1. Introduction

Let \(\Sigma\) denote the class of functions of the form \[f(z)=\frac{1}{z}+\sum_{k=1}^\infty a_kz^{k-1},\] which are analytic in the punctured unit disc \( \mathbb E_0=\mathbb E\setminus\{0\},\) where \(\mathbb E = \{z\in\mathbb C:|z|< 1\}. \) A function \(f \in \Sigma \) is said to be meromorphic starlike of order \(\alpha \) if \(f(z) \neq 0 \) for \(z \in \mathbb E_0 \) and

\[-\Re\left(\frac{zf'(z)}{f(z)}\right)>\alpha,\hspace{1.7cm}(\alpha< 1;z \in \mathbb E).\] The class of such functions is denoted by \(\mathcal {MS}^*(\alpha)\) and write \(\mathcal {MS}^*=\mathcal {MS}^*(0)\)-the class of meromorphic starlike functions.

In 1959, Sakaguchi [1] introduced and studied the class of starlike functions with respect to symmetric points in \(\mathbb E\). Further investigations into the class of starlike functions with respect to symmetric points can be found in [2,3].

Recently, Ghaffar et al., [4] introduced and investigated a class of meromorphic starlike functions with respect to symmetric points which satisfies the condition

\[-\frac{2zf'(z)}{f(z)-f(-z)}\prec \frac{1+Az}{1+Bz}, ~z\in\mathbb E_0\] and \(-1\leq B< A\leq1.\) We denote the above class by \(\mathcal {MS}^{s*}[A,B].\) If \(f\) is analytic and \(g\) is analytic univalent in open unit disk \(\mathbb E\), we say that \(f(z)\) is subordinate to \(g(z)\) in \(\mathbb E\) and written as \(f(z)\prec g(z)\) if \(f(0) = g(0)\) and \(f( \mathbb E) \subset g(\mathbb E)\). To derive certain sandwich-type results, we use the dual concept of differential subordination and superordination.

Let \(~\Phi:\mathbb C^2 \times \mathbb E\longrightarrow \mathbb C\) (\(\mathbb C\) is the complex plane) and \(h\) be univalent in \(\mathbb E\). If \(p\) is analytic in \(\mathbb E\) and satisfies the differential subordination

then \(p\) is called a solution of the differential subordination (1). The univalent function \(q\) is called a dominant of differential subordination (1) if \(p\prec q\) for all \(p\) satisfying \(1.\) A dominant \(\tilde q \prec q \) for all dominants \(q\) of \(1,\) is said to be the best dominant of \(1.\)

Let \(\Psi:\mathbb C^2 \times \mathbb E\longrightarrow \mathbb C\) \((\mathbb C\) is the complex plane) be analytic and univalent in domain \(\mathbb C^2 \times \mathbb E, h\) be analytic in \(\mathbb E,\) \(p\) is analytic and univalent in \(\mathbb E,\) with \((p(z), zp'(z); z)\in \mathbb C^2 \times \mathbb E \) for all \(z\in \mathbb E.\) Then \(p\) is called a solution of first order differential superordination if it satisfies

An analytic function \(q\) is called a subordinant of differential superordination (2) if \(q\prec p\) for all \(p\) satisfying \((2).\) A univalent subordinant \(\tilde q\) that satisfies \(q \prec \tilde q \) for all subordinants \(q\) for (2), is said to be the best subordinant of (2).

In this paper we study the concepts of subordination and superordination to obtain meromorphic starlikeness with respect to symmetric points.On the basis of the theory we also investigate some important sandwich results of symmtric meromorphic functions.

2. Preliminaries

We shall use the following lemmas to prove our result.

Lemma 1. [5] Let \(q\) be univalent in \(\mathbb E\) and let \(\theta\) and \(\phi\) be analytic in a domain \(\mathbb D\) containing \(q(\mathbb E)\), with \(\phi(w)\neq 0\), when \(w\in q (\mathbb E)\). Set \(Q_1(z)=zq'(z)\phi[q(z)],~h(z)=\theta[q(z)]+Q_1(z)\) and suppose that either

• (i) \(h\) is convex, or
• (ii) \(Q_1\) is starlike.

In addition, assume that

• (iii) \(\displaystyle \Re\left(\frac{zh'(z)}{Q_1(z)}\right)>0\) for all \(z\) in \(\mathbb E\).

If \(p\) is analytic in \(\mathbb E\), with \(p(0)=q(0),~p(\mathbb E)\subset \mathbb D\) and \[\theta[p(z)]+zp'(z)\phi[p(z)]\prec\theta[q(z)]+zq'(z)\phi[q(z)],~z\in\mathbb E,\] then \(p(z)\prec q(z)\) and \(q\) is the best dominant.

Definition 1. We denote by \(Q\) the set of functions \(p\) that are analytic and injective on \(\overline{\mathbb E}\setminus\mathbb B(p),\) where \[\mathbb B(p)=\left\{\zeta\in \partial{\mathbb E}: \lim_{z\rightarrow\zeta} {p(z)=\infty} \right\},\] and are such that \(p'(\zeta)\neq 0\) for \(\zeta \in \partial \mathbb E\setminus\mathbb B(p).\)

Lemma 2. [6] Let \(q\) be the univalent in \(\mathbb E\) and let \(\theta\) and \(\phi\) be analytic in a domain \(\mathbb D\) containing \(q(\mathbb E).\) Set \(Q_1(z)=zq'(z)\phi[q(z)],~h(z)=\theta[q(z)]+Q_1(z)\) and suppose that (i) \( Q_1(z)\) is starlike in \(\mathbb E;\) and (ii) \( \Re\left(\frac{\theta'(q(z))}{\phi(q(z))}\right)> 0,\) for \(z \in \mathbb E.\) If \(p\in \mathcal H[q(0),1]\cap Q,\) with \(p(\mathbb E)\subset \mathbb D\) and \(\theta[p(z)]+zp'(z)\phi[p(z)]\) is univalent in \(\mathbb E\) and \[\theta[q(z)]+zq'(z)\phi[q(z)]\prec \theta[p(z)]+zp'(z)\phi[p(z)],z\in \mathbb E,\] then \(q(z)\prec p(z)\) and \(q\) is the best subordinant.

Lemma 3. [7] The function \(\displaystyle q(z)=\frac{1}{(1-z)^{2ab}}\) is univalent in \(E\) if and only if \(|2ab-1|\leq1\) or \(|2ab+1|\leq1.\)

3. Subordination results

Theorem 1. Let \(q\) be univalent in \(\mathbb E,\) with \(q(0)=1,\) and let

where \(\gamma\in\mathbb C^*:=\mathbb C\setminus\{0\}.\) If \(f\in\Sigma\) satisfy the condition and then \[\frac{-2zf'(z)}{f(z)-f(-z)}\prec q(z),\] and \(q\) is the best dominant of (5).

Proof. Setting \[p(z)=\frac{-2zf'(z)}{f(z)-f(-z)},~z\in\mathbb E,\] from assumption (4) it follows that \(p\) is analytic in \(\mathbb E,\) with \(p(0)=1.\) A simple computation shows that \[\frac{-2(1+\lambda)zf'(z)}{f(z)-f(-z)}+\frac{2\lambda z^2 f'(z)(f'(z)+f'(-z))}{(f(z)-f(-z))^2}-\frac{2\lambda z^2 f''(z)}{f(z)-f(-z)}=p(z)+\lambda zp'(z),\] hence, the subordination (5) is equivalent to \(p(z)+\lambda zp'(z)\prec q(z)+\lambda zq'(z)\).

Now, in order to prove our result we will use Lemma 1. Consider the functions \(\theta(w)=w\) and \(\phi(w)=\gamma\) analytic in \(\mathbb C,\) and set

\begin{equation} Q(z)=zq'(z) \phi(q(z))=\gamma zq'(z) \  \  and \  \  h(z)=\theta(q(z))+Q(z)=q(z)+\gamma z q'(z). \end{equation} Since \(Q(0)=0\) and \(Q'(0)=\gamma q'(0)\neq0\), the assumption (3) implies that \(Q\) is starlike in \(\mathbb E\) and \[\Re\left(\frac{zh'(z)}{Q(z)}\right)=\Re \left(1+\frac{zq''(z)}{q'(z)}+\frac{1}{\gamma}\right)>0, ~z \in \mathbb E.\] Therefore, Lemma 1 and assumption (5) imply \(p(z)\prec q(z)\) and the function \(q\) is the best dominant of (5).

If we take \(\displaystyle q(z)=\frac{1+Az}{1+Bz},-1\leq B< A\leq 1\) in Theorem 1 then the condition (3) reduces to

It is easy to see that the function \(\displaystyle \chi(\zeta)=\frac{1-\zeta}{1+\zeta}\) with \(|\zeta|< |B|\) is convex in \(E\) and since \(\chi(\overline{\zeta}) =\overline{\chi(\zeta) }\) for all \(|\zeta|< |B|\), it follows that \(\chi(\mathbb E)\) is a convex domain symmetric with respect to the real axis. Hence Thus, the inequality (6) is equivalent to \(\displaystyle\Re\frac {1}{\lambda}\geq \frac{|B|-1}{|B|+1}\), hence we deduce the following corollary:

Corollary 1. Let \(\lambda\in\mathbb C^*\) and \(-1\leq B< A\leq 1\) with \(\displaystyle\frac{1-|B|}{1+|B|}\geq max\left\{0;~-\Re \frac{1}{\lambda}\right\}\). If \(f\in\Sigma\) satisfy the condition (4) and

then \[\frac{-2zf'(z)}{f(z)-f(-z)}\prec \frac{1+Az}{1+Bz},\] i.e., \(f\in\mathcal{MS}^{s*}[A,B].\) Moreover, the function \(\displaystyle\frac{1+Az}{1+Bz}\) is the best dominant of (8).

For \(A=1\) and \(B=-1\), the above corollary reduces to the next special case:

Remark 1. Let \(\lambda\in\mathbb C^*\) with \(\Re \frac{1}{\lambda}\geq0\). If \(f\in\Sigma\) satisfy the condition (4) and

then \[\frac{-2zf'(z)}{f(z)-f(-z)}\prec \frac{1+z}{1-z},\] i.e., \(f\in\mathcal{MS}^{s*},\) or \(f\) is meromorphic starlike with respect to symmetrical points in \(\mathbb E\). Moreover, the function \(\displaystyle\frac{1+z}{1-z}\) is the best dominant of (9).

Theorem 2. Suppose that \(q\) be univalent in \(\mathbb E\) with \(q(0)=1\) and \(q(z)\neq0\) for all \(z\in\mathbb E\) such that

where \(\gamma,\mu\in\mathbb C^*\) and \(\nu,\eta\in\mathbb C\) with \(\nu+\eta\neq 0\) and let \(f\in\Sigma\) satisfy the conditions and If then \[-\left[\frac{2(\nu+\eta) zf'(z)}{\nu z(f'(z)+ f'(-z)) +\eta (f(z)- f(-z))}\right]^\mu\prec q(z)\] and \(q\) is the best dominant of (13) (the power is the principal one).

Proof. Define a function

According to the assumptions (11) and (12), the multivalued power function \(p\) has an analytic branch in \(\mathbb E\) with \(p(0)=1\) and from the (14) it follows that \[\mu\left[1-\frac{\nu z[z(f'(z)+ f'(-z))]' +\eta z(f'(z)+ f'(-z))}{\nu z(f'(z)+ f'(-z)) +\eta (f(z)- f(-z))}+\frac{zf''(z)}{f'(z)}\right]=\frac{zp'(z)}{p(z)}.\] To prove our desired result we will use Lemma 1. Thus, let the functions \(\theta(w)=1\) and \(\phi(w)=\displaystyle\frac{\gamma}{w}\). Also, if we let \( Q(z)=zq'(z) \phi(q(z))=\gamma zq'(z)\) and \(h(z)=\theta(q(z))+Q(z)=1+\gamma\displaystyle\frac{z q'(z)}{q(z)}.\)

Since \(Q(0)=0\) and \(Q'(0)=\gamma q'(0)\neq0\), the assumption (10) implies that \(Q\) is starlike in \(\mathbb E\) and

\[\Re\left(\frac{zh'(z)}{Q(z)}\right)=\Re \left(1+\frac{zq''(z)}{q'(z)}+\frac{z q'(z)}{q(z)}\right)>0, ~z \in \mathbb E.\] Therefore, using Lemma 1 and assumption (13) implies \(p(z)\prec q(z)\) and the function \(q\) is the best dominant of (13).

In particular, taking \(\nu=0, \eta=\gamma=1\) and \(\displaystyle q(z)=\frac{1+Az}{1+Bz}\) in the above theorem, it is easy to check that the inequality (10) holds whenever \(-1\leq B< A\leq 1\). Hence, we deduce the following corollary:

Corollary 2. Let \(-1\leq B< A\leq 1\) and \(\mu\in\mathbb C^*\). Let \(f\in\Sigma\) satisfy the conditions (4) and

If then \[-\left[\frac{2zf'(z)}{f(z)-f(-z)}\right]^\mu\prec \frac{(1+Az)}{(1+Bz)}, \] and the function \( \displaystyle\frac{(1+Az)}{(1+Bz)}\) is the best dominant of (16) (the power is principal one).

Using \(\nu=0, \eta=1, \gamma=\displaystyle \frac{1}{ab}\) with \(a,b\in\mathbb C^*, \mu=a\) and \( q(z)=\displaystyle \frac{1}{(1-z)^{2ab}}\) in Theorem 2, then using this result together with Lemma 3, we deduce the following result:

Corollary 3. Let \(a,b\in\mathbb C^*\) such that \[|2ab-1|\leq1 \hspace{1cm}or \hspace{1cm}|2ab+1|\leq1\] and suppose that \(f\in\Sigma\) satisfy the conditions (4) and (15). If \[1+\frac{1}{b}\left[1-\frac{z(f'(z)+f'(-z))}{f(z)-f(-z)}+\frac{zf''(z)}{f'(z)}\right]\prec\frac{1+z}{1-z},\] then

and the function \(\displaystyle \frac{1}{(1-z)^{2ab}}\) is the best dominant of (17) (the power is the principal one).

By using \(\nu=0, \eta=\gamma=1\) and \( q(z)= \displaystyle(1+Bz)^{\frac{\mu(A-B)}{B}}\), \(-1\leq B< A\leq 1\), \(B\neq0\) in Theorem 2, then using Lemma 3 we have the next result:

Corollary 4. Let \(-1\leq B< A\leq 1\) with \(B\neq0\), and suppose that \[\left|\frac{\mu(A-B)}{B-1}\right|\leq1 \hspace{1cm}or \hspace{1cm}\left|\frac{\mu(A-B)}{B+1}\right|\leq1,\] where \(\mu\in\mathbb C^*\). If \(f\in\Sigma\) satisfy the conditions (4), (15) and

then \[-\left[\frac{2zf'(z)}{f(z)-f(-z)}\right]^\mu\prec (1+Bz)^{\frac{\mu(A-B)}{B}} \] and the function \((1+Bz)^{\frac{\mu(A-B)}{B}}\) is the best dominant of (18) (the power is the principal one).

By taking \(\nu=0, \eta=1\), \(\gamma=\frac{e^{\iota\lambda}}{ab \cos \lambda}, a,b\in\mathbb C^*, |\lambda|< \displaystyle\frac{\pi}{2}, \mu=a\) and \(q(z)=\displaystyle\frac{1}{(1-z)^{2abe^{-\iota\lambda} \cos \lambda}}\) in Theorem 2, we obtain the following:

Corollary 5. Let \(a,b\in\mathbb C^*\) and \(|\lambda|< \frac{\pi}{2}\) and suppose that \[ |2abe^{-\iota\lambda} \cos \lambda-1|\leq1 \hspace{1cm} \text{or} \hspace{1cm} |2abe^{-\iota\lambda} \cos \lambda+1|\leq1 .\] If \(f\in\Sigma\) satisfy the conditions (4), (15) and \[1+\frac{e^{\iota\lambda}}{b \cos \lambda}\left[1-\frac{z(f'(z)+f'(-z))}{f(z)-f(-z)}+\frac{zf''(z)}{f'(z)}\right]\prec\frac{1+z}{1-z},\] then

and the function \(\displaystyle \frac{1}{(1-z)^{2abe^{-\iota\lambda} \cos \lambda}}\) is the best dominant of (19) (the power is the principal one).

Theorem 3. Suppose that \(q\) be univalent in \(\mathbb E\) with \(q(0)=1\) and \(q(z)\neq0\) for all \(z\in\mathbb E\), such that

where \(\gamma,\mu\in\mathbb C^*\) and \(\delta,\nu,\eta\in\mathbb C\) with \(\nu+\eta\neq 0\) and let \(f\in\Sigma\) satisfy the conditions (11), (12) and \[\left[\delta+\gamma\mu\left(1-\frac{\nu z[z(f'(z)+ f'(-z))]' +\eta z(f'(z)+ f'(-z))}{\nu z(f'(z)+ f'(-z)) +\eta (f(z)-f(-z))}+\frac{zf''(z)}{f'(z)}\right)\right]\,.\] If then \[-\left[\frac{2(\nu+\eta) zf'(z)}{\nu z(f'(z)+ f'(-z)) +\eta (f(z)- f(-z))}\right]^\mu\prec q(z)\] and \(q\) is the best dominant of (13) (the power is the principal one).

Proof. Define a function

From the assumptions (11) and (12) it follows that the multivalued power function \(p\) has an analytic branch in \(\mathbb E\) with \(p(0)=1\) and from the (23) it follows that \[\mu p(z) \left[1-\frac{\nu z[z(f'(z)+ f'(-z))]' +\eta z(f'(z)+ f'(-z))}{\nu z(f'(z)+ f'(-z)) +\eta (f(z)- f(-z))}+\frac{zf''(z)}{f'(z)}\right]=zp'(z).\] Consider the functions \(\theta(w)=\delta w\) and \(\phi(w)=\gamma\) that are analytic in \(\mathbb C^*\). Also, if we let \( Q(z)=zq'(z) \phi(q(z))=\gamma zq'(z)\) and \(h(z)=\theta(q(z))+Q(z)=\delta q(z)+\gamma\frac{z q'(z)}{q(z)}.\)

Since \(Q(0)=0\) and \(Q'(0)=\gamma q'(0)\neq0\), the assumption (20) implies that \(Q\) is starlike in \(\mathbb E\) and

\[\Re\left(\frac{zh'(z)}{Q(z)}\right)=\Re \left(\frac{\delta}{\gamma}+1+\frac{zq''(z)}{q'(z)}+\frac{z q'(z)}{q(z)}\right)>0, ~z \in \mathbb E.\] Therefore, using Lemma 1 and assumption (22) implies \(p(z)\prec q(z)\) and the function \(q\) is the best dominant of (22).

Taking \(\displaystyle q(z)=\frac{1+Az}{1+Bz}\) in Theorem 3, where \(-1\leq B< A\leq1\), according to (7) the condition (20) becomes \[max\left\{0;~-\Re \frac{\delta}{\gamma}\right\}\leq\frac{1-|B|}{1+|B|}.\] Hence, for the particular cases \(\eta=0\) and \(\mu=\gamma=1\), we have the following result:

Corollary 6. Let \(-1\leq B< A\leq 1, \mu\in\mathbb C^*\) and \(\delta\in\mathbb C\) with \[max\left\{0;~-\Re \frac{\delta}{\gamma}\right\}\leq\frac{1-|B|}{1+|B|}.\] If \(f\in\Sigma\) satisfy the conditions (4), (15) and

then \[ -\left[\frac{2f'(z)}{f'(z)+ f'(-z)}\right]^\mu \prec \frac{1+Az}{1+Bz}\] and the function \(\displaystyle \frac{1+Az}{1+Bz}\) is the best dominant of (24) (all the powers are principal ones).

Taking \(\eta=\gamma=1,\mu=0\) and \(q(z)=\displaystyle\frac{1+z}{1-z}\) in Theorem 3, we get the following corollary:

Corollary 7. Let \(\mu\in\mathbb C^*\) and \(\delta\in\mathbb C\) with \(\Re \delta \geq 0.\) If \(f\in\Sigma\) satisfy the conditions (4), (15) and

then \[ -\left[\frac{2f'(z)}{f'(z)+ f'(-z)}\right]^\mu \prec \frac{1+z}{1-z},\] and the function \(\displaystyle \frac{1+z}{1-z}\) is the best dominant of (25) (all the powers are principal ones).

4. Superordination and sandwich theorems

Theorem 4. Let \(q\) be convex in \(\mathbb E\) with \(q(0)=1\) and \(\lambda\in\mathbb C\) with \(\Re~ \lambda>0\). Let \(f\in\Sigma\) satisfy the condition (4) such that \(\displaystyle-\frac{2zf'(z)}{f(z)-f(-z)}\in\mathbb Q\) and suppose that the function \[ \frac{-2(1+\lambda)zf'(z)}{f(z)-f(-z)}+\frac{2\lambda z^2 f'(z)(f'(z)+f'(-z))}{(f(z)-f(-z))^2}-\frac{2\lambda z^2 f''(z)}{f(z)-f(-z)}\] is univalent in \(\mathbb E\). If

then \[q(z)\prec-\frac{2zf'(z)}{f(z)-f(-z)}\] and \(q\) is the best subordinant of (26).

Proof. Setting \[p(z)=\frac{-2zf'(z)}{f(z)-f(-z)},~z\in\mathbb E,\] then \(p\) is analytic in \(\mathbb E\) with \(p(0) = 1.\) Taking logarithmic differentiation of the above relation with respect to \(z,\) we have \[\frac{zp'(z)}{p(z)}=1-z\left(\frac{f'(z)+f'(-z)}{f(z)-f(-z)}-\frac{f''(z)}{f'(z)}\right)\] and a simple calculation yields that the assumption (26) is equivalent to \[ q(z)+\lambda zq'(z)\prec p(z)+\lambda zp'(z).\] Now, in order to prove our result we will use Lemma 2. Consider the functions \(\theta(w)=w\) and \(\phi(w)=\lambda\) analytic in \(\mathbb C\) and set \[h(z)=zq'(z)\phi(q(z))=\lambda z q'(z).\] Since \(h(0)=0\), \(h'(0)=\lambda q'(0)\neq0\) and \(q\) is convex in \(\mathbb E\), it follows that \(h\) is starlike in \(\mathbb E\) and \[\Re \frac{\theta'q(z)}{\phi(q(z))}=\Re \frac{1}{\gamma}>0, ~z \in \mathbb E.\] Therefore, Lemma 2 and assumption (26) imply \(q(z)\prec p(z)\) and the function \(q\) is the best subordinant of (26).

Taking \(\displaystyle q(z)=\frac{1+Az}{1+Bz}\) in Theorem 4, where \(-1\leq B< A\leq1\), we get the following corollary:

Corollary 8. Let \(q\) be convex in \(\mathbb E\) with \(q(0)=1\) and \(\lambda\in\mathbb C\) with \(\Re~ \lambda>0\). Let \(f\in\Sigma\) satisfy the condition (4) such that \(\displaystyle-\frac{2zf'(z)}{f(z)-f(-z)}\in\mathbb Q\) and suppose that the function \[ \frac{-2(1+\lambda)zf'(z)}{f(z)-f(-z)}+\frac{2\lambda z^2 f'(z)(f'(z)+f'(-z))}{(f(z)-f(-z))^2}-\frac{2\lambda z^2 f''(z)}{f(z)-f(-z)}\] is univalent in \(\mathbb E\). If

then \[\displaystyle\frac{1+Az}{1+Bz}\prec-\frac{2zf'(z)}{f(z)-f(-z)}\] and \(q\) is the best subordinant of (27).

Using the same techniques as in proof of Theorem 3 and then applying Lemma 2, we could prove the next theorem:

Theorem 5. Let \(\gamma,\mu\in\mathbb C^*\) and \(\delta,\nu,\eta\in\mathbb C\) with \(\nu+\eta\neq 0\) and \(\Re\left (\displaystyle\frac{\delta}{\gamma}\right)>0\). Suppose that \(q\) is convex in \(\mathbb E\), with \(q(0)=1\) and let \(f\in\Sigma\) satisfy the conditions (11), (12) and \[-\left[\frac{2(\nu+\eta) zf'(z)}{\nu z(f'(z)+ f'(-z)) +\eta (f(z)- f(-z))}\right]^\mu\in\mathbb Q. \] If the function \(\phi \) given by (21) is univalent in \(\mathbb E\) and

then \[q(z)\prec -\left[\frac{2(\nu+\eta) zf'(z)}{\nu z(f'(z)+ f'(-z)) +\eta (f(z)- f(-z))}\right]^\mu\] and \(q\) is the best subordinant of (28) (the power is the principal one).

If we combine Theorem 1 with Theorem 4 and Theorem 3 with Theorem 5, we deduce the following sandwich results, respectively:

Theorem 6. Let \(q_1\) and \(q_2\) be two convex functions in \(\mathbb E\) with \(q_1(0)=q_2(0)=1\) and \(\lambda\in\mathbb C\) with \(\Re~ \lambda>0\). Let \(f\in\Sigma\) satisfy the condition (4), such that \(\displaystyle-\frac{2zf'(z)}{f(z)-f(-z)}\in\mathbb Q\) and suppose that the function \[ \frac{-2(1+\lambda)zf'(z)}{f(z)-f(-z)}+\frac{2\lambda z^2 f'(z)(f'(z)+f'(-z))}{(f(z)-f(-z))^2}-\frac{2\lambda z^2 f''(z)}{f(z)-f(-z)}\] is univalent in \(\mathbb E\). If \[ q_1(z)+\lambda zq_1'(z)\prec \frac{-2(1+\lambda)zf'(z)}{f(z)-f(-z)}+\frac{2\lambda z^2 f'(z)(f'(z)+f'(-z))}{(f(z)-f(-z))^2}-\frac{2\lambda z^2 f''(z)}{f(z)-f(-z)}\,.\]

then \[q_1(z)\prec\frac{-2zf'(z)}{f(z)-f(-z)}\prec q_2(z),\] where \(q_1\) and \(q_2\) are respectively the best subordinant and the best dominant of (29).

Theorem 7. Let \(q_1\) and \(q_2\) be two convex functions in \(\mathbb E\) with \(q_1(0)=q_2(0)=1\) and let \(\gamma,\mu\in\mathbb C^*\) and \(\nu,\eta\in\mathbb C\) with \(\nu+\eta\neq 0\) and \(\Re\displaystyle \frac{\delta}{\gamma}>0\). Let \(f\in\Sigma\) satisfy the condition (11), (12) and such that \(\displaystyle-\frac{2zf'(z)}{f(z)-f(-z)}\in\mathbb Q\) and \[-\left[\frac{2(\nu+\eta) zf'(z)}{\nu z(f'(z)+ f'(-z)) +\eta (f(z)-f(-z))}\right]^\mu\in\mathbb Q. \] If the function \(\phi \) given by (21) is univalent in \(\mathbb E\) and

then \[q_1(z)\prec -\left[\frac{2(\nu+\eta) zf'(z)}{\nu z(f'(z)+ f'(-z)) +\eta (f(z)- f(-z))}\right]^\mu\prec q_2(z)\] and \(q_1\) and \(q_2\) are respectively the best subordinant and the best dominant of (29) (the power is the principal one).

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflicts of Interest

''The authors declare no conflict of interest.''

1. Introduction

Throughout this paper, let \(\Gamma\) be a nonempty closed subset of a real Banach space \(\Psi\), \(\mathbb{N}\) the set of all natural numbers, \(\Re\) the set of all real numbers and \(C([d,e])\) denotes the set of all continuous real-valued functions defined on \([d,e]\subset\Re\).

A mapping \(T:\Gamma\to \Gamma\) is called

• \(\bullet\) contraction if there exists a constant \(\delta\in[0,1)\) such that
  \begin{eqnarray} \|T\psi-T\zeta\|\leq \delta\|\psi-\zeta\|,\,\,\forall\,\psi,\zeta\in \Gamma; \end{eqnarray}

  (1)
• \(\bullet\) nonexpansive if
  \begin{eqnarray} \|T\psi-T\zeta\|\leq \|\psi-\zeta\|,\,\,\forall\,\psi,\zeta\in \Gamma. \end{eqnarray}

  (2)

Clearly, every contraction map is a nonexpansive map for \(\delta=1\).

For some decades now, fixed point theory has been developed into a basic and an essential tool for different branches of both applied and pure mathematics. Particularly, it may be seen as an essential subject of nonlinear functional analysis. Moreover, fixed point theory is one of the useful tools to solve many problems in applied sciences and engineering such as: the existence of solutions to integral equations, differential equations, matrix equations, dynamical system, models in economy, game theory, fractals, graph theory, optimization theory, approximation theory, computer science and many other subjects.

It is well known that several mathematical problems are naturally formulated as fixed point problem,

where \(T\) is some suitable mapping, which may be nonlinear. For example, given a mapping \(\varphi:[d,e]\subset\Re\to \Re\) and \(k:[d,e]\times[d,e]\times\Re\to \Re\), then the solution to the following nonlinear integral equation: where \(\psi\in C([d,e])\), is tantamount to finding the fixed point of the integral operator \(T:C([d,e])\to C([d,e])\) defined by for all \(\psi \in C([d,e])\).

A solution \(\psi\) of the problem (3) is called a fixed point of the mapping \(T\). We will denote the set of all fixed points of \(T\) by \(F(T)\), i.e., \(F(T)=\{\psi\in \Gamma: T\psi=\psi\}\).

On the other hand, once the existence of some fixed points of a given mapping is guaranteed, then finding such fixed point of the mapping is cumbersome in some cases. To surmount this difficulty, iterative algorithm is usually employed for approximating them. An efficient iterative should at least be \(T\)-stable, converges faster than a number of some existing iterative scheme in the literature, converges to a fixed point of an operator or should be data dependent (see [1]).

The Picard iterative algorithm

is one of the first iterative algorithms which has been widely used for approximating the fixed points of contraction mappings. However, the success of Picard iterative algorithm has not been carried over to the more general classes of operators such as nonexpansive mappings even when a fixed exists.

For example, the mapping \(T:[0,1]\to[0,1]\) defined by \(T\psi=1-\psi\) for all \(\psi\in [0,1]\) is a nonexpansive mapping with a unique fixed point \(\psi=\frac{1}{2}\). Notice that for \(\psi_0\in [0,1]\), \(\psi_0\neq\frac{1}{2}\), the Picard iterative algorithm (6) generates the sequence \(\{1-\psi_0,\psi_0,1-\psi_0,...\}\) for which fails to converge to the fixed point \(z\) of \(T\).

To overcome the failure recorded by Picard iterative algorithm, many researchers in nonlinear analysis got busy with constructing several new iterative algorithms for approximating the fixed points of nonexpansive mappings and other mappings more general than the classes of nonexpansive mappings.

Some notable iterative algorithms in the existing literature includes: Mann [2], Ishikawa [3], Noor [4], Argawal et al., [5], Abbas and Nazir [6], SP [7], S* [8], CR [9], Normal-S [10], Picard-S [11], Thakur et al., [12], M [13], M* [14], Garodia and Uddin [15], Two-Step Mann [16] iterative algorithms and so on.

In 2007, the following iterative algorithm which is known as S iteration was introduced by Argawal et al., [5]:

where \(\{\mu_s\}\) and \(\{r_s\}\) are sequences in \([0,1].\)

In 2014, the following iterative algorithm known as Picard-S iteration was introduced by Gursoy and Karakaya [11]:

where \(\{\mu_s\}\) and \(\{r_s\}\) are sequences in \([0,1].\) The authors showed with the aid of an example that Picard-S iterative algorithm (8) converges at a rate faster than all of Picard, Mann, Ishikawa, Noor, SP, CR, S, S*, Abbas and Nazir, Normal-S and Two-Step Mann iteration processes for contraction mappings.

In 2016, Thakur et al., [12] introduced the following three steps iterative algorithm:

where \(\{\mu_s\}\) and \(\{r_s\}\) are sequences in \([0,1].\) With the help of numerical example, they proved that (9) is faster than Picard, Mann, Ishikawa, Agarwal, Noor and Abbas iterative algorithm for Suzuki generalized nonexpansive mappings.

In 2018, Ullah and Arshad [13] introduced the M iterative algorithm as follows:

where \(\{r_s\}\) is a sequence in \([0,1].\) Numerically they show that M iterative algorithm (10) converges faster than S iterative algorithm (7) and Picard-S iteration process (8) for Suzuki generalized nonexpansive mapping.

Recently, Garodia and Uddin [15] introduced the following three steps iterative algorithm:

where \(\{r_s\}\) is a sequence in \([0,1].\) The authors showed both analytically and numerically that their iterative algorithm (11) converges faster than M iterative algorithm (10). Also, they showed that the iterative algorithm (11) converges faster than all of S, Abbas and Nazir, Thakur New, M, Noor, Picard-S, Thakur, M* iterative algorithms for contractive-like mappings and Suzuki generalized nonexpansive mappings.

Clearly, from the performance of the iterative algorithm (11), we note that it is one of the leading iterative schemes.

Problem 1. Is it possible to construct a four-steps iterative algorithm which have better rate of converges than the three-step iterative algorithm (11) for contraction mappings?

To solve the above problem, we introduce the following iterative algorithm, called AU iterative algorithm: where \(\{r_s\}\) is a sequence in \([0,1]\).

The aim of this paper is to prove the strong convergence of AU iterative algorithm (12) to the fixed points of contraction mappings. We will prove analytically that AU iterative algorithm (12) converges faster than the iterative algorithm (11). With the aid of an example, we show numerically that AU iterative algorithm has a better rate of convergence than the iterative algorithm (11) and several other leading iterative algorithms existing in the literature. Also, we will show that AU iterative algorithm is \(T\)-stable and data dependent.

Additionally, we will use AU iterative algorithm (12) to find the unique solutions of a nonlinear integral equation.

2. Preliminaries

The following definitions and lemmas will be useful in proving our main results.

Definition 2.[17] Let \(\{a_s\}\) and \(\{b_s\}\) be two sequences of real numbers that converge to \(a\) and \(b\) respectively and assume that there exists \begin{eqnarray*} \ell=\lim\limits_{s\to\infty}\frac{\|a_s-a\|}{\|b_s-b\|}. \end{eqnarray*}Then,

• (\(R_1\)) if \(\ell=0\), we say that \(\{a_s\}\) converges faster to \(a\) than \(\{b_s\}\) does to \(b\).
• (\(R_2\)) If \(0< \ell< \infty\), we say that \(\{a_s\}\) and \(\{b_s\}\) have the same rate of convergence.

Definition 3. [17] Let \(\{\omega_s\}\) and \(\{\mu_s\}\) be two fixed point iteration processes that converge to the same point \(z\), the error estimates \begin{eqnarray*} \|\omega_s-z\|&\leq& a_s, \,\,s \in\mathbb{N},\\ \|\mu_s-z\|&\leq& b_s, \,\,s \in\mathbb{N}, \end{eqnarray*}are available where \(\{a_s\}\) and \(\{b_s\}\) are two sequences of positive numbers converging to zero. Then we say that \(\{\omega_s\}\) converges faster to \(z\) than \(\{\mu_s\}\) does if \(\{a_s\}\) converges faster than \(\{b_s\}\).

Definition 4.[17] Let \(T\), \(\tilde{T}:\Gamma\to \Gamma\) be two operators. We say that \(\tilde{T}\) is an approximate operator for \(T\) if for some \(\epsilon>0\), we have

Definition 5.[18] Let \(\{\zeta_s\}\) be any sequence in \(\Gamma\). Then, an iterative algorithm \(\psi_{s+1}=f(T,\psi_s)\), which converges to fixed point \(z\), is said to be \(T\)-stable if for \(\varepsilon_s=\|\zeta_{s+1}-f(T,\zeta_s)\|\), \(\forall\,s\in\mathbb{N}\), we have

Lemma 6. [19] Let \(\{ \mathfrak{\theta}_s\}\) and \(\{\lambda_s\}\) be nonnegative real sequences satisfying the following inequalities:

where \(\sigma_s\in (0,1)\) for all \(s\in \mathbb{N}\), \(\sum\limits_{s=0}^{\infty}\sigma_s=\infty\) and \(\lim\limits_{s\to\infty}\frac{ \mathfrak{\lambda}_s}{\sigma_s}=0\), then \(\lim\limits_{s\to\infty} \mathfrak{\theta}_s=0\).

Lemma 7. [20] Let \(\{ \mathfrak{\theta}_s\}\) be a nonnegative real sequence and there exits an \(s_0\in \mathbb{N}\) such that for all \(s\geq s_0\) satisfying the following condition: \begin{eqnarray*} \mathfrak{\theta}_{s+1}\leq(1-\sigma_s) \mathfrak{\theta}_s+\sigma_s \lambda_s, \end{eqnarray*}where \(\sigma_s\in (0,1)\) for all \(s\in \mathbb{N}\), \(\sum\limits_{s=0}^{\infty}\sigma_s=\infty\) and \(\lambda_s\geq0\) for all \(s\in \mathbb{N}\), then \begin{eqnarray*} 0\leq\limsup\limits_{s\to\infty} \mathfrak{\theta}_s\leq\limsup\limits_{s\to\infty}\lambda_s. \end{eqnarray*}

3. Convergence result

In this section, we prove the strong convergence of AU iterative algorithm (12) for contraction mappings.

Theorem 8. Let \(\Gamma\) be a nonempty closed convex subset of a real Banach space \(\Psi\) and \(T:\Gamma\to \Gamma\) be a contraction mapping such that \(F(T)\neq\emptyset\). Let \(\{\psi_s\}\) be the sequence iteratively generated by (12) with a real sequence \(\{r_s\}\) in \([0,1]\) satisfying \(\sum\limits_{s=0}^{\infty}r_s=\infty\). Then \(\{\psi_s\}\) converges strongly to a unique fixed point of \(T\).

Proof. Given \(z\in F(T)\), then from (12) we have

Now, from (12) and (16) we obtain Also, from (12) and (17) we get Finally, from (12) and (18) we have From (19), the following inequality is obtained: From (20), we have Since for all \(s\in \mathbb{N}\), \(\{r_s\}\in [0,1]\) and \(\delta\in (0,1)\), it follows that \((1-(1-\delta)r_s)< 1\). From classical analysis we know that \(1-\psi< \exp^{-\psi}\) for all \(\psi\in [0,1]\). Then from (21), we have If we take the limits of both sides of (22), we get \(\lim\limits_{s\to\infty}\|\psi_s-z\|=0\). Hence, \(\{\psi_s\}\) converges strongly to the fixed point of \(T\) as required.

4. Stability result

Theorem 9. Let \(\Gamma\) be a nonempty closed convex subset of a Banach space \(\Psi\) and \(T:\Gamma \to \Gamma\) be a contraction mapping. Let \(\{\psi_s\}\) be an iterative algorithm defined by (12) with a real sequence \(\{r_s\}\) in [0,1] satisfying \(\sum\limits_{s=0}^{\infty}r_s=\infty\). Then the iterative algorithm (12) is \(T\)-stable.

Proof. Let \(\{\zeta_s\}\subset \Psi\) be an arbitrary sequence in \(\Gamma\) and suppose that the sequence iteratively generated by (12) is \(\psi_{s+1}=f(T,\psi_s)\), which converges to a unique point \(z\) and that \(\varepsilon_s=\|\zeta_{s+1}-f(T,\zeta_s)\|\). To prove that (12) is T-stable, we have to show that \(\lim\limits_{s\to\infty}\varepsilon_s=0\Leftrightarrow \lim\limits_{s\to\infty}\zeta_s=z\).

Let \(\lim\limits_{s\to\infty}\varepsilon_s=0\). Then from (12) and the demonstration above, we have

Putting (24) into (23) we obtain Substituting (26) into (25), we obtain Substituting (28) into (27), we get For all \(s\in \mathbb{N}\), put \begin{eqnarray*} \theta_s&=&\|\zeta_s-z\|,\\ \sigma_s&=&(1-\delta)r_s\in (0,1),\\ \lambda_s&=&\varepsilon_s. \end{eqnarray*}Since \(\lim\limits_{s\to\infty}\varepsilon_s=0\), this implies that \(\frac{\lambda_s}{\sigma_s}=\frac{\varepsilon_s}{(1-\delta)r_s}\to 0\) as \(s\to\infty\). Apparently, all the conditions of Lemma 6 are fulfilled. Hence, we have \(\lim\limits_{s\to\infty}\zeta_s=z\).

Conversely, let \(\lim\limits_{s\to\infty}\zeta_s=z\). The we have

From (30), it follows that \(\lim\limits_{s\to\infty}\varepsilon_s=0\). Hence, our new iterative algorithm (12) is stable with respect to \(T\).

5. Rate of convergence

In this section, we show that AU iterative algorithm (12) converges faster than Garodia and Uddin iterative algorithm (11) for contraction mappings.

Theorem 10. Let \(\Gamma\) be a nonempty closed convex subset of a Banach space and \(T:\Gamma\to\Gamma\) be a contraction mapping with fixed point \(z\). For any \(w_0=\psi_0\in\Gamma\), let \(\{w_s\}\) and \(\{\psi_s\}\) be two sequences iteratively generated by (11) and (12) respectively, with real sequence \(\{r_s\}\in [0,1]\) such that \(r\leq r_s< 1\), for some \(r>0\) and \(s\in \mathbb{N}\). Then \(\{\psi_s\}\) converges faster to \(z\) than \(\{w_s\}\) does.

Proof. Recalling the inequality (21), we have

Following our assumption that \(r\leq r_s< 1\), for some \(r>0\) and together with \(s\in \mathbb{N}\), then from (31) we obtain Let Similarly, for any \(z\in F(T)\), it follows from (11) that Using (11) and (34), we obtain Again, from (11) and (35) we get From (36), we obtain the following inequalities: From (37), we have Then from (38) we obtain that Put Now, from (33) and (40) we have that Since \(\lim\limits_{s\to\infty}\frac{\theta_{s+1}}{\theta_{s}}=\lim\limits_{s\to\infty}\frac{\delta^{s+2}}{\delta^{s+1}}=\delta< 1\), so from ratio test we know that \(\sum\limits_{s=0}^{\infty}\theta_s< \infty\). Hence, from (41) we have From the above demonstrations, it implies that \(\{\psi_s\}\) converges at a rate faster than \(\{w_s\}\). Hence, our new iterative algorithm (12) converges faster than Garodia and Uddin iterative algorithm (11). To support the analytical proof of Theorem 10 and to illustrate the efficiency of AU iterative algorithm (12), we will consider the following numerical example.

Example 1. Let \(\Psi=\Re\) and \(\Gamma=[1,50]\). Let \(T:\Gamma\to\Gamma\) be a mapping defined by \(T\psi=\sqrt[3]{2\psi+4}\) for all \(\psi\in \Gamma\). Clearly, \(T\) is contraction and \(z=2\) is a fixed point of \(T\). Take \(\mu_s=r_s=\frac{1}{2}\), with an initial value of 30.

By using the above example, we will show that AU iterative algorithm (12) converges faster a number of leading iterative algorithms in existing literature.

Table 1.A comparison of the different iterative algorithm.

Step	S	MANN	M	NOOR	AU
1	30.000000000	30.000000000	30.000000000	30.000000000	30.000000000
2	3.6809877034	17.000000000	2.2052183845	16.671526557	2.0055900713
3	2.1987214827	10.180987703	2.0032795388	9.7690754673	2.0000025151
4	2.0258396187	6.5399580891	2.0000531287	6.1528499922	2.0000000011
5	2.0034028155	4.5576312697	2.0000008609	4.2363659316	2.0000000000
6	2.0004488682	3.4579473856	2.0000000139	3.2106233814	2.0000000000
7	2.0000592237	2.8381224183	2.0000000002	2.6575629271	2.0000000000
8	2.0000078142	2.4845256350	2.0000000000	2.3578892859	2.0000000000
9	2.0000010310	2.2811112127	2.0000000000	2.1950165817	2.0000000000
10	2.0000001360	2.1634532374	2.0000000000	2.1063366837	2.0000000000
11	2.0000000179	2.0951662880	2.0000000000	2.0580036416	2.0000000000
12	2.0000000024	2.0554515931	2.0000000000	2.0316457914	2.0000000000
13	2.0000000003	2.0323255723	2.0000000000	2.0172673331	2.0000000000
14	2.0000000000	2.0188493597	2.0000000000	2.0094223922	2.0000000000
15	2.0000000000	2.0109929989	2.0000000000	2.0051417581	2.0000000000
16	2.0000000000	2.0064117448	2.0000000000	2.0028058861	2.0000000000

Table 2. A comparison of the different Iterative methods.

Step	ISHIKAWA	GARODIA	ABBASS	THAKUR	AU
1	30.000000000	30.000000000	30.000000000	30.000000000	30.000000000
2	16.680987703	2.0287550076	2.8050437150	2.2052183845	2.0055900713
3	9.7832248288	2.0000774455	2.0456798729	2.0032795388	2.0000025151
4	6.1683753023	2.0000002091	2.0027147499	2.0000531287	2.0000000011
5	4.2509176634	2.0000000006	2.0001617881	2.0000008609	2.0000000000
6	3.2228294683	2.0000000000	2.0000096435	2.0000000139	2.0000000000
7	2.6669712534	2.0000000000	2.0000005748	2.0000000002	2.0000000000
8	2.3646881630	2.0000000000	2.0000000343	2.0000000000	2.0000000000
9	2.1996958627	2.0000000000	2.0000000020	2.0000000000	2.0000000000
10	2.1094405974	2.0000000000	2.0000000001	2.0000000000	2.0000000000
11	2.0600055172	2.0000000000	2.0000000000	2.0000000000	2.0000000000
12	2.0329091715	2.0000000000	2.0000000000	2.0000000000	2.0000000000
13	2.0180511628	2.0000000000	2.0000000000	2.0000000000	2.0000000000
14	2.0099021119	2.0000000000	2.0000000000	2.0000000000	2.0000000000
15	2.0054321205	2.0000000000	2.0000000000	2.0000000000	2.0000000000
16	2.0029800349	2.0000000000	2.0000000000	2.0000000000	2.0000000000

Table 3.A comparison of the different Iterative methods.

Step	SP	CR	AU
1	30.0000000000	30.0000000000	30.000000000
2	6.5399580891	2.9645498633	2.0055900713
3	2.8381224183	2.0693639599	2.0000025151
4	2.1634532374	2.0053107041	2.0000000011
5	2.0323255723	2.0004085862	2.0000000000
6	2.0064117448	2.0000314469	2.0000000000
7	2.0012725149	2.0000024204	2.0000000000
8	2.0002525810	2.0000001863	2.0000000000
9	2.0000501359	2.0000000143	2.0000000000
10	2.0000099517	2.0000000011	2.0000000000
11	2.0000019754	2.0000000001	2.0000000000
12	2.0000003921	2.0000000000	2.0000000000
13	2.0000000778	2.0000000000	2.0000000000
14	2.0000000154	2.0000000000	2.0000000000
15	2.0000000031	2.0000000000	2.0000000000
16	2.0000000006	2.0000000000	2.0000000000
17	2.0000000001	2.0000000000	2.0000000000
18	2.0000000000	2.0000000000	2.0000000000

From the above Tables and Figures, it is clear that AU iterative algorithm converges faster than a number of existing iterative algorithms.

6. Data dependence result

In this section, our focus is on the prove of data dependence result for fixed points of contraction mappings by utilizing AU iterative algorithm (12).

Theorem 11. Let \(\tilde{T}\) be an approximate solution of a contraction mapping \(T\). Let \(\{\psi_s\}\) be a sequence iteratively generated by AU iterative algorithm (12) and define an iterative algorithm \(\{\tilde{\psi_s}\}\) as follows:

where \(\{r_s\}\) is a sequence in \([0,1]\) satisfying the following conditions:

• (i) \(\frac{1}{2}\leq r_s,\,s\in\mathbb{N}\),
• (ii) \(\sum\limits _{s=0}^{\infty}r_s=\infty\).

If \(Tz=z\) and \(\tilde{T}\tilde{z}=\tilde{z}\) such that \(\lim\limits_{s\to\infty}\tilde{\psi}_s=\tilde{z}\), we have where \(\epsilon>0\) is a fixed number.

Proof. From (12) and (43), we have

Putting (45) into (46), we obtain Substituting (47) into (48), we have Putting (49) into (50), we obtain \begin{eqnarray} \nonumber\|\psi_{s+1}-\tilde{\psi}_{s+1}\|&\leq&\delta^4(1-(1-\delta)r_s)\|\psi_s-\tilde{\psi}_s\|+\delta^4 r_s\epsilon+\delta^3\epsilon+\delta^2\epsilon+\delta\epsilon+\epsilon. \label{e7}\end{eqnarray} Since \(r_s\in[0,1]\) and \(\delta\in[0,1)\), it implies that and using our assumption (i), we have \begin{eqnarray*} 1-r_s\leq r_s \Rightarrow 1=1-r_s+r_s\leq r_s+r_s=2r_s. \end{eqnarray*}From (51), we have Let \(\theta_s=\|\psi_{s}-\tilde{\psi}_{s}\|, \,\sigma_s=(1-\delta)r_s,\,\lambda_s=\frac{9\epsilon}{(1-\delta)}\), then from Lemma 7 and (52), we obtain Recalling Theorem 8, we have \(\lim\limits_{s\to\infty}\psi_s=z\) and from the assumption that \(\lim\limits_{s\to\infty}\tilde{\psi_s}=\tilde{z}\) together with (53) we have Hence, AU iterative algorithm (12) is data dependent. This completes the proof of Theorem 6.1

7. Application to a Volterra-Fredholm functional integral equation

In this section, we will use AU iterative algorithm (12) to find the solutions of a nonlinear integral equation.

Many problems of mathematical physics, applied sciences and engineering are reduced to Volterra-Fredholm integral equations (see for example, [21,22] and the references therein).

In 2011, Cracium and Serbian [23] considered and studied the following mixed-type Volterra-Fredholm functional nonlinear integral equation:

where \([u_1;v_1]\times\cdots\times[u_m;v_m]\) is an interval in \(\mathbb{\Re}^m\), \(K,H:[u_1;v_1]\times\cdots\times[u_m;v_m]\times[u_1;v_1]\times\cdots\times[u_m;v_m]\times \mathbb{\Re}\to \mathbb{\Re}\) continuous functions and \(F:[u_1;v_1]\times\cdots\times[u_m;v_m]\times\mathbb{\Re}^3\to\mathbb{\Re}\).

Recently, many authors in nonlinear analysis have constructed some iterative algorithms for approximating the unique solution of the mixed-type Volterra-Fredholm functional nonlinear integral equation (55) in Banach spaces (see for example, [24,25,26] and the references therein).

In this paper, we will prove the strong convergence of AU iterative algorithm (12) to the unique solution of the problem (55). The following theorem which was given by Cracium and Serbian [23] will be of great importance in proving our main results.

Theorem 12.[23] We assume that the following conditions are satisfied:

• (\(B_1\)) \(K,H\in \Gamma([u_1;v_1]\times\cdots\times[u_m;v_m]\times[u_1;v_1]\times\cdots\times[u_m;v_m]\times \mathbb{\Re})\);
• (\(B_2\)) \(F\in([u_1;v_1]\times\cdots\times[u_m;v_m]\times\mathbb{\Re}^3)\);
• (\(B_3\)) there exists nonnegative constants \(\alpha,\beta,\gamma\) such that \begin{eqnarray*} |F(t,f_1,\xi_1,h_1)-F(t,f_2,\xi_2,h_2)|\leq\alpha|f_1-f_2|+\beta|\xi_1-\xi_2|+\gamma|h_1-h_2|, \end{eqnarray*}for all \(t\in[u_1;v_1]\times\cdots\times[u_m;v_m]\), \(f_1,\xi_1,h_1,f_2,\xi_2,h_2\in\mathbb{\Re}\);
• (\(B_4\)) there exist nonnegative constants \(L_K\) and \(L_H\) such that \begin{eqnarray*} |K(t,\rho,f)-K(t,\rho,\xi)|\leq L_K|f-\xi|,\\ |H(t,\rho,f)-H(t,\rho,\xi)|\leq L_H |f-\xi|, \end{eqnarray*}for all \(t,\rho\in [u_1;v_1]\times\cdots\times[u_m;v_m],f,\xi\in\mathbb{\Re}\);
• (\(B_5\)) \(\alpha+(\beta L_K+\gamma L_H)(v_1-u_1)\cdots(v_m-u_m)< 1\). Then, the nonlinear integral equation (55) has a unique solution \(z\in C ([u_1;v_1]\times\cdots\times[u_m;v_m]).\)

We are now ready to prove our main result.

Theorem 13. Assume that all the conditions \((B_1)-(B_5)\) in Theorem 12 are satisfied. Let \(\{\psi_n\}\) be defined by AU iterative algorithm (12) with real sequence \(r_s\in[0,1]\), satisfying \(\sum\limits_{s=1}^{\infty}r_s=\infty\). Then (55) has a unique solution and the AU iterative algorithm (12) converges strongly to the unique solution of the mixed type Volterra-Fredholm functional nonlinear integral equation (55), say \(z\in C([u_1;v_1]\times\cdots\times[u_m;v_m])\).

Proof. We now consider the Banach space \(\Psi=C([u_1;v_1]\times\cdots\times[u_m;v_m],\|\cdot\|_C)\), where \(\|\cdot\|_C\) is the Chebyshev's norm. Let \(\{\psi_n\}\) be the iterative sequence generated by AU iterative algorithm (12) for the operator \(A:\Psi\to \Psi\) define by

Our intention now is to prove that \(\psi_s\to z\) as \(s\to \infty\). Now, by using (12), (55), (56) and the assumptions (\(B_1\))-\((B_5)\), we have that \begin{eqnarray} \nonumber \|g_s-z\|&=&\|A((1-r_s)\psi_s+r_sA\psi_s)-z\|\\ \nonumber &=&|A[(1-r_s)\psi_s+r_sA\psi_s](t)-A(z)(t)|\\ \nonumber&=&|F(t,[(1-r_s)\psi_s+r_sA\psi_s](t),\int_{u_1}^{q_1}\dots\\\nonumber&&\int_{u_m}^{q_m}K(t,\rho,[(1-r_s)\psi_s+r_sA\psi_s](\rho))d\rho,\int_{u_1}^{v_1}\dots\\\nonumber&&\int_{u_m}^{v_m}H(t,\rho,[(1-r_s)\psi_s+r_sA\psi_s](\rho))d\rho)\\&&\nonumber- F\left(t,z(t),\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}K(t,\rho,z(\rho))d\rho,\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}H(t,\rho,z(\rho))d\rho\right)|\\\nonumber \nonumber \end{eqnarray} \begin{eqnarray} \nonumber&\leq&\alpha|[(1-r_s)\psi_s+r_sA\psi_s](t)-z(t)|+\beta\int_{u_1}^{q_1}\dots\\\nonumber&&\int_{u_m}^{q_m}|K(t,\rho,[(1-r_s)\psi_s+r_sA\psi_s](\rho))-K(t,\rho,z(\rho))|d\rho\\&&\nonumber +\gamma\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}|H(t,\rho,[(1-r_s)\psi_s+r_sA\psi_s](\rho))-H(t,\rho,z(\rho))|d\rho \\ \nonumber&\leq&\alpha|[(1-r_s)\psi_s+r_sA\psi_s](t)-z(t)|+\beta\int_{u_1}^{q_1}\dots\\\nonumber&&\int_{u_m}^{q_m}L_K|[(1-r_s)\psi_s+r_sA\psi_s](\rho)-z(\rho)|ds\\&&\nonumber+\gamma\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}L_H|[(1-r_s)\psi_s+r_sA\psi_s](\rho)-z(\rho)|ds\\ \nonumber&\leq&\alpha\|(1-r_s)\psi_s+r_sA\psi_s-z\|+\beta \prod_{i=1}^{m}(v_i-u_i)L_K\|(1-r_s)\psi_s+r_sA\psi_s-p\|\\&&\nonumber+\gamma \prod_{i=1}^{m}(v_i-u_i)L_H\|(1-r_s)\psi_s+r_sA\psi_s-z\| \end{eqnarray} \begin{eqnarray} \nonumber \|(1-r_s)\psi_s&+&r_sA\psi_s-z\|\\\nonumber&\leq&(1-r_n)|\psi_s(t)-z(t)|+r_n|A(\psi_s)(t)-A(p)(t)|\\ \nonumber&=&(1-r_s)|\psi_s(t)-z(t)|\\&&\nonumber+r_s\left|F\left(t,\psi_s(t),\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}K(t,\rho,\psi_s(\rho))d\rho,\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}H(t,\rho,\psi_s(\rho))d\rho\right)\right.\\&&\left.\nonumber- F\left(t,\rho(t),\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}K(t,\rho,z(\rho))d\rho,\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}H(t,\rho,z(\rho))d\rho\right)\right| \end{eqnarray} \begin{eqnarray} \nonumber&\leq&(1-r_s)|\psi_s(t)-z(t)|+r_n\alpha|\psi_s(t)-z(t)|+r_s\beta\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}L_K|\psi_s(\rho)-z(\rho)|d\rho\\&&\nonumber+r_s\gamma\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}L_H|\psi_s(\rho)-z(\rho)|d\rho \end{eqnarray} Putting (59) into (58), we obtain From (12) and (60), we have \begin{eqnarray*} \nonumber \|k_{s}-z\|&=&\|Ag_s-z\|=|A(g_s)(t)-A(z)(t)|\\ \nonumber&=&\left|F\left(t,g_s(t),\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}K(t,\rho,g_s(\rho))d\rho,\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}H(t,\rho,g_s(\rho))d\rho\right)\right.\\&&\nonumber\left.- F\left(t,z(t),\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}K(t,\rho,z(\rho))d\rho,\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}H(t,\rho,z(\rho))d\rho\right)\right| \end{eqnarray*} \begin{eqnarray} \nonumber&\leq&\alpha|g_s(t)-z(t)|+\beta|\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}K(t,\rho,g_s(\rho))d\rho-\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}K(t,\rho,z(\rho))d\rho|\\&&\nonumber +\gamma|\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}H(t,\rho,g_s(\rho))d\rho-\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}H(t,\rho,z(\rho))d\rho|\\ \nonumber&\leq&\alpha|g_s(t)-z(t)|+\beta\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}|K(t,\rho,g_s(\rho))-K(t,\rho,z(\rho))|d\rho\\&&\nonumber +\gamma\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}|H(t,\rho,g_s(\rho))-H(t,\rho,z(\rho))|d\rho \\ \nonumber&\leq&\alpha|g_s(t)-z(t)|+\beta\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}L_K|g_s(\rho)-z(\rho)|d\rho+\gamma\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}L_H|g_s(\rho)-z(\rho)|d\rho \end{eqnarray} \begin{eqnarray} \nonumber&\leq&\alpha\|g_s-z\|+\beta \Pi_{i=1}^{m}(v_i-u_i)L_K\|g_s-z\|+\gamma \Pi_{i=1}^{m}(v_i-u_i)L_H\|g_s-z\|\\ \nonumber &=&[\alpha+(\beta L_K+\gamma L_H)\prod_{i=1}^{m}(v_i-u_i)]\|g_s-z\| \end{eqnarray} From (12) and (61), we have \begin{eqnarray} \nonumber \|\eta_{s}-z\|&=&\|Ak_s-z\|\\ \nonumber &=&|A(k_s)(t)-A(z)(t)|\\ \nonumber&=&|F\left(t,k_s(t),\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}K(t,\rho,k_s(\rho))d\rho,\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}H(t,\rho,k_s(\rho))d\rho\right)\\&&\nonumber- F\left(t,z(t),\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}K(t,\rho,z(\rho))d\rho,\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}H(t,\rho,z(\rho))d\rho\right)|\\\nonumber &\leq&\alpha|k_s(t)-z(t)|+\beta|\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}K(t,\rho,k_s(\rho))d\rho-\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}K(t,\rho,z(\rho))d\rho|\\&&\nonumber +\gamma|\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}H(t,\rho,k_s(\rho))d\rho-\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}H(t,\rho,z(\rho))d\rho|\\ \nonumber&\leq&\alpha|k_s(t)-z(t)|+\beta\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}|K(t,\rho,k_s(\rho))-K(t,\rho,z(\rho))|d\rho\\&&\nonumber +\gamma\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}|H(t,\rho,k_s(\rho))-H(t,\rho,z(\rho))|d\rho \\ \nonumber&\leq&\alpha|k_s(t)-z(t)|+\beta\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}L_K|k_s(\rho)-z(\rho)|d\rho+\gamma\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}L_H|k_s(\rho)-z(\rho)|d\rho \\ \nonumber&\leq&\alpha\|k_s-z\|+\beta \Pi_{i=1}^{m}(v_i-u_i)L_K\|k_s-z\|+\gamma \Pi_{i=1}^{m}(v_i-u_i)L_H\|k_s-z\|\\ \nonumber &=&[\alpha+(\beta L_K+\gamma L_H)\prod_{i=1}^{m}(v_i-u_i)]\|k_s-z\|\\ \end{eqnarray} Again, from (12) and (62), we obtain \begin{eqnarray*} \nonumber \|\eta_{s}-z\|&=&\|Ak_s-z\|=|A(k_s)(t)-A(z)(t)|\\ \nonumber \nonumber&=&\left|F\left(t,k_s(t),\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}K(t,\rho,k_s(\rho))d\rho,\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}H(t,\rho,k_s(\rho))d\rho\right)\right.\\&&\nonumber\left.- F\left(t,z(t),\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}K(t,\rho,z(\rho))d\rho,\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}H(t,\rho,z(\rho))d\rho\right)\right|\\\nonumber &\leq&\alpha|k_s(t)-z(t)|+\beta\left|\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}K(t,\rho,k_s(\rho))d\rho-\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}K(t,\rho,z(\rho))d\rho\right|\end{eqnarray*} \begin{eqnarray}&&\nonumber +\gamma\left|\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}H(t,\rho,k_s(\rho))d\rho-\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}H(t,\rho,z(\rho))d\rho\right|\\ \nonumber&\leq&\alpha|k_s(t)-z(t)|+\beta\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}|K(t,\rho,k_s(\rho))-K(t,\rho,z(\rho))|d\rho\\&&\nonumber +\gamma\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}|H(t,\rho,k_s(\rho))-H(t,\rho,z(\rho))|d\rho\\ \nonumber&\leq&\alpha|k_s(t)-z(t)|+\beta\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}L_K|k_s(\rho)-z(\rho)|d\rho+\gamma\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}L_H|k_s(\rho)-z(\rho)|d\rho \\ \nonumber&\leq&\alpha\|k_s-z\|+\beta \Pi_{i=1}^{m}(v_i-u_i)L_K\|k_s-z\|+\gamma \Pi_{i=1}^{m}(v_i-u_i)L_H\|k_s-z\|\\ \nonumber &=&[\alpha+(\beta L_K+\gamma L_H)\prod_{i=1}^{m}(v_i-u_i)]\|k_s-z\| \end{eqnarray} Finally, from (12) and (63), we obtain \begin{eqnarray} \nonumber \|\psi_{s+1}-z\|&=&\|A\eta_s-z\|\\ \nonumber &=&|A(\eta_s)(t)-A(z)(t)|\\ \nonumber \nonumber&=&\left|F\left(t,\eta_s(t),\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}K(t,\rho,\eta_s(\rho))d\rho,\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}H(t,\rho,\eta_s(\rho))d\rho\right)\right.\\&&\nonumber\left.- F\left(t,z(t),\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}K(t,\rho,z(\rho))d\rho,\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}H(t,\rho,z(\rho))d\rho\right)\right|\\\nonumber &\leq&\alpha|\eta_s(t)-z(t)|+\beta\left|\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}K(t,\rho,\eta_s(\rho))d\rho-\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}K(t,\rho,z(\rho))d\rho\right|\\&&\nonumber +\gamma\left|\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}H(t,\rho,\eta_s(\rho))d\rho-\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}H(t,\rho,z(\rho))d\rho\right|\\ \nonumber&\leq&\alpha|\eta_s(t)-z(t)|+\beta\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}|K(t,\rho,\eta_s(\rho))-K(t,\rho,z(\rho))|d\rho\\&&\nonumber +\gamma\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}|H(t,\rho,\eta_s(\rho))-H(t,\rho,z(\rho))|d\rho\\ \nonumber&\leq&\alpha|\eta_s(t)-z(t)|+\beta\int_{u_1}^{q_1}\dots\int_{u_m}^{q_m}L_K|\eta_s(\rho)-z(\rho)|d\rho+\gamma\int_{u_1}^{v_1}\dots\int_{u_m}^{v_m}L_H|\eta_s(\rho)-z(\rho)|d\rho \\ \notag&\leq&\alpha\|\eta_s-z\|+\beta \Pi_{i=1}^{m}(v_i-u_i)L_K\|\eta_s-z\|+\gamma \Pi_{i=1}^{m}(v_i-u_i)L_H\|\eta_s-z\| \\ \nonumber &=&[\alpha+(\beta L_K+\gamma L_H)\prod_{i=1}^{m}(v_i-u_i)]\|\eta_s-z\| \end{eqnarray} Since from condition (\(B_5\)) we have \([\alpha+(\beta L_K+\gamma L_H)\prod\limits_{i=1}^{m}(v_i-u_i)]< 1\), it follows that \( ([\alpha+(\beta L_K+\gamma L_H)\prod\limits_{i=1}^{m}(v_i-u_i)])^4 < 1\). Thus, (64) reduces to From (65), we have the following inequalities: From (66), we have Since \(r_n\in [0,1]\) for all \(n\in \mathbb{N}\) and recalling from assumption \((B_5)\) that \([\alpha+(\beta L_K+\gamma L_H)\prod_{i=1}^{m}(v_i-u_i)]< 1\), then we have From classical analysis, it is that \(1-\psi\leq e^{-\psi}\) for all \(\psi\in [0,1]\), thus from (67), we have Taking the limit of both sides of the above inequalities, we have \( \lim\limits_{s\to \infty}\|\psi_s-z\|=0\). Hence, (12) converges strongly to the unique solution of the mixed type Volterra-Fredholm functional nonlinear integral equation (55).

7. Conclusion

In this paper, we have proved that our new iterative algorithm (12) outperforms several well known iterative algorithms in the literature in terms of rate of convergence. The stability result of AU iterative algorithm has also been obtained. We have also shown that AU iterative algorithm (12) is data dependent. Finally, to illustrate the efficiently of AU iterative algorithm (12), we have proved approximated the unique solution of a nonlinear integral equation. Hence, our results are generalization and improvements of several well known results in the existing literature.

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflicts of Interest

''The authors declare no conflict of interest.''

1. Introduction

Suppose \(g:[a_1,a_2]\to \mathbb{R}\) is a four times continuously differentiable mapping on \((a_1,a_2)\) and \(||g^{(4)}||_{\infty}=\displaystyle\sup_{x\in (a_1,a_2)}|g^{(4)}(x)|< \infty\). Then the following inequality

\begin{equation*} \begin{aligned} &\Big|\frac{1}{3}\Big[\frac{g(a_1)+g(a_2)}{2}+2g\Big(\frac{a_1+a_2}{2}\Big)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\Big|\le\frac{1}{2880}||g^{(4)}||_{\infty}(a_2-a_1)^4 \end{aligned} \end{equation*} holds, and it is well known in the literature as Simpson's inequality, named after the English mathematician Thomas Simpson. If the mapping \(g\) is neither four times differentiable nor is the fourth derivative \(g^{(4)}\) bounded on \((a_1, a_2)\), then we cannot apply the classical Simpson quadrature formula. In [1] Dragomir et al., proved some recent developments on Simpson's inequality for which the remainder is expressed in terms of lower derivatives than the fourth. For recent refinements, counterparts, generalizations and new Simpson's type inequalities, see [1,2,3,4].

The classical convexity of functions is a fundamental notions in mathematics, they have widely applications in many branches of mathematics and physics. A function \(g:I\subset \mathbb{R}\to \mathbb{R}\) is said to be a convex in the classical sense, if \[g(ta_2+(1-t)a_1)\le tg(a_2)+(1-t)g(a_1)\] for all \(a_1,a_2 \in I\) and \(t \in [0,1]\). We call \(g\) concave if the inequality is reversed. A somewhat generalization of the above definition is given by Awan et al., [5] as follows:

Definition 1([5]). A function \(g:I\subset \mathbb{R}\to \mathbb{R}\) is called exponentially convex, if

for all \(a_1, a_2 \in I\), \(t\in [0,1]\), and \(\alpha \in \mathbb{R}\). If (1) holds in the reversed sense, then \(g\) is said to be exponentially concave.

For example, the function \(g:\mathbb{R} \to \mathbb{R}\), defined by \(g(x)=-x^2\) is a concave function, thus this function is exponentially convex for all \(\alpha < 0\). An exponentially convex function on a closed interval is bounded, it also satisfies the Lipschitzian condition on any closed interval \([a_1,a_2]\subset \mathring{I}\) (interior of I). Therefore an exponentially convex function is absolutely continuous on \([a_1,a_2]\subset \mathring{I}\) and continuous on \(\mathring{I}\).

Recently, Nie et al., [6] introduced the notion of exponentially quasi-convex as thus:

Definition 2([6]). Let \(\alpha \in \mathbb{R}\). Then a mapping \(g:I\subset \mathbb{R}\to \mathbb{R}\) is said to be exponentially quasi-convex if \begin{eqnarray*}\label{ec 1.4} g(ta_2+(1-t)a_1)\le \max\left\{\frac{g(a_2)}{e^{\alpha a_2}}, \frac{g(a_1)}{e^{\alpha a_1}}\right\} \end{eqnarray*} for all \(a_1, a_2 \in I\), and \(t\in [0,1]\).

Remark 1. Note that if \(\alpha =0\), then the classes of exponentially convex and quasi-convex functions reduce to the classes of classical convex and quasi-convex function.

Inspired by the work of Sarikaya et al., [7] and Vivas et al., [8], we aim to establish new Simpson's type results for the class of functions whose derivatives in absolute value at certain powers are exponentially convex and exponentially quasi-convex. By taking \(\alpha=0\), we recapture some already established results in the literature. The main results are framed and justified in section symbol §2, followed by applications of our results to some special means in section symbol §3, and Simpson's quadrature in section symbol §4.

2. Main results

For the proof of our main results, the following lemma will be useful.

Lemma 1([7]). Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g''\in L_1[a_1,a_2]\), where \(a_1,a_2\in I\) with \(a_1< a_2\). Then the following equality holds: \begin{equation*} \begin{aligned} &\frac{1}{6}\left[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\right]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)\,dx \\ &=(a_2-a_1)^2\int_{0}^{1}k(t)g''(ta_2+(1-t)a_1)\,dt, \end{aligned} \end{equation*} where \begin{equation*} k(t)= \begin{cases} \frac{t}{2}\left(\frac{1}{3}-t\right)\quad \mbox if \quad t\in \left[0,\frac{1}{2}\right)\\[1ex] (1-t)\left(\frac{t}{2}-\frac{1}{3}\right) \quad \mbox if \quad t\in \left[\frac{1}{2},1\right]. \end{cases} \end{equation*}

2.1. Simpson's inequality for exponentially convex

We now give a new refinement of Simpson's inequality for twice differentiable functions:

Theorem 1. Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g''\in L_1[a_1,a_2]\), where \(a_1,a_2 \in I\) with \(a_1< a_2\). If \(|g''|\) is exponentially convex on \([a_1,a_2]\), then the following inequality holds: \begin{equation*}\label{4} \begin{aligned} &\left|\frac{1}{6}\Big[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\right|\le \frac{(a_2-a_1)^2}{162}\left[\frac{\left|g''(a_2)\right|}{e^{\alpha a_2}}+ \frac{\left|g''(a_1)\right|}{e^{\alpha a_1}}\right]. \end{aligned} \end{equation*}

Proof. Using Lemma 1 and since \(|g''|\) is exponentially convex, we have \begin{align*} \Big|\frac{1}{6}\Big[g(a_1)&+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\Big|\\ \leq& (a_2-a_1)^2\int_0^1|k(t)|\left|g''(ta_2+(1-t)a_1)\right|dt \\ \le& (a_2-a_1)^2\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|\Big(t\frac{\left|g''(a_2)\right|}{e^{\alpha a_2}}+ (1-t)\frac{\left|g''(a_1)\right|}{e^{\alpha a_1}} \Big)dt\\ &+ (a_2-a_1)^2\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|\Big(t\frac{\left|g''(a_2)\right|}{e^{\alpha a_2}}+ (1-t)\frac{\left|g''(a_1)\right|}{e^{\alpha a_1}} \Big)dt\\ =& (a_2-a_1)^2(I_1+I_2), \end{align*} where \begin{align*} I_1=&\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|\left(t\frac{\left|g''(a_2)\right|}{e^{\alpha a_2}}+ (1-t)\frac{\left|g''(a_1)\right|}{e^{\alpha a_1}} \right)\,dt\\ =&\frac{\left|g''(a_2)\right|}{e^{\alpha a_2}}\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|tdt+\frac{\left|g''(a_1)\right|}{e^{\alpha a_1}}\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|(1-t)\,dt\\ =&\frac{59}{31104}\frac{\left|g''(a_2)\right|}{e^{\alpha a_2}}+\frac{133}{31104}\frac{\left|g''(a_1)\right|}{e^{\alpha a_1}}, \end{align*} and \begin{align*} I_2=&\int_{\frac{1}{2}}^{1}\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|\Big(t\frac{\left|g''(a_2)\right|}{e^{\alpha a_2}}+ (1-t)\frac{\left|g''(a_1)\right|}{e^{\alpha a_1}} \Big)dt\\ =&\frac{\left|g''(a_2)\right|}{e^{\alpha a_2}}\int_{\frac{1}{2}}^{1}\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|tdt+\frac{\left|g''(a_1)\right|}{e^{\alpha a_1}}\int_{\frac{1}{2}}^{1}\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|(1-t)dt\\ =&\frac{133}{31104}\frac{\left|g''(a_2)\right|}{e^{\alpha a_2}}+\frac{59}{31104}\frac{\left|g''(a_1)\right|}{e^{\alpha a_1}}, \end{align*} which completes the proof.

Corollary 1. Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g''\in L_1[a_1,a_2]\), where \(a_1,a_2 \in I\) with \(a_1< a_2\). If \(g(a_1)=g\Big(\frac{a_1+a_2}{2}\Big)=g(a_2)\) and \(|g''|\) is exponentially convex on \([a_1,a_2]\), then the following inequality holds: \begin{eqnarray*} \Big|\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx-g\Big(\frac{a_1+a_2}{2}\Big)\Big| \le \frac{(a_2-a_1)^2}{162}\left[\frac{\left|g''(a_2)\right|}{e^{\alpha a_2}}+ \frac{\left|g''(a_1)\right|}{e^{\alpha a_1}}\right]. \end{eqnarray*}

Remark 2. In Theorem 1, by letting \(\alpha=0\) we get [7,Theorem 2.2].

Theorem 2. Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g''\in L_1[a_1,a_2]\), where \(a_1,a_2 \in I\) with \(a_1< a_2\). If \(|g''|^q\) is exponentially convex on \([a_1,a_2]\) and \(q\ge 1\), then the following inequality holds: \begin{equation*}\label{5} \begin{aligned} &\left|\frac{1}{6}\Big[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\right|\\ &\le (a_2-a_1)^2\left(\frac{1}{162}\right)^{1-\frac{1}{q}}\left[\left(\frac{59}{31104}\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^q+\frac{133}{31104}\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^q\right)^{\frac{1}{q}}\right.\\ &\quad\quad\left.+ \left(\frac{133}{31104}\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^q+\frac{59}{31104}\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^q\right)^{\frac{1}{q}}\right]. \end{aligned} \end{equation*}

Proof. Suppose that \(q\ge 1\). From Lemma 1, we have \begin{equation*} \begin{aligned} &\Big|\frac{1}{6}\Big[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+f(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\Big|\\ &\le (a_2-a_1)^2\int_0^1|k(t)|\left|g''(ta_2+(1-t)a_1)\right|dt \\ &=(a_2-a_1)^2\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|\left|g''(ta_2+(1-t)a_1)\right|dt\\ &\quad+ (a_2-a_1)^2\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|\left|g''(ta_2+(1-t)a_1)\right|dt. \end{aligned} \end{equation*} Using the Hölder's inequality for functions \[\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|^{1-\frac{1}{q}}\] and \[\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|^{\frac{1}{q}}\left|g''(ta_2+(1-t)a_1)\right|,\] for the first integral, and the functions \[ \left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|^{1-\frac{1}{q}}\] and \[ \left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|^{\frac{1}{q}}\left|g''(ta_2+(1-t)a_1)\right|,\] for the second integral, from the above relation we get the inequality: \begin{equation*} \begin{aligned} &\Big|\frac{1}{6}\Big[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\Big|\\ &\leq (a_2-a_1)^2\Big(\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|dt\Big)^{1-\frac{1}{q}}\\ &\quad\times\left(\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|\left|g''(ta_2+(1-t)a_1)\right|^qdt\right)^{\frac{1}{q}}\\ &\quad+ (a_2-a_1)^2\Big(\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|\,dt\Big)^{1-\frac{1}{q}}\\ &\quad\times \Big(\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|\left|g''(ta_2+(1-t)a_1)\right|^qdt\Big)^{\frac{1}{q}}. \end{aligned} \end{equation*} Using the fact that \[\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|dt=\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|dt=\frac{1}{162}\] and the exponentially convexity of \(|g''|^q\), we have

and From (2) and (3), we have \begin{equation*} \begin{aligned} &\Big|\frac{1}{6}\Big[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+\Psi(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\Big|\\ &\le(a_2-a_1)^2\Big(\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|dt\Big)^{1-\frac{1}{q}}\left(\frac{59}{31104}\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^q+\frac{133}{31104}\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^q\right)^{\frac{1}{q}}\\ &\quad+ (a_2-a_1)^2\Big(\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|dt\Big)^{1-\frac{1}{q}} \left(\frac{133}{31104}\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^q+\frac{59}{31104}\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^q\right)^{\frac{1}{q}}\\ &=(a_2-a_1)^2\left(\frac{1}{162}\right)^{1-\frac{1}{q}}\left[\left(\frac{59}{31104}\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^q+\frac{133}{31104}\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^q\right)^{\frac{1}{q}}\right.\\ &\left.\quad+ \left(\frac{133}{31104}\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^q+\frac{59}{31104}\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^q\right)^{\frac{1}{q}}\right]. \end{aligned} \end{equation*} This completes the proof.

Corollary 2. Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g''\in L_1[a_1,a_2]\), where \(a_1,a_2 \in I\) with \(a_1< a_2\). If \(g(a_1)=g\Big(\frac{a_1+a_2}{2}\Big)=g(a_2)\) and \(|g''|^q\) is exponentially convex on \([a_1,a_2]\) and \(q\ge 1\), then the following inequality holds: \begin{equation*} \begin{aligned} &\Big|\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx-g\Big(\frac{a_1+a_2}{2}\Big)\Big| \\ &\le (a_2-a_1)^2\left(\frac{1}{162}\right)^{1-\frac{1}{q}}\left[\left(\frac{59}{31104}\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^q+\frac{133}{31104}\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^q\right)^{\frac{1}{q}}\right.\\ &\left.\quad+ \left(\frac{133}{31104}\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^q+\frac{59}{31104}\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^q\right)^{\frac{1}{q}}\right]. \end{aligned} \end{equation*}

Remark 3. By setting \(\alpha =0\) in Theorem 2, we recapture [7,Theorem 2.5].

Corollary 3. Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g''\in L_1[a_1,a_2]\), where \(a_1, a_2 \in I\) with \(a_1< a_2\). If \(g(a_1)=g\Big(\frac{a_1+a_2}{2}\Big)=g(a_2)\) and \(|g''|^2\) is exponentially convex on \([a_1,a_2]\), then the following inequality holds: \begin{equation*} \begin{aligned} &\left|\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)\,dx-g\Big(\frac{a_1+a_2}{2}\Big)\right| \\ &\le (a_2-a_1)^2\left(\frac{1}{162}\right)^{\frac{1}{2}}\left[\left(\frac{59}{31104}\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^2+\frac{133}{31104}\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^2\right)^{\frac{1}{2}}\right.\\ &\left.\quad+ \left(\frac{133}{31104}\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^2+\frac{59}{31104}\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^2\right)^{\frac{1}{2}}\right]. \end{aligned} \end{equation*}

2.2. Simpson's inequality for exponentially quasi-convex

Theorem 3. Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g''\in L_1[a_1,a_2]\), where \(a_1,a_2 \in I\) with \(a_1< a_2\). If \(|g''|\) is exponentially quasi-convex on \([a_1,a_2]\), then the following inequality holds: \begin{equation*} \begin{aligned} &\left|\frac{1}{6}\Big[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\right| \\ &\le \frac{(a_2-a_1)^2}{81}\max\left\{\frac{|g''(a_2)|}{e^{\alpha a_2}},\frac{|g''(a_1)|}{e^{\alpha a_1}}\right\}. \end{aligned} \end{equation*}

Proof. From Lemma 1 and by using the exponentially quasi-convexity of \(|g''|\), we get \begin{equation*} \begin{aligned} &\Big|\frac{1}{6}\Big[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\Big|\\ &\leq (a_2-a_1)^2\int_0^1|k(t)|\left|g''(ta_2+(1-t)a_1)\right|dt \\ &\le (a_2-a_1)^2\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|\max\left\{\frac{|g''(a_2)|}{e^{\alpha a_2}},\frac{|g''(a_1)|}{e^{\alpha a_1}}\right\}dt\\ &\quad+ (a_2-a_1)^2\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|\max\left\{\frac{|g''(a_2)|}{e^{\alpha a_2}},\frac{|g''(a_1)|}{e^{\alpha a_1}}\right\}dt\\ &=(a_2-a_1)^2(I_1+I_2), \end{aligned} \end{equation*} where \begin{equation*} \begin{aligned} I_1&=\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|\max\left\{\frac{|g''(a_2)|}{e^{\alpha a_2}},\frac{|g''(a_1)|}{e^{\alpha a_1}}\right\}dt\\ &=\max\left\{\frac{|g''(a_2)|}{e^{\alpha a_2}},\frac{|g''(a_1)|}{e^{\alpha a_1}}\right\}\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|dt\\ &=\frac{1}{162}\max\left\{\frac{|g''(a_2)|}{e^{\alpha a_2}},\frac{|g''(a_1)|}{e^{\alpha a_1}}\right\}, \end{aligned} \end{equation*} and \begin{equation*} \begin{aligned} I_2&=\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|\max\left\{\frac{|g''(a_2)|}{e^{\alpha a_2}},\frac{|g''(a_1)|}{e^{\alpha a_1}}\right\}dt\\ &=\max\left\{\frac{|g''(a_2)|}{e^{\alpha a_2}},\frac{|g''(a_1)|}{e^{\alpha a_1}}\right\}\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|dt\\ &=\frac{1}{162}\max\left\{\frac{|g''(a_2)|}{e^{\alpha a_2}},\frac{|g''(a_1)|}{e^{\alpha a_1}}\right\}, \end{aligned} \end{equation*} which completes the proof.

Corollary 4. Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g''\in L_1[a_1,a_2]\), where \(a_1,a_2 \in I\) with \(a_1< a_2\). If \(g(a_1)=g\Big(\frac{a_1+a_2}{2}\Big)=g(a_2)\) and \(|g''|\) is exponentially quasi-convex on \([a_1,a_2]\), then the following inequality holds: \begin{equation*} \begin{aligned} \left|\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx-g\Big(\frac{a_1+a_2}{2}\Big)\right| \le \frac{(a_2-a_1)^2}{81}\max\left\{\frac{|g''(a_2)|}{e^{\alpha a_2}},\frac{|g''(a_1)|}{e^{\alpha a_1}}\right\}. \end{aligned} \end{equation*}

Theorem 4. Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g''\in L_1[a_1,a_2]\), where \(a_1, a_2 \in I\) with \(a_1< a_2\). If \(|g''|^q\) is exponentially quasi-convex on \([a_1,a_2]\) and \(q\ge 1\), then the following inequality holds: \begin{equation*} \begin{aligned} &\left|\frac{1}{6}\Big[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\right| \\ &\le \frac{(a_2-a_1)^2}{81}\left(\max\left\{\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^q,\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^q\right\}\right)^{\frac{1}{q}}. \end{aligned} \end{equation*}

Proof. Suppose that \(q\ge 1\). From Lemma 1, we have \begin{equation*} \begin{aligned} &\left|\frac{1}{6}\Big[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\right|\\ &\le(a_2-a_1)^2\int_0^1|k(t)|\left|g''(ta_2+(1-t)a_1)\right|dt \\ &=(a_2-a_1)^2\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|\left|g''(ta_2+(1-t)a_1)\right|dt\\ &\quad+ (a_2-a_1)^2\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|\left|g''(ta_2+(1-t)a_1)\right|dt. \end{aligned} \end{equation*} Using the Hölder's inequality for functions \[\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|^{1-\frac{1}{q}}\] and \[\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|^{\frac{1}{q}}\left|g''(ta_2+(1-t)a_1)\right|\] for the first integral and the functions \[ \left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|^{1-\frac{1}{q}}\] and \[ \left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|^{\frac{1}{q}}\left|g''(ta_2+(1-t)a_1)\right|,\] for the second integral, from the above relation we get the inequalities: \begin{equation*} \begin{aligned} &\left|\frac{1}{6}\Big[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\right|\\ &\le (a_2-a_1)^2\Big(\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|dt\Big)^{1-\frac{1}{q}}\\ &\quad\times\Big(\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|\left|g''(ta_2+(1-t)a_1)\right|^qdt\Big)^{\frac{1}{q}}\\ &\quad+(a_2-a_1)^2\Big(\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|dt\Big)^{1-\frac{1}{q}}\\ &\quad\times\Big(\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|\left|g''(ta_2+(1-t)a_1)\right|^qdt\Big)^{\frac{1}{q}}. \end{aligned} \end{equation*} Since \(|g''|^q\) is exponentially quasi-convex, therefore we have

and From (4) and (5), we have \begin{equation*} \begin{aligned} &\left|\frac{1}{6}\Big[g(a_1)+4g\Big(\frac{a_1+a_2}{2}\Big)+g(a_2)\Big]-\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx\right|\\ &\le (a_2-a_1)^2\Big(\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|dt\Big)^{1-\frac{1}{q}}\left(\frac{1}{162}\max\left\{\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^q,\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^q\right\}\right)^{\frac{1}{q}}\\ &\quad+ (a_2-a_1)^2\Big(\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|dt\Big)^{1-\frac{1}{q}} \left(\frac{1}{162}\max\left\{\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^q,\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^q\right\}\right)^{\frac{1}{q}}\\ &= 2(a_2-a_1)^2\left(\frac{1}{162}\right)^{1-\frac{1}{q}}\left(\frac{1}{162}\right)^{\frac{1}{q}}\left(\max\left\{\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^q,\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^q\right\}\right)^{\frac{1}{q}}\\ &= \frac{(a_2-a_1)^2}{81}\left(\max\left\{\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^q,\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^q\right\}\right)^{\frac{1}{q}}, \end{aligned} \end{equation*} where we used the fact that \[\int_0^{\frac{1}{2}}\left|\frac{t}{2}\left(\frac{1}{3}-t\right)\right|dt=\int_{\frac{1}{2}}^1\left|(1-t)\left(\frac{t}{2}-\frac{1}{3}\right)\right|dt=\frac{1}{162}.\] The proof is complete.

Corollary 5. Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g''\in L_1[a_1,a_2]\), where \(a_1,a_2 \in I\) with \(a_1< a_2\). If \(g(a_1)=g\Big(\frac{a_1+a_2}{2}\Big)=g(a_2)\) and \(|g''|^q\) is exponentially quasi-convex on \([a_1,a_2]\) and \(q\ge 1\), then the following inequality holds: \begin{equation*} \begin{aligned} &\left|\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx-g\Big(\frac{a_1+a_2}{2}\Big)\right| \le \frac{(a_2-a_1)^2}{81}\left(\max\left\{\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^q,\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^q\right\}\right)^{\frac{1}{q}}. \end{aligned} \end{equation*}

Corollary 6. Let \(g:I\subset \mathbb{R}\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g''\in L_1[a_1,a_2]\), where \(a_1,a_2 \in I\) with \(a_1< a_2\). If \(g(a_1)=g\Big(\frac{a_1+a_2}{2}\Big)=g(a_2)\) and \(|g''|^2\) is exponentially quasi-convex on \([a_1,a_2]\), then the following inequality holds: \begin{equation*} \begin{aligned} &\Big|\frac{1}{a_2-a_1}\int_{a_1}^{a_2}g(x)dx-g\Big(\frac{a_1+a_2}{2}\Big)\Big| \le \frac{(a_2-a_1)^2}{81}\left(\max\left\{\left|\frac{g''(a_2)}{e^{\alpha a_2}}\right|^2,\left|\frac{g''(a_1)}{e^{\alpha a_1}}\right|^2\right\}\right)^{\frac{1}{2}}. \end{aligned} \end{equation*}

3. Applications to special means

We now consider the following special means for positive real numbers \(a_1\) and \(a_2\).

1. The arithmetic mean: \(\mathcal{A}=\mathcal{A}(a_1,a_2)=\frac{a_1+a_2}{2}.\)
2. The harmonic mean: \[ \mathcal{H}=\mathcal{H}(a_1,a_2)=\frac{2a_1a_2}{a_1+a_2}.\]
3. The logarithmic mean: \[\mathcal{L}=\mathcal{L}(a_1,a_2)=\left\{ \begin{array}{lcl} a_1 & \mbox{ if } & a_1=a_2 \\ & & \\ \frac{a_2-a_1}{\ln a_2-\ln a_1} & \mbox{ if } & a_1\neq a_2. \end{array} \right. \]
4. The \(p\)-logarithmic mean: \[\mathcal{L}_p=\mathcal{L}_p(a_1,a_2)=\left\{ \begin{array}{lcl} \Big[\frac{a_2^{p+1}-a_1^{p+1}}{(p+1)(a_2-a_1)}\Big]^{\frac{1}{p}} & \mbox{ if } & a_1\neq a_2 \\ & & \\ a_1 & \mbox{ if } & a_1=a_2 \end{array} \right. , \ \ p\in \mathbb{R}\setminus\{-1,0\}.\]

It is well know that \(\mathcal{L}_p\) is monotonic nondecreasing over \(p\in \mathbb{R}\) with \(\mathcal{L}_{-1}=\mathcal{L}\). In particular, we have the following inequalities \[ \mathcal{H}\le \mathcal{L}\le \mathcal{A}\] Some new inequalities are derived for the above means.

Proposition 1. Let \(a_1, a_2 \in \mathbb{R}\), \(0< a_1< a_2\). Then, we have \begin{eqnarray*}\label{eq 12} \left|\frac{1}{3}\mathcal{A}(a_1^4,a_2^4)+\frac{2}{3}\mathcal{A}^4(a_1,a_2)-\mathcal{L}_4^4(a_1,a_2)\right|\le \frac{2(a_2-a_1)^2}{27}\left[\frac{a_2^2}{e^{\alpha a_2}}+\frac{a_1^2}{e^{\alpha a_1}}\right]. \end{eqnarray*}

Proof. The assertion follows from Theorem 1 and a simple computation applied to \(g(x)=\frac{x^4}{12}, \ \ x\in [a_1, a_2]\), where \(|g''|\) is exponentially convex mapping.

Proposition 2. Let \(a_1, a_2 \in \mathbb{R}\), \(0< a_1< a_2\). Then, we have \begin{eqnarray*} \left|\frac{1}{3}\mathcal{A}(a_1^{r+1},a_2^{r+1})+\frac{2}{3}\mathcal{A}^{r+1}(a_1, a_2)-\mathcal{L}_{r+1}^{r+1}(a_1, a_2)\right|\le \frac{r(r+1)(a_2-a_1)^2}{81}\max\left\{\frac{a_2^{r-1}}{e^{\alpha a_2}},\frac{a_1^{r-1}}{e^{\alpha a_1}}\right\}. \end{eqnarray*}

Proof. This time we use Theorem 3 and a simple computation applied to \(g(x)=\frac{x^{r+1}}{r+1}, \ r\ge 1, \ \ x\in [a_1,a_2]\). Here, the function \(|g''(x)|=rx^{r-1}\) is increasing and exponentially quasi-convex.

4. Applications to Simpson's formula

Let \(g:[a_1,a_2]\to \mathbb{R}\) and \(\mathbf{P}\) be a partition of the interval \([a_1,a_2]\); i.e. \[\mathbf{P}: a_1=s_0< s_1< \cdots < s_{n-1}< s_n=a_2;\quad\quad h_i=\frac{(s_{i+1}-s_i)}{2}.\] Now, for the given Simpson's quadrature: \begin{eqnarray*}\label{eq 16} S(g,\mathbf{P})=\sum_{i=0}^{n-1}\frac{g(s_i)+4g(s_i+h_i)+g(s_{i+1})}{3}h_i, \end{eqnarray*} it is well known that if \(g\) is differentiable such that \(g^{(4)}(x)\) exist on \((a_1,a_2)\) and \(K=\displaystyle\max_{x\in [a_1,a_2]}|g^{(4)}(x)|< \infty\), then where the approximation error \(E_s(g,\mathbf{P})\) of the integral \(I\) by Simpson's formula \(S(g,\mathbf{P})\) satisfies It is clear that if the function \(g\) is not four times differentiable or the fourth derivative is not bounded on \((a_1,a_2)\), then (7) cannot be applied.

Theorem 5. Let \(g:I\subset [0,\infty)\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g''\in L_1[a_1,a_2]\), where \(a_1,a_2\in I\) with \(a_1< a_2\). If \(|g''|\) is exponentially convex on \([a_1,a_2]\), then in (6) for every division \(\mathbf{P}\) of \([a_1,a_2]\), the following holds: \begin{eqnarray*} |E_s(g,\mathbf{P})|\le \frac{1}{162}\sum_{i=0}^{n-1}(s_{i+1}-s_i)^3\left[\frac{\left|g''(s_{i+1})\right|}{e^{\alpha s_{i+1}}}+ \frac{\left|g''(s_{i})\right|}{e^{\alpha s_i}}\right]. \end{eqnarray*}

Proof. Applying Theorem 1 on the subintervals \([s_i,s_{i+1}],\ (i=0,1,2,\cdots, n-1)\) of the division \(\mathbf{P}\), we get \begin{eqnarray*} &&\left|\frac{(s_{i+1}-s_i)}{6}\left[g(s_i)+4g\left(\frac{s_{i+1}-s_i}{2}\right)+g(s_{i+1})\right]-\int_{s_i}^{s_{i+1}}g(s)ds\right|\\ &&\;\;\;\;\le \frac{(s_{i+1}-s_i)^3}{162}\left[\frac{\left|g''(s_{i+1})\right|}{e^{\alpha s_{i+1}}}+ \frac{\left|g''(s_{i})\right|}{e^{\alpha s_i}}\right]. \end{eqnarray*} Adding over \(i\) for \(0\) to \(n-1\) and taking into account that \(|g''|\) is exponentially convex, we have: \begin{eqnarray*} \left|S(g,\mathbf{P})-\int_{a_1}^{a_2}g(s)ds\right|\le \sum_{i=0}^{n-1}\frac{(s_{i+1}-s_i)^3}{162}\left[\frac{\left|g''(s_{i+1})\right|}{e^{\alpha s_{i+1}}}+ \frac{\left|g''(s_{i})\right|}{e^{\alpha s_i}}\right], \end{eqnarray*} which completes the proof.

Corollary 7. If \(\alpha =0\), we get \begin{eqnarray*} |E_s(g,\mathbf{P})|\le \frac{1}{162}\sum_{i=0}^{n-1}(s_{i+1}-s_i)^3\left[g''(s_i)+ g''(s_{i+1})\right]. \end{eqnarray*}

Theorem 6. Let \(g:I\subset [0,\infty)\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g''\in L_1[a_1,a_2]\), where \(a_1, a_2\in I\) with \(a_1< a_2\). If \(|g''|\) is exponentially quasi-convex on \([a_1,a_2]\), then in (6) for every division \(\mathbf{P}\) of \([a_1,a_2]\), the following holds: \begin{eqnarray*} \left|E_s(g,\mathbf{P})\right|\le \frac{1}{81}\sum_{i=0}^{n-1}(s_{i+1}-s_i)^3\max\left\{\frac{|g''(s_{i+1})|}{e^{\alpha s_{i+1}}},\frac{|g''(s_i)|}{e^{\alpha s_i}}\right\}. \end{eqnarray*}

Proof. Applying Theorem 3 and proceeding as in the proof of Theorem 5, we obtain the desired result.

Proposition 3. Let \(g:I\subset [0,\infty)\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g''\in L_1[a_1,a_2]\), where \(a_1,a_2\in I\) with \(a_1< a_2\). If \(|g''|^q\) is exponentially convex on \([a_1,a_2]\) and \(q\ge 1\), the following holds: \begin{eqnarray*} |E_s(g,\mathbf{P})|\le \left(\frac{1}{162}\right)^{1-\frac{1}{q}}\sum_{i=0}^{n-1}(s_{i+1}-s_i)^3\Big[M_{\alpha}^q(g''(s_i), g''(s_{i+1}))\Big], \end{eqnarray*} where \begin{align*} M_{\alpha}^q\left(g''(s_i), g''(s_{i+1})\right)&=\left(\frac{59}{31104}\left|\frac{g''(s_{i+1})}{e^{\alpha s_{i+1}}}\right|^q+\frac{133}{31104}\left|\frac{g''(s_i)}{e^{\alpha s_i}}\right|^q\right)^{\frac{1}{q}}\\ &\quad+ \left(\frac{133}{31104}\left|\frac{g''(s_{i+1})}{e^{\alpha s_{i+1}}}\right|^q+\frac{59}{31104}\left|\frac{g''(s_i)}{e^{\alpha s_i}}\right|^q\right)^{\frac{1}{q}}. \end{align*}

Proof. The proof is immediate by using Theorem 2.

Proposition 4. Let \(g:I\subset [0,\infty)\to \mathbb{R}\) be a twice differentiable mapping on \(I^{\circ}\) such that \(g''\in L_1[a_1,a_2]\), where \(a_1,a_2\in I\) with \(a_1< a_2\). If \(|g''|^q\) is exponentially quasi-convex on \([a_1,a_2]\) and \(q\ge 1\), the following holds: \begin{eqnarray*} |E_s(g,\mathbf{P})|\le \frac{1}{81}\sum_{i=0}^{n-1}(s_{i+1}-s_i)^3\left(\max\left\{\left|\frac{g''(s_{i+1})}{e^{\alpha s_{i+1}}}\right|^q,\left|\frac{g''(s_i)}{e^{\alpha s_i}}\right|^q\right\}\right)^{\frac{1}{q}}. \end{eqnarray*}

Proof. The proof follows by applying Theorem 4.

5. Conclusion

Several inequalities of the Simpson's type for exponentially convex and exponentially quasi-convex functions are hereby established. For \(\alpha=0\), we recapture results in [7]. Furthermore, we presented some applications to special means and to Simpson's formula. We look forward to further investigation in this direction.

Acknowledgments

Many thanks to the referee for his/her comments. Thanks to the Dirección de investigación from Pontificia Universidad Católica del Ecuador for technical support to our research project entitled: ``Algunas generalizaciones de desigualdades integrales".

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflicts of Interest

''The authors declare no conflict of interest.''

1. Introduction

One of the main problems in the theory of ordinary differential equations is the study of their limit cycles, their existence, their number and their stability, these properties of limit cycles were studied extensively by mathematicians and physicists, and more recently also by chemists, biologists, economists, etc. A limit cycle of a differential equation is a periodic orbit in the set of all isolated periodic orbits of the differential equation.

The second part of the \(16\)th Hilbert's problem [1] is related to the least upper bound on the number of limit cycles of polynomial vector fields having a fixed degree. The study of differential equations or planar differential systems has been considered by several authors. In [2] the authors studied the limit cycles for a variant of a generalized Riccati equation. Mathieu, in [3] considered the second order differential equation

\begin{equation*} \ddot{x}+b(1+\cos t)x=0, \end{equation*} where b is a real constant. It is called Mathieu equation, which is the simplest mathematical model of an excited system on a parameter. We also recall the Ermakov-Pinney equation which is the Mathieu-Duffing type equation \begin{equation*} \ddot{x}+b(1+\cos t)x-x^\beta=0, \end{equation*} where \(\beta\) is an integer and \(b>0\). The last two equations modeled the dynamics of a system with harmonic parametric excitation and a nonlinear term corresponding to a restoring force, see [4,5].

There are several methods exist to study the number of limit cycles that bifurcate from the periodic orbits such as the integrating factor, the abelian integral method, the Poincaré-Melnikov integral method, Poincaré return map and averaging theory. The study of limit cycles for differential equations or planar differential systems by applying the averaging method has been considered by several authors see for instance [6,7,8].

In [9], the authors studied the limit cycles of the second-order differential equations

\begin{equation*} \ddot{x}+\varepsilon(1+\cos^m \theta)Q(x,y)+x=0, \end{equation*} where \(Q(x,y)\) is an arbitrary polynomial of degree \(n\), and for each integer non-negative \(m\).

In this paper, our goal is to study the maximum number of limit cycles of a differential planar system bifurcating from the periodic orbits of the linear center

\begin{equation*} \begin{cases} \dot{x}=-y\\ \dot{y}=x, \end{cases} \end{equation*} given by where \(|\varepsilon|>0\) is a small parameter, \(m,n\) are non-negative integers, \(P(x,y)\) and \(Q(x,y)\) are polynomials of degree \(n_{1}\) and \(n_{2}\) respectively. Our main result is the following theorem .

Theorem 1. For all polynomials \(P(x,y)\) and \(Q(x,y)\) have degree \(n_{1}\) and \(n_{2}\) respectively, \(n\) and \(m\) are non-negative integers, then for \(|\varepsilon|>0\) sufficiently small, the maximum number of limit cycles of the differential systems (1) bifurcating from the periodic orbits of the linear center \(\dot{x}=-y, \dot{y}=x\) using averaging theory of first order

• (1) If \(m\) odd and \(n\) odd \begin{equation*} \max \left\{ n_{1},n_{2}\right\}, \end{equation*}
• (2) If \(m\) even and \(n\) even \begin{equation*} \max\left\{\left[ \frac{n_{1}}{2}\right] ,\left[ \frac{n_{2}}{2}\right] \right\}, \end{equation*}
• (3) If \(m\) odd and \(n\) even \begin{equation*} \max \left\{ n_{2},n_{2}+\left[ \frac{n_{1}}{2}\right] -\left[ \frac{n_{2}}{2 }\right] \right\} , \end{equation*}
• (4) If \(m\) even and \(n\) odd \begin{equation*} \max \left\{ n_{1},n_{1}+\left[ \frac{n_{2}}{2}\right] -\left[ \frac{n_{1}}{2 }\right] \right\} , \end{equation*} where \([.]\) denotes the integer part function.

The statements of Theorem 1 is proved in §3. In §2 we recall the averaging theory of first order.

2. The averaging theory of first order

The averaging theory of first for studying periodic orbits was developed in [10,11].

Theorem 2. We Consider the differential system

where \(H:\mathbb{R}\times D \rightarrow \mathbb{R}^n, R:\mathbb{R} \times D \times (-\varepsilon_{f},\varepsilon_{f}) \rightarrow \mathbb{R}^n\) are continuous functions, \(T\)-periodic in the first variable, and \(D\) is an open subset of \(\mathbb{R}^n\). We define \(h:D\rightarrow \mathbb{R}^n\) as and assume that

• (i) \(H\) and \(R\) are locally Lipschitz with respect to x,
• (ii) For \(a\in D\) with \(h(a)=0\), there exists a neighborhood \(V\) of a such that \(h(z)\neq0\) for all \(z\in \bar V \setminus\{ a\}\) and \(d_{B}(h,V,0)\neq 0\).

Then for \(|\varepsilon|>0\) sufficiently small there exists an isolated \(T\)-periodic solution \(\varphi(\cdot, \varepsilon)\) of system (2) such that \(\varphi(\cdot, \varepsilon)\to a\) as \(\varepsilon\to 0\).

Here we will need some facts from the proof of Theorem 2. Hypothesis (i) assures the existence and uniqueness of the solution of each initial value problem on the interval \(\left[0,T\right]\). Hence, for each \(z\in D\), it is possible to denote by \(x(\cdot,z,\varepsilon)\) the solution of (2) with the initial value \(x(0,z,\varepsilon)=z\).

We consider also the function \(\zeta:D\times(-\varepsilon_{f},\varepsilon_{f})\to\mathbb{R}^n\) defined by

From (2) it follows for every \(z\in D\) that The function \(\zeta\) can be written in the form where \(h\) is given by (3), then for \(|\varepsilon|>0\) sufficiently small satisfies that \(z_{\varepsilon}=x(0,\varepsilon)\) tends to be an isolated zero of \(\zeta (.,\varepsilon )\) when \(\varepsilon\to 0\). Of course, due to (5) the function \(\zeta\) is a displacement function for system (2), and its fixed points are initial conditions for the T-periodic solution of system (2).

For additional information on the averaging theory, see the books [12,13].

Theorem 3.(Discartes Theorem). Consider the real polynomial \( p(x)=a_{i_{1}}x^{i_{1}}+a_{i_{2}}x^{i_{2}}+......+a_{i_{k}}x^{i_{k}}\) with \( 0\leq i_{1}< i_{2}< ...< i_{k}\) and \(a_{i_{j}}\neq 0\) real constants for \(j\in \{1,2,...,k\}.\) When \(a_{i_{j}}a_{i_{j+1}}< 0,\) we say that \(a_{i_{j}}\) and \( a_{i_{j+1}}\) have a variation of sign. If the number of variations of signs is \(m\), then \(p(x)\) has at most \(m\) positive real roots. Moreover, it is always possible to choose the coefficients of \(p(x)\) in such a way that \(p(x) \) has exactly \(k-1\) positive real roots.

3. Proof Theorem 1

We need the first order averaging theory in to prove of Theorem 1. In order to apply first order averaging method we write system (1), in polar coordinates \((r,\theta)\) where \(x=r\cos(\theta), y=rsin(\theta)\) \(r>0\). If we take \begin{equation*} \left\{ \begin{array}{l} P(x,y)=\displaystyle\sum_{i+j=0}^{n_{1}}a_{ij}x^{i}y^{j}, \\ Q(x,y)=\displaystyle\sum_{i+j=0}^{n_{2}}b_{ij}x^{i}y^{j}, \end{array} \right. \end{equation*} system (1) can be written as follows \begin{equation*} \left\{ \begin{array}{l} \dot{r}=\varepsilon \left(\displaystyle\sum_{i+j=0}^{n_{1}}a_{ij}\left(\cos ^{i+2}\theta \sin ^{j}\theta +\cos ^{i+2}\theta \sin ^{j+n}\theta \right)r^{i+j+1} \right.\\ \;\;\;\;\;\;\;\;\;\;\;\left.+\displaystyle\sum_{i+j=0}^{n_{2}}b_{ij}\left(\cos ^{i}\theta \sin ^{j+2}\theta +\cos ^{i+m}\theta \sin ^{j+2}\theta \right)r^{i+j+1}\right), \\ \dot{\theta}=1+\varepsilon \left(\displaystyle\sum_{i+j=0}^{n_{2}}b_{ij}\left(\cos ^{i+1}\theta \sin ^{j+1}\theta +\cos ^{i+m+1}\theta \sin ^{j+1}\theta \right)r^{i+j} \right.\\ \;\;\;\;\;\;\;\;\;\;\;\left.-\displaystyle\sum_{i+j=0}^{n_{1}}a_{ij}\left(\cos ^{i+1}\theta \sin ^{j+1}\theta +\cos ^{i+1}\theta \sin ^{j+n+1}\theta \right)r^{i+j}\right). \end{array} \right. \end{equation*} Taking \(\theta\) as the new independent variable system, (1) can be written as \begin{align*} \frac{dr}{d\theta } =&\varepsilon \left(\sum_{i+j=0}^{n_{1}}a_{ij}\left(\cos ^{i+2}\theta \sin ^{j}\theta +\cos ^{i+2}\theta \sin ^{j+n}\theta \right)r^{i+j+1} \right.\\ &\left.+\sum_{i+j=0}^{n_{2}}b_{ij}\left(\cos ^{i}\theta \sin ^{j+2}\theta +\cos ^{i+m}\theta \sin ^{j+2}\theta \right)r^{i+j+1}\right)+O(\varepsilon ^{2}) \end{align*}\begin{align*} =&\varepsilon F\left(r,\theta \right)+O(\varepsilon ^{2}), \end{align*} where \begin{eqnarray*} F(r,\theta ) &=&F_{1}(r,\theta )+F_{2}(r,\theta ), \\ F_{1}(r,\theta ) &=&\sum_{i+j=0}^{n_{1}}a_{ij}\left(\cos ^{i+2}\theta \sin ^{j}\theta +\cos ^{i+2}\theta \sin ^{j+n}\theta \right)r^{i+j+1}, \\ F_{2}(r,\theta ) &=&\sum_{i+j=0}^{n_{2}}b_{ij}\left(\cos ^{i}\theta \sin ^{j+2}\theta +\cos ^{i+m}\theta \sin ^{j+2}\theta \right)r^{i+j+1}. \end{eqnarray*} Let \(F_{10}\) be the averaging equation of first order associated with system (1), using the notation introduced in Theorem 2 we compute \(F_{10}\) by integrating \(F_{1}\) with respect to \(\theta\), In order to calculate the exact expression of \(F_{10}\) we use the following formulas \begin{eqnarray*} \int_{0}^{2\pi }\sin ^{p}\theta \cos ^{2q}\theta d\theta &=&\frac{(2q-1)!!}{ (2q+p)(2q+p-2)..(p+2)}\int_{0}^{2\pi }\sin ^{p}\theta d\theta ,\\ \end{eqnarray*} \begin{eqnarray*} \int_{0}^{2\pi }\cos ^{p}\theta \sin ^{2q}\theta d\theta &=&\frac{(2q-1)!!}{ (2q+p)(2q+p-2)..(p+2)}\int_{0}^{2\pi }\cos ^{p}\theta d\theta .\\ \end{eqnarray*} These formulas are applicable for arbitrary real \(p\) and arbitrary positive integer \(q\), except for the following negative even integers \(p=-2,-4,...,-2n\).

If \(p\) is a natural number and \(q=0\) we have

\begin{eqnarray*} &&\int_{0}^{2\pi }\sin ^{2l}\theta d\theta =\frac{(2l-1)!!}{2^{l}l!}2\pi ,\\ &&\int_{0}^{2\pi }\sin ^{2l+1}\theta d\theta =0,\\ &&\int_{0}^{2\pi }\cos ^{2l}\theta d\theta =\frac{(2l-1)!!}{2^{l}l!}2\pi, \\&&\int_{0}^{2\pi }\cos ^{2l+1}\theta d\theta =0. \end{eqnarray*} We have also \begin{eqnarray*} \int_{0}^{2\pi }\sin ^{p}\theta \cos ^{2q+1}\theta d\theta &=&0,\\ \int_{0}^{2\pi }\cos ^{p}\theta \sin ^{2q+1}\theta d\theta &=&0. \end{eqnarray*} These last formulas are applicable for arbitrary real \(p\) and non-negative integer \(q\), except the following negative odd integers \(p=-1,-3,...,-(2n+1)\). For more details of these integrals and other, see [14].

Now we determine \(\frac{1}{2\pi }\int_{0}^{2\pi }F_{1}(r,\theta )d\theta\), in the following cases

• (1) If \(n\) odd and \(n_{1}\) even \begin{eqnarray*} f_{1}(r) &=&\frac{1}{2\pi }\int_{0}^{2\pi }F_{1}(r,\theta )d\theta \\ &=&\frac{1}{2\pi }\int_{0}^{2\pi }\sum_{i+j=0}^{n_{1}}\left[a_{i,j}\left(\sin ^{j}\theta +\sin ^{j+n}\theta \right)\cos ^{i+2}\theta \right]r^{i+j+1}d\theta \end{eqnarray*}\begin{eqnarray*} &=&\frac{1}{2\pi }\int_{0}^{2\pi }\sum_{2q+j=2}^{n_{1}+2}\left[a_{2q-2,j}(\sin ^{j}\theta +\sin ^{j+n}\theta )\cos ^{2q}\theta \right]r^{2q+j-1}d\theta \\ &=&\frac{1}{2\pi }\left[\sum_{2q+2l+1=3}^{n_{1}+1}a_{2q-2,2l+1}r^{2q+2l} \int_{0}^{2\pi }\sin ^{2l+n+1}\theta \cos ^{2q}\theta d\theta \right. \\ &&\left. +\sum_{2l+2q=2}^{n_{1}+2}a_{2q-2,2l}r^{2q+2l-1}\int_{0}^{2\pi }\sin ^{2l}\theta \cos ^{2q}\theta d\theta \right] \\ &=&\sum_{l+q=1}^{n_{1}/2}a_{2q-2,2l+1}r^{2l+2q}\frac{(2q-1)!!}{ (2q+2l+1)(2q+2l-1)...(2l+3)}\frac{(2l+n)!!}{2^{\tfrac{2l+n+1}{2}}(\frac{ 2l+n+1}{2})!} \\ &&+\sum_{l+q=1}^{(n_{1}+2)/2}a_{2q-2,2l}r^{2l+2q-1}\frac{(2q-1)!!}{ (2q+2l)(2q+2l-2)...(2l+2)}\frac{(2l-1)!!}{2^{l}l!}\\ &=&\sum_{l+q=1}^{n_{1}/2}a_{2q-2,2l+1}r^{2q+2l}\frac{ (2l+n)!!(2q-1)!!}{2^{\tfrac{2l+n+1}{2}}(\frac{2l+n+1}{2} )!(2q+2l+1)(2q+2l-1)...(2l+3)} \\ &&+\sum_{l+q=1}^{(n_{1}+2)/2}a_{2q-2,2l}r^{2q+2l-1}\frac{ (2q-1)!!(2l-1)!!}{2^{l+q}l!(q+l)(q+l-1)...(l+1)} \\ &=&\sum_{k=1}^{n_{1}+1}A_{k}r^{k}. \end{eqnarray*}

• (2)If \(n\) odd and \(n_{1}\) odd \begin{eqnarray*} f_{2}(r) &=&\frac{1}{2\pi }\int_{0}^{2\pi }F_{1}(r,\theta )d\theta \\ &=&\frac{1}{2\pi }\int_{0}^{2\pi }\sum_{i+j=0}^{n_{1}}\left[a_{i,j}(\sin ^{j}\theta +\sin^{j+n}\theta )\cos ^{i+2}\theta \right]r^{i+j+1}d\theta \\ &=&\frac{1}{2\pi }\int_{0}^{2\pi }\sum_{2q+j=2}^{n_{1}+2}\left[a_{2q-2,j}(\sin ^{j}\theta +\sin ^{j+n}\theta )\cos ^{2q}\theta \right]r^{2q+j-1}d\theta \\ &=&\frac{1}{2\pi }\left[\sum_{2q+2l+1=3}^{n_{1}+2}a_{2q-2,2l+1}r^{2q+2l} \int_{0}^{2\pi }\sin ^{2l+n+1}\theta \cos ^{2q}\theta d\theta \right. \\ &&\left.+\sum_{2l+2q=2}^{n_{1}+1}a_{2q-2,2l}r^{2q+2l-1}\int_{0}^{2\pi }\sin ^{2l}\theta \cos ^{2q}\theta d\theta \right]\\ &=&\sum_{l+q=1}^{(n_{1}+1)/2}a_{2q-2,2l+1}r^{2q+2l}\frac{ (2l+n)!!(2q-1)!!}{2^{\tfrac{2l+n+1}{2}}(\frac{2l+n+1}{2} )!(2q+2l+1)(2q+2l-1)...(2l+3)} \\ &&+\sum_{l+q=1}^{(n_{1}+1)/2}a_{2q-2,2l}r^{2q+2l-1}\frac{ (2q-1)!!(2l-1)!!}{2^{l+q}l!(q+l)(q+l-1)...(l+1)} \\ &=&\sum_{k=1}^{n_{1}+1}\tilde{A}_{k}r^{k}. \end{eqnarray*}
• (3) If \(n\) even and \(n_{1}\) even \begin{eqnarray*} f_{3}(r) &=&\frac{1}{2\pi }\int_{0}^{2\pi }F_{1}(r,\theta )d\theta \\ &=&\frac{1}{2\pi }\int_{0}^{2\pi }\sum_{i+j=0}^{n_{1}}[a_{i,j}(\sin ^{j}\theta +\sin ^{j+n}\theta )\cos ^{i+2}\theta ]r^{i+j+1}d\theta \end{eqnarray*}\begin{eqnarray*} &=&\frac{1}{2\pi }\int_{0}^{2\pi }\sum_{2q+j=2}^{n_{1}+2}[a_{2q-2,j}(\sin ^{j}\theta +\sin ^{j+n}\theta )\cos ^{2q}\theta ]r^{2q+j-1}d\theta \\ &=&\frac{1}{2\pi }\left[\sum_{2l+2q=2}^{n_{1}+2}a_{2q-2,2l}r^{2q+2l-1}\int_{0}^{2 \pi }(\sin ^{2l}\theta +\sin ^{2l+n}\theta )\cos ^{2q}\theta d\theta \right]\\ &=&\sum_{l+q=1}^{(n_{1}+2)/2}a_{2q-2,2l}r^{2q+2l-1}\frac{(2q-1)!!}{ 2^{q}(q+l)(q+l-1)...(l+1)}\left[\frac{(2l-1)!!}{2^{l}l!}+\frac{(2l+n-1)!!}{2^{ \tfrac{2l+n}{2}}(\frac{2l+n}{2})!}\right] \\ &=&\sum_{\substack{ k=1 \\ k\text{ }odd}}^{n_{1}+1}\bar{A}_{k}r^{k}. \end{eqnarray*}
• (4) If \(n\) even and \(n_{1}\) odd \begin{eqnarray*} f_{4}(r) &=&\frac{1}{2\pi }\int_{0}^{2\pi }F_{1}(r,\theta )d\theta \\ &=&\frac{1}{2\pi }\int_{0}^{2\pi }\sum_{i+j=0}^{n_{1}}[a_{i,j}(\sin ^{j}\theta +\sin ^{j+n}\theta )\cos ^{i+2}\theta ]r^{i+j+1}d\theta \\ &=&\frac{1}{2\pi }\int_{0}^{2\pi }\sum_{2q+j=2}^{n_{1}+2}[a_{2q-2,j}(\sin ^{j}\theta +\sin ^{j+n}\theta )\cos ^{2q}\theta ]r^{2q+j-1}d\theta \\ &=&\frac{1}{2\pi }\left[\sum_{2l+2q=2}^{n_{1}+1}a_{2q-2,2l}r^{2q+2l-1}\int_{0}^{2 \pi }(\sin ^{2l}\theta +\sin ^{2l+n}\theta )\cos ^{2q}\theta d\theta \right]\\ &=&\sum_{l+q=1}^{(n_{1}+1)/2}a_{2q-2,2l}r^{2q+2l-1}\frac{(2q-1)!!}{ 2^{q}(q+l)(q+l-1)...(l+1)}\left[\frac{(2l-1)!!}{2^{l}l!}+\frac{(2l+n-1)!!}{2^{ \tfrac{2l+n}{2}}(\frac{2l+n}{2})!}\right]\\ &=&\sum_{\substack{ k=1 \\ k\text{ }impair}}^{n_{1}}\bar{A}_{k}r^{k}. \end{eqnarray*}

And we determine \(\frac{1}{2\pi }\int_{0}^{2\pi }F_{2}(r,\theta )d\theta\) in the following cases

• (5) If \(m\) odd and \(n_{2}\) even \begin{eqnarray*} f_{5}(r) &=&\frac{1}{2\pi }\int_{0}^{2\pi }F_{2}(r,\theta )d\theta \\ &=&\frac{1}{2\pi }\int_{0}^{2\pi }\sum_{i+j=0}^{n_{2}}\left[b_{i,j}(\cos ^{i}\theta +\cos ^{i+m}\theta )\sin ^{j+2}\theta \right]r^{i+j+1}d\theta \\ &=&\frac{1}{2\pi }\int_{0}^{2\pi }\sum_{i+2q=2}^{n_{2}+2}\left[b_{i,2q-2}(\cos ^{i}\theta +\cos ^{i+m}\theta )\sin ^{2q}\theta \right]r^{i+2q-1}d\theta \\ &=&\frac{1}{2\pi }\left[\sum_{2l+1+2q=3}^{n_{2}+1}b_{2l+1,2q-2}r^{2l+2q} \int_{0}^{2\pi }\cos ^{2l+m+1}\theta \sin ^{2q}\theta d\theta \right. \\ &&\left.+\sum_{2l+2q=2}^{n_{2}+2}b_{2l,2q-2}r^{2l+2q-1}\int_{0}^{2\pi }\cos ^{2l}\theta \sin ^{2q}\theta d\theta \right]\\ &=&\sum_{l+q=1}^{n_{2}/2}b_{2l+1,2q-2}r^{2l+2q}\frac{(2q-1)!!}{ (2q+2l+1)(2q+2l-1)...(2l+3)}\frac{(2l+m)!!}{2^{\tfrac{2l+m+1}{2}}(\frac{ 2l+m+1}{2})!} \\ &&+\sum_{l+q=1}^{(n_{2}+2)/2}b_{2l,2q-2}r^{2l+2q-1}\frac{(2q-1)!!}{ (2q+2l)(2q+2l-2)...(2l+2)}\frac{(2l-1)!!}{2^{l}l!} \end{eqnarray*} \begin{eqnarray*} &=&\sum_{l+q=1}^{n_{2}/2}b_{2l+1,2q-2}r^{2l+2q}\frac{(2l+m)!!(2q-1)!! }{2^{\tfrac{2l+m+1}{2}}(\frac{2l+m+1}{2})!(2q+2l+1)(2q+2l-1)...(2l+3)} \\ &&+\sum_{l+q=1}^{(n_{2}+2)/2}b_{2l,2q-2}r^{2l+2q-1}\frac{ (2l-1)!!(2q-1)!!}{2^{l+q}l!(q+l)(q+l-1)...(l+1)} \\ &=&\sum_{k=1}^{n_{2}+1}B_{k}r^{k}. \end{eqnarray*}
• (6) If \(m\) odd and \(n_{2}\) odd \begin{eqnarray*} f_{6}(r) &=&\frac{1}{2\pi }\int_{0}^{2\pi }F_{2}(r,\theta )d\theta \\ &=&\frac{1}{2\pi }\int_{0}^{2\pi }\sum_{i+j=0}^{n_{2}}\left[b_{i,j}(\cos ^{i}\theta +\cos ^{i+m}\theta )\sin ^{j+2}\theta \right]r^{i+j+1}d\theta \\ &=&\frac{1}{2\pi }\int_{0}^{2\pi }\sum_{i+2q=2}^{n_{2}+2}\left[b_{i,2q-2}(\cos ^{i}\theta +\cos ^{i+m}\theta )\sin ^{2q}\theta \right]r^{i+2q-1}d\theta \\ &=&\frac{1}{2\pi }\left[\sum_{2l+1+2q=3}^{n_{2}+2}b_{2l+1,2q-2}r^{2l+2q} \int_{0}^{2\pi }\cos ^{2l+m+1}\theta \sin ^{2q}\theta d\theta \right. \\ &&\left.+ \sum_{2l+2q=2}^{n_{2}+1}b_{2l,2q-2}r^{2l+2q-1}\int_{0}^{2\pi }\cos ^{2l}\theta sin ^{2q}\theta d\theta \right] \\ &=&\sum_{l+q=1}^{(n_{2}+1)/2}b_{2l+1,2q-2}r^{2l+2q}\frac{ (2l+m)!!(2q-1)!!}{2^{\tfrac{2l+m+1}{2}}(\frac{2l+m+1}{2} )!(2q+2l+1)(2q+2l-1)...(2l+3)} \\ &&+\sum_{l+q=1}^{(n_{2}+1)/2}b_{2l,2q-2}r^{2l+2q-1}\frac{ (2l-1)!!(2q-1)!!}{2^{l+q}l!(q+l)(q+l-1)...(l+1)} \\ &=&\sum_{k=1}^{n_{2}+1}\tilde{B}_{k}r^{k}. \end{eqnarray*}
• (7) If \(m\) even and \(n_{2}\) even \begin{eqnarray*} f_{7}(r) &=&\frac{1}{2\pi }\int_{0}^{2\pi }F_{2}(r,\theta )d\theta \\ &=&\frac{1}{2\pi }\int_{0}^{2\pi }\sum_{i+j=0}^{n_{2}}\left[b_{i,j}(\cos ^{i}\theta +\cos ^{i+m}\theta )\sin ^{j+2}\theta \right]r^{i+j+1}d\theta \\ &=&\frac{1}{2\pi }\int_{0}^{2\pi }\sum_{i+2q=2}^{n_{2}+2}\left[b_{i,2q-2}(\cos ^{i}\theta +\cos ^{i+m}\theta )\sin ^{2q}\theta \right]r^{i+2q-1}d\theta \\ &=&\frac{1}{2\pi }\left[\sum_{2l+2q=2}^{n_{2}+2}b_{2l,2q-2}r^{2l+2q-1}\int_{0}^{2 \pi }(\cos ^{2l}\theta +\cos ^{2l+m}\theta )\sin ^{2q}\theta d\theta \right] \\ &=&\sum_{l+q=1}^{(n_{2}+2)/2}b_{2l,2q-2}r^{2l+2q-1}\frac{(2q-1)!!}{ 2^{q}(q+l)(q+l-1)...(l+1)}\left[\frac{(2l-1)!!}{2^{l}l!}+\frac{(2l+m-1)!!}{2^{ \tfrac{2l+m}{2}}(\frac{2l+m}{2})!}\right]\\ &=&\sum_{\substack{ k=1 \\ k\text{ }odd}}^{n_{2}+1}\bar{B}_{k}r^{k}. \end{eqnarray*}
• (8) If \(m\) even and \(n_{2}\) odd \begin{eqnarray*} f_{8}(r)&=&\frac{1}{2\pi }\int_{0}^{2\pi }F_{2}(r,\theta )d\theta \end{eqnarray*}\begin{eqnarray*} &=&\frac{1}{2\pi }\int_{0}^{2\pi }\sum_{i+j=0}^{n_{2}}\left[ b_{i,j}(\cos ^{i}\theta +\cos ^{i+m}\theta )\sin ^{j+2}\theta \right]r^{i+j+1}d\theta \\ &=&\frac{1}{2\pi }\int_{0}^{2\pi }\sum_{i+2q=2}^{n_{2}+2}\left[b_{i,2q-2}(\cos ^{i}\theta +\cos ^{i+m}\theta )\sin ^{2q}\theta \right]r^{i+2q-1}d\theta \\ &=&\frac{1}{2\pi }\left[\sum_{2l+2q=2}^{n_{2}+1}b_{2l,2q-2}r^{2l+2q-1}\int_{0}^{2 \pi }(\cos ^{2l}\theta +\cos ^{2l+m}\theta )\sin ^{2q}\theta d\theta \right] \\ &=&\sum_{l+q=1}^{(n_{2}+1)/2}b_{2l,2q-2}r^{2l+2q-1}\frac{(2q-1)!!}{ 2^{q}(q+l)(q+l-1)...(l+1)}\left[\frac{(2l-1)!!}{2^{l}l!}+\frac{(2l+m-1)!!}{2^{ \tfrac{2l+m}{2}}(\frac{2l+m}{2})!}\right]\\ &=&\sum_{\substack {k=1 \\ k\text{ }odd}}^{n_{2}}\bar{B}_{k}r^{k}. \end{eqnarray*}

Going back to the Equation (7), and we distinguish the following cases and subcases

• (a) If \(m\) odd and \(n\) odd
  • (a.1) n\(_{1}\) even et n\(_{2}\) even \begin{equation*} F_{10}(r)=\sum_{k=1}^{n_{1}+1}A_{k}r^{k}+\sum_{k=1}^{n_{2}+1}B_{k}r^{k}, \end{equation*}
  • (a.2) n\(_{1}\) odd et n\(_{2}\) odd \begin{equation*} F_{10}(r)=\sum_{k=1}^{n_{1}+1}\tilde{A}_{k}r^{k}+\sum_{k=1}^{n_{2}+1}\tilde{B }_{k}r^{k}, \end{equation*}
  • (a.3) n\(_{1}\) odd et n\(_{2}\) even \begin{equation*} F_{10}(r)=\sum_{k=1}^{n_{1}+1}\tilde{A}_{k}r^{k}+ \sum_{k=1}^{n_{2}+1}B_{k}r^{k}, \end{equation*}
  • (a.4) n\(_{1}\) even et n\(_{2}\) odd \begin{equation*} F_{10}(r)=\sum_{k=1}^{n_{1}+1}A_{k}r^{k}+\sum_{k=1}^{n_{2}+1}\tilde{B} _{k}r^{k}. \end{equation*}

We have that \(F_{10}\) is the polynomial in the variable \(r\), then by Descartes Theorem \(F_{10}\) has most \(\max \left\{ n_{1},n_{2}\right\}\) limit cycles, this completes the proof of statement \((1)\) of Theorem 1.

• (b) If \(m\) even and \(n\) even.
  • (b.1) n\(_{1}\) even et n\(_{2}\) even \begin{equation*} F_{10}(r)=\sum_{\substack{ k=1 \\ k\text{ }odd}}^{n_{1}+1}\bar{A} _{k}r^{k}+\sum_{\substack{ k=1 \\ k\text{ }odd}}^{n_{2}+1}\bar{B}_{k}r^{k}, \end{equation*}
  • (b.2) n\(_{1}\) odd et n\(_{2}\) odd \begin{equation*} F_{10}(r)=\sum_{\substack{ k=1 \\ k\text{ }odd}}^{n_{1}}\bar{A} _{k}r^{k}+\sum_{\substack{ k=1 \\ k\text{ }odd}}^{n_{2}}\bar{B}_{k}r^{k}, \end{equation*}
  • (b.3) n\(_{1}\) odd et n\(_{2}\) even \begin{equation*} F_{10}(r)=\sum_{\substack{ k=1 \\ k\text{ }odd}}^{n_{1}}\bar{A} _{k}r^{k}+\sum_{\substack{ k=1 \\ k\text{ }odd}}^{n_{2}+1}\bar{B}_{k}r^{k}, \end{equation*}
  • (b.4) n\(_{1}\) even et n\(_{2}\) odd \begin{equation*} F_{10}(r)=\sum_{\substack{ k=1 \\ k\text{ }odd}}^{n_{1}+1}\bar{A} _{k}r^{k}+\sum_{\substack{ k=1 \\ k\text{ }odd}}^{n_{2}}\bar{B}_{k}r^{k}. \end{equation*}

We have that \(F_{10}\) is the polynomial in the variable \(r^2\), then by Descartes Theorem \(F_{10}\) has most \(\max\{\left[ \frac{n_{1}}{2}\right] ,\left[ \frac{n_{2}}{2}\right] \}\) limit cycles, this completes the proof of statement (2) of Theorem 1.

• (c) If \(m\) odd and \(n\) even.
  • (c.1) n\(_{1}\) even et n\(_{2}\) even \begin{equation*} F_{10}(r)=\sum_{\substack{ k=1 \\ k\text{ }odd}}^{n_{1}+1}\bar{A} _{k}r^{k}+\sum_{k=1}^{n_{2}+1}B_{k}r^{k}, \end{equation*}
  • (c.2) n\(_{1}\) odd et n\(_{2}\) odd \begin{equation*} F_{10}(r)=\sum_{\substack{ k=1 \\ k\text{ }odd}}^{n_{1}}\bar{A} _{k}r^{k}+\sum_{k=1}^{n_{2}+1}\tilde{B}_{k}r^{k}, \end{equation*}
  • (c.3) n\(_{1}\) odd et n\(_{2}\) even \begin{equation*} F_{10}(r)=\sum_{\substack{ k=1 \\ k\text{ }odd}}^{n_{1}}\bar{A} _{k}r^{k}+\sum_{k=1}^{n_{2}+1}B_{k}r^{k}, \end{equation*}
  • (c.4) n\(_{1}\) even et n\(_{2}\) odd \begin{equation*} F_{10}(r)=\sum_{\substack{ k=1 \\ k\text{ }odd}}^{n_{1}+1}\bar{A} _{k}r^{k}+\sum_{k=1}^{n_{2}+1}\tilde{B}_{k}r^{k}. \end{equation*}

We have that \(F_{10}\) is the sum of two polynomials, one in the variable \(r\) and the other in \(r^2\), then by Descartes Theorem \(F_{10}\) has most \(\max \left\{ n_{2},n_{2}+\left[ \frac{n_{1}}{2}\right] -\left[ \frac{n_{2}}{2 }\right] \right\}\) limit cycles, this completes the proof of statement \((3)\) of Theorem 1.

• (d) If \(n\) odd and \(m\) even.
  • (d.1) n\(_{1}\) even et n\(_{2}\) even \begin{equation*} F_{10}(r)=\sum_{k=1}^{n_{1}+1}A_{k}r^{k}+\sum_{\substack{ k=1 \\ k\text{ } odd}}^{n_{2}+1}\bar{B}_{k}r^{k}, \end{equation*}
  • (d.2) n\(_{1}\) odd et n\(_{2}\) odd \begin{equation*} F_{10}(r)=\sum_{k=1}^{n_{1}+1}\tilde{A}_{k}r^{k}+\sum_{\substack{ k=1 \\ k \text{ }odd}}^{n_{2}}\bar{B}_{k}r^{k}, \end{equation*}
  • (d.3) n\(_{1}\) odd et n\(_{2}\) even \begin{equation*} F_{10}(r)=\sum_{k=1}^{n_{1}+1}\tilde{A}_{k}r^{k}+\sum_{\substack{ k=1 \\ k \text{ }odd}}^{n_{2}+1}\bar{B}_{k}r^{k}, \end{equation*}
  • (d.4) n\(_{1}\) even et n\(_{2}\) odd \begin{equation*} F_{10}(r)=\sum_{k=1}^{n_{1}+1}A_{k}r^{k}+\sum_{\substack{ k=1 \\ k\text{ } odd}}^{n_{2}}\bar{B}_{k}r^{k}. \end{equation*}

We have that \(F_{10}\) is the sum of two polynomials, one in the variable \(r\) and the other in \(r^2\), then by Descartes Theorem \(F_{10}\) has most \(\max \left\{ n_{1},n_{1}+\left[ \frac{n_{2}}{2}\right] -\left[ \frac{n_{1}}{2 }\right] \right\}\) limit cycles, this completes the proof of statement \((4)\) of Theorem 1.

4. Example

We consider the system By doing the change of variables \(x=r\cos\theta, y=r\sin\theta\) and taking \(\theta\) as a new independent variable, we get \begin{eqnarray*} \dot{r}&=&\varepsilon\left(\left( \frac{1}{84}r^{3}\cos ^{4}\theta -\frac{23}{240}r\cos ^{2}\theta\right ) \left( 1+\sin ^{2}\theta \right) +\left(\frac{1}{8}r^{5}\sin ^{4}\theta \cos ^{2}\theta -\frac{23}{18}r^{4}\sin ^{4}\theta \cos \theta \right.\right.\\ &&\left.\left.+\frac{1}{12}r^{3}\sin ^{4}\theta -\frac{13}{48}r^{2}\sin ^{2}\theta \cos \theta +\frac{1}{8}r\sin ^{2}\theta \right)\left(1+\cos ^{3}\theta \right)\right),\end{eqnarray*}\begin{eqnarray*} \dot{\theta}&=& 1+\varepsilon\left( \left(-\frac{1}{84}r^{2}\cos ^{3}\theta \sin \theta + \frac{23}{240}\sin \theta \cos \theta \right)\left(1+\sin ^{2}\theta \right)+ \left(-\frac{ 23}{18}r^{3}\cos ^{2}\theta \sin ^{3}\theta \right.\right.\\ &&\left.\left.+\frac{1}{12}r^{2}\sin ^{3}\theta \cos \theta +\frac{1}{8}r^{4}\cos ^{3}\theta \sin ^{3}\theta - \frac{13}{48}r\cos \theta +\frac{1}{8}\cos \theta \sin \theta \right)\right). \end{eqnarray*} Taking \(\theta\) as the new independent variable, we get \begin{eqnarray*} \frac{dr}{d\theta }&=&\varepsilon F(r,\theta )+O(\varepsilon ^{2}), \end{eqnarray*} where \begin{eqnarray*} F(r,\theta ) &=&\left( \frac{1}{84}r^{3}\cos ^{4}\theta -\frac{23}{240}r\cos ^{2}\theta \right) \left( 1+\sin ^{2}\theta \right) +\left(\frac{1}{8}r^{5}\sin ^{4}\theta \cos ^{2}\theta -\frac{23}{18}r^{4}\sin ^{4}\theta \cos \theta \right.\\ &&\left.+\frac{1}{12}r^{3}\sin ^{4}\theta -\frac{13}{48}r^{2}\sin ^{2}\theta \cos \theta +\frac{1}{8}r\sin ^{2}\theta \right)\left.(1+\cos ^{3}\theta \right). \end{eqnarray*} The function of averaging theory of first order is \[F_{10}=\frac{1}{768}r\left(6r^{4}-23r^{3}+28r^{2}-13r+2\right),\] that has exactly four positive zeros which are \(r_{1}=\dfrac{1}{3},r_{2}=\dfrac{1}{2},r_{3}=1\), and \(r_{4}=2.\) Which satisfy \begin{align*}&\frac{dF_{10}(r)}{dr}\left. {}\right\vert _{r=r_{1}}=-\frac{5}{10368}\neq 0,\\ &\frac{dF_{10}(r)}{dr}\left. {}\right\vert _{r=r_{2}}=\frac{1}{2048}\neq 0,\\ &\frac{dF_{10}(r)}{dr}\left. {}\right\vert _{r=r_{3}}=-\frac{1}{384}\neq 0,\\ &\frac{dF_{10}(r)}{dr}\left. {}\right\vert _{r=r_{4}}=\frac{5}{128}\neq 0,\end{align*} then we conclude that the system (8) has two stable limit cycles for \(r_{1}=\dfrac{1}{3}\) and \(r_{3}=1\), and two unstable limit cycles for \(r_{2}=\dfrac{1}{2}\) and \(r_{4}=2\) (see Figure 1). \begin{figure}[H] \centering \includegraphics[scale=0.5]{limit.png} \caption{Four limit cycles for \(\varepsilon=10^{-3}\)} \label{figure2} \end{figure}

5. Conclusion

In the present paper, by using the averaging theory of the first order we show that the maximum number of the limit cycles bifurcating from linear center \( \dot{x}=-y, \dot{y}=x,\) for a generalized planar differential system.

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflicts of Interest

"The authors declare no conflict of interest."

1. Introduction

Fractional calculus is the extension of integer derivatives to real or complex number derivatives. Therefore, it has gained an important place in terms of examining the sensitivity of many phenomena in natural sciences. Most of the problems in scientific fields such as physics, biology, astronomy, medical sciences, optical fibers, chemical biology, radiology can be modeled by differential equations [1,2,3,4,5,6]. The fractional orders of the differential equations enable the investigations to be more comprehensive and obtain optimal solutions. That's why fractional differential equations have been the focus of various researchers over the past decades. Articles and books have been written to demonstrate the existence of solutions to such equations or to develop new solution methods [7,8,9].

Many people have studied the existence and positive solutions or multiplicity of solutions for nonlinear fractional boundary value problems using fixed point theorems such as the Schauder fixed-point theorem, the Guo-Krasnosel'skii fixed-point theorem, and the Leggett-Williams fixed-point theorem. In recent studies, boundary value problems involving multi point boundary condition and fractional differential equations have attracted attention.

In order to guarantee the existence of positive solutions for boundary value problems, the nonlinearity is usually nonnegative. If the nonlinearity changing sign, it will bring much more difficulties to the study of the problem. Because of this, there is only a few study on such problems [10,11,12,13,14].

In [15], Ahmad consider the two point Liouville-Caputo boundary value problem

\begin{eqnarray*}\label{1} \left. \begin{array}{ll} ^{c}D^\beta x(t)=-f(t,x(t)), \quad t \in (a,b)\\ x(a)=\delta_1, \quad x(b)=\delta_2, \quad \delta_i \in \mathbb{R}, i=1,2 \end{array} \right. \end{eqnarray*} where \(^{c}D^\beta\) denotes the Caputo fractional derivative of order \(\beta\) with \(1< \beta< 2\) and \(f\) is a continuous function.

In this paper, we investigate the existence result for the following m-point fractional boundary value problem with changing sign nonlinearity,

where \(D_a^\beta\) is the standard Caputo fractional derivatives of fractional order \(\beta\) with \(1< \beta\leq 2\) , \(\xi_i \quad (1\leq i\leq m-2)\) are positive real constants with \( 0< \sum_{i=1}^{m-2} \xi_i< 1\), \(\eta_i \in(a,b)\) and a continuous function \(f:[a,b]\times \mathbb{R}\rightarrow \mathbb{R}\) may change sign.

The paper is organized as follows: In Section 2, present some background materials and preliminaries. In Section 3, we give an existence result and also an example is given to exemplify main result.

2. Preliminaries

In this section we gather some preliminary definitions, theorems and fundamental results of fractional calculus theory that will be used in subsequent section. Basic properties of the fractional calculus and its applications can be found in Kilbas, Srivastava and Trujillo [4] and Podlubny [6].

Definition 1. For a function \(f\) given on the interval \([a,b]\), the Caputo derivative of fractional order \(r\) is defined as \begin{equation*} D^rf(t)=\frac{1}{\Gamma(n-r)}\int_0^t(t-s)^{n-r-1}f^{(n)}(s)ds, \quad n=[r]+1, \end{equation*} where \([r]\) denotes the integer part of \(r\).

Definition 2. The Riemann-Liouville fractional integral of order \(r\) for a function \(f\) is defined as \begin{equation*} I^rf(t)=\frac{1}{\Gamma(r)}\int_0^t (t-s)^{r-1}f(s)ds, \quad r>0, \end{equation*} where \([r]\) denotes the integer part of \(r\).

Lemma 1. Let \(r>0\). Then the differential equation \(D^rx(t)=0\) has solutions \begin{equation*} x(t)=c_0+c_1t+c_2t^2+\cdots+c_{n-1}t^{n-1}, \end{equation*} where \(c_i \in \mathbb{R}\), \(i=0,1,2, . . .,n\), \(n=[r]+1\).

Lemma 2. Let \(r>0\). Then \begin{equation*} I^r(D^rx)(t)=x(t)+c_0+c_1t+c_2t^2+ . . . +c_{n-1}t^{n-1}, \end{equation*} where \(c_i \in \mathbb{R}\), \(i=0,1,2, . . ., n\), \(n=[r]+1\).

For finding a solution of the problem (1)-(2) , we first consider the following fractional differential equation with the boundary condition (2).

Let we define \( \Delta:=b-a-\sum_{i=1}^{m-2}\xi_{i}(\eta_i-a)\).

Lemma 3. Let \(\beta \in (1, 2]\) and \(t\in [a,b]\). The boundary value problem (3)-(2) has a unique solution \(u\) in the form \begin{eqnarray*} u(t)&=&\left[1+\frac{ \sum_{i=1}^{m-2} \xi_{i}-1}{\Delta}(t-a)\right]\delta_1+\frac{t-a}{\Delta}\delta_2 +\int_a^b G(t,s)h(s)ds, \end{eqnarray*} where

\(i=1,2,\ldots m-2.\)

Proof. The equation \(D^\beta u(t)=-h(t)\) has a unique solution

where \(c_0, c_1 \in \mathbb{R}\).

By \(u(a)=\delta_1\) and \( u(b)=\sum\limits_{i=1}^{m-2}\xi u(\eta_i)+\delta_2\), we have \(c_0=\delta_1\) and

\[ c_1=\frac{1}{\Delta}\left(-\frac{1}{\Gamma(\beta)}\sum_{i=1}^{m-2} \xi_{i} \int_a^{\eta_i} (\eta_i-s)^{\beta-1}h(s)ds+\frac{1}{\Gamma(\beta)}\int_a^b(b-s)^{\beta-1}h(s)ds+\delta_1 \left[\sum_{i=1}^{m-2}\xi_{i}-1\right] +\delta_2 \right).\] Substituting \(c_0, c_1\) into equation (5) we find, \begin{align*} u(t)=&\left[1+\frac{ \sum_{i=1}^{m-2}\xi_i -1}{\Delta}(t-a)\right]\delta_1+\frac{t-a}{\Delta}\delta_2 -\frac{1}{\Gamma(\beta)}\left(\int_a^t (t-s)^{\beta -1}h(s)ds+\frac{t-a}{\Delta} \sum_{i=1}^{m-2} \xi_i \int_a^{\eta_i} (\eta_i-s)^{\beta-1}h(s)ds\right.\\ &\left.-\frac{t-a}{\Delta}\int_a^b (b-s)^{\beta-1} h(s)ds\right)\\ =& \left[1+\frac{ \sum_{i=1}^{m-2}\xi_i -1}{\Delta}(t-a)\right]\delta_1+\frac{t-a}{\Delta}\delta_2+\int_a^b G(t,s)h(s)ds, \end{align*} where \(G(t,s)\) is defined by (4). The proof is completed.

Lemma 4. If \( 0< \sum_{i=1}^{m-2}\xi_i < 1\), then

• i) \(\Delta>0\),
• ii) \( (b-s)^{\beta-1}- \sum\limits_{j=i}^{m-2}\xi_j(\eta_j - s)^{\beta -1 }>0\).

Proof.

• i) We can easily see that \begin{align*} &\eta_i < b ,\\ &\xi_i(\eta_i-a)<\xi_i(b-a),\\ & -\sum_{i=1}^{m-2}\xi_i(\eta_i-a)>-\sum_{i=1}^{m-2}\xi_i(b-a),\\ &b-a-\sum_{i=1}^{m-2}\xi_i(\eta_i-a)>b-a-\sum_{i=1}^{m-2}\xi_i(b-a)=(b-a)\left[1-\sum_{i=1}^{m-2}\xi_{i}\right]. \end{align*} If \( 1-\sum_{i=1}^{m-2}\xi_i>0\), then \( b-a- \sum_{i=1}^{m-2}\xi_i (\eta_i -a)>0\). So we have \(\Delta>0\).
• ii) Since \( 0< \beta-1 \leq 1\), we have \(b-a>b-s>(b-s)^{\beta-1}\) and \(\eta_i - a>\eta_i -s > (\eta_i -s)^{\beta -1}\). Also we know that \((\eta_j-s)^{\beta-1}< (b-s)^{\beta -1}\). Thus we have \begin{eqnarray*} \sum_{j=i}^{m-2}\xi_j(\eta_j-s)^{\beta-1}< \sum_{j=i}^{m-2}\xi_j(b-s)^{\beta-1} \leq (b-s)^{\beta-1}\sum_{i=1}^{m-2}\xi_i < (b-s)^{\beta -1} \end{eqnarray*} and so \( (b-s)^{\beta -1}- \sum\limits_{j=i}^{m-2}\xi_j(\eta_j-s)^{\beta-1}>0. \)

Remark 1. Let \(G(t,s)\) be the Green's function for the problem (1)-(2). It is easy to find that

Remark 2. Let \(p(t)\in L^1[a,b]\) and \(w(t)\) be a solution of the problem

then \( w(t)=\int_a^b G(t,s)p(s)ds\).

The following fixed point theorem is fundamental and important to the proof of our main result.

Theorem 1. [14][ Schauder-Tychonoff Fixed Point Theorem] Let \( X \) be a Banach space. Assume that \( K \) is a closed, bounded, convex subset of \( X \). If \( T:K\longrightarrow K \) is compact, then \( T \) has a fixed point in \(K.\)

3. Existence Result

We also make the following assumption throughout this paper.

• (H1) There exists a nonnegative function \(p \in L^1[a,b]\) and \( \int_a^b p(t)dt>0\) such that \(f(t,u)\geq - p(t)\) for all \((t,u)\in [a,b]\times \mathbb{R}\).
• (H2) \(f(t,u)\neq 0\), for \((t,u)\in[a,b]\times \mathbb{R}\).

Let \(B=\mathbb{C}([a,b],\mathbb{R})\) denote the Banach space of all continuous function from \([a,b]\) into \(\mathbb{R}\) endowed with the usual norm \(\Vert u \Vert = sup\{\vert u(t) \vert: t\in[a,b]\}\).

First we shall show that the following fractional equation

with the boundary condition (2) has a solution, where \(F:[a,b]\times \mathbb{R}\rightarrow \mathbb{R}\) and \(u^*(t)=max\{(u-w)(t),0\}\) such that \(w\) is the unique solution of the problem (7). We define an operator \(T:B \rightarrow B\) associated with the problem (8)-(2) as where \(G(t,s)\) is given by (4). The existence of a fixed point for the operator \(T\) implies the existence of a solution for the problem (8)-(2).

Theorem 2. Assume that (H1)-(H2) are satisfied. If \( K > 0 \) satisfies \(\left[1+\frac{ \sum_{i=1}^{m-2}\xi_i -1}{\Delta}(b-a)\right]\delta_1+ \frac{b-a}{\Delta}\delta_2 + L M \leq K \) where \(L\geq max\{\left|F(t,u) \right|:t \in [a,b], \ \left|u \right|\leq K \} \) and \(M\) is given in (6) then the problem (8)-(9) has a solution \(u(t)\).

Proof. Let we define \( P:=\{u\in B:\left\|u \right\|\leq K \}\). It can be easily seen that \( P \) is a closed, bounded and convex subset of \( B \) to which Schauder fixed point theorem is applicable. Define \( T:P\longrightarrow B\) by (10). It is easily seen that \( T:P\longrightarrow B\) is continuous. Claim \(T:P \longrightarrow P\). Let \( u \in P \). Consider \(u^*(t) \leq u(t) \leq K\) for all \(t \in [a, b]\). Then \begin{align*} \left|Tu(t) \right|&=\left|\left[1+\frac{ \sum_{i=1}^{m-2}\xi_i -1}{\Delta}(t-a)\right]\delta_1+\frac{t-a}{\Delta}\delta_2+\int_a^b G(t,s)F(s,u^*(s))ds \right|\\ &\leq \left[1+\frac{ \sum_{i=1}^{m-2}\xi_i -1}{\Delta}(b-a)\right]\delta_1+\frac{b-a}{\Delta}\delta_2+ L M \leq K \end{align*} for all \( t \in [a,b].\) This implies that \( \left\|Tu\right\|\leq K \). Hence \( T:K\longrightarrow K.\) Using the Arzela-Ascoli theorem it can be shown that \( T:K\longrightarrow K \) is a compact operator. Hence \( T \) has a fixed point \( u \) in \(P\) by the Schauder-Tychonov theorem. This implies that \(u\) is a solution of the problem (8)-(2).

Lemma 5. \(u^*(t)\) is the solution of the boundary value problem (1)-(2) with \(u(t)>w(t)\) for all \(t\in[a,b]\) if and only if \(u=u^*+w\) is the positive solution of the boundary value problem (8)-(2).

Proof. Let \(u(t)\) is the solution of the boundary value problem (8)-(2). Then \begin{align*} u(t)&=\left[1+\frac{ \sum_{i=1}^{m-2}\xi_i-1}{\Delta}(t-a)\right]\delta_1+\frac{t-a}{\Delta}\delta_2 + \frac{1}{\Gamma(\beta)}\int_a^b G(t,s)F(s,u^*(s))ds\\ &=\left[1+\frac{ \sum_{i=1}^{m-2}\xi_i-1}{\Delta}(t-a)\right]\delta_1+\frac{t-a}{\Delta}\delta_2 + \frac{1}{\Gamma(\beta)}\int_a^b G(t,s)(f(s,u^*(s))+p(s))ds\\ &=\left[1+\frac{ \sum_{i=1}^{m-2}\xi_i-1}{\Delta}(t-a)\right]\delta_1+\frac{t-a}{\Delta}\delta_2 + \frac{1}{\Gamma(\beta)}\int_a^b G(t,s)f(s,(u-w)(s))ds+\frac{1}{\Gamma(\beta)}\int_a^b G(t,s)p(s)ds\\ &=\left[1+\frac{ \sum_{i=1}^{m-2}\xi_i-1}{\Delta}(t-a)\right]\delta_1+\frac{t-a}{\Delta}\delta_2 + \frac{1}{\Gamma(\beta)}\int_a^b G(t,s)f(s,(u-w)(s))ds+w(t) \end{align*} or \begin{eqnarray*} u(t)-w(t)=&\left[1+\frac{ \sum_{i=1}^{m-2}\xi_i-1}{\Delta}(t-a)\right]\delta_1+\frac{t-a}{\Delta}\delta_2 + \frac{1}{\Gamma(\beta)}\int_a^b G(t,s)f(s,(u-w)(s))ds \end{eqnarray*} and hence we get \begin{eqnarray*} u^*(t)&= \left[1+\frac{ \sum_{i=1}^{m-2}\xi_i-1}{\Delta}(t-a)\right]\delta_1+\frac{t-a}{\Delta}\delta_2+\frac{1}{\Gamma(\beta)}\int_a^b G(t,s)f(s,u^*(s))ds. \end{eqnarray*} In other words, if \(u^*\) is a solution of the boundary value problem (1)-(2) then we get \begin{eqnarray*} D_a^\beta (u^*(t)+w(t))=D_a^\beta u^*(t)+ D_a^\beta w(t) = -f(t,u^*(t))-p(t)= -[f(t,u^*(t))+p(t)]= - F(t,u^*(t)), \end{eqnarray*} which implies that \begin{equation*} D_a^\beta u(t)=-F(t,u^*(t)). \end{equation*} Also from the boundary conditions, we easily see that \begin{equation*} u^*(a)=u(a)-w(a)=u(a)-0=\delta_1, \end{equation*} i.e., \( u(0)=\delta_1\) and \begin{align*} &u^*(b)=\sum_{i=1}^{m-2}\xi_i u^*(\eta_i)+\delta_2\\ &u(b)-w(b)= \sum_{i=1}^{m-2}\xi_i (u(\eta_i)-w(\eta_i))+\delta_2=\sum_{i=1}^{m-2}\xi_i u(\eta_i)-\sum_{i=1}^{m-2}\xi_i w(\eta_i)+\delta_2, \end{align*} i.e., \[u(b)= \sum_{i=1}^{m-2}\xi_i u(\eta_i)+\delta_2.\] Therefore \(u(t)\) is a solution of the boundary value problem (8)-(9).

Example 1. Consider the following fractional boundary value problem

with the function \(f(t,u(t)) = \frac{e^t}{1+t^2}\sin(u(t))\).

Choosing \(p(t)= e^t\) we get \(\int_{0}^{1} e^t dt =e-1 > 0\), so it is easy to check that the assumptions (H1)-(H2) hold. Calculating \(\Delta = \frac{29}{40}, M\cong2.9\) and seeing \(|F(t,u)| < 2 e = L\) such that \(|u|\leq K\) where \(K=20\), we can easily verify that

\[\left[1+\frac{ \sum_{i=1}^{m-2}\xi_i -1}{\Delta}(b-a)\right]\delta_1+ \frac{b-a}{\Delta}\delta_2 + L M \cong 15,3 \leq 20.\] Then applying Theorem 2 the problem (11)-(12) has a solution \(u(t)\).

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflicts of Interest

''The authors declare no conflict of interest.''

1. Introduction

Herpes simplex virus (HSV) is one of the most highly widespread sexually transmitted infections [1]. Lowestein confirmed the infectious nature of the HSV experimentally in 1919. In the years 1920s and 1930s, it was found that the HSV also infects the central nervous system [2]. The two strains of the disease are, HSV1 mostly known as cold sores and HSV2 also known as genital herpes. Both strains are transmitted sexually, however, HSV1 can also be transmitted non-sexually through contact with the fluid of an infected person. Looker et al., [3] estimated that about 417 million individuals with 267 million women inclusive age \(15-49\) have HSV2 infections worldwide, with the highest prevalence of 87% in Africa. Although South-East Asia and western pacific regions have a low prevalence of HSV2, they contribute significant amount to the global prevalence due to high population size. Fisman et al., [4] in their study on future dimensions and cost of HSV2 in the United States found that the prevalence of infection among individuals from age \(15-39\) years was projected to increase \(39%\) in men and \(49%\) in women by 2025.

In order to mitigate the spread and global socio-economic burden, mathematical modeling is an important tool that can provide insight into the long-term dynamics of a disease. It helps to simplify complex disease systems and has been a catalyst for decision making. Indeed, modeling has also been useful to present possible future outcomes of current trends and potential decision by policy makers. Some individuals infected with HSV are undiagnosed or do not display any physical symptoms. Latency is one of the major characteristics of the HSV [5,6,7]. Immunocompromised individuals with HSV stand a higher risk of acquiring HIV [8,9], and it has been found that HSV2 epidemics can more than double the peak of HIV incidence [8]. In fact, a key characteristic of HIV infection is poor control of herpes virus infections, which reactivate from latency and cause opportunistic infections in immunocompetent individuals [10].

Symptoms associated with HSV1 are tingling, itching or pains and sore throat which lead to blisters appearing on the face leading to sores, while in the case of HSV2, it appears on the genital areas [11,12] with associated symptoms such as headache, nerve pains, itching, lower abdominal pains, urinary difficulties, yeast infections, vaginal discharge, fever and open sores. Though HSV symptoms are mostly mild, there are other severity associated with such as ocular herpes which affects the eye leading to blindness; encephalitis also comes about as a result of being infected in the brain leading to death [13]. There is a reactivation after the latent infection to cause one or more rounds of the disease. And this reactivation can occur when infected individuals become sexually active again, weak immune system and inadequate or lack of treatment. Because there is currently neither complete treatment nor HSV vaccine, HSV which is a a life-long sexually-transmitted infection has no complete cure and an infected individual would live with it until death [14]. However, there are type-specific serology testing in the absence of symptoms which helps to determine the particular strain of the virus [15], and palliative treatment is administered to infected individuals to help get rid of the sores, reduce the risk of transmission as well as minimize the number and intensity of within host outbreaks. A study by the American College of Obstetricians and Gynecologists currently recommended the use of suppressive therapy to decrease transmission in discordant couples [16]. A comparison of two of these suppression drugs noted their effectiveness in decreasing both symptomatic and sub-clinical viral shedding [17,18].

Mathematical modeling is a useful tool to explore complex real life issue and guide experimental strategies and decision making [19]. Mathematical models of multiple strains of diseases such as HIV/AIDS, influenza and malaria have received much attention compared to HSV [20,21,22,23]. Nuno et al., [24] investigated the dynamics of a two-strain influenza with isolation and determined threshold conditions for the co-existence of the two strains. Because HSV treatement is only palliative, Schiffer et al., [25] developed a mathematical model to help optimize drug dose selection in clinical practice. To the best of our knowledge, a model that investigates conditions under which one strain of the virus would dominate or persist alone has not yet been considered. It is expected that this study will help fill the gap on HSV strains co-dynamics and provide a platform to further investigate if one strain could dominate and potentially drive the other to extinction.

The rest of this paper is organized as follows: The model formulation and the underlying assumptions are provided in \(§\)2. Theoretical analysis of the model is provided in \(§\)3. In \(§\)4, graphical representations generated (using the python programming language) to support the theoretical results are provided. \(§\)5 is the conclusion.

2. Model Formulation

We formulate a simple HSV transmission dynamics model of the 2 HSV strains in the presence of treatment. Our model partitions the total population at any time \(t\) denoted by \( N(t)\) into seven epidemiological states depending on individuals disease status. The fully susceptible class denoted by \(S(t)\). The class of individuals who have come in contact with the HSV virus strain 1 and are infectious is denoted by \(I_{1}(t)\), and those in contact with the HSV2 virus and infectious is denoted by \(I_{2}(t)\). Individuals affected with the two strains are grouped in the \(I_{12}(t)\) class. The \(I_{1}(t)\) class of individuals who are on palliative treatment move to the \(T_1(t)\) class, and those in \(I_2(t)\) under palliative treatment move to the \(T_2(t)\) class. Individuals co-infected with both HSV1 and HSV2 strain \(I_{12}(t)\) under treatment move to the \(T_{12}(t)\) class.

Individuals are recruited into the susceptible compartment at a constant rate \(\pi\). Individuals in each compartment die naturally at a rate \(\mu\). Since there is no cure for HSV, treatment is just for relief and to reduce individual infectiousness. We assume no simultaneous infections by both strains. Susceptible individuals progress into \(I_{1}\) class at a transmission rate of \(\beta_1\) and to \(I_{2}\) class at a transmission rate of \(\beta_2\). Individuals in the \(I_{1}\) move to the \(T_{1}\) class and those in the \(I_{2}\) move to the \(T_{2}\) class at a treatment rate of \(q_{1}\) and \(q_{2}\) respectively or enter into the dual infection class at the rate of \(\beta_{2}\) and \(\beta_{1}\) respectively. The infected individuals with both diseases move to the \(T_{12}\) class at a rate of \(q_{12}\). Because treatment is not permanent, relapse is common, thus from the treatment class, the disease can re-activate at a rate \(r_{1}\), \(r_{2}\) and \(r_{12}\) into the \(I_{1}\), \(I_{2}\) and \(I_{12}\) class respectively. The description of the model variables and parameters are summarized in Table 1. Figure 1 gives a graphical interpretation of our proposed model based on the above description and assumptions.

Table 1. Description of model variables and parameters.

	Description	Value	Reference
Parameter
\(\beta_1\)	Transmission  rate  for individuals with HSV1	\(0.007(0.001-0.03)yr^{-1}\)	Assumed
\(\beta_2\)	Transmission  rate  for individuals with HSV2	\(0.001(0.001-0.03)yr^{-1}\)	[26]
\(\mu\)	natural death rate	\(0.019(0.015-0.02)yr^{-1}\)	[27]
\(r_1\)	Reactivation  rate  of  HSV1	\(0.6\)	Assumed
\(r_2\)	Reactivation  rate  of  HSV2	\(0.6\)	Assumed
\(r_{12}\)	Reactivation  rate  of  both  HSV1 and HSV2	\(0.6\)	Assumed
\(q_{12}\)	Treatment     rate  of  both HSV1 and HSV2	\(0.45(0-1.0)yr^{-1}\)	Assumed
\(q_{1}\)	Treatment     rate  of  HSV1	\(0.45(0-1.0)yr^{-1}\)	Assumed
\(q_{2}\)	Treatment     rate  of  HSV2	\(0.45(0-1.0)yr^{-1}\)	[28]
\(\pi\)	Recruitment   rate	\(0.3yr^{-1}\)	Assumed
Variable
\(I_1(t)\)	Individuals infected with HSV1
\(I_2(t)\)	Individuals infected with HSV2
\(I_{12}(t)\)	Individuals infected with both HSV1 and HSV2
\(T_1(t)\)	Individuals infected with HSV1 and receiving treatment
\(T_2(t)\)	Individuals infected with HSV2 and receiving treatment
\(T_{12}(t)\)	Co-infected individuals receiving treatment

From the aforementioned, we established the following non-linear ordinary differential equations given by system (1)

with initial conditions \( S(0) > 0 , I_1 (0)\geq 0, I_2 (0)\geq 0, I_{12}(0)\geq 0, T_{1}(0)\geq 0, T_{2}(0)\geq 0, T_{12}(0)\geq 0.\) All the model parameters, their description, values and sources are presented in the Table 1.

3. Model analysis

The total non constant population is given by \(N(t) = S(t) + I_{1}(t) + I_{2}(t) + I_{12}(t) + T_{1}(t) + T_{2}(t) + T_{12}(t)\).

Since model 1 describe the dynamics of a human population, all state variables should be positive for the model to be epidemiological meaningful. Thus the following Lemma holds.

Lemma 1. The feasible set of model system (1) is given by \[\Omega = \left\lbrace \left( S, I_{1}, I_{2}, I_{12}, T_{1}, T_{2}, T_{12} \right) \in R^{7}_{+} : S+I_{1}+I_{2}+I_{12}+T_{1}+T_{2}+T_{12} \leq \frac{\pi}{\mu} \right\rbrace, \] which is bounded, positively invariant and attracting for all \(t\geq 0.\)

The disease-free equilibrium of model system (1) is given by \begin{align*} E_0 &= \left( \frac{\pi}{\mu}, 0, 0, 0, 0, 0, 0 \right). \end{align*} Using the next generation matrix method of van den Driessche and Watmough [29], compute the basic reproduction number, which represents the expected number of secondary cases produced by a typical infected individual during its entire period of infectiousness in a completely susceptible population when a single infected individual is introduced [29].

First, re-arrange the equations according to the infected compartments \(I_{1}(t), I_{2}(t), I_{12}(t), T_1,T_2,T_{12}\) .

Next, we compute the spectral radius (dominant eigenvalue) of the matrix \(\rho\left( \mathbf{F}\mathbf{V}^{-1}\right) = R_0\) from the arranged system (2). The matrices \(\mathbf{F}\) and \(\mathbf{V}\) are given by \begin{align*} \mathbf{F}&= \left(\begin{array}{rrrrrr} \frac{\beta_{1} \pi}{\mu} & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac{\beta_{2} \pi}{\mu} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right),\;\;\;\;\text{and}\;\;\;\;\; \mathbf{V}=\left(\begin{array}{rrrrrr} \mu + q_{1} & 0 & 0 & -r_{1} & 0 & 0 \\ 0 & \mu + q_{2} & 0 & 0 & -r_{2} & 0 \\ 0 & 0 & \mu + q_{12} & 0 & 0 & -r_{12} \\ -q_{1} & 0 & 0 & \mu + r_{1} & 0 & 0 \\ 0 & -q_{2} & 0 & 0 & \mu + r_{2} & 0 \\ 0 & 0 & -q_{12} & 0 & 0 & \mu + r_{12} \end{array}\right). \end{align*} The eigenvalues \(\lambda_{1,2}\) of \(\mathbf{F} \mathbf{V^{-1}}\) are solutions of the equation \(|\mathbf{F} \mathbf{V^{-1}} - \lambda I| = 0\), given by From Equation (3) we obtain, Let \(R_1\) and \(R_2\) be the reproduction number for HSV1 and HSV2 respectively, that are \begin{align*} R_1&= \frac{\beta_{1} \mu \pi + \beta_{1} \pi r_{1}}{\mu^{3} + \mu^{2} q_{1} + \mu^{2} r_{1}},\\ R_2 &= \frac{\beta_{2} \mu \pi + \beta_{2} \pi r_{2}}{\mu^{3} + \mu^{2} q_{2} + \mu^{2} r_{2}}. \end{align*} Thus, the basic reproduction number for the model system (1) is the spectral radius that is, the dominant eigenvalue of the next generation matrix \(\rho(\mathbf{F} \mathbf{V^{-1}})= R_0 = \max\left\lbrace R_1, R_2 \right\rbrace.\)

3.1. Local Stability of the Disease-Free Equilibrium

Theorem 1. The disease-free equilibrium of our system (1) is locally asymptotically stable if \(R_0 < 1\), and unstable if \(R_0 > 1\).

Proof. The proof is investigated by the linearization method. The Jacobian matrix associated with the model system (1) at the disease-free equilibrium is given by

The eigenvalues of \(J_{E_0}\) are \(\lambda_1 = -\mu - q_{12} - r_{12}\), \(\lambda_2 =-\mu\) repeated, while the other four eigenvalues are the solutions of the following equations: and From Equation (6) the coefficients \(\mathbb{A}, \mathbb{B}\) and \(\mathbb{C}\) are given by \begin{align*} \mathbb{A} &= 1,\\ \mathbb{B} & = \left(2 \, \mu - \frac{\beta_{2} \pi}{\mu} + q_{2} + r_{2}\right), \\ \mathbb{C} & = \mu^{2} - \beta_{2} \pi + \mu q_{2} + \mu r_{2} - \frac{\beta_{2} \pi r_{2}}{\mu} = [\mu^2 + \mu q_2 + \mu r_2 - \beta_2 \pi] [1-R_2]. \end{align*} From Equation (7), the coefficients \(\mathbb{D}, \mathbb{E}\) and \(\mathbb{F}\) are given by \begin{align*} \mathbb{D} &= 1, \\ \mathbb{E} &= \left(2 \, \mu - \frac{\beta_{1} \pi}{\mu} + q_{1} + r_{1}\right), \\ \mathbb{F} &= \mu^{2} - \beta_{1} \pi + \mu q_{1} + \mu r_{1} - \frac{\beta_{1} \pi r_{1}}{\mu}=[\mu^2 + \mu q_1 + \mu r_1 - \beta_1 \pi][1-R_1]. \end{align*} Next, we apply the Routh-Hurwitz criteria [30] which states that for the polynomial \(P(\lambda)\) to be negative or have negative real part, all coefficients must be strictly positive, which is a necessary but not a sufficient condition. Obviously, if \(R_0 < 1\), then all the eigenvalues of \(\mathbb{E}_0\) are negative and we can therefore conclude based on Routh-Hurwitz criterion that the disease-free equilibrium of model system (1) is locally stable.

3.2. Global Stability of the Disease-Free Equilibrium

To prove the global stability of the disease-free equilibrium, we employ the approach by Castillo Chavez [31]. From the model (1), the system of equations can be rewritten as \begin{align*} X'(t) &= F(X,Y),\\ Y'(t) &= G(X,Y),\;\;\;\;\;\; G(X,0) = 0, \end{align*} where \(X = (S)\) and \(Y = (I_{1}, I_{2}, I_{12}, T_{1}, T_{2}, T_{12})\) with \(X \in \mathbb{R}_{+}\) denoting (its components) the number of uninfected individuals and \(Y \in \mathbb{R}_{+}^{6}\) denoting (its components) the number of infected individuals. The disease-free equilibrium is now denoted by \(E_0 = (X_0,0)\) where \(X_0 = \dfrac{\pi}{\mu}\). The conditions for global stability of the disease-free equilibrium are given by

• (H1) For \(X'(t) = F(X^*,0),\)   \( X^*\) is globally asymptotically stable.
• (H2) \( G(X,Y) = AY-\hat{G}(X,Y),\)   \(\hat{G}(X,Y)\geq 0\) for \((X,Y)\in \Omega.\)

This implies that \(A=D_I G(X,0)\) is an \(M\)-matrix, that is the off diagonal entries of \(A\) are non-negative and \(\Omega\) is the region where the system of equations of the model makes epidemiological meaning. If the conditions above are satisfied using our model system (1) the following theorem holds:

Theorem 2. The fixed point \(E_0\) is a globally stable point of model system (1) provided \(R_0< 1\).

Proof. Consider \(F(X,0)=[\pi - \mu S]\), \begin{align*} A = \begin{bmatrix} \frac{\beta_1 \pi}{\mu}-\beta_1 I_2 - (q_{1}+\mu) & -\beta_1 I_1 & 0 & r_1 &0 &0\\ -\beta_2 I_1 & \frac{\beta_2 \pi}{\mu}-\beta_2 I_1 - (q_2 + \mu) & 0 & r_2 & 0 & 0 \\ \beta_1 I_2 + \beta_2 I_2 & \beta_1 I_1 + \beta_2 I_1& -q_{12} + \mu & 0 & 0 & r_{12}\\ q_1 & 0 & 0 & -(r+\mu) & 0 & 0 \\ 0 & q_2 & 0 & 0 & -(r_2 + \mu) & 0\\ 0 & 0 & q_{12} & 0 & 0 & -(r_{12} + \mu) \end{bmatrix} \end{align*} and \begin{align*} \hat{G}(X,Y) = \begin{bmatrix} \beta_1 I_1 \bigg(\dfrac{\pi}{\mu} - \dfrac{S}{N} \bigg)\\ \beta_2 I_2 \bigg(\dfrac{\pi}{\mu} - \dfrac{S}{N} \bigg) \end{bmatrix}\geq 0 \end{align*} and \(0\) for \(I_{12}, T_1, T_2, T_{12}\) respectively. Therefore \(\hat{G}(X,Y)\geq 0\) for all \((X,Y)\in \Omega\) implies that \(\mathbb{E}_0\) is globally asymptotically stable for \(R_0 < 1\).

3.3. Endemic Equilibria

When there is no infection with strain \(2\), that is when \(I_2^* = 0\), there is an equilibrium \[E_1 = (s*, i_1^*, 0,0,t_1^*,0,0),\] where \( s^* = \dfrac{\pi}{\mu} \dfrac{1}{R_1},\)   \( t_1^* = \dfrac{q_1 \mu^2 [R_1 - 1]}{\beta_1\mu^3(\mu + r_1)}\) and \(i^*_1 = \dfrac{\mu[R_1-1]}{\beta_1}.\)

This equilibrium makes biological sense only when \(R_1 > 1\). Note that \(E_1\) partitions the population into parts, that is \(\dfrac{1}{R_1}\) uninfected which we observed from the \(s^*\) term. Also, a portion of the population represented by \(\dfrac{\pi}{\mu}\) remains in that class until death. The other parts are \(\dfrac{\mu}{\beta_1}\) and \(\dfrac{q_1}{\beta_1 \mu(\mu+r_1)}\).

When there is no infection with strain 1, that is when \(I_1^* = 0\), a second single-strain equilibrium is given by \[E_2 = (s^*, 0, i^*_2, 0, 0, t^*_2,0),\] where \(s^* = \dfrac{\pi}{\mu} \dfrac{1}{R_2},\)   \(\dfrac{\mu [R_2 - 1]}{\beta_2}\) and \(t^*_2 = \dfrac{q_2 \mu^2[R_2-1]}{\beta_2\mu^3(\mu+r_2)}.\) This equilibrium makes biological sense only when \(R_2>1\).

3.4. Stability analysis using invasion method

Using the approach in [32], we construct the invasion reproduction number \(\tilde{R}_1\), where \(E_2\) is considered the disease-free equilibrium. Hence, we only consider those equations representing the classes infected with strain 1; that is the \(\mathcal{F}_1\) matrix is given by the new infection terms in the equations \(I_1^{\prime}, I_{12}^{\prime}, T_1^{\prime}, T_{12}^{\prime}\) and the \(\mathcal{V}_1\) matrix consists of the remainder of the terms in those equations. \begin{align*} \mathcal{F}_1 - \mathcal{V}_1 = \begin{pmatrix} \dfrac{\beta_1 I_1 S}{N} \\[5pt] \dfrac{\beta_1 I_1 I_2}{N} \\[5pt] 0 \\[5pt] 0 \end{pmatrix} - \begin{pmatrix} \dfrac{\beta_2 I_1(\tilde{I_2})}{N} + q_1 I_1 + \mu I_1 - r_1 T_1 \\[5pt] -\dfrac{\beta_2 \tilde{I}_2 I_1}{N} - r_{12}T_{12} + (q_{12}+\mu)I_{12}\\[5pt] -q_1I_1 + (r_1+\mu)T_1 \\[5pt] -q_{12}I_{12} + (r_{12}+\mu)T_{12} \end{pmatrix}. \end{align*} Computing the Jacobian of each matrix at the equilibrium in strain 1, we obtain the following matrices \(F_1\) and \(V_1\): \begin{align*} F_1 & = \begin{pmatrix} \beta_1 S^{*} & \beta_1 S^{*} & 0 & 0 \\[5pt] \beta_1 I^{*}_{2} & \beta_1 I_2^{*} & 0 & 0 \\[5pt] 0 & 0 & 0 & 0 \\[5pt] 0 & 0 & 0 & 0 \end{pmatrix}\quad \text{and} \quad V_1 = \begin{pmatrix} \beta_2 I_2^{*} + q_1 + \mu & 0 & -r_1 & 0 \\[5pt] -\beta_2 I_{2}^{*} & (q_{12} + \mu) & 0 & -r_{12}\\[5pt] -q_1 & 0 & r_1+\mu & 0 \\[5pt] 0 & -q_{12} & 0 & r_{12}+\mu \end{pmatrix}. \end{align*} Next, we compute the dominant eigenvalue of the matrix \(F_1V_1^{-1}\) which represents the invasion reproductive number \(\tilde{R}_1\) where the assumption that \(R_2>1\) implicitly is made and is given by \begin{align*} \tilde{R}_1 =\dfrac{\beta_1}{\mu} \bigg[ \frac{\mathbf{A} + \mathbf{B} + {\left(\mathbf{C} + \mathbf{D} + {\left(\beta_{1} \mu^{2} + \beta_{1} \mu q_{12}\right)} r_{1} + {\left(\beta_{1} \mu^{2} + \beta_{1} \mu r_{1}\right)} r_{12}\right)} S}{\mu^{3} + \mu^{2} q_{1} + \mathbf{E} + {\left(\mu^{2} + \mu q_{1}\right)} q_{12} + {\left(\mu^{2} + \mu q_{12}\right)} r_{1} + {\left(\mu^{2} + \mu q_{1} + \mu r_{1}\right)} r_{12}} \bigg], \end{align*} where \begin{align*} \mathbf{A} &= {\left( \beta_{2} \mu^{2} + \beta_{2} \mu r_{1} + {\left( \beta_{2} \mu + \beta_{2} r_{1}\right)} r_{12}\right)} I_{2}^{2}, \\[5pt] \mathbf{B} &= {\left( \mu^{3} + \mu^{2} q_{1} + \mu^{2} r_{1} + {\left( \mu^{2} + \mu q_{1} + \mu r_{1}\right)} r_{12}\right)} I_{2},\\[5pt] \mathbf{C} &= \mu^{3} + \mu^{2} q_{12},\\[5pt] \mathbf{D} &= {\left( \beta_{2} \mu^{2} + \beta_{2} \mu r_{1} + {\left( \beta_{2} \mu + \beta_{2} r_{1}\right)} r_{12}\right)} I_{2},\\[5pt] \mathbf{E} &= {\left(\beta_{2} \mu^{2} + \beta_{2} \mu q_{12} + {\left(\beta_{2} \mu + \beta_{2} q_{12}\right)} r_{1} + {\left(\beta_{2} \mu + \beta_{2} r_{1}\right)} r_{12}\right)} I_{2}. \end{align*} After some algebraic manipulations, we obtain \begin{align*} \tilde{R}_1 =& R_1 \dfrac{\mu(a)}{\pi(\mu+r_1)}\\&\times\dfrac{s^*\mu^3 + \mu(a)(\mu+r_{12})i^*_2+s^*\mu\beta_1(\mu r_1+(\mu+r_1)r_{12})+s^*\mu q_{12}(\mu+r_1\beta_1)+(\mu+r_1)(\mu+r_{12})i^*_2(s^*+i^*_2)\beta_2}{\mu(a)(\mu+q_{12}+r_{12})+(q_{12}(\mu+r_1)+\mu(\mu+2r_1)+(\mu+r_1)r_{12}i^*_2\beta_2}, \end{align*} where \(a = \mu+q_1 +r_1\) and \(s^*, i^*_2\) defined in \(E_2\). Note that \(\tilde{R}_1\) is essentially \(R_1\) multiplied by a term representing an altered vulnerability to infection with strain 1.

We further consider the invasion reproductive number \(\tilde{R}_2\). It is the ability of strain \(2\) to invade the susceptible population at \(E_1\). To determine \(\tilde{R}_2\), we follow a similar approach using the next generation matrix method. The \(\mathcal{F}_2\) matrix is given by the new infection terms in the equations \(I^{\prime}_{2}, I^{\prime}_{12}, T^{\prime}_{2}, T^{\prime}_{12}\) and the \(\mathcal{V}_2\) matrix consists of the remainder of the terms in those equations. Thus, we obtain

\begin{align*} \mathcal{F}_2 - \mathcal{V}_2 = \begin{pmatrix} \dfrac{\beta_2 \tilde{I_2} S}{N} \\[5pt] \dfrac{\beta_2\tilde{I_2} I_1}{N} \\[5pt] 0 \\[5pt] 0 \end{pmatrix} - \begin{pmatrix} \dfrac{\beta_1 I_2\tilde{I}_1}{N} + (q_2 + \mu)I_2 - r_2 T_2 \\[5pt] -\dfrac{\beta_1 \tilde{I}_1 I_2}{N} - r_{12}T_{12} + (q_{12}+\mu)I_{12}\\[5pt] -q_2I_2 + (r_2+\mu)T_2 \\[5pt] -q_{12}I_{12} + (r_{12}+\mu)T_{12} \end{pmatrix}. \end{align*} We then compute the Jacobian of the following \(F_2\) and \(V_2\) at strain \(2\) to obtain \begin{align*} F_2 & = \begin{pmatrix} \beta_2 S^{*} & \beta_2 S^{*} & 0 & 0 \\[5pt] \beta_2 I^{*}_{1} & \beta_2 I_1^{*} & 0 & 0 \\[5pt] 0 & 0 & 0 & 0 \\[5pt] 0 & 0 & 0 & 0 \end{pmatrix}\quad \text{and} \quad V_2 = \begin{pmatrix} \beta_1 I_1^{*} + q_2 + \mu & 0 & -r_2 & 0 \\[5pt] -\beta_1 I_{1}^{*} & (q_{12} + \mu) & 0 & -r_{12}\\[5pt] -q_2 & 0 & r_2+\mu & 0 \\[5pt] 0 & -q_{12} & 0 & r_{12}+\mu \end{pmatrix}. \end{align*} The dominant eigenvalue of the matrix which is determined by \(F_1V_1^{-1}\) and is the invasion reproductive number \(\tilde{R}_2\) is given by \begin{align*} \tilde{R}_2 = \frac{\beta_2}{\mu} \bigg[ \frac{\textbf{M} + \textbf{N} + {\left(\textbf{O} + \textbf{P} + {\left(\beta_{2} \mu^{2} + \beta_{2} \mu q_{12} + \beta_{2} \mu r_{12}\right)} r_{2}\right)} S}{\mu^{3} + \mu^{2} q_{12} + \textbf{Q} + {\left(\mu^{2} + \mu q_{12}\right)} q_{2} + {\left(\mu^{2} + \mu q_{2}\right)} r_{12} + {\left(\mu^{2} + \mu q_{12} + \mu r_{12}\right)} r_{2}} \bigg], \end{align*} where \begin{align*} \textbf{M} & = {\left(\beta_{1} \mu^{2} + \beta_{1}\mu r_{12} + {\left(\beta_{1}\mu + \beta_{1} r_{12}\right)} r_{2}\right)} I_{1}^{2}, \\[5pt] \textbf{N} &= {\left(\mu^{3} + \mu^{2} q_{2} + {\left(\mu^{2} +\mu q_{2}\right)} r_{12} + {\left(\mu^{2} + \mu r_{12}\right)} r_{2}\right)} I_{1}, \\[5pt] \textbf{O} & = \mu^{3} + \mu^{2} q_{12} + \mu^{2} r_{12}, \\[5pt] \textbf{P} & = {\left(\beta_{1} \mu^{2} + \beta_{1} \mu r_{12} + {\left(\beta_{1}\mu + \beta_{1}r_{12}\right)} r_{2}\right)} I_{1}, \\[5pt] \textbf{Q} & = {\left(\beta_{1} \mu^{2} + \beta_{1} \mu q_{12} + \beta_{1} \mu r_{12} + {\left(\beta_{1} \mu + \beta_{1} q_{12} + \beta_{1} r_{12}\right)} r_{2}\right)} I_{1}. \end{align*} After some algebraic manipulations, \(\tilde{R}_2\) can further be expressed as \begin{align*} \tilde{R}_2 = R_2\dfrac{\mu(a_1)}{\pi(\mu+r_2)}\dfrac{s^*\mu q_{12}(\mu+r_2\beta_2)+(\mu+r_{12})(\mu+r_2)i^{*^2}_1\beta_1+i^*_1(\mu q_2 +(\mu+r_2)(\mu+s^*\beta_1))+s^*\mu(\mu+r_2\beta_2))}{(\mu+q_{12}+r_{12})(\mu q_2+(\mu+r_2)(\mu+i^*\beta_1))} \end{align*} where \(a_1 = \mu+q_2+r_2\) and \(s^*, i^*_1\) is defined in \(E_1\). Similarly, \(\tilde{R}_2\) is essentially \(R_2\) multiplied by a term representing an altered vulnerability to infection with strain \(2\).

4. Numerical simulations

To support the analytical results, numerical simulations of the model system 1 are provided. Model parameter values used for the numerical simulations are listed in Table 1. For the purpose of illustration, we assumed heuristic model parameter values within realistic range for some of the model parameter values, and the following initial values: \(S = 30\), \(I_{1} = 8\), \(I_{2} = 8\), \(I_{12} = 8\) , \(T_{1} = 3\), \(T_{2} = 3\), \(T_{12} = 3\).

Figure 2 depicts the graphical representation of infectious individuals with HSV1 and HSV when either \(R_1>1>R_2\) Figure 2(a) or \(R_2>1>R_1\) Figure 2(b). In Figure 2(a), strain 1 dominates while in Figure 2(b), strain 2 dominates.

Next, we illustrate the effects of increasing treatment rates on the dynamics of population with HSV1 and HSV2 in Figure 3.

Figure 3 illustrates the effects of increasing treatment as a control strategy in a given population. In Figure 3(a), it is observed that when treatment rates with respect to HSV1 is small, the infected individuals increases. In Figure 3(b), the treatment rates for HSV1 infection is increased to a very reasonable value (\(0.65\)) and it is observed that although there is a reduction in the number of infected individuals, the impact is minimal. The treatment rate for the HSV1 is further increased to a reasonably high value (\(0.95\)) and it is observed that the infection persists, but at a lower rate. Thus, treatment only minimizes the rate of transmitting HSV strain 1, but does not eradicate it, which agrees with what is know about this disease that treatment is only palliative.

Simulations in Figure 4 illustrates the effects of increasing treatment as a control strategy in a given population. In Figure 4(a), it is observed that when treatment rates with respect to HSV2 is small, the infected individuals increases. In Figure 4(b) the treatment rate for HSV2 infection was increased to a very reasonable value (\(0.65\)) and it is observed that although there is a reduction in the number of infected individuals the impact is not that great. The treatment rate for the HSV2 was further increased to a reasonably high value (\(0.95\)) and it is observed that the infection persists but at a lower rate. Again, treatment only minimizes the rate of transmitting HSV strain 2, but does not eradicate it, which agrees with what is know about HSV that treatment only palliative.

4.1. Sensitivity Analysis

Some of the model parameter values used herein were obtained from various sources, while the rest were assumed for illustrative purposes. Because errors could occur while collecting data or estimating model parameter values, it is important to investigate the sensitivity of the model parameters. In general, sensitivity analysis is to determine which model input parameters exert the most influence on the model results [33]. This information could then be used to tailor disease control strategy on the most sensitive parameters. Thus, we aim to investigate the relative importance of each parameter on the transmission dynamic of the disease, using the normalized forward sensitivity method [34]. This approach states that sensitivity indices are determined when a change in parameter allow us to measure the relative change in a state variable. It is important to note that while some sensitivity methods are mathematically elegant and comprehensive, their results in many cases are comparable to those obtained from simpler techniques [35].

Using sage programming language, we derive the following:

\begin{align*} \beta_1 &=\newcommand{\Bold}[1]{\mathbf{#1}}\frac{{\left(\mu \pi + \pi r_{1}\right)} \beta_{1}}{\beta_{1} \mu \pi + \beta_{1} \pi r_{1}},\\ r_1 &=\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{{\left(\mu^{3} + \mu^{2} q_{1} + \mu^{2} r_{1}\right)} {\left(\frac{{\left(\beta_{1} \mu \pi + \beta_{1} \pi r_{1}\right)} \mu^{2}}{{\left(\mu^{3} + \mu^{2} q_{1} + \mu^{2} r_{1}\right)}^{2}} - \frac{\beta_{1} \pi}{\mu^{3} + \mu^{2} q_{1} + \mu^{2} r_{1}}\right)} r_{1}}{\beta_{1} \mu \pi + \beta_{1} \pi r_{1}},\\ q_1 &=\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{\mu^{2} q_{1}}{\mu^{3} + \mu^{2} q_{1} + \mu^{2} r_{1}},\\ \\ \mu &=\newcommand{\Bold}[1]{\mathbf{#1}}\frac{\mu{\left(\mu^{3} + \mu^{2} q_{1} + \mu^{2} r_{1}\right)} {\left(\frac{\beta_{1} \pi}{\mu^{3} + \mu^{2} q_{1} + \mu^{2} r_{1}} - \frac{{\left(\beta_{1} \mu \pi + \beta_{1} \pi r_{1}\right)} {\left(3 \, \mu^{2} + 2 \, \mu q_{1} + 2 \, \mu r_{1}\right)}}{{\left(\mu^{3} + \mu^{2} q_{1} + \mu^{2} r_{1}\right)}^{2}}\right)}}{\beta_{1} \mu \pi + \beta_{1} \pi r_{1}},\\ \pi &=\newcommand{\Bold}[1]{\mathbf{#1}}\frac{{\left(\beta_{1} \mu + \beta_{1} r_{1}\right)} \pi}{\beta_{1} \mu \pi + \beta_{1} \pi r_{1}},\\ \beta_2 &=\newcommand{\Bold}[1]{\mathbf{#1}}\frac{{\left(\mu \pi + \pi r_{2}\right)} \beta_{2}}{\beta_{2} \mu \pi + \beta_{2} \pi r_{2}},\\ r_2 &=\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{{\left(\mu^{3} + \mu^{2} q_{2} + \mu^{2} r_{2}\right)} {\left(\frac{{\left(\beta_{2} \mu \pi + \beta_{2} \pi r_{2}\right)} \mu^{2}}{{\left(\mu^{3} + \mu^{2} q_{2} + \mu^{2} r_{2}\right)}^{2}} - \frac{\beta_{2} \pi}{\mu^{3} + \mu^{2} q_{2} + \mu^{2} r_{2}}\right)} r_{2}}{\beta_{2} \mu \pi + \beta_{2} \pi r_{2}},\\ q_2 &=\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{\mu^{2} q_{2}}{\mu^{3} + \mu^{2} q_{2} + \mu^{2} r_{2}}. \end{align*} Then, using the parameter values \(\beta_1 = 0.007\), \(\beta_2 = 0.001\), \(r_1= 0.6\), \(r_2= 0.6\), \(q_1 = 0.45\), \(q_1 = 0.45\), \(\mu = 0.019\), \(\pi = 0.3\), we compute the sensitivity indices of \(R_1\) and \(R_2\) which are summarized in Tables 2 and 3.

Table 2. Sensitivity indices of \(R_1\).

Parameter	Sensitivity index
\(\beta_1\)	\(1.00000\)
\(r_1\)	\(0.408033 \)
\(q_1\)	\(-0.42095\)
\(\mu \)	\(-1.98708\)
\(\pi \)	\(1.00000\)

Table 3. Sensitivity indices of \(R_2\).

Parameter	Sensitivity index
\(\beta_2\)	\(1.00000\)
\(r_2\)	\(0.408033 \)
\(q_2\)	\(-0.42095\)
\(\mu \)	\(-1.98708\)
\(\pi \)	\(1.00000\)

From Tables 2 and 3, the sensitivity index with positive sign indicate that the value of the reproduction numbers \(R_1\) and \(R_2\) increase when the corresponding parameters increase while the parameters with negative signs indicate that, for an increase in the corresponding parameters, there is a decrease in the value of the reproduction numbers \(R_1\) and \(R_2\). Using the model parameter values in Table 1, we graphically represent the sensitivity index profile of the reproduction numbers \(R_1\) and \(R_2\).

From Figure 5(a) and Figure 5(b), it can be observed that \(\beta_1\), \(\beta_2\) and \(\pi\) have the highest influence on the reproduction number \(R_0\) followed by \(\mu\), \(q_1\) and \(q_2\). This implies that an increase in the contact rates \(\beta_1\), \(\beta_2\) and recruitment/inflow rate \(\pi\) will lead to a corresponding increase in the basic reproduction number. On the other hand, \(\mu\), \(q_2\) and \(q_1\) correlates negatively with the basic reproduction number as an increase in such parameters will lead to a corresponding decrease in the basic reproduction number.

5. Conclusion

We formulated and analyzed a mathematical model for the transmission of HSV infection with palliative treatment. Individuals from the susceptible compartments could either be infected with strain 1 or strain 2 of the disease. Infected individuals could then either go for treatment or get infected with the other strain (co-infection) and also receive treatment. Since this disease has no cure, the treatment only reduces the intensity of the disease but does not cure it. The model is then analyzed qualitatively with numerical simulations provided to support the theoretical results. The model basic reproduction number is computed for both strains independently with \(R_0=max \lbrace R_1, R_2 \rbrace \), and used to investigate the stability of the model equilibria. The disease-free equilibrium of the proposed model is locally-asymptotically stable when the basic reproduction number \(R_0=max\lbrace R_1, R_2 \rbrace \) is less than unity. Numerical simulations indicate that the two HSV strains co-exist, with HSV1 dominating but not driving out HSV2 whenever \(R_1 > R_2 > 1\) and vice versa. If infection with one strain confers incomplete immunity against the other, the model 1 exhibits the phenomenon of competitive exclusion, where the first strain HSV1 could drive out the second strain HSV2 when \(R_1 > 1 > R_2\). Using the method in [32], we establish existence and local stability of single strain equilibria through invasion reproductive numbers. In order to determine the relative importance of model parameters to initial disease transmission, sensitivity analysis is carried out. The reproduction number is most sensitive respectively to the contact rates \(\beta_1\) and \(\beta_2\) as well as the recruitment rate \( \pi \). The application of control measures such as palliative treatment has an impact on the infection dynamics, but does not completely eradicate the disease. This study is not exhaustive as there are a number of limitations. Treatment could significantly help to minimize transmission of the disease but not eradicate it (because it is only a palliative measure), development of treatment resistance is to be expected [36]. Also, while there is no definitive vaccine, studies have shown that even an imperfect prophylactic HSV-2 vaccine could have an important public health benefits on HSV-2 incidence [14]. Future studies accounting for heterogeneity in infection rates such as by age, sex and sexual activity and incorporating the above limitations are viable.

Acknowledgments

JK thanks the African Institute for Mathematical Science (AIMS-Tanzania) for financial support towards her M.Sc. in mathematical sciences.

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflicts of Interest

''The authors declare no conflict of interest.''

1. Introduction

Let \(\mathcal{U}=\{z:z\in\mathbb{C}, |z|< 1\}\) be a unit disk and let \(\mathcal{A}\) denote the class of analytic functions of the form

normalized by the conditions \(f(0)=f'(0)-1=0\). Let \(\mathcal{S}\subset\mathcal{A}\) be the class of analytic and univalent functions in \(\mathcal{U}\).

Let \(\mathcal{W}\) denote the class of functions

\begin{equation*}\label{funSchwarz} w(z)=w_1z + w_2z^2 + w_3z^3+\cdots\quad (z\in\mathcal{U}), \end{equation*} such that \(w(0)=0\) and \(|w(z)|< 1\). The class \(\mathcal{W}\) is known as the class of Schwarz functions.

By [1], let \(j(z)\), \(J(z)\in\mathcal{A}\), then \(j(z)\prec J(z)\), \(z\in\mathcal{U}\), if \(\exists w(z)\) analytic in \(\mathcal{U}\), such that \(w(0)=0\), \(|w(z)|< 1\) and \(j(z)=J(w(z))\). If the function \(J(z)\) is univalent in \(\mathcal{U}\), then \(j(z)\prec J(z)\implies j(0) = J(0)\) and \(j(\mathcal{U})\subset J(\mathcal{U})\).

Let \(\mathcal{P}\) denote the class of functions

which are analytic in \(\mathcal{U}\) such that \(\mathcal{R}e(p(z))>0\) and \(p(0)=1\). It is known that functions in classes \(\mathcal{P}\) and \(\mathcal{W}\) are related such that In [2], Ma and Minda defined a function \(\phi\in\mathcal{P}\ (z\in\mathcal{U})\) such that \(\phi(0)=1\), \(\phi'(0)>0\) and \(\phi(\mathcal{U})\) is starlike with respect to 1 and symmetric with respect to the real axis. Such function \(\phi\) can be expressed as Fekete and Szegö [3] investigated the coefficient functional \begin{equation*}\label{funFSFunctional} g_\rho(f)=|a_3 - \rho a_2^2|, \end{equation*} which arose from the disproof of Littlewood-Parley conjecture (see [1]) that says modulus of coefficients of odd univalent functions are less than 1. This functional has been investigated by many researchers, see for instance [4,5].

Historically, Lewin [6] introduced a subclass of \(\mathcal{A}\) called the class of bi-univalent functions and established that \(|a_2|\leq 1.51\) for all bi-univalent functions. Also, the Koebe 1/4 theorem (see [1]) states that the range of every function \(f\in\mathcal{S}\) contains the disk \(D=\{\omega:|\omega|< 0.25\}\subseteq f(\mathcal{U})\). This implies that \(\forall f\in\mathcal{S}\) has an inverse function \(f^{-1}\) such that

\begin{align*} f^{-1}(f(z)) &= z \quad (z\in\mathcal{U}), \end{align*} and \begin{align*} f(f^{-1}(\omega)) &= \omega\quad (\omega:|\omega| < r_0(f);\; r_0(f)\geq 0.25), \end{align*} where \(f^{-1}(\omega)\) is expressed as Thus, a function \(f\in\mathcal{A}\) is said to be bi-univalent in \(\mathcal{U}\) if both \(f(z)\) and \(F(\omega)\) are univalent in \(\mathcal{U}\). Let \(\mathcal{B}\) denote the class of analytic and bi-univalent functions in \(\mathcal{U}\).

Some functions \(f\in\mathcal{B}\) includes \(f(z)=z\), \(f(z)=z/(1-z)\), \(f(z)=-\log(1-z)\) and \(f(z)=\frac{1}{2}\log[(1+z)/(1-z)]\). Observe that some familiar functions \(f\in\mathcal{S}\) such as the Koebe function \(K(z)=z/(1-z)^2\), its rotation function \(K_\sigma(z)=z/(1-e^{i\sigma}z)^2\), \(f(z)=z - z^2/2\) and \(f(z)=z/(1-z^2)\) are nonmembers of \(\mathcal{B}\). See [4,5,7,8,9,10,11] for more details.

Jackson [12] (see also [8,13,14]) introduced the concept of \(q\)-derivative operator. For functions \(f\in\mathcal{A}\), the \(q\)-derivative of \(f\) can be defined by

where \(\mathcal{D}_q f(0)=f'(0)\) and \(\mathcal{D}_qf(z)z=\mathcal{D}_q(\mathcal{D}_qf(z))\). From (1) and (6) we get where \([n]_q=\frac{1-q^n}{1-q}\), \([n-1]_q=\frac{1-q^{n-1}}{1-q}\), \(\lim\limits_{q\uparrow 1}[n]_q = n\) and \(\lim\limits_{q\uparrow 1}[n-1]_q = n-1\).

For instance, if \(\alpha\) is a constant, then for the function \(f(z)=\alpha z^n\),

\[\mathcal{D}_qf(z)=\mathcal{D}_q(\alpha z^n)=\frac{1-q^n}{1-q}\alpha z^{n-1}=[n]_q\alpha z^{n-1}\,,\] and note that \[\lim\limits_{q\uparrow 1}\mathcal{D}_qf(z)=\lim\limits_{q\uparrow 1}[n]_q\alpha z^{n-1}=n\alpha z^{n-1}=:f'(z)\,,\] where \(f'(z)\) is the classical derivative.

In this study, the \(q\)-derivative operator and the subordination principle are used to define and generalize a subclass of bi-univalent functions. Afterwards, some coefficient bounds and some Fekete-Szegö estimates were investigated. Some of our results generalised that of Srivastava and Bansal in [10] and some new results are added.

Definition 1. Let \(0< q< 1\), \(\tau \in\mathbb{C}\setminus\{0\}\), \(0\leq \lambda\leq 1\) and \(\phi\) is defined in (4). A function \(f\in\mathcal{B}\) is said to be in the class \(\mathcal{B}_q(\tau,\lambda,\phi)\) if the subordination conditions

and where \(F(\omega)=f^{-1}(\omega)\) are satisfied.

Remark 1. Let \(q\uparrow 1\) in (8) and (9), then \(\mathcal{B}_q(\tau,\lambda,\phi)\) becomes the class \(\mathcal{B}(\tau, \lambda, \phi)\) investigated by Srivastava and Bansal [10].

2. Preliminary Lemmas

To establish our results, we shall need the following lemmas. Let \(p(z)\) be as defined in (2).

Lemma 2 ([1]). If \(p(z)\in\mathcal{P}\), then \(|p_n|\leq 2\ (n\in\mathbb{N}). \) The result is sharp for the well-known Möbius function.

Lemma 3 ([15,16]). If \(p(z)\in\mathcal{P}\), then \(2p_2 = p^2_1 + (4-p^2_1)x \) for some \(x\) and \(|x|\leq 1\).

3. Main Results

Unless otherwise mentioned in what follows, we assume throughout this work that \(0< q< 1\), \(\tau\in\mathbb{C}\setminus\{0\}\), \(0\leq \lambda \leq 1\), \(\phi\) is as defined in (4) and \(f\in\mathcal{B}\), hence our results are as follows:

Theorem 4. Let \(f\in\mathcal{B}_q(\tau,\lambda,\phi)\), then

where \(\beta_1>0\) and \(\beta_n\ (n\in\mathbb{N})\) are coefficients of \(\phi(z)\) in (4).

Proof. Let \(f(z)\in\mathcal{B}\) and \(F(\omega)=f^{-1}(\omega)\), then there exists the analytic functions \(u(z), v(\omega)\in\mathcal{W}\), \(z,\omega\in\mathcal{U}\) such that \(u(0)=0=v(0)\), \(|u(z)|< 1\), \(|v(\omega)|< 1\) so that they satisfy the subordination conditions:

and By substituting (7) into LHS of (12) we respectively get and following the same process for \(F(\omega)\) in (5) gives Now to expand and in series form, let \(\delta_1(z)=1+b_1z+b_2z^2+\dots\), \(\delta_2(\omega)=1+c_1\omega+c_2\omega^2+\dots\in\mathcal{P}\), then by (3), and following the same process Substituting (18) into (16) as expressed by (4) we get and substituting (19) into (17) as expressed by (4) we get Now comparing the coefficients in (14) and (20) we get and comparing the coefficients in (15) and (21) gives Now adding (22) and (24) and simplifying we get Also from (22) and (24) we get and adding (23) and (25) and using (26) we get From (27) and using (26) we get So that by substituting for \(b_1^2\) in (28) we get and applying Lemma 2 gives (10).

Again by subtracting (23) from (25), using (26) and simplifying we get

Thus, from (27), using (26) and simplifying we get and applying Lemma 2 gives (11).

Let \(q\uparrow 1\), then Theorem 4 becomes

Corollary 5. Let \(f(z)\in\mathcal{B}_q(\tau, \lambda, \phi)\), then as \(q\uparrow 1\), \begin{align*} |a_2| &\leq \frac{|\tau|\beta_1^{3/2}} {\sqrt{|\tau[3]_q \beta_1^2 +[2]_q^2 (\beta_1-\beta_2)|}}\,,\\ |a_3| &\leq \frac{|\tau|^2 \beta_1^2}{[2]^2_q} + \frac{ |\tau| \beta_1}{[3]_q}\,. \end{align*} which is the result of Srivastava and Bansal [10].

Theorem 6.( Fekete-Szegö Estimate, \(\varrho\in\mathbb{R}\)). If \(f\in\mathcal{B}_q(\tau,\lambda,\phi)\) and \(\varrho\in\mathbb{R}\), then \[ \mbox{\(|a_3 - \varrho a_2^2|\)}\leq \left\{ \begin{array}{rl} \frac{|\tau|\beta_1}{[3]_q(1+[2]_q\lambda)} & \mbox{for \(0\leq |h(\varrho)|\leq\frac{1}{[3]_q(1+[2]_q\lambda)}\);}\\ |\tau|\beta_1|h(\rho)| & \mbox{for \(|h(\varrho)|\geq \frac{1}{[3]_q(1+[2]_q\lambda)}\),} \end{array}\right. \] where

Proof. From (30) and (31), \begin{align*} |a_3 - \varrho a_2^2| &= \left|\frac{\tau \beta_1 (b_2 - c_2)}{4[3]_q(1+[2]_q\lambda)} + (1 - \varrho)a^2_2\right|\\ &= \left|\frac{\tau \beta_1}{4}\left\{\frac{(b_2 - c_2)}{[3]_q(1+[2]_q\lambda)} + (b_2+c_2)h(\varrho)\right\}\right|\,, \end{align*} where \(h(\varrho) \) is given in (33), so that by applying triangle inequality, (4), Lemma 2 and simplifying complete the proof.

Theorem 7( Fekete-Szegö Estimate, \(\rho\in\mathbb{C}\)). If \(f\in\mathcal{B}_q(\tau, \lambda, \phi)\) and \(\rho\in\mathbb{C}\), then

where \begin{equation*} \xi = \frac{[2]^2_q(1+[1]_q\lambda)^2}{|\tau|\beta_1[3]_q(1+[2]_q\lambda)}\,. \end{equation*}

Proof. From (27) and (31) and using (26),

From Lemma 3 and (26) for some \(x, y,|x|\leq 1,|y|\leq 1\) and \(|b_1|\in[0,2]\). Thus using (36) in (35) simplifies to \begin{equation*} a_3 - \rho a^2_2 = (1-\rho)\frac{\beta_1^2 b_1^2 \tau^2}{4[2]^2_q(1+[1]_q\lambda)^2} + \frac{\beta_1\tau (4-b^2_1)}{8[3]_q(1+[2]_q\lambda)}(x-y). \end{equation*} For \(\delta(z)=1+b_1z + b_2z^2 + \cdots\in\mathcal{P}\), \(|b_1| \leq 2\) by Lemma 2. Letting \(b= b_1\), we may assume without any restriction that \(b\in [0,2]\). Now using triangle inequality, letting \(X=|x|\leq 1\) and \(Y = |y|\leq 1\), then we get \begin{align*} |a_3 - \rho a^2_2|&\leq |1-\rho|\frac{\beta_1^2 b^2|\tau|^2}{4[2]^2_q(1+[1]_q\lambda)^2} + \frac{\beta_1|\tau|(4-b^2)}{8[3]_q(1+[2]_q\lambda)}(X + Y) = H(X,Y). \end{align*} For \(X,Y\in [0,1]\); \begin{equation*} \max\{H(X,Y)\} = H(1,1) = \frac{\beta_1^2 |\tau|^2}{4[2]^2_q(1+[1]_q\lambda)^2}\left\{ |1-\rho| - \frac{[2]_q^2(1+[1]_q\lambda)^2}{\beta_1|\tau|[3]_q(1+[2]_q\lambda)}\right\}b^2 + \frac{\beta_1|\tau|}{[3]_q(1+[2]_q\lambda)} = G(b)\label{G(t)}. \end{equation*} For \(b\in[0,2]\); which implies that there is a critical point at \(G'(b)=0\), that is at \(b=0\). Hence for \[G'(b)< 0; \; |1-\rho|\in \left[0,\frac{[2]_q^2(1+[1]_q\lambda)^2}{\beta_1|\tau|[3]_q(1+[2]_q\lambda)}\right)\,,\] thus, \(G(b)\) is strictly a decreasing function of \(|1-\rho|\), therefore from (3), \[\max\{G(b):b\in[0,2]\}=G(0)=\frac{\beta_1|\tau|}{[3]_q(1+[2]_q\lambda)}.\] Also for \[G'(b)\geq 0; \; |1-\rho|\in \left[\frac{[2]_q^2(1+[1]_q\lambda)^2}{\beta_1|\tau|[3]_q(1+[2]_q\lambda)}, 0\right)\,,\] thus, \(G(b)\) is an increasing function of \(|1-\rho|\), therefore from (3), \[\max\{G(b):b\in[0,2]\}=G(2)=\frac{|1-\rho|\beta_1^2|\tau|^2}{[2]_q^2(1+[1]_q\lambda)^2}.\] So that by putting the results together leads to (34).

4. Conclusion

In this work, we were able to establish the first two coefficient bounds and also solve the Fekete-Szegö problem for the class \(\mathcal{B}_q(\tau,\lambda,\phi)\) of analytic and bi-univalent functions in \(\mathcal{U}\). The results in the first theorem generalized that of Srivastava and Bansal [10].

Acknowledgments

The authors thank the referees for their valuable suggestions to improve the paper.

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflicts of Interest

The authors declare no conflict of interest.

1. Introduction

Studies have been conducted by different authors on continuity and its aspects. Many results have so far been obtained. Most of these results have been successfully obtained by use of separation criteria. This can be done by choosing a topological space that one may wish to use in testing a property of either topological or bitopological space. Separation criteria involve axioms that guide the separation of topological and bitopological spaces. Therefore, separation axioms are restrictions that are often made depending on the kind of topological or bitopological spaces that we intended to consider. Separation axioms involve the use of spaces which distinguish disjoint sets and distinct points. These separation axioms are also called Tychonoff separation axioms.

Fora [1] states that spaces that can be topologically distinguished are said to be separable spaces. Abu-Donia and El-Tantawy [2] conducted a study to show some classes of sets of a bitopological space \((X, \tau_{1}, \tau_{2})\) which included the infra topologies and supra topologies. These classes introduce new bitopological properties and new types of continuous functions between bitopological spaces. They did some work which indicated that bitopological separation properties are preserved under some types of continuous functions. The properties \(T_{\frac{1}{2}},\) \(T_{b},\) \(\alpha T_{b},\) \(T_{d},\) \(\alpha T_{d}\) are exhibited by topological spaces and finally extended to bitopological spaces. Patil and Nagashree [3] effected some study to show new separation axioms in binary soft topological spaces alongside their properties and characterizations. They have also introduced the notions of binary separation axioms as binary \(T_{0},\) binary \(T_{1},\) binary \(T_{2}\) spaces.

According to [3, Theorem 3.4], every \(n-T^{\ast}_{0}\) is binary soft \(n-T_{0}.\) This result implies that if we have \((x_{1}, y_{1})\) and \((x_{1}, y_{2})\) as two distinct points of a binary soft \(n-T^{\ast}_{0}\) space \((U_{1}, U_{2}, \tau_{b}, E)\) then there exists at least one binary soft open set \((F, E)\) or \((G, E)\) such that \((x_{1}, y_{2})\in (F, E)\), \((x_{2}, y_{2})\in (F, E)\prime\) or \((x_{2}, y_{2}) \in (G, E),\) \((x_{1}, y_{1}) \in (G, E).\) So it implies that \((x_{1}, y_{1})\in (F, E),\) \((x_{2}, y_{2})\) does not exist in \((F, E)\) or \((x_{2}, y_{2})\in (G, E),\) \((x_{1}, y_{1})\) does not exist \((G, E).\) In our study we have considered \(T_{2\frac{1}{2}}\)-spaces as binary soft bitopological space.

According to [3, Theorem 3.11], a binary soft topological space \((U_{1}, U_{2}, \tau_{b}, E)\) is a binary soft \(n-T^{\ast}_{1}\) space if for any pair of distinct points \((x_{1}, y_{1}),\) \((x_{2}, y_{2})\in U_{1} \times U_{2}.\) There exists \((F, E),\) \((G, E)\tau_{b}\) such that \((x_{1}, y_{1})\in (F, E)^{'}\) and \((x_{2}, y_{2}) \in (G, E),\) \((x_{1}, y_{1}) \in (G, E)^{'}.\)

Binary soft property is also seen to be hereditary to some separation axioms as observed in the next result. In [3, Theorem 3.20], it was proved that the property of binary soft \(n-T_{2}\) is hereditary. It illustrates that when \((x_{1}, y_{1})\) \((x_{2}, y_{2})A\times B\) is a pair of distinct points. Then \((x_{1}, y_{1})\), \((x_{2}, y_{2})\in U_{1} \times U_{2}\) are distinct. The fact that \((U_{1}, U_{2}, \tau_{b}, E)\) is binary soft \(n-T_{2}\) there exists disjoint binary soft open sets \((F, E)\) and \((G, E)\) in \((U_{1}, U_{2},\tau_{b}, E)\) such that \((x_{1}, y_{1})\in (F, E)\) and \((x_{2}, y_{2})\in (G, E).\) Therefore, we have disjoint binary soft open sets \(^{Y}(F, E)\) and \(^{Y}(G, E)\) in \((Y, \tau_{bY}, E)\) such that \((x_{1}, y_{1})\in ^{Y}(F, E)\) and also \((x_{2}, y_{2})\in ^{Y}(G, E).\)

Selvanayaki and Rajesh [4] introduced another type of separation axioms in their research of quasi \(T_{\frac{1}{2}^\ast}\) space. A space \((X, \tau_{1}, \tau_{2})\) is said to be quasi \(T_{0}\) if for any two distinct points \(x\) and \(y\) of \(X\) there exists \(A\in QO(X, \tau_{1}, \tau_{2})\) such that \(x\in X\) where \(y\) does not exists in \(A\) or \(y\in A\) where \(x\) does not exist in \(A.\) A space \((X, \tau_{1}, \tau_{2})\) is said to be \(T_{1}\) if for any distinct points \(x, y\) of \(X\) there exists \(A, B\in QO(X, \tau_{1}, \tau_{2})\) such that \(x\ A,\) \(y\) does not exist in \(A\) and \(y\in B,\) \(x\) does not exist in \(B.\)

A space \((X, \tau_{1}, \tau_{2})\) is said to be quasi \(T_{2}\) if for any two distinct points \(x, y\) of \(X\) there exists two disjoint sets \(A, B\in QO(X, \tau_{1}, \tau_{2}) = QGC(X, \tau_{1}, \tau_{2}).\) Therefore, it is true that \(T_{2} \rightarrow T_{1} \rightarrow T_{\frac{1}{2}} \rightarrow T_{0}.\) A point \(x\in X\) can be said to be a limit point of a subset \(A\) of a bitopological space \((X, \tau_{1}, \tau_{2})\). In [4, Theorem 2.3] it was proved that every quasi \(T_{{\frac{1}{2}^\ast}}\) is quasi \(T_{0}.\)

In our study we are aiming at establishing separation axioms that can be used in aspects of continuity through \(ij\) notion. For a bitopological space \((x, \delta, \tau)\) can be referred to as \(T_{0}\) space if for all \(x, y \in X\) with \(x \neq y\) then there exists in \(\delta \cup \tau\) such that \(x\in U\) whereas \(y\) is not a members of \(U\) on the other hand when \(x\) is not a cardinality of \(U\) then \(y\in U.\) The result is indicated in below: According to [5, Theorem 2.76], a bitopological space \((X, \delta, \tau)\) is called \(T_{1}\) space if for all \(x, y \in X\) with \(x\neq y\) then there exists \(U \in \delta\) and \(V\in \tau\) such that \(x \in U\) where \(y\) does not exists in \( V\) and \(x\) does not exists in \(U\) and \(y \in U.\) In [5, Theorem 2.7], it was proved that a bitopological space \((X, \delta, \tau)\) is called a \(T_{2}\) space if for all \(x, y\in X\) with \(x\neq y\) then there exists \(U\in \delta,\) \(V\in \tau\) such that \(x\in U\) where \(y\in V\) and \(U\cap V=\phi.\)

Rupaya and Hossan [5] have shown some of the results of heredity property exhibited by some separation axioms as given below: In [5, Theorem 3.1], it was proved that if \((X, \tau_{1}, \tau_{2})\) is a bitopological space then \(T_{0}\) is considered to have hereditary property. This result illustrates that \((X, \delta, \tau)\) is a \(T_{0}\) space and \(A\subset X\) shows that \((A, \delta, \tau)\) is also \(T_{0}\) space. Having \(x, y\in A\) with \(x\neq y\) and \(x, y\in X\) with \(x\neq y.\) Hence if \((X, \delta, \tau)\) is \(T_{0}\) space then there exists \(U \in \delta \cup \tau\) such that \(x\in U\) where \(y\) does not exist in \(U\) or \(x\) also does not exist in \(U\) but \(y\in U\) so \(U\in \delta \cup \tau.\) This imply that \(U\in \delta\) or \(U\in \tau\) then \(U\cap A \in \delta_{A}\) or \(U\cap A \in \tau_{A}.\) Therefore, \(U\cap A \in \delta \cup \tau.\) Then \(x, y \in A\) and \(x\in U\cap A\) where \(y\) does not exist in \(U\cap A\) or \(x\) does not exist in \(U\cap A\) where \(y\in U\cap A.\) Hence \((A, \delta_{A}, \tau_{A})\) is also \(T_{0}.\) Separation axioms also act even on the bitopological spaces that exhibit the homeomorphic property. Homeomorphic image of a particular separation axioms is still that axiom. This has been shown by Rupaya and Hossain [5] in the study of properties of separation axioms in bitopological spaces.

Some result shows that a function \(f:(X, \delta, \tau) \rightarrow (Y, W, Z)\) be a homeomorphism and \((X, \delta, \tau)\) is \(T_{0}\) space. Then \((Y, W, Z)\) can be shown to be a \(T_{0}\) space. This work shows that by letting \(f(X, \delta, \tau)\rightarrow (Y, W, Z)\) to be a homeomorphism and \((X, \delta, \tau)\) is \(T_{1}\) space. Then \((Y, W, Z)\) is also \(T_{1}\) space. Hence \(y_{1}, y_{2}\in Y\) with \(y_{1}\neq y_{2}\). Since \(f\) is onto then there exists \(x_{1}, x_{2} \in X\) with \(f(x_{1})= y_{1}\) and \(f(x_{2}= y_{2}.\) A function \(f\) is one with \(y_{1}\neq y_{2} \rightarrow f(x_{1})\neq f(x_{2})\) this also implies that \(x_{1}\neq x_{2}.\) Then \(x_{1}, x_{2}\in X\) with \(x_{1}\neq x_{2}.\) Again since \((X, \delta, \tau)\) is \(T_{1}\) space then there exists \(U\in \delta\) and \(V\in \tau \) such that \(x_{1}\in U\) and \(x_{2}\) does not exists in \(U\) and \(x_{1}\) does not in \(V\) and \(x_{2}\in V.\) Since \(f\) is open then \(f(U)\in W\) and \(f(V)\in Z.\) In summary, the notion of bitopological spaces was first introduced by Kelly [6] in 1963. Kelly stated that a bitopological space is endowed by two independent topological structures say \(\tau_{1}\) and \(\tau_{2}.\) So a bitopological space that is equipped by two topologies is denoted by \((X, \tau_{1}, \tau_{2}).\) A non empty set \(X\) is equipped by two structures \(\tau_{1}\) and \(\tau_{2}.\) By Uryshon's lemma, the results show that existence of quasi-metrics is related to the existence of real-valued functions. Does other properties of bitopological spaces exhibit two topologies a bitopological space?.

In 1984, Ivanov [7] conducted a study on the structures of bitopological spaces. Results obtained show that some properties of a bitopological space as a set \(X\) on which there are given two structures \(\tau_{1}\) and \(\tau_{2}\) which are defined on the same set as \((X, \tau_{1}, \tau_{2}).\) The result was obtained by the use of \(T_{0}\) separation axioms as research methodology. However, from the open question that asks whether all the properties of bitopological spaces inherited by these structures. In 1988, Coy [8] carried out a study on some properties of bitopological spaces such as normality, separability and compactness. Results indicate that they are inherited by a topology in a topological space and as well can be extended to structures in bitopological spaces by the use of conditions of normality and compactness. Does continuity property apply in all spaces?. We consider this question in our study and seek to determine if aspects of continuity can apply to all types of topological spaces. In 2005, Abu-Donia and El-Tantaway [2] gave out generalized separation axioms in bitopological spaces where it was proved that some of the introduced bitopological separation properties are preserved under some types of continuous functions. The extension of semi-continuity, pre-continuity, \(\alpha\)-continuity have also been extended to bitopological space. Does it follow that aspects of continuity can be expressed bilaterally as \(ij\)-continuity in bitopological spaces and N-topological spaces? We consider this question and try to characterize \(ij\)-continuity.

In 2006, Noiri [9] defines semi-open sets and semi-continuity in bitopological spaces. A subset \(A\) of \(X\) is said to be semi-open if it is \((1, 2)\)-semi open and \((2, 1)\)-semiopen. It is also continuous if it is \(ij\)-continuous. Is the continuity determined by all the separation axioms?. We also focus on this question and try to find out if other separation axioms apart from \(T_{0}\) apply. In 2007, Orihuela [10] gave a brief account of topological open problems in the area of renormings of Banach spaces. From question four in open problems, let \(X\) be a weakly Cech analytic Banach space where every norm open set is a countable union of sets which are differences of closed sets for the weak topology. Does it follow that the identity map \(Id : (X, \sigma(X,X^{*})) \rightarrow (X, \| � \|)\) is \(\sigma\)-continuous?. We also consider this question and seek to determine if it holds true when \(X\) is a bitopological space. In 2010, Kohli [11] in the study of between strong and almost continuity gave an account that strong continuity and weak continuity lie strictly between strong continuity of Lavine and strong continuity of Singal. If \(X\) is endowed with a partition topology then, every continuous function \(f: X\rightarrow Y\) is perfectly continuous and hence completely continuous. Is the inverse of a real value function \(f\) continuous?. This question form basis of our research to determine if inverse functions of \(n\)-topologies are also continuous. In 2018, Caldas [12] carried out a study on some topological concepts in bitopological spaces among the concepts given account on are normality, compactness separability among others. From question two in open problems, let function \(f\) be continuous if it maps spaces \((X, \tau_{1}, \tau_{2})\) to \((Y, \tau_{1}, \tau_{2}).\) Does most aspects of continuity hold?. We consider this question and try to determine characterization of aspects of continuity.

Birman [13] conducted a study in continuity of bitopological spaces in 2018. Results obtained by using criterion for continuity as research methodology show that most properties including continuity can be induced in a bitopological space \((X, \tau_{1}, \tau_{2}).\) The open question is that can these properties be extended to \(n\)-topological spaces \(A_{1} \cup A_{2} \cup A_{3} \cup ...\cup A_{n}\)?. We also look into possibility of extending properties of bitopological spaces to \(n\)-topologies. Currently in 2019, Rupaya [5] introduced some concepts of separation axioms in bitopological spaces which satisfy topological and hereditary properties. Since a topological space does not imply bitopological space and vice versa. Open question is that do these properties hold in general for both topological and bitopological space?.

In this paper we characterize separability criteria for bitopological spaces.

2. Preliminaries

In this section, we outline the basic concepts which are useful in the sequel.

Definition 1. [14, Definition 1.1.1] Let \(X\) be a nonempty set. A collection of subsets of \(X\) denoted by \(\tau\) is said to be a topology on \(X\) if and only if the following properties are satisfied:

• (i). \(X\) and \(\emptyset\) belong to \(\tau.\)
• (ii). Any arbitrary union of members of \(\tau\) belongs to \(\tau.\)
• (iii). Any finite intersection of any two members of \(\tau\) belongs to \(\tau.\)

The ordered pair \((X, \tau)\) is called a topological space.

Definition 2. [15, Definition 2.1] Let \(X\) be a nonempty set and \(\tau_{1},\) \(\tau_{2}\) be topologies on \(X.\) Then \((X, \tau_{1}, \tau_{2})\) is said to be a bitopological space.

Definition 3. [16, Definition 1.7] Let \((X, \tau)\) be a topological space and \((Y, \tau_{X})\) be a subspace of \(X\) which is equipped with a topology induced or inherited from that of \(X.\)

Definition 4.[17, Definition 2.4] Let \((X, \tau_{1}, \tau_{2}, E)\) be a soft bitopological space over \(X\) and \(Y\) be a nonempty subset of \(X.\) Then \(\tau_{1Y} = \{(Y, F, E): (F, E)\in \tau_{1}\}\) and \(\tau_{2Y} = \{(Y, G, E) \in \tau_{2}\}\) are said to be relatively topologies on \(Y.\) Therefore, \(\{Y, \tau_{1Y}, \tau_{2Y}, E\}\) is called a relatively soft bitopological space of \((X, \tau_{1}, \tau_{2}).\)

Example 1. [18, Example 1.1.2] Let \(X = \{a, b, c\}.\) Let the collection of subsets of \(X\) is denoted by topology \(\tau.\) Then \(\tau = \{\{a\}, \{b\}, \{c\},\{a,b\}, \{b, c\}, \{a, c\}, \{X\}, \emptyset\}.\) Then \(\tau\) is a topology on \(X\) if it satisfies conditions (i), (ii) and (iii) in Definition 1.

For condition (i) both \(X\) and \(\phi\) are in \(\tau.\)

For condition (ii) \(\{ a, c\} \cup \{ b,c\} = \{ a, b, c \} = X,\) which is a member of \(\tau.\)

For condition (iii) \(\{a\} \cap \{a, c\} = \{a\},\) which is a member of \(\tau.\)

Definition 5.[19, Definition 2.4] A mapping \(f: (X, \tau_{1}, \tau_{2})\rightarrow (X_{1}, \tau_{3}, \tau_{4})\) is called \(P\)-continuous (respectively open, \(P\)-closed) if the induced mapping \(f:( X, \tau_{1})\rightarrow (X_{1}, \tau_{3})\) and \(f: (X, \tau_{2})\rightarrow (X_{1}, \tau_{4})\) are continuous (respectively open, closed).

Definition 6. [20, Definition 4.1] Let \(X\) and \(Y\) be \(N\)-topological space. A function \(f: X\rightarrow Y\) is said to be \(N^{\ast}\)-continuous on \(X\) if the inverse image of every \(N\tau\)-open set in \(Y\) is a \(N\tau\)-open set in \(X.\)

Definition 7. [5, Definition 2.5] A bitopological space \((X, \tau_{1}, \tau_{2})\) is called \(T_{0}\) space if \(\forall\) \(x, y \in X\) with \(x\neq y.\) Then there exists \(U\in \tau_{1} \cup \tau_{2}\) such that \(x\in U,\) \(y\) does not exists in \(U\) or \(x\) does not exists in \(Y,\) \(y\in U.\)

Example 2.[14, Example 4.1.2] Consider \(X = \{a, b, c, e, f\}\), \(Y=\{b, c, e\}\) and \(\tau = \{ X, \phi, \{a\}, \{c, d\}, \{a, c, d\}, \{b, c, d, e, f\}\}\) and \(Y = \{b, c,e\}.\) Therefore, \((Y, \tau)\) is a subspace of \((X, \tau).\) Since topology \(\tau\) on \(Y\) is induced from \(X\) then all properties of a topological space \((X, \tau)\) has are also in a subspace \((Y, \tau).\) This subspace is denoted as \(\tau_{Y} = Y \mid \tau _{X}.\)

Definition 8.[6, Definition 1.2] A bitopological space \((X, \tau_{1}, \tau_{2})\) is a space that is endowed with two independent topologies say \(\tau_{1}\) and \(\tau_{2}\) denoted as \((X, \tau_{1}, \tau_{2}).\)

Example 3. [21, Example 3.2] Let \(X=\{a, b,\}, \tau=\{\emptyset, \{a\}, \{c\}, \{a, c\}, X\}\) and \(\tau_{\beta}=\{\emptyset, X, \{a\}, \{c\}, \{a, c\},\\ \{b, c\}, \{a, b\}\}\) then \(\{a, c\}\) is connected.

Definition 9.[5, Definition 2.6] A bitopological space \((X, \tau_{1}, \tau_{2})\) is called \(T_{1}\) space if \(\forall\) \(x, y \in X\) with \(x\ne y.\) Then there exists \(U\in \tau_{1}\) and \(V\in \tau_{2}\) such that \(x\in U,\) \(y\) does not exists in \(U\) and \(x\) does not exists in \(U\) and \(y\in U.\)

Example 4.[19, Example 1] Consider \(X=\{a, b, c\}\) with topologies \(P=\{\emptyset, \{b\}, \{c\}, \{b, c\}, X\}\) and \(Q=\emptyset, \{a\}, \{b\}, \{a, b\}, \{b, c\}, x\}\) defined on \(X.\) Therefore, \(P\)-closed subsets of \(X\) are \(\emptyset, \{a, c\}, \{a, b\}, \{a\}\) and \(X\). \(Q\)-closed subsets of \(X\) are \(\emptyset, \{b, c\}, \{a, c\}, \{c\}, \{a\}\) and \(X.\) It follows that \((X, P, Q)\) does satisfy the condition \(P\)-normal. Hence \((X, P, Q)\) is \(P\)-normal.

Definition 10.[22, Definition 3.19] A bitopological space \((X, \tau_{1}, \tau_{2})\) is said to be:

• (i). Pairwise \(S\)-closed if every pairwise regular, pairwise closed cover of \((X, \tau_{1}, \tau_{2})\) has a finite subcover.
• (ii). Pairwise \(S\)-closed if every pairwise countable cover of \((X, \tau_{1}, \tau_{2})\) by pairwise regular closed sets with respect to \(\tau_{1}\) and \(\tau_{2}\) has a finite subcover.
• (iii). Pairwise \(S\)-Lindelof if every pairwise cover of \((X, \tau_{1}, \tau_{2})\) by pairwise regular closed sets with respect to \(\tau_{1}\) and \(\tau_{2}\) has a finite pairwise countable subcover.
• (iv). Nearly pairwise compact if every pairwise regular open cover of \((X, \tau_{1}, \tau_{2})\) has a finite subcover.
• (v). Pairwise countably nearly pairwise compact if every pairwise countable cover of \((X, \tau_{1}, \tau_{2})\) by pairwise regular open sets with respect to \(\tau_{1}\) and \(\tau_{2}\) has a finite subcover.
• (vi). Nearly pairwise Lindelof if every pairwise regular open cover of \((X, \tau_{1}, \tau_{2})\) has a pairwise countable subcover.

Definition 11.[23, Definition 3.14] Let \((X, \tau_{1}, \tau_{2})\) and \((Y, \sigma_{1}, \sigma_{2})\) be two bitopological spaces. A function \(f: (X, \tau_{1}, \tau_{2})\rightarrow (Y, \sigma_{1}, \sigma_{2})\) is called \((1, 2)-\gamma\) semi-continuous if the inverse image of each \(\sigma_{1}\)-open set in \(Y\) is \((1, 2)-\gamma\) -semi-open set in \(X.\)

Remark 1.[24, Remark 5] Each of \(ij\)-irresoluteness \(ij\)-almost \(s\)-continuity and \(i\)-continuity is independent of one another and \(ij\)-almost \(s\)-continuity is not a generalization of \(i\)-continuity. Each of \(ij\)-quasi irresoluteness \(ij\)-semi-continuity and \(ij\)-almost continuity is independent of one another.

Definition 12.[25, Definition 2.1] A function \(f:X\rightarrow Y\) from a topological space \(X\) to A topological space \(Y\) is said to be:

• (i). Strongly continuous if \(f (\overline{A}) \subset A\) for all \(A \subset X.\)
• (ii). Perfectly continuous if \(f^{-1}(V)\) is clopen in \(X\) for every open set \(V\subset Y.\)
• (iii). \(\sigma\)-perfectly continuous if for each \(\sigma\)-open set \(V\) in \(Y\), \(f^{-1}(V)\) is a clopen set.

Definition 13.[26, Definition 1.4.1] A function \(f\) is continuous at some point \(x\in X\) if and only if for any neighborhood \(V\) of \(f(X)\) hence there is a neighborhood \(U\) of \(x\) such that \(f(U)\subseteq V.\)

Definition 14.[11, Definition 3.2] Let \((X, \tau_{1}, \tau_{2})\) be a topological space and \(N\) be a subset of \(X\) and \(p\) a point in \(N.\) Then \(N\) is said to be a neighborhood of the point \(p\) if there exists an open set \(U\) such that \(p\in U\subseteq N.\)

Example 5.[27, Example 2.3] Let \(X=\{1, 2, 3\}\), \(P=\{\emptyset, X, \{1, 2\}, \{3\}\}\) and \(Q=\{\emptyset, X, \{1\}, \{2, 3\}\}\). It is easy to check that \((X, P, Q)\) is pairwise \(T_{0}\) but not weak pairwise \(T_{1}\) considering the points 1, 2.

Definition 15.[28, Definition 2] The intersection (resp. union) of all (i, j) -\(\delta\)-closed (resp. (i, j)-\(\delta\)-open sets \(X\) containing (resp. contained in) \(A\subset X\) is called the \((i, j)-\delta\)-closure (resp. \((i, j)-\delta\)-interior of an \(A.\)

Definition 16.[29, Definition 3] A subset of a bitopological space \((X, \tau_{1}, \tau_{2})\) is said to be (i, j)-\(\delta\)-b-open. If \(A\subset j Cl (i Int-{\delta} (A))\cup i Int(j Cl_{\delta} (A))\), where \(i \neq j i, j =1, 2.\) The complement of an \((i, j)-\delta-b\)-open set is called an \((i, j)-\delta- b\)-closed set.

Example 6. [30, Example 7] Suppose that intersection (resp.union) for all \((i, j)-\delta-b\)-closed \((resp. (i, j)-\delta-b\)-open) sets of \(X\) containing (resp. contained in) \(A\subset X\) is called the \((i, j)-\delta-b\)-closure \((resp. (i, j)-\delta-b\)-interior) of \(A\) and is denoted by \((i, j)-b Cl_{\delta} (A) (resp. (i, j)-b Int_{\delta} (A))\) The intersection of all \((i, j) -b-\delta\)-open sets of \(X\) containing \(A\) is called \((i, j)-b-\delta\)-kernel of \(A\) and is denoted by \((i, j)-b Ker_{\delta}(A).\)

Example 7.[19, Example 2] Consider \(X=\{a, b, c, d\}\) with topologies \(P=\{\emptyset, \{a, b\}, X\}\) and \(Q=\{\emptyset, \{a\}, \{b, c, d\}, X\}\) defined on \(X.\) Observe that \(P\)-closed subsets of \(X\) are \(\emptyset, \{c, d\}\) and \(X.\) \(Q\)-closed subsets of \(X\) are \(\emptyset, \{b, c, d\}, \{a\}\) and \(X.\) Hence \((X, P, Q)\) is \(P1\)-normal as we can check since the only pairwise closed sets of \(X\) are \(\emptyset\) and \(X.\) However, \((X, P, Q)\) is not \(P\)-normal since the \(P\)-closed set \(A=\{c, d\}\) and \(Q\)-closed set \(B=\{a\}\) satisfy \(A\cap B=\emptyset\) but do not exists the \(Q\)-open set \(U\) and \(P\)-open set \(V\) such that \(A\subseteq U, B\subseteq V\) and \(U\cup V=\emptyset.\)

Definition 17.[31, Definition 4.2] A space \((X, \tau_{1}, \tau_{2})\) is said to be pairwise \(R_{0}\) if for every \(T_{i}\)-open set \(G\), \(x\in G\) implies \(T_{i}-cl\{x\}\subset G\) i,j = 1, 2, \(i\neq j.\)

Remark 2.[24, Remark 1] Every \(ij\)-regular open set is \(ji\) semiregular.

Definition 18.[32, Definition 4.3] A space \((X, \tau_{1}, \tau_{2})\) is said to be pairwise \(R_{0}\) if for every \(T_{i}\)-open set \(G,\) \(x\in G\) implies that \(T_{j}-\delta cl \{x\}\subset G\), i, j = 1, 2, \(i\neq j.\)

Example 8. [33, Example 3.12] Let \(X=\{a, b, c\},\) \(\tau_{1}=\{X, \emptyset, \{a\}\}\) and \(\tau_{2}=\{X, \emptyset, \{a\}, \{a, b\}, \{a, c\}\}.\) Clearly, \((X, \tau_{1})=(X, \tau_{2})=(X, \tau_{1}, \tau_{2})= \tau_{2}.\) Thus \((X, \tau_{1}, \tau_{2})\) is a a quasi-\(b-T_\frac{1}{2}\) space that is not quasi-\(b-T_{1}.\)

Definition 19.[19, Definition 3.6] A space \((X, \tau_{1}, \tau_{2})\) is called:

• (i). Pairwise pre-\(T_{0}\) (resp. pairwise pre-\(T_{1}\)) if for any pair of distinct points \(x\) and \(y\) in \(X,\) there exists a \(T_{i}\)-preopen set which contains one of them but not the other \(i= 1\) or \(2\) resp. there exists \(T_{i}\)-preopen set \(U\) and \(T_{j}\)-preopen set \(V\) such that \(x\in U,\) \(y\in V\) and \(U\cap V = \emptyset i, j =1, 2 , i\neq j.\)
• (ii). A space \((X, \tau_{1}, \tau_{2})\) is said to be pairwise pre-\(T_{2}\) if for any pair of distinct points \(x\) and \(y\) in \(X,\) there exists \(\tau_{i}\)-preopen set \(U\) and \(\tau_{j}\)-preopen set \(V\) such that \(x\in U,\) \(y\in V\) and \(U\cap V \neq \emptyset\) i, j= 1, 2, where \(i\neq j.\)

Example 9.[34, Example 2] Let \(X=\{a, b, c, d, e\}, \tau_{1}=\{\emptyset, X\}, \tau_{2}=\{\emptyset, X, \{b, e\}\},\) \( Y=\{a, b, c, d\}, \sigma_{1}=\{\emptyset, Y, \{c\}\}\) and \(\sigma_{2}=\{\emptyset, Y, \{b, d\}\}.\) Suppose \(f:(X, \tau_{1}, \tau_{2})\rightarrow (Y, \sigma_{1}, \tau_{2})\) is defined by \(f(a)=f(c)=f(d), f(b)=f(c), f(b)=f(e)=b.\) Then the map \(f\) is \(12\)-preweak continuous but not \(12\)-preweak semicontinuous since \(f^{-1}(\{c\})=\{a, c, d\}\) which is not 1-open set in the subspace \(2-cl(1-int(2-cl(f^{-1}(2-cl(\{c\}))))).\)

Definition 20. [35, Definition 3.19] A bitopological space \((X, \tau_{1}, \tau_{2})\) is said to be:

• (i). Pairwise \(S\)-closed if every pairwise regular, pairwise closed cover of \((X, \tau_{1}, \tau_{2})\) has a finite subcover.
• (ii). Pairwise \(S\)-closed if every pairwise countable cover of \((X, \tau_{1}, \tau_{2})\) by pairwise regular closed sets with respect to \(\tau_{1}\) and \(\tau_{2}\) has a finite subcover.
• (iii). Pairwise \(S\)-Lindelöf if every pairwise cover of \((X, \tau_{1}, \tau_{2})\) by pairwise regular closed sets with respect to \(\tau_{1}\) and \(\tau_{2}\) has a finite pairwise countable subcover.
• (iv). Nearly pairwise compact if every pairwise regular open cover of \((X, \tau_{1}, \tau_{2})\) has a finite subcover.
• (v). Pairwise countably nearly pairwise compact if every pairwise countable cover of \((X, \tau_{1}, \tau_{2})\) by pairwise regular open sets with respect to \(\tau_{1}\) and \(\tau_{2}\) has a finite subcover.
• (vi). Nearly pairwise Lindelöf if every pairwise regular open cover of \((X, \tau_{1}, \tau_{2})\) has a pairwise countable subcover.

Definition 21.[36, Definition 3.1] Let \((X, \tau_{1}, \tau_{2})\) be a bitopological space. Then, we say that a subset \(A\) of \(X\) is \(\tau_{1}\) semi open with respect to \(\tau_{2}\) if and only if there exists a \(\tau_{2}\) open set \(O\) such that \(O \subset A\subset \tau_{2}\) closure \(O.\)

Definition 22.[37, Definition 2.4] The quasi \(b\)-closure of a subset \(A\) of \((X, \tau_{1}, \tau_{2})\) is defined to be \(qbCl(A)=\cap {F: F\in QBC(X, \tau_{1}, \tau_{2}), A\subseteq F}.\) A subset \(A\) of \((X, \tau_{1}, \tau_{2})\) is quasi-b-generalized closed (simply, quasi-bg-closed) if \(qbCl(A)\subseteq U.\)

Definition 23.[38, Definition 2.3] A bitopological space \((X, \tau_{1}, \tau_{2})\) is said to be \(i\)-Lindelöf if both the topological space \((X, \tau_{i})\) is Lindeloöf. \(X\) is called Lindelöf if it is both 1-Lindelöf and 2-Lindelöf. Equivalently, \((X, \tau_{1}, \tau_{2})\) is Lindelöf if every \(i\)-open cover of \(X\) has a countable subcover for each \(i = 1, 2\).

Definition 24.[17, Definition 5] The union of two soft sets of \((F, A)\) and \((G, B)\) over the common universe \(U\) is the soft set \((H, C)\), where \(C= A\cup B\) and for all \(e\in E.\) This can be denoted as \((F, A) \cup (G, B)= (H, C).\)

Example 10.[4, Definition 3.1] A space \((X, \tau_{1}, \tau_{2})\) is quasi-\(b-T_{0}\) if for every two distinct points \(x, y\) of \(X\), there exists \(A\in QBO(X, \tau_{1}, \tau_{2})\) such that \(x \in A\) and \(y\) not in \(A\) or \(x\) not in \(A\) and \(y\in A.\)

Definition 25. [3, Definition 3.2] A binary soft topological space \((U_{1}, U_{2}, \tau_{b}, E)\) is said to be binary soft \(n-T_{0}\) if for any pair \((x_{1}, y_{1})\), \((x_{1}, y_{2})\in U_{1} \times U_{2}\) of distinct points that is, \(x_{1}\neq x_{2}\) and \(y_{1} \neq y_{2}\) there exists at least one binary soft open set \((F, E)\) or \((G, E)\) such that \((x_{1}, y_{1})\in (F, E)\), \((x_{2}, y_{2})\) does not exists in \((G, E)\) or \((x_{2}, y_{2})\in (G, E)\) and \((x_{1}, y_{1})\) does not exists in \((G, E).\)

Definition 26. [33, Definition 3.4] A space \((X, \tau_{1}, \tau_{2})\) is quasi-\(b-T_{2}\) if for every two distinct points \(X,\) \(y\) of \(X,\) there exists disjoint sets \(A, B\in QBO(X, \tau_{1}, \tau_{2})\) such that \(x\in A\) and \(y\in B.\)

Definition 27. [17, Definition 3] The intersection \((H, C)\) of two sets \((F, A)\) and \((G, B)\) over a common universe \(U\). Denoted as \((F, A)\cap (G, B)\), is defined as \(C= A\cap B.\) Also \(H(e)= F(e)\cap G(e)\) for all \(e\in C.\)

Definition 28. [39, Definition 4.5] A space \(X\) is normal on a subset \(Y.\) Then if every two disjoint closed subsets \(F\) and \(G\) of \(X\) satisfying \(F=\overline{F \cap Y}\) and \(G=\overline{G \cap Y}\) can be separated in \(X\) by disjoint open sets. Suppose that we have a separable space \(X\) then the following conditions are equivalent:

• (i). \(X\) is said to be normal on every dense countable subset.
• (ii). Any two separable disjoint subspaces if \(X\) can be represented by disjoint open set.

Remark 3. [40, Definition 2.1] A bitopological space \((X, P, Q)\) is weak pairwise \(T_{0}\) if and only if each pair of distinct points, there is a set which is either \(p\)-open or \(q\)-open containing one but not the other.

Definition 29.[4, Definition 3.1] Let \((X, \tau_{1}, \tau_{2})\) be a topological space and \(A\subset X.\) Then \(A\) is said to be weakly quasi separated from set \(B\) if there exists a quasi open set \(G\) such that \(A\subset G\) and \(G\cap B=\emptyset\) or \(A\cap qcl(B)=\emptyset.\)

Definition 30. [41, Definition 2.2] Let \(U_{1},\) \(U_{2}\) be two initial universe sets with powers \(P(U_{1})\) and \(P(U_{2})\) respectively and \(E\) be a set of parameters. A pair \((F, E)\) is said to be a binary soft set over \(U_{1}\) and \(U_{2}\) where \(F\) is defined as \(F: E\rightarrow P(U_{1}) \times P(U_{2}),\) \(F(e)= (X, Y)\) for each \(e\in E\) such that \(X\subset U_{1},\) \(Y\subset U_{2}.\)

3. Main results

Proposition 1. Let \((X, \tau_{1}, \tau_{2})\) be a \(T_{0}\) space then the property of \(T_{0}\) is hereditary and topological.

Proof. We prove that \(T_{0}\) has hereditary property. Let \((X, \tau_{1}, \tau_{2})\) be a \(T_{0}\) space and let \(D\subseteq X.\) Then we show that a bitopological subspace \((D, \tau_{D1}, \tau_{D2})\) is also a \(T_{0}\) space. Since \((D,s \tau_{D1}, \tau_{D2})\) has induced properties from \((X, \tau_{1}, \tau_{2})\) then it implies that \(a, b\in D\) with \(a\neq b,\) then \(a, b\in X\) with \(a \neq b\) as in Definition 7. Since \((X, \tau_{1}, \tau_{2})\) is a \(T_{0}\) space then \(\exists U\in \tau_{1} \cup \tau_{2}\) such that \(a\in U,\) \(b\) does not exists in \(U\) or \(a\) is not a member of \(U\) but \(b\in U.\) Then it follows that \(U\in \tau_{1} \cup \tau_{2}\) such that \(U\in \tau_{1}\) or \(U\in \tau_{2}.\) Therefore, \(U\cap D\in \tau_{D1}\) or \(U\cap D\in \tau_{D2}\) and so \(U\cap D\in \tau_{D1} \cap \tau_{D2}.\) Again since \(a, b\in D\) then \(a\in U\cap D,\) \(b\) does not exists in \(U\cap D\) or \(a\) does not exists in \(U\cap D,\) and \(b\in U\cap D.\) Hence \((D, \tau_{D1}, \tau_{D2})\) is also a \(T_{0}\) space. Secondly, we prove that \(T_{0}\) has topological property. Let \(\chi: (X, \tau_{1}, \tau_{2})\rightarrow (Y, \tau_{3}, \tau_{4})\) be a homeomorphism and let \((X, \tau_{1}, \tau_{2})\) be a \(T_{0}\) space we therefore show that \((Y, \tau_{3}, \tau_{4})\) is also a \(T_{0}\) space. By Definition, \(\chi:(X, \tau_{1})\rightarrow (Y, \tau_{3})\) and \(\chi: (X, \tau_{2})\rightarrow (Y, \tau_{4})\) are continuous (open, closed, homeomorphism respectively). Let \(b_{1}, b_{2} \in Y\) with \(b_{1}\neq b_{2},\) since \(\chi\) is an onto function then \(\exists a_{1}, a_{2}\in X\) with \(\chi (a_{1}) = \chi(b_{1})\) and \(\chi(a_{2}) = b_{2}\) as Definition 5. Since \(\chi\) is an injective function with \(b_{1} \neq b_{2}\) therefore it implies that \(\chi(a_{1}) \neq \chi(a_{2})\) hence \(a_{1} \neq a_{2}.\) Since \((X, \tau_{1}, \tau_{2})\) is \(T_{0}\) space and \(a_{1}, a_{2}\in X\) where \(a_{1} \neq a_{2}\) then it implies that there exists \(U\in \tau_{1} \cup \tau_{2}\) such that \(a_{1}\in U,\) \(a_{1}\) does not exists in \(U\) or \(a_{1}\) does not exists in \(U,\) \(a_{2}\in U\) or \(a_{1}\in U,\) \(a_{2}\) does not exists in \(U.\) Then \(U\in \tau_{1}\cup \tau_{2}\) follows that \(\chi (U)\in \chi(\tau_{1} \cup \tau_{2})\) since \(\chi\) is open and continuous. By Tychonoff separation axiom, it implies that \(\chi (U)\in\chi(\tau_{1})\cup\chi(\tau_{2})\in \tau_{3} \cup \tau_{4}.\) Also \(a_{1}\in U\) which implies that \(\chi(a_{1})\in \chi (U)\) or \(b_{1}\in \chi(U)\) and \(a_{2}\) does not exists in \(U\) which imply that \(\chi(a_{2})\) does not exists in \(\chi(U)\) or \(b_{2}\) does not exists \(\chi(U).\) For any \(b_{1}, b_{2}\in Y\) with \(b_{1} \neq b_{2},\) \(\chi(U)\in \tau_{3} \cup \tau_{4}\) is obtained such that \(b_{1}\in \chi (U),\) \(b_{2}\) does not exists in \(\chi(U).\) Therefore, \((Y, \tau_{3}, \tau_{4})\) is a \(T_{0}\) space. Every homeomorphic image of \(T_{0}\) space implies that it is topological property.

Proposition 2. Let \((X, \tau_{1}, \tau_{2})\) be a \(T_{1}\) space then the property of \(T_{1}\) is topological and hereditary.

Proof. Suppose that \(T_{1}\) has hereditary property then \((X, \tau_{1}, \tau_{2})\) is also \(T_{1}\) space. Let \(D\subseteq X\) and hence \((D, \tau_{D1}, \tau_{D2})\) is also \(T_{1}\) space. Let \(a, b\in D\) with \(a\neq b\) and \(a, b\in X\) with \(a \neq b.\) Since \((X, \tau_{1}, \tau_{2})\) is a \(T_{1}\) space then \(\exists U\in \tau_{1} \cup \tau_{2}\) such that \(a\in U,\) \(b\) does not exists in \(U\) and \(a\) does not exists in \(V\) but \(b\in U.\) By Proposition 1, we have \(U\in \tau_{1} \cup \tau_{2}.\) Then \(U\in \tau_{1}\) or \(U\in \tau_{2}\) with \(U\cap D\in \tau_{D1}\) or \(U\cap D\in \tau_{D2}\) also \(U\cap D\in \tau_{D1} \cap \tau_{D2}.\) Since \(a, b\in D\) hence \(a\in U\cap D,\) \(b\) does not exists in \(U\cap D\) or \(a\) does not exists in \(U\cap D,\) \(b\in U\cap D.\) Therefore, \((D, \tau_{D1}, \tau_{D2})\) is also a \(T_{1}\) space. Secondly, we show that \(T_{1}\) space has a topological property. Let \(\chi: (X, \tau_{1}, \tau_{2})\rightarrow (Y, \tau_{3}, \tau_{4})\) be a homeomorphism and \((X, \tau_{1}, \tau_{2})\) be \(T_{0}\) space. Then by hypothesis \((Y, \tau_{3}, \tau_{4})\) is a \(T_{1}\) space. Let \(b_{1}, b_{2} \in Y\) where \(b_{1}\neq b_{2}.\) Suppose that \(\chi\) is surjective then it follows that there exists \(a_{1}, a_{2}\in X\) with \(\chi (a_{1}) = \chi(b_{1})\) and \(\chi(a_{2}) = b_{2}.\) Hence \(\chi\) is also one to one function with \(b_{1} \neq b_{2}\) this implies that \(\chi(a_{1}) \neq \chi(a_{2})\) hence \(a_{1} \neq a_{2}.\) Since \((X, \tau_{1}, \tau_{2})\) is a \(T_{1}\) space and \(a_{1}, a_{2}\in X,\) with \(a_{1} \neq a_{2}.\) Then \(\exists U\in \tau_{1} \cup \tau_{2}\) such that \(a_{1}\in U,\) \(a_{1}\) does not exists in \(U\) or \(a_{1}\) does not exists in \(U,\) \(a_{2}\in U.\) Since \(a_{1}\in U,\) \(a_{2}\) does not exists in \(U\) then it implies that \(U\in \tau_{1}\cup\tau_{2}.\) Therefore, \(\chi (U)\in \chi(\tau_{1} \cup \tau_{2}).\) By Tychonoff Theorem, \(\chi\) is open and continuous then \(\chi (U)\in\chi(\tau_{1})\cup\chi(\tau_{2})\in \tau_{3} \cup \tau_{4}.\) Similarly, \(a_{1}\in U\) hence it follows that \(\chi(a_{1})\in \chi (U)\) also \(b_{1}\in \chi(U)\) and \(a_{2}\) does not exists in \(U\) which implies that \(\chi(a_{2})\) does not exists in \(\chi(U),\) and so \(b_{2}\) does not exists \(\chi(U).\) Then for any \(b_{1}, b_{2}\in Y\) with \(b_{1} \neq b_{2}\) and \(\chi(U)\in \tau_{3} \cup \tau_{4}\) is obtained such that \(b_{1}\in \chi (U),\) \(b_{2}\) does not exists in \(\chi(U).\) Therefore, \((Y, \tau_{3}, \tau_{4})\) is also a \(T_{1}\) space. Hence \(\chi\) is continuous (open, closed and homeomorphism) if and only if the maps \(\chi:(X, \tau_{1})\rightarrow (Y, \tau_{3})\) and \(\chi: (X, \tau_{2})\rightarrow (Y, \tau_{4})\) are continuous (open, closed and homeomorphism respectively). By hypothesis, every homeomorphism image of \(T_{1}\) space imply \(T_{0}\) space. Therefore, a \(T_{1}\) space is a topological property.

Proposition 3. Suppose that \((X, \tau_{1}, \tau_{2})\) is a \(T_{2}\) space then the property of \(T_{2}\) is topological and hereditary.

Proof. Let \((X, \tau_{1}, \tau_{2})\) and \((Y, \tau_{1}, \tau_{2})\) be two bitopological spaces. If \((X, \tau_{1}, \tau_{2})\) is a \(T_{2}\) space then it exhibits topological properties. Let \(\chi: (X, \tau_{1}, \tau_{2})\rightarrow (Y, \tau_{3}, \tau_{4})\) be a homeomorphism and \((X, \tau_{1}, \tau_{2})\) is also a \(T_{2}\) space. Then we show that \((Y, \tau_{3}, \tau_{4})\) is also \(T_{2}\) space. By Definition 27, let \(b_{1}, b_{2}\in Y\) with \(y_{1} \neq y_{2}\). Since all elements in \(Y\) are images of elements in \(X\) then \(\chi\) is a surjective function. Then there exists \(a_{1} ,a_{2} \in X\) with \(\chi (a_{1}) = b_{1}\) and \(\chi(a_{2}) = b_{2}.\) Again since \(\chi\) is an injective function then \(b_{1} \neq b_{2}.\) This implies that \(\chi(a_{1}) \neq \chi(a_{2}),\) and \(a_{1} \neq a_{2}.\) Therefore, \(a_{1}, a_{2} \in X\) with \(a_{1} \neq a_{2}.\) Consequently, since \((X, \tau_{1}, \tau_{2})\) is a \(T_{2}\) space then it shows that \(\exists U\in \tau_{1}\) and \(V\in \tau_{2}\) such that \(a_{1}\in U,\) \(a_{2}\in V\) then \(U\cap V\neq \emptyset.\) Suppose that \(\chi\) is open then \(\chi(U)\in \tau_{3}\) and \(\chi (V)\in \tau_{4}.\) Therefore, \(\chi(U)\cap \chi(V)\neq \emptyset\) then there exists \(c\in X\) such that \(c\in \chi(U) \cap \chi(V).\) This shows that \(c\in \chi(U)\) and \(c\in \chi(V)\) then \(\exists p_{1}\in U\) and \(p_{2}\in V\) such that \(c= \chi(p_{1})\) and \(c\in \chi(p_{2})\) with \(\chi(p_{1}) = \chi(p_{2})\) and \(p_{1}= p_{2}\) since \(\chi\) is a one to one function and so \(p_{1}\in U\) and \(p_{1}\in V.\) hence \(p_{1}\in U\cap V\neq \emptyset\) which is by contradiction. Suppose that \(U\cap V= \emptyset\) which implies that \(\chi(U) \cap \chi(V)=\emptyset.\) Therefore, for any \(b_{1}, b_{2}\in Y\) with \(b_{1} \neq b_{2}\) hence \(\chi(U) =\tau_{3}\) and \(\chi (V)\in c\) is obtained such that \(b_{1} \in \chi(U),\) \(b_{2} \in \chi(V)\) and so \(\chi(U) \cap \chi(V)\neq \emptyset.\) Hence \((Y, \tau_{3}, \tau_{4})\) is a \(T_{2}\) space. So it implies that every homeomorphism image of a \(T_{2}\) is a \(T_{2}\) space. Then \(T_{2}\) is a topological property. Let \((X, \tau_{1}, \tau_{2})\) be a \(T_{2}\) space then it has hereditary property. Let \((X, \tau_{1}, \tau_{2})\) also be \(T_{2}\) space. Since \(D\subseteq X,\) we prove that \((D, \tau_{D1}, \tau_{D2})\) is also \(T_{2}\) space. Let \(a, b\in D\) with \(a\neq b\) and \(a, b\in X\) and also \(a \neq b.\) By Definition 9, it follows that \(\exists U\in \tau_{1} \cup \tau_{2}\) such that \(a\in U,\) \(b\) does not exists in \(U\) and also \(a\) does not exists in \(U\) but \(b\in U.\) So \(U\in \tau_{1} \cup \tau_{2},\) it implies that \(U\in \tau_{1}\) or \(U\in \tau_{2}\) where \(U\cap D\in \tau_{D1}\) or \(U\cap D\in \tau_{D2}.\) By Tychonoff theorem, \(U\cap D\in \tau_{D1} \cap \tau_{D2}.\) Again since \(a, b\in D\) then \(a\in U\cap D,\) \(b\) does not exists in \(U\cap D\) or \(a\) does not exists in \(U\cap D\), \(b\in U\cap D.\) Therefore, \((D, \tau_{D1}, \tau_{D2})\) is also a \(T_{2}\) space and has a topological property.

Proposition 4. Let \((X, \tau_{1}, \tau_{2})\) be a \(T_\frac{5}{2}\) space then the property of \(T_\frac{5}{2}\) is topological and hereditary.

Proof. Let \(\chi: (X, \tau_{1}, \tau_{2})\rightarrow (Y, \tau_{3}, \tau_{4}).\) By hypothesis, \(T_{1}\) space imply \(T_{2}\) space which also implies \(T_\frac{5}{2}\) space. Therefore, we prove that \(T_\frac{5}{2}\) space has hereditary property. Let \((X, \tau_{1}, \tau_{2})\) be a \(T_\frac{5}{2}\) space and let \(K\subseteq X.\) Then \((K, \tau_{K1}, \tau_{K2})\) is a \(T_\frac{5}{2}\) space since it is bitopological subspace of \((X, \tau_{1}, \tau_{2}).\) Then it implies that \((K, \tau_{K1}, \tau_{K2})\) is a \(T_\frac{5}{2}\) space. Let \(m, n\in K\) with \(m\neq n\) then if \((X, \tau_{1}, \tau_{2})\) is a \(T_\frac{5}{2}\) space then \(\exists A\in \tau_{1}\) and \(B\in \tau_{2}\) such that \(m\in A,\) \(n\in B\) such that the intersection of \(A\) and \(B\) is empty that is, \(A\cap B = \emptyset.\) Hence \(A\in \tau_{1},\) \(B\in \tau_{2}\) then it follows that \(A\cap K\in \tau_{K1}\) and \(B\cap K\in \tau_{K2}.\) Therefore, \(m, n\in K\) then \(m\in A\cap K\), \(n\in B\cap K.\) So \((A\cap K)\cap (B\cap K)= (A\cap K)\cap K=\emptyset \cap K=\emptyset,\) hence \((K, \tau_{K1}, \tau_{K2})\) is \(T_\frac{5}{2}\) space. Then \((X, \tau_{1}, \tau_{2})\) is also a \(T_\frac{5}{2}\) space so it also has a topological property. By hypothesis \(\chi: (X, \tau_{1}, \tau_{2})\rightarrow (Y, \tau_{3}, \tau_{4})\) and \(\chi\) is a homeomorphic function then it follows that \((Y,\tau_{3}, \tau_{4})\) is also a \(T_\frac{5}{2}\) space. Therefore, \(n_{1}, n_{2} \in Y\) with \(n_{1}\neq n_{2}.\) Since \(\chi\) is onto function then it implies that \(\exists m_{1}, m_{2}\in X\) with \(\chi (m_{1}) = \chi(n_{1})\) and \(\chi(m_{2}) = n_{2}.\) Suppose \(\chi\) is injective with \(n_{1} \neq n_{2}\) then it implies that \(\chi(m_{1}) \neq \chi(m_{2})\) and so \(m_{1} \neq m_{2}.\) Hence \((X, \tau_{1}, \tau_{2})\) is \(T_\frac{5}{2}\) space then \(m_{1}, m_{2}\in X,\) with \(m_{1} \neq m_{2}\) and \(\exists A\in \tau_{1} \cup \tau_{2}\) such that \(m_{1}\in A,\) \(m_{1}\) does not exists in \(A\) or \(m_{1}\) does not exists in \(A,\) \(a_{2}\in A.\) Similarly, \(m_{1}\in A\), \(m_{2}\) does not exists in \(U\) therefore it implies that \(A\in \tau_{1} \cup \tau_{2}\) such that \(\chi (A)\in \chi(\tau_{1} \cup \tau_{2}).\) By condition for separation axioms, \(\chi (A) \in \chi(\tau_{1}) \cup \chi(\tau_{2})\in \tau_{3} \cup \tau_{4},\) \(m_{1}\in A\) such that \(\chi(m_{1})\in \chi (A)\) hence \(n_{1}\in \chi(A)\) and \(m_{2}\) does not exists in \(A\) and \(\chi(m_{2})\) does not exists in \(\chi(A)\), this implies that \(n_{2}\) does not exists \(\chi(A)\) for any \(n_{1}, n_{2}\in Y\) with \(n_{1} \neq n_{2},\) \(\chi(A)\in \tau_{3} \cup \tau_{4}\) is obtained such that \(n_{1}\in \chi (A),\) \(n_{2}\) does not exists in \(\chi(A).\) Therefore, \((Y, \tau_{3}, \tau_{4})\) is also \(T_\frac{5}{2}\) space. Each homeomorphic image of \(T_\frac{5}{2}\) space is also \(T_\frac{5}{2}\) space and hence it has topological property.

Lemma 1. Let \((X, \tau_{1}, \tau_{2})\) be a normal space then the property of normality imply topological and hereditary.

Proof. Let \((X, \tau_{1}, \tau_{2})\) be a normal bitopological space. Then there exist two disjoint closed sets \(x\) and \(y\) with \(x\neq y\) and two disjoint open sets say \(U\) and \(V\) such that \(x\subset U\) and \(y\subset V.\) By Definition, two disjoint closed sets \(x, y \in X\) this therefore implies that \(x\in U,\) \(y\) does not exists in \(U\) and \(x\) does not exists in \(V\) but \(y\in V.\) Since normal bitopological space implies \(T_{2}\) space then we have \(x, y\in X\) with \(x\neq y\) then \(\exists U\in \tau_{1}\cup \tau_{2}\) such that \(x\in U,\) \(y\) does not exists in \(U\) and also \(x\) does not exists in \(V\) but \(y\in V\) therefore normal spaces have topological property. Secondly, we prove that normality imply hereditary property. By Proposition 2, \(\chi: (X, \tau_{1}, \tau_{2})\rightarrow (Y, \tau_{3}, \tau_{4})\) and \(\chi\) is homeomorphism if and only if it is a bijective function. Let \(A\subseteq X\) and let \((X, \tau_{1}, \tau_{2})\) be a normal space then this implies that \(A\) is also normal. By conditions for normality, \((X, \tau_{1}, \tau_{2})\) is a normal space then \((A, \tau_{A1}, \tau_{A2})\) is also normal since disjoints closed sets \(x,y\in A\) and \(x, y\in X\) with \(x\neq y.\) Therefore, considering disjoint open sets \(U\) and \(V\) we have \(U\in \tau_{1}\) and \(V\in \tau_{2}\) such that \(x\in U,\) \(y\in V\) and \(U\cap V=\emptyset.\) Therefore, \(U\in \tau_{1}\) and \(V\in \tau_{2}\) then it implies that \(U\cap A\in \tau_{A1}\) and \(V\cap A\in \tau_{A2}\) hence it follows that \(x\in \cap AU\), \(y\in V\cap A.\) Then it implies that \((U\cap A) \cap (V\cap A) \cap A = \emptyset \cap A = \emptyset.\) By hypothesis, \((A, \tau_{A1}, \tau_{A2})\) is normal and has topological property induced from \((X, \tau_{1}, \tau_{2}).\)

Proposition 5. A bitopological space \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-T_{\lambda}\) if and only if it is \(ij-\pi_{\lambda}\)-symmetric.

Proof. Assume that \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-T_{\lambda}\). Let \(x\in ij-Cl\pi_{\lambda} (\{y\})\) and \(U\) to be any \(ij-\pi_{\lambda}\)-open set such that \(y\in U\) and \(x\in U.\) Therefore, this implies that every \(ij-\pi_{\lambda}\)-open set that is containing \(y\) also contains \(x\) with \(x\neq y\) hence \(y\in ij-Cl\pi_{\lambda} (\{x\}).\) Conversely, let \(U\) be \(ij-\pi_{\lambda}\)-open in \(X\) hence it implies that \(x\in U,\) \(y\) does not exist in \(U,\) and \(x\) does not exists in \(ij-Cl\pi_{\lambda} (\{y\}).\) Then if \(y\) does not exist in \(ij-Cl\pi_{\lambda} (\{x\})\) then \(ij-Cl\pi_{\lambda}(\{x\}) \subseteq U.\) Hence it implies that both \(U\) and \(ij-Cl\pi_{\lambda} (\{y\})\) are disjoint open sets where \(x\in U\) and \(y\) does not exists in \(U\) or \(y\in ij-Cl\pi_{\lambda}(\{x\})\) and \(x\) does not exist in \(ij-Cl\pi_{\lambda}(\{x\}).\) By Proposition 1, \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}- T_{0}\) since it has topological property. Therefore, \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-T_{\lambda}.\)

Proposition 6. A bitopological space \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-T_{0}\) and \(ij-\pi_{\lambda}-T_{1}\) if and only if it is \(ij-\pi_{\lambda}-T_{\lambda}\) symmetric.

Proof. Let \((X, \tau_{1}, \tau_{2})\) be \(ij-\pi_{\lambda}-T_{\lambda}\) symmetric and let \(x, y\in X\) with \(x\neq y.\) Since \((X, \tau_{1}, \tau_{2})\) \(ij-\pi_{\lambda}-T_{\lambda}\) symmetric we need prove that it is also \(ij-\pi_{\lambda}-T_{0}.\) Let \(x\) and \(y\) be two disjoint closed sets in \(X.\) Then \(U\) and \(ij-\pi_{\lambda} (\{y\})\) be any two disjoint open sets. By Definition 9, two disjoint closed sets \(x\) and \(y\) are both members of open sets either \(U\) or \(ij-\pi_{\lambda} (\{y\}).\) Hence suppose that each \(ij-\pi_{\lambda}\)-open set contains \(x\) and \(y\) then \(y\in U\) and \(x\in U.\) Since \(U\) is a member of \(ij-\pi_{\lambda}\)-open set then it implies that \(x\in U\) and \(y\) does not exists in \(U.\) By Tychonoff theorem, it follows that \(ij-\pi_{\lambda} (\{x\})\subseteq U\) hence \(y\) does not exists \(U.\) Hence it implies that \(y\) does not exists in \(ij-\pi_{\lambda} (\{x\})\) thus by assumption \(x\) does not exists in \(ij-\pi_{\lambda} (\{x\}).\) Since \(ij-\pi_{\lambda}(\{x\})\subseteq U\) then \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-T_{0}.\) Now, it implies that \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-T_{\lambda}\) symmetric. Therefore, every \(ij-\pi_{\lambda}-T_{\lambda}\) symmetric imply \(ij-\pi_{\lambda}-T_{1}.\) Since \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}- T_{0}\) we assume without any loss of generality that \(x\in K\subset X\{y\}\) for \(ij-\pi_{\lambda}\)-open set \(K\) where \(x\) does not exists in \(ij-Cl\pi _{\lambda}(\{y\})\) and \(y\) does not exists in \(ij-Cl\pi (\{x\}).\) Therefore, \(X\setminus ij-Cl\pi(\{x\})\) is an \(ij-\pi_{\lambda}\)-open set containing \(y\) but not \(x.\) Hence \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-T_{1}.\)

Lemma 2. A bitopological space \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-T_{0}\) if and if for any \(x, y\in X\) and \(ij-Cl\pi_{\lambda}(\{x\})= ij-Cl\pi_{\lambda} (\{y\})\) then implies that \(ij-Cl\pi_{\lambda}(\{x\}) \cap ij-Cl\pi_{\lambda} (\{y\})=\emptyset.\)

Proof. Let \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-T_{0}.\) Then we have disjoint closed sets \(x\) and \(y\) where \(x\in X\) such that \(ij-Cl\pi_{\lambda} (\{y\}) \neq ij-Cl\pi _{\lambda}(\{x\}).\) Therefore, there exists \(x\in ij-Cl\pi_{\lambda}(\{x\})\) such that \(x\) does not exists in \(ij-Cl\pi_{\lambda}(\{y\})\) this implies that \(y\in ij-Cl\pi_{\lambda}(\{y\})\) and \(n\) does not exists in \(ij-Cl\pi_{\lambda}(\{x\}).\) Since \(x\) is not a member of \(ij-Cl\pi_{\lambda}(\{y\})\) therefore there exists \(V\in ij-B\lambda O (X, x)\) such that \(y\) does not exists in \(V.\) However, \(x\in ij-Cl\pi_{\lambda}(\{x\})\) hence \(x\in V.\) Therefore, this follows that \(x\) is not a member of \(ij-Cl\pi_{\lambda}(\{y\}).\) Then it implies that \(x\in X\setminus ij-Cl\pi_{\lambda}(\{y\}) \in ij-B\lambda O (X).\) Since \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-T_{0}\) then \(ij-Cl\pi_{\lambda}(\{x\})\subset X\setminus ij-Cl\pi_{\lambda}(\{y\}).\) By Proposition 6, we have \(ij-Cl\pi_{\lambda}(\{x\})\cap ij-\pi Cl_{\lambda}(\{y\}) = \emptyset.\) Conversely, let \(V\in ij-B\lambda O(X, x).\) We show that \(ij-Cl\pi_{\lambda}(\{x\})\subset V.\) Let \(y\) not to be an element of \(V\) then it follows that \(y\in X\setminus V\) hence \(y=x\) and \(x\) does not exists in \(ij-Cl\pi_{\lambda}(\{y\}).\) This shows that \(ij-Cl\pi_{\lambda} (\{y\}) \neq ij-Cl\pi_{\lambda}(\{x\}).\) By assumption \(ij-Cl\pi_{\lambda} (\{y\}) \cap ij-Cl\pi_{\lambda}(\{x\})=\emptyset.\) By hypothesis, \(y\) does not exists in \(ij-Cl\pi_{\lambda}(\{x\})\) and so \(ij-Cl\pi_{\lambda}(\{x\}) \subseteq V.\) Therefore, \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-T_{0}.\)

Theorem 1. A bitopological space \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-T_{1}\) if and only if for any points \(x\) and \(y\) in \(X,\) \(ij-Ker\pi_{\lambda} (\{x\}) =ij-Ker\pi _{\lambda} (\{y\})\) implying that \(ij- Ker\pi_{\lambda} (\{x\}) \cap ij-Ker\pi _{\lambda} (\{y\}) = \emptyset.\)

Proof. Let \((X, \tau_{1}, \tau_{2})\) be \(ij-\pi_{\lambda}-T_{1}.\) Then let disjoint closed sets \(x, y\in X.\) By hypothesis, \(ij- Ker\pi_{\lambda} (\{x\}) \neq ij-Ker\pi_{\lambda} (\{y\})\) then it follows that \(ij-\pi Cl_{\lambda} (\{y\}) \neq ij-\pi Cl_{\lambda}(\{x\}).\) Suppose that \(z\in ij-Ker\pi_{\lambda} (\{x\}) \cap ij-Ker\pi_{\lambda}(\{y\})\) then it implies that \(z\in ij- Ker\pi_{\lambda} (\{x\}).\) Therefore, it follows that \(z\in ij-Cl\pi_{\lambda}(\{z\}).\) Thus by Lemma 2, we have \(ij-Cl\pi_{\lambda}(\{x\}) = ij-Cl\pi_{\lambda}(\{z\})\) this is by contraction. Hence \(ij-Cl\pi_{\lambda} (\{y\}) = ij-Cl\pi_{\lambda} (\{z\}) = ij-Cl\pi_{\lambda}(\{x\}).\) Therefore, \(ij-Ker\pi_{\lambda}(\{x\})\cap ij-\pi Ker_{\lambda} (\{y\}) = \emptyset.\) Conversely, let \((X, \tau_{1}, \tau_{2})\) be a bitopological space such that disjoint closed points \(x\) and \(y\) are members of \(X.\) Then it implies that \(ij- Ker\pi_{\lambda} (\{x\}) \neq ij-Ker\pi_{\lambda} (\{y\})\) hence \(ij-Ker\pi_{\lambda} (\{x\}) \cap ij-Ker\pi_{\lambda} (\{y\}) = \emptyset.\) Therefore, \(ij-Ker\pi_{\lambda} (\{x\}) \neq ij-Cl\pi_{\lambda} (\{y\})\) then \(\cap ij- Ker\pi_{\lambda} (\{y\})=\emptyset.\) By assumption we have \(z\in ij-Ker\pi_{\lambda}(\{x\})\) and \(x\in ij-Ker\pi_{\lambda} (\{y\})\) therefore \(ij- Ker\pi_{\lambda} (\{x\}) \cap ij-Ker\pi_{\lambda} (\{z\}) = \emptyset.\) Therefore, \(ij- \pi Ker_{\lambda} (\{x\}) = ij-\pi Ker_{\lambda} (\{z\}).\) Thus it follows that \(z\in ij-Cl\pi_{\lambda}(\{x\}) = ij-Cl\pi_{\lambda}(\{y\}).\) Then \(ij- Ker\pi_{\lambda} (\{x\}) = ij-Ker\pi_{\lambda} (\{z\})= ij-Ker\pi_{\lambda} (\{y\})\) this is by contradiction. Therefore, \(ij-Ker\pi_{\lambda} (\{x\}) \neq ij-Ker\pi_{\lambda}(\{y\}).\) Then this implies that \(ij-Ker\pi_{\lambda} (\{x\}) \cap ij-Ker\pi_{\lambda}(\{y\}) = \emptyset.\) Therefore, \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-T_{1}.\)

Proposition 7. For bitopological space \((X, \tau_{1}, \tau_{2})\) the following are equivalent:

• (i). \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-T_{1}.\)
• (ii). For each \(x, y\in X\) then \(U\) is \(ij-\pi_{\lambda}-T_{1}\)-open then \(x\in U\) if and only if \(y\in U\). \(x\in U\) and \(y\in V.\)
• (iii). If \(x, y\in X\) such that \(ij-\pi-Cl_{\lambda}(\{x\}) \neq ij-\pi-Cl_{\lambda}(\{y\}),\) then there exists closed sets \(F_{1}\) and \(F_{2}\) whereby \(x\in F_{1},\) \(y\) does not exists in \(F_{1},\) \(y\in F_{2}\) \(x\) does not exists in \(F_{2}\) and \(X= F_{1} \cup F_{2}.\)

Proof. Proving for \((i)\Rightarrow (ii).\) Let \(x, y\) be two closed disjoint sets in \(X.\) By Theorem 1, \(ij-Cl\pi_{\lambda} (\{x\} = ij-Cl\pi_{\lambda} (\{y\}\) or \(ij-Cl\pi_{\lambda} (\{x\} \neq ij-Cl\pi_{\lambda} (\{y\}.\) Let \(U\) be \(ij-\pi_{\lambda}\)-open therefore \(x\in U\) such that \(y\in ij-Cl\pi_{\lambda} (\{x\}\subset U.\) Since \(x\in U\) then \(x\in ij-\pi-Cl_{\lambda} (\{y\}\subset U.\) Therefore, \(ij-Cl\pi_{\lambda} (\{x\} \neq ij-Cl\pi_{\lambda} (\{y\}.\) This implies that there exists disjoint \(ij-\pi_{\lambda}\)-open sets \(U\) and \(V\) such that \(x\in ij-Cl\pi_{\lambda} (\{x\}\subset U\) and \(y\in ij-Cl\pi_{\lambda} (\{y\}\subset V.\)

For \((ii). \Rightarrow (iii).\) Let \(x, y\in X\) such that \(ij-Cl\pi_{\lambda} (\{x\} \neq ij-Cl\pi_{\lambda} (\{y\}.\) Since \(x\) is not a member of \( ij-Cl\pi_{\lambda} (\{y\}\) and \(y\) is not a member of \(ij-Cl\pi_{\lambda} (\{x\}\) then \(x\) does not belongs to \(ij-Cl\pi_{\lambda} (\{y\}\). Therefore, there exists \(ij-\pi_{\lambda}\)-open set \(A\) such that \(x\in A\), \(y\) does not exists in \(A.\) By \((ii)\) there exists disjoints \(ij-\pi_{\lambda}\)-open sets \(U\) and \(V\) such that \(x\in U,\) \(y\in V.\) Then it implies that \(F_{1} = X\setminus V\) and \(F_{2} = X\setminus U\) are \(ij-\pi_{\lambda}\)-closed sets such that \(x\in F_{1},\) \(y\) does not exists in \(F_{1},\) and \(y\in F_{2}\), \(x\) does not exists in \(F_{2}\) hence it follows that \(X= F_{1}\cap F_{2}.\)

For \((iii) \Rightarrow (i).\) We need to show that \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-T_{0}\) space. Let \(U\) be an \(ij-\pi_{\lambda}\)-open set such that \(x\in U.\) Then \(ij-\pi-Cl_{\lambda} (\{x\}\subset U\) this implies that \(y\in ij-Cl\pi_{\lambda} (\{x\}\cap (X\setminus U).\) By (i) it implies that \(ij-Cl\pi_{\lambda} (\{x\} \neq ij-Cl\pi_{\lambda} (\{y\} = ij-Cl\pi_{\lambda} (\{y\})\) then \(y\in U.\) By \((iii),\) there exists \(ij-\pi_{\lambda}\)-closed sets \(F_{1}\) and \(F_{2}\) such that \(x\in F_{1},\) \(y\) does not exists \(F_{1}\) but \(y\in F_{2},\) \(x\) does not exists in \(F_{2}\) such that \(X= F_{1}\cup F_{2}.\) Hence \(y\in F_{2}\setminus F_{1}\setminus X \setminus F_{1}\in ij-B\lambda O(X)\) and \(x\) does not exists in \(X\setminus F_{1}\) this is by contradiction. Therefore, \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}- T_{0}\) space. Let \(p, q \in X\) such that \(ij-Cl\pi_{\lambda} (\{p\} \neq ij-Cl\pi_{\lambda} (\{q\},\) then there are \(ij-\pi_{\lambda}\)-closed sets \(H_{1}\) and \(H_{2}\) such that \(p\in H_{1},\) \(q\in H_{1},\) \(q\) is not a member of \(H_{1}\) and \(q\in H_{2}\) while \(p\in H_{2}\) and \(X=H_{1}\cap H_{2}.\) Therefore, \(p\in H_{1}\setminus H_{2}\) and \(q\in H_{2}\setminus H_{1}.\) Then \(H_{1}\setminus H_{2}\) and \(H_{2}\setminus H_{1}\) are disjoint \(ij-\pi_{\lambda}\)-open sets. Hence \(ij-Cl\pi_{\lambda} (\{p\}) \subset H_{1} \setminus H_{2}\) and \(ij-Cl\pi_{\lambda}(\{q\}) \subset H_{2}\setminus H_{1}.\)

Theorem 2. Every normal \(ij-\pi_{\lambda}-T_{2}\) bitopological space \((X, \tau_{1}, \tau_{2})\) is Hausdorff space.

Proof. Let \((X, \tau_{1}, \tau_{2})\) be a normal bitopological space then we have disjoint closed sets \(x\) and \(y\) with \(x\neq y.\) Let \(U\) and \(V\) be disjoint open sets such that \(x\subset U\) and \(y\subset V.\) By Definition 24, if there exist two disjoint closed sets \(x, y \in X\) then it implies that \(x\in U,\) \(y\) is not a member of \(U\) and \(x\) does not exists in \(V\) but \(y\in V.\) By hypothesis, normal bitopological spaces are also \(T_{2}\) spaces. Since \(x, y\in X\) with \(x\neq y\), then \(\exists U\in \tau_{1}\cup \tau_{2}\) such that \(x\in U,\) \(y\) does not exists in \(U\) and \(x\) does not exists in \(V\) but \(y\in V.\) By Lemma 1, if \((X, \tau_{1}, \tau_{2})\) is a normal space then \((A, \tau_{A1}, \tau_{A2})\) is also normal since \(A\subseteq X.\) Then there exists open disjoint sets \(U\) and \(V\) where \(U\in \tau_{1}\) and \(V\in \tau_{2}\) such that \(x\in U,\) \(y\in V\) and \(U\cap V=\emptyset.\) Consequently, by conditions for normality it implies that \(U\in \tau_{1}\) and \(V\in \tau_{2}\) then \(U\cap A\in \tau_{A1}\) therefore, \(V\cap A\in \tau_{A2},\) and \(x\in \cap AU,\) \(y\in V\cap A.\) Then \((U\cap A) \cap (V\cap A) \cap A = \emptyset \cap A=\emptyset.\) Since \((A, \tau_{A1}, \tau_{A2})\) is a bitopological subspace then it implies that it has a topological property. Let \(T_{2}\) space be Hausdorff space then we have two closed sets \(x\) and \(y\) with \(x\neq y.\) By hypothesis, we have two disjoint open sets \(U\) and \(V\) such that \(x\in U,\) \(y\) does not exists in \(V\) and \(x\) does not exists in \(V\) but \(y\in V\) Then it implies that \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-T_{2}.\) Therefore, \((X, \tau_{1}, \tau_{2})\) is a Hausdorff space and every normal \(ij-\pi_{\lambda}-T_{2}\) space is also Hausdorff space.

Corollary 1. The property of \(ij-\pi_{\lambda}-T_{2}\) in bitopological space \((X, \tau_{1}, \tau_{2})\) is hereditary.

Proof. By hypothesis, \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-T_{2}.\) Let \(X\) be any set and \(A\subset X\) then \((A, \tau_{A1}, \tau_{A2})\) is also \(T_{2}\) space. Since \((A, \tau_{A1}, \tau_{A2})\) is a bitopological subspace therefore it inherits properties from \((X, \tau_{1}, \tau_{2}).\) Let \(m, n\in A\) with \(m\neq n\) and \(m, n\in X\) with \(m \neq n.\) Then it follows that we have disjoint open sets \(U\) and \(V.\) Therefore, it implies that there exists \(U\in \tau_{1} \cup \tau_{2}\) such that \(m\in U,\) \(n\) does not exists in \(V\) and \(m\) does not exists in \(V\) but \(m\in V.\) By Theorem 2, we have \(U\in \tau_{1} \cup \tau_{2}.\) This implies that \(U\in \tau_{1}\) or \(U\in \tau_{2}\) such that \(U\cap A\in \tau_{A1}\) or \(U\cap A\in \tau_{A2}.\) Hence by separation axioms it suffices that \(U\cap A\in \tau_{A1} \cap \tau_{A2}.\) Then \(m, n\in A\) and \(m\in U\cap A,\) where \(n\) is not a member of \(U\cap A\) or \(m\) is not a member of \(U\cap A,\) \(n\in U\cap A.\) Hence \((A, \tau_{A1}, \tau_{A2})\) implies \(T_{2}\) space. Since \((A, \tau_{A1}, \tau_{A2})\) is a \(T_{2}\) space then it implies that it is also \(ij-\pi_{\lambda}-T_{2}.\) Therefore, \((A, \tau_{A1}, \tau_{A2})\) is also \(ij-\pi_{\lambda}-T_{2}\) space and so it is hereditary.

Corollary 2. The property of \(ij-\pi_{\lambda}-T_{2}\) in bitopological space \((X, \tau_{1}, \tau_{2})\) is topological.

Proof. Let \(\chi: (X, \tau_{1}, \tau_{2})\rightarrow (Y, \tau_{1}, \tau_{2}).\) By hypothesis, \((X, \tau_{1}, \tau_{2})\) is a \(T_{2}\) space. Let \(\chi: (X, \tau_{1}, \tau_{2})\rightarrow (Y, \tau_{3}, \tau_{4})\) be a homeomorphism. We have disjoint open sets \(y_{1}, y_{2}\in Y\) with \(y_{1} \neq y_{2}.\) By Definition 11, all elements in \(Y\) are images of elements in \(X\) and so \(\chi\) is a bijective function. Therefore, it implies that \(\exists x_{1} ,x_{2} \in X\) with \(\chi (x_{1}) = y_{1}\) and \(\chi(x_{2}) = y_{2}.\) Suppose \(\chi\) is a one to one function with \(y_{1} \neq y_{2}\) therefore this implies that \(\chi(x_{1}) \neq \chi(x_{2}),\) then it implies that \(x_{1} \neq x_{2}\) hence \(x_{1}, x_{2} \in X\) with \(x_{1} \neq x_{2}.\) Since \((X, \tau_{1}, \tau_{2})\) is a \(T_{2}\) space then it implies that \(\exists U\in \tau_{1}\) and \(V\in \tau_{2}\) such that \(x_{1}\in U,\) \(x_{2}\in V\) and so \(U\cap V\neq \emptyset.\) Since \(\chi\) is open hence \(\chi(U)\in \tau_{3}\) and \(\chi (V)\in \tau_{4}.\) By Tychonoff theorem, \(\chi(U)\cap \chi(V)\neq \emptyset\) suppose that there exists \(c\in X\) therefore \(c\in \chi(U) \cap \chi(V).\) This shows that \(c\in \chi(U)\) and \(c\in \chi(V)\) then \(\exists p_{1}\in U\) and \(p_{2}\in V\) such that \(c= \chi(p_{1})\) and \(c\in \chi(p_{2})\) with \(\chi(p_{1} = \chi(p_{2}).\) Suppose that \(p_{1}= p_{2}\) then it implies that \(p_{1}\in U\) and \(p_{1}\in V\) so \(p_{1}\in U\cap V\neq \emptyset.\) by contradiction if \(U\cap V= \emptyset\) then \(\chi(U) \cap \chi(V) = \emptyset.\) For any \(y_{1}, y_{2}\in Y\) with \(y_{1} \neq y_{2}\) then \(\chi(U) =\tau_{3}\) and \(\chi (V)\in c\) is obtained such that \(y_{1} \in \chi(U),\) \(y_{2} \in \chi(V)\) and \(\chi(U) \cap \chi(V)\neq \emptyset.\) Therefore, \((Y, \tau_{3}, \tau_{4})\) is a \(T_{2}\) space. Hence \((X, \tau_{1}, \tau_{2})\) and \((Y, \tau_{3}, \tau_{4})\) are \(ij-\pi_{\lambda}-T_{2}\) spaces and so they are topological.

Corollary 3. The property of \(ij-\pi_{\lambda}-T_\frac{5}{2}\) in bitopological space \((X, \tau_{1}, \tau_{2})\) is both topological and heredity.

Proof. Let \((X, \tau_{1}, \tau_{2})\) be a \(T_\frac{5}{2}\) space and let \(M\subseteq X.\) By hypothesis, it implies that a bitopological subspace \((M, \tau_{M1}, \tau_{M2})\) is also a \(T_\frac{5}{2}\) space. By Proposition 4, we have \(x\in M\) with \(x\neq y.\) Since \((X, \tau_{1}, \tau_{2})\) is a \(T_\frac{5}{2}\) space it implies that there exists \(A\in \tau_{1}\) and \(B\in \tau_{2}\) such that \(x\in A,\) \(y\in B\) and \(A\cap B=\emptyset\) hence it implies that \(A\in \tau_{1}\) and \(B\in \tau_{2}.\) By separation axioms technique, \(A\cap M\in \tau_{M1}\) and \(B\cap M\in \tau_{M2}\) therefore \(x, y\in M\) hence \(x\in A\cap M\) and \(y\in B\cap M.\) Then it follows that \((A\cap M)\cap (B\cap M)= (A\cap M)\cap M=\emptyset \cap M=\emptyset.\) Therefore, \((M, \tau_{M1}, \tau_{M2})\) is \(ij-\pi_{\lambda}-T_\frac{5}{2}.\) Since \((X, \tau_{1}, \tau_{2})\) is an \(ij-\pi_{\lambda}-T_\frac{5}{2}\) space then it has topological property. Let \(\chi: (X, \tau_{1}, \tau_{2})\rightarrow (Y, \tau_{3}, \tau_{4})\) and \(\chi\) is homeomorphic. Since \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-T_\frac{5}{2}\) then it implies that \((Y,\tau_{3}, \tau_{4})\) is also an \(ij-\pi_{\lambda}-T_\frac{5}{2}\) space. By hypothesis, \(y_{1}, y_{2} \in Y\) with \(n_{1}\neq y_{2}.\) Suppose that \(\chi\) is onto function then there exists \(x_{1}, x_{2}\in X\) such that \(\chi (x_{1}) = \chi(y_{1})\) and \(\chi(y_{2}) = x_{2}.\) Also if \(\chi\) is an injective function with \(y_{1} \neq y_{2}\) it follows that \(\chi(x_{1}) \neq \chi(x_{2})\) and \(x_{1} \neq x_{2}.\) Therefore, since \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-T_\frac{5}{2}\) space then it implies that \(x_{1}, x_{2}\in X,\) with \(x_{1} \neq x_{2}\) and so \(\exists A\in \tau_{1} \cup \tau_{2}\) hence \(x_{1}\in A,\) \(x_{1}\) does not exists in \(A\) or \(x_{1}\) does not exists in \(A,\) \(x_{2}\in A\) similarly, \(x_{1}\in A\), \(x_{2}\) does not exists in \(A\) hence \(A\in \tau_{1} \cup \tau_{2}.\) Then \(\chi (A)\in \chi(\tau_{1} \cup \tau_{2})\) since \(\chi\) is open we have \(\chi (A) \in \chi(\tau_{1}) \cup \chi(\tau_{2})\in \tau_{3} \cup \tau_{4}.\) Since \(x_{1}\in A\) then it implies that \(\chi(x_{1})\in \chi (A)\) so \(y_{1}\in \chi(A)\) and \(x_{2}\) does not exists in \(A\) which implies that \(\chi(x_{2})\) does not exists in \(\chi(A)\) and \(y_{2}\) does not exists \(\chi(A).\) For any \(y_{1}, y_{2}\in Y\) with \(y_{1} \neq y_{2},\) \(\chi(A)\in \tau_{3} \cup \tau_{4}\) is obtained such that \(y_{1}\in \chi (A)\) and \(y_{2}\) does not exists in \(\chi(A).\) Hence \((Y, \tau_{3}, \tau_{4})\) is a \(ij-\pi_{\lambda}-T_\frac{5}{2}\) space. Therefore, \(ij-\pi_{\lambda}-T_\frac{5}{2}\) space is both topological and hereditary.

Theorem 3. A bitopological space \((X, \tau_{1}, \tau_{2})\) is pairwise \(\lambda T_{0}\) if and only if there is either closed distinct point of \(X\) either \(\tau_{1}-\eta\) or \(\tau_{2}-\eta.\)

Proof. Let \(x, y\in X\) be two distinct points in \(X.\) By hypothesis there exist two disjoint open sets \(U\) and \(V.\) Then \(U\) is a \(\tau_{1}-\eta\)-open set containing \(x\) but not \(y.\) Similarly, \(V\) is a \(\tau_{2}-\eta\)-open set that contains \(y\) but not \(x.\) Thus by Definition 17, \(y\in \tau_{1}-\eta cl \{y\}\subset X-U\) and so it implies that \(x\) does not belongs to \(\tau_{1}-\eta cl \{y\}.\) By Proposition 7, it follows that \(\tau_{1}-\eta cl\{x\}\neq\tau_{1}-\eta cl \{y\}\) as \(x\), \(y\) are distinct points in \(X.\) Therefore \(\tau_{1}-\eta cl \{x\}\neq \tau_{1}-\eta cl \{y\}\) and \(\tau_{2}-\eta cl \{x\}\neq \tau_{2}-\eta cl \{y\}.\) Suppose that \(p\) is a point of \(X\) such that \(p\in \tau_{1}-\eta cl \{y\}\) then \(p\) does not belongs to \(\tau_{1}-\eta cl \{x\}.\) This implies that \(y\) does not belongs to \(\tau_{1}-\eta cl \{x\}.\) Hence if \(y\in\tau_{1}-\eta cl \{x\}\) then \(\tau_{1}-\eta cl \{y\}\subset \tau_{1}-\eta cl \{x\}\) and \(p\in \tau_{1}-\eta cl \{y\}\subset \tau_{1}-\eta cl \{x\}.\) By contradiction \(p\) does not belongs to \(\tau_{1}-\eta cl \{x\}\) and so \(p\in\tau_{1}-\eta cl \{y\}\subset \tau_{1}-\eta cl \{x\}.\) This contradicts the fact that \(p\) is not a member of \(\tau_{1}-\eta cl \{x\}\) hence \(y\) also does not belongs to \(\tau_{1}-\eta cl \{x\}.\) Therefore, \(U= X - \tau_{1}- \eta cl \{x\}\) is a \(\tau_{1}-\eta\)-open set containing \(y\) but not \(x.\) This implies that \(\tau_{2}-\eta cl \{x\}\neq \tau_{2}-\eta cl \{y\}.\) Hence \(\tau_{1}-\eta\) and \(\tau_{2}-\eta\) closures are distinct points of \(X.\)

Theorem 4. A bitopological space \((X, \tau_{1}, \tau_{2})\) is pairwise \(\lambda T_{0}\) if either \((X, \tau_{1})\) or \((X, \tau_{2})\) is \(\lambda T_{0}.\)

Proof. Let \((X, \tau_{1}, \tau_{2})\) be a bitopological space. Then \((X, \tau_{1}, \tau_{2})\) is pairwise \(\lambda T_{0}\) if and only if either \((X, \tau_{1})\) or \((X, \tau_{2})\) is \(\lambda T_{0}.\) By hypothesis, there are two disjoint closed sets \(x\) and \(y\) which are members of \(X.\) Similarly, there are two disjoint open sets \(U\) and \(V.\) By Theorem 3, \(U\) is a \(\tau_{1}-\eta\)-open set containing \(x\) but not \(y\) hence \(y\in \tau_{1}-\eta cl \{y\}\subset X-U\) and \(x\) does not belongs to \(\tau_{1}-\eta cl\{y\}.\) Since topological spaces imply bitopological spaces we have \(\tau_{1}-\eta cl \{x\}\neq \tau_{1}-\eta cl \{y\}\)\(\tau_{1}-\eta\) and \(\tau_{2}-\eta\) closures are distinct points. Conversely, this need not to be true in general. Let \(X= \{a, b, c\},\) \(\tau_{1} = \{X, \emptyset, \{a\}, \{b, c\}\}\) and \(\tau_{2} = \{X, \emptyset, \{c\}, \{a, b\}\}\). Therefore, it shows that a bitopological space \((X, \tau_{1}, \tau_{2})\) is pairwise \(\lambda T_{0}\) when neither \((X, \tau_{1})\) nor is \((X, \tau_{2})\) is \(\lambda T_{0}.\)

Theorem 5. A bitopological space \((X, \tau_{1}, \tau_{2})\) is \(ij-\lambda T_{0}\) if it is normal.

Proof. Let \((X, \tau_{1}, \tau_{2})\) be a normal space. By hypothesis, it follows that there exists two disjoint closed sets \(a\) and \(b\) where \(a\neq b.\) Similarly, this implies that there exists two disjoint open sets \(M\) and \(N\) such that \(a\subset M\) and \(b\subset N.\) Then \(a, b \in X\) and so \(a\in M\) where \(a\) does not exists in \(N\) but \(b\in N\) for all \(a, b\in X\) with \(a\neq b.\) By conditions for normality, there exists \(M\in \tau_{1}\cup \tau_{2}\) such that \(a\in M,\) \(b\) does not exists in \(U\) also \(a\) does not exists in \(N\) but \(b\in N.\) This shows that \((X, \tau_{1}, \tau_{2})\) is a normal space. Since \((X, \tau_{1}, \tau_{2})\) is \(\lambda T_{0}\) then it implies that it is \(ij-\lambda T_{0}\)-normal. Thus, by Definition 18, \(M\) is a \(\tau_{1} -\eta\)-open set containing \(a\) but does not contain \(b.\) Therefore, it implies that \(b\in \tau_{1}-\eta cl \{b\}\subset X-M\) and so \(a\) does not belongs to \(\tau_{1}-\eta cl \{b\}.\) By Tychonoff theorem, we have \(\tau_{1}-\eta cl \{a\}\neq \tau_{1}-\eta cl \{b\}.\) Since \(a\), \(b\) are two distinct points of \(X\) then neither \(\tau_{1}-\eta cl \{a\}\neq \tau_{1}-\eta cl \{b\}\) nor \(\tau_{2}-\eta cl \{a\}\neq \tau_{2}-\eta cl \{b\}.\) Let \(c\) to be any point of \(X\) such that \(c\in \tau_{1}-\eta cl \{b\},\) \(c\) does not belongs to \(\tau_{1}-\eta cl \{a\}\) and \(b\) does not belong to \(\tau_{1}-\eta cl \{a\}.\) If \(b\in \tau_{1}-\eta cl\{a\}\) then \(\tau_{1}-\eta cl \{b\}\subset \tau_{1}-\eta cl \{a\}\) and \(c\in \tau_{1}-\eta cl \{b\}\subset \tau_{1}-\eta cl \{a\}.\) By contradiction, since \(c\) does not belongs to \(\tau_{1}-\eta cl \{a\}\) then it implies that \(b\) also does not exists in \(\tau_{1}-\eta cl \{a\}\) thus \(M=X-\tau_{1}- \eta cl \{a\}\) is a \(\tau_{1}-\eta\)-open set that contains \(b\) but not \(x.\) Hence it implies that \(\tau_{2}-\eta cl \{a\}\neq \tau_{2}-\eta cl \{b\}.\) Therefore, \((X, \tau_{1}, \tau_{2})\) is \(ij-\lambda T_{0}\) and it implies that is a normal space.

Corollary 4. Every \(ij-\pi_{\lambda}-{\lambda} T_{2}\) is \(ij-\pi_{\lambda}-\lambda T_{1}\) and \(ij-\pi_{\lambda}-\lambda T_{0}.\)

Proof. Let \((X, \tau_{1}, \tau_{2})\) be \(ij-\pi_{\lambda}- \lambda T_{2}.\) By hypothesis, \((X, \tau_{1}, \tau_{2})\) is pairwise \(\lambda T_{0}.\) Let \(G\) be any \(T_{i}-\pi_{\lambda}\)-open set and hence \(x\in G\) such that each point \(y\in X-G,\) \(T_{j}-\pi Cl \{y\}.\) Then it implies that there exists \(T_{i}-\pi_{\lambda}\) open set \(U_{y}\) and any \(T_{j}-\pi_{\lambda}\)-open set \(V_{y}\) such that \(x\in U_{y},\) \(y\in V_{y}.\) Therefore, by hypothesis we have \(U_{y} \cap V_{y}=\emptyset.\) If \(A=\bigcup \{V_{y}:y\in X-G\}\) then \(X-G\subset A\) and \(x\) does not exists in \(A.\) Therefore, \(T_{j}-\pi_{\lambda}\) openness of \(A\) implies that \(T_{j}-\pi Cl \{x\}\subset X-A\subset G.\) Hence \(X\) is \(\lambda T_{0}\) and \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi{\lambda}-T_{0}.\) By hypothesis, \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}- \lambda T_{0}\) and there exists closed disjoint sets \(x\) and \(y\) with \(x\neq y\) and \(x\in ij-\pi Cl_{\lambda} (\{y\}).\) By assumption \(y\) does not exists in \(ij-\pi Cl_{\lambda} (\{x\})\) then \(ij-\pi Cl_{\lambda} (\{x\})\subseteq U.\) Thus this implies that \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-\lambda T_{0}.\) Since \((X, \tau_{1}, \tau_{2})\) is \(ij-\pi_{\lambda}-\lambda T_{0}\) then it is \(ij-\pi_{\lambda}-{\lambda} T_{2}.\) Therefore, \(ij-\pi_{\lambda}-{\lambda} T_{2}\) imply \(ij-\pi_{\lambda}-\lambda T_{1}\) which also imply \(ij-\pi_{\lambda}-\lambda T_{0}.\)

Proposition 8. A space \((X, \tau_{1}, \tau_{2})\) is pairwise \(\Omega-T_{1}\) if and only if \((X, \tau_{1})\) and \((X, \tau_{2})\) are \(\Omega-T_{1}.\)

Proof. Let \((X, \tau_{1}, \tau_{2})\) be pairwise \(\Omega-T_{1}\) space. Let \(x\) and \(y\) be a pair of distinct points of set \(X.\) By Definition 19, there exists a \(\tau_{i}\)-preopen set which is containing \(x\) but does not contain \(y.\) Since \(x\) and \(y\) are closed sets in \(X\) then it implies that \(T_{i}-\Omega Cl\{x\} \neq T_{j}-\Omega Cl \{y\},\) \(i, j = 1, 2\) with \(i\neq j.\) Therefore, there exists a \(T_{i}-\Omega\)-open set \(U\) and a \(T_{j}-\Omega\)-open set \(V\) such that \(x\in V,\) \(y\in U\) and \(U\cap V=\emptyset.\) Therefore, it follows that \(i= 1\) or \(2\) then there exist \(\tau_{i}\)-preopen set \(U\) and \(\tau_{j}\)-preopen set \(V.\) This therefore implies that \(T_{i}-\Omega Cl \{x\}\subset V\) and \(y\in U\) implies \(T_{j}-\Omega \{y\} \subset U,\) with \(U\cap V=\emptyset\) and \(i, j = 1, 2\) where \(i\neq j\). Therefore, \((X, \tau_{1})\) and \((X, \tau_{2})\) are \(\Omega-T_{1}\) and hence \((X, \tau_{1}, \tau_{2})\) is pairwise \(\Omega-T_{1}.\)

Theorem 6. Every bisemiopen subset of a pairwise \(p\)-regular space is pairwise \(p\)-regular.

Proof. Let \((X, \tau_{1}, \tau_{2})\) be pairwise \(p\)-regular space and let \(Y\subset X\) be a bisemiopen set. We show that subspace \((Y, \tau_{1Y}, \tau_{2Y})\) is pairwise \(p\)-regular. Let \(F\) be any \(\tau_{iY}\)-closed set and so \(x\) does not belongs to \(F.\) Then there exists a \(\tau_{i}\)-closed set \(A\) such that \(F= A\cap Y.\) Since \((X, \tau_{1}, \tau_{2})\) is a pairwise \(p\)-regular space and \(x\in A\) then there exists \(U\in PO (X, \tau_{j})\) and \(V\in PO(X, \tau_{i})\) hence we have \(A\subset U,\) \(x\in V\) and \(U\cap V=\emptyset\) where \(i, j = 1, 2\) with \(i\neq j.\) Now, it implies that we have \(L=U\cap Y\) and \(M=V\cap Y.\) By separation axioms, it follows that \(L\in PO(X, \tau_{i})\), \(M\in PO(Y, \tau_{iY}).\) Hence \(F\subset L,\) \(x\in M\) and so it implies that the intersection of \(L\) and \(M\) is emptyset that is, \(L\cap M=\emptyset.\) Therefore, \((Y, \tau_{1Y}, \tau_{2Y})\) is pairwise \(p\)-regular space.

Theorem 7. Every quasi \(T_\frac{7}{2}\) space is quasi \(T_{0}.\)

Proof. Let \((X, \tau_{1}, \tau_{2})\) be a quasi \(T_\frac{7}{2}\) space. We have \(x, y\in X\) with \(x\neq y.\) Hence it follows that \(qKer(\{x\}) \cap qKer(\{y\})=\emptyset\) since \(\{x\}\) and \(\{y\}\) are disjoint open sets with \(\{x\}\neq\{y\}.\) If \(qKer (\{x\}) \cap qKer(\{y\}) =\emptyset\) then it implies that \(T_\frac{7}{2}\) is a quasi space. Since \(\{x\}\neq \{y\}\) then by separation axioms it suffices that \(qKer (\{x\}) \neq qKer(\{y\}).\) Therefore, it implies that \((X, \tau_{1}, \tau_{2})\) is quasi \(T_{0}.\) Since \((X, \tau_{1}, \tau_{2})\) is quasi \(T_{0}\) then it is also quasi \(T_\frac{7}{2}.\)

Theorem 8. A bitopological space \((X, \tau_{1}, \tau_{2})\) is quasi \(T_\frac{7}{2}\) if and only if

\((qcl(\{y\}))=\cap \{y\} \bigcup(qcl(\{y\})\cap \{x\})\) is degenerate.

Proof. Let \(X\) be quasi \(T_\frac{7}{2}.\) Therefore, it shows that we have any of the two cases either \(x\) is weakly quasi separated from \(y\) or \(y\) weakly quasi separated from \(x.\) Suppose that \(x\) is weakly quasi separated from \(y\) then it implies that we have \(\{x\}\cap qcl\{y\}=\emptyset\) and \(\{y\}\cap qcl(\{x\})\) is called a degenerated set. Similarly, if \(y\) is weakly quasi separated from \(x\) then \(\{y\}\cap qcl(\{x\})=\emptyset\) and \(\{x\}\cap qcl(\{y\})\) is also a degenerated set. By Definition 29, it suffices that \((qcl(\{x\})\cap \{y\})\bigcup (qcl(\{x\})\cap \{y\})\) is a degenerate set. Hence \(qcl(\{x\})\cap \{y\})\bigcup (qcl(\{x\})\cap \{y\})\) is a degenerate set. By separation axioms, it implies that it is either an empty or singleton set. Then suppose that it is a singleton then its value is either \(\{x\}\) or \(\{y\}.\) Therefore, if it is \(\{x\}\) then \(y\) is weakly quasi separated from \(x.\) Also if it is \(\{y\}\) it is then \(x\) is weakly separated from \(y.\) Hence \((X, \tau_{1}, \tau_{2})\) is quasi \(T_\frac{7}{2}.\)

4. Conclusion

In this paper, we have given necessary conditions and characterized separation criteria for bitopological spaces via \(ij\)-continuity. We have shown that if a bitopological space is a separation axiom space, then that separation axiom space exhibits both topological and heredity properties. For instance, let \((X, \tau_{1}, \tau_{2})\) be a \(T_{0}\) space then, the property of \(T_{0}\) is topological and hereditary. Similarly, when \((X, \tau_{1}, \tau_{2})\) is a \(T_{1}\) space then the property of \(T_{1}\) is topological and hereditary. Lastly, we have shown that separation axiom \(T_{0}\) implies separation axiom \(T_{1}\) which also implies separation axiom \(T_{2}\) and the converse is true.

Acknowledgments

The authors are grateful to the reviewers for their useful comments.

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflicts of Interest

The authors declare no conflict of interest.

1. Introduction

The fixed point theory is a very interesting research area in due to its wide range of applicability, to resolve diverse problems emanating from the theory of nonlinear differential equations and integral equations.

Wardowski [1], generalized the famous Banach theorem [2] for \(F-\)contraction on metric spaces, several mathematicians extended this new notion for contraction on metric spaces [3,4,5,6].

The concept of a rectangular metric space was introduced by Branciari in [7]. After that, several interesting results about the existence of fixed points in rectangular metric spaces have been obtained [8,9,10,11]. Recently, Kari et al., [12], obtained some results for generalized \(\theta-\phi-\)expansive mapping in rectangular metric spaces.

In 1984, Wang et al., [13], presented some interesting work on expansion mappings in metric spaces. Recently, Kumar et al., [14], introduced a new concept of \((\alpha, \psi)-\)expansive mappings and established some fixed point theorems for such mapping in complete rectangular metric spaces.

In this paper, inspired by the idea of \(F-\)contraction introduced by Wardowski [1] in metric spaces, we presented generalized \(F-\)expansive mapping and establish various fixed point theorems for such mappings in complete rectangular metric spaces. Our theorems extend, generalize and improve many existing results.

2. Preliminaries

By an expansion mappings [13] on a metric space \((X,d)\), we understand a mapping \(T: X \to X\) satisfying for all \(x, y\in X\): \begin{equation*} d(Tx, Ty)\geq kd(x, y), \end{equation*} where \(k\) is a real in \(]1, +\infty[\).

In 2000 Branciari [7] introduced the concept of rectangular metric spaces.

Definition 1. [7] Let \(X\) be a non-empty set and \(\ d:X\times X\rightarrow \mathbb{R}^{+}\) be a mapping such that for all \(x,y\) \(\in X\) and for all distinct points \(u,v\in X\), each of them different from \(x\) and \(y,\) on has

• (i) \(d(x,y)=0\) if and only if \(x=y\);
• (ii) \(d(x,y)=d(y,x)\) for all distinct points \(x,y\in X\);
• (iii) \(d(x,y)\leq d(x,u)+d(u,v)+d(v,y)\)( the rectangular inequality).

Then \(\left( X,d\right) \) is called an rectangular metric space.

Definition 2. [15] Let \(T:X\rightarrow X\) and \(\alpha,\eta \) :\(X\times X\rightarrow \left[ 0,+\infty \right[.\) We say that \(T\) is a triangular \(\left( \alpha ,\eta \right)\)-admissible mapping if

• \(\left( T_{1}\right) \) \(\alpha \left( x,y\right) \geq 1\) \(\Rightarrow \) \(\alpha \left( Tx,Ty\right) \geq 1,\) \(x,y\in X\);
• \(\left( T_{2}\right) \) \(\eta \left( x,y\right) \leq 1\) \(\Rightarrow \) \(\eta\left( Tx,Ty\right) \leq 1,\) \(x,y\in X\);
• \(\left( T_{3}\right) \) \(\left\{ \begin{array}{c}\alpha \left( x,y\right) \geq 1\\ \alpha \left( y,z\right) \geq 1\end{array}\right. \Rightarrow \alpha \left( x,z\right) \geq 1\) for all \(x,y,z\in X\);
• \(\left( T_{4}\right) \) \( \left\{ \begin{array}{c}\eta \left( x,y\right) \leq 1 \\ \eta \left( y,z\right) \leq 1\end{array}\right. \Rightarrow \eta \left( x,z\right) \leq 1\) for all \(x,y,z\in X\).

Definition 3. [15] Let \(\left( X,d\right) \) be a rectangular metric space and let \(\alpha ,\eta \) :\(X\times X\rightarrow \left[ 0,+\infty \right[ \) be two mappings. Then

• (a) T is \(\alpha -\)continuous mapping on \(\left( X,d\right) ,\) if for given point \(x\in X\) and sequence \(\lbrace x_{n}\rbrace \) in \(X,\) \(x_{n}\rightarrow x\) and \(\alpha \left( x_{n},x_{n+1}\right) \geq 1\) for all \(n\in \mathbb{N},\) imply that \(Tx_{n}\rightarrow Tx\).
• (b) T is \(\eta \) sub\(-\)continuous mapping on \(\left( X,d\right) ,\) if for given point \(x\in X\) and sequence \(\lbrace x_{n}\rbrace \) in \(X,\) \(x_{n}\rightarrow x\) and \(\eta \left( x_{n},x_{n+1}\right) \leq 1\) for all \( n\in \mathbb{N},\) imply that T\(x_{n}\rightarrow Tx\).
• (c) T is \(\left( \alpha ,\eta \right) \) \(-\)continuous mapping on \(\left(X,d\right) ,\) if for given point \(x\in X\) and sequence \(\lbrace x_{n}\rbrace \) in \(X,\) \(x_{n}\rightarrow x\) and \(\alpha \left( x_{n},x_{n+1}\right) \geq 1\) or \( \eta \left( x_{n},x_{n+1}\right) \leq 1\) for all \(n\in \mathbb{N},\) imply that \(Tx_{n}\rightarrow Tx\).

Recently Hussain et al., gives the following definition [16]:

Definition 4. [16] Let \(d\left( X,d\right) \) be a rectangular metric space and let \(\alpha ,\eta \) :\(X\times X\rightarrow \left[ 0,+\infty \right[ \) be two mappings. The space \( X \) is said to be

• (a) \(\alpha -\)complete\(,\) if every Cauchy sequence \(\lbrace x_{n}\rbrace \) in \(X\) with \(\alpha \left( x_{n},x_{n+1}\right) \geq 1\) for all \(n\in \mathbb{N},\) converges in \(X.\)
• (b) \(\eta -\sup -\)complete\(,\) if every Cauchy sequence \(\lbrace x_{n}\rbrace \) in \(X\) with \(\eta \left( x_{n},x_{n+1}\right) \leq 1\) for all \(n\in \mathbb{N},\) converges in \(X.\)
• (c) \(\left( \alpha ,\eta \right)-\)complete\(,\) if every Cauchy sequence \(\lbrace x_{n}\rbrace \) in \(X\) with \(\alpha \left( x_{n},x_{n+1}\right) \geq 1 \) or \(\eta \left( x_{n},x_{n+1}\right) \leq 1\) for all \(n\in \mathbb{N},\) converges in \(X.\)

Definition 5. [16] Let \(\left( X,d\right) \) be a rectangular metric space and let \(\alpha ,\eta \) :\(X\times X\rightarrow \left[ 0,+\infty \right[ \) be two mappings. The space \( (X,d) \) is said to be

• (a) \(\left( X,d\right) \) is \(\alpha \) -regular, if \(x_{n}\rightarrow x\), where \(\alpha \left( x_{n},x_{n+1}\right) \geq 1\) for all \(n\in \mathbb{N},\) implies \(\alpha \left( x_{n},x\right) \geq 1\) for all \(n\in \mathbb{N}.\)
• (b) \(\left( X,d\right) \) is \(\eta -\)sub -regular, if \(x_{n}\rightarrow x\), where \(\eta \left( x_{n},x_{n+1}\right) \leq 1\) for all \(n\in \mathbb{N},\) implies \(\eta \left( x_{n},x\right) \leq 1\) for all \(n\in \mathbb{N}.\)
• (c) \(\left( X,d\right) \) is \(\left( \alpha ,\eta \right) \)-regular, if \(x_{n}\rightarrow x\), where \(\alpha \left( x_{n},x_{n+1}\right) \geq 1\) or \( \eta \left( x_{n},x_{n+1}\right) \leq 1\) for all \(n\in \mathbb{N},\) imply that \(\alpha \left( x_{n},x\right) \geq 1\) or \(\eta \left( x_{n},x\right) \leq 1\) for all \(n\in \mathbb{N}.\)

The following definition introduced by Wardowski [1]:

Definition 6. [1] Let \(\mathbb{F} \) be the family of all functions \(F\colon \) \(\mathbb{R}^{+}\rightarrow \mathbb{R}\) such that

• (i) \(F\) is strictly increasing;
• (ii) for each sequence \(\left( x_{n}\right) _{n\in \mathbb{N} }\) of positive numbers \( \lim_{n\rightarrow\infty}x_{n}=0,\,\,\,\text{ if and only if }\,\,\,\lim_{n\rightarrow \infty }F\left( x_{n}\right) =-\infty; \)
• (iii) there exists \(k\in \left] 0,1\right[ \) such that \(\lim_{x\rightarrow 0}x^{k}F\left( x\right) =0.\)

Recently, Piri and Kuman [4] extended the result of Wardowski [1] by changing the condition (iii) in the Definition 6 as follow:

Definition 7. [4] Let \(\Gamma \) be the family of all functions \( F\colon \) \( \mathbb{R}^{+}\rightarrow \mathbb{R}\) such that

• (i) \(F\) is strictly increasing;
• (ii) for each sequence \(\left( x_{n}\right) _{n\in \mathbb{N}}\) of positive numbers \( \lim_{n\rightarrow\infty}x_{n}=0,\,\,\,\text{ if and only if } \,\,\,\lim_{n\rightarrow \infty }F\left( x_{n}\right) =-\infty; \)
• (iii) \(F\) is continuous.

3. Fixed point theorem on rectangular metric spaces

We introduce a new notion of generalized \( F- \)expansive mapping in the context of rectangular metric spaces as follows:

Definition 8. Let \( (X,d) \) be a rectangular metric space and \( T:X\rightarrow X \) be a given mapping. \(T\) is said to be generalized \(F-\)expansive mapping if there exists \( F\in\mathbb{F}\) and \( \tau >0 \) such that

where \(M(x,y )=\min\left\lbrace d(x,y),d(x,Tx),d(y,Ty),d(x,Ty) \right\rbrace. \)

Theorem 1. Let \( (X,d )\) be a \( (\alpha-\eta) \)-complete generalized metric space, and \( T:X\rightarrow X \) be a bijective, generalized \(F\)-expansive mapping satisfying following conditions

• (i) \(T^{-1}\) is a triangular \(\left( \alpha ,\eta \right) -\)admissible mapping;
• (ii) there exists \(x_{0}\in X\) such that \(\alpha \left(x_{0},T^{-1}x_{0}\right) \geq 1\) or \(\eta \left( x_{0},T^{-1}x_{0}\right)\leq 1;\)
• (iii) \(T\) is a \(\left( \alpha ,\eta \right) -\)continuous.

Then \(T\) has a fixed point. Moreover, \(T\) has a unique fixed point when \(\alpha\left( z,u\right) \geq 1\) or \(\eta \left(z,u\right) \leq 1\) for all \(z,u\in Fix(T).\)

Proof. Let \(x_{0} \in X \) such that \(\alpha \left(x_{0},T^{-1}x_{0}\right) \geq 1\) or \(\eta \left( x_{0},T^{-1}x_{0}\right)\leq 1.\) We define the sequence \( \left\lbrace x_n \right\rbrace\) in \( X \) by \( x_{n}=Tx_{n+1} \), for all \( n\in \mathbb{N}. \)

Since \(T^{-1}\) is an triangular \(\left( \alpha ,\eta \right) -\)admissible mapping, then

\[\alpha \left( x_{0},x_{1}\right) =\alpha \left( x_{0},T^{-1}x_{0}\right) \geq 1\Rightarrow \alpha \left( T^{-1}x_{0},T^{-1}x_{1}\right) =\alpha \left( x_{1},x_{2}\right) \geq 1,\] or \[\eta \left( x_{0},x_{1}\right) =\eta \left( x_{0},T^{-1}x_{0}\right) \leq 1\Rightarrow \eta \left( T^{-1}x_{0},T^{-1}x_{1}\right) =\eta \left( x_{1},x_{2}\right)\leq 1 .\] Continuing this process we have \[\alpha \left( x_{n-1},x_{n}\right) \geq 1,\] or \[\eta \left( x_{n-1},x_{n}\right) \leq 1,\] for all \(n\in \mathbb{N}.\) By \(\left( T_{3}\right) \) and \(\left(T_{4}\right) ,\) one has. Suppose that there exists \(n_{0}\) \(\in \mathbb{N}\) such that \(x_{n_{0}}=Tx_{n_{0}}.\) Then \(x_{n_{0}}\) is a fixed point of \(T\) and the prove is finished. Hence, we assume that \(x_{n}\neq Tx_{n}\), i.e., \(d\left( x_{n-1},x_{n}\right) >0\) for all \(\ n\in \mathbb{N}.\)

Step \( 1 \): We shall prove

\[\lim\limits_{n \rightarrow +\infty}d( x_{n}, x_{n+1})=0. \] Applying inequality (1) with \( x=x_{n}\) and \(y= x_{n+1} \), we obtain where \begin{align*} M\left( x_{n}, x_{n+1}\right)&=\min\left\lbrace d( x_{n}, x_{n+1}),d( x_{n},T x_{n}),d( x_{n+1},T x_{n+1}),d(x_{n},T x_{n+1})\right\rbrace \\ &=\min\left\lbrace d( x_{n}, x_{n+1}),d( x_{n},x_{n-1}),d( x_{n+1},x_{n}),d(x_{n}, x_{n})\right\rbrace \\ &=\min\left\lbrace d( x_{n}, x_{n+1}),d( x_{n},x_{n-1})\right\rbrace. \end{align*} If for some \( n, \) \( M\left( x_{n}, x_{n+1}\right)=d( x_{n},x_{n-1})\), then the inequality (3), we get It is a contradiction. Hence \( M \left( x_{n}, x_{n+1}\right)=d( x_{n},x_{n+1}).\) Therefore Thus, Continuing this process, we get Now, by (6) and the condition \(F_3 \) of Definition 2, we deduce that Taking the limit as \(n\rightarrow\infty \) in (7) and using the condition \(F_2 \), we get

Step 2: Now, we shall prove

On the contrary, assume that \(x_{n}=x_{m}\) for some \(n=m+k>m\). Indeed, suppose that \(x_{n}=x_{m}\), so we have \[x_{n}=Tx_{n+1}=Tx_{m+1}=x_{m}.\] Denote \( d_{n}=d\left( x_{n},x_{n+1}\right).\) By the inequality (8), we have \begin{equation*} d_{n}< d_{n-1}. \end{equation*} Continuing this process, we get Which is a contradiction. Thus (10) hold.

Step 3: We prove shall

Applying inequality (1) with \( x=x_{n} \), \( y= x_{n+2} \), we obtain where \begin{align*} M \left( x_{n}, x_{n+2}\right)&=\min\left\lbrace d( x_{n}, x_{n+2}),d( x_{n},T x_{n}),d( x_{n+2},T x_{n+2}),d(x_{n},T x_{n+2})\right\rbrace \\ &=\min\left\lbrace d( x_{n}, x_{n+2}),d( x_{n},x_{n-1}),d( x_{n+2},x_{n+1}),d(x_{n}, x_{n+1})\right\rbrace \\ &=\min\left\lbrace d( x_{n}, x_{n+2}),d( x_{n+1},x_{n+2})\right\rbrace. \end{align*} Take \(a_{n}=d\left( x_{n},x_{n+2}\right) \) and \(b_{n}=d\left(x_{n+1},x_{n+2}\right).\) Thus, by (13), one can write \begin{align*} F\left(a_{n-1}\right)&=F\left( d\left( x_{n-1},x_{n+1}\right)\right)=F\left( d\left( Tx_{n},Tx_{n+2}\right)\right) >F\left( d\left( Tx_{n},Tx_{n+2}\right)\right)-\tau \geq F\left[ M\left( x_{n},x_{n+2}\right)\right] =F \left( \min\lbrace a_{n}, b_{n}\rbrace \right). \end{align*} Therefore, \begin{equation*} a_{n-1}\geq \min\left\lbrace a_{n}, b_{n}\right\rbrace. \end{equation*} Again, by (8) \begin{equation*} b_{n-1}\geq b_{n}\geq \min \left\{ a_{n},b_{n}\right\}. \end{equation*} Which implies that \begin{equation*} \min \left\{ a_{n},b_{n}\right\} \leq \min \left\{ a_{n-1},b_{n-1}\right\},\text{ }\forall n\in \mathbb{N}. \end{equation*} Then the sequence \( \min \left\{ a_{n},b_{n}\right\}_{n\in \mathbb{N}} \) is monotone non increasing. Thus, there exists \(\lambda \geq 0\) such that \begin{equation*} \lim_{n\rightarrow \infty }\min \left\{ a_{n},b_{n}\right\} =\lambda. \end{equation*} Assume that \(\lambda >0\). By (9), we have \begin{equation*} \lim_{n\rightarrow \infty }\sup a_{n}=\lim_{n\rightarrow \infty }\sup \min\left\{ a_{n},b_{n}\right\} =\lim_{n\rightarrow \infty }\min \left\{a_{n},b_{n}\right\}=\lambda. \end{equation*} Taking the \(\limsup_n\rightarrow \infty \) in (13), and using \(( F_3) \), we obtain \begin{equation*} F(\lambda)=F \left( \lim_{n\rightarrow \infty }\sup \lbrace a_{n-1}\rbrace \right)\geq F \left(\lim_{n\rightarrow \infty }\sup \left\{ a_{n}\right\} \right)> F \left(\lim_{n\rightarrow \infty }\sup \left\{ a_{n}\right\} \right)-\tau \geq F \left(\lim_{n\rightarrow \infty }\min \left\{ a_{n}, b_{n}\right\} \right), \end{equation*} which implies that \begin{equation*} F(\lambda) > F \left(\lambda \right)-\tau \geq F(\lambda). \end{equation*} Therefore, \begin{equation*} F \left( \lambda\right) < F \left(\lambda\right). \end{equation*} By \(\left( F _{1}\right) \), we get \begin{equation*} \lambda < \lambda. \end{equation*} It is a contradiction, then

Step 4: We shall prove that \( \left\lbrace x_n \right\rbrace \) is a Cauchy sequence in \( (X,d) \), that is

If otherwise there exists an \(\varepsilon \) \(>0\) for which we can find sequence of positive integers \(\left\lbrace x_{n_{\left( k\right) }}\right\rbrace _k \) and \(\left\lbrace x_{m_{\left( k\right) }}\right\rbrace _k \) of \(\left\lbrace x_{n}\right\rbrace \) such that, for all positive integers \(k\), \(n_{\left( k\right) }>m_{\left( k\right) }>k,\) Now, using (9), (14), (16) and the rectangular inequality, we find \begin{align*} \varepsilon & \leq d\left( x_{m_{\left( k\right) }},x_{n_{\left( k\right)}}\right)\\ & \leq d\left( x_{m_{\left( k\right) }},x_{m_{\left( k\right)+1}}\right) +d\left( x_{m_{\left( k\right) +1}},x_{m_{\left( k\right)-1}}\right) +d\left( x_{m_{\left( k\right) -1}},x_{n_{\left( k\right) }}\right)\\ & < d\left( x_{m_{\left( k\right) }},x_{m_{\left( k\right) +1}}\right)+d\left( x_{m_{\left( k\right) +1}},x_{m_{\left( k\right) }-1}\right)+\varepsilon. \end{align*} Then Now, by rectangular inequality, we have Letting \(k\rightarrow \infty \) in the above inequalities, using (9), (16) and (17), we obtain and On the other hand \begin{align*} M\left( x_{m_{\left( k\right) }},x_{n_{\left( k\right) }}\right)& =\min\left\{d\left( x_{m_{\left( k\right) }},x_{n_{\left( k\right)}}\right),d\left( x_{m_{\left( k\right) }},Tx_{m_{\left( k\right)}}\right) ,d\left( x_{n_{\left( k\right) }},Tx_{n_{\left(k\right)}}\right),d\left( x_{m_{\left( k\right) }},Tx_{n_{\left( k\right)}}\right)\right\}\\ &=\min\left\{d\left( x_{m_{\left( k\right) }},x_{n_{\left( k\right)}}\right),d\left( x_{m_{\left( k\right) }},x_{m_{\left( k\right)-1}}\right) ,d\left( x_{n_{\left( k\right) }},x_{n_{\left(k\right)-1}}\right),d\left( x_{m_{\left( k\right) }},x_{n_{\left( k\right)-1}}\right)\right\}. \end{align*} Letting \(k\rightarrow \infty \) in the above inequalities and using (9), (17) and (22), we get that By (21), let \( A=\frac{\varepsilon}{2}> 0 \), from the definition of the limit, there exists \(n_{0}\) \(\in \mathbb{N}\) such that \begin{equation*} \vert d\left( x_{m_{\left( k\right) +1}},x_{n_{\left( k\right) +1}}\right)-\varepsilon\vert \leq A \text{ }\forall n\geq n_{0}. \end{equation*} This implies that \begin{equation*} d\left( x_{m_{\left( k\right) +1}},x_{n_{\left( k\right) +1}}\right) \geq A > 0 \text{ }\forall n\geq n_{0}, \end{equation*} and by (23), let \( B=\frac{\varepsilon}{2}> 0 \), from the definition of the limit, there exists \(n_{1}\) \(\in \mathbb{N}\) such that \begin{equation*} M\left( x_{m_{\left( k\right)}},x_{n_{\left( k\right)}}\right) \geq B> 0 \text{ }\forall n\geq n_{1}. \end{equation*} Applying (1) with \( x=x_{m_{\left( k\right)}} \) and \( y= x_{n_{\left( k \right)}}\), we obtain \begin{equation*} F \left( d( x_{m_{\left( k\right)+1}}, x_{n_{\left( k\right)+1}})\right)> F \left( d( x_{m_{\left( k\right)+1}},x_{n_{\left( k\right)+1}})\right)-\tau \geq F\left( M (d( x_{m_{\left( k\right)}},x_{n_{\left( k \right)}}))\right) \end{equation*} Letting \( k\rightarrow\infty \) the above inequality and using \(\left( F _{3}\right)\), we obtain \begin{equation*} F \left(\lim_{k\rightarrow \infty }d\left( x_{m_{\left( k\right)+1}},x_{n_{\left( k\right) +1}}\right)\right)> F \left(\lim_{k\rightarrow \infty }d\left( x_{m_{\left( k\right)+1}},x_{n_{\left( k\right)+1 }}\right)\right) -\tau \geq F \left ( \lim_{k\rightarrow \infty }M\left( x_{m_{\left( k\right)}},x_{n_{\left( k\right) }}\right) \right), \end{equation*} Therefore, \[ F(\varepsilon)< F(\varepsilon). \] It is a contradiction. Then \begin{equation*} \lim_{n,m\rightarrow \infty }d\left( x_{m},x_{n}\right) =0. \end{equation*} It follows that \(\left\lbrace x_{n}\right\rbrace \) is a Cauchy sequence in X. Since \(\left( X,d\right)\) is \( (\alpha,\eta )\)-complete and \[\alpha(x_{n-1},x_{n})\geq 1 \ or \ \eta(x_{n-1},x_{n})\leq 1, \] for all \( n\in\mathbb{N} \), the there exists \(z\in X\) such that \begin{equation*} \lim_{n\rightarrow \infty }d\left( x_{n},z\right) =0. \end{equation*}

Step 5: We show that \(d\left( Tz,z\right) =0\) arguing by contradiction, we assume that

\begin{equation*} d\left( Tz,z\right)>0. \end{equation*} By rectangular inequality we get, By letting \(n\rightarrow \infty \) in inequality (24) and (25) we obtain \begin{equation*} d\left( z,Tz\right) \leq \lim_{n\rightarrow \infty }d\left(Tx_{n},Tz\right) \leq d\left( z,Tz\right). \end{equation*} Therefore, Since \( T \) is \( (\alpha,\eta) \)-continuous, then \( Tx_{n}\rightarrow Tz \) i.e \(\lim_{n\rightarrow \infty }d\left( Tx_{n},Tz\right)= 0 \). Hence \( d(Tz,z)=0 \), so \(Tz=z\).

Step 6: (Uniqueness) Now, suppose that \(z,u\in X\) are two fixed points of \(T\) such that \(u\neq z\) and \( \alpha(z,u)\geq 1 \) or \( \eta(z,u)\leq 1 \). Therefore, we have

\begin{equation*} d\left( Tz,Tu\right) =d\left( z,u\right) >0. \end{equation*} Applying (1) with \( x=z \) and \( y=u \), we have \begin{equation*} F \left( d\left( Tu,Tz\right)\right)-\tau \geq F \left( M\left( z,u\right)\right), \end{equation*} where \begin{equation*} M\left( z,u\right) =min \left\{d\left( z,u\right),d\left(z,Tz\right),d\left( u,Tu\right),d\left( z,Tu\right)\right\} =d\left(z,u\right). \end{equation*} Therefore, we have \begin{equation*} F \left( d\left( z,u\right)\right) >F\left( d\left( u,z\right)\right)-\tau \geq F \left( d\left( z,u\right)\right). \end{equation*} It is a contradiction. Therefore \(u=z\).

Theorem 2. Let \(\alpha ,\eta \) : \(X\times X\rightarrow \mathbb{R}^{+}\) be two function and let \(\left( X,d\right) \) be a \(\left( \alpha,\eta \right) -\)complete rectangular metric space. Let \(T:X\rightarrow X\) be a bijective mapping satisfying the following conditions:

• (i) \(T^{-1}\) is a triangular \(\left( \alpha ,\eta \right) \)-admissible mapping;
• (ii) \(T\) is a generalized \(\left( \alpha ,\eta \right) -F\)-expansive mapping ;
• (iii) \(\alpha \left( z,T^{-1}z\right) \geq 1\) or \(\eta \left(z,T^{-1}z\right) \leq 1,\) for all \(z\in \) Fix \(\left( T\right) .\)

Then \(\ T\) has a fixed point.

Proof. Let \(z\) \(\in \) Fix \(\left( T^{n}\right) \) for some fixed \(n>1\). As \(\alpha \left( z,T^{-1}z\right) \geq 1\) or \(\eta \left( z,T^{-1}z\right) \leq 1\) and \(T^{-1}\) is a triangular \(\left( \alpha ,\eta \right) \)-admissible mapping, then \begin{equation*} \alpha \left( T^{-1}z,T^{-2}z\right) \geq 1 \text{ or } \eta \left( T^{-2}z,T^{-1}z\right)\leq 1. \end{equation*} Continuing this process, we have \begin{equation*} \alpha \left( T^{-n}z,T^{-n-1}z\right) \geq 1\text{ or }\eta \left( T^{-n}z,T^{-n-1}z\right) \leq 1, \end{equation*} for all \(n\in \mathbb{N}\). By \(\left( T3\right)\) and \(\left( T4\right),\) we get \begin{equation*} \alpha \left( T^{-m}z,T^{-n}z\right) \geq 1\text{ or }\eta \left( T^{-m}z,T^{-n}z\right) \leq 1,\text{ }\forall \text{ }m,n\in \mathbb{N},\text{ }n\neq m. \end{equation*} Since \( T \) is a bijective mapping, then \( T^{-n}z=z=T^{n}z \) for all \(n\in\mathbb{N} \) and \( z\in \) Fix \(\left( T\right)\). Therefore, \begin{equation*} \alpha \left( T^{m}z,T^{n}z\right) \geq 1\text{ or }\eta \left( T^{m}z,T^{n}z\right) \leq 1,\text{ }\forall \text{ }m,n\in \mathbb{N},\text{ }n\neq m. \end{equation*} Assume that \(z\notin \) Fix \(\left( T\right) ,\) i.e. \(d\left( z,Tz\right) >0.\) Then, we have \begin{equation*} d\left( z,Tz\right) =d\left( T^{n}z,Tz\right) =d\left( TT^{n-1}z,Tz\right). \end{equation*} Applying (1) with \(x=z \) and \(y= T^{n-1}z \), we obtain \begin{equation*} F\left( d\left( z,Tz\right)\right)-\tau= F\left( d\left( TT^{n-1}z,Tz\right)\right)-\tau \geq F \left(M\left( T^{n-1}z,z\right)\right), \end{equation*} where

Letting \( n\rightarrow\infty \) in (27), we obtain \begin{equation*} \lim\limits_{n \rightarrow +\infty}M\left( T^{n-1}z,z\right) =d\left( z,Tz\right) \end{equation*} Now, using \(\left( F_{3}\right)\), we get \begin{equation*} F\left( d\left( z,Tz\right) \right)-\tau \geq F\left( d\left( z,Tz\right) \right). \end{equation*} It is a contradiction. Then \( z\in Fix(T)\).

Example 1. Let \( X=\left[1,+\infty \right[ \) and \( d:X\times X\rightarrow \left[0,+\infty \right[\) define by \begin{equation*} d\left( x,y\right) = \vert x-y\vert. \end{equation*} Then \( (X,d) \) is a metric space and rectangular metric space. Define mapping \(T:X\rightarrow X\) and \(\alpha ,\eta :X\times X\rightarrow \left[ 0,+\infty \right[ \) by \begin{equation*} T(x)=x^{2} \end{equation*} and \[ \alpha \left( x,y\right) =\frac{x+y}{\max \left\{ x,y\right\}+1 },\] \ \[% \eta \left( x,y\right) =\frac{\left\vert x-y\right\vert }{\max \left\{x,y\right\}+1 }.\] Then, \(T\) is an \(\left( \alpha ,\eta \right) -\)continuous triangular \(\left(\alpha ,\eta \right) -\)admissible mapping and \( T \) is a bijective mapping.

Let \(F \left( t\right) =ln(t),\) \(\tau =ln(2).\) Evidently, \(\left( \alpha \left( x,y\right) \geq 1\text{ or }\left( x,y\right) \leq 1\right) \) and \(\min\lbrace d\left( x,y\right),d\left( x,Tx\right),d\left( y,Ty\right),d\left( y,Tx\right)\rbrace >0\) are when \( x\neq y\neq 1 \).

Now, consider the following two cases:

Case 1: \(( x>y>1 )\)

As \[d(Tx,Ty)=x^{2}-y^{2},\ F(d(Tx,Ty))= ln(x^{2}-y^{2})= ln(x-y)+ln(x+y).\] Thus, \[ F( d(Tx,Ty))-\tau =ln(x^{2}-y^{2})-ln(2)= ln(x-y)+ln(x+y)-ln(2).\] We have \[ F(d(x,y)) =ln(x-y).\] On the other hand \begin{align*} F( d(x,y))- F( d(Tx,Ty))+\tau& =ln(x^{2}-y^{2})= ln(x-y)-ln(x-y)-ln(x+y)+ln(2)=ln(x-y)+ln(2). \end{align*} Since \( x,y\in \left] 1,+\infty \right[ \), then \begin{align*} -ln(x+y)+ln(2)\leq 0. \end{align*} Which implies that \begin{align*} F( d(Tx,Ty))-\tau &\geq F( d(x,y)) \geq F \left[ \min\lbrace d\left( x,y\right),d\left( x,Tx\right),d\left( y,Ty\right),d\left( y,Ty\right)\right] . \end{align*}

Case 2: \( (y>x>1) \)

As \[d(Ty,Tx)=y^{2}-x^{2},\ F(d(Ty,Tx))= ln(y^{2}-x^{2})= ln(y-x)+ln(y+x),\] Thus, \[ F( d(Ty,Tx))-\tau =ln(y^{2}-x^{2})-ln(2)= ln(y-x)+ln(y+x)-ln(2).\] We have \[ F(d(x,y)) =ln(y-x).\] On the other hand \begin{align*} F( d(y,x))- F( d(Ty,Tx))+\tau&= ln(y-x)-ln(y-x)-ln(y+x)+ln(2)=-ln(x-y)+ln(2). \end{align*} Since \( y,x\in \left] 1,+\infty \right[\), then \begin{align*} ln(y-x)+ln(2)\leq 0. \end{align*} Which implies that \begin{align*} F( d(Ty,Tx))-\tau &\geq F( d(y,x)) \geq F \left[ \min\lbrace d\left( y,x\right),d\left( y,Ty\right),d\left( x,Tx\right),d\left( x,Ty\right)\right] . \end{align*} Hence, the condition (1) is satisfied. Therefore, \( T \) has a unique fixed point \( z=1 \).

Theorem 3. Let \(\alpha,\eta :\) \(X\times X\rightarrow \mathbb{R} ^{+}\) be two functions and let \(d\left( X,d\right) \) be a \(\left( \alpha,\eta \right) -\)complete rectangular metric space. Let \(T:X\rightarrow X\) be a bijective mapping satisfying the following assertions:

• (i) \(T^{-1}\) is triangular \(\left( \alpha ,\eta \right) -\)admissible;
• (ii) \(T\) is a generalized \(\left( \alpha ,\eta \right) -F-\)expansive mapping;
• (iii) there exists \(x_{0}\in X\) such that \(\alpha \left(x_{0},T^{-1}x_{0}\right) \geq 1\) or \(\eta \left( x_{0},T^{-1}x_{0}\right) \leq 1;\)
• (iv) \(\left( X,d\right) \) is a \(\left( \alpha ,\eta \right) \)-regular rectangular metric space.

Then \(T\) has a fixed point. Moreover, \(T\) has a unique fixed point whenever \(\alpha \left( z,u\right) \geq 1\) or \(\eta \left( z,u\right) \leq 1\) for all \(z,u\in Fix\left( T\right).\)

Proof. Let x\(_{0}\in X\) such that \(\alpha \left(x_{0},T^{-1}x_{0}\right) \geq 1\) or \(\eta \left( x_{0},T^{-1}x_{0}\right) \leq 1\). Similar to the proof of Theorem 1, we can conclude that \begin{equation*} \left( \alpha \left( x_{n},x_{n+1}\right) \geq 1\text{ or }\eta \left(x_{n},x_{n+1}\right) \leq 1\right),\text{ and }x_{n}\rightarrow z\text{ as }n\rightarrow \infty, \end{equation*} and from inequality (26), we have \[\lim_{n\rightarrow \infty }d(Tx_{n},Tz)=d(z,Tz). \] From (iv) \( \alpha \left( x_{n},z\right) \geq 1\text{ or }\eta \left( x_{n},z\right)\leq 1 \), hold for \(n\in \mathbb{N}.\)

Suppose that \(Tz=x_{n_{0-1}}=Tx_{n_{0}}\) for some \(n_{0}\in \mathbb{N^{*}}.\) From Theorem 1 we know that the members of the sequence \(\lbrace x_{n}\rbrace \) are distinct. Hence, we have \(Tz\neq Tx_{n},\) i.e. \(d\left(Tz,Tx_{n}\right) >0\) for all \(n>n_{0}.\) Thus, we can apply (1) to \(x_{n}\) and \(z\) for all \(n>n_{0}\) to get

\begin{equation*}%\label{3.28} F \left( d\left( Tz,Tx_{n}\right)\right)-\tau \geq F\left( M\left( z,x_{n}\right) \right),\text{ }\forall n\geq n_0, \end{equation*} where \begin{align*} M\left( z,x_{n}\right) &=\min \left\{ d\left( z,x_{n}\right) ,d\left( z,Tz\right),d\left( x_{n},Tx_{n}\right),d\left( z,Tx_{n}\right)\right\}\\ &=\min \left\{ d\left( z,x_{n}\right) ,d\left( z,Tz\right),d\left( x_{n},x_{n-1}\right),d\left( z,x_{n-1}\right)\right\}. \end{align*} Therefore, By letting \(n\rightarrow \infty \) in inequality (28), we obtain \begin{equation*}%\laabel{3.30} \lim_{n\rightarrow \infty }F\left(d\left( Tz,Tx_{n}\right) \right)> \lim_{n\rightarrow \infty } F\left(d\left( Tz,Tx_{n}\right) \right) -\tau \geq \lim_{n\rightarrow \infty }F\left( \min \left\{ d\left( z,x_{n}\right) ,d\left( z,Tz\right),d\left( x_{n},x_{n-1}\right),d\left( z,x_{n-1}\right)\right\} \right). \end{equation*} Since \( F \) is continuous function and \(\lim\limits_{n \rightarrow +\infty} M(z,x_{n})=d(z,Tz) \), we conclude that \begin{equation*} F\left(d(z,Tz) \right) > F\left(d(z,Tz) \right), \end{equation*} which implies that \begin{equation*} d(z,Tz)< d(z,Tz). \end{equation*} It is a contradiction. Hence \(Tz=z\). The proof of the uniqueness is similarly to that of Theorem 1.

Corollary 1. Let \( \alpha,\eta :X\times X\rightarrow \left[0,+\infty \right[ \) be two functions, \( (X,d) \) be a \( ( \alpha,\eta) \)-complete rectangular metric space and \( T:X\rightarrow X \) be a bijective mapping. Suppose that for all \( x,y\in X \) with \( \alpha(x,y)\geq 1 \) or \(\eta(x,y)\leq 1 \) and \( M(x,y) > 0\) we have \begin{equation*} F \left( d\left( Tx,Ty\right)\right)-\tau \geq F \left( d\left( x,y\right)\right). \end{equation*} Then \( T \) has a fixed point, if

• (i) \(T^{-1}\) is a triangular \(\left( \alpha ,\eta \right) -\)admissible mapping;
• (ii) there exists \(x_{0}\in X\) such that \(\alpha \left(x_{0},T^{-1}x_{0}\right) \geq 1\) or \(\eta \left( x_{0},T^{-1}x_{0}\right)\leq 1;\)
• (iii) \(T\)is a \(\left( \alpha ,\eta \right) -\)continuous; or
• (iv) \(\left( X,d\right) \) is an \(\left( \alpha ,\eta \right) \)-regular rectangular metric space.

Moreover, \(T\) has a unique fixed point when \(\alpha\left( z,u\right) \geq 1\) or \(\eta \left(z,u\right) \leq 1\) for all \(z,u\in Fix(T).\)

4. Fixed point theorem on rectangular metric spaces endowed with a partial order

Definition 9. [16] Let \( (X,d,\preceq) \) be an ordered rectangular metric space and \( T: X\rightarrow X \) be a mapping. Then

• 1) \( (X,d) \) is said to be O-complete, if every Cauchy \( \left\lbrace n_n \right\rbrace \) in \( X \) with \( x_{n}\preceq x_{n+1} \) for all \( n\in \mathbb{N} \) or \(x_{n}\succeq x_{n+1} \) for all \( n\in \mathbb{N} \), converges in \( X \).
• 2) \( (X,d) \) is said to be O-regular, if for each sequence \( \left\lbrace n_n \right\rbrace \) in \( X \) \( \left\lbrace x_n \right\rbrace \rightarrow x \) and \( x_{n}\preceq x_{n+1} \) for all \( n\in \mathbb{N} \) or \(x_{n}\succeq x_{n+1} \) for all \( n\in \mathbb{N} \) imply that \( \left\lbrace n_n \right\rbrace\preceq x \) or \( \left\lbrace n_n \right\rbrace \succeq x \) respectively.
• 3) \(T \) is said to be O-continuous, if for given \( x\in X \) and sequence \( \left\lbrace n_n \right\rbrace \) with \( x_{n}\succeq x_{n+1} \) or \( x_{n}\preceq x_{n+1} \) for all \( n\in \mathbb{N} \), \( \left\lbrace n_n \right\rbrace \rightarrow x \Rightarrow Tx_n \rightarrow Tx \).

Definition 10. Let \( (X,d,\preceq) \) be an ordered rectangular metric spaces and \( T: X\rightarrow X \) be a mapping. We say that \(T\) be an ordered \( F\)- expansive mapping, if for all \( x,y \in X \) with \( x\preceq y \) or \( x\succeq y \) such that \[ M(x,y)> 0\Rightarrow F\left( d\left( Tx,Ty\right) \right)-\tau \geq F\left( M(x,y)\right),\] where \(M(x,y )=\min\left\lbrace d(x,y),d(x,Tx),d(y,Ty),d(x,Ty) \right\rbrace \).

Theorem 4. Let \( (X,d,\preceq) \) be an O-complete partially ordered rectangular metric space. Let \(T:X\rightarrow X\) be a bijective self mapping on \( X \) satisfying the following assertions:

• (i) \(T^{-1}\) is monotone ;
• (ii) \( T \) is an ordered \(F\)- expansive mapping;
• (iii) there exists \(x_{0}\in X\) such that \( x_{0}\preceq T^{-1}x_{0}\) or \( x_{0}\succeq T^{-1}x_{0}\)
• (iv) either \( T \) is O-continuous; or
• (v) \(( X,d) \) is O-regular.

Then \(T\) has a fixed point. Moreover, \(T\) has a unique fixed point whenever \( z\preceq u\) or \(z\succeq u\) for all \(z,u\in Fix\left( T\right).\)

Proof. Define the mapping \( \alpha :X\times X\rightarrow \left[ 0,+\infty\right[ \) by \[\alpha(x, y)=\left\{\begin{array}{ll} 1 & \text { if } x\preceq y \\ 0 & \text { otherwise } \end{array}\right. \] and the mapping \( \eta :X\times X\rightarrow \left[ 0,+\infty\right[ \) by \[\eta(x, y)=\left\{\begin{array}{ll} 1 & \text { if } x\succeq y \\ 0 & \text { otherwise } \end{array}\right. \] Using condition (iii) we have \[ x_{0}\preceq T^{-1}x_{0}\Rightarrow \alpha( x_{0},T^{-1}x_{0})\geq 1 ,\] or \[x_{0}\succeq T^{-1}x_{0}\Rightarrow \eta( x_{0},T^{-1}x_{0})\leq 1 .\] Owing to the monotonicity of \( T^{-1} \), we get \[\alpha(x,y)\geq 1 \Rightarrow x\preceq y \Rightarrow T^{-1}x \preceq T^{-1}y\Rightarrow \alpha(T^{-1}x , T^{-1}y)\geq 1,\] or \[\eta(x,y)\leq 1 \Rightarrow x\succeq y \Rightarrow T^{-1}x \succeq T^{-1}y \Rightarrow \eta(T^{-1}x , T^{-1}y)\leq 1.\] Therefore, \( (T_1) \) and \( (T_2) \) hold.

On the other hand, if

\[ \left\{\begin{array}{ll} & \alpha(x,y)\geq 1\\ & \alpha(x,y)\geq 1 \end{array}\right. \Rightarrow \left\{\begin{array}{ll} & x\preceq y\\ & y\preceq z \end{array}\right. \] or \[ \left\{\begin{array}{ll} & \eta(x,y)\leq 1\\ & \eta(x,y)\leq 1 \end{array}\right. \Rightarrow \left\{\begin{array}{ll} & x\succeq y\\ & y\succeq z \end{array}\right. \] Since \( (X,d) \) be an O-complete partially ordered rectangular metric space, we conclude that \[ x\preceq z \ or \ x\succeq z \Rightarrow \alpha(x,z)\geq 1 \ or \ \eta(x,z)\leq 1 .\] Thus, \(( T_3 )\) and \( (T_4) \) hold. This shows that \( T^{-1} \) is a triangular \(\left( \alpha ,\eta \right)-\)admissible mapping then \begin{equation*} \left( \alpha \left( x_{n},x_{n+1}\right) \geq 1\text{ or }\eta \left(x_{n},x_{n+1}\right) \leq 1\right). \end{equation*} Now, if \(T\) is O-continuous, then \( x_{n}\preceq x_{n+1} \) or \( x_{n}\succeq x_{n+1} \) \( \Rightarrow \) \(\alpha(x_{n},x_{n+1})\geq 1 \) or \(\eta(x_{n},x_{n+1})\leq 1\) and \( x_{n}\rightarrow z \) as \( n\rightarrow \infty \) with \( z\in X \Longrightarrow Tx_{n}\rightarrow Tx. \) The existence and uniqueness of a fixed point follows from Theorem 1.

Now, suppose that follow \( (X,d,\preceq) \) is O-regular. Let \( \left\lbrace x_n \right\rbrace \) be a sequence such that

\[ \left\lbrace n_n \right\rbrace\preceq x \ or \ \left\lbrace n_n \right\rbrace \succeq x ,\] which implies that \begin{equation*} \left( \alpha \left( x_{n},x\right) \geq 1\text{ or }\eta \left(x_{n},x\right) \leq 1\right), \end{equation*} for all \( n \) and \( x_n \) \( \rightarrow x\) as \( n\rightarrow\infty .\) This shows that \((X,d) \) is \(\left( \alpha ,\eta\right)-\) regular. Thus, the existence and uniqueness of fixed point from Theorem 3.

Corollary 2. Let \( (X,d,\preceq) \) be an O-complete partially ordered rectangular metric spaces. Further, let \(T:X\rightarrow X\) be a bijective self mapping on \( X \) be such that \( T^{-1} \) is a monotone mapping and there exist \( k \in \left]0,1 \right[ \) such that \( kd(Tx,Ty)\geq d(x,y) \), for all \( x,y \in X \) with \( x\preceq y \) or \( x\succeq y \). Also, suppose that the following conditions hold:

• (i) there exists \(x_{0}\in X\) such that \( x_{0}\preceq T^{-1}x_{0}\) or \( x_{0}\succeq T^{-1}x_{0}\);
• (ii) either \( T \) is O-continuous; or
• (iii) X is O-regular.

Then \(T\) has a fixed point. Moreover, \(T\) has a unique fixed point whenever \( z\preceq u\) or \(z\succeq u\) for all \(z,u\in Fix\left( T \right).\)

Proof. Let \(F(t)=ln(t)\) for all \(t \in \) \(\left] 0,+\infty \right[ \) and \( \tau =\frac{1}{ln(k)}\). Clearly \( F\in\mathbb{F}\) and \( \tau >0 \). We prove that T is a generalized \(F\)-expansive mapping. Indeed, Since \begin{align*} kd(Tx,Ty)&\geq d(x,y). \end{align*} We have \begin{align*} ln\left[ k .d(Tx,Ty)\right] &=ln\left[ d(Tx,Ty)\right] +ln(k)\\ &=ln\left( d(Tx,Ty)\right)-\frac{1}{ln(k)}\\ & \geq ln\left[ d(x,y)\right] \\ &\geq ln\left[\min\lbrace d(x,y),d(x,Tx),d(y,Ty),d(Tx,y)\rbrace \right]. \end{align*} As in the proof of Theorems 1 and 4, \(T\) has a unique fixed point \(x\in X\).

Corollary 3. Let \( (X,d,\preceq) \) be an O-complete partially ordered rectangular metric spaces. Further, let \(T:X\rightarrow X\) be a bijective self mapping on \( X \) such that \( T^{-1} \) is a monotone mapping and there exist \( \alpha \in \left]0,\frac{1}{2} \right[ \) such that \[ \alpha d(Tx,Ty)\geq \frac{ d(x,Tx)+d(y,Ty)}{2} \] for all \( x,y \in X \) with \( x\preceq y \) or \( x\succeq y \). Also suppose that the following conditions hold:

• (i) there exists \(x_{0}\in X\) such that \( x_{0}\preceq T^{-1}x_{0}\) or \( x_{0}\succeq T^{-1}x_{0}\);
• (ii) either \( T \) is O-continuous; or
• (iii) X is O-regular.

Then \(T\) has a fixed point. Moreover, \(T\) has a unique fixed point whenever \( z\preceq u\) or \(z\succeq u\) for all \(z,u\in Fix\left( T \right).\)

Proof. Let \(F(t)=ln(t)\) for all \(t \in \) \(\left] 0,+\infty \right[ \) and \( \tau =\frac{1}{ln(2\alpha)}\). Clearly \( F\in\mathbb{F}\) and \( \tau >0 \). We prove that \(T\) is a generalized \(F\)-expansive mapping. Indeed, since \begin{align*} \alpha d(Tx,Ty)&\geq \frac{d(x,Tx)+d(y,Ty)}{2}. \end{align*} We have \begin{align*} ln\left[ 2\alpha .d(Tx,Ty)\right] &=ln\left[ d(Tx,Ty)\right] +ln(2\alpha)\\ &=ln\left( d(Tx,Ty)\right)-\frac{1}{ln(2\alpha)}\\ & \geq ln\left[d(x,Tx)+d(y,Ty)\right] \\ &\geq ln\left[\min\lbrace d(x,y),d(x,Tx),d(y,Ty),d(Tx,y)\rbrace \right]. \end{align*} As in the proof of Theorems 1 and 4, \(T\) has a unique fixed point \(x\in X\).

Corollary 4. Let \( (X,d,\preceq) \) be an O-complete partially ordered rectangular metric spaces. Further, let \(T:X\rightarrow X\) be a bijective self mapping on \( X \), such that \( T^{-1} \) is a monotone mapping and there exist \( \lambda \in \left]0,\frac{1}{3} \right[ \) such that \[ \alpha d(Tx,Ty)\geq \frac{ d(x,y)+d(x,Tx)+d(y,Ty)}{3} \] for all \( x,y \in X \) with \( x\preceq y \) or \( x\succeq y \). Also suppose that the following conditions hold:

• (i) there exists \(x_{0}\in X\) such that \( x_{0}\preceq T^{-1}x_{0}\) or \( x_{0}\succeq T^{-1}x_{0}\);
• (ii) either \( T \) is O-continuous; or
• (iii) X is O-regular.

Then \(T\) has a fixed point. Moreover, \(T\) has a unique fixed point whenever \( z\preceq u\) or \(z\succeq u\) for all \(z,u\in Fix\left( T \right).\)

Proof. Let \(F(t)=ln(t)\) for all \(t \in \) \(\left] 0,+\infty \right[ \), and \( \tau =\frac{1}{ln(3\alpha)}\). Clearly \( F\in\mathbb{F}\) and \( \tau >0 \). We prove that \(T\) is a \(F\)-expansive mapping. Indeed, since \begin{align*} \lambda d(Tx,Ty)&\geq \frac{d(x,y)+d(x,Tx)+d(y,Ty)}{3} . \end{align*} We have \begin{align*} ln\left[ 3\lambda .d(Tx,Ty)\right] &=ln\left[ d(Tx,Ty)\right] +ln(3\lambda)\\ &=ln\left( d(Tx,Ty)\right)-\frac{1}{ln(3\lambda)}\\ & \geq ln\left[d(x,y)+d(x,Tx)+d(y,Ty)\right] \\ &\geq ln\left[\min\lbrace d(x,y),d(x,Tx),d(y,Ty),d(Tx,y)\rbrace \right]. \end{align*} As in the proof of Theorems 1 and 4, \(T\) has a unique fixed point \(x\in X\).

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflicts of Interest

The authors declare no conflict of interest.

1. Introduction

In this paper, we use standard notations from the value distribution theory of meromorphic functions (see [1,2,3]). We suppose that \(f\) is a meromorphic function in whole complex plane \(\mathbb{C}\). In addition, we denote the order of growth of \(f\) by \(\rho (f)\), and use the notation \(\rho _{2}(f)\) to denote the hyper-order of \(f\), defined by \begin{equation*} \rho _{2}(f)=\underset{r\rightarrow +\infty }{{\lim \sup }}\frac{\log \log T(r,f)}{\log r}, \end{equation*} where \(T(r,f)\) is the Nevanlinna characteristic function of \(f\).

To give the precise estimate of fixed points, we denote the exponent of convergence of fixed points by \(\tau (f)\), which is defined by

\begin{equation*} \tau (f)=\lambda(f-z)=\underset{r\rightarrow +\infty }{{\lim \sup }}\frac{\log N\left( r,\frac{1}{f-z}\right) }{\log r} \end{equation*} and the hyper-exponent of convergence of fixed points and distinct fixed points are denoted by \(\tau _{2}(f)\) and \(\overline{\tau }_{2}(f)\) and are defined by \begin{equation*} \tau _{2}(f)=\lambda_{2}(f-z)=\underset{r\rightarrow +\infty }{{\lim \sup }}\frac{\log \log N\left( r,\frac{1}{f-z}\right) }{\log r}, \end{equation*} and \begin{equation*} \overline{\tau }_{2}(f)=\overline{\lambda}_{2}(f-z)=\underset{r\rightarrow +\infty }{{\lim \sup }}\frac{\log \log \overline{N}\left( r,\frac{1}{f-z}\right) }{\log r}, \end{equation*} respectively, where \(N\left( r,\frac{1}{f-z}\right) \) and \(\overline{N}\left( r,\frac{1}{f-z}\right) \) are respectively the integrated counting function of fixed points and distinct fixed points of \(f\). We denote the exponent of convergence of zeros (distinct zeros) of \(f\) by \(\lambda (f)\) \((\overline{\lambda }(f))\) and the hyper-exponent of convergence of zeros (distinct zeros) of \(f\) by \(\lambda _{2}(f)\) \((\overline{\lambda }_{2}(f))\).

Consider the second-order homogeneous linear differential equation

where \(P(w)\) and \(Q(w)\) are not constants polynomials in \(w=e^{z}\) \((z\in\mathbb{C})\). It's well-known that every solution of Equation (1) is entire.

Suppose \(f\not\equiv 0\) is a solution of (1). If \(f\) satisfies the condition

\begin{equation*} \underset{r\rightarrow +\infty }{{\lim \sup }}\frac{\log T(r,f)}{r}=0, \end{equation*} then we say that \(f\) is a nontrivial subnormal solution of (1), and if \(f\) satisfies the condition [4], \begin{equation*} \underset{r\rightarrow +\infty }{{\lim \sup }}\frac{\log T(r,f)}{r^{n}}=0, \end{equation*} then we say that \(f\) is a nontrivial \(n\)-subnormal solution of (1). In [5], Wittich investigated the subnormal solution of (1), and obtained the form of all subnormal solutions in the following theorem:

Theorem 1. [5] If \(f\not\equiv 0\) is a subnormal solution of (1), then \(f\) must have the form \begin{equation*} f(z)=e^{cz}(a_{0}+a_{1}e^{z}+\cdots +a_{m}e^{mz}), \end{equation*} where \(m\geq 0\) is an integer and \(c\), \(a_{0}\), \(a_{1}\),...,\(a_{m}\) are constants with \(a_{0}a_{m}\neq 0\).

Gundersen and Steinbert [6] refined Theorem 1 and got the following theorem:

Theorem 2. [6] Under the assumption of Theorem 1, the following statements hold:

• (i) If \(\deg P>\deg Q\) and \(Q\not\equiv 0\), then any subnormal solution \(f\not\equiv 0\) of (7) must have the form \begin{equation*} f(z)=\sum_{k=0}^{m}h_{k}e^{-kz}, \end{equation*} where \(m\geq 1\) is an integer and \(h_{0}\), \(h_{1} \), ..., \(h_{m}\) are constants with \(h_{0}\neq 0\) and \(h_{m}\neq 0\).
• (ii) If \(\deg P\geq 1\) and \(Q\equiv 0\), then any subnormal solution of Equation (7) must be constant.
• (iii) If \(\deg P< \deg Q\), then the only subnormal solution of (7) is \(f\equiv 0.\)

Chen and Shon [7] investigated more general equation than (7), and got the following theorem: Set where \(d_{j}\geq 0\) \((j=0,\cdots ,n),\) \(m_{k}\geq 0\) \((k=0,\cdots ,s)\) are integers, \(a_{jd_{j}},...,a_{j0};\) \(b_{km_{k}},...,b_{k0}\) are complex constants such that \(a_{jd_{j}}\neq 0,\) \(b_{km_{k}}\neq 0\).

Theorem 3. [7] Let \(a_{n}(z),...,a_{1}(z),\) \(a_{0}\left( z\right) ,\) \(b_{s}(z),...,b_{1}(z),\) \(b_{0}\left( z\right) \) be polynomials and satisfy (2) and (3), and \(a_{n}(z)b_{s}(z)\neq 0\). Suppose that \(P^{\ast}(e^{z})=a_{n}(z)e^{nz}+\cdots +a_{1}(z)e^{z}+a_{0}(z)\), \(Q^{\ast}(e^{z})=b_{s}(z)e^{sz}+\cdots +b_{1}(z)e^{z}+b_{0}(z)\). If \(n< s\), then every solution \(f\) \((\not\equiv 0)\) of equation \begin{equation*} f^{\prime \prime }+P^{\ast }(e^{z})f^{\prime }+Q^{\ast }(e^{z})f=0 \end{equation*} satisfies \(\rho _{2}(f)=1\).

Many authors investigated the growth of solutions and the existence of subnormal solutions for some class of higher order linear differential equations (see [4,7,8,9,10,11,12,13]). For the higher-order linear homogeneous differential equation

where \(P_{j}(e^{z})\) \((j=0,\cdots ,k-1)\) are polynomials in \(z\), Yang and Li [11] generalized the result of Theorem 2 to the higher order and obtained the following results: Set where \(d_{jm_{j}}\geq 0\) \((j=0,\cdots,k-1)\) are integers, \(a_{jm_{j}d_{jmj}},...,a_{jm_{j}0}\) are complex constants, \(a_{jm_{j}d_{jmj}}\neq 0\).

Theorem 4. [11] Let \(a_{jm_{j}}\left(z\right) \) be polynomials and satisfy (5). Suppose that \begin{equation*} P_{j}(e^{z})=a_{jm_{j}}\left( z\right) e^{m_{j}z}+\cdots +a_{j1}\left(z\right) e^{z}, \end{equation*} where \(a_{jm_{j}}\left( z\right) \not\equiv 0\). If there exists an integer \(s\) \((s\in \{0,\cdots ,k-1\})\) satisfying \begin{equation*} m_{s}>\max \left\{ m_{j}:j=0,\cdots ,s-1,s+1,\cdots ,k-1\right\} =m, \end{equation*} then every solution \(f\not\equiv 0\) of Equation (4) satisfies \(\rho _{2}\left( f\right) =1\) if one of the following condition holds:

• (i) \(s=0\) or \(1\).
• (ii) \(s\geq 2\) and \(\deg a_{0j}\left( z\right) >\deg a_{ij}\left( z\right) \) \(\left( i\neq 0\right) \).

Theorem 5. [11] Under the assumption of Theorem 4, if \(zP_{0}(e^{z})+P_{1}(e^{z})\) \(\not\equiv0\), then we have every solution \(f\not\equiv 0\) of Equation (4) satisfies \begin{equation*} \tau _{2}(f)=\overline{\tau }_{2}(f)=\rho _{2}\left( f\right) =1. \end{equation*}

In particular, they also investigated the exponents of convergence of the fixed points of solutions and their first derivatives for a second order Equation (1) and obtained the following theorem:

Theorem 6. [11] Let \(a_{n}(z)\),..., \(a_{1}(z)\), \(b_{s}(z)\),..., \(b_{1}(z)\) be polynomials and satisfy (2) and (3), and \(a_{n}(z)b_{s}(z)\neq 0\). Suppose that \(P(e^{z})=a_{n}(z)e^{nz}+\cdots +a_{1}(z)e^{z}\), \(Q(e^{z})=b_{s}(z)e^{sz}+\cdots +b_{1}(z)e^{z}\). If \(n\neq s\), then every solution \(f\) \((\not\equiv 0)\) of Equation (1) satisfy \(\lambda (f-z)=\lambda (f^{\prime}-z)=\rho \left( f\right) =\infty \) and \(\lambda _{2}(f-z)=\lambda_{2}(f^{\prime }-z)=\rho _{2}\left( f\right) =1\).

Thus, it is natural to ask what will happen if we change \(\exp \{z\}\) in the coefficients of (4) into \(\exp \{A(z)\}\)? In this paper, we consider the above problem to Theorems 3, 4, 5 and 6, we obtain the following results: We set

\begin{equation*} A(z)=c_{n}z^{n}+c_{n-1}z^{n-1}+\cdots +c_{1}z+c_{0}, \end{equation*} where \(n\geq 1\) is an integer and \(c_{0},...,c_{n}\) are complex constants such that \(\mathit{Re}c_{n}>0\), throughout the rest of this paper.

Theorem 7. Let \(a_{jm_{j}}\left( z\right) \) be polynomials and satisfy (5). Suppose that

where \(a_{jm_{j}}\left( z\right) \not\equiv 0\). If there exists an integer \(s\) \((s\in \{0,\cdots ,k-1\})\) satisfying \begin{equation*} m_{s}>\max \left\{ m_{j}:j=0,\cdots ,s-1,s+1,\cdots ,k-1\right\} =m, \end{equation*} then every solution \(f\not\equiv 0\) of equation satisfies \(\rho \left( f\right) =\infty \) and \(\rho_{2}\left( f\right) =n\) if one of the following condition holds:

• (i) \(s=0\) or \(1\).
• (ii) \(s\geq 2\) and \(\deg a_{0j}\left( z\right) >\deg a_{ij}\left( z\right) \) \(\left( i\neq 0\right) \).

Example 1. Let \(f=e^{e^{z^2}}\) be a solution of the equation \begin{equation*} f^{(4)}-2ze^{z^2}f^{(3)}-12z^2e^{z^2}f^{\prime \prime}-24z^3e^{z^2}f^{\prime} -[24z^2e^{3z^2}+(96z^2+12)e^{2z^2}+(16z^{4}+48z^{2}+12)e^{z^2}]f=0. \end{equation*} Set \begin{align*}P_3(e^{A(z)})&=a_{3,1}(z)e^{A(z)}=-2ze^{z^2},\\ P_2(e^{A(z)})&=a_{2,1}(z)e^{A(z)}=-12z^2e^{z^2},\\ P_1(e^{A(z)})&=a_{1,1}(z)e^{A(z)}=-24z^3e^{z^2}, \\ P_0(e^{A(z)})&=a_{0,3}(z)e^{3A(z)}+a_{0,2}(z)e^{2A(z)}+a_{0,1}(z)e^{A(z)} =-24z^2e^{3z^2}-(96z^2+12)e^{2z^2}-(16z^{4}+48z^{2}+12)e^{z^2}. \end{align*} We remark that \(s=0\) and \(m_{0}=3>\max \left\{ m_{j}:j=1,2,3\right\} =m=1\). Obviously, the conditions of Theorem 7 are satisfied, we see that \(\rho \left( f\right) =\infty \) and \(\rho_{2}\left( f\right)=n =2.\)

Remark 1. Very recently, Li et al., [4] have investigated \(n\) subnormal solutions of the Equation (7) with \begin{equation*} P_{j}(e^{A(z)})=a_{jm_{j}}e^{m_{j}A(z)}+\cdots +a_{j1}e^{A(z)}\text{ }\left( j=0,...,k-1\right) , \end{equation*} where \(a_{jm_{j}},\cdots ,a_{j1}\) \(\left( j=0,...,k-1\right) \) are complex constants instead of polynomials and obtained some results concerning their growth.

Corollary 1. Under the assumption of Theorem 7, if \(zP_{0}(e^{A(z)})+P_{1}(e^{A(z)})\not\equiv 0\), then we have every solution \(f\not\equiv 0\) of Equation (4) satisfies \begin{equation*} \tau (f)=\overline{\tau }(f)=\rho \left( f\right) =\infty \text{ and }\tau _{2}(f)=\overline{\tau }_{2}(f)=\rho _{2}\left( f\right) =n. \end{equation*}

In particular, we also investigate the exponents of convergence of the fixed points of solutions and their first derivatives for a second order equation and we obtain the following theorems:

Theorem 8. Let \(a_{p}(z)\),..., \(a_{1}(z)\), \(b_{s}(z)\),..., \(b_{1}(z)\) be polynomials and satisfy (2) and (3), and \(a_{p}(z)b_{s}(z)\neq 0\). Suppose that \(P(e^{A(z)})=a_{p}(z)e^{pA(z)}+\cdots +a_{1}(z)e^{A(z)}\), \(Q(e^{A(z)})=b_{s}(z)e^{sA(z)}+\cdots +b_{1}(z)e^{A(z)}\). If \(p\neq s \), then every solution \(f\) \((\not\equiv 0)\) of Equation (8) satisfies \(\lambda (f-z)=\lambda (f^{\prime }-z)=\rho \left( f\right) =\infty \) and \(\lambda_{2}(f-z)=\lambda _{2}(f^{\prime }-z)=\rho _{2}\left( f\right) =n\).

Example 2. Let \(f=e^{e^{z^2}}\) be a solution of the equation \begin{equation*} f^{\prime \prime }-3ze^{z^2}f^{\prime }+[2z^2e^{2z^2}-(4{z^2}+2)e^{z^2}]f=0. \end{equation*} Set \begin{align*}P(e^{A(z)})&=a_{1}(z)e^{A(z)}=-3ze^{z^2},\\ Q(e^{A(z)})&=b_{2}(z)e^{2A(z)}+b_{1}(z)e^{A(z)}=2z^2e^{2z^2}-(4z^2+2)e^{z^2}.\end{align*} It is clear that the conditions of Theorem 8 are satisfied with \(p=1\neq s=2\), we see that \(\lambda (e^{e^{z^2}}-z)=\lambda (2ze^{z^2}e^{e^{z^2}}-z)=\rho \left( f\right) =\infty \) and \(\lambda _{2}(e^{e^{z^2}}-z)=\lambda _{2}(2ze^{z^2}e^{e^{z^2}}-z)=\rho _{2}\left( f\right)=n =2\).

Remark 2. If \(p=s,\) then the conclusions of Theorem 8 does not hold. For instance, consider the following equation

We can easily see that (9) has solution \(f\left( z\right) =z \) which satisfies \(\rho \left( f\right) =0\neq \infty \) and \(\rho _{2}\left( f\right) =0\neq n=3.\ \) In this example, we have \(p=s=2, \) \( A(z)=\left( 1+5i\right) z^{3}+z,\) \(a_{2}(z)=z^{4}+2iz,\) \(a_{1}(z)=-z^{2}+ \left( 2-i\right) z,\) \(b_{2}(z)=-\left( z^{3}+2i\right) \) and \(% b_{1}(z)=-\left( -z+2-i\right).\)

Theorem 9. Let \(a_{p}(z),...,a_{1}(z),a_{0}\left( z\right) ,b_{s}(z),...,b_{1}(z),b_{0}\left( z\right) \) be polynomials and satisfy (2) and (3), and \(a_{p}(z)b_{s}(z)\neq 0\). Suppose that \[P^{\ast }(e^{A(z)})=a_{p}(z)e^{pA(z)}+\cdots +a_{1}(z)e^{A(z)}+a_{0}(z),\] \[Q^{\ast}(e^{A(z)})=b_{s}(z)e^{sA(z)}+\cdots +b_{1}(z)e^{A(z)}+b_{0}(z).\] If \(p< s\), then every solution \(f\) \((\not\equiv 0)\) of equation

satisfies \(\rho \left( f\right) =\infty \) and \(\rho_{2}\left( f\right) =n\).

Example 3. Let \(f=e^{z} e^{e^z}\) be a solution of the equation \begin{equation*} f^{\prime \prime }+(e^{z+1}-3)f^{\prime }+[(-e^{-2}-e^{-1})e^{2(z+1)}-e^{z+1}+2]f=0. \end{equation*} Set \begin{align*} P^{\ast }(e^{A(z)})&=a_1(z)e^{A(z)}+a_0(z)=e^{z+1}-3,\\ Q^{\ast}(e^{A(z)})&=b_2(z)e^{2A(z)}+b_1(z)e^{A(z)}+b_0(z)=(-e^{-2}-e^{-1})e^{2(z+1)}-e^{z+1}+2.\end{align*} It is clear that the conditions of Theorem 9 are satisfied with \(p=1< s=2\), here we have \(\rho \left( f\right) =\infty \) and \(\rho _{2}\left( f\right) =n=1\).

Remark 3. If \(p\geq s,\) then the conclusions of Theorem 9 does not hold. For instance, consider the following equation

It is easy to see that (11) has solution \(f\left( z\right)=z\) which satisfies \(\rho \left( f\right) =0\neq \infty \) and \(\rho_{2}\left( f\right) =0\neq n=2. \) In this example, we have \(p=s=1, \) \(A(z)=\left( 1-i\right) z^{2}+2z+i,\) \(a_{1}(z)=-\left( 2z^{2}+3z\right) ,\) \(a_{0}(z)=-\left( iz^{3}-z^{2}+\left( 1+i\right) z\right) ,\) \(b_{1}(z)=2z+3\) and \(b_{0}(z)=iz^{2}-z+1+i.\)

Remark 4. Setting \(c_{n}=1\), \(c_{n-1}=\cdots =c_{0}=0\) and \(n=1,\) in Theorem 7, Corollary 1, Theorem 8 and Theorem 9, we obtain Theorem 4, Theorem 5, Theorem 6 and Theorem 3 respectively.

2. Auxiliary Lemmas

Recall that \begin{equation*} A(z)=c_{n}z^{n}+c_{n-1}z^{n-1}+\cdots +c_{0},,c_{l}=\alpha _{l}e^{i\theta _{l}},\text{ }z=re^{i\theta },\text{ }\mathit{Re}c_{n}>0, \end{equation*} we set \(\delta _{l}(A,\theta )=\mathit{Re}(c_{l}(e^{i\theta })^{l})=\alpha_{l}\cos (\theta _{l}+l\theta )\), and \(H_{l,0}=\{\theta \in \lbrack 0,2\pi):\delta _{l}(A,\theta )=0\}\), \(H_{l,+}=\{\theta \in \lbrack 0,2\pi ):\delta_{l}(A,\theta )>0\}\), \(H_{l,-}=\{\theta \in \lbrack 0,2\pi ):\delta_{l}(A,\theta )< 0\}\), for \(l=1,\cdots ,n\), throughout the rest of this paper. Obviously, if \(\delta _{n}(A,\theta )\neq 0\), as \(r\rightarrow+\infty \), we get

Lemma 1. [3] Let \(f_{j}(z)\) \((j=1,\cdots ,n)\) \((n\geq 2)\) be meromorphic functions, \(g_{j}(z)\) \((j=1,\cdots ,n)\) be entire functions, and satisfy

• (i) \( \sum_{j=1}^{n}e^{g_{j}(z)}\equiv 0;\)
• (ii) when \(1\leq j\leq k\leq n\), then \(g_{i}(z)-g_{k}(z)\) is not a constant;
• (iii) when \(1\leq j\leq n\), \(1\leq h\leq k\leq n\),

then \begin{equation*} T(r,f_{j})=o\{T(r,e^{g_{h}-g_{k}})\}\hspace{0.5cm}(r\rightarrow +\infty ,r\not\in E), \end{equation*} where \(E\subset (1,\infty )\) is of finite linear measure or logarithmic measure. Also, \(f_{j}(z)\equiv 0\) \((j=1,\cdots ,n)\).

Lemma 2. Let \(A(z)\), \(P_{j}(e^{A(z)}) \), \(m_{j}\), \(m_{s}\), \(m\) and \(a_{ij}(z)\) satisfy the hypotheses of Theorem 7, then Equation (7) has no constant polynomial solution.

Proof. Suppose that \(f_{0}\left( z\right) =b_{l}z^{l}+\cdots +b_{1}z+b_{0}\) \(\left(l\geq 1\right) \) is a nonconstant polynomial solution of (7), where \(b_{l}\neq 0,\cdots ,b_{0} \) are complex constants.

If \(l\geq s\), then \(f^{(s)}\not\equiv 0\). Taking \(z=r\), we have

Substituting \(f_{0}\) into (7) and using (13), we conclude that where \(d=\max \{d_{jm_{j}}:j=0,\cdots ,s-1,s+1,\cdots ,k-1\}\) and \(M_{0}>0\) is some constant. Since \(m_{s}>m\), we see that (3) is a contradiction. Obviously, when \(s=0\) or \(1\), we can get that the Equation (7) has nonconstant polynomial solution from the above process. If \(l< s\), then Set \(\max \{m_{j}:j=0,\cdots ,l\}=h\). If \(m_{j}< h\), then we can rewrite \begin{equation*} P_{j}(e^{A(z)})=a_{jh}\left( z\right) e^{hA(z)}+\cdots +a_{j(m_{j}+1)}\left( z\right) e^{(m_{j}+1)A(z)} +a_{jm_{j}}\left( z\right) e^{m_{j}A(z)}+\cdots +a_{j1}\left( z\right) e^{A(z)} \end{equation*} for \(j=0,\cdots ,l\), where \(a_{jh}\left( z\right) =\cdots=a_{j(m_{j}+1)}\left( z\right) =0\). Thus, we conclude by (15) that Set Since \(f_{0}\) and \(a_{ij}(z)\) are polynomials, we see that By Lemma 1 and (16)-(18), we conclude that Since \(\deg f_{0}>\deg f_{0}^{\prime }>\cdots >\deg f_{0}^{(l)}\) and \(\deg a_{0j}(z)>\deg a_{ij}(z)\) \((i\neq 0)\), so by (16) and (19), we get a contradiction.

Lemma 3. [14,15] Let \(f\left( z\right) \) be an entire function and suppose that \(|f^{(k)}(z)|\) is unbounded on some ray \(\arg z=\theta \). Then, there exists an infinite sequence of points \(z_{m}=r_{m}e^{i\theta }\) \(\left( m=1,2,\cdots \right) \), where \(r_{m}\rightarrow +\infty\) such that \(f^{(k)}(z_{m})\rightarrow \infty \) and \begin{equation*} \left\vert \frac{f^{\left( j\right) }\left( z_{m}\right) }{f^{\left( k\right) }\left( z_{m}\right) }\right\vert \leq \left\vert z_{m}\right\vert ^{k-j}(1+o(1))\hspace{0.5cm}\left( j=0,\cdots ,k-1\right) . \end{equation*}

Lemma 4. [16] Let \(f\left(z\right) \) be a transcendental meromorphic function of finite order \(\rho .\) Let \(\Gamma =\left\{ \left( k_{1},j_{1}\right) ,\left( k_{2},j_{2}\right) ,\cdots ,\left( k_{m},j_{m}\right) \right\}\) denote a set of distinct pairs of integers satisfying \(k_{i}>j_{i}\geq 0\) \(\left( i=1,2,\cdots ,m\right) \) and let \(\varepsilon >0\) be a given constant. Then, there exists a set \(E_{1}\subset \left[ 0,2\pi \right) \) that has linear measure zero such that if \(\theta \in \left[ 0,2\pi \right) \diagdown E_{1}\), then there is a constant \(R_{1}=R_{1}\left( \theta \right) >1\) such that for all \(z\) satisfying \(\arg z=\theta \) and \(\left\vert z\right\vert \geq R_{1}\) and for all \(\left( k,j\right) \in \Gamma \), we have \begin{equation*} \left\vert \frac{f^{\left( k\right) }\left( z\right) }{f^{\left( j\right)}\left( z\right) }\right\vert \leq \left\vert z\right\vert ^{\left(k-j\right) \left( \rho -1+\varepsilon \right) }. \end{equation*}

Lemma 5. [17] Let \(f(z)\) be an entire function with \(\rho \left( f\right) =\rho < \infty \). Suppose that there exists a set \(E_{2}\subset \lbrack 0,2\pi )\) that has linear measure zero, such that for any ray \(\arg z=\theta _{0}\in \lbrack 0,2\pi )\diagdown E_{2}\), \(|f(re^{i\theta_{0}})|\leq Mr^{k}\) \((M=M(\theta _{0})>0\) is a constant and \(k>0\) is a constant independent of \(\theta _{0}) \). Then \(f(z)\) is a polynomial with \(\deg f\leq k\).

Lemma 6. [16] Let \(f\) be a transcendental meromorphic function, and \(\alpha >1\) be a given constant. Then, there exists a set \(E_{3}\subset (1,\infty )\) with finite logarithmic measure and a constant \(C>0\) that depends only on \(\alpha \) and \(i\), \(j\) \((i,j\in \mathbb{N})\), such that for all \(z\) satisfying \(|z|=r\not\in E_{3}\cup \lbrack 0,1]\),

Remark 5. From the proof of Lemma 6 ([16, Theorem 3]), we can see that the exceptional set \(E_{4}\) equals \(\{|z|:z\in (\cup _{n=1}^{+\infty }O(a_{n}))\}\), where \(a_{n}(n=1,2,\cdots )\) denote all zeros and poles of \(f^{(i)}\), and \(O(a_{n})\) denote sufficiently small neighborhoods of \(a_{n}\). Hence, if \(f(z) \) is a transcendental entire function and \(z\) is a point that satisfies \(|f(z)|\) to be sufficiently large, then the point \(z\not\in E_{4}\) satisfies (20). For details, see , [9, Remark 2.10].

Lemma 7. [10,18] Let \(A_{0}\), \(\cdots \), \(A_{k-1}\) be entire functions of finite order. If \(f(z)\) is a solution of equation \begin{equation*} f^{\left( k\right) }+A_{k-1}f^{\left( k-1\right) }+\cdots +A_{0}f=0, \end{equation*} then \(\rho _{2}\left( f\right) \leq \max \{\rho (A_{j}):j=0,\cdots ,k-1\}\).

Lemma 8. [19] Let \(g(z)\) be an entire function of infinite order with the hyper-order \(\rho_{2}(g)=\rho \), and let \(\nu (r)\) be the central index of \(g\). Then, \begin{equation*} \underset{r\rightarrow +\infty }{{\lim \sup }}\frac{\log \log \nu (r)}{\log r}=\rho _{2}(g)=\rho . \end{equation*}

Lemma 9. [7] Let \(f(z)\) be an entire function that satisfies \(\rho \left( f\right) =\rho (n< \rho < \infty )\); or \(\rho \left( f\right) =\infty \) and \(\rho _{2}=0\); or \(\rho _{2}=\alpha (0< \alpha < \infty )\), and a set \(E_{5}\subset \lbrack 1,\infty )\) has a finite logarithmic measure. Then, there exists a sequence \(\{z_{k}=r_{k}e^{i\theta _{k}}\}\) such that \(|f(z_{k})|=M(r_{k},f)\), \(\theta _{k}\in \lbrack 0,2\pi )\), \(\lim_{k\rightarrow \infty }\theta _{k}=\theta _{0}\in \lbrack 0,2\pi )\), \(r_{k}\not\in E_{5}\), and \(% r_{k}\rightarrow \infty \), such that

• (i) if \(\rho \left( f\right) =\rho \) \((n< \rho < \infty )\), then for any given \(\varepsilon_{1}(0< \varepsilon _{1}< \frac{\rho -n}{2})\), \begin{equation*} {r_{k}}^{\rho -\varepsilon _{1}}< \nu (r_{k})< {r_{k}}^{\rho +\varepsilon_{1}}; \end{equation*}
• (ii) if \(\rho \left( f\right) =\infty \) and \(\rho _{2}\left( f\right) =0\), then for any given \(\varepsilon _{2}(0< \varepsilon _{2}< \frac{1}{2})\), and for any large \(M\) \((>0)\), we have, as \(r_{k}\) is sufficiently large, \begin{equation*} {r_{k}}^{M}< \nu (r_{k})< \exp \{{r_{k}}^{\varepsilon _{2}}\}; \end{equation*}
• (iii) if \(\rho _{2}\left( f\right) =\alpha (0< \alpha < \infty )\), then for any given \(\varepsilon _{3}(0< \varepsilon _{3}< \alpha )\), \begin{equation*} \exp \{{r_{k}}^{\alpha -\varepsilon _{3}}\}< \nu (r_{k})< \exp \{{r_{k}}^{\alpha +\varepsilon _{3}}\}. \end{equation*}

Lemma 10. [20] Let \(g\) be a non-constant entire function, and let \(0< \delta < 1\). There exists a set \(E_{6}\subset \lbrack 1,\infty )\) of finite logarithmic measure with the following property. For \(r\in \lbrack 1,\infty )\diagdown E_{6}\), the central index \(\nu (r)\) of \(g\) satisfies \begin{equation*} \nu (r)\leq (\log M(r,g))^{1+\delta }. \end{equation*}

Lemma 11. [21,22] Let \(A_{0}\), ..., \(A_{k-1}\), \(F\not\equiv 0\) be finite order meromorphic functions. If \(f\) is a meromorphic solution of the equation \begin{equation*} f^{(k)}+A_{k-1}f^{(k-1)}+\cdots +A_{0}f=F, \end{equation*} with \(\rho \left( f\right) =+\infty \) and \(\rho_{2}\left( f\right) =\rho \), then \(f\) satisfies \(\overline{\lambda }(f)=\lambda (f)=\rho \left( f\right) =\infty \) and \(\overline{\lambda }_{2}(f)=\lambda _{2}(f)=\rho _{2}\left( f\right) =\rho \).

Lemma 12. [14] Let \(\varphi :\left[ 0,+\infty \right) \rightarrow\mathbb{R}\) and \(\psi :\left[ 0,+\infty \right) \rightarrow\mathbb{R}\) be monotone non-decreasing functions such that \(\varphi \left(r\right) \leq \psi \left( r\right) \) for all \(r\notin E_{7}\cup \left[ 0,1\right] \), where \(E_{7}\subset \left( 1,+\infty \right) \) is a set of finite logarithmic measure. Let \(\gamma >1\) be a given constant. Then there exists a \(r_{1}=r_{1}\left( \gamma \right) >0\) such that \(\varphi \left( r\right) \leq \psi \left( \gamma r\right) \) for all \(r>r_{1}.\)

3. Proofs of the results

Proof of Theorem 7. Suppose that \(f\not\equiv 0\) is a solution of (7), then \(f\) is an entire function. By Lemma 2, we see that \(f\) is transcendental.

First step. We prove that \(\rho (f)=\infty \).

Suppose, to the contrary, that \(\rho (f)=\rho < \infty \). By Lemma 4, for any given \(\varepsilon >0\), there exists a set \(E_{1}\subset \lbrack 0,2\pi ) \) with linear measure zero, such that if \(\theta \in \lbrack 0,2\pi )\diagdown E_{1}\), then there exists a constant \(R_{1}=R_{1}(\theta )>1\), such that for all \(z\) satisfying \(\arg z=\theta \) and \(|z|=r>R_{1}\), we have

Case 1. Take a ray \(\arg z=\theta \in H_{n,+}\diagdown E_{1}\), then \(\delta _{n}(A,\theta )>0\). We assume that \(|f^{(s)}(re^{i\theta })|\) is bounded on the ray \(\arg z=\theta \). If \(|f^{(s)}(re^{i\theta })|\) is unbounded on the ray \(\arg z=\theta \), then by Lemma 3, there exists a sequence \(\{z_{t}=r_{t}e^{i\theta }\}\) such that as \(r_{t}\rightarrow +\infty \), \(f^{(s)}(z_{t})\rightarrow \infty \) and

By (7), we get For \(r_{t}\rightarrow +\infty \), we have and where \(d=\max \{d_{jm_{j}}:j=0,\cdots ,s-1,s+1,\cdots ,k-1\}\). Substituting (21), (22), (24), (25) into (23), we obtain that for sufficiently large \(r_{t}\) where \(C_{0}>0\) is a constant. From (26), we can get a contradiction by \(m_{s}>m\) and \(\delta _{n}(A,\theta )>0\), so on the ray \(\arg z=\theta \in H_{n,+}\diagdown E_{1}\).

Case 2. Now, we take a ray \(\arg z=\theta \in H_{n,-}\), then \(\delta _{n}(A,\theta )< 0\). If \(|f^{(k)}(re^{i\theta })|\) is unbounded on the ray \(\arg z=\theta \), then by Lemma 3, there exists a sequence \( \{z_{t}=r_{t}e^{i\theta }\}\) such that as \(r_{t}\rightarrow +\infty \), \( f^{(s)}(z_{t})\rightarrow \infty \) and

By (7), we get For \(r_{t}\rightarrow +\infty \), we have Substituting (28) and (30) into (29), we obtain that for sufficiently large \(r_{t}\) Since \(\delta _{n}(A,\theta )< 0\), when \(r_{t}\rightarrow +\infty \), by (31), we get \(1\leq 0\), this is a contradiction. Hence on the ray \(\arg z=\theta \in H_{n,-}\diagdown E_{1}\). From Lemma 5, (27) and (32), we know that \(f(z)\) is a polynomial, which contradicts the assertion that \(f(z)\) is transcendental. Therefore, \(\rho (f)=\infty \).

Step 2. We prove that \(\rho _{2}(f)=n\). By Lemma 7 and \(\rho (P_{j}(e^{A(z)}))=n\) \((j=0,\cdots ,k-1)\), we see that \(\rho _{2}(f)\leq \max \{\rho (P_{j}(e^{A(z)}))\}=n\).

Now, we suppose that there exists a solution \(f_{0}\) satisfies \(\rho _{2}(f_{0})=\alpha < n\). Then we have

By Lemma 6, we see that there exists a subset \(E_{3}\subset (1,\infty )\) having finite logarithmic measure such that for all \(z\) satisfying \(% |z|=r\not\in E_{3}\cup \lbrack 0,1]\), where \(C(>0)\) is some constant. From the Wiman-Valiron theory, there is a set \(E_{8}\subset (1,\infty )\) having finite logarithmic measure, such that we can choose a \(z\) satisfying \(|z|=r\not\in \lbrack 0,1]\cup E_{8}\) and \(% |f_{0}(z)|=M(r,f_{0})\), then we get where \(\nu (r)\) is the central index of \(f_{0}(z)\). By Lemma 9, we see that there exists a sequence \(\{z_{t}=r_{t}e^{i\theta _{t}}\}\) such that \(% |f_{0}(z_{t})|=M(r_{t},f_{0})\), \(\theta _{t}\in \lbrack 0,2\pi )\), with \(% r_{t}\not\in \lbrack 0,1]\cup E_{5}\cup E_{8}\), \(r_{t}\rightarrow +\infty \) and for any sufficiently large \(M_{3}(>2k+3)\)

Case 1. Suppose \(\theta _{0}\in H_{n,+}\). Since \(\delta _{n}(A,\theta )=\alpha _{n}\cos (\theta _{n}+n\theta )\) is a continuous function of \(\theta \), by \(\theta _{t}\rightarrow \theta _{0}\) we get \(% \lim_{t\rightarrow \infty }\delta _{n}(A,\theta _{t})=\delta _{n}(A,\theta _{0})>0\). Therefore, there exists a constant \(N(>0)\) such that as \(t>N\),

\begin{equation*} \delta _{n}(A,\theta _{t})\geq \frac{1}{2}\delta _{n}(A,\theta _{0})>0. \end{equation*} By (33), for any given \(\varepsilon _{1}(0< \varepsilon _{1}< % \frac{1}{2^{n+1}(k+1)}\delta _{n}(A,\theta _{0}))\), and \(t>N\), By (34), (35) and (37), we have By (7), we get Substituting (24), (25) and (35) into (39), we get for sufficiently large \(r_{t}\), By (36), (38) and (40), we get \begin{align} |a_{sm_{s}d_{sm_{s}}}|&{r_{t}}^{d_{sm_{s}}}e^{m_{s}\delta _{n}(A,\theta _{t}){% r_{t}}^{n}(1+o(1))}(1+o(1))\notag\\ &\leq 2\left( \frac{\nu (r_{t})}{r_{t}}\right) ^{k-s}(1+o(1)) +\sum_{j=0,j\neq s}^{k-1}4|a_{{jm_{j}}{d_{jm_{j}}}}|{r}_{t}^{d}e^{m\delta _{n}(A,\theta ){r}_{t}^{n}(1+o(1))}\left( \frac{\nu (r_{t})}{r_{t}}\right) ^{j-s}(1+o(1))\notag\\ &\leq 2\left( \frac{\nu (r_{t})}{r_{t}}\right) ^{k-s}(1+o(1)) +\sum_{j=0}^{s-1}4|a_{{jm_{j}}{d_{jm_{j}}}}|{r}_{t}^{d}e^{m\delta _{n}(A,\theta ){r}_{t}^{n}(1+o(1))}\left( \frac{\nu (r_{t})}{r_{t}}\right) ^{j-s}(1+o(1))\notag\\ &\;\;\;\;+\sum_{j=s+1}^{k-1}4|a_{{jm_{j}}{d_{jm_{j}}}}|{r}_{t}^{d}e^{m\delta _{n}(A,\theta ){r}_{t}^{n}(1+o(1))}\left( \frac{\nu (r_{t})}{r_{t}}\right) ^{j-s}(1+o(1))\notag\end{align} \begin{align} &\leq C_{2}{r_{t}}^{d}e^{m\delta _{n}(A,\theta ){r}_{t}^{n}(1+o(1))}\left( \frac{\nu (r_{t})}{r_{t}}\right) ^{k-s}(1+o(1)),\notag\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \end{align} where \(C_{2}>0\) is a constant. From this inequality and (38), it follows that Since \(m_{s}-m\geq 1>\frac{1}{2}\) and \(\delta (A,\theta _{t})\geq \frac{1}{2}% \delta _{n}(A,\theta _{0})>0\), we see that (41) is a contradiction.

Case 2. Suppose \(\theta _{0}\in H_{n,-}\). Since \(\delta _{n}(A,\theta )\) is a continuous function of \(\theta \), by \(\theta _{t}\rightarrow \theta _{0}\) we get \(\lim_{t\rightarrow \infty }\delta _{n}(A,\theta _{t})=\delta _{n}(A,\theta _{0})< 0\). Therefore, there exists a constant \(N(>0)\) such that as \(t>N\),

\begin{equation*} \delta _{n}(A,\theta _{t})\leq \frac{1}{2}\delta _{n}(A,\theta _{0})< 0. \end{equation*} By (7), we can write From (6) and \(\delta _{n}(A,\theta _{t})< 0\), we get where \(C_{3}>0\) is a constant. Substituting (35) and (43) into (42), we get where \(C_{4}>0\) is a constant. By substituting (36) into (44), we have Since \(\delta (A,\theta _{t})\) \(\leq \frac{1}{2}\delta _{n}(A,\theta _{0})< 0\), we see (45) is also a contradiction.

Case 3. Suppose \(\theta _{0}\in H_{n,0}\). Since \(\theta _{t}\rightarrow \theta _{0}\), for any given \(\varepsilon _{2}\) \(% (0< \varepsilon _{2}< \frac{1}{10n})\), there exists as integer \(N\) \((>0)\), such that as \(t>N\), \(\theta _{t}\in \lbrack \theta _{0}-\varepsilon _{2},\theta _{0}+\varepsilon _{2}]\), and

\begin{equation*} z_{t}=r_{t}e^{i\theta _{t}}\in \overline{\Omega }=\{z:\theta _{0}-\varepsilon _{2}\leq \arg z\leq \theta _{0}+\varepsilon _{2}\}. \end{equation*} By Lemma 6, we se that there exist a subset \(E_{3}\subset (1,\infty )\) having logarithmic measure \(lmE_{3}< \infty \), and a constant \(C>0\) such that for all \(z\) satisfying \(|z|=r\not\in E_{3}\cup \lbrack 0,1]\), Now, we consider the growth of \(f_{0}(re^{i\theta })\) on a ray \(\arg z=\theta \in \overline{\Omega }\diagdown \{\theta _{0}\}\). By the properties of cosine function, we suppose without loss of generality that \(\delta _{n}(A,\theta )>0\) for \(\theta \in \lbrack \theta _{0}-\varepsilon _{2},\theta _{0})\) and \(\delta _{n}(A,\theta )< 0\) for \(\theta \in (\theta _{0},\theta _{0}+\varepsilon _{2}]\).

Subcase 3.1 For a fixed \(\theta \in \lbrack \theta _{0}-\varepsilon _{2},\theta _{0})\), we have \(\delta _{n}(A,\theta )>0\). Since \(\rho _{2}(f_{0})< n\), we get that \(f_{0}\) satisfies (33). From \(T(r,{f_{0}}^{(s)})< (s+1)T(r,f_{0})+S(r,f_{0}),\) where \(S(r,f)=o(T(r,f))\), as \(r\rightarrow +\infty \) outside of a possible exceptional set of finite logarithmic measure, we get that \({f_{0}}^{(s)}\) also satisfies (33). So for any given \(\varepsilon _{2}\) satisfying \( 0< \varepsilon _{2}< \frac{1}{2^{n+1}(k-s+1)}\delta _{n}(A,\theta ) \), we have

We assert that \(|f_{0}^{(s)}(re^{i\theta })|\) is bounded on the ray \(\arg z=\theta \in \lbrack \theta _{0}-\varepsilon _{2},\theta _{0})\). If \(% |f^{(s)}(re^{i\theta })|\) is unbounded on the ray \(\arg z=\theta \), then, by Lemma 3, there exists a sequence \(\{y_{j}=R_{j}e^{i\theta }\}\) such that as \(R_{j}\rightarrow \infty \), \(f_{0}^{(s)}(y_{j})\rightarrow \infty \) and By Remark 5 and \(f_{0}^{(s)}(y_{j})\rightarrow \infty \), we know that \(% |y_{j}|=R_{j}\not\in E_{4}\). By (46) and (47), we have for sufficiently large \(j\), Substituting (24), (25), (48) and (49) into (23) where \(C_{5}>0\) is a constant, which yields a contradiction by \(m_{s}-m\geq 1>\frac{1}{2}\) and \(\delta _{n}(A,\theta )>0\). Hence \(|f_{0}^{(s)}(re^{i \theta })|\) is bounded on the ray \(\arg z=\theta \), so on the ray \(\arg z=\theta \in \lbrack \theta _{0}-\varepsilon _{4},\theta _{0})\).

Subcase 3.2 For a fixed \(\theta \in (\theta _{0},\theta _{0}+\varepsilon _{2}]\), we have \(\delta _{n}(A,\theta )< 0\). Using a reasoning similar to that in Subcase 3.1, we obtain

on the ray \(\arg z=\theta \in (\theta _{0},\theta _{0}+\varepsilon _{4}]\). By (51) and (52), we see that on the ray \(\arg z=\theta \in \overline{\Omega }\diagdown \{\theta _{0}\}\), But since \(\rho (f_{0}(re^{i\theta }))=\infty \) and \(\{z_{t}=r_{t}e^{i\theta _{t}}\}\) satisfies \(|f_{0}(z_{t})|=M(r_{t},f_{0})\), we see that, for any large \(M_{6}(>k)\), as \(t\) is sufficiently large, Since \(z_{t}\in \overline{\Omega }\), by (53) and (54), we see that \(\theta _{t}=\theta _{0}\) as \(t\rightarrow \infty \). Therefore, \(\delta _{n}(A,\theta _{t})=0\) as \(t\rightarrow \infty \). Thus, for sufficiently large \(t\), where \(j=0,\cdots ,k-1\) and \(C_{6}>0\) is a constant. By (7), (35) and (55), we get that \begin{equation*} |-(\frac{\nu (r_{t})}{z_{t}})^{k}(1+o(1))|=|-\frac{f_{0}^{(k)}(z_{t})}{% f_{0}(z_{t})}|\leq C_{7}r^{d}(\frac{\nu (r_{t})}{z_{t}})^{k-1}(1+o(1)), \end{equation*} i.e., where \(C_{7}>0\) is a constant. Substituting (36) into (56), we obtain also a contradiction. So we have \(\rho_{2}(f)=n\).

Proof of Corollary 1. From Theorem 7, we get \(\rho (f)=\infty \) and \(\rho _{2}(f)=n\). Let \(% g=f-z\), then \(f=g+z\). Substituting it into (7), we have \begin{equation*} g^{(k)}+P_{k-1}(e^{A(z)})g^{(k-1)}+\cdots +P_{0}(e^{A(z)})g=-zP_{0}(e^{A(z)})-P_{1}(e^{A(z)}). \end{equation*} Since \(-zP_{0}(e^{A(z)})-P_{1}(e^{A(z)})\not\equiv 0\), from Lemma 11, \( \rho (g)=\infty \) and \(\rho _{2}(g)=n\) we conclude \(\overline{\lambda }% (g)=\lambda (g)=\rho (g)=\infty \) and \(\overline{\lambda }_{2}(g)=\lambda _{2}(g)=\rho _{2}(g)=n\). So \(\overline{\tau }(f)=\tau (f)=\rho (f)=\infty \) and \(\overline{\tau }_{2}(f)=\tau _{2}(f)=\rho _{2}(f)=n\).

Proof of Theorem 8. From Theorem 7, we get \(\rho (f)=\infty \) and \(\rho _{2}(f)=n\).

• (i) Let \(g=f-z\), then \(f=g+z\). Substituting it into (8), we have \begin{equation*} g^{\prime \prime }+P(e^{A(z)})g^{\prime }+Q(e^{A(z)})g=-P(e^{A(z)})-zQ(e^{A(z)}). \end{equation*} Since \(p\neq s\), we get \(-P(e^{A(z)})-Q(e^{A(z)})z\not\equiv 0\). From Lemma 11, we obtain \({\lambda }(g)=\rho (g)=\rho (f)=\infty \) and \(\lambda _{2}(g)=\rho _{2}(g)=\rho _{2}(f)=n\). So \(\lambda (f-z)=\infty \) and \(\lambda _{2}(f-z)=n\).
• (ii) Differentiating both sides of (8), we get that
  \begin{equation} f^{\prime \prime \prime }+P(e^{A(z)})f^{\prime \prime }+[(P(e^{A(z)}))^{\prime }+Q(e^{A(z)})]f^{\prime }+(Q(e^{A(z)}))^{\prime }f=0. \label{3.36a} \end{equation}

  (57)
  By (8), we have
  \begin{equation} f=-\frac{f^{\prime \prime }+P(e^{A(z)})f^{\prime }}{Q(e^{A(z)})}. \label{3.37a} \end{equation}

  (58)
  Substituting (58) into (57), we get
  \begin{equation} f^{\prime \prime \prime }+[(P(e^{A(z)}))^{\prime }-\frac{(Q(e^{A(z)}))^{% \prime }}{Q(e^{A(z)})}]f^{\prime \prime } +[(P(e^{A(z)}))^{\prime }+Q(e^{A(z)})-\frac{(Q(e^{A(z)}))^{\prime }}{% Q(e^{A(z)})}P(e^{A(z)})]f^{\prime }=0. \label{3.38a} \end{equation}

  (59)
  Let \(g=f^{\prime }-z\), then \(f^{\prime }=g+z\), \(f^{\prime \prime }=g^{\prime }+1\), \(f^{\prime \prime \prime }=g^{\prime \prime }\). Substituting these into (59), we get that
  \begin{align} g^{\prime \prime }&+[(P(e^{A(z)}))^{\prime }-\frac{(Q(e^{A(z)}))^{\prime }}{% Q(e^{A(z)})}]g^{\prime }+[(P(e^{A(z)}))^{\prime }+Q(e^{A(z)})-\frac{% (Q(e^{A(z)}))^{\prime }}{Q(e^{A(z)})}P(e^{A(z)})]g\notag\\ &=-P(e^{A(z)})+\frac{(Q(e^{A(z)}))^{\prime }}{Q(e^{A(z)})} -[(P(e^{A(z)}))^{\prime }+Q(e^{A(z)})-\frac{(Q(e^{A(z)}))^{\prime }}{% Q(e^{A(z)})}P(e^{A(z)})]z\notag\\&=h(z). \label{3.39} \end{align}

  (60)
  Next, we prove that \(h(z)\not\equiv 0\). If \(h(z)\equiv 0\), then \begin{equation*} -P(e^{A(z)})+\frac{(Q(e^{A(z)}))^{\prime }}{Q(e^{A(z)})}\equiv \lbrack (P(e^{A(z)}))^{\prime }+Q(e^{A(z)})-\frac{(Q(e^{A(z)}))^{\prime }}{% Q(e^{A(z)})}P(e^{A(z)})]z. \end{equation*} Since \(Q(z)\not\equiv 0\), we have
  \begin{equation} (Q(e^{A(z)}))^{\prime }-(Q(e^{A(z)}))^{2}z\equiv P(e^{A(z)})Q(e^{A(z)}) +[(P(e^{A(z)}))^{\prime }Q(e^{A(z)})-(Q(e^{A(z)}))^{\prime }P(e^{A(z)})]z. \label{3.40a} \end{equation}

  (61)
  Suppose \(p>s.\) By taking \(z=r\), we have \begin{equation*} P(e^{A(r)})=a_{p}(r)e^{pA(r)}+\cdots +a_{1}(r)e^{A(r)},\;\;\;\;\;\text{and}\;\;\;\;\;\; Q(e^{A(r)})=b_{s}(r)e^{sA(r)}+\cdots +b_{1}(r)e^{A(r)}. \end{equation*} We get \begin{align*} (P(e^{A(r)}))^{\prime }&=\overset{p}{\underset{j=1}{\sum }}(a_{j}^{\prime }(r)+jA^{\prime }(r)a_{j}(r))e^{jA(r)}\\ &=(a_{p}^{\prime }(r)+pA^{\prime }(r)a_{p}(r))e^{pA(r)}+\cdots +(a_{1}^{\prime }(r)+A^{\prime }(r)a_{1}(r))e^{A(r)} \end{align*} and \begin{align*} (Q(e^{A(r)}))^{\prime }&=\overset{s}{\underset{j=1}{\sum }}(b_{j}^{\prime }(r)+jA^{\prime }(r)b_{j}(r))e^{jA(r)}\\ &=(b_{s}^{\prime }(r)+sA^{\prime }(r)b_{s}(r))e^{sA(r)}+\cdots +(b_{1}^{\prime }(r)+A^{\prime }(r)b_{1}(r))e^{A(r)}. \end{align*} So, we obtain \begin{align*} &|P(e^{A(r)})Q(e^{A(r)})+(P(e^{A(r)}))^{\prime }Q(e^{A(r)})r-(Q(e^{A(r)}))^{^{\prime }}P(e^{A(r)})r|\\ &=|a_{p}(r)b_{s}(r)+(p-s)rA^{\prime }(r)a_{p}(r)b_{s}(r) +(a_{p}^{\prime }(r)b_{s}(r)-a_{p}(r)b_{s}^{\prime }(r))r|e^{(p+s)\mathit{Re}% c_{n}r^{n}\left( 1+o\left( 1\right) \right) }(1+o(1)). \end{align*} Since \(a_{p}(r)\), \(b_{s}(r)\) and \(A(r)\) are polynomials and \(p>s\), we get \begin{equation*} \deg ((p-s)rA^{\prime }(r)a_{p}(r)b_{s}(r))>\deg [a_{p}(r)b_{s}(r)+(a_{p}^{\prime }(r)b_{s}(r)-a_{p}(r)b_{s}^{\prime }(r))r]. \end{equation*} So, we have \begin{equation*} |(p-s)rA^{\prime }(r)a_{p}(r)b_{s}(r)+a_{p}(r)b_{s}(r)+(a_{p}^{\prime }(r)b_{s}(r)-a_{p}(r)b_{s}^{\prime }(r))r| =Mr^{d_{1}}(1+o(1))\not\equiv 0, \end{equation*} where \(M>0\) and \(d_{1}>0\) are some constants. It follows that \begin{equation*} |P(e^{A(r)})Q(e^{A(r)})+(P(e^{A(r)}))^{\prime }Q(e^{A(r)})r-(Q(e^{A(r)}))^{^{\prime }}P(e^{A(r)})r| =Mr^{d_{1}}e^{(p+s)\mathit{Re}c_{n}r^{n}\left( 1+o\left( 1\right) \right) }(1+o(1)). \end{equation*} From (61), we have \begin{align*} Mr^{d_{1}}e^{(p+s)\mathit{Re}c_{n}r^{n}\left( 1+o\left( 1\right) \right) }(1+o(1)) &=|P(e^{A(r)})Q(e^{A(r)})+(P(e^{A(r)}))^{^{\prime }}Q(e^{A(r)})r-(Q(e^{A(r)}))^{^{\prime }}P(e^{A(r)})r|\\ &=|(Q(e^{A(r)}))^{^{\prime }}-(Q(e^{A(r)}))^{2}r|\leq M_{1}r^{d_{2}}e^{2s% \mathit{Re}c_{n}r^{n}\left( 1+o\left( 1\right) \right) }(1+o(1)), \end{align*} where \(M_{1}>0\) and \(d_{2}>0\) are some constants, which is a contradiction. So we have \(h(z)\not\equiv 0.\) If \(p< s,\) by (61) for \(z=r\) we have \begin{align*} M_{2}r^{d_{3}}e^{2s\mathit{Re}c_{n}r^{n}\left( 1+o\left( 1\right) \right) }(1+o(1)) &=[(Q(e^{A(r)}))^{2}+(P(e^{A(r)}))^{\prime }Q(e^{A(r)})-(Q(e^{A(r)}))^{\prime }P(e^{A(r)})]r\\ &=\left\vert (Q(e^{A(r)}))^{\prime }-P(e^{A(r)})Q(e^{A(r)})\right\vert\\& \leq M_{3}r^{d_{4}}e^{\left( p+s\right) \mathit{Re}c_{n}r^{n}\left( 1+o\left( 1\right) \right) }(1+o(1)), \end{align*} where \(M_{2}>0,\) {\(d_{3}>0,\) }\(M_{3}>0\) and \(d_{4}>0\) are some constants. This is a contradiction. So, we obtain \(h(z)\not\equiv 0.\) Hence, if \(% p\neq s\) we have \(h(z)\not\equiv 0.\) From Lemma 11, we get \(\lambda (g)=\rho (g)=\rho (f^{\prime }-z)=\rho (f)=\infty \) and \(\lambda _{2}(g)=\rho _{2}(g)=\rho _{2}(f^{\prime }-z)=\rho _{2}(f)=n\).

Proof of Theorem 9. Suppose that \(f\not\equiv 0\) is a solution of (10). Since \(\rho (P^{\ast })=\rho (Q^{\ast })=n,\) then by Lemma 7, we see that

By Lemma 6, we se that there exist a subset \(E_{3}\subset (1,\infty )\) having logarithmic measure \(lmE_{3}< \infty \), and a constant \(C>0\) such that for all \(z\) satisfying \(|z|=r\not\in E_{3}\cup \lbrack 0,1]\), Taking \(z=r\), in (2) and (3), we obtain that for sufficiently large \(r\) and Substituting (63)-(65) into (10), we deduce that for all \(z\) satisfying \(|z|=r\not\in E_{3}\cup \lbrack 0,1]\) By (66), we deduce that for all \(z\) satisfying \(|z|=r\not\in E_{3}\cup \lbrack 0,1]\) Since \(s-p>0\), by (67) and Lemma 12, we get From (62) and (68) we obtain \(\rho (f)=+\infty \) and \(\rho _{2}(f)=n.\)

Acknowledgments

This paper was supported by Directorate-General for Scientific Research and Technological Development(DGRSDT).

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflicts of Interest

The authors declare no conflict of interest.