1. Introduction

The object of this work is to study the global solution to the following boundary value problem for the Moore-Gibson-Thompson equation

where \(\Omega \) is a bounded domain in \(R^n(n\ge 1)\) with sufficiently smooth boundary \(\partial \Omega\), \(u_0(x),u_1(x)\) and \(u_2(x)\) are given functions and \(f\) is a given nonlinear function. All the parameters \(\alpha,\beta,c^2,r\) are assumed to be positive constants.

In recent years, increasing attention has been paid to the well-posedness and asymptotic behavior of the Moore-Gibson-Thompson (MGT) equation, see [1,2,3,4,5,6,7]. The MGT model is considered through third-order (in time), strictly hyperbolic partial differential equation as follows

it is one of the nonlinear acoustic models describing the propagation of acoustics wave in gases and liquid, it has a wide range of applications in medical and industry. In the physical context of the acoustic waves, \(u \) is the velocity potential of the acoustic phenomena, \(\alpha\) denotes the thermal relaxation time, \(c\) denotes the speed of sound, \(\beta\) denotes friction, and \(b\) denotes a parameter of diffusivity.

It is often convenient to write MGT equation as an abstract form

and it has been shown [8,9] that the linear part of Eq. (5) generates a strongly continuous semigroup as long as \(r>0\). In [10], the authors provided a brief overview of well-posedness results, both local and global, pertinent to various configurations of MGT equations. Especially, the authors in [11] considered the following model with nonlinear control feedback where the parameter \(\beta >0\), \(p(u)\) denotes an active force and the operator \(A\) is strictly positive. By semigroup method, it was proved in [11] we that (6) with initial data of arbitrary size in \(H\) is locally and globally well-posed under the following assumption: \(p\in C^1(R)\) and its derivative satisfies \(-\delta\le p'(s)\le m \) for some positive constants \(\delta \) and \(m\). Kaltenbacher et al., [12] established the well-posedness by Galerkin approximations and then employ fixed-point arguments for well-posedness of the Jordan-Moore-Gibson-Thompson (JMGT) equation More recently, Boulaaras et al., [13] proved the existence and uniqueness of the weak solution of the Moore-Gibson-Thompson equation with the integral condition by applying the Galerkin method.

In this paper, we extend the results in [11] to Problem (1)-(3) by applying the Galerkin method and compact method. The contents of this paper are organized as follows; In §2, we prepare some materials needed for our proof. Finally, in §3, we give the main result and the proof.

2. Preliminaries

Throughout this paper, the domain \(\Omega\) is assumed to be sufficiently smooth to admit integration by parts and second-order elliptic regularity. We use \(C\) to denote a universal positive constant that may have different values in different places. \(W^{m,2}(\Omega)=H^m(\Omega)\) and \(W^{m,2}_0(\Omega)=H^m_0(\Omega)\) denote the well-known Soblev space. We denote by \(||.||_p\) the \(L^p(\Omega)\) norm and by \(||\nabla .||\) the norm in \(H^1_0(\Omega)\). In particular, we denote \(||.||=||.||_2\)

By a weak solution \(u(x,t)\) of Problem (1)-(3) on \(\Omega\times [0,T]\) for any \(T>0\), we mean \(u\in L^{\infty}((0,T);H^2(\Omega)\cap H^1_0(\Omega))\cap W^{1,\infty}((0,T);H^1_0(\Omega))\cap W^{2, \infty}((0,T);L^2(\Omega))\), \(\Delta u_t, u_{ttt}\in L^{\infty}((0,T); H^{-1}(\Omega))\) such that \(u(x,0)=u_0(x)\) a.e. in \(\Omega\), \(u_t(x,0)=u_1(x)\) a.e. in \(\Omega\), \(u_{tt}(x,0)=u_2(x)\) a.e. in \(\Omega\), and

\begin{align} & \alpha (u_{ttt}+\beta u_{tt}- c^2\Delta u-r \Delta u_t+f(u),v)=0\nonumber \end{align} for any \(v\in H_0^1(\Omega)\), a.e. \(t\in[0,T]\).

In this paper, we assume \(\alpha,\beta,c^2,r>0\) and

Lemma 1.[14] Let \(\Omega \in R^n\) be a bounded domain and \(w_j\) be a base of \(L^2(\Omega)\). Then for any \(\epsilon>0\) there exist a positive constant \(N_{\epsilon}\), such that \[||u||\le (\sum_{j=1}^{N_{\epsilon}}(u,w_j))^{\frac{1}{2}}+\epsilon ||u||_{1,p}\] for any \(u\in W_0^{1,p}(\Omega)(2\le p< \infty)\), where \(N_{\epsilon}\) is independent on \(u\).

Lemma 2.[15] Let \(G(z_1,z_2,...z_h)\) be the function of the variables \(z_1,z_2,...z_h\) and suppose that \(G\) is continuous differentiable for k-times \((k\ge 1)\) with respect to every variable. Let \(z_i(x,t)\in L^\infty([0,T];H^k(\Omega))(i=1,2,...h)\), then the estimation \[\int_\Omega|D_x^kG(z_1(x,t),z_2(x,t),...,z_h(x,t))|^2dx< C(M,k,h)\sum_{i=1}^h||z_i||_{H^k(\Omega)}\] holds, where \(D_x=\frac{\partial}{\partial x}, M=\max_{i=1,2,...,h}\max_{0\le t\le T,x\in \Omega}|z_i(x,t)|\).

3. Solvability of the problem

In this section, by using Galerkin's method and compactness method, we shall prove the existence of global solutions of Problem (1)-(3).

Let \(\{w_j(x)\}_{j\in N}\) be the eigenfunctions of the following boundary problem

corresponding to the eigenvalue \(\lambda_j(j=1,2,3,...)\). Then \(\{w_j(x)\}_{j\in N}\) can be normalized to from an orthogonal basis of \(H^2(\Omega)\cap H_0^1(\Omega)\) and to be orthnormal with respect to the \(L^2(\Omega)\) scalar product.

Now, we seek an approximate solution of Problem (1)-(3) in the form of

where the constants \(T_{jN}\) are defined by the conditions \(T_{jN}(t)=(u^N(x, t),w_j (x))\) and can be determined from the relation

Lemma 3. Assume (8) holds, \(u_0 \in H^2(\Omega)\cap H_0^1(\Omega)\), \(u_1 \in H_0^1(\Omega)\), and \(u_2 \in L^2(\Omega)\), then for any \(T>0\), Problem (11)-(12) possesses a solution \(u^N\) on \([0,T]\), and the following estimate holds in the class

Proof. Problem (11)-(12) leads to a system of ODEs for unknown functions \(T_{jN} (t)\). Based on standard existence theory for ODE, one can obtain functions \( T_{jN} (t):[0, t_k) \rightarrow R , j = 1, 2, ..., k,\) which satisfy approximate Problem (11)-(12) in a maximal interval \([0, t_k), t_k\in(0, T]\). This solution is then extended to the closed interval \([0, T]\) by using the estimate below.

Multiplying (11) by \(T_{jNtt}(t)\), summing up the products for \(j=1,2,...,N\) and integrating by parts, we get

Integrating (14) with respect to \(t\) from 0 to \(t\), we obtain We observe that and Adding \(2[(u^N,u^N_t)+(u^N_{t},u^N_{tt})+( \nabla u^N, \nabla u^N_{t})]\) to both sides of (15) and a substitution of the equalities (16) and (17) in (15) gives Then, by Hölder inequality and the fact \(|f(s)|=|\int_0^tf'(s)ds|\le C_1|s|\) by (A1), we arrive at Taking into account that \[||u^N_{tt}(0)||^2+||\nabla u^N_t(0)||^2+||\nabla u^N(0)||^2\rightarrow ||u_2||^2 +||\nabla u_0||^2+||\nabla u_1||^2\] and \[||u^N(0)||^2+ ||u^N_t(0)||^2\rightarrow ||u_0||^2+||u_1||^2\] as \(N \rightarrow \infty\), then applying the Gronwall inequality to (19) and then integrating from 0 to \(t\) appears that Multiplying (11) by \(\lambda_j T_{jN}(t)\) summing up the products for \(j=1,2,...N\), integrating by parts and integrating with respect to \(t\), we get Combining Cauchy inequality, the fact \(||\Delta u^N(0)||^2\rightarrow ||\Delta u_0||^2\), and \(|f(s)|\le C_1|s|\), and making use of the following inequality \begin{align} \int_0^t(u^N_{ttt},\Delta u^N)d\tau &=(u^N_{tt},\Delta u^N)|_0^t -\int_0^t(u^N_{tt},\Delta u^N_t)d\tau\nonumber\\ &=(u^N_{tt},\Delta u^N)-(u^N_{tt}(0),\Delta u^N(0)) +\frac{1}{2}||\nabla u^N_t||^2-\frac{1}{2}||\nabla u^N_t(0)||^2, \nonumber \end{align} we have Choosing \(\epsilon_1\) sufficiently small and \(\epsilon_2\) sufficiently large such that \(\epsilon_2>2c^2\), then it follows from (22) and (20) that Thus, applying Gronwall's inequality to (23), we deduce Combining (20) and (24), we get Furthermore, by (25), we have that (11)-(12) possesses a global solution.

Theorem 1. Assume (8) holds, \(u_0 \in H^2(\Omega)\cap H_0^1(\Omega)\), \(u_1 \in H_0^1(\Omega)\), and \(u_2 \in L^2(\Omega)\), then for any \(T>0\), Problem (1)-(3) possesses a unique global solution.

Proof. For any \(v\in H_0^1(\Omega)\), it follows that

Thus, using Lemma 3, it follows that Similarly, we have From Lemma 3, (27) and (28), there exist a subsequence of \(\{u^N\}\), still denoted by \(\{u^N\}\), and a function \(u,\xi,\eta\), such that and for any \(t\in [0,T]\) Since \(f\in C^1\) and \(||f(u^N)||\le C||u^N||\le C\), for any \(v\in H_0^1(\Omega)\) and any \(t\in [0,T]\), we have as \(N\rightarrow \infty\). Then we get \(\xi=f(u), \eta=\nabla u_t\), combining this with (35)-(40), we have \[u\in L^{\infty}((0,T);H^2(\Omega)\cap H^1_0(\Omega))\cap W^{1,\infty}((0,T);H^1_0(\Omega))\cap W^{2, \infty}((0,T);L^2(\Omega)),\] \[\Delta u_t, u_{ttt}\in L^{\infty}((0,T); H^{-1}(\Omega)).\] By using Lemma 3 and (27), we observe that where \(u^N_{t^k}=\frac{\partial^k u^N}{\partial t^k}\). Then, by Ascoli-Arcela theorem, we can select from \(\{u^N\}\) a subsequence, still denoted by \(\{u^N\}\), such that as \(N\rightarrow \infty\), the subsequence In particular, we take \(t=0\) and we note that \(\{w_j(x)\}_{j\in N}\) are an orthogonal basis of \(L^2(\Omega)\), we know that By (29)-(34),(44) and Lemma 2.1, we have Thanks to (29)-(42), letting \(N\rightarrow \infty\) in (11), leads to for any \(v\in H_0^1(\Omega)\). Altogether, we conclude that \(u\) is a solution of the initial boundary Problem (1)-(3).

Now, suppose that there exist two different solutions \(u_1,u_2\) for Problem (1)-(3), then the difference \(w=u_1-u_2\) satisfies

Integrating (48) for \(t\) from 0 to \(t\), we have Multiplying the Eq. (51) by \(w_t\), integrating over \(\Omega\), adding up \((w,w_t)\), we obtain where we have used mean value theorem and \(|\theta|\le 1\). By applying Gronwall inequality, we deduce that This implies that \(w=0\) for all \(t\in[0,T]\). Thus the uniqueness is proved.

Acknowledgments :

The authors would like to thank the referee for his/her valuable comments that resulted in the present improved version of the article.

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 on norm-attainable operators have been considered by several mathematicians (see [1,2,3,4,5]). Of interest has been the norm-attainability conditions for these operators in different algebras. For instance, the author in [4] considered these conditions in Banach spaces while others worked in Hilbert spaces particularly. However, norm and structural characterizations of these operators have not been done in detail, particularly in other classes of algebras [2]. Spectraloid cones of these operators have not been given attention, particularly in general Banach space setting [3]. In [6], the author studied cones on involutive stereotype tubes and showed that the unit balls of these tubes could be perturbed by the identity when calculating the distance between the identity and the commutant of these operators. Moreover, when the identity has an algebraically connected component, then [7] established that these operators become spectraloid if and only if the spectral radius is algebraically stereotyped with a converging sequence of eigenvalues to an algebraic multiplicity of degree \(n\). This research considers the class of norm-attainable operators on Involutive Stereotype Tubes (IST) with Algebraically Connected Component of the Identity (ACCI). Certain interesting algebraic features are exhibited by the algebra of norm-attainable operators on involutive stereotype tubes with an algebraically connected component of the identity when perturbed by infinitesimal summands of orthogonal isometries idempotents, orthogonal projections, co-isometries among other classes of operators [8]. However, in this paper, we restrict ourselves to IST with ACCI and study them in a general Banach space setting. We outline new characterizations of the set of norm-attainable operators on involutive stereotype tubes with algebraically connected components of the identity in terms of eigenvalues and the corresponding eigenvectors in (IST). In particular, we prove reflexivity, boundedness, and compactness properties when the cones contain unit balls with involution for the tubes when they are of the stereotype category.

2. Preliminaries

We give some definitions and make some important remarks. Consider \(H\) as a complex Hilbert space and \(B(H)\) be the algebra oh all bounded linear operators on \(H\). We state the following definition;

Definition 1.([9], Definition 2.3) An operator \(A \in B(H)\) is called a scalar operator of order \(m\) if it possesses a spectral distribution of order \(m\), i.e., if there exists a continuous unital morphism \(\phi : C^{m} _{0} (\mathbb{C})\rightarrow B(H)\) such that \(\phi(z) = A,\) where \(z\) stands for the identity function on \(\mathcal{C}\) and \(C^{m} _{0} (\mathbb{C})\) for the space of compactly supported functions on \(\mathbb{C}\) continuously differentiable of order \(m\), \(0 \leq m \leq \infty.\) An operator \(A_{0} \in B(H)\) is called subscalar if it is similar to the restriction of a scalar operator to an invariant subspace of \(H\).

Definition 2.([1], Definition 1.1) An operator \(A\in B(H)\) is said to be norm-attainable if there exists a unit vector \(x_{0}\in H\) such that \(\|Ax_{0}\|=\|A\|.\) The set of all norm-attainable operators on a Hilbert space \(H\) is denoted by \(NA(H).\)

Remark 1. Consider ISTs with ACCI denoted by \(X\) and \(Y,\) and let \(\mathcal{L}(X,Y)\) be the set of all norm-attainable operators \(T : X \longrightarrow Y\) endowed with the usual operator norm \( \|T\|_{\mathcal{L}(X,Y)} = \sup_{\|x\|_{X}=1} \|Tx\|_{Y} = \sup_{\|x\|_{X}=1, \|y^{\star}\|_{Y^{\star}}=1} |\langle Tx,y^{\star} \rangle_{Y,Y^{\star}}|,\) in which \(T \in \mathcal{L}(X,Y),\) then \(\mathcal{L}(X,Y)\) is an IST with ACCI. In general, we denote the set of all norm attaining operators by \(NA(H)\) and the spectraloid cone by \(H^{\infty}(\mathbb{B}_{n},X).\) Let \(b:\mathbb{B}_{n} \longrightarrow \mathcal{L}(\overline{X},Y)\) and consider \(b \in \mathcal{H}(\mathbb{B}_{n}, \mathcal{L}(\overline{X},Y)).\) The norm-attainable operator with operator-valued symbol \(b,\) given as \(h_b\) is well defined for \(z \in \mathbb{B}_n\) as \(\displaystyle h_{b}f(z) := \int_{\mathbb{B}_n}\dfrac{b(w)\overline{f(w)}}{(1- \langle z,w\rangle)^{n+1+\alpha}}\mathrm{d}\nu_{\alpha}(w), \) \( f \in H^{\infty}(\mathbb{B}_{n},X).\) We note that \(b\) satisfies the condition \( \displaystyle \int_{\mathbb{B}_n} \dfrac{\|b(w)\|_{\mathcal{L}(\overline{X},Y)}}{|1-\langle z,w \rangle|^{n+1+\alpha}}\mathrm{d}\nu_{\alpha}(w) < \infty, \; \mbox{for every}~~z \in \mathbb{B}_{n},\label{hypo1} \) unless stated otherwise in the sequel. The next section forms the key part of this work in which new characterizations of the set of norm-attainable operators on IST with ACCI are unveiled.

3. Main results

Here we give the main results of this work. We start by some auxilliary proposition.

Proposition 1. Consider an orthonormal sequence \(\lbrace a_k \rbrace\) of non-negative scalars. Let \( M_{k} := (a_{0}I+N) \circ (a_{1}I+N) \circ \ldots \circ (a_{k-1}I+N)\) be a norm-attainable operator for some constant \(k>0.\) Then a function \(f\) is in \(\Gamma_{\gamma}(\mathbb{B}_{n},X)\) if and only if \(k > \gamma\) is such that \(\sup_{z \in \mathbb{B}_{n}}(1-|z|^2)^{k-\gamma}\|M_{k}f(z)\|_{X} < \infty.\)

Proof. Without loss of generality let \(f \in \Gamma_{\gamma}(\mathbb{B}_{n},X).\) From the statement of the proposition consider \(k > \gamma \) and constant \(C>0\) such that \( \|N^{k}f(z)\|_{X} \leq C (1 - |z|^2)^{\gamma-k},\) for all \(z \in \mathbb{B}_{n}.\) A simple and straightforward manipulation together with some substitutions give \( \phi'(r) = \dfrac{1}{r^{a+1}}\int_{0}^{r} s^{a}\psi'(s)\mathrm{d}s.\) Since \(k > \gamma,\) we get the desired result.

For ISTs with ACCI, we consider at this point two classes of norm-attainable operators \(D_{k},\) and \(L_{k},\) having an equivalence relation with \(a_{j} = n+\alpha+j+1.\) Now for simplicity we denote \( H^{\infty}(\mathbb{B}_{n},X)\) and \( H^{\infty}(\mathbb{B}_{n},Y^{\star})\) by \(\mathfrak{S}_{X}\) and \(\mathfrak{S}_{Y}\) respectively.

Proposition 2. Let \(\mathfrak{T}\) be an IST with ACCI and let \(f \in \mathfrak{S}_{X}\) and \(g \in \mathfrak{S}_{Y^{*}}.\) Given that \(b \in \mathcal{H}(\mathbb{B}_{n}, \mathcal{L}(\overline{X},Y)),\) then we have \( \displaystyle \langle h_{b}f,g \rangle_{\alpha,Y} = \int_{\mathbb{B}_n} \langle b(z)\overline{f(z)}, g(z) \rangle_{Y,Y^{\star}} \mathrm{d}\nu_{\alpha}(z). \)

Proof. Let \(\mathfrak{T}\) be an IST with ACCI and consider \(f \in \mathfrak{S}_{X}\) and \(g \in \mathfrak{S}_{Y^{*}}.\) From Hahn-Banach theorem for norm-attainable operators, the inner product property \(\langle \cdot,\cdot \rangle_{\alpha,Y},\) Proposition 1 and properties of reproducing kernels, we obtain \begin{eqnarray*} \langle h_{b}(f),g \rangle_{\alpha,Y} & = & \displaystyle \int_{\mathbb{B}_n} \langle b(w)\overline{f(w)}, g(w) \rangle_{Y,Y^{\star}} \mathrm{d}\nu_{\alpha}(w). \end{eqnarray*} Using an analogy of Hahn-Banch theorem and Parseval's equality, we have double integral \( \displaystyle \int_{\mathbb{B}_{n}} \int_{\mathbb{B}_{n}} \left|\dfrac{g(z)\left( b(w)(\overline{f(w)})\right)}{(1 - \langle z,w \rangle)^{n+1+\alpha}} \right|\mathrm{d}\nu_{\alpha}(w)\mathrm{d}\nu_{\alpha}(z) \) giving the finite strict inequality \(\displaystyle \int_{\mathbb{B}_{n}} \|b(w)\|_{\mathcal{L}(\overline{X},Y)} \log \left( \dfrac{1}{1-|w|^2} \right) \mathrm{d}\nu_{\alpha}(w) < \infty. \) This completes the proof.

Lemma 1. Let \(\mathfrak{T}\) be an IST with ACCI and let \(f \in \mathfrak{S}_{X}\) and \(z \in \mathbb{B}_n.\) For some \(b \in \mathcal{H}(\mathbb{B}_{n}, \mathcal{L}(\overline{X},Y))\) we have the map \(g_{z}(w)\) in \(\mathfrak{S}_{X}\) and the condition \( \displaystyle h_{b}(f)(z)\) is true for any integer \(k>0\) and \(C_{k}>0.\)

Proof. Since \(g_{z} \in \mathfrak{S}_{X},\) then by the reproducing kernel property and norm-attainability property we have \begin{eqnarray*} h_b(f)(z) & = & c_{k}^{-1} \int_{\mathbb{B}_n}L_{k}\left( \int_{\mathbb{B}_n} \dfrac{b(w)(\overline{g_{z}(\zeta)})}{(1 - \langle \zeta,w \rangle)^{n+1+\alpha}} \mathrm{d}\nu_{\alpha}(w) \right) \mathrm{d}\nu_{\alpha+k}(\zeta)\\ & = & c_{k}^{-1} \int_{\mathbb{B}_n}L_{k}\left( b(\zeta)(\overline{g_{z}(\zeta)})\right)\mathrm{d}\nu_{\alpha+k}(\zeta). \end{eqnarray*} Clearly, the conditions for Tonelli's theorem are satisfied. In fact, by Proposition 2 we have that \[\displaystyle \int_{\mathbb{B}_n} \left\|\int_{\mathbb{B}_n} \dfrac{b(w)(\overline{g_{z}(\zeta)})}{(1 - \langle \zeta,w \rangle)^{n+1+\alpha+k}}\mathrm{d}\nu_{\alpha}(w)\right\|_{Y}\mathrm{d}\nu_{\alpha+k}(\zeta)< \infty.\] This completes the proof of this lemma.

At this juncture, we make some assumptions on symbol \(b\) by postulating that \( \displaystyle \int_{\mathbb{B}_{n}} \|b(z)\|_{\mathcal{L}(\overline{X},Y)} \log\left( \dfrac{1}{1 - |z|^2} \right) \mathrm{d}\nu_{\alpha}(z) < \infty\label{hypimp} \) holds. We use this postulate in the next theorem.

Theorem 1. Let \(\mathfrak{T}\) be an IST with ACCI and let \(f \in \mathfrak{S}_{X}\) and \(g \in \mathfrak{S}_{X^{*}}.\) Let \(z_{n}\) be an orthonormal sequence of a complex Hilbert space \(H\) converging to \(z\) as \(n\rightarrow\infty\) and consider \(0 < p \leq 1.\) The class \((A^p_\alpha(\mathfrak{S}_{X}))^{\star}\) has equivalence relation with \(\Gamma_{\gamma}(\mathfrak{S}_{X}^{\star})\) having \(\gamma = (n+1+\alpha)\left( \frac{1}{p}-1\right)\) and for any class \(D_{k}\) which is dense, reflexive we have the coupling \( \displaystyle \langle f,g \rangle_{\alpha,X} = c_{k}\int_{\mathbb{B}_n} \langle f(z),D_{k}g(z) \rangle_{X,X^{\star}}(1 - |z|^2)^{k} \mathrm{d}\nu_{\alpha}(z),\label{intpair} \) Furthermore, \(\|g\|_{\Gamma_{\gamma}(\mathfrak{S}_{X^{\star}})} \simeq \sup_{\|f\|_{A^p_\alpha(\mathfrak{S}_{X})} =1} |\langle f,g \rangle_{\alpha,X}|.\)

Proof. Consider \(g \in \Gamma_{\gamma}( \mathfrak{S}_{Y^{*}}^{\star})\) having \(\gamma = (n+1+\alpha)\left( \frac{1}{p}-1\right) .\) For \(\alpha > 0\), let \(\displaystyle \wedge_{g}: A^p_{\alpha}(\mathfrak{S}_{X}) \longrightarrow \mathbb{C},\;\) \(f \mapsto \wedge_{g}(f) = c_{k}\int_{\mathfrak{S}_{X}} \langle f(z),D_{k}g(z) \rangle_{X,X^{\star}}(1 - |z|^2)^{k}\mathrm{d}\nu_{\alpha}(z)\) be a be positive linear functional where \(k > \gamma,\) and \(c_k>0\) a constant. We show that \(\wedge_{g}\) is well defined. To see this, let \(f \in A^p_{\alpha}(\mathfrak{S}_{X}).\) By Proposition 1, we have \( |\wedge_{g}(f)| \lesssim \|g\|_{\Gamma_{\gamma}(\mathfrak{S}_{X} ^{\star})}\|f\|_{p,\alpha,X}. \) Hence, the boundedness property of \(\wedge_{g}\) on \(A^p_{\alpha}(\mathfrak{S}_{X} )\) is satisfied.

Conversely, consider \(\wedge\) as above on \(A^p_{\alpha}(\mathfrak{S}_{X} ).\) It suffices show the existence of the function \(g \in \Gamma_{\gamma}(\mathfrak{S}_{X} ^{\star}),\) with \(\gamma = (n+1+\alpha)\left(\frac{1}{p}-1\right)\) in which \(\wedge = \wedge_{g}.\) By straightforward computation, by Lemma 1 and a simplified manipulation we have \( \wedge(f) = \displaystyle\dfrac{c_{\alpha}c_{k}}{c_{\alpha+k}}(1-|w|^2)^{k-\gamma} \big\langle x,D_{k}g(w)\big\rangle_{X,X^{\star}}. \) Now, \(f \in A^p_{\alpha}(\mathfrak{S}_{X},X).\) By reflexivity of \(\mathfrak{S}_{X},\) \( \|D_{k}g(w)\|_{X^{\star}} \lesssim \dfrac{\|\wedge\|}{(1-|w|^2)^{k-\gamma}}. \) Applying separability of the IST with ACCI and the density of \(\mathfrak{S}_{X}\) the proof is complete.

Theorem 2. Let \(\mathfrak{T}\) be an IST with ACCI and let \(f \in \mathfrak{S}_{X}\) and \(g \in \mathfrak{S}_{X^{*}}.\) Let \(z_{n}\) be an orthonormal sequence of a complex Hilbert space \(H\) converging to \(z\) as \(n\rightarrow\infty.\) Then the norm-attainable operator \(h_{b}: A^p_{\alpha}( \mathfrak{S}_{X},X) \longrightarrow A^{1,\infty}_{\alpha}( \mathfrak{S}_{X},Y)\) is continuous, linear and bounded.

Proof. The proofs for continuity and linearity are trivial so we only need to prove boundedness. Using Lemma 2 and the reproducing kernel property gives \( |\langle h_{b}(f),g \rangle_{\alpha,Y}| = \displaystyle \frac{(1-|w|^2)^{k-\gamma}}{c_{k}} \left| \big\langle L_{k}\left( b(w)\left( \overline{x}\right) \right) , y^{\star} \big\rangle_{Y,Y^{\star}} \right|. \)

Now, we use Lemma 1 to obtain \( \|h_{b}f(z)\|_{Y} = \|b\|_{\Gamma_{\gamma}( \mathfrak{S}_{X},\mathcal{L}(\overline{X},Y))}P^{+}_{\alpha} g(z), \) but by parallelogram law the property of the reproducing kernel holds and we have that \(\displaystyle P^{+}_{\alpha} g(z) = \int_{ \mathfrak{S}_{X}} \dfrac{(1-|w|^2)^{\gamma}\|f(w)\|_{X}}{|1 - \langle z,w \rangle|^{n+1+\alpha}} \mathrm{d}\nu_{\alpha}(w)\) is norm-attainable. We obtain the set \(\nu_{\alpha} (\lbrace z \in \mathfrak{S}_{X} : \|h_{b}f(z)\|_{Y} > \lambda \rbrace ) \leq \nu_{\alpha}(\lbrace z \in \mathfrak{S}_{X} : c_{k}\|b\|_{\Gamma_{\gamma}( \mathfrak{S}_{X},\mathcal{L}(\overline{X},Y))}P^{+}_{\alpha}g(z) > \lambda \rbrace ).\) By positivity of \(P^{+}_{\alpha} \), it is bounded and this completes the proof.

The following consequence comes immediately;

Corollary 1. Let \(\mathfrak{T}\) be an IST with ACCI and let \(f \in \mathfrak{S}_{X}\) and \(g \in \mathfrak{S}_{X^{*}}.\) Let \(z_{n}\) be an orthonormal sequence of a complex Hilbert space \(H\) converging to \(z\) as \(n\rightarrow\infty\) and consider \(0 < p \leq 1,\) and \(\alpha > -1.\) The norm-attainable operator \(h_{b}\) has an extension via \(A^p_{\alpha}( \mathfrak{S}_{X},X)\) into \(A^q_{\alpha}( \mathfrak{S}_{X},Y).\)

Proof. This is direct analogously from the proof of Theorem 2 and the fact that the sets \(NA(H)\) and \(\mathfrak{S}_{X}\) are desnse and satisfy the duality condition for reflexive spaces.

Theorem 3. Let \(\mathfrak{T}\) be an IST with ACCI and let \(f \in \mathfrak{S}_{X}\) and \(g \in \mathfrak{S}_{X^{*}}.\) Let \(z_{n}\) be an orthonormal sequence of a complex Hilbert space \(H\) converging to \(z\) as \(n\rightarrow\infty.\) and consider \(0 < p \leq 1,\) \(\alpha >-1\) and \(\gamma = (n+1+\alpha)\left( \frac{1}{p}-1\right).\) Then the norm-attainable operator is extendable and we have \( \|N^{k}b(w)\|_{\mathcal{L}(\overline{X},Y)} \leq \dfrac{C}{(1-|w|^2)^{k-\gamma}}\left(\log \dfrac{1}{1-|w|^2} \right)^{-1}.\)

Proof. Utilizing Corollary 1, using Lemma \(1\) for every \(f \in H^{\infty}(\mathfrak{S}_{X},X),\) and invoking the conditions of Lemma \(1\) and Proposition \(1\) we obtain \( \|h_{b}f\|_{A^{1}_{\alpha}(\mathfrak{S}_{X},Y)} \lesssim \|f\|_{p,\alpha,X}. \) Conversely, let \(h_{b}\) be extendable via \(A^p_{\alpha}(\mathfrak{S}_{X},X)\) to \(A^{1}_{\alpha}(\mathfrak{S}_{X},Y).\) With some manipulations involving \(f\) and \(g\) we get \( \langle h_{b}f,g \rangle_{\alpha,Y} = \displaystyle (1-|w|^2)^{k-\gamma}\log(1-|w|^2) \langle L_{k}(b(w)(\overline{x})),y^{\star} \rangle_{Y,Y^{\star}} + \displaystyle \langle \int_{\mathfrak{S}_{X}} b(z)(\overline{\varphi(z)}) \mathrm{d}\nu_{\alpha}(z),y^{\star} \rangle_{Y,Y^{\star}}. \) Now \(I_{2}\) is estimated as \(\displaystyle |I_{2}| \leq \|h_b\|\|\varphi\|_{p,\alpha,X}\|y^{\star}\|_{Y^{\star}} \lesssim \|h_b\|\|x\|_{X}\|y^{\star}\|_{Y^{\star}}.\) But by the fact that \(I_{1} = \langle h_{b}f,g \rangle_{\alpha,Y} - I_{2},\) Proposition \(1\) and prior estimation on \(I_{2}\) gives \(|I_{1}| \leq |\langle h_{b}f,g \rangle_{\alpha,Y}| + |I_{2}| \lesssim \|h_b\|\|x\|_{X}\|y^{\star}\|_{Y^{\star}}.\) But \(x \in X,\) \(y^{\star} \in Y^{\star}\) are not fixed and so invoking Corollary 1 completes the proof.

Next, we consider reflexivity and characterize symbols \(b\) for compact norm-attainable operators with regard to IST with ACCI. We state the following proposition.

Proposition 3. Let \(\mathfrak{T}\) be an IST with ACCI and let \(f \in \mathfrak{S}_{X}\) and \(g \in \mathfrak{S}_{X^{*}}.\) Let \(z_{n}\) be an orthonormal sequence of a complex Hilbert space \(H\) converging to \(z\) as \(n\rightarrow\infty\). For an integer \(N>0\) and \(\alpha\) we have an integro-differential norm-attainable operator (IDNAO) of order \(N\) with polynomial coefficients defined by \(\displaystyle R^{\alpha,N}f(z) = \sum_{m \in \mathbb{N}^{n},|m| \leq N} p_{m}(z)\dfrac{\partial^{|m|} f}{\partial z^{m}}(z),\) for some polynomial \(p_{m}.\)

Proof. Let \(x \in X\) and \(w \in \mathfrak{S}_{X}.\) By integrability of multinomial formula for IDNAO we have \(\langle z,w \rangle^{k} = \displaystyle \sum_{|m| = k} \dfrac{k!}{m!}z^{m}\overline{w}^{m},\) and a simple calculation follows immediately that \( R^{\alpha,N} = \displaystyle \sum_{k=0}^{N}\sum_{|m| = k}c_{mk}z^{m}\dfrac{\partial^{k} }{\partial z^{m}}.\) This completes the proof.

Remark 2. Next, we characterize integro-differential norm-attainable operator (IDNAO) of order \(N\) with polynomial coefficients defined in the infinite dimensional case by \(\displaystyle R^{\alpha,N}f(z) = \sum_{m \in \mathbb{N}^{n},|m| \leq N} p_{m}(z)\dfrac{\partial^{|m|} f}{\partial z^{m}}(z),\) for some polynomial \(p_{m}\) in a general setting for spectraloid cones of \(NA(H).\) Moreover, we consider monomiality for IDNAO. We characterize integro-differential norm-attainable operator of order \(N\) with existing and unique polynomial coefficients well defined by \(\displaystyle R^{\alpha,N}f(z) = \sum_{m \in \mathbb{N}^{n},|m| \leq N} p_{m}(z)\dfrac{\partial^{|m|} f}{\partial z^{m}}(z),\) for some polynomial \(p_{m}\) in a general setting for spctraloid cones of \(NA(H).\) We take into consideration the duality and reflexivity of the Banach spaces here. We state the following proposition.

Proposition 4. Let \(\mathfrak{T}\) be an IST with ACCI and let \(f \in \mathfrak{S}_{X}\) and \(g \in \mathfrak{S}_{X^{*}}.\) Let \(z_{n}\) be an orthonormal sequence of a complex Hilbert space \(H\) converging to \(z\) as \(n\rightarrow\infty\). Let \(1 < p < \infty\) and for all \(x^{\star} \in X^{\star}\) and \(z \in \mathfrak{S}_{X},\) suppose that \( e_{z,x^{\star}}(w) = \dfrac{x^{\star}}{(1-\langle w,z \rangle)^{n+1+\alpha}},\; w \in \mathfrak{S}_{X}.\) Then \(e_{z,x^{\star}} \in A^{p'}_{\alpha}(\mathfrak{S}_{X},X^{\star})\) and \(e_{z,x^{\star}}\) generates a separable subspectraloid cube in \(A^{p'}_{\alpha}(\mathfrak{S}_{X},X^{\star}).\)

Proof. Consider \(\phi \in A^p_{\alpha}(\mathfrak{S}_{X},X)\) with \(\langle \phi,e_{z,x^{\star}} \rangle_{\alpha,X} = 0,\) for every \(z \in \mathfrak{S}_{X}\) and \(x^{\star} \in X^{\star}.\) Suppose that \(f^{\star} \in A^{p'}_{\alpha}(\mathfrak{S}_{X},X^{\star}).\) By Radon-Nikodym theorem, it is enough to show that \(\langle \phi,f^{\star} \rangle_{\alpha,X} = 0.\) By the reproducing kernel formula for IDNAO, we have \begin{eqnarray*} 0 & = & \langle \phi,e_{z,x^{\star}} \rangle_{\alpha,X} = \displaystyle \int_{\mathfrak{S}_{X}} \langle \phi(w),e_{z,x^{\star}}(w) \rangle_{X,X^{\star}}\mathrm{d}\nu_{\alpha}(w)\\ & = & \displaystyle \int_{\mathfrak{S}_{X}} \langle \phi(w),\dfrac{x^{\star}}{(1-\langle w,z \rangle)^{n+1+\alpha}} \rangle_{X,X^{\star}}\mathrm{d}\nu_{\alpha}(w)\\ & = & \displaystyle \int_{\mathfrak{S}_{X}} \langle \dfrac{\phi(w)}{(1-\langle z,w \rangle)^{n+1+\alpha}},x^{\star}\rangle_{X,X^{\star}}\mathrm{d}\nu_{\alpha}(w)\\ & = & \langle \phi(z),x^{\star} \rangle_{X,X^{\star}}. \end{eqnarray*} Hence, for every \(x^{\star} \in X^{\star},\) we get \(\langle \phi(z),x^{\star} \rangle_{X,X^{\star}} = 0.\) Clearly, \(f^{\star} \in A^{p'}_{\alpha}(\mathfrak{S}_{X},X^{\star}),\) and so \(\langle \phi,f^{\star} \rangle_{\alpha,X} = \int_{\mathfrak{S}_{X}}\langle \phi(z),f^{\star}(z) \rangle_{X,X^{\star}}\mathrm{d}\nu_{\alpha}(z) = 0.\)

Lemma 2.Let \(\mathfrak{T}\) be an IST with ACCI and let \(f \in \mathfrak{S}_{X}\) and \(g \in \mathfrak{S}_{X^{*}}.\) Let \(z_{n}\) be an orthonormal sequence of a complex Hilbert space \(H\) converging to \(z\) as \(n\rightarrow\infty\) and let \(\beta_{0} \in \mathbb{N}^{n}\) and \(\lbrace f_j \rbrace\) converge strongly to \(0.\) Then \(\lbrace x_{j} \rbrace\) converges strongly to \(0\) in \(A^p_{\alpha}(\mathfrak{S}_{X},X).\)

Proof. We have that \(\lbrace f_j \rbrace\) is strongly bounded in \(X\) because \(f_j \rightarrow 0\) strongly in \(X\) as \(j \rightarrow \infty.\) Now, for all \(g \in A^{p'}_{\alpha}(\mathfrak{S}_{X},X^{\star}),\) we get \begin{eqnarray*} \langle x_{j},g \rangle_{\alpha,X} & = & \displaystyle \int_{\mathfrak{S}_{X}} \langle x_{j}(z),g(z) \rangle_{X,X^{\star}} \mathrm{d}\nu_{\alpha}(z)\\ & = & \displaystyle \int_{\mathfrak{S}_{X}} \langle z^{\beta_{0}}f_{j},g(z) \rangle_{X,X^{\star}}\mathrm{d}\nu_{\alpha}(z)\\ & = & \displaystyle \int_{\mathfrak{S}_{X}} z^{\beta_{0}} \langle f_{j},g(z) \rangle_{X,X^{\star}}\mathrm{d}\nu_{\alpha}(z), \end{eqnarray*} for \begin{eqnarray*} \displaystyle \left| z^{\beta_{0}} \langle f_{j},g(z) \rangle_{X,X^{\star}} \right| & \leq & \displaystyle |z^{\beta_{0}} \langle f_{j},g(z) \rangle_{X,X^{\star}}|\\ & \leq & \|f_{j}\|_{X}\|g(z)\|_{X^{\star}} \\ & \leq & C\|g(z)\|_{X^{\star}}, \end{eqnarray*} and \[ \displaystyle \int_{\mathfrak{S}_{X}} \|g(z)\|_{X^{\star}}\mathrm{d}\nu_{\alpha}(z) \leq \left( \int_{\mathfrak{S}_{X}} \|g(z)\|^{p'}_{X^{\star}}\mathrm{d}\nu_{\alpha}(z)\right)^{1/p'} < \infty.\] Applying Fatou's lemma and Lebegue's Dominated convergence theorem gives \begin{eqnarray*} \displaystyle \limsup_{j\longrightarrow \infty}\langle x_{j},g \rangle_{\alpha,X} & = & \displaystyle \int_{\mathfrak{S}_{X}} z^{\beta_{0}} \lim_{j \longrightarrow \infty} \langle f_{j},g(z) \rangle_{X,X^{\star}} \mathrm{d}\nu_{\alpha}(z) = 0. \end{eqnarray*} This completes the proof as required.

Proposition 5. Let \(\mathfrak{T}\) be an IST with ACCI and let \(f \in \mathfrak{S}_{X}\) and \(g \in \mathfrak{S}_{X^{*}}.\) Let \(z_{n}\) be an orthonormal sequence of a complex Hilbert space \(H\) converging to \(z\) as \(n\rightarrow\infty\) and let \(1 < p \leq q < \infty,\) \(0 \leq r < 1\) and \(\gamma \in \mathbb{N}^{n}.\) If \(a_{\gamma} \in \mathcal{K}(\overline{X},Y),\) then the nor-attainable \(h_{g^{\gamma}_{r}} : A^{p}_{\alpha}(\mathfrak{S}_{X},X) \rightarrow A^{q}_{\alpha}(\mathfrak{S}_{X},Y)\) is compact, in which \(g^{\gamma}_{r}(z) = a_{\gamma}(rz)^{\gamma}\) for all \(z \in \mathfrak{S}_{X}.\)

Proof. A simple manipulation and by Theorem 2 gives the desired result.

Theorem 4. Let \(\mathfrak{T}\) be an IST with ACCI and let \(f \in \mathfrak{S}_{X}\) and \(g \in \mathfrak{S}_{X^{*}}.\) Let \(z_{n}\) be an orthonormal sequence of a complex Hilbert space \(H\) converging to \(z\) as \(n\rightarrow\infty\) and \(1 < p \leq q < \infty.\) The norm-attainable operator \(h_{b}: A^p_{\alpha}(\mathfrak{S}_{X},X) \rightarrow A^{q}_{\alpha}(\mathfrak{S}_{X},Y)\) is bounded if and only if \(b \in \Lambda_{\gamma_{0}}(\mathfrak{S}_{X},\mathcal{L}(\overline{X},Y)),\) in which \(\gamma_{0} = (n+1+\alpha)\left(\frac{1}{p} - \frac{1}{q}\right).\) Furthermore, \(\|h_{b}\|_{A^p_{\alpha}(\mathfrak{S}_{X},X) \rightarrow A^{q}_{\alpha}(\mathfrak{S}_{X},Y)} \simeq \|b\|_{ \Lambda_{\gamma_{0}}(\mathfrak{S}_{X},\mathcal{L}(\overline{X},Y))}.\)

Proof. Since \(h_{b}\) is norm-attainable from \(A^p_{\alpha}(\mathfrak{S}_{X},X)\) to \(A^q_{\alpha}(\mathfrak{S}_{X},Y)\) and having the norm \(\|h_b\| = \|h_b\|_{A^p_{\alpha}(\mathfrak{S}_{X},X) \rightarrow A^q_{\alpha}(\mathfrak{S}_{X},Y)}.\) Let \(z \in \mathfrak{S}_{X}\) then for the function \( f(w) = \dfrac{x}{(1-\langle w,z \rangle)^{k}},\; w \in \mathfrak{S}_{X}\), we have that \( h_{b}f(z) = R^{\alpha,k}b(z)(\overline{x}). \) and a straightforward calculation gives \( \|R^{\alpha,k}b(z)(\overline{x})\|_{Y} = \|h_{b}f(z)\|_{Y}. \) Now for all \(x \in X\) and \(\|x\|_{X} = \|\overline{x}\|_{\overline{X}}\) we obtain \( \|R^{\alpha,k}b(z)\|_{\mathcal{L}(\overline{X},Y)} \lesssim \dfrac{\|h_b\|}{(1-|z|^2)^{k-\gamma_{0}}}.\) Therefore, \( \sup_{z \in \mathfrak{S}_{X}} (1-|z|^2)^{k-\gamma_{0}}\|R^{\alpha,k}b(z)\|_{\mathcal{L}(\overline{X},Y)} \lesssim \|h_b\|.\) This implies that the symbol \(b \in \Lambda_{\gamma_{0}}(\mathfrak{S}_{X},\mathcal{L}(\overline{X},Y))\) and \(\|b\|_{\Lambda_{\gamma_{0}}(\mathfrak{S}_{X},\mathcal{L}(\overline{X},Y))} \lesssim \|h_{b}\|.\) For the reverse inclusion suppose that \(b \in \Lambda_{\gamma_{0}}(\mathfrak{S}_{X},\mathcal{L}(\overline{X},Y)).\) Let \(f \in A^p_{\alpha}(\mathfrak{S}_{X},X),\) \(g \in A^{q'}_{\alpha}(\mathfrak{S}_{X},Y^{\star})\) and \(k > \gamma_{0}.\) Then we obtain \(b \in \Lambda_{\gamma_{0}}(\mathfrak{S}_{X},\mathcal{L}(\overline{X},Y)) \subset A^{p'}_{\alpha}(\mathfrak{S}_{X},\mathcal{L}(\overline{X},Y)).\) The rest is clear from Proposition 1 and Theorem 3.

4. Conclusion

In this work, we have studied the class of norm-attainable operators on involutive stereotype tubes with an algebraic connected component of the identity. We give characterizations of spectraloid cones of norm-attainable operators on involutive stereotype tubes with algebraic connected components of identity in terms of eigenvalues and the corresponding eigenvectors in involutive stereotype tubes. In particular, we have proven reflexivity, boundedness, and compactness properties when the cones contain unit balls with involution for the tubes when they are of the stereotype category.

Acknowledgments :

The first author is grateful for the Chebyshev grant (ICM No.: 15729326) to attend the International Congress of Mathematicians (ICM-2022) in Saint Petersburg, Russia.

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.''

Data Availability:

All data required for this research is included within this paper.

Funding Information:

No funding is available for this research.

1. Introduction

Consider the following initial-boundary value problem for the Navier-Stokes equations in two dimensions with non-local viscosity. It means, find a vector function

\[ {u}: \Omega \times [0,T] \rightarrow \mathbb{R}^2,\] and a scalar function \[p: \Omega \times [0,T] \rightarrow \mathbb{R},\] satisfying where \(\Omega\) is a domain sufficiently regular, \(\partial \Omega \) its boundary well regular, and we have that \(c(l(u_{1}(x,t)),l(u_{2}(x,t)))\) satisfies these hypotheses: Given \(x=(x_1,x_2), y=(y_1,y_2) \in \mathbb{R}^2,\)

• (A1) \(0< c_- \leq c(x_1,x_2) \leq c_+,\)
• (A2) \(|c(x) - c(y)| \leq A_1 |x_1-y_1| + A_2 | x_2 - y_2|, \mbox{ for some } A_1,A_2>0,\)

and \(l:L^2(\Omega) \rightarrow \mathbb{R}\) is a continuous linear functional, defined by \(u \mapsto \int_\Omega u d\Omega.\)

We mentioned that the existence, uniqueness and exponential decay of the solution to the problem (1)-(4) were studied by Ferreira, Shahrouzi, Andrade and Panni in [1].

The motivation to study this kind of problem is we can describe motion of fluids which viscosity depends of time and satisfies the hypotheses \((A1)-(A2)\), and, when \(c(l(u_{1}(x,t)),l(u_{2}(x,t)))=\mu\), constant, we obtain the regular Navier-Stokes equations in two dimensions. This non-local term was introduced by Chipot [2], and it arrives naturally when we study the growth of a bacteria population, one kind of this problem was suggested by Ladyzhenskaya [3] where \(c(t):= \mu_0 + \mu_1 \| {u}(t)\|^2\) when \(\mu_0\) and \(\mu_1\) are positive constants.

The rest of the paper is organized as follows. In §2, we recall some notations, the weak formulation, lemmas and theorems. In §3, we study the existence of strong solutions to the problem (1)-(4). In §4, we introduce the existence of periodic solutions using Brouwer's fixed point theorem. The conclusions of the paper are presented in §5.

2. Preliminaries

In this section we introduce some notations, achieve weak formulation and enunciate some important results.

2.1. Notations

Let \(\Omega \subset \mathbb{R}^2\) be a regular domain with \(\partial \Omega\) a well regular boundary. We denote the inner product in \(H^1_0(\Omega)\) by \(((\cdot,\cdot))\) and \((\cdot,\cdot)\) in \(L^2(\Omega)\), and norms respectively by \(\|\cdot\|\) and \(|\cdot|\). By \( {H}^1_0(\Omega)\) we denote \((H_0^1(\Omega))^2\) and, \( {L}^2(\Omega)\) by \((L(\Omega))^2\). The set \( {\mathfrak{V}}\) is the set of all distributions \( {u}: (\mathcal{D}(\Omega))^2 \mapsto \mathbb{R}^2\), which its divergent is null, in other words, \( {\mathfrak{V}}(\Omega):= \{ {u} \in (\mathcal{D}(\Omega))^2 ; div( {u}) =0\}\). Also, we denote the closure of \( {\mathfrak{V}}(\Omega)\) in \( {H}^1_0(\Omega)\) by \( {V}\) and the closure of \( {\mathfrak{V}}(\Omega)\) in \((L(\Omega))^2\) by \(H\).

A well known propriety of non-local term, see [4,5,6,7,8], is that this term commutes with spatial integral sign

\[\int_\Omega c(l(u_1),l(u_2)) {u} d\Omega = c(l(u_1),l(u_2)) \int_\Omega {u}d\Omega.\]

2.2. Weak formulation

Consider \( {v} \in {V}\). Doing inner-product in \(L^2(\Omega)\) with Eq. (1) we get, \begin{align*} &\left(\dfrac{d {u}(x,t)}{dt}, {v}(x)\right)- c(l(u_1),l(u_2))(\nabla {u}(x,t), {v}(x) ) + (( {u}(x,t) \cdot \nabla ) {u}(x,t), {v}(x))+ (\nabla p(x), {v}(x)), = ( {f}(x,t), {v}). \end{align*} By green first identity and integration by parts, \begin{align*} &\dfrac{d}{dt}\left( {u}(x,t), {v}(x)\right) + c(l(u_1),l(u_2))(( {u}(x,t), {v}(x) ))+ (( {u}(x,t) \cdot \nabla ) {u}(x,t), {v}(x)) = ( {f}(x,t), {v}(x)). \end{align*} Now we define a bilinear form \(a( {u}, {v}):=(( {u}, {v} ))\) and a trilinear form \(b( {u}, {v}, {w}):= (( {u} \cdot \nabla ) {v}, {w})\), and then we obtain the weak form of Eq. (1), \begin{equation*} \dfrac{d}{dt}\left( {u}, {v}\right) + c(l(u_1),l(u_2))a ( {u}, {v} ) + b( {u}, {u}, {v}) = ( {f}, {v}). \end{equation*}

2.3. Some results

Lemma 1.[9] Let \( {u} \in {L}^2(0,T; {V})\), then the function \(B {u}\) defined by, \[ \langle B {u}(t), {v} \rangle := b( {u}, {v}, {w}), \forall {v} \in {V}, \text{for a.e. } t \in [0,T],\] belongs to \( {L}^1(0,T; {V}').\)

Lemma 2. Let \(\Omega \subset \mathbb{R}^n\) be an bounded Lipschitz open set in \(\mathbb{R}^2\).

1. If a distribution \(p\) has all its first-order derivatives \(D_i p\) in \(L^2(\Omega),\) then \(p \in L^2(\Omega)\) and \[\|p\|_{L^2(\Omega) \backslash \mathbb{R} } \leq c(\Omega) |\nabla p|_{L^2(\Omega)}.\]
2. If a distribution \(p\) has all its first-order derivatives in \(H^{-1}(\Omega).\) Then \(p \in L^2(\Omega)\) and \[\|p\|_{L^2(\Omega) \backslash \mathbb{R}} \leq c \|\nabla p\|_{H^{-1}(\Omega)},\]

where \(L^2(\Omega)\backslash \mathbb{R}:= \left\{ p \in L^2(\Omega) \big\rvert \int_\Omega p(x) dx =0 \right\}.\)

Problem 1. For \( {f}\) and \( {u}_0\) given, with

to find \( {u}\) satisfying, \begin{align*} {u} \in L^2(0,T; {V}), {u}' \in L^1(0,T; {V}'), \end{align*} \begin{align*} &\dfrac{d}{dt}\left( {u}(x,t), {v}(x)\right) + c(l(u_1),l(u_2))(( {u}(x,t), {v}(x) )) + (( {u}(x,t) \cdot \nabla ) {u}(x,t), {v}(x)) = ( {f}(x,t), {v}(x)), \end{align*} for any \( {v} \in {V}\).

3. Existence of strong solutions

Suppose the existence of weak solutions to the problem (1)-(4). Our goal in this section is recover the pressure and prove the existence of strong solutions.

Theorem 2. Given \( {f}\) and \( {u}_0\) satisfying (5) and (6). Suppose that \( {u}\) is a solution of the Problem 1 and \[ {f}-c(l(u_1),(u_2))A {u} -B {u} - {u}' \in L^2(0,T;V'),\] then the solution \( {u}\) is also strong.

Proof. Let, \[ {U}(t):=\int_0^t {u}(s)ds, {F}(t):=\int_0^t {f}(s) ds \hspace{0.2cm} \mbox{ and } \beta(t):=\int_0^t B( {u}(s), {u}(s))ds \in V'.\] Since \( {u}, {f},B {u} \in L^2(0,T;V')\) then,

Integrating \(c(l(u_1),(u_2))A {u} +B {u}+ {u}' = {f}\), and, by (7), we get \[ {u}(t) - {u}(0) + c(l(u_1),l(u_2))\int_0^t A {u}(s)ds + \int_0^t B {u}(s)ds = \int_0^t {f}(s) ds \mbox{ in V'.}\] Then, \[ {u}(t)- {u}_0 + c(l(u_1),l(u_2))A {U}(t) + \beta(t) = {F}(t) \mbox{ in }V', \forall t \in [0,T].\] So, for each \(\phi \in {\mathfrak{V}}\), Define, For each \(t \in [0,T]\) it is possible to extend \( {S}(t)\) on a functional \( {T}(t)\in H^{-1}(\Omega)\) such as, But, from (8) and (10) we can conclude that, \[\langle {T}(t), {\phi} \rangle =0, \forall {\phi} \in \mathfrak{V}.\] From Lemma 2 results that \(\exists P(t) \in L^2(\Omega)\) satisfying, So, from (10) and (11) we get, Replacing (12) in (9), \[ {u}(t) - {u}_0 + c(l(u_1),l(u_2))A {U}(t) + \beta(t) - {F}(t) = \nabla P(t) \mbox{ in }V', \forall v \in [0,T].\] As the expression on the left belongs to the space \( C ^ 0 (0, T; V ') \) we have \( \nabla P \in C ^ 0 (0, T; V') \), and hence we can derive the above equation in the sense of distributions, with this: \[ {u}' + c(l(u_1),l(u_2))A {u} - {f} + B {u}= \nabla \dfrac{\partial P}{\partial t} \mbox{ in } L^2(0,T;V').\] Therefore is possible to say that equality above is given a.e. in \( (0, T) \). Setting \(p(x,t) = - \dfrac{\partial P}{\partial t},\) results in, \[ {u}' + c(l(u_1),l(u_2)) A {u} +B {u} = {f} -\nabla p \in L^2(0,T;V').\]

4. Existence of periodic solutions

The purpose of this section is to prove the existence of periodic solutions to the Navier-Stokes equations.

Theorem 3. Let \(\Omega \subset \mathbb{R}^2\) a bounded open set with boundary \(\partial \Omega\) well regular and \(Q:=[0,T] \times \Omega\). Consider the following problem,

where \( {f} \in L^2(0,T;V')\). This problem admits weak solution in \( {u}:Q \rightarrow \mathbb{R}^2,\) \( {u} \in L^2(0,T;H)\cap L^\infty(0,T;H)\) and \( {u}' \in L^2(0,T;V').\)

Proof. The weak formulation of (13) is given by,

Consider \(\{ {w}_1,\cdots, {w}_n,\cdots\}\) a base of \(V\). We truncate the series in \( m \)-th term, which leads to the approximate solution space \( V_m \). Setting \( {u}_m(t):=g_{im}(t) {w}_i,\) where \(j=1,\cdots, m\).

The approximate system above has a global solution, since by similar procedure to the case of the existence of solutions [1], we obtain the following inequality,

\[| {u}(t)|^2 + \int_0^t \| {u}_m(s)\|^2ds \leq | {v}| + \frac{1}{c_-}\| {f}\|_{L^2(0,T;V')}\leq c(m),\] as \(m\) is fixed, we can extend \( {u}(t)\) in \([0,T].\) Our goal is to show that, among all solutions of the approximate equation, there is at least one \( {u}_m \) solution that satisfies periodicity, \[ {u}_m(0) = {u}_m(T).\] To do this, just prove that for every \( m \in \mathbb{N}, \) the application, \begin{align} & {\tau}_m: V_m \rightarrow V_m \nonumber \\ & {v} \mapsto {\tau}_m( {v}) = {u}_m(T), \nonumber \end{align} has a single fixed point, because in this case there will be a single function \( {v} \in V_m \) such that Thus (16) we have a \( ( {u}_m) \) sequence of approximate solutions such that they all satisfy the periodicity condition.

Lemma 3. Exists \(\rho_0 >0\) such as \( {\tau}_m \overline{(B_{\rho_0}(0))} \subset \overline{B_{\rho_0}(0)}.\)

Proof. Using the \(H\) induced topology in \(V_m\), it suffices to prove that \begin{align*} \exists\;\;\;\; \rho_0 > 0 \mbox{ such that } | {\tau}_m( {v})|_H \leq \rho_0; \forall v \in V_m, \mbox{ where } | {v}|_H \leq \rho_0. \end{align*} Applying the energy method, \begin{align*} \frac{1}{2} \dfrac{d}{dt}| {u}_m(t)|^2 + c_- \| {u}_m(t)\|^2 &+ b( {u}_m(t), {u}_m(t), {u}_m(t))\\ &\leq \frac{1}{2} \dfrac{d}{dt}| {u}_m(t)|^2 + c(l(u_1),l(u_2)) \| {u}_m(t)\|^2 + b( {u}_m(t), {u}_m(t), {u}_m(t))\\ &=\langle {f}(t), {u}_m(t) \rangle\\ &\leq\| {f}(t)\|_{V'} | {u}_m(t)|, \end{align*} implies that, \begin{align*} \frac{1}{2} \dfrac{d}{dt}| {u}_m(t)|^2 + c_- \| {u}_m(t)\|^2 \leq \frac{1}{2c_-}\| {f}(t)\|^2_{V'} + \frac{c_-}{2}\| {u}_m(t)\|^2, \end{align*} then,

As \(V \hookrightarrow H,\) exists \(c_0 >0\) such as, Thus from (17) and (18) we get, \[\dfrac{d}{dt}| {u}_m(t)|^2 + c_0^2 c_- | {u}_m(t)|^2 \leq \frac{1}{c_-} \| {f}(t)\|^2_{V'}.\] Multiplying both sides by \(e^{c_0^2c_-t}\): \begin{align*} \dfrac{d}{dt}(| {u}_m(t)|^2e^{c_0^2c_-t}) \leq \frac{1}{c_-}\| {f}(t)\|_{V'}^2 e^{c_0^2c_-t}. \end{align*} Integrating from \(0\) to \(T\) we get, \begin{align*} | {u}_m(t)|^2e^{c_0^2c_-t} \leq | {u}_m(0)|^2 + \dfrac{1}{c_-}\int_0^T \| {f}(t)\|^2_{V'}e^{c_0^2c_-t}dt, \end{align*} which means, \begin{equation*} | {u}_m(t)|^2 \leq e^{-c_0^2c_-T}| {u}_m(0)|^2 + \frac{1}{c_-}\int_0^T \| {f}(t)\|^2_{V'}dt, \end{equation*} then, \begin{equation*} | {u}_m(t)|^2\leq e^{-c_0^2 c_- T}| {u}_m(0)| + \frac{1}{c_-}\| {f}\|^2_{L^2(0,T;V')}. \end{equation*} Denoting \(\theta = e^{-c_0^2 c_- T}\) and \(c= \frac{1}{c_-}\| {f}\|^2_{L^2(0,T;V')},\) we can write \[| {u}_m(t)|^2 \leq \theta| {u}_m(0)|^2 +c,\] so, \[| {\tau}_m( {v})|^2 \leq \theta | {v}|^2 + c, \hspace{0.2cm} \forall v \in V_m.\] Now, how \(0 < \theta < 1\) then \(0 < 1-\theta< 1.\) That way there is a \(\rho_0 >0\), big enough that \(c< (1-\theta)\rho^2_0.\) So if \(| {v}| < \rho_0\) then, \[\theta | {v}|^2 + c \leq \theta \rho_0^2 + (1-\theta)\rho^2_0 = \rho^2_0\,,\] where, \[| {\tau}_m( {v})|^2 \leq \rho_0^2, \hspace{0.2cm}\forall m \in \mathbb{N},\] which proves this lemma.

Lemma 4. The application \( {\tau}_m: V_m \mapsto V_m\) defined in (15) is continuous.

Proof. Let \( {v}_1, {v}_2 \in V_m\) and \( {u}_m, {z}_m\) solutions of the approximate problem with initial data \( {v}_1\) and \( {v}_2\), respectively. Our goal is to show that the solutions are Lipschitz-continuous, \(| {\tau}_m( {v}_1) - {\tau}_m( {v}_2)|\leq c_m | {v}_1- {v}_2|\) for some \(c_m>0.\) \begin{align*} &( {u}'_m(t), {w}_j) + c(l(u_1),l(u_2))(( {u}_m(t), {w}_j)) + b( {u}_m(t), {u}_m(t), {w}_j) = \langle {f}(t), {w}_j \rangle,\\ &( {z'}_m(t), {w}_j) + c(l(z_1),l(z_2))(( {z}_m(t), {w}_j)) + b( {z}_m(t), {z}_m(t), {w}_j) = \langle {f}(t), {w}_j \rangle. \end{align*} Doing the difference between these equations and defining \( {\eta}_m = {z}_m - {u}_m,\) \begin{align*} ( {\eta}_m, {w}_j) &+ c(l(u_1),l(u_2))(( {u}_m(t), {w}_j)) -a(l(z_1),l(z_2))(( {z}_m(t), {w}_j))\\ &+b( {u}_m(t), {u}_m(t), {w}_j)-b( {z}_m(t), {z}_m(t), {w}_j) =0, \end{align*} we proceed as in [1] \begin{equation*} \frac{d | {\eta}_m|^2}{dt} - | {\eta}_m|^2\left(\frac{2}{c_-}\|u_{2m}(t)\|^2 + \frac{K^2}{c_-}\| {z}_m \|^2\right)\leq 0. \end{equation*} Defining \(\theta_m(t) = \left(\frac{2}{c_-}\|u_{2m}(t)\|^2 + \frac{K^2}{c_-}\| {z}_m \|^2\right)\), we get, \begin{equation*} \frac{d | {\eta}_m|^2}{dt} - | {\eta}_m|^2\theta_m(t)\leq 0. \end{equation*} Multiplying both sides of inequality \(e^{-\int_0^t \theta_m(s)ds},\) \begin{equation*} \dfrac{d}{dt}\left( | {\eta}_m(t)|^2e^{-\int_0^t \theta_m(s)ds} \right) \leq 0. \end{equation*} Integrating the inequality from \(0\) to \(T\), \[| {\eta}_m(T)|^2e^{-\int_0^t \theta_m(s)ds} - | {\eta}_m(0)|^2\leq 0.\] Defining \(c_m = e^{-\int_0^t \theta_m(s)ds},\) \[| {\eta}_m(T)|^2 \leq c_m| {\eta}_m(0)|^2.\] By other hand, \[ {\eta}_m(s) = {u}_m(s) - {z}_m(s),\] so, \[| {u}_m(T)- {z}_m(T)|^2 \leq c_m | {u}_m(0) - {z}_m(0)|^2.\] Then, \[| {\tau}_m( {v}_1) - {\tau}_m( {v}_2)| \leq c_m| {v}_1 - {v}_2|,\] which is what we want to prove.

The hypotheses of Brouwer's fixed point theorem are satisfied by virtue of Lemmas 3 and 4, so we have \[ {\tau}_m: \overline{B_{\rho_0}(0)} \rightarrow \overline{B_{\rho_0}(0)},\] admits a fixed point, which means, there is a \( {v} \in \overline{B_{\rho_0}(0)}\) such as \( {\tau}_m( {v}) = {v},\) so, \( {u}_m(0) = {u}_m(T).\)

Then, for each \(m\in \mathbb{N},\) there is a least one \( {u}_m(t)\) such as \( {u}_m(0) \in \overline{B_{\rho_0}(0)}\) and, \(\forall j=1,\cdots,m\),

\begin{align*} \begin{cases} ( {u}'_m(t), {w}_j)+ c(l(u_1),l(u_2)) (( {u}_m(t), {w}_j)) + b( {u}_m(t), {u}_m(t), {w}_j) = \langle {f}(t), {w}_j \rangle, \\ {u}_m(0) = {u}_m(t). \end{cases} \end{align*} From the fact that \( {u}_m(0) \in \overline{B_{\rho_0}(0)}\) we can repeat the estimates getting a subsequence \(( {u}_{\nu})\) of \(( {u}_m)\) such as From the convergence results (19) - (21), by passing the limit in the approximate equation desired in (14). Similarly to the proof of the initial condition in the previous case, we prove that \( {u}(0) = {u}(T), \) which concludes the statement.

5. Conclusions

We studied the Navier-Stokes equations with non-local viscosity, considering a bounded domain \(\Omega \subset \mathbb{R}^2\) with smooth boundary \(\partial \Omega\). Using Faedo-Galerkin's method and Brouwer's fixed point theorem, we proved the strong solutions and periodic solutions.

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.''

Data Availability:

All data required for this research is included within this paper.

Funding Information:

The third author was supported by FCT - Fundação para a Ciência e a Tecnologia, through Centro de Matemática e Aplicações - Universidade da Beira Interior, under the Grant Number UI/BD/150794/2020, and also supported by MCTES, FSE and UE.

1. Introduction

Viscoelastic materials exhibit an instantaneous elasticity effect and creep characteristics at the same time. The importance of the viscoelastic properties of materials has been realized because of the rapid developments in rubber and plastics industry. The modeling of the dynamics of physical phenomena such as heat flow in conductors with memory, hereditary polarization in dielectrics, population dynamics, viscolasticity can be described by an abstract integro-differential equation of the form

where \('\) represents a derivative with respect to time \(t\), \(A: \mathcal{D}(A)\subset H \longrightarrow H\) is a positive definite self-adjoint operator on \(H\), \(g\) is the relaxation function (convolution kernel), \(\alpha \in [0,1]\), \(u_0,\, u_1\) are given history function and initial data respectively.

The study of viscoelastic problems has attracted the attention of many authors and several decay and blow up results have been established. We start with the pioneer work [6,7,] where Dafermos considered a one-dimensional viscoelastic problem and established various existence results and then proved, for smooth monotone decreasing relaxation functions, that the solutions go to zero as \(t\) goes to infinity. After that, many results dealing with the existence, uniqueness, regularity and asymptotic behavior of many systems of the form (1) have been studied; see, for example, [1,8,9,10,11]. In the case of finite memory, that is, \(u_0(t)=0\) for \(t< 0\), see [12,13,14,15,16,17,18]. In particular, Rivera et al., [15] considered the interpolating cases \(\alpha \in (0,1)\) and a relaxation function \(g\) which decays exponentially to zero at infinity, that is,

They showed that the energy decays polynomially at the rate of \(\frac{1}{t}\). Recently, Hassan and Messaoudi [19] considered and established a new general decay rate result for which the relaxation function \(g\) satisfies condition For case of infinite memory, see [20,21,22,23,24,25]. In particular, Guesmia [1] considered and introduced a new ingenuous approach for proving a more general decay result based on the properties of convex functions and the use of the generalized Young inequality. He used a larger class of infinite history kernels satisfies the following condition such that where \(G:\mathbb{R}_+\to \mathbb{R}_+\) is an increasing strictly convex function. Al-Mahdi and Al-Gharabli [2] considered the following viscoelastic problem and they established decay results with using a relaxation function \(g\), satisfying the condition Very recently, Guesmia [26] considered two models of wave equations with infinite memory and established an explicit and general decay rate results where the relaxation function satisfying the condition (4).

Motivated by the above works, we intend to study the following class of viscoelastic equations of the form

where \(A: \mathcal{D}(A)\subset H \longrightarrow H\) is a positive definite self-adjoint operator on \(H\) such that the embedding \(\mathcal{D} (A^\beta) \hookrightarrow \mathcal{D} (A^\sigma)\) is compact for any \(\beta > \sigma \geq 0\) and \( \alpha \in (0,1) \).

Remark 1. The assumption \(\mathcal{D} (A^\beta) \hookrightarrow \hookrightarrow \mathcal{D} (A^\sigma)\) for any \(\beta > \sigma \geq 0\) guarantees the existence of some constants \(\omega,\, \omega_0,\, \omega_1\) such that

and

2. Our main objectives

We intend to establish a two fold objective:

1. improve many earlier works such as the ones in [11,15,19] from finite memory to infinite memory;
2. prove a general decay estimate for the solution of Problem (10) with a wider class of relaxation functions than the ones considered in [1,2,3,4,5] by getting a better decay rate with imposing a weaker assumption on the boundedness of initial data than the one considered in the earlier papers such as the one in [1,2,3,4,5].

The paper is organized as follows: We present some assumptions and remarks in §3. We state and prove some technical lemmas in §4. The main result, its proof and some examples are presented in §5.

3. Assumptions

In this section, we state some assumptions needed in the proof of our main decay result. The strictly decreasing differentiable relaxation (kernel) function \( g : [0,\infty)\longrightarrow (0,\infty)\) satisfies the following assumptions:

• ( A.1) \(\qquad\qquad\qquad \displaystyle g(0)>0 \qquad\mathrm{and\qquad} 1- _0\int_0^{+\infty}g(s)ds=l>0. \)
• ( A.2) There exists a non-increasing differentiable function \(\xi\!:\mathbb{R}_+\longrightarrow(0,\infty)\) and a \(\boldsymbol{C}^1\) function \(G:[0,+\infty)\longrightarrow[0,+\infty)\) which is linear or it is strictly increasing and strictly convex \(C^2\) function on \((0,r]\), with \(G(0)=G^{\prime}(0)=0\), such that
  \begin{equation}\label{a4} g'(t)\leq-\xi(t)G(g(t)),\qquad\forall\,t \geq 0, \end{equation}

  (14)
  where \(\xi\) is satisfying \(\int_0^{+ \infty}\xi(s)ds = + \infty.\)
• ( A.3) We assume that \[\int_{0}^{+\infty}g(s)\vert\vert A^{\alpha/2}u_0(s)\vert\vert^{2} ds < +\infty,\] and \[\int_{0}^{+\infty}g(s)\vert\vert A^{1/2}u_0(s)\vert\vert^{2} ds < +\infty.\]

Remark 2.The class of relaxation functions satisfying ( A.1)-( A.2) in the present paper is larger than the ones satisfying (6) and (7) used in some earlier papers such as the one in [1]. In fact, the boundedness of the sup in (6) use in [1], can be interpreted as the inequality in ( A.2) in the present paper (with \(\xi=1\)). The conditions (6) and (7) used in [1] ask also the boundedness of the integral. So, it is better to consider the relaxation functions satisfy ( A.1)-( A.2) used in the present paper than the one used in [1].

Remark 3. Hypothesis ( A.3) is needed for proving the existence and stability results. For the stability, if ( A.3) holds, then the functions \(h_0\) and \(h_1\) defined in Lemma 4 well be defined. Moreover, Hypothesis ( A.3) is weaker than the one used in [1,2,3,4,5] that is, there exists a positive constant \(M\) such that \[\vert\vert \nabla A^{\alpha/2} u_{0}(s)\vert\vert^{2} \leq M,\] and \[\vert\vert \nabla A^{1/2} u_{0}(s)\vert\vert^{2} \leq M.\]

Remark 4. As is in Mustafa [14], if \(G\) is a strictly increasing and strictly convex \(\boldsymbol{C}^2\) function on \((0,r]\), with \(G(0)=G'(0)=0\), then there is a strictly convex and strictly increasing \(\boldsymbol{C}^2\) function \(\overline{G}:[0,+\infty)\longrightarrow[0,+\infty)\) which is an extension of \(G\). For instance, we can define \(\overline{G}\), for any \(t>r\), by \[\overline{G}(t):=\frac{G''(r)}{2}t^2+\big(G'(r)-G''(r)r\big)t+\left(G(r)+\frac{G''(r)}{2}r^2-G'(r)r\right).\]

We state the existence, regularity and uniqueness theorem whose proof is in [15].

Theorem 1([15]). Suppose that \((u_0(\cdot, 0),u_1) \in \mathcal{D}(A) \times \mathcal{D}(A^{1/2})\) and ( A.1- A.3) hold. Then, Problem (10) has a unique global solution satisfying \[u \in \boldsymbol{C} \left(\mathbb{R}_+;\mathcal{D}(A)\right) \cap \boldsymbol{C}^1 \left(\mathbb{R}_+;\mathcal{D}\left(A^{1/2}\right)\right) \cap \boldsymbol{C}^2 \left(\mathbb{R}_+;H\right).\] Moreover, if \((u_0(\cdot, 0),u_1) \in \mathcal{D}\left(A^{\sigma+1/2}\right) \times \mathcal{D}\left(A^{\sigma}\right)\) for \(\sigma \geq 0\), then the solution satisfies \[u \in \boldsymbol{C} \left(\mathbb{R}_+;\mathcal{D}\left(A^{\sigma+1/2}\right)\right) \cap \boldsymbol{C}^1 \left(\mathbb{R}_+;\mathcal{D}\left(A^{\sigma}\right)\right) \cap \boldsymbol{C}^2 \left(\mathbb{R}_+;\mathcal{D}\left(A^{\sigma-1/2}\right)\right).\]

The "modified" energy functionals associated to our problem are given by for any \(t \geq 0\), where for \(v \in L^2_{loc}(\mathbb{R}_+;H)\), \[(g \circ v)(t):=\int_0^{\infty} g(s)\|v(t)-v(t-s)\|^2 ds.\]

Remark 5. The positiveness of the energy functionals comes from inequalities (12) and (13).

Lemma 1([15]). For any initial data \((u_0,u_1) \in \mathcal{D}(A) \times \mathcal{D}\left(A^{1/2}\right)\), the energy functionals associated to Problem (10) satisfy, for any \(t \geq 0\), the identities

As in Jin et al., [27], we set, for any \(0< \varepsilon< 1\), \[C_\varepsilon:=\int_0^\infty\frac{g^2(s)}{\varepsilon g(s)-g'(s)}ds \qquad\mathrm{and}\qquad h_\varepsilon(t):=\varepsilon g(t)-g'(t).\]

Lemma 2([27]). Assume that the condition ( A.1) holds. Then, for any \(v \in L^2_{loc} \big([0,+\infty);L^2(0,L)\big)\), we have

Lemma 3(Jensen's inequality). Let \(F:[a,b]\longrightarrow\mathbb{R}\) be a convex function. Assume that the functions \(f: \longrightarrow[a,b]\) and \(h:\Omega\longrightarrow\mathbb{R}\) are integrable such that \(h(x)\geq0\), for any \(x\in\Omega\) and \(\displaystyle\int_\Omega h(x)dx=k>0\). Then, \[F\left(\frac{1}{k}\int_\Omega f(x)h(x)dx\right) \leq \frac{1}{k}\int_\Omega F(f(x))h(x)dx.\]

4. Technical lemmas

In this section, we state and prove some Lemmas that are useful in the proof of Theorem 2. Through out this work we use \(c>1\) to represent a generic constant, which is independent of \(t\) and the initial data.

Lemma 4. Assume that ( A.1- A.3) hold. Then, there exist two positive constants \(M_0, M_1\) such that

and where \(h_0(t)=\int_{0}^{+\infty}g(t+s)\left(1+\vert \vert A^{\alpha/2}u_0(s)\vert\vert^{2} \right) ds,\) and \(h_1(t)=\int_{0}^{+\infty}g(t+s)\left(1+\vert \vert A^{1/2}u_0(s)\vert\vert^{2} \right) ds.\)

Proof. Indeed, we have

where \(M_0=\max\big\{2, \frac{4 \omega_0 E(0)}{1-l} \big\}\).

The proof of (21) can be established similarly to the proof of (20).

Lemma 5. Assume that conditions ( A.1- A.3) hold. Then, for any \(0< \delta< 1\), the functional \(I_1\) defined by \[I_1(t):=-\left\langle u'(t), \int_0^{\infty} g(s)(u(t)-u(t-s))ds\right\rangle\] satisfies, along the solution of (10), the estimate

Proof. Differentiating \(I_1\) and exploiting the differential equation in Problem (10), we get

Next, we estimate the terms in the right-hand side of the above identity. Using the Cauchy-Schwarz, Young and Hölder inequalities, Lemma 2 and inequalities (11) and (12), it follows that, for any \(0< \delta< 1\), \begin{align*} &\left\langle A^{1/2}u(t), \int_0^\infty g(s)A^{1/2}(u(t)-u(t-s))ds \right\rangle \\ &\leq \frac{\delta}{2} \|A^{1/2}u(t)\|^2 + \frac{c}{\delta}C_\varepsilon \left( h_\varepsilon \circ A^{1/2}u \right)(t), -\left\langle \int_0^\infty g(s)A^{\alpha/2}u(t-s)ds,\right. \left.\int_0^\infty g(s)A^{\alpha/2}(u(t)-u(t-s))ds \right\rangle \\ & = \left\| \int_0^\infty g(s)A^{\alpha/2}(u(t)-u(t-s))ds \right\|^2 - \left\langle\left(\frac{1-l}{\omega_0}\right)A^{\alpha/2}u(t), \int_0^\infty g(s)A^{\alpha/2}(u(t)-u(t-s))ds\right\rangle \\ & \leq \frac{\delta}{2\omega_0}\left\|A^{\alpha/2}u(t)\right\|^2 + \frac{c}{\delta}C_\varepsilon \left( h_\varepsilon \circ A^{\alpha/2}u\right)(t) \\ & \leq \frac{\delta}{2}\left\|A^{1/2}u(t)\right\|^2 + \frac{c}{\delta}C_\varepsilon \left( h_\varepsilon \circ A^{1/2}u\right)(t), \end{align*} and \begin{align*} &\left\langle u'(t),\int_0^\infty g'(s)(u(t)-u(t-s))ds \right\rangle \\ &= \left\langle u'(t),\varepsilon\int_0^\infty g(s)(u(t)-u(t-s))ds \right\rangle - \left\langle u'(t),\int_0^\infty h_\varepsilon(s)(u(t)-u(t-s))ds \right\rangle \end{align*}\begin{align*} &\leq \frac{\delta}{2} \|u'(t)\|^2 + \frac{\varepsilon^2}{2\delta}\left\|\int_0^\infty g(s)(u(t) - u(t-s)) ds \right\|^2 + \frac{\delta}{2} \|u'(t)\|^2 + \frac{1}{2\delta}\left\| \int_0^\infty h_\varepsilon(s)(u(t)- u(t-s)) ds \right\|^2 \\ &\leq \delta \|u'(t)\|^2 + \frac{c}{\delta}(C_\varepsilon + 1)\left( h_\varepsilon \circ A^{1/2}u \right)(t). \end{align*} Plugging the above estimates in (24), we obtain the desired estimate.

Lemma 6. Under the conditions ( A.1- A.3), the functional \(I_2\) defined by \[I_2(t):=\langle u'(t),u(t)\rangle\] satisfies, along the solution of (10), the estimate

Proof. Differentiating \(I_2\), using the equation in (10), and repeating the above computations, we get \begin{align*} I_2'(t)&=\|u'(t)\|^2 - \left\|A^{1/2}u(t)\right\|^2 + \left( \frac{1-l}{\omega_0} \right)\left\|A^{\alpha/2}\right\|^2 + \left\langle \int_0^\infty g(s)A^{\alpha/2}(u(s)-u(t-s))ds, A^{\alpha/2}u(t)\right\rangle \\ &\leq \|u'(t)\|^2 -l \left\|A^{1/2}u(t)\right\|^2 + \frac{l}{2\omega_0}\left\| A^{\alpha/2}u(t) \right\|^2 + \frac{\omega_0}{2l}\left\| \int_0^\infty g(s) A^{\alpha/2}((u(t) - u(t-s)) ds \right\|^2 \\ &\leq \|u'(t)\|^2 - \frac{l}{2}\left\|A^{1/2}u(t)\right\|^2 + cC_\varepsilon\left( h_\varepsilon \circ A^{1/2}u \right)(t), \qquad\forall\, t \geq 0. \end{align*}

Lemma 7. Assume that ( A.1- A.3) hold. Then, the functionals \(J_1\) and \(J_2\) defined by \[ J_1(t) := \int_0^t p(t-s)\left\| A^{\alpha/2} u(s) \right\|^2 ds \] and \[ J_2(t) := \int_0^t p(t-s)\left\| A^{1/2} u(s) \right\|^2 ds \] with \( \displaystyle p(t) := \int_t^\infty g(s) ds \) satisfy, along the solution of (10), the estimates \begin{equation*}\label{e8s3} J_1'(t) \leq 3(1-l)\|A^{1/2} u(t) \|^2 - \frac{1}{2}(g \circ A^{\alpha/2}u)(t)+ \frac{1}{2}\int_t^{\infty} g(s) \|A^{\alpha/2} u(t)-A^{\alpha/2} u(s)\|_2^2ds, \end{equation*} and \begin{equation*}\label{e9s3} J_2'(t) \leq \frac{3}{ _0}(1-l)\|A^{1/2} u(t)\|^2 - \frac{1}{2}(g \circ A^{1/2}u)(t)+ \frac{1}{2}\int_t^{\infty} g(s) \|A^{1/2} u(t)-A^{1/2} u(s)\|_2^2ds, \end{equation*} for any \( t \geq 0 \).

Proof. Exploiting Young's inequality, ( A.1- A.3), inequality (12) and the fact that \( p(t) \leq p(0) = \frac{1-l}{ _0} \), we obtain, for any \( t \geq 0 \),

Similarly, differentiating \(J_2\) and repeating the above computations, we get, for any \( t \geq 0 \),

Lemma 8. Assume ( A.1- A.3) hold. Then, the functional \(\mathcal{L}\) defined by \[ \mathcal{L} (t):=N(E(t) + \mathcal E (t)) + \varepsilon_1 I_1(t) + \varepsilon_2 I_2(t)\] satisfies, for a suitable choice of \(N, \varepsilon_1, \varepsilon_2 > 0\),

and the estimate

Proof. It is straightforward to establish the equivalence (28). To prove (29), we start by exploiting relations (17), (18), (23) and (25) to get

Now, we set \(\beta:=\frac{1-l}{\omega_0}\) and choose \(\delta\) small enough so that \[d< \min\left\{\frac{1}{2}\beta, \, \frac{l}{8}\beta \right\}.\] Consequently, for \( \varepsilon_1 = \frac{16(1-l)}{l\beta}\left( 4 + \frac{3}{2 _0} \right) \), we pick \( \varepsilon_2 = \frac{3}{8}\beta\varepsilon_1 \) satisfying \[\frac{1}{4}\beta\varepsilon_1 < \varepsilon_2 < \frac{1}{2}\beta\varepsilon_1.\] Then, \[ (\beta-d)\varepsilon_1 - \varepsilon_2 > \frac{1}{2}\beta\varepsilon_1-\varepsilon_2 = \frac{1}{8}\beta\varepsilon_1 = \frac{2}{l}(1 - l)\left( 4 + \frac{3}{2 _0} \right) \] and \[ \frac{l}{2}\varepsilon_2 - \delta\varepsilon_1 > \frac{l}{2}\left(\varepsilon_2 - \frac{1}{4}\beta\varepsilon_1 \right) = \frac{l}{16}\beta\varepsilon_1 = (1 - l)\left( 4 + \frac{3}{2 _0} \right). \] From \(\displaystyle \frac{\varepsilon g^2(s)}{\varepsilon g(s)-g'(s)}< g(s)\) and the Lebesgue Dominated Convergence Theorem, we deduce \[\lim_{\varepsilon\rightarrow0^+}\varepsilon C_\varepsilon=\lim_{\varepsilon\rightarrow0^+}\int_0^\infty\frac{\varepsilon g^2(s)}{\varepsilon g(s)-g'(s)}ds=0.\] So there exists \(0 < \varepsilon_0 < 1\) such that if \(\varepsilon < \varepsilon_0\), then \[\varepsilon C_\varepsilon < \frac{1}{\frac{8c}{\delta}\left( \varepsilon_1 + \varepsilon_2 \right)}.\] Now, we choose \(N\) large enough so that \( \mathcal{L} \sim E + \cal E \) and \[ N > \max \left\lbrace \frac{4c}{\delta}(\varepsilon_1 + \varepsilon_2),\,\,\frac{1}{2\varepsilon_0} \right\rbrace. \] For \(\varepsilon=\frac{1}{2N}\), we have \[ \frac{N}{4} - \frac{c}{\delta}(\varepsilon_1 + \varepsilon_2) > 0 \qquad\mathrm{and}\qquad \varepsilon< \varepsilon_0.\] This gives \[ \frac{N}{2} - \frac{c}{\delta}(\varepsilon_1 + \varepsilon_2) - \frac{c}{\delta}C_\varepsilon(\varepsilon_1 + \varepsilon_2) > \frac{N}{2} - \frac{c}{\delta}(\varepsilon_1 + \varepsilon_2) - \frac{1}{8\varepsilon} = \frac{N}{4} - \frac{c}{\delta}(\varepsilon_1 + \varepsilon_2) > 0. \] Thus estimate (30) becomes \begin{eqnarray*} \mathcal{L}'(t) &\leq & - \frac{2}{l}(1 - l)\left( 4 + \frac{3}{2 _0} \right)\|u'(t)\|^2 - (1 - l)\left( 4 + \frac{3}{2 _0} \right)\|A^{1/2}u(t)\|^2 \\ &&+ \frac{1}{4}\left( g \circ A^{\alpha/2}u + g \circ A^{1/2}u \right)(t), \qquad \forall \, t \geq 0.\end{eqnarray*}

Lemma 9. Assume that ( A.1- A.3) hold. Then, the energy functional satisfies, for all \(t\in \mathbb{R}^+\) and for some positive constant \(\tilde{m}\), the following estimate

where \(f(t)=1+\int_{0}^{t} h(s)ds\) and \(h=h_0+h_1\) and \(h_0\), \(h_1\) are defined in (20) and (21).

Proof. Let \(F(t)=\mathcal{L}(t)+J_1(t)+ \frac{1}{2}J_2(t)\), then we obtain, for all \(t\in \mathbb{R}_+\),

Estimates (18) and (32) yield, for some positive constant \(\lambda\) and for all \(t\in \mathbb{R}_+\), \begin{align*} F^{\prime}(t)\le& - \lambda E(t)+\frac{1}{2}\int_{t}^{+\infty} g(s)\Vert A^{\alpha/2} u(t)-A^{\alpha/2} u(t-s) \Vert^{2} ds dx\\&+\frac{1}{2}\int_{t}^{+\infty} g(s)\Vert A^{1/2} u(t)-A^{1/2} u(t-s) \Vert^{2} ds dx. \end{align*} Therefore, using (20) and (21) and integrating both sides of the last inequality, over \((0,t)\), we arrive at Hence, we get where \(\tilde{m}=\max \big\{\frac{F(0)}{\lambda},\frac{M_0}{2\lambda}\big\}\).

Corollary 1. There exists \(0< q_0 < 1\) such that, for all \(t \ge 0\), we have the following estimate:

where \(G\) is defined in Remark 4 and \(f(t)\) is defined in (31).

Proof. We introduce a functional \(\eta\) defined by \[\eta(t) := q(t)\int_{0}^t\left(\left\| A^{\alpha/2}(u(t)-u(t-s))\right\|^2+\left\| A^{1/2}(u(t)-u(t-s))\right\|^2\right)ds,\quad\forall\, t \geq 0,\] and observe, from inequality (12), that

Use of (15), (17) and (38) yields \begin{align*} \eta(t) & \leq 2q(t)\int_{0}^t\left(\left\| A^{\alpha/2}u(t)\right\|^2 + \left\|A^{\alpha/2}u(t-s)\right\|^2+\left\| A^{1/2}u(t)\right\|^2 +\left\|A^{1/2}u(t-s)\right\|^2\right)ds \\ & \leq \frac{4q(t)}{l}(1+ _0) \int_{0}^t \big( E(t) + E(t-s) \big)ds \\ & \leq \frac{8q(t)}{l}(1+ _0) \int_{0}^t E(s) ds, \qquad\forall\, t \geq 0. \end{align*}Thanks to (31), we can pick \(0< q_0 < \min\big\{1, \frac{l}{8\tilde{m}(1+\omega_0)}\big\} \) so that To prove (35), we define another functional \(\mu\) by Also, the strict convexity of \(G\) and the fact that \(G(0)=0\) entail that \[G(s\tau) \leq sG(\tau),\qquad \mathrm{for} \quad 0 \leq s \leq 1 \quad \mathrm{and} \quad \tau \in (0,r].\] Combining this with the hypothesis ( A.2), Jensen's inequality and (39), we obtain, for any \(t\geq0\), \begin{align*} \mu(t)&=-\frac{1}{\eta(t)}\int_{0}^t\eta(t)g'(s)\left(\left\| A^{\alpha/2}(u(t)-u(t-s))\right\|^2+\left\| A^{1/2}(u(t)-u(t-s))\right\|^2\right)ds\\ &\geq\frac{1}{\eta(t)}\int_{0}^t\eta(t)\xi(s)G(g(s))\left(\left\| A^{\alpha/2}(u(t)-u(t-s))\right\|^2+\left\| A^{1/2}(u(t)-u(t-s))\right\|^2\right)ds\\ &\geq\frac{\xi(t)}{\eta(t)}\int_{0}^t G(\eta(t)g(s))\left(\left\| A^{\alpha/2}(u(t)-u(t-s))\right\|^2+\left\| A^{1/2}(u(t)-u(t-s))\right\|^2\right)ds\\ &\geq\frac{\xi(t)}{q(t)}\overline{G}\left( q(t)\int_{0}^t g(s)\left(\left\| A^{\alpha/2}(u(t)-u(t-s))\right\|^2+\left\| A^{1/2}(u(t)-u(t-s))\right\|^2\right)ds\right), \end{align*} where \(\overline{G}\) is a \(\boldsymbol{C}^2\) extension of \(G\) which is strictly increasing and strictly convex on \((0,\infty)\). For simplicity, in the rest of this paper, we use \(G\) instead of \(\overline{G}\). Then we have for any \(t \geq 0\), \[ \int_{0}^t g(s)\left(\left\| A^{\alpha/2}(u(t)-u(t-s))\right\|^2+\left\| A^{1/2}(u(t)-u(t-s))\right\|^2\right)ds \leq \frac{1}{q(t)}{G}^{-1}\left(\frac{q(t) \mu(t)}{\xi(t)}\right).\]

5. The main result

In this section, we state and prove our decay result. We introduce the following functions: It is not difficult to show that the above functions are convex and increasing on \((0,r]\). Now we state our main result.

Theorem 2. Assume that hypotheses ( A.1)-( A.3) hold and the initial data satisfy \[(u_0,u_1)\in \left[\mathcal{D}\left(A^{1-\alpha/2}\right) \times \mathcal{D}\left(A^{(1-\alpha)/2}\right)\right] \cap \left[\mathcal{D}\left(A^{1/2}\right) \times H\right].\] Then, for all \(0 \leq s \leq t\) and for strictly positive constant \(C\), we have the following decay results

where \(q\) is defined in (37), \(h=h_0+h_1\) where \(h_0\), \(h_1\) are defined in (20) and (21) and the functions \(G_2(s)\) and \(G_{4}(s)\) are defined in (41).

Proof. We start by combining (15), (20), (21), (29) and (35); then, for some \(m>0\) and for any \(t\geq 0\), we have

Let \( 0 < \varepsilon_0 < r\), then define a functional \(\mathcal{F}\) by \[\mathcal{F}(t) := G'\left( \frac{\varepsilon_0 q(t)E(t)}{E(0)} \right)\mathcal{L}(t), \qquad \forall\, t \geq 0.\] Using the facts that \(E'\leq0\), \(G'>0\) and \(G''>0\), we get for any \(t \geq 0\), \begin{align} \mathcal{F}'(t) &= \frac{\varepsilon_0 q(t) E'(t)}{E(0)} G''\left(\frac{\varepsilon_0 q(t)E(t)}{E(0)}\right)\mathcal{L}(t) + G'\left(\frac{\varepsilon_0 q(t)E(t)}{E(0)}\right)\mathcal{L}'(t)\notag\end{align} Let \(G^{*}\) be the convex conjugate of \(G\) in the sense of Young (see [28]), then and it satisfies the following generalized Young inequality So, with \(A=G^{\prime}\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)\), \(B=G^{-1}\left(\frac{q(t) \mu(t)}{\xi(t)}\right)\), and using (17), (18), and (44)-(46), we arrive at So, multiplying (47) by \(\xi(t)\) and using (40) and the fact that \(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}< r\), we obtain \begin{equation*} \begin{aligned} \xi(t) \mathcal{F}^{\prime}(t)\le& -m \xi(t) E(t)G^{\prime}\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)+c \xi(t)\varepsilon_{0}\frac{E(t)}{E(0)}G^{\prime}\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)\\&+c \mu(t)q(t)+c \xi(t)h(t)G^{\prime}\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)\\ \le&-\varepsilon_{0}(\frac{m E(0)}{\varepsilon_{0}} -c ) \xi(t)\frac{E(t)}{E(0)}G^{\prime}\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)-c\big(E'(t) + \mathcal E'(t)\big)(t)\\&+c \xi(t) h(t)G^{\prime}\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right). \end{aligned} \end{equation*}Consequently, recalling the definition of \(G_2\) and choosing \(\varepsilon_{0}\) so that \(k=(\frac{m E(0)}{\varepsilon_{0}} -c)>0\), we obtain, for all \(t \in \mathbb{R}_+\), where \(\mathcal{F}_{1}=\xi \mathcal{F}+c (E +\mathcal E)\). Since \(G^{\prime}_{2}(t)=G^{\prime}(t)+t G^{\prime\prime}(t),\) then, using the strict convexity of \(G\) on \((0,r],\) we find that \(G_{2}^{\prime}(t), G_{2}(t)>0\) on \((0,r].\)

Using the general Young inequality (46) for the last term in (48) with \(A=G^{\prime}\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)\) and \(B=[\frac{c}{d}h (t)]\), we have for any \(d>0\),

where \(G_2\), \(G_3\) and \(G_4\) are given in (41). Now, combining (48) and (49) and choosing \(d\) small enough so that \(k_1=(k-d)>0\), we arrive at Since \(E'< 0\) and \(q'< 0\), then \((qE)(t)\) is decreasing function. Using this fact and since \(G_2\) is increasing, we have, for \(0 \leq t\leq T\), Combining (50) with (51) and multiplying by \(q(t)\), we get Since \(q'< 0\), then for all \(0 \leq t\leq T\), Integrating (53) over \([0, T]\) and using the fact \(q(0)=q_0\), we have Hence, Thus which yields where \(C=\max\big\{1, \frac{\mathcal{F}_{1}(0)}{c}, \frac{c}{d}, \frac{1}{\varepsilon_0} \big\}\).

Example 1. Let \(g(t)=\frac{a}{(1+t)^\nu}\), where \(\nu >1\) and \(0< a< \nu-1\). In this case \(\xi(t)=\nu a^{\frac{-1}{\nu}}\) and \(G(t)=t^{\frac{\nu+1}{\nu}}\). Then \(G'(t)=a_0 t^{\frac{1}{\nu}}\). We will discuss two cases:

Case 1: if \(m_0 \leq 2+\vert \vert A^{\alpha/2}u_{0}+A^{1/2}u_{0}\vert \vert2 \leq m_1\). Then we have the following:

Then Thus for \(\nu \geq 2\) or \(\sqrt{2} < \nu < 2\) we have \(\lim_{T\rightarrow +\infty} E(T)=0\).

Case 2: if \(m_0 (1+t)^{r} \leq 2+\vert \vert A^{\alpha/2}u_{0}+A^{1/2}u_{0}\vert \vert^2 \leq m_1 (1+t)^{r}\), where \(0 < r < \nu-1\), then we have the following:

Then, Thus for \(\nu -r\geq 2\) or \(\frac{1}{2}\big(r+\sqrt{r^2+4r+8}\big) < \nu < r+2\) we have \(\lim_{T\rightarrow +\infty} E(T)=0\).

Acknowledgments :

The authors thank University of King Fahd University of Petroleum and Minerals (KFUPM) and the referee for his/her very careful reading and valuable comments. This work is sponsored by KFUPM under Project No. SB201012.

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.''

Data Availability:

All data required for this research is included within this paper.

Funding Information:

No funding is available for this research.

1. Introduction

In this paper, we investigate the existence of global and decay of solutions for the p-biharmonic parabolic equation with logarithmic nonlinearity

where \(\Omega \) is bounded domain \( \mathbb{R} ^{n}\) with smooth boundary \(\partial \Omega ,\) \(p,\) \(q\) are positive constants, \(2< p< q< p\left( 1+\frac{4}{n}\right) ,\) and \(u_{0}\in \left( W_{0}^{1,p}\left( \Omega \right) \cap W^{2,p}\left( \Omega \right) \right) \backslash \left\{ 0\right\} .\) The term \(\Delta \left( \left\vert \Delta u\right\vert ^{p-2}\Delta u\right) \) is called a \(p\)-biharmonic operator.

Studies of logarithmic nonlinearity have a long history in physics as it occurs naturally in different areas of physics such as supersymmetric field theories, inflationary cosmology, nuclear physics, optics and quantum mechanics [1,2]. Peng and Zhou [3] studied the following heat equation with logarithmic nonlinearity

\begin{equation*} u_{t}-\Delta u_{t}=\left\vert u\right\vert ^{p-2}u\ln \left\vert u\right\vert . \end{equation*} They obtained the global existence and blow-up of solutions. Also, they discussed the upper bound of blow-up time under suitable conditions. Nhan and Truong [4] studied the following nonlinear pseudo-parabolic equation \begin{equation*} u_{t}-\Delta u_{t}-div\left( \left\vert \nabla u\right\vert ^{p-2}\nabla u\right) =\left\vert u\right\vert ^{p-2}u\ln \left\vert u\right\vert . \end{equation*} They obtained results as regard the existence or non-existence of global solutions. Also, He et al., [5] proved the decay and the finite time blow-up for weak solutions of the equation. Cao and Liu [6] studied the following nonlinear evolution equation with logarithmic source \begin{equation*} u_{t}-\Delta u_{t}-div\left( \left\vert \nabla u\right\vert ^{p-2}\nabla u\right) -k\Delta u_{t}=\left\vert u\right\vert ^{p-2}u\ln \left\vert u\right\vert . \end{equation*} They established the existence of global weak solutions. Moreover, they considered global boundedness and blowing-up at \(\infty \).

Wang and Liu [7] considered the following p-biharmonic parabolic equation with the logarithmic nonlinearity

\begin{equation*} u_{t}+\Delta \left( \left\vert \Delta u\right\vert ^{p-2}\Delta u\right) =\left\vert u\right\vert ^{q-2}u\ln \left\vert u\right\vert \end{equation*} They studied existence of weak solutions by potential well method, blow up at finite time by concative method.

Recently some authors studied the hyperbolic and parabolic equation with logarithmic source term (see [8,9,10,11,12,13,14,15,16,17,18,19,20]). This paper is organized as follows: In the §2, we introduce some lemma which will be needed later. In §3, under some conditions, we obtain the unique global weak solution of the problem (1). Meanwhile, we find that the solution is decay polynomially.

It is necessary to note that prence of the logarithmic nonlinearity causes some difficulties in deploying the potantial well method. In order to handle this situation we need the following logarithmic Sobolev inequality which was introduced by ([4,21,22]).

Proposition 1. Let \(u\) be any function in \(H^{1}\left( \mathbb{R} ^{n}\right) \) and \(\mu >0\) be any number. Then

where \begin{equation*} \mathcal{L} _{p}=\frac{p}{n}\left( \frac{p-1}{e}\right) ^{p-1}\pi ^{-\frac{p }{2}}\left[ \frac{\Gamma \left( \frac{\pi }{2}+1\right) }{\Gamma \left( n \frac{p-1}{p}+1\right) }\right] ^{\frac{p}{n}}. \end{equation*}

2. Preliminaries

For simplicity, we denote \begin{equation*} \text{ }\left\Vert u\right\Vert _{s}=\left\Vert u\right\Vert _{L^{s}(\Omega )},\text{ }\left\Vert u\right\Vert _{W_{0}^{2,p}\left( \Omega \right) }=\left\Vert u\right\Vert _{2,s}=\left( \left\Vert \Delta u\right\Vert _{s}^{s}+\left\Vert \nabla u\right\Vert _{s}^{s}+\left\Vert u\right\Vert _{s}^{s}\right) ^{\frac{1}{s}}, \end{equation*} for \(1< s< \infty \) (see [23,24], for details). We also use notation \(X_{0}\) to denote \(\left( W_{0}^{1,p}\left( \Omega \right) \cap W^{2,p}\left( \Omega \right) \right) \backslash \left\{ 0\right\} \) and \(W^{-2,p^{\prime }}\left( \Omega \right) \) to denote the dual space of \(W^{2,s}\left( \Omega \right) \), where \( s^{\prime }\) is Hölder conjugate functional of \(s>1.\)

Let us introduce the energy functional \(J\) and Nehari functional \(I\) defined on \(X_{0}\) as follow

and By (3) and (4), we get Let \begin{equation*} \mathcal{N}=\{u\in X_{0}:I(u)=0\}, \end{equation*} be the Nehari manifold. Thus, we may define \(d\) is positive and is obtained by some \(u\in \mathcal{N}.\) Then it is obvious that \begin{equation*} M=\frac{1}{p^{2}}\left( \frac{p^{2}e}{n\mathcal{L} _{p}}\right) ^{\frac{n}{p} }. \end{equation*} From [4], we know \(d\geq M.\)

The local existence of the weak solutions can be obtained via the standard parabolic theory. It is easy to obtain the following equality

Lemma 1. Let \(u\in X_{0}\). Then we possess

• (i) \(\lim_{\lambda \to 0^{+}}j(\lambda )=0\) and \(\lim_{\lambda \to +\infty }j(\lambda )=-\infty ;\)
• (ii) there is a unique \(\lambda ^{\ast }>0\) such that \(j^{\prime }(\lambda ^{\ast })=0;\)
• (iii) \(j(\lambda )\) is increasing on \((0,\lambda ^{\ast }),\) decreasing on \( (\lambda ^{\ast },+\infty )\) and attains the maximum at \(\lambda ^{\ast };\)
• (iv) \(I(\lambda u)>0\) for \(0< \lambda < \lambda ^{\ast },\) \(I(\lambda u)< 0\) for \(\lambda ^{\ast }< \lambda < +\infty \) and \(I(\lambda ^{\ast }u)=0.\)

Proof. For \(u\in X_{0},\) by the definition of \(j,\) we get

It is clear that (i) holds due to \(\int\nolimits_{\Omega }\left\vert u\right\vert ^{q}dx\neq 0.\) We have \begin{eqnarray*} \frac{d}{d\lambda }j(\lambda ) &=&\lambda ^{p-1}\left\Vert \Delta u\right\Vert _{p}^{p}-\lambda ^{q-1}\int\nolimits_{\Omega }\left\vert u\right\vert ^{q}\ln \left\vert u\right\vert dx-\lambda ^{q-1}\ln \lambda \left\Vert u\right\Vert _{q}^{q}, \\ &=&\lambda ^{p-1}\left( \left\Vert \Delta u\right\Vert _{p}^{p}-\lambda ^{q-p}\int\nolimits_{\Omega }\left\vert u\right\vert ^{q}\ln \left\vert u\right\vert dx-\lambda ^{q-p}\ln \lambda \left\Vert u\right\Vert _{q}^{q}\right) . \end{eqnarray*} Since \(\lambda >0,\) let \(\varphi \left( \lambda \right) =\lambda ^{1-p}j^{\prime }(\lambda ),\) through direct calculation, we get \begin{equation*} \varphi ^{\prime }(\lambda )=-\lambda ^{q-p-1}\left( \left( q-p\right) \int\nolimits_{\Omega }\left\vert u\right\vert ^{q}\ln \left\vert u\right\vert dx+\left( q-p\right) \ln \lambda \left\Vert u\right\Vert _{q}^{q}+\left\Vert u\right\Vert _{q}^{q}\right) . \end{equation*} Hence, there exists a \begin{equation*} \lambda ^{\ast }=\exp \left( \frac{\left( p-q\right) \int\nolimits_{\Omega }\left\vert u\right\vert ^{q}\ln \left\vert u\right\vert dx-\left\Vert u\right\Vert _{q}^{q}}{\left( q-p\right) \left\Vert u\right\Vert _{q}^{q}} \right) >0, \end{equation*} such that \(\varphi ^{\prime }(\lambda )>0\) on \((0,\lambda ^{\ast }),\) \( \varphi ^{\prime }(\lambda )< 0\) on \((\lambda ^{\ast },+\infty )\) and on \( \varphi ^{\prime }(\lambda )=0.\) So, \(\varphi (\lambda )\) is increasing on \( (0,\lambda ^{\ast }),\) decreasing on \((\lambda ^{\ast },+\infty ).\) Since \( \lim_{\lambda \to 0^{+}}\) \(\varphi (\lambda )=\left\Vert \nabla u\right\Vert ^{2}>0,\) \(\lim_{\lambda \to +\infty }\) \(\varphi (\lambda )=-\infty ,\) there exists a unique \(\lambda ^{\ast }>0\) such that \( \varphi (\lambda ^{\ast })=0,\) i.e., \(j^{\prime }(\lambda ^{\ast })=0.\) So (ii) holds. Then, \(j^{\prime }(\lambda )=\lambda \varphi (\lambda )\) is positive on \((0,\lambda ^{\ast }),\) negative on \((\lambda ^{\ast },+\infty ).\) Thus, \(j(\lambda )\) is increasing on \((0,\lambda ^{\ast }),\) decreasing on \((\lambda ^{\ast },+\infty )\) and attains the maximum at \( \lambda ^{\ast }.\) So (iii) holds. The last property, (iv) , is only a simple corallary of the fact that \begin{eqnarray*} I(\lambda u) &=&\left\Vert \Delta \left( \lambda u\right) \right\Vert _{p}^{p}-\int\nolimits_{\Omega }\left\vert \lambda u\right\vert ^{q}\ln \left\vert \lambda u\right\vert dx \\ &=&\lambda ^{p}\left\Vert \Delta u\right\Vert _{p}^{p}-\lambda ^{q}\int\nolimits_{\Omega }\left\vert u\right\vert ^{q}\ln \left\vert u\right\vert dx-\lambda ^{q}\ln \lambda \left\Vert u\right\Vert _{q}^{q} \\ &=&\lambda j^{\prime }(\lambda ). \end{eqnarray*} Thus, \(I(\lambda u)>0\) for \(0< \lambda < \lambda ^{\ast },\ I(\lambda u)< 0\) for \(\lambda ^{\ast }< \lambda < +\infty \) and \(I(\lambda ^{\ast }u)=0.\) So (iv) holds. The proof is complete.

Next we denote

\begin{equation*} R:=\left( \frac{p^{2}e}{n\mathcal{L} _{p}}\right) ^{n/p^{2}}. \end{equation*}

Lemma 2.

• (i) if \(I(u)>0\) then \(0< \left\Vert u\right\Vert _{p}< R,\)
• (ii) if \(I(u)< 0\) then \(\left\Vert u\right\Vert _{p}>R,\)
• (iii) if \(I(u)=0\) then \(\left\Vert u\right\Vert _{p}\geq R.\)

Proof. By the definition of \(I(u)\), we get \begin{eqnarray*} I(u) &=&\left\Vert \Delta u\right\Vert _{p}^{p}-\int\nolimits_{\Omega }\left\vert u\right\vert ^{q}\ln \left\vert u\right\vert dx \\ &\geq &\left\Vert \Delta u\right\Vert _{p}^{p}-\int\nolimits_{\Omega }\left\vert u\right\vert ^{p}\left( \ln \frac{\left\vert u\right\vert }{ \left\Vert u\right\Vert _{p}}+\ln \left\Vert u\right\Vert _{p}\right) dx \\ &\geq &\left( 1-\frac{\mu }{p}\right) \left\Vert \Delta u\right\Vert _{p}^{p}+\left( \frac{n}{p^{2}}\ln \left( \frac{p\mu e}{n\mathcal{L} _{p}} \right) -\ln \left\Vert u\right\Vert _{p}\right) \left\Vert u\right\Vert _{p}^{p}. \end{eqnarray*} Choosing \(\mu =p,\) we have \begin{equation*} I(u)\geq \left( \frac{n}{p^{2}}\ln \left( \frac{p^{2}e}{n\mathcal{L} _{p}} \right) -\ln \left\Vert u\right\Vert _{p}\right) \left\Vert u\right\Vert _{p}^{p}. \end{equation*} (i) if \(I(u)>0,\) then \begin{equation*} \ln \left\Vert u\right\Vert _{p}< \ln \left( \frac{p^{2}e}{n\mathcal{L} _{p}} \right) ^{\frac{n}{p^{2}}}, \end{equation*} that's mean \begin{equation*} \left\Vert u\right\Vert _{p}< \left( \frac{p^{2}e}{n\mathcal{L} _{p}}\right) ^{\frac{n}{p^{2}}}=R, \end{equation*} and (ii) if \(I(u)< 0,\) we obtain \begin{equation*} \left\Vert u\right\Vert _{p}>\left( \frac{p^{2}e}{n\mathcal{L} _{p}}\right) ^{\frac{n}{p^{2}}}=R, \end{equation*} property (iii) we can argue similarly the proof of (ii).

The proof of lemma is complete.

Lemma 3. [25] For any \(u\in W_{0}^{1,p}(\Omega )\), \(p\geq 1\), \(r\geq 1\) and \(p_{\ast } =\frac{np}{n-p}\), the inequality \begin{equation*} \left\Vert u\right\Vert _{q}\leq C\left\Vert \nabla u\right\Vert _{p}^{\theta }\left\Vert u\right\Vert _{r}^{1-\theta }, \end{equation*} is valid, where \begin{equation*} \theta =\left( \frac{1}{r}-\frac{1}{q}\right) \left( \frac{1}{n}-\frac{1}{p} + \frac{1}{r}\right) ^{-1}, \end{equation*} and for \(p\geq n=1,\) \(r\leq q\leq \infty ;\) for \(n>1\) and \(p< n,\) \(q\in \lbrack r,p_{\ast }]\) if \(r< p_{\ast }\) and \(q\in \lbrack p_{\ast },r]\) if \( r\geq p_{\ast }\) for \(p=n>1,\) \(r\leq q\leq \infty ;\) for \(p>n>1,\) \(r\leq q\leq \infty .\)

Here, the constant \(C\) depends on \(n,p,q\) and \(r.\)

Lemma 4. [26] Let \(f:R^{+}\to R^{+}\) be a nonincreasing function and \(\sigma \) is a nonnegative constant such that \begin{equation*} \int\nolimits_{t}^{+\infty }f^{1+\sigma }(s)ds\leq \frac{1}{\omega } f^{\sigma }(0)f(t),\text{ }\forall t\geq 0. \end{equation*} Hence

• (a) \(f(t)\leq f(0)e^{1-\omega t},\) for all \(t\geq 0,\) whenever \(\sigma =0,\)
• (b) \(f(t)\leq f(0)\left( \frac{1+\sigma }{1+\omega \sigma t}\right) ^{\frac{1 }{\sigma }},\) for all \(t\geq 0,\) whenever \(\sigma >0.\)

3. Main results

Now as in ([4]), we introduce the follows sets: \begin{eqnarray*} \mathcal{W}_{1} &=&\{u\in X_{0}:J(u)< d\},\text{ }\mathcal{W}_{2}=\{u\in X_{0}:J(u)=d\},\text{ }\mathcal{W}=\mathcal{W}_{1}\cup \mathcal{W}_{2}, \\ \mathcal{W}_{1}^{+} &=&\{u\in \mathcal{W}_{1}:I(u)>0\},\text{ }\mathcal{W} _{2}^{+}=\{u\in \mathcal{W}_{2}:I(u)>0\},\text{ }\mathcal{W}^{+}=\mathcal{W }_{1}^{+}\cup \mathcal{W}_{2}^{+}, \\ \mathcal{W}_{1}^{-} &=&\{u\in \mathcal{W}_{1}:I(u)< 0\},\text{ }\mathcal{W} _{2}^{-}=\{u\in \mathcal{W}_{2}:I(u)< 0\},\text{ }\mathcal{W}^{-}=\mathcal{W }_{1}^{-}\cup \mathcal{W}_{2}^{-}. \end{eqnarray*}

Definition 1. (Maximal Existence Time). Assume that \(u\) be weak solutions of problem (1). We define the maximal existence time \(T_{\max }\) as follows \begin{equation*} T_{\max }=\sup \{T>0:u(t)\text{ exists on }[0,T]\}. \end{equation*} Then

• (i) If \(T_{\max }< \infty ,\) we say that \(u\) blows up in finite time and \( T_{\max }\) is the blow-up time;
• (ii) If \(T_{\max }=\infty ,\) we say that \(u\) is global.

Definition 2. (Weak solution). We define a function \(u\in L^{\infty }(0,T;X_{0})\) with \( u_{t}\in L^{2}(0,T;H_{0}^{1}(\Omega ))\) to be a weak solution of problem (1) over \([0,T],\) if it satisfies the initial condition \( u(0)=u_{0}\in X_{0},\) and \begin{equation*} \left\langle u_{t},w\right\rangle +\left\langle \left\vert \Delta u\right\vert ^{p-1},\Delta w\right\rangle +\left\langle \nabla u_{t},\nabla w\right\rangle=\int\nolimits_{\Omega }u\left\vert u\right\vert ^{q-2}\ln \left( \left\vert u\right\vert \right) wdx,\text{ } \end{equation*} for all \(w\in X_{0},\) and for a.e. \(t\in \lbrack 0,T].\)

Theorem 1. (Global Existence). Let \(u_{0}\in \) \(\mathcal{W}^{+},\) \(0< J(u_{0})< M\) and \( I(u)>0.\) Then there is a unique global weak solution \(u\) of (1) satisfying \(u(0)=u_{0}.\) We have \(u(t)\in \mathcal{W}^{+}\)holds for all \( 0\leq t< +\infty ,\) and the energy estimate \begin{equation*} \int\nolimits_{0}^{t}\left\Vert u_{s}(s)\right\Vert _{H_{0}^{1}(\Omega )}^{2}ds+J(u(t))\leq J(u_{0}),\text{ }0\leq t\leq +\infty . \end{equation*} Also, the solution decay polynomially provided \(u_{0}\in \mathcal{W} _{1}^{+}. \)

Proof. The Faedo-Galerkin's methods is used. In the space \(W_{0}^{1,p}\left( \Omega \right) \cap W^{2,p}\left( \Omega \right) ,\) we take a bases \( \{w_{j}\}_{j=1}^{\infty }\) and define the finite orthogonal space \begin{equation*} V_{m}=span\{w_{1},w_{2},...,w_{m}\}. \end{equation*} Let \(u_{0m}\) be an element of \(V_{m}\) such that

as \(m\to \infty .\) We construct the following approximate solution \( u_{m}(x,t)\) of the problem (1) where the coefficients \(a_{mj}\) \((1\leq j\leq m)\) satisfy the ordinary differential equations for \(i\in \{1,2,...,m\},\) with the initial condition We multiply both sides of (11) by \(a_{mi}^{\prime },\) sum for \( i=1,...,m\) and integrating with respect to time variable on \([0,t],\) we get where \(T_{\max }\) is the maximal existence time of solution \(u_{m}(t).\) We shall prove that \(T_{\max }=+\infty .\) From (9), (13) and the continuity of \(J\), we obtain Thanks to \(J(u_{0})< d\) and the continuity of functional \(J,\) it follows from (14) that \begin{equation*} J(u_{0m})< d,\text{ for sufficiently large }m. \end{equation*} And therefore, from (13), we obtain for sufficiently large \(m.\) Next, we will study for sufficiently large \(m.\) We assume that (16) does not process and think that there exists a sufficiently small time \(t_{0}\) such that \( u_{m}(t_{0})\notin \mathcal{W}_{1}^{+}.\) Then, by continuity of \( u_{m}(t_{0})\in \partial \mathcal{W}_{1}^{+}.\) So, we get and Nevertheless, by definition of \(d,\) we see that (17) could not consist by (15) while if (18) holds then, we get \begin{equation*} J(u_{m}(t_{0}))\geq \underset{u\in \mathcal{N}}{\inf }J(u)=d, \end{equation*} which also contradicts with (15). Moreover, we have (16), i.e., \(J(u_{m}(t))< d,\) and \(I(u_{m}(t))>0,\) for any \(t\in \lbrack 0,T_{\max }),\) for sufficiently large \(m.\) Then, from (5), we obtain \begin{eqnarray*} d &>&J(u_{m}(t)) \\ &=&\frac{1}{q}I(u_{m})+\left( \frac{1}{p}-\frac{1}{q}\right) \left\Vert \Delta u_{m}\right\Vert _{p}^{p}+\frac{1}{q^{2}}\left\Vert u_{m}\right\Vert _{q}^{q} \\ &\geq &\left( \frac{1}{p}-\frac{1}{q}\right) \left\Vert \Delta u_{m}\right\Vert _{p}^{p}+\frac{1}{q^{2}}\left\Vert u_{m}\right\Vert _{q}^{q}, \end{eqnarray*} which gives and Since \(u_{m}(x,t)\in \mathcal{W}_{1}^{+}\) for \(m\) large enough, it follows from (5) that \(J(u_{m})\geq 0\) for \(s\) large enough. So, by (15 ) it follows for \(m\) large enough By (20), we know that \begin{equation*} T_{\max }=+\infty . \end{equation*} It follows from (19) and (21) that there exist a function \( X_{0}\) and a subsequence of \(\{u_{m}\}_{j=1}^{\infty }\) is indicated by \( \{u_{m}\}_{j=1}^{\infty }\) such that \begin{equation*} \left\vert \Delta u\right\vert ^{p-2}\Delta u\to \chi \text{ weakly in }L^{\infty }\left( 0,\infty ;W^{-2,p^{\prime }}\left( \Omega \right) \right). \end{equation*} By (22), (23) and Aubin-Lions compactness theorem, we obtain \begin{equation*} u_{m}\to u\text{ strongly in }C([0,+\infty ];L^{2}(\Omega )). \end{equation*} This yields that Moreover, since \begin{equation*} \alpha ^{r-1}\ln \alpha =-(e\left( r-1\right) )^{-1}\text{ for }\alpha >1, \end{equation*} and \begin{equation*} \ln \alpha =2\ln \left( \alpha ^{\frac{1}{2}}\right) \leq 2\alpha ^{\frac{1}{ 2}}\text{ for }\alpha >0. \end{equation*} By (19), we have where \begin{equation*} \Omega _{1}=\{x\in \Omega :\left\vert u_{m}(t)\right\vert \leq 1\},\text{ and }\Omega _{2}=\{x\in \Omega :\left\vert u_{m}(t)\right\vert \geq 1\}. \end{equation*} Hence, it follows from (24) and (25) that \begin{equation*} u_{m}\left\vert u_{m}\right\vert ^{q-2}\ln \left\vert u_{m}\right\vert \to u\left\vert u\right\vert ^{q-2}\ln \left\vert u\right\vert \text{ weakly* in }L^{\infty }(0,+\infty ;L^{\frac{2q}{2q-1}}(\Omega ))\text{ }. \end{equation*} Then integrating (11) respect to \(t\) for \(0\leq t< \infty ,\) we obtain \begin{equation*} \left\langle u_{t},w\right\rangle +\left\langle \chi (t),\Delta w\right\rangle +\left\langle \nabla u_{t},\nabla w\right\rangle =\int\nolimits_{\Omega }u\left\vert u\right\vert ^{p-2}\ln \left( \left\vert u\right\vert \right) wdx, \end{equation*} for all \(w\in W_{0}^{2,p}\left( \Omega \right) \) and for almost every \(t\in \left[ 0,\infty \right] .\) Finally, well known arguments of the theory of monotone operators implied \begin{equation*} \chi =\left\vert \Delta u\right\vert ^{p-2}\Delta u, \end{equation*} which yields \begin{equation*} \left\langle u_{t},w\right\rangle +\left\langle \left\vert \Delta u\right\vert ^{p-1},\Delta w\right\rangle +\left\langle \nabla u_{t},\nabla w\right\rangle =\int\nolimits_{\Omega }u\left\vert u\right\vert ^{p-2}\ln \left\vert u\right\vert wdx. \end{equation*} for all \(w\in W_{0}^{2,p}\left( \Omega \right) \) and for a.e. \(t\in \left[ 0,\infty \right] .\)

Finally, we discuss the decay results.

Thanks to \(u(t)\in \mathcal{W}_{1}^{+},\) we deduce from (13) that

\begin{equation*} \left( \frac{1}{p}-\frac{1}{q}\right) \left\Vert \Delta u\right\Vert _{p}^{p}+\frac{1}{q^{2}}\left\Vert u\right\Vert _{q}^{q}\leq J(u(t))\leq J(u_{0}),\text{ }t\in \lbrack 0,T]. \end{equation*} By using (5) and Proposition 1, we put \(p\left( \frac{J(u_{0})}{M} \right) ^{\frac{p}{n}}< \mu < p,\) we know \begin{eqnarray*} I(u(t)) &\geq &\left( 1-\frac{\mu }{p}\right) \left\Vert \Delta u\right\Vert _{p}^{p}+\left( \frac{n}{p^{2}}\ln \left( \frac{p\mu e}{n\mathcal{L} _{p}} \right) -\ln \left\Vert u(t)\right\Vert _{p}\right) \left\Vert u(t)\right\Vert _{p}^{p} \\ &\geq &\left( 1-\frac{\mu }{p}\right) \left\Vert \Delta u\right\Vert _{p}^{p}+\frac{1}{p}\ln \left( \frac{M}{J(u_{0})}\left( \frac{\mu }{p} \right) ^{\frac{n}{p}}\right) \left\Vert u(t)\right\Vert _{p}^{p} \\ &=&C_{1}\left\Vert u(t)\right\Vert _{W^{2,p}\left( \Omega \right) }^{p}. \end{eqnarray*} Integrating the \(I(u(s))\) with respect to \(s\) over \((t,T)\), we obtain where \(C_{2}\) stand by the best constant in the embedding \(W^{2,p}\left( \Omega \right) \hookrightarrow \to H_{0}^{1}\left( \Omega \right) \) From (26), we have Let \(T\to +\infty \) in (27), we can get \begin{equation*} \int\nolimits_{t}^{\infty }\left\Vert u(t)\right\Vert _{W^{2,p}\left( \Omega \right) }^{p}ds\leq \frac{1}{\omega }\left\Vert u(t)\right\Vert _{W^{2,p}\left( \Omega \right) }^{2}. \end{equation*} From Lemma 5, we have \(f(t)=\left\Vert u(t)\right\Vert _{W^{2,p}\left( \Omega \right) }^{2},\) \(\sigma =\frac{p}{2}-1,\) \(f(0)=1\) \begin{equation*} \left\Vert u(t)\right\Vert _{W^{2,p}\left( \Omega \right) }\leq \left\Vert u_{0}\right\Vert _{W^{2,p}\left( \Omega \right) }\left( \frac{p}{2+\omega \left\Vert u_{0}\right\Vert _{W_{0}^{2,p}\left( \Omega \right) }^{p-2}\left( p-2\right) t}\right) ^{\frac{1}{p-2}},\text{ }t\geq 0. \end{equation*} The above inequality implies that the solution \(u\) decays polynomially.

Acknowledgments :

The author would like to thank Prof. Charles N. Moore of Washington State University, USA for his valuable suggestions on this article.

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 declares no conflict of interest.''

Data Availability:

All data required for this research is included within this paper.

Funding Information:

No funding is available for this research.

1. Introduction

The Henstock integral of real-valued functions was first defined by Henstock[1,2]. It is a direct generalization of the Riemann integral because it uses the concept of the tagged partition and the Riemann sum. In Henstock integral, the concept of the norm of a tagged partition in the Riemann integral is replaced by the positive gauge function. Therefore, the definition of the Henstock integral is as simple as the definition of the Riemann integral. On the other hand, to introduce the Lebesgue integral, a good amount of measure theory is required. It is one of the reasons why the Henstock integral is simpler than the Lebesgue integral. However, the Henstock integral is also a generalization of the Lebesgue integral. Every Lebesgue integrable function is Henstock integrable, and both integrals are the same. One of the Lebesgue integral deficits is that not every continuous function that is differentiable everywhere, possible except for a countable number of points, is recovered from its derivative by the Lebesgue integrable. In this sense, we say that the Lebesgue integral does not recover a function from its derivative. On the other hand, the Henstock integral overcomes this drawback: every continuous function \(F:[a,b] \to R\) that is differentiable everywhere except for countable number of points on \([a,b]\) can be recovered from its derivative by the Henstock integral, and \(\int_a^x F' = F(x) - F(a)\).

The approximate Henstock integral (AP-Henstock integral) [3] further generalizes the Henstock integral by using the concept of the approximate derivative [4], and the gauge function in the Henstock integral is generalized to the choice in the AP-Henstock integral. Every Henstock integrable function is AP-Henstock integrable, and the integrals are the same. Furthermore, the AP-Henstock integral recovers an approximate continuous function from its approximate derivative:

Theorem 1.[4] If \(F:[a, b] \to R\) is approximately continuous on \([a,b]\) and approximately differentiable everywhere except for countable number of points on \([a,b]\), then the approximate derivative of \(F\), denoted by \(F'_{ap}\), is AP-Henstock integrable, and \(\int_a^x F'_{ap} = F(x) - F(a)\) for any \(x\in [a,b]\).

For the detailed introduction of the AP-Henstock integral, the reader is referred to [5,6,7,8,9].

Although the space of integrable functions is closed under the addition and the scalar multiplication, the product of two integrable functions is not necessarily integrable. Therefore, it is an important question of what kind of properties of an integrable function guarantees the integrability of the product of two integrable functions. We call those properties the multiplier properties and the related theorems the multiplier theorems. For example, in the case of the Lebesgue, Denjoy, and Henstock integral, the product of an integrable function \(f: [a, b ]\to \mathbb{R}\) and a function of bounded variation(which is integrable in any sense mentioned above) \(G: [a, b] \to \mathbb{R}\) is integrable, and

\begin{equation*} \int_a^b fG = F(b)G(b) - \int_a^b F \mathrm{d}G, \end{equation*} where \(F(x) = \int_a^x f\), and the last integral is the Riemann-Stieltjes integral. For the Perron integral, a restricted condition on the function of bounded variation is required, see [6]. In this paper, we develop the same kind of multiplier properties that ensure the \(AP\)-Henstock integrability of the product of two \(AP\)-Henstock integrable functions.

On the other hand, the Henstock integral can be defined on unbounded intervals [10]. It is now known as Hake's theorem that there is no such thing as an ``improper integral'' for the Henstock integral. By modifying the definition of the Henstock integral on unbounded intervals, we define the \(AP\)-Henstock integrals on unbounded intervals. In this setting, we investigate some properties for the AP-Henstock integral and prove that, under some additional conditions, the product of an AP-integrable function and a function of bounded variation. Furthermore, the product of an AP-integrable function and a regulated function is AP-Henstock integrable.

We state the mean value theorem for the Riemann-Stieltjes integral which will be used in the main body of our work.

Theorem 2.[1] Let \(f\) be a continuous functions on \([a,b]\) and let \(\varphi\) be a bounded increasing function on \([a,b]\). Then there exists \(\xi\) in \([a,b]\) such that \begin{equation*} \int_a^b f d\varphi = f(\xi) (\varphi(b) - \varphi(a)). \end{equation*}

2. Definition and basic properties

An approximate neighborhood(or ap-nbd) of \( x \in [a,b]\) is a measurable set \( S_x \subset [a,b] \) containing \(x\) as a point of density. For every \( x \in E \subset [a,b] \), choose an ap-nbd \(S_x \subset [a,b] \) of \(x\). Then we say that \(\mathcal{S}= \{ S_x : x \in E \} \) is a choice on \(E\). A tagged interval \( ([u,v],x )\) is said to fine to the choice \(\mathcal{S}=\{S_x\}\) if \(u,v\in S_{x}\). Let \(\mathcal{P}=\{([x_{i-1},x_{i}],t_{i}):1 \leq i \leq n \}\) be a finite collection of non-overlapping tagged intervals. If \(([x_{i-1},x_{i}],t_i)\) is fine to the choice \(\mathcal{S}\) for each \(i=1,\cdots, n\), then we say that \(\mathcal{P}\) is \(\mathcal{S}\)-fine. Let \(E \subset [a,b]\). If \(\mathcal{P}\) is \(\mathcal{S}\)-fine and \(t_{i} \in E\) for each \(i=1,\cdots, n\), then \(\mathcal{P}\) is said to be \(\mathcal{S}\)-fine on \(E\). If \(\mathcal{P}\) is \(\mathcal{S}\)-fine and \([a,b]=\cup_{i=1}^{n} [x_{i-1}, x_{i}]\), then we say that \({\mathcal{P}} \) is \(\mathcal{S}\)-fine partition of \([a,b]\).

Definition 1.[4] A function \(f : [a,b] \to \mathbb{R}\) is said to be approximate Henstock integrable (AP-Henstock integrable) on \([a,b]\) if there exists a real number \(A\) such that for each \( \epsilon > 0\) there is a choice \(\mathcal{S}\) on \([a,b]\) such that \[ \left| \sum_{i=1}^n f( t_i )(x_i - x_{i-1} ) - A \right| < \epsilon, \] for each \(\mathcal{S}\)-fine partition \(\mathcal{P}=\{([x_{i-1},x_{i}],t_{i}):1 \leq i \leq n \}\) of \([a,b]\). In this case, \(A\) is called the AP-Henstock integral of \(f\) on \([a,b]\), and we write \(A= \int_a^b f.\) We denote \(\sum_{i=1}^n f(t_i)(x_i - x_{i-1})=S(\mathcal{P}; f)\) and the collection of all functions that are AP-Henstock integrable on an interval \(I\) by \(AH(I)\).

Theorem 3.[4] Let \(f\) and \(g\) be AP-Henstock integrable functions on \([a,b]\). then for any real numbers \(\alpha\) and \(\beta\), \(\alpha f\) + \(\beta g\) is AP-Henstock integrable on \([a,b]\) and \( \int_a^b ( \alpha f + \beta g ) = \alpha \int_a^b f + \beta \int_a^b g \).

Theorem 4. Let \(f:[a,b] \to \mathbb{R}\). If \(f=0\) almost everywhere on \([a,b]\), then \(f\in AH([a,b])\) and \(\int_a^b f = 0.\)

Proof. The facts that \(f\) is Henstock integrable on \([a,b]\) and the integral is \(0\) are proved in [6]. Since every Henstock integrable function is \(AP\)-Henstock integrable and the integrals are the same, \(f\in AH([a,b])\) and \(\int_a^b f = 0.\)

Corollary 1. Let \(f\in AH([a,b])\) and \(g : [a,b] \to \mathbb{R}\). If \(g = f\) almost everywhere on \([a,b]\), then \(g\in AH([a,b])\) and \(\int_a^b g = \int_a^b f\).

Proof. Since \(g-f = 0\) almost everywhere on \(I\), by the theorem, \(g-f \in AH([a,b])\) and \(\int_a^b (g-f) = 0\). Therefore, \(g = (g -f) + f \in AH(I)\) and \(\int _a^b g = \int_a^b (g-f) + \int_a^b f = \int_a^b f.\)

Theorem 5.[4] Let \(f:[a,b] \to \mathbb{R}\) be \(AP\)-Henstock integrable on \([a,b]\) and let \(F(x) = \int_a^x f\) for each \(x \in [a,b]\). Then

1. the function \(F\) is measurable;
2. the function \(F\) is approximately continuous on \([a,b]\);
3. the function \(F\) is approximately differentiable almost everywhere on \([a,b]\) and \(F'_{ap} = f\) almost everywhere on \([a,b]\); and
4. the function f is measurable.

3. Integral of the translate of a function

In this section, we prove that the translate of an AP-Henstock integrable function is AP-Henstock integrable.

Let \(I:= [a,b]\) and let \(r \in \mathbb{R}\). We define the \(r\)-additive translate of \(I\) to be the interval \(I_r := [a+r, b+r]\), and the \(r\)-additive translate of \(f\) to be the function \(f_r (y): = f(y-r)\) for all \(y \in I_r\). Similarly, if \(r>0\), we define the \(r\)-multiplicative translate of \(I\) to be the interval \(I_{(r)} := [ar, br]\), and the \(r\)-multiplicative translate of \(f\) to be the function \(f_{(r)} (z) := f(z/r ) \) for all \(z \in I_{(r)}\).

Theorem 6.

• (a) If \(f\) is AP-Henstock integrable on \(I\), then \(f_r\) is AP-Henstock integrable on \(I_r\) and \(\int_{I_r } f_r = \int_I f\).
• (b) If \(f\) is AP-Henstock integrable on \(I\), then \(f_{(r)}\) is AP-Henstock integrable on \(I_{(r)}\) and \( \int_{I_{(r)}} f_{(r)} = r \int_I f\).

Proof.

• (a) Let \(\epsilon > 0\). Since \(f \in AH(I)\), there exist a choice \(\mathcal{S} = \{ S_x : x \in I \} \) on \(I\) such that if \( \mathcal{P}_1\) is a \(\mathcal{S}\)-fine partition of \(I\), then \( | S(f; \mathcal{P}_1 ) - \int_I f | \leq \epsilon\). Now, we define \( \mathcal{S}_{\epsilon} : = \{S_{y-r} + r : S_{y-r} \in \mathcal{S}, ~y \in I_r \} \). Suppose that \(\mathcal{Q}: = \{ ([ y_{i-1} , y_i] , s_i ) \}_{i=1}^n \) is a \(\mathcal{S}_{\epsilon}\)-fine partition of \(I_r \). If we let \(x_i := y_i -r \) and \(t_i:= s_i - r\), then \(x_{i-1} \leq t_i \leq x_i\), \(x_{i-1} , x_i \in S_{t_i}\), and \(t_i \in I\), whence \(\mathcal{P}: = \left\{([x_{i-1} , x_i], t_i) \right\}_{i=1}^n\) is a \(\mathcal{S}\)-fine partition of \(I\). Since \(S( f_r ;\mathcal{Q}) = S(f; \mathcal{P})\), we infer that \[ \left| S(f_r ; \mathcal{Q}) - \int_I f \right| = \left| S(f;\mathcal{P}) - \int_I f \right| \leq \epsilon. \] Since \(\epsilon > 0\) is arbitrary, we conclude that \( f_r \in AH(I_r ) \) and \( \int_{I_r } f_r = \int_I f.\)
• (b) Let \(\epsilon > 0\). Since \(f \in AH(I)\), there exists a choice \(\mathcal{S} = \{ S_x : x \in I \} \) of \(I\) such that if \( \mathcal{P}_1 \) is a \(\mathcal{S}\)-fine partition on \(I\), then \(|S(f; \mathcal{P}_1 ) - \int_I f | \leq {\epsilon}/r \). Now, we define \( \mathcal{S}_{\epsilon} : = \{ rS_{y/r} : S_{y/r} \in \mathcal{S}, ~y \in I_{(r)} \} \), then \(\mathcal{S}_{\epsilon} \) is a choice on \(I_{(r)}\). Suppose that \(\mathcal{Q}: = \{ ([ y_{i-1} , y_i] , s_i ) \}_{i=1}^n \) is a \(S_{\epsilon}\)-fine partition of \(I_{(r)}\). If we let \(x_i := y_i/r \) and \(t_i:= {s_i}/r\), then \(x_{i-1} \leq t_i \leq x_i\), \( x_{i-1} , x_i \in S_{t_i}\), and \(t_i \in I\), whence \(\mathcal{P}: = \{([x_{i-1} , x_i ], t_i ) \}_{i=1}^n \) is a \(\mathcal{S}\)-fine partition of \(I\). Since \(S( f_{(r)} ;\mathcal{Q}) =\sum_{i=1}^n f_{(r)}(s_i)(y_{i}-y_{i-1}) =r \sum_{i=1}^n f(t_i)(x_i - x_{i-1}) = r S(f; \mathcal{P})\), we have \[ \left| S(f_{(r)} ; \mathcal{Q}) - r \int_I f \right| = r \left| S(f;\mathcal{P}) - \int_I f\right| \leq \epsilon. \] Since \(\epsilon > 0\) is arbitrary, we conclude that \( f_{(r)} \in AH(I_{(r)} ) \) and \(\int_{I_{(r)} } f_{(r)} = r\int_I f.\)

4. Multiplier Properties on bounded intervals

It is well known that the product of two AP-Henstock integrable functions is not necessarily AP-Henstock integrable, even when one of them is bounded or continuous. In this section, we provide some conditions under which the product of two AP-Henstock integrable functions is AP-Henstock integrable. We also establish the Mean Value Theorems. We start by showing the squeeze theorem for the AP-Henstock integral.

Theorem 7. A function \(f\) belongs to \(AH(I:=[a,b])\) if and only if for every \( \epsilon > 0\) there exist functions \(\varphi_{\epsilon} \) and \( \psi_{\epsilon} \) in \(AH(I)\) with \(\varphi_{\epsilon} (x) \leq f(x) \leq \psi_{\epsilon} (x) \) for all \(x \in I \), and such that \[ \int_I (\psi_{\epsilon} - \varphi_{\epsilon} ) \leq \epsilon. \]

Proof. Let \(f \in AH(I)\) and let \(\epsilon > 0 \). We can take \(\varphi_{\epsilon}: = \psi_{\epsilon}: = f \). Conversely, assume that for every \(\epsilon>0\) there exist functions \(\varphi_{\epsilon}\) and \(\psi_{\epsilon}\) in \(AH(I)\) with \(\varphi_{\epsilon} (x) \leq f(x) \leq \psi_{\epsilon} (x) \) for all \(x \in I \), and such that \(\int_I (\psi_{\epsilon} - \varphi_{\epsilon} ) \leq \epsilon\). Then for each \(\epsilon > 0 \), it follows that \( S(\varphi_{\epsilon}; \mathcal{P}) \leq S(f; \mathcal{P}) \leq S(\psi_{\epsilon}; \mathcal{P})\) for any tagged partition \(\mathcal{P}\) of \(I\). Since \(\varphi_{\epsilon} \in AH(I)\), there exists a choice \(\mathcal{S}_1 \) on \(I\) such that if \(\mathcal{P} \) is a \(\mathcal{S}_1\)-fine partition of \(I\), then \(| S(\varphi_{\epsilon}; \mathcal{P}) - \int_I \varphi_{\epsilon} | \leq \epsilon\), whence it follows that \(\int_I \varphi_{\epsilon} - {\epsilon} \leq S(\varphi_{\epsilon}; \mathcal{P})\). Similarly there exists a choice \(\mathcal{S}_2 \) on \(I\) such that if \(\mathcal{P} \) is a \(\mathcal{S}_2\)-fine partition of \(I\), then \( S(\psi_{\epsilon}; \mathcal{P}) \leq \int_I \psi_{\epsilon} + \epsilon \). Now let \( \mathcal{S}: = \{ S_1 \cap S_2 : S_1\in \mathcal{S}_1, S_2 \in \mathcal{S}_2 \}\) and let \( \mathcal{P} \) be a \(\mathcal{S}\)-fine tagged partition of \(I\). Then we have \[ \int_I \varphi_{\epsilon} - \epsilon \leq S(f; \mathcal{P}) \leq \int_I \psi_{\epsilon} + \epsilon, \] and if \(\mathcal{Q} \) is a \(\mathcal{S}\)-fine partition of \(I\), then \[ - \int_I \psi_{\epsilon} - \epsilon \leq - S(f; \mathcal{Q}) \leq - \int_I \varphi_{\epsilon} + \epsilon. \] Adding these inequalities, we obtain \[ - \int_{I} (\psi_{\epsilon} - \varphi_{\epsilon}) - 2\epsilon \leq S(f; \mathcal{P}) - S(f; \mathcal{Q}) \leq \int_{I} (\psi_{\epsilon} - \varphi_{\epsilon}) + 2\epsilon. \] Hence we conclude that \[ \mid S(f; \mathcal{P}) - S(f; \mathcal{Q}) \mid \leq \int_{I} (\psi_{\epsilon} - \varphi_{\epsilon}) + 2\epsilon \leq 3\epsilon. \] Since \(\epsilon > 0\) is arbitrary, \(f\) satisfies the Cauchy Criterion([4], Theorem 16.6) for the AP-Henstock integral. Therefore \(f\in AH(I)\).

Definition 2.[1] Let \(I:= [a,b]\). A function \(f : I \to \mathbb{R}\) is said to be regulated on \(I\) if for every \( \epsilon > 0\) there exists a step function \( s_{\epsilon} : I \to \mathbb{R}\) such that \[ \mid f(x) - s_{\epsilon} (x) \mid \leq \epsilon \] for all \( x \in I \).

It easy to see that a function \(f\) is regulated on \(I\) if and only if there is a sequence \(\left\{ s_n \right\}_{n=1}^{\infty} \) of step functions on \(I\) that converges uniformly to \(f\) on \(I\).

Theorem 8. Let \(I:= [a,b]\). If \(f : I \to \mathbb{R}\) is regulated on \(I\), then \(f\) is AP-Henstock integrable on \(I\).

Proof. Let \(f : I \to \mathbb{R}\) be regulated on \(I\) and let \(\epsilon > 0\). Then there exists a step function \( s_{\epsilon} : I \to \mathbb{R}\) such that \[ \mid f(x) - s_{\epsilon} (x) \mid \leq \epsilon \] for all \( x \in I \). Therefore, we have \( s_{\epsilon} (x) - \epsilon \leq f(x) \leq s_{\epsilon}(x) + \epsilon \) for all \(x \in [a,b]\). If we let \( \varphi_{\epsilon} (x):= s_{\epsilon} (x) - \epsilon \) and \(\psi_{\epsilon} (x) : = s_{\epsilon} (x) + \epsilon\) for all \(x \in I\), then the functions \(\varphi_{\epsilon}\) and \( \psi_{\epsilon}\) are AP-Henstock integrable on \(I\) and \(\varphi_{\epsilon} (x) \leq f(x) \leq \psi_{\epsilon} (x) \) for \(x \in I \). Moreover, since \[ \int_I ( \psi_{\epsilon} - \varphi_{\epsilon} ) \leq 2(b-a) \epsilon , \] it follows from Theorem 7 that \(f\) is AP-Henstock integrable on \(I\).

Theorem 9. Let \(f\in AH(I:=[a,b])\) be bounded below and \(g\) be regulated on \(I\). Then the product \(fg \) belongs to \(AH(I)\).

Proof. Assume that \(f(x) \geq 0\) for \( x \in I\). It is clear that if \(s\) is a step function, then \(sf\) belongs to \(AH(I)\). Let \( A > \int_I f \geq 0\) and let \( \epsilon > 0\). Since \(g\) is regulated on \(I\), there exists a step function \(s_{\epsilon} \) on \(I\) such that \(| g(x) - s_{\epsilon}(x) | \leq \frac{\epsilon}{2A}\) for all \(x \in I\). Now, we define \(\varphi_{\epsilon} (x) := f(x) \left( s_{\epsilon}(x) - \frac{\epsilon}{2A} \right)\) and \(\psi_{\epsilon} := f(x) \left( s_{\epsilon} (x) + \frac{\epsilon}{2A} \right) \) for all \(x \in I\), then \(\varphi_{\epsilon}\), \(\psi_{\epsilon} \in AH(I)\) and it follows that \(\varphi_{\epsilon}(x) \leq f(x) g(x) \leq \psi_{\epsilon} (x)\) for all \(x \in I\), and that \[ \int_I ( \psi_{\epsilon} - \varphi_{\epsilon} ) = \frac{\epsilon}{A} \int_I f \leq \epsilon. \] Therefore, it follows from Theorem 7 that \(fg\in AH(I)\). Now, let \(f(x) \geq M\) on \(I\). Then by writing \(fg = (f-M)g + M g\), we see that \(fg\in AH(I)\).

Theorem 10. Let \(f\) and \(|f|\) be AP-Henstock integrable on \(I:=[a,b]\) and let \(g\) be a bounded, measurable function on \(I\). Then the product \(fg \) is AP-Henstock integrable on \(I\).

Proof. Let \( h: = fg\). Since \(h\) is measurable, there exists a sequence \(\left\{s_n \right\}_{n=1}^{\infty}\) of step functions such that \(\left\{s_n \right\}_{n=1}^{\infty}\) converges to \(h\) almost everywhere on \(I\). Let \( \overline{s_n }\) be the middle function of \(-M|f|, ~s_n\), and \(M |f|\). Then \(-M|f| \leq \overline{s_n } \leq M |f|\) and \(\left\{\overline{s_n }\right\}\) converges to \(h\) almost everywhere on \(I\). Therefore, it follows from the Dominated Convergence Theorem for the AP-Henstock integral [9] that \(h\in AH(I)\).

Theorem 11. Let \(I:=[a,b]\), \(f\in AH(I)\), \(\varphi\) is of bounded variation on \(I\), and \(F(x) := \int_a^x f\) on \(I\). If \(F\) is Riemann-Stieltjes integrable with respect to \( \varphi\) on \(I\), then the product \(f \varphi \) belongs to \(AH(I)\) and \begin{equation*} \int_I f \varphi = \int_I \varphi dF = F(b) \varphi (b) - \int_I F d \varphi, \end{equation*} where the second and third integrals are the Riemann-Stieltjes integrals.

Proof. Since \(F\) is Riemann-Stieltjes integrable with respect to \(\varphi\), the third integral exists, and the existence of the second integral and the validity of the second equality follows from the well-known integration by part formula for the Riemann-Stieljes integral ([4], Theorem 12.14). Therefore, we only need to show the first equality. To this end, let \(\epsilon > 0\). Since \(\varphi\) is Riemann-Stieltjes integrable with respect to \(F\), there exist \(\delta > 0\) such that if \( \mathcal{P} = \{ ([x_{i-1} , x_i ], t_i ) \}_{i=1}^n\) is any tagged partition of \(I\) with norm less than \(2\delta\), then \[ \left\lvert \sum_{i=1}^n \varphi (t_i ) (F(x_i ) - F(x_{i-1}) ) - \int_I \varphi dF \right\rvert \leq \epsilon . \] Let \(|\varphi(x)| \leq M \) for all \(x \in I\). Since \(f \in AH(I)\), there exist a choice \(\mathcal{S}\) on \(I\) such that if \( \mathcal{P} = \{ ([x_{i-1} , x_i ], t_i ) \}_{i=1}^n\) is \(\mathcal{S}\)-fine partition of \(I\), then \[ \left\lvert \sum_{i=1}^n \left\{ f(t_i ) ( x_i - x_{i-1}) -( F(x_i ) - F(x_{i-1})) \right\} \right\rvert \leq {\epsilon}/{2M}, \] and it follows from the Saks-Henstock Lemma for the AP-Henstock integral that \begin{equation*} \sum_{i=1}^n \left| f(t_i ) ( x_i - x_{i-1}) -(F(x_i) - F(x_{i-1} ))\right| \leq {\epsilon}/M. \end{equation*} Define \( \mathcal{S}': = \{S_x \cap (x-\delta, x+ \delta) : S_x \in \mathcal{S} \}\) and let \( \mathcal{P} = \{ ([x_{i-1} , x_i ], t_i ) \}_{i=1}^n\) be \(\mathcal{S}'\)-fine partition of \(I\). Then, \begin{align} &\left\lvert \sum_{i=1}^n f( t_i ) \varphi (t_i ) (x_i - x_{i-1}) - \int_I \varphi dF \right\rvert \nonumber\\ &~~\leq \left\vert \sum_{i=1}^n f( t_i ) \varphi (t_i ) (x_i - x_{i-1}) - \sum_{i=1}^n \varphi (t_i ) \left(F(x_i ) - F(x_{i-1} ) \right) \right\vert+ \left\vert \sum_{i=1}^n \varphi (t_i ) \left(F(x_i ) - F(x_{i-1})\right) - \int_I \varphi dF\right\vert \nonumber\end{align}\begin{align} &~~\leq M \sum_{i=1}^n \left\lvert f(t_i) (x_i - x_{i-1}) - \left( F(x_i) - F(x_{i-1}) \right) \right\rvert + \epsilon \leq 2\epsilon. \nonumber \end{align} Since \(\epsilon > 0 \) is arbitrary, \(f \varphi \in AH(I) \) and \(\int_I f\varphi = \int_I \varphi dF\).

Note that the indefinite AP-Henstock integral is only approximately continuous, not necessarily continuous on \(I\). It is shown in [6] (Exercise 12.10) that if two bounded functions \(F\) and \(\varphi\) on \(I\) share a common point of discontinuity in \(I\), then \(F\) is not Rieman-Stieltjes integrable with respect to \(\varphi\) on \(I\). Therefore, the condition in the above theorem that \(F\) is Reimann-Stiltjes integrable with respect to \(\varphi\) cannot be removed. On the other hand, it is well-known fact that if \(F\) is continuous and \(\varphi\) is of bounded variation on \(I\), then \(F\) is Riemann-Stieltjes integrable with respect to \(\varphi\) [6]. Therefore, the following corollary follows.

Corollary 2. Let \(f\) be AP-Henstock integrable on \(I:= [a,b]\), \(\varphi\) be bounded variation on \(I\), and \( F(x) := \int_a^x f\) on \(I\). If \(F\) is continuous on \(I\), then \(f \varphi \in AH(I)\) and \begin{equation*} \int_I f \varphi = \int_I \varphi dF = F(b) \varphi (b) - \int_I F d \varphi, \end{equation*} where the second and third integrals are the Riemann-Stieltjes integrals.

In addition to the multiplier theorem above, we provide a version of integration by parts theorem.

Theorem 12. Let \(I:=[a,b]\) and let \(F, G : I \to \mathbb{R}\) be approximately continuous on I. If \(f, g \in AH(I)\), and \(F'_{ap} = f\), \(G'_{ap}=g\) except for countably many points in \(I\), then \(Fg + fG \in AH(I)\) and \begin{equation*} \int_I (Fg + fG) = F(b)G(b) - F(a)G(a). \end{equation*} Moreover, \(Fg \in AH(I)\) if and only if \(fG \in AH(I)\), in which case \begin{equation*} \int_I Fg = F(b)G(b) - F(a)G(a) - \int_I fG. \end{equation*}

Proof. By the hypothesis, there exist countable sets \(C_f\) and \(C_g\) such that \(F'_{ap}(x) = f(x)\) for \(x \in I - C_f\) and \(G'_{ap}(x) = g(x)\) for \(x\in I - C_g\). Let \(C := C_f \cup C_g\) be a countable set. For \(x \in I - C\), \((FG)'_{ap} = F'_{ap}G + F G'_{ap} = fG + Fg\). Also, by the hypothesis, \(FG\) is approximately continous on \(I\). Therefore, by Theorem 1, \((FG)'_{ap} \in AH(I)\) and \(\int_I (FG)'_{ap} = F(g)G(g) - F(a)G(a)\). Since \(C\) is a countable set, \(fG + Fg \in AH(I)\) and \(\int_I fG + Fg = \int_I (FG)' _{ap}\). Moreover, if \(Fg\in AH(I),\) since \(fG = (Fg + fG) - Fg\), \(fG\in AH(I)\).

We now establish the Mean Value Theorems for the AP-Henstock integral.

Theorem 13.(First Mean Value Theorem). If \(f\) is continuous on \(I:=[a,b]\), and if \( p \in AH(I)\) does not change sign on \(I\), then there exists \( \xi \in I\) such that \[ \int_I fp = f( \xi) \int_I p. \]

Proof. Since \(f\) is continuous, \(f\) is bounded on \(I\). We invoke the fact that a nonnegative AP-Henstock integrable function is Lebesgue integrable and that the product of a Lebesgue integrable function and a bounded, Lebesgue integrable function is Lebesgue integrable. Therefore, \(p\) is Lebesgue integrable and \(fp\) is Lebesgue integrable on \(I\). If \(p \geq 0\), then \(mp \leq fp \leq Mp,\) where \(m:= \min \{ f(x) : x \in I \} \) and \(M: = \max \{f(x) : x \in I \},\) and \[ m \int_I p \leq \int_I fp \leq M \int_I p. \] If \( \int_I p = 0,\) then the result is trivial; if not, it follows from the Bolzano Intermediate Value Theorem in \(\mathbb{R}\). If \(p \leq 0\), then the argument is similar.

Theorem 14.(Second Mean Value Theorem). If \(f \in AH(I:=[a,b])\), \(F(x) = \int_a^x f \) is continuous on \(I\), and if \(g\) is monotone on \(I\), then there exists \( \xi \in I\) such that \[ \int_I fg = g(a) \int_a^{\xi} f + g(b) \int_{\xi}^b f. \]

Proof. Since \(g\) is of bounded variation, it follows from Theorem 11 that \( fg \in AH(I)\) and \[ \int_I fg = \int_I g dF = g(b)F(b) - g(a)F(a) -\int_I F dg, \] where the third integral is the Riemann-Stieltjes integral. Then, by Theorem 1.2, there exists \(\xi \in I\) such that the last two terms equal \begin{eqnarray*} g(b)F(b) - g(a)F(a) - F(\xi)(g(b) - g(a)) &=& g(a)( F(\xi) - F(a)) + g(b)(F(b) - F(\xi))\\ &=& g(a)\int_a^{\xi}f + g(b)\int_{\xi}^b f. \end{eqnarray*}

5. Multiplier Properties on unbounded intervals.

In this section, we define the \(AP\)-Henstock integral on unbounded intervals and investigate some properties of the integral including some multiplier properties.

We extend any function \(f: [a, \infty) \to \mathbb{R}\) to a function defined on \([a,\infty]\) in the extended real numbers \( \mathbb{R}^* := \mathbb{R} \cup \{\infty, - \infty \}\) by defining \(f(\infty) = 0\). We then take a tagged partition of the interval \([a, \infty]\) :

\[ \mathcal{P} : = \{([x_0 , x_1], t_i ) , \cdots , ([x_{n-1} , x_n ] , t_n), ([x_n , x_{n+1}], t_{n+1} ) \}, \] so that \(x_0 =a\) and \(x_{n+1} = \infty\). A choice \(\mathcal{S} = \left\{ S_x : x\in [a,\infty] \right\}\) on \([a, \infty]\) is a set of ap-nbd \(S_x \subset [a,\infty]\) that contains \(x\) as a point of density. We require that \(S_x\) is bounded for each \(x \in \mathbb{R}\) and \(S_{\infty} = [d, \infty]\) for some \(d>a\). We say that the tagged partition \(\mathcal{P}\) is \(\mathcal{S}\)-fine if \(x_{i-1}, x_{i} \in S_{t_i}\) for \(i = 1, \cdots, n+1\), and \(d\leq x_n\). Because \(S_x\) is bounded for \(x\in \mathbb{R}\), \(t_{n+1} = \infty\). Define \( 0\cdot \infty = 0\) so that the contribution of the final term in \(\mathcal{P}\) to the Riemann sum is \(f(\infty)\cdot \infty = 0\). Now, we give the definition of the \(AP\)-integral of a function \(f:[a,\infty] \to \mathbb{R}\).

Definition 3. A function \(f : [a,\infty] \to \mathbb{R}\) is AP-Henstock integrable on \([a,\infty)\), or on \([a,\infty]\) if there exists a real number \(A\) such that for each \( \epsilon > 0\) there is a choice \(\mathcal{S}\) on \([a,\infty]\) such that \[ \left| \sum_{i=1}^n f( \xi_i )(x_i - x_{i-1} ) - A \right| < \epsilon \] for each \(\mathcal{S}\)-fine partition \(\mathcal{P}=\{([x_{i-1},x_{i}],t_{i}):1 \leq i \leq n+1 \}\) of \([a,\infty]\). In this case, \(A\) is called the AP-Henstock integral of \(f\) on \([a,\infty)\) and we write \(A= \int_a^{\infty} f\).

The collection of all functions that are AP-Henstock integrable on an interval \([a,\infty)\) will be denoted by \(AH([a,\infty))\). The next theorem is the Cauchy Criterion for the \(AP\)-Henstock integral on unbounded intervals.

Theorem 15. Let \(I:=[a,\infty]\) and let \(f: I \to \mathbb{R}\). Then, \(f\in AH(I)\) if and only if for any \(\epsilon>0\) there exists a choice \(\mathcal{S}_{\epsilon}\) of \(I\) such that if \(\mathcal{P}\) and \(\mathcal{Q}\) are any partitions of \(I\) that are \(\mathcal{S}\)-fine, then \(|S(f;\mathcal{P}) - S(f;\mathcal{Q})|\leq \epsilon\).

Proof. Let \(f \in AH([a,\infty])\) with \(\tilde{A}:= \int_a^{\infty} f\). Let \(\tilde{\mathcal{S}}_{\epsilon}\) be a choice on \(I\) such that if \(\mathcal{P}, \mathcal{Q}\) are \(\tilde{\mathcal{S}}_{\epsilon}\)-fine partitions of \(I\), then \(|S(f;\mathcal{P}) - A| \leq \epsilon/2\) and \(|S(f;\mathcal{Q}) - A| \leq \epsilon/2\), which follows that \[|S(f;\mathcal{P}) - S(f;\mathcal{Q})| \leq |S(f;\mathcal{P}) - \tilde{A}| + | S(f;\mathcal{Q}) -\tilde{A}| \leq \epsilon.\] Now, suppose that for any \(\epsilon > 0\) there exists a choice \(\mathcal{S}_{\epsilon}\) on \(I\) such that if \(\mathcal{P}\) and \(\mathcal{Q}\) are any partitions of \(I\) that are \(\mathcal{S}_{\epsilon}\)-fine of \(I\), then \(|S(f;\mathcal{P}) - S(f;\mathcal{Q})| \leq \epsilon\). For each \(n\in \mathbb{N}\), let \(\mathcal{S}_n = \{S_{n,x} : x\in I\}\) be a choice on \(I\) such that if \(\mathcal{P}\) and \(\mathcal{Q}\) are \(\mathcal{S}_n\)-fine, then \(|S(f;\mathcal{P}) - S(f;\mathcal{Q})| \leq 1/n\). We may assume that \(S_{n+1,x} \subset S_{n,x}\) for all \(x\in I\), \(n \in \mathbb{N}\). For each \(n \in \mathbb{N}\), let \(\mathcal{P}_n\) be a \(\mathcal{S}_n\)-fine partition of \(I\). If \(m>n\), then \(\mathcal{P}_m\) and \(\mathcal{P}_n\) are \(\mathcal{S}_n\)-fine. Therefore, for \(m>n\), \(|S(f;\mathcal{P}_n) - S(f;\mathcal{P}_m) | \leq 1/n\), and it follows that \(\{S(f;\mathcal{P}_n)\}_{n=1}^{\infty}\) is a Cauchy sequence. Let \(A:=\lim_{n\to\infty} S(f;\mathcal{P}_n)\). By taking \(m\to \infty\), we have \(|S(f;\mathcal{P}_n) - A| \leq 1/n\). Now, for any given \(\epsilon>0\), let \(K \in \mathbb{N}\) be such that \(1/K \leq \epsilon/2\). If \(\mathcal{Q}\) be a \(\mathcal{S}_K\)-fine partition of \(I\), then \[|S(f;\mathcal{Q}) - A| \leq |S(f; \mathcal{Q}) - S(f; \mathcal{P}_K)| + |S(f;\mathcal{P}_K)-A|\leq 2/K \leq \epsilon.\]

The following theorem is the additive property of the \(AP\)-Henstock integral of a function on \([a,\infty]\).

Theorem 16. Let \(I := [a, \infty]\), \(f : I\to \mathbb{R}\) and let \(c >a\). Then \(f\in AH(I)\) if and only if the restriction of \(f\) to \([a,c]\) and \([c,\infty]\) are both integrable. In this case we have \begin{equation*} \int_a^{\infty} f = \int_a^c f + \int_c^{\infty} f. \end{equation*}

Proof. Let \(I_1 := [a,c]\) and \(I_2 := [c, \infty]\). Suppose that \(f \in AH(I_1)\) and \(f\in AH(I_2)\). Let \(f_1\) be the restriction of \(f\) to \(I_1\) and let \(f_2\) be the restriction of \(f\) to \(I_2\). Let \(A_1 := \int_{I_1} f_1\) and let \(A_2:= \int_{I_2}f\). Given \(\epsilon>0\), let \(\mathcal{S}'_{\epsilon} : = \{S'_{\epsilon,x} : x\in I_1\}\) be a choice on \(I_1\) and let \(\mathcal{S}''_{\epsilon}:= \{S''_{\epsilon,x} : x\in I_2\}\) be a choice on \(I_2\) such that if \(\mathcal{P}_1\) is a \(\mathcal{S}'_{\epsilon}\)-fine partition of \(I_1\) and \(\mathcal{P}_2\) is a \(\mathcal{S}''_{\epsilon}\)-fine partition of \(I_2\), then \begin{equation} |S(f_1; \mathcal{P}_1) - A_1| \leq \frac{1}{2}\epsilon ~~ and ~~|S(f_2; \mathcal{P}_2) - A_2| \leq \frac{1}{2}\epsilon. \end{equation} We define a choice \(\mathcal{S}_{\epsilon} :=\{S_{\epsilon,x} : x\in I\}\) on \(I\) by \begin{equation*} S_{\epsilon,x} = \begin{cases} S'_{\epsilon,x} \cap [a, c) & \text{if} ~~x\in[a,c) \\ (S'_{\epsilon,c} \cap [a,c]) \cup (S_{\epsilon,c}'' \cap [c, \infty)) & \text{if} ~~x = c \\ S''_{\epsilon,x} \cap (c, \infty) & \text{if} ~~x\in(c,\infty)\\ S''_{\epsilon, \infty} & \text{if} ~~ x = \infty. \end{cases} \end{equation*} Let \(\mathcal{P}\) be a \(S_{\epsilon}\)-fine partition of \(I\) and suppose that each tag occurs only once. Then the point \(c\) must be a tag of an subinterval in \(\mathcal{P}\). Let \(([u,v],c)\) be the tagged interval in \(\mathcal{P}\) of which the tag is \(c\). Then \(\mathcal{P}\) is of the form \(\mathcal{P}_a \cup ([u,v],c) \cup \mathcal{P}_b\) where the tags of \(\mathcal{P}_a\) are less than \(c\) and the tags of \(\mathcal{P}_b\) are greater than \(c\). Let \(\mathcal{P}_1:= \mathcal{P}_a \cup ([u,c],c)\) and let \(\mathcal{P}_2 := \mathcal{P}_b \cup ([c,v],c)\). Then \(\mathcal{P}_1\) is a \(\mathcal{S}'_{\epsilon}\)-fine partition of \(I_1\) and \(\mathcal{P}_2\) is a \(\mathcal{S}''_{\epsilon}\)-fine partition of \({I_2}\). Therefore, \[|S(f;\mathcal{P}) - A_1 - A_2| \leq |S(f_1;\mathcal{P}_1) - A_1 |+ |S(f_2;\mathcal{P}_2) - A_2|.\] Since \(\epsilon>0\) is arbitrary, \(f\) is integrable on \(I\) to \(\int_{I_1} f + \int_{I_2} f\).

Now, suppose that \(f\in AH([a,\infty])\). For each \(\epsilon > 0\), let \(\tilde{\mathcal{S}}_{\epsilon}:=\{\tilde{S}_{\epsilon,x} : x \in I\}\) be a choice on \(I\) that satisfies the Cauchy Criterion (Theorem 15). Let \(f_1\) denote the restriction of \(f\) to \(I_1\) and let \(\tilde{\mathcal{S}}'_{\epsilon} := \{\tilde{S}_{\epsilon,x}\cap I_1 : x \in I_1\}\) be the restriction of \(\tilde{\mathcal{S}}_{\epsilon}\) to \(I_1\). Let \(\mathcal{P}_1, \mathcal{Q}_1\) be \(\tilde{S}'_{\epsilon}\)-fine partitions of \(I_1\). By adjoining the same tagged partition of \(I_2\), extend \(\mathcal{P}_1,\mathcal{Q}_1\) to partitions \(\mathcal{P}, \mathcal{Q}\) of \(I\) that are \(\tilde{S}_{\epsilon}\)-fine . Then,

\[|S(f_1; \mathcal{P}_1) - S(f_1;\mathcal{Q}_1)| = |S(f; \mathcal{P}) - S(f;\mathcal{Q})|\leq \epsilon.\] Therefore, by Theorem 15, \(f_1\) is integrable on \(I_1\). In the same way, the restriction of \(f\) to \(I_2\) is integrable on \(I_2\)

Corollary 3. If \(f\in AH([a,\infty])\) and if \([c,d] \subset [a, \infty]\), then the restriction of \(f\) to \([c,d]\) is integrable.

Proof. Let \(f \in AH([a,\infty])\) and \([c,d]\subset [a, \infty]\). Then it follows from the theorem that \(f \in AH([c, \infty])\), which follows that \(f\in AH([c,d])\).

Theorem 17. Let \(I:=[a, \infty]\) and let \(f : I \to \mathbb{R}\). Then \(f\in AH(I)\) if and only if \(f\in AH([a,c])\) for every \(c \geq a\) and there exists \(A\in \mathbb{R}\) such that \begin{equation*} \lim_{c\to \infty}\int_a^c f = A. \end{equation*} In this case, \(\int_a^{\infty} f = A.\)

Proof. Let \(f \in AH(I)\), \(\int_a^{\infty} f = A,\) and let \(\epsilon>0\). Then there exists a choice \(\mathcal{S} := \left\{S_x : x\in I \right\}\) on \(I\) such that if \(\mathcal{P} = \left\{ ([x_{i-1}, x_i], t_i)\right\}_{i = 1}^{n+1}\) is a \(\mathcal{S}\)-fine partition of \(I\), then \(\left|S(f;\mathcal{P}) - A \right| \leq \frac{1}{2}\epsilon\). Let \(c\geq x_n\). Since \(f\in AH([a,c])\) by Theorem 16, there exists a choice \(\mathcal{S}_c := \left\{S_{c,x}: x\in [a,c] \right\}\) on \([a,c]\) such that if \(\mathcal{P}_c\) is a \(S_c\)-fine partition of \([a,c]\), then \(\left| S(f; \mathcal{P}_c) - \int_a^c f\right|\) \(\leq \frac{1}{2}\epsilon\). We may assume that \(S_{c,x} \subset S_x\) for all \(x\in[a,c]\). Let \(\mathcal{P}_c^* := \mathcal{P}_c \cup ( [c, \infty],\infty)\), then \(\mathcal{P}_c^*\) is a \(\mathcal{S}\)-fine partition of \([a,\infty]\) such that \(S(f;\mathcal{P}_c) = S(f;\mathcal{P}^*_c)\). Therefore, \begin{equation*} \left| \int_a^c f - A\right| \leq \left| \int_a^c f - S(f;\mathcal{P}_c) \right| + \left| S(f; \mathcal{P}_c^*) - A\right| \leq \epsilon. \end{equation*} Since \(\epsilon >0\) is arbitrary, \(\lim_{c\to \infty} \int_a^c f = A\).

Now, suppose that \(f\in AH([a,c])\) for every \(c \geq a\) and that there exists \(A\in \mathbb{R} \) such that \(\lim_{c \to \infty} \int_a^c f = A\). Take a strictly increasing unbounded sequence \(\{c_k\}_{k=0}^{\infty}\) with \(c_0 = a\). Given \(\epsilon >0 \), let \(N \in \mathbb{N}\) be such that if \(b\geq c_{N}\), then \(\left| \int_a^b f - A\right| \leq \epsilon\). Since \(f \in AH(I_k:= [c_{k-1}, c_k])\) for each \(k\in \mathbb{N}\), let \(\mathcal{S}_k := \left\{S_{k,x} : x\in I_k \right\}\) be a choice on \(I_k\) such that if \(\mathcal{P}_k\) is a \(\mathcal{S}_k\)-fine partition of \(I_k\), then \(\left| S(f; \mathcal{P}_k) -\int_{I_k} f \right| \leq \epsilon/2^k.\) We may assume that

1. \(S_{1,c_0} \subset \left[c_0, \frac{c_0 + c_1}{2}\right]\),

   and if \(k\geq 1\), that

2. \(S_{k+1, c_k} \subset S_{k, c_k} \cap \left(\frac{c_{k-1}+c_k}{2}, \frac{c_k + c_{k+1}}{2} \right)\), and
3. \(S_{k,x} \subset \left( \frac{c_{k-1} + x}{2}, \frac{x+c_k}{2}\right)\) for \(x\in(c_{k-1},c_k)\).

Now, in order to define a choice on \(I\), we assign a measurable set \(S_x\) to each \(x \in I\) by \begin{equation*} S_x = \left\{ \begin{array}{ll} S_{k,x} & \text{if} ~~x \in [c_{k-1}, c_k), ~k\in \mathbb{N} \\ \left[c_{N}, \infty \right] & \text{if} ~~ x = \infty,\\ \end{array} \right. \end{equation*} so that \(\mathcal{S}^* = \left\{S^*_x : x\in I \right\}\) be a choice on \(I\). Let \(\mathcal{P} = \left\{ ([x_{i-1}, x_i], t_i)\right\}_{i=1}^{n+1}\) be a \(\mathcal{S}^*\)-fine partition of \(I\). By the definition of \(\mathcal{S}^*\), the tag for the unbounded subinterval \([x_n, \infty]\) in \(\mathcal{P}\) must be \(\infty\) and \(c_{N} \leq x_n\). Now let \(s\in \mathbb{N}\) be the smallest positive integer such that \(x_n \leq c_s\) so that \(N \leq s\). Again, by the the condition (3), for \(k=1, \cdots, s-1\), the point \(c_k\) must be the tag for any subinterval in \(\mathcal{P}\) that contains \(c_k\), and we may assume that \(c_k\) appears as an end point to the intervals. We let \begin{equation*} \mathcal{Q}_1 :=\mathcal{P} \cap [c_0, c_1], \cdots , \mathcal{Q}_{s-1} := \mathcal{P} \cap [c_{s-2}, c_{s-1}], \mathcal{Q}_s := \mathcal{P} \cap [c_{s-1}, x_n]. \end{equation*} Then, \(\mathcal{Q}_k(k = 1, \cdots, s-1)\) is \(\mathcal{S}_k\)-fine partition of \(I_k\). Therefore, we have \begin{equation*} \left| S(f;\mathcal{Q}_k) - \int_{I_k} f \right| \leq \frac{\epsilon}{2^k}. \end{equation*} Also, since \(\mathcal{Q}_s\) is a \(\mathcal{S}_s\)-fine subpartion of \(I_s\), by the Saks-Henstock Lemma, \begin{equation*} \left| S(f; \mathcal{Q}_s) - \int_{c_{s-1}}^{x_n} f \right| \leq \frac{\epsilon}{2^s}. \end{equation*} Let \(\mathcal{Q}_{\infty} := \left\{ ([x_n, \infty], \infty) \right\}\) so that \(S(f;\mathcal{Q}_{\infty}) = 0\). Now, since \(\mathcal{P} = \mathcal{Q}_1 \cup \cdots \cup \mathcal{Q}_s \cup \mathcal{Q}_{\infty}\), we have \begin{align} &\left| S(f; \mathcal{P}) - A \right| = \left| \sum_{i = 1} ^s S(f; \mathcal{Q}_i)+S(f;Q_{\infty}) - A \right| \leq \left| \sum_{i=1}^s S(f; \mathcal{Q}_i) - \int_a^{x_n}f \right| + \left| S(f;\mathcal{Q}_{\infty}) \right| + \left| \int_a^{x_n} f - A \right| \leq 2 \epsilon \nonumber \end{align} Since \(\epsilon>0\) is arbitrary, \(f \in AH(I)\) and \(\int_I f = A\).

We give a different version of Cauchy Criterion for \(f \in AH([a,\infty])\).

Theorem 18. Let \(f : [a,\infty] \to \mathbb{R}\) be such that \( f \in AH([a, c])\) for all \(c \geq a\). Then \( f \in AH([a, \infty])\) if and only if for every \( \epsilon > 0 \) there exists \(K(\epsilon) \geq a\) such that if \(q > p \geq K(\epsilon )\), then \( \mid \int_p^q f \mid \leq \epsilon. \)

Proof. Suppose that \(f\in AH([a,\infty])\). Let \(\epsilon > 0.\) By the previous theorem, there exists \(K(\epsilon)>0\) such that \(\left| \int_a^c f - \int_a^{\infty} f \right| < \epsilon/2\) for all \(c \geq K(\epsilon)\). Let \(q>p > K(\epsilon)\), then \begin{equation*} \left|\int_p^q f \right| = \left| \int_a^q f - \int_a^p f \right| = \left| \int_a^q f - \int_a^{\infty}f \right| + \left| \int_a^p f - \int_a^{\infty}f \right| < \epsilon. \end{equation*} Conversely, suppose that for any given \(\epsilon>0,\) there exists \(K(\epsilon)>0\) such that if \(q > p \geq K(\epsilon),\) then \(\left|\int_p^q f \right| \leq \epsilon\). Let \(\left\{x_n\right\}\) be an unbounded increasing sequence with \(x_0 \geq a\). Since for any \(x_m\geq x_n \geq K(\epsilon)\), \(\left|\int_a^{x_m} f - \int_a^{x_n} f \right| = \left| \int_{x_n}^{x_m} f \right| \leq \epsilon\), the sequence \(\left\{ \int_a^{x_n} f\right\}_{n=1}^{\infty}\) is a Cauchy sequence. Let \(\lim_{n\to\infty} \int_a^{x_n} f := A\) and \(N\) be an integer such that \(x_N \geq K(\epsilon)\) and \(\left|\int_a^{x_n}f -A \right| < \epsilon\) whenever \(n\geq N\). If \(c> x_N\), then \begin{equation} \left| \int_a^c f - A\right| = \left| \int_a^{x_N} f - A\right|+\left|\int_{x_N}^c f \right| < 2\epsilon. \nonumber \end{equation} Since \(\epsilon >0\) is arbitrary, \(\lim_{c \to \infty}\int_a^c f = A\), and by Theorem 17, \(f\in AH(I)\).

We now consider the multiplier properties for the AP-Henstock integral on unbounded intervals.

Theorem 19. Let \(f \in AH([a,\infty)) \) be bounded below and let \(g\) be a regulated function on \([a,\infty)\). Then the product \(fg\in AH([a,\infty))\).

Proof. Assume that \(f(x) \geq 0\) on \([a,\infty]\). By Corollary 3 and Theorem 9, \(fg \in AH([p,q])\) for any \(q >p \geq a\). Let \(s\) be a step function such that \(\left| g(x)-s(x)\right|< 1\) for all \(x\in[a,\infty]\). Let \(\epsilon >0\). By Theorem 18, there exists \(K(\epsilon) > d\) such that \(|\int_p^q f|< \epsilon\) whenever \(q> p\geq K(\epsilon)\). If \(q\geq x \geq p \geq K(\epsilon)\), then \(|g(x)|< M\) for some \(M>0\) and \(|f(x)g(x)|< Mf(x)\). Since \(|fg|\) and \(Mf\) are measurable on \([p,q]\), \(|fg|\) is Lebesgue integrable and hence \(|fg| \in AH([p,q])\). It follows that \(\left|\int_p^q fg \right| \leq \int_p^q |fg| < \int_p^q f < \epsilon.\) Therefore, by Theorem 18, \(fg \in AH([a,\infty])\). Now, if \(f(x) > \alpha\) on \([a,\infty]\) for some \(\alpha < 0\), then since \((f-\alpha)g\), \(\alpha g \in AH([a,\infty])\), the result follows from \(fg = (f-\alpha)g + \alpha g\).

Theorem 20. Let \(I:=[a, \infty)\) and let \( f, \varphi : I \to \mathbb{R}\). Suppose that \(f \in AH(I)\), \(F(x) := \int_a^x f \) is continuous on \(I\), and that \(\varphi\) is bounded and monotone on \(I\). Then the product \(f \varphi \in AH(I)\).

Proof. Let \(\epsilon>0\). Since \(\varphi\) is bounded on \(I\), there exists \(M>0\) such that \(| \varphi (x) | \leq M \) for all \(x \in I\). By Theorem 18, there exists \(K(\epsilon) \geq a\) such that if \(q>p \geq K(\epsilon),\) then \(\mid \int_p^q f \mid \leq {\epsilon}/{2M}.\) Since \(\varphi \) is monotone, it follows from Corollary 2 that \(f \varphi \in AH([p,q])\) and from Theorem 14 that there exists \( \xi \in[p,q]\) such that \[ \int_p^q f \varphi = \varphi(p) \int_p^{\xi} f + \varphi (q) \int_{\xi}^q f. \] Thus, if \(q>p \geq K(\epsilon)\), then \(\mid \int_p^q f {\varphi } \mid \leq M ( {\epsilon}/{2M} ) + M({\epsilon}/ {2M} ) = {\epsilon}.\) Since \(\epsilon >0\) is arbitrary, by Theorem 18, \( f \varphi \) is AP-Henstock integrable on \(I\).

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.''

Data Availability:

All data required for this research is included within this paper.

Funding Information:

No funding is available for this research.

1. Introduction

Perturbation theory comprises methods for finding an approximate solution to a problem; in perturbation theory, the solution is expressed as a power series in a small parameter \(\varepsilon \). The first term is the known solution to the solvable problem. Successive terms in the series at higher powers of \(\varepsilon\) usually become smaller. Assume \(X\) is a Banach space, \(X_n\subseteq X\) is a finite set, \(T(t)\) the \(C_{0}\)-semigroup, \(\omega-OCP_n\) the \(\omega\)-order preserving partial contraction mapping, \(M_{m}\) be a matrix, \(L(X)\) be a bounded linear operator on \(X\), \(P_n\) a partial transformation semigroup, \(\rho(A)\) a resolvent set, \(\sigma(A)\) a spectrum of \(A\) and \(A\in \omega-OCP_n\) is a generator of \(C_{0}\)-semigroup. This paper consists of results of \(\omega\)-order preserving partial contraction mapping generating a one-parameter semigroup.

Akinyele et al., [1] introduced perturbation of the infinitesimal generator in the semigroup of the linear operator. Batty [2] established some spectral conditions for stability of one-parameter semigroup and also in [3] Batty et al., revealed some asymptotic behavior of semigroup of the operator. Balakrishnan [4] obtained an operator calculus for infinitesimal generators of the semigroup. Banach [5] established and introduced the concept of Banach spaces. Chill and Tomilov [6] deduced some resolvent approaches to stability operator semigroup. Davies [7] obtained linear operators and their spectra. Engel and Nagel [8] introduced a one-parameter semigroup for linear evolution equations. R\(\ddot{a}\)biger and Wolf [9] deduced some spectral and asymptotic properties of the dominated operator. Rauf and Akinyele [10] introduced \(\omega\)-order preserving partial contraction mapping and established its properties, also in [11], Rauf et al., deduced some results of stability and spectra properties on semigroup of a linear operator. Vrabie [12] proved some results of \(C_{0}\)-semigroup and its applications. Yosida [13] established and proved some results on differentiability and representation of one-parameter semigroup of linear operators.

In this paper, we show that adding a bounded linear operator \(B\) to an infinitesimal generator \(A\) of a semigroup of the linear operator does not destroy A's property. Furthermore, \(A\) is the generator of a one-parameter semigroup, and \(B\) is a small perturbation so that \(A+B\) is also the generator of a one-parameter semigroup.

2. Preliminaries

Definition 1.(\(C_0\)-Semigroup) [8] A \(C_0\)-Semigroup is a strongly continuous one parameter semigroup of bounded linear operator on Banach space.

Definition 2.(\(\omega\)-\(OCP_n\))[11] A transformation \(\alpha\in P_{n}\) is called \(\omega\)-order preserving partial contraction mapping if \(\forall x,y \in~ \) Dom \(\alpha:x\le y~~\implies~~ \alpha x\le \alpha y\) and at least one of its transformation must satisfy \(\alpha y=y\) such that \(T(t+s)=T(t)T(s)\) whenever \(t,s>0\) and otherwise for \(T(0)=I\).

Definition 3.(Perturbation) [1] Let \(A : D(A) \subseteq X \to X\) be the generator of a strongly continuous semigroup \((T(t))_{t\geq 0}\) and consider a second operator \(B : D(B) \subseteq X \to X\) such that the sum \(A + B\) generates a strongly continuous semigroup \((S(t))_{t\geq 0}\). We say that \(A\) is perturbed by operator \(B\) or that \(B\) is a perturbation of \(A\).

Definition 4.(Analytic Semigroup) [12] We say that a \(C_0\)-semigroup \(\{T(t); t \geq 0\}\) is analytic if there exists \(0 < \theta \leq \pi\), and a mapping \(S : \bar{\mathbb{C}}_{\theta} \to L(X)\) such that:

1. \(T(t) = S(t)\) for each \(t \geq 0\);
2. \(S(z_{1} + z_{2}) = S(z_{1})S(z_{2})\) for \(z_{1},z_{2} \in \bar{\mathbb{C}}_{\theta}\);
3. \(\lim_{z_{1} \in \bar{\mathbb{C}}_{\theta},z_{1} \to 0} S(z_{1})x = x\) for \(x \in X\); and
4. the mapping \(z_{1} \to S(z_{1})\) is analytic from \(\bar{\mathbb{C}}_{\theta}\) to \(L(X)\). In addition, for each \(0 < \delta < \theta \), the mapping \(z_{1} \to S(z_{1})\) is bounded from \( \mathbb{C}_{\delta}\) to \(L(X)\), then the \(C_{0}\)-Semigroup \(\{T(t);t \geq 0\}\) is called analytic and uniformly bounded.

Definition 5.(Perturbation class) [7] We say that operator \(B\) is a class \(P\) perturbation of the generator \(A\) of the one-parameter semigroup \(T(t)\) if:

Note that \(BT(t)\) is bounded for all \(t > 0\) under conditions (1)\(_1\) and (1)\(_2\) by the closed graph theorem.

Example 1(\(2\times 2\) matrix \({M_m(\mathbb{N} \cup\{0\})}\)). Suppose \[ A=\begin{pmatrix} 2&0\\ 1&2 \end{pmatrix} \] and let \(T(t)=e^{t A}\), then \[ e^{t A}=\begin{pmatrix} e^{2t}&e^{I}\\ e^{t}& e^{2t} \end{pmatrix} .\]

Example 2(\(3\times 3\) matrix \({M_m(\mathbb{N} \cup \{0\})}\)). Suppose \[ A=\begin{pmatrix} 2&2&3\\2& 2&2\\1& 2&2 \end{pmatrix} \] and let \(T(t)=e^{t A}\), then \[ e^{t A}=\begin{pmatrix} e^{2t}&e^{2t}&e^{3t}\\e^{2t}&e^{2t}&e^{2t}\\e^{t}& e^{2t}&e^{2t} \end{pmatrix}. \]

Example 3(\(3\times 3\) matrix \({M_m(\mathbb{C})}\)). Since we have for each \(\lambda>0\) such that \(\lambda\in \rho(A)\) where \(\rho(A)\) is a resolvent set on \(X\). Suppose we have \[ A=\begin{pmatrix} 2&2&3\\ 2&2&2\\ 1&2&2 \end{pmatrix} \] and let \(T(t)=e^{t A_\lambda}\), then \[ e^{t A_\lambda}=\begin{pmatrix} e^{2t\lambda}&e^{2t\lambda}&e^{3t\lambda}\\ e^{2t\lambda}&e^{2t\lambda}&e^{2t\lambda}\\ e^{t\lambda}&e^{2t\lambda}&e^{2t\lambda}\end{pmatrix} .\]

3. Main results

This section present results of one-parameter semigroup generated by \(\omega\)-\(OCP_{n}\) using perturbation theory.

Theorem 1. Let \(A\in\omega-OCP_n\) be the generator of a one-parameter semigroup \(T(t)_{z\geqslant 0}\) on the Banach space \(X\) and suppose that \[ \|T(t)\|\leqslant Me^{at} \] for all \(t\geqslant 0\). If \(B\) is a bounded operator on \(X\), then \((A+B)\) is the generator of a one-parameter semigroup \(S(t)_{t\geqslant 0}\) on \(X\) such that \[ \|S(t)\|\leqslant Me^{(a+M\|B\|)t} \] for all \(t\geqslant 0\) and \(B\in\omega-OCP_n\).

Proof. We define the operators \(S(t)\) by

The \(nth\) term is an \(n\)-fold integral whose integrand is a norm continuous function of the variables. It is easy to verify that the series is norm convergent and that for all \(f\in X\), \(t\geqslant 0\) and \(B\in\omega-OCP_n\).

Since \(S(s)S(t)=S(s+t)\) and if \(f\in X\), then

\[ \lim\limits_{t\rightarrow 0}\|s(t)f-f\|\leqslant\lim\limits_{t\rightarrow 0}\left\{\|T(t)f-f\|+\sum_{n=1}^{\infty}Me^{at}\|f\|(tM\|B\|)^n/n!\right\}\geqslant 0 \,,\] so that \(s(t)\) is a one-parameter semigroup. If \(f\in X\) and \(B\in\omega-OCP_n\), then It follows that \(f\) lies in the domain of the generator \(Y\) of \(S(t)\) if and only if it lies in the domain of \(A\), and that for such \(f\).

As well as being illuminating in its own right, (2) easily leads to the identities

Hence the proof is complete.

Theorem 2. Suppose \(B\) is a class \(P\) perturbation of the generator \(A\), then \[ Dom(B)\supseteq Dom(A). \] If \(\varepsilon>0\) and \(A,B\in\omega-OCP_n\), then

for all large enough \(\lambda>0.\) Hence \(B\) has relative bound \(0\) with respect to \(A\).

Proof. Combining (1) with the bound \[ \|BT(t)\|\leqslant\|BT(t)\|Me^{a(t-1)}\,, \] valid for all \(t\geqslant 1\), we then see that \[ \int_{0}^{\infty}\|BT(t)\|e^{-\lambda t}dt< \infty\,,\] for all \(\lambda>a\). Suppose \(\varepsilon>0\) and \(A,B\in\omega-OCP_n\), then for all large enough \(\lambda\) we have \[ \int_{0}^{\infty}\|BT(t)\|e^{-\lambda t}dt\leqslant\varepsilon. \] Now, \[ \int_{0}^{\infty}T(t)e^{-\lambda t}fdt = R(\lambda,A)f \,,\] for all \(f\in X\), so by the closedness of \(B\), we see that \(R(\lambda,A)f\in Dom(B)\) and \[ \|BR(\lambda,A)f\|\leqslant\varepsilon\|f\| \,,\] as required to prove (7).

If \(g\in Dom(A)\) and we put \(f:=(\lambda I-A)g\), then we deduce from (7) that

for all large enough \(\lambda>0\). This implies the last statement of the theorem and hence the proof is complete.

Theorem 3. Assume \(B\) is a class \(P\) perturbation of the generator \(A\) of the one-parameter semigroup \(T(t)\) on \(X\), then \(B+A\) is the generator of a one-parameter semigroup \(S(t)\) on \(X\) and \(A,B\in\omega-OCP_n\).

Proof. Let \(a\) be small enough that

We may define \(S(t)\) by the convergent series (2) for \(0\leqslant t\leqslant 2a\), and verify as in the proof of Theorem 1 that \(S(s)S(t)=S(s+t)\) for all \(s,t\geqslant 0\) such that \(s+t\leqslant 2a\). We now extend the definition of \(S(t)\) inductively for \(t\geqslant 2a\) by putting if \(n\in\mathbb{N}\) and \(na< t\leqslant(n+1)a\). It is straight forward to verify that \(S(t)\) is a semigroup. Now suppose that \(\|T(t)\|\leqslant N\) for \(0\leqslant t\leqslant a\).

Assume \(f\in X\) and \(B\in\omega-OCP_n\), then

\[ \|S(t)f-f\|\leqslant\|T(t)f-f\|+\sum_{n=1}^{\infty}N\left(\int_{0}^{t}\|BS(t)\|ds\right)^n\|f\|, \] so that and \(S(t)\) is a one-parameter semigroup on \(X\). It is an immediate consequence of the definition that for all \(f\in X\), \(B\in\omega-OCP_n\) and all \(0\leqslant t\leqslant a\). Suppose that this holds for all \(t\) such that \(0\leqslant t\leqslant na\). If \(na\leqslant u\leqslant(n+1)a\), then By induction, (12) holds for all \(t\geqslant 0\).

We finally have to identify the generator \(Y\) of \(S(t)\). The subspace

\[ D:=\underset{t>0}{\bigcup}T(t)\{Dom(A)\} \,,\] is contained in \(Dom(A)\) and is invariant under \(T(t)\) and so is a core for \(A\). If \(f\in D\), then there exists \(g\in Dom(A)\) where \(A\in\omega-OCP_n\) and \(\varepsilon>0\) such that \(f=T(\varepsilon)g\). Hence, Therefore, \(Dom(Y)\) contains \(D\) and \(Yf(B+A)\) for all \(f\in D\) and \(A,B\in\omega-OCP_n\). If \(f\in Dom(A)\), then there exists a sequence \(f_n\in D\) such that \(\|f_n-f\|\rightarrow 0\) and \(\|Af_n-Af\|\rightarrow 0\) as \(n\rightarrow\infty\). It follows by Theorem 2 that \(\|Bf_n-Bf\|\rightarrow 0\) and hence that \(Yf_n\) converges. Since \(Y\) is a generator that is closed, then we deduce that \[ Yf=(B+A)f \,,\] for all \(f\in Dom(A)\) and \(A,B\in\omega-OCP_n\). Multiplying (12) by \(e^{-\lambda t}\) and integrating over \((0,\infty)\), we see as in the proof of Theorem 2 that if \(\lambda>0\) is large enough, then \[ R(\lambda,Y)f=R(\lambda,A)f+R(\lambda,Y)BR(\lambda,A)f \,,\] for all \(f\in Y\) and \(A,B\in\omega-OCP_n\).

If \(\lambda\) is also large enough that

\[ \|BR(\lambda,A)\|< 1 \,,\] we deduce that \[ R(\lambda,Y)=R(\lambda,A)(I-BR(\lambda,A))^{-1}. \] Hence, \[ Dom(Y)=Ran(R(\lambda,Y))=Ran(R(\lambda,A))=Dom(A) \,,\] and \(Y=A+B\), and this achieve the proof.

Theorem 4. Let \(A:=-H\) where \(H=(-\Delta)^n\geqslant 0\) acts in \(L^2(\mathbb{R}^N)\). Also let \(B\) be a lower order perturbation of the form \[ (Bf)(x):=\sum_{|\alpha|< 2n}a_\alpha(x)(D^\alpha f)(x). \] If \(a_\alpha\in L^{P_\alpha}(\mathbb{R}^N)+L^\infty(\mathbb{R}^N)\) for each \(\alpha\), where \(P_\alpha\geqslant 2\) and \(P_\alpha>N/(2n-|\alpha|)\), the \(A+B\) is the generator of a one-parameter semigroup and \(B\) has relative bound \(0\) with respect to \(A\) where \(A,B\in\omega-OCP_n\).

Proof. Suppose \(A\in\omega-OCP_n\) is the generator of holomorphic semigroup \(T(t)\) such that \[ \|T(t)\|\leqslant c_1,\quad \|AT(t)\leqslant c_2/t\| \,,\] for all \(t\in(0,1)\). And also the operator \(B\in\omega-OCP_n\) has domain containing \(Dom(A)\) and there exists \(\alpha\in(0,1)\), such that

for all \(f\in Dom(a)\) and \(0< \varepsilon\leqslant 1\). Then for all \(t\in(0,1)\) so that \(B\) is a class \(P\) perturbation of \(A\) and by Theorem 3 under the stated conditions on \(t\) and \(\varepsilon\), we have \begin{align*} \|BT(t)f\|&\leqslant\varepsilon\|AT(t)f\|+c_3\varepsilon^{-\alpha/(1-\alpha)}\|T(t)f\|\\ &\leqslant(\varepsilon c_2t^{-1}+c_1c_3\varepsilon^{-\alpha/(1-\alpha)})\|f\|. \end{align*} By putting \(\varepsilon=t^{1-\alpha}\), then we obtain (16).

Assume \(\alpha\in(0,1)\), \(H\) is a non-negative self-adjoint operator on \(P\) and \(B\) is a linear operator with \(Dom(B)\geqslant(H)\), we have

\[ \|Bf\|\leqslant\varepsilon\|Af\|+c_3\varepsilon^{-\alpha/(1-\alpha)}\|f\| \,,\] for all \(\varepsilon>0\) if and only if there is a constant \(c_4\) such that \[ \|Bf\|\leqslant c_4\|Af\|^\alpha\|f\|^{1-\alpha} \,,\] for all \(f\in Dom(A)\) and \(A,B\in\omega-OCP_n\).

By Theorem 3, it is sufficient to prove that for each \(\alpha\) there exists \(\beta< 1\) for which

\[ X_\alpha:=a_\alpha(\cdot)D^\alpha(H+1)^{-\beta} \] is bounded.

Let \(X_\alpha=a_\alpha(Q)b_\alpha(P)\), where

\[ b_\alpha(\varepsilon)=\frac{i^{|\alpha|}\varepsilon^\alpha}{(|\varepsilon|^{2n}+1)^\beta}. \] If \(a_\alpha\in L^\infty(\mathbb{R}^N)\), then \(\|X\|\leqslant\|a_\alpha\|_\infty\|b_\alpha\|_\infty< \infty\) provided \(|\alpha|/2n< \beta< 1\). On the other hand, if \(a_\alpha\in L^P(\mathbb{R}^N)\) where \(P\geqslant 2\) and \(P>N/(2n-|\alpha|)\), then there exists \(\beta\) such that \[ \frac{N+|\alpha|P}{2np}< \beta< 1. \] This implies that \((|\alpha|-2n\beta)p+N< 0\) and hence \(b_\alpha\in L^p(\mathbb{R}^N)\).

4. Conclusion

In this paper, it has been established that \(\omega\)-order preserving partial contraction mapping generates a one-parameter semigroup using a perturbation theory on Banach space by showing that the semigroup of a linear operator is bounded, that \(B\) has a relative bound \(0\) with respect to \(A\), and also that \(B + A\) is a generator of the one-parameter semigroup.

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.''

Data Availability:

All data required for this research is included within this paper.

Funding Information:

No funding is available for this research.

1. Introduction and Main Results

This paper deals with 1D and 2D exactly solvable models for the Schrödinger operators. We study the existence of negative eigenvalues lying under the continuous spectrum of the Schrödinger operator with \(\delta\) potential. The main goal is to find the threshold value of the coupling constant for which such eigenvalues exist. In one dimensional case, our result's are similar to the consequences of the more general results concerning the Schrödinger operators with a finite number of \(\delta-\)interactions

on the line [1]. In [1], Albeverio and Nizhnik have provided the effective algorithm for determining the number of negative eigenvalues of such operators in terms of the intensities \(\alpha_k\) and the distances \(d_k = x_{k+1}-x_k\) between the interactions. In this paper, we consider the eigenvalue problem on the positive half-axis with Dirichlet, Neumann, and Robin type boundary conditions. For all solvable models under consideration, there exists a critical value \(\beta_{cr}\) such that the operator possesses the negative eigenvalues if \(\beta>\beta_{cr}\) and has no eigenvalues if \(\beta>\beta_{cr}\). The goal of this paper is to present and estimate the exact values of the beta critical for Eq. (2) considering appropriate boundary condition. We define beta critical as a critical value of the coupling constant denoted by \(\beta_{cr},\) the value of \(\beta\) such that Eq. (2) does not have negative eigenvalues for \(\beta< \beta_{cr}\) and has them if \(\beta>\beta_{cr}\). The delta potential allows solutions for both the bound states \(\lambda < 0\) and scattering states \(\lambda >0.\) There has been a considerable interest in the study of coupling constant and the problem was investigated by several researchers like S.Albeverio and others in [2], P.Exner and K.Pankrashkin in [3], Barry Simon in [4], Martin Klaus in [5], and Cranston, Koralov, Molchanov and Vainberg in [7]. It is known that the spectrum of \(-\Delta-\beta V(x)\) consists of the absolutely continuous part \([0, \infty)\) and at most a finite number of of negative eigenvalues The classification of the spectrum into discrete and continuous parts usually corresponds to a classification of the dynamics into localized (bound) states and locally decaying states when time increases (scattering) respectively. The lower bound, \(0\), of the absolutely continuous spectrum is called the ionization threshold. This follows from the fact that the particle is no longer localized, but moves freely when \(\lambda>0\). This classification is related to the space-time behaviour of solutions of the corresponding Schrödinger equation.

It is known that [7] \(\beta_{cr}>0\) in the case of Dirichlet boundary condition and \(\beta_{cr}=0\) in the case of the Neumann boundary condition in the dimension 1 and 2 . It was shown that the choice \(\beta_{cr}>0\) or \(\beta_{cr}=0\) depends on whether the truncated resolvent is bounded or goes to infinity when \(\lambda\to 0^-\). In fact, \(\beta_{cr}\) was expressed through truncated resolvent operator and depend on the boundary condition and dimension [7]. The main results of this paper in 1D is stated as follows.

Theorem 1. Consider one dimensional eigenvalue problem

then the \(\beta_{cr}=\frac{1}{a}>0\) in the case of Dirichlet Boundary condition and \(\beta_{cr}=0\) in the case of Neumann and Robin boundary condition.

Proof. The solution of the problem (4) is given by

where \(P\) and \(Q\) are constants and \(\lambda= -k^2 < 0.\) One can determine the value of constant P and Q by using the Dirichlet condition \(\ y(0)=0\) with \(\ y(a)=1\). Then we split the solution of the problem (4) in to two different regions. \[ \begin{cases} y_1(x)= \frac{\sinh kx}{\sinh ka},& \text{if } 0\leq x\leq a\\ y_2(x)= e^{k(a-x)},& \text{if } a\leq x. \end{cases} \] Integrate the Eq. (4) with respect to \(x\) over a small interval \(\Delta \epsilon\) at the point \(x=a\), The integral of the second derivative is just the first derivative function and the integral over the function in the right hand side goes to zero. This yields, When \(\epsilon \xrightarrow{}0\), we get \[ k+k \coth ka =\beta.\] This implies, \( \frac{k}{\beta}=\frac{1}{1+ \coth ka}\) and \( \frac{ka}{\beta a}=\frac{1}{1+ \coth ka}\). Let \(ka=A, \ \beta a= B,\) then \(e^{-2A}=1-\frac{2A}{B}.\) Again let \(2A=z, \) then we have, From Eq. (8), \[1-\frac{z}{B}\geq 0.\] By solving the inequality, we get \[\frac{\beta}{2}\geq k \] and ultimately, \(\frac{\beta^2}{4}\geq k^2.\) This leads to Hence, if \(\beta=0\) then there is no possibility of having negative eigenvalues. Therefore, \(\beta_{cr}\) must be greater than zero to produce negative eigenvalues. We notice that we will not have a solution of the Eq. (8) if \(\frac{1}{B}\geq 1.\) That means, there is no negative eigenvalues when \( \frac{1}{a}\geq \beta. \) However, for \(\frac{1}{B}\leq 1\), there is a solution of the Eq. (8). That means, when \(\frac{1}{ a}\leq \beta,\) we will have the negative eigenvalues. As we defined \(\beta_{cr},\) the value of \(\beta\) such that Eq. (4) does not have negative eigenvalues for \(\beta< \beta_{cr}\) and has them if \(\beta>\beta_{cr}\), we conclude that \(\beta_{cr}=\frac{1}{a}\) for the Eq. (4) with Dirichlet boundary condition.

If we consider the delta potential located at \(x=a_n\) then the \(\beta_{cr} \xrightarrow{} \infty \) as \( a_n \xrightarrow{}0.\) That means, if we approach the potential towards to the boundary, i.e., \(a_n \xrightarrow{}0\) then \(\beta_{cr} \xrightarrow{} \infty .\)

We consider the Neumann boundary condition then the Eq. (2) becomes

Similar to Dirichlet problem, we divide the solution of Neumann problem in to two different regions: \[ \begin{cases} y_1(x)= \frac{\cosh kx}{\cosh ka},& \text{if } 0\leq x\leq a\\ y_2(x)= e^{k(a-x)},& \text{if } a\leq x. \end{cases} \] Integrate the Eq. (10) with respect to \(x\) over a small interval and take \(\epsilon \xrightarrow{}0\) yields, \[ k+k \tanh ka =\beta.\] After simplification, we get \[ \frac{k}{\beta}=\frac{1}{1+ \tanh ka} \] and then we multiply and divide by \(a\) for the right hand side, \[\frac{ka}{\beta a}=\frac{1}{1+ \tanh ka}.\] Let \(ka=A,\ \beta a= B\), then \[e^{-2A}=\frac{2A}{B}-1.\] Following the same analysis as Dirichlet case, we get \(\frac{\beta^2}{4}\leq k^2.\) This implies that \(\lambda =-k^2 \leq \frac{-\beta^2}{4}\) and concludes that if \(\beta=0\) then there is still a possibility of having a negative eigenvalues. Hence, \(\beta_{cr}=0.\)

Now we consider the Robin boundary condition. The Eq. (2) with Robin boundary condition is given by

As above, we divide the solution of this problem in to two different regions: \[ \begin{cases} y_1(x)= \frac{k\cosh kx-\sinh kx}{k\cosh ka-\sinh ka},& \text{if } 0\leq x\leq a\\ y_2(x)= e^{k(a-x)},& \text{if } a\leq x. \end{cases} \] We integrate the Eq. (11) with respect to \(x\) over a small interval \(\Delta \epsilon\), \[\int_{a-\epsilon}^{a+\epsilon}(-y^{''}-\beta \delta(x-a) y \ )dx= \int_{a-\epsilon}^{a+\epsilon}(\lambda y)dx.\] After integration and taking \( \epsilon \xrightarrow{}0,\) we get, \[ k+k \bigg(\frac{k-\coth ka}{k\coth ka-1}\bigg) =\beta.\] After simplification, \[ \frac{k}{\beta}=\frac{1}{1+ \bigg(\frac{k-\coth ka}{k\coth ka-1}\bigg)} ,\] and \[ \frac{ka}{\beta a}=\frac{1}{1+ \bigg(\frac{k-\coth ka}{k\coth ka-1}\bigg)}.\] Let \(ka=A, \beta a= B\), then \[e^{-2A}=\frac{2A^2-2Aa-AB+Ba}{AB+Ba}.\] Observe that \(\frac{2A^2-2Aa-AB+Ba}{AB+Ba}\) must be \(\geq 0.\) After solving this inequality, we get \[\frac{\beta}{2}\leq k. \] Now, \(\lambda =-k^2 \leq \frac{-\beta^2}{4}.\) Hence, if \(\beta=0\) then there is a possibility of having negative eigenvalues so \(\beta_{cr}\) must be zero.

Two Dimensional eigenvalue problem

The rotational invariance suggests that the two dimensional Laplacian should take a particularly simple form in polar coordinates. We use polar coordinates \((r, \theta)\) and look for solutions depending only on r. For \(d=2,\) we do not consider the Neumann boundary condition since \(\beta_{cr}\) is always zero in this case.

Theorem 2. Consider two dimensional Dirichlet problem with delta potential on the circle,

then the \(\beta_{cr}>0\) and \(\beta\in (0,1/2)\) .

Proof. The Eq. (2) takes the following form for \(d=2\) with delta potential on the circle,

We divide the solution of Eq. (13) into two different regions: region (I) with \(1\leq r < 1+a\) and region (II) with \(1+a< r\), \[ \begin{cases} y_1(r)= \frac{Y_0(k)J_0(kr)-J_0(k)Y_0(kr)}{Y_0(k)J_0(k(1+a))-J_0(k)Y_0(k(1+a))},& \text{if } 1\leq r\leq 1+a\\ y_2(r)= \frac{K_0(kr)}{K_0(k(1+a))},& \text{if } 1+a\leq r. \end{cases} \] Using the same argument as in the one dimensional problem, we get \[ -y^{'}\bigg|_{1+a-\epsilon}^{1+a+\epsilon}-\beta y(1+a)=0.\] After simplification, we have \[ -\big(y_2^{'}|_{1+a+\epsilon}-y_1^{'}|_{1+a-\epsilon}\big)=\beta.\] When \(\epsilon \xrightarrow{}0,\) we get

where \(Y_0\) and \(Y_1\) are Bessel function of second kind and and \(J_0\) and \(J_1\) are Bessel function of first kind. Similarly, \(K_0\) and \(K_1\) are a modified Bessel function of second kind. We define \[ g(k,a)=\frac{g_1(k,a)}{g_2(k,a)}=\frac{-Y_0(k)J_1(k(1+a)+J_0(k)Y_1(k(1+a))}{Y_0(k)J_0(k(1+a))-J_0(k)Y_0(k(1+a))}.\] We notice that \(g(k,a)=-1\) for all the values of \(a\) as shown in the Figures 1-4.

From Eq. (14), we get

We will use the following fact from [8] to prove that \(\beta_{cr}>0.\)

Lemma 1([8]). Let \(p, q\geq 0.\) Then the double inequalities

hold for all \(x>0\) if only if \(p\geq 1/4\) and \(q=0.\)

Now, from Eqs (15) and (16) we get, \[\frac{1}{2(x+p)}< \frac{K_1(x)}{K_0(x)}-1< \frac{1}{2(x+q)}.\] When \(x=k(a+1)>0,\) \[\frac{1}{2(k(a+1)+p)}< \frac{K_1(k(a+1))}{K_0(k(a+1))}-1< \frac{1}{2(k(a+1)+q)}.\] \[\frac{1}{2(k(a+1)+p)}< \frac{\beta}{k}< \frac{1}{2(k(a+1)+q)}.\] \[\frac{1}{2((a+1)+\frac{p}{k})}< \beta < \frac{1}{2((a+1)+\frac{q}{k})}.\] Since \(a>0, p\geq \frac{1}{4}, k>0\), which tells us that \(\beta >0\) and hence \(\beta_{cr}>0\) and \(\beta \in (0,\frac{1}{2}).\)

Conflicts of Interest:

The author declares no conflict of interest.

Data Availability:

All data required for this research is included within this paper.

Funding Information:

No funding is available for this research.

1. Introduction

Kolmogorov's law of the iterated logarithm (LIL) for the sequence of independent random variables is in the words of K. L. Chung, ``a crowning achievement in classical probability theory". We first begin with Kolmogorov's celebrated law of the iterated logarithm.

Theorem 1(Kolmogorov [1]). Let \(S_{m}=\sum_{k=1}^{m} X_{k}\) where \(\{X_{k}\} \) is a sequence of real valued independent random variables. Let \(s_{m}\) be the variance of \(S_{m}.\) Suppose \(s_{m}\rightarrow \infty\) and \(|X_{m}|^{2}\leq \dfrac{K_{m} s_{m}^{2}} {\log\log{(e^{e}+s_{m}^{2})}}\) for some sequence of constants \(K_{m}\rightarrow 0.\) Then, almost surely, \[ \limsup_{m\rightarrow \infty}\frac{S_{m}(\omega)}{\sqrt{2s_{m} \log\log{s_{m}^{2}}}}=1.\]

This beautiful law of the iterated logarithm result of Kolmogorov was first proved by Khintchine [2] for Bernoulli random variables. Khintchine obtained this result while improvising the efforts of Hausdorff \((1913)\), Hardy and Littlewood \((1914)\) and Steinhaus \((1922)\) to obtain the exact rate of convergence in Borel's Theorem on normal numbers. Over the years, people have obtained the analog of the Kolmogorov's result in various settings in analysis. Some of the existing settings are lacunary trigonometric series, martingales, harmonic functions, Bloch functions etc. Readers are referred to a survey article by Bingham [3] which has more than \(400\) references on the law of the iterated logarithm. Salem and Zygmund [4] obtained the analogue of Kolmogorov's LIL in the context of lacunary trigonometric series and their result is the first LIL in analysis. Moreover, Salem and Zygmund [4] also introduced a law of the iterated logarithm for the tail sums of lacunary trigonometric series, known as tail LIL. The tail LIL of lacunary series was then completed by Ghimire and Moore [5].

In \(1970\), Stout [6] obtained a martingale version of Kolmogorov's LIL where he used the probabilistic approach. In this paper, we prove one side of the same law of the iterated logarithm for dyadic martingales using a different approach. Precisely, we use the harmonic analysis approach and easily obtain the upper bound. In the proof, we make the use of a subgaussian type estimate and Borel-Cantelli lemma. Our main result is:

Theorem 2. If \(\{f_{n}\}_{n=0}^{\infty}\) is a dyadic martingale on \([0,1)\), then \[ \limsup_{n\rightarrow \infty} \frac{|f_{n}(x)|} {S_{n}f(x)\sqrt{2\log\log {S_{n}f(x)}}}\leq 1\] almost everywhere on the set where \(\{f_{n}\}_{n=0}^{\infty}\) is unbounded.

2. Preliminaries

We first fix some notations, give some definitions and state some lemmas which will be used in the course of the proof.

Let \(\mathcal{D}_n\) denote the family of dyadic subintervals of the unit interval \([0,1)\) of the form \(\left[\left.\frac{j}{2^{n}}, \frac{j+1}{2^{n}}\right)\right.\), where \(n=0,1,2 \cdots\) and \(j=0,1, \cdots 2^{n}-1.\)

Definition 1.(Dyadic martingale) A dyadic martingale is a sequence of integrable functions, \(\{f_{n}\}_{n=0}^{\infty}\) with \(f_{n}:[0,1)\rightarrow \mathbb{R} \) such that,

• (i) for every \(n\), \(f_{n}\) is \(\mathfrak{F}_{n}-\) measurable where \({\mathfrak{F}}_{n}\) is the \(\sigma-\)algebra generated by dyadic intervals of the form \([\frac{j}{2^{n}}, \frac{j+1}{2^{n}}),\) \(j\in \{0, 1, 2, \cdots 2^{n}-1\}\);
• (ii) and the following conditional expectation condition holds \[ \mathbb{E}(f_{n+1}|\mathfrak{F}_{n})=f_{n},\] where \(\mathbb{E}(f_{n+1}|\mathfrak{F}_{n})(x)= \frac{1}{|Q_{n}|}\int_{Q_{n}}f_{n+1}(y)dy\), for \(Q_n \in \mathcal{D}_n\) and \(x\in Q_n\).

Definition 2. For a dyadic martingale, \(\{f_{n}\}_{n=0}^{\infty},\) we define

• (i) the increments: \(d_{k}=f_{k}-f_{k-1}.\) So \(f_{n}(x)=\sum_{k=1}^{n}d_{k}(x)+f_{0}.\)
• (ii) the quadratic characteristics or square function: \(S_{n}^{2}f(x)=\sum_{k=1}^{n} d_{k}^{2}(x).\)
• (iii) the limit function: \(S^{2}f(x)= \lim_{n\rightarrow \infty} S_{n}^{2}f(x) = \sum_{k=1}^{\infty} d_{k}^{2}(x).\)

Next, we define Hardy-Littlewood maximal function:

Definition 3.(Hardy-Littlewood maximal function) Let \(f\in L^p(\mathbb{R}^n),\) \(1\leq p < \infty.\) Let \[Mf(x)=\sup_{r>0} \frac{1}{|B(x,r)|} \int_{B(x,r)} |f(y)| dy.\] Then \(Mf\) is called the Hardy-Littlewood maximal function of \(f.\) Here \(|B(x,r)|\) denotes the measure of the ball centered at \(x\) and of radius \(r.\)

Let \(m\) denote the Lebesgue measure on \(\mathbb{R}\).

Lemma 1(Borel-Cantelli [7]). Let \(\{E_{k}\}_{k=1}^{\infty}\) be a countable collections of measurable sets for which \(\sum_{k=1}^{\infty}m(E_{k}) < \infty.\) Then almost all \(x\in \mathbb{R}\) belong to at most finitely many of the sets \(E_{k}'s.\)

Next, we obtain an estimate for the sequence of dyadic martingales. This estimate will be used in the proof of a lemma. The estimate is stated as a lemma below:

Lemma 2. For a dyadic martingale \(\{f_{n}\}_{n=0}^{\infty}\), with \(f_{0}=0\) \[ \int_{0}^{1} \exp \left( f_{n}(x)-\frac{1}{2}S^{2}_{n}f(x) \right) dx \leq 1.\]

This estimate was originally obtained by Chang et al., [8] using the probabilistic approach. Recently, S. Ghimire also obtained the same estimate using the analytic approach. Please refer [9] for the detail.

Remark 1. Note that if we rescale the sequence \(\{f_{n}\}\) by \(\lambda,\) then Lemma 2 gives, \[\int_{0}^{1} \exp \left( \lambda f_{n}(x)-\frac{1}{2}\lambda^{2}S^{2}_{n}f(x) \right) dx \leq 1.\] This shows that the above inequality is inhomogeneous type. We will make the use of this form in the proof the lemma that follows.

With the help of Lemma 2, we now obtain a subgaussian type estimate related to dyadic martingales. This estimate plays the central role in the proof of our main result. The proof of the estimate can be found in [9].

We also revisit the same proof here. The estimate is given as a lemma below;

Lemma 3. For a dyadic martingale \(\{f_{n}\}\) and \(\lambda > 0\), we have \[ \left|\left\{ x\in [0,1): \sup_{m\geq1} |f_{m}(x)|>\lambda \right\} \right| \leq 6 \exp \left( \dfrac{-\lambda^{2}}{2|| Sf||^{2}_{\infty}} \right).\]

Proof. Fix \(n.\) Let \(\lambda >0,\) \(\gamma >0.\) Then for every \(m\leq n,\) \[ f_{m}(x)= \frac{1}{|Q_{m}|} \int_{Q_{m}} f_{n} (y) dy, \quad x\in Q_{m},\quad |Q_{m}|=\frac{1}{2^{m}}.\] Fix \(x.\) Then \( \sup_{1\leq m \leq n}|f_{m}(x)|\leq M|f_{n}|(x)\), where \(Mf_{n}\) is the Hardy-Littlewood maximal function of \(f_{n}.\) Then using Jensen's inequality, we have \begin{align*} \exp(\gamma |f_{m}(x)|)&= \exp \left( \gamma \left|\int_{Q_{m}} f_{n}(y) d\left( \dfrac{y}{|Q_{m}|}\right)\right|\right)\\ &\leq \dfrac{1}{|Q_{m}|} \int_{Q_{m}} \exp(\gamma |f_{n}(y)|)dy\\ &\leq M (e^{\gamma |f_{m}(x)|})(x). \end{align*} Using the Hardy-Littlewood maximal estimate, we have \begin{align*} \left|\left\{ x\in [0,1): \sup_{1 \leq m\leq n} |f_{m}(x)|>\lambda \right\}\right| &= \left|\left\{ x\in [0,1): \sup_{1 \leq m\leq n} e^{\gamma |f_{m}(x)|}>e^{\gamma \lambda} \right\}\right|\\ &\leq \left|\left\{ x\in [0,1): M (e^{\gamma |f_{m}|})(x)>e^{\gamma \lambda} \right\}\right|\\ &\leq \frac{3}{e^{\gamma \lambda}} \int_{0}^{1} \exp(\gamma |f_{n}(y)|)dy\\ &\leq \frac{3}{e^{\gamma \lambda}}\exp\left(\frac{\gamma^{2}}{2}||S_{n}f||^{2}_{\infty}\right) \int_{0}^{1} \exp\left(\gamma |f_{n}(y)|-\frac{\gamma^{2}}{2}S_{n}^{2}f(y)\right)dy. \end{align*} Using Lemma 2, we have \begin{align*} \int_{0}^{1} &\exp\left(\gamma |f_{n}(y)|-\frac{\gamma^{2}}{2}S_{n}^{2}f\right)dy\\ &= \int_{\{y:f_{n}(y)\geq 0\}} \exp\left(\gamma f_{n}(y)-\frac{\gamma^{2}}{2}S_{n}^{2}f(y)\right)dy + \int_{\{y:f_{n}(y) < 0\}} \exp\left(-\gamma f_{n}(y)-\frac{\gamma^{2}}{2}S_{n}^{2}f(y)\right)dy\\ &\leq \int_{0}^{1} \exp\left(\gamma f_{n}(y)-\frac{\gamma^{2}}{2}S_{n}^{2}f(y)\right)dy + \int_{0}^{1} \exp\left(-\gamma f_{n}(y)-\frac{\gamma^{2}}{2}S_{n}^{2}f(y)\right)dy\leq 2. \end{align*} So, \[ \left|\left\{ x\in [0,1): \sup_{1\leq m\leq n} |f_{m}(x)|>\lambda \right\} \right| \leq \frac{6}{e^{\gamma \lambda}}\exp\left(\frac{\gamma^{2}}{2}||S_{n}f||^{2}_{\infty}\right).\] Choose \(\gamma=\dfrac{\lambda}{||S_{n}f||^{2}_{\infty}}.\) With this \(\gamma\), the above inequality becomes, \[\left|\left\{ x\in [0,1): \sup_{1\leq m\leq n} |f_{m}(x)|>\lambda \right\} \right| \leq 6 \exp\left(\dfrac{-\lambda^{2}}{2||S_{n}f||^{2}_{\infty}} \right).\] Note that for the dyadic martingale \(\{f_{n}\}\), \[S_{n}^{2}f(x)=\sum_{k=1}^{n}d_{k}^{2}(x) \longrightarrow S^{2}f(x)=\sum_{k=1}^{\infty}d_{k}^{2}(x).\] Consequently, \[\dfrac{-1}{2||S_{n}f||^{2}_{\infty}} \leq \dfrac{-1}{2||Sf||^{2}_{\infty}}.\] Recall the continuity property of Lebesgue measure, if \(\{E_{n}\}\) is a sequence of sets with \(E_{n}\subset E_{n+1}\) for all \(n\) and \(E= \bigcup_{n=1}^{\infty}E_{n}\), then \(|E|= \lim _{n\rightarrow \infty}|E_{n}|\). Using this we get, \[ \left|\left\{ x\in [0,1): \sup_{m\geq1} |f_{m}(x)|>\lambda \right\} \right| \leq 6 \exp \left( \dfrac{-\lambda^{2}}{2|| Sf||^{2}_{\infty}} \right). \] This completes the proof of the lemma.

Burkholder and Gundy [10] obtained the asymptotic behavior of dyadic martingale. They showed that the sets \(\{ x: Sf(x)< \infty\}\) and \(\{x:\lim f_{n} \quad \text{exists}\}\) are equal almost everywhere (a.e.) where a.e. equal means that the measure of the set where they are not equal is zero. From the result of Burkholder and Gundy, we see that the sequence of dyadic martingales \(\{f_n\}\) behave asymptotically well on the set \(\{ x: Sf(x)< \infty\}\). How does the dyadic martingale behave on the set \(\{ x: Sf(x)=\infty\}\), which is the complement of the set \(\{ x: Sf(x)< \infty\}\)? The behavior of dyadic martingales is quite pathological on the set \(\{ x: Sf(x)=\infty\}\). Precisely, it is unbounded a.e. on this set. Even though, one can obtain the rate of growth of \(|f_n|\) on the set \(\{ x: Sf(x)=\infty\}\). The rate of growth of \(|f_n|\) can be obtained by the martingale analogue of Kolmogorov's law of the iterated logarithm. Stout [6] obtained the law of the iterated logarithm for dyadic martingales. Here we obtain the same upper bound in the law of the iterated logarithm for dyadic martingales using the estimates obtained in Lemma 3 and Borel-Cantelli Lemma (Lemma 1).

3. Proof of main result

Proof of Theorem 2. Let \(\theta >1\) and \(\delta >0.\) We note that for every \(x \in [0,1)\), we have either \(S_{n}f(x) > \theta^{k}\) for some \(n\) or \(S_{n}f(x)\leq \theta^{k},\) for every \(n\), and thus, \(Sf(x) \leq \theta^{k}.\) We define stopping time as; \begin{equation*} \gamma_{k}(x)=\left\{ \begin{array}{ll} \min \left(n: S_{n+1}f(x) > \theta^{k}\right); \\ \infty, \quad if \quad \hbox{\(Sf(x) \leq \theta^{k}\).} \end{array} \right. \end{equation*} So by stopping time, \(\gamma_{k}\) is the smallest index such that \(S_{\gamma_{k}+1}f(x) > \theta^{k}.\) This means \(S_{\gamma_{k}}f(x) \leq \theta^{k}.\) Define, \begin{equation*} \tilde{f_{n}}(x)=f_{n\wedge \gamma_{k}}(x)=\left\{ \begin{array}{ll} f_{1}(x), f_{2}(x),\ldots, f_{\gamma _{k}}(x), f_{\gamma _{k}}(x), \ldots, & \hbox{for \(\gamma_{k}\neq \infty\),} \\ f_{1}(x), f_{2}(x), f_{3}(x),\ldots,& \hbox{if \(\gamma_{k}=\infty.\)} \end{array} \right. \end{equation*} We first show that \(S\tilde{f}\leq \theta^{k}.\) So for \(n< \gamma_{k}(x),\) we have \(S\tilde{f_{n}}(x)=Sf_{n}(x)\leq Sf_{\gamma_{k}}(x) \leq \theta^{k}.\) Again if \(n \geq \gamma_{k}(x) ,\) then \(S\tilde{f_{n}}(x)=Sf_{\gamma_{k}}(x) \leq \theta^{k}.\) Thus, \(\forall n\) \(S\tilde{f_{n}}(x)\leq \theta^{k}.\) Then, \( \lim_{n\rightarrow \infty} S\tilde{f_{n}}(x) \leq \theta^{k}.\) So we have \(S\tilde{f}\leq \theta^{k}.\) Choose \(\lambda=(1+\delta) \theta^{k} \sqrt{2\log\log\theta^{k}}.\) Then using Lemma 3 for the dyadic martingale \(\{\tilde{f_{n}}\}\) with the chosen \(\lambda\), we get \begin{align*} \left|\left\{ x\in [0,1): \sup_{n\geq1} |\tilde{f_{n}}(x)|>(1+\delta) \theta^{k} \sqrt{2\log\log\theta^{k}}\right\} \right| & \leq 6 \exp \left( \dfrac{-(1+\delta)^{2}\theta^{2k}2\log\log\theta^{k}} {2|| Sf||^{2}_{\infty}} \right)\\ &\leq 6 \exp \left( \dfrac{-(1+\delta)^{2}\theta^{2k}2\log\log\theta^{k}} {2 \theta^{2k}} \right)\\ &= \dfrac{6}{(k\log\theta)^{(1+\delta)^{2}}}. \end{align*} Summing over all \(k,\) we have \begin{align*} \sum_{k=1}^{\infty}\left|\left\{ x\in [0,1): \sup_{n\geq1} |\tilde{f_{n}}(x)|>(1+\delta) \theta^{k} \sqrt{2\log\log\theta^{k}}\right\} \right| &\leq \dfrac{6}{(\log\theta)^{(1+\delta)^{2}}} \sum_{k=1}^{\infty} \frac{1}{k^{(1+\delta)^{2}}} < \infty. \end{align*} Then by Borel-Cantelli Lemma 2, we have for a.e. \(x\), \[\sup_{n\geq1} |\tilde{f_{n}}(x)|\leq (1+\delta) \theta^{k} \sqrt{2\log\log\theta^{k}}\] for sufficiently large \(k\), say, \(k\geq M\), \(M\) depends on \(x.\) Thus for a.e. \(x,\) we have, \[\sup_{n\geq1} |f_{n\wedge \gamma_{k}}(x)(x)|\leq (1+\delta) \theta^{k} \sqrt{2\log\log\theta^{k}}\] for sufficiently large \(k\geq M\). We choose \(x\) such that \(f_{n}(x)\) is unbounded. Then from [10] we have, \[\{x:Sf(x)< \infty \}\overset{a.e.}{=}\{ x:f_{n}(x)\quad \text{converges}\}.\] So we have \(Sf(x)=\infty\). Then \(\gamma_{1}(x)\leq \gamma_{2}(x) \leq \gamma_{3}(x)\leq \ldots \) i.e. for every \(i,\) \( \gamma_{i}(x)< \infty.\)

Let \(n \geq \gamma_{M}.\) Then choose \(k\) such that \(\gamma_{k}(x)< n \leq \gamma_{k+1}(x).\) Here, \(\gamma_{k}(x)< n\) gives \(\gamma_{k}(x)\leq n-1.\) Thus, \(S_{n}f(x)=S_{n-1+1}f(x)> \theta^{k}.\) Using this, we have

\begin{align*} |f_{n}(x)|&\leq \sup_{1\leq m\leq \gamma_{k+1}}|f_{m\wedge\gamma_{k+1}}(x)|\\ &\leq \sup_{m\geq 1}|f_{m\wedge\gamma_{k+1}}(x)|\\ & \leq (1+\delta)\theta^{k+1}\sqrt {2\log\log\theta^{k+1}}\\ &< (1+\delta) S_{n}f(x) \theta \sqrt{2\log(\log S_{n}f(x) +\log\theta).} \end{align*} So, \[ \limsup_{n\rightarrow \infty} \dfrac{|f_{n}(x)|} {S_{n}f(x) \sqrt{2\log(\log S_{n}f(x))}} < (1+\delta) \theta \limsup_{n\rightarrow \infty} \sqrt{\dfrac{2 \log(\log S_{n}f(x)+\log\theta)}{2\log(\log S_{n}f(x))}}. \] Clearly, \[\limsup_{n\rightarrow \infty} \sqrt{\dfrac{ \log(\log S_{n}f(x)+\log\theta)}{\log(\log S_{n}f(x))}}=1.\] Therefore for a.e. \(x,\) \[ \limsup_{n\rightarrow \infty} \dfrac{|f_{n}(x)|} {S_{n}f(x) \sqrt{2\log\log S_{n}f(x)}} < (1+\delta) \theta. \] Letting \(\theta \searrow 1\) we get, \[ \limsup_{n\rightarrow \infty} \dfrac{|f_{n}(x)|} {S_{n}f(x) \sqrt{2\log\log S_{n}f(x)}} \leq 1+ \delta. \] This can be done for every \(\delta>0.\) Hence we have for a.e. \(x,\) \[ \limsup_{n\rightarrow \infty} \dfrac{|f_{n}(x)|} {S_{n}f(x) \sqrt{2\log\log S_{n}f(x)}} \leq 1. \]

4. Conclusion

The upper bound of the law of the iterated logarithm in the context of dyadic martingale using analytic approach has been obtained where we made the use of some subgaussian type estimates and Borel-Cantelli Lemma. We look forward to obtain the lower bound result using the similar approach.

Acknowledgments :

The author would like to thank Prof. Charles N. Moore of Washington State University, USA for his valuable suggestions on this article.

Conflicts of Interest:

''The author declares no conflict of interest.''

Data Availability:

All data required for this research is included within this paper.

Funding Information:

No funding is available for this research.