1. Introduction

Let \(z\in \mathbb{C}\) and assume that \(\Re z > 0\). It is well-known that

where \(J(z)\) can be written as The analytic function \(J(z)\) is known as Binet's function and has several equivalent expressions; see for example, Henrici [1, (8.5-7)].

Binet's function has an asymptotic expansion

or more precisely, for non-negative integers \(k\), where and It may be shown that where \(B_{2k+2}\) is a Bernoulli number (\(B_2 = 1/6, B_4 = -1/30\), etc.). Proofs of these results are given in Henrici's book [1, §11.1]\(^1\).

Substituting (4) into (1) gives an asymptotic expansion for \(\ln\Gamma(z)\) that is usually named after James Stirling, although some credit is due to Abraham de Moivre. For the history and early references, see Dutka [2]. It is interesting to note that de Moivre started (about 1721) by trying to approximate the central binomial coefficient \(\binom{2n}{n}\), not the factorial (or Gamma) function- see Dutka [2,pg.227].

It is easy to see from (5) and (6) that

where Suppose now that \(z\) is real and positive. Since \(z^2/(z^2+\eta^2) \in (0,1)\) and the logarithmic factors in (9) are positive for all \(\eta\in(0,\infty)\), we see that Thus, the remainder \(r_k(z)\) given by (8) has the same sign as the next term \((-1)^k\beta_k/z^{2k+1}\) in the asymptotic series, and is smaller in absolute value. In the terminology used by Pólya and Szegö [3, Ch.4], the asymptotic series for \(\ln\Gamma(z)\) strictly envelops\(^2\) the function \(\ln\Gamma(z)^3\).

§2 shows that we can deduce a strictly enveloping asymptotic series for \(\ln(\Gamma(2z+1)/\Gamma(z+1)^2)\) or equivalently, if \(z=n\) is a positive integer, for the logarithm of the central binomial coefficient \(\binom{2n}{n}\). The series itself is well known, but we have not found the enveloping property or the resulting error bound mentioned in the literature. Henrici was aware of it, since in his book [1, §11.2, Problem 6] he gave the special case \(k=3\) as an exercise, along with a hint for the solution. Hence, we do not claim any particular originality. Our purpose is primarily to make some useful asymptotic series and their associated error bounds readily accessible. Related results and additional references may be found, for example, in [4,5,6].

In §2 we consider the central binomial coefficient and its generalisation to a complex argument. Then, in §3, we consider some closely related asymptotic series that we can prove to be strictly enveloping. In §4 we make some remarks on asymptotic series that are not enveloping. An Appendix gives numerical values of the coefficients appearing in three of the asymptotic series.

Finally, we remark that it is possible to give asymptotic series related to \(\Gamma(z+\frac12)/\Gamma(z)\) and \(\binom{2n}{n}\), but in general these series do not alternate in sign. See, for example, [7], [8], [9, ex.9.60 and pg.602], [10], and [11].

2. Asymptotic series for central binomial coefficients

Define \begin{align*} \widetilde{\Gamma}(z) :=& \frac{\Gamma(2z+1)}{\Gamma(z+1)^2}\,,\\ \widetilde{J}(z) :=& J(2z) - 2J(z),\\ \widetilde{r}_k(z) :=& r_k(2z) - 2r_k(z), \end{align*} and As noted above, the central binomial coefficient \(\binom{2n}{n}\) is simply \(\widetilde{\Gamma}(n)\).

Using elementary properties of the Gamma function, we have

Thus, from (1) and the same equation with \(z \mapsto 2z\), we have Also, from (4) and the definition of \(\widetilde{J}(z)\), we have an asymptotic series for \(\widetilde{J}(z)\), namely: Since \(\binom{2n}{n} = \widetilde{\Gamma}(n)\), equations (13)-(14) give an asymptotic series for \(\ln\binom{2n}{n}\). Lemma 1 shows that the remainder \(\widetilde{r}_k(z)\) can be expressed as an integral analogous to the integral (6) for \(r_k(z)\).

Lemma 1. For \(z\in\mathbb{C}, \Re z > 0\), and \(k\) a non-negative integer,

and

Proof. Making the change of variables \(z \mapsto 2z\) and \(\eta \mapsto 2\eta\) in (6), we obtain \[ r_k(2z) = \frac{(-1)^k}{\pi z^{2k-1}}\int_0^\infty \frac{\eta^{2k}}{z^2+\eta^2} \ln\left(\frac{1}{1-e^{-4\pi\eta}}\right)\,d\eta\,. \] Now \begin{equation*} \ln\left(\frac{1}{1-e^{-4\pi\eta}}\right) - 2\ln\left(\frac{1}{1-e^{-2\pi\eta}}\right) = \ln\left(\frac{1-e^{-2\pi\eta}}{1+e^{-2\pi\eta}}\right) = \ln\tanh(\pi\eta), \end{equation*} so (16)-(17) follow from the definitions of \(\widetilde{r}_k(z)\) and \(\widetilde{J}(z)\). The proof of (15) is similar.

Corollary 1 gives a result analogous to Equations (8)-(9).

Corollary 1. For \(z\in\mathbb{C}, \Re z > 0\), and \(k\) a non-negative integer,

where

Proof. This is straightforward from Equations (15)-(16) of Lemma 1.

Corollary 2 gives a result analogous to the bound (10).

Corollary 2. If \(z\) is real and positive, then \(\widetilde{\theta}_k(z) \in (0,1)\).

Proof. We write (19) as

Observe that \(\coth(y) = \cosh(y)/\sinh(y) > 1\) for \(y\in(0,\infty)\), so \(\ln\coth(y) > 0\) for \(y = {\pi\eta} > 0\). Since \(z^2/(z^2+\eta^2) \in (0,1)\) for real positive \(z\) and \(\eta\), the result follows.

Corollary 3. If \(z\) is real and positive, then the asymptotic series (14) for \(\widetilde{J}(z)\) is strictly enveloping.

Proof. This is immediate from Corollary 2.

Remark 1. We may compare Corollary 2 with (the proof of) Lemma 2.7 of [12]. The latter, after allowing for different notation, gives the bound \[ \frac{-1}{4^{k+1}-1} < \widetilde{\theta}_k(z) < \frac{4^{k+1}}{4^{k+1}-1}. \] This is clearly weaker than the bound of Corollary 2, and not sufficient to prove Corollary 3.

3. Some related asymptotic series

Lemma 2. If \(z\in\mathbb{C}, \Re z > 0\), then \[ \widetilde{J}(z) = \ln\left(\frac{\Gamma(z+\frac12)}{z^{1/2}\Gamma(z)}\right) . \]

Proof. This follows from Equations (12)-(13) and the duplication formula \(\Gamma(z)\Gamma(z+\frac{1}{2}) = 2^{1-2z}\pi^{1/2}\Gamma(2z)\).

From Lemma 2 and (14) we immediately obtain an asymptotic expansion which is strictly enveloping if \(z\) is real and positive.

Define

Using the asymptotic expansion for \(\ln\Gamma(z)\) given by Equations (1) and (3), we see from (21) that \(\ln\Gamma(z+\frac12)\) has an asymptotic expansion In fact, the expansion (23) was already obtained by Gauss [13, Eqn.[59] of Art.29] in 1812. However, Gauss did not explicitly bound the truncation error. Writing (23) as we have an unsurprising result for the truncation error \(\widehat{r}_k(z)\): the error is of the same sign as the first neglected term \((-1)^{k+1}\widehat{\beta}_k/z^{2k+1}\), and is bounded in magnitude by this term. This is shown in Lemma 3 and Corollaries 4-5 below.

Lemma 3. For \(z\in\mathbb{C}, \Re z > 0\), and \(k\) a non-negative integer,

and

Proof. This is similar to the proof of Lemma 1.

Corollary 4. For \(z\in\mathbb{C}, \Re z > 0\), and \(k\) a non-negative integer,

where

Proof. This is a straightforward consequence of Lemma 3.

Corollary 5. If \(z\) is real and positive, then the asymptotic expansion for \(\ln\Gamma(z+\frac{1}{2})\) given in (24) is strictly enveloping.

Proof. From (28) we have \(\widehat{\theta}_k(z) \in (0,1)\).

Remark 2. If we make the change of variables \(z \mapsto n+\frac12\) in (23), and assume that \(n\) is a positive integer, we obtain an asymptotic series for \(n!\) in negative powers of \((n+\frac{1}{2})\):

In fact, (29) was stated (without proof) by de Moivre [14,15] as early as 1730, see Dutka [1,(5), pg.233].

4. Non-enveloping asymptotic series

Lest the reader has gained the impression that all "naturally occurring" asymptotic series are enveloping (for real positive arguments), we give two classes of examples to show that this is not the case. In fact, enveloping series are the exception, not the rule. Our first class of examples is given by the following Lemma.

Lemma 4. Let \(x\in(0,+\infty)\) and \(f(x) := J(x) + \exp(-bx)\) for some constant \(b \in (0, 2\pi)\). Then \(f(x)\) has an asymptotic series

However, the series (30) does not envelop \(f\).

Proof. For all \(k \ge 0\), \(\exp(-bx) = O(x^{-2k-1})\) as \(x \to +\infty\). Thus, it follows from (4) that \(f(x)\) has the claimed asymptotic series (in fact the same series as the Binet function \(J\).) This proves the first claim.

To prove the final claim, suppose, by way of contradiction, that the series (30) envelops \(f\). For each integer \(k > 0\), define \(x_k := k/\pi\). From (7), the \(\beta_k\) grow like \((2k)!/(2\pi)^{2k}\), and from Stirling's approximation we see that

Since the same series envelops both \(f\) and \(J\), (31) implies that \[ |f(x_k)-J(x_k)| = O(\exp(-2\pi x_k)) \;\text{ as }\; k \to \infty. \] Since \(\exp(-2\pi x) = o(\exp(-bx))\), it follows that, for sufficiently large \(k\), \[ |f(x_k)-J(x_k)| < \exp(-bx_k). \] This contradicts the definition of \(f\), so the assumption that the series (30) envelops \(f\) must be false.

Remark 3. Lemma 4 can be generalised. For example, the conclusion holds if \(f(x) = J(x) + g(x)\), where \(g(x) = O(x^{-k})\) for all positive integers \(k\), but \(g(x) \ne O(\exp(-2\pi x))\). Also, we can replace the function \(J(x)\) by a different function that has an enveloping asymptotic series whose terms grow at the same rate as those of \(J(x)\).

Our second class of examples involves asymptotic expansions where all (or all but a finite number) of the terms are of the same sign (assuming a positive real argument \(x\)). Such series can not be strictly enveloping [3, Ch.4]. As examples, we mention the Bessel function \(I_0(x)\) (see Olver and Maximon [16,§10.40.1]), the product of two Bessel functions \(I_0(x)K_0(x)\) (see [16, §10.40.6] and [17, Lemma 3.1]), and the Riemann-Siegel theta function (see [18, §6.5]). In all these examples the terms have constant sign, so the remainder changes monotonically as the number of terms increases with the argument \(x\) fixed. Eventually the remainder changes sign and starts increasing in absolute value. Often the point where the remainder changes sign is close to where the terms are smallest in absolute value, but this is not always true- see for example [19, §§4-5].

5. Concluding remarks

We have considered three different but related asymptotic series that can all be proved to be strictly enveloping. Our proofs depend on the fact that the three relevant functions \(-\ln(1-e^{-2\pi\eta})\), \(\ln\coth(\pi\eta)\), and \(\ln(1+e^{-2\pi\eta})\) are positive for all \(\eta\in(0,\infty)\). We remark that these three functions are linearly dependent, since \[ \coth(\pi\eta) = \frac{1+e^{-2\pi\eta}}{1-e^{-2\pi\eta}}\,. \] It follows that the sequences \((\beta_k)_{k\ge 0}\), \((\widetilde{\beta}_k)_{k\ge 0}\) and \((\widehat{\beta}_k)_{k\ge 0}\) are linearly dependent. In fact, \(\widetilde{\beta}_k = \beta_k + \widehat{\beta}_k\) for all \(k\ge 0\). A table of numerical values is given in the Appendix.

Acknowledgments: We thank an anonymous referee for simplifying the proof of Lemma 4 and for noting that the domain of validity of (2) is the right half-plane \(\Re z > 0\) (although the Binet function \(J(x)\) may be continued into the left half-plane by analytic continuation, see [1, Thm.8.5a]).

Conflicts of Interest:

"The author declares no conflict of interest".

1. Introduction

It is well known that the CEV model is one of very popular models in finance. The dynamic of this model is described by the following Itô stochastic differential equation

where \(X_0,a,b,\sigma\) are positive constants, \(\alpha\in (\frac{1}{2},1)\) and \(B=(B_t)_{0\leq t\leq T}\) is a standard Brownian motion.

The solution \((X_t)_{0\leq t\leq T}\) to the model (1) is a Markov process without memory. However, in the last few decades, there are many observations showing that an asset price or an interest rate is not always a Markov process since it has long-range aftereffects. Many studies have pointed out that the dynamics driven by fractional Brownian motion are a suitable choice to model such objects, see [2] and the references therein. Hence, it is important to take into account the effect of fractional noise to the model (1). We recall that a fractional Brownian motion (fBm) of Hurst parameter \(H\in (0,1)\) is a centered Gaussian process \(B^H=(B^H_t)_{0\leq t\leq T}\) with covariance function

\[R_H(t,s):=E\left[B^H_t B^H_s\right]=\frac{1}{2}\left(t^{2H}+s^{2H}-|t-s|^{2H}\right).\] For \(H>1/2\), \(B^H_t\) admits the so-called Volterra representation (see [3] pp. 277-279) where \((W_t)_{t\geq 0}\) is a standard Brownian motion, \[ K(t,s):=c_H\,s^{\frac{1}{2}-H}\int_s^t (u-s)^{H-\frac{3}{2}}u^{H-\frac{1}{2}}du,\quad\text{\(s\leq t\)} \] and \( c_H=\sqrt{\frac{H(2H-1)}{\beta(2-2H,H-\frac{1}{2})}},\) \(\text{where \(\beta\) is the Beta function.}\)

In this paper, we consider the mixed-fractional CEV model that is defined as the stochastic differential equations of the form

where the initial condition \(X_0\) and \(a,b,\sigma,\sigma_H\) are positive constants, \(\frac{1}{2}< \alpha< 1,\mbox{ and } B_t^H \) is fBm with \(H>\frac{1}{2}.\) The stochastic integral with respect to \(B\) is the Itô integral. Meanwhile, the stochastic integral with respect to \(B^H\) is interpreted as a pathwise Stieltjes integral, which has been frequently used in the studies related to fBm. We refer readers to Zähle's paper [4] for a detailed presentation of this integral.

Recently, the applications in finance of the mixed-fractional CEV model have been extensively discussed, see [5] and references therein. In the present paper, our aim is to study the tail distribution of solutions to (3). This problem is important because the probability distribution function is one of the most natural features for any random variable. In fact, in the last decade, the tail distribution estimates for various random variables have been investigated by many authors, see e.g. [1,6,7] and references therein. In the present paper, we will focus on providing explicit estimates for the probability distribution of \(X_t,\) see Theorem 1 below.

The volatility coefficient of the model (3) violates the Lipschitz continuous condition which is traditionally imposed in the study of stochastic differential equations. This causes some mathematical difficulties which make the study of the model (3) particularly interesting. In order to be able to handle such difficulties, our tools are the techniques of Malliavin calculus and a result established recently in [1].

The rest of the paper is organized as follows: In §2, we recall some fundamental concepts of Malliavin calculus. The main results of the paper are stated and proved in §3.

2. Preliminaries

This paper is strongly based on techniques of Malliavin calculus. For the reader's convenience, let us recall some elements of Malliavin calculus. We refer to [3] for a more complete treatment of this topic. We assume that two-dimensional Browian motion \(w=(B,W)\) is defined in a complete probability space \((\Omega ,\mathcal{F},P)\) and the \(\sigma \)-field \(\mathcal{F}\) is generated by \(w\). Let us denote by \(H\) the Hilbert space \(L^{2}([0,T];\mathbb{R}^{2}),\) and for any function \(h=\left(h^B,h^W\right)\in H\) we set \begin{equation*} w(h)=\int_{0}^Th^{B}(t)dB_{t}+\int_{0}^Th^{W}(t)dW_{t}. \end{equation*} Let \(\mathcal{S}\) be the class of smooth and cylindrical random variables of the form \begin{equation*} F=f(w(h_{1}),\ldots ,w(h_{n})), \end{equation*} where \(n\geq 1\), \(h_{1},\ldots ,h_{n}\in H\), and \(f\) is an infinitely differentiable function such that together with all its partial derivatives have at most polynomial growth order. The derivative operator of the random variable \(F\) is defined as \begin{align*} &D_{t}^{B}F=\sum_{j=1}^{n}\frac{\partial f}{\partial x_{j}}(w(h_{1}),\ldots,w(h_{n}))h_{j}^{B}(t),\\ &D_{t}^{W}F=\sum_{j=1}^{n}\frac{\partial f}{\partial x_{j}}(w(h_{1}),\ldots,w(h_{n}))h_{j}^{W}(t), \end{align*} where \(t\in [0,T]\). In this way, we interpret \(DF\) as a random variable with values in the Hilbert space \(H\). The derivative is a closable operator on \(L^{2}(\Omega )\) with values in \(L^{2}(\Omega ;H)\). We denote by \(\mathbb{D}^{1,2}\) the Hilbert spaced defined as the completion of \(\mathcal{S}\) with respect to the scalar product \begin{equation*} \left\langle F,G\right\rangle _{1,2}=E[FG]+E\left[\int_{0}^TD_{t}^{B}FD_{t}^{B}Gdt+\int_{0}^TD_{t}^{W}FD_{t}^{W}Gdt\right] . \end{equation*} A random variable \(F\) is said to be Malliavin differentiable if it belongs to \(\mathbb{D}^{1,2}.\) We have the following general estimate for tail probabilities.

Lemma 1. Let \(Z\) be a centered random variable in \(\mathbb{D}^{1,2}.\) Assume there exists a non-random constant \(L\) such that

Then following estimate for tail probabilities holds

Proof. The proof is similar to that of Lemma 2.2 in [1]. By Clark-Ocone formula we have \[Z=\int_0^T E\left[D^B_rZ|\mathcal{F}_r\right]dB_r+\int_0^T E\left[D^W_rZ|\mathcal{F}_r\right]dW_r.\] Hence, for any \(\lambda\in \mathbb{R},\) we obtain \begin{align*}Ee^{\lambda Z}=&E\exp\left(\lambda\int_0^T E\left[D^B_rZ|\mathcal{F}_r\right]dB_r+\lambda\int_0^T E\left[D^W_rZ|\mathcal{F}_r\right]dW_r\right)\\ =&E\exp\left(\lambda\int_0^T E\left[D^B_rZ|\mathcal{F}_r\right]dB_r-\frac{\lambda^2}{2}\int_0^T \left(E\left[D^B_rZ|\mathcal{F}_r\right]\right)^2dr+\frac{\lambda^2}{2}\int_0^T \left(E\left[D^B_rZ|\mathcal{F}_r\right]\right)^2dr\right)\\ &\times E\exp\left(\lambda\int_0^T E\left[D^W_rZ|\mathcal{F}_r\right]dB_r-\frac{\lambda^2}{2}\int_0^T \left(E\left[D^W_rZ|\mathcal{F}_r\right]\right)^2dr+\frac{\lambda^2}{2}\int_0^T \left(E\left[D^W_rZ|\mathcal{F}_r\right]\right)^2dr\right)\\ \leq& e^{\frac{\lambda^2}{2}M^2}EN_T, \end{align*} where \((N_t)_{t\in[0,T]}\) is a stochastic process defined by \[N_t:=\exp\left(\lambda\int_0^t E\left[D^B_rZ|\mathcal{F}_r\right]dB_r+\lambda\int_0^t E\left[D^W_rZ|\mathcal{F}_r\right]dW_r-\frac{\lambda^2}{2}\int_0^t \left(\left(E\left[D^B_rZ|\mathcal{F}_r\right]\right)^2+\left(E\left[D^B_rZ|\mathcal{F}_r\right]\right)^2\right)dr\right).\] By using Itô formula, we obtain \[N_T=1+\lambda \int_0^TN_rE\left[D^B_rZ|\mathcal{F}_r\right]dB_r+\lambda \int_0^TN_rE\left[D^W_rZ|\mathcal{F}_r\right]dW_r,\] which implies that \(EN_T=1.\) Thus we get \[Ee^{\lambda Z}\leq e^{\frac{\lambda^2}{2}L^2}EN_T= e^{\frac{\lambda^2}{2}L^2}.\] This, together with Markov's inequality, gives us \[P\left(Z\geq x\right)=P\left(e^{\lambda Z}\geq e^{\lambda x}\right)\leq e^{\frac{\lambda^2}{2}L^2-\lambda x},\,\,\lambda>0,x\in \mathbb{R}.\] When \(x>0,\) we choose \(\lambda=x/L^2,\) and we get \[P\left(Z\geq x\right)\leq e^{-\frac{x^2}{2L^2}},\quad x>0.\] So we can finish the proof of Lemma.

3. The main results

We first show that the equation (3) has a unique solution. Following the method used in [8], we consider the following equation the initial value \(V_0:=X_0^{1-\alpha}> 0.\) We put \[g(x)=(1-\alpha)\left(ax^{\frac{-\alpha}{1-\alpha}}-bx-\frac{\alpha\sigma^2}{2x}\right),\,\,\,x>0,\] and rewrite the Equation (6) as follows \[ V_t=V_0+\int_0^tg(V_s)ds+\sigma(1-\alpha)B_t+\sigma_H(1-\alpha)B_t^H,\,\,\,t\geq 0.\]

Lemma 2. We have

where \(x_0\in (0,\infty)\) such that \( x_0^{\frac{1}{1-\alpha}-2}=\frac{a}{(1-\alpha)^2\sigma^2}.\)

Proof. We have \[ g'(x)=-a\alpha x^{\frac{-1}{1-\alpha}} -b(1-\alpha)+\frac{\alpha(1-\alpha)\sigma^2}{2x^2} \] and \begin{align*} g''(x)&=x^{\frac{-1}{1-\alpha}-1}\left(\frac{a\alpha}{1-\alpha}-\alpha(1-\alpha)\sigma^2x^{\frac{1}{1-\alpha}-2}\right). \end{align*} We note that \(\frac{1}{2}< \alpha< 1\) and so \(\frac{1}{1-\alpha}-2>0\). Hence, it is easy to see that \(g''(x_0)=0\) and \(\sup\limits_{x>0}g'(x)=g'(x_0).\) We thus obtain the relation (7).

Proposition 1. The Equation (6) admits a unique positive solution. Moreover, \(V_t>0\mbox{ } a.s.\) for any \(t\geq 0.\)

Proof. We observe that the function \(g(x)=(1-\alpha)\left(ax^{\frac{-\alpha}{1-\alpha}}-bx-\frac{\alpha\sigma^2}{2x}\right)\) is Lipschitz continuous on the neighborhood of \(V_0>0.\) Hence, there exists a local solution \(V_t\) on the interval \([0,\tau],\) where \(\tau\) is the stopping time such that \(\tau=\inf\left\{t>0:V_t=0\right\}.\) Assume that \(\tau< \infty.\)

For all \(t\in [0,\tau),\) we have

We note that \[g(x)x^{\frac{\alpha}{1-\alpha}}=(1-\alpha)\left(a-bx^{\frac{1}{1-\alpha}}-\frac{\alpha}{2}\sigma^2x^{\frac{2\alpha-1}{1-\alpha}}\right).\] Because \(\frac{1}{2}< \alpha< 1\mbox{ we have }\frac{1}{1-\alpha}>0 \mbox{ and }\frac{2\alpha-1}{1-\alpha}>0.\) Therefore, \[\lim\limits_{x\to 0^+}g(x)x^{\frac{\alpha}{1-\alpha}}=a(1-\alpha)>0.\] Hence, there exists \(\varepsilon >0\) such that \[ g(x)>\frac{a(1-\alpha)}{2x^{\frac{\alpha}{1-\alpha}}},\mbox{ }\forall x\in (0,\varepsilon). \] Since \(V_t\) is continuous, and \(V_{\tau}=0,\) there exists \(t_0\) such that \(V_t\in (0,\varepsilon), \mbox{ }\forall t\in [t_0,\tau) \) which implies that Recall that the paths of Brownian motion are \(\beta\)-Hölder continuous for any \(\beta< \frac{1}{2}\) and the paths of fBm are \(\beta\)-Hölder continuous for any \(\beta< H.\) So, fixed \(\beta< \frac{1}{2}\) then there exists a finite random variable \(C_{\beta}(\omega )\) such that \[ \left|\sigma(1-\alpha)(B_{\tau}-B_t)+\sigma_H(1-\alpha)\left(B_{\tau}^H-B_t^H\right)\right|\le C_{\beta}(\omega)\left|\tau-t\right|^{\beta}. \] This, combined with (8), gives us \begin{align*} 0< V_t&=-\int_t^{\tau}g(V_s)ds-\sigma(1-\alpha)(B_{\tau}-B_t)-\sigma_H(1-\alpha)\left(B_{\tau}^H-B_t^H\right)\\ &< \left|\sigma(1-\alpha)(B_{\tau}-B_t)+\sigma_H(1-\alpha)\left(B_{\tau}^H-B_t^H\right)\right|\\ &< C_{\beta}(\omega)\left(\tau-t\right)^{\beta},\mbox{ }\forall t\in [t_0,\tau), \end{align*} and \[ 0\leq \int_t^{\tau}g(V_s)ds< C_{\beta}(\omega)(\tau-t)^{\beta},\mbox{ }\forall t\in [t_0,\tau). \] As a consequence, it follows from (9) that \[C_{\beta}(\omega)(\tau-t)^{\beta}>\int_t^{\tau}g(V_s)ds> \int_t^{\tau}\frac{a(1-\alpha)}{2V_s^{\frac{\alpha}{1-\alpha}}}ds>\int_t^{\tau} \frac{a(1-\alpha)}{2\left[C_{\beta}(\omega)(\tau-s)^{\beta}\right]^{\frac{\alpha}{1-\alpha}}}ds,\mbox{ }\forall t\in [t_0,\tau).\] Therefore, it holds that or equivalently \[ 2\left[C_{\beta}(\omega)\right]^{\frac{1}{1-\alpha}}\frac{1}{a(1-\alpha)}>(\tau-t)^{1-\frac{\beta}{1-\alpha}},\mbox{ }\forall t\in [t_0,\tau). \] We choose \(\beta\) such that \(\frac{1}{2}>\beta>1-\alpha\) then \(1-\frac{\beta}{1-\alpha}< 0.\) We get a contradiction beacause the right hand side of (10) tends to \(\infty\) as \(t\to\tau.\) We conclude that \(\tau=\infty.\) Thus, the Equation (6) exists global solution with \(V_0>0.\)

The uniqueness of the solutions can be verified as follows. Let \(V_t\) and \(V_t^*\) be two solutions of (6) with the same initial condition \(V_0.\) We have

\[V_t-V_t^*=\int_0^t \left[g(V_s)-g(V_s^*)\right] ds,\,\,\,0\leq t\leq T,\] and hence, \[ \left(V_t-V_t^*\right)^2=2\int_0^t\left(V_s-V_s^*\right)\left[g(V_s)-g(V_s^*)\right] ds,\,\,\,t\geq 0.\] By using Lagrange's theorem, there exists a random variable \(\theta \) lying between 0 and 1 such that \[ \left(V_t-V_t^*\right)^2=2\int_0^t g'\left(V_s+\theta (V_s^*-V_s)\right)\left(V_s-V_s^*\right)^2 ds,\,\,\,t\geq 0.\] By Lemma 2, we deduce \[ \left(V_t-V_t^*\right)^2\le 2M\int_0^t \left(V_s-V_s^*\right)^2 ds\le \varepsilon +2M\int_0^t \left(V_s-V_s^*\right)^2 ds,\mbox{ }\forall \varepsilon >0.\] We use Gronwall's lemma to get \[ \left(V_t-V_t^*\right)^2\le \varepsilon e^{2Mt}\le \varepsilon e^{2MT},\mbox{ }\forall t\geq 0, \mbox{ }\forall \varepsilon >0.\] The right hand converges to \(0 \) as \(\varepsilon\to 0,\) hence, \(V_t=V_t^*,\mbox{ }\forall t\in [0,T].\) The proof of Proposition is complete.

Proposition 2. The Equation (3) has a unique solution given by \(X_t=V_t^{\frac{1}{1-\alpha}},\,\,0\leq t\leq T,\) where \(V_t\) is the solution of (6).

Proof. The proof is similar to that of Theorem 2.1 in [8]. So we omit it.

Next, we will prove the solution \(V_t\) of (6) is Malliavin differentiable. By Volterra expression of fBm, we can rewrite (6) by the following equation

Proposition 3. Let \((V_t)_{0\leq t\leq T}\) be the solution of the Equation (6). Then, for each \(t\in(0,T],\) the random variable \(V_t\) is Malliavin differentiable. Moreover, we have \begin{align*} D_s^B V_t &=\sigma(1-\alpha)\exp\left(\int_s^tg'(V_r)dr\right) \mathbb{I}_{[0,t]}(s)\\ D_s^W V_t &=\sigma_H(1-\alpha)\int_s^tK_1(v,s)\exp\left(\int_v^tg'(V_r)dr\right)dv\mathbb{I}_{[0,t]}(s) \end{align*} where \( K_1(v,s) = \frac{\partial}{\partial v}K(v,s)= c_{H}(v-s)^{H- \frac{3}{2}}v^{H-\frac{1}{2}} r^{\frac{1}{2}-H}.\)

Proof. Fix \(t\in(0,T].\) Let us compute the directional derivative \(\langle D^BV_t,h\rangle_{L^2[0,T]}\) with \(h\in L^2[0,T]:\) \[\langle D^BV_t,h\rangle_{L^2[0,T]} = \frac{dV^\varepsilon_t}{d\varepsilon}|_{\varepsilon =0},\] where \(V^\varepsilon_t\) solves the following equation \[ V_t^{\varepsilon } =V_0+\int_0^t g\left(V_s^{\varepsilon }\right)ds+\sigma(1-\alpha)\left(B_t+\varepsilon\int_0^t h_sds\right)+\sigma_H(1-\alpha)dB_t^H, t\in[0,T],\varepsilon\in(0,1).\] By using Lagrange's theorem, we get

for some random variables \(\xi_s\) lying between 0 and 1. The solution of (12) is given by \[ V_t^{\varepsilon }-V_t= \sigma(1-\alpha)\varepsilon\int_0^t h_s\left(\exp\int_s^tg'\left(V_r+\xi _r(V_r^{\varepsilon}-V_r)\right)dr\right)ds,\,\, t\in[0,T],\] which implies that \[ \frac{V_t^{\varepsilon }-V_t}{\varepsilon}=\sigma(1-\alpha)\int_0^t h_s\left(\exp\int_s^tg'\left(V_r+\xi _r(V_r^{\varepsilon}-V_r)\right)dr\right)ds.\] We recall that \(g'(x)\leq M< \infty,\mbox{ }\forall x>0.\) Hence, by the dominated convergence theorem, we obtain \begin{align*} \lim_{\varepsilon\to 0^+}\frac{V_t^{\varepsilon }-V_t}{\varepsilon}&= \sigma(1-\alpha)\int_0^t h_s\exp\left(\int_s^tg'(V_r)dr\right)ds\\ &=\sigma(1-\alpha)\int_0^T h_s\exp\left(\int_s^tg'(V_r)dr\right)\mathbb{I}_{[0,t]}ds\\ &= \Big < h,\sigma(1-\alpha)\exp\left(\int_s^tg'(V_r)dr\right)\mathbb{I}_{[0,t]}\Big >_{L^2[0,T]}, t\in[0,T], \end{align*} where the limit holds in \(L^2(\Omega ).\) According to the results of Sugita [9], we can conclude that \(V_t\) is Malliavin differentiable with respect to \(B\) and its derivative is given by \begin{align*} D_s^B V_t=&\sigma(1-\alpha)\exp\left(\int_s^tg'(V_r)dr\right)\mathbb{I}_{[0,t]}(s). \end{align*} In a same way, we compute the directional derivative \(\langle D^WV_t,h\rangle= \frac{dV^\theta_t}{d\varepsilon}|_{\theta =0}\), where \(V_t^{\theta }\) satisfies \begin{align*} V_t^{\theta } &=V_0+\int_0^t g\left(V_s^{\theta }\right)ds+\sigma(1-\alpha)B_t+\sigma_H(1-\alpha)\int _0^t K(t,s)d\left(W_s+\theta \int_0^s h_udu\right)\end{align*}\begin{align*} &=V_0+\int_0^t g\left(V_s^{\theta }\right)ds+\sigma(1-\alpha)B_t+\sigma_H(1-\alpha)\int _0^t K(t,s)\left(dW_s+\theta h_sds\right), t\in[0,T],\theta \in[0,1). \end{align*} Using Lagrange's theorem again, we have where \(\zeta_s \) is a random variable between 0 and 1. The solution of (13) is represented by \[ V_t^{\theta } -V_t=\theta\sigma_H(1-\alpha)\int_0^t \left(\int_0^sK_1(s,u)h_udu\right)\exp\left(\int_s^tg'\left(V_r+\zeta _r\left(V_r^{\theta }-V_r\right) \right)dr\right)ds. \] Hence, \begin{align*} \lim_{\theta\to 0^+}\frac{V_t^{\theta } -V_t}{\theta}&=\sigma_H(1-\alpha)\int_0^t \left(\int_0^sK_1(s,u)h_udu\right)\exp\left(\int_s^tg'\left(V_r \right)dr\right)ds.\\ &= \Big < h_s,\sigma_H(1-\alpha)\int_s^tK_1(v,s)\exp\left(\int_v^tg'(V_r)dr\right)dv\mathbb{I}_{[0,t]}(s)\Big >_{L^2[0,T]}. \end{align*} Thus \(V_t\) is Malliavin differentiable with respect to \(W\) and we have \begin{align*} D_s^W V_t&=\sigma_H(1-\alpha)\int_s^tK_1(v,s)\exp\left(\int_v^tg'(V_r)dr\right)dv\mathbb{I}_{[0,t]}(s). \end{align*} The proof of Proposition is complete.

We now are in a position to state and prove the main result of this paper.

Theorem 1. Let \((X_t)_{0\leq t\leq T}\) be the unique solution of the Equation (3). Then, for each \(t\in(0,T],\) the tail distribution of \(X_t\) satisfies \[ P(X_t\ge x)\leq\exp\left(-\frac{\left(x^{1-\alpha}-\mu_t^{1-\alpha}\right)^2}{2\left(\frac{\sigma^2\left(1-\alpha\right)^2}{2M}\left(e^{2Mt} -1\right) +\sigma_H^2\left(1-\alpha\right)^2e^{2Mt}t^{2H}\right)}\right),\,\,\,x> \mu_t,\] where \(\mu_t:=E[X_t]\) and \(M\) is defined by (7).

Proof. Recalling Proposition 3 we get \begin{align*} 0&\leq D_r^B V_t\le\sigma(1-\alpha)e^{M(t-r)},\\ 0&\leq D_r^W V_t = \sigma_H(1-\alpha)\int_r^tK_1(v,r)e^{\left(\int_v^tg'(V_r)dr\right)}dv\\ &\le \sigma_H(1-\alpha)\left( K(t,r)+ M\int_r^t K(v,r)e^{M(t-v)}dv\right) , \mbox{ }0\le r\le t\le T. % & = \sigma_H(1-\alpha) \end{align*} Because the function \(v \rightarrow K(v,r)\) is non-decreasing, this implies \begin{align*} D_r^W V_t&\le \sigma_H(1-\alpha)\left( K(t,r)+ MK(t,r)\int_r^t e^{M(t-v)}dv\right)\\ &=\sigma_H(1-\alpha)K(t,r)e^{M(t-r)}, \mbox{ }0\le r\le t\le T. \end{align*} We have \begin{align*} \int_0^T\left(E\left[D_r^BV_t|\mathcal{F}_s\right]\right)^2dr &= \int_0^t\left(E\left[D_r^BV_t|\mathcal{F}_s\right]\right)^2dr\\&\le \int_0^t\left(\sigma(1-\alpha)e^{M(t-r)}\right)^2dr\\ & = \frac{\sigma^2(1-\alpha)^2}{2M}\left(e^{2Mt} -1\right) \end{align*} and \begin{align*} \int_0^T\left(E\left[D_r^WV_t|\mathcal{F}_r\right]\right)^2dr &= \int_0^t\left(E\left[D_r^WV_t|\mathcal{F}_r\right]\right)^2dr\\ &\le \int_0^t\left(\sigma_H\left(1-\alpha\right)K(t,r)e^{M(t-r)}\right)^2dr \\&= \sigma_H^2(1-\alpha)^2\int_0^tK^2(t,r)e^{2M(t-r)}dr\\ & \le \sigma_H^2\left(1-\alpha\right)^2e^{2Mt}\int_0^tK^2(t,r)dr, \mbox{ }0\le r\le t\le T. \end{align*} Since \(\int_0^tK^2(t,s)ds=E|B_t^H|^2=t^{2H}\) we have \[ \int_0^T\left(E\left[D_r^WV_t|\mathcal{F}_r\right]\right)^2dr \le \sigma_H^2(1-\alpha)^2e^{2Mt}t^{2H}. \] Fixed \(t\in (0,T], \) put \(F=V_t-E[V_t]\) then \(EF=0\) and \(D_s^BF=D_s^BV_t,D_s^WF=D_s^WV_t\). We obtain the following estimate \begin{align*} \int_0^T\left(E\left[D_s^BF|\mathcal{F}_s\right]\right)^2ds + \int_0^T\left(E\left[D_s^WF|\mathcal{F}_s\right]\right)^2ds&=\int_0^T\left(E\left[D_s^BV_t|\mathcal{F}_s\right]\right)^2ds + \int_0^T\left(E\left[D_s^WV_t|\mathcal{F}_s\right]\right)^2ds\\ & \le \frac{\sigma^2(1-\alpha)^2}{2M}\left(e^{2Mt} -1\right) + \sigma_H^2(1-\alpha)^2e^{2Mt}t^{2H}. \end{align*} We observe that, by Lyapunov's inequality, \(E\left[X_t^{1-\alpha}\right]\leq \left(E\left[X_t\right]\right)^{1-\alpha}=\mu_t^{1-\alpha}.\) Hence, by applying Lemma 1 to \(F,\) we obtain \begin{align*} P(X_t\ge x)&=P\left(V_t\ge x^{1-\alpha}\right)\\ &=P\left(V_t-E\left[V_t\right]\ge x^{1-\alpha}-E\left[V_t\right] \right)\\ &=P\left(F\ge x^{1-\alpha}-E\left[X_t^{1-\alpha}\right]\right)\\ &\leq P\left(F\ge x^{1-\alpha}-\mu_t^{1-\alpha}\right)\\ &\le \exp\left(-\frac{\left(x^{1-\alpha}-E\left[X_t^{1-\alpha}\right]\right)^2}{2\left(\frac{\sigma^2(1-\alpha)^2}{2M}\left(e^{2Mt} -1\right) +\sigma_H^2(1-\alpha)^2e^{2Mt}t^{2H}\right)}\right),\,\,x>\mu_t. \end{align*} The proof of Theorem is complete.

Remark 1. In [5], Araneda obtained an analytical expression for the transition probability density function of solutions to the Equation (3). However, the stochastic integral with respect to \(B^H\) considered there is interpreted as a Wick-Itô integral. This integral is different from the pathwise Stieltjes integral using in our work (the relation between two integrals can be found in §5.6 of [10]). In particular, unlike the Wick-Itô integral, the pathwise Stieltjes integral has non-zero expectation. We therefore think that it is not easy to extend the method developed in [5] to the setting of pathwise Stieltjes integrals. That is why we have to employ a different method to investigate the tail distributions as in Theorem 1.

Remark 2. The transition probability density and tail distribution can be used to compute the price of options. In the setting of the mixed-fractional CEV model using pathwise Stieltjes integrals, to the best of our knowledge, the option pricing formula is still an open problem. Solving this problem is beyond the scope of the present paper. However, if such a formula exists then the tail distribution estimates obtained in Theorem 1 will be useful to provide an upper bound for the price of options.

4. Conclusion

In this paper, we used the techniques of Malliavin calculus to estimate the tail distribution of the mixed-frational CEV model. Our contribution is that we are able to obtain an explicit estimate for the tail distributions. Our work provides one more fundamental property of CEV models. In this sense, we partly enrich the knowledge of CEV models.

Acknowledgments:

The authors would like to thank the anonymous referees for their valuable comments.

Author Contributions

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

Conflicts of Interest

"The authors declare no conflict of interest".

1. Introduction

Mathematical trees (and their natural counterparts) have been studied for a very long time. In this work, we consider labelled \(t\)-ary trees, i.e. these are rooted trees embedded in the plane such that each vertex has a degree of at most \(t\) and the vertex set is \([n]:=\{1,2,\ldots,n\}\). Here, a degree of a vertex is the number of edges that come out of a vertex if the edges are oriented away from the root. Trees treated here have local orientation [1], i.e. the edges are oriented from a vertex of lower label towards a vertex of higher label. A vertex \(v\) is said to be reachable from a vertex \(i\) if there is an oriented path from vertex \(i\) to vertex \(v\) and we say that a path \(p\) has length \(\ell\) if there are \(\ell\) edges on the path. All paths are oriented. A vertex in which there is no edge that is oriented away from it is called a sink.

We note that in any labelled tree on \(n\) vertices, only one vertex (i.e itself) is reachable from vertex \(n\) since it is always a sink. It is also clear that many vertices can be reached from vertex \(1\) than from any other vertex. The vertices with the same parent are called siblings. Since the siblings are linearly ordered, they are always drawn in a left-right pattern where the leftmost sibling is referred to as first child. At level \(\ell\), the left most child of all the siblings is the eldest child. A left most path refers to a sequence of edges joining eldest children at each level in a plane tree. The corresponding results for ordinary trees and noncrossing trees were obtained in the PhD thesis [2]. Using the aforementioned statistics, Nyariaro and Okoth [3] worked on the corresponding results for ordered trees. From now henceforth, \(t\)-ary trees will be referred to as trees.

The paper is organized as follows: In §2, we obtain the number of these trees as enumerated by path lengths, number of sinks and number of leaf sinks. Enumeration by left most paths and first children is considered in §3. Finally, we use non-first children and non-leaves to enumerate the trees in §4. We derive most of our results by means of generating functions and Lagrange Inversion Formula [4].

2. Enumeration by path lengths, sinks and leaf sinks

The aim of this section is to find the average number of vertices, sinks and leaf sinks in labelled trees of order \(n\) that are reachable from a root of label \(i\).

Proposition 1. There are

labelled trees on \([n]\) such that vertex \(j\), of degree \(d\) and at length \(\ell\geq 0\), is reachable from root \(i\).

Proof. Let \(T(x)\) be the generating function of unlabelled trees such that \(x\) marks the number of non-root vertices. It follows that \(T(x)=(1+xT(x))^t\). Consider a tree rooted at \(i\) and having a path of length \(\ell\) terminating at a vertex of degree \(d\). There are \(\ell+1\) vertices in this path. The trees with such a path can be decomposed as shown in Figure 1.

We therefore have the generating function for these trees as \[(xT(x)^2)^{\ell}x(xT(x))^d=x^{\ell+d+1}T(x)^{2\ell +d}.\] We need to extract the coefficient of \(x^{n}\) in the generating function. Set \(xT(x)=S(x)\) so that the generating function becomes \(x^{-\ell+1}S(x)^{2\ell+d}\). We also have \(S(x)=x(1+S(x))^t.\)

By Lagrange Inversion Formula, we obtain

\begin{align*} [x^n]x^{\ell+d+1}T(x)^{2\ell +d}&=[x^n]x^{-\ell+1}S(x)^{2\ell +d} =[x^{n+\ell-1}]S(x)^{2\ell +d}\\ &=\frac{2\ell+d}{n+\ell-1}[s^{n-\ell-d-1}](1+s)^{t(n+\ell-1)}\\ &=\frac{2\ell+d}{n+\ell-1}[s^{n-\ell-d-1}]\sum_{i\geq 0}{t(n+\ell-1)\choose i}s^{i}\\ &=\frac{2\ell+d}{n+\ell-1}{t(n+\ell-1)\choose n-\ell-d-1}. \end{align*}

This formula counts the number of unlabelled trees with a path of length \(\ell\) starting at the root and ending at a vertex of degree \(d\). There are \({t\choose d}\) possible positions for the children of the terminating vertex. Lets consider a path of length \(\ell\) starting at root \(i\) and ending at vertex \(j.\) There are \(\binom{j-i-1}{\ell-1}\) such paths. Once the \(\ell+1\) vertices on the path are labelled, there are \((n-\ell-1)!\) choices for labels of the vertices which are not on the path. Putting everything together, we obtain the required formula.

We remark that Equation (1) also gives the number of vertices of label \(j\) with degree \(d\) that are reachable from root \(i\) in \(\ell\) steps in all labelled trees of order \(n\). A number of corollaries follow for example by summing over all \(d\), we obtain the following corollary:

Corollary 1. The total number of labelled trees on \(n\) vertices such that vertex \(j\) is reachable from root \(i\) in \(\ell\geq 0\) steps is given by \begin{align*} \dfrac{(2\ell+1)(n-\ell-1)!}{n+\ell}{j-i-1\choose \ell-1} {t(n+\ell)\choose n-\ell-1}. \end{align*}

Corollary 2.The total number of labelled trees on \(n\) vertices such that vertex \(j\) is reachable from the root in \(\ell\geq 0\) steps is given by

Proof. The result follows by summing over all \(i\) in Equation (1).

Similarly, summing over all \(j\) in Equation (1), we obtain:

Corollary 3. There are

paths of length \(\ell\geq 0\) starting at the root in labelled trees of order \(n\).

By substituting for \(\ell=1\) and \(i=1\) in Equation (3), we obtain:

Corollary 4. There are a total of \(\frac{3n!}{n^2+n}{tn+t\choose n-2}\) children of root \(1\) in all labelled trees of order \(n.\)

By summing over all \(j\) in Equation (2) or by summing over all \(i\) in Equation (3) we arrive at the total number of paths of length \(\ell\) in labelled trees on \([n]\) whose formula is given by If \(\ell=0\) in Equation (4), we recover the formula, \((n-1)!{tn\choose n-1}\), for the number of labelled trees on \(n\) vertices. By substituting for \(\ell=1\) in Equation (4), we obtain:

Corollary 5. There are a total of \(\frac{3n!}{2n+2}{tn+t\choose n-2}\) children of the root in all labelled trees of order \(n.\)

Corollary 6. The average number of vertices, at length \(\ell\geq 0\), that are reachable from the root in a random labelled tree is \((2\ell+1)/(t^{\ell}(\ell+1)!)\).

Proof. By dividing Equation (4) by \((n-1)!{tn\choose n-1}\), the number of labelled trees on \(n\) vertices, we obtain

as the average number of vertices that can be reached from the root in labelled trees of order \(n\). Now, let \[A=\dfrac{2\ell+1}{(\ell+1)!}\cdot\dfrac{(t(n+\ell)-1)(t(n+\ell)-2)\cdots (t(n+\ell)-(n-\ell-2))n(n-1)\cdots (n-\ell)}{n(tn-1)(tn-2)\cdots (tn-n+2)},\] so that \begin{align*} \lim_{n\rightarrow \infty}A&=\dfrac{2\ell+1}{(\ell+1)!}\cdot\lim_{n\rightarrow \infty}\dfrac{t^{n-\ell-2}n^{n-\ell-2}n^{\ell+1 }+\cdots}{t^{n-2}n^{n-1}+\cdots}=\dfrac{2\ell+1}{(\ell+1)!}\cdot\lim_{n\rightarrow \infty}\dfrac{n^{n-1}+\cdots}{t^{\ell}n^{n-1}+\cdots}\\ &=\dfrac{2\ell+1}{(\ell+1)!}\cdot\dfrac{1}{t^{\ell}}\cdot\lim_{n\rightarrow \infty}\dfrac{1+\cdots}{1+\cdots}=\dfrac{2\ell+1}{(\ell+1)!}\cdot\dfrac{1}{t^{\ell}}. \end{align*} This completes the proof.

From Equation (4), we get the total number of reachable vertices in these trees of order \(n\) as \begin{align*} n!\sum_{\ell=0}^{n-1}\dfrac{(2\ell+1)}{(\ell+1)!(n+\ell)} {t(n+\ell)\choose n-\ell-1}. \end{align*} We now enumerate the trees by sinks:

Proposition 2. The number of labelled trees on \([n]\) such that a sink \(j\), of degree \(d\) and at length \(\ell\geq 0\), is reachable from root \(i\) is given by,

Proof. From the proof of Proposition 1, it follows that there are \begin{align*} \frac{2\ell+d}{n+\ell-1}{t(n+\ell-1)\choose n-\ell-d-1} \end{align*} unlabelled trees with a path of length \(\ell\) starting at the root such that the terminating vertex has degree \(d\). Lets consider a path of length \(\ell\) starting at root \(i\) and ending at vertex \(j.\) There are \(\binom{j-i-1}{\ell-1}\) such paths. Since vertex \(j\) is a sink of degree \(d\), the labels of the \(d\) vertices must be less than \(j\). Thus there are \({j-\ell-1\choose d}\) choices for these labels. There are \({t}\choose {d}\) possible positions for children of final vertex. Once the \(\ell+1\) vertices on the path and the \(d\) children of \(j\) are labelled, there are \((n-\ell-d-1)!\) choices for other labels in the tree. Collecting everything, we arrive at the required formula.

It is worthwhile to note Equation (6) also gives the number of sinks of label \(j\) with degree \(d\) that are reachable from root \(i\) in \(\ell\) steps in all labelled trees of order \(n\). Again a number of corollaries follow. By summing over all \(d\), we obtain that

Corollary 7. The total number of sinks of degree \(d\) that are reachable from vertex \(i\) in \(\ell\) steps in labelled trees of order \(n\) is given by \begin{align*} {(n-\ell-d-1)!}\frac{2\ell+d}{n+\ell-1}\sum_{j=\ell+i}^{n}{\begin{pmatrix}{j-i-1}\\{\ell-1}\end{pmatrix}}\begin{pmatrix}{j-\ell-1}\\{d}\end{pmatrix}{{t}\choose {d}}\begin{pmatrix}{t(n+\ell-1}\\{n-d-\ell-1}\end{pmatrix}. \end{align*}

Proof. We obtain the formula by summing over all \(j\) in Equation (6).

Corollary 8. The total number of labelled trees with \(n\) vertices such that root \(i\) is a sink of degree \(d\) is given by:

Proof. The result follows by setting \(\ell=0\) and \(j=i\) in Equation (6).

Corollary 9. The total number of labelled trees of order \(n\) with root sinks of degree \(d\) is given by:

Proof. The result follows by summing over all \(i\) in Equation (7).

The following result is immediate by setting \(\ell=1\) in Equation (6).

Corollary 10. Irrespective of the label of the root, the number of children of the root labelled \(j\) and having degree \(d\), in labelled trees on \(n\) vertices is given by:

Summing over all \(j\) in Equation (9), we obtain the number of children of the root which are sinks of degree \(d\), in trees of order \(n\) as \begin{equation*} {(n-d-2)!}\frac{2+d}{n}\begin{pmatrix}{n-d-1}\\{d+1}\end{pmatrix}{{t}\choose{d}}\begin{pmatrix}{tn}\\{n-d-2}\end{pmatrix}. \end{equation*}

Corollary 11. The average number of root sinks of degree \(d\) in a random tree of order \(n\) is

Proof. Diving the total number of labelled trees of order \(n\) with root sinks of degree \(d\) (Equation (8)), by the total number of labelled trees we get, \begin{align*} \frac{{(n-d-1)!}\frac{d}{n-1}\begin{pmatrix}{n}\\{d+1}\end{pmatrix}\displaystyle{t\choose d}\begin{pmatrix}{t(n-1)}\\{n-d-1}\end{pmatrix}}{\frac{n!}{n}\begin{pmatrix}{tn}\\{n-1}\end{pmatrix}}, \end{align*} as the average number of root sinks of degree \(d\) in a random plane tree on \(n\) vertices. The result follows by simplification and tending \(n\) to infinity.

Setting \(d=0\) in Equation (10) we get that the average number of root sinks of degree \(0\) is zero. This implies that there is no leaf sink which is also a root.

For the remainder of this section, we enumerate the trees by leaf sinks. The following result follows by setting \(d=0\) in Equation (6). However, to show the decomposition of trees with a leaf sink, we shall construct the proof.

Corollary 12. The number of labelled trees on \([n]\) such that vertex \(j\) is a leaf sink reachable from vertex \(i\) in \(\ell \) steps is given by

Proof. Let \(T(x)\) be the generating function of unlabelled trees such that \(x\) marks the number of non-root vertices. Consider a tree rooted at vertex \(i\) and having a path of length \(\ell\) starting at \(i\) and ending at \(i+\ell\). Let vertex \(i+\ell\) be a leaf sink. The path decomposes the tree into right an left subtrees up to length \(\ell\). Such trees are pictorially represented in Figure 2.

The generating function for these trees is thus \((xT(x)^{2})^{\ell}x=x^{\ell+1}T(x)^{2\ell}.\) We set \(xT(x)=S(x)\) and use Lagrange Inversion Formula to obtain

\begin{align*} [x^n]x^{\ell+1}T(x)^{2\ell}=[x^{n+\ell-1}]S(x)^{2\ell}&=\frac{2\ell}{n+\ell-1}[s^{n-\ell-1}](1+s)^{t(n+\ell-1)}\\ &=\frac{2\ell}{n+\ell-1}[s^{n-\ell-1}]\sum_{i\geq{0}}{t(n+\ell-1)\choose i}s^i\\ &=\frac{2\ell}{n+\ell-1}{t(n+\ell-1)\choose n-\ell-1}. \end{align*} Consider a path of length \(\ell\) starting from vertex \(i\) and terminating at vertex \(j\). There are \({j-i-1 \choose \ell-1}\) such paths. Once \(\ell+1\) vertices on the path have been labelled, there are \((n-\ell-1)!\) choices for labels of the vertices which are not on the path. Therefore, putting everything together we obtain the desired formula.

Summing over all \(j\) in Equation (11), we obtain the result below:

Corollary 13. There are

paths of length \( \ell\geq{0}\) starting at root \(i\) and ending at a leaf sink in a labelled tree of order \(n\).

Corollary 14. The total number of labelled trees on \(n\) vertices such that vertex \(j\) is a leaf sink reachable from the root in \(\ell\geq{0}\) steps is given by

Proof. We obtain the desired result by summing over all \(i\) in Equation (12).

Moreover, summing over all \(\ell\) in Equation (13) we obtain the following result:

Corollary 15. The number of leaf sinks in a labelled tree of order \(n\) that are reachable from the root is given by \begin{align*} n!\sum_{\ell=0}^{n-1}\frac{2\ell}{(\ell+1)!(n+\ell-1)}{t(n+\ell-1)\choose n-\ell-1}. \end{align*}

Corollary 16. The average number of leaf sinks that are reachable from the root in \(\ell\) steps in a random tree of order \(n\) is given by: \begin{align*} \frac{2\ell}{(\ell+1)!t^\ell}\,. \end{align*}

Proof. The result follows by dividing Equation (13) by \((n-1)!{tn\choose n-1}\), and tending \(n\to\infty\).

3. Enumeration by left most paths and first children

In this section, we obtain the number of trees with a given number of first children and also with a left most path of a given length.

Proposition 3. The number of labelled trees of order \(n\) in which there is a left most path of length \(\ell\) from root \(i\) to vertex \(j\) is given by

Proof. Let \(T(x)\) be the generating function for unlabelled trees, where vertex \(x\) marks the number of non-root vertices. Consider a tree rooted at vertex \(i\) and having a path of length \(\ell\) terminating at vertex \(\ell+1\). There are \(\ell+1\) vertices on the path. These trees are represented pictorially as shown in Figure 3.

The generating function for the trees is therefore expressed as \(x^{\ell+1}T(x)^{\ell+1}.\) Since \(T(x)=(1+xT(x))^{t}\), we set \(xT(x)=S(x)\), and making use of Lagrange Inversion formula, we get

\begin{align*} [x^{n}]x^{\ell+1}T(x)^{\ell+1}=[x^{n}]S(x)^{\ell+1}&=\frac{\ell+1}{n}{[s^{n-\ell-1}]}(1+s)^{tn} =\frac{\ell+1}{n}[s^{n-\ell-1}]\sum_{i\geq{0}} {tn \choose i} s^{i} =\frac{\ell+1}{n}{tn \choose n-\ell-1}. \end{align*} As before, there are \(\displaystyle{j-i-1 \choose \ell-1}\) choices for paths of length \(\ell\) between vertices \(i\) and \(j\). Once \(\ell+1\) vertices on the path have been labelled, there are \((n-\ell-1)!\) ways of labelling the remaining vertices which are not on the path. Thus, having everything altogether we obtain Formula (14).

Corollary 17. The number of labelled trees on \(n\) vertices with a left most path of length \(\ell\) from vertex \(i\) is given by

Proof. The desired formula is obtained by summing over all \(j\) in Equation (14).

The following result follows by summing over all \(i\) in Equation (15).

Corollary 18. The number of labelled trees of order \(n\) in which there is a left most path of length \(\ell\) from the root is given by

Corollary 19. The number of labelled trees of order \(n\) in which there is a left most path starting form the root is given by \begin{align*} (n-1)!\sum_{\ell=0}^{n-1}\frac{1}{\ell!}{tn\choose n-\ell-1}. \end{align*}

Proof. We obtain the required formula by summing over all \(\ell\) in Equation (16).

Corollary 20. The average number of trees on \(n\) vertices with a left most path of length \(\ell\) from the root is given by \(\frac{1}{\ell! t^{\ell}}\).

Proof. The result follows by dividing Equation (16) by \((n-1)!{tn\choose n-1}\), and tending \(n\to\infty\).

Proposition 4. The number of labelled trees of order \(n\) rooted at vertex \(i\) in which there is a left most path of length \(\ell\) such that the final vertex \(j\) is a leaf sink is given by

Proof. Let \(T(x)\) be the generating function of a trees where \(x\) marks non-root vertices. Now, consider a tree rooted at vertex \(i\) such that there is a left most path of length \(\ell\) starting at root \(i\) and ending at vertex \(i+\ell\) which is also a leaf sink. The path decomposes the tree into right subtrees only up to length \(\ell\). The final vertex, being a leaf sink, has no subtree attached to it. See Figure 4 for the decomposition.

Therefore, the generating function of the above tree is given by \(x^{\ell+1}T(x)^{\ell}.\) As already seen in this paper, we set \(xT(x)=S(x)\) so that \(x^{\ell+1}T(x)^{\ell}= xS(x)^{\ell} \) and we use Lagrange Inversion Formula to extract the coefficient of \(x^n\) in the generating function:

\begin{align*} [x^{n}]x^{\ell+1}T(x)^{\ell}=[x^{n-1}]S(x)^{\ell}&=\frac{\ell}{n-1}[s^{n-\ell-1}](1+s)^{t(n-1)} =\frac{\ell}{n-1}[s^{n-\ell-1}]\sum_{i\geq{0}} {t(n-1) \choose i}s^{i} =\frac{\ell}{n-1}{t(n-1) \choose n-\ell-1}. \end{align*} The result follows by choosing the labels for the vertices on the path and for the vertices which are not on the path.

By summing of all \(j\) in Equation (17), we obtain the following result.

Corollary 21. There are

labelled trees of order \(n\) with a left most path of length \(\ell\) with vertex \(i\) as a root and final vertex \(j\) as a leaf sink.

Also, summing over all \(i\) in Equation (18), we obtain the following result.

Corollary 22. The number of labelled trees of order \(n\) in which there is a left most path of length \(\ell\) from the root and a final vertex as a leaf sink is given by

Corollary 23. The number of labelled trees of order \(n\) in which there is a left most path starting from the root and ending at a leaf sink is given by \begin{align*} \frac{n!}{n-1}\sum_{\ell=0}^{n-1} \frac{\ell}{(\ell+1)!} {t(n-1)\choose n-\ell-1}. \end{align*}

Proof. The result is immediate by summing over all \(\ell\) in Equation (19).

Proposition 5. The number of labelled trees on \(n\) vertices with vertex \(i\) as a root and vertex \(j\) as a first child at level \(\ell\) is given by

Proof. Let \(T(x)\) be the generating function of trees where \(x\) marks non-root vertices. Consider tree with a path of length \(\ell\) starting from vertex \(i\) and terminating at vertex \(i+\ell\) which is also a first child. The path decomposes the tree into left and right subtrees up to length \(\ell-1\). Since vertex \(i+\ell\) is a first child then its parent has no left subtrees. Moreover, vertex \(i+\ell\) has a subtree attached to it, which can possibly be empty. The tree is represented as in Figure 5.

The generating function for the number of these tree is thus \((xT(x)^{2})^{\ell-1}(xT(x))^{2}=x^{\ell+1}T(x)^{2\ell}\). We set \(xT(x)=S(x)\) and apply Lagrange Inversion Formula to obtain

\begin{align*} [x^{n}]x^{\ell+1}T(x)^{2\ell}=[x^{n+\ell-1}]S(x)^{2\ell}&=\frac{2\ell}{n+\ell-1}[s^{n-\ell-1}](1+s)^{t(n+\ell-1)}\\ &=\frac{2\ell}{n+\ell-1}[s^{n-\ell-1}]\sum_{i\geq{0}}{t(n+\ell-1) \choose i}s^{i}\end{align*} \begin{align*}&=\frac{2\ell}{n+\ell-1}{t(n+\ell-1)\choose {n-\ell-1}}. \end{align*} The number of choices for a path of length \(\ell\) between vertex \(j\) and vertex \(i\) is \(j-i-1\choose \ell-1\). Once the \(\ell+1\) vertices along the path have been labelled, then the remaining vertices can be labelled in \((n-\ell-1)!\) ways. Therefore, by putting all the terms together we obtain the desired closed formula.

Corollary 24. The number of first children at level \(\ell\) that are reachable from vertex \(i\) in a labelled tree of order \(n\) is given by

Proof. We sum over all \(j\) in Equation (20) to obtain the desired result.

Corollary 25. The number of first children at level \(\ell\) that are reachable from the root in a labelled tree on \(n\) vertices is

Proof. The required formula follows by summing over all \(i\) in Equation (21).

By summing over all \(\ell\) in Equation (22), we obtain the following result.

Corollary 26. There are \begin{align*} n!\sum_{\ell=0}^{n-1}\frac{2\ell}{(\ell+1)!(n+\ell-1)}{t(n+\ell-1)\choose n-\ell-1} \end{align*} first children in labelled tree of order \(n\) that are reachable from the root.

Corollary 27. The average number of first children at level \(\ell\) in a random labelled tree is given by \(\frac{2\ell}{(\ell+1)!t^{\ell}}.\)

Proof. The required formula is arrived at by dividing Equation (22) by \((n-1)!{tn \choose {n-1}}\) and tending \(n\) to infinity.

Remark 1. It is worth noting that the generating function for the number of trees with a leaf sink at length \(\ell\) from the root is the same as the generating function for the trees with a first child at length \(\ell\). Thus the results are quite the same.

4. Enumeration by non-first children and non-leaves

In this section, we enumerate trees by number of reachable non-first children and non-leaves.

Proposition 6. The number of labelled trees on \([n]\) in which vertex \(i\) is a root such that there is a path of length \(\ell\) starting at \(i\) and terminating at vertex \(j\) which is also a non-first child is given by

Proof. Consider a tree rooted at vertex \(i\) with a path of length \(\ell\) from the root to vertex \(i+\ell\) which is non-first child. This path decomposes the tree into left and right subtrees up to step \(\ell\). Moreover, since vertex \(i+\ell\) is a non-first child, there must be an elder sibling of \(i+\ell\). Vertex \(i+\ell\) can have a subtree attached to it, which is possibly empty. This decomposition is represented by Figure 6.

The generating function for these trees is thus \((xT(x)^{2})^{\ell}xT(x)xT(x)=x^{\ell+2}T(x)^{2\ell+2}\), where \(T(x)\) is the generating function for unlabelled rooted trees with \(x\) marking non-root vertices. We set \(xT(x)=S(x)\) and apply Lagrange Inversion Formula to obtain

\begin{align*} [x^{n}]x^{\ell+2}T(x)^{2\ell+2}=[x^{n+\ell}]S(x)^{2\ell+2}&=\frac{2\ell+2}{n+\ell}[s^{n-\ell-2}(1+s)^{t(n+\ell)}\end{align*}\begin{align*} &=\frac{2\ell+2}{n+\ell}[s^{n-\ell-2}]\sum_{i\geq{0}}{t(n+\ell)\choose i}s^{i}\\ &=\frac{2\ell+2}{n+\ell}{t(n+\ell)\choose n-\ell-2}. \end{align*} There are \(j-i-1 \choose \ell-1\) choices for paths of length \(\ell\) between vertex \(i\) to vertex \(j\). Once the \(\ell+1\) vertices on the path have been labelled, the remaining vertices can be labelled in \((n-\ell-1)!\) ways. Thus putting everything together we obtain the desired formula.

The next result follows by summing over all \(j\) in Equation (23).

Corollary 28. The total number of labelled trees of order \(n\) such that a non-first child at length \(\ell\) is reachable from a root \(i\) is given by

Summing over all \(i,\) in (24) we obtain:

Corollary 29. The total number of labelled trees on \(n\) vertices such that a non-first child at length \(\ell\) is reachable from the root is given by

Similarly, by summing over all \(\ell\) in Equation (25), we find that

Corollary 30. There are \begin{align*} 2n!\sum_{\ell=0}^{n-1}\frac{1}{\ell!(n+\ell)}{t(n+\ell)\choose n-\ell-2} \end{align*} labelled trees with a non-first child at level \(\ell\) that is reachable from the root.

Corollary 31. The average number of non-first children at level \(\ell\) in a random tree is given by \(\frac{2}{\ell!t^{\ell}}\).

Proof. Dividing Equation (25) by \((n-1)! {tn \choose {n-1}}\), we obtain the average number of non-first children that are reachable from the root at length \(\ell\) in trees of order \(n\). Now, simplifying the average and tending \(n\) to infinity we arrive at the required formula.

We now enumerate the trees by the number of non-leaves.

Proposition 7. The number of labelled trees on \([n]\) in which vertex \(i\) is a root such that there is a path of length \(\ell\) starting at \(i\) and terminating at a non-leaf vertex \(j\) is given by the formula,

Proof. Consider a tree rooted at vertex \(i\) with a path of length \(\ell\) from the root to vertex \(i+\ell\) which is non-leaf. This path decomposes the tree into left and right subtrees up to step \(\ell\). Moreover, since vertex \(i+\ell\) is a non-leaf, there must be a subtree of \(i+\ell\) which may be empty and a subtree, rooted at a child of \(i+\ell\). This subtree may also be empty. The decomposition is therefore given by Figure 7.

The generating function for the trees is thus \((xT(x)^{2})^{\ell}xT(x)xT(x)=x^{\ell+2}T(x)^{2\ell+2}\), where \(T(x)\) is the generating function for unlabelled rooted trees with \(x\) marking non-root vertices. The required result therefore follows by applying Lagrange Inversion formula, upon setting \(xT(x)=S(x)\), and giving the number of choices for the labels on the path and those that are not on the path.

Remark 2. The number of trees with a non-leaf vertex \(j\) which is reachable at length \(\ell\) from a root \(i\) have similar generating functions (though different decompositions) as for the case of non-first children. The results in the case of non-first children therefore hold for non-leaf vertices.

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

Since Ostrowski first time proved his inequality in 1938, after that many researchers did a lot of work on it. Some monographs presented by Barnet et al., [1] and Dragomir et al. [2] on Ostrowski's type inequalities. In the past years, many researchers [3,4,5,6,7] did efforts to obtain tighter error bounds of Ostrowski type inequalities. Inspired and motivated by the work of above famous Mathematician [8,9] and [2,10,11], we started our work to extend and produce new and generalized Ostrowski's integral inequalities.

In this paper, we introduced some new generalized different types of Kernels, development of new identities and new error bounds of Ostrowski's type inequalities for first and second derivable mappings. By utilizing our obtained results, previous famous results are recaptured as special cases.

2. Results for Quadratic mapping

Theorem 1. Let \(J\subseteq \mathbb{R} \) such that \(c,d\in J, \) and \(c< d. \) If \(s: J\rightarrow\mathbb{R} \) is a derivable function such that \(\gamma \leq s^{\prime }\left( t\right) \leq \Gamma , \) and \(\varphi ,\psi ,\gamma ,\Gamma\in\mathbb{R}, \) then we get

for all \(t\in \left[ c,d\right]. \)

Proof. Define a new peano Kernel \(L\left( x,t\right) :\left[ c,d\right] \rightarrow\mathbb{R} \) by

for all \( x\in \left[ c,d\right]. \) By using (2), we get Again, by using (2), we get Using (3) and (4), we get On the other hand Let \(C=\frac{\Gamma +\gamma }{2} \), then, \( \underset{t\in \left[ c,d\right] }{\max }\left\vert s^{\prime \prime }\left( t\right) -C\right\vert \leq \frac{\Gamma -\gamma }{2}. \) Thus (6) becomes Using (5) in (8), we get our required result (1).

Remark 1. By putting \(\varphi =\psi \) in (1), we get

Corollary 2. By putting \(x=\frac{c+d}{2} \) in (9), we get mid point inequality: \begin{align*} & \left\vert \int\limits_{c}^{d}s\left( t\right) dt-\left( d-c\right) s\left( \frac{c+d}{2}\right) -\frac{1}{48}\left( \Gamma +\gamma \right) \left( d-c\right) ^{3}\right\vert \leq \frac{1}{48}\left( \Gamma -\gamma \right) \left( d-c\right) ^{3}. \end{align*}

3. Applications in numerical integration

Using [5], we suppose that \(J_{n}:c=x_{0}< x_{1}< x_{2} < ....< x_{n-1}< x_{n}=d \) a partition of \([c,d] \), \(\xi_{i}\in\left[ x_{i}+\delta\frac{\varrho_{i}}{2},x_{i+1}-\delta\frac{\varrho_{i}}{2}\right] , \) \(\left( i=0,1,.....,n-1\right) \) and \(\varrho_{i}=x_{i+1}-x_{i} \) \(,\left( i=0,1,.....,n-1\right), \) then following theorem exist:

Theorem 3. Let \(s:\left[ c,d\right] \rightarrow\mathbb{R} \) be continuous on \(\left[ c,d\right] \) and derivable on \(\left(c,d\right) , \) then following formula exist:

where and remainder satisfies the estimation for all \(\xi _{i}\in \left[ x_{i},x_{i+1}\right] . \)

Proof. By using Theorem 1 on \(\left[ x_{i},x_{i+1}\right] ,\xi _{i}\in \left[ x_{i},x_{i+1}\right] , \) to get:

or \begin{align*} & \left\vert \sum\limits_{i=0}^{n-1}\frac{1}{2\left( \varphi +\psi \right) } \left( \varphi \left( \xi _{i}-x_{i}\right) ^{2}-\psi \left( \xi _{i}-x_{i+1}\right) ^{2}\right) s^{\prime }\left( \xi _{i}\right) \right. \left. -\sum\limits_{i=0}^{n-1}\frac{1}{\varphi +\psi }\left( \varphi \left( \xi _{i}-x_{i}\right) +\psi \left( \xi _{i}-x_{i+1}\right) \right) s\left( \xi _{i}\right) \right. \\ & \left. +\sum\limits_{i=0}^{n-1}\frac{1}{\varphi +\psi }\left( \varphi \int\limits_{c}^{x}s\left( t\right) dt+\psi \int\limits_{x}^{d}s\left( t\right) dt\right) \right. \left. -\frac{\Gamma +\gamma }{2}\sum\limits_{i=0}^{n-1}\frac{\varrho _{i} }{6\left( \varphi +\psi \right) }\left( \varphi \left( \xi _{i}-x_{i}\right) ^{3}-\psi \left( \xi _{i}-x_{i+1}\right) ^{3}\right) \right\vert \\ & \leq \frac{\Gamma -\gamma }{2}\sum\limits_{i=0}^{n-1}\frac{\varrho _{i}}{ 6\left( \varphi +\psi \right) }\left( \varphi \left( \xi _{i}-x_{i}\right) ^{3}-\psi \left( \xi _{i}-x_{i+1}\right) ^{3}\right) . \end{align*} With the help of generalized triangular inequality, we get the desired estimation.

4. Results for generalized linear mapping

Theorem 4. Let \(r:I\rightarrow\mathbb{R} \) and \(v,w\in I, \) \(v< w. \) If \(g^{\prime }:I\rightarrow\mathbb{R} \), such that \(\gamma \leq r^{\prime }\left( t\right) \leq \Gamma ,\ \forall \) \(t\in \left[ v,w\right] \) and \(\varphi ,\psi ,\ \gamma ,\Gamma \in\mathbb{R}. \) We have

Proof. First we define the mapping \(L\left( x,t\right) \ :\left[ v,w\right] \rightarrow\mathbb{R} \) by

By using (15), we get and We put \(C=\frac{\Gamma +\gamma }{2} \) and using (16) and (17), we get Let \begin{equation*} C=\frac{\Gamma +\gamma }{2}. \end{equation*} Then Now Using (19) and (21), we have Using (18) and (22), we get our required result (14).

Remark 2. By putting \(\varrho =0 \) in (14), we get \begin{align*} & \left\vert \frac{1}{\varphi +\psi }\left[ \left( \varphi \left( x-v\right) -\psi \left( x-w\right) \right) r\left( x\right) -\left( \varphi \int\limits_{v}^{x}r\left( t\right) dt+\psi \int\limits_{x}^{w}r\left( t\right) dt\right) \right. \right. \\ & \left. \left. -\frac{1}{4}\left( \Gamma +\gamma \right) \left( \varphi \left( x-v\right) ^{2}-\psi \left( x-w\right) ^{2}\right) \right] \right\vert \\ & \leq \frac{\Gamma -\gamma }{2}\left[ \frac{\varrho ^{2}}{8}\left( w-v\right) ^{2}+\frac{1}{2\left( \varphi +\psi \right) }\left( \varphi \left( x-v\right) ^{2}+\psi \left( x-w\right) ^{2}\right) \right] . \end{align*}

5. Results for generalized Quadratic mapping

Theorem 5. Let \(z:I\subseteq \mathbb{R}, \) and \( c,d\in I,\ c< d. \) If \(z:I\rightarrow \mathbb{R} \) is a derivable function such that \(\gamma \leq z^{\prime }\left( t\right) \leq \Gamma ,\ \forall \ t\in \left[ c,d\right] , \) the constants \(\varphi ,\psi ,\ \gamma ,\Gamma \in \mathbb{R} . \) Then, we get

Proof. Let us define the mapping

By using (24), we get and Using (25) and (26), we get But on the other side, Now, again by using (24), we get and Also Using (28) and (29), we get Using (27) and (32), we get our required result (23).

6. Conclusion

In this paper, we proved the results by using quadratic mapping, generalized linear mapping and generalized quadratic mapping. We developed application for numerical integration also.

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

Polylogarithm function

The polylogarithm function \[ \text{Li}_{s}(z)= \sum_{n=1}^{\infty}\frac{z^{n}}{n^{s}} = z+\frac{z^{2}}{2^{s}}+\frac{z^{3}}{3^{s}}+\cdots, \quad s, z\in\mathbb{C}, |z|< 1 \] plays a significant role in many areas of number theory; its origin, the dilogarithm \(\text{Li}_{2}(z)\), dates back to Abel, Euler, Kummer, Landen and Spence etc. See Kirillov [1], Lewin [2], Zagier [3] for more details. The main theme of this article is to better understand the relation between the dilogarithm, trilogarithm \(\text{Li}_{3}(z)\) functions and zeta values \(\zeta(2)\), \(\zeta(3)\) (Apéry constant), \(\zeta(4)\) in terms of new integral representations.

Main results

First, we wish to briefly explain work of Boo Rim Choe (1987) [4], Ewell (1990) [5] and Williams-Yue (1993) [6, p.1582-1583] which motivated us. Their common idea is that, from Maclaurin series involving \(\sin^{-1}x\), they each derived certain infinite sums related to \(\zeta(2)\) and \(\zeta(3)\) with termwise Wallis integral. Figure 1 gives summary of this.

Table 1. Summary of Boo Rim Choe, Ewell and Williams-Yue's work.

Boo Rim Choe	\(\sin^{-1}x=\sum\limits_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!}\frac{x^{2n+1}}{2n+1}\)	\(\rightarrow\)	\(\sum\limits_{n=0}^{\infty}\frac{1}{(2n+1)^{2}}=\frac{3}{4}\zeta(2)=\frac{\pi^{2}}{8}\)
Ewell	\(\frac{\sin^{-1}x}{x}=\sum\limits_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!}\frac{x^{2n}}{2n+1}\)	\(\rightarrow\)	\(\sum\limits_{n=0}^{\infty}\frac{1}{(2n+1)^{3}}=\frac{7}{8}\zeta(3)\)
Williams-Yue	\(\frac{(\sin^{-1}x)^{2}}{x}=\frac{1}{2}\sum\limits_{n=1}^{\infty}\frac{(2n)!!}{(2n-1)!!}\frac{x^{2n-1}}{n^{2}}\)	\(\rightarrow\)	\(\frac{\pi}{8}\sum\limits_{n=1}^{\infty}\frac{1}{n^{3}}=\frac{\pi}{8}\zeta(3)\)

In this article, we reformulate their ideas introducing Wallis operator and naturally extend their results.

• We find new integral representations for \(\text{Li}_{2}(t)\), \(\text{Li}_{3}(t)\), Legendre \(\chi\) functions of order 2, 3 and even for Apéry, Catalan constants (Theorems 2, 5, Corollaries 3, 6).
• We give a lower bound for \(\text{Li}_{2}(t)\) on the unit interval (Theorem 7).
• Making use of \((\sin^{-1}x)^{3}\) and \((\sin^{-1}x)^{4}\), we prove new Euler sums (Theorem 8).

Notation

Throughout this paper, \(n\) denotes a nonnegative integer. Let \begin{align*} (2n)!!&=2n(2n-2)\cdots 4\cdot 2, \\(2n-1)!!&=(2n-1)(2n-3)\cdots 3\cdot 1. \end{align*} In particular, we understand that \((-1)!!=0!!=1\). Moreover, let \[ w_{n}=\frac{(n-1)!!}{n!!}. \] Notice the relation \(w_{2n}w_{2n+1}=\frac{1}{2n+1}\) as we will see in the sequel.

Remark 1.

1. The sequence \(\{w_{n}\}\) appears in Wallis integral as \[ \int_{0}^{\pi/2}{\sin^n x} = \begin{cases} \frac{\pi}{2} w_{n}& \text{\(n\) even,}\\ w_{n}& \text{\(n\) odd.}\\ \end{cases} \]
2. It also appears in the literature in the disguise of central binomial coefficients as \[ w_{2n}= \frac{(2n-1)!!}{(2n)!!}=\frac{1}{2^{2n}}\binom{2n}{n}. \] See Apéry [7], van der Poorten [8], for example.

Unless otherwise specified, \(t, u, x, y\) are real numbers. By \(\sin^{-1} x\) and \(\cos^{-1} x\), we mean the real inverse sine and cosine functions (\(\arcsin x, \arccos x\)), that is,

\[ \begin{array}{ccl} y=\sin^{-1}x&\iff & x=\sin y, \quad-\frac{\,\pi\,}{2}\le y\le \frac{\,\pi\,}{2},\\ y=\cos^{-1}x&\iff & x=\cos y, \quad0\le y\le \pi. \end{array} \]

Fact 1. (Gradshteyn-Ryzhik [9, p.60, 61])

Further, \(\sinh^{-1}x=\log(x+\sqrt{x^{2}+1})\) \((x\in\mathbb{R})\) denotes the inverse hyperbolic sine function (some authors write \(\text{arsinh  } x,\text{arcsinh  } x\) or \(\text{argsinh  } x\) for this one).

2. Dilogarithm function

2.1. Definition

Definition 1. For \(0\le t\le 1\), the dilogarithm function is \[ \text{Li}_2(t)= \sum_{n=1}^{\infty}\frac{t^{n}}{n^{2}}. \]

In particular, \(\text{Li}_2(1)=\zeta(2)=\frac{\pi^{2}}{6}.\)

It is possible to describe its even part by \(\text{Li}_2\) itself since

\[ \sum_{n=1}^{\infty}\frac{t^{2n}}{(2n)^{2}} =\frac{1}{4}\sum_{n=1}^{\infty}\frac{(t^2)^{n}}{n^{2}} =\frac{1}{4}\text{Li}_2(t^{2}). \] Its odd part is called the Legendre \(\chi\) function of order 2: \[ \chi_2(t)=\sum_{n=1}^{\infty}\frac{t^{2n-1}}{(2n-1)^{2}}. \] Here is a fundamental relation of these two parts.

Observation 1. \[ \text{Li}_2(t)=\chi_2(t)+\frac{1}{4}\text{Li}_2(t^{2}). \]

Definition 2. Define \[ \text{Ti}_{2}(t)= \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n-1)^{2}}t^{2n-1} \] as a signed analog of \(\chi_2(t)\).

This is also called the inverse tangent integral of order 2 because of the integral representations \[ \text{Ti}_{2}(t)= \int_{0}^{t}{\frac{\tan^{-1}x}{x}}\,dx. \]

2.2. Wallis operator

Let \(\mathbb{R}[[t]]\) denote the set of power series in \(t\) over real coefficients. Set \[ F(t)= \{f\in \mathbb{R}[[t]]\;\;\left| \text{\(f(t)\) is convergent for \(|t|\le 1\)}\right. \}. \]

Definition 3. For \(f\in F(t)\), define \(W:F(t)\rightarrow F(t)\) by \[ Wf(t)= \int_0^1f(tu)\frac{du}{\sqrt{1-u^2}}. \] Call \(W\) the Wallis operator.

Remark 2. [9, p.17] Power series may be integrated and differentiated termwise inside the circle of convergence without changing the radius of convergence. In the sequel, we will frequently use this without mentioning explicitly.

It is now helpful to understand \(W\) coefficientwise.

Lemma 1. Let \(f(t)= \sum_{n=0}^{\infty}a_{n}t^{n}\in F(t)\). Then \[ Wf(t)= \sum_{n=0}^{\infty}a_{2n}\left(\frac{\pi}{2}w_{2n}\right) t^{2n}+ \sum_{n=0}^{\infty}a_{2n+1}w_{2n+1}t^{2n+1}. \]

Proof. \begin{align*} Wf(t)&=\int_{0}^{1}{f(tu)}\,\frac{du}{\sqrt{1-u^{2}}} \\&=\int_{0}^{1}{ \left(\sum_{n=0}^{\infty}a_{2n}t^{2n}u^{2n} + \sum_{n=0}^{\infty}a_{2n+1}t^{2n+1}u^{2n+1}\right) }\,\frac{du}{\sqrt{1-u^{2}}} \\&= \sum_{n=0}^{\infty}a_{2n}t^{2n} \int_{0}^{1}{u^{2n}}\,\frac{du}{\sqrt{1-u^{2}}} + \sum_{n=0}^{\infty}a_{2n+1}t^{2n+1} \int_{0}^{1}{u^{2n+1}}\,\frac{du}{\sqrt{1-u^{2}}} \end{align*}\begin{align*}&= \sum_{n=0}^{\infty}a_{2n}\left(\frac{\pi}{2}w_{2n}\right) t^{2n}+ \sum_{n=0}^{\infty}a_{2n+1}w_{2n+1}t^{2n+1}. \end{align*}

Observe that \(W\) is linear in the sense that \(W(f+g)=W(f)+W(g)\) and \(W(cf)=cW(f)\) for \(f, g\in F(t), c\in\mathbb{R}\).

2.3. Main Theorem 1

Lemma 2. All of the following are convergent power series for \(|t|\le1\).

Proof. We already saw (3) and (4) in Introduction. (5) is \((3)+\frac{2}{\pi}(4)\). (6) and (7) follow from (3), (4) and \(\sinh^{-1}z=\frac{1}{i}\sin^{-1}(iz)\) (for all \(z\in\mathbb{C}\)) [9, p.56].

Theorem 2. For \(0\le t\le 1\), all of the following hold;

Proof. Note that these are equivalent to the following statements:

With Lemmas 1 and 2, we can verify (13)-(16) by checking coefficients of those series. For example, \begin{align*} W(\sin^{-1}t)&=W\left( \sum_{n=0}^{\infty}w_{2n}\frac{t^{2n+1}}{2n+1} \right) =\sum_{n=0}^{\infty}w_{2n}w_{2n+1}\frac{t^{2n+1}}{2n+1} =\sum_{n=0}^{\infty}\frac{t^{2n+1}}{(2n+1)^{2}}=\chi_2(t). \end{align*} It remains to show (17). \begin{align*} W\left( {\frac{\,1\,}{2} (\sinh^{-1}t)^{2}}\right) &=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{w_{2n}} \left(w_{2n}\frac{\pi }{2}\right) \frac{t^{2n}}{(2n)^{2}} \\&=\frac{\pi }{2} \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n)^{2}} t^{2n} \\&=\frac{\pi }{2} \left( \sum_{n=1}^{\infty}\frac{1}{(2n)^{2}} t^{2n} - 2 \sum_{n=1}^{\infty}\frac{1}{(4n)^{2}} t^{4n} \right) \\&= \frac{\pi}{2} \left(\frac{1}{4}\text{Li}_{2}(t^{2})-\frac{\,1\,}{8}\text{Li}_{2}(t^{4})\right). \end{align*}

Corollary 3.

Proof. These are \(\chi_{2}(1), \frac{1}{4} \text{Li}_2(1^{2}), \text{Li}_2(1), \text{Ti}_{2}(1)\) and \(\frac{\pi}{2} \left(\frac{1}{4}\text{Li}_{2}(1^{2})-\frac{\,1\,}{8}\text{Li}_{2}(1^{4})\right)\).

3. Trilogarithm function

3.1. Definition

Definition 4. The trilogarithm function for \(0\le t\le 1\) is \[ \text{Li}_{3}(t)= \sum_{n=1}^{\infty}\frac{t^{n}}{n^{3}}. \]

Its odd part is the Legendre \(\chi\) function of order 3: \[ \chi_3(t)= \sum_{n=1}^{\infty}\frac{t^{2n-1}}{(2n-1)^{3}}. \] In particular, \(\text{Li}_{3}(1)=\zeta(3)\) and \(\chi_{3}(1)=\frac{7}{8}\zeta(3)\).

Observation 4. \[ \text{Li}_3(t)=\chi_3(t)+\frac{1}{8}\text{Li}_3(t^{2}). \]

Further, a signed analog of \(\chi_3(t)\) is \[ \text{Ti}_{3}(t)= \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n-1)^{3}}t^{2n-1}. \]

3.2. Main Theorem 2

Lemma 3.

Proof. We can derive all of these by integrating (3)-(7) termwise.

As a consequence, we obtain the equalities below (cf. (13)-(17)). In this way, the five functions above come to possess double integral representations. For example, \[ \chi_3(t)= \int_{0}^{1}{ \left( \int_{0}^{tu}{\frac{\sin^{-1}y}{y}}\,dy \right) }\,\frac{du}{\sqrt{1-u^{2}}}. \] We can indeed simplify such integrals to single ones by exchanging order of integrals.

Theorem 5.

Proof. We give a proof altogether. For \(t=0\), all the equalities hold as \(0=0\). Suppose \(0< t\le 1\). Let \[ f(y)\in\left\{ \sin^{-1}y, \frac{1}{\pi}(\sin^{-1}y)^{2}, \sin^{-1}y+\frac{1}{\pi}(\sin^{-1}y)^{2}, \sinh^{-1}y, \frac{1}{2}(\sinh^{-1}y)^{2}\right\}. \] Then \begin{align*} W\left( \int_{0}^{t}{\frac{f(y)}{y}}\,dy\right) &= \int_{0}^{1}{ \int_{0}^{tu}{\frac{f(y)}{y}}\,dy }\,\frac{du}{\sqrt{1-u^{2}}} \\&= \int_{0}^{t}{ \int_{y/t}^{1}{\frac{f(y)}{y}} \frac{1}{\sqrt{1-u^{2}}} }\,dudy \\&= \int_{0}^{t} {\frac{f(y)}{y}}\cos^{-1}{\frac{y}{t}} \,dy \\&= \int_{0}^{1} {\frac{f(tx)}{x}}\cos^{-1}{x} \,dx. \end{align*}

Corollary 6.

Proof. These are \(\chi_{3}(1), \frac{1}{8}\text{Li}_3(1^{2}), \text{Li}_3(1), \text{Ti}_{3}(1)\) and \(\frac{\pi}{2}\left(\frac{1}{8}\text{Li}_3(1^{2})-\frac{1}{32}\text{Li}_3(1^{4})\right)\).

4. Applications

4.1. Inequalities

It is easy to see from the definitions \(\text{Li}_2(t)=\sum_{n=1}^{\infty}\frac{t^{n}}{n^{2}}\) and \(\chi_2(t)=\sum_{n=1}^{\infty}\tfrac{t^{2n-1}}{(2n-1)^{2}}\) (\(0\le t\le 1\)) that \[ 0\le \text{Li}_2(t)\le \frac{\pi^{2}}{6} \quad \text{and}\quad 0\le \chi_2(t)\le \frac{\pi^{2}}{8}. \] In fact, we can improve these inequalities a little more. For upper bounds, it is immediate that \begin{align*} &\text{Li}_2(t)= \sum_{n=1}^{\infty}\frac{t^{n}}{n^{2}} \le \sum_{n=1}^{\infty}\frac{t}{n^{2}} =\frac{\pi^{2}}{6}t, \\&\chi_{2}(t)= \sum_{n=1}^{\infty}\frac{t^{2n-1}}{(2n-1)^{2}} \le \sum_{n=1}^{\infty}\frac{t}{(2n-1)^{2}} =\frac{\pi^{2}}{8}t. \end{align*} We next prove nontrivial lower bounds for these functions and also \(\text{Ti}_{2}(t)\).

Theorem 7. For \(0\le t\le 1\),

Before the proof, we need a lemma. It provides another integral representation of \(\text{Li}_2(t)\) which seems interesting itself.

Lemma 4. For \(0\le t\le 1\),

\( \left(\text{cf.} \quad \text{Li}_3(t)=\frac{8}{\pi} \int_{0}^{1}\frac{(\sin^{-1}(\sqrt{t}x))^{2}\cos^{-1}x}{x}dx, \quad \text{\(t\mapsto \sqrt{t}\) in (34).} \right)\)

Proof. If \(t=0\), then both sides are \(0\). For \(0< t\le 1\), \begin{align*} \text{RHS}&= \frac{8}{\pi} \int_{0}^{\sqrt{t}}{\frac{\sin^{-1}y}{\sqrt{1-y^{2}}}}\cos^{-1}\frac{y}{\sqrt{t}}\,dy \\&= \frac{8}{\pi} \int_{0}^{\sqrt{t}}{\frac{\sin^{-1}y}{\sqrt{1-y^{2}}}} \int_{y/\sqrt{t}}^{1}{\frac{1}{\sqrt{1-u^{2}}}}\,dudy \\&=\frac{8}{\pi} \int_{0}^{1} \int_{0}^{\sqrt{tu}} {\frac{\sin^{-1}y}{\sqrt{1-y^{2}}}}dy \frac{1}{\sqrt{1-u^{2}}}\,du \\&=\frac{8}{\pi} W\left( \int_{0}^{\sqrt{t}} {\frac{\sin^{-1}y}{\sqrt{1-y^{2}}}} dy\right) \\&=\frac{8}{\pi} W\left(\frac{1}{2}\left(\sin^{-1}\sqrt{t}\right) ^{2}\right) \\&=\frac{8}{\pi} \left(\frac{\pi}{2} \frac{1}{4} \text{Li}_2\left((\sqrt{t})^{2}\right) \right) =\text{Li}_2(t). \end{align*}

Proof of Theorem 7. If \(t=0\), then all of (43)-(45) hold as \(0\ge0\). Suppose \(t>0\). Since \(\sin^{-1}\) is increasing on \([0, 1]\), \(\sin^{-1}(tx)\le \sin^{-1}(\sqrt{t}x)\) for all \(0< t, x\le 1\). Then \begin{align*} \text{Li}_2(t) &= \frac{8\sqrt{t}}{\pi} \int_{0}^{1}{ \frac{\sin^{-1}(\sqrt{t}x)}{\sqrt{1-tx^{2}}} }\cos^{-1}x\,dx \\&\ge \frac{8t}{\pi} \int_{0}^{1}{ \frac{\sin^{-1}(tx)}{\sqrt{1-t^{2}x^{2}}} }\cos^{-1}x\,dx \\&= \frac{8{t}}{\pi} \int_{0}^{1}{ \frac{1}{t}\left(\frac{1}{2}(\sin^{-1}tx)^{2}\right) }\cos^{-1}x\,dx \\&= \frac{8}{\pi} \left( \underbrace{\left[ {\frac{1}{2}(\sin^{-1}tx)^{2}}\cos^{-1}x \right]^{1}_{0}}_{0} - \int_{0}^{1}{\frac{1}{2}(\sin^{-1}tx)^{2}\frac{-1}{\sqrt{1-x^{2}}}}\,dx \right) \\&\ge \frac{4}{\pi} \int_{0}^{1}{ \frac{(\sin^{-1}tx)^{2}}{\sqrt{1-t^{2}x^{2}}}} \,dx \\&=\frac{4}{\pi}\left[\frac{1}{3t}(\sin^{-1}tx)^{3}\right]^{1}_{0} =\frac{4}{3\pi}\frac{(\sin^{-1}t)^{3}}{t}. \end{align*} Next, we prove (44). Note that \[ \frac{\sin^{-1}(tx)}{\sqrt{1-t^{2}x^{2}}}\le \frac{\sin^{-1}(tx)}{\sqrt{1-x^{2}}} \] for \(0< t, x< 1\). Integrate these from \(0\) to \(1\) in \(x\) so that \begin{align*} \int_{0}^{1}\frac{\sin^{-1}(tx)}{\sqrt{1-t^{2}x^{2}}} \,dx &\le \int_{0}^{1} \frac{\sin^{-1}(tx)}{\sqrt{1-x^{2}}} \, dx, \\ \left[\frac{(\sin^{-1}(tx))^{2}}{2t}\right]^{1}_{0}&\le \chi_2(t), \\ \frac{(\sin^{-1}t)^{2}}{2t} &\le \chi_2(t). \end{align*} Quite similarly, for \(0< t, x< 1\), it also holds that \begin{align*} \frac{\sinh^{-1}(tx)}{\sqrt{1+t^{2}x^{2}}}&\le \frac{\sinh^{-1}(tx)}{\sqrt{1-x^{2}}}, \\ \int_{0}^{1}{ \frac{\sinh^{-1}(tx)}{\sqrt{1+t^{2}x^{2}}}}\,dx &\le \int_{0}^{1}{ \frac{\sinh^{-1}(tx)}{\sqrt{1-x^{2}}} }\,dx = \text{Ti}_{2}(t). \end{align*} The left hand side is \[ \left[\frac{\sinh^{-1}(tx)^{2}}{2t}\right]^{1}_{0}=\frac{(\sinh^{-1}t)^{2}}{2t}. \]

4.2. Euler sums

Definition 5. A harmonic number is \(H_{n}=\sum_{k=1}^{n} \frac{1}{k}\). More generally, for \(m, n\ge 1\), an \((m, n)\)- harmonic number is \[ H_{n}^{(m)}= \sum_{k=1}^{n}\frac{1}{k^{m}}. \]

In particular, \(H_{n}^{(1)}=H_{n}\). Any series involving such numbers is called an Euler sum.

Valean [10, p.292-293] presents truly remarkable Euler sums such as

\begin{align*} \sum_{n=1}^{\infty}\frac{H_{n}^{2}}{n^{2}}&=\frac{17}{4}\zeta(4), \\\sum_{n=1}^{\infty}\frac{H_{n}^{2}}{n^{3}}&=\frac{7}{2}\zeta(5)-\zeta(2)\zeta(3), \\\sum_{n=1}^{\infty}\frac{H_{n}^{2}}{n^{4}}&= \frac{97}{24}\zeta(6)-2\zeta^{2}(3), \\\sum_{n=1}^{\infty}\frac{H_{n}^{2}}{n^{5}}&=6\zeta(7)-\zeta(2)\zeta(5)-\frac{5}{2}\zeta(3)\zeta(4), \\\sum_{n=1}^{\infty}\frac{H_{n}H_{n}^{(2)}}{n^{2}}&=\zeta(2)\zeta(3)+\zeta(5). \end{align*} There are many ideas to prove such formulas; Borwein and Bradley [11] gives thirty two proofs for \[ \sum_{n=1}^{\infty}\frac{H_{n-1}}{n^{2}}= \zeta(3)= 8\sum_{n=1}^{\infty}\frac{(-1)^{n}H_{n-1}}{n^{2}} \] by integrals, polylogarithm functions, Fourier series and hypergeometric functions etc. Here, as an application of our main idea, Wallis operators, we prove two new Euler sums. Let \[ {O}_{n}^{(2)}=H_{2n-1}^{(2)}-\frac{1}{4}H_{n-1}^{(2)} = \sum_{k=1}^{2n-1}\frac{1}{k^{2}}- \sum_{k=1}^{n-1}\frac{1}{(2k)^{2}}= \sum_{k=0}^{n-1} \frac{1}{(2k+1)^{2}}. \]

Theorem 8.

For the proof, we make use of less-known Maclaurin series for \((\sin^{-1}t)^{3}\) and \((\sin^{-1}t)^{4}\); thus we can interpret this result as a natural subsequence of Boo, Ewell and Williams-Yue's work.

Lemma 5.

\( \left( \text{cf.}\quad \sin^{-1}t= \sum\limits_{n=0}^{\infty} w_{2n} \frac{t^{2n+1}}{2n+1}, \quad (\sin^{-1}t)^{2}= \frac{\,1\,}{2} \sum\limits_{n=1}^{\infty} \frac{1}{w_{2n}}\frac{t^{2n}}{n^{2}} \right) \).

Proof. First, write \((\sin^{-1}t)^{3}= \sum_{n=0}^{\infty}A_{n}t^{2n+1}\), \(A_{n}\in\mathbb{R}\) and let \(a_{n}=\frac{2n+1}{w_{2n}}A_{n}\, (n\ge0)\). It is enough to show that \(a_{n}=6O_{n}^{(2)}\). Since the series \((\sin^{-1}t)^{3}=(t+\frac{t^{3}}{6}+\cdots)^{3}\) starts from the \(t^{3}\) term, \(A_{0}=a_{0}=0\). For convenience, set \[ f_{n}(x)=\frac{\sin^{2n+1}x}{(2n+1)!}. \] Then \[ f_{n}'(x)=\frac{\sin^{2n}x}{(2n)!}\cos x, \] \[ f_{n}''(x)=\frac{1}{(2n)!} \left(2n\sin^{2n-1}x(1-\sin^{2}x)-\sin^{2n+1}x\right) =f_{n-1}(x)-(2n+1)^{2}f_{n}(x). \] Now let \(x=\sin^{-1}t\) (\(-\frac{\,\pi\,}{2}\le x\le \frac{\,\pi\,}{2}\)), \(b_{n}=(2n-1)!!\). Recall that \[ \sin^{-1}t= \sum_{n=0}^{\infty} w_{2n} \frac{t^{2n+1}}{2n+1}. \] In terms of \(x\), \(b_{n}, f_{n}(x)\), this is \[ x= \sum_{n=0}^{\infty} w_{2n} \frac{\sin^{2n+1}x}{2n+1} = \sum_{n=0}^{\infty} \frac{(2n-1)!!}{(2n)!!}(2n)! \frac{\sin^{2n+1}x}{(2n+1)!} = \sum_{n=0}^{\infty}b_{n}^{2} f_{n}(x). \] Thus, \begin{align*} x^{3}&=\sum_{n=0}^{\infty} A_{n}\sin^{2n+1}x =\sum_{n=0}^{\infty} a_{n} \left(w_{2n}\frac{\sin^{2n+1}x}{2n+1}\right) =\sum_{n=0}^{\infty} a_{n}b_{n}^{2} f_{n}(x). \end{align*} Differentiate both sides twice in \(x\): \begin{align*} 6x&= \sum_{n=0}^{\infty}a_{n}b_{n}^{2}f_{n}''(x) \\&= \sum_{n=0}^{\infty}a_{n}b_{n}^{2}(f_{n-1}(x)-(2n+1)^{2}f_{n}(x)) \\&= \sum_{n=0}^{\infty} (a_{n+1}b_{n+1}^{2}f_{n}(x)- a_{n}b_{n}^{2}(2n+1)^{2}f_{n}(x)), \\ 6 \sum_{n=0}^{\infty}b_{n}^{2}f_{n}(x) &= \sum_{n=0}^{\infty} (a_{n+1}b_{n+1}^{2}f_{n}(x)- a_{n}b_{n}^{2}(2n+1)^{2}f_{n}(x)). \end{align*} Equating coefficients of \(f_{n}(x)\) yields \[ 6b_{n}^{2}=a_{n+1}b_{n+1}^{2}- a_{n}b_{n}^{2}(2n+1)^{2}, \quad n\ge0. \] Since \(b_{n+1}=(2n+1)b_{n}\) and \(b_{n}\ne0\), we must have \[ a_{n+1}-a_{n}=\frac{6}{(2n+1)^{2}}. \] With \(a_{0}=0\), we now arrive at \[ a_{n}= \sum_{k=0}^{n-1}\frac{6}{(2k+1)^{2}}=6O_{n}^{(2)}, \] as required.

The proof for (50) proceeds along the same line. Write \((\sin^{-1}t)^{4}= \frac{1}{2}\sum_{n=0}^{\infty}C_{n}t^{2n}\), \(C_{n}\in\mathbb{R}\) and let \(c_{n}=C_{n}w_{2n}n^{2}\, (n\ge1)\). It is enough to show that \(c_{n}=3H_{n-1}^{(2)}\). Since the series \((\sin^{-1}t)^{4}\) starts from the \(t^{4}\) term, \(C_{1}=c_{1}=0\). For convenience, set

\[ g_{n}(x)=\frac{\sin^{2n}x}{(2n)!}. \] Then \[ g_{n}'(x)=\frac{\sin^{2n-1}x}{(2n-1)!}\cos x, \] \[ g_{n}''(x)=\frac{1}{(2n-1)!} \left((2n-1)\sin^{2n-2}x(1-\sin^{2}x)-\sin^{2n}x\right) =g_{n-1}(x)-(2n)^{2}g_{n}(x). \] Now let \(x=\sin^{-1}t\) (\(-\frac{\,\pi\,}{2}\le x\le \frac{\,\pi\,}{2}\)), \(d_{n}=2^{n}(n-1)!\). Recall that \[ (\sin^{-1}t)^{2}= \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{w_{2n}} \frac{t^{2n}}{n^{2}}. \] In terms of \(x\), \(d_{n}, g_{n}(x)\), this is \[ x^{2}= \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{w_{2n}} \frac{\sin^{2n}x}{n^{2}} = \frac{1}{2} \sum_{n=1}^{\infty} \frac{(2n)!!}{(2n-1)!!} \frac{(2n)!}{n^{2}} \frac{\sin^{2n}x}{(2n)!} = \frac{1}{2}\sum_{n=1}^{\infty}d_{n}^{2} g_{n}(x). \] Thus, \begin{align*} x^{4}= \frac{1}{2}\sum_{n=1}^{\infty} c_{n} \left(\frac{1}{w_{2n}}\frac{\sin^{2n}x}{n^{2}}\right)= \frac{1}{2}\sum_{n=1}^{\infty} c_{n} d_{n}^{2}g_{n}(x). \end{align*} Differentiate both sides twice in \(x\): \begin{align*} 12x^{2}&= \frac{1}{2}\sum_{n=1}^{\infty}c_{n}d_{n}^{2}g_{n}''(x) \\&= \frac{1}{2} \sum_{n=1}^{\infty}c_{n}d_{n}^{2}(g_{n-1}(x)-(2n)^{2}g_{n}(x)) \end{align*}\begin{align*}&= \frac{1}{2} \sum_{n=0}^{\infty} c_{n+1}d_{n+1}^{2}g_{n}(x)- \frac{1}{2} \sum_{n=1}^{\infty} c_{n}d_{n}^{2}(2n)^{2}g_{n}(x) \\&= \frac{1}{2} \sum_{n=1}^{\infty} c_{n+1}d_{n+1}^{2}g_{n}(x)- \frac{1}{2} \sum_{n=1}^{\infty} c_{n}d_{n}^{2}(2n)^{2}g_{n}(x) \quad\text{(\(c_{1}=0\))} \\ &= \frac{1}{2} \sum_{n=1}^{\infty} (c_{n+1}d_{n+1}^{2}-c_{n}d_{n}^{2}(2n)^{2})g_{n}(x),\\ \frac{12}{2}\sum_{n=1}^{\infty}d_{n}^{2} g_{n}(x) &= \frac{1}{2} \sum_{n=1}^{\infty} (c_{n+1}d_{n+1}^{2}-c_{n}d_{n}^{2}(2n)^{2})g_{n}(x). \end{align*} Equating coefficients of \(g_{n}(x)\) yields \[ \frac{12}{2}d_{n}^{2}= \frac{1}{2} (c_{n+1}d_{n+1}^{2}-c_{n}d_{n}^{2}(2n)^{2}), \quad n\ge 1. \] Since \(d_{n+1}=2n d_{n}\) and \(d_{n}\ne0\), we must have \[ c_{n+1}-c_{n}=\frac{12}{(2n)^{2}}. \] With \(c_{1}=0\), we conclude that \[ c_{n}= \sum_{k=1}^{n-1}\frac{12}{(2k)^{2}} = \sum_{k=1}^{n-1}\frac{3}{k^{2}}=3H_{n-1}^{(2)}. \]

Proof of Theorem 8. Note that \begin{align*} W\left(\frac{1}{6}(\sin^{-1}t)^{3}\right)&= \sum_{n=1}^{\infty}\frac{\widetilde{O}_{n}^{(2)}w_{2n}}{2n+1}w_{2n+1}t^{2n+1} \\&= \sum_{n=1}^{\infty}\frac{\widetilde{O}_{n}^{(2)}}{(2n+1)^{2}}t^{2n+1}. \end{align*} Clearly, \(t=1\) gives the sum for (47). Therefore, \[ W\left(\frac{1}{6}(\sin^{-1}t)^{3}\right) \Biggr|_{t=1} = \int_{0}^{1}{ \frac{1}{6}(\sin^{-1}u)^{3} }\,\frac{du}{\sqrt{1-u^{2}}} = \left[\frac{1}{24}(\sin^{-1} u)^{4}\right]^{1}_{0}= \frac{\pi^{4}}{384}. \] Similarly, we have \[ W\left(\frac{2}{3}(\sin^{-1}t)^{4}\right) = \frac{\pi}{2} \sum_{n=1}^{\infty}\frac{H_{n-1}^{(2)}}{n^{2}}t^{2n} \] so that \begin{align*} W\left(\frac{2}{3}(\sin^{-1}t)^{4}\right) \Biggr|_{t=1} &= \int_{0}^{1}{ \frac{2}{3}(\sin^{-1}u)^{4} }\,\frac{du}{\sqrt{1-u^{2}}} = \left[\frac{2}{15}(\sin^{-1} u)^{5}\right]^{1}_{0}= \frac{\pi^{5}}{240}. \end{align*} We conclude that \[ \sum_{n=1}^{\infty}\frac{H_{n-1}^{(2)}}{n^{2}} = \frac{2}{\pi} \left(\frac{\pi^{5}}{240}\right)=\frac{\pi^{4}}{120}. \]

Remark 3.

1. (47) is a variation of De Doelder's formula \(\sum_{n=1}^{\infty}\frac{O_{n}^{(2)}}{n^{2}}=\frac{\pi^{4}}{32}\) [12, p.1196 (13)] and (48) gives another proof of \(\sum_{n=1}^{\infty}\frac{H_{n}^{(2)}}{n^{2}}=\frac{7}{4}\zeta(4)\) [10, p.286] because \[ \sum_{n=1}^{\infty}\frac{H_{n}^{(2)}}{n^{2}} = \sum_{n=1}^{\infty} \left(\frac{H_{n-1}^{(2)}}{n^{2}} +\frac{1}{n^{4}} \right) = \frac{3}{4}\zeta(4)+\zeta(4) = \frac{7}{4}\zeta(4). \]
2. After preparation of the manuscript, Christophie Vignat kindly told me that recently Guo-Lim-Qi (2021) [13] described Maclaurin series of integer powers of arcsin. In fact, it was the result from J.M. Borwein-Chamberland (2007) [14].

4.3. Integral evaluation

As byproduct of our discussions, we find evaluation of many integrals with known special values of \(\text{Li}_2(t), \text{Li}_3(t)\). Here, we record several examples. Let \(\phi=\frac{1+\sqrt{5}}{2}\) be the golden ratio. Observe that \[ \phi^{-1}=\frac{\sqrt{5}-1}{2}, \quad \phi^{-2}=\frac{3-\sqrt{5}}{2}. \] We write \(\log^{2}x\) for \((\log x)^{2}\). Note that \[ \log^{2}(\phi^{-1})= (\log(\phi^{-1}))^{2}= (-\log(\phi))^{2}= (\log(\phi))^{2}= \log^{2}(\phi). \]

Fact 2. ([2]).

Corollary 9.

Proof. \begin{align*} \int_{0}^{1}{ \frac{\sin^{-1} (\phi^{-1}u)}{\sqrt{1-u^{2}}} \,{du}} &= \chi_{2}(\phi^{-1})=\text{Li}_2(\phi^{-1})-\frac{1}{4}\text{Li}_2(\phi^{-2}) \\ &= \left(-\log^{2}(\phi)+\frac{\pi^{2}}{10}\right) -\frac{\,1\,}{4} \left(-\log^{2}(\phi)+\frac{\pi^{2}}{15}\right) = -\frac{3}{4}\log^{2}(\phi)+\frac{\pi^{2}}{12}. \end{align*} \begin{align*} W\left(\frac{1}{2}(\sin^{-1}t)^{2}\right) \bigr|_{t=1/\sqrt{2}}&= \frac{\pi}{8}\text{Li}_2\left(\frac{1}{2}\right) = \frac{\pi}{8}\left(\frac{\pi^{2}}{12}-\frac{\,1\,}{2}\log^{2}2\right). \end{align*} (34) for \(t=\phi^{-1}\) with (53) gives (58). \[ W(\frac{1}{2}(\sinh^{-1}t)^{2}) \bigr|_{t=\phi^{-1/2}}= \frac{\pi}{2}\left( \frac{1}{4}\text{Li}_2(\phi^{-1})- \frac{1}{8}\text{Li}_2(\phi^{-2}) \right) = \frac{\pi}{2}\left(- \frac{\,1\,}{8}\log^{2}\phi+ \frac{\pi^{2}}{60}\right). \] Finally, (34) for \(t=1/\sqrt{2}\) with (55) gives (60).

5. Concluding remarks

Here, we record several remarks for our future research.

1. For \(0\le \alpha\le 1\), define a generalized Wallis operator \[ W_{\alpha}f(t)= \int_{0}^{\alpha}{f(tu)}\,\frac{du}{\sqrt{1-u^{2}}} \] so that we can deal with more general integrals. Study \(W_{\alpha}\), particularly for \(\alpha=1/2, \sqrt{2}/2, \sqrt{3}/2\).
2. Can we show any inequality for \(\text{Li}_3(t), \chi_3(t)\) and \(\text{Ti}_{3}(t)\) in a similar way?
3. Discuss \((\sinh^{-1}t)^{3}\), \((\sinh^{-1}t)^{4}\) and related Euler sums.
4. Wolfram alpha [15] says that \begin{align*} & \int_{0}^{1}{\frac{(\sin^{-1}x)^{3}}{x}}\,dx = \int_{0}^{\pi/2}{u^{3}\cot u}\,du =\frac{\pi^{3}}{8}\log2-\frac{9}{16}\pi\zeta(3), \\ & \int_{0}^{1}{\frac{(\sin^{-1}x)^{4}}{x}}\,dx = \int_{0}^{\pi/2}{u^{4}\cot u}\,du = \frac{1}{32}\left(-18\pi^{2}\zeta(3)+93\zeta(5)+2\pi^{4}\log2\right). \end{align*} It should be possible to describe such integrals as certain infinite sums with or without numbers \(w_{2n}\). We plan to study those details in subsequent publication.
5. It is interesting that (38) happens to be quite similar to \[ \int_{0}^{1}{\frac{\tan^{-1}x\cot^{-1}x}{x}}\,dx =\frac{7}{8}\zeta(3). \] Not often this result appears in this form in the literature, though. Now, let us see how we evaluate this integral. Let \begin{align*} &I= \int_{0}^{1}{\frac{\tan^{-1}x\cot^{-1}x}{x}}\,dx, \\ &I_{1}= \int_{0}^{1}{\frac{\tan^{-1}x}{x}}\,dx, \\ &I_{2}= \int_{0}^{1}{\frac{(\tan^{-1}x)^{2}}{x}}\,dx. \end{align*} Then \begin{align*} I&= \int_{0}^{1}{\frac{\tan^{-1}x\cot^{-1}x}{x}}\,dx \\&= \int_{0}^{1}{\frac{\tan^{-1}x\left( \frac{\pi}{2}-\tan^{-1}x\right) }{x}}\,dx = {\frac{\pi}{2}I_{1}-I_{2}}. \end{align*} We can compute \(I_{1}\) and \(I_{2}\) as follows. \begin{align*} I_{1}&= \int_{0}^{1}{\frac{\tan^{-1}x}{x}}\,dx = \int_{0}^{1}{ \sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}x^{2n} }\,dx \\&=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1} \int_{0}^{1}{x^{2n}}\,dx= \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{2}}= G. \end{align*} For \(I_{2}\), recall from Fourier analysis that \[ \log\left(\tan\frac{y}{2}\right)=-2 \sum_{n=0}^{\infty}\frac{1}{2n+1}\cos(2n+1)y, \quad 0< y< \pi. \] It follows that \begin{align*} I_{2}&= \int_{0}^{1}{\frac{(\tan^{-1}x)^{2}}{x}}\,dx =_{[y=2\tan^{-1}x]} \frac{\,1\,}{4} \int_{0}^{\pi/2}{\frac{y^{2}}{\sin y}}\,dy \\&=\frac{\,1\,}{4} \left(\left[y^{2} \log\left(\tan \frac{y}{2}\right) \right]^{\pi/2}_{0} - \int_{0}^{\pi/2}{ 2y\log\left(\tan \frac{y}{2}\right) }\,dy \right) \\&=-\frac{1}{2} \int_{0}^{\pi/2}{ y \left(-2 \sum_{n=0}^{\infty}\frac{1}{2n+1} \cos(2n+1)y \right) }\,dy \\&=\sum_{n=0}^{\infty} \frac{1}{2n+1} \int_{0}^{\pi/2}{y\cos(2n+1)y}\,dy \\&=\sum_{n=0}^{\infty} \frac{1}{2n+1} \left(\left[ y\frac{\sin(2n+1)y}{2n+1} \right]^{\pi/2}_{0}- \int_{0}^{\pi/2}{ \frac{\sin(2n+1)y}{2n+1} }\,dy \right) \\&=\sum_{n=0}^{\infty} \left(\frac{\,\pi\,}{2}\frac{(-1)^{n}}{(2n+1)^{2}} -\frac{1}{(2n+1)^{3}}\right) =\frac{\pi G}{2}-\frac{7}{8}\zeta(3). \end{align*} Finally, we see \[ I= { \frac{\pi G}{2}- \left(\frac{\pi G}{2}-\frac{7}{8}\zeta(3) \right) }=\frac{7}{8}\zeta(3). \]

Open Question What if we replace \(\tan^{-1}\) by \(\tanh^{-1}\)?

In this article, we encountered many integral representations for dilogarithm, trilogarithm and hence \(\zeta(2)\), the Catalan constant \(G\) and \(\zeta(3)\) as a reformulation of Boo Rim Choe (1987) [4], Ewell (1990) [5] and Williams-Yue (1993) [6] on the inverse sine function. As an application, we also proved new Euler sums. Indeed, there are subsequent results on multiple zeta and \(t\)-values \(\zeta(3, 2, \cdots, 2)\), \(t(3, 2, \cdots, 2)\) as Hoffman and Zagier discussed in [16,17]. We will write them with more details at another opportunity.

Conflicts of Interest: 

"The author declares no conflict of interest".

1. Introduction

This paper is divided into three parts. The first part deals with some technical issues concerning Jefimenko's equations, which are perhaps not completely clear from [1]. Namely that we can obtain solutions to Maxwell's equations from a given charge and current configuration \(\left(\rho,\overline{J}\right)\), satisfying the continuity equation, with \(\overline{J}\) vanishing at infinity, using Jefimenko's equation to define the electric and magnetic fields \(\left\{\overline{E},\overline{B}\right\}\). In particular, this is the case for localised charge and current configurations, or when the charge and current decays rapidly at infinity, a condition true for the Schwartz class of functions which we have denoted by \(S\left(\overline{R}^{3}\right)\) and we have referred to as smoothly decaying. This result is applied in Lemma 5, where we construct \(\left(\rho,\overline{J},\overline{E},\overline{B}\right)\) satisfying Maxwell's equations and various other relations, just from the assumptions that \(\rho\) satisfies the wave equation, has a restriction on the initial condition, and belongs to the Schwartz class.

The second part of the paper is mainly concerned with deriving these relations, and proving a converse in the context of special relativity, that these relations are characterised by a no radiation condition in all inertial frames. In Lemma 3, we characterise the smoothly decaying solutions of the wave equation \(\square^{2}\left(\overline{E}\right)=\overline{0}\) for electric fields, as well as the smoothly decaying electromagnetic solutions \(\left(\overline{E},\overline{B}\right)\) to Maxwell's equations in free space. This requires a careful Fourier analysis, and, in particular, we require the smoothly decaying hypothesis to apply the inversion theorem. We conclude the Lemma by proving that if \(\square^{2}\left(\overline{E}\right)=\overline{0}\), then we can find a free space solution \(\left(\overline{E}_{0},\overline{B}_{0}\right)\), such that \(\bigtriangledown\times\left(\overline{E}-\overline{E}_{0}\right)=\overline{0}\). This result is required in Lemma 5.

In Lemma 4, we characterise \(\square^{2}\left(\overline{E}\right)\) and \(\square^{2}\left(\overline{B}\right)\) in terms of the quantities \(\left(\rho,\overline{J}\right)\). Combining these last two Lemmas, we obtain the result of Lemma 5 mentioned above, that we can obtain a number of conditions on \(\left(\rho,\overline{J},\overline{E},\overline{B}\right)\) from essentially the assumption that \(\rho\) satisfies the wave equation. In particularly, we can arrange for the magnetic field \(\overline{B}\) and Poynting vector \(\overline{E}\times\overline{B}\) to be both zero. In Lemma 6, we strengthen this result, to show that if we can obtain the relations of Lemma 5 for \(\left(\rho,\overline{J},\overline{E},\overline{B}\right)\) in the rest frame \(S\), then for any inertial frame \(S'\), we can extend the transformed charge and current \(\left(\rho',\overline{J}'\right)\) to \(\left(\rho',\overline{J}',\overline{E}',\overline{B}'\right)\), satisfying the same relations. This requires the transformation rules for quantities in \(S\), and we rely on the fact that the transformed quantities are bounded, which follows from the smoothly decaying hypothesis, to apply Liouville's theorem.

These last two Lemmas are the basis for the non-radiating condition, which we formulate in Definition 2, that in any inertial frame \(S'\) we can arrange for a solution \(\left(\rho',\overline{J}',\overline{E}',\overline{B}'\right)\), extending the transformed quantities \(\left(\rho',\overline{J}'\right)\), with \(\overline{B}'=\overline{0}\). This certainly ensures that the transformed system doesn't radiate, according to the definition in [1], as the Poynting vector \(\overline{E}'\times \overline{B}'\) is zero. Interestingly, in Lemma 7, a converse is proved, namely that for any system satisfying the non-radiating condition, in the rest frame \(S\), \(\rho\) and \(\overline{J}\) satisfy the wave equation, and we can obtain all the relations proved in Lemma 5. This suggests that the no-radiation condition might be enough to consider using wave equations for \(\left(\rho,\overline{J}\right)\) in atomic systems, which should not radiate according to Rutherford's observation.

However, there are weaker definitions of non-radiation, which we give in Definition 3. Some of these we are able to exclude, while supporting our hypothesis, which we do in Lemmas 8 and 9, and, others exclusions we leave as conjectures. Certainly, a successful proof of these conjectures, which we believe might be possible with a careful analysis of the stress energy tensor, see [2], would be a compelling argument for the use of the wave equation in quantum theory, and we should make some comparisons. Although Schrodinger's equation has a relativistic formulation in quantum field theory, through the use of the Klein-Gordon equation or the Dirac equation, it still lacks the relativistic invariance of the relations which we were able to derive, and, indeed, uses the possibility of radiating atomic systems to account for the Lamb shift in the hydrogen spectrum. We believe this shift might be accounted for in the discussion of the final part of the paper, and still believe that radiative behaviour would be a strange property. For example, the radiated energy would have to oscillate, if the system is to remain stable, and we do not observe this behaviour in the Lamb shift. Another important point to make here is that, if energy is radiated locally in the system, \(div\left(\overline{E}\times\overline{B}\right)\neq 0\), then, assuming that locally work is done on electrons in the system, the condition \(\left(\overline{E},\overline{J}\right)\neq 0\), we can obtain a transfer of heat from electrons in one part of the system to another. If we impose a condition of thermal equilibrium and expect that the system does not radiate to infinity, necessary to counter Rutherford's observation, we can obtain a violation of the second law of thermodynamics. We save this point of view for another occasion, the reader might be interested in [3], which considers cavity radiation on a fairly qualitative level. A successful resolution of these issues is clearly required and we believe that we have made a step in the right direction.

In the last part of this paper, we deal with a common criticism of classical electromagnetism, in it's failure to explain quantised phenomena, in particular the behaviour of the Balmer series or the results of the Franck-Hertz experiment. In Lemma 10, we start with the relations of Lemma 5, and characterise these relations in terms of the coefficients governing the Fourier expansions of \(\rho\) and \(\overline{J}\), a coefficient relation which we later refer to as a radial transform condition. In Lemma 11, we show that these relations can be obtained while also imposing the condition that \(\overline{J}|_{S(r_{0})}=\overline{0}\), a natural requirement for an atomic system. This introduces a discreteness phenomena in terms of the zeroes of Bessel functions vanishing at \(r_{0}\) and, in Lemma 12, we prove a technical result on how these functions, combined with the spherical harmonics, provide an orthogonal basis for smooth functions on the ball \(B(r_{0})\), vanishing at the boundary.

In Lemma 13, we compute the energy stored in the electromagnetic field, when restricted to \(B(r_{0})\), and find, that when the total charge confined to the ball is non-zero, we are able to remove the one continuous degree of freedom, to obtain quantisation of the energy values, in line with the Balmer series, see [4]. The ionisation condition, on the total charge \(Q\) confined to the ball, seems quite interesting, and is observed in the Frank-Hertz experiment. Again, some comparative remarks should be made, namely the success of Schrodinger's equation, together with Bohr's atomic model, in accounting for the Balmer series. We were not able to verify the value of the Rydberg constant in this paper, due to computational issues, but we emphasise that we were able to get the \({1\over n^{2}}\) dependence, where \(n\) is an integer. Possibly, further computations of Bessel functions and spherical harmonics can be done to resolve this issue, and, again, we believe that we have gone some way to answering this potential criticism of the use of the wave equation. We believe there is some scope for developing the classical theory here, without relying on Planck's photonic theory of radiation, in how the system switches between electromagnetic energy levels, and in how light of certain energies, which we propose would come from the differences in these energy levels, is emitted.

We make some observations which are not part of this paper, but instead direct the reader to further work. A principle justification for using the Schrodinger equation is its reliance on wave-particle duality. This requires that an electron can simultaneously be both a particle and a guiding wave, modelled on the square of the wave function \(|\Psi|^{2}\). The particle nature of the electron can be observed in experiments such as Millikan's deflection or Compton scattering, while the wave nature, can be seen in the interference patterns of electron diffraction, see [4]. However, it seems that a dual nature which explains these experiments is a failure to find a better explanation, with the cost of developing a rather paradoxical theory. Instead, a wave function, modelling the behaviour of charge and current, and linking directly with classical electromagnetism, has been proposed here.

The author believes that, using nonstandard analysis, one can develop the idea of the wave being composed of individual infinitesimal entities, which behave as particles, and propagate independently, with trajectories determined by a charge and current \(\left(\rho,\overline{J}\right)\) satisfying the continuity equation. Some successes for this viewpoint can be found in [5], where the heat equation solution is modelled as a random diffusion, and [6], where the nonstandard wave equation is shown to perfectly approximate the wave equation at standard values. Instead of considering individual velocities and momenta, one can instead compute the distribution of these quantities, and some unfinished progress in this direction has been made for the diffusion equation in [7]. A similar idea is present in Boltzmann's derivation of the distribution of molecular speeds for ideal gases. In the case of the wave equation satisfying additional conditions which we have examined, we expect a sharp peak in the distribution function, as would be expected from a single particle. Similar consideration might apply to photons.

We finally mention the theory of black body radiation, developed by Planck, and explained concisely in [8]. A criticism of classical electromagnetism has been its failure to explain the spectrum of black body radiation, by postulating that the radiated energy is independent of frequency, often referred to as the ultraviolet catastrophe. A central concept used in the experimentally successful resolution of this problem, was Planck's use of the quantisation of particular energies of molecular systems. The author hopes that the present paper goes some way to assuage these criticisms, without abandoning the main component of Planck's argument in quantum statistical mechanics.

2. Jefimenko's Equations

We begin this paper by first clarifying a technical issue surrounding Jefimenko's equations.

Lemma 1. Suppose that \(\left\{\overline{E},\overline{B}\right\}\) satisfy Maxwell equations, for given charge and current \(\left\{\rho,\overline{J}\right\}\), and the potentials \(\left\{\overline{A},V\right\}\), with; \[\overline{E}=-\bigtriangledown\left(V\right)-{\partial\overline{A}\over \partial t}\] and \[\overline{B}=\bigtriangledown\times\overline{A}\] are chosen to satisfy the Lorentz gauge condition; \[\bigtriangledown\centerdot\overline{A}+\mu_{0}\epsilon_{0}{\partial V\over \partial t}=0.\] Suppose that \(\left\{\overline{A}',V'\right\}\) also satisfy the Lorentz gauge condition, with the additional equations;

Then the corresponding fields \(\left\{\overline{E}',\overline{B}'\right\}\) still satisfy Maxwell's equations for the same charge and current \(\left\{\rho,\overline{J}\right\}\).

Proof. It follows from [1] that \(\left\{\overline{A},\overline{V}\right\}\) satisfy the Equations (1) and (2) from the lemma. Let \(V''=V-V'\), \(\overline{A}''=\overline{A}-\overline{A}'\), then \(\left\{\overline{A}'',V''\right\}\) satisfy the equations;

and the Lorentz gauge condition. Let \(\left\{\overline{E}'',\overline{B}''\right\}\) be the corresponding fields. Then, it is sufficient to show that they satisfy Maxwell's equation in vacuum; For (5), we have, using the definition of \(\overline{E}''\), the Lorentz gauge condition, and condition (3) that; \begin{align*}\bigtriangledown\centerdot\overline{E}''&=\bigtriangledown\centerdot\left(-\bigtriangledown\left(V''\right)-{\partial \overline{A''}\over \partial t}\right)\\ &=-\bigtriangledown^{2}\left(V''\right)-{\partial\left(\bigtriangledown\centerdot\overline{A''}\right)\over \partial t}\\ &=-\mu_{0}\epsilon_{0}{\partial^{2}V''\over \partial t^{2}}-{\partial \left(\bigtriangledown\centerdot\overline{A''}\right)\over \partial t}\\ &=-\mu_{0}\epsilon_{0}{\partial^{2}V''\over \partial t^{2}}+\mu_{0}\epsilon_{0}{\partial^{2}V''\over \partial t^{2}}=0.\end{align*} For (6), we have, using vector analysis, see [9]; \[\bigtriangledown\centerdot\overline{B}''=\bigtriangledown\centerdot\left(\bigtriangledown \times\overline{A''}\right)=0.\] For (7), we have, using the definition of \(\overline{E}''\), \(\overline{B}''\) and vector analysis, see [9], that; \[\bigtriangledown\times \overline{E}''=\bigtriangledown\times\left(-\bigtriangledown\left(V''\right)-{\partial \overline{A''}\over \partial t}\right) =-(\bigtriangledown \times\left(\bigtriangledown (V'')\right)-{\partial \left(\bigtriangledown\times\overline{A}''\right)\over\partial t} =-{\partial \overline{B}''\over \partial t}.\] For (8), we have, using the definition of \(\overline{B}''\) and condition (4) that; \[\bigtriangledown\times \overline{B}''=\bigtriangledown\times\left(\bigtriangledown\times\overline{A}''\right).\] A simple calculation shows that; \[\bigtriangledown\times\left(\bigtriangledown\times\overline{A}''\right)=(d_{1},d_{2},d_{3})=(c_{1},c_{2},c_{3}),\] where; \begin{align*} d_{1}&={\partial^{2}a_{2}\over \partial x \partial y}+{\partial^{2}a_{3}\over \partial x \partial z}+\left(-{\partial^{2}a_{1}\over \partial y^{2}}-{\partial^{2}a_{1}\over \partial z^{2}}\right),\\ d_{2}&={\partial^{2}a_{1}\over \partial x \partial y}+{\partial^{2}a_{3}\over \partial y \partial z}+\left(-{\partial^{2}a_{2}\over \partial x^{2}}-{\partial^{2}a_{2}\over \partial z^{2}}\right),\\ d_{3}&={\partial^{2}a_{1}\over \partial x \partial z}+{\partial^{2}a_{2}\over \partial y \partial z}+\left(-{\partial^{2}a_{3}\over \partial x^{2}}-{\partial^{2}a_{3}\over \partial y^{2}}\right),\\ c_{1}&={\partial^{2}a_{2}\over \partial x \partial y}+{\partial^{2}a_{3}\over \partial x \partial z}+\left({\partial^{2}a_{1}\over \partial x^{2}}-\mu_{0}\epsilon_{0}{\partial^{2}a_{1}\over \partial t^{2} }\right),\\ c_{2}&={\partial^{2}a_{1}\over \partial x \partial y}+{\partial^{2}a_{3}\over \partial y \partial z}+\left({\partial^{2}a_{2}\over \partial y^{2}}-\mu_{0}\epsilon_{0}{\partial^{2}a_{2}\over \partial t^{2} }\right),\\ c_{3}&={\partial^{2}a_{1}\over \partial x \partial z}+{\partial^{2}a_{2}\over \partial y \partial z}+\left({\partial^{2}a_{3}\over \partial z^{2}}-\mu_{0}\epsilon_{0}{\partial^{2}a_{3}\over \partial t^{2}}\right),\end{align*} whereas, using the definition of \(\overline{E}''\); \[\mu_{0}\epsilon_{0}{\partial \overline{E}''\over \partial t} =\mu_{0}\epsilon_{0}{\partial \left(-\bigtriangledown \left(V''\right)-{\partial \overline{A}'' \over \partial t}\right)\over \partial t} =-\mu_{0}\epsilon_{0}\bigtriangledown \left({\partial V''\over\partial t}\right)-\mu_{0}\epsilon_{0}{\partial^{2}\overline{A}''\over \partial t^{2}}.\] It is, therefore, sufficient to prove that; \[-\mu_{0}\epsilon_{0}\bigtriangledown\left({\partial V''\over \partial t}\right)=\left(e_{1},e_{2},e_{3}\right),\] where; \begin{align*} e_{1}&={\partial^{2}a_{2}\over \partial x \partial y}+{\partial^{2}a_{3}\over \partial x \partial z}+{\partial^{2}a_{1}\over \partial x^{2}},\\ e_{2}&={\partial^{2}a_{1}\over \partial x \partial y}+{\partial^{2}a_{3}\over \partial y \partial z}+{\partial^{2}a_{2}\over \partial y^{2}},\\ e_{3}&={\partial^{2}a_{1}\over \partial x \partial z}+{\partial^{2}a_{2}\over \partial y \partial z}+{\partial^{2}a_{3}\over \partial z^{2}}.\end{align*} This holds, using the Lorentz gauge condition, as; \(-\mu_{0}\epsilon_{0}\bigtriangledown\left({\partial V''\over \partial t}\right) =-\mu_{0}\epsilon_{0}\bigtriangledown \left(-{\bigtriangledown\centerdot \overline{A}''\over \mu_{0}\epsilon_{0} }\right) =\bigtriangledown \left(\bigtriangledown\centerdot \overline{A}''\right) =\bigtriangledown \left({\partial a_{1}\over \partial x}+{\partial a_{2}\over \partial y}+{\partial a_{3}\over \partial z}\right) =(e_{1},e_{2},e_{3}).\)

Lemma 2. Let the potentials \(\{\overline{A}',V'\}\) be defined be given as retarded potentials; \[V'\left(\overline{r},t\right)={1\over 4\pi\epsilon_{0}}\int {\rho\left(\overline{r}',t_{r}\right)\over \mathfrak{r}}d\tau',\] \[\overline{A}'\left(\overline{r},t\right)={\mu_{0}\over 4\pi}\int {\overline{J}\left(\overline{r}',t_{r}\right)\over \mathfrak{r}}d\tau',\] then, assuming \(\left\{\rho, \overline{J}\right\}\) satisfy the continuity equation and \(\overline{J}\) vanishes at infinity, these potentials satisfy the Lorentz gauge condition and the equations in the hypotheses (1) and (2) from Lemma 1. In particular, the corresponding fields \(\left\{\overline{E}',\overline{B}'\right\}\), given by Jefimenko's Equations; \[\overline{E}'\left(\overline{r},t\right)={1\over 4\pi\epsilon_{0}}\int \left[{\rho\left(\overline{r'},t_{r}\right)\over \mathfrak{r}^{2}}\hat{\mathfrak{\overline{r}}}+{\dot{\rho}\left(\overline{r'},t_{r}\right)\over c\mathfrak{r}}\hat{\mathfrak{\overline{r}}}-{\dot{\overline{J}}\left(\overline{r'},t_{r}\right)\over c^{2}\mathfrak{r}}\right]d\tau',\] \[\overline{B}'\left(\overline{r},t\right)={\mu_{0}\over 4\pi}\int \left[{\overline{J}\left(\overline{r'},t_{r}\right)\over \mathfrak{r}^{2}}+{\dot{\overline{J}}\left(\overline{r'},t_{r}\right)\over c\mathfrak{r}}\right]\times \hat{\mathfrak{\overline{r}}} d\tau',\] satisfy Maxwell's equations.

Proof. The first part of the claim, under the stated hypotheses, is proved in [1], p424, see also footnote 2 of that page and Exercise 10.8, with the solutions given in [10]. For the final part of the claim, one can assume the existence of a solution \(\left\{\rho,\overline{J},\overline{E},\overline{B}\right\}\) to Maxwell's equations, then construct potentials \(\left\{V,\overline{A}\right\}\) abstractly, satisfying the Lorentz gauge condition, as is done in [1]. Applying the result of Lemma 1, we obtain the result. Alternatively, one can verify Maxwell's equations directly, for \(\left\{\overline{E}',\overline{B}'\right\}\), using the method of Lemma 1, just replacing the conditions (3) and (4) on \(\left\{V',\overline{A}'\right\}\), in the proof, with their non-homogeneous versions. The fact that the fields \(\left\{\overline{E}',\overline{B}'\right\}\) are given by Jefimenko's equations is proved in [1], p427-428.

Remark 1. There is an alternative strategy to construct explicit solutions for \(\left(\overline{A},V\right)\), given the corresponding fields \(\left\{\overline{E},\overline{B}\right\}\), satisfying Maxwell's equations, and the potentials satisfying the Lorentz gauge condition. Namely, one can use the explicit formulas, suitably rescaled, for solutions to the homogeneous and inhomogeneous wave equations given in [11], (p73 formula (22) and p82 formula (44)) respectively, with the initial conditions given by \(\left\{V_{0},V_{0,t},\overline{A}_{0},\overline{A}_{0,t}\right\}\) and the driving terms given by \(\left\{-{\rho\over\epsilon_{0}},-\mu_{0}\overline{J}\right\}\). In the context of Lemma 2, one can easily compute the initial conditions, and, in principle derive new formulas for \(\left(\overline{A},V \right)\), replacing the retarded potentials, and for \(\left\{\overline{E},\overline{B}\right\}\), replacing Jefimenko's equations.

3. A No Radiation Condition

We now clarify some results about smoothly decaying solutions to Maxwell's equations in free space.

Definition 1. By Maxwell's equations in free space, we mean;

• (i) \(div\left(\overline{E}\right)=0\),
• (ii) \(div\left(\overline{B}\right)=0\),
• (iii) \(\bigtriangledown\times\overline{E}=-{\partial \overline{B}\over \partial t}\),
• (iv) \(\bigtriangledown\times\overline{B}=\mu_{0}\epsilon_{0}{\partial \overline{E}\over \partial t}\).

We abbreviate the operator \(\bigtriangledown^{2}-\mu_{0}\epsilon_{0}{\partial^{2}\over \partial t^{2}}\) by \(\square^{2}\).

Lemma 3. The smoothly decaying solutions of the wave equation \(\square^{2}\left(\overline{E}\right)=\overline{0}\) are given by; \[\overline{E}\left(\overline{x},t\right)=\int_{\mathcal{R}^{3}}\overline{A}\left(\overline{k}\right)e^{i\left(\overline{k}\centerdot\overline{x}-\omega\left(\overline{k}\right)t\right)}d\overline{k}+\int_{\mathcal{R}^{3}}\overline{B}\left(\overline{k}\right)e^{i\left(\overline{k}\centerdot\overline{x}+\omega\left(\overline{k}\right)t\right)}d\overline{k},\] where \(A,B\subset S\left(\mathcal{R}^{3}\right)\), while the smoothly decaying solutions of Maxwell's equations in free space, are given by;

where \(k=|\overline{k}|\), \(\omega\left(\overline{k}\right)=c |\overline{k}|={k\over \sqrt{\mu_{0}\epsilon_{0}}}\) and \(\{G,H\}\subset \mathcal{S}(M)\) where \(S_{\overline{k}}=\left(S^{2}\left(\overline{k},1\right)\cap P_{\overline{k}}\right)\), \(P_{\overline{k}}=\left\{\overline{n}:\left(\overline{n}-\overline{k}\right)\centerdot\overline{k}=0\right\}\), \(d\overline{S}_{\overline{k}}\left(\overline{n}\right)=\left(\overline{n}-\overline{k}\right)dS_{\overline{k}}\), \(M=\left\{\left(\overline{k},\overline{n}\right)\in\mathcal{R}^{6}:\left(\overline{n}-\overline{k}\right)\centerdot\overline{k}=0,|\overline{n}-\overline{k}|=1\right\}\) and \(\mathcal{S}(M)=\left\{f\in C(M):\int_{S_{\overline{k}}}fd\overline{S}_{\overline{k}}\in\mathcal{S}\left(\mathcal{R}^{3},\mathcal{R}_{\geq 0},\mathcal{C}\right)\right\}\), where \(k=|\overline{k}|\), \(\omega\left(\overline{k}\right)=c|\overline{k}|={k\over \sqrt{\mu_{0}\epsilon_{0}}}\), \(\overline{M}\left(\overline{k},\overline{n}\right)={G\left(\overline{k},\overline{n}\right)\over\omega\left(\overline{k}\right)}\left(\overline{k}\times \overline{n}\right)\), \(\overline{N}\left(\overline{k},\overline{n}\right)={-H\left(\overline{k},\overline{n}\right)\over\omega\left(\overline{k}\right)}\left(\overline{k}\times \overline{n}\right)\).

Finally, if \(\overline{E}\) is a smoothly decaying solutions of the wave equation \(\square^{2}\left(\overline{E}\right)=\overline{0}\), there exists a pair \(\left(\overline{E}_{0},\overline{B}_{0}\right)\) which is a smoothly decaying solution of Maxwell's equations in free space, such that;

\[\bigtriangledown\times\left(\overline{E}-\overline{E}_{0}\right)=\overline{0}.\]

Proof. It is easily checked that the solutions (9) and (10) satisfy (i)-(iv) of Definition 1. Conversely, let \(\{\overline{E},\overline{B}\}\) be smooth solutions of (i)-(iv). We have, using (i), (iii) and (iv), that; \[\left(\bigtriangledown\times\overline{E}\right)=-{\partial \overline{B}\over \partial t}\] and, hence; \[\bigtriangledown\times\left(\bigtriangledown\times\overline{E}\right)=-\left(\bigtriangledown\times {\partial \overline{B}\over \partial t}\right),\] \[-\bigtriangledown^{2}\overline{E}=-{\partial\over \partial t}\left(\bigtriangledown\times\overline{B}\right)=-\mu_{0}\epsilon_{0}{\partial^{2}\overline{E}\over \partial t^{2}},\] and \[\bigtriangledown^{2}\overline{E}={1\over c^{2}}{\partial^{2}\overline{E}\over \partial t^{2}},\,\,\,\,\,\,\,\, \left(c={1\over \sqrt{\mu_{0}\epsilon_{0}}}\right).\] If \(E_{i}\in \mathcal{S}_{\left(\mathcal{R}^{3},\mathcal{R}\right)}(\mathcal{R}^{4})\) solve the wave equation, \(\bigtriangledown^{2}E_{i}={1\over c^{2}}{\partial^{2}E_{i}\over \partial t^{2}}\), where \(\mathcal{S}_{(\mathcal{R}^{3},\mathcal{R})}(\mathcal{R}^{4})=\{f\in C^{\infty}(\mathcal{R}^{4}):f_{t}\in\mathcal{S}(\mathcal{R}^{3})\}\), for \(t\in\mathcal{R}\), then; \[E_{i}\left(\overline{x},t\right)=\left({1\over 2\pi}\right)^{3}\int_{\mathcal{R}^{3}}\hat{E_{i}}\left(\overline{k},t\right)e^{i\overline{k}\centerdot\overline{x}}d\overline{k},\] by the inversion theorem of Fourier analysis, see [11] and [12]. Hence \[\bigtriangledown^{2}E_{i}=-\left({1\over 2\pi}\right)^{3}\int_{\mathcal{R}^{3}}|\overline{k}|^{2}\hat{E_{i}}\left(\overline{k},t\right)e^{i\overline{k}\centerdot\overline{x}}d\overline{k},\] \[{\partial^{2}E_{i}\over \partial t^{2}}=\left({1\over 2\pi}\right)^{3}\int_{\mathcal{R}^{3}}{\partial^{2}\hat{E_{i}}\over \partial t^{2}}\left(\overline{k},t\right)e^{i\overline{k}\centerdot\overline{x}}d\overline{k},\] and \[\bigtriangledown^{2}E_{i}-{1\over c^{2}}{\partial^{2}E_{i}\over \partial t^{2}}=\left({1\over 2\pi}\right)^{3}\int_{\mathcal{R}^{3}}\left(-|\overline{k}|^{2}\hat{E_{i}}-{1\over c^{2}}{\partial^{2}\hat{E_{i}}\over \partial t^{2}}\right)\left(\overline{k},t\right)e^{i\overline{k}\centerdot\overline{x}}d\overline{k}=0.\] So that \(|\overline{k}|^{2}\hat{E_{i}}+{1\over c^{2}}{\partial^{2}\hat{E_{i}}\over \partial t^{2}}=0\) using the inversion formula again. \begin{align*}\hat{E_{i}}\left(\overline{k},t\right)=A_{i}\left(\overline{k}\right)e^{-i|\overline{k}|ct}+B_{i}\left(\overline{k}\right)e^{i\left|\overline{k}\right|ct},\end{align*} as \(\hat{E_{i}}\left(\overline{k},t\right)\in\mathcal{S}_{\left(\mathcal{R}^{3},\mathcal{R}\right)}(\mathcal{R}^{4}).\) \begin{align*} E_{i}\left(\overline{x},t\right)&=\left({1\over 2\pi}\right)^{3}\left(\int_{\mathcal{R}^{3}}\left(A_{i}\left(\overline{k}\right)e^{-i|\overline{k}|ct}\right)e^{i\overline{k}\centerdot\overline{x}}d\overline{k}+\int_{\mathcal{R}^{3}}\left(B_{i}\left(\overline{k}\right)e^{i|\overline{k}|ct}\right)e^{i\overline{k}\centerdot\overline{x}}d\overline{k}\right)\\ &=\left({1\over 2\pi}\right)^{3}\left(\int_{\mathcal{R}^{3}}A_{i}\left(\overline{k}\right)e^{i\left(\overline{k}\centerdot\overline{x}-\omega\left(\overline{k}\right)t\right)}d\overline{k}+\int_{\mathcal{R}^{3}}B_{i}\left(\overline{k}\right)e^{i\left(\overline{k}\centerdot\overline{x}+\omega\left(\overline{k}\right)t\right)}d\overline{k}\right),\end{align*} and \begin{align*}\overline{E}(\overline{x},t)=\int_{\mathcal{R}^{3}}\overline{A}\left(\overline{k}\right)e^{i\left(\overline{k}\centerdot\overline{x}-\omega\left(\overline{k}\right)t\right)}d\overline{k}+\int_{\mathcal{R}^{3}}\overline{B}\left(\overline{k}\right)e^{i\left(\overline{k}\centerdot\overline{x}+\omega\left(\overline{k}\right)t\right)}d\overline{k},\end{align*} where \(A,B\subset S(\mathcal{R}^{3})\), as required for the first part. Using \((i)\), we have that \[\int_{\mathcal{R}^{3}}\left(\overline{A}\left(\overline{k}\right)\centerdot i\overline{k}\right)e^{i(\overline{k}\centerdot\overline{x}-\omega(\overline{k})t)}d\overline{k}+\int_{\mathcal{R}^{3}}\left(\overline{B}\left(\overline{k}\right)\centerdot i\overline{k}\right)e^{i(\overline{k}\centerdot\overline{x}+\omega(\overline{k})t)}d\overline{k}=0.\] At \(t=0\) and \(t=1\), we obtain that \[\int_{\mathcal{R}^{3}}\left(\overline{A}\left(\overline{k}\right)\centerdot i\overline{k}\right)e^{i(\overline{k}\centerdot\overline{x})}d\overline{k}+\int_{\mathcal{R}^{3}}\left(\overline{B}\left(\overline{k}\right)\centerdot i\overline{k}\right)e^{i(\overline{k}\centerdot\overline{x})}d\overline{k}=0,\] which implies \[\int_{\mathcal{R}^{3}}\left(\overline{A}\left(\overline{k}\right)\centerdot i\overline{k}\right)e^{i(\overline{k}\centerdot\overline{x}-ck)}d\overline{k}+\int_{\mathcal{R}^{3}}\left(\overline{B}\left(\overline{k}\right)\centerdot i\overline{k}\right)e^{i(\overline{k}\centerdot\overline{x}+ck)}d\overline{k}=0.\] As this holds for all \(\overline{x}\in\overline{R}^{3}\), using the inversion formula, we obtain \[(\overline{A}(\overline{k})+\overline{B}(\overline{k}))\centerdot \overline{k}=0,\] and \[(\overline{A}(\overline{k})e^{-ck}+\overline{B}(\overline{k})e^{ck})\centerdot \overline{k}=0.\] Assuming \(k\neq 0\), we obtain \(\overline{A}(\overline{k})\centerdot \overline{k}=\overline{B}(\overline{k})\centerdot \overline{k}=0\), so that \(A(\overline{k})=\int_{S_{\overline{k}}}G(\overline{k},\overline{n})d\overline{S}_{\overline{k}}(\overline{n})\),    \(B(\overline{k})=\int_{S_{\overline{k}}}H(\overline{k},\overline{n})d\overline{S}_{\overline{k}}(\overline{n})\) and the first part of the result (9) follows.

Using the same argument, we obtain that

Using (iii); where \(\overline{M}(\overline{k},\overline{n})={G(\overline{k},\overline{n})\over\omega(\overline{k})}(\overline{k}\times \overline{n})\), \(\overline{N}(\overline{k},\overline{n})={-H(\overline{k},\overline{n})\over\omega(\overline{k})}(\overline{k}\times \overline{n})\), and \(\overline{\theta}(\overline{x})=0\), using (11). This proves the second part of the result. For the final part, we can use the first part to write where \(\overline{A},\overline{B}\subset S(\mathcal{R}^{3})\). For \(\overline{k}\neq\overline{0}\), let \[\overline{A}_{2}(\overline{k})={(\overline{A}(\overline{k})\centerdot\overline{k})\over |\overline{k}|^{2}}\overline{k}=A_{3}(\overline{k})\overline{k},\] and \[\overline{A}_{1}(\overline{k})=\overline{A}(\overline{k})-\overline{A}_{2}(\overline{k}).\] Then \(\overline{A}_{1}(\overline{k})\centerdot\overline{k}=0\), and we can write \(\overline{A}_{1}(\overline{k})=A_{2}(\overline{k})(\overline{n}-\overline{k})\) with \(\overline{n}\in S_{\overline{k}}\) and \(A_{2}\in S(\mathcal{R}^{3})\) so that \(\overline{A}_{1}(\overline{k})=\int_{S_{\overline{k}}}G(\overline{k},\overline{n})d\overline{S_{\overline{k}}}(\overline{n})\) with \(G\in S(M)\).

It follows that we can write

\[\int_{\mathcal{R}^{3}}\overline{A}(\overline{k})e^{i(\overline{k}\centerdot\overline{x}-\omega(\overline{k})t)}d\overline{k} =\int_{\overline{k}\in \mathcal{R}^{3}}\int_{S_{\overline{k}}}G(\overline{k},\overline{n})e^{i(\overline{k}\centerdot\overline{x}-\omega(\overline{k})t)}d\overline{S_{\overline{k}}}(\overline{n})d\overline{k}+\int_{\overline{k}\in \mathcal{R}^{3}}A_{3}(\overline{k})\overline{k}e^{i(\overline{k}\centerdot\overline{x}-\omega(\overline{k})t)}\overline{k}d\overline{k}.\] Similarly, repeating the procedure for \(\overline{B}(\overline{k})\), we can write \[\int_{\mathcal{R}^{3}}\overline{B}(\overline{k})e^{i(\overline{k}\centerdot\overline{x}+\omega(\overline{k})t)}d\overline{k} =\int_{\overline{k}\in \mathcal{R}^{3}}\int_{S_{\overline{k}}}H(\overline{k},\overline{n})e^{i(\overline{k}\centerdot\overline{x}+\omega(\overline{k})t)}d\overline{S_{\overline{k}}}(\overline{n})d\overline{k}+\int_{\overline{k}\in \mathcal{R}^{3}}B_{3}(\overline{k})\overline{k}e^{i(\overline{k}\centerdot\overline{x}+\omega(\overline{k})t)}\overline{k}d\overline{k}.\] We set \[\overline{E}_{0}(\overline{x},t)=\int_{\overline{k}\in \mathcal{R}^{3}}\int_{S_{\overline{k}}}G(\overline{k},\overline{n})e^{i(\overline{k}\centerdot\overline{x}-\omega(\overline{k})t)}d\overline{S_{\overline{k}}}(\overline{n})d\overline{k}+\int_{\overline{k}\in \mathcal{R}^{3}}\int_{S_{\overline{k}}}H(\overline{k},\overline{n})e^{i(\overline{k}\centerdot\overline{x}+\omega(\overline{k})t)}d\overline{S_{\overline{k}}}(\overline{n})d\overline{k}.\] We need to check that if \(\overline{E}\) is real, then so is \(\overline{E}_{0}\). If \(\overline{E}={\overline{E}}^{*}\), then, using (13) and equating coefficients, we have \(\overline{A}(\overline{k})^{*}=\overline{B}(-\overline{k})\) and \(\overline{B}(\overline{k})^{*}=\overline{A}(-\overline{k})\). It follows that \[\overline{A}_{2}(\overline{k})^{*}={(\overline{A}(\overline{k})^{*}\centerdot\overline{k})\over |\overline{k}|^{2}}\overline{k} ={(\overline{B}(-\overline{k})\centerdot\overline{k})\over |\overline{k}|^{2}}\overline{k} =-{(\overline{B}(-\overline{k})\centerdot -\overline{k})\over |-\overline{k}|^{2}}\overline{k} =\overline{B}_{2}(-\overline{k}),\] and, similarly \(\overline{B}_{2}(\overline{k})^{*}=\overline{A}_{2}(-\overline{k})\), so that \(\overline{E}-\overline{E}_{0}\) is real, and, therefore, \(\overline{E}_{0}\) is real.

Using the second part, we can find \(\overline{B}_{0}\) such that \((\overline{E}_{0},\overline{B}_{0})\) is a smoothly decaying solution of Maxwell's equations in free space. Finally, we compute

\begin{align*} \bigtriangledown\times (\overline{E}-\overline{E}_{0})& =\bigtriangledown\times \left(\int_{\overline{k}\in \mathcal{R}^{3}}A_{3}(\overline{k})e^{i(\overline{k}\centerdot\overline{x}-\omega(\overline{k})t)}\overline{k}d\overline{k}+\int_{\overline{k}\in \mathcal{R}^{3}}B_{3}(\overline{k})e^{i(\overline{k}\centerdot\overline{x}+\omega(\overline{k})t)}\overline{k}d\overline{k}\right)\\ &=\int_{\overline{k}\in \mathcal{R}^{3}}A_{3}(\overline{k})e^{i(\overline{k}\centerdot\overline{x}-\omega(\overline{k})t)}(-i\overline{k}\times\overline{k})d\overline{k} +\int_{\overline{k}\in \mathcal{R}^{3}}B_{3}(\overline{k})e^{i(\overline{k}\centerdot\overline{x}+\omega(\overline{k})t)}(-i\overline{k}\times\overline{k})d\overline{k}=\overline{0},\end{align*} as required.

Lemma 4. Let \((\rho,\overline{J},\overline{E},\overline{B})\) satisfy Maxwell's equations, then \[\square^{2}(\overline{E})={\bigtriangledown(\rho)\over\epsilon_{0}}+\mu_{0}{\partial\overline{J}\over \partial t},\] and \[\square^{2}(\overline{B})=-\mu_{0}(\bigtriangledown\times\overline{J}).\]

Proof. As is done in [1], we can choose potentials \((V,\overline{A})\) such that \[\overline{E}=-\bigtriangledown (V)-{\partial \overline{A}\over \partial t},\] \[\overline{B}=\bigtriangledown\times \overline{A}\] and \((V,\overline{A})\) satisfy the Lorentz gauge condition. It follows, see Lemma 1, that \((V,\overline{A})\) satisfy the equations \[\square^{2}(V)=-{\rho\over\epsilon_{0}},\] and \[\square^{2}(\overline{A})=-\mu_{0}\overline{J}.\] It is an easy exercise in vector calculus to show that \(\square^{2}\) commutes with the gradient operator \(\bigtriangledown\), the curl operator \(\bigtriangledown\times\) and partial differentiation \({\partial\over \partial t}\). Therefore, we compute \[\square^{2}(\overline{E})=\square^{2}\left(-\bigtriangledown (V)-{\partial \overline{A}\over \partial t}\right) =-\bigtriangledown(\square^{2}(V))-{\partial (\square^{2}(\overline{A}))\over \partial t} =-\bigtriangledown\left(-{\rho\over\epsilon_{0}}\right)-{\partial (-\mu_{0}\overline{J}) \over \partial t} =\bigtriangledown\left({\rho\over\epsilon_{0}}\right)+\mu_{0}{\partial\overline{J} \over \partial t},\] and \[\square^{2}(\overline{B})=\square^{2}(\bigtriangledown\times \overline{A}) =\bigtriangledown\times \square^{2}(\overline{A}) =\bigtriangledown\times (-\mu_{0}\overline{J}) =-\mu_{0}(\bigtriangledown\times \overline{J}).\]

Lemma 5. Let \(\rho\) satisfy the wave equation \(\square^{2}(\rho)=0\), with the initial conditions \({\partial \rho\over \partial t}|_{t=0}\in S(\overline{R}^{3})\) and \(\rho|_{t=0}\in S(\overline{R}^{3})\), then there exists \(\overline{J}\) such that \((\rho,\overline{J})\) satisfies the continuity equation, \(\overline{J}\) satisfies the wave equation \(\square^{2}(\overline{J})=\overline{0}\) with \(\bigtriangledown\times\overline{J}=\overline{0}\), and \((\overline{E},\overline{B})\) such that \((\rho,\overline{J},\overline{E},\overline{B})\) satisfy Maxwell's equations, with \(\square^{2}(\overline{E})=\overline{0}\) and \(\overline{B}=\overline{0}\). In particular, the Poynting vector \(\overline{E}\times \overline{B}=\overline{0}\).

Proof. Define \(\overline{J}\) by \(\overline{J}(\overline{x},t)={-1\over \epsilon_{0}\mu_{0}}\int_{0}^{t}\bigtriangledown(\rho)ds+\overline{h}(\overline{x})\). Then, by the fundamental theorem of calculus, we have \({\partial \overline{J}\over\partial t}=-{\bigtriangledown(\rho)\over \epsilon_{0}\mu_{0}}\). In particular,

Applying the divergence operator, differentiating under the integral sign, using the wave equation for \(\rho\), we have \begin{align*} div(\overline{J})&={-1\over \epsilon_{0}\mu_{0}}\int_{0}^{t}div(\bigtriangledown(\rho))ds+div(\overline{h})\\ &={-1\over \epsilon_{0}\mu_{0}}\int_{0}^{t}\bigtriangledown^{2}(\rho)ds+div(\overline{h})\\ &=-\int_{0}^{t}{\partial^{2}\rho\over\partial s^{2}}ds+div(\overline{h})\\ &=-{\partial \rho\over \partial t}+{\partial \rho\over \partial t}|_{t=0}+div(\overline{h}).\end{align*} Choose \(\overline{h}\) so that \(div(\overline{h})=-{\partial \rho\over \partial t}|_{t=0}\), then \(div(\overline{J})=-{\partial \rho\over \partial t}\), so that \((\rho,\overline{J})\) satisfy the continuity equation. By the initial conditions, that \(\rho_{0}\) and \({\partial \rho\over \partial t}|_{t=0}\) belongs to the Schwartz class, the general solution of the homogeneous wave equation with initial conditions, see [11], and an appropriate choice of \(\overline{h}\), we have that \(\overline{J}\) vanishes at infinity. It follows, applying the result of Lemma 2, that we can find a pair \((\overline{E},\overline{B})\), such that \((\rho,\overline{J},\overline{E},\overline{B})\) satisfy Maxwell's equations, and, by the result of Lemma 4 and the condition (14), we have that \(\square^{2}(\overline{E})=\overline{0}\), so that \(\overline{E}\) also satisfies the wave equation. Moreover, by the explicit formulas in Jefimenko's equations, one can check that \(\overline{E}\) is also smoothly decaying. Then, using the result of Lemma 3, we can find \((\overline{E}_{0},\overline{B}_{0})\), which are smoothly decaying solutions of Maxwell's equations in free space and with \(\bigtriangledown\times(\overline{E}-\overline{E}_{0})=\overline{0}\), (13). Clearly, \((\rho,\overline{J},\overline{E}-\overline{E}_{0},\overline{B}-\overline{B}_{0})\) still satisfy Maxwell's equations, and, by (13) and Maxwell's equations, we have \[-{\partial (\overline{B}-\overline{B}_{0})\over \partial t} =\bigtriangledown\times(\overline{E}-\overline{E}_{0})=\overline{0}\] that is the field \(\overline{B}-\overline{B}_{0}\) is magnetostatic. Set \(\mu_{0}\overline{J}_{0}=\bigtriangledown\times (\overline{B}-\overline{B}_{0})\), then \((0,\overline{J}_{0},\overline{0},\overline{B}-\overline{B}_{0})\) satisfy Maxwell's equations, so that, subtracting solutions, \((\rho,\overline{J}-\overline{J}_{0},\overline{E}-\overline{E}_{0},\overline{0})\) also satisfies Maxwell's equations. We must have then that \((\rho,\overline{J}-\overline{J}_{0})\) satisfies the continuity equation. As \((\overline{E}_{0},\overline{B}_{0})\) were solutions to Maxwell's equations in free space, we have \(\square^{2}(\overline{E}_{0})=\overline{0}\), so that, as \(\square^{2}(\overline{E})=\overline{0}\), we must have \(\square^{2}(\overline{E}-\overline{E}_{0})=\overline{0}\) as well. By Lemma 4, we have By elementary vector calculus, (15), the continuity equation, the fact that \(\square^{2}(\overline{E}-\overline{E}_{0})=\overline{0}\), and the first result of Lemma 4, we have \begin{align*} \bigtriangledown^{2}(\overline{J}-\overline{J}_{0})&=\bigtriangledown(div(\overline{J}-\overline{J}_{0}))-\bigtriangledown\times (\bigtriangledown\times (\overline{J}-\overline{J}_{0})) =\bigtriangledown(div(\overline{J}-\overline{J}_{0}))\\ &=-\bigtriangledown\left({\partial \rho\over \partial t}\right) =-{\partial\left(-\epsilon_{0}\mu_{0}{\partial(\overline{J}-\overline{J}_{0})\over \partial t}\right)\over \partial t} ={1\over c^{2}}{\partial^{2}(\overline{J}-\overline{J}_{0})\over \partial t^{2}},\end{align*} so that \(\square^{2}(\overline{J}-\overline{J}_{0})=\overline{0}\). This proves the main claim. The fact that the Poynting vector is zero follows trivially from the fact that the magnetic field vanishes.

We now strengthen this result.

Lemma 6. Let \((\rho,\overline{J},\overline{E},\overline{B})\) satisfy the conclusions of Lemma 5, then in any inertial frame \(S'\) moving with velocity vector \(\overline{v}\) relative to \(S\), if \((\rho',\overline{J}')\) are the transformed charge distribution and current, there exists a pair \((\overline{E}',\overline{B}')\) such that \((\rho',\overline{J}',\overline{E}',\overline{B}')\) satisfy Maxwell's equations in \(S'\), and \(\square'^{2}(\overline{E}')=\overline{0}\) with \(\overline{B}'=\overline{0}\) in \(S'\). In particular, the Poynting vector \(\overline{E}'\times\overline{B}'=\overline{0}\). Moreover, the pair \((\rho',\overline{J}')\) still satisfy the wave equations \(\square'^{2}(\rho')=0\) and \(\square'^{2}(\overline{J}')=\overline{0}\), with \(\bigtriangledown\times \overline{J}'=\overline{0}\).

Proof. Let \((\rho',\overline{J}',\overline{E}'',\overline{B}'')\) be the transformed quantities in \(S'\), corresponding to \((\rho,\overline{J},\overline{E},\overline{B})\) in \(S\). The transformation rule for the electric field, see [13], is given by \[\overline{E}''=\overline{E}_{||}+\gamma(\overline{E}_{\perp}+\overline{v}\times\overline{B}),\] where \(||\) and \(\perp\) denote the parallel and perpendicular components respectively. Note that \(\overline{E}_{||}\) is defined in the equation by \((\overline{E}\centerdot\overline{v}){\overline{v}\over |\overline{v}|^{2}}\), and \(\overline{E}_{\perp}\) by \(\overline{E}-\overline{E}_{||}\). As \(\overline{v}\times\overline{B}=\overline{0}\), from the assumption that \(\overline{B}=\overline{0}\) in \(S\), we have that an observer in \(S'\) sees the electric field \[\overline{E}''=\overline{E}_{||}+\gamma\overline{E}_{\perp}.\] Let \(\square'^{2}\) be the d'Alembertian operators in \(S'\), then using the Lorentz invariance of the d'Alembertian operator, the obvious fact that it commutes with parallel and perpendicular components, the above transformation rule, and the fact that \(\square^{2}(\overline{E})=\overline{0}\), we have \[\square'^{2}(\overline{E}'')=\square'^{2}(\overline{E}_{||}+\gamma\overline{E}_{\perp}) =\square^{2}(\overline{E}_{||}+\gamma\overline{E}_{\perp}) =(\square^{2}(\overline{E}))_{||}+\gamma(\square^{2}\overline{E})_{\perp}=\overline{0}.\] Similarly the transformation rule for the magnetic field, see [13], is given by \[\overline{B}''=\overline{B}_{||}+\gamma\left(\overline{B}_{\perp}-{\overline{v}\times\overline{E}\over c^{2}}\right),\] which, using the fact that \(\overline{B}=\overline{0}\) in \(S\) again, becomes \[\overline{B}''=-{\gamma\over c^{2}}(\overline{v}\times\overline{E}).\] Using, a similar argument to the above, this time using the fact that the d'Alembertian commutes with taking a cross product with \(\overline{v}\), we have \[\square'^{2}(\overline{B}'')=\square'^{2}\left(-{\gamma\over c^{2}}(\overline{v}\times\overline{E})\right) =\square^{2}\left(-{\gamma\over c^{2}}(\overline{v}\times\overline{E})\right) =-{\gamma\over c^{2}}(\overline{v}\times\square^{2}(\overline{E}))=\overline{0}.\] As in Lemma 5, and using the last part of the result of Lemma 3, with the fact that \(\square'^{2}(\overline{E}'')=\overline{0}\), we can find a pair \((\overline{E}_{0}'',\overline{B}_{0}'')\) which are smoothly decaying solutions of Maxwell's equation in free space, and with

Then clearly, we still have that \[\square'^{2}(\overline{E}''-\overline{E}_{0}'')=\square'^{2}(\overline{B}''-\overline{B}_{0}'')=\overline{0},\] and, moreover, by Maxwell's equations in \(S'\) and (16) \[\bigtriangledown'\times (\overline{E}''-\overline{E}_{0}'') =-{\partial\over \partial t'}(\overline{B}''-\overline{B}_{0}'')=\overline{0}.\] so that \(\overline{B}''-\overline{B}_{0}''\) is magnetostatic. However, we then have \[\square'^{2}(\overline{B}''-\overline{B}_{0}'')=\bigtriangledown'^{2}(\overline{B}''-\overline{B}_{0}'')=\overline{0},\] so that \(\overline{B}''-\overline{B}_{0}''\) satisfies Laplace's equation and is harmonic. Using the fact that \(\overline{B}''-\overline{B}_{0}''\) is bounded, we can use Liouville's theorem to conclude that \(\overline{B}''-\overline{B}_{0}''\) is constant, and using the fact that \(\overline{B}''-\overline{B}_{0}''\) vanishes at infinity, that in fact \(\overline{B}''-\overline{B}_{0}''=\overline{0}\). Setting \(\overline{E}'=\overline{E}''-\overline{E}_{0}''\) and \(\overline{B}'=\overline{B}''-\overline{B}_{0}''\) then gives the first result. The result about the Poynting vector is clear. Finally, we have the transformation rules for current and charge, see [13], given by \[\rho'=\gamma\left(\rho-{vJ_{||}\over c^{2}}\right),\] and \[\overline{J}'=\gamma\left(\overline{J}_{||}-\rho\overline{v}\right)+\overline{J}_{\perp},\] where \(\overline{v}=v\hat{\overline{v}}\) and \(\overline{J}_{||}=J_{||}\hat{\overline{J}_{||}}\). We then compute, using the usual commutation rules, the transformation rules just given, and the fact that \(\square^{2}(\rho)=0\) and \(\square^{2}(\overline{J})=\overline{0}\) in \(S\), that \[\square'^{2}(\rho')=\square'^{2}\left(\gamma\left(\rho-{vJ_{||}\over c^{2}}\right)\right) =\square^{2}\left(\gamma\left(\rho-{vJ_{||}\over c^{2}}\right)\right) =\gamma\left(\square^{2}(\rho)-{v\square^{2}(J_{||})\over c^{2}}\right)=0,\] and \[\square'^{2}(\overline{J}')=\square'^{2}(\gamma(\overline{J}_{||}-\rho\overline{v})+\overline{J}_{\perp}) =\square^{2}(\gamma(\overline{J}_{||}-\rho\overline{v})+\overline{J}_{\perp}) =(\gamma(\square^{2}(\overline{J})_{||}-\square^{2}(\rho)\overline{v})+\square^{2}(\overline{J})_{\perp})=\overline{0}.\] Finally, the fact \(\bigtriangledown'\times \overline{J}'=\overline{0}\) follows from the result \(\square'^{2}(\overline{B}')=\overline{0}\) and the second result in Lemma 4.

We now prove a kind of converse to this result. We first require the following definition:

Definition 2. Let \((\rho,\overline{J})\) be a charge distribution and current, satisfying the continuity equation in the rest frame \(S\). Then, we say that \((\rho,\overline{J})\) is non-radiating if in any inertial frame \(S'\), with velocity vector \(\overline{v}\), for the transformed current and charge \((\rho',\overline{J}')\), there exist electric and magnetic fields \((\overline{E}',\overline{B}')\) in \(S'\) such that \((\rho',\overline{J}',\overline{E}',\overline{B}')\) satisfy Maxwell's equations in \(S'\) and with \(\overline{B}'=\overline{0}\).

Lemma 7. Let \((\rho,\overline{J})\), as in Definition 2, be non-radiating, then \((\rho,\overline{J})\) satisfy the wave equations \(\square^{2}(\rho)=0\) and \(\square^{2}(\overline{J})=\overline{0}\).

Proof. By the definition of non-radiating, there exist fields \((\overline{E},\overline{B})\) in the rest frame \(S\) such that \((\rho,\overline{J},\overline{E},\overline{B})\) satisfy Maxwell's equations and \(\overline{B}=\overline{0}\). We then have that \(\square^{2}(\overline{B})=\overline{0}\) and, by the second result in Lemma 4, that \(\bigtriangledown\times \overline{J}=\overline{0}\). By the same argument, and using the definition of non-radiating, we must have that \(\bigtriangledown'\times \overline{J}'=\overline{0}\) for the transformed current and charge \((\rho',\overline{J}')\) in any inertial frame \(S'\) with velocity vector \(\overline{v}\). We now compute \(\bigtriangledown'\times\overline{J}'\). We have, as above, the transformation rule for \(\overline{J}'\) given by \[\overline{J}'=\gamma(\overline{J}_{||}-\rho\overline{v})+\overline{J}_{\perp},\] so that, using elementary vector calculus; \[\bigtriangledown'\times\overline{J}'=\gamma(\bigtriangledown'\times \overline{J}_{||})+\gamma(\overline{v}\times \bigtriangledown'(\rho))+\bigtriangledown'\times \overline{J}_{\perp}.\] We also have, see [13], the transformation rule for \(\bigtriangledown'\); \[\bigtriangledown'=\gamma\left(\bigtriangledown_{||}+{\overline{v}\over c^{2}}{\partial\over\partial t}\right)+\bigtriangledown_{\perp}.\] Taking \(\overline{v}=v\hat{x}\), we have \(\bigtriangledown_{||}=\left({\partial\over \partial x},0,0\right)\), \(\bigtriangledown_{\perp}=\left(0,{\partial\over \partial y},{\partial\over \partial z}\right)\), \({\overline{v}\over c^{2}}{\partial\over\partial t}=\left({v\over c^{2}}{\partial\over\partial t},0,0\right)\), \(\overline{J}_{||}=(J_{1},0,0)\) and \(\overline{J}_{\perp}=(0,J_{2},J_{3})\) so that \[\bigtriangledown'=\left({\partial\over \partial x'},{\partial\over \partial y'},{\partial\over \partial z'}\right) =\gamma\left({\partial\over \partial x},0,0\right)+\gamma\left({v\over c^{2}}{\partial\over\partial t},0,0\right)+\left(0,{\partial\over \partial y},{\partial\over \partial z}\right) =\left(\gamma{\partial\over \partial x}+{\gamma v\over c^{2}}{\partial\over\partial t},{\partial\over \partial y},{\partial\over \partial z}\right),\] while \begin{align*} &\gamma(\bigtriangledown'\times \overline{J}_{||})=\gamma\left(0,{\partial J_{1}\over \partial z'},-{\partial J_{1}\over \partial y'}\right) =\left(0,\gamma{\partial J_{1}\over \partial z},-\gamma{\partial J_{1}\over \partial y}\right),\\ &\bigtriangledown'\times \overline{J}_{\perp}=\left({\partial J_{3}\over\partial y'}-{\partial J_{2}\over\partial z'},-{\partial J_{3}\over\partial x'},{\partial J_{2}\over\partial x'}\right) =\left({\partial J_{3}\over\partial y}-{\partial J_{2}\over\partial z},-\gamma{\partial J_{3}\over\partial x}-{\gamma v\over c^{2}}{\partial J_{3}\over\partial t},\gamma{\partial J_{2}\over\partial x}+{\gamma v\over c^{2}}{\partial J_{2}\over\partial t}\right),\end{align*} and \[\bigtriangledown'(\rho)=\left({\partial\rho\over \partial x'},{\partial\rho\over \partial y'},{\partial\rho\over \partial z'}\right) =\left(\gamma{\partial\rho\over \partial x}+{\gamma v\over c^{2}}{\partial\rho\over \partial t},{\partial\rho\over \partial y},{\partial\rho\over \partial z}\right)\\ \gamma(\overline{v}\times\bigtriangledown'(\rho))=\left(0,-\gamma v{\partial\rho\over \partial z},\gamma v{\partial\rho\over \partial y}\right).\] Combining these results, it follows that \begin{align*} \bigtriangledown'\times\overline{J}'&=\left(0,\gamma{\partial J_{1}\over \partial z},-\gamma{\partial J_{1}\over \partial y}\right)+\left(0,-\gamma v{\partial\rho\over \partial z},\gamma v{\partial\rho\over \partial y}\right) +\left({\partial J_{3}\over\partial y}-{\partial J_{2}\over\partial z},-\gamma{\partial J_{3}\over\partial x}-{\gamma v\over c^{2}}{\partial J_{3}\over\partial t},\gamma{\partial J_{2}\over\partial x}+{\gamma v\over c^{2}}{\partial J_{2}\over\partial t}\right)\\ &=\left({\partial J_{3}\over\partial y}-{\partial J_{2}\over\partial z},\gamma{\partial J_{1}\over \partial z}-\gamma{\partial J_{3}\over\partial x}-{\gamma v\over c^{2}}{\partial J_{3}\over\partial t}-\gamma v{\partial\rho\over \partial z},-\gamma{\partial J_{1}\over \partial y}+\gamma{\partial J_{2}\over\partial x}+{\gamma v\over c^{2}}{\partial J_{2}\over\partial t}+\gamma v{\partial\rho\over \partial y}\right).\end{align*} As we have seen, \(\bigtriangledown\times\overline{J}=\overline{0}\) in \(S\). In coordinates, this implies that \[{\partial J_{3}\over\partial y}-{\partial J_{2}\over\partial z}={\partial J_{1}\over\partial z}-{\partial J_{3}\over\partial x}={\partial J_{2}\over\partial x}-{\partial J_{1}\over\partial y}=0.\] It follows that \[\bigtriangledown'\times\overline{J}'=\left(0,-{\gamma v\over c^{2}}{\partial J_{3}\over\partial t}-\gamma v{\partial\rho\over \partial z},{\gamma v\over c^{2}}{\partial J_{2}\over\partial t}+\gamma v{\partial\rho\over \partial y}\right) =\gamma\left(\overline{v}\times \left(\bigtriangledown(\rho)+{1\over c^{2}}{\partial\overline{J}\over \partial t}\right)\right).\] As \(\overline{v}\) is arbitrary and \(\bigtriangledown'\times\overline{J}'=\overline{0}\), we conclude that

Taking the divergence \(div\) of (17) and using the continuity equation, we obtain that \[div(\bigtriangledown(\rho))+{1\over c^{2}}{\partial div(\overline{J})\over \partial t} =\bigtriangledown^{2}(\rho)-{1\over c^{2}}{\partial^{2}\rho\over \partial t^{2}}=0,\] so that \(\square^{2}(\rho)=0\). We can conclude as in Lemma 5, using (17), the continuity equation and the fact that \(\bigtriangledown\times\overline{J}=\overline{0}\), that \(\square^{2}(\overline{J})=\overline{0}\) as required.

We make a further definition;

Definition 3. Let \((\rho,\overline{J})\) be a charge distribution and current, satisfying the continuity equation in the rest frame \(S\). Then, we say that \((\rho,\overline{J})\) is strongly non-radiating if in any inertial frame \(S'\), with velocity vector \(\overline{v}\), for the transformed current and charge \((\rho',\overline{J}')\), there exist electric and magnetic fields \((\overline{E}',\overline{B}')\) in \(S'\) such that \((\rho',\overline{J}',\overline{E}',\overline{B}')\) satisfy Maxwell's equations in \(S'\) and with \(\overline{E}'=\overline{0}\). We say that \((\rho,\overline{J})\) is mixed non-radiating if in any inertial frame \(S'\), with velocity vector \(\overline{v}\), for the transformed current and charge \((\rho',\overline{J}')\), there exist electric and magnetic fields \((\overline{E}',\overline{B}')\) in \(S'\) such that \((\rho',\overline{J}',\overline{E}',\overline{B}')\) satisfy Maxwell's equations in \(S'\) and with either \(\overline{E}'=\overline{0}\) or \(\overline{B}'=\overline{0}\). We say that \((\rho,\overline{J})\) is Poynting non-radiating if in any inertial frame \(S'\), with velocity vector \(\overline{v}\), for the transformed current and charge \((\rho',\overline{J}')\), there exist electric and magnetic fields \((\overline{E}',\overline{B}')\) in \(S'\) such that \((\rho',\overline{J}',\overline{E}',\overline{B}')\) satisfy Maxwell's equations in \(S'\) and with the Poynting vector \(\overline{E}'\times\overline{B}'=\overline{0}\). We say that \((\rho,\overline{J})\) is surface non-radiating if in any inertial frame \(S'\), with velocity vector \(\overline{v}\), for the transformed current and charge \((\rho',\overline{J}')\), there exist electric and magnetic fields \((\overline{E}',\overline{B}')\) in \(S'\) such that \((\rho',\overline{J}',\overline{E}',\overline{B}')\) satisfy Maxwell's equations in \(S'\) and with \(div(\overline{E}'\times\overline{B}')=0\).

We note the following;

Lemma 8. Let \((\rho,\overline{J})\) be strongly non-radiating, then \((\rho,\overline{J})\) is trivial, that is \(\rho=0\) and \(\overline{J}=\overline{0}\).

Proof. In the rest frame \(S\), we can find a pair \((\overline{E},\overline{B})\) such that \((\rho,\overline{J},\overline{E},\overline{B})\) satisfy Maxwell's equations and with \(\overline{E}=\overline{0}\). By Maxwell's equations, we have that \(div(\overline{E})={\rho\over\epsilon_{0}}=0\), so that

In an inertial frame \(S'\), with velocity vector \(\overline{v}\), we have, by the transformation rules and (18), that; Using the fact that we can find \((\overline{E}',\overline{B}')\) such that \((\rho',\overline{J}',\overline{E}',\overline{B}')\) satisfy Maxwell's equations and with \(\overline{E}'=\overline{0}\), we can conclude again, that \(\rho'=0\). By (19), we then have that \(J_{||}=0\), so that \((\overline{J},\overline{v})=0\), for every velocity vector \(\overline{v}\). This clearly implies that \(\overline{J}=\overline{0}\) as required.

Lemma 9. Let \((\rho,\overline{J})\) be mixed non-radiating, but not non-radiating, then \((\rho,\overline{J})\) is trivial, that is \(\rho=0\) and \(\overline{J}=\overline{0}\).

Proof. Without loss of generality, using the result of Definition 3, we can assume that in the rest frame \(S\), there exists a pair \((\overline{E},\overline{B})\) such that \((\rho,\overline{J},\overline{E},\overline{B})\) satisfy Maxwell's equations and with \(\overline{E}=\overline{0}\), and that there exists an inertial frame \(S'\), with velocity vector \(\overline{v}\), such that, for the transformed charge and current \((\rho',\overline{J}')\), there exists a pair \((\overline{E}',\overline{B}')\) such that \((\rho',\overline{J}',\overline{E}',\overline{B}')\) satisfy Maxwell's equations and with \(\overline{B'}=\overline{0}\). Working in the rest frame \(S\), we have that, by Maxwell's equations, \(div(\overline{E})={\rho\over\epsilon_{0}}\), so that \(\rho=0\). As \(\square^{2}(\overline{E})=\overline{0}\), we have by Lemma 4, that; \[\mu_{0}{\partial\overline{J}\over \partial t}={\bigtriangledown(\rho)\over\epsilon_{0}}+\mu_{0}{\partial\overline{J}\over \partial t}=\overline{0},\] so that \(\overline{J}\) is static. Again by Maxwell's equations, we have that; \[\bigtriangledown\times\overline{E}=-{\partial\overline{B}\over \partial t}\\ \bigtriangledown\times\overline{B}=\mu_{0}\overline{J}+\mu_{0}\epsilon_{0}{\partial\overline{E}\over \partial t},\] so that, as \(\overline{E}=\overline{0}\), \(\overline{B}\) is static, and

Switching to the frame \(S'\), using the fact that \(\overline{B}'=\overline{0}\) and the second result of Lemma 4, we have that \(\bigtriangledown'\times\overline{J}'=\overline{0}\). Repeating the calculation of Lemma 7, and using the fact that \(\square^{2}(\overline{E})=\overline{0}\), we have that; \[\bigtriangledown'\times\overline{J}'=(\bigtriangledown\times\overline{J})_{||}+\gamma((\bigtriangledown\times\overline{J})_{\perp})+\gamma(\overline{v}\times\square^{2}(\overline{E})) =(\bigtriangledown\times\overline{J})_{||}+\gamma((\bigtriangledown\times\overline{J})_{\perp})=\overline{0}.\] It follows that; \[((\bigtriangledown\times\overline{J}),\overline{v}) =(\bigtriangledown\times\overline{J})_{||},\overline{v}) =-\gamma((\bigtriangledown\times\overline{J})_{\perp},\overline{v})=\overline{0},\] so that; \[(\bigtriangledown\times\overline{J})_{||}=-\gamma(\bigtriangledown\times\overline{J})_{\perp}=\overline{0}\] and \[(\bigtriangledown\times\overline{J})=(\bigtriangledown\times\overline{J})_{||}+(\bigtriangledown\times\overline{J})_{\perp}=\overline{0}.\] By the second result of Lemma 4, we obtain that \(\square^{2}(\overline{B})=\overline{0}\), but \(\overline{B}\) is static, so in fact \(\bigtriangledown^{2}(\overline{B})=\overline{0}\). Applying Liouville's theorem, and using the fact that \(\overline{B}\) is bounded and vanishing at infinity, we obtain that \(\overline{B}=\overline{0}\). From (20), we must have that \(\overline{J}=\overline{0}\) as well, proving the claim.

Remark 2. We conjecture that if \((\rho,\overline{J})\) are Poynting or surface non-radiating, but not non-radiating, then \((\rho,\overline{J})\) are trivial. Given these conjectures, if an electromagnetic system fails to satisfy the wave equation outline above, then in some inertial frame, without loss of generality, we would have that \(div(\overline{E}\times\overline{B})>0\) on some open \(U\). By the divergence theorem, this would imply an energy flux through the boundary of \(U\). This imposes strong restrictions on the nature of this flux, as if the total energy \(V\) of the system were to reduce to zero, or even decrease then, we can consider Rutherford's observation, that, in an atomic system, the orbiting electrons would spiral into the nucleus.

4. The Balmer Series

We now consider flows satisfying the wave equation.

Lemma 10. Let \((\rho,\overline{J})\) be a pair, satisfying the continuity equation, with \(\rho\in S(\mathcal{R}^{3},\mathcal{R})\), \(\overline{J}\in S((\mathcal{R}^{3},\mathcal{R}^{3}))\) and the wave equations \(\square^{2}(\rho)=0\) and \(\square^{2}(\overline{J})=\overline{0}\), with the additional equation;

Then if \[\rho(\overline{x},t)=\int_{\mathcal{R}^{3}}f(\overline{k})e^{i(\overline{k}\centerdot\overline{x}-\omega(\overline{k})t)}d\overline{k}+\int_{\mathcal{R}^{3}}g(\overline{k})e^{i(\overline{k}\centerdot\overline{x}+\omega(\overline{k})t)}d\overline{k},\] and \[\overline{J}(\overline{x},t)=\int_{\mathcal{R}^{3}}\overline{F}(\overline{k})e^{i(\overline{k}\centerdot\overline{x}-\omega(\overline{k})t)}d\overline{k}+\int_{\mathcal{R}^{3}}\overline{G}(\overline{k})e^{i(\overline{k}\centerdot\overline{x}+\omega(\overline{k})t)}d\overline{k},\] we have that, for \(\overline{k}\neq\overline{0}\); \[\overline{F}(\overline{k})={cf(\overline{k})\overline{k}\over |\overline{k}|},\;\;\; f(\overline{k})={(\overline{k},F(\overline{k}))\over c|\overline{k}|},\;\;\; \overline{G}(\overline{k})=-{cg(\overline{k})\overline{k}\over |\overline{k}|},\;\;\; g(\overline{k})=-{(\overline{k},G(\overline{k}))\over c|\overline{k}|}.\] If \(\overline{J}\) is tangential, that is for \(\overline{x}\neq\overline{0}\), and \(t\in\mathcal{R}_{\geq 0}\), \((\overline{x},\overline{J}(\overline{x},t))=0\), then the pair \((\rho,\overline{J})\) is trivial, that is \(\rho=0\) and \(\overline{J}=\overline{0}\).

Proof. By the first part of Lemma 3, using the fact that \(\rho\) and \(\overline{J}\) satisfy the wave equation, we can write; \[\rho(\overline{x},t)=\int_{\mathcal{R}^{3}}f(\overline{k})e^{i(\overline{k}\centerdot\overline{x}-\omega(\overline{k})t)}d\overline{k}+\int_{\mathcal{R}^{3}}g(\overline{k})e^{i(\overline{k}\centerdot\overline{x}+\omega(\overline{k})t)}d\overline{k},\] and \[\overline{J}(\overline{x},t)=\int_{\mathcal{R}^{3}}\overline{F}(\overline{k})e^{i(\overline{k}\centerdot\overline{x}-\omega(\overline{k})t)}d\overline{k}+\int_{\mathcal{R}^{3}}\overline{G}(\overline{k})e^{i(\overline{k}\centerdot\overline{x}+\omega(\overline{k})t)}d\overline{k},\] where \(f,g\subset S(\mathcal{R}^{3},\mathcal{R})\), \(F,G\subset S(\mathcal{R}^{3},\mathcal{R})\) and \(\omega(\overline{k})=c|\overline{k}|\).

We have that;

\[\bigtriangledown(\rho)(\overline{x},t)=\int_{\mathcal{R}^{3}}f(\overline{k})i\overline{k}e^{i(\overline{k}\centerdot\overline{x}-\omega(\overline{k})t)}d\overline{k}+\int_{\mathcal{R}^{3}}g(\overline{k})i\overline{k}e^{i(\overline{k}\centerdot\overline{x}+\omega(\overline{k})t)}d\overline{k},\] while; \[{\partial\overline{J}\over \partial t}(\overline{x},t)=\int_{\mathcal{R}^{3}}-i\omega(\overline{k})\overline{F}(\overline{k})e^{i(\overline{k}\centerdot\overline{x}-\omega(\overline{k})t)}d\overline{k}+\int_{\mathcal{R}^{3}}i\omega(\overline{k})\overline{G}(\overline{k})e^{i(\overline{k}\centerdot\overline{x}+\omega(\overline{k})t)}d\overline{k},\] so that, equating coefficients, using the Inversion Theorem, and (21), we have that; We have that; \[{\partial\rho\over \partial t}(\overline{x},t)=\int_{\mathcal{R}^{3}}-i\omega(\overline{k})f(\overline{k})e^{i(\overline{k}\centerdot\overline{x}-\omega(\overline{k})t)}d\overline{k}+\int_{\mathcal{R}^{3}}i\omega(\overline{k})g(\overline{k})e^{i(\overline{k}\centerdot\overline{x}+\omega(\overline{k})t)}d\overline{k},\] and \[div(\overline{J})(\overline{x},t)=\int_{\mathcal{R}^{3}}i(\overline{k},\overline{F}(\overline{k}))e^{i(\overline{k}\centerdot\overline{x}-\omega(\overline{k})t)}d\overline{k}+\int_{\mathcal{R}^{3}}i(\overline{k},\overline{G}(\overline{k}))e^{i(\overline{k}\centerdot\overline{x}+\omega(\overline{k})t)}d\overline{k},\] so that, equating coefficients again, and using the continuity equation \({\partial\rho\over \partial t}+div(\overline{J})=0\), we have; Now suppose that \(\overline{J}\) is tangential. We then have, applying the Fourier transform \(\mathcal{F}\), see [11]; \[\mathcal{F}(x_{1}J_{1}+x_{2}J_{2}+x_{3}J_{3}) =-i\left({\partial\mathcal{F}(J_{1})\over \partial k_{1}}+{\partial\mathcal{F}(J_{2})\over \partial k_{2}}+{\partial\mathcal{F}(J_{3})\over \partial k_{3}}\right) =-i(div(\mathcal{F}(\overline{J})(\overline{k},t)))=0,\] so that \(div(\mathcal{F}(\overline{J})(\overline{k},t))=0\) which implies, equating coefficients, that \(div(\overline{F})(\overline{k})=div(\overline{G})(\overline{k})=0\). Using the formula (22), we have; \[div(\overline{F})(\overline{k})=div\left({cf(\overline{k})\overline{k}\over |\overline{k}|}\right) =c\left(\bigtriangledown(f)(\overline{k}),{\overline{k}\over |\overline{k}|}\right)+cf(\overline{k})div\left({\overline{k}\over |\overline{k}|}\right) =c\left(\bigtriangledown(f)(\overline{k}),{\overline{k}\over |\overline{k}|}\right)+cf(\overline{k}){2\over |\overline{k}|}=0,\] so that; \[(\bigtriangledown(f),\overline{k})=-2f,\;\;\text{ for }\;\;\overline{k}\neq \overline{0}.\] In coordinates, this would imply that; \[{\partial f\over \partial k_{1}}={\partial f\over \partial k_{2}}={\partial f\over \partial k_{3}}={\partial f\over \partial k_{1}}+{\partial f\over \partial k_{2}}+{\partial f\over \partial k_{3}}=-2f,\] so that \(-6f=-2f\) and \(f=0\). Similarly, we conclude that \(g=0\), and, using the equations (22), that \(\overline{F}=\overline{G}=\overline{0}\). This implies that \(\rho=0\) and \(\overline{J}=\overline{0}\) as required.

Lemma 11. We can find a pair \((\rho,\overline{J})\) satisfying the hypotheses of Lemma 10, with the additional requirement that \(\overline{J}|_{S(r_{0})}=\overline{0}\).

Proof. We convert to spherical polar coordinates, \(x=rcos(\phi)sin(\theta)\), \(y=rsin(\phi)sin(\theta)\), \(z=rcos(\theta)\), for \(0\leq\theta\leq\pi\), \(-\pi\leq\phi\leq\pi\), writing the Laplacian;

where \(R_{r}(u)={\partial^{2}u\over\partial r^{2}}+{2\over r}{\partial u\over \partial r}\) and \(A_{\theta,\phi}(u)={1\over\sin(\theta)}{\partial\over\partial\theta}\left(sin(\theta){\partial u\over\partial\theta}\right)+{1\over sin^{2}\theta}{\partial^{2}u\over\partial\phi^{2}}\) are the radial and angular components respectively. The eigenvectors of the operator \(A_{\theta,\phi}\) are the spherical harmonics, defined by; \[Y_{l,m}(\theta,\phi)=(-1)^{m}\left({2l+1\over 4\pi}{(l-m)!\over (l+m)!}\right)^{1\over 2}P_{l,m}(cos(\theta))e^{im\phi}, \;\;\;\;(l\geq 0,0\leq m\leq l),\] and \[\overline{Y_{l,m}}(\theta,\phi)=(-1)^{m}Y_{l,-m},\;\;\;\;\;(l\geq 0,0\leq m\leq l),\] where \(P_{l,m}(x)=(1-x^{2})^{m\over 2}{d^{m}\over dx^{m}}(P_{l}(x))\), \((l\geq 0,0\leq m\leq l)\) and \(P_{l}(x)={1\over 2^{l}l!}{d^{l}\over dx^{l}}((x^{2}-1)^{l})\), \((l\geq 0)\), see the appendix of [14]. We have that \(\{Y_{l,m}:l\geq 0,-l\leq m\leq l\}\) forms a complete orthonormal basis of \(L^{2}(S(1))\), and, moreover \[A_{\theta,\phi}(Y_{l,m})=-l(l+1)Y_{l,m},\;\;\; (l\geq 0,-l\leq m\leq l),\] see the appendix of [14] again. We look for eigenvectors of \(\bigtriangledown^{2}\) of the form \(\psi_{l,m,E}(r,\theta,\phi)=Y_{l,m}(\theta,\phi)\chi_{l,E}(r)\). We have, using (24), that \[\bigtriangledown^{2}(\psi_{l,m,E})=\left(R_{r}+{1\over r^{2}}A_{\theta,\phi}\right)(Y_{l,m}(\theta,\phi)\chi_{l,E}(r)) =Y_{l,m}(\theta,\phi)R_{r}(\chi_{l,E}(r))-{l(l+1)\over r^{2}}Y_{l,m}(\theta,\phi)\chi_{l,E}(r)),\] so that \(\bigtriangledown^{2}(\psi_{l,m,E})=E\psi_{l,m,E}\) iff \(\chi_{l,E}(r)\) satisfies the radial equation; We can solve (26), using the method of Frobenius, see [15], but the solutions are only bounded for \(E< 0\). Explicitly, taking \(E=-k^{2}\), with \(k>0\), and making the change of variables \(s=kr\), the radial equation reduces to the spherical Bessel equation; which, as noted in [14] has a unique bounded solution (up to scalar multiplication) on \((0,\infty)\) defined by; \[j_{l}(s)=\left({\pi\over 2s}\right)^{1\over 2}J_{l+{1\over 2}}(s),\] where \(J_{l+{1\over 2}}(s)\) denotes the ordinary (of the first kind) Bessel function of order \(l+{1\over 2}\). As is shown in [14] again, see also [16], the functions; \[k\left({2\over \pi}\right)^{1\over 2}Y_{l,m}(\theta,\phi)j_{l}(kr),\;\;\;\;(k\in (0,\infty))\] form a complete orthonormal set in \(C(R^{3})\). Moreover, we have the explicit representations; \begin{align*} &J_{1\over 2}(s)=\left({2\over \pi s}\right)^{1\over 2}sin(s),\\ &J_{l+{1\over 2}}(s)=\left({2\over \pi s}\right)^{1\over 2}\left(P_{l}\left({1\over s}\right)sin(s)-Q_{l-1}\left({1\over s}\right)cos(s)\right),\;\;\;\;(l\in\mathcal{Z}_{\geq 1}),\end{align*} where \(\{P_{l},Q_{l}\}\subset \mathcal{Q}[x]\) are polynomials of degree \(l\), with the property that \(P_{l}(-1)=(-1)^{l}P_{l}(1)\) and \(Q_{l}(-1)=(-1)^{l}Q_{l}(1)\), for \(l\geq 0\), see [17].

We set

\[\chi_{l,E}(r)=\tau_{l,k}(r)=k\left({2\over \pi}\right)^{1\over 2}j_{l}(kr),\] and \[\gamma_{l,m,k}(r,\theta,\phi)=Y_{l,m}(\theta,\phi)\tau_{l,k}(r),\] where \(E=-k^{2}\) for \(k>0\). By what has been shown \(\{\gamma_{l,m,k}:k\in (0,\infty),l\geq 0,-l\leq m\leq l\}\) forms a complete orthonormal set, (9), and \(\bigtriangledown^{2}(\gamma_{l,m,k})=-k^{2}\gamma_{l,m,k}\). It follows easily, that we can write a general solution for the charge \(\rho\) and current contributions \(\overline{J}\) in the wave equation using the forms; We say that \(\overline{J}\) satisfies the radial transform condition, if, in the notation of Lemma 10, we have that, for \(\overline{k}\neq\overline{0}\); \begin{align*} &\overline{F}(\overline{k})=\alpha(k)\overline{k},\\ &\overline{G}(\overline{k})=\beta(k)\overline{k},\end{align*} for some \(\{\alpha,\beta\}\subset S(\mathcal{R}_{>0})\). As is easily shown, if \(\overline{J}\) satisfies the radial transform condition, then if we define \(\rho\) according to the second pair of equations in Lemma 10, we automatically have that \(\overline{J}\) satisfies the first pair, and all the assumptions of Lemma 10 are met. By considering the representation of \(\overline{J}\) in Lemma 10, equating coefficients, and applying the inversion theorem, we see that; for \(\overline{k}\neq\overline{0}\). We compute these integrals using the representation of \(\overline{J}\) in (28) and the representation; \[e^{i\overline{k}\centerdot\overline{x}}=4\pi\sum_{l=0}^{\infty}\sum_{m=-l}^{l}i^{l}j_{l}(kx)Y_{l,m}(\hat{\overline{k}})Y_{l,m}(\hat{\overline{x}}) =4\pi\sum_{l=0}^{\infty}\sum_{m=-l}^{l}{i^{l}\over k}\left({\pi\over 2}\right)^{1\over 2}Y_{l,m}(\hat{\overline{k}})\gamma_{l,m,k},\] given in [14], where \(k=|\overline{k}|\). We have, using the property (9), that; \begin{align*} \int_{\overline{R}^{3}}\overline{J}_{0}&e^{-i\overline{k}\centerdot\overline{x}}d\overline{x}=\int_{\overline{R}^{3}}\left(\sum_{l\geq 0}\sum_{-l\leq m\leq l}\int_{k>0}(\overline{U}(l,m,k)\gamma_{l,m,k}+\overline{V}(l,m,k)\gamma_{l,m,k})dk\right)e^{-i\overline{k}\centerdot\overline{x}}d\overline{x}\\ &=4\pi\int_{\overline{R}^{3}}\left(\sum_{l\geq 0}\sum_{-l\leq m\leq l}\int_{k>0}(\overline{U}(l,m,k)\gamma_{l,m,k}+\overline{V}(l,m,k)\gamma_{l,m,k})dk\right) \left(\sum_{l=0}^{\infty}\sum_{m=-l}^{l}{i^{-l}\over k}\left({\pi\over 2}\right)^{1\over 2}Y_{l,m}^{*}(\hat{\overline{k}})\gamma_{l,m,k}^{*}\right)d\overline{x}\\ &=4\pi\sum_{l=0}^{\infty}\sum_{m=-l}^{l}(\overline{U}(l,m,k)+\overline{V}(l,m,k)){i^{-l}\over k}\left({\pi\over 2}\right)^{1\over 2}Y_{l,m}^{*}(\hat{\overline{k}}).\end{align*} A similar calculation shows that; \[\int_{\overline{R}^{3}}\left({\partial\overline{J}\over \partial t}\right)_{0}e^{-i\overline{k}\centerdot\overline{x}}d\overline{x} =4\pi\sum_{l=0}^{\infty}\sum_{m=-l}^{l}(-ick\overline{U}(l,m,k)+ick\overline{V}(l,m,k)){i^{-l}\over k}\left({\pi\over 2}\right)^{1\over 2}Y_{l,m}^{*}(\hat{\overline{k}}).\] It follows from (29) that; We can compute \(\overline{k}\) in spherical harmonics by; \[\overline{k}=\overline{k}^{*}=k(cos(\phi)sin(\theta),sin(\phi)sin(\theta),cos(\theta)) =\sum_{l=0}^{\infty}\sum_{m=-l}^{l}k\overline{W}(l,m)^{*}Y_{l,m}^{*}(\hat{\overline{k}}).\] noting that, by orthonormality of the spherical harmonics; Equating coefficients, the radial transform condition is satisfied setting; where \(\{\alpha,\beta\}\subset S(\mathcal{R}_{>0})\), and \(l\geq 0\), \(-l\leq m\leq l\), \(k>0\).

We now impose the boundary condition, that \(\overline{J}|_{S(r_{0})}=\overline{0}\). We can achieve this by requiring that

\(\gamma_{l,m,k}|_{S(r_{0})}=0\), or equivalently, that \(\tau_{k,l}(r_{0})=0\), or \(j_{l}(kr_{0})=0\). The positive zeros \(S_{l}\) of \(j_{l}\) form a discrete set and we require that \(k\in {S_{l}\over r_{0}}\). Using the asymptotic approximation; \[j_{l}(s)\sim_{s\rightarrow\infty}{sin(s-{l\pi\over 2})\over s}\] given in [14], we have \[k\sim {\pi\over r_{0}}\left(n+{l\over 2}\right), \;\;\;\;n\in\mathcal{Z}_{>0},\] for large values of \(k\). Using (28), we have that \(\overline{J}\) takes the form; where we have that; for some nonstandard infinite \(\eta\) and the coefficients \(\{\overline{U},\overline{V}\}\) are chosen to satisfy (32) at the discrete eigenvalues.

Remark 3. Technically, the calculation (30) requires smoothness of the coefficients \(\{\overline{U},\overline{V}\}\) in the continuous variable \(k\), so that we can invoke the Riemann-Lebsegue lemma, to eliminate the orthogonal terms \(k\neq k'\). When passing to a discrete sum, we lose this property, and, an argument involving equating coefficients is required. We have sketched over this by involving a nonstandard element \(\eta\), but, if the reader is unfamiliar with this circle of ideas, we are essentially using distributions. As this is primarily a Physics paper, we leave the technical details for another occasion.

Lemma 12. Let \(s_{l,k}(r)={\tau_{l,k}(r)\over c_{l,k}}\), where \(c_{l,k}={k^{1\over 2}r_{0}\over \sqrt{2}}J_{l+{3\over 2}}(kr_{0})\), then, for \(k\in {S_{l}\over r_{0}}\), \(l\geq 0\), \(l\) fixed, \(s_{k,l}\) forms a complete orthonormal system in \(C_{0,2}((0,r_{0}))\), consisting of continuous functions on the interval \((0,r_{0})\), vanishing at \(r_{0}\), with respect to the measure \(r^{2}dr\). Moreover, the functions \(\delta_{l,m,k}(r,\theta,\phi)=Y_{l,m}(\theta,\phi)s_{l,k}(r)\), for \(l\geq 0\),\(-l\leq m\leq l\), \(k\in {S_{l}\over r_{0}}\) form a complete orthonormal system in \(C_{0,2}(B(r_{0}))\), consisting of continuous functions on the ball \(B(r_{0})\) of radius \(r_{0}\), vanishing at the boundary \(S(r_{0})\), with respect to the standard measure \(dB\).

Proof. Let \(M_{l,r}={d^{2}\over dr^{2}}+{2\over r}{d\over dr}-{l(l+1)\over r^{2}}\) and \(L_{l,r}=r^{2}M_{l,r}=r^{2}{d^{2}\over dr^{2}}+2r{d\over dr}-l(l+1)\) so that \(L_{l,r}(f)=-(-r^{2}f')'-l(l+1)f\) for \(f\in C^{2}(0,r_{0})\). By Lagrange's identity, see [15], we have that

and, with notation as above, we have \(\tau_{l,k}={k^{1\over 2}\over r^{1\over 2}}J_{l+{1\over 2}}(kr)\). As \(M_{l,r}(\tau_{l,k})=-k^{2}\tau_{l,k}\), we have applying (35), that \begin{align*} (k'^{2}-&k^{2})\int_{0}^{r_{0}}\tau_{l,k}\overline{\tau_{l,k'}}r^{2}dr =(r^{2}(\tau_{l,k}'\tau_{l,k'}-\tau_{l,k}\tau_{l,k'}'))|_{0}^{r_{0}}\\ =&r_{0}^{2}\left(\left({-k^{1\over 2}\over 2r_{0}^{3\over 2}}J_{l+{1\over 2}}(kr_{0})+{k^{3\over 2}\over r_{0}^{1\over 2}}J'_{l+{1\over 2}}(kr_{0})\right){k'^{1\over 2}\over r_{0}^{1\over 2}}J_{l+{1\over 2}}(k'r_{0}) -\left({-k'^{1\over 2}\over 2r_{0}^{3\over 2}}J_{l+{1\over 2}}(k'r_{0})+{k'^{3\over 2}\over r_{0}^{1\over 2}}J'_{l+{1\over 2}}(k'r_{0})\right){k^{1\over 2}\over r_{0}^{1\over 2}}J_{l+{1\over 2}}(kr_{0})\right)\\ =&{r_{0}^{2}\over r_{0}}\left(k^{3\over 2}k'^{1\over 2}J'_{l+{1\over 2}}(kr_{0})J_{l+{1\over 2}}(k'r_{0})-k'^{3\over 2}k^{1\over 2}J'_{l+{1\over 2}}(k'r_{0})J_{l+{1\over 2}}(kr_{0})\right).\end{align*} Clearly, if \(\{k,k'\}\subset {S_{l}\over r_{0}}\) are distinct, this proves that \(\tau_{l,k}\) and \(\tau_{l,k'}\) are orthogonal with respect to the measure \(r^{2}dr\). We then have, using l'Hospital's rule, assuming that \(k\in {S_{l}\over r_{0}}\) and the recurrence relation for Bessel functions, see [14]; \begin{align*} ||\tau_{l,k}||^{2}_{r^{2}dr}&=lim_{k'\rightarrow k}{r_{0}\left(k^{3\over 2}k'^{1\over 2}J'_{l+{1\over 2}}(kr_{0})J_{l+{1\over 2}}(k'r_{0})-k'^{3\over 2}k^{1\over 2}J'_{l+{1\over 2}}(k'r_{0})J_{l+{1\over 2}}(kr_{0})\right)\over (k'+k)(k'-k)}\\ &={r_{0}\over 2}lim_{k'\rightarrow k}{\left(kJ'_{l+{1\over 2}}(kr_{0})J_{l+{1\over 2}}(k'r_{0})-k'J'_{l+{1\over 2}}(k'r_{0})J_{l+{1\over 2}}(kr_{0})\right)\over (k'-k)}\\ &={r_{0}\over 2}\left(kr_{0}J'_{l+{1\over 2}}(kr_{0})J'_{l+{1\over 2}}(kr_{0})-J'_{l+{1\over 2}}(kr_{0})J_{l+{1\over 2}}(kr_{0})-kr_{0}J''_{l+{1\over 2}}(kr_{0})J_{l+{1\over 2}}(kr_{0})\right)\\ &={kr_{0}^{2}\over 2}\left[J'_{l+{1\over 2}}(kr_{0})\right]^{2} ={kr_{0}^{2}\over 2}\left[J_{l+{3\over 2}}(kr_{0})\right]^{2}.\end{align*} It follows immediately, that, for fixed \(l\geq 0\), the \(s_{l,k}\) form an orthonormal system. The proof that the \(s_{l,k}\) form a complete system is sketched in [18]. As \(\{Y_{l,m}:l\geq 0,-l\leq m\leq l\}\) forms an orthogonal system on \(S(1)\), we have that; \begin{align*} \int_{S(r_{0})}\delta_{l,m,k}\overline{\delta_{l',m',k'}}dB& =\int_{S(r_{0})}Y_{l,m}s_{l,k}\overline{Y_{l',m'}s_{l',k'}}dB\\ &=\int_{0}^{r_{0}}\int_{S(1)}Y_{l,m}\overline{Y_{l',m'}}(\theta,\phi)s_{l,k}\overline{s_{l',k'}}(r)r^{2}dS(1)dr\\ &=\delta_{l,m}\delta_{l,k}\end{align*} proving that the \(\delta_{l,m,k}\) form an orthonormal system. Completeness then follows easily from completeness of the \(Y_{l,m}\) and the \(s_{l,k}\).

Lemma 13. For the fundamental electric field solutions \(\overline{E}_{l_{0},k_{0}}^{\alpha,\beta}\), as defined below, the corresponding time averaged energies \(< U_{em,l_{0},k_{0}}^{Q}>\), determined by the conserved quantity \(Q\neq 0\), defined below, are quantised and display the properties of the Balmer series. Moreover, for a general bounded electric field solution \(\overline{E}\), determined by \((\rho,\overline{J})\), satisfying the hypotheses of Lemma 11, the corresponding energy \(U_{em}\) can be computed in terms of the fundamental energies.

Proof. We compute the electric field \(\overline{E}\), assuming the magnetic field \(\overline{B}\) vanishes. By Maxwell's equations; \[{\partial \overline{E}\over \partial t}=-{1\over\epsilon_{0}}\overline{J},\] so that, integrating (33) of Lemma 11, requiring the boundedness condition, using the result of Lemma 12, and the relations, (34) of Lemma 11, we have;

From here, we rely on the fact, proved in [18], that for \(\{l_{1},l_{2}\}\subset\mathcal{Z}_{\geq 0}\) distinct, the Bessel functions \(J_{l_{1}+{1\over 2}}\) and \(J_{l_{2}+{1\over 2}}\) have no common zeros. We define the fundamental solutions \(\overline{E}_{l_{0},k_{0}}^{\alpha,\beta}\), \(l_{0}\geq 0\), \(k_{0}\in S_{l_{0}}\) by requiring that \(\alpha\) and \(\beta\) are both supported at a single point \(k_{0}\in S_{l_{0}}\) of the discrete union \(\bigcup_{l\geq 0}S_{l}\), so that; and both \(\overline{U}(l_{0},m,k_{0})\) and \(\overline{V}(l_{0},m,k_{0})\) are defined by (32), in Lemma 11. By Poynting's Theorem, see [1], using the facts \((*)\), (32) of Lemma 11, and the coefficient relations in Lemmas 10 and 12, the total energy stored in the electric field \(\overline{E}_{l_{0},k_{0}}^{\alpha,\beta}\), restricted to \(B(r_{0})\), is given by; where \(\beta_{l_{0}}=\sum_{-l_{0}\leq m\leq l_{0}}|\overline{W}(l_{0},m)|^{2}\).

Now let \(Q_{t}^{\alpha,\beta}=\int_{B(r_{0})}\rho_{t}^{\alpha,\beta}d\overline{x}\). Note that \(Q_{t}^{\alpha,\beta}\) is conserved, as, by the continuity equation, the divergence theorem, and the vanishing of \(\overline{J}_{t}^{\alpha,\beta}\) on \(S(r_{0})\);

\[{dQ_{t}^{\alpha,\beta}\over dt}=\int_{B(r_{0})}{\partial \rho_{t}^{\alpha,\beta}\over \partial t}d\overline{x} =\int_{B(r_{0})}-div(\overline{J}_{t}^{\alpha,\beta})d\overline{x} =\int_{S(r_{0})}-\overline{J}_{t}^{\alpha,\beta}\centerdot d\overline{S}_{r_{0}}=0.\] Using the relations \(f(\overline{k})={\alpha(k)k\over c}\) and \(g(\overline{k})=-{\beta(k)k\over c}\) from Lemma 10 and the radial condition, we have, using the integral representation in Lemma 10, that for a fundamental solution; \begin{align*} &\rho_{l_{0},k_{0}}^{\alpha,\beta}={1\over \sqrt{\eta}}\int_{\overline{k}\in S(k_{0})}{\alpha(k_{0})k_{0}\over c}e^{i(\overline{k}\centerdot\overline{x}-k_{0}t)}-{\beta(k_{0})k_{0}\over c}e^{i(\overline{k}\centerdot\overline{x}+k_{0}t)}dS_{k_{0}},\\ &{\partial \rho_{l_{0},k_{0}}^{\alpha,\beta}\over \partial t}={1\over \sqrt{\eta}}\int_{\overline{k}\in S(k_{0})}{-i\alpha(k_{0})k_{0}^{2}\over c}e^{i(\overline{k}\centerdot\overline{x}-k_{0}t)}-{i\beta(k_{0})k_{0}^{2}\over c}e^{i(\overline{k}\centerdot\overline{x}+k_{0}t)}dS_{k_{0}},\end{align*} We can then use this representation, the result in [19], together with the conservation property, to obtain; \begin{align*} Q&=\int_{B(r_{0})}\rho_{l_{0},k_{0}}^{\alpha,\beta}d\overline{x}\\ &={1\over \sqrt{\eta}}\int_{\overline{k}\in S(k_{0})}{\alpha(k_{0})k_{0}\over c}\left({2\pi r_{0}\over k_{0}}\right)^{3\over 2}J_{3\over 2}(k_{0}r_{0})-{\beta(k_{0})k_{0}\over c}\left({2\pi r_{0}\over k_{0}}\right)^{3\over 2}J_{3\over 2}(k_{0}r_{0})dS_{k_{0}}\\ &={(\alpha(k_{0})-\beta(k_{0}))\over \sqrt{\eta}}{(2\pi r_{0})^{3\over 2}\over ck_{0}^{3\over 2}}k_{0}4\pi k_{0}^{2}J_{3\over 2}(k_{0}r_{0})\\ &={(\alpha(k_{0})-\beta(k_{0}))\over \sqrt{\eta}}{(2\pi r_{0})^{3\over 2}\over c}4\pi k_{0}^{3\over 2}J_{3\over 2}(k_{0}r_{0}),\\ 0&=\int_{B(r_{0})}{\partial \rho_{l_{0},k_{0}}^{\alpha,\beta}\over \partial t}d\overline{x}\\ &={-i(\alpha(k_{0})+\beta(k_{0}))\over \sqrt{\eta}}{(2\pi r_{0})^{3\over 2}\over c}4\pi k_{0}^{5\over 2}J_{3\over 2}(k_{0}r_{0}),\end{align*} so that, rearranging \(\alpha(k_{0})=-\beta(k_{0})\),    \(f(k_{0})=g(k_{0})\), and, for \(l_{0}\neq 1\); \begin{align*} &\alpha(k_{0})={Q\sqrt{\eta}c\over 8\pi(2\pi r_{0})^{3\over 2}k_{0}^{3\over 2}J_{3\over 2}(k_{0}r_{0})},\\ &f(k_{0})={Q\sqrt{\eta}\over 8\pi(2\pi r_{0})^{3\over 2}k_{0}^{1\over 2}J_{3\over 2}(k_{0}r_{0})}.\end{align*} Now we can substitute in (38), to obtain; \[U_{em,l_{0},k_{0}}^{Q}={r_{0}^{2}k_{0}J^{2}_{l_{0}+{3\over 2}}(k_{0}r_{0})\beta_{l_{0}}\over 32\epsilon_{0}\pi^{3}\eta}\left({2Q^{2}\eta(1+cos(2ck_{0}t))\over 64\pi^{2}(2\pi r_{0})^{3}k_{0}J^{2}_{3\over 2}(k_{0}r_{0})}\right) ={Q^{2}\beta_{l_{0}}(1+cos(2ck_{0}t))\over 1024\pi^{8}\epsilon_{0}r_{0}}{J^{2}_{l_{0}+{3\over 2}}(k_{0}r_{0})\over J^{2}_{3\over 2}(k_{0}r_{0})},\] and, taking the average over a cycle; \[< U_{em,l_{0},k_{0}}^{Q}>={Q^{2}\beta_{l_{0}}\over 1024\pi^{8}\epsilon_{0}r_{0}}{J^{2}_{l_{0}+{3\over 2}}(k_{0}r_{0})\over J^{2}_{3\over 2}(k_{0}r_{0})}.\] By the explicit representation of Bessel functions in Lemma 11, we have that; \[{J^{2}_{l_{0}+{3\over 2}}(k_{0}r_{0})\over J^{2}_{3\over 2}(k_{0}r_{0})}={(P_{l_{0}+1}({1\over k_{0}r_{0}})sin(k_{0}r_{0})-Q_{l_{0}}({1\over k_{0}r_{0}})cos(k_{0}r_{0}))^{2}\over (P_{1}({1\over k_{0}r_{0}})sin(k_{0}r_{0})-Q_{0}cos(k_{0}r_{0}))^{2}},\] and, using the asymptotic description of \({S_{l_{0}}\over r_{0}}\) for large values of \(k_{0}\), in Lemma 11, we have that; \begin{align*} cos(k_{0}r_{0})\simeq\begin{cases} (-1)^{n_{0}}(-1)^{l_{0}\over 2},&\text{if}\ l_{0}\ \text{even},\\ 0,&\text{if}\ l_{0}\ \text{odd}.\end{cases}\end{align*} \begin{align*}sin(k_{0}r_{0})\simeq\begin{cases} 0,&\text{if}\ l_{0}\ \text{even},\\ (-1)^{n_{0}}(-1)^{l_{0}-1\over 2},&\text{if}\ l_{0}\ \text{odd}.\end{cases}\end{align*} So that, for \(l_{0}\) even; \[{J^{2}_{l_{0}+{3\over 2}}(k_{0}r_{0})\over J^{2}_{3\over 2}(k_{0}r_{0})}\simeq {Q_{l_{0}}^{2}\left({1\over k_{0}r_{0}}\right)\over Q_{0}^{2}}={Q_{l_{0},0}^{2}\over Q_{0}^{2}}+{2Q_{l_{0},0}Q_{l_{0},2}\over Q_{0}^{2}k_{0}^{2}r_{0}^{2}}+O\left({1\over k_{0}^{4}r_{0}^{4}}\right),\] and, for \(l_{0}\) odd; \[{J^{2}_{l_{0}+{3\over 2}}(k_{0}r_{0})\over J^{2}_{3\over 2}(k_{0}r_{0})}\simeq {P_{l_{0}+1}^{2}\left({1\over k_{0}r_{0}}\right)\over P_{1}^{2}\left({1\over k_{0}r_{0}}\right)}={P_{l_{0}+1,1}^{2}\over P_{1,1}^{2}}+{2P_{l_{0}+1,1}P_{l_{0}+1,3}\over P_{1,1}^{2}k_{0}^{2}r_{0}^{2}}+O\left({1\over k_{0}^{4}r_{0}^{4}}\right).\] It follows that, for \(l_{0}\) even, and large \(\{k_{0},k_{1}\}\); \[< U_{em,l_{0},k_{0}}^{Q}>-< U_{em,l_{0},k_{1}}^{Q}>\simeq {2Q^{2}Q_{l_{0},0}Q_{l_{0},2}\beta_{l_{0}}\over 1024\pi^{8}\epsilon_{0}Q_{0}^{2}r_{0}^{3}(k_{0}^{2}-k_{1}^{2})} \simeq {Q^{2}Q_{l_{0},0}Q_{l_{0},2}\beta_{l_{0}}\over 128\pi^{10}\epsilon_{0}Q_{0}^{2}r_{0}(m_{0}^{2}-m_{1}^{2})},\] and for \(l_{0}\) odd, and large \(\{k_{0},k_{1}\}\); \[< U_{em,l_{0},k_{0}}^{Q}>-< U_{em,l_{0},k_{1}}^{Q}>\simeq {2Q^{2}P_{l_{0}+1,1}P_{l_{0}+1,3}\beta_{l_{0}}\over 1024\pi^{8}\epsilon_{0}P_{1,1}^{2}r_{0}^{3}(k_{0}^{2}-k_{1}^{2})} \simeq {Q^{2}P_{l_{0}+1,1}P_{l_{0}+1,3}\beta_{l_{0}}\over 128\pi^{10}\epsilon_{0}P_{1,1}^{2}r_{0}(m_{0}^{2}-m_{1}^{2})},\] where \(k_{0}\simeq {\pi\over r_{0}}\left(n_{0}+{l_{0}\over 2}\right)\) and \(m_{0}=2n_{0}+l_{0}\), \(k_{1}\simeq {\pi\over r_{0}}\left(n_{1}+{l_{0}\over 2}\right)\) and \(m_{1}=2n_{1}+l_{0}\), with \(\{m_{1},m_{2}\}\subset{\mathcal{Z}}_{\geq 1}\), which agrees closely with the Balmer series as claimed. Observe that for distinct \((l_{0},k_{0})\) and \((l_{1},k_{1})\), using the representation (37) and the orthogonality of the series \(\delta_{l,m,k}\), that for \(\{\alpha_{0},\beta_{0}\}\) and \(\{\alpha_{1},\beta_{1}\}\) supported on \(k_{0}\in S_{l_{0}}\) and \(k_{1}\in S_{l_{1}}\) respectively, that; For any \(\overline{E}\) represented as in (36) we have that; where \(\alpha_{k}\) and \(\beta_{k}\) are the restrictions of \(\alpha\) and \(\beta\) to \(k\in S_{l}\). It follows from (39) and (40), that; \[U_{em}=\int_{B(r_{0})}\left|\overline{E}\right|^{2}d\overline{x} =\sum_{l\geq 0}\sum_{k\in S_{l}}\int_{B(r_{0})}\left|\overline{E}_{k,l}^{\alpha_{k},\beta_{k}}\right|^{2}d\overline{x} =\sum_{l\geq 0}\sum_{k\in S_{l}}U_{em,k,l}^{Q_{k,l}^{\alpha_{k},\beta_{k}}},\] where \(Q_{k,l}^{\alpha_{k},\beta_{k}}=\int_{B(r_{0})}\rho_{k,l}^{\alpha_{k},\beta_{k}}d\overline{x}\).

Remark 4. Note that the condition \(Q=0\) places no restriction on the values of \(\alpha\) and \(\beta\), when \(l_{0}=1\). As the values of \(\alpha\) and \(\beta\) can vary continuously, this suggests that the quantisation phenomenon, observed in the previous lemma, occurs only when the atom is ionised, in which case \(Q\neq 0\) and we observe the behaviour of the Balmer series. This point of view is supported by the results of the Franck-Hertz experiment.

Conflicts of Interest: 

"The author declares no conflict of interest".

1. Introduction

Kamps [1] introduced the concept of generalized order statistics \((gos)\) as follows: Let us note \(n\in N\), \(k\geq1\), and \(\tilde{m}=(m_1,m_2,\ldots,m_{n-1})\in\mathfrak{R}^{n-1},\) \(1\leq r\leq {n-1}\), such that

\begin{equation*} \gamma_r=k+n-r+\sum_{j=r}^{n-1} m_{j}>0~~\  for \  ~~1\leq r \leq n-1. \end{equation*} The random variables \(X(1,n,\tilde{m},k),X(2,n,\tilde{m},k), \ldots,X(n,n,\tilde{m},k)\) are said to be \(gos\) from a continuous population with cumulative distribution function (\(cdf\)) \(F(x)\) and probability distribution function (\(pdf\)) \(f(x)\) if their joint \(pdf\) is of the form defined on the cone \(F^{-1}(0+)< x_1\leq x_2 \leq \ldots \leq x_n< F^{-1}(1)\) of \( \mathfrak{R}^{n}\), where \(\bar{F}(x)=1-F(x)\).

The model of \(gos\) contains special cases such as ordinary order statistics \((\gamma_i=n-i+1; i=1,2,\dots,n ~i.e. ~ m_{1}=m_{2}=\dots=m_{n-1}=0, \,k=1)\), \(k\)-th record values \((\gamma_{i}=k ~i.e.,~ m_{1}=m_{2}=\dots=m_{n-1}=-1,\, k\in N)\), sequential order statistics \((\gamma_{i}=(n-i+1)\alpha_{i};\, \alpha_{1},\alpha_{2},\dots,\alpha_{n}> 0)\), order statistics with non-integer sample size \((\gamma_{i}=\alpha-i+1; \,\alpha > 0)\), Pfeifer's record values \((\gamma_{i}=\beta_{i};\, \beta_{1},\beta_{2},\dots,\beta_{n} > 0)\) and progressive type II censored order statistics \((\gamma_r=n-r+1+\sum_{i=r}^{l} m_{i}, 1\leq r\leq l \leq n,\, m_{i}\in N,\, k=m_n+1)\), see [1,2,3].

Here we shall obtain the results for \(\gamma_i\ne\gamma_j\) and then deduce the results for \(\gamma_i=\gamma_j\) (\(m_{1}=m_{2}=\dots=m_{n-1}=m\ne-1\)).

Therefore, we will consider two cases:

Case I: \(\gamma_i=\gamma_j\,(m_{1}=m_{2}=\dots=m_{n-1}=m\ne-1)\) [1].

Case II: \(\gamma_i\ne\gamma_j,\,i\ne j\, i,j=1,2,\dots{,{n-1}}\) [2].

Case I: The \(pdf\) of \(r-\)th \(gos\) \(X(r,n,m,k),\) is given by

and the joint \(pdf\) of \(X(r,n,m,k)\) and \(X(s,n,m,k)\), \(1\leq r< s\leq n\), is given by where \begin{equation*} C_{r-1}=\prod_{i=1}^r\gamma_i ,\, \,~~~\gamma_i=k+(n-i) (m+1), \end{equation*} \begin{equation*} h_m(x)=\begin{cases} -\displaystyle\frac{1}{m+1}\,(1-x)^{m+1}&,~~~ m\neq -1\\ -\ln(1-x)&,~~~m=-1 \end{cases} \end{equation*} and \begin{equation*} g_m(x)~=~h_m(x)-h_m(0)=\int_0^x(1-t)^mdt,~~x\in [0,1). \end{equation*} Case II: The \(pdf\) of \(r-\)th \(gos\) \(X(r,n,\tilde{m},k),\) is given by with the joint \(pdf\) of \(X(r,n,\tilde{m},k)\) and \(X(s,n,\tilde{m},k)\), \(1\leq r< s\leq n\), where \begin{align*} &C_{r-1}=\prod_{i=1}^r\gamma_i,\\ &\gamma_r=k+n-r+\sum_{j=r}^{n-1} m_{j},\\ &a_{i}(r)~=~\prod_{j=1}^r\frac{1}{(\gamma_j-\gamma_i)},\ \ \ \ j\ne i,\ \ \ \ \gamma_j\ne\gamma_i, \ \ \ \ 1\leq i\leq r\leq n,\\ &a_{i}^{(r)}(s)~=~\prod_{j=r+1}^n\frac{1}{(\gamma_j-\gamma_i)},\ \ \ \ j\ne i,\ \ \ \ \gamma_j\ne\gamma_i, \ \ \ \ r+1\leq i\leq s\leq n. \end{align*} For \(m_{1}=m_{2}=\dots=m_{n-1}=m\ne -1\), it can be shown that [3]: and In this paper we are interested in a situation when a random variable \(X\) follows the Pareto-Rayleigh(P-R) distribution with \(pdf\) and with \(df\) In view of (8) and (9), Pareto-Rayleigh distribution can be seen as a member of Transformed-Transformer family (or T-X family) of distributions proposed by Alzaatreh et al., [4]. This distribution is recognized as a good model for fitting various lifetime data, see Jebeli and Deiri [5]. This is also confirmed in [6] were a comparative study on the performance of Pareto-Rayleigh distribution against biased Lomax distribution was conducted. Further, for more details on Pareto-Rayleigh distribution one can see [7,8,9].

Exact moments expressions of gos for different distributions have been obtained by literature. Some examples are exponentiated Log-logistic distribution, Burr type XII distribution, linear exponential distribution, Erlang-truncated exponential distribution, Burr distribution, power function distribution, type II exponentiated Log-logistic distribution, extended exponential distribution, generalized Pareto distribution, q-Weibull distribution; see, respectively, Athar and Nayabuddin [10], Khan et al., [11], Ahmad [12], Khan et al., [13], Khan and Khan [3], Kumar and Khan [14], Kumar [15], Kumar and Dey [16], Malik and Kumar [17], Singh et al., [18] and Kumar et al., [19].

In this paper, we have derived explicit expression for single and product moments of Pareto-Rayleigh distribution based on gos.

2. Relations for Product Moments

In this section, we derive the exact expressions for product moments of generalized order statistics in the following theorems. Before coming to the main result, the following lemma is proved.

Lemma 1. For the Pareto-Rayleigh distribution with \(cdf\) \((1.9)\) next relations holds

where \begin{equation*}\label{2.2} \Phi_{j,l}(a,b)=\int_{0}^{\infty}\int_{0}^{y}\frac{x^{j+1}}{\big(1+\frac{x^{2}}{2\sigma^2}\big)^{a\,\alpha+1}}\,\frac{y^{l+1}}{\big(1+\frac{y^{2}}{2\sigma^2}\big)^{\alpha\,b+1}}dx\,dy \end{equation*} and \begin{equation*} _{p}F_{q}[a_{1},\dots,a_{p};b_{1},\dots,b_{q};x]=\sum_{r=0}^{\infty}\Big[\prod_{j=1}^{p}\frac{\Gamma(a_{j}+r)}{\Gamma(a_{j})}\Big]\Big[\prod_{j=1}^{q}\frac{\Gamma(b_{j})}{\Gamma(b_{j}+r)}\Big]\frac{x^r}{r!}, \end{equation*} for \(p=q+1\) and \(\sum_{j=1}^{q}b_{j}-\sum_{j=1}^{p}a_{j}\,>0\).

Proof. We have

Let Substituting \(1-u=\frac{1}{\big(1+\frac{x^{2}}{2\sigma^2}\big)}\) in (14), we get \begin{align*} B(y)&=\frac{(2\sigma^2)^{\big(1+\frac{j}{2}\big)}}{2}\,\int_{0}^{\frac{\frac{y^{2}}{2\sigma^2}}{\big(1+\frac{y^{2}}{2\sigma^2}\big)}}u^{\frac{j}{2}}\,(1-u)^{a\,\alpha-\frac{j}{2}-1}\,du\\ &=\frac{(2\sigma^2)^{\big(1+\frac{j}{2}\big)}}{2}\,B_{\frac{\frac{y^{2}}{2\sigma^2}}{\big(1+\frac{y^{2}}{2\sigma^2}\big)}}\Big(\frac{j}{2}+1,\,a\alpha-\frac{j}{2} \Big). \end{align*} From (13), we have where \( B_{x}(p,q)=\,\int_{0}^{x}u^{p-1}\,(1-u)^{q-1}\,du. \) We know that and Substituting (16) and (17) in (15), we get Setting \(t={\frac{\frac{y^{2}}{2\sigma^2}}{1+\frac{y^{2}}{2\sigma^2}}}\) in \((18)\), we get \begin{align*} \Phi_{j,l}(a,b)=&\frac{(2\sigma^2)^{\left(\frac{j+l}{2}+2\right)}}{2\,(j+2)}\,\int_{0}^{1}\,t^{\frac{j+l}{2}+1}\,(1-t)^{\alpha\,b-\frac{l}{2}-1}\,_2{F}_1\left[ \frac{j}{2}+1,\,1-a\,\alpha+\frac{j}{2},;\,\frac{j}{2}+2;\,t\right]\,dt \\&=\frac{(2\sigma^2)^{\left(\frac{j+l}{2}+2\right)}}{2\,(j+2)}\,B\left(\frac{j+l}{2}+2,\,\alpha\,b-\frac{l}{2}\right) _3{F}_2\left(\frac{j}{2}+1,\,1-a\,\alpha+\frac{j}{2},\,\frac{j+l}{2}+2;\,\frac{j}{2}+2,\,\frac{j}{2}+\alpha\,b+2;1\right). \end{align*}

Lemma 2. Setting \(j=0\) or \(l=0\) in Lemma 1, we obtain

and where \[ \Phi_{j}(a)=\int_{0}^{\infty}\frac{x^{j+1}}{\left(1+\frac{x^{2}}{2\sigma}\right)^{a\,\alpha+1}}\,\ \ dx=\frac{(2\sigma^2)^{\left(1+\frac{j}{2}\right)}}{2}\,B\left(a\alpha-\frac{j}{2}, 1+\frac{j}{2}\right). \]

Proof. Substituting \(j=0\) in (13), we get \begin{align*} \Phi_{0,l}(a,b)&=\int_{0}^{\infty}\,\frac{y^{l+1}}{\left(1+\frac{y^{2}}{2\sigma^2}\right)^{\alpha\,b+1}}\left[\int_{0}^{y}\frac{x}{\left(1+\frac{x^{2}}{2\sigma^2}\right)^{a\,\alpha+1}}dx\right]\,dy\\ &=\frac{\sigma^2}{a\,\alpha}\int_{0}^{\infty}\,\frac{y^{l+1}}{\left(1+\frac{y^{2}}{2\sigma^2}\right)^{\alpha\,b+1}}\left[1-\frac{1}{\left(1+\frac{y^{2}}{2\sigma^2}\right)^{a\,\alpha}}\right]\,dy\\ &=\frac{\sigma^2}{a\,\alpha}\,[\Phi_{j}(b)-\Phi_{l}(a+b)]. \end{align*} Similarly, we get (20) by noting that \begin{equation*} _3{F}_2(a,\,b,\,c;\,c,\,d;1)=_2{F}_1(a,\,b;\,d;1)=\frac{\Gamma (d)\,\,\Gamma(d-a-b)}{\Gamma(d-a)\,\,\Gamma(d-b)}. \end{equation*}

Theorem 1. Generalized product moments for Pareto-Rayleigh distribution are given as

Proof. We have \begin{equation*} \mu^{(j,\,l)}_{r,s,n,\tilde{m},k}=~C_{s-1}\,\int_{0}^{\infty}\int_{0}^{y}x^j\,y^{l} \left[\sum_{i=r+1}^{s}a_{i}^{(r)}(s)\left\{\frac{\bar{F}(y)}{\bar{F}(x)}\right\}^{\gamma_{i}}\right] \left(\sum_{i=1}^{r}a_{i}(r)\left\{{\bar{F}(x)}\right\}^{\gamma_{i}}\right)\frac{f(x)}{\bar{F}(x)}\,\frac{f(y)}{\bar{F}(y)}\,dx\,dy. \end{equation*} which yields (21).

Corollary 2. Product moment for Pareto-Rayleigh distribution, when \(m_1=m_2=\dots=m_{n-1}=m\ne -1\) is given as

Remark 1. Setting \( m_{1}=m_{2}=\dots=m_{n-1}=0\) and \(k=1\) in (22), we get the result as the product moment of order statistics as

Corollary 3. Single moments of the Pareto-Rayleigh distribution are of the form

Proof. Putting \(j=0\) in (21) and using (19), we get \begin{align*} \mu^{(l)}_{r,s,n,\tilde{m},k}= &{C_{s-1}}\left(\frac{\alpha}{\sigma^2}\right) \left[\sum_{t=r+1}^{s}\frac{a_{t}^{(r)}(s)}{(\gamma_{i}-\gamma_{t})}\,\left(\sum_{i=1}^{r}\,a_{i}(r)\,\left\{\Phi_{l}(\gamma_{t})-\Phi_{l}(\gamma_{i})\right\}\right)\right].\\ \mu^{(l)}_{s,n,\tilde{m},k}=&{C_{s-1}}\left(\frac{\alpha}{\sigma^2}\right) \left[\sum_{t=r+1}^{s}\,{a_{t}^{(r)}(s)}\,\Phi_{l}(\gamma_{t})\,\left(\sum_{i=1}^{r}\,\frac{a_{i}(r)}{(\gamma_{i}-\gamma_{t})}\right)\right] \\ &+~{C_{s-1}}\left(\frac{\alpha}{\sigma^2}\right)\left[\sum_{i=1}^{r}\,a_{i}(r)\,\Phi_{l}(\gamma_{i})\,\left(\sum_{t=r+1}^{s}\,\frac{a_{t}^{(r)}(s)}{(\gamma_{i}-\gamma_{t})}\right)\right]. \end{align*} Now using the results found in [20] we obtain \begin{equation*} \sum_{i=1}^{r}\frac{a_{i}(r)}{(\gamma_i-\gamma_j)}~=~\prod_{j=1}^r\frac{1}{(\gamma_i-\gamma_j)},\quad j\ne i,\quad \gamma_j\ne\gamma_i, \quad 1\leq i\leq r\leq n, \end{equation*} and \begin{equation*} \sum_{i=r+1}^{s}\frac{a_{i}^{(r)}(s)}{(\gamma_i-\gamma_j)}~=~\prod_{j=r+1}^s\frac{1}{(\gamma_i-\gamma_j)},\quad j\ne i,\quad \gamma_j\ne\gamma_i, \quad r+1\leq i\leq s\leq n. \end{equation*} Hence, \begin{equation*} \mu^{(l)}_{s,n,\tilde{m},k}=~{C_{s-1}}\Big(\frac{\alpha}{\sigma^2}\Big)\Bigg[\sum_{t=r+1}^{s}\,{a_{t}^{(r)}(s)}\,\Phi_{l}(\gamma_{t})\,\Bigg(\prod_{j=1}^r\frac{1}{(\gamma_i-\gamma_j)}\Bigg)\Bigg] \end{equation*} \begin{equation*} \hspace{1.7cm}+~{C_{s-1}}\Big(\frac{\alpha}{\sigma^2}\Big)\Bigg[\sum_{i=1}^{r}\,a_{i}(r)\,\Phi_{l}(\gamma_{i})\,\Bigg(\prod_{j=r+1}^s\frac{1}{(\gamma_i-\gamma_j)}\Bigg)\Bigg], \end{equation*} which yields (24).

Corollary 4. Single moments of gos for Pareto-Rayleigh distribution, when \(m_1=m_2=\dots=m_{n-1}=m\ne -1\), are given as

Proof. Setting \( m_{1}=m_{2}=\dots=m_{n-1}=m\ne -1\) in (24) and using (7) we get the result as the single moment.

Remark 2. Setting \( m_{1}=m_{2}=\dots=m_{n-1}=0\) and \(k=1\) in (25), we get the result as the single moment from order statistics

Remark 3. Setting \( j=0\) and \(l=0\) in \((21)\) we get

and setting \(l=0\) in (24) we obtain Combining (27) and (28), we get another identity, When \( m_{1}=m_{2}=\dots=m_{n-1}=m\ne -1\), (29) reduces to another identity which is obtained in [3].

Remark 4. Setting \(\gamma_r=k+n-r+\sum_{i=r}^{l}m_{j}\), \(1\leq r\leq l\leq n\), \(m_i \in N\), in (21), then the product moments of progressive type II censored order statistics of Pareto-Rayleigh distribution can be obtained.

3. Numerical Computations

Here we have calculated means and variances for order statistics (Table 1 & 2), and generalized order statistics \((gos)\) (Table 3 & 4). All computations here we obtained using Mathematica. Mathematica like other algebraic manipulation packages allow for arbitrary precisions, so the accuracy of the given values is not an issue. In case of order statistics, the relation \( \sum_{r=1}^{n}\mu^{j}_{r,n,0,1}=n\,\mu^{j}_{1,1,0,1}, \qquad j=1,2, \) is used to evaluate the means and variancess, see [21]. It is observed that when the sample size \(n\) is fixes, increasing the value of \(r\) directly increases the means and variances, whereas, for fixed \(r\), the opposite occurs in the case when the sample size \(n\) increases.

Table 1. Means of order statistics from Pareto-Rayleigh distribution (\(\alpha\)=2, \(\sigma\)=1).

	n
r	1	2	3	4	5	6	7	8
1	1.1107	0.6942	0.5469	0.4653	0.4120	0.3736	0.3443	0.3209
2		1.5272	0.9892	0.7907	0.6786	0.6040	0.5496	0.5078
3			1.7962	1.1877	0.9589	0.8279	0.7398	0.6752
4				1.9991	1.3403	1.0900	0.9453	0.8474
5					2.1638	1.4654	1.1984	1.0433
6						2.3035	1.5722	1.2916
7							2.4253	1.6658
8								2.5339

Table 2. Variances of order statistics from Pareto-Rayleigh distribution (\(\alpha\)=2, \(\sigma\)=1).

	n
r	1	2	3	4	5	6	7	8
1	0.7663	0.1847	0.1011	0.0692	0.0525	0.0422	0.0353	0.0303
2		1.0009	0.2214	0.1176	0.0792	0.0594	0.0475	0.0395
3			1.1735	0.2464	0.1281	0.0852	0.0635	0.0504
4				1.3180	0.2672	0.1366	0.0900	0.0666
5					1.4450	0.2855	0.1441	0.0942
6						1.5599	0.3021	0.1509
7							1.6655	0.3175
8								1.7639

Table 3. Means of gos from Pareto-Rayleigh distribution (\(\alpha\)=2, \(\sigma\)=1, \(m\)=1, \(k\)=2).

	n
r	1	2	3	4	5	6	7	8
1	0.3471	0.2327	0.1868	0.1605	0.1428	0.1300	0.1200	0.1121
2		0.2308	0.1622	0.1329	0.1155	0.1036	0.0947	0.0878
3			0.1325	0.0957	0.0795	0.0697	0.0628	0.0577
4				0.0724	0.0532	0.0447	0.0394	0.0357
5					0.0386	0.0288	0.0243	0.0216
6						0.0203	0.0153	0.0130
7							0.0105	0.0080
8								0.0054

Table 4. Variances of gos from Pareto-Rayleigh distribution (\(\alpha\)=2, \(\sigma\)=1, \(m\)=1, \(k\)=2).

	n
r	1	2	3	4	5	6	7	8
1	0.2128	0.0887	0.0560	0.0409	0.0322	0.0266	0.0226	0.0197
2		0.2086	0.0971	0.0641	0.0481	0.0385	0.0321	0.0275
3			0.1480	0.0733	0.0499	0.0380	0.0308	0.0259
4				0.0914	0.0471	0.0327	0.0253	0.0207
5					0.0527	0.0280	0.0197	0.0154
6						0.0292	0.0158	0.0113
7							0.0158	0.0087
8								0.0084

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

Consider the problem of solving equation

where \(F:\Omega \subset B_1\longrightarrow B_2\) is continuously Fréchet differentiable, \(X, Y\) are Banach spaces and \(\Omega\) is a nonempty convex set.

In this paper we study the local convergence of a family of sixth order iterative methods using assumptions only on the first derivative of \(F.\) Usually the convergence order is obtained using Taylor expansions and conditions on high order derivatives not appearing on the methods [1,2,3,4,5,6,7,8,9,10,11,12,13]. These conditions limit the applicability of the methods.

For example, let \( X=Y=\mathbb{R}, \,D= [-\frac{1}{2}, \frac{3}{2}].\) Define \(f\) on \(D\) by

\[f(s)=\left\{\begin{array}{cc} s^3\log s^2+s^5-s^4& if\,\,s\neq0\\ 0& if\,\, s=0. \end{array}\right. \] Then, we have \(x_*=1,\) and \[f'(s)= 3s^2\log s^2 + 5s^4- 4s^3+ 2s^2 ,\] \[f''(s)= 6x\log s^2 + 20s^3 -12s^2 + 10s,\] \[f'''(s) = 6\log s^2 + 60s^2-24s + 22.\] Obviously \(f'''(s)\) is not bounded on \(D.\) So, the convergence of these methods is not guaranteed by the analysis in these papers.

The family of methods we are interested are:

where \(B_n=b_1I+b_2C(x_n,y_n)+b_3C(y_n,x_n)+b_4C(x_n,y_n)^2,\) \(\gamma=\frac{2}{3},\,\ \ a_1=\frac{5-8a_2}{8},\,\ \ a_3=\frac{a_2}{3}\), \(a_4=\frac{9-8a_2}{24},\,\ \ b_2=\frac{3+8b_1}{8},\,\ \ b_3=\frac{15-8b_1}{14},\,\ \ b_4=\frac{9+4b_1}{12}\) with \(a_2,\ \ b_1\) and \(\gamma\) free.

The efficiency and convergence order was given in [14] when \(X=Y=\mathbb{R}^k.\) The convergence was shown using the seventh derivative. We include error bounds on \(\|x_n-x_*\|\) and uniqueness results not given in [14]. Our technique is so general that it can be used to extend the usage of other methods [1,2,3,4,5,6,7,8,9,10,11,12,13].

The article contains local convergence analysis in Section 2 and the numerical examples in Section 3.

2. Local convergence

We develop some real parameters and functions. Set \(S=[0, \infty).\) Suppose function:

• (i) \( \omega_0(t)-1 \) has a least zero \(R_0\in S-\{0\}\) for some function \(\omega:S\longrightarrow S\) continuous and nondecreasing. Set \(S_0=[0, R_0).\)
• (ii) \( \varphi_{1}(t)-1=0 \) has a least zero \(r_1\in S_0-\{0\}\) for some functions \(\omega:S_0\longrightarrow S, \omega_1:S_0\longrightarrow S\) continuous and nondecreasing with \(\varphi_1:S_0\longrightarrow S\) defined by \[\varphi_1(t)=\frac{\int_0^1\omega((1-\theta)t)d\theta+|1-\gamma|\int_0^1\omega_1(\theta t)d\theta}{1-\omega_0(t)}.\]
• (iii) \(\varphi_2(t)-1\) has a least zero \(r_2\in S_0-\{0\}\) for some function \(\zeta:S_0\longrightarrow S\) with \(\varphi_2:S_0\longrightarrow S\) defined by \[\varphi_{2}(t)=\frac{\int_0^1\omega((1-\theta)t)d\theta+\zeta(t)\int_0^1\varphi_1(\theta t)d\theta}{1-\omega_0(t)},\] where \(\zeta(t)=|a_1-1|+\frac{\omega_1(t)}{1-\omega_0(\varphi_1(t)t)}+\frac{|a_3|\omega_0(\varphi_{1}(t)t)}{1-\omega_0(t)}+|a_4|\left(\frac{\omega_1(t)}{1-\omega_0(\varphi_1(t)t)}\right)^2.\)
• (iv) \(\omega_0(\varphi_1(t)t)-1\) has a least zero \(R_1\in S_0-\{0\}.\) Set \(R=\min\{R_0, R_1\}\) and \(S_1=[0, R).\)
• (v) \(\varphi_3(t)-1\) has a least zero \(r_3\in S_1-\{0\}\) for some function \(\psi:S_1\longrightarrow S\) defined by \begin{align*} \varphi_3(t)=&\left[\frac{\int_0^1\omega((1-\theta)\varphi_2(t)t)}{1-\omega_0(\varphi_2(t)t)}\right.+\left.\frac{(\omega_0(\varphi_{2}(t)t)+\omega_0(\varphi_1(t)t))\int_0^1\omega_1(\theta \varphi_2(t)t)d\theta}{(1-\omega_0(\varphi_2(t)t))(1-\omega_0(\varphi_1(t)t))}\right.\\&+\left.\frac{\psi(t)\int_0^1\omega_1(\theta\varphi_2(t)t)d\theta}{1-\omega_0(\varphi_2(t)t)}\right]\varphi_2(t) \end{align*} where \(\psi(t)=|b_1-1|+|b_2|\frac{\omega_1(\varphi_1(t)t)}{1-\omega_0(t)}+|b_3|\frac{\omega_1(t)}{1-\omega_0(\varphi_1(t)t)}+|b_4|\left(\frac{\omega_1(\varphi_1(t)t)}{1-\omega_0(t)}\right)^2.\)

Define parameter \(r\) by

It shall be shown that \(r\) is a convergence radius for method (2). Set \(S_2=[0,r).\) Notice that for each \(t\in S_2\) the following hold and By \( \bar{T}(x,\delta)\) we denote the closure of the open ball \(T(x,\delta)\) with center \(x\in X\) and of radius \(\delta > 0.\)

Our local convergence analysis uses hypotheses (H) provided that the functions ``\(\omega\)`` are as previously given, and \(x_*\) is a simple zero of \(F.\) Suppose:

• (H1) \(\|F'(x_*)^{-1}(F'(u)-F'(x_*))\|\leq \omega_0(\|u-x_*\|)\) for each \(u\in \Omega.\) Set \(\Omega_0=\Omega\cap T(x_*,R_0)\);
• (H2) \(\|F'(x_*)^{-1}(F'(u)-F'(v))\|\leq \omega(\|u-v\|)\) and \(\|F'(x_*)^{-1}F'(u)\|\leq \omega_1(\|u-x_*\|)\) for each \(u,v\in \Omega_0\);
• (H3) \(\bar{T}(x_*,r)\subset \Omega;\) and
• (H4) There exists \(\beta\geq r\) satisfying \(\int_0^1\omega_0(\theta \beta)d\theta < 1.\) Set \(\Omega_1=\Omega\cap \bar{T}(x_*,\beta).\)

Next, the local convergence analysis follows for method (2) utilizing hypotheses (H).

Theorem 1. Under hypotheses (H) choose starting point \(x_0\in T(x_*,r)-\{x_*\}.\) Then, sequence \(\{x_n\}\) generated by method (2) for any starting point \(x_0\) is well defined in \(T(x_*,r),\) remains in \(T(x_*,r)\) and \(\lim_{n\longrightarrow \infty}x_n=x_*,\) which is the only zero of \(F\) in the set \(\Omega_1\) given in (H4).

Proof. The following assertions shall be shown using induction

and where the radius \(r\) is defined in (3) and the \(\varphi_m\) functions are as previously given. Let \(x\in T(x_*,r)-\{x_*\}.\) Using (3), (4), and (H1), we get so by a Lemma due to Banach [15,16,17,18,19] on invertible operators \(F'(x)\) is invertible and Notice also \(y_0\) exists by the first substep of method (2) from which we can write By (3), (6) (for \( m=1\)), (11) (for \(x=x_0\)), (H2) and (12), we have showing (7) for \(n=0\) and \(y_0\in T(x_*,r).\) Then, we also have that (11) holds for \(x=y_0\) and \(F'(y_0)\) is invertible. Hence, \(z_0\) exists by the second substep of method (2) from which we can also write By (3), (6) (for \(m=2\)), (11) (for \(x=x_0,y_0\)), (13) and (14), we have showing (8) for \(n=0\) and \(z_0\in T(x_*,r),\) where we also used the estimate Similarly, we have that \(x_1\) exists and we can write by the third substep of method (2) Then, by (3), (6)( for \(m=3\)), (11) (for \(x=z_0, y_0\)), (13), (15) and (17), we get showing (9) for \(n=0\) and \(x_1\in T(x_*,r),\) where we also used Exchange \(x_0,y_0, z_0, x_1\) by \(x_n, y_n, z_n, x_{n+1}\) in the preceding calculations to complete the induction for (7)-(9). Then, from the estimation where \(p=\varphi_3(\|x_0-x_*\|)\in [0,1),\) we get \(\lim_{n\longrightarrow\infty}x_n=x_*,\) and \(x_{n+1}\in T(x_*,r).\)

Set \(M=\int_0^1F'(x_*+\theta(q-x_*))d\theta\) for some \(q\in \Omega_1\) with \(F(q)=0.\) Using (H1) and (H4) \[\|F'(x_*)^{-1}(M-F'(x_*))\|\leq \int_0^1\omega_0(\theta\|q-x_*)\|d\theta \leq \int_0^1\omega_0(\theta \beta)d\theta < 1,\] so \(q=x_*\) is implied by the identity \(0=F(q)-F(x_*)=M(q-x_*)\) and the invertability of \(M.\)

Remark 1.

• 1. In view of (H2) and the estimate \begin{eqnarray*} \|F'(x^\ast)^{-1}F'(x)\|&=&\|F'(x^\ast)^{-1}(F'(x)-F'(x^\ast))+I\|\\ &\leq& 1+\|F'(x^\ast)^{-1}(F'(x)-F'(x^\ast))\| \leq 1+\varphi_0(\|x-x^\ast\|) \end{eqnarray*} the second condition in (H3) can be dropped and \(\varphi_1\) can be replaced by \(\varphi_1(t)=1+\varphi_0(t)\) or \(\varphi_1(t)=1+\varphi_0(R_0),\) since \(t\in [0, R_0).\)
• 2. The results obtained here can be used for operators \(F\) satisfying autonomous differential equations [15] of the form \(F'(x)=P(F(x))\) where \(P\) is a continuous operator. Then, since \(F'(x^\ast)=P(F(x^\ast))=P(0),\) we can apply the results without actually knowing \(x^\ast.\) For example, let \(F(x)=e^x-1.\) Then, we can choose: \(P(x)=x+1.\)
• 3. Let \(\varphi_0(t)=L_0t,\) and \(\varphi(t)=Lt.\) In [15,16] we showed that \(r_A=\frac{2}{2L_0+L}\) is the convergence radius of Newton's method:
  \begin{equation} x_{n+1}=x_n-F'(x_n)^{-1}F(x_n)\,\,\,\  for \  each \  \,\,\,n=0,1,2,\cdots \end{equation}

  (21)
  under the conditions (H1) - (H3). It follows from the definition of \(\alpha,\) that the convergence radius \(r\) of the method (2) cannot be larger than the convergence radius \(r_A\) of the second order Newton's method (21). As already noted in [15,16] \(r_A\) is at least as large as the convergence radius given by Rheinboldt [10]
  \begin{equation} r_R=\frac{2}{3L},\end{equation}

  (22)
  where \(L_1\) is the Lipschitz constant on \(D.\) The same value for \(r_R\) was given by Traub [13]. In particular, for \(L_0 < L_1\) we have that \(r_R < r_A\) and \(\frac{r_R}{r_A}\rightarrow \frac{1}{3}\,\,\, as\,\,\, \frac{L_0}{L_1}\rightarrow 0.\) That is the radius of convergence \(r_A\) is at most three times larger than Rheinboldt's.
• 4. We can compute the computational order of convergence (COC) defined by \(\xi= \frac{\ln\left(\frac{d_{n+1}}{d_n}\right)}{\ln\left(\frac{d_n}{d_{n-1}}\right)}, \) where \(d_n=\|x_n-x^\ast\|\) or the approximate computational order of convergence \(\xi_1= \frac{\ln\left(\frac{e_{n+1}}{e_n}\right)}{\ln\left(\frac{e_n}{e_{n-1}}\right)}, \) where \(e_n=\|x_n-x_{n-1}\|.\)

3. Numerical Examples

Example 1. Consider the kinematic system \[F_1'(x)=e^x,\, F_2'(y)=(e-1)y+1,\, F_3'(z)=1\] with \(F_1(0)=F_2(0)=F_3(0)=0.\) Let \(F=(F_1,F_2,F_3).\) Let \({B}_1={B}_2=\mathbb{R}^3, D=\bar{B}(0,1), p=(0, 0, 0)^t.\) Define function \(F\) on \(D\) for \(w=(x,y, z)^t\) by \[ F(w)=(e^x-1, \frac{e-1}{2}y^2+y, z)^t. \] Then, we get \[F'(v)=\left[ \begin{array}{ccc} e^x&0&0\\ 0&(e-1)y+1&0\\ 0&0&1 \end{array}\right], \] so \( \omega_0(t)=(e-1)t, \omega(t)=e^{\frac{1}{e-1}}t, \omega_1(t)=e^{\frac{1}{e-1}}.\) Then, the radii are \[r_{1}=0.154407,\, r_2=0.367385,\, r_3=0.323842.\]

Example 2. Consider \({B}_1={B}_2=C[0,1],\) \(D=\overline{B}(0,1)\) and \(F:D\longrightarrow B_2\) defined by

We have that \[F'(\phi(\xi))(x)=\xi(x)-15\int_0^1x\theta\phi(\theta)^2\xi(\theta)d\theta,\,\,\,\  for \  each \  \,\,\, \xi \in D.\] Then, we get that \(x^* =0,\) so \( \omega_0(t)=7.5t, \omega(t)=15t\) and \(\omega_1(t)=2.\) Then, the radii are \[r_{1}=0.02222,\, r_2=0.091401,\, r_3=0.0656309.\]

Example 3. By the academic example of the introduction, we have \(\omega_0(t)=\omega(t)=96.6629073 t\) and \(\omega_1(t) =2.\) Then, the radii are \[r_{1}=0.00229894,\, r_2=0.0065021,\, r_3=0.0905654.\]

Author Contributions

All authors contributed equally.

Conflicts of Interest

The authors declare no conflict of interest.

1. Introduction

The concept of an intuitionistic fuzzy matrix (IFM) was introduced by Khan et al. [1] and Im et al. [2] to generalize the concept of Thomason's fuzzy matrix [3]. Each element in an IFM is expressed by an ordered pair \(\left\langle \mu_{a_{ij}},\nu_{a_{ij}}\right\rangle\) with \(\mu_{a_{ij}},\nu_{a_{ij}}\in [0,1]\) and \(0\leq \mu_{a_{ij}}+\nu_{a_{ij}}\leq 1\). Since the IFS was proposed, it has received a lot of attention in many fields, such as pattern recognition, medical diagnosis, and so on. But if the sum of the membership degree and the nonmembership degree is greater than 1, the IFM is no longer applicable. Khan and Pal [4] defined some basic operations and relations of IFMs including maxmin, minmax, complement, algebraic sum, algebraic product etc. and proved equality between IFMs. After the introduction of IFM theory, many researchers attempted the important role in IFM theory [5,6,7,8,9,10,11,12,13,14].

Yager [15] introduced the concept of a Pythagorean fuzzy set (PFS) and developed some aggregation operations for PFS. Zhang and Xu [16] studied various binary operations over PFS and also proposed a decision making algorithm based on PFS. Recently, Yager [17] proposed the concept of the q-ROFS, in which MD u and NMD satisfy \(\mu^q+\nu^q\leq 1 (q\geq 1)\). We can see that the IFS and PFS are special cases of q-ROFS. As q-rung increases, the range of processing fuzzy information increases. In recent years, the topic of information aggregation has attracted a lot of attention and is one of the key research issues in the problems of MAGDM. As far as q-ROFS is concerned, different aggregation operators have been introduced and applied, such as q-ROFWA and q-ROFWG operator [18]. After the introduction of q-ROFS theory, many researchers attempted the important role in PFS and q-ROFS theory [19,20,21,22,23,24,25].

Using the theory of PFS ans q-ROFS, Silambarasan and Sriram [26] defined the Pythagorean fuzzy matrix (PFM) theory and its algebraic operations. Each element in an PFM is expressed by an ordered pair \(\left\langle \mu_{a_{ij}},\nu_{a_{ij}}\right\rangle\) with \(\mu_{a_{ij}},\nu_{a_{ij}}\in [0,1]\) and \(0\leq \mu^2_{a_{ij}}+\nu^2_{a_{ij}}\leq 1\). Also,they constructed \(nA\) and \(A^n\) of a Pythagorean fuzzy matrix \(A\) and using these operations. Further, they defined the commutative monoid on Pythagorean fuzzy matrices and proved that the set of all PFMs forms a commutative monoid [27]. After the introduction of PFM theory, many researchers worked in PFM and Fermatean fuzzy matrix theory [28]. Since the PFM was brought up, it has been widely applied in FM operations on q-ROFMs and prove their desirable properties. In Section 5, we define necessity and possibility on q-ROFMs and proved some algebraic properties of these operations. In Section 6, we define a new operation(@) on q-ROFMs and investigated their algebraic properties. We write the conclusion of the paper in the last section.

2. Preliminaries

In this section, some basic concepts related to the intuitionistic fuzzy matrix (IFM) and Pythagorean fuzzy matrix (PFM) have been given.

Definition 1.[1] An intuitionistic fuzzy matrix (IFM) is a pair \(A=\left[\left\langle \mu_{a_{ij}}, \nu_{a_{ij}}\right\rangle\right]\) of a non negative real numbers \(\mu_{a_{ij}}, \nu_{a_{ij}}\in [0,1]\) satisfying \(0\leq \mu_{a_{ij}}+\nu_{a_{ij}}\leq 1\) for all \(i,j.\)

Definition 2.[26] A Pythagorean fuzzy matrix (PFM) is a pair \(A=\left[\left\langle \mu_{a_{ij}}, \nu_{a_{ij}}\right\rangle\right]\) of non negative real numbers \(\mu_{a_{ij}}, \nu_{a_{ij}}\in [0,1]\) satisfying the condition \(0\leq\mu^2_{a_{ij}}+\nu^2_{a_{ij}}\leq 1\), for all \(i,j\). Where \(\mu_{a_{ij}}\in[0,1]\) is called the degree of membership and \(\nu_{a_{ij}}\in[0,1]\) is called the degree of non-membership.

3. q-rung orthopair fuzzy matrices (q-ROFMs)

In this section, we briefly introduce the q-rung orthopair fuzzy matrices and give examples.

Definition 3. A q-rung orthopair fuzzy matrix (q-ROFM) is a pair \(A=\left[\left\langle \mu_{a_{ij}}, \nu_{a_{ij}}\right\rangle\right]\) of non negative real numbers \(\mu_{a_{ij}}, \nu_{a_{ij}}\in [0,1]\) satisfying the condition \(0\leq\mu^q_{a_{ij}}+\nu^q_{a_{ij}}\leq 1 (q\geq 1)\), for all \(i,j\). Where \(\mu_{a_{ij}}\in[0,1]\) is called the degree of membership and \(\nu_{a_{ij}}\in[0,1]\) is called the degree of non-membership.

For understanding the q-ROFM better, we give an instance to illuminate the understandability of the q-ROFM: We can definitely get \(0.9+0.6 > 1\), and, therefore, it does not follow the condition of intuitionistic fuzzy matrices. Also, we can get \((0.9)^2+(0.6)^2 =0.81 + 0.36 = 1.17 > 1\), which does not obey the constraint condition of Pythagorean fuzzy matrices. However, we can get \((0.9)^q+(0.6)^q \leq 1 ~(q\geq 1)\), which is good enough to apply the q-ROFM to control it.

Theorem 1. The q-ROFMs is larger than the set of PFMs and IFMs.

Proof.

Any intuitionistic fuzzy matrix \((\mu_{a_{ij}},\nu_{a_{ij}})\) that is an IFM is also a PFM and a q-ROFM. For any two fuzzy matrices \(A, B\in [0,1]\), we get \(\mu^q_{a_{ij}}\leq \mu^2_{a_{ij}}\leq \mu_{a_{ij}}\) and \(\nu^q_{a_{ij}}\leq \nu^2_{a_{ij}}\leq \nu_{a_{ij}}\). Thus \(\mu_{a_{ij}}+\nu_{a_{ij}}\leq 1 \Rightarrow \mu^2_{a_{ij}}+\nu^2_{a_{ij}}\leq 1 \Rightarrow \mu^q_{a_{ij}}+\nu^q_{a_{ij}}\leq 1.\) Consider a point \((0.9,0.6)\), we see that \((0.9)^q+(0.6)^q\leq 1, (q \geq 1)\) thus this is an q-ROFM. Since \((0.9)^2+(0.6)^2 =0.81+0.36=1.17\geq 1\) and \(0.9+0.6\geq 1\), therefore \((0.9,0.6)\) is neither a PFM nor an IFM.

This development can be evidently recognized in Figure 1. Here we notice that IFMs are all points beneath the line \(\mu_{a_{ij}}+\nu_{a_{ij}}\leq 1,\) the PFMs are all points with \(\mu^2_{a_{ij}}+\nu^2_{a_{ij}}\leq 1,\) and the q-ROFMs are all points with \(\mu^q_{a_{ij}}+\nu^q_{a_{ij}}\leq 1\). We see then that the q-ROFMs enable for the presentation of a bigger body of nonstandard membership function then IFMs and PFMs. Here \(Q_{m\times n}\) denote the set of all the q-ROFMs.

4. PFM operations on q-ROFMs

In this section we propose the definition of q-rung orthopair fuzzy matrix (q-ROFM) and introduce some operations on q-ROFM. Also, we prove some algebraic properties, such as commutativity, associativity, identity, distributivity and De Morgan's laws over complement.

Definition 4. Let \(A=\left[\left\langle \mu_{a_{ij}},\nu_{a_{ij}}\right\rangle\right]\) and \(B=\left[\left\langle \mu_{b_{ij}},\nu_{b_{ij}}\right\rangle\right]\) be two q-ROFMs of the same size. Then

• (i) \(A\vee_{q} B=\left[\left\langle \max\left\{\mu_{a_{ij}},\mu_{b_{ij}}\right\}\min\left\{\nu_{a_{ij}},\nu_{b_{ij}}\right\}\right\rangle\right]\),
• (ii) \(A\wedge_{q} B=\left[\left\langle \min\left\{\mu_{a_{ij}},\mu_{b_{ij}}\right\}\max\left\{\nu_{a_{ij}},\nu_{b_{ij}}\right\}\right\rangle\right]\),
• (iii) \(A^C=\left[\left\langle \mu_{a_{ij}},\nu_{a_{ij}}\right\rangle\right]\).

Definition 5. Let \(A=\left[\left\langle \mu_{a_{ij}},\nu_{a_{ij}}\right\rangle\right]\) and \(B=\left[\left\langle \mu_{b_{ij}},\nu_{b_{ij}}\right\rangle\right]\) be two q-ROFMs of the same size. Then

• (i) \( A\oplus_{q} B=\left[\left\langle \left(\mu_{a_{ij}}^q+\mu_{b_{ij}}^q-\mu_{a_{ij}}^q\mu_{b_{ij}}^q\right)^{1/q},\nu_{a_{ij}}\nu_{b_{ij}}\right\rangle\right]\),
• (ii) \( A\otimes_{q} B=\left[\left\langle \mu_{a_{ij}}\mu_{b_{ij}}, \left(\nu^q_{a_{ij}}+\nu^q_{b_{ij}}-\nu^q_{a_{ij}}\nu^q_{b_{ij}}\right)^{1/q}\right\rangle\right]\),
• (iii) \( nA=\left[\left\langle \left(1-(1-\mu_{a_{ij}}^q)^{n}\right)^{1/q},(\nu_{a_{ij}})^n\right\rangle\right]\),
• (iv) \( A^{n}=\left[\left\langle \mu_{a_{ij}}^{n},\left(1-(1-\mu_{a_{ij}}^q)^{n}\right)^{1/q}\right\rangle\right]\),

where \(+,-\) and \(.\) are ordinary addition, subtraction and multiplication respectively.

Theorem 2. For \(A, B\in Q_{m\times n}\), we have

• (i) \(A\oplus_{q} B=B\oplus_{q} A,\)
• (ii) \(A\otimes_{q} B=B\otimes_{q} A\),
• (iii) \(n(A\oplus_{q} B)=nA\oplus_{q} nB, n>0\),
• (iv) \((n_{1}+n_{2})A=n_{1}A\oplus_{q} n_{2}A, n_{1},n_{2}>0\),
• (v) \((A\oplus_{q} B)^{n}=A^{n}\otimes_{q} B^{n}, n>0,\)
• (vi) \(A^{n_{1}}\otimes_{q} A^{n_{2}}=A^{(n_{1}+n_{2})}, n_{1},n_{2}>0\).

Proof.

• (i) \begin{align*}A\oplus_{q} B&=\left[\left\langle \left(\mu_{a_{ij}}^q+\mu_{b_{ij}}^q-\mu_{a_{ij}}^q\mu_{b_{ij}}^q\right)^{1/q},\nu_{a_{ij}}\nu_{b_{ij}}\right\rangle\right]\\ &=\left[\left\langle \left(\mu_{b_{ij}}^q+\mu_{a_{ij}}^q-\mu_{b_{ij}}^q\mu_{a_{ij}}^q\right)^{1/q},\nu_{b_{ij}}\nu_{a_{ij}}\right\rangle\right]\\ &=B\oplus_{q} A.\end{align*}
• (ii) \begin{align*} A\otimes_{q} B&=\left[\left\langle \mu_{a_{ij}}\mu_{b_{ij}}, \left(\nu^q_{a_{ij}}+\nu^q_{b_{ij}}-\nu^q_{a_{ij}}\nu^q_{b_{ij}}\right)^{1/q}\right\rangle\right]\\ &=\left[\left\langle \mu_{b_{ij}}\mu_{a_{ij}}, \left(\nu^q_{b_{ij}}+\nu^q_{a_{ij}}-\nu^q_{b_{ij}}\nu^q_{a_{ij}}\right)^{1/q}\right\rangle\right]\\ &=B\otimes_{q} A.\end{align*}
• (iii) \begin{align*} n(A\oplus_{q} B)&= n \left[\left\langle \left(\mu_{a_{ij}}^q+\mu_{b_{ij}}^q-\mu_{a_{ij}}^q\mu_{b_{ij}}^q\right)^{1/q},\nu_{a_{ij}}\nu_{b_{ij}}\right\rangle\right]\\ &=\left[\left\langle \left(1-\left[1-(\mu_{a_{ij}}^q+\mu_{b_{ij}}^q-\mu_{a_{ij}}^q\mu_{b_{ij}}^q)\right]^n\right)^{1/q},(\nu_{a_{ij}}\nu_{b_{ij}})^{n}\right\rangle\right]\\ &=\left[\left\langle \left(1-(1-\mu_{a_{ij}}^q)^{n}(1-\mu_{b_{ij}}^q)^{n}\right)^{1/q},(\nu_{a_{ij}}\nu_{b_{ij}})^{n}\right\rangle\right]\\ nA\oplus_{q} nB&=\left[\left\langle \left(1-(1-\mu_{a_{ij}}^q)^{n}\right)^{1/q},(\nu_{a_{ij}})^n \right\rangle \oplus_{q} \left\langle \left(1-(1-\mu_{a_{ij}}^q)^{n}\right)^{1/q},(\nu_{a_{ij}})^n\right\rangle\right]\\ &=\left[\left\langle \left(1-(1-\mu_{a_{ij}}^q)^{n}(1-\mu_{a_{ij}}^q)^{n}\right)^{1/q},(\nu_{a_{ij}}\nu_{b_{ij}})^n\right\rangle\right]\\ &=n(A\oplus_{q} B).\end{align*}
• (iv) \begin{align*} (n_{1}+n_{2})A&=\left[\left\langle \left(1-(1-\mu_{a_{ij}}^q)^{n_{1}+n_{2}}\right)^{1/q},(\nu_{a_{ij}})^{n_{1}+n_{2}}\right\rangle\right]\\ &=\left[\left\langle \left(1-(1-\mu_{a_{ij}}^q)^{n_{1}}(1-\mu_{a_{ij}}^q)^{n_{2}}\right)^{1/q},(\nu_{a_{ij}}\nu_{b_{ij}})^{n_{1}+n_{2}}\right\rangle\right]\\ &=\left[\left\langle \left(1-(1-\mu_{a_{ij}}^q)^{n_{1}}\right)^{1/q},(\nu_{a_{ij}})^{n_{1}} \right\rangle \oplus_{q} \left\langle \left(1-(1-\mu_{a_{ij}}^q)^{n_{2}}\right)^{1/q},(\nu_{a_{ij}})^{n_{2}}\right\rangle\right]\\ &=n_{1}A\oplus_{q} n_{2}A.\end{align*}
• (v) \begin{align*} (A\otimes_{q} B)^{n}&=\left[\left\langle \mu_{a_{ij}}\mu_{b_{ij}}, \left(\nu^q_{a_{ij}}+\nu^q_{b_{ij}}-\nu^q_{a_{ij}}\nu^q_{b_{ij}}\right)^{1/q}\right\rangle\right]^n\\ &=\left[\left\langle \left(\mu_{a_{ij}}\mu_{b_{ij}}\right)^n, \left(1-\left(1-\nu^q_{a_{ij}}-\nu^q_{b_{ij}}+\nu^q_{a_{ij}}\nu^q_{b_{ij}}\right)^n\right)^{1/q}\right\rangle\right]^n\\ &=\left[\left\langle \left(\mu_{a_{ij}}\right)^n\left(\mu_{b_{ij}}\right)^n, \left(1-(1-\mu_{a_{ij}}^q)^{n}(1-\mu_{a_{ij}}^q)^{n}\right)^{1/q} \right\rangle\right]\\ &=\left[\left\langle \mu_{a_{ij}}^{n},\left(1-(1-\mu_{a_{ij}}^q)^{n}\right)^{1/q}\right\rangle\ \otimes_{q} \left\langle \mu_{b_{ij}}^{n},\left(1-(1-\mu_{b_{ij}}^q)^{n}\right)^{1/q}\right\rangle\right]\\ &=A^{n}\otimes_{q} B^{n}.\end{align*}

• (vi) \begin{align*} A^{n_{1}}\otimes_{q} A^{n_{2}}&=\left[\left\langle \mu_{a_{ij}}^{n_{1}},\left(1-(1-\mu_{a_{ij}}^q)^{n_{1}}\right)^{1/q}\right\rangle\ \otimes_{q} \left\langle \mu_{a_{ij}}^{n_{2}},\left(1-(1-\mu_{a_{ij}}^q)^{n_{2}}\right)^{1/q}\right\rangle\right]\\ &=\left[\left\langle \mu_{a_{ij}}^{n_{1}+n_{2}},\left(1-(1-\mu_{a_{ij}}^q)^{n_{1}+n_{2}}\right)^{1/q}\right\rangle\right]\\ &=A^{(n_{1}+n_{2})}. \end{align*}

Theorem 3. For \(A, B\in Q_{m\times n}\), we have

• (i) \(A\wedge_{q} B=B\wedge_{q} A,\)
• (ii) \( A\vee_{q} B=B\vee_{q} A,\)
• (iii) \( A\wedge_{q}(B\wedge_{q} C)=(A\wedge_{q} B)\wedge_{q} C,\)
• (iv) \( A\vee_{q}(B\vee_{q} C)=(A\vee_{q} B)\vee_{q} C,\)
• (v) \( n(A\wedge_{q} B)=nA\wedge_{q} nB,\)
• (vi) \( n(A\vee_{q} B)=nA\vee_{q} nB,\)
• (vii) \((A\wedge_{q} B)^n=A^n\wedge_{q} B^n,\)
• (viii) \( (A\vee_{q} B)^n=A^n\vee_{q} B^n\).

Proof. Here we prove (i), (iii) and (vi). The remaining are similar.

• (i) \begin{align*}(A\wedge_{q} B)&=\left(\min\left\{\mu_{a_{ij}},\mu_{b_{ij}}\right\}, \max\left\{\nu_{a_{ij}},\nu_{b_{ij}}\right\}\right) \\ &=\left(\min\left\{\mu_{b_{ij}},\mu_{a_{ij}}\right\}, \max\left\{\nu_{b_{ij}},\nu_{a_{ij}}\right\}\right)\\ &=B\wedge_{q} A.\end{align*}
• (iii) \begin{align*} A\wedge_{q}(B\wedge_{q} C) &=\left(\mu_{a_{ij}},\nu_{b_{ij}}\right)\wedge_{q} \left(\min\left\{\mu_{b_{ij}},\mu_{c_{ij}}\right\}, \max\left\{\nu_{b_{ij}},\nu_{c_{ij}}\right\} \right)\\ &=\left(\min\left\{\mu_{a_{ij}},\min\left\{\mu_{b_{ij}},\mu_{c_{ij}}\right\}\right\}, \max\left\{\nu_{a_{ij}},\max\left\{\nu_{b_{ij}},\nu_{c_{ij}}\right\}\right\}\right)\\ &=\left(\min\left\{\min\left\{\mu_{a_{ij}},\mu_{b_{ij}}\right\},\mu_{c_{ij}}\right\}, \max\left\{\max\left\{\nu_{a_{ij}},\nu_{b_{ij}}\right\}, \nu_{c_{ij}}\right\}\right)\\ &=\left(\min\left\{\mu_{a_{ij}},\mu_{b_{ij}}\right\}, \max\left\{\nu_{a_{ij}},\nu_{b_{ij}}\right\} \right)\wedge_{q} \left(\mu_{c_{ij}},\nu_{c_{ij}}\right)\\ &=(A\wedge_{q} B)\wedge_{q} C.\end{align*}
• (vi) \begin{align*} n(A\vee_{q} B)&=nA\vee_{q} nB\\ &=n\left(\min\left\{\mu_{a_{ij}},\mu_{b_{ij}}\right\}, \max\left\{\nu_{a_{ij}},\nu_{b_{ij}}\right\} \right)\\ &=\left[\left\langle \left(1-(1-\max\left\{\mu_{a_{ij}}^q,\mu_{b_{ij}}^q\right\})^{n}\right)^{1/q},\min\left\{(\nu_{a_{ij}})^n,(\nu_{b_{ij}})^n\right\}\right\rangle\right]\\ nA\vee_{q} nB&=\left[\left\langle \left(\left(1-(1-\mu_{a_{ij}}^q)^{n}\right)^{1/q},{\nu_{a_{ij}}}^{n}\right) \vee \left( \left(1-(1-\mu_{a_{ij}}^q)^{n}\right)^{1/q},(\nu_{a_{ij}})^n\right)\right\rangle\right]\\ &=\left[\left\langle \max\left\{\left(1-(1-\mu_{a_{ij}}^q)^{n}\right)^{1/q},\left(1-(1-\mu_{b_{ij}}^q)^{n}\right)^{1/q}\right\},\min\left\{(\nu_{a_{ij}})^n,(\nu_{b_{ij}})^n\right\}\right\rangle\right]\\ &=\left[\left\langle \left(1-(1-\max\left\{\mu_{a_{ij}}^q,\mu_{b_{ij}}^q\right\})^{n}\right)^{1/q},\min\left\{(\nu_{a_{ij}})^n,(\nu_{b_{ij}})^n\right\}\right\rangle\right]\\ &=n(A\vee_{q} B).\end{align*}

Theorem 4. For \(A, B\in Q_{m\times n}\), we have

• (i) \((A\wedge_{q} B)^C=A^C\vee_{q} B^C,\)
• (ii) \((A\vee_{q} B)^C=A^C\wedge_{q} B^C,\)
• (iii) \((A\oplus_{q} B)^C=A^C\otimes_{q} B^C,\)
• (iv) \((A\otimes_{q} B)^C=A^C\oplus_{q} B^C,\)
• (v) \((A^C)^n=(nA)^C,\)
• (vi) \( n(A^C)=(A^n)^C.\)

Proof. Here we prove (i), (iii) and (iv). The remaining are similar.

• (i) \begin{align*} (A\wedge_{q} B)^C&=\left[\left\langle \left(\min\left\{\mu_{a_{ij}},\mu_{b_{ij}}\right\},\max\left\{\nu_{a_{ij}},\nu_{b_{ij}}\right\}\right)^C\right\rangle\right]\\ &=\left[\left\langle \max\left\{\nu_{a_{ij}},\nu_{b_{ij}}\right\},\min\left\{\mu_{a_{ij}},\mu_{b_{ij}}\right\}\right\rangle\right]\\ &=\left(\nu_{a_{ij}},\mu_{a_{ij}}\right)\vee_{q} \left(\nu_{b_{ij}},\mu_{b_{ij}}\right)\\ &=A^C\vee_{q} B^C.\end{align*}
• (iii) \begin{align*}(A\oplus_{q} B)^C&=\left[\left\langle\left( \left(\mu_{a_{ij}}^q+\mu_{b_{ij}}^q-\mu_{a_{ij}}^q\mu_{b_{ij}}^q\right)^{1/q},\nu_{a_{ij}}\nu_{b_{ij}}\right)^C\right\rangle\right]\\ &=\left[\left\langle \mu_{a_{ij}}\mu_{b_{ij}},\left(\nu_{a_{ij}}^q+\nu_{b_{ij}}^q-\nu_{a_{ij}}^q\nu_{b_{ij}}^q\right)^{1/q}\right\rangle\right]\\ &=(\nu_{a_{ij}}\mu_{b_{ij}}) \otimes (\nu_{b_{ij}}\mu_{b_{ij}})\\ &=A^C\otimes_{q} B^C.\end{align*}
• (v) \begin{align*} (A^C)^{n}&=(\nu_{a_{ij}},\mu_{a_{ij}})^n\\ &=\left[\left\langle \nu^n_{a_{ij}},\left(1-(1-\mu_{a_{ij}}^q)^n\right)^{1/q}\right\rangle\right]\\ &=\left[\left\langle \left(\left(1-(1-\mu_{a_{ij}}^q)^n\right)^{1/q}, \nu^n_{a_{ij}}\right)^C\right\rangle\right]\\ &=(nA)^C.\end{align*}

Theorem 5. For \(A, B, C\in Q_{m\times n}\), we have

• (i) \((A\vee_{q} B)\wedge_{q} C=(A\wedge_{q} C)\vee_{q} (B\wedge_{q} C),\)
• (ii) \((A\wedge_{q} B)\vee_{q} C=(A\vee_{q} C)\wedge_{q} (B\vee_{q} C),\)
• (iii) \((A\vee_{q} B)\oplus_{q} C=(A\oplus_{q} C)\vee_{q} (B\oplus_{q} C),\)
• (iv) \((A\wedge_{q} B)\oplus_{q} C=(A\oplus_{q} C)\wedge_{q} (B\oplus_{q} C),\)
• (v) \((A\vee_{q} B)\otimes_{q} C=(A\otimes_{q} C)\vee_{q} (B\otimes_{q} C),\)
• (vi) \((A\wedge_{q} B)\otimes_{q} C=(A\otimes_{q} C)\wedge_{q} (B\otimes_{q} C).\)

Proof. Here we prove(i), (iii) and (v). The remaining can be proved analogously.

(i)

\begin{align*} (A\vee_{q} B)\wedge_{q} C &=\left[\left\langle \min\left\{\max\left\{\mu_{a_{ij}},\mu_{b_{ij}}\right\},\mu_{c_{ij}}\right\},\max\left\{\min\left\{\nu_{a_{ij}},\nu_{b_{ij}}\right\}\nu_{c_{ij}}\right\}\right\rangle\right]\\ &=\left[\left\langle \max\left\{\min\left\{\mu_{a_{ij}},\mu_{b_{ij}}\right\},\min\left\{\mu_{a_{ij}},\mu_{c_{ij}}\right\}\right\},\min\left\{\max\left\{\nu_{a_{ij}},\nu_{b_{ij}}\right\},\max\left\{\nu_{b_{ij}},\nu_{c_{ij}}\right\}\right\}\right\rangle\right]\\ &=\left[\left\langle\left\{\min\left\{\mu_{a_{ij}},\mu_{c_{ij}}\right\},\max\left\{\nu_{a_{ij}},\nu_{c_{ij}}\right\}\right\}\bigvee \left\{\min\left\{\nu_{b_{ij}},\nu_{c_{ij}}\right\},\max\left\{\nu_{b_{ij}},\nu_{c_{ij}}\right\}\right\}\right\rangle\right]\\ &=(A\wedge_{q} C)\vee_{q} (B\wedge_{q} C).\end{align*} Hence, \((A\vee_{q} B)\wedge_{q} C=(A\wedge_{q} C)\vee_{q} (B\wedge_{q} C).\)

(iii)

\begin{align*} (A\vee_{q} B)\oplus_{q} C&=\left(\max\left\{\mu_{a_{ij}},\mu_{b_{ij}}\right\},\min\left\{\nu_{a_{ij}},\nu_{b_{ij}}\right\}\right)\oplus \left(\mu_{c_{ij}}, \nu_{c_{ij}} \right)\\ &=\left[\left\langle \left(\max\left\{\mu_{a_{ij}}^q,\mu_{b_{ij}}^q\right\}+\mu_{c_{ij}}^q-\max\left\{\mu_{a_{ij}}^q,\mu_{b_{ij}}^q\right\}\mu_{c_{ij}}^q\right)^{1/q} ,\min\left\{\nu_{a_{ij}},\nu_{b_{ij}}\right\}\nu_{c_{ij}}\right\rangle\right]\\ &=\left[\left\langle \left((1-\mu_{c_{ij}}^q)\max\left\{\mu_{a_{ij}}^q,\mu_{b_{ij}}^q\right\}+\mu_{c_{ij}}^q\right)^{1/q},\min\left\{\nu_{a_{ij}}\nu_{c_{ij}},\nu_{b_{ij}}\nu_{c_{ij}}\right\}\right\rangle\right].\end{align*} \begin{align*}(A\oplus_{q} C)\vee_{q} (B\oplus_{q} C)&=\left[\left\langle \max\left\{\left(\mu_{a_{ij}}^q+\mu_{c_{ij}}^q-\mu_{a_{ij}}^q\mu_{c_{ij}}^q\right)^{1/q},\left(\mu_{b_{ij}}^q+\mu_{c_{ij}}^q-\mu_{b_{ij}}^q\mu_{c_{ij}}^q\right)^{1/q}\right\},\min\left\{\nu_{a_{ij}}\nu_{c_{ij}},\nu_{b_{ij}}\nu_{c_{ij}}\right\}\right\rangle\right]\\ &=\left[\left\langle \max\left\{\left((1-\mu_{c_{ij}}^q)\mu_{a_{ij}}^q+\mu_{c_{ij}}^q\right)^{1/q},\left((1-\mu_{c_{ij}}^q)\mu_{b_{ij}}^q+\mu_{c_{ij}}^q\right)^{1/q}\right\},\min\left\{\nu_{a_{ij}}\nu_{c_{ij}},\nu_{b_{ij}}\nu_{c_{ij}}\right\}\right\rangle\right]\\ &=\left[\left\langle \left((1-\mu_{c_{ij}}^q)\max\left\{\mu_{a_{ij}}^q,\mu_{b_{ij}}^q\right\}+\mu_{c_{ij}}^q\right)^{1/q},\min\left\{\nu_{a_{ij}}\nu_{c_{ij}},\nu_{b_{ij}}\nu_{c_{ij}}\right\}\right\rangle\right]\\ &=(A\vee_{q} B)\oplus_{q}C.\end{align*} Hence, \((A\vee_{q} B)\oplus_{q} C=(A\oplus_{q} C)\vee_{q} (B\oplus_{q} C)\).

(v)

\begin{align*} (A\vee_{q} B)\otimes_{q}C&=\left[\left\langle \max\left\{\mu_{a_{ij}},\mu_{b_{ij}}\right\}\mu_{c_{ij}},\left(\min\left\{\nu^q_{a_{ij}},\nu^q_{b_{ij}}\right\}+\nu^q_{c_{ij}}-\min\left\{\nu^q_{a_{ij}},\nu^q_{b_{ij}}\right\}\nu^q_{c_{ij}}\right)^{1/q}\right\rangle\right]\\ &=\left[\left\langle \max\left\{\mu_{a_{ij}},\mu_{b_{ij}}\right\}\mu_{c_{ij}},\left((1-\nu^q_{c_{ij}})\min\left\{\nu^q_{a_{ij}},\nu^q_{b_{ij}}\right\}+\nu^q_{c_{ij}}\right)^{1/q}\right\rangle\right].\end{align*} \begin{align*}(A\otimes_{q} C)\vee_{q} (B\otimes_{q} C) &=\left[\left\langle \max\left\{\mu_{a_{ij}}\mu_{c_{ij}},\mu_{b_{ij}}\mu_{c_{ij}}\right\},\min\left\{\left(\nu^q_{a_{ij}}+\nu^q_{c_{ij}}-\nu^q_{a_{ij}}\nu^q_{c_{ij}}\right)^{1/q},\left(\nu^q_{b_{ij}}+\nu^q_{c_{ij}}-\nu^q_{b_{ij}}\nu^q_{c_{ij}}\right)^{1/q}\right\}\right\rangle\right]\\ &=\left[\left\langle \max\left\{\mu_{a_{ij}}\mu_{c_{ij}},\mu_{b_{ij}}\mu_{c_{ij}}\right\},\min\left\{\left((1-\nu_{c_{ij}}^q)\nu_{a_{ij}}^q+\nu_{c_{ij}}^q\right)^{1/q},\left((1-\nu_{c_{ij}}^q)\nu_{b_{ij}}^q+\nu_{c_{ij}}^q\right)^{1/q}\right\}\right\rangle\right]\\ &=\left[\left\langle \max\left\{\mu_{a_{ij}},\mu_{b_{ij}}\right\}\mu_{c_{ij}},\left((1-\nu^q_{c_{ij}})\min\left\{\nu^q_{a_{ij}},\nu^q_{b_{ij}}\right\}+\nu^q_{c_{ij}}\right)^{1/q}\right\rangle\right]\\ &=(A\vee_{q} B)\otimes_{q} C.\end{align*} Hence, \((A\vee_{q} B)\otimes_{q} C=(A\otimes_{q} C)\vee_{q} (B\otimes_{q} C) \).

Theorem 6. For any q-ROFM \(A\), we have

• (i) \((A\oplus_{q} O)=(O\oplus_{q} A)=A,\)
• (ii) \((A\otimes_{q} J)=(J\otimes_{q} A)=A.\)

Proof.

• (i) \( A\oplus_{q} O =\left\langle \mu_{a_{ij}}, \nu_{a_{ij}}\right\rangle \oplus_{q}\left\langle 0,1\right\rangle =\left[\left\langle \left(\mu^q_{a_{ij}}+0-\mu^q_{a_{ij}}.0 \right)^{1/q}, \nu_{a_{ij}}.1\right\rangle\right] =\left[\left\langle \mu_{a_{ij}}, \nu_{a_{ij}}\right\rangle\right] =A.\)

  Similarly, we can prove \(O\oplus_{q} A=A\).

• (ii) \( A\otimes_{q} J=\left\langle \mu_{a_{ij}}, \nu_{a_{ij}}\right\rangle \otimes_{q}\left\langle 1,0\right\rangle =\left[\left\langle \mu_{a_{ij}}.1,\left({\nu_{a_{ij}}}^q+0-{\nu_{a_{ij}}}^q.0\right)^{1/q}\right\rangle\right] =\left[\left\langle \mu_{a_{ij}}, \nu_{a_{ij}}\right\rangle\right] =A.\)

  Similarly, we can prove \(J\otimes_{q} A=A\).

Theorem 7. For any q-ROFM \(A\), we have

• (i) \((A\oplus_{q} J)=(J\oplus_{q} A)=J,\)
• (ii) \((A\otimes_{q} O)=(O\otimes_{q} A)=O.\)

Proof.

• (i) \((A\oplus_{q} J)=\left\langle \mu_{a_{ij}},\nu_{a_{ij}}\right\rangle \oplus_{q} \left\langle 1,0\right\rangle =\left[\left\langle \left(\mu^q_{a_{ij}}+1-\mu^q_{a_{ij}}.1\right)^{1/q},a_{ij}.0\right\rangle\right] =\left\langle 1,0\right\rangle =J.\)

  Similarly, we can prove \(J\oplus_{q} A=J\).

• (ii) \((A\otimes_{q} O) =\left\langle \mu_{a_{ij}}, \nu_{a_{ij}}\right\rangle \otimes_{q} \left\langle 0,1\right\rangle =\left[\left\langle \mu_{a_{ij}}.0,\left(\nu^q_{a_{ij}}+1-\nu^q_{a_{ij}}.1\right)^{1/q}\right\rangle\right] =\left\langle 0,1\right\rangle=O\).

  Similarly, we can prove \(O\otimes_{q} A=O\).

5. Necessity and Possibility operators on q-ROFMs

In this section, we define necessity and possibility operators for q-ROFMs and proved their algebraic properties.

Definition 6. For every q-ROFM \(A\), the necessity \((\Box)\) and possibility \((\Diamond)\) operators are defined as follows: \[\Box A=\left[\left\langle \mu_{a_{ij}},\left(1-\mu_{a_{ij}}^q\right)^{1/q}\right\rangle\right],\] \[\Diamond A=\left[\left\langle \left(1-\nu^q_{a_{ij}}\right)^{1/q},\nu_{a_{ij}}\right\rangle\right].\]

Theorem 8. For \(A, B\in Q_{m\times n}\), we have

• (i) \(\Box(A\oplus_{q} B)=\Box A\oplus_{q} \Box B,\)
• (ii) \(\Diamond (A\oplus_{q} B)=\Diamond A\oplus_{q} \Diamond B\).

Proof. (i) \begin{align*} \Box(A\oplus_{q} B)&=\left[\left\langle \left(\mu_{a_{ij}}^q+\mu_{b_{ij}}^q-\mu_{a_{ij}}^q\mu_{b_{ij}}^q\right)^{1/q},\left(1-(\mu_{a_{ij}}^q+\mu_{b_{ij}}^q-\mu_{a_{ij}}^q\mu_{b_{ij}}^q)\right)^{1/q}\right\rangle\right]\\ \Box A\oplus_{q} \Box B&=\left[\left\langle \left(\mu_{a_{ij}}^q+\mu_{b_{ij}}^q-\mu_{a_{ij}}^q\mu_{b_{ij}}^q\right)^{1/q},\left(1-\mu_{a_{ij}}^q\right)^{1/q} \left(1-\mu_{b_{ij}}^q\right)^{1/q}\right\rangle\right]\\ &=\left[\left\langle \left(\mu_{a_{ij}}^q+\mu_{b_{ij}}^q-\mu_{a_{ij}}^q\mu_{b_{ij}}^q\right)^{1/q},\left(1-(\mu_{a_{ij}}^q+\mu_{b_{ij}}^q-\mu_{a_{ij}}^q\mu_{b_{ij}}^q)\right)^{1/q}\right\rangle\right]. \end{align*} Hence, \(\Box(A\oplus_{q} B)=\Box A\oplus_{q} \Box B\).

(ii)

\begin{align*} \Diamond (A\oplus_{q} B)&=\left[\left\langle \left(1-\nu^q_{a_{ij}}\nu^q_{b_{ij}}\right)^{1/q},\nu_{a_{ij}}\nu_{b_{ij}}\right\rangle\right]\\ \Diamond A\oplus_{q} \Diamond B &=\left[\left\langle \left((1-\nu^q_{a_{ij}})+(1-\nu^q_{b_{ij}})-(1-\nu^q_{a_{ij}})(1-\nu^q_{b_{ij}})\right)^{1/q},\nu_{a_{ij}}\nu_{b_{ij}}\right\rangle\right] &=\left[\left\langle \left(1-\nu^q_{a_{ij}}\nu^q_{b_{ij}}\right)^{1/q},\nu_{a_{ij}}\nu_{b_{ij}}\right\rangle\right].\end{align*} Hence, \(\Diamond (A\oplus_{q} B)=\Diamond A\oplus_{q} \Diamond B\).

Theorem 9. For \(A, B\in Q_{m\times n}\), we have

• (i) \(\Box(A\otimes_{q} B)=\Box A\otimes_{q} \Box B,\)
• (ii) \(\Diamond (A\otimes_{q} B)=\Diamond A\otimes_{q} \Diamond B\).

Proof.

• (i) \begin{align*} \Box(A\otimes_{q} B)&=\left[\left\langle \mu_{a_{ij}}\mu_{b_{ij}},\left(1-\mu_{a_{ij}}^q\mu_{b_{ij}}^q\right)^{1/q}\right\rangle\right]\\ \Box A\otimes_{q} \Box B&=\left[\left\langle \mu_{a_{ij}}\mu_{b_{ij}},\left((1-\mu_{a_{ij}}^q)+(1-\mu_{b_{ij}}^q)-(1-\mu_{a_{ij}}^q)(1-\mu_{b_{ij}}^q)\right)^{1/q}\right\rangle\right]\\ &=\left[\left\langle \mu_{a_{ij}}\mu_{b_{ij}},\left(1-\mu_{a_{ij}}^q\mu_{b_{ij}}^q\right)^{1/q}\right\rangle\right].\end{align*} Hence, \(\Box(A\otimes_{q} B)=\Box A\otimes_{q} \Box B\).
• (ii) It can be proved analogously.

Theorem 10. For \(A, B\in Q_{m\times n}\), we have

• (i) \(\left(\Box(A^C\oplus_{q} B^C)\right)^{C}=\Diamond A\otimes_{q} \Diamond B,\)
• (ii) \(\left(\Box(A^C\otimes_{q} B^C)\right)^{C}=\Diamond A\oplus_{q} \Diamond B\).

Proof.

• (i) \[(A^C\oplus_{q} B^C) =\left[\left\langle \left(\nu^q_{a_{ij}}+\nu^q_{b_{ij}}-\nu^q_{a_{ij}}\nu^q_{b_{ij}}\right)^{1/q},\mu_{a_{ij}}\mu_{b_{ij}}\right\rangle\right],\] \begin{align*}\Box(A^C\oplus_{q} B^C) &=\left[\left\langle \left(\nu^q_{a_{ij}}+\nu^q_{b_{ij}}-\nu^q_{a_{ij}}\nu^q_{b_{ij}}\right)^{1/q}, \left(1-(\nu^q_{a_{ij}}+\nu^q_{b_{ij}}-\nu^q_{a_{ij}}\nu^q_{b_{ij}})\right)^{1/q}\right\rangle\right],\\ \left(\Box(A^C\oplus_{q} B^C)\right)^{C} &=\left[\left\langle \left(1-(\nu^q_{a_{ij}}+\nu^q_{b_{ij}}-\nu^q_{a_{ij}}\nu^q_{b_{ij}})\right)^{1/q},\left(\nu^q_{a_{ij}}+\nu^q_{b_{ij}}-\nu^q_{a_{ij}}\nu^q_{b_{ij}}\right)^{1/q}\right\rangle\right] =\Diamond A\otimes_{q} \Diamond B.\end{align*}
• (ii) It can be proved analogously.

Theorem 11. For \(A, B\in Q_{m\times n}\), we have

• (i) \(\left(\Diamond(A^C\oplus_{q} B^C)\right)^{C}=\Box A\otimes_{q} \Box B,\)
• (ii) \(\left(\Diamond(A^C\otimes_{q} B^C)\right)^{C}=\Box A\oplus_{q} \Box B\).

Proof.

• (i) \begin{align*} \Diamond(A^C\oplus_{q} B^C)&=\left[\left\langle \left(1-\mu_{a_{ij}}^q\mu_{b_{ij}}^q\right)^{1/q},\mu_{a_{ij}}\mu_{b_{ij}}\right\rangle\right],\\ \left(\Diamond(A^C\oplus_{q} B^C)\right)^{C}&=\left[\left\langle \mu_{a_{ij}}\mu_{b_{ij}},\left(1-\mu_{a_{ij}}^q\mu_{b_{ij}}^q\right)^{1/q}\right\rangle\right] =\Box A\otimes_{q} \Box B.\end{align*}
• (ii) It can be proved similarly.

6. New operation (@) on q-ROFMs

In this section, we define the @ operation on q-ROFMs and present their algebraic properties. We discuss the Distributivity law in the case the operation of Algebraic sum and Algebraic product, \(\vee_{q}\) and \(\wedge_{q}\) are combined each other.

Definition 7. Let \(A=\left[\left\langle \mu_{a_{ij}}, \nu_{a_{ij}}\right\rangle\right],\) and \(B=\left[\left\langle\mu_{b_{ij}}, \nu_{b_{ij}}\right\rangle\right]\) be any two q-ROFMs. The new operation of q-ROFM is defined by \[A@B=\left[\left\langle\left(\dfrac{\mu^q_{a_{ij}}+\mu^q_{b_{ij}}}{2}\right)^{1/q},\left(\dfrac{\nu^q_{a_{ij}}+\nu^q_{b_{ij}}}{2}\right)^{1/q}\right\rangle\right].\]

Theorem 12. For any q-ROFM \(A\), we have \(A@A=A.\)

Proof. \begin{align*} A@A&=\left[\left\langle \left(\dfrac{\mu^q_{a_{ij}}+\mu^q_{a_{ij}}}{2}\right)^{1/q},\left(\dfrac{\nu^q_{a_{ij}}+\nu^q_{a_{ij}}}{2}\right)^{1/q}\right\rangle\right]\\ &=\left[\left\langle \left(\dfrac{2\mu^q_{a_{ij}}}{2}\right)^{1/q}, \left(\dfrac{2\nu^q_{a_{ij}}}{2}\right)^{1/q}\right\rangle\right]\\ &=\left[\left\langle \mu^q_{a_{ij}}, \nu^q_{a_{ij}}\right\rangle\right]\\ &=\left[\left\langle \mu_{a_{ij}}, \nu_{a_{ij}}\right\rangle\right]\\ &=A.\end{align*}

Theorem 13. For \(A, B\in Q_{m\times n}\), we have

• (i) \((A\oplus_{q} B)\wedge_{q} (A\otimes_{q} B)=A\otimes_{q} B,\)
• (ii) \((A\oplus_{q} B)\vee_{q} (A\otimes_{q} B)=A\oplus_{q} B,\)
• (iii) \((A\oplus_{q} B)\wedge_{q} (A@B)=A@B,\)
• (iv) \((A\oplus_{q} B)\vee_{q} (A@B)=A\oplus_{q} B,\)
• (v) \((A\otimes_{q} B)\wedge_{q} (A@B)=A\otimes_{q} B,\)
• (vi) \((A\otimes_{q} B)\vee_{q} (A@B)=A@B.\)

Proof. (i) \begin{align*} (A\oplus_{P} B)\wedge_{q} (A\otimes_{q} B) &=\left[\left\langle \min\left\{ \left(\mu^q_{a_{ij}}+\mu^q_{b_{ij}}-\mu^q_{a_{ij}}\mu^q_{b_{ij}}\right)^{1/q}, \mu_{a_{ij}}\mu_{b_{ij}}\right\},\max\left\{ \nu_{a_{ij}}\nu_{b_{ij}},\left(\nu^q_{a_{ij}}+\nu^q_{b_{ij}}-\nu^q_{a_{ij}}\nu^q_{b_{ij}}\right)^{1/q}\right\}\right\rangle\right]\\ &=\left[\left\langle \mu_{a_{ij}}\mu_{b_{ij}}, \left(\nu^q_{a_{ij}}+\nu^q_{b_{ij}}-\nu^q_{a_{ij}}\nu^q_{b_{ij}}\right)^{1/q}\right\rangle\right]\\ &=A\otimes_{q} B.\end{align*} Hence, \((A\oplus_{q} B)\wedge_{q} (A\otimes_{q} B)=A\otimes_{q} B\).

(ii)

\begin{align*} (A\oplus_{q} B)\vee_{q} (A\otimes_{q} B) &=\left[\left\langle \max\left\{ \left(\mu^q_{a_{ij}}+\mu^q_{b_{ij}}-\mu^q_{a_{ij}}\mu^q_{b_{ij}}\right)^{1/q},\mu_{a_{ij}}\mu_{b_{ij}}\right\},\min\left\{ \nu_{a_{ij}}\nu_{b_{ij}},\left(\nu^q_{a_{ij}}+\nu^q_{b_{ij}}-\nu^q_{a_{ij}}\nu^q_{b_{ij}}\right)^{1/q}\right\} \right\rangle\right]\\ &=\left[\left\langle \left(\mu_{a_{ij}}^q+\mu_{b_{ij}}^q-\mu_{a_{ij}}^q\mu_{b_{ij}}^q\right)^{1/q},\nu_{a_{ij}}\nu_{b_{ij}}\right\rangle\right]\\ &=A\oplus_{q} B.\end{align*} Hence, \((A\oplus_{q} B)\vee_{q} (A\otimes_{q} B)=A\oplus_{q} B.\)

(iii)

\begin{align*} (A\oplus_{q} B)\wedge_{q} (A@B) &=\left[\left\langle\min\left\{\left(\mu^q_{a_{ij}}+\mu^q_{b_{ij}}-\mu^q_{a_{ij}}\mu^q_{b_{ij}}\right)^{1/q},\left(\dfrac{\mu^q_{a_{ij}}+\mu^q_{b_{ij}}}{2}\right)^{1/q}\right\},\max\left\{\nu_{a_{ij}}\nu_{b_{ij}},\left(\dfrac{\nu^q_{a_{ij}}+\nu^q_{b_{ij}}}{2}\right)^{1/q}\right\}\right\rangle\right]\\ &=\left[\left\langle\left(\dfrac{\mu^q_{a_{ij}}+\mu^q_{b_{ij}}}{2}\right)^{1/q},\left(\dfrac{\nu^q_{a_{ij}}+\nu^q_{b_{ij}}}{2}\right)^{1/q}\right\rangle\right]\\ &=A@B.\end{align*} Hence, \((A\oplus_{q} B)\wedge_{q} (A@B)=A@B.\)

(iv)

\begin{align*} (A\oplus_{q} B)\vee_{q} (A@B) &=\left[\left\langle \max \left\{\left(\mu^q_{a_{ij}}+\mu^q_{b_{ij}}-\mu^q_{a_{ij}}\mu^q_{a_{ij}}\right)^{1/q},\left(\dfrac{\mu^q_{a_{ij}}+\mu^q_{b_{ij}}}{2}\right)^{1/q}\right\},\min\left\{\nu_{a_{ij}}\nu_{b_{ij}},\left(\dfrac{\nu^q_{a_{ij}}+\nu^q_{b_{ij}}}{2}\right)^{1/q}\right\}\right\rangle\right]\\ &=\left[\left\langle \left(\mu_{a_{ij}}^q+\mu_{b_{ij}}^q-\mu_{a_{ij}}^q\mu_{b_{ij}}^q\right)^{1/q},\nu_{a_{ij}}\nu_{b_{ij}}\right\rangle\right]\\ &=A\oplus_{q} B.\end{align*} Hence, \((A\oplus_{q} B)\vee_{q} (A@B)=A\oplus_{q} B.\)

(v)

\begin{align*} (A\otimes_{q} B)\wedge_{q} (A@B) &=\left[\left\langle\min\left\{\mu_{a_{ij}}\mu_{b_{ij}},\left(\dfrac{\mu^q_{a_{ij}}+\mu^q_{b_{ij}}}{2}\right)^{1/q}\right\},\max\left\{\left(\nu^q_{a_{ij}}+\nu^q_{b_{ij}}-\nu^q_{a_{ij}}\nu^q_{b_{ij}}\right)^{1/q},\left(\dfrac{\nu^q_{a_{ij}}+\nu^q_{b_{ij}}}{2}\right)^{1/q}\right\}\right\rangle\right]\\ &=\left[\left\langle \mu_{a_{ij}}\mu_{b_{ij}}, \left(\nu^q_{a_{ij}}+\nu^q_{b_{ij}}-\nu^q_{a_{ij}}\nu^q_{b_{ij}}\right)^{1/q}\right\rangle\right]\\ &=A\otimes_{q} B.\end{align*} Hence, \((A\otimes_{q} B)\wedge_{q} (A@B)=A\otimes_{q} B.\)

(vi)

\begin{align*} (A\otimes_{q} B)\vee_{q} (A@B) &=\left[\left\langle\max\left\{\mu_{a_{ij}}\mu_{b_{ij}},\left(\dfrac{\mu^q_{a_{ij}}+\mu^q_{b_{ij}}}{2}\right)^{1/q}\right\},\min\left\{\left(\nu^q_{a_{ij}}+\nu^q_{b_{ij}}-\nu^q_{a_{ij}}\nu^q_{b_{ij}}\right)^{1/q},\left(\dfrac{\nu^q_{a_{ij}}+\nu^q_{b_{ij}}}{2}\right)^{1/q}\right\}\right\rangle\right]\\ &=\left[\left\langle \left(\dfrac{\mu^q_{a_{ij}}+\mu^q_{b_{ij}}}{2}\right)^{1/q},\left(\dfrac{\nu^q_{a_{ij}}+\nu^q_{b_{ij}}}{2}\right)^{1/q}\right\rangle\right]\\ &=A@B.\end{align*} Hence, \((A\otimes_{q} B)\vee_{q} (A@B)=A@B.\)

Remark 1. The q-rung orthopair fuzzy matrix forms a commutative monoid, associativity, commutativity and identity under the q-rung orthopair fuzzy matrix operation of algebraic sum and algebraic product. The distributive law also holds for \(\oplus_{q}, \otimes_{q}\) and \(\wedge_{q}, \vee_{q}, @\) are combined each other.

7. Application

The formation of q-ROFMs is commutative monoid structure, q-rung orthopair fuzzy matrix and algebraic structure on this matrix, the results are applicable.

8. Conclusion

Generalized orthopair fuzzy matrices are extensions of intuitionistic fuzzy matrices and Pythagorean fuzzy matrices. Each element is expressed as an ordered pair of values, the former indicating the support for membership and the latter support against membership. The restriction on the memberships is that the sum of the \(q^{th}\) powers of the support for and support against is equal to or less than one. Thus it greatly increases the modelers' ability to capture their judgment of the appropriate orthopair membership grade. In this paper, we proposed q-rung orthopair fuzzy matrices and its algebraic operations are defined. Then we proved some algebraic properties of q-ROFMs, such as associativity, commutativity, identity, distributivity and De Morgan's laws over complement. Furthermore, we defined necessity and possibility operators on q-ROFMs and investigated their algebraic properties. Finally, a new operation(@) on q-ROFMs are defined and discussed distributive laws in the case where the operations of \(\oplus_{q}, \otimes_{q}, \wedge_{q}\) and \(\vee_{q}\) are combined each other. For the development of q-rung orthopair fuzzy commutative monoid structure and its algebraic property the results of this paper would be helpful.

Conflicts of Interest

The author declares no conflict of interest.

1. Introduction and Preliminaries

Fractional derivative was defined for responding to a question 'what does it mean \( \frac{d^{\alpha}f}{dt^{\alpha}} \) if \( \alpha=\frac{1}{2} \)' in 1695. Following that, finding the right definition of fractional derivative has attracted significant attention of researcher and in the last few years it has seen significantly progress in mathematical and non-mathematical journals (see [1,2,3,4,5,6,7,8,9,10]). In fact, there are articles which in few months have gained hundreds of citations. In particular in past three years several definitions of fractional derivative have been proposed (see [3,4,10,11,12,13,14,15,16,17,18,19,20,21,22,23]). Since some of previous definitions do not satisfy the classical formulas of the usual derivative, it has been proposed an ad hoc algebra associated to each definition. To unify that diversity, we propose a version of fractional derivative that has the advantages that generalized the already existing in the literature and where the different algebras are unified under the notion of fractional differential ring.

The present paper is organized as follows: In the Section 2 we give the previous definitions of fractional derivative and our generalized fractional derivative (GFD) definition, in the Section 3 we introduce a fractional differential ring, in the Section 4 we give some result of GFD, in the Section 5 we give a definition of partial fractional differential derivative, in the Section 6 we give a definition of GFD when \( \alpha\in(n,n+1] \).

2. Fractional derivative

Let \( \alpha\in (0,1] \) be a fractional number, we want to give a definition of generalized fractional derivative of order \( \alpha\) for a differentiable function \( f \). We denote \( \alpha-\)th derivative of \( f\) by \( D^{\alpha}(f) \) and we denote the first derivative of \( f\) by \( D(f) \).

We begin the present section listing previous definition of fractional derivative; later we present our proposal of generalized one showing how it contains the once already described. We finish the section providing some examples.

• 1. The Caputo fractional derivative was defined by Michele Caputo in [24]:
  \begin{equation}\label{caputo} D^{\alpha}(f)=\frac{1}{\Gamma(1-\alpha)}\int_{a}^{t}\frac{f'(x)}{(t-x)^{\alpha}}dx. \end{equation}

  (1)
• 2. The conformable fractional derivative was defined by Khalil, Al Horani, Yousef and Sababheh in [11]:
  \begin{equation}\label{arabe} D^{\alpha}f(t)=\lim_{\varepsilon\rightarrow 0}\frac{f(t+ t^{1-\alpha}\varepsilon)-f(t)}{\varepsilon}. \end{equation}

  (2)
• 3. The conformable fractional derivative was defined by Anderson and Ulness in [17]:
  \begin{equation}\label{gringos} D^{\alpha}f(t)=(1-\alpha)\mid t \mid^{\alpha}f(t)+\alpha \mid t\mid^{1-\alpha}Df. \end{equation}

  (3)
• 4. The fractional derivative was defined by Udita N.Katugampola in [17]:
  \begin{equation}\label{algeriano} D^{\alpha}f(t)=\lim_{\varepsilon\rightarrow 0}\frac{f(te^{\varepsilon t^{-\alpha}})-f(t)}{\varepsilon}. \end{equation}

  (4)
• 5. The fractional derivative was defined by Guebbai and Ghiat in [20] for an increasing and positive function \( f \):
  \begin{equation} D^{\alpha}f(t)=\lim_{\varepsilon\rightarrow 0}\left(\frac{f(t+f(t)^{\frac{1-\alpha}{\alpha}}\varepsilon)-f(t)}{\varepsilon}\right)^{\alpha}. \end{equation}

  (5)
• 6. The conformable ratio derivative was defined by Camrud in [14] for a function \(f (t)\geq 0\) with \(Df(t)\geq 0\):
  \begin{equation} D^{\alpha}f(t)=\lim_{\varepsilon\rightarrow 0}f(t)^{1-\alpha}\left(\frac{f(t+\varepsilon)-f(t)}{\varepsilon}\right)^{\alpha}. \end{equation}

  (6)

From all these definitions, we propose a definition that unifies almost all of them.

Definition 1. Given a differentiable function \(f :[0,\infty) \rightarrow \mathbb{R}\), the generalized fractional derivative(GFD) for \( \alpha \in (0,1] \) at point \( t \) is defined by: \begin{array}{llll} D^{\alpha}f(t)= \lim_{\varepsilon\rightarrow 0}\frac{f(t+w_{t,\alpha} t^{1-\alpha}\varepsilon)-f(t)}{\varepsilon}, \end{array} where \(w_{t,\alpha}\) is a function that may depend on \( \alpha \) and \( t \).

Remark 1. As a consequence of Definition 1 we can see \[ D^{\alpha}f(t)=w_{t,\alpha} t^{1-\alpha}Df(t). \]

Definition 2. A differentiable function \( f: [0,\infty) \rightarrow \mathbb{R}\) is said to be \(\alpha-\)generalized fractional differentiable function over \( [0,\infty) \) if it exists \( D^{\alpha}(f)(t) \) for all \( t\in [0,\infty) \) for \( \alpha\in (0,1] \).

We denote \( C^{\alpha}[0,\infty) \) the set of \( \alpha- \)generalized differentiable functions with real values in the interval \([0,\infty)\) in variable \(t\). The set \( (C^{\alpha}[0,\infty),+,.)\) is a ring. In the following we want to see the relation between GFD and the others definitions:

• 1. The fractional derivative of Khalil, Al Horani, Yousef and Sababheh in [11] is a particular case of GFD where \( w_{t,\alpha}=1\).
• 2. The fractional derivative of Anderson and Ulness in [25] is a particular case of GFD where \[ w_{t,\alpha}=\frac{(1-\alpha)t^{\alpha}f(t)+\alpha t^{1-\alpha}Df}{\alpha t^{1-\alpha}}. \] In this fractional derivative \( w_{t,\alpha} \) depends on \( \alpha \) and \(t\).
• 3. The fractional derivatives of Guebbai and Ghiat in [20] and Camrud in [14] are particular cases of GFD where \( w_{t,\alpha}=\left(\frac{tDf}{f} \right)^{\alpha-1}\).

We are particularly interested in discussing GFD where \(w_{t,\alpha}=g(t,\alpha)\tau^{\alpha-1}\) such that \( g:[0,\infty)\times(0,1]\rightarrow \mathbb{R}\) is a function and \( \tau\) is the characteristic of system with the properties \[ w_{t,\alpha} =1 \quad \textit{if and only if} \quad \alpha=1. \]

If the system is periodic with period \( T\), then we have \( \tau=T \). In the quantum systems \( \tau \) is the Bohr radius and in astronomy \( \tau \) is the light year. The characteristic of system \( \tau \) depends on the systems and the derivative. If \( t \) is time, \( \tau \) is time too. If \( t \) is space, \( \tau \) is space too. In fact the unit of \( t \) is \( \tau \), i.e., \( t=c\tau \) where \( c \) is a constant. In the general \( \tau =1\).

Example 1. Let \( \alpha,\beta \in (0,1] \). Let \( f,h \) be two functions in \(C^{\alpha}[0,\infty) \). We suppose \(w_{t,\alpha}= g(t,\alpha)\tau^{\alpha-1} \) with \( \tau=1\).

• 1. If \(g(t,\alpha)\) is a function with \( g(0,0)=0,\) then \( \lim_{\alpha \rightarrow 0} D^{\alpha}(f)=0\).
• 2. If \( g(t,\alpha)=\alpha \), we have the chain rule \[ D^{\alpha}(f\circ h) =\frac{t^{\alpha-1}}{\alpha} D^{\alpha}\left( f(h)\right) D^{\alpha}(h).\]

Example 2. We want to present the corresponding figure to the generalized fractional derivatives for \( \alpha=\frac{3}{4}\) for a trigonometric, using all the fractional derivative definitions that we have already mentioned in this article. It can be seen from all the figures that in principle these definitions do not find a reason to discard them. That is, they have a fairly reasonable behavior. We consider \( f(t)=\sin(2t) \), the graph of \(f\) can be seen in Figure 1.

3. Generalized fractional differential ring

In this section we want to stress out that instead of defining a new derivative, we focus on the notion of differentiable operator and the ring that it carries with.

Definition 3. Let \(R\) be a commutative ring with unity. A derivation on \(R\) is a map \(d : R \rightarrow R\) that satisfies \(d(a + b) = d(a) + d(b)\) and \(d(ab) = d(a)b + ad(b),\forall a, b \in R\). The pair \((R, d)\) is called a differential ring (see [26]).

Theorem 1. Let \( \alpha\in (0,1] \), then the ring \( C^{\alpha}[0,\infty)\) with operator \(D^{\alpha}: C^{\alpha}[0,\infty)\rightarrow C^{\alpha}[0,\infty)\) is a differential ring.

Proof. Since \( C^{\alpha}[0,\infty)\) is a commutative ring with unity \( f(t)=1 \) and the derivation \(D^{\alpha}\) for \( \alpha\in (0,1] \) satisfies following properties from Remark 1

• 1. \( D^{\alpha}(af_{1}+bf_{2})=aD^{\alpha}(f_{1})+bD^{\alpha}(f_{2}), \) \( \forall f_{1},f_{2} \in C^{\alpha}[0,\infty), \forall a,b\in \mathbb{R}\),
• 2. \( D^{\alpha}(f_{1}f_{2})=f_{1}D^{\alpha}(f_{2})+f_{2}D^{\alpha}(f_{1})\), \( \forall f_{1},f_{2} \in C^{\alpha}[0,\infty) \).

Let \( \alpha\in (0,1]\) be a fractional number, and \( f_{1},f_{2} \in C^{\alpha}[0,\infty)\) be two functions, then GFD has the following properties:

• 1. \(D^{\alpha}(\frac{f_{1}}{f_{2}})=\frac{f_{2}D^{\alpha}f_{1}-f_{1}D^{\alpha}f_{2}}{f_{2}^2}.\)
• 2. \( D^{\alpha}(f_{1}\circ f_{2}) =\frac{t^{\alpha-1}}{w_{t,\alpha}} D^{\alpha}\left( f_{1}(f_{2})\right)D^{\alpha}(f_{2}).\)
• 3. \(D^{\alpha+\beta}(f_{1})=\frac{w_{t,\alpha}w_{t,\beta}t}{w_{t,\alpha+\beta}}D^{\alpha}D^{\beta}(f_{1}).\)

It is easy to see these properties from Remark 1 that if \( w_{t,\alpha}=t^{1-\alpha} \) we have the equality \[ D^{\alpha}(f_{1}\circ f_{2}) =D^{\alpha}\left( f_{1}(f_{2})\right) D^{\alpha}(f_{2}).\] If \( \frac{w_{t,\alpha+\beta}}{w_{t,\alpha}w_{t,\beta}}=t\) we have the equality \[ D^{\alpha+\beta}(f_{1})= D^{\alpha}D^{\beta}(f_{1}), \forall \alpha,\beta\in (0,1].\] Parts 4 and 5 of the properties imply that we can create function spaces with different algebras using different expressions for \(w_{t,\alpha}\).

By considering previous properties of GFD we called \( C^{\alpha}[0,\infty)\) a \(w_{t,\alpha} -\)generalized fractional differential ring of functions and we denote it by \((C^{\alpha}[0,\infty) ,D^{\alpha},w_{t,\alpha}) \). Let \( I \subset C^{\alpha}[0,\infty)\) be an ideal. If \( D^{\alpha}(I)\subset I \) then the ideal \( I \) is called a \(w_{t,\alpha} -\)generalized fractional differential ideal. By using the previous properties we can see the following result:

Theorem 2. Let \( \alpha\in(0,1] \), then, associated to any \( \alpha \) and any \( w_{t,\alpha} \) there exists a fractional differential ring.

4. Some results of Generalized Fractional Derivative

Let \( \alpha\in (0,1] \) be a fractional number and \( t\in [0,\infty) \) then GFD has the following properties:

• 1. \( D^{\alpha}(\frac{t^{\alpha}}{\alpha w_{t,\alpha}})=1 \),
• 2. \( D^{\alpha}(\sin(\frac{t^{\alpha}}{\alpha w_{t,\alpha}}))=\cos(\frac{t^{\alpha}}{\alpha w_{t,\alpha}})\),
• 3. \( D^{\alpha}(\cos(\frac{t^{\alpha}}{\alpha w_{t,\alpha}}))=-\sin(\frac{t^{\alpha}}{\alpha w_{t,\alpha}})\),
• 4. \( D^{\alpha}(e^{(\frac{t^{\alpha}}{\alpha w_{t,\alpha}})})=e^{(\frac{t^{\alpha}}{\alpha w_{t,\alpha}})}\).

Theorem 3. (Rolle's Theorem for \( \alpha- \)Generalized Fractional Differentiable Functions)

Let \( a>0 \) and \( f:[a,b]\rightarrow \mathbb{R} \) be a function with the properties that

• 1. \( f \) is continuous on \( [a,b] \),
• 2. \(f\) is \( \alpha- \)generalized fractional differentiable on \((a, b)\) for some \( \alpha\in(0,1] \),
• 3. \( f(a)=f(b) . \)

Then, there exist \( c\in (a,b) \) such that \( D^{\alpha}f(c)=0 \).

Proof. Since \(f \) is continuous on \( [a,b]\) and \( f(a)=f(b) \), then the function \( f \) has a local extreme in a point \( c\in (a,b) \) and \[ D^{\alpha} f(c)=\lim_{\varepsilon\rightarrow 0^{+}}\frac{f(c+w_{t,\alpha} c^{1-\alpha}\varepsilon)-f(c)}{\varepsilon}=\lim_{\varepsilon\rightarrow 0^{-}}\frac{f(c+w_{t,\alpha} c^{1-\alpha}\varepsilon)-f(c)}{\varepsilon}.\] But two limits have different signs, so \(D^{\alpha}f(c)=0 . \)

Theorem 4. (Mean Value Theorem for \( \alpha- \)Generalized Fractional Differentiable Functions) Let \( a>0 \) and \( f:[a,b]\rightarrow \mathbb{R} \) be a function with the properties that

• 1. \( f \) is continuous on \( [a,b] \),
• 2. \(f\) is \( \alpha-\)Generalized fractional differentiable on \((a,b)\) for some \( \alpha\in (0,1] \).

Then, there exists \( c\in (a,b) \) such that \( D^{\alpha}f(c)=\frac{\alpha w_{t,\alpha}(f(b)-f(a))}{b-a}\).

Proof. Consider function \[ h(t)=f(t)-f(a)-\frac{\alpha w_{t,\alpha}(f(b)-f(a))}{b-a}\left(\frac{t^{\alpha}}{\alpha w_{t,\alpha}} -\frac{a^{\alpha}}{\alpha w_{t,\alpha}}\right). \] Then, the function \( h \) satisfies the conditions of the fractional Rolle’s Theorem. Hence, there exists \( c\in (a,b) \) such that \( D^{\alpha}h(c)=0. \) We have the result since \[ D^{\alpha}h(c)= D^{\alpha}f(c)-\frac{\alpha w_{t,\alpha}(f(b)-f(a))}{b-a}(1)=0. \]

5. Generalized Partial Fractional Derivative

In this section we introduce a partial fractional derivative of first and second order. Also we introduce a partial fractional differential ring.

Definition 4. Let \(f(t_{1},\cdots,t_{n}):[0,\infty)^n\rightarrow \mathbb{R}\) be a function with \( n \) variables such that \(\forall i\), then there exists the partial derivative of \( f \) respect to \(t_{i}\). Let \( \alpha\in (0,1]\) be a fractional number. We define \( \alpha-\)generalized partial fractional derivative(GPFD) of \( f \) with respect to \( t_{i} \) at point \( t=(t_{1},\ldots,t_{n}) \) \begin{array}{llll} \frac{\partial^{\alpha}f(t)}{\partial t_{i}^{\alpha}}= \lim_{\varepsilon\rightarrow 0}\frac{f(t_{1},\ldots,t_{i}+w_{t_{i},\alpha} t_{i}^{1-\alpha}\varepsilon,\ldots,t_{n})-f(t)}{\varepsilon}, \end{array} where \(w_{t_{i},\alpha}\) can be a function depend on \( \alpha\) and \( t_{i} \).

Remark 2. As a consequence of Definition 2 we can see for \( \alpha\in(0,1] \) and \( 1\leq i\leq n \): \[ \frac{\partial^{\alpha}f}{\partial t_{i}^{\alpha}}(t)=w_{t_{i},\alpha}( t_{i})^{1-\alpha} \frac{\partial f}{\partial t_{i}}(t). \]

Let \( \alpha\in (0,1] \) and \( 1\leq i\leq n \). A partial differentiable function \( f: [0,\infty)^n \rightarrow \mathbb{R}\) is said to be a \( \alpha- \)generalized partial fractional differentiable function respect to \( t_{i} \) over \( [0,\infty) \) if there exists \( \frac{\partial^{\alpha}f(t)}{\partial t_{i}^{\alpha}} \) for all \( t\in [0,\infty) \). We denote by \( C_{i}^{\alpha}[0,\infty)^n \) the set of \( \alpha-\)generalized partial fractional differentiable functions respect to \( t_{i} \) with real values in the interval \([0,\infty)^n\) in variable \(t=(t_{1},\ldots,t_{n})\). The set \( (C_{i}^{\alpha}[0,\infty)^n,+,.)\) is a ring.

Theorem 5. Let \( \alpha\in(0,1] \) and \( 1\leq i\leq n \). The ring \( C_{i}^{\alpha}[0,\infty)^n\) with operator \[ \frac{\partial^{\alpha}}{\partial t_{i}^{\alpha}}: C_{i}^{\alpha}[0,\infty)^n\rightarrow C_{i}^{\alpha}[0,\infty)^n\] is a differential ring.

Proof. Since the ring \( C_{i}^{\alpha}[0,\infty)^n\) is a commutative ring with unity \( f(t_{1},\ldots,t_{n})=1 \) and the derivation \(\frac{\partial^{\alpha}}{\partial t_{i}^{\alpha}}\) for \(\alpha\in (0,1] \) satisfies the following properties from Remark 2;

• 1. \(\frac{\partial^{\alpha}(f_{1}+f_{2})}{\partial t_{i}^{\alpha}} =\frac{\partial^{\alpha}(f_{1})}{\partial t_{i}^{\alpha}}+\frac{\partial^{\alpha}(f_{2})}{\partial t_{i}^{\alpha}}\quad f_{1},f_{2}\in C^{\alpha}[0,\infty)^n,\)
• 2. \(\frac{\partial^{\alpha}(f_{1}f_{2})}{\partial t_{i}^{\alpha}} =f_{1}\frac{\partial^{\alpha}f_{2}}{\partial t_{i}^{\alpha}}+f_{2}\frac{\partial^{\alpha}f_{1}}{\partial t_{i}^{\alpha}}\quad f_{1},f_{2}\in C^{\alpha}[0,\infty)^n\).

Let \( \alpha\in (0,1]\) and \( 1\leq i\leq n \). Further let \( f_{1},f_{2} \in C^{\alpha}[0,\infty)^n\) be two functions, then GPFD has the following properties from Remark 2:

• 1. \(\frac{\partial^{\alpha}(\frac{f_{1}}{f_{2}})}{\partial t_{i}^{\alpha}}=\frac{f_{2} \frac{\partial^{\alpha}f_{1}}{\partial t_{i}^{\alpha}}-f_{1} \frac{\partial^{\alpha}f_{2}}{\partial t_{i}^{\alpha}}}{f_{2}^2},\)
• 2. \( \frac{\partial^{\alpha}f_{1}\circ f_{2}}{\partial t_{i}^{\alpha}}= \frac{t_{i}^{\alpha-1}}{w_{t_{i},\alpha}} \frac{\partial^{\alpha}( f_{1}(f_{2}))}{\partial t_{i}^{\alpha}}\frac{\partial^{\alpha}(f_{2})}{\partial t_{i}^{\alpha}}, \)
• 3. \( \frac{\partial^{\alpha+\beta}( f_{1})}{\partial t_{i}^{\alpha}}=\frac{w_{t_{i},\alpha}w_{t_{i},\beta}t_{i}}{w_{t_{i},\alpha+\beta}} \frac{\partial^{\alpha}}{\partial t_{i}^{\alpha}}\frac{\partial^{\alpha}(f_{1})}{\partial t_{i}^{\alpha}}.\)

By considering previous properties of GPFD we called the ring \( C_{i}^{\alpha}[0,\infty)^n\) a \( w_{t_{i},\alpha}- \) generalized partial fractional differential ring. We denote it by \( (C_{i}^{\alpha}[0,\infty)^n,\frac{\partial^{\alpha}}{\partial t_{i}^{\alpha}} ,w_{t_{i},\alpha}) \).

We can see the following result by using the previous properties:

Theorem 6. Let \( \alpha\in(0,1] \) and \( 1\leq i \leq n \). Associated to any \( \alpha \) and any \( w_{t_{i},\alpha} \) there is a partial fractional differential ring.

Example 3. Let \( f(t_{1},t_{2}) =t_{1}^3\sin(t_{2})\) and \( \alpha\in [0,1)\) then we have \[ \frac{\partial^{\alpha}f}{\partial t_{1}^{\alpha}}= w_{t,\alpha}(t_{1})^{1-\alpha}(3t_{1}^2)\sin(t_{2}).\]

Definition 5. Let \( \alpha\in(0,1]\) be a fractional number. We define \( \alpha- \)generalized partial fractional derivative of second order with respect to \(t_{i}\) and \( t_{j} \) at point \( t=(t_{1},\cdots,t_{n}) \) is \begin{array}{llll} \frac{\partial^{\alpha^2}f(t)}{\partial t_{j}^{\alpha} \partial t_{i}^{\alpha}}=\frac{\partial^{\alpha}}{\partial t_{j}^{\alpha}}(\frac{\partial^{\alpha}f(t)}{\partial t_{i}^{\alpha}})= \lim_{\varepsilon\rightarrow 0}\frac{ \frac{\partial^{\alpha}f}{\partial t_{i}^{\alpha}}(t_{1},\ldots,t_{j}+w_{t_{j},\alpha} t_{j}^{1-\alpha}\varepsilon,\ldots,t_{n})-\frac{\partial^{\alpha}f(t)}{\partial t_{i}^{\alpha}}}{\varepsilon}. \end{array}

Remark 3. As a consequence of Definition 5 we can see for \( \alpha\in(0,1] \) and \( 1\leq i,j\leq n \); \[ \frac{\partial^{\alpha^2}f(t)}{\partial t_{j}^{\alpha} \partial t_{i}^{\alpha}}=w_{t_{j},\alpha}w_{t_{i},\alpha}(t_{j} t_{i})^{1-\alpha} \frac{\partial }{\partial t_{j}}( \frac{\partial f}{\partial t_{i}}(t)). \]

A partial differentiable function of second order \( f: [0,\infty)^n \rightarrow \mathbb{R}\) is said to be a \( \alpha- \)generalized fractional partial differentiable function of second order respect to \( t_{i}\) and \(t_{j} \) over \( [0,\infty) \) if there exists \( \frac{\partial^{\alpha^2}f(t)}{\partial t_{j}^{\alpha}\partial t_{i}^{\alpha}}\) for all \( t\in [0,\infty) \). We denote \( C_{i,j}^{\alpha^2}[0,\infty)^n \) the set of \( \alpha-\)generalized partial fractional differentiable functions of second order respect to \( t_{i}\) and \(t_{j} \) with real values in the interval \([0,\infty)^n\) in variable \(t=(t_{1},\ldots,t_{n})\). The set \( (C_{i,j}^{\alpha^2}[0,\infty)^n,+,.)\) is a ring.

6. Generalized Fractional Derivative for \( \alpha\in (n,n+1] \)

In this section we define a fraction differential derivative for \( \alpha\in (n,n+1]. \)

Definition 6. Let \( \alpha\in (n,n+1]\) be a fractional number for \( n\in \mathbb{N} \) and \( f:[0.\infty) \rightarrow \mathbb{R}\) be a \( n -\) differentiable. The generalized fractional derivative of order \( \alpha \) is defined by \begin{array}{llll} D^{\alpha}f(t)= \lim_{\varepsilon\rightarrow 0}\frac{f^{[\alpha]-1}(t+w_{t,\alpha} t^{[\alpha]-\alpha}\varepsilon)-f^{[\alpha]-1}(t)}{\varepsilon}, \end{array} where \([\alpha]\) is the smallest integer greater than or equal to \( \alpha. \)

As a consequence of Definition 6 we can see

\[ D^{\alpha}(f)=w_{t,\alpha}t^{[\alpha]-\alpha}D^{[\alpha]}(f), \] where \( \alpha\in (n,n+1] \).

Let \( n< \alpha\leq n+1 \). A function \( f: [0,\infty) \rightarrow \mathbb{R}\) is said to be \(\alpha-\)generalized differentiable over \( [0,\infty)\) if there exists \( D^{\alpha}(f)(t) \) for all \( t\in [0,\infty). \) We denote \( C^{\alpha}[0,\infty) \) the set of \( \alpha- \)generalized fractional differentiable functions with real values in the interval \([0,\infty)\) in variable \(t\). The set \( (C^{\alpha}[0,\infty),+,.)\) is a ring.

Theorem 7. The ring \( (C^{\alpha}[0,\infty),+,.)\) with operator \( D^{\alpha} \) is not a differential ring for fractional number \( \alpha\in (n,n+1] \).

Proof. Since \( D^{\alpha}(fg)\neq fD^{\alpha}g+gD^{\alpha}(f) \) for every \( f,g\in C^{\alpha}[0,\infty) \).

7. \(\alpha\)-Fractional Taylor Series

There are some articles about fractional Taylor series see ([12,27,28,29]). In this section we use GFD to define a fractional taylor series for a function \( f\in C^{r}[0,\infty)\) for every fractional number \( r \).

Let \(0 < \alpha< 1 \), we define the \(\alpha\)-fractional taylor series of \( f \) at real number \( x_{0} \)

\[ f(x)=f(x_{0})+\sum_{i=1}^{\infty} \frac{D^{i}f(x_{0})}{w_{x,\alpha}\overline{(\alpha+i-1)!}} (x-x_{0})^{\alpha+i-1} , \] where \(\overline{(\alpha+i-1)!}= \alpha(\alpha+1)\cdots(\alpha+i-1) \).

Let \( 1< \alpha\leq 2 \), we define the \(\alpha\)-fractional taylor series of \( f \) at real number \( x_{0}\)

\[ f(x)=f(x_{0})+Df(x_{0})(x-x_{0})+\sum_{i=2}^{\infty}\frac{D^{i}f(x_{0})(x-x_{0})^{\alpha+i-2}}{w_{x,\alpha}\overline{(\alpha+i-2)!}}, \] where \(\overline{(\alpha+i-2)!}= (\alpha-1)\alpha(\alpha+1)\cdots(\alpha+i-2) .\)

Let \( n< \alpha\leq n+1 \) such that \( \alpha=n+A \) with \( 0< A< 1 \). We define the \(\alpha\)-fractional taylor series of \( f \) at real number \( x_{0}\),

\[ f(x)=f(x_{0})+\sum_{i=1}^{n}\frac{D^if(x_{0})(x-x_{0})^{i}}{i!}+\sum_{i=n+1}^{\infty}\frac{D^{i}f(x_{0})}{w_{x,\alpha}\overline{(A+i-1)!}}(x-x_{0})^{A+i-1}, \] where \( \alpha=n+A \), \(\overline{(A+i-1)!}= A(A+1)\cdots(A+i-1) \).

8. Application to differential equations

There are some articles for applications of fractional differential derivative such as [3,6,7,9]. In this section we solve some (partial) fractional differential equations by using our definitions. Firstly we solve the fractional differential equations with the form where \(y=f(t)\) be a differentiable function and \(0< \alpha< 1 \).

By substituting GFD in the Equation (7) we have

\[ aw_{t,\alpha}t^{1-\alpha}Dy+by=c\Longrightarrow Dy+\frac{bt^{\alpha-1}}{aw_{t,\alpha}}y=\frac{t^{\alpha-1}c}{aw_{t,\alpha}}, \] the solutions of this equation have the form \(y(t)= \frac{c}{b}+c_{1}e^{(-\frac{bt^{\alpha}}{aw_{t,\alpha}\alpha})}. \)

Example 4. We consider the partial fractional differential equation with boundary conditions

where \( u(x,t)\) be a differentiable function respect to \( x \) and \( t \), \( u(x,t)\) be a \( \frac{1}{3}- \)partial fractional differentiable function of first order respect to \(x \), \(u_{t}=\frac{\partial u}{\partial t } \) and \(\sqrt[3]{u_{x}}=\frac{\partial^{\frac{1}{3}} u}{\partial x^{\frac{1}{3}} }. \) For \(w_{t,\frac{1}{3}}=\frac{1}{3}\), by using Remark 2 we can write \[u_{t}+2\sqrt[3]{xu_{x}}+u=x^{2} \Longrightarrow u_{t}+2\sqrt[3]{x} w_{t,\alpha} x^{1-\alpha}u_{x}+u=x^{2} \Longrightarrow u_{t}+\frac{2}{3}xu_{x}+u=x^{2}. \] We solve this equation by taking Laplace transform of equation with respect to \( t \). We denote by \(U(x,s)\) the Laplace of \( u(x,t) \) with respect to \( t \), we have the following equation Then \[ U_{x}+\frac{3+3s}{2x}U=\frac{3x}{2s}\Longrightarrow U(x,s)=\frac{3x^2}{s(3s+7)}+c(s)x^{\frac{-3-3s}{2}}.\] By substituting \(U(0,x)=0\), we have \( c(s)=0 \), then \( U(x,s)=x^2(\frac{3}{7s}+\frac{9}{7(3s+7)})\). The solution of equation is \(u(x,t)=\frac{x^2}{7}(1-e^{\frac{-7}{3}}) .\)

Example 5. We consider the partial fractional differential equation of second order;

where \( u(x,t) \) be a \( \frac{1}{5}- \)fractional partial differentiable function of second order respect to \( t,x \). For \( w_{x,\frac{1}{5}}=x^2, w_{t,,\frac{1}{5}}=\frac{1}{\sqrt[3]{t}} \) by using remark 3, we have We consider a solution of this differential equation of the form \( u(x,t)=f(x)g(t) \) such that \( f \) a function depends on \( x \) and \( g \) a function depends on \( t \). By substituting \( u(x,t) \) in the Equation (11) we have We can write Equation (12) in a form that divide the functions of \( t \) and \( x \): two sides of the Equality (13) is a constant \( k \). We have The solution of Equation (10) has the form \[ u(x,t)= \exp( k\sqrt[15]{x}+c_{1})\exp(\frac{-2\sqrt[15]{t^8}}{k}+c_{2})=c\exp(k\sqrt[15]{x}+\frac{-2\sqrt[15]{t^8}}{k}).\]

9. Conclusion

We defined a generalized fractional derivative (GFD). We showed that the previous derivatives are particular cases. We also showed how it is possible to have infinite fractional derivatives with their algebra. We present the fractional differential ring, the fractional partial derivatives and their applications.

Conflicts of Interest

The authors declare no conflict of interest.

Data Availability

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Funding Information

No funding is available for this research.

Acknowledgments

This paper was supported by PRODEP of Mexico for Zeinab Toghani postdoctoral position.