Schrödinger equation with potential function vanishing exponentially fast

Abstract

Explicitly oscillating solutions of differential equation $y^{″}+\left(\lambda +20{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^{2}x\right)y=0$ and its eigenpairs are obtained by calculating complex residues. Eigenpairs, spectral function and eigenfunction expansions are also reported for this specific Pöschl–Teller differential equation.

Keywords:

• Schrödinger equation
• Pöschl–Teller potential
• complex residues
• explicit solution
• eigenpair

1. Introduction

In this paper, we consider the following second-order differential equation: (1) $y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+(\lambda -q\left(x\right))y\left(x\right)=0,$(1) with specific potential function $q\left(x\right)=-20{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^{2}x$. Theory related to (1(1) $y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+(\lambda -q\left(x\right))y\left(x\right)=0,$(1) ) is elaborately given in [1]. One may say that the only case to solve (1(1) $y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+(\lambda -q\left(x\right))y\left(x\right)=0,$(1) ) explicitly for all λ is the case $q\left(x\right)=0$ where the solutions are oscillating. In reality, this is not correct. There are indeed some examples which are not reported yet. One such example is (2) $y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+\left(\lambda +20{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^{2}x\right)y\left(x\right)=0,$(2) which is a special case of $y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+\left(\lambda +n(n+1){\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^{2}x\right)y\left(x\right)=0$ with n=4. This example is slightly different from what Titchmarsh does in the following example where its solutions involving Legendre functions $\begin{array}{rl}& y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+\left(\lambda +\left({\nu }^2-\frac{1}{4}\right){\sec }^{2}x\right)y\left(x\right)=0,\\ & x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right).\end{array}$ For similar examples and results, see [1–8].

It is worth mentioning here how our interest in this title arises. There is a relationship between the Schrödinger equation and Sturm–Liouville differential equation (1(1) $y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+(\lambda -q\left(x\right))y\left(x\right)=0,$(1) ). The Schrödinger equation arises from several partial differential equation appearing in physics by separation of variables [9]. For instance, for a single quantum-mechanical particle of mass m moving in one space dimension in a potential $V\left(x\right)$, the time-dependent Schrödinger equation is $i\hslash {\mathrm{\Psi }}_t=-\frac{{\hslash }^2}{2m}{\mathrm{\Psi }}_{xx}+V\left(x\right)\mathrm{\Psi }.$ Looking for separable solutions $\mathrm{\Psi }=\varphi \left(x\right){\mathrm{e}}^{-iEt/\hslash }$ where E is energy eigenstates. Then one finds that $-\frac{{\hslash }^2}{2m}{\varphi }^{\mathrm{\prime }\mathrm{\prime }}+V\left(x\right)\varphi =E\varphi .$ After normalization, one obtains Equation (1(1) $y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+(\lambda -q\left(x\right))y\left(x\right)=0,$(1) ) in the subject line above where the potential function $q=\left(2m/{\hslash }^2\right)V\left(x\right)$, the eigenvalue parameter $\lambda =\left(2m/{\hslash }^2\right)E$ and $y=\varphi$.

Therefore, if one considers $x\to \mathrm{\infty }$, then ${\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^{2}x=\frac{4}{{\mathrm{e}}^{2x}+2+{\mathrm{e}}^{-2x}}\sim 4{\mathrm{e}}^{-2x}.$ Hence, one may see that why we called title in the subject line above. It is also important to note that Equation (2(2) $y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+\left(\lambda +20{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^{2}x\right)y\left(x\right)=0,$(2) ) describes the diatomic molecular vibration and such potential also arises in the solutions of the Korteweg–de Vries equation [10, 11].

In general, it is not easy to obtain the solutions of Equation (1(1) $y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+(\lambda -q\left(x\right))y\left(x\right)=0,$(1) ) as we obtain in our case by employing residues. Conventionally, solutions to Equation (1(1) $y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+(\lambda -q\left(x\right))y\left(x\right)=0,$(1) ) are written either in terms of hypergeometric functions or as series. The purpose of this article is to provide analytical solutions of Equation (2(2) $y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+\left(\lambda +20{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^{2}x\right)y\left(x\right)=0,$(2) ) by using local complex residues.

2. Preliminaries

We now give some preliminaries. To get the expansion of an arbitrary function $f\left(x\right)$ in terms of eigenfunctions one actually needs to know the following definitions and lemmas taken from [1, 2]. If $\theta (x,\lambda )$ and $\phi (x,\lambda )$ are the solutions of (1(1) $y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+(\lambda -q\left(x\right))y\left(x\right)=0,$(1) ), with α is real, satisfying (3) $\begin{array}{rl}& \phi (0,\lambda )=\sin \alpha ,{\phi }^{\mathrm{\prime }}(0,\lambda )=-\cos \alpha ,\\ & \theta (0,\lambda )=\cos \alpha ,{\theta }^{\mathrm{\prime }}(0,\lambda )=\sin \alpha .\end{array}$(3) Then Wronskian $W_x(\theta ,\phi )={\sin }^2\alpha +{\cos }^2\alpha =1$. It is well known that the general solution of (1(1) $y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+(\lambda -q\left(x\right))y\left(x\right)=0,$(1) ) is $\psi (x,\lambda )=\theta (x,\lambda )+m\left(\lambda \right)\phi (x,\lambda )\in L^2(0,\mathrm{\infty }).$ The definition of the non-decreasing spectrum function is given by $k\left(\lambda \right)=-\underset{\delta \to 0}{lim}{\int }_0^{\lambda }\mathrm{I}\mathrm{m}(m(u+i\delta ))\mathrm{d}u.$

Lemma 2.1

For proof, see [1]

For suitable class of functions on $(0,\mathrm{\infty })$ one has $\begin{array}{rl}f\left(x\right)& =\frac{1}{\pi }{\int }_0^{\mathrm{\infty }}\phi (x,\lambda )\mathrm{d}k\lambda {\int }_0^{\mathrm{\infty }}\phi (t,\lambda )f\left(t\right)\mathrm{d}t\\ & +\sum _{N=0}^{\mathrm{\infty }}{\phi }_N(x,\lambda ){\int }_0^{\mathrm{\infty }}{\phi }_N(t,\lambda )f\left(t\right)\mathrm{d}t.\end{array}$ If $m\left(\lambda \right)$ does not have poles, then the summative terms in the expansion of $f\left(x\right)$ disappear.

Lemma 2.2

For proof, see [1]

Let $q\left(x\right)$ be given. (If $q\left(x\right)$ is an even function, then $m_1\left(\lambda \right)=-m_2\left(\lambda \right)$). Hence, for a suitable class of functions on $(-\mathrm{\infty },\mathrm{\infty })$ one has (4) $\begin{array}{rl}f\left(x\right)& =\frac{1}{\pi }({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\theta (x,\lambda )\mathrm{d}\xi {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\theta (y,\lambda )f\left(y\right)\mathrm{d}y\\ & +{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\phi (x,\lambda )\mathrm{d}\zeta {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\phi (y,\lambda )f\left(y\right)\mathrm{d}y)\\ & +\sum _{N=0}^{\mathrm{\infty }}{\phi }_N(x,\lambda ){\int }_0^{\mathrm{\infty }}{\phi }_N(t,\lambda )f\left(t\right)\mathrm{d}t,\end{array}$(4) where $\begin{array}{rl}& {\psi }_1(x,\lambda )=\theta (x,\lambda )+m_1\left(\lambda \right)\phi (x,\lambda )\in L^2(-\mathrm{\infty },0),\\ & {\psi }_2(x,\lambda )=\theta (x,\lambda )+m_2\left(\lambda \right)\phi (x,\lambda )\in L^2(0,\mathrm{\infty }),\\ & \xi \left(\lambda \right)=-\underset{\delta \to 0}{lim}{\int }_0^{\lambda }\mathrm{I}\mathrm{m}(\frac{1}{m_1(u+i\delta )-m_2(u+i\delta )})\mathrm{d}u,\\ & {\xi }^{\prime }\left(\lambda \right)\underset{\mathrm{q}\left(\mathrm{x}\right)\mathrm{i}\mathrm{s}\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}{=}\mathrm{I}\mathrm{m}(\frac{1}{2m_2\left(\lambda \right)}),\\ & \zeta \left(\lambda \right)=-\underset{\delta \to 0}{lim}{\int }_0^{\lambda }\mathrm{I}\mathrm{m}(\frac{m_1(u+i\delta )m_2(u+i\delta )}{m_1(u+i\delta )-m_2(u+i\delta )})\mathrm{d}u,\\ & {\zeta }^{\prime }\left(\lambda \right)\underset{\mathrm{q}\left(\mathrm{x}\right)\mathrm{i}\mathrm{s}\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}{=}\mathrm{I}\mathrm{m}(\frac{-1}{2m_2\left(\lambda \right)}).\end{array}$ If $m\left(\lambda \right)$ does not have poles, then the summative terms in (4(4) $\begin{array}{rl}f\left(x\right)& =\frac{1}{\pi }({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\theta (x,\lambda )\mathrm{d}\xi {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\theta (y,\lambda )f\left(y\right)\mathrm{d}y\\ & +{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\phi (x,\lambda )\mathrm{d}\zeta {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\phi (y,\lambda )f\left(y\right)\mathrm{d}y)\\ & +\sum _{N=0}^{\mathrm{\infty }}{\phi }_N(x,\lambda ){\int }_0^{\mathrm{\infty }}{\phi }_N(t,\lambda )f\left(t\right)\mathrm{d}t,\end{array}$(4) ) disappear.

3. Main results

We are now aiming to deal with the solution of (2(2) $y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+\left(\lambda +20{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^{2}x\right)y\left(x\right)=0,$(2) ) which is given by with $s^2=-\lambda$ (5) $y\left(x\right)={\cosh }^{5}x{\oint }_C\frac{\cosh \left(zs\right)}{(\sinh z-\sinh x{)}^5}\mathrm{d}z,$(5) where C is just including the point z=x and excluding the other zeros of $\sinh z-\sinh x$.

Theorem 3.1

Case for any n, see [2, 4]

The integral (5(5) $y\left(x\right)={\cosh }^{5}x{\oint }_C\frac{\cosh \left(zs\right)}{(\sinh z-\sinh x{)}^5}\mathrm{d}z,$(5) ) solves Equation (2(2) $y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+\left(\lambda +20{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^{2}x\right)y\left(x\right)=0,$(2) ).

Proof.

$\begin{array}{rl}{y}^{\prime }\left(x\right)& =5\tanh \left(x\right)y\left(x\right)\\ & +5{\cosh }^{6}x{\oint }_C\frac{\cosh \left(zs\right)}{(\sinh z-\sinh x{)}^6}\mathrm{d}z,\\ y^{″}\left(x\right)& =25y\left(x\right)-20{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^2\left(x\right)y\left(x\right)\\ & +55{\cosh }^{5}x\sinh x{\oint }_C\frac{\cosh \left(zs\right)}{(\sinh z-\sinh x{)}^6}\mathrm{d}z\\ & +30{\cosh }^{7}x{\oint }_C\frac{\cosh \left(zs\right)}{(\sinh z-\sinh x{)}^7}\mathrm{d}z.\end{array}$Therefore,

(6) $\begin{array}{rl}& y^{″}\left(x\right)+20{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^2\left(x\right)y\left(x\right)\\ & =5{\cosh }^{5}x{\oint }_C\frac{\cosh \left(zs\right)(5{\sinh }^{2}z+\sinh z\sinh x+6)}{(\sinh z-\sinh x{)}^7}\mathrm{d}z.\end{array}$(6) Applying two times partial integration to (5(5) $y\left(x\right)={\cosh }^{5}x{\oint }_C\frac{\cosh \left(zs\right)}{(\sinh z-\sinh x{)}^5}\mathrm{d}z,$(5) ) successively, one obtains $\begin{array}{rl}y\left(x\right)& =\frac{5{\cosh }^{5}x}{s}{\oint }_C\frac{\sinh \left(zs\right)\cosh z}{(\sinh z-\sinh x{)}^6}\mathrm{d}z\\ & =\frac{5{\cosh }^{5}x}{s^2}{\oint }_C\frac{\cosh \left(zs\right)\left(5{\sinh }^{2}z+\sinh z\sinh x+6\right)}{(\sinh z-\sinh x{)}^7}\mathrm{d}z.\end{array}$ By using $s^2=-\lambda$, (7) $\lambda y\left(x\right)=-5{\cosh }^{5}x{\oint }_C\frac{\cosh \left(zs\right)\left(5{\sinh }^{2}z+\sinh z\sinh x+6\right)}{(\sinh z-\sinh x{)}^7}\mathrm{d}z.$(7) Comparing (6(6) $\begin{array}{rl}& y^{″}\left(x\right)+20{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^2\left(x\right)y\left(x\right)\\ & =5{\cosh }^{5}x{\oint }_C\frac{\cosh \left(zs\right)(5{\sinh }^{2}z+\sinh z\sinh x+6)}{(\sinh z-\sinh x{)}^7}\mathrm{d}z.\end{array}$(6) ) and (7(7) $\lambda y\left(x\right)=-5{\cosh }^{5}x{\oint }_C\frac{\cosh \left(zs\right)\left(5{\sinh }^{2}z+\sinh z\sinh x+6\right)}{(\sinh z-\sinh x{)}^7}\mathrm{d}z.$(7) ), we see that $y^{″}\left(x\right)+(\lambda +20{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^{2}x)y\left(x\right)=0.$

Corollary 3.1

The factor $\cosh \left(zs\right)$ in (5(5) $y\left(x\right)={\cosh }^{5}x{\oint }_C\frac{\cosh \left(zs\right)}{(\sinh z-\sinh x{)}^5}\mathrm{d}z,$(5) ) plays little part in the argument. By replacing $\cosh \left(zs\right)$ by $\sinh \left(zs\right),$ the other linearly independent solution to Equation (2(2) $y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+\left(\lambda +20{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^{2}x\right)y\left(x\right)=0,$(2) ) is given by with $s^2=-\lambda$ (8) $y\left(x\right)={\cosh }^{5}x{\oint }_C\frac{\sinh \left(zs\right)}{(\sinh z-\sinh x{)}^5}\mathrm{d}z.$(8)

Theorem 3.2

Case for any n, see [2, 4]

The integral (8(8) $y\left(x\right)={\cosh }^{5}x{\oint }_C\frac{\sinh \left(zs\right)}{(\sinh z-\sinh x{)}^5}\mathrm{d}z.$(8) ) solves Equation (2(2) $y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+\left(\lambda +20{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^{2}x\right)y\left(x\right)=0,$(2) ).

Proof.

The proof is the same as Theorem 3.1. Hence, it is excluded.

4. Eigenpairs obtained by calculating the relevant residues

As a result of very laborious calculations, one obtains that the residue at $z=x$ for (5(5) $y\left(x\right)={\cosh }^{5}x{\oint }_C\frac{\cosh \left(zs\right)}{(\sinh z-\sinh x{)}^5}\mathrm{d}z,$(5) ) is $\begin{array}{rl}& \frac{\cosh \left(xs\right)(9+(15+45s^2-105){\tanh }^{2}x+s^4-10s^2+105{\tanh }^{4}x)}{4!(z-x{)}^5{\cosh }^{5}x}\\ & +\frac{s\sinh \left(xs\right)((55-10s^2)\tanh x-105{\tanh }^{3}x)}{4!(z-x{)}^5{\cosh }^{5}x}.\end{array}$ Hence, by using $s^2=-\lambda$, one solution is $\begin{array}{rl}{y}_1& =\cos \left(x\sqrt{\lambda }\right)(9+(-90-45\lambda ){\tanh }^{2}x+{\lambda }^2\\ & +10\lambda +105{\tanh }^{4}x)\\ & -\sqrt{\lambda }\sin \left(x\sqrt{\lambda }\right)((55+10\lambda )\tanh x-105{\tanh }^{3}x).\end{array}$ Similarly, by examining the residue of (8(8) $y\left(x\right)={\cosh }^{5}x{\oint }_C\frac{\sinh \left(zs\right)}{(\sinh z-\sinh x{)}^5}\mathrm{d}z.$(8) ), one obtains second solution as $\begin{array}{rl}{y}_2& =\sin \left(x\sqrt{\lambda }\right)(9+(-90-45\lambda ){\tanh }^{2}x+{\lambda }^2\\ & +10\lambda +105{\tanh }^{4}x)\\ & +\sqrt{\lambda }\cos \left(x\sqrt{\lambda }\right)((55+10\lambda )\tanh x-105{\tanh }^{3}x).\end{array}$ If $Y(x,s)=y_1\left(x\right)-y_2\left(x\right)$, where $y_1\left(x\right)$ and $y_2\left(x\right)$ are given by (5(5) $y\left(x\right)={\cosh }^{5}x{\oint }_C\frac{\cosh \left(zs\right)}{(\sinh z-\sinh x{)}^5}\mathrm{d}z,$(5) ) and (8(8) $y\left(x\right)={\cosh }^{5}x{\oint }_C\frac{\sinh \left(zs\right)}{(\sinh z-\sinh x{)}^5}\mathrm{d}z.$(8) ) respectively, then $Y(x,s)={\cosh }^{5}x{\oint }_C\frac{{\mathrm{e}}^{-sz}}{(\sinh z-\sinh x{)}^5}\mathrm{d}z.$

Theorem 4.1

Case for any n, see [2, 3]

$\begin{array}{rl}Y(0,s)& ={\oint }_C\frac{{\mathrm{e}}^{-sz}}{{\sinh }^{5}z}\mathrm{d}z\\ & =\frac{i\pi }{12}(3+s)(1+s)(s-1)(s-3),\end{array}$where the contour C is excluding all zeros of $\cos z$ except $z=\pi /2$. If $Y(0,s)=0,$ then $\lambda =-1,-9$.

Proof.

If $z=i(z-\pi /2)$, then $\begin{array}{rl}Y(0,s)& =-{\mathrm{e}}^{is\pi /2}{\oint }_C\frac{{\mathrm{e}}^{-isz}}{{\cos }^{5}z}\mathrm{d}z\\ & =I_1+I_2,\end{array}$ where

$\begin{array}{rl}{I}_1& =-{\mathrm{e}}^{is\pi /2}{\oint }_C\frac{\cos \left(sz\right)}{{\cos }^{5}z}\mathrm{d}z\\ & =-{\mathrm{e}}^{is\pi /2}{f}_5\left(s\right)\\ & =-{\mathrm{e}}^{is\pi /2}{\oint }_C\frac{\cos ((s-1)z+z)}{{\cos }^{5}z}\mathrm{d}z\\ & =-{\mathrm{e}}^{is\pi /2}{\oint }_C\frac{\cos (s-1)z\cos \left(z\right)-\sin (s-1)z\sin \left(z\right)}{{\cos }^{5}z}\mathrm{d}z,\end{array}$integration by parts $\begin{array}{rl}& =-{\mathrm{e}}^{is\pi /2}(1+\frac{s-1}{4})f_4(s-1)\\ & =-{\mathrm{e}}^{is\pi /2}\frac{3+s}{4}{f}_4(s-1),\end{array}$ by continuing this argument one gets (9) $I_1=-{\mathrm{e}}^{is\pi /2}\frac{3+s}{4}\frac{1+s}{3}\frac{s-1}{2}\frac{s-3}{1}{f}_1(s-4),$(9) where $\begin{array}{rl}{f}_1(s-4)& ={\oint }_C\frac{\cos (s-4)z}{\cos z}\mathrm{d}z\\ & =-2\pi i\cos \frac{(s-4)\pi }{2}.\end{array}$ After all, $\begin{array}{rl}{I}_1& =\frac{2i{\mathrm{e}}^{is\pi /2}\pi }{4!}(3+s)(1+s)(s-1)(s-3)\cos \frac{(s-4)\pi }{2}\end{array}$ and $\begin{array}{rl}{I}_2& =i{\mathrm{e}}^{is\pi /2}{\oint }_C\frac{\sin \left(sz\right)}{{\cos }^{5}z}\mathrm{d}z\\ & =i{\mathrm{e}}^{is\pi /2}{\hat{f}}_5\left(s\right)\\ & =i{\mathrm{e}}^{is\pi /2}{\oint }_C\frac{\sin ((s-1)z+z)}{{\cos }^{5}z}\mathrm{d}z\\ & =i{\mathrm{e}}^{is\pi /2}{\oint }_C\frac{\sin (s-1)z\cos \left(z\right)+\sin \left(z\right)\cos (s-1)z}{{\cos }^{5}z}\mathrm{d}z,\end{array}$ partial integral implies that $I_2=i{\mathrm{e}}^{is\pi /2}(1+\frac{s-1}{4}){\hat{f}}_4(s-1)$ keep continuing this argument, $I_2=i{\mathrm{e}}^{is\pi /2}\frac{3+s}{4}\frac{1+s}{3}\frac{s-1}{2}\frac{s-3}{1}{\hat{f}}_1(s-4),$ where $\begin{array}{rl}{\hat{f}}_1(s-4)& ={\oint }_C\frac{\sin (s-4)z}{\cos z}\mathrm{d}z\\ & =-2i\pi \sin \frac{(s-4)\pi }{2}.\end{array}$ That is, $I_2=\frac{2{\mathrm{e}}^{is\pi /2}\pi }{4!}(3+s)(1+s)(s-1)(s-3)\sin \frac{(s-4)\pi }{2}.$ By combining $I_1$ and $I_2$, one gets the expected result. $\begin{array}{rl}{I}_1+I_2& =\frac{2i{\mathrm{e}}^{is\pi /2}\pi }{4!}(3+s)(1+s)(s-1)(s-3)\cos \frac{(s-4)\pi }{2}\\ & +\frac{2{\mathrm{e}}^{is\pi /2}\pi }{4!}(3+s)(1+s)(s-1)(s-3)\sin \frac{(s-4)\pi }{2}\\ & =\frac{2i\pi {\mathrm{e}}^{is\pi /2}}{4!}{\mathrm{e}}^{-i(s-4)\pi /2}(3+s)(1+s)(s-1)(s-3)\\ & =\frac{i\pi }{12}{\mathrm{e}}^{2i\pi }(3+s)(1+s)(s-1)(s-3)\\ & =\frac{i\pi }{12}(3+s)(1+s)(s-1)(s-3).\end{array}$ If $Y\left(0\right)=0$, then $(3+s)(1+s)(s-1)(s-3)=0$. Hence, $s^2=1$ or $\lambda =-1$ and $s^2=9$ or $\lambda =-9$. That completes the proof.

Theorem 4.2

Case for any n, see [2, 3]

$\begin{array}{rl}{Y}^{\prime }(0,s)& =5{\oint }_C\frac{{\mathrm{e}}^{-sz}}{{\sinh }^{6}z}\mathrm{d}z\\ & =\frac{-i\pi }{12}(4+s)(2+s)s(s-2)(s-4)\end{array}$where the contour C is excluding all zeros of $\cos z$ except $z=\pi /2$. If $Y^{\prime }(0,s)=0,$ then $\lambda =-4,-16$.

Proof.

Proof is the same as Theorem 4.1. Hence, it is excluded. If $Y^{\prime }(0,s)=0$, then $s^2=4,16$ or $\lambda =-4,-16$. That completes the proof.

Theorem 4.3

Case for any n, see [2, 3]

$Y(x,s)\to 0\text{as}x\to \mathrm{\infty }$.

Proof.

Set z=iw and x=iy. So $\begin{array}{rl}Y(y,s)& ={\cos }^{5}y\int \frac{{\mathrm{e}}^{-isw}}{(\sin w-\sin y{)}^5}\mathrm{d}w\\ & =2^{-5}{\cos }^{5}y\int \frac{{\mathrm{e}}^{-isw}}{{\left(\sin \frac{w-y}{2}\cos \frac{w+y}{2}\right)}^5}\mathrm{d}w\\ & =\int \frac{{\mathrm{e}}^{-isw}}{(w-y{)}^5}\mathrm{d}w.\end{array}$ Let w−y=u and isu=v. Then one obtains $\begin{array}{rl}Y(y,\lambda )& =s^4{\mathrm{e}}^{-isy}\int {\mathrm{e}}^{-v}{v}^{-5}\mathrm{d}v\\ & =s^4{\mathrm{e}}^{-sx}\int {\mathrm{e}}^{-v}{v}^{-5}\mathrm{d}v.\end{array}$ Therefore, this integral approaches to zero as x goes to ∞.

5. Eigenfunction expansions

If $\theta (x,\lambda )$ and $\phi (x,\lambda )$ are the solutions of (2(2) $y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+\left(\lambda +20{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^{2}x\right)y\left(x\right)=0,$(2) ) and satisfying (3(3) $\begin{array}{rl}& \phi (0,\lambda )=\sin \alpha ,{\phi }^{\mathrm{\prime }}(0,\lambda )=-\cos \alpha ,\\ & \theta (0,\lambda )=\cos \alpha ,{\theta }^{\mathrm{\prime }}(0,\lambda )=\sin \alpha .\end{array}$(3) ). Then one finds (10) $\begin{array}{rl}\phi (x,\lambda )& =\frac{y_1(x,\lambda )\sin \alpha }{{\lambda }^2+10\lambda +9}-\frac{y_2(x,\lambda )\cos \alpha }{\sqrt{\lambda }({\lambda }^2+20\lambda +64)},\\ \theta (x,\lambda )& =\frac{y_1(x,\lambda )\cos \alpha }{{\lambda }^2+10\lambda +9}+\frac{y_2(x,\lambda )\sin \alpha }{\sqrt{\lambda }({\lambda }^2+20\lambda +64)}.\end{array}$(10) One needs to find spectral function $k\left(\lambda \right)$. To get desired result, we have to find the asymptotics of (10(10) $\begin{array}{rl}\phi (x,\lambda )& =\frac{y_1(x,\lambda )\sin \alpha }{{\lambda }^2+10\lambda +9}-\frac{y_2(x,\lambda )\cos \alpha }{\sqrt{\lambda }({\lambda }^2+20\lambda +64)},\\ \theta (x,\lambda )& =\frac{y_1(x,\lambda )\cos \alpha }{{\lambda }^2+10\lambda +9}+\frac{y_2(x,\lambda )\sin \alpha }{\sqrt{\lambda }({\lambda }^2+20\lambda +64)}.\end{array}$(10) ) as $x\to \mathrm{\infty }$. If $\mathrm{I}\mathrm{m}\left(\sqrt{\lambda }\right)>0$, then the asymptotics are (11) $\phi (x,\lambda )\sim \frac{{\mathrm{e}}^{-ix\sqrt{\lambda }}{M}_1\left(\lambda \right)}{2\sqrt{\lambda }({\lambda }^2+10\lambda +9)({\lambda }^2+20\lambda +64)},$(11) where $\begin{array}{rl}{M}_1\left(\lambda \right)& =\sqrt{\lambda }({\lambda }^2+20\lambda +64)({\lambda }^2-35\lambda +24)\sin \alpha \\ & -\sqrt{\lambda }(10\lambda -50)({\lambda }^2+10\lambda +9)\cos \alpha \\ & +i\{-\lambda (10\lambda -50\left)\right({\lambda }^2+20\lambda +64)\sin \alpha \\ & -({\lambda }^2-35\lambda +24)({\lambda }^2+10\lambda +9)\cos \alpha \}\end{array}$ and (12) $\theta (x,\lambda )\sim \frac{{\mathrm{e}}^{-ix\sqrt{\lambda }}M\left(\lambda \right)}{2\sqrt{\lambda }({\lambda }^2+10\lambda +9)({\lambda }^2+20\lambda +64)},$(12) where $\begin{array}{rl}M\left(\lambda \right)& =\sqrt{\lambda }({\lambda }^2+20\lambda +64)({\lambda }^2-35\lambda +24)\cos \alpha \\ & +\sqrt{\lambda }(10\lambda -50)({\lambda }^2+10\lambda +9)\sin \alpha \\ & +i\{-\lambda (10\lambda -50\left)\right({\lambda }^2+20\lambda +64)\cos \alpha \\ & +({\lambda }^2-35\lambda +24)({\lambda }^2+10\lambda +9)\sin \alpha \}.\end{array}$ After arranging that the linear combination of (11(11) $\phi (x,\lambda )\sim \frac{{\mathrm{e}}^{-ix\sqrt{\lambda }}{M}_1\left(\lambda \right)}{2\sqrt{\lambda }({\lambda }^2+10\lambda +9)({\lambda }^2+20\lambda +64)},$(11) ) and (12(12) $\theta (x,\lambda )\sim \frac{{\mathrm{e}}^{-ix\sqrt{\lambda }}M\left(\lambda \right)}{2\sqrt{\lambda }({\lambda }^2+10\lambda +9)({\lambda }^2+20\lambda +64)},$(12) ), then the terms ${\mathrm{e}}^{-ix\sqrt{\lambda }}$ cancel out. $m\left(\lambda \right)=\left\{\begin{array}{ll}-\frac{M\left(\lambda \right)}{M_1\left(\lambda \right)}& \mathrm{w}\mathrm{h}\mathrm{e}\mathrm{n}\lambda >0\\ 0& \mathrm{w}\mathrm{h}\mathrm{e}\mathrm{n}\lambda <0.\end{array}\right.$ Hence, the spectrum is $\begin{array}{rl}k\left(\lambda \right)& =-\mathrm{I}\mathrm{m}(m\left(\lambda \right))\\ & =\left\{\begin{array}{ll}\frac{\sqrt{\lambda }({\lambda }^2+10\lambda +9)({\lambda }^2+20\lambda +64)}{\lambda ({\lambda }^2+20\lambda +64{)}^2{\sin }^2\alpha +({\lambda }^2+10\lambda +9{)}^2{\cos }^2\alpha },& \lambda >0\\ 0,& \lambda <0.\end{array}\right.\end{array}$ In particular, if $\alpha =0$ and $\lambda >0$ then one gets the continuous spectrum as $\begin{array}{rl}k\left(\lambda \right)& =-\mathrm{I}\mathrm{m}(m\left(\lambda \right))\\ & =\left\{\begin{array}{ll}\frac{\sqrt{\lambda }(\lambda +4)(\lambda +16)}{(\lambda +1)(\lambda +9)},& \lambda >0\\ 0,& \lambda <0.\end{array}\right.\end{array}$ Hence, with $\alpha =0$, there exist two eigenvalues occurring at $\lambda =-1$ and $\lambda =-9$ and their associated eigenfunctions are (13) $\begin{array}{rl}\phi (x,-1)& =\frac{\begin{array}{c}\sinh x\{45{\tanh }^{2}x-105{\tanh }^{4}x\}\\ -\cosh x\{45\tanh x-105{\tanh }^{3}x\}\end{array}}{45},\\ \phi (x,-9)& =\sinh \left(3x\right)\{3{\tanh }^{2}x+{\tanh }^{4}x\}\\ & -\cosh \left(3x\right)\{\tanh x+3{\tanh }^{3}x\},\end{array}$(13) where $\phi (x,\lambda )$ is (10(10) $\begin{array}{rl}\phi (x,\lambda )& =\frac{y_1(x,\lambda )\sin \alpha }{{\lambda }^2+10\lambda +9}-\frac{y_2(x,\lambda )\cos \alpha }{\sqrt{\lambda }({\lambda }^2+20\lambda +64)},\\ \theta (x,\lambda )& =\frac{y_1(x,\lambda )\cos \alpha }{{\lambda }^2+10\lambda +9}+\frac{y_2(x,\lambda )\sin \alpha }{\sqrt{\lambda }({\lambda }^2+20\lambda +64)}.\end{array}$(10) ). As a result of that, by using Lemma 2.1, one writes expansion of $f\left(x\right)$ on $(0,\mathrm{\infty })$ with $\lambda <0$ as $\begin{array}{rl}f\left(x\right)& ={\int }_0^{\mathrm{\infty }}\phi (x,\lambda )\mathrm{d}k\left(\lambda \right){\int }_0^{\mathrm{\infty }}\phi (y,\lambda )f\left(y\right)\mathrm{d}y\\ & +c_1\phi (x,-1)+c_2\phi (x,-9).\end{array}$ In particular, if $\alpha =\pi /2$, then $k\left(\lambda \right)=-\mathrm{I}\mathrm{m}(m\left(\lambda \right))=\left\{\begin{array}{ll}\frac{(\lambda +1)(\lambda +9)}{\sqrt{\lambda }(\lambda +4)(\lambda +16)},& \lambda >0\\ 0,& \lambda <0.\end{array}\right.$ Hence, the eigenvalues are occurring at $\lambda =-4$ and $\lambda =-16$ and their associated eigenfunctions are (14) $\begin{array}{rl}\phi (x,-4)& =-\cosh \left(2x\right)\{6{\tanh }^{2}x-1+7{\tanh }^{4}x\}\\ & -2\sinh \left(2x\right)\{\tanh x-7{\tanh }^{3}x\},\\ \phi (x,-16)& =\cosh \left(4x\right)\{1+6{\tanh }^{2}x+{\tanh }^{4}x\}\\ & -4\sinh \left(4x\right)\{\tanh x+{\tanh }^{3}x\}.\end{array}$(14) So we once again conclude that if $\phi (x,\lambda )$ is (10(10) $\begin{array}{rl}\phi (x,\lambda )& =\frac{y_1(x,\lambda )\sin \alpha }{{\lambda }^2+10\lambda +9}-\frac{y_2(x,\lambda )\cos \alpha }{\sqrt{\lambda }({\lambda }^2+20\lambda +64)},\\ \theta (x,\lambda )& =\frac{y_1(x,\lambda )\cos \alpha }{{\lambda }^2+10\lambda +9}+\frac{y_2(x,\lambda )\sin \alpha }{\sqrt{\lambda }({\lambda }^2+20\lambda +64)}.\end{array}$(10) ) and $\alpha =\pi /2$, by using Lemma 2.1, then $\begin{array}{rl}f\left(x\right)& ={\int }_0^{\mathrm{\infty }}\phi (x,\lambda )\mathrm{d}k\left(\lambda \right){\int }_0^{\mathrm{\infty }}\phi (y,\lambda )f\left(y\right)\mathrm{d}y\\ & +c_3\phi (x,-4)+c_4\phi (x,-16).\end{array}$ Thus, one has formally proved the following theorem.

Theorem 5.1

If $\alpha =0$, then there exist only two eigenvalues occurring at $\lambda =-1$ and $\lambda =-9$ and their associated eigenfunctions $\phi (x,-1)$ and $\phi (x,-9)$ are given in (13(13) $\begin{array}{rl}\phi (x,-1)& =\frac{\begin{array}{c}\sinh x\{45{\tanh }^{2}x-105{\tanh }^{4}x\}\\ -\cosh x\{45\tanh x-105{\tanh }^{3}x\}\end{array}}{45},\\ \phi (x,-9)& =\sinh \left(3x\right)\{3{\tanh }^{2}x+{\tanh }^{4}x\}\\ & -\cosh \left(3x\right)\{\tanh x+3{\tanh }^{3}x\},\end{array}$(13) ).

If $\alpha =\pi /2$, then there exist only two eigenvalues occurring at $\lambda =-4$ and $\lambda =-16$ and their associated eigenfunctions $\phi (x,-4)$ and $\phi (x,-16)$ are given in (14(14) $\begin{array}{rl}\phi (x,-4)& =-\cosh \left(2x\right)\{6{\tanh }^{2}x-1+7{\tanh }^{4}x\}\\ & -2\sinh \left(2x\right)\{\tanh x-7{\tanh }^{3}x\},\\ \phi (x,-16)& =\cosh \left(4x\right)\{1+6{\tanh }^{2}x+{\tanh }^{4}x\}\\ & -4\sinh \left(4x\right)\{\tanh x+{\tanh }^{3}x\}.\end{array}$(14) ).

Consider the interval $(-\mathrm{\infty },\mathrm{\infty })$. Since $q\left(x\right)$ is an even function, we use definition of ${\xi }^{\prime }\left(\lambda \right)$ and ${\zeta }^{\prime }\left(\lambda \right)$ from Lemma 2.2, one gets (15) $\begin{array}{rl}& {\xi }^{\prime }\left(\lambda \right)=\left\{\begin{array}{ll}\frac{2^{-1}\left\{\sqrt{\lambda }\right({\lambda }^2+20\lambda +64\left)\right({\lambda }^2+10\lambda +9\left)\right\}}{\lambda ({\lambda }^2+20\lambda +64{)}^2{\cos }^2\alpha +({\lambda }^2+10\lambda +9{)}^2{\sin }^2\alpha },& \lambda >0,\\ 0,& \lambda <0,\end{array}\right.\\ & {\zeta }^{\prime }\left(\lambda \right)=\left\{\begin{array}{ll}\frac{2^{-1}\left\{\sqrt{\lambda }\right({\lambda }^2+20\lambda +64\left)\right({\lambda }^2+10\lambda +9\left)\right\}}{\lambda ({\lambda }^2+20\lambda +64{)}^2{\sin }^2\alpha +({\lambda }^2+10\lambda +9{)}^2{\cos }^2\alpha },& \lambda >0,\\ 0,& \lambda <0.\end{array}\right.\end{array}$(15) If so, by using Lemma 2.2, one gets $\begin{array}{rl}f\left(x\right)& =\frac{1}{\pi }\{{\int }_0^{\mathrm{\infty }}\theta (x,\lambda )\mathrm{d}\xi \left(\lambda \right){\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\theta (y,\lambda )f\left(y\right)\mathrm{d}y\\ & +{\int }_0^{\mathrm{\infty }}\phi (x,\lambda )\mathrm{d}\zeta \left(\lambda \right){\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\phi (y,\lambda )f\left(y\right)\mathrm{d}y\}\\ & +c_1\phi (x,-1)+c_2\phi (x,-9)\\ & +c_3\phi (x,-4)+c_4\phi (x,-16),\end{array}$

where $c_1$, $c_2$, $c_3$ and $c_4$ are constants. $\phi (x,\lambda )$, $\theta (x,\lambda )$, ${\xi }^{\prime }\left(\lambda \right)$ and ${\zeta }^{\prime }\left(\lambda \right)$ are given by (10(10) $\begin{array}{rl}\phi (x,\lambda )& =\frac{y_1(x,\lambda )\sin \alpha }{{\lambda }^2+10\lambda +9}-\frac{y_2(x,\lambda )\cos \alpha }{\sqrt{\lambda }({\lambda }^2+20\lambda +64)},\\ \theta (x,\lambda )& =\frac{y_1(x,\lambda )\cos \alpha }{{\lambda }^2+10\lambda +9}+\frac{y_2(x,\lambda )\sin \alpha }{\sqrt{\lambda }({\lambda }^2+20\lambda +64)}.\end{array}$(10) ) and (15(15) $\begin{array}{rl}& {\xi }^{\prime }\left(\lambda \right)=\left\{\begin{array}{ll}\frac{2^{-1}\left\{\sqrt{\lambda }\right({\lambda }^2+20\lambda +64\left)\right({\lambda }^2+10\lambda +9\left)\right\}}{\lambda ({\lambda }^2+20\lambda +64{)}^2{\cos }^2\alpha +({\lambda }^2+10\lambda +9{)}^2{\sin }^2\alpha },& \lambda >0,\\ 0,& \lambda <0,\end{array}\right.\\ & {\zeta }^{\prime }\left(\lambda \right)=\left\{\begin{array}{ll}\frac{2^{-1}\left\{\sqrt{\lambda }\right({\lambda }^2+20\lambda +64\left)\right({\lambda }^2+10\lambda +9\left)\right\}}{\lambda ({\lambda }^2+20\lambda +64{)}^2{\sin }^2\alpha +({\lambda }^2+10\lambda +9{)}^2{\cos }^2\alpha },& \lambda >0,\\ 0,& \lambda <0.\end{array}\right.\end{array}$(15) ) respectively.

6. Conclusion

The explicit solutions of the Pöschl–Teller differential equation (2(2) $y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+\left(\lambda +20{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^{2}x\right)y\left(x\right)=0,$(2) ) and its eigenpairs are obtained by calculating complex residues. Eigenfunction expansions for arbitrary function $f\left(x\right)$ satisfying suitable conditions are also obtained. We concluded that differential equation (2(2) $y^{\mathrm{\prime }\mathrm{\prime }}\left(x\right)+\left(\lambda +20{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}}^{2}x\right)y\left(x\right)=0,$(2) ) has oscillating solutions which are not reported in the literature.

Disclosure statement

No potential conflict of interest was reported by the author.

ORCID

Tanfer Tanriverdi  http://orcid.org/0000-0003-4686-1263