Validating solution to PDE using integral transforms

I'm trying to obtain the analytical solution of a Fokker-Planck PDE, which the solution is a probability density function, and then use this to find the mean of some quantity in the paper. The paper has a solution which they say can be found via Mehler-Fock transform. Their solution which I am trying to obtain reads

$$P_{\varepsilon}(L,u) = \frac{e^{-\varepsilon^2sL/4}}{2\sqrt{2\pi}(\varepsilon^2sL)^{\frac32}}\int_{u}^{\infty}\frac{xe^{-x^2/(\varepsilon^2sL)}}{\sqrt{\cosh{(x)}-\cosh{(u)}}}dx.$$

I obtain a different solution to this one.

$$\textbf{My attempt:}$$ The equation in the paper reads \begin{align} \frac{\partial P_\varepsilon}{\partial L} = \varepsilon^2s\frac{\partial}{\partial u}\bigg[(u^2-1)\frac{\partial P_\varepsilon}{\partial u}\bigg], \quad P_\varepsilon(L=0,u) = \delta(u-1). \end{align} This is a $$1d$$ Fokker-Planck equation, where $$P_\varepsilon(L,u)$$ is a probability density for some diffusion Markov process. To solve this equation, note that the right hand side is simply a Legendre differential equation. Denote the legendre function of the first kind via \begin{align} \frac{d}{du}(u^2-1)\frac{d}{du}P_{-\frac12+i\mu}(u) = -\bigg(\mu^2+\frac14\bigg)P_{-\frac12+i\mu}(u), \end{align} which has an integral representation \begin{align} P_{-\frac12+i\mu}(u) = \frac{\sqrt{2}}{\pi}\cosh{(\pi\mu)}\int_{0}^{\infty}\frac{\cos{(\mu\tau)}}{\sqrt{\cosh{(\tau)}+u}}d\tau. \end{align} Now use the Mehler-Fock transform. The Mehler-Fock transform of an integrble function $$f$$ defined on $$[1,\infty)$$ is the function $$\check f$$ defined on $$[0,\infty)$$ where \begin{align} \hat{f}(\mu) = \int_{1}^{\infty}f(u) P_{-\frac12+i\mu}(u)\,du, \end{align} with inverse transform \begin{align} f(u) = \int_{0}^{\infty}\check{f}(\mu)\mu\tanh{(\mu\pi)}P_{-\frac12+i\mu}(u)\,d\mu. \end{align} Applying the Mehler-Fock transform to the PDF $$P_\varepsilon(L,u)$$ gives \begin{align} \check p(L,\mu) = \int_{1}^{\infty}p(L,u)P_{-\frac12+i\mu}(u)du. \end{align} Taking a partial derivative in $$L$$ gives \begin{align} \frac{\partial\check p}{\partial L}(L,\mu) = \varepsilon^2s\int_{1}^{\infty}\frac{\partial}{\partial u}\bigg[(u^2-1)\frac{\partial p}{\partial u}(L,u)\bigg]P_{-\frac12+i\mu}(u)du. \end{align} Integrating twice more by parts gives \begin{align} \frac{\partial\check p}{\partial L}(L,\mu) = \varepsilon^2s\int_{1}^{\infty}p(L,u)\frac{\partial}{\partial u}\bigg[(u^2-1)\frac{\partial P_{-\frac12+i\mu}}{\partial u}(u)\bigg]du. \end{align} Using the ODE $$(2)$$ which satisfies the Legendre function, the Mehler-Fock transform satisfies the ODE \begin{align} \frac{\partial \check p}{\partial L}(L,\mu) = -\varepsilon^2s\bigg(\mu^2+\frac14\bigg)\check p(L,\mu), \quad \check p(L=0,\mu) = 1. \end{align} Then \begin{align} \check p(L,\mu) = \exp{\bigg(-\bigg(\mu^2+\frac14\bigg)L\varepsilon^2s\bigg)}. \end{align} Hence, the solution to $$(1)$$ is \begin{align}\nonumber P_\varepsilon(L,u) &= \int_{0}^{\infty}\mu\tanh{(\mu\pi)}P_{-\frac12+i\mu}(u)\exp{\bigg(-\bigg(\mu^2+\frac14\bigg)L\varepsilon^2s\bigg)}d\mu \\ &= \int_{0}^{\infty}\mu\tanh{(\mu\pi)}P_{-\frac12+i\mu}(u) = \frac{\sqrt{2}}{\pi}\cosh{(\pi\mu)}\int_{0}^{\infty}\frac{\cos{(\mu\tau)}}{\sqrt{\cosh{(\tau)}+u}} \\ &\times\exp{\bigg(-\bigg(\mu^2+\frac14\bigg)L\varepsilon^2s\bigg)}d\tau d\mu. \end{align}

This paper claims to have a solution

$$P_{\varepsilon}(L,u) = \frac{e^{-\varepsilon^2sL/4}}{2\sqrt{2\pi}(\varepsilon^2sL)^{\frac32}}\int_{u}^{\infty}\frac{xe^{-x^2/(\varepsilon^2sL)}}{\sqrt{\cosh{(x)}-\cosh{(u)}}}dx.$$ Here I have computed my solution vs theirs, for varying parameter $$L\in[1,10]$$ and fixed $$\varepsilon^2s$$

Clearly they do not agree.

$$\textbf{My second question}$$ is how they use this solution to find a mean value via

\begin{align}\mathbb{E(R)} &= \int_{1}^{\infty}\bigg(\frac{u-1}{u+1}\bigg)P_{\varepsilon}(L,u)du \\ &= \frac{e^{(-\varepsilon^2sL/4)}}{2\sqrt{2\pi}(\varepsilon^2Ls)^{\frac32}}\int_{1}^{\infty}\bigg(\frac{u-1}{u+1}\bigg)\int_{u}^{\infty}\frac{xe^{-x^2/(4\varepsilon^2sL)}}{\sqrt{\cosh{(x)}-\cosh{(u)}}}dxdu \end{align}

$$\textbf{AND then reduce this expression to read}$$

$$\mathbb{E(R)}=1-\frac{4}{\sqrt{\pi}}e^{-\varepsilon^2sL}\int_{0}^{\infty}\frac{x^2e^{-x^2}}{\cosh{(\sqrt{\varepsilon^2sL}x)}}dx,$$ since it's easy to see asymptotically that they behave differently. I'm looking for help in either of these questions. Thanks for the help in advance.

• Any suggestions? I can link the paper if that would help but this is all the information given. Feb 24, 2020 at 15:55
• I would be interested to see the link to the paper. Feb 24, 2020 at 18:38
• @ThomasL I've added a link. The equation in question starts at (3.9) on page 5. Feb 24, 2020 at 18:50
• @ThomasL I've added some numerical plots. The solutions don't agree. Something strange is going on here. It's highly cited so I'm not sure what to make of it. Feb 25, 2020 at 12:21
• thanks for your info. I tried a couple of transformations (substitutions), but could not get any closer to the required formula. I noticed that the function is symmetrical to 0 in regards to x, but it does not really help. I highly doubt this is a one step integration. Based on your feedback, I give up at that point, thanks. Feb 25, 2020 at 12:38