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I have derived two nonlinear parabolic equation as $\begin{align*} \frac{\partial S}{\partial t}&=a\exp\left(\frac{x-b}{c}\right)^2\frac{\partial^2 S}{\partial x^2} \tag{1}\\ \frac{\partial S}{\partial t}&=a\exp\left(\frac{x-b}{c}\right)^2\frac{\partial}{\partial x}\left(S\frac{\partial S}{\partial x}\right) \tag{2}\\ \end{align*}$ I wonder if anyone can give a detail analysis for exact solution. Thank you!

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    For (1), in fact it is better to use separation of variables rather than the "integral kernel method", so I make large edits of my answer now. Sorry for making you misleading for many months.2012-11-30

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For (1), note that it is a linear PDE.

First have a "warm-up" by using separation of variables:

Let $ S(x,t)=X(x)T(t) $, Then $X(x)T'(t)=ae^{\left(\frac{x-b}{c}\right)^2}X''(x)T(t)$ $\dfrac{T'(t)}{T(t)}=\dfrac{ae^{\left(\frac{x-b}{c}\right)^2}X''(x)}{X(x)}=f(s)$

$ \begin{cases}\dfrac{T'(t)}{T(t)}=f(s) \\ae^{\left(\frac{x-b}{c}\right)^2}X''(x)-f(s)X(x)=0 \end{cases} $ Therefore $ae^{\left(\frac{x-b}{c}\right)^2}\dfrac{\partial^2K(x,s)}{\partial x^2}-s K(x,s)=0 $ For complying the conditions $S(0,t)=0$ and $\dfrac{\partial S}{\partial x}(L,t)=0$ , You should take the solution as $S(x,t)=\sum\limits_sC_1(s)e^{tf(s)}X_1(x,s)$, where $X_1(x,s)$ is some or all solutions of $ae^{\left(\frac{x-b}{c}\right)^2}X''(x)-f(s)X(x)=0$ that satisfies $X(0)=0$ and $X'(L)=0$ .

But to solve $ae^{\left(\frac{x-b}{c}\right)^2}X''(x)-f(s)X(x)=0$ is just like to solve second-order linear ODE with general variable coefficients and is very complicated. I provide this article to you to have deep investigation on this issue.

For (2), I have no idea.

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    I mean to simplify the final formula to make it feasible for practical use. So how to difine the values of s maybe a key question. By the way, do you have some references about the integral and summation kernel method?2012-06-30