Q:

how to solve a nonlinear parabolic equations?

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!

A:

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.

