how to solve a nonlinear parabolic equations?

Question

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!

Answer 1

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 $X(x)T'(t)=ae^{\left(\frac{x-b}{c}\right)^2}X''(x)T(t)$

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

$\begin{cases}\dfrac{T'(t)}{T(t)}=f(s)\\ae^{\left(\frac{x-b}{c}\right)^2}X''(x)-f(s)X(x)=0\end{cases}$

$\therefore ae^{\left(\frac{x-b}{c}\right)^2}\dfrac{\partial^2K(x,s)}{\partial x^2}-sK(x,s)=0$

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.