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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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Okay, twice now you've taken the material I tried to make legible, and messed it up. Perhaps you can decide what it is you want to write and, if you don't know how to typeset, just let others do it for you? –  Arturo Magidin Jun 8 '12 at 3:35
    
Is the second line supposed to be just a restatement of the first one? If so, there is a difference between $S_{xx}$, which means $\frac{\partial^2}{\partial x^2}S$, and $\frac{\partial}{\partial x}\left(S\frac{\partial S}{\partial x}\right)$. If they are not meant to be the same equation, then you are asking that $S_xx = \frac{\partial}{\partial x}(S\frac{\partial S}{\partial x})$, which seems rather weird. –  Arturo Magidin Jun 8 '12 at 3:41
    
Be aware that multiple editing will turn your question into a Community Wiki question. You are skirting close to the limit already. –  Arturo Magidin Jun 8 '12 at 4:02
    
For question (2), because of the evil coefficient $ae^{\left(\frac{x-b}{c}\right)^2}$ exist, eqworld.ipmnet.ru/en/solutions/npde/npde1209.pdf might be the slim chance, although this is already the closest case for question (2) basing on eqworld.ipmnet.ru/en/solutions/npde/npde-toc1.htm and the whole EqWorld site, as it suggested to use combination of variables. –  doraemonpaul Jun 15 '12 at 5:25
    
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. –  doraemonpaul Nov 30 '12 at 1:59

1 Answer 1

up vote 1 down vote accepted

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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Thank you for your reply to my question. But I have little knowledge about kernel method and I wish you can provide me a reference for your derivation. You use a integration to subsititue S(x,t) and suggest possible ways for solving the K(x,s)'s ODE equation. My question is how to solve the one dimentional ODE equation with two variables. The boundary and intial conditions for my original equations are S=0 at x=0 & t>=0, dS/dx=0 at x=L & t>=0, S=D at x=L & t=0. –  qaz Jun 8 '12 at 18:22
    
why m and n are chosen arbitrarily in this case? –  qaz Jun 12 '12 at 16:10
    
The reason is simple. Since the integral kernel in this case is for transformation of $t$ only. As the coefficient of the terms of PDE in this case does not contain functions of $t$ , integration by parts are not involved in the processes. That's why $m$ and $n$ are chosen arbitrarily in this case. –  doraemonpaul Jun 13 '12 at 22:33
    
Thank you so much. –  qaz Jun 14 '12 at 15:01
    
I have derived the similar equations according to your suggestions. While S in my case is decreasing with time, so can I substitute exp(ts) into exp(-ts) in the integral formula. I am not sure about the integral kernel method and the summation kernel method. Wish you can recommend some reference about these two methods. In addition, can I adopt m=n=s in my case? –  qaz Jun 29 '12 at 20:53

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