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There is a well established theory for the uniformly parabolic equations, i.e. for equations of the form $$u_t=a(x,t)u_{xx}, x \in D, t\in (0,T], u(x,0)=u_0$$ when $a(x,t)\geq a_0 >0$. In fact, if write the fundamental solution for that (i.e. the Green's function) in the special case of $a(x,t)\equiv a_0$, I can see that $a_0$ and $t$ both are present in the denominator(or in several dimension it is a determinant). However, if $a(x,t)$ takes value $0$ for some $t>0$, for example, if $a(x,t)=x(x-1)$, then it doesn't fall into the class of uniformly parabolic p.d.e. Thus, I have two questions I am confused about:

  1. Does the fundamental solution in fact in this case has a similar form as in the case of constant coefficients and similar to heat density with minimum of $a(x,t)$ being in the denominator? Because if it is the case, then the fundamental solution becomes a delta function for any $t>0$. Is there a problem with a fundamental solution then? Or existence of it?

  2. Might be a bit of a general question but can someone provide an example or simple explanation why "degenerate" equations of high importance. Is that because the tools used for proving existence and uniqueness in the case of uniformly parabolic equations are not valid or what's the reason behind a separate theory for those equations? What are the tipical issues with those equations?

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2 Answers 2

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1.) Without some conditions on the coefficient $a(x,t)$, the question is too general. Let's assume for simplicity that $a$ depends only on $x$. If $a$ is allowed to take negative values, then, on the region $\{a<0\}$ the equation is not well posed. If $a\ge0$, something might be said, but as far as I know, there is no general theory. If $a=0$ oh an interval $(a,b)$, then the solution $u$ is constant on $(a,b)\times(0,T]$. If $a(x)=|x|^n$, the equation has different behavior for different values of $n$.

2.) Another reason this type of problem is interesting are semi-linear equations like $u_t=f(u)\,u_{xx}$. The equation becomes degenerate where $f(u)$ vanishes. Even more interesting are equations in divergence form $u_t=(f(u)\,u_x)_x$.

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I am interested in a linear case, for example when $a(x,t)=x^2$. Yes, this is trivial and the solution can be found by the change of variables but the point is to start with a simple example. That is, the solution is symmetric on $(-\infty,0)$ and $(0,\infty)$, so instead of considering the whole where we construct the solution we can look only at the $[0,\infty)$. So, my question is still the same: how do we construct the solution there? Do we follow the same arguments as in uniformly parabolic case or if not why we can't? –  Medan Nov 15 '12 at 1:49
    
All the usual estimates for parabolic equations will fail where $a=0$. –  Julián Aguirre Nov 15 '12 at 8:17
    
However, what if $a(x)=x^2,\; D=(\epsilon,\infty)$ I have boundedness of the $a$ in the domain $D$, in fact it is $\inf a = \epsilon^2$, so my equation is uniformly parabolic there and I can find a solution in a usual way, so the problem is to extending the domain to $0$ and finding the solution on the extended domain or it is not only that?? –  Medan Nov 17 '12 at 23:02
    
Assuming $a\ge0$, a better approach is the following. Given $\epsilon>0$ let $a_\epsilon=a(x,t)+\epsilon$. The equation $v_t=a_\epsilon v_{xx}$ is uniformly parabolic, and has a unique solution $v_\epsilon$ such that $v_\epsilon(x,0)=u_0(x)$ (this way you do not have to worry about boundary values at $x=\epsilon$.) Now you would like to prove that $v_\epsilon$ (or a subsequence) converges in some sense to a solution $u$ of the original problem. For this you usually need uniform estimates on $v_\epsilon$, which may be lacking if $a$ vanishes somewhere. –  Julián Aguirre Nov 18 '12 at 11:38

In fact the Green's function of $u_t=a(x,t)u_{xx}$ is the solution that satisfy $u_t=a(x,t)u_{xx}$ with the condition $u(x,0)=\delta(x)$ .

The form of $a(x,t)$ will greatly influence the difficulties of finding the Green's function.

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