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Look at $u_t=a(x)u_{xx}$ if I have $a(x)\geq a_0>0$ then I can see in all books that $C^{2,1}$ solution exist and it is unique. However, if $a(x)\geq0$, that is degenerate, I see in Friedman's book the construction of $K_{\epsilon}$: sequence convergent uniformly to $K$ which is the solution of the degenerate equation. But he doesn't state the property of the latter. What are the problems of those equations? Looks like I have problem constructing a weak solution because I lack coercivity property, so I have little hope to have it $C^{2,1}$. But what class a solution of the degenerate equation belongs to? thnaks!

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    Certainly you cannot have $C^{2,1}$ property, still you can expect some continuity for the solution(or weak solution). There is some result for it.[link](http://math.tkk.fi/~ttkuusi/articles/Cauchy_Kuusi_Parviainen.pdf)2012-05-08
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    thanks for the link, I will look into it. They consider a Cauchy problem, so if I put smooth boundary condition and make IVBP, would that make much difference?2012-05-08
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    I don't think it will help much, think about 1-dim case.2012-05-08
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    the reason I am asking is that I have a pde and I solve it numerically, so it is IBVP. And I see it converges to something and I claim it is a solution. However, I need to justify that there is one and that degeneracy is causing problems for me because it is very hard to show...2012-05-08
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    It depends on your $a(x,t)$, imagine if $a(x)=0$ on a interval, then $u_t=0$ on this interval, there is no uniqueness for our solution, even provided IVBP.2012-05-08
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    no, $a(x,t)=x^2$, so the degeneracy only when $x=0$, so it is a point on the boundary...2012-05-08
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    There is a book called "degenerate parabolic equations", maybe it can help, but the equations require some regularity on $a(x,t,u,Du)$, which $a(x)=x^2$ does not satisfy.....there are also some results about your problem when $a(x)=x^\alpha$, however, they required $\frac{1}{\sqrt{a}}\in L^1(0,1)$.2012-05-08
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    @Yimin how come there is no uniqueness if $a(x) = 0$? This just means that $u(t,x) = f(x)$ where $f$ is the initial condition or am I missing something? Also, I think the example shows that best you can hope for is to preserve the regularity of the initial condition...2012-05-08
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    @PZZ yes! if $a(x)=0$ on an interval, then the temperature $u(t,x)$ will be unchanged on the interval.It is unique in this case if the domain is finite.2012-05-08
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    @PZZ I still believe the uniqueness for this degenerated case need conditions on $a(x)$, even $a(x)$ is smooth enough. But I cannot find a counterexample.2012-05-08

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This is not a complete answer, but is too long to be a comment. It is an analysis of the pure Cauchy problem in the case $a(x)=x^2$. Using semigroup theory, it can be proved that the problem is well posed on the space $$ X=\Bigr\{u\colon\mathbb{R}\to\mathbb{R}:\int_{-\infty}^\infty|u(x)|^2\frac{dx}{x^2}<\infty\Bigl\}. $$ The fact that the initial value $u_0(x)=u(x,0)\in X$ implies that $u_0$ must vanish to a certain order at $x=0$.

The equation $u_t-x^2u_{xx}=0$ can be transformed into the heat equation by means of a change of variable. Consider first the region $x>0$. The change $x=e^{-z}$ transforms the equation into $u_t-u_z-u_{zz}=0$, $u(z,0)=u_0(e^{-z})$. Now if $v(z,t)=u(z-t,t)$, then $v_t-v_{zz}=0$ and $v(z,0)=u_0(e^{-z})$. A similar computation can be caried out for $x<0$. This allows to prove for instance that if $$ \lim_{x\to0}x\,u_0'(x)=\lim_{x\to0}x^2\,u_0''(x)=0,\tag{1} $$ then there exists a unique classical solution.

Some remarks are in order. We have $u(0,t)=u_0(0)$ for all $t>0$. Moreover, $x=0$ acts as a barrier to diffusion. If $u_0$ has compact support contained in $(0,\infty)$, then $u(x,t)=0$ for all $x<0$ and $t>0$. This is in sharp contrast with the heat equation. The equation is not regularizing at $x=0$. The solution is $C^{2,1}$ 0n $(\mathbb{R}\setminus\{0\})\times(0,\infty)$, but only as regular as $u_o$ at $x=0$. What happens if $u_0$ does not satisfy (1)? The solution can still be constructed, but in general it will be a solution only in a weak sense. For instance, if $u_0(x)=\sin(-\log|x|)$, then $u(x,t)=e^{-t} \sin(-\log|x|-t)$, which is discontinuous at $x=0$ for all $t>0$.

Finally, can something be said for the more general equation $u_t-|x|^\alpha u_{xx}=0$, $\alpha\ge0$? If $0\le\alpha<2$, then the operator $|x|^\alpha u''$ has some compactnes properties, and it is possible to obtain some results. If $\alpha>2$, very little is known.

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    Julian, thanks for your answer. I got some idea and keep looking for the theory behind that. You mentioned semigroup theory for stating the space where the problem is well-posed. Is there a paper or a book you might suggest I can find such analysis? I am working on an algorithm for solving 2d equations of type $u_t-x^2u_{xx}-xu_x-u_y=0$, which can be converted to $u_t-u_{zz}-u_y=0$ so it is pure degenerate and I find the solution numerically on $[0,x_{max}]x[0,y_{max}]$, but I have to work thorough the well-posedeness over that interval. Thus, any reference would be of a great help!2012-05-11
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    There is this paper: THE SHAPE OF BLOW-UP FOR A DEGENERATE PARABOLIC EQUATION, J. Aguirre & J. Giacomoni,Differential and Integral Equations, Volume 14, Number 5, May 2001, Pages 589–604, and the references therein. In the years since I worked on this there might be new results. I do not think it is available on-line. If you send me an email, I can send you a copy.2012-05-11
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    Dear Julian can I have your email address, I cannot find it on the web and I would like a copy of this paper2014-08-31
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    @math101 julian.aguirre@ehu.es2014-09-02