Consider the nonlinear heat equation $u_{t} = \Delta(u^{4})$ in $\{x \in \mathbb{R}^{3}: |x| < 1\}$ with $u = 0$ on $\{x \in \mathbb{R}^{3}: |x| = 1\}$. The problem I am working on is to show that all solutions to this PDE tend to zero (pointwise) as $t \rightarrow \infty$.

The equation looks like the porous medium equation so I thought about looking at how the Barenblatt solution is derived, but the issue is that this doesn't cover all solutions, so I'm thinking that there is a general argument to prove the above result.


Let $v = u^4$ and rewrite the PDE in terms of $v$ to get: $$ v^{-3/4} v_t = \Delta v. $$ Let $\;v = A(t) B(x,y,z)\;$ and separate variables to get: $$ A' = 4 A^{7/4} c \quad \mbox{and} \quad B^{3/4} \Delta B = c \quad \mbox{for a constant } c . $$ The DE in $A$ has the solution $\;A(t)^{3/4} = -\frac{4/3}{4 c t + k}\;$ for a constant $k$. So when $t$ goes to infinity $A$ goes to $0$ and therefore also $v$ and $u$.

  • $\begingroup$ You should format your answer using $\rm\LaTeX$. It's a bit hard to read as is. $\endgroup$ – Cameron Williams Apr 22 '15 at 3:52
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    $\begingroup$ You've done a good job on the separable solutions; I upvoted. But what about (possible) non-seperable ones? Cheers! $\endgroup$ – Robert Lewis Apr 22 '15 at 4:16
  • $\begingroup$ By the way, are you near Berkeley, Ca? I know some people at Berkeley Chocolate . . . $\endgroup$ – Robert Lewis Apr 22 '15 at 5:24
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    $\begingroup$ @berkeleychocolate: I'm curious why the non-separable case is the sum of separable solutions? $\endgroup$ – user187437 Apr 23 '15 at 3:09
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    $\begingroup$ To user187437: I have no idea why and likely this is not even true since the DE is non-linear and so sums of solutions are not solutions. One needs a completely different argument for the non-separable case. $\endgroup$ – berkeleychocolate Apr 26 '15 at 1:33

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