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In Evans' PDE, he discusses how to find a scaling invariant solution of the Porous medium equation for $\gamma > 1$. On page 186 (2nd edition), he assumes that $w, w^{\prime} \rightarrow 0$ as $r$ goes to infinity. We find that $w = \left(b - \frac{\gamma - 1}{2\gamma}\beta r^2\right)^{+ \frac{1}{\gamma -1}}$ (where $b$ is a constant, $+$ denotes the positive part of the expression in parenthesis, and $\beta = \frac{1}{n(\gamma - 1) + 2}$). It's not apparent to me why $w \rightarrow 0$ as $r \rightarrow \infty$. If anyone can explains why, that would be greatly appreciated.

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  • $\begingroup$ Isn't $w$ compactly supported? As $r\to \infty$ the quantity in parenthesis tends to $-\infty$ so the positive part vanishes. $\endgroup$ – Jose27 Nov 1 '15 at 18:20
  • $\begingroup$ Evans doesn't assume anything on $w$ other than the limit mentioned above. It does seem reasonable if $w$ has compact support to conclude $w \rightarrow 0$. $\endgroup$ – MathNewbie Nov 1 '15 at 21:39
  • $\begingroup$ The statement $w,w'\to 0$ is not a conclusion, but an assumption. In fact it's not enough to assume that, we need that $w,w'$ decay fast enough (for instance $w=o(r^{-n})$ and $w'=o(1)$ is enough). See Evans' errata for the book. $\endgroup$ – Jose27 Nov 3 '15 at 3:28
  • $\begingroup$ @Jose27, I was referring to the formula I got for $w$. I understand that $w, w^{\prime} \rightarrow 0$ is an assumption as explained in the errata. I actually got it. I forgot that $+$ meant the max of $0$ and the expression in parenthesis. $\endgroup$ – MathNewbie Nov 3 '15 at 4:41

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