0
$\begingroup$

A probably simple question: why can a function $u \in \mathcal C^2(\mathbb R^n \times \mathbb R)$ which is positive for $|x| \geq 1$ and for $|t| \geq 1$ and with $u(0,0) = -1$ be not a solution of the heat equation?

My guess is one has to apply the maximum principle (or the comparision principle) but I do not see how.

  • 0
    Could you clarify when the function is positive? Do you really mean its positive for $|x| \geq 1$ or when $t \geq 1?$ – 2017-02-24
  • 0
    I edited: I mean its positive for all $|x| \geq 1$ and for all $t \geq 1$. – 2017-02-24
  • 0
    If that is the case, it *feels* wrong. Consider $u(x,0) = 10000$ for all $x$ except in an $\epsilon-$neighborhood of $x=0,$ where it smoothly but rapidly decreases to meet the condition that $u(0,0) = -1.$ The heat equation damps out high frequencies first, thus this rapid transition neighborhood would get smoothed out quickly, and I'd suspect that the entire function would be positive by the time $t = 1$ comes around. I could be wrong, but the problem statement at least *feels* wrong. – 2017-02-24
  • 0
    Hm. You could be right. Maybe it meant that either one or the other is the case. Would the statement be then true? – 2017-02-24
  • 0
    The same example would be positive for all $|x| \geq 1$ for all time, if it behaved as I previously mentioned. Also, it would be positive for $t > 1...$ so the same example covers either case still. If one wishes to actually test my hypothesis, try letting $u(x,0) = 10000$ on $|x| > \epsilon$ and $u(x,0) = 10001x^2/ \epsilon^2 - 1$ for $|x| \leq \e.$ Its $C^2$ almost everywhere. Then, do the convolution $u(x,t) = \frac{1}{\sqrt{4\pi t}} \int_R u(\xi,0) exp(\frac{(x-\xi)^2}{4t}) \, d\xi.$ See if $u(1,1)$ is positive. If it is, then you have a counterexample. The IC could be made $C^2$ easily. – 2017-02-24

0 Answers 0