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I am studying the following basic heat equation. (All notations follows Evans book) \begin{align} u_t -\Delta u =0 &\text{ in }\mathbb R^2\times(0,\infty)\\ u=u_0 &\text{ on }\mathbb R^2\times\{t=0\} \end{align} i.e., the Cauchy problem. Here we assume $u_0\in C_c^\infty(\mathbb R^2)$ and $u_0$ is non-negative.

Let $u(x,t)$ denote the standard solution so that $$ u(x,t):=\frac{1}{4\pi t}\int_{\mathbb R^2}e^{-\frac{|x-y|^2}{4t}}u_0(y)dy \tag 1 $$

Now given a function $a(x)\in C_c^\infty(\mathbb R^2)$ so that $$ \int_{\mathbb R^2}a(x)\,dx=0,\,\,\int_{\mathbb R^2}a(x)u_0(x)\,dx=C = \int_{\mathbb R^2}(a(x))^2dx, $$ where $C>0$ is a constant. (Note the above equation does not mean $u_0=a$)


Now I need to study the following equation: $$ f(t):=\int_{\mathbb R^2} [u(x,t)-u_0(x)]\cdot\partial_t u(x,t)\,dx - \int_{\mathbb R^2} \partial_tu(x,t)\cdot a(x)\,dx. \tag 2 $$ Notice that $$f(t) = \frac{d}{dt} \frac{1}{2}\int (u(x,t)-u_0+a)^2dx$$

My questions:

  1. Assume that there exist $t_0>0$ so that $f(t_0)=0$. Prove that for all $0<t<t_0$, $f(t)<0$; and for all $t>t_0$, $f(t)>0$. And I expect $f(t)\to 0$ as $t\to\infty$.

  2. Give a condition on $u_0$ so that there exists such $t_0>0$. (the condition can not be $u_0$ is a constant.)


What I tried so far is put $(1)$ into $(2)$ and see what's going on. But so far I got nothing... I suspect $$ 0\leq \int_{\mathbb R^2}a(x)u(x,t_2)\,dx\leq \int_{\mathbb R^2}a(x)u(x,t_1)\,dx\leq \int_{\mathbb R^2}a(x)u_0(x)\,dx $$ for $0<t_1<t_2$ but I can not prove it.

Any help is really welcome!

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  • $\begingroup$ Is Q1 equivalent to 'f is nondecreasing' ? $\endgroup$ – BCLC Nov 17 '15 at 21:30
  • $\begingroup$ @BCLC not exactly. What I want is $f(t)<0$ as $t<t_0$, $f(t)>0$ as $t>t_0$; so the graph of $f(t)$ looks like a spoon. $\endgroup$ – spatially Nov 17 '15 at 21:34
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    $\begingroup$ @BCLC: I restated the problem. Thanks for pointing out! $\endgroup$ – spatially Nov 17 '15 at 21:35
  • $\begingroup$ tankonetoone, sooooo w/o your edit, Q1 is equivalent to 'f is nondecreasing' ? $\endgroup$ – BCLC Nov 17 '15 at 21:36
  • $\begingroup$ @BCLC: Yes. But I actually expect $f$ decreasing first and than increasing $\endgroup$ – spatially Nov 17 '15 at 21:38

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