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Let $u$ denote to the solution of the heat equation

$$\begin{cases} u_t(x,t)-\Delta u(x,t) & = & 0 & t>0 \\ u(x,0) & = & g(x) \end{cases}$$

where $x\in\mathbb{R}^n$.

I want to show that

  1. if $||g||_\infty<\infty$ then $u$ tends to some constant as $t\to\infty$
  2. if $\text{supp}(g) \Subset \mathbb{R}$ (that is compact) then this constant is $0$.

I started from $$u(x,t) = (g*K_t)(x) \ \ \ \ \ \ \ \ \ (\diamond)$$ where $K_t$ is the heat kernel and tried to prove 1. by showing $||u'(x,t)||_{\infty}\to 0$ as $t\to\infty$. Unfortunately this might be the wrong way since formula $(\diamond)$ solves the heat equation (with initial data) only if $g \in L^p$ (what is not clear as we only claim $g$ to be bounded).

Who can help?

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Do you really only have $g$ being bounded? No continuity or local integrability? (I am wondering what would happen if $g$ were not measurable!) –  Willie Wong Jul 6 '12 at 15:51
@WillieWong I'm assuming condition 1. should mean $g \in L^\infty$, that is $g$ is measurable with finite essential supremum. –  user12014 Jul 6 '12 at 17:25

1 Answer 1

up vote 3 down vote accepted

Without additional assumptions on $u$, I believe this is not true.

There exist nontrivial solutions to the heat equation with initial condition 0. Indeed, one can find nontrivial $v(x,t)$ such that $v_t - \Delta v = 0$ for all $(x,t) \in \mathbb{R}^{n} \times \mathbb{R}$ and $v(x,t) = 0$ for all $x$ and all $t \le 0$. See for example

P. C. Rosenbloom and D. V. Widder. A temperature function which vanishes initially. Amer. Math. Monthly, 65:607–609, 1958.

The relevant example is at the very end.

I did not check whether this $v$ converges as $t \to \infty$. If it does not, we are done. If it does, we could take something like $u(x,t) = \sum_{n=1}^\infty v(x, t-n)$ so that disturbances keep appearing and prevent convergence.

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Note that uniqueness is available if one requires $u$ to not grow too fast as $|x|\to \infty$. In particular, IIRC, requiring that $u$ is also uniformly bounded in $x$ at each $t$ should suffice. –  Willie Wong Jul 7 '12 at 15:39

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