# Uniqueness of a nonlinear heat equation?

Let $U$ be a bounded open set with smooth boundary $\partial U$. Show that $C^2$ solutions of the initial-boundary value problem:

$u_t-\Delta u+ \cos (u)=0$ in $U\times\mathbb R^+$

$u=0$ on $\partial U\times\mathbb R^+$

$u(x,0)=u_0(x)$ in $U$

are unique.

I believe I should use the maximum principle to solve the problem. But the nonlinear term $\cos(u)$ can have different sign. It make it hard to apply the theorem. Anyone know how to deal with that term? Please help, thanks!!

-
Cosine is a Lipschitz function: $|\cos u_1-\cos u_2|\le |u_1-u_2|$. This implies that the difference of two solutions, $v=u_1-u_2$, satisfies the inequality $|v_t-\Delta v|\le |v|$. Try to see what this means for its energy function $E(t)=\int_U v(x,t)^2dx$. –  user53153 Jan 6 '13 at 20:40
If you get stuck, feel free to comment. The basic idea is that $E$ will be identically zero by Gronwall's inequality. –  user53153 Jan 6 '13 at 23:11
Oh, I see! $\frac{d}{dt}(\int v^2 dx)=2\int v v_t dx\le 2\int v(\Delta v+|v|)dx\le 2\int |v|^2 dx$ then you use the Gronwall's inequality to get that $E(t)$ is identically zero! Great! Thanks! –  Siming HE Jan 6 '13 at 23:28
Actually, the first inequality in your chain isn't quite right because $v$ could be negative. I corrected this in the answer. –  user53153 Jan 7 '13 at 0:45
OK, thank you very much for the help. Yeah, I made a mistake there. And the remark after the answer is very interesting. –  Siming HE Jan 7 '13 at 4:30

Copied from comments (with a small fix): $$\frac{d}{dt}\left(\int v^2 dx\right)=2\int v v_t dx \le 2\int v \Delta v dx + 2\int |v| |v_t-\Delta v| dx \le 2\int |v|^2 dx$$ then you use the Gronwall's inequality to get that $E(t)$ is identically zero.