Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!!

share|cite|improve this question
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
up vote 3 down vote accepted

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.

If one is so inclined, parabolic problems can be phrased as an ODE in a Hilbert space. In this formulation the appearance of the Lipschitz condition and the application of Gronwall's inequality are things to be expected. But introducing an energy function one gets the benefit of ODE methods without the cost of function-spaces machinery.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.