2
$\begingroup$

Let's consider the following wave equation in $B\times (0,\infty)$ in which $B$ is the open unit ball in $\Bbb R^n$: $$ \begin{align} u_{tt} - \Delta u = f,&\quad (x,t)\in B\times (0,\infty)\\ u = u_0,\,\, u_t=u_1,&\quad (x,t)\in B\times\{t=0\}\\ u = 0,&\quad (x,t)\in \partial B\times (0,\infty) \end{align} $$

Suppose $u_0$ is $C^2$, $u_1$ is $C^1$, $f$ is continuous, and $u$ is a $C^2$ solution. Prove the following energy estimate: $$E(t)\le 2E(0) + 2(\int_0^t \|f(s,\cdot)\|_{L^2}ds)^2,\quad t\ge 0.$$ in which $E(t) = \|u_t\|^2_{L^2}+\|\nabla u\|_{L^2}^2$. The $L^2$-norm is taken over $B$.

Usually we will consider $E'(t)$, upper bound it and then integrate. But since the RHS has $2E(0)$ instead of $E(t)$, this doesn't look very hopeful.

Any thoughts?

$\endgroup$
4
$\begingroup$

Compute as per the homogeneous case: $$ \frac{d}{dt} E(t) = 2\int_B u_t u_{tt} + \nabla u \cdot \nabla u_t = 2 \int_B (u_{tt}-\Delta u) u_t = 2 \int_B f u_t \le 2 \Vert u_t \Vert_{L^2} \Vert f \Vert_{L^2} \le 2 \sqrt{E(t)} F(t) $$ for $F(t) = \Vert f(\cdot,t)\Vert_{L^2}$. Note that the boundary term vanishes when we IBP because $u=0$ there, and so $u_t=0$ there as well. Then $$ \frac{d}{dt} \sqrt{E(t)} = \frac{1}{2 \sqrt{E(t)}} \frac{d}{dt}E(t) \le F(t), $$ and so upon integrating we see that $$ \sqrt{E(t)} \le \sqrt{E(0)} + \int_0^t F(s) ds. $$ Now square both sides and apply Cauchy's inequality, $2ab \le a^2 + b^2$: $$ E(t) \le E(0) + 2\sqrt{E(0)} \int_0^t F(s)ds + \left(\int_0^t F(s)ds \right)^2 \\ \le 2E(0) + 2\left(\int_0^t F(s)ds \right)^2. $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks! For what is worth I was having trouble with the boundary term that came out of $\int_B \Delta u\cdot \Delta u_t $. I really should have realised that $u = 0$ on all of $\partial B\times (0,\infty)$ clearly implies $u_t = 0$ on $\partial B\times (0,\infty)$ and so the boundary term vanishes :) $\endgroup$ – Vim Jun 23 '18 at 4:15
  • $\begingroup$ On the second line, it should be $1/(2\sqrt{E(t)})$ I believe. $\endgroup$ – Chee Han Jun 23 '18 at 9:28
  • $\begingroup$ @CheeHan Thanks: fixed. $\endgroup$ – Glitch Jun 23 '18 at 11:52
  • $\begingroup$ @Vim You're welcome. For the sake of future readers, I added a line to the main text pointing out that $u_t$ vanishes on the boundary. $\endgroup$ – Glitch Jun 23 '18 at 11:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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