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?


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, $2ab \le a^2 + b^2$: $$ E(t) \le E(0) + 2\sqrt{E(0)} \int_0^t F(s)ds + \left(\int_0^t F(s)ds \right)^2 \\ \le 2E(0) + 2\left(\int_0^t F(s)ds \right)^2. $$

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  • $\begingroup$ 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

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