Estimate $\Vert \Delta u(t)\Vert_{2}$ in term of energy

Question

We consider the wave equation $$\left\{ \begin{array}{ll} u_{tt}(x,t)-\Delta u(x,t)=0, x \in \Omega, t>0\\ u=0, \quad u \in \partial \Omega, t>0 \\ u(0,x)=u_{0}(x), \quad u_{t}(0,x)=u_{1}(x), x \in \Omega \end{array} \right. $$ the energy functional $E(t)$ is \begin{equation} E(t)=\dfrac{1}{2}\Vert u_{t}(t)\Vert_{2}^{2}+\dfrac{1}{2}\Vert \nabla u(t)\Vert_{2}^{2} \end{equation} How can I estimate $\Vert \Delta u(t)\Vert_{2}$ in term of $E(t)$ or can we find a constant $c>0$ such that $\Vert \Delta u(t)\Vert_{2}^{2}\leq cE(t) $?

Answer 1

You cannot.

Let $v$ be a Dirichlet eigenfunction of $-\Delta$ on the domain $\Omega$ with eigenvalue $\lambda > 0$. The function

$$ u(t,x) = \sin(\sqrt{\lambda}t) v(x) $$

solves the wave equation. The energy functional evaluates to

$$ E(t) = \frac12 \lambda \|v\|_{L^2}^2 $$

But

$$ \|\Delta u\|_{L^2}^2 = \lambda^2 \sin^2(\sqrt{\lambda}t) \|v\|_{L^2}^2 $$

Take $\lambda\nearrow \infty$ you see that no bound of the type you hoped for is possible.