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

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 $$$$E(t)=\dfrac{1}{2}\Vert u_{t}(t)\Vert_{2}^{2}+\dfrac{1}{2}\Vert \nabla u(t)\Vert_{2}^{2}$$$$ How can I estimate $$\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)$$?

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.