Integral inequality (Divergence theorem)

Question

I'm trying to prove the following inequality:

$$2 \int_U |\nabla \phi|^2 dx \leq \int_U \phi^2 dx + \int_U |\Delta \phi|^2 dx$$

where $U \subset \mathbb{R}^n$ is bounded and open and $\phi \in C^\infty_c(U)$. I actually think I have managed to prove this just using one of Green's identities $$\int_UD \phi \cdot D \phi dx = -\int_U \phi \Delta \phi dx + \int_{\partial U}\frac{\partial \phi}{\partial \nu} \phi dx$$ (which comes from the divergence theorem), and then using the fact that $\phi = 0$ on $\partial U$. This then gives:

$$\int_U |\nabla \phi|^2 dx = -\int_U \phi \Delta \phi dx \leq \int_U |\phi| |\Delta \phi| dx \leq \int_U \frac{\phi^2}{2}+\frac{|\Delta \phi|^2}{2}$$ where the last inequality comes from Cauchy's inequality $ab \leq a^2/2 + b^2/2$.

Does this seem correct? The problem is that I have been given a hint which says to use the fact that $\nabla \cdot(\phi \nabla \phi) = |\nabla \phi|^2 + \phi \Delta \phi$ and I'm not sure how to use this hint. Would this just lead to an alternative proof of the inequality?

Answer 1

Divergence theorem is $$ \int_{U} \nabla \cdot F\,dx = \int_{\partial U} F\cdot \nu\,dS. $$ The hint you have is: $$\nabla \cdot(\phi \nabla \phi) = |\nabla \phi|^2 + \phi \Delta \phi.$$ So we let $F = \phi \nabla \phi$: $$ \int_{U}(|\nabla \phi|^2 + \phi \Delta \phi)\,dx = \int_{U} \nabla \cdot (\phi \nabla \phi)\,dx = \int_{\partial U} \phi \nabla \phi\cdot \nu\,dS = \int_{\partial U} \phi \frac{\partial \phi}{\partial \nu}\,dS . $$ And here we have the Green's identity and the rest of the proof goes like you wrote. Also I want to point out that the integration on boundary is with respect to the surface measure $dS$, not the volume measure $dx$ anymore, you may wanna modify the Green's identity in your question a little bit. :)

Answer 2

You already used it :) That's where the Green's identity came from!