Divergence inequality for a function that is vanishing on boundary

Question

How to prove:

$$2\int_U |\nabla \varphi|^2 dx \leq \epsilon \int_U \varphi^2 dx + \frac{1}{\epsilon} \int_U |\Delta \varphi|^2 dx$$

for every $\epsilon >0$? How to use divergence theorem here?

Answer 1

By Green's identity, we have $$\int_U|\nabla \varphi|^2+\int_U \varphi \Delta\varphi=\int_{\partial U}\varphi\frac{\partial\varphi}{\partial\nu}.$$ Since $\varphi$ vanishes on the boundary, i.e. $\varphi=0$ on $\partial U$, we have $$\tag{1}\int_U|\nabla \varphi|^2=-\int_U \varphi \Delta\varphi.$$

As John pointed out, we can apply the Cauchy-Schwarz inequality $2ab \leq \epsilon a^2 + \frac{1}{\epsilon} b^2$ with $a=-\varphi$ and $b=\Delta\varphi$, we have $$-2\varphi \Delta\varphi\leq \epsilon\varphi^2+\frac{1}{\epsilon}(\Delta\varphi)^2.$$ Integrating it over $U$, we get $$\tag{2}-2\int_U\varphi \Delta\varphi\leq \epsilon\int_U\varphi^2+\frac{1}{\epsilon}\int_U(\Delta\varphi)^2.$$ Now the required inequality follows from combining $(1)$ and $(2)$.