How the calculate the variation of this functional?

Question

given $L=\int\frac{1}{2}|\nabla\phi|^2dx$I now wanna to calculate : $\delta L(\phi)=\frac{1}{2}\int\delta\{\nabla(\phi\nabla\phi)-\phi\nabla^2\phi\}dx $ $=\int\frac{-1}{2}\delta(\phi\nabla^2\phi) dx$, and $\delta(\phi\nabla^2\phi)=\phi\nabla^2(\delta \phi)+(\nabla^2\phi)(\delta\phi)$. I wonder what is the next process?

Answer 1

Let $\varphi_s$ be a one-parameter family of functions. Then $$\frac{1}{2}\frac{d}{ds}|\nabla\varphi_s|^2=\langle\nabla\varphi_s,\nabla\dot{\varphi}_s\rangle=\operatorname{div}\big(\dot{\varphi}_s\nabla\varphi_s\big)-\dot{\varphi}_s\Delta\varphi_s.$$ So if $\varphi_s$ are defined on, say, a Riemannian manifold $M$ with boundary $\partial M$, then $$\frac{1}{2}\frac{d}{ds}\int_M |\nabla\varphi_s|^2\,d\mu_g=\int_{\partial M}\dot{\varphi}_s\frac{\partial\varphi_s}{\partial\nu}\,d\mu_{\partial M}-\int_M\dot{\varphi}_s\Delta\varphi_s\,d\mu_g.$$ This is the first variation formula. You can read off that the Euler-Lagrange equation is $\Delta\varphi=0$.