Calculus of Variation, Need help finding Euler Lagrange Equation for Klein Gordon

Question

I have been given the following Question:
Find the Euler-Lagrange equation for the variational problem with the fundamental Integral: $$ \int_G\left[\frac{1}{2}\left(A_{\alpha \beta}\frac{\partial u}{\partial t_{\alpha}}\frac{\partial u}{\partial t_{\beta}}\right)+\frac{1}{2}m^2u^2\right]dt_1...dt_4 $$ where $m$ is a constant and $$ A=\begin{bmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}\\ \alpha,\beta=1,2,3,4 $$ I am not sure where to start here. Can anyone help me please.

Answer 1

Write $$ I(u) = \frac{1}{2}\int_G \langle A\nabla u,\nabla u\rangle + m^2u^2. $$ If $u$ is a critical point of $I$, then $$ 0=\frac{d}{d\epsilon}\bigg|_{\epsilon =0}I(u+\epsilon v) $$ for all functions $v$ of compact support. (This condition says that the first derivative of $I$ vanishes at $u$.) The right hand side of this equation can be computed via integration by parts and the symmetry of $A$: \begin{align} 0 &=\frac{d}{d\epsilon}\bigg|_{\epsilon =0}I(u+\epsilon v) \\ &=\frac{1}{2}\int_G \langle A\nabla v,\nabla u\rangle +\langle A\nabla u,\nabla v\rangle +2m^2uv \\ &=\int_G (-\mathrm{div}(A\nabla u)+m^2 u)v. \end{align} Hence $0=\int_G (\mathrm{div}(A\nabla u)+m^2 u)v$ for all compactly supported $v$. This can only be true if the integrand is zero, i.e. if $$ 0=-\mathrm{div}(A\nabla u)+m^2 u = - A_{\alpha\beta}\frac{\partial^2 u}{\partial t_{\alpha} \partial t_{\beta}} +m^2u. $$ This is the Euler-Lagrange equation, which is just the Klein-Gordon equation.