From Hamiltonian to equation of motion

Question

It is said that from the Hamiltonian $$ H=\int\left[\frac{\Phi_t^2}{2}+\frac{\Phi_r^2}{2}+V(\Phi)\right]\, dx~~(1) $$ follows the equation of motion $$ \Phi_{tt}-\Phi_{xx}+\frac{dV(\Phi)}{d\Phi}=0.~~(2) $$

How to get from (1) to (2)?

Answer 1

Formally. For a time-independent Hamiltonian, we have that conservation of energy \begin{align} \frac{d}{dt} \int\left[ \frac{\Phi_t^2}{2}+\frac{\Phi_x^2}{2}+ V(\Phi)\right]\ dx=0 \end{align} which means \begin{align} \int \Phi_t\Phi_{tt} + \Phi_x\Phi_{xt} + \nabla V(\Phi)\Phi_t\ dx =0. \end{align} Applying integration by parts on the middle term yields \begin{align} \int [\Phi_{tt} - \Phi_{xx} + \nabla V(\Phi)]\Phi_t\ dx =0. \end{align}

In short, we know that the pde \begin{align} \Phi_{tt} - \Phi_{xx} + \nabla V(\Phi)=0 \end{align} will give raise to the above Hamiltonian.