Functional differentiation gives
$$
\mu=f'(\phi)-\kappa\nabla^2\phi.
$$
For $\Pi=\mu\phi-\mathbb F$ and $\Sigma_{ij}=-\Pi\delta_{ij}-\kappa(\nabla_i\phi)(\nabla_j\phi)$,
$$
\begin{aligned}
\nabla_i\Sigma_{ij}
={}&-\nabla_j(\mu\phi-\mathbb F)
-\kappa\nabla_i[(\nabla_i\phi)(\nabla_j\phi)]\\
={}&-\phi\nabla_j\mu
+[f'(\phi)-\mu-\kappa\nabla^2\phi]\nabla_j\phi.
\end{aligned}
$$
The bracket vanishes by the expression for $\mu$, while the two mixed second-derivative terms cancel. Therefore
$$
\boxed{\nabla_i\Sigma_{ij}=-\phi\nabla_j\mu.}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 344 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-344/1/b/solution