The Lagrangian is
$$
F(X,Y,\nabla X,\nabla Y)=X^{-2}(|\nabla X|^2+|\nabla Y|^2).
$$
The <Euler-Lagrange equation> for $X$ is
$$
-2X^{-3}(|\nabla X|^2+|\nabla Y|^2)
-\operatorname{div}(2X^{-2}\nabla X)=0.
$$
Expanding the divergence and simplifying gives
$$
\boxed{\Delta X=\frac{|\nabla X|^2-|\nabla Y|^2}{X}}.
$$
The equation for $Y$ is
$$
\boxed{\operatorname{div}\left(\frac{\nabla Y}{X^2}\right)=0}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 107 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-107/2/a/solution