The first variation of the graph area functional is
$$
\frac d{dt}A(u+t\phi)\bigg|_{t=0}
=\int_{B_1}\frac{Du\cdot D\phi}{\sqrt{1+|Du|^2}}.
$$
Thus the <minimal surface equation for a graph> is
$$
\operatorname{div}\frac{Du}{\sqrt{1+|Du|^2}}=0,
$$
or, in nondivergence form,
$$
\boxed{\left(\delta_{ij}-\frac{D_iuD_ju}{1+|Du|^2}\right)D_{ij}u=0.}
$$
Differentiate the divergence equation with respect to $x_k$. Then $w=D_ku$ satisfies
$$
D_i(a^{ij}(Du)D_jw)=0,
$$
where
$$
\boxed{a^{ij}(p)=\frac{\delta_{ij}}{\sqrt{1+|p|^2}}
-\frac{p_ip_j}{{(1+|p|^2)}^{3/2}}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 107 / 3 / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-107/3/b/i/solution