For a compactly supported variation $v+t\phi$, differentiation of the area functional at $t=0$ gives the <first variation>
$$
0=\left.\frac d{dt}\right|_{t=0}\int\sqrt{1+|D(v+t\phi)|^2}
=\int\frac{Dv\mathbin\cdot D\phi}{\sqrt{1+|Dv|^2}}.
$$
After <integration by parts>, this is the <minimal surface equation for a graph>
$$
D_i\left(\frac{D_iv}{\sqrt{1+|Dv|^2}}\right)=0.
$$
For $v_R(x)=R^{-1}v(Rx)$ one has $Dv_R(x)=Dv(Rx)$. The equation is invariant under this scaling, so $v_R$ solves it on $B_1$ for every $R>0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 107 / 3 / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-107/3/b/i/solution