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

The first variation of the graph area functional is

\frac{d}{d t} A (u + tϕ)_{t = 0} = \int_{B_{1}} \frac{D u \cdot D ϕ}{1 + ∣ D u ∣ ^{2}} .

Thus the minimal surface equation for a graph is

div \frac{D u}{1 + ∣ D u ∣ ^{2}} = 0,

or, in nondivergence form,

(δ_{ij} - \frac{D _{i} u D _{j} u}{1 + ∣ D u ∣ ^{2}}) D_{ij} u = 0.

Differentiate the divergence equation with respect to

x_{k}

. Then

w = D_{k} u

satisfies

D_{i} (a^{ij} (D u) D_{j} w) = 0,

where

a^{ij} (p) = \frac{δ _{ij}}{1 + ∣ p ∣ ^{2}} - \frac{p _{i} p _{j}}{( 1 + ∣ p ∣ ^{2} ) ^{3/2}} .

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