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

Table of contents
- Solution i

Solution

For a compactly supported variation

v + tϕ

, differentiation of the area functional at

t = 0

gives the first variation

0 = \frac{d}{d t}_{t = 0} \int 1 + ∣ D (v + tϕ) ∣^{2} = \int \frac{D v \cdot D ϕ}{1 + ∣ D v ∣ ^{2}} .

After integration by parts, this is the minimal surface equation for a graph

D_{i} (\frac{D _{i} v}{1 + ∣ D v ∣ ^{2}}) = 0.

For

v_{R} (x) = R^{- 1} v (R x)

one has

D v_{R} (x) = D v (R x)

. The equation is invariant under this scaling, so

v_{R}

solves it on

B_{1}

for every

R > 0

.

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