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

Solution

Put

h = lo g X

. The first equation becomes

Δ h = \frac{Δ X}{X} - \frac{∣\nabla X ∣ ^{2}}{X ^{2}} = - \frac{∣\nabla Y ∣ ^{2}}{X ^{2}},

while the second remains

div (X^{- 2} \nabla Y) = 0.

The assumed

C^{0, α}

regularity and the bounds away from zero make

X^{- 2}

C^{0, α}

uniformly elliptic coefficient. The divergence-form

C^{1, α}

estimate from part (d) first gives

Y \in C_{loc}^{1, α}

. Hence the right side of the equation for

h

C^{0, α}

h

, and therefore

X

, in

C_{loc}^{2, α}

Expanding the

Y

Δ Y - 2\nabla h \cdot \nabla Y = 0.

The higher-order Schauder estimates now alternate between the equations for

h

and

Y

, gaining derivatives at each step. Induction yields

(X, Y) \in C^{\infty} (Ω^{'}; H)

for every

Ω^{'} ⋐ Ω

There are no discussions about this article yet.

There are currently no matching articles.