Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 107 / 3 / i

Table of contents
- Solution i

Solution

Here

w

is the harmonic replacement of

u

in

B_{r} = B (x_{0}, r)

, so

v = u - w \in H_{0}^{1} (B_{r})

. Subtract the weak equations and test with

v

:

\int_{B_{r}} ∣\nabla v ∣^{2} = \frac{1}{2} \int_{B_{r}} x_{1}^{2} u_{x_{1}} v_{x_{1}} + \int_{B_{r}} f v .

Because

B_{r} \subset B (0, 1)

,

∣ x_{1} ∣ \leq 1

. The Sobolev inequality

∥ v ∥_{6} \leq C ∥\nabla v ∥_{2}

and the Holder inequality give

∥\nabla v ∥_{2}^{2} \leq \frac{1}{2} ∥\nabla u ∥_{2} ∥\nabla v ∥_{2} + C ∥ f ∥_{6/5} ∥\nabla v ∥_{2} .

After division and squaring,

\int_{B_{r}} ∣\nabla v ∣^{2} \leq C_{1} \int_{B_{r}} ∣\nabla u ∣^{2} + C_{2} (\int_{B_{r}} ∣ f ∣^{6/5})^{5/3} .

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