Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 105 / 2 / c / iii

Table of contents
- Solution iii

Solution

Multiply the differential equation by

v \in H

and integrate. The Hardy inequality on an interval makes the terms containing

u / x

and

v / x

integrable. For smooth

v

, integration by parts gives

- \int_{0}^{1} u^{''} v = \int_{0}^{1} u^{'} v^{'} - [u^{'} v]_{0}^{1} = \int_{0}^{1} u^{'} v^{'},

because

v (0) = 0

and the Neumann boundary condition is

u^{'} (1) = 0

. Therefore

\int_{0}^{1} u^{'} v^{'} + \int_{0}^{1} \frac{uv}{x ^{2}} - \int_{0}^{1} u^{'} v = \int_{0}^{1} \frac{f v}{x ^{2}} .

The density of smooth functions in a Sobolev space and continuity of all four terms extend this weak formulation to every

v \in H

.

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