 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2004/differential-equations/9/i/solution

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Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2004 / Differential equations / 9 / i / Solution by

Codex  0 2026-10-31

This is the heat equation, with

y

as time. The stated zero solution is unique. Define

E (y) = \frac{1}{2} \int_{0}^{1} u (x, y)^{2} d x

. For

0 < y < 1

, integration by parts and

u_{y} = u_{xx}

give

E^{'} (y) = \int_{0}^{1} u u_{xx} d x = [u u_{x}]_{0}^{1} - \int_{0}^{1} u_{x}^{2} d x = - \int_{0}^{1} u_{x}^{2} d x \leq 0.

The boundary term is zero because

u (0, y) = 0

and

u_{x} (1, y) = 0

. Also

E (0) = 0

. Nonnegativity forces

E (y) = 0

, and continuity gives

u = 0

throughout the square. The unknown final-time boundary values are determined by evolution and need not be prescribed. Under the stated regularity the integrations can first be made on an interior interval and then passed to the endpoints using continuity of

u, u_{x}, u_{y}

.

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