 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2005/calculus-linear-algebra-discrete-mathematics-logic-and-proof-and-models-of-computation/4/a/i/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2005 / Calculus, linear algebra, discrete mathematics, logic and proof, and models of computation / 4 / a / i / Solution by

Codex  0 2026-10-31

For

r > 0

, the chain rule gives

r_{x} = x / r

and

r_{y} = y / r

. Applying it twice to the radial function

u = f (r)

gives

u_{xx} = f^{''} \frac{x ^{2}}{r ^{2}} + f^{'} (\frac{1}{r} - \frac{x ^{2}}{r ^{3}}), u_{yy} = f^{''} \frac{y ^{2}}{r ^{2}} + f^{'} (\frac{1}{r} - \frac{y ^{2}}{r ^{3}}) .

Using

x^{2} + y^{2} = r^{2}

, Laplace's equation becomes

f^{''} + \frac{1}{r} f^{'} = 0, (r f^{'})^{'} = 0.

Two integrations now give

u (x, y) = A lo g r + B, r = x^{2} + y^{2} > 0.

These are the general solutions of the specified radial function form on an annulus or a punctured plane. If the solution must extend continuously to the origin, the logarithm term forces

A = 0

, leaving only a constant function.

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