Past exam of the mathematics course of the University of Cambridge / 2009 / ia / Paper 2 / 8C / a

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- Solution a

Solution

The chain rule gives

\partial_{x} = \partial_{u} + \partial_{v}

and

\partial_{t} = \partial_{u} - \partial_{v}

. Hence

\partial_{x}^{2} - \partial_{t}^{2} = 4 \partial_{u} \partial_{v}

, and the forced wave equation becomes

y_{uv} = 1

. Integrating twice gives

y = uv + F (u) + G (v)

. At

t = 0

the initial conditions are

x^{2} + F (x) + G (x) = sin x, F^{'} (x) - G^{'} (x) = 0.

Thus

F - G

is constant, which cancels from the final expression. Combining the remaining sum yields

y = uv + \frac{1}{2} (sin u + sin v - u^{2} - v^{2}) = sin x cos t - 2 t^{2} .

Differentiation checks the forcing and both initial conditions.

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