Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2003 / Paper 2 / 10 / b

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- Solution b

Solution

Write

c = cos θ

,

s = sin θ

. Since

θ

is constant, the multivariable chain rule gives

\partial_{x} = c \partial_{u} - s \partial_{v}

and

\partial_{y} = s \partial_{u} + c \partial_{v}

. Applying these operators twice to a twice continuously differentiable function gives

g_{xx} = c^{2} g_{uu} - 2 cs g_{uv} + s^{2} g_{vv}, g_{yy} = s^{2} g_{uu} + 2 cs g_{uv} + c^{2} g_{vv} .

The mixed terms cancel and

c^{2} + s^{2} = 1

, so

g_{xx} + g_{yy} = g_{uu} + g_{vv} .

This is rotational invariance of the Laplacian: an orthogonal change of coordinates preserves the trace of the Hessian matrix. A nonconstant angle would introduce additional derivatives and is not covered by this constant rotation calculation.

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