 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-1/19/a/solution

Assume

c \neq = 0

, so

(ξ, η) = (x - c t, x + c t)

is an invertible coordinate change. The chain rule gives

\partial_{x} = \partial_{ξ} + \partial_{η}, \partial_{t} = c (- \partial_{ξ} + \partial_{η}) .

For a twice continuously differentiable solution the mixed partial derivatives commute. Squaring the two operators therefore gives

\partial_{x}^{2} - c^{- 2} \partial_{t}^{2} = 4 \partial_{ξ} \partial_{η} .

ψ_{ξ η} = 0.

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