 Solution

ID: past-exam-of-the-differential-equations-and-applications-course-of-the-university-of-oxford/2011/a/2/b/i/solution

Write

v = L y

v (x) = cos x \int_{x}^{π} sin t y (t) d t + sin x \int_{0}^{x} cos t y (t) d t .

Differentiating twice, the moving-limit terms cancel at first order and yield

v^{''} + v = y

at second order. Also

v^{'} (0) = 0

and

v (π) = 0

. Thus this integral operator is the inverse of

D^{2} + 1

For

κ_{n} = n + 1/2

y_{n} (x) = cos (κ_{n} x)

satisfies those conditions and

(D^{2} + 1) y_{n} = (1 - κ_{n}^{2}) y_{n}

. The corresponding mixed homogeneous equation has no nonzero solution, so uniqueness gives

L y_{n} = λ_{n} y_{n}, λ_{n} = \frac{1}{1 - κ _{n}^{2}} = - \frac{4}{4 n ^{2} + 4 n - 3} .

This verifies the asserted eigenvalues and eigenfunctions for every

n \geq 0

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