 Solution

ID: past-exam-of-the-differential-equations-and-applications-course-of-the-university-of-oxford/2008/a/2/d/solution

Expand a forcing in the complete orthonormal basis

ϕ_{n}

. Since

L ϕ_{n} = - λ_{n} ϕ_{n}

, its inverse multiplies the

n

- 1/ λ_{n}

. Consequently

g (x, t) = - n = 0 \sum \infty \frac{ϕ _{n} ( x ) ϕ _{n} ( t )}{λ _{n}} = - 2 n = 0 \sum \infty \frac{sin ( k _{n} x ) sin ( k _{n} t )}{k _{n}^{2} - 1} .

The negative sign follows from the eigenvalue convention

L y = - λ y

. This kernel satisfies the mixed endpoint conditions in its first argument and

L_{x} g = δ (x - t)

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