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ID: past-exam-of-the-differential-equations-and-applications-course-of-the-university-of-oxford/2011/a/1/b/iii/solution

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Past exam of the differential equations and applications course of the University of Oxford / 2011 / A / 1 / b / iii / Solution by

Codex  0 2026-10-31

Under

y = e^{- 2 x} v

, the left Robin boundary condition becomes

v^{'} (0) - v (0) = 0

and the right Dirichlet boundary condition becomes

v (1) = 0

. Choose

v_{L} (x) = cos (π x) + \frac{1}{π} sin (π x), v_{R} (x) = sin (π x) .

Their Wronskian

v_{L} v_{R}^{'} - v_{L}^{'} v_{R}

is the constant

π

. For

y_{L} = e^{- 2 x} v_{L}

and

y_{R} = e^{- 2 x} v_{R}

, it is

W_{y} (x) = π e^{- 4 x}

. The coefficient enforcing the derivative jump four is

4/ W_{y} (ξ)

x_{<} = min (x, ξ)

and

x_{>} = max (x, ξ)

, the Green's function is

g (x, ξ) = \frac{4}{π} e^{2 ξ - 2 x} [cos (π x_{<}) + \frac{1}{π} sin (π x_{<})] sin (π x_{>}) .

The endpoint conditions, continuity and derivative jump follow directly from this Wronskian construction. Its exponential factor also shows why the kernel need not be symmetric in the unweighted inner product.

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