 Solution

ID: past-exam-of-the-differential-equations-2-course-of-the-university-of-oxford/2023/1/a/ii/solution

Use the unweighted inner product

⟨ w, y ⟩ = \int_{0}^{a} \overline{w} y d x

. Two applications of integration by parts give

\int_{0}^{a} \overline{w} (y^{''} + 2 y^{'} + 2 y) d x = \int_{0}^{a} (\overline{w}^{''} - 2 \overline{w}^{'} + 2 \overline{w}) y d x + [\overline{w} y^{'} - \overline{w}^{'} y + 2 \overline{w} y]_{0}^{a} .

y = 0

at both ends, but leave

y^{'} (0)

and

y^{'} (a)

independently variable. The remaining boundary term vanishes for every admissible

y

precisely when

w (0) = w (a) = 0

L^{*} w = w^{''} - 2 w^{'} + 2 w, w (0) = w (a) = 0 .

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