Use the normalization from part (ii), and writeThe Sturm-Liouville eigenfunction expansion coefficient is . Although has zero endpoint values, generally does not. Retain the endpoint term in the Lagrange identity for a Sturm-Liouville operator:The positive sign comes from subtracting the contribution at . The differential equation also gives . Substitution into the preceding identity yieldsHence, away from resonance in a differential equation,This is the effect of boundary data in a Sturm-Liouville eigenfunction expansion: omitting the endpoint contribution would discard the dependence on .
If for a positive integer , the Fredholm solvability condition isWhen it fails, there is no solution. When it holds, is arbitrary and the other coefficients retain the displayed formula, giving infinitely many solutions differing by a multiple of . Sufficiency follows by subtracting the smooth lift , which has the required endpoint values and satisfies , and applying the Fredholm alternative to the resulting homogeneous Dirichlet boundary conditions.
Finally, for , the displayed Sturm-Liouville eigenfunction expansion represents the solution on and in the weighted inner product sense. Every term is zero at , so that series cannot give the nonzero prescribed value by substitution at the endpoint. The boundary trace must be imposed separately. Equivalently, expand with homogeneous Dirichlet boundary conditions and add back afterwards.
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