Solution

ID: past-exam-of-the-differential-equations-2-course-of-the-university-of-oxford/2021/1/c/ii/solution

Putting both data at zero makes this an initial value problem. The uniqueness theorem for ordinary differential equations applies, and one can also prove uniqueness directly: the difference of two solutions is a homogeneous solution whose value and first derivative at zero are both zero, forcing . In fact, for every prescribed continuous function ,
satisfies both initial data and . Consequently non-uniqueness is impossible for every , including . In this setting the kernel is trivial. The boundary conditions at a single endpoint do not give the same self-adjoint operator domain as the two separated Dirichlet boundary conditions, so the earlier exceptional lengths do not apply.

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