Past exam of the differential equations 2 course of the University of Oxford / 2021 / 1 / c

At $l=\pi$, the Fredholm solvability condition isby orthogonality of the Fourier sine basis. Thus a non-unique solution exists exactly when $a=0$, with any real $b$. Since $(d^2/d{x}^2+1)\sin 2x=-3\sin 2x$, the complete family isIf $a\ne 0$ there is no solution, for any $b$. In particular, when $a=b=0$, the family reduces to the kernel $C\sin x$; it is still non-unique. There is no choice of $a,b$ producing a unique solution for this length.

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 $A\cos x+B\sin x$ whose value and first derivative at zero are both zero, forcing $A=B=0$. In fact, for every prescribed continuous function $f$,satisfies both initial data and $y^{\prime \prime }+y=f$. Consequently non-uniqueness is impossible for every $l>0$, including $l=m\pi$. 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.