Past exam of the differential equations and applications course of the University of Oxford / 2011 / A / 2 / b

Write $v=Ly$ asDifferentiating twice, the moving-limit terms cancel at first order and yield $\boxed{v^{\prime \prime }+v=y}$ at second order. Also $v^{\prime }(0)=0$ and $v(\pi )=0$. Thus this integral operator is the inverse of $D^2+1$ with mixed Neumann boundary condition and Dirichlet boundary condition.

For ${\kappa }_n=n+1/2$, $y_n(x)=\cos ({\kappa }_{n}x)$ satisfies those conditions and $(D^2+1)y_n=(1-{\kappa }_n^2)y_n$. The corresponding mixed homogeneous equation has no nonzero solution, so uniqueness givesThis verifies the asserted eigenvalues and eigenfunctions for every $n\ge 0$.

Since ${\lambda }_1=-4/5$, the kernel of a linear map $L+\frac{4}{5}I$ is the span of $\psi (x)=\cos (3x/2)$. The self-adjoint Fredholm alternative requires $⟨f,\psi ⟩=0$ for solvability. If it holds, an arbitrary multiple of $\psi$ can be added, so a solution is not unique.

For $f=\cos x$,so case (A) has no solution. For $f=\sin x\sin (3x/2)$,by sine-function orthogonality. Thus case (B) has solutions, and they are nonunique.

For $\alpha \ne 0$, the Fredholm alternative gives a unique solution exactly when $\alpha \ne -{\lambda }_n$ for all $n\ge 0$, unless a resonant problem could exceptionally be compatible. Here none is: with ${\kappa }_n=n+1/2$,At every resonance the compatibility condition therefore fails.

The case $\alpha =0$ must be checked separately because $L$ is a compact first-kind integral operator. For every admissible $y$, $Ly$ is continuous and $(Ly)(\pi )=0$, whereas $\cos \pi =-1$. Hence $Ly=\cos x$ has no solution. Combining these cases,In the nonresonant nonzero case the shifted eigenvalues stay away from zero, since ${\lambda }_n\to 0$ and $\alpha \ne 0$, so inversion is bounded on the Hilbert space.