Past exam of the differential equations and applications course of the University of Oxford / 2008 / A / 3

Write the integral part as $Ky=12b⟨a,y⟩-30a⟨b,y⟩$, where $a(x)=e^{-x}$ and $b(x)=e^{-2x}$. The kernel is real, so its adjoint transposes the two arguments. HenceThis follows by exchanging the two integrations in the $L^2(0,\infty )$ inner product; the exponential factors make the finite-rank operators bounded.

Every eigenvector with eigenvalue different from one lies in $E=\mathrm{span}\{a,b\}$, because $(\lambda -1)y=Ky\in E$. The elementary overlaps are $⟨a,a⟩=1/2$, $⟨a,b⟩=1/3$, $⟨b,b⟩=1/4$. ThusIts characteristic polynomial is $\lambda (\lambda +4)$, so the finite-multiplicity eigenpairs areEach eigenspace has dimension one. On $E^{\perp }$, both integrals defining $K$ vanish, so $Ly=y$. Thus $\boxed{\lambda =1}$ is also an eigenvalue, of infinite multiplicity, and there are no others.

The adjoint also leaves $E$ invariant, with $L^{\ast }a=5a-15b$ and $L^{\ast }b=3a-9b$. By adjoint in weighted coordinates, its matrix in the same basis is $\left(\begin{array}{cc}5& 3\\ -15& -9\end{array}\right)$. Solving the two eigenvector equations givesThe basis $a,b$ is not orthonormal, so one must not simply transpose the previous numerical matrix without accounting for its Gram matrix. Directly transposing the integral kernel avoids that error.

The nonzero kernel of $L$ rules out a unique solution for every value of $c$. By the Fredholm alternative, solvability requires orthogonality of the forcing to $\ker L^{\ast }=\mathrm{span}\{3a-5b\}$. Its pairing with $6a+cb$ isThus only $c=-4$ is compatible. In that case $6a-4b=2y_{-4}$ and a particular solution is $-y_{-4}/2$. All solutions areTherefore infinitely many solutions exist for $c=-4$, no solution exists for $c\ne -4$, and no parameter value gives a unique solution.