Past exam of the differential equations and applications course of the University of Oxford / 2009 / A / 2 / B

Use the complex $L^2$ inner product $⟨f,g⟩={\int }_0^{\pi }\overline{f}g$, linear in its second argument. Exchanging the order of integration givesSince the kernel is real, the adjoint operator is thereforeor, renaming variables, $L^{\ast }f(x)={\int }_0^{\pi }k(t,x)f(t)dt$. Continuity of the kernel on the compact integration square justifies these integrals and the exchange.

Symmetry $k(x,t)=k(t,x)$ makes the adjoint kernel equal to the original kernel, so $L^{\ast }=L$. For a nonzero eigenfunction $Ly=\lambda y$,Thus $\boxed{\lambda \in \mathbb{R}}$. The argument includes the zero eigenvalue and does not require normalizing the eigenfunction.

Antisymmetry $k(x,t)=-k(t,x)$ instead gives $L^{\ast }=-L$, so $L$ is skew-adjoint. Its quadratic form satisfies $⟨y,Ly⟩=-\overline{⟨y,Ly⟩}$. Hence an eigenpair satisfiesNonzero eigenvalues are purely imaginary in the complexified function space; zero is allowed as well. On a strictly real vector space this means there are no nonzero real eigenvalues.

Write $c(x)=\cos 2x$. The separated kernel givesThus its image is $E=\mathrm{span}\{1,c\}$. Since $L1=\pi c$ and $Lc=\pi /2$, its matrix on this basis is $\left(\begin{array}{cc}0& \pi /2\\ \pi & 0\end{array}\right)$. ThereforeBoth eigenspaces have dimension one, and there are no further nonzero eigenvalues because every such eigenfunction belongs to the image. The zero eigenspace is $E^{\perp }$, which is infinite-dimensional. One explicit nonzero example is $\boxed{y(x)=\sin 2x}$, orthogonal to both $1$ and $\cos 2x$.

The image of $L$ consists only of $a+b\cos 2x$. If $\sin x$ were such a function, its endpoint values would give $a+b=0$ and its midpoint value $a-b=1$, forcing $a=1/2$, $b=-1/2$. That combination is ${\sin }^{2}x$, which differs from $\sin x$, for example at $x=\pi /6$. Continuity makes the same argument valid for equality almost everywhere. Thus the forcing is outside the image and no solution exists.

Here ${\sin }^{2}x=(1-\cos 2x)/2$ belongs to the image. Matching the two coefficients using $L(A+B\cos 2x)=B\pi /2+A\pi \cos 2x$ gives $A=-1/(2\pi )$, $B=1/\pi$. ConsequentlyThere are infinitely many solutions, since any element of the infinite-dimensional kernel may be added.

The eigenvalues of $L$ are $\pi /\sqrt{2}$, $-\pi /\sqrt{2}$ and zero, none equal to one. On $E=\mathrm{span}\{1,\cos 2x\}$ the map $L-I$ is an invertible two-dimensional operator; on $E^{\perp }$ it is $-I$. Hence it is invertible on all of $L^2(0,\pi )$. Thus $(L-I)y=\sin x$ has exactly one solution. This is the finite-rank Fredholm alternative; no range obstruction survives the nonresonant identity term.