Past exam of the differential equations and applications course of the University of Oxford / 2010 / A / 2

Set $t=\log x$, so $0<t<\pi$, and write $y(x)=x^{-2}w(t)$. Direct differentiation of this Cauchy-Euler equation givesThe eigenvalue equation is therefore $-w^{\prime \prime }=(\lambda -3)w$ with $w(0)=w(\pi )=0$. For $\lambda -3\le 0$, the linear/hyperbolic solutions cannot satisfy both endpoint conditions except trivially. For $\lambda -3>0$, the sine solution vanishes at both endpoints exactly when its frequency is a positive integer. ThusEach associated eigenspace is one-dimensional.

The displayed differential operator is not self-adjoint for the unweighted inner product. Let $A=x^2{\partial }_x^2+5x{\partial }_x+1$. Integration by parts on compactly supported smooth functions gives its formal adjointwhich differs from $A$. One can see an explicit failure of symmetry without boundary terms: take a nonzero real compactly supported smooth $f$ in the interval and $g=xf$. ThenBoth test functions obey the Dirichlet endpoint conditions. Changing the sign of $A$ for the eigenproblem does not repair this failure. A different, weighted inner product with weight $x^3$ does give the Sturm-Liouville form $x^{3}Ay=(x^5{y}^{\prime }{)}^{\prime }+x^{3}y$, explaining why a real discrete spectrum is nevertheless natural.

Use the cosine difference identity to separate the kernel:Thus $L$ has image $E=\mathrm{span}\{c,s\}$, where $c(x)=\cos 2x$, $s(x)=\sin 2x$. Their squared norms in $L^2(0,\pi )$ are $\pi /2$ and they are orthogonal. ConsequentlyEvery nonzero eigenvalue must belong to this two-dimensional image, so the complete list of eigenvalues of finite multiplicity isEach has multiplicity one. On $E^{\perp }$ the operator vanishes, giving the zero eigenvalue infinite multiplicity. For example $\boxed{y(x)=1}$ is a nonzero element of the kernel because both relevant integrals over the interval vanish.

The forcing is $\frac{1}{2}\sin 2x$, which lies in the image of $L$. Since the eigenvalue on $\sin 2x$ is $\pi /4$, a particular solution is $y_p=(2/\pi )\sin 2x$. The full family isThus solutions exist and are not unique: adding any element of the infinite-dimensional kernel leaves $Ly$ unchanged. For example every constant can be added.

Every image of $L$ is a linear combination $a\cos 2x+b\sin 2x$. The function $\cos x$ is not such a combination. Indeed these functions are continuous, so equality almost everywhere would imply equality everywhere; evaluation at the endpoints would require simultaneously $a=1$ and $a=-1$. Therefore the forcing is outside the image and there is no solution. Orthogonality to the kernel is the equivalent image test for this self-adjoint finite-rank operator.

Rewrite the equation as $(L-I)y=\cos x$. On $E=\mathrm{span}\{\cos 2x,\sin 2x\}$, the operator $L-I$ has eigenvalues $\pi /2-1$ and $\pi /4-1$, both nonzero. On $E^{\perp }$ it equals $-I$. The orthogonal decomposition $L^2(0,\pi )=E\oplus E^{\perp }$ therefore makes $L-I$ bijective with bounded inverse. In particular there is exactly one solution for this forcing. This is also the finite-rank Fredholm alternative: $1$ is not an eigenvalue of $L$, so adding the identity term removes the image obstruction present in $Ly=f$.