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

The characteristic polynomial is $r^2+2r+2=(r+1{)}^2+1$, with distinct complex roots $r=-1\pm i$. Taking the real and imaginary parts of the corresponding exponential functions gives the real fundamental system of solutions $e^{-x}\cos x$ and $e^{-x}\sin x$. Consequently the general real solution isTheir Wronskian is $e^{-2x}\ne 0$, so they span the two-dimensional real solution space.

For each $0<\xi <2\pi$, the Green's function is a distribution in $x$ satisfyingEquivalently, $g$ is a classical homogeneous solution on each side of $\xi$, satisfies the same endpoint conditions, and obeys the Green function jump conditionsTo see equivalence, let $[g]$ and $[g_x]$ denote its jumps. The singular part of $L_{x}g$ is $[g]{\delta }^{\prime }(x-\xi )+([g_x]+2[g])\delta (x-\xi )$. Matching a unit Dirac delta distribution with no derivative first forces continuity, then the unit derivative jump. Conversely these conditions produce exactly that distribution. The solution for a regular forcing is $y(x)={\int }_0^{2\pi }g(x,\xi )f(\xi )d\xi$. The homogeneous endpoint problem has only the zero solution, which ensures uniqueness of this kernel.

Choose a left-end solution $u(x)=e^{-x}\sin x$ and a right-end solution $v(x)=e^{-x}(\sin x+\cos x)$. They satisfy $u(0)=0$ and $v^{\prime }(2\pi )=0$. Their Wronskian, with the order used here, isThe Green's function is $u(x_{<})v(x_{>})/W(\xi )$, where $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. Explicitly,The two expressions agree at the source. Their derivative difference is $[u(\xi )v^{\prime }(\xi )-u^{\prime }(\xi )v(\xi )]/W(\xi )=1$, checking both normalization and sign. The first-order derivative term means the kernel need not be symmetric under exchange of $x$ and $\xi$.

A distribution is a continuous linear functional on the space of test functions $\mathcal{D}(\mathbb{R})=C_c^{\infty }(\mathbb{R})$. A precise useful form of continuity is: for every compact set $K\subset \mathbb{R}$ there exist an integer $m\ge 0$ and a constant $C_K$ such thatEquivalently, if a sequence of test functions has supports in one fixed compact set and every derivative tends uniformly to zero, then its images under $T$ tend to zero. Keeping a common compact support is part of the test function topology; mere pointwise convergence is insufficient.

Linearity follows from linearity of differentiation and point evaluation. For a test function supported in a fixed compact set $K$, its value and derivative vanish at any point outside $K$. ThereforeThis proves the continuity estimate with $m=1$ and $C_K=2$ for every $K$. Thus the stated mapping is a distribution of order at most one, supported in $\{0,1\}$. In particular, a sequence tending to zero in the test function topology has both ${\varphi }_n^{\prime }(0)\to 0$ and ${\varphi }_n(1)\to 0$.

The distributional derivative is defined by moving the derivative onto a test function and changing the sign:Since $⟨{\delta }_a,\varphi ⟩=\varphi (a)$, one has $⟨{\delta }_0^{\prime },\varphi ⟩=-{\varphi }^{\prime }(0)$. HenceThe minus sign in front of ${\delta }_0^{\prime }$ is essential: it makes its action the positive derivative evaluation required here.

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 $Ly=f$.

Normalize the Laguerre differential equation as $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$, where $P(x)=1/x-1$ and $Q(x)=n/x$. At $x=0$, the products $xP=1-x$ and $x^{2}Q=nx$ are analytic. Thus zero is a regular singular point, with indicial equationAt $x=1$ the coefficients $P,Q$ are analytic, so it is an ordinary point. No singular-point indicial equation is needed there.

For infinity put $z=1/x$ and $Y(z)=y(1/z)$. Then the equation becomesAt $z=0$, the coefficient $z(z^{-1}+z^{-2})=1+1/z$ is not analytic. Therefore infinity is an irregular singular point, including when $n=0$. A terminating polynomial solution does not change the classification of the differential equation.

Substitute $y={\sum }_{k\ge 0}{a}_k{x}^k$ and compare the coefficient of $x^k$. The two derivative terms give $(k+1{)}^2{a}_{k+1}$, while the remaining terms give $(n-k)a_k$. ThusIterating the recurrence givesIndeed the recurrence first vanishes at $k=n$ and then remains zero. Consequently the Laguerre polynomial iswith exact degree $n$ and leading coefficient $(-1{)}^n/n!$. This is the regular power-series branch normalized at zero; the second branch at the repeated indicial root is not an additional analytic solution with this normalization.

Assume first that $n\ge m$. Substituting Rodrigues' formula for $L_n$ givesHere integration by parts was performed $n$ times. Each boundary term vanishes: at infinity a polynomial times $e^{-x}$ tends to zero; at zero every derivative of $x^n{e}^{-x}$ of order less than $n$ vanishes. If $n>m$, the derivative $L_m^{(n)}$ is zero, proving orthogonality. The case $m>n$ follows by exchanging the two factors.

For $m=n$, the leading coefficient found above gives $L_n^{(n)}=(-1{)}^n$. ThereforeThe same conclusion includes $n=0$ directly. Thus the full Laguerre orthonormality relation is