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

A distribution is a linear functional on the space of test functions that is continuous in its test-function topology. Equivalently, for every compact $K\subset \mathbb{R}$ there are an integer $m$ and a constant $C_K$ such that $|⟨T,\varphi ⟩|\le C_K{\max }_{0\le j\le m}\Vert {\varphi }^{(j)}{\Vert }_{\infty }$ whenever $\varphi$ has support in $K$.

Linearity of $T_n$ follows from linearity of the integral and derivative. Since the integration interval has length $2/n$,This is a test-function seminorm estimate of order one, valid for every $K$ and even uniformly in $n$. It proves that each $T_n$ is a distribution.

The function $\varphi -{\varphi }^{\prime }$ is continuous at zero, so its average over $[-1/n,1/n]$ tends to its value there. ConsequentlyThe Dirac delta satisfies $⟨\delta ,\varphi ⟩=\varphi (0)$, and its distributional derivative satisfies $⟨{\delta }^{\prime },\varphi ⟩=-{\varphi }^{\prime }(0)$. Hence the distributional convergence is

The characteristic polynomial after multiplication by four is $r^2+4r+4+{\pi }^2$, with roots $-2\pm i\pi$. Thus the general real solution of this constant-coefficient ordinary differential equation isEquivalently the change of dependent variable $y=e^{-2x}v$ gives $Ly=\frac{1}{4}{e}^{-2x}(v^{\prime \prime }+{\pi }^{2}v)$.

In the distribution definition, the Green's function is characterized, for fixed $0<\xi <1$, byIn the classical definition, solve $L_{x}g=0$ separately for $0<x<\xi$ and $\xi <x<1$, with those endpoint conditions, and imposeA jump in $g$ would introduce a distributional derivative of the Dirac delta, absent from the right side. Continuity eliminates it; integrating the equation across $\xi$ leaves one quarter of the derivative jump equal to one. These Green function jump conditions make the definitions equivalent. The forced solution is $y(x)={\int }_0^{1}g(x,\xi )f(\xi )d\xi$.

Under $y=e^{-2x}v$, the left Robin boundary condition becomes $v^{\prime }(0)-v(0)=0$ and the right Dirichlet boundary condition becomes $v(1)=0$. ChooseTheir Wronskian $v_L{v}_R^{\prime }-v_L^{\prime }{v}_R$ is the constant $\pi$. For $y_L=e^{-2x}{v}_L$ and $y_R=e^{-2x}{v}_R$, it is $W_y(x)=\pi e^{-4x}$. The coefficient enforcing the derivative jump four is $4/W_y(\xi )$. Writing $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$, the Green's function isThe endpoint conditions, continuity and derivative jump follow directly from this Wronskian construction. Its exponential factor also shows why the kernel need not be symmetric in the unweighted inner product.

Put $t=\log x$ and $Y(t)=y(e^t)$. Then $x{y}^{\prime \prime }+y^{\prime }=Y^{\prime \prime }/x$, so the eigenvalue problem reduces toA nonzero Dirichlet eigenfunction therefore has $4+\lambda =n^2$ for an integer $n\ge 1$. The complete family isThe alternatives $4+\lambda \le 0$ give only the zero solution under both endpoint conditions.

The operator itself is self-adjoint with the given unweighted inner product, on the usual regular Dirichlet domain. Indeed $My=(x{y}^{\prime }{)}^{\prime }+4y/x$, and integration by parts givesfor functions satisfying the endpoint conditions. This is a regular Sturm-Liouville operator with positive leading coefficient $x$, whose Dirichlet realization is self-adjoint. The displayed generalized eigenproblem contains the additional weight $1/x$; its eigenfunctions are accordingly orthogonal with respect to $\int fg/x$, which is separate from the self-adjointness assertion about $M$.

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.

After division by $x$, the ordinary differential equation has coefficients $P(x)=(m+1)/x-1$ and $Q(x)=n/x$. At zero, $xP\to m+1$ and $x^{2}Q\to 0$, making zero a regular singular point. The indicial equation isso the indicial exponents are $\boxed{0,-m}$, coincident when $m=0$.

At infinity put $t=1/x$ and $Y(t)=y(1/t)$. The transformed equation isThe coefficient of $Y^{\prime }$ already fails the regular-singularity test at $t=0$. Thus $\boxed{x=\infty \text{ is an irregular singular point}}$, including when $n=0$.

There is a normalization conflict in the source: the printed Rodrigues' formula gives the standard Generalized Laguerre polynomial $P_n=L_n^{(m)}$ with $P_n(0)=\left(\genfrac{}{}{0ex}{}{n+m}{n}\right)$, which is generally not one. The norm requested here corresponds to that Rodrigues normalization. A solution normalized to one at zero is ${\hat{P}}_n=P_n/\left(\genfrac{}{}{0ex}{}{n+m}{n}\right)$.

Set $w=x^m{e}^{-x}$ and $p=x^{m+1}{e}^{-x}$. The equation becomes $(p{P}_n^{\prime }{)}^{\prime }+nw{P}_n=0$. Multiply the equations for $P_n,P_k$ by the opposite polynomial, subtract and integrate. The boundary term $p(P_n^{\prime }{P}_k-P_k^{\prime }{P}_n)$ vanishes at zero and infinity, soThis proves orthogonality for $k\ne n$ under either normalization.

For the squared norm, use the printed Rodrigues' formula and integrate by parts $n$ times:All boundary terms vanish: the required derivatives have a positive power of $x$ at zero and exponential decay at infinity. The leading coefficient is $(-1{)}^n/n!$, so $P_n^{(n)}=(-1{)}^n$. The Gamma function integral givesFor the source's alternative unit-at-zero normalization the correct answer instead isBoth cannot equal the printed norm when $m,n>0$. This distinction is the normalization of generalized Laguerre polynomials.

For $m=0$ the two normalizations agree, and the Laguerre polynomials are orthonormal with weight $e^{-x}$. Take real $-1<h<1$. Their expansion coefficients for the requested function areSubstitute Rodrigues' formula and integrate by parts $n$ times. Differentiating $e^{-xh/(1-h)}$ transfers a factor $[h/(1-h){]}^n$ after the two sign factors cancel. ThereforeThe integral is $n!(1-h{)}^{n+1}$ by the Gamma function. Boundary terms vanish throughout this range.

The resulting generating function for Laguerre polynomials isOne can also verify convergence and equality directly: Rodrigues' formula gives $L_n(x)={\sum }_{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)(-x{)}^j/j!$, and summing ${\sum }_{n\ge j}\left(\genfrac{}{}{0ex}{}{n}{j}\right)h^n=h^j/(1-h{)}^{j+1}$ reduces the double series to the exponential on the left. The absolute-value double series converges for fixed $x$ and $|h|<1$, justifying the interchange; the formula then also holds for complex $h$ in that disk.