Past exam of the differential equations and applications course of the University of Oxford / 2011 / A / 1

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.