Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 327 / 2

The space of test functions is $\mathcal{D}(\mathbb{R})=C_c^{\infty }(\mathbb{R})$. A sequence ${\phi }_j$ converges to $\phi$ when its supports eventually lie in a single compact set and every derivative converges uniformly. The distribution space ${\mathcal{D}}^{\prime }(\mathbb{R})$ is its continuous linear dual; weak convergence means $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every test function. We use $⟨u^{\prime },\phi ⟩=-⟨u,{\phi }^{\prime }⟩$ and $⟨au,\phi ⟩=⟨u,a\phi ⟩$ for multiplication of a distribution by a smooth function.

If a test function $\phi$ has integral zero, thenis another compactly supported smooth test function with ${\psi }^{\prime }=\phi$. Hence $v^{\prime }=0$ implies $⟨v,\phi ⟩=-⟨v^{\prime },\psi ⟩=0$ for every such $\phi$.

Choose a test function $\eta$ with integral one. For arbitrary $\phi$, the function $\phi -\eta \int \phi$ has zero integral, soThus $v$ is the constant distribution $⟨v,\eta ⟩$. This proves a distribution with zero derivative is constant.

The principal-value reciprocal distribution isThe limit exists because the constant part of the numerator cancels symmetrically at zero. Multiplying by $x$ gives $x\mathrm{pv}(1/x)=1$.

If $v$ is any other solution, $w=v-\mathrm{pv}(1/x)$ obeys $xw=0$. To identify this kernel of multiplication by a coordinate, choose a cutoff function $\eta$ equal to one near zero. Every test function has the form $\phi =\phi (0)\eta +x\psi$ with $\psi \in \mathcal{D}$. Thus $⟨w,\phi ⟩=\phi (0)⟨w,\eta ⟩$, proving $w=c{\delta }_0$. ThereforeFor real distributions the constants are real.

The Leibniz rule rewrites the left side as $(xu{)}^{\prime }$. By a distribution with zero derivative is constant,The Dirac delta multiplication identity $x{\delta }_0^{\prime }=-{\delta }_0$ supplies a particular solution, and the preceding division result supplies all solutions of $xu=c$. HenceMultiplying by $x$ and then differentiating verifies the equation. The coordinate-multiplication kernel and the constant-derivative kernel show that no additional solutions are missing.

The distributional derivative of the Heaviside step function gives $H^{\prime }={\delta }_0$, while $((x-1)H(x-1){)}^{\prime \prime }={\delta }_1$. A particular solution is therefore $H(x)-2(x-1)H(x-1)$.

The difference from any other solution has second derivative zero. Applying a distribution with zero derivative is constant twice shows that difference is affine. ThusThe ramp function is continuous at $x=1$, and its first derivative jumps there, producing the indicated delta source in its second derivative.

Since $(x^2-1)(-{\delta }_0)={\delta }_0$, write $u^{\prime }=-{\delta }_0+w$, where $(x^2-1)w=0$. The polynomial has two simple zeros. The distribution annihilated by a function with simple zeros lemma givesIntegrating with the Heaviside step function yieldsThis includes every solution: its derivative has exactly the required delta terms, and any two primitives differ only by a constant distribution.