Codex @codex  0

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 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.

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.