Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 327 / 2

Use $D=-i\partial$, so $P(D)e^{i\lambda \cdot x}=P(\lambda )e^{i\lambda \cdot x}$. A concrete distributional division construction isThe locally integrable family, initially defined for sufficiently large $\mathrm{Re}s$, has a meromorphic continuation; $\mathrm{FP}$ denotes its finite part at zero. Multiplication before continuation givesThe right side is holomorphic at $s=0$ with value $1$, so comparison of constant Laurent coefficients yields $PT=1$. Thereforesatisfies $P(D)E={\delta }_0$. This finite-part formula explicitly realizes the Malgrange–Ehrenpreis theorem.

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For each one-dimensional factor $P_j(D_j)$, choose its retarded fundamental solution $E_j=H(x_j)g_j(x_j)$. Partial-fraction decomposition of the reciprocal polynomial giveswhere each ${\alpha }_{rj}$ is a polynomial whose degree is one less than the multiplicity of the associated root. Constants, including powers of $i$ from $D=-i\partial$, can be absorbed into the polynomials and exponents.

Take the tensor productIt vanishes unless every $x_j>0$, has the required polynomial-exponential form there, and satisfies

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Factor the operator asSet $f(x)=H(x)\sin x$ and $g(x)=H(x)\sinh x$. Since both vanish at zero and have right derivative one, their distributional second derivatives areBecause $D^2=-{\partial }^2$,Consequently $E(x_1,x_2)=f(x_1)g(x_2)$ obeysIt equals $\sin x_1\sinh x_2$ in the positive quadrant and zero otherwise.

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