Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 5 / b

Transform only the time variable. Two integrations by parts give ${\mathcal{F}}_t[y_{tt}]=(i\omega {)}^2\tilde{y}=-{\omega }^2\tilde{y}$, when the ordinary transform boundary terms vanish; the same derivative rule holds for tempered distributions. Spatial derivatives and the fixed spatial delta commute with the transform. ThusorThe spatial Dirichlet boundary conditions transform to $\tilde{y}(\omega ,0)=\tilde{y}(\omega ,L)=0$.

Multiply the transformed equation by $\sin (k_{n}x)$ and integrate from zero to $L$. Two integrations by parts, using the zero endpoint values of both $\tilde{y}$ and the sine, giveThe delta source contributes $\tilde{F}(\omega )\sin (k_n{x}_0)$. Multiplying by $2/L$ and using the coefficient definition therefore yieldsAway from the resonant frequencies this givesAt $\omega =\pm k_n$ the quotient must be interpreted with a pole prescription if the source excites that mode. In distributional form, arbitrary delta terms at those frequencies may also be added; they are the free homogeneous oscillations. This is the frequency-space issue addressed explicitly in part (iii).

Expand in the complete Fourier sine series $\tilde{y}(\omega ,x)={\sum }_{n\ge 1}{b}_n(\omega )\sin (k_{n}x)$ and apply Fourier inversion. Substitution of part (ii) givesTo check the source normalization, apply ${\partial }_t^2-{\partial }_x^2$ mode by mode. It cancels the denominator; the remaining inverse transform gives $2\pi F(t)$, and completeness givesThe endpoints vanish termwise. These identities establish the equation in the distributional sense, with classical validity wherever the series and derivatives converge suitably.

The displayed real-axis integrals in the paper are not ordinary convergent integrals at their poles. One definite choice is the Cauchy principal value at each $\omega =\pm k_n$. If a causal solution is wanted, use $[k_n^2-(\omega -i0{)}^2{]}^{-1}$ instead. Its inverse transform is $H(t)\sin (k_{n}t)/k_n$, so the causal solution iswhen the past integral exists. The principal-value choice instead uses the time-symmetric kernel $\sin (k_n|t|)/(2k_n)$. No temporal initial data are given, so homogeneous sine and cosine modes remain permissible additions. This is the causal Fourier resolvent of a point-driven fixed string.