Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2011 / Paper 2 / 5 / Solution

First prove the convolution theorem with the stated negative-exponential Fourier transform. Assuming absolute integrability sufficient for Fubini's theorem, change variables $u=t-s$:

For the differential equation, integration by parts givesThe boundary terms vanish by the assumed decay of $y,y^{\prime }$. As with the convolution step, the transforms can first be justified for integrable functions and derivatives; the resulting kernel gives the desired decaying response. ThusThe convolution theorem and Fourier inversion theorem imply $y=f\ast G$.

For the inverse transform of $\tilde{G}$, useThe poles are at $ib,ia$. For $t>0$, close the inverse-transform contour above; the residues of $e^{ikt}\tilde{G}(k)$ there are $-i{e}^{-bt}/(a-b)$ and $i{e}^{-at}/(a-b)$. The inverse-transform factor $1/(2\pi )$ makes the answer $i$ times their sum. For $t<0$, close below, enclosing no poles. The large arcs vanish: the complete rational factor is $O(|k{|}^{-2})$ and the exponential is bounded on the appropriate semicircle. Therefore the causal Green function of an overdamped oscillator isIt is continuous, with $G(0)=0$ and $G^{\prime }(0^{+})=1$, so $(D_t+a)(D_t+b)G=\delta (t)$. The response can also be written $y(t)={\int }_{-\infty }^{t}G(t-s)f(s)ds$.

For the bilateral exponential forcing, split the real integral at zero:ConsequentlyFor negative time the lower half-plane contains only $k=-i$. The residue of $e^{ikt}\tilde{y}$ there is $i{e}^t/[(a+1)(b+1)]$. Clockwise closure makes the inverse transform $-i$ times that residue. For positive time the three upper residues, at $i,ia,ib$ respectively, areMultiplication of their sum by $i$ gives the bilateral exponential forcing of an overdamped oscillator:Both expressions have $y(0)=y^{\prime }(0)=1/[(a+1)(b+1)]$. This verifies the matching that prevents an extra Dirac delta function at zero. Direct substitution gives the forcing on either side, and all terms decay at their corresponding infinities. A causal Green function does not imply zero response before $t=0$ when the forcing has already been acting at negative times. The assumptions $a,b\ne 1$ avoid repeated poles; their limiting cases would contain polynomial times exponential terms.