Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 5 / b

Substitution givesThe inverse Fourier transform is $(2\pi {)}^{-1}{\int }_{\mathbb{R}}\tilde{y}(\omega )e^{i\omega t}d\omega$. For $t<0$, close clockwise in the lower half-plane, where there are no poles; hence $y(t)=0$.

For $t>0$, close counterclockwise in the upper half-plane. The three simple-pole residues of the full inversion integrand areMultiplying their sum by $i$ and combining the conjugate exponential contributions yieldsThe large arcs vanish because the rational factor is $O({\omega }^{-3})$ and the exponential is bounded on the chosen half-plane. At zero, the right-hand value and first derivative are both zero, matching the negative-time branch, so no unwanted delta or delta derivative appears in the second-order equation.

For an independent dynamical check, this is the exponential forcing of an underdamped oscillator with damping $\gamma =1$, natural frequency $\sqrt{5}$, damped frequency $2$, exponent $\beta =2$ and force amplitude $-1$. For positive time its substitution gives $y^{\prime \prime }+2y^{\prime }+5y=-e^{-2t}$.

The result of part (i) vanishes identically on the open interval $t<0$, so its first and second derivatives also vanish there. Substituting into the equation of motion immediately givesThus the force has no negative-time prehistory, consistent with the causal Green function of a damped oscillator. In fact, direct inversion or the positive-time substitution above identifies $f(t)=-e^{-2t}H(t)$; the given numerator $+i$ corresponds to a negative exponential force with this Fourier convention. Here $H$ is the Heaviside step function; its value exactly at the jump does not affect the transform.