Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 5 / Solution

The one-sided Laplace transform is $\bar{f}(p)={\int }_0^{\infty }{e}^{-pt}f(t)dt$. Its Bromwich inversion formula is $f(t)=(2\pi i{)}^{-1}{\int }_{\gamma -i\infty }^{\gamma +i\infty }{e}^{pt}\bar{f}(p)dp$, with an upward vertical contour to the right of all singularities and within the appropriate convergence half-plane.

To derive the convolution theorem for Laplace transforms, transform $x(t)={\int }_0^{t}f(\tau )g(t-\tau )d\tau$. With $u=t-\tau \ge 0$, the triangular integration region becomes a quadrant. Fubini's theorem gives $\bar{x}(p)={\int }_0^{\infty }{\int }_0^{\infty }{e}^{-p(u+\tau )}f(\tau )g(u)dud\tau =\bar{f}(p)\bar{g}(p)$, under the usual absolute convergence hypotheses. Inversion gives the required convolution formula.

For the differential equation, the transformed initial conditions give $(p^2-1)\bar{y}=\bar{g}+p+1$, hence $\bar{y}=\bar{g}/(p^2-1)+1/(p-1)$. Since $\mathcal{L}[\sinh t]=1/(p^2-1)$ for $\mathrm{Re}p>1$, the convolution yieldsThe integral and its first derivative vanish at zero, so the homogeneous term supplies both initial values. Differentiating twice gives the forcing through the unit derivative of the hyperbolic-sine kernel at zero.