Past exam of the short options course of the University of Oxford / 2015 / 5 / b

Let $K(p)=\mathcal{L}[k](p)$ and $F(p)=\mathcal{L}[f](p)$. Apply the Laplace transform of a derivative twice and use the two zero initial conditions. The differential equation becomesThe factor $1/(p^2+1)$ is the Laplace transform of $\sin x$. The convolution theorem for Laplace transforms therefore givesThis is a causal convolution: the value at $x$ uses only the forcing up to $x$. For ordinary continuous $k$, direct differentiation gives $f^{\prime }(x)={\int }_0^x\cos (x-t)k(t)dt$ and $f^{\prime \prime }(x)=k(x)-f(x)$, verifying both the differential equation and the initial conditions. For an impulsive $k$, the same formula is interpreted using distributions.

An impulse at $a$ shifts the causal sine response. Substituting the Dirac delta in the convolution from part (i), or using the Laplace transform time-shift rule, giveswhere $H$ is the Heaviside step function. The answer is continuous at $a$, while its first derivative jumps from $0$ to $1$. Consequently its second distributional derivative contains exactly one unit Dirac delta at $a$, giving $f^{\prime \prime }+f=\delta (x-a)$. Since $a>0$, both initial conditions vanish.