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

Use the unilateral Laplace transform convention $\overline{f}(p)={\int }_0^{\infty }{e}^{-px}f(x)dx$. The Laplace transform of a derivative isThe formula holds in the common Laplace convergence region under the usual endpoint and growth hypotheses. The initial value is the right-hand value at zero.

With the same unilateral Laplace transform convention, the standard sine formula is

The defining evaluation rule of the Dirac delta givesThe impulse lies strictly inside the integration domain, so there is no endpoint convention to settle. This Laplace transform exists for every complex $p$.

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.

Put $G(p)=\mathcal{L}[J_0](p)$, initially for real $p>0$. Besides the Laplace transform of a derivative, use multiplication by the original variable in the formTransforming the Bessel differential equation, with its given initial conditions, yieldsThe constant terms cancel, leaving $(1+p^2)G^{\prime }+pG=0$. This first-order ordinary differential equation integrates toThe initial value theorem for Laplace transforms gives ${\lim }_{p\to \infty }pG(p)=J_0(0)=1$, so $C=1$. Thus the Laplace transform of the Bessel function J0 isFor complex $p$ with $\mathrm{Re}p>0$, choose the analytic square-root branch agreeing with the positive square root on the positive real axis.

For the forcing $k=J_0$, part (b) now gives $F(p)=(1+p^2{)}^{-3/2}$. On the other hand, the same multiplication rule givesTaking an integral from zero divides the Laplace transform by $p$. By uniqueness of the Laplace transform,A direct check also explains the cancellation. This expression has $f^{\prime }=x{J}_0$ and $f^{\prime \prime }=J_0+x{J}_0^{\prime }$. The Bessel differential equation says $(x{J}_0^{\prime }{)}^{\prime }=-x{J}_0$, so integration from zero gives $x{J}_0^{\prime }=-{\int }_0^{x}t{J}_0(t)dt=-f$. Hence $f^{\prime \prime }+f=J_0$, and both initial conditions hold.