Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 4 / 14A (Complex Methods)

For $a>0$, the Heaviside step function vanishes before $a$ and is one afterwards. For $\mathrm{Re}p>0$,The value assigned at the single jump point does not affect this Laplace transform. The positive real part ensures convergence at infinity.

Write $F(p)=\mathcal{L}[f](p)$. Assume $f$ and $f^{\prime }$ have right limits at zero and sufficient regularity and exponential-order bounds for the following integration by parts; take $\mathrm{Re}p$ large enough to kill the boundary terms at infinity. First,Apply the same calculation to $f^{\prime }$:This is the Laplace transform of a derivative formula with both initial-data terms retained.

Transform in time and use the zero interior initial conditions. The Laplace transform of a derivative formula turns the wave equation into $c^2{Y}_{xx}=p^{2}Y$. Its transformed endpoint conditions are $Y(0,p)=a/p$ and $Y(L,p)=0$. A homogeneous solution satisfying the right condition is a multiple of $\sinh [p(L-x)/c]$; the left condition fixes that multiple. Thus, for $\mathrm{Re}p>0$,Here $L,c>0$ and the denominator has no zeros in this half-plane. The initial-boundary corner at $(0,0)$ is deliberately discontinuous: the applied step launches propagating fronts, so one should not demand a globally smooth solution across them.

Rewrite the ratio of hyperbolic sines asFor $\mathrm{Re}p>0$, expand the denominator by the geometric series. Part (a) inverts each delayed exponential:At each finite time only finitely many terms are nonzero, so this reflected-step solution of the wave equation is locally a finite sum. The first front travels from $x=0$; the second is its sign-reversed reflection from the fixed endpoint, with repeated round trips. At $x=L$ the delays coincide and cancel. At $x=0$ the sum telescopes to $a$ for $t>0$.

An equivalent standing-wave Fourier series isIts static ramp has exactly the sine coefficients $2a/(n\pi )$, so cancellation gives the zero initial displacement in the interior. Both representations solve the equation away from the fronts and in the distributional sense across them; the convention $H(0)=1/2$ agrees with Fourier midpoint values.

Introduce $\tau =ct/L$. At the midpoint, the positive fronts arrive at $\tau =2m+1/2$ and the negative fronts at $\tau =2m+3/2$. HenceThe midpoint displacement is a square wave, initially zero, with period $2L/c$ and a phase delay $L/(2c)$. At a front, choosing $H(0)=1/2$ gives value $a/2$. The dotted vertical segments below denote jumps; they are not continuous changes through intermediate displacements.