Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 5

The Fourier transform of $u_{xx}$ is $-k^2\tilde{u}$: two integrations by parts have no boundary terms under the stated decay assumption. Time differentiation commutes with the spatial transform. ThereforeFor every fixed spatial wavenumber $k$, this is a constant-coefficient ordinary differential equation in time.

The characteristic polynomial is $(r+\gamma {)}^2+c^2{k}^2$, with roots $r=-\gamma \pm ick$. For $ck\ne 0$, the transformed general solution isThe repeated-root case $ck=0$ is $e^{-\gamma t}[A_0+B_{0}t]$, rather than two independent copies of the same exponential. An alternative basis regular at $k=0$ is $e^{-\gamma t}[C(k)\cos (ckt)+D(k)\sin (ckt)/(ck)]$, using the limit $t$ for the last factor at zero.

For $c\ne 0$, the Fourier shift theorem turns the two exponential factors into translations. Equivalently, set $v=e^{\gamma t}u$; then $v_{tt}=c^2{v}_{xx}$. The D'Alembert formula gives the general formwhere $F,G$ are arbitrary functions with the required regularity and decay. The arbitrary profiles can accommodate a spatially transformed zero-frequency velocity by their integral components; the isolated repeated root must not be used to discard that degree of freedom. If $c=0$, the equation instead has $u=e^{-\gamma t}[F(x)+tG(x)]$.

This is the exponentially conjugated wave equation. Positive $\gamma$ produces damping, but the algebra also covers negative $\gamma$, which gives exponential amplification. The sign of $c$ is immaterial because the differential equation contains only $c^2$.

The initial profile is $f(x)=e^{-|x|}$, with the absolute value confirmed in the original PDF. Its Fourier transform isUsing ${\tilde{u}}_t(k,0)=0$ in the regular cosine-sine form giveswith the quotient interpreted as $t$ when $ck=0$. Put $v_0=|c|>0$. Inverting, or applying the D'Alembert formula to $e^{\gamma t}u$, whose initial velocity is $\gamma f$, givesFor an explicit expression at $t\ge 0$, let $r=|x|$ and $s=v_{0}t$. The integral is $2e^{-r}\sinh s$ for $r\ge s$ and $2(1-e^{-s}\cosh r)$ for $r\le s$. ThereforeThe branches agree at $r=v_{0}t$. At zero time, the integral vanishes and the translated profiles coincide, recovering $f$. Their time derivatives cancel, while the integral contributes $\gamma f$, cancelling the derivative of the prefactor, so $u_t(x,0)=0$.

For $c=0$ the limiting answer is $u=(1+\gamma t)e^{-\gamma t}{e}^{-|x|}$. The cusp in $f$ means the solution is understood as a weak solution, classically smooth away from the characteristic fronts $x=\pm v_{0}t$; it is not globally twice differentiable. No assumption that $\gamma$ is positive is needed for this initial-value solution.