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

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.