Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2 / 14C / ii / Solution

Split the real contour at zero and rotate its positive part to angle $\theta \in (0,\pi /4)$ and its negative part to angle $\pi -\theta$. Rotate the two rays of $\partial D^{+}$ to these same angles, preserving orientation. The initial-transform pole at $k=i$ is not crossed by either real-contour rotation, while ${\hat{u}}_0(-k)$ has its pole at $-i$, outside the upper half-plane. The apparent poles in ${\tilde{g}}_0$ are removable.

On the rotated rays, both $\mathrm{Im}k>0$ and $\mathrm{Re}{k}^2>0$. The initial-data factor has Gaussian decay $|e^{ikx-k^{2}t}|=e^{-x\mathrm{Im}k-t\mathrm{Re}{k}^2}$. For the boundary term, retain the integral formwhich is $O(|k{|}^{-2})$ there. Multiplication by $e^{ikx}$ gives exponential decay as $|k|\to \infty$ for every fixed $x>0$. Thus all integrands are exponentially damped on the deformed contours; the endpoint boundary value at $x=0$ is taken afterwards, rather than claiming uniform exponential decay there.