Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2 / 14C (Further Complex Methods)

Use the half-line transform $\hat{u}(k,t)={\int }_0^{\infty }{e}^{-ikx}u(x,t)dx$ and let $g_j(t)={\partial }_x^{j}u(0,t)$. Integrating the heat equation twice by parts givesHere ${\hat{u}}_0(k)=(1+ik{)}^{-2}$ andwith removable singularities at the apparent denominator zeros. Let $D^{+}=\{\mathrm{Im}k>0,\mathrm{Re}{k}^2<0\}$, and orient its boundary from infinity on the $3\pi /4$ ray to zero and then out on the $\pi /4$ ray. Transform inversion and contour deformation giveTo eliminate the unknown boundary derivative, evaluate the transform identity at $-k$: ${\tilde{g}}_1={\hat{u}}_0(-k)+ik{\tilde{g}}_0-e^{k^{2}t}\hat{u}(-k,t)$. The resulting extra contour integral of $e^{ikx}\hat{u}(-k,t)$ vanishes by analyticity and decay in the upper half-plane. This proves that the boxed expression involves only the prescribed data. It is intended for $x,t>0$, with boundary and initial values obtained as limits.

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.