Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 1 / 14E (Further Complex Methods)

Extend $q$ by zero to $x<0$ and apply the assumed Fourier inversion theorem. For sufficiently regular, decaying data and $x>0$ this gives $q(x)=(2\pi {)}^{-1}{\int }_{\mathbb{R}}{e}^{ikx}\hat{q}(k)dk$, with $\hat{q}(k)={\int }_0^{\infty }{e}^{-ikx}q(x)dx$. If $\mathrm{Im}k\le 0$, the exponential in this last integral is bounded in magnitude by one; it defines a holomorphic transform in the lower half-plane, with its boundary values on the real axis.

The PDF orients $L$ from $i\infty$ down to zero, then from zero to $+\infty$. The function $e^{ikx}\hat{q}(-k)$ is holomorphic in the upper half-plane, and for $x>0$ its exponential decays there. Close the two rays through a large first-quadrant arc. Cauchy integral theorem and Jordan lemma giveFor example, if $q,q^{\prime }$ are integrable, integration by parts gives $\hat{q}(-k)=O(1/k)$ in that quadrant, which suffices for the arc estimate. Standard limiting arguments extend the identity to the admissible Fourier-inversion class. Adding any constant multiple of this zero integral to ordinary inversion provesThe extra integral is a reflected, zero-supported contribution, not an extra datum.

Set $\Theta =e^{-ikx+i{k}^{2}t}$. Expanding the proposed local relation givesThus its vanishing is exactly $i{q}_t+q_{xx}=0$. Integrating over $x>0$, using decay at infinity, yieldsIntegrate in time and use the initial data to obtain the global relationHere ${\tilde{g}}_j(z,t)={\int }_0^t{e}^{iz\tau }{g}_j(\tau )d\tau$, with $g_0=q(0,\cdot )$ and $g_1=q_x(0,\cdot )$. In particular these finite-time transforms are entire in $z$.

Write $F(k)=e^{ikx-i{k}^{2}t}$ and $B_{\pm }(k)=k{\tilde{g}}_0(k^2,t)\pm i{\tilde{g}}_1(k^2,t)$. Substitution into part (i), including replacing $k$ by $-k$ in the reflected transform, givesFor the boundary integrands, rotate the negative real ray to the positive imaginary ray through the second quadrant. There are no singularities, and for $k=a+ib$ with $a\le 0,b\ge 0$,After writing $F{B}_{+}$ as its finite time integral, this estimate and the usual smooth-data arc bounds justify the rotation. The positive real ray stays fixed and the new imaginary ray is directed downwards, givingThe same argument applies separately to the ${\tilde{g}}_0$ and ${\tilde{g}}_1$ terms. Thus the net boundary integral in the inversion formula isFor Dirichlet boundary conditions choose $c=-1$, eliminating the unknown normal derivative. For Neumann boundary conditions choose $c=+1$, eliminating the unknown boundary value. The resulting unified transform representations areEach contains only the initial transform and the prescribed boundary transform. The signs depend on the PDF's orientation of $L$, checked above.