Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 4 / 15D (Complex Methods)

Use the Fourier transform convention $\hat{u}(k,y)={\int }_{\mathbb{R}}u(x,y)e^{-ikx}dx$ and inverse factor $1/(2\pi )$. Laplace equation gives ${\hat{u}}_{yy}-k^2\hat{u}=0$. Selecting its decaying Fourier branch gives $\hat{u}(k,y)={\hat{u}}_0(k)e^{-|k|y}$, the standard bounded solution for bounded continuous boundary data. The inverse transform of the multiplier is the Poisson kernel for the upper half-plane:The convolution theorem therefore gives the Poisson integralFor suitably regular boundary data, this kernel is an approximate identity as $y\downarrow 0$, recovering $u_0$.

For $a>0$, contour integration gives ${\hat{u}}_0(k)=(\pi /a)e^{-a|k|}$. Multiplication by $e^{-y|k|}$ and inversion yieldThe stated form assumes $a>0$; if $a<0$, replace it by $|a|$ because the boundary function only depends on $a^2$. The case $a=0$ has singular boundary data and is outside this regular calculation.

The transform argument requires a growth condition. A zero-boundary harmonic function with pointwise vertical decay shows that the printed conditions alone do not ensure uniqueness:vanishes at $y=0$ and tends to zero as $y\to \infty$ for every fixed $x$. It can be added without changing those conditions, but its growth in $x$ excludes it from the bounded solution class. The displayed Poisson integral is the standard bounded solution for regular bounded boundary data; pointwise vertical decay by itself cannot justify a uniqueness claim.