Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / 4 / Solution

Split the Fourier transform integral at zero:By Fourier inversion applied to this pair, or by the residue theorem, the other transform isFor example, inversion with $\lambda =\mu$ gives $\int e^{ikx}(k^2+{\mu }^2{)}^{-1}dk=(\pi /\mu )e^{-\mu |x|}$; evenness supplies the desired sign convention.

Take the spatial Fourier transform of the Laplace equation. For $y>0$ it gives ${\hat{\psi }}_{yy}-k^2\hat{\psi }=0$. Removing the exponentially growing solution and matching the boundary givesAt $k=0$ this formula is interpreted by continuity; a single transform value does not alter the inverse integral. The previous transform pair identifiesthe Poisson kernel for the upper half-plane. The convolution theorem therefore gives the Poisson integralThis is the bounded harmonic solution for bounded decaying data, or the corresponding solution in an appropriate Fourier class. Such a growth condition matters if vertical decay is interpreted only pointwise: the nonzero harmonic function $e^{x^2-y^2}\sin (2xy)=\mathrm{Im}{e}^{(x+iy{)}^2}$ vanishes on the real axis and tends to zero for fixed $x$ as $y\to \infty$, but grows rapidly in $x$ and has no ordinary spatial Fourier transform. It can be excluded by requiring boundedness; the stated pointwise limits alone would not establish uniqueness.

For the interval data, direct integration gives the Poisson extension of interval boundary dataAs $y\downarrow 0$ it approaches $c$ for $|x|<a$, zero for $|x|>a$, and $c/2$ at $x=\pm a$. At those jump points the prescribed isolated values cannot be attained by a continuous extension; they do not affect the integral. The solution decays vertically, with $\psi \sim 2ac/(\pi y)$ for fixed $x$ and large $y$.