Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 2 / 14A (Methods)

Direct evaluation of the integral givesThe continuous value at $k=0$ is zero, in agreement with the vanishing integral of the odd function $f$. Thus

Because $e^{-\lambda |k|}$ is an even function, Fourier inversion reduces to a cosine integral:Therefore

Extend the boundary data oddly to the whole real axis. The extended function is precisely the function $f$ from part (a). Taking the Fourier transform in $x$, the Laplace equation becomesDecay as $y\to \infty$ selectsPart (b), together with the convolution theorem, says that the inverse transform is convolution with the Poisson kernelFor $x,y>0$ this givesThe odd extension enforces $u(0,y)=0$, while the Poisson integral takes the prescribed boundary values at every continuity point and decays at infinity.

By the superposition principle, interchange $x$ and $y$ in the solution from part (c) and add the two solutions:The first term supplies the required data on the positive $x$-axis and vanishes on the positive $y$-axis; the second does the reverse. Explicitly,It is harmonic in the quarter-plane, has the stated boundary values, and decays in both unbounded directions.