Past exam of the short options course of the University of Oxford / 2017 / 5

Use the angular-frequency Fourier transform conventionFor $f,f^{\prime }\in L^1(\mathbb{R})$ with the usual local absolute-continuity hypothesis, the Fourier transform of a derivative isFor example, this applies when $f$ is $C^1$ and both displayed functions are integrable. These assumptions make integration by parts valid and the endpoint terms vanish.

Split the integral at the origin and reflect the negative half:Thus the Fourier transform of a two-sided exponential is

With this convention, the Fourier inversion theorem states that if $f,\hat{f}\in L^1(\mathbb{R})$, thenalmost everywhere and at every continuity point of $f$. Apply this to the continuous function in part (ii):The denominator is even. Replacing $x$ by $-t$ and renaming the integration variable givesBoth integrable members of this transform pair satisfy the inversion hypotheses.

Complete the square: $x^2+2x+2=(x+1{)}^2+1$. In the Fourier transform, the substitution $u=x+1$ givesThe phase $e^{is}$ is important. Taking the real part at $s=a$ gives the cosine integral, since changing the sign in the exponential does not change its real part:

For integrable functions $u,v$, the convolution theorem in the chosen convention isLet $F(s,y)$ be the Fourier transform of $f(\cdot ,y)$ in the horizontal coordinate. Transforming Laplace's equation givesFor $s\ne 0$, exclude the growing exponential and obtain $F(s,y)=2e^{-|s|y}/(1+s^2)$. At $s=0$, continuity in frequency gives $F(0,y)=2$; a constant zero-frequency transform is compatible with spatial decay of $f$.

By the transform pair in part (a), the inverse transform of $e^{-|s|y}$ is the Poisson kernel for the upper half-plane, $P_y(x)=y/[\pi (x^2+y^2)]$. ThusThe Poisson extension of exponential boundary data is harmonic, approaches the continuous boundary data and decays at spatial infinity. For uniqueness, apply the maximum principle for harmonic functions to the difference of two solutions on increasingly large half-discs: it is zero on the real boundary and uniformly small on the distant semicircle.

Use the squaring map from a quadrant to the upper half-plane:By conformal invariance of harmonicity, the answer isOn the horizontal axis it gives $f(x^2,0)=e^{-x^2}$, and on the vertical axis it gives $f(-y^2,0)=e^{-y^2}$. Also $|(x+iy{)}^2|=x^2+y^2$, so spatial infinity maps to spatial infinity and the required decay follows. The two boundary values agree at the origin.