Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 4 / 15A (Complex Methods)

With the Fourier transform convention of 5A, use the printed Fourier inversion formula directly. The possible choice of values at $s=\pm 1$ has no effect on the integral. Integration over the support givesCombining the two fractions yieldsIn particular $g$ is imaginary-valued and odd, as expected from the real odd transformed function with this inverse-transform sign.

Take the Fourier transform in $x$ of the Laplace equation. Writing $U(s,y)$ for the transformed solution gives $U_{yy}-s^{2}U=0$. Its boundary values givewith the multiplier at $s=0$ understood as $1-y$. The particular data simplify this to $U(s,y)=2\sinh (s(1-y))$ for $|s|<1$, and zero outside. Thus, with $a=1-y$, the Fourier solution of the strip Dirichlet problem isThe displayed closed expression applies in the open strip. The integral defines its continuous boundary values, including the removable point $(x,a)=(0,0)$. It gives $u(x,0)=g(x)$ and $u(x,1)=0$. Differentiation under this finite integral verifies $u_{xx}+u_{yy}=0$ directly, since each integrand has opposite second derivatives in $x$ and $y$.

The boundary conditions alone do not specify a unique solution without a growth restriction: adding $C{e}^{\pi x}\sin (\pi y)$ preserves both boundary values and harmonicity. The constructed solution is bounded, and it is unique among bounded solutions continuous on the closed strip. To see this, apply the maximum principle for harmonic functions to a bounded real or imaginary component $w$ of the difference. For $0<\delta <\pi$, the positive harmonic function $B(x,y)=\cosh (\delta x)\cos (\delta (y-1/2))$ has a positive minimum factor in $y\in [0,1]$. For any $\epsilon >0$, the functions $\pm w-\epsilon B$ are nonpositive on the horizontal sides of a sufficiently wide rectangle and on its vertical sides, because $w$ is bounded and $B$ grows there. The maximum principle gives $|w|\le \epsilon B$ at every fixed point. Letting $\epsilon \downarrow 0$ proves uniqueness in this natural Fourier solution class.