Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2010 / B / 6

Glue the two analytic functions on their overlap, where their values agree. Since $\alpha <\beta$, the two half-planes cover the whole complex plane. The gluing therefore defines an entire function. It is bounded outside a sufficiently large disk by the two given infinity bounds, and bounded on the remaining compact disk by continuity. The Liouville theorem then gives one common constant.

For the standard canonical Wiener-Hopf factorization, first assume the nonzero kernel has Index of a Wiener-Hopf kernel zero and a continuous analytic logarithm $\ell =\log F$ tending to zero at both ends of the strip, with sufficient decay for the Cauchy projections. Choose $\alpha <a<b<\beta$. Let ${\Gamma }_{+}$ be the line $\mathrm{Im}\zeta =a$ oriented left to right, with its factor used above that line; let ${\Gamma }_{-}$ be the line $\mathrm{Im}\zeta =b$ oriented left to right, with its factor used below that line.

For $a<\mathrm{Im}z<b$, closing the rectangle between the two lines and using the Cauchy integral formula givesThus $F=F_{+}/F_{-}$. Multiplying $F{w}_{+}=w_{-}$ gives $F_{+}{w}_{+}=F_{-}{w}_{-}$ in the overlap. The canonical factors are analytic, nonzero, and normalized at infinity in their respective half-planes, so part (a) yieldsContours may be shifted within the strip without changing their factors on the appropriate side. When necessary, a subtracted Cauchy kernel fixes convergence and the common normalization constant.

The printed hypotheses omit the index and factor-growth qualifications; $F\to 1$ alone does not ensure them. An explicit counterexample is $F(z)=(z+i)/(z-i)$ on $-1/2<\mathrm{Im}z<1/2$, with $w_{+}=(z-i)/(z+i)$ and $w_{-}=1$. Both $w$ functions are analytic and bounded at infinity in their prescribed half-planes, and the relation holds. If admissible analytic factors made their common product a nonzero constant, $w_{-}=1$ would force $F_{-}$ constant and hence $F_{+}$ proportional to $F$, impossible because $F$ has a pole at $i$ in the plus half-plane. The natural factors $F_{+}=z+i$, $F_{-}=z-i$ instead give the nonconstant common polynomial $z-i$. Thus the boxed constant conclusion is the intended index-zero result, not a consequence of the literal assumptions alone.

Let $k=s+i\eta$. On the negative half-line, the integrand magnitude is bounded at infinity by a constant times $e^{(\alpha -\eta )x}$, which is integrable as $x\to -\infty$ when $\eta <\alpha$. On the positive half-line the corresponding bound is $e^{(\beta -\eta )x}$, integrable when $\eta >\beta$. ThusOn compact subsets strictly inside these half-planes, the same exponential domination absorbs every power of $|x|$. Differentiation under the integral is therefore valid and proves holomorphy, rather than only absolute convergence. The bars here denote the printed one-sided Fourier transforms, not complex conjugation.

Use the Fourier convention $\hat{u}(k,y)={\int }_{\mathbb{R}}u(x,y)e^{ikx}dx$. Transforming the differential equation givesDefine $k_{+}(k)=\sqrt{k}$ analytic above the downward cut from zero, and $k_{-}(k)=\sqrt{k-i}$ analytic below the upward cut from $i$. More explicitly take $\arg k\in (-\pi /2,3\pi /2)$ and $\arg (k-i)\in (-3\pi /2,\pi /2)$. In the strip $0<\mathrm{Im}k<1$, put $q=k_{+}{k}_{-}$. Then $q^2=k^2-ik$ and $\mathrm{Re}q>0$; at $k=i/2$, $q=1/2$, fixing the choice throughout the strip. The decaying $y$ mode is $e^{-qy}$.

The positive-half-line boundary value has transform ${\int }_0^{\infty }{e}^{ikx}dx=i/k$ for $\mathrm{Im}k>0$. Thus $\hat{u}(k,0)=\overline{f}+i/k$ and ${\hat{u}}_y(k,0)=\overline{g}$. The transformed boundary relation isLet $c=k_{-}(0)=e^{-i\pi /4}$. Splitting off the pole at zero givesThe left side is analytic in the lower domain up to $\mathrm{Im}k<1$: its apparent pole at zero is removable. The right side is analytic for $\mathrm{Im}k>0$. By the supplied boundedness of $k\overline{f}$ and $\overline{g}$, both sides tend to zero at infinity in their half-planes. The gluing and Liouville theorem argument therefore makes their common value zero. ConsequentlyChoose the inversion contour $\Gamma =\{s+i\eta :s\in \mathbb{R}\}$ oriented left to right, with any fixed $0<\eta <1$. It passes above the pole at zero and below the branch point at $i$. Fourier inversion gives exactlyThus the appropriate value is $\sqrt{-i}=e^{-i\pi /4}$. The other proposed value reverses the residue of the boundary step and violates its prescribed positive value.

There is also a far-field qualification in the source. The formula is the intended bounded mixed-boundary solution, but it does not tend to zero along every path to infinity. This can be checked explicitly: in parabolic coordinates $x=a^2-b^2$, $y=2ab$, with $a,b>0$, the mixed-boundary convection-diffusion solution in parabolic coordinates isIndeed $\Delta u=u_{bb}/[4(a^2+b^2)]$ and $u_x=-b{u}_b/[2(a^2+b^2)]$; the identity $u_{bb}=-2b{u}_b$ proves the differential equation. At $b=0$ the boundary value is one, and at $a=0$ its normal derivative is zero. Its traces are $f(x)=\mathrm{erfc}\sqrt{-x}$ for $x<0$ and $g(x)=-1/\sqrt{\pi x}$ for $x>0$, whose one-sided transforms are the expressions derived above.

At a fixed angle away from the positive $x$ axis, $b\to \infty$ and $u\to 0$. But at fixed $y>0$ and $x\to +\infty$, $b\sim y/(2\sqrt{x})$ and $u\to 1$. Hence the radial decay statement cannot hold uniformly as printed; the intended decay is away from the downstream boundary layer. The exact expression also identifies its thickness as $O(\sqrt{x})$.