= Solution
{c}
Use the Fourier convention $\widehat u(k,y)=\int_{\mathbb R}u(x,y)e^{ikx}dx$. Transforming the <differential equation> gives
$$
\widehat u_{yy}-(k^2-ik)\widehat u=0.
$$
Define $k_+(k)=\sqrt k$ 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 $0<\operatorname{Im}k<1$, put $q=k_+k_-$. Then $q^2=k^2-ik$ and $\operatorname{Re}q>0$; 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 $\operatorname{Im}k>0$. Thus $\widehat u(k,0)=\overline f+i/k$ and $\widehat u_y(k,0)=\overline g$. The transformed boundary relation is
$$
\overline g=-k_+k_-(\overline f+i/k).
$$
Let $c=k_-(0)=e^{-i\pi/4}$. Splitting off the <pole> at zero gives
$$
k_-\overline f+i\frac{k_--c}{k}
=-\frac{\overline g}{k_+}-\frac{ic}{k}.
$$
The left side is analytic in the lower domain up to $\operatorname{Im}k<1$: its apparent <pole> at zero is removable. The right side is analytic for $\operatorname{Im}k>0$. 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. Consequently
$$
\overline g=-\frac{ic}{\sqrt k},\qquad
\overline f=\frac{i(c-k_-)}{kk_-},\qquad
\widehat u(k,0)=\frac{ic}{k\sqrt{k-i}}.
$$
Choose 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 exactly
$$
\boxed{u(x,y)=\frac1{2\pi}\int_\Gamma
\frac{i e^{-i\pi/4}}{k\sqrt{k-i}}\,e^{-\sqrt{k^2-ik}\,y}e^{-ikx}dk.}
$$
Thus the appropriate value is \b[$\sqrt{-i}=e^{-i\pi/4}$]. 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> is
$$
u(x,y)=\operatorname{erfc}b,\qquad
b=\sqrt{\frac{\sqrt{x^2+y^2}-x}{2}}.
$$
Indeed $\Delta u=u_{bb}/[4(a^2+b^2)]$ 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 $f(x)=\operatorname{erfc}\sqrt{-x}$ 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 $u\to0$. But at fixed $y>0$ and $x\to+\infty$, $b\sim y/(2\sqrt x)$ and $u\to1$. Hence \b[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 $O(\sqrt x)$.
\Image[/past-exam-of-the-perturbation-methods-and-applied-complex-variables-course-of-the-university-of-oxford/2010-downstream-layer.png]
{title=The exact mixed-boundary solution decays away from the downstream boundary layer}
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