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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.

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 $O(\sqrt{x})$.

A meromorphic function on the plane with $|f(z)|\le C|z{|}^n$ outside a disc has no poles there and only finitely many poles inside. Subtract their finite principal parts to obtain an entire function. For $n\ge 0$, Cauchy estimates make every derivative of order above $n$ vanish, so the remainder is a polynomial. For $n<0$ the remainder tends to zero at infinity and is zero by Liouville theorem. Adding back the principal parts gives a rational function in either case.

A conformal map from ${\mathbb{C}}^{\ast }$ into the punctured unit disc would be bounded near its isolated singularity at zero, hence extend to a bounded entire function. The Liouville theorem would make it constant.

For generic fixed $w$, regard the right side as an elliptic function of $z$. The apparent pole at zero cancels, giving limiting value $\wp (w)$. At $z=w$ the slope ratio is finite. At $z=-w$, its squared slope has the same double principal part as $\wp (z+w)$ and no simple pole term. These exhaust the possible poles because the Weierstrass elliptic function has degree two. The difference is therefore an entire elliptic function, hence constant by Liouville theorem, and its value at zero is zero. Meromorphic continuation extends the identity to the exceptional parameters.

The Weierstrass elliptic function satisfiesWriting $G_{2r}={\sum }_{\omega \in \Lambda \setminus \{0\}}{\omega }^{-2r}$, the Laurent coefficients of the Weierstrass elliptic function giveThus $g_2=60G_4$ and $g_3=140G_6$ cancel every nonremovable term in the Laurent series of ${\wp }^{\prime 2}-4{\wp }^3+g_2\wp +g_3$ at each point of the period lattice. The resulting elliptic function is an entire function and is bounded, hence is zero by Liouville theorem.

Let $h_i$ represent a nonzero two-torsion point of a complex torus, and write $e_i=\wp (h_i)$. The product $(\wp (z+h_i)-e_i)(\wp (z)-e_i)$ is an elliptic function whose apparent poles cancel against double zeros. It is therefore constant by Liouville theorem. Its value at $z=0$ is ${\wp }^{\prime \prime }(h_i)/2$. The Weierstrass elliptic differential equation gives ${\wp }^{\prime \prime }(h_i)/2=3e_i^2-g_2/4=(e_i-e_j)(e_i-e_l)$, proving the formula.