The <Wirtinger derivative> $u_z=(u_x-iu_y)/2$ is <holomorphic>, since $\partial_{\bar z}u_z=\Delta u/4=0$ by the <Laplace equation>. The <conformal map> $w=z^2$ takes the quadrant to the <complex upper half-plane>. Integrating the prescribed <tangential boundary derivatives>, choose a common corner value $c$ and define the real boundary function
$$
f(s)=\begin{cases}
c+\displaystyle\int_0^{\sqrt s}g_2(r)\,dr,&s>0,\\
c+\displaystyle\int_0^{\sqrt{-s}}g_1(r)\,dr,&s<0.
\end{cases}
$$
The decaying data are bounded under the usual regularity at the corner, so $f(s)=O(1+\sqrt{|s|})$. Thus its <Poisson integral> converges. A <holomorphic> function whose <real part> is that <Poisson integral> is supplied by the regularized <Schwarz integral formula>:
$$
W(w)=\frac1{\pi i}\int_{-\infty}^{\infty}f(s)\left(\frac1{s-w}-\frac{s}{1+s^2}\right)ds.
$$
For $u_*(z)=\operatorname{Re}W(z^2)$, the <chain rule> gives $(u_*)_z=zW'(z^2)$. Apply <integration by parts> to obtain
$$
W'(w)=\frac1{\pi i}\int_{\mathbb R}\frac{f'(s)}{s-w}\,ds,
\qquad
f'(s)=\begin{cases}
g_2(\sqrt s)/(2\sqrt s),&s>0,\\
-g_1(\sqrt{-s})/(2\sqrt{-s}),&s<0.
\end{cases}
$$
Substituting $s=r^2$ and $s=-r^2$ gives the requested particular <integral representation>:
$$
\boxed{(u_*)_z(z)=\frac{z}{\pi i}\left[\int_0^\infty\frac{g_2(r)}{r^2-z^2}\,dr+\int_0^\infty\frac{g_1(r)}{r^2+z^2}\,dr\right].}
$$
Both integrals converge absolutely for an interior point of the quadrant. The <Sokhotski–Plemelj formula> gives $\operatorname{Re}(u_*)_z(x)=g_2(x)/2$ on the horizontal edge and $\operatorname{Im}(u_*)_z(iy)=-g_1(y)/2$ on the vertical edge, verifying the prescribed derivatives.

The printed conditions alone do not determine $u_z$ uniquely. The general answer adds a <holomorphic> function $H$ satisfying
$$
\operatorname{Re}H(x)=0,\qquad \operatorname{Im}H(iy)=0,
\qquad x,y>0.
$$
Equivalently, write $H(z)=ih(z^2)/z$, where $h$ is <holomorphic> in the <complex upper half-plane> with real boundary values on the nonzero real axis. An <antiderivative> of $2H$ has a <harmonic> <real part> with zero <tangential boundary derivatives>. For example, $u=xy$ contributes $H=-iz/2$ without changing either datum. If the two edge constants differ by $c_2-c_1$, the angular <harmonic function> contributes $i(c_2-c_1)/(\pi z)$. The boxed expression selects the <Poisson integral> representative with equal edge constants; regularity and suitable growth conditions can be used to select that representative.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 328 / 1 / Solution

{c}
{title2=$\partial_z,\partial_{\bar z}$}
{wiki=Wirtinger_derivatives}

For $z=x+iy$, the Wirtinger derivatives are $\partial_z=(\partial_x-i\partial_y)/2$ and $\partial_{\bar z}=(\partial_x+i\partial_y)/2$. The <Cauchy-Riemann equations> become $\partial_{\bar z}F=0$ for a <holomorphic function> $F$, and $4\partial_z\partial_{\bar z}=\Delta$. Thus $u_z$ is <holomorphic> whenever $u$ is a twice continuously <differentiable> <harmonic function>.

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