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Two bounded harmonic functions continuous to the quadrant edges with the same Dirichlet boundary conditions are equal. Their difference is bounded and zero on both axes. Odd reflection across each axis gives a bounded harmonic extension away from the corner; the removable singularity for a bounded harmonic function fills in that corner. The harmonic Liouville theorem makes that extension constant, and oddness makes the constant zero. This does not exclude unbounded examples such as the zero-boundary harmonic growth in a quadrant.

Interpret the malformed printed domain inequality as the first quadrant $x>0,y>0$, as indicated by the two axis boundary conditions. There is also a genuine missing hypothesis: decay of the boundary data does not require the harmonic function itself to be bounded. For zero data, both $q=0$ and $q=xy$ satisfy the Laplace equation and both edge values, but the Wirtinger derivative of the latter isThus the data do not determine an arbitrary solution's derivative without a growth restriction. We derive the integral representation for the bounded solution, which is also the standard decaying solution when the continuous edge data decay. Any other admissible unbounded solution can differ by a zero-boundary harmonic function.

The conformal map $w=z^2$ sends the quadrant bijectively to the complex upper half-plane: the positive real axis maps to $s>0$, and the positive imaginary axis maps to $s<0$. Define its real-axis boundary function byThe compatibility at the corner makes $f$ continuous there. The Dirichlet Poisson integral in a quadrant follows from the upper-half-plane Poisson kernel:It is bounded by the supremum of the bounded edge data, and the kernel's approximate identity property supplies both Dirichlet boundary conditions. More explicitly, for $z^2=a+ic$, where $a=x^2-y^2$ and $c=2xy>0$, this integral isIt displays directly which part of the real-line boundary corresponds to each original axis.

To obtain its derivative, first take real data and use the Schwarz integral formula. A holomorphic function whose real part is the Poisson solution in the $w$-plane isThe regularizing subtraction is independent of $w$ and affects only the imaginary normalization; it permits bounded data without an artificial integrability assumption. Then $q=\mathrm{Re}F(z^2)$. By the chain rule and the Wirtinger derivative,The factor is $z$, not $2z$: taking the real part contributes the factor $1/2$ in the complex derivative. Differentiation under the integral sign is justified uniformly on compact interior subsets because the denominator stays away from the real boundary and decays quadratically at infinity. Splitting the real integral at zero and substituting $s=r^2$ or $s=-r^2$ gives the holomorphic derivative of a quadrant Dirichlet solution:This converges for bounded data, and in particular for the stated sufficiently decaying data. No interior pole is encountered since $\mathrm{Im}(z^2)=2xy>0$. For complex-valued data the same formula holds by applying the real-data construction separately to their real and imaginary parts and using linearity.

An equivalent form, often convenient when the data's derivatives decay, comes from integration by parts. Sincethe two endpoint terms are $-g_2(0)/z^2$ and $+g_1(0)/z^2$. They cancel by the specified compatibility, leavingThis is the harmonic quadrant solution from tangential boundary derivatives applied to the actual derivatives $g_1^{\prime },g_2^{\prime }$; the supplied $g_1,g_2$ in this question are boundary values, not derivatives. Omitting those primes would solve a different boundary problem.

For completeness, bounded Dirichlet uniqueness in a quadrant selects the solution just constructed. The difference of two bounded solutions is zero on both axes. Odd reflection across each axis gives a bounded harmonic extension to the punctured plane; the removable singularity for a bounded harmonic function fills the corner. The harmonic Liouville theorem makes the extension constant, and oddness makes it zero. Decaying data give a decaying Poisson extension: split the real-axis integral into a fixed compact interval and a tail with arbitrarily small data; the compact contribution tends to zero as $|z^2|\to \infty$, while the tail is bounded by that small supremum.

As literally printed without boundedness or a growth condition, the most one can assert isThe zero-boundary harmonic growth in a quadrant includes $v=cxy$ and more generally $v=\mathrm{Im}(z^{2m})$. This is an actual nonuniqueness of the printed problem, not a defect in the integral for the normalized bounded solution.