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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.

If a harmonic function on one side of a straight boundary has zero normal derivative there, even reflection extends it harmonically across the boundary. Locally take the boundary to be $x=0$ and assume $u$ and its first derivatives extend continuously to its open segment. Define $\tilde{u}(x,y)=u(|x|,y)$. The zero value of $u_x$ makes the first derivatives match across the join. Applying Green's first identity on the two sides against a smooth compactly supported test function leaves no interface term, so the reflected function is weakly harmonic and hence harmonic by Weyl lemma.

Reflecting across both edges of a right-angle corner gives a harmonic function on a punctured neighbourhood when both normal derivatives vanish. If the original function is bounded at the corner, the removable singularity for a bounded harmonic function makes the extension harmonic there too. Thus a maximum on such an edge or corner can be tested using the interior strong maximum principle for harmonic functions.

A harmonic function on ${\mathbb{R}}^2\setminus \{0\}$ bounded on one side is constant. Shift and change sign to make it nonnegative. Its circular mean has the form $a\log r+b$; positivity for every $r>0$ forces $a=0$. For $|x|=r$, the mean value property for harmonic functions bounds its value by the mean integral over $r/2<|z|<3r/2$, giving at most $8b$. A removable singularity for a bounded harmonic function fills the origin, and the harmonic Liouville theorem finishes the proof. The dimensional distinction matters: $|x{|}^{2-n}$ is a positive nonconstant harmonic function on punctured ${\mathbb{R}}^n$ for $n\ge 3$.

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.

The conclusion holds in dimension two. Shift and change sign to obtain a nonnegative harmonic function $v$. Its circular mean satisfiesHere the angular second derivative in the polar-coordinate Laplacian integrates to zero. Thus $m(r)=a\log r+b$; nonnegativity for every $r>0$ forces $a=0$. For $|x|=r$, the mean value property for harmonic functions and nonnegativity giveThe removable singularity for a bounded harmonic function fills in the origin, and the harmonic Liouville theorem makes the extension constant.

The conclusion fails for $n\ge 3$: $\boxed{u(x)=|x{|}^{2-n}}$ is positive and nonconstant, and its radial Laplacian $u^{\prime \prime }+(n-1)u^{\prime }/r$ vanishes away from zero. For $n=1$, $\boxed{u(x)=|x|}$ is a nonconstant function bounded below whose second derivative vanishes on both components of the punctured line.

No such function exists. Continuity on the compact closed unit disk makes $u$ bounded near the origin. The removable singularity for a bounded harmonic function extends $u$ harmonically across the origin. The extended function is continuous on the closed disk and vanishes on its boundary, so the maximum principle for harmonic functions, applied to both $u$ and $-u$, forces $u=0$ throughout the disk. This contradicts $u(0,0)=1$.

Let $w$ be the difference of two solutions. It is a harmonic function in the open rectangle, continuous on its closure, withInterpret these Neumann boundary conditions classically on the open edges, with the corresponding first derivatives extending there. We use the strong maximum principle for harmonic functions together with even reflection across a homogeneous Neumann boundary. This avoids imposing differentiability at the two bottom corners, where the original data fail a corner compatibility condition for mixed boundary data.

By continuity on the compact rectangle, $w$ attains a maximum $M$. Suppose $M>0$. It cannot occur on the bottom edge, where $w=0$. If it occurs in the interior, the strong maximum principle for harmonic functions makes $w$ constant, contradicting its bottom value. If it occurs on an open vertical or top edge, the zero normal derivative allows an even reflection across that edge. The reflected harmonic function then has an interior maximum, giving the same contradiction.

The only remaining possibilities are the two top corners. Reflect evenly across both meeting edges. This gives a bounded harmonic function on a punctured neighbourhood of the corner. The removable singularity for a bounded harmonic function fills the centre with its continuous value. The corner is now an interior maximum, again impossible. Therefore $w\le 0$. Applying the identical argument to $-w$ gives $w\ge 0$.

Thus $\boxed{w=0}$, and the solution is unique among the continuous classical solutions with the stated edge conditions.