Poisson integral

The Poisson integral convolves boundary data with the Poisson kernel to produce a harmonic function in a half-space. In the upper half-plane,

The conformal map $w=z^2$ sends $x,y>0$ to the complex upper half-plane. For bounded continuous boundary values $g_2$ on the horizontal axis and $g_1$ on the vertical axis with equal corner values, set $f(s)=g_2(\sqrt{s})$ for $s>0$ and $f(s)=g_1(\sqrt{-s})$ for $s<0$. The bounded harmonic function with those Dirichlet boundary conditions isEquivalently, if $z^2=a+ic$ with $c>0$, this is $\frac{2c}{\pi }{\int }_0^{\infty }r\left[g_2(r)/((r^2-a{)}^2+c^2)+g_1(r)/((r^2+a{)}^2+c^2)\right]dr$. The Poisson kernel is a positive approximate identity, so the prescribed edge values are attained. Boundedness is essential for uniqueness on this unbounded domain.

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.

The harmonic function $v(x,y)=xy$ vanishes on both axes but is nonzero in the quadrant, and $v_z=-iz/2$. More generally $\mathrm{Im}{z}^{2m}$, for a positive integer $m$, vanishes on both axes and is harmonic. Adding these unbounded functions leaves all edge Dirichlet boundary conditions unchanged. Thus decay of the boundary data does not imply decay of every solution, and an edge-data-only formula cannot determine an arbitrary solution's Wirtinger derivative. A boundedness or suitable growth condition must select the solution.

For the bounded Dirichlet Poisson integral in a quadrant, the Wirtinger derivative isThis follows by differentiating the Schwarz integral formula composed with $z^2$, remembering that for real boundary data $u_z$ is half the derivative of a holomorphic function whose real part is $u$. For complex boundary data, apply the construction to the real and imaginary parts separately and use linearity. If the data and their derivatives decay sufficiently, integration by parts givesThe endpoint terms cancel exactly when $g_1(0)=g_2(0)$. Interior denominators do not vanish because $\mathrm{Im}{z}^2>0$.

The Schwarz integral formula reconstructs a holomorphic function from boundary values of its real part. For a real boundary function $f$ on the complex upper half-plane satisfying ${\int }_{\mathbb{R}}|f(s)|/(1+s^2)ds<\infty$, the regularized form isIts real part is the Poisson integral of $f$. The subtraction makes the integral converge and changes only the imaginary normalization. Integration by parts, when the boundary terms vanish, gives $W^{\prime }(w)=(\pi i{)}^{-1}\int f^{\prime }(s)/(s-w)ds$; jumps of $f$ are included as Dirac delta function terms in its distributional derivative.

For real boundary data $u$, the displayed holomorphic integral has real part equal to the disk Poisson integral of $u$ and imaginary part its normalized harmonic conjugate. If $u\ge 0$ is not identically zero, the real part is strictly positive inside the disk. A holomorphic logarithm then has imaginary part between $-\pi /2$ and $\pi /2$, while its real part records the logarithm of the magnitude. This separates uniform imaginary-part control from large real parts on small boundary sets.

For a harmonic function in the quadrant with boundary derivatives $u_y(0,y)=g_1(y)$ and $u_x(x,0)=g_2(x)$, integrating the data to equal corner values and applying the Schwarz integral formula after the conformal map $w=z^2$ givesThese are tangential boundary derivatives, so uniqueness requires an additional normalization or growth restriction. A homogeneous contribution has the form $ih(z^2)/z$ with $h$ holomorphic in the complex upper half-plane and real on its boundary away from zero. For example, $u=xy$ has zero prescribed derivatives on both edges.