Dirichlet Green function

A Dirichlet Green function for a differential operator satisfies homogeneous Dirichlet boundary conditions in one variable. It converts source and boundary data into an integral representation of the solution.

For the gap $a<r<b$, put $u_{\ell }(r)=r^{\ell }-a^{2\ell +1}{r}^{-\ell -1}$, $v_{\ell }(r)=r^{-\ell -1}-r^{\ell }/b^{2\ell +1}$ and $d_{\ell }=(2\ell +1)[1-(a/b{)}^{2\ell +1}]$. With orthonormal spherical harmonics, the Dirichlet Green function isFor a short verification, $u_{\ell }(a)=v_{\ell }(b)=0$ and both solve the radial Laplace equation. Their Wronskian is $u_{\ell }{v}_{\ell }^{\prime }-u_{\ell }^{\prime }{v}_{\ell }=-d_{\ell }/r^2$. Thus the radial derivative jump is $-1/r^{\prime 2}$; completeness of the spherical harmonics gives the stated Dirac delta function normalization. A point charge has electric potential $qG/(4\pi {ϵ}_0)$, and the inner-wall surface charge density is $-q{\partial }_{r}G{|}_{a^{+}}/(4\pi )$, with the angular source coordinates held fixed.

For $D=\{x>0,z>0\}\subset {\mathbb{R}}^3$, take the Dirichlet Green function convention $-{\Delta }_{P}G(P,Q)={\delta }_Q(P)$. Reflect $Q=(\xi ,\eta ,\zeta )$ to $Q_{ij}=((-1{)}^i\xi ,\eta ,(-1{)}^j\zeta )$. The displayed image sum vanishes on both faces, has exactly one positive unit source inside $D$, and decays at infinity. Pairwise cancellation proves the boundary conditions, while the three image terms are harmonic inside $D$. Its inward source-normal derivative gives the Poisson kernel of a three-dimensional quarter-space. The source sign must be kept consistent with Green's second identity.

For the face $z=0$ of $x>0,z>0$, set $r_{\mp }^2=(x\mp \xi {)}^2+(y-\eta {)}^2+z^2$, with source boundary coordinates $\xi >0$, $\eta \in \mathbb{R}$. The Poisson kernel for data on that face and zero data on $x=0$ is the displayed positive difference. It follows by differentiating the Dirichlet Green function for a three-dimensional quarter-space in the inward source direction. Integrating it against bounded continuous face data gives the bounded harmonic solution away from compatibility issues at the edge. Without a boundedness or suitable infinity condition, uniqueness fails because the harmonic function $xz$ vanishes on both faces.

Its integral over this one face is $2\arctan (x/z)/\pi$, the harmonic measure of that face. To calculate it, integration in $\eta$ first gives $z\{[(x-\xi {)}^2+z^2{]}^{-1}-[(x+\xi {)}^2+z^2{]}^{-1}\}/\pi$; the remaining half-line integral is an elementary arctangent difference. This mass lies between zero and one: the other face carries the remaining boundary influence.

Let $G$ vanish on the boundary of a domain and satisfy $\Delta G(\cdot ,y)={\delta }_y$. Green's second identity givesThus the Poisson kernel is $+{\partial }_{n}G$ for this convention. If instead $-\Delta K=\delta$, then $K=-G$ and the same kernel is $-{\partial }_{n}K$. At the boundary of $x>0$, the outward normal is $-e_x$, so the first convention gives $P=-{\partial }_{x}G$. Changing the delta-source sign or replacing the outward derivative by an inward coordinate derivative changes a sign; changing both leaves the physical kernel unchanged.

For $L=d^2/d{x}^2+4x^{-1}d/dx-4x^{-2}$ on $(0,1)$, regularity at zero selects the homogeneous solution $u=x$, while the right Dirichlet boundary condition selects $v=x^{-4}-x$. Their Wronskian is $-5x^{-4}$. Thus the displayed Green function, with $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$, is continuous, has unit first-derivative jump and satisfies $L_{x}G=\delta (x-\xi )$. For bounded forcing $f$, $y(x)={\int }_0^{1}G(x,\xi )f(\xi )d\xi$ is the unique solution continuous at zero with zero endpoint values. The singular homogeneous branch $x^{-4}$ is excluded by that regularity. The kernel uses ordinary source measure $d\xi$, not the measure for the multiplied self-adjoint operator.

With the convention ${\Delta }_z{G}_D=\delta (z-\zeta )$, the displayed Dirichlet Green function vanishes on the unit circle. Indeed $|z-\zeta |=|1-z\bar{\zeta }|$ when $|z|=1$. For $\zeta \ne 0$, the second logarithm corresponds to an opposite image source at $\zeta /|\zeta {|}^2$, outside the unit disc; it is a harmonic function throughout the interior. At $\zeta =0$ the formula becomes $(2\pi {)}^{-1}\log |z|$. Its outward normal derivative in the boundary variable is $(2\pi {)}^{-1}(1-|\zeta {|}^2)/|z-\zeta {|}^2$, the Poisson kernel divided by $2\pi$. Green's second identity therefore gives the Poisson integral on the unit disk.

For $\kappa \ge 0$ in the upper half-space, let $r^{\prime \ast }$ be the reflection of the interior source $r^{\prime }$ across its boundary plane. The displayed decaying Green function solves $(\Delta -{\kappa }^2)G_D=\delta (r-r^{\prime })$ with zero Dirichlet boundary condition. The direct term supplies the source; the image source lies outside the domain and its opposite sign makes the two terms cancel on the plane. The minus sign relative to the positive static Yukawa Green function comes from the operator convention. The case $\kappa =0$ is the Coulomb limit. An infinity condition is needed for uniqueness on an unbounded domain.

For $D=d/dx$ and $L_a=(D-a{)}^2+b^2$ on $(0,L)$, with $b>0$ and $\sin (bL)\ne 0$, the substitution $y=e^{ax}v$ gives $L_{a}y=e^{ax}(v^{\prime \prime }+b^{2}v)$. Here $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. The displayed Green function vanishes at the endpoints, is continuous at $x=\xi$, and its first derivative has jump one, so $L_{a,x}G=\delta (x-\xi )$. It is the usual sine Dirichlet Green function conjugated by the exponential factor. At $bL\in \pi \mathbb{Z}\setminus \{0\}$ there is a homogeneous zero mode, so the inverse kernel does not exist without a solvability condition and a normalization.

For a three-dimensional slab $0<x_1<L$ and source $y=(y_1,y_2,y_3)$, define image positions $y_n^{+}=(y_1+2nL,y_2,y_3)$ and $y_n^{-}=(-y_1+2nL,y_2,y_3)$. The displayed method of images series uses convention ${\Delta }_x{G}_D=+\delta (x-y)$. Reflection in $x_1=0$ pairs the plus image $n$ with the minus image $-n$, while reflection in $x_1=L$ pairs it with the minus image $1-n$. Both boundary values therefore vanish. Only $y_0^{+}=y$ lies in the slab. Each bracket is $O(n^{-2})$, giving absolute convergence of the paired series away from sources. This selects the usual Dirichlet Green function that decays at large transverse separation.

For the three-dimensional ball $|x|<a$, the displayed Dirichlet Green function has convention ${\Delta }_x{G}_D=+\delta (x-y)$. Its method of images uses the exterior inverse point $y^{\ast }=a^{2}y/|y{|}^2$. On $|x|=a$, the distance identity $|x-y^{\ast }|=(a/|y|)|x-y|$ makes the boundary value zero. The image lies outside the ball and therefore adds no source singularity to the domain. At a central source the weighted image term tends to $1/a$, giving $G_D(x,0)=-(4\pi |x|{)}^{-1}+(4\pi a{)}^{-1}$. Spherical inversion with the required radial weight is the Kelvin transform.

For the upper half of a three-dimensional ball, let $\overline{y}$ be the reflection of an interior source $y$ in the equatorial plane. The Dirichlet Green function inside a sphere, evaluated at the source and its reflection and then subtracted, vanishes on both the spherical cap and flat equatorial disc. Reflection makes the source distances and image distances agree on the plane. Only the original source is in the upper half-ball; the reflected source is in the lower half-ball, and both inversion images are outside the ball. Thus the difference has precisely the required Dirac delta distribution and Dirichlet boundary conditions.

For the exterior of the unit disc in two dimensions, with the sign convention ${\Delta }_{x}G=+\delta (x-y)$, this is the Dirichlet Green function that vanishes on the circle and is bounded at infinity. The method of images places an opposite logarithmic source at the inverse point $y/|y{|}^2$. The constant $-(2\pi {)}^{-1}\log |y|$ enforces the zero boundary value. The inversion in a circle identity $|x-y|=|y||x-y/|y{|}^2|$ for $|x|=1$ verifies that condition directly.

The outward normal of the exterior domain is $-x$ on the unit circle. Its normal derivative gives the positive exterior Poisson kernelFor bounded harmonic functions this kernel integrates boundary data to the interior value. Boundedness at infinity is essential to uniqueness: without it, a multiple of $\log |y|$ can be added without changing the circle's Dirichlet boundary condition.

Let $0<a<b$ and impose zero Dirichlet boundary conditions at $a,b$. With the convention $L_{x}G(x,s)=\delta (x-s)$ for the Dirac delta distribution, the Green's function is continuous and its first derivative has jump one. For $n\ne 0$ choose $\ell (x)=x^{-n}-a^{-n}$ and $r(x)=x^{-n}-b^{-n}$. Their Wronskian is $n(a^{-n}-b^{-n})s^{-n-1}$, soFor $n=0$, take the continuous limit or use $\ell =\log (x/a)$ and $r=\log (x/b)$ to obtainThe solution is $u(x)={\int }_a^{b}G(x,s)f(s)ds$. These kernels use ordinary measure $ds$. The factor $s^{n+1}$ comes from the nonconstant Wronskian and must not be discarded by confusing the original operator with its multiplied divergence form $(x^{n+1}{u}^{\prime }{)}^{\prime }$.

For ${\partial }_x^2+k^2$ on $(0,\ell )$ with zero endpoint values and $\sin (k\ell )\ne 0$, let $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. The displayed Green function is continuous at $x=\xi$ and its first derivative has jump one, giving unit Dirac delta forcing. The formula follows from the left solution $\sin (kx)$ and right solution $\sin (k(\ell -x))$, whose Wronskian is $-k\sin (k\ell )$. At $k=0$ its limit is $-x_{<}(\ell -x_{>})/\ell$, although zero is excluded by a condition that $k$ is not an integer in the length-$\pi$ version.

The density of expected Brownian occupation time before exit from a planar domain. For planar Brownian motion with infinitesimal generator $\frac{1}{2}\Delta$, it obeys $-\frac{1}{2}{\Delta }_w{G}_D(z,w)={\delta }_z(w)$ and has singularity $-{\pi }^{-1}\log |w-z|$. A conformal bijection preserves this Dirichlet Green function: the squared derivative in the conformal Brownian clock cancels the area Jacobian determinant.

For the positive Laplacian convention $\Delta G=\delta$, the method of images in the first quadrant reflects an interior pole $(a,b)$ across each axis with negative sign and across both with positive sign:The Dirichlet Green function vanishes on both axes. Its outward normal derivative on the horizontal axis isand the vertical-axis kernel follows by exchange of coordinates. Green second identity converts these derivatives into the quadrant Poisson integral for boundary data.

For the canonical Dirichlet Green function with logarithmic singularity, its harmonic correction satisfies $G_z(z)=-\log {\mathrm{crad}}_D(z)$. If $\varphi$ maps the unit disc to $D$ with $\varphi (0)=z$, thenThe quotient extends at zero to ${\varphi }^{\prime }(0)$. On unbounded domains the canonical Green function, rather than boundary values alone, fixes the harmonic correction.