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The velocity is $\mathbf{u}=\nabla \varphi =q{\mathbf{e}}_r/(2\pi r)$. Its divergence in polar coordinates is zero for $r>0$, while a circle about the origin has fluxBy contour deformation the same holds for any enclosing contour. This is a two-dimensional point source of strength $q$.

For the second potential,which vanishes at $y=0$. In the full plane this is a pair of equal sources of strength $q$ at $(0,\pm a)$. In $y>0$, it is one source at $(0,a)$ beside an impermeable slip wall, with the lower source serving as its image.

Remove a small disc about $x$ and apply Green second identity to $u(x_0)$ and $G_0(x;x_0)$. Since ${\nabla }_{x_0}^2{G}_0=\delta (x_0-x)$ and ${\nabla }^{2}u=-\rho$, the shrinking inner boundary contributes $u(x)$ and gives

For the unit disc, the method of images gives the Dirichlet Green functionwhich vanishes on the boundary by the hinted identity. The boundary representation with $u(1,{\theta }_0)=\delta ({\theta }_0-\alpha )$ evaluates the normal derivative of $G_D$ at ${\theta }_0=\alpha$, giving the Poisson kernel

The method of images replaces the earthed plane by an image charge $-q$ at $(-a,0,0)$. The two Coulomb potentials cancel at $x=0$, so uniqueness for the Dirichlet problem makes the resulting field the physical field in $x>0$. With $\mathbf{r}=(x,y,z)$ and ${\mathbf{e}}_x=(1,0,0)$,and

The electric multipole expansion from part (a) gives, for $r\gg a$,Thus the leading field is that of an electric dipole.

On the plane,Taking the plane normal to be $+{\mathbf{e}}_x$ and using polar coordinates,With the outward normal of the region $x>0$, the sign is reversed. This is consistent with Gauss's law: all electric flux from the real charge terminates on the grounded conductor.

The electrostatic boundary condition gives the induced surface charge densityand its integral isIn the plane $z=0$, the field lines leave the positive charge, meet the conductor normally, and are the right-half-plane portions of the field lines joining the real charge to its negative image.

The flat four-dimensional retarded Green function isFor a Dirichlet boundary condition at $z=0$, the method of images givesThe boundary-to-bulk kernel for $\psi$ is its inward boundary derivative. For a source at the spacetime origin,Multiplying by the Weyl factor $z$ gives the requested kernel for the original bulk field:It is causal and supported on the future bulk light cone of the boundary insertion. Its distributional boundary limit satisfies $\varphi /z\to {\delta }^3(x)$, while the source-free initial term vanishes for the CFT vacuum in this response calculation.

Reflect the source point ${\mathbf{r}}_0=(x_0,y_0,z_0)$ across the plane to ${\mathbf{r}}_0^{\ast }=(x_0,y_0,-z_0)$. Equal-sign source and image make the normal derivatives cancel on the plane. Thus the method of images gives the Neumann Green function

Expand the divergence of $\Phi \nabla \Psi -\Psi \nabla \Phi$. The cross terms cancel, leaving $\Phi \Delta \Psi -\Psi \Delta \Phi$. The divergence theorem gives Green's second identity,Use the Green function convention ${\Delta }_{x}G(x,x_0)=\delta (x-x_0)$. Integrating over a bounded domain containing $x_0$ yields $\boxed{{\int }_S{\partial }_{n}GdS=1}$. Thus zero Neumann boundary data would be incompatible with this unit Dirac delta source. A suitable Neumann Green function with constant boundary flux instead has ${\partial }_{n}G=1/|S|$, where $|S|$ is the boundary area. Substituting $\Psi =G$, $\Delta \Phi =\sigma$, ${\partial }_n\Phi =f$ givesThe last term is an undetermined constant. The data must obey the Neumann Poisson problem compatibility condition ${\int }_V\sigma dV={\int }_{S}fdS$, obtained by integrating the original equation. Under this condition an additive constant in $G$ cancels from the first two terms.

For the upper half-space, reflect the source to $x_0^{\ast }=(x_{0,1},x_{0,2},-x_{0,3})$. The method of images with equal-sign sources givesThe reflected singularity lies outside the domain; on $z=0$ the two normal derivatives cancel, so ${\partial }_{n}G=-{\partial }_{z}G=0$. Its total unit flux passes through a large hemisphere at infinity, where $G\sim -1/(2\pi R)$. This is why the bounded-domain condition on the boundary flux cannot be applied to the infinite plane alone.

With zero interior source and the given point boundary source, the boundary representation is $\Phi (x_0)=-G((0,0,0),x_0)$. The terms at infinity vanish for this resulting $1/R$ field. ThereforeTo check the singular boundary condition, $-{\partial }_z\Phi =z/[2\pi (x^2+y^2+z^2{)}^{3/2}]$ is the Poisson kernel for the upper half-space. Its integral over a horizontal plane is one and its mass concentrates at the origin as $z\downarrow 0$, giving ${\partial }_n\Phi =\delta (x)\delta (y)$ in the distributional sense. This is the Neumann point source on a half-space boundary.

Choose the Green function convention ${\Delta }_{\mathbf{r}}G(\mathbf{r};{\mathbf{r}}_0)={\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}_0)$, with decay at infinity and no extra singularities. For sufficiently decaying source data,Away from its source, this fundamental solution of the Laplace equation is harmonic. Its radial derivative is $1/(4\pi {\rho }^2)$, so the flux of its gradient through a small source-centred sphere is one. The divergence theorem therefore supplies exactly the unit Dirac delta function required by the equation. Decay at infinity excludes an added globally regular harmonic term.

For the half-space, reflect the source to ${\mathbf{r}}_0^{\ast }=(-x_0,y_0,z_0)$. The method of images gives the zero-boundary Dirichlet Green functionThe two distances coincide at $x=0$, and also cancel if the source lies on the plane. The image singularity lies outside the physical domain.

The sign conventions for a Dirichlet Green function require care here. With $\Delta \tilde{G}=+\delta$, Green's second identity gives $\Phi ({\mathbf{r}}_0)={\int }_{\partial \Omega }f{\partial }_n\tilde{G}dS$. At this wall the outward normal is $-{\mathbf{e}}_x$, so ${\partial }_n\tilde{G}=-{\partial }_x\tilde{G}$. Direct differentiation gives ${\partial }_x\tilde{G}{|}_{x=0}=-x_0/(2\pi d^3)$, where $d^2=x_0^2+(y-y_0{)}^2+(z-z_0{)}^2$. Thus the printed intermediate formula with $+{\partial }_x\tilde{G}$ needs a minus sign for its stated Green convention. The final Poisson kernel for the upper half-space is positive:Its boundary integral is one for constant data, providing an independent sign check.

Let ${\mathbf{x}}_0^{-}=(x_0,y_0,-z_0)$ be the reflection of the source across the plane. The method of images gives the Dirichlet Green functionThe two source distances agree on $z=0$, so the boundary value vanishes. The reflected source lies outside the upper half-space, leaving only the required Dirac delta distribution inside. Both terms vanish at infinity.

Write $R=|x-x_0|$. A radial function harmonic away from the source satisfies $(R^2{G}^{\prime }{)}^{\prime }=0$, hence $G=A+B/R$. Decay at infinity forces $A=0$. Integrating $\Delta G=\delta (x-x_0)$ over a small sphere, the divergence theorem requires flux one:Therefore the fundamental solution in this sign convention isAdding point sources outside the domain changes only a harmonic function inside it. The method of images selects their positions and strengths to cancel boundary values. The sign reverses for the alternative convention $-\Delta G=\delta$.

For the grounded wall, use an image charge $-q$ at $(2b,0)$. Define the Euclidean distances $r=\sqrt{x^2+y^2}$ and $r_{\ast }=\sqrt{(x-2b{)}^2+y^2}$. The method of images givesThe real source has the required Dirac delta function coefficient, and the image is outside the domain. On $x=b$, $r=r_{\ast }$, so $V_D=0$ exactly. This verifies the Dirichlet boundary condition rather than only its derivative.

Reflect $r^{\prime }=(x^{\prime },y^{\prime },z^{\prime })$ to $r^{\prime \ast }=(x^{\prime },y^{\prime },-z^{\prime })$. The method of images gives the Dirichlet image kernel for the three-dimensional Helmholtz equation:The two distances coincide on the plane, so its boundary value is zero; the image source is outside the domain. It therefore has the required positive delta source inside and the selected outgoing behaviour.

Let $h(R)=e^{ikR}/R$. At a boundary point $r=(\xi ,\eta ,0)$, the distances to the real and reflected sources have $z$ derivatives $-z^{\prime }/R$ and $+z^{\prime }/R$. ThusThe outward unit normal of the upper half-space is $-e_z$, and $z^{\prime }/R=\cos \theta$. HenceUsing the decay assumption to discard the distant hemisphere in the boundary representation givesThe qualification at infinity concerns the whole Green boundary term, not just the pointwise limit of $\varphi$. Sufficient decay of the field and its derivatives, as assumed, makes it vanish; compatible radiation conditions can also do so. As a sign check, taking $k\to 0$ in the kernel gives $z^{\prime }/(2\pi R^3)$, the positive Poisson kernel for the upper half-space.

Let the evaluation point be ${\mathbf{r}}_0=(x,y,z)$, $z>0$, and reflect it to ${\mathbf{r}}_0^{\ast }=(x,y,-z)$. The method of images constructsThe two Euclidean distances coincide on the plane, so $G=0$ there. The negative image source lies outside the domain; inside, only the real source contributes to $-{\nabla }_{\mathbf{r}}^{2}G={\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}_0)$. The difference also decays at infinity.

On the plane write $\mathbf{r}=(\xi ,\eta ,s)$ and set $s=0$. The outward unit normal of the upper half-space is $n=-e_z$. Direct differentiation givesSubstituting into the boundary representation from part (c) yields the Poisson kernel for the upper half-space:Its sign is positive, as expected from the maximum principle for nonnegative boundary data. Its normalization can be checked in polar coordinates:As $z\downarrow 0$ the kernel is an approximate identity, recovering $f$ at its continuity points. Boundary data must have the decay and regularity compatible with the stated condition at infinity.

The domain is the three-dimensional upper half-space. Reflect the source point to ${\mathbf{r}}^{\prime \ast }=(x^{\prime },y^{\prime },-z^{\prime })$. The method of images places an opposite source at this reflected point, giving the Dirichlet image kernel for the three-dimensional modified Helmholtz equationOn $z=0$, the two Euclidean distances are equal, so the terms cancel exactly. The image lies outside the domain, so its contribution satisfies the homogeneous modified Helmholtz equation everywhere inside; only the original source contributes a Dirac delta. Both terms decay at infinity, including for $k=0$, and the homogeneous Dirichlet boundary condition follows.

For $x\ne \xi$, the homogeneous solutions are $e^x,e^{-x}$. The left boundary condition selects $\sinh x$, while boundedness at infinity selects $e^{-x}$ on the right. Thus write $G=A(\xi )\sinh x$ for $x<\xi$ and $G=B(\xi )e^{-x}$ for $x>\xi$.

A unit Dirac delta distribution in the second derivative requires continuity of $G$ and a derivative jump of one:The second equation and the first give $-A{e}^{\xi }=1$, so the Dirichlet half-line Green function for d2 minus 1 isThe second expression is also a method of images construction. Applying the Green's function to the forcing givesFor bounded forcing this is bounded, vanishes at zero and solves the equation; a bounded homogeneous solution vanishing at zero is zero, proving uniqueness. More general forcing is allowed when the integrals converge and the resulting function has the required growth. Arbitrary unbounded forcing need not admit a bounded solution; this distinction is relevant in part (b).

For $\Delta G=+{\delta }^{(3)}$, the fundamental solution of the Laplace equation in three dimensions isIts sign follows from the flux: on a small sphere of radius $\epsilon$, its outward radial derivative is $1/(4\pi {\epsilon }^2)$, with total flux one.

Let $s=|{\mathbf{r}}^{\prime }|\in (0,1)$ and put the image point at ${\mathbf{r}}^{\prime \ast }={\mathbf{r}}^{\prime }/s^2$. The method of images gives the Dirichlet Green function inside a sphere:The image lies outside the ball, so its potential is harmonic inside. On $|\mathbf{r}|=1$, the identity $|\mathbf{r}-{\mathbf{r}}^{\prime }|=s|\mathbf{r}-{\mathbf{r}}^{\prime \ast }|$ makes the two terms cancel. The original source retains unit strength. For a source at the centre, take the limit:

A Dirichlet Green function satisfies ${\Delta }_{\mathbf{x}}{G}_D(\mathbf{x},\mathbf{y})={\delta }^{(2)}(\mathbf{x}-\mathbf{y})$ in the domain and $G_D=0$ when $\mathbf{x}$ lies on its boundary. For a source $\mathbf{y}\ne 0$ in the unit disc, put its opposite image at ${\mathbf{y}}^{\ast }=\mathbf{y}/|\mathbf{y}{|}^2$. The method of images givesThe image lies outside the unit disc, so its logarithm is a harmonic function inside. The original logarithm has precisely the unit source strength found in part (a). On $|\mathbf{x}|=1$,so the boundary value is zero. Equivalently, in complex coordinates, the Dirichlet Green function inside the unit disk isThis also covers a source at the origin, where $G_D(z,0)=(2\pi {)}^{-1}\log |z|$.

Reflect the source across the plane, defining ${\mathbf{r}}^{\prime \ast }=(x^{\prime },y^{\prime },-z^{\prime })$ for an interior point $z^{\prime }>0$. Superpose the direct Green function from part (b) and an image of opposite sign:The direct term supplies the delta source in the upper half-space; the image singularity lies outside that domain and contributes no source there. On $z=0$ the two distances are equal, so the terms cancel, giving homogeneous Dirichlet boundary conditions. Both decay at infinity, which is the additional condition selecting this half-space solution. This is the method of images realization of the Dirichlet image kernel for the three-dimensional modified Helmholtz equation. It includes the unscreened case $k=0$.

The three-dimensional fundamental solution of the Laplace equation, with $\Delta F=+\delta$, isIt is harmonic away from the source, and its gradient has total outward flux one through a small source-centred sphere.

For $s=|{\mathbf{r}}^{\prime }|\in (0,1)$, place the image at ${\mathbf{r}}^{\ast }={\mathbf{r}}^{\prime }/s^2$, outside the unit ball. On $|\mathbf{r}|=1$, expansion of squared distances gives $|\mathbf{r}-{\mathbf{r}}^{\ast }|=|\mathbf{r}-{\mathbf{r}}^{\prime }|/s$. The method of images therefore gives the Dirichlet Green function inside a sphereThe boundary value is zero by the distance identity, and the image term is harmonic throughout the physical ball. Thus $\Delta G_0=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$ there. At the central source, take the finite limit of the weighted image term:

Reflect the physical source across $z=0$, writing $\overline{{\mathbf{r}}^{\prime }}=(x^{\prime },y^{\prime },-z^{\prime })$. Its radius is still $s=|{\mathbf{r}}^{\prime }|$, and its spherical image is $\overline{{\mathbf{r}}^{\ast }}$. The Dirichlet Green function inside a hemisphere isExplicitly this four-source method of images isBoth $G_0$ terms vanish on the spherical cap. On the flat equatorial disc, reflection equates the two physical-source distances and also the two image-source distances, so both pairs cancel. The reflected physical source lies in the lower half-ball and the two inverted images lie outside the sphere. Consequently the only source in the upper half-ball is ${\mathbf{r}}^{\prime }$, with the correct coefficient:The Dirichlet uniqueness for the Poisson equation then identifies this as the required Green function.

Write $s=|{\mathbf{r}}^{\prime }|$ and ${\mathbf{r}}^{\prime \prime }=a^2{\mathbf{r}}^{\prime }/s^2$. If $|\mathbf{r}|=a$, expansion of squared Euclidean distances givesConsequently $|\mathbf{r}-{\mathbf{r}}^{\prime \prime }|=(a/s)|\mathbf{r}-{\mathbf{r}}^{\prime }|$, and the two terms in $H$ cancel. This method of images gives the Dirichlet Green function inside a sphere:The image point is outside the sphere, so only the physical source contributes a Dirac delta distribution inside. As $s\to 0$, the image position tends to infinity and its coefficient also diverges, but their ratio has the finite limitThusThe apparent singularity of the image construction at a central source is removable; the physical source singularity at $\mathbf{r}=0$ remains.