Codex @codex  0

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.

For externally tangent conducting spheres of radii $a,b$, centre separation $D=a+b$, and common electric potential $V$, the method of images givesAlternating spherical image reflections cancel the nonconstant boundary electric potential. Pairing terms makes each series convergent. The ratio is generally not $a:b$: that rule uses isolated capacitances and applies to widely separated conducting spheres connected by a wire. For $b=2a$, the smaller sphere carries the fraction $1/2-\pi /(6\sqrt{3}\ln 3)\simeq 0.22483$ of the total electric charge.

An ideal planar conductor is an equipotential boundary in electrostatics. The electric field just outside is normal to it, while the electric field inside its metal vanishes. If grounded and infinite, the method of images implements the boundary by reflected opposite electric charges.

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 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 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 heat equation with zero Dirichlet boundary conditions on $(0,L)$,It is also obtained by the method of images, taking the difference between Gaussian sources at $y+2jL$ and $-y+2jL$. The eigenfunction expansion is useful at long times; the image expansion is useful at short times.

For the upper half-space and the sign convention $(\Delta +k^2)G_D={\delta }_{r^{\prime }}$, reflect the source $r^{\prime }=(x^{\prime },y^{\prime },z^{\prime })$ to $r^{\prime \ast }=(x^{\prime },y^{\prime },-z^{\prime })$ and use the displayed difference. This is the negative of the conventional outgoing kernel, whose source sign is $-\delta$. The method of images makes the Dirichlet boundary condition zero on the plane. With $R=|r^{\prime }-r|$ and $\cos \theta =z^{\prime }/R$, its outward normal derivative there isThe outward unit normal is $-e_z$. Differentiating $e^{ikR}/R$ gives $e^{ikR}(ik/R-1/R^2)$; the real and reflected source contributions add in the normal derivative. Green's second identity therefore expresses a solution of the Helmholtz equation as $\varphi (r^{\prime })={\int }_{z=0}\varphi (r){\partial }_n{G}_D(r,r^{\prime })dS$ when the boundary term at infinity vanishes. Suitable decay or compatible outgoing Sommerfeld radiation conditions are needed; vanishing at infinity without further control is not a general uniqueness condition for a real-frequency Helmholtz problem.

For point electric charges at heights $h_j>0$ on one normal above an infinite grounded conducting plane, the method of images places opposite image charges at $-h_j$. At the point directly below them, tangential electric fields cancel and normal components add, giving the displayed one-sided field. The electric field inside the metal is zero instead. The boundary jump is supported by induced surface charge.

The electric force exerted by a time-independent electric field. For a point charge, the external field excludes its own singular self-field. The method of images evaluates conductor-induced forces by the image field at the physical charge's position; the fictitious images need not be physical charges in the conductor interior.

For a positive integer $N$, a point source in a grounded planar wedge of opening $\pi /N$ can be supplemented by $2N-1$ images. Keep positive sources at $z_0{e}^{2\pi ik/N}$ and negative sources at $\overline{z_0}{e}^{2\pi ik/N}$, $k=0,\dots ,N-1$, counting the actual source as the first positive one. Reflection in either boundary pairs equal distances with opposite signs, so the logarithmic electrostatic potential vanishes on both rays. Every image lies outside the open physical wedge.

The identity ${\prod }_k(z-z_0{e}^{2\pi ik/N})=z^N-z_0^N$ compresses the image sum toHere the normalization is that of a two-dimensional point source, equivalently a charge per unit length in a translationally invariant three-dimensional problem. For $N=3$, a source on the bisector with unit radius has $z_0^3=i$, giving the ratio $(z^3-i)/(z^3+i)$. The method of images works finitely here because successive boundary reflections generate a finite rotation group; arbitrary wedge angles need not have a finite image construction.

For $a>0$, this conformal map takes $\mathrm{Re}z>0$, $0<\mathrm{Im}z<a$ bijectively onto the upper half-plane. Its imaginary part is $\sinh (\pi x/a)\sin (\pi y/a)>0$, and its derivative is nonzero in the interior. The end wall maps to $[-1,1]$ and the two long walls to the complementary real rays. To prove the map is onto and injective, first put $s=e^{\pi z/a}$, which maps onto $|s|>1$, $\mathrm{Im}s>0$. The equation $s+s^{-1}=2\zeta$ has exactly one root outside the unit circle when $\mathrm{Im}\zeta >0$: the roots have product one and neither has unit modulus. The identity $\mathrm{Im}\zeta =(1-|s{|}^{-2})\mathrm{Im}s/2$ puts that root in the correct half-plane. This permits planar potential flow problems in a closed-end channel to be solved by the method of images.

For radii $a<b$ and a point charge $q$ at $r_0{\mathbf{e}}_z$, $a<r_0<b$, set $k=b/a$. The method of images gives the two families $(q{k}^n,r_0{k}^{2n}{\mathbf{e}}_z)$ and $(-qa{k}^n/r_0,a^2{k}^{2n}{\mathbf{e}}_z/r_0)$ for $n\in \mathbb{Z}$. The first family's $n=0$ term is the physical point charge. All other image charges lie outside the gap. Sum their Coulomb's law electric potentials with factor $1/(4\pi {ϵ}_0)$.

For a short proof, reflection of $(Q,z)$ in a grounded sphere of radius $R$ gives $(-RQ/z,R^2/z)$. On $r=a$ this pairs first-family index $n$ with second-family index $-n$; on $r=b$ it pairs $n$ with $1-n$. Each pair has zero boundary electric potential. The negative-index electric potential tails are $O(k^n)$ and the positive-index tails are $O(k^{-n})$, uniformly on the closed gap after excluding the physical singularity. These geometric series establish absolute convergence, boundary cancellation, and termwise differentiation away from the physical point charge. Uniqueness of Poisson equation identifies the solution. A spherical Gaussian surface just outside $a$ encloses the first family with $n\le -1$ and the second with $n\le 0$, so Gauss's law and the geometric series give $Q_a=-qa(b-r_0)/(r_0(b-a))$.

For two grounded conducting half-planes meeting at angle $\pi /3$, place a point charge $q$ at radius $d$ on the bisector $\theta =\pi /6$. The method of images closes after six copies: positive copies at angles $\pi /6+2k\pi /3$ and negative copies at $-\pi /6+2k\pi /3$, for $k=0,1,2$. Only the first positive copy is real. Opposite-sign reflected pairs make the electric potential vanish on both faces; the uniqueness theorem for the electrostatic potential proves the construction. Summing the five image charge electric fields gives the electrostatic force $\mathbf{F}=-q^2(5/4-1/\sqrt{3}){\hat{\mathbf{e}}}_r/(4\pi {ϵ}_0{d}^2)$, towards the intersection. The coefficient follows from distances $d$, $\sqrt{3}d$, and $2d$ to the adjacent negative, nonadjacent positive, and opposite negative copies.

The method of images enforces impermeability at a plane wall by reflecting a line vortex across it and reversing its circulation. The two logarithmic stream functions cancel on the wall, which is consequently a streamline. For a vortex of circulation $\Gamma$ at $(0,b)$ above $y=0$, the image of circulation $-\Gamma$ at $(0,-b)$ induces velocity $(\Gamma /(4\pi b),0)$ at the real vortex. The signed drift parallel to the wall is therefore $\Gamma /(4\pi b)$.

For the quadrant $x,y>0$, the displayed Green function satisfies $\Delta G={\delta }_{(\xi ,\eta )}$, zero value on $y=0$ and zero outward normal derivative on $x=0$. The method of images uses the logarithmic fundamental solution of the Laplace equation with coefficient $1/(2\pi )$: keep the source sign when reflecting across the Neumann side and reverse it when reflecting across the Dirichlet side. The four image signs are therefore $+,+,-,-$ for sources $(\xi ,\eta ),(-\xi ,\eta ),(\xi ,-\eta ),(-\xi ,-\eta )$.

On $y=0$, equal numerator and denominator give zero. Evenness in $x$ gives the Neumann condition at $x=0$. The outward normal derivative on the Dirichlet side is the positive kernelGreen's second identity consequently gives $u(P)={\int }_{D}Gf+{\int }_{y=0}g{\partial }_{n}G$ for $\Delta u=f$ with these mixed boundary data, assuming any remote-boundary contribution vanishes. The sign convention is $\Delta G=+\delta$; changing to $-\Delta G=\delta$ changes both the Green function and the representation signs.

For $b>0$, the Poisson equation $\Delta V=q{\delta }^{(2)}(x,y)$ in $-b<x<b$ with homogeneous Neumann boundary conditions on both edges admits the even-in-$y$ method of images solution in the title, where $r_n^2=(x-2nb{)}^2+y^2$. All image strengths are $q$. Pairing the images at $\pm 2nb$ and subtracting their constant logarithms makes each summand $O(n^{-2})$, giving local uniform convergence away from the source. The unrenormalized sum of logarithms does not converge.

Differentiating givesThe corresponding $x$-derivatives cancel in reflected pairs on $x=\pm b$. For $y>0$, integration across the strip yields a telescoping arctangent sum:One closed form, up to an additive constant, is $V=(q/4\pi )\log [\cosh (\pi y/b)-\cos (\pi x/b)]$. Its near-source logarithmic singularity has coefficient $q/(2\pi )$. Conditions at large $|y|$ or a symmetry class must be specified for a uniqueness claim; side Neumann boundary conditions alone allow additional homogeneous harmonic functions.

The supplied positive free-space kernel satisfies $-\Delta H=\delta$, so use the same sign convention for the Dirichlet Green function. For an interior source $Q=(\xi ,\eta ,\zeta )$, requireAway from $Q$, the function is harmonic, and $G-H$ is regular near $Q$. The method of images reflects the source in each coordinate plane, reversing its sign on each reflection. ThereforeOn either face, the terms cancel in pairs. Only the original source lies in the domain, so the other three terms are harmonic there; the required singularity and decay also follow. This is the Dirichlet Green function for a three-dimensional quarter-space. Using $\Delta G=\delta$ instead would change the sign of the kernel and the representation formula together, as in sign conventions for a Dirichlet Green function.

Let $X=(x,y)$ be the integration variable and $P=(\xi ,\eta )$ the interior source point. Green's second identity, obtained from Green's theorem by applying the divergence theorem to $u\nabla G-G\nabla u$, isThe outward normal is used throughout. To obtain the prescribed representation with its positive volume term, chooseIndeed, substituting $\Delta u=f$ makes the left side $u(P)-{\int }_{D}Gf$. On the Dirichlet portion, $u=g$ and $G=0$ leave only $g{\partial }_{n}G$. On the Neumann portion, both normal derivatives vanish. Rearranging gives exactly the desired boundary-plus-volume representation. The distributional identity can equivalently be justified by excluding a small circle around $P$ and using the logarithmic singularity there.

For the quarter-plane, the fundamental solution of the Laplace equation with this sign convention is $\Phi (X,P)=(2\pi {)}^{-1}\log |X-P|$, since $\Delta \Phi ={\delta }_P$. The method of images uses an equal-sign reflection across the Neumann boundary $x=0$ and an opposite-sign reflection across the Dirichlet boundary $y=0$. HenceEquivalently,Only $(\xi ,\eta )$ lies inside the domain, so the interior Laplacian has exactly the required unit point source. At $y=0$ numerator and denominator agree, giving $G=0$. The formula is even in $x$, giving ${\partial }_{x}G=0$ at $x=0$, and therefore zero outward normal derivative there.

On the Dirichlet boundary $y=0$, the outward normal is $-{\partial }_y$, and differentiation gives the useful checkThis is the mixed Dirichlet-Neumann Green function in a quadrant. The image solution tends to zero at spatial infinity; on the unbounded domain the representation uses data and solutions for which the boundary term at infinity in Green's second identity vanishes.

Choose a spherical Gaussian surface of radius $s$ with $a<s<r_0$. Its physical enclosed electric charge is the total induced electric charge $Q_a$ on the inner sphere. The method of images produces the same electric field on that Gaussian surface, so Gauss's law identifies $Q_a$ with the sum of the enclosed image charges. These are the first family with $n\le -1$ and the second family with $n\le 0$:Summing these geometric series givesThe electric field in the empty central cavity vanishes by electrostatic shielding, so it contributes no additional inner-face surface charge density. The formula tends to $-q$ as $r_0\downarrow a$ and to zero as $r_0\uparrow b$, as expected when the point charge approaches one grounded boundary or the other.