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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 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.

The method of images places an image charge $-q$ at the reflected position below the plane. The candidate potential in the physical upper half-space isAt $z=0$ the two terms cancel, so $\Phi =0$ and $E_x=E_y=0$. Direct differentiation givesThus the upper-side field satisfies exactly the conductor boundary conditions from part (i). The fictitious pair does have a field below the plane, but that part of its solution is not used for the physical conductor.

In $z>0$, the potential has the correct real point source, the correct zero Dirichlet boundary condition, and the required decay at infinity. By Uniqueness of the Dirichlet problem, it is the physical potential there. This is why the artificial charge can represent the induced conductor response without being a second physical charge.

Let $A_s$ denote the surface area of an n-sphere for the unit sphere $S^s$. In this notation $s$ is the intrinsic dimension of the sphere. Cartesian factorization of the Gaussian integral gives $I_N={\pi }^{N/2}$. In hyperspherical coordinates,The substitution $t={\rho }^2$ supplies the defining Gamma function integral. Scaling the sphere radius multiplies its surface volume by $r^{N-1}$, henceThis is the volume of the sphere itself, not of its enclosed ball.

Take $D$ to denote spacetime dimension and $d=D-1$ spatial dimensions. In uncompactified flat space, Gauss's law for a charge $q$ and permittivity ${\epsilon }_d$ gives $E_r{A}_{d-1}{r}^{d-1}=q/{\epsilon }_d$. Since $E_r=-d\phi /dr$, the electric potential for $d>2$, normalized to vanish at infinity, isThus the four-dimensional electric potential falls as $1/r$, and the five-dimensional one as $1/r^2$. If a convention counts only spatial dimensions as $D$, replace $d$ by $D$ in the first formula. For $d=2$, the integral instead gives $\phi =-q\log (r/r_0)/(2\pi {\epsilon }_2)$; for $d=1$, it gives $\phi =-q|x|/(2{\epsilon }_1)$ up to a constant. Neither case has a finite zero at infinity.

For a Newtonian gravitational potential ${\Phi }_g$, define the flux-normalized coupling ${\kappa }_d$ by ${\nabla }^2{\Phi }_g={\kappa }_{d}m{\delta }^{(d)}(\mathbf{x})$ and acceleration $-\nabla {\Phi }_g$. Exactly the same flux integration gives the attractive potentialIn $d=3$, ${\kappa }_3=4\pi G_4$ reproduces $-G_{4}m/r$. If $G_D$ is defined by the $D$-dimensional Einstein-Hilbert action $S=(16\pi G_D{)}^{-1}\int \sqrt{-g}R$, the weak static Einstein field equations give ${\kappa }_d=8\pi G_D(D-3)/(D-2)$ for $D>3$. ThereforeIndeed, $R_{00}\simeq {\nabla }^2{\Phi }_g$ and $T_{00}-g_{00}T/(D-2)=(D-3)\rho /(D-2)$ in signature $(-+\cdots +)$. A Poisson-defined Newtonian theory in two spatial dimensions has a logarithmic potential, but pure three-dimensional Einstein gravity has no analogous local Newtonian point-mass force; the relativistic normalization must not be extrapolated to that exceptional dimension.

For an unwarped product with a fixed extra-dimensional metric, normalize the Einstein-Hilbert action by $\frac{1}{2}{M}_{\ast }^{n+2}\int d^{4+n}x\sqrt{-g}R$. Integrating the part containing the four-dimensional curvature over the internal space gives $\frac{1}{2}{M}_{\ast }^{n+2}\mathrm{vol}(S_R^n)\int d^{4}x\sqrt{-g_4}{R}_4$. Thus, using the reduced Planck mass,This is a volume relation for compactification; a curved sphere additionally requires a mechanism supporting and stabilizing its background. It is not by itself a proof that a sphere times flat spacetime solves a vacuum gravitational theory.

With $M_{\ast }=10^3\mathrm{GeV}$, $M_{\mathrm{Pl}}=2.435\times 10^{18}\mathrm{GeV}$ and $1{\mathrm{GeV}}^{-1}=1.97327\times 10^{-16}\mathrm{m}$, the spherical-volume factors $A_1=2\pi$, $A_2=4\pi$, $A_6=16{\pi }^3/15$ giveThe first estimate is astronomical; the second is roughly a tenth of a millimetre. If both Planck scales are instead defined by $S=M_D^{D-2}\int \sqrt{-g}R/(16\pi )$, the same algebraic volume relation uses the unreduced $M_{\mathrm{planck}}=1.221\times 10^{19}\mathrm{GeV}$. Holding that convention's $M_{\ast }=1\mathrm{TeV}$ gives radii larger by $(8\pi {)}^{1/n}$: approximately $4.69\times 10^{12}\mathrm{m}$, $6.80\times 10^{-4}\mathrm{m}$, $2.54\times 10^{-14}\mathrm{m}$. These are different conventions for what is held fixed at one TeV, not different scaling laws.

Finally, compactify the fifth coordinate on a circle of circumference $L=2\pi R$. For a source and observer at the same compact coordinate, the method of images turns a five-dimensional static potential $C/{\rho }^2$ intoOne can derive this image sum without assuming a summation formula. The Fourier transform of $1/(r^2+y^2)$ is $\pi e^{-r|k|}/r$; Poisson summation then gives $\pi [1+2{\sum }_{\ell \ge 1}{e}^{-2\pi \ell r/L}]/(Lr)$, which is the displayed hyperbolic cotangent. ThereforeThe short-distance limit is $C/r^2$; the long-distance limit is the four-dimensional $1/r$ potential. These exponential corrections are the nonzero Kaluza-Klein modes, with masses $2\pi |\ell |/L$. For electrostatics $C=q/(4{\pi }^2{\epsilon }_4)$, so the effective permittivity is ${\epsilon }_3=L{\epsilon }_4$. For a Poisson-normalized gravitational potential, $C=-{\kappa }_{4}m/(4{\pi }^2)$ gives ${\kappa }_3^{\mathrm{eff}}={\kappa }_4/L$. A massless scalar associated with the circle radius also contributes in an unstabilized gravitational compactification; recovering pure four-dimensional Einstein gravity with $G_4=G_5/L$ requires its stabilization or removal. The $1/r$ crossover itself is independent of this tensor-versus-scalar normalization issue.

Use unfolding a reflected interval trajectory: reflect the interval across each wall so that a bouncing trajectory becomes a free trajectory in an image interval. A return path starting at $q$ ends at $q+2rL$ after an even number of reflections, or at $-q+2rL$ after an odd number. Each hard-wall reflection contributes phase $-1$, giving the positive direct images and negative reflected images of the Dirichlet heat kernel on an interval.

The method of images therefore givesHere $\tau$ has inverse-energy units, since $\beta =1/(k_{B}T)$. In physical imaginary time $u=\hslash \tau$, the same dimensionless action is $\mathcal{A}=S_E/\hslash$, with $S_E={\int }_0^{\hslash \beta }(m/2)(dq/du{)}^{2}du$. Thus the question's convention $S=(m/(2\hslash )){\int }_0^{\beta }({\partial }_{\tau }q{)}^{2}d\tau$ has exactly the weight $e^{-S/\hslash }$.

The method of images uses the real line charge at the origin and the opposite image charge at $(2d,0)$. LetSuperposing their logarithmic electrostatic potentials gives, in the physical region $x<d$,At $x=d$, $r_1=r_2$, so $\varphi =0$. Its tangential derivatives also vanish on that plane, giving $E_y=E_z=0$, exactly the electrostatic boundary conditions at a conductor. The image lies outside the physical domain, so it creates no extra physical source there. Together with the appropriate behaviour at infinity, uniqueness of Poisson equation identifies this field with the grounded-conductor solution.

The method of images for the Neumann boundary conditions extends $bx$ evenly through zero and then with period $2\pi$. Define the triangular waveThe formulas agree at their endpoints. It is even, $2\pi$-periodic, and agrees with $bx$ on $[0,\pi ]$. Its Fourier cosine series has the coefficients in part (a). The cosine product identity then gives the D'Alembert formulaIt is the average of two oppositely traveling reflected waves. For $b>0$, the sketch rises from zero at even multiples of $\pi$ to $b\pi$ at odd multiples; for $b<0$ it is reflected vertically, and for $b=0$ it is zero.

Write the pole as $(x_0,y_0)$ with $x_0,y_0>0$. The method of images uses three reflected poles with signs $-,-,+$. ThusThe logarithms cancel in pairs on each axis. Only the original pole is in the interior, so this is the required Dirichlet Green function in a quadrant.

On the horizontal boundary $(t,0)$ the outward normal is $-{\mathbf{e}}_y$. Differentiating the four logarithms givesMultiply by the horizontal Dirichlet boundary condition $e^{-t^2}$ and integrate to obtainBy exchanging the axes, the vertical boundary contributes $F(y_0,x_0)$. The representation in part (a) consequently givesFor the unbounded-domain condition, the same expression is the Poisson integral after the conformal change of variables $w=(x+iy{)}^2$ to the upper half-plane, with real-axis data $e^{-|s|}$. This continuous boundary function tends to zero at both ends. Its Poisson integral tends to zero as $|w|\to \infty$: split the data into a compact part, whose kernel integral tends to zero, and a tail uniformly smaller than any prescribed positive number. Hence the constructed function has the required decay, and its two boundary traces agree at the corner. The maximum principle for harmonic functions on expanding quarter-discs proves uniqueness among decaying solutions. The corner value is obtained by continuity as $1$.

Use the Bromwich inversion formula in the real variable $\omega$, with $p=\gamma +i\omega$:When the relevant Laplace transforms and spatial integral transforms are explicit, every sample of this integrand can be evaluated directly, without a time-stepping approximation to the partial differential equation. Split the initial condition integral at $y=x$: the two pieces of $R_q$ are linear combinations of $e^{\pm qy}$, so truncated spatial exponential transforms suffice. For real data the negative-frequency half is the complex conjugate of the positive-frequency half.

A practical numerical integration is to truncate to $|\omega |\le \Omega$, apply an adaptive quadrature rule, and independently increase $\Omega$ and refine the quadrature. Choose $\gamma >0$ large enough to remain to the right of all singularities, but avoid an unnecessarily large $e^{\gamma t}$. Evaluate hyperbolic function ratios in scaled form; for example,This prevents overflow when $\mathrm{Re}qL$ is large. The boundary contributions at an interior point are damped by $e^{-x\mathrm{Re}q}$ or $e^{-(L-x)\mathrm{Re}q}$, with $\mathrm{Re}q\sim \sqrt{|\omega |/2}$. Near an endpoint more frequencies are needed; at the endpoint itself use the prescribed Dirichlet boundary condition. The initial condition resolvent has a generally only $O(1/\omega )$ tail. Treat its oscillatory tail accurately, subtract a known transform with the same leading term, or evaluate that contribution separately with the heat kernel.

An equally useful check is the eigenfunction expansion. DefineThe causal modal formula isIt follows either from the heat kernel formula or directly from integration by parts against a sine eigenfunction. Exponential-transform formulas evaluate the time integral explicitly when available. The initial condition contribution has a Gaussian function cutoff in $n$ for each positive time. Nonzero endpoint forcing produces a more slowly convergent sine series tail. A boundary lifting $\ell (x,t)=(1-x/L)g_0(t)+(x/L)e^{aL}h(t)$ gives $w=\ell +z$, where $z$ has zero Dirichlet boundary conditions and forcing $-{\ell }_t-a^2\ell$; expanding $z$ gives better convergence and imposes the endpoint values exactly. At very small times the method of images is often more efficient than retaining many modes.

Check convergence by increasing both the frequency cutoff and quadrature resolution, and compare with a separately truncated causal modal or image-kernel evaluation. Causality in finite-time Laplace contour inversion means a leftward contour deformation must respect both resolvent operator poles and growth of the transformed data. In particular, the transform of data extended by zero after $T$ can grow like $e^{-pT}$ in the left half-plane: for $t<T$ one cannot close that contour indiscriminately and discard its large arc. The modal forcing integral with upper limit $t$ avoids this causality error.

Put $a=\alpha /2$ and apply the Dirichlet gauge transform for constant drift $w(x,t)=e^{ax}u(x,t)$. Taking derivatives directly removes the advection term and gives the damped heat equation $w_t=w_{xx}-a^{2}w$. Thus $w_0(y)=e^{ay}{u}_0(y)$, while the two Dirichlet boundary conditions become $w(0,t)=g_0(t)$ and $w(L,t)=e^{aL}h(t)$.

The Dirichlet heat kernel on an interval and its damped version areThe method of images gives an alternative, often better at short times:To determine the boundary signs, multiply the equation for $w$ by the backward heat kernel $K_a(x,y,t-s)$ and use integration by parts in $y$. Since the kernel vanishes at $y=0,L$, the surviving boundary expression is $K_{a,y}(x,0,t-s)g_0(s)-K_{a,y}(x,L,t-s)e^{aL}h(s)$. Consequently an integral representation containing only the given data isThis is a Dirichlet boundary-forcing heat-kernel formula. For $0<x<L$ the short-time Gaussian decay makes the forcing integral well defined. The endpoint values are interior limits of the complete formula: evaluating a termwise sine series at an endpoint before taking the time integral loses the nonzero boundary values. At the initial corners a continuous classical solution requires $u_0(0)=g_0(0)$ and $u_0(L)=h(0)$; otherwise the same formula describes the solution away from those corners.

Here is a second integral representation, useful when the integral transforms are explicit. Extend $g_0,h$ by zero after $T$; the values of the extension cannot affect a solution at $t<T$. Write their Laplace transforms as $G(p),H(p)$ and set $q=\sqrt{p+a^2}$ using the principal square root. The transformed Dirichlet Green function for $-{\partial }_x^2+q^2$ isIt vanishes at both endpoints and its first derivative in $x$ jumps by $-1$, which verifies the sign of its unit source. The resolvent kernel for Dirichlet advection-diffusion on an interval gives the transformed solutionThe Bromwich inversion formula therefore yieldsThe square-root notation creates no genuine branch singularity here: all three kernels are even in $q$. Their only spatial resolvent operator poles are $p=-a^2-q_n^2$. This transformed formula and the causal heat kernel formula represent the same solution in the usual smooth-data class.

The insulated boundary imposes the Neumann boundary condition $T_{x_1}(0,x_2,t)=0$. Use the method of images by reflecting the initial data evenly across $x_1=0$. At the origin, the original sector and its reflection contribute equally. Using the given two-dimensional heat kernel,Differentiation givesThis vanishes exactly when $e^{3/(4Kt)}=4$. Its sign is positive before that time and negative afterwards. Moreover $T\to 0$ as $t\downarrow 0$ and as $t\to \infty$. Therefore this is the unique global maximum:The factor two from the insulated-boundary reflection is essential; applying the full-plane heat kernel only to the unreflected sector would give half the temperature.

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.