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The trigonometric identityshows that only angular Fourier modes $1$ and $3$ are required. Differentiating the harmonic expansion on a disk at $r=1$ multiplies the coefficient of each nonconstant Fourier mode by $n$. Thus $B_1=3/4$ and $3B_3=-1/4$, givingEach term is a regular harmonic function, and the radial derivative isThe Neumann compatibility condition for Laplace's equation holds becauseas is necessary by the divergence theorem applied to the gradient of a harmonic function.

The additive constant is undetermined. This Neumann boundary condition is unchanged by $T\mapsto T+C$, so the paper's blanket uniqueness assertion requires a normalization in this case. The uniqueness of the Neumann problem gives uniqueness up to a constant; prescribing $T(0)=0$, or zero mean temperature, selects the displayed answer with $C=0$.

Take $v=|\mathbf{x}{|}^2$ in Green's second identity. Since ${\nabla }^{2}u=0$, ${\nabla }^{2}v=6$, $v=1$ on the unit sphere, and ${\partial }_{n}v=2$ there,The final term vanishes by the divergence theorem, because it equals ${\iiint }_B{\nabla }^{2}udV=0$. ThereforeWe assume sufficient boundary regularity for the stated boundary values and integrals, or use the corresponding trace/limit formulation. This is the equality of volume and surface averages for a harmonic function, since the unit ball and sphere have volume $4\pi /3$ and area $4\pi$.

Treat the boundary data as $2\pi$-periodic in the angular variable. Denote the outer data by $g(\theta )$, and set $h(\theta )=f(a,\theta )$ for the inner data. Define their mean and nonconstant Fourier coefficients byThe outer data have reflection symmetry, so all their sine Fourier coefficients vanish. Their mean is $1/2$, and their cosine Fourier coefficients are

The angular Fourier harmonics for a single-valued harmonic function in an annulus are $\cos (n\theta )$ and $\sin (n\theta )$, with integer $n\ge 1$. Their radial factors are independent combinations of $r^n$ and $r^{-n}$; the zero angular mode has radial factor $A_0+B_0\log r$. For arbitrary inner data, the sine and cosine modes need independent radial coefficients: forcing them to share one radial factor, as in the product form supplied in the paper, would impose an extra restriction. For either sine or cosine mode, if its outer and inner coefficients are $v$ and $w$, the radial coefficients $\alpha ,\beta$ obeySolving this linear system givesIt is convenient to collect the result into the annular harmonic interpolation factorsThey satisfy $H_n(1)=1$, $H_n(a)=0$, $K_n(1)=0$ and $K_n(a)=1$. The zero mode satisfies $A_0=1/2$ and $B_0=(p_0-1/2)/\log a$. Therefore the required separation of variables solution isAll dependence on $f(a,\theta )$ is given by the displayed integrals. Substituting $r=a$ leaves its Fourier series, and substituting $r=1$ leavesthe Fourier series of the specified outer step. The radial factors satisfy $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$, so each nonconstant term satisfies Laplace's equation; $1$ and $\log r$ do also. For integrable boundary data the radial factors give exponential decay on every closed subannulus strictly inside $a<r<1$, justifying the necessary termwise partial derivatives there.

There is a qualification at the two outer jump discontinuities. By pointwise convergence of a piecewise smooth Fourier series, the outer series gives $1/2$ at $\theta =\pm \pi /2$, whereas the printed data assign $1$ at those two points. Thus the displayed solution attains the outer data at their continuity points, and has radial limit $1/2$ at the jumps. One may separately assign the boundary value $1$ at the two exceptional points without changing the interior harmonic function, but no continuous extension can match the discontinuous data there. For piecewise smooth inner data, the analogous boundary trace agrees with $h$ at continuity points and with its midpoint at jumps. Some regularity of $h$ is necessary for a classical interpretation; the problem imposes none explicitly.

Put $u(r,\theta )=R(r)\Theta (\theta )$ in Laplace's equation and multiply by $r^2/(R\Theta )$ on intervals where the factors are nonzero. Separation of variables givesand henceThe angular Dirichlet boundary conditions are $\Theta (0)=\Theta (\pi /2)=0$. Multiplication of the angular equation by $\Theta$, followed by integration by parts, givesFor a nonzero angular eigenfunction this forces $\lambda >0$: equality would make $\Theta$ constant and then zero. Write $\lambda ={\mu }^2$. The first angular condition removes the cosine term, and the second requires $\sin (\mu \pi /2)=0$. ThusThe radial Cauchy-Euler equation has the two solutions $r^{2n}$ and $r^{-2n}$. By superposition, the general separated series is thereforeThe coefficients are arbitrary subject to convergence with enough partial derivatives on compact subsets to give a harmonic function and to the prescribed zero values on the two open radial edges. Completeness of the sine eigenfunctions gives the usual series representation for solutions smooth up to those edges on closed radial subintervals. There is no constant or logarithmic zero mode because its angular factor would be linear in $\theta$ and would vanish at both ends only if it were zero.

The negative powers must be retained: the stated domain omits $r=0$, and no boundedness at that vertex is prescribed. For example, $r^{-2}\sin (2\theta )$ is a harmonic function satisfying both edge conditions. An additional boundedness condition as $r\to 0$ would remove all $B_n$; an additional value on the circular arc would then determine the remaining $A_n$. With only the conditions in the paper, neither restriction is available and the solution is not unique.

Let $u_1,u_2$ be two solutions with the same data, and write $w=u_1-u_2$ on the rectangle $\Omega =(0,\pi )\times (-1,1)$. By linearity,The normal derivative is $-w_x$ on the left edge and $w_x$ on the right edge, so it vanishes on both. The energy method applies Green's first identity, which follows by integratingIt givesThe boundary term vanishes on each horizontal edge because $w=0$, and on each vertical edge because ${\partial }_{n}w=0$. The interior term vanishes because $w$ is a harmonic function. Since the continuous integrand $|\nabla w{|}^2$ is nonnegative, a zero integral forces $\nabla w=0$. The rectangle is connected, so $w$ is constant; its zero value on either horizontal edge makes that constant zero. HenceThis proof assumes enough boundary regularity for Green's first identity. It also works in the finite Dirichlet energy class using the weak identity with test function $w$, so it remains applicable when corner derivatives are not continuous. It proves uniqueness conditional on existence and does not impose a compatibility condition for arbitrary $f,g$.

Set $u=X(x)Y(y)$ in Laplace's equation. Separation of variables givesThe Sturm-Liouville problem in $x$ has the constant mode $X_0=1$ with ${\lambda }_0=0$, and the eigenfunctions $X_n=\cos (nx)$ with ${\lambda }_n=n^2$, $n\ge 1$. Indeed, integration by parts makes $\lambda$ nonnegative; for $\lambda >0$ the first Neumann boundary condition removes the sine term and the second requires $\sqrt{\lambda }$ to be an integer. For $\lambda =0$, both derivative conditions make $X$ constant.

Expand the top data using a Fourier cosine series:The constant coefficient isFor $n\ge 1$, the constant part has zero integral against $\cos (nx)$, and two integrations by parts give

Since the bottom value is zero, a convenient factor depending on $y$ for the nonconstant modes is $\sinh (n(y+1))$, which vanishes at $y=-1$. Its value at $y=1$ is $\sinh (2n)$. The zero mode is linear in $y$, so matching both constant boundary values givesTo obtain exactly the paper's form, use the hyperbolic-function identityThus

Each product of hyperbolic sine or hyperbolic cosine with $\cos (nx)$ has $u_{xx}+u_{yy}=0$. The zero mode is also a harmonic function. On $-1\le y\le 1$, the factor $\sinh (n(y+1))/\sinh (2n)$ lies between zero and one. Since $\sum n^{-2}$ converges, the Weierstrass M-test gives uniform convergence on the closed rectangle. The series therefore gives zero at the bottom and the continuous Fourier cosine series of ${\pi }^2-x^2$ at the top, including its endpoints. On compact subsets with $y<1$, the factors decay exponentially with $n$, justifying termwise partial derivatives. In particular $u_x=0$ on both open vertical edges because every differentiated cosine factor vanishes there. The solution has finite Dirichlet energy: Fourier orthogonality gives the $n$th nonconstant mode's energy as $(\pi /2)n{g}_n^2\coth (2n)=O(n^{-3})$, whose sum is finite. Here $\coth$ is the hyperbolic cotangent. Thus a proves uniqueness in that class.

There is a corner compatibility condition for mixed boundary data in the printed wording. The top data have $g^{\prime }(\pi )=-2\pi$, but the paper writes $u_x(\pi ,y)=0$ even at $y=1$. The series solves the interior equation and all the open-edge boundary conditions; the literal derivative condition at $(\pi ,1)$ is incompatible with the top data. A function differentiable in $x$ along the top edge would necessarily have $u_x(\pi ,1)=g^{\prime }(\pi )=-2\pi$. Hence it cannot simultaneously satisfy the printed zero derivative at that corner, and there is no solution with all those literal corner requirements. The standard interpretation of this mixed boundary condition imposes the Neumann boundary condition on the open sides and the Dirichlet boundary condition continuously on the horizontal edges. Differentiating the series term by term on the top edge would be invalid at this corner; the uniform convergence of the undifferentiated series does not license that operation.

The zero value at $x=0$ removes the even factor in $x$, and the zero value at $y=0$ removes the even factor in $y$. Up to multiplicative constants, the nonzero separated modes areThe zero-separation case also supplies $\boxed{xy}$. Each is a harmonic function and vanishes on both axes.

The substitution gives $x^2-y^2=u^2\cos 2v$. For this function $f_{uu}=2\cos 2v$, $f_u/u=2\cos 2v$, and $f_{vv}/u^2=-4\cos 2v$, whose sum is zero. Thus it satisfies the polar form of Laplace's equation for $u>0$. The apparent coordinate singularity at $u=0$ does not make the original Cartesian polynomial singular; it is a smooth harmonic function through the origin.

For a separated harmonic function $V=F(x)G(y)$, Laplace's equation gives $F^{\prime \prime }/F=-G^{\prime \prime }/G=-k^2$. Bounded periodic dependence in $x$ has $F=A\cos (kx)+B\sin (kx)$; decay on each half-plane selects $G=e^{-k|y|}$ for $k>0$, when the potential is matched continuously at the sheet. These are the separated decaying modes. General decaying solutions can be superpositions of them; decay alone does not force a single value of $k$.

The prescribed cosine density selects $k=1$ and $V=C\cos x{e}^{-|y|}$. Its normal derivative jump is $-2C\cos x$. Matching the stated jump $2\pi {\Sigma }_0\cos x$ givesThis is the gravitational potential of a sinusoidal sheet in the normalization of the question. It is continuous at the sheet and decays on both sides. Since ${\Sigma }_0\cos x$ changes sign, it can physically describe a Fourier component of a density perturbation rather than an everywhere positive isolated mass sheet.

Simplify the real component to $u=(1-x^2-y^2)/D$. Thus $u=0$ is the unit circle with $(1,0)$ excluded from the mapping's domain. Its upper semicircle has $v>0$ and its lower semicircle has $v<0$; the point $(-1,0)$ maps to $u=v=0$. Inside the disk, $u>0$, so the principal arctangent is unambiguous. Setting $\alpha =V$ givesThe Laplace equation follows from (b). Approaching an open upper boundary arc makes the ratio tend to $+\infty$, giving $+V$; the lower arc gives $-V$. This solution is $V(2F-1)$ in terms of the harmonic measure of a semicircle in a disk $F$.

The boundary data jump at $(\pm 1,0)$, so no path-independent continuous boundary value can be imposed at those two junctions. The displayed bounded harmonic function is the solution with the specified limits on the open arcs; those endpoint discontinuities are part of the idealized capacitor geometry.

For the nonconstant boundary data, use separation of variables in polar coordinates and put $\varphi =R(r)\cos 2\theta$. Laplace's equation reduces toThis Cauchy-Euler equation has independent solutions $r^2$ and $r^{-2}$. Unlike a disk containing the origin, the annulus permits both. The zero inner boundary value requires $R=A(r^2-a^4{r}^{-2})$, and the outer value requires $A=V{b}^2/(b^4-a^4)$. HenceSubstituting $r=a$ gives zero, while $r=b$ gives $V\cos 2\theta$. This is the Dirichlet problem on an annulus for one Fourier mode; no other angular mode is needed.

For the changed inner boundary value, use superposition to add a radial harmonic function that is $-V$ at $a$ and zero at $b$. The radial zero-mode equation is $(r{R}^{\prime }{)}^{\prime }=0$, giving $R=C+D\log r$. Imposing the two constant boundary values givesTherefore the second answer isThe logarithmic term is independent of $\theta$, so it changes only the angular average of the electric potential. Both answers are unique for these Dirichlet boundary conditions, by uniqueness for Laplace's equation.

Green identity. With $n$ the outward unit normal, the product rule gives $\nabla \cdot (u\nabla v)=u{\nabla }^{2}v+\nabla u\cdot \nabla v$. Integrating over $V$ and applying the divergence theorem yields Green's first identity:

Uniqueness for nonzero mass. Suppose two real solutions have the same prescribed boundary data, and let $w$ be their difference. The spatial equation called the Klein-Gordon equation in the question is the modified Helmholtz equation, so ${\nabla }^{2}w=m^{2}w$. With $u=v=w$, Green's first identity becomesFor identical Dirichlet boundary conditions, $w=0$ on $S$; for identical Neumann boundary conditions, ${\partial }_{n}w=0$ there. Either makes the right-hand side zero. When $m$ is real and nonzero, both integrands on the left are nonnegative, forcing $w=0$. ThusThis is the required uniqueness assertion; existence is understood for admissible smooth data on a suitable bounded region. Complex-valued solutions obey the same result by treating their real and imaginary parts separately.

Zero mass. For $m=0$, the argument gives $\nabla w=0$, hence $w=C$ on the connected region. Dirichlet boundary conditions still force $C=0$, but Neumann boundary conditions leave an arbitrary additive constant:The prescribed normal data in the zero-mass case must also satisfy ${\int }_S{\partial }_n\Phi dS=0$, by the divergence theorem. For nonzero $m$, that flux can instead equal $m^2{\int }_V\Phi dV$.

Energy minimization. Let $\eta =w-v$ for the two functions in the last part. They share their boundary values, so $\eta =0$ on $S$. Since $v$ is a harmonic function, Green's first identity givesExpand the square in the Dirichlet energy of $w=v+\eta$:The cross term vanishes, soEquality requires $w-v$ to be constant, and the shared boundary values then make it zero. This is the Dirichlet principle: among functions with fixed boundary values, the harmonic function minimizes Dirichlet energy.

Harmonic parts. Write the holomorphic function as $f=u+iv$. The Cauchy-Riemann equations say $u_x=v_y$ and $u_y=-v_x$. Equality of mixed partial derivatives therefore givesThus both the real and imaginary parts are harmonic functions and satisfy two-dimensional Laplace's equation.

Single-valued logarithm. A convenient choice isusing the principal logarithm for each term. Its branch cuts in the $z$ plane are $(-\infty ,-1]$ and $[1,\infty )$. The branch points are $-1$ and $1$: going once around the first adds $2\pi i$, while going once around the second subtracts $2\pi i$. Removing the two exterior rays makes this a consistent branch of a multivalued function and in particular leaves the whole open unit disk available.

Potential inside the cylinder. Put $z=r{e}^{i\theta }$ with $r<1$. Multiplication by $1-\overline{z}$ givesThe denominator is positive and the real part of the numerator is positive. Thus its argument lies strictly between $-\pi /2$ and $\pi /2$, andAs $r\uparrow 1$, this tends to $\pi /2$ on the upper open semicircle and $-\pi /2$ on the lower open semicircle. Scale the imaginary part to obtainIt is a harmonic function, has the required boundary values, and gives $\Phi (0)=0$. The two joining points of the boundary arcs have discontinuous boundary data, so no common continuous boundary value is demanded there. Along the real diameter, including radial approaches to either joining point, the displayed solution is zero. This is the disk potential with two semicircular boundary values.

For a second check, the Taylor series of the logarithms gives $f(z)=2{\sum }_{j\ge 0}{z}^{2j+1}/(2j+1)$. Taking the imaginary part yields the Fourier serieswhose boundary limit is exactly the required alternating pair of constant semicircle values.

Transform convention and derivatives. Use the Fourier transform pairFor sufficiently decaying $f,f^{\prime }$ with integrable second derivative, two applications of integration by parts giveThe identity also holds for tempered distributions, which is the setting needed for the Dirac delta function boundary data.

Upper half-plane solution. Take the Fourier transform in $x$ only, writing $\Phi (k,y)=\tilde{\varphi }(k,y)$. Laplace's equation becomes ${\Phi }_{yy}-k^2\Phi =0$. For $k\ne 0$, discard the exponentially growing solution and use $\Phi (k,0)=1$, the Fourier transform of the Dirac delta function. This gives $\Phi (k,y)=e^{-|k|y}$. The value at $k=0$ can be set to its continuous limit one. Pointwise decay of $\varphi$ as $y\to \infty$ does not require this individual transformed value to decay: the total integral in $x$ remains one while the profile spreads.

The inverse Fourier transform now yields, for $y>0$,Evaluate the last integral as the real part of ${\int }_0^{\infty }{e}^{-(y-ix)k}dk=(y-ix{)}^{-1}$:This is the Poisson kernel for the upper half-plane. Its boundary condition is a distributional convergence, rather than an ordinary value at $(0,0)$: for a smooth compactly supported test function $g$,The dominated convergence theorem applies because $g$ is bounded and $(1+t^2{)}^{-1}$ is integrable. Also $0\le \varphi (x,y)\le 1/(\pi y)$, proving the required decay as $y\to \infty$.

Direct verification. Differentiating the rational expression givesTheir sum is zero throughout $y>0$, so the solution is a harmonic function there.

Compute the complex derivative along horizontal and vertical increments. Along the real direction, $f^{\prime }=u_x+i{v}_x$; along the imaginary direction, $f^{\prime }=(u_y+i{v}_y)/i=v_y-i{u}_y$. Equality gives the Cauchy-Riemann equationsThe real and imaginary parts of a holomorphic function are smooth. Equality of their mixed partial derivatives then gives $u_{xx}+u_{yy}=v_{yx}-v_{xy}=0$ and $v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0$, proving that both are harmonic functions.

For the first real part, $u_x=2x$ and $u_y=-2y$. The Cauchy-Riemann equations require $v_y=2x$ and $v_x=2y$, so its harmonic conjugate is $v=2xy+C$. ThereforeFor the second real part, $u=x/(x^2+y^2)$ is the real part of $1/z$. Its harmonic conjugate is $-y/(x^2+y^2)+C$, givingThese are the most general answers on a connected open domain: the difference of two candidates has zero real part, so the Cauchy-Riemann equations make its imaginary part constant. The second function cannot extend holomorphically through zero.

For a separated harmonic function $R(r)\Theta (\theta )$ in polar coordinates, single-valuedness forces integer angular frequencies. The radial factors for frequency $m\ge 1$ are $r^m,r^{-m}$, while frequency zero has a constant and $\ln r$. Finiteness at the origin excludes $r^{-m}$ and $\ln r$.

The source's angular factor is $\cos \theta$ in every term, as printed, rather than $\cos (n\theta )$. The power-law angular mode for the planar Poisson equation givesHence, assuming the series permits termwise differentiation, the general finite, single-valued solution isThe arbitrary harmonic series must itself converge to a finite function on the prescribed disk. If ${\alpha }_0\ne 0$, the source has a direction-dependent limit at the origin; the displayed potential still extends finitely there, but the equation at that point is understood weakly rather than as a smooth classical source.

For the special source, a particular potential is $r^3\cos \theta /8$. Its boundary cosine term is canceled by $-a^{2}r\cos \theta /8$, and the boundary sine term is supplied by $(r/a{)}^2\sin 2\theta$. The resulting Dirichlet boundary conditions has solutionIt is finite at zero, has the required boundary values, and its Laplacian is $r\cos \theta$.

Put $\Phi (r,\theta )=R(r)Y(\theta )$ in Laplace's equation, multiply by $r^2/(RY)$, and separate the two variables:With $s=\cos \theta$, the angular equation becomes $(1-s^2)Y_{ss}-2s{Y}_s+\lambda Y=0$. Requiring a smooth axisymmetric harmonic function at both poles selects $\lambda =n(n+1)$ and the Legendre polynomial $Y=P_n(s)$, for nonnegative integers $n$. The radial Cauchy-Euler equation has solutions $r^n$ and $r^{-n-1}$. Superposition over the angular eigenfunctions therefore givesThis expansion assumes regularity on each whole angular sphere; without it, singular angular solutions would also be possible. The boundary data in the shell involve only the $n=0,1$ modes. Define the spherical-shell harmonic interpolation weightsEach $p_n$ equals one at the inner boundary and zero at the outer one, and $q_n$ does the reverse. Hence the steady temperature isEach term is a harmonic function, and substitution verifies both prescribed boundary values. Uniqueness follows from the maximum principle for harmonic functions.

Let $s(\theta )$ be the $2\pi$-periodic ramp represented by $\theta$ on $(-\pi ,\pi )$. The Fourier series of a periodic sawtooth givesBoth boundary data are multiples of this odd function, so their angular means and cosine coefficients vanish. Consequently $A_0=C_0=A_n=C_n=0$. Put $h_n=2(-1{)}^{n+1}/n$. Matching the sine coefficients at the two radii gives $B_n{a}^n+D_n{a}^{-n}=h_n{\mu }_a$ and $B_n{b}^n+D_n{b}^{-n}=h_n{\mu }_b$. Solving this two-by-two linear system yieldsEquivalently, two-boundary annular harmonic interpolation expresses the same solution asEvery term is a harmonic function, and the radial factors enforce the specified boundary values. For a radius strictly between $a$ and $b$, their geometric decay makes the series smooth. At the boundary discontinuity $\theta =\pi$, the limiting Fourier value is the mean of the two one-sided values, zero. The boundary data are imposed away from that exceptional point, as stated.

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 $-\Delta G=\delta$.

Split the Fourier transform integral at zero:By Fourier inversion applied to this pair, or by the residue theorem, the other transform isFor example, inversion with $\lambda =\mu$ gives $\int e^{ikx}(k^2+{\mu }^2{)}^{-1}dk=(\pi /\mu )e^{-\mu |x|}$; evenness supplies the desired sign convention.

Take the spatial Fourier transform of the Laplace equation. For $y>0$ it gives ${\hat{\psi }}_{yy}-k^2\hat{\psi }=0$. Removing the exponentially growing solution and matching the boundary givesAt $k=0$ this formula is interpreted by continuity; a single transform value does not alter the inverse integral. The previous transform pair identifiesthe Poisson kernel for the upper half-plane. The convolution theorem therefore gives the Poisson integralThis is the bounded harmonic solution for bounded decaying data, or the corresponding solution in an appropriate Fourier class. Such a growth condition matters if vertical decay is interpreted only pointwise: the nonzero harmonic function $e^{x^2-y^2}\sin (2xy)=\mathrm{Im}{e}^{(x+iy{)}^2}$ vanishes on the real axis and tends to zero for fixed $x$ as $y\to \infty$, but grows rapidly in $x$ and has no ordinary spatial Fourier transform. It can be excluded by requiring boundedness; the stated pointwise limits alone would not establish uniqueness.

For the interval data, direct integration gives the Poisson extension of interval boundary dataAs $y\downarrow 0$ it approaches $c$ for $|x|<a$, zero for $|x|>a$, and $c/2$ at $x=\pm a$. At those jump points the prescribed isolated values cannot be attained by a continuous extension; they do not affect the integral. The solution decays vertically, with $\psi \sim 2ac/(\pi y)$ for fixed $x$ and large $y$.