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Every boundary point of $D$ satisfies the exterior sphere condition. For a point on the inner spherical boundary, use a smaller ball inside the removed ball and tangent at that point; for a point on the cube, use a ball in a supporting exterior half-space. If the exterior ball has centre $y$ and radius $r$, a local positive harmonic barrier isfor $n\ge 3$, while in two dimensions use $w(x)=\log (|x-y|/r)$. Adding a sufficiently large positive multiple of a global superharmonic function extends the local barrier across the bounded domain. Hence every boundary point is regular by part (v), and the Perron method for the Dirichlet problem produces a harmonic function attaining the prescribed continuous boundary data. The maximum principle for harmonic functions gives uniqueness.

For the stated flat metric and orientation,Because the metric coefficients and the coordinate one-forms are constant, the Hodge Laplacian acts coefficientwise:Thus a harmonic one-form $\alpha ={\sum }_i{\alpha }_{i}d{x}^i$ has harmonic coefficient functions. Every harmonic function on the compact connected torus is constant by the maximum principle for harmonic functions. HenceThe Hodge decomposition theorem now gives

No such function exists. Continuity on the compact closed unit disk makes $u$ bounded near the origin. The removable singularity for a bounded harmonic function extends $u$ harmonically across the origin. The extended function is continuous on the closed disk and vanishes on its boundary, so the maximum principle for harmonic functions, applied to both $u$ and $-u$, forces $u=0$ throughout the disk. This contradicts $u(0,0)=1$.

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.

Set $\Phi =R(r)P(\theta )$ and multiply the Laplace equation by $r^2/(RP)$. Separation of variables givesWith $\xi =\cos \theta$, the angular equation is $[(1-{\xi }^2)P_{\xi }{]}^{\prime }+\lambda P=0$. Regularity at both poles selects the Legendre polynomials $P_n$ and $\lambda =n(n+1)$. The radial Cauchy-Euler equation has solutions $r^n,r^{-n-1}$. Expanding the angular data in this complete orthogonal system therefore giveswith the usual convergence required of a harmonic expansion on the region considered. Angular Legendre functions singular at the poles are excluded.

For $0<a<b$, the outer data have expansion $b^2{\cos }^2\theta =b^2/3+(2b^2/3)P_2(\cos \theta )$, while the inner data are pure $P_0$. Thus only degrees zero and two are needed. A radial interpolation matching each pair of boundary values givesThe first term is a linear combination of $1,r^{-1}$ and takes values $a^2,b^2/3$ at the two radii. The second is a combination of $r^2,r^{-3}$ and takes values zero and $2b^2/3$. These checks establish both boundary conditions; each term is harmonic. The maximum principle for harmonic functions gives uniqueness. This is an example of axisymmetric harmonic interpolation between concentric spheres.

For any interior point $\mathbf{p}$, choose a sufficiently small circle centred on $\mathbf{p}$ and wholly inside the disk. Translating the origin does not change the Laplace equation, so part (b) givesSuppose a real harmonic function, continuous on the closed disk, attains its global maximum $M$ at an interior point. Every value on each such circle is at most $M$, and their average is $M$. By continuity, every value on the circle must therefore equal $M$. Varying its radius shows that a neighbourhood of the point also has value $M$.

Thus the set of points attaining $M$ is open in the disk, and it is closed by continuity. The disk is connected, so this set must be the whole disk. The function is constant and has the same maximum on the boundary. Otherwise the maximum cannot be interior. Compactness of the closed disk guarantees existence of a maximum, proving the maximum principle for harmonic functions:Continuity up to the boundary is the standard hypothesis for this attained-maximum formulation.

The Fourier orthogonality relation on this angular interval isProjecting the outer boundary condition onto each angular eigenfunction givesThe solution is unique in the class of bounded harmonic functions with the prescribed boundary trace. For continuous boundary data, the difference of two solutions is zero on the boundary and bounded at the vertex, so the maximum principle for harmonic functions makes it zero throughout the sector. Equivalently all its Fourier sine series coefficients vanish. The usual convergence interpretation applies if the data are only piecewise smooth.

Set $u=\cos \theta$. The boundary data simplify to $u+2u^2=\frac{2}{3}{P}_0(u)+P_1(u)+\frac{4}{3}{P}_2(u)$. The axisymmetric harmonic extension from a sphere keeps only decaying radial powers outside and regular powers inside. HenceandEach Legendre polynomial coefficient agrees at $r=a$. The exterior solution tends to zero, and the interior solution is finite at $r=0$. The maximum principle for harmonic functions proves uniqueness in these respective classes.

For integrable functions $u,v$, the convolution theorem in the chosen convention isLet $F(s,y)$ be the Fourier transform of $f(\cdot ,y)$ in the horizontal coordinate. Transforming Laplace's equation givesFor $s\ne 0$, exclude the growing exponential and obtain $F(s,y)=2e^{-|s|y}/(1+s^2)$. At $s=0$, continuity in frequency gives $F(0,y)=2$; a constant zero-frequency transform is compatible with spatial decay of $f$.

By the transform pair in part (a), the inverse transform of $e^{-|s|y}$ is the Poisson kernel for the upper half-plane, $P_y(x)=y/[\pi (x^2+y^2)]$. ThusThe Poisson extension of exponential boundary data is harmonic, approaches the continuous boundary data and decays at spatial infinity. For uniqueness, apply the maximum principle for harmonic functions to the difference of two solutions on increasingly large half-discs: it is zero on the real boundary and uniformly small on the distant semicircle.

For a spatial initial slice with vanishing extrinsic curvature, the vacuum momentum constraint is automatic and the Hamiltonian constraint requires zero Ricci scalar. In three spatial dimensions, the indicated conformally flat metric has $\mathcal{R}=-8{\psi }^{-5}\Delta \psi$, with a nonzero conformal factor. Thus a harmonic function supplies the data on its domain. A positive example is $\psi =1+a/r$ on a punctured or exterior region. There is no nonconstant globally smooth example on all of Euclidean three-space tending to one at infinity, by the maximum principle for harmonic functions. Time symmetry of an initial slice does not imply that its entire evolved spacetime is static.