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For boundary data $f(u)$ on the sphere $r=a$, with $u=\cos \theta$, define its Legendre polynomial coefficients by $f_{\ell }=(2\ell +1){\int }_{-1}^{1}f(u)P_{\ell }(u)du/2$. Separation of variables in the Laplace equation gives radial factors $r^{\ell }$ and $r^{-\ell -1}$. Regularity at the origin selects the first for the interior Dirichlet problem; decay at infinity selects the second for the exterior problem. Matching each boundary coefficient gives the displayed series. For continuous boundary data the resulting harmonic functions attain those data, with convergence interpreted through their harmonic extension rather than requiring uniform convergence of the unweighted boundary series. The maximum principle for harmonic functions proves uniqueness in the bounded interior and decaying exterior classes.

Suppose the inner and outer boundary values on spheres of radii $0<a<b$ have convergent Legendre polynomial expansions with coefficients $u_n,v_n$. For each degree set $F_n(r)=r^n-a^{2n+1}{r}^{-n-1}$ and $H_n(r)=r^n-b^{2n+1}{r}^{-n-1}$. ThenBoth radial terms solve $(r^2{R}_n^{\prime }{)}^{\prime }=n(n+1)R_n$, so the displayed angular expansion is harmonic wherever the series and required derivatives converge. The two radial factors take endpoint values $(1,0)$ and $(0,1)$, respectively, proving the boundary matching. For continuous data with the usual harmonic-extension convergence, the maximum principle for harmonic functions gives uniqueness. The construction separates the angular orthogonality from the two-boundary radial interpolation.

If $u,v$ are harmonic functions on a bounded domain and continuous on its closure, their difference is a harmonic function. Its minimum is attained on the boundary by the maximum principle for harmonic functions applied to its negative. Nonnegative boundary difference therefore implies nonnegative interior difference. Applying the comparison in both directions proves uniqueness for the Dirichlet problem whenever a classical solution exists. This is an order statement and does not itself establish existence.

The difference of two classical Poisson equation solutions with the same Dirichlet boundary conditions is a harmonic function vanishing on the boundary. Under sufficient boundary smoothness, Green's first identity gives ${\int }_V|\nabla ({\varphi }_1-{\varphi }_2){|}^2=0$. More generally the maximum principle for harmonic functions proves the same uniqueness for functions continuous up to the boundary and twice continuously differentiable in the bounded interior.

For a point electric charge in a charge-free neighbourhood, a static electric potential obeys Laplace's equation. The maximum principle for harmonic functions excludes a strict interior minimum or maximum of that electric potential, hence a strict minimum of the charge's potential energy. A static electric field alone therefore cannot provide stable three-dimensional confinement under these assumptions.

For the Laplace equation on $0<y<1$, prescribe bottom data $g$ and zero top data. A horizontal Fourier transform reduces the equation to ${\hat{u}}_{yy}-s^2\hat{u}=0$, giving the displayed multiplier and its limit $1-y$ at $s=0$. Integrable transformed data give a bounded inverse-transform solution. The bounded solution is unique by the maximum principle for harmonic functions, using barriers $\epsilon \cosh (\delta x)\cos (\delta (y-1/2))$ for $0<\delta <\pi$ on expanding rectangles. Without a growth condition, one can add $e^{\pi x}\sin (\pi y)$, so the two boundary traces alone do not determine an arbitrary harmonic function on an infinite strip.

A harmonic function on one side of a hyperplane that is continuous up to an open portion of the hyperplane and zero there extends harmonically by odd reflection. One local proof takes a ball centered on the boundary and assigns odd continuous boundary data on its sphere. The Poisson integral on the ball is odd, so it vanishes on the equatorial disk. Uniqueness by the maximum principle for harmonic functions identifies it with the original function on the half-ball. Thus the odd extension is harmonic near each boundary point. A globally bounded zero-boundary harmonic function on a half-space is consequently zero by the Liouville theorem for harmonic functions.

For a compact H-hull, $v(z)=\mathrm{Im}{g}_K(z)$ is harmonic and satisfies $0<v(z)\le \mathrm{Im}z$. Local boundedness and the interior continuity of $g_K^{-1}$ show that $v$ tends to zero at every finite boundary point. The maximum principle for harmonic functions applied to $v-\mathrm{Im}z$, with an exhaustion controlling infinity through the Laurent expansion, proves the inequality without boundary smoothness.

An ion trap confines an ion using electromagnetic fields, allowing repeated interrogation of the same particle. The name also covers related traps for an electron. A Paul trap uses an oscillating electric quadrupole field; a Penning trap combines a static electric quadrupole field with a static magnetic field. Confinement creates time to measure an energy level, frequency or magnetic dipole moment; it does not itself cool the particle or identify its internal quantum state.

The obstacle is Earnshaw's theorem. In a charge-free neighbourhood the static electric potential $\Phi$ obeys Laplace's equation, ${\nabla }^2\Phi =0$. For a particle of fixed nonzero electric charge $Q$, the potential energy $Q\Phi$ is also a harmonic function. The maximum principle for harmonic functions excludes a strict interior minimum. For a quadratic example, the Hessian matrix has zero trace; it cannot have three positive eigenvalues. A static purely electric field therefore cannot supply three-dimensional stable confinement of a free point electric charge. Time variation in a Paul trap, and the velocity-dependent Lorentz force in a Penning trap, change the assumptions of this argument.

A maximum principle bounds a solution through its boundary or initial data, often forbidding nonconstant solutions from attaining an interior extremum. Weak and strong versions depend on the differential operator and hypotheses. The maximum principle for harmonic functions and its strong counterpart are examples.

For a harmonic function, $\Delta u=0$ and $\Delta (-u)=0$. Apply the preceding minimum argument to $u$ and to $-u$:These are the weak minimum and maximum principle for harmonic functions.

The lower formula has a smooth continuation as a function of $(x,y)$, and on $y=0$ it has trace $u(x,0)=0$. Its second tangential derivative iswhich vanishes at $y=0$. Hence the continued lower formula satisfies the equation restricted to the interface, where the formal equation is $u_{xx}=0$.

Assuming the solution itself is continuous across the interface, the upper region therefore has the Dirichlet boundary value problemThe zero function satisfies these data. The maximum principle for harmonic functions, applied to both $u$ and $-u$, proves uniqueness among harmonic functions continuous on the closed rectangle. Thus $\boxed{u=0\text{ for }y\ge 0}$ is the continuous upper continuation. Satisfaction of the degenerate interface equation does not impose continuity of the normal derivative; that is the issue in part (d).

Let $\zeta =\zeta (z,t)$ be the inverse conformal map. With $Q$ real and nonzero, the function $Q\log |\zeta |$ is harmonic away from the source, vanishes on the moving boundary, and has the required logarithmic singularity. Indeed $F(\zeta ,t)=a_1(t)\zeta +O({\zeta }^2)$ with $a_1\ne 0$, so $\log |\zeta |=\log |z|-\log |a_1|+o(1)$ near the origin. Subtracting this candidate from the physical velocity potential leaves a bounded harmonic function near the source. Its singularity is removable, and the difference has zero boundary values. The maximum principle for harmonic functions makes it identically zero. ThereforeOnly the real part is needed, so the multivalued imaginary part of the logarithm causes no ambiguity in $\varphi$.

Differentiate $F(\zeta (z,t),t)=z$ while holding the physical point $z$ fixed:Also the local analytic complex potential of a two-dimensional potential flow has derivativeThe distinction between differentiating at fixed $z$ and fixed $\zeta$ is essential because the boundary map depends on time. Substituting into the boundary evolution law and multiplying by $|\zeta F_{\zeta }{|}^2/Q$ givesTaking the conjugate inside a real part leaves it unchanged, so the desired conformal map equation for planar Darcy growth isAs a normalization check, parameterize the boundary by $\zeta =e^{i\theta }$. Its outward normal vector times arclength is $\zeta F_{\zeta }d\theta$, so its area growth rate is ${\int }_0^{2\pi }\mathrm{Re}(F_t\overline{\zeta F_{\zeta }})d\theta =2\pi Q$, exactly the specified source flux. For an expanding circular map $F=R(t)\zeta$, the equation reduces to $R\dot{R}=Q$.

Choose the positive Green function conventionAway from its pole, $G$ is harmonic, and the maximum principle for harmonic functions makes it positive inside $D$. The sign of its logarithmic singularity is important: it is the convention for $-\Delta$, whereas a Green function for $+\Delta$ has the opposite sign.

Green's second identity, justified by removing a small disk about the pole and taking its radius to zero, givesSince $\Delta u=-f$, $\Delta G=-{\delta }_{\xi }$ and $G$ vanishes on the outer boundary, this rearranges toThe local logarithmic singularity is integrable, so continuous $f$ gives a well-defined volume integral.

In complex coordinates $z=x+iy$, $\zeta =\xi +i\eta$ for the unit disk, useThe numerator has no zeros inside the disk, so its logarithm is harmonic there. On $|z|=1$, the numerator and denominator have equal modulus, giving the required zero boundary value. Differentiating in the outward radial direction at $z=e^{i\theta }$ givesWith zero forcing the representation therefore becomes the Poisson integralFinally replace $\eta$ by $-\eta$ and change variable $\theta \mapsto 2\pi -\theta$. Cosine stays the same, sine changes sign and the assumed boundary symmetry leaves $g$ unchanged. The integral is therefore invariant, so $\boxed{u(\xi ,-\eta )=u(\xi ,\eta )}$. Equivalently, reflection produces a harmonic function with the same Dirichlet data, and uniqueness identifies it with the original solution.

Use separation of variables in Laplace's equation: writing $\varphi =R(r)\Theta (\theta )$ givesSingle-valuedness requires $2\pi$-periodicity, so the nonconstant angular Fourier modes have integer $n\ge 1$ and are $\cos n\theta ,\sin n\theta$. Their radial solutions are $r^n,r^{-n}$. The zero angular mode has radial solutions $1,\log r$. These give the harmonic Fourier expansions in planar concentric domains used below.

For the annulus with $0<a<b$, constant boundary data suggest a radial harmonic function $A+B\log r$. Enforcing both boundary values givesThis has the required boundary values and satisfies $(r{\varphi }_r{)}_r=0$. It is unique: the difference of two continuous solutions is harmonic in the annulus and zero on both boundary circles, so the maximum principle for harmonic functions, applied to the difference and its negative, makes it zero.

For a continuously differentiable vector field on a volume and its boundary, the divergence theorem equates ${\int }_V\nabla \cdot FdV$ to the outward flux ${\int }_{\partial V}F\cdot ndS$. For the given field, $\nabla \cdot F=5(x^2+y^2+z^2)=5r^2$, soOn the sphere, $n=(x,y,z)/R$, andThus the surface flux is also $4\pi R^5$, verifying the theorem by explicit calculation.

Apply the product rule to $\nabla \cdot (\varphi \nabla \psi )=\varphi \Delta \psi +\nabla \varphi \cdot \nabla \psi$, then apply the divergence theorem. This proves Green's first identity:Swap $\varphi$ and $\psi$ and subtract. The gradient products cancel, giving Green's second identity:For uniqueness, set $w={\varphi }_1-{\varphi }_2$. It is a harmonic function with zero Dirichlet boundary conditions. Under the stated smoothness assumptions, Green's first identity with $\varphi =\psi =w$ gives ${\int }_V|\nabla w{|}^{2}dV=0$. Thus $w$ is constant on each connected component, and its boundary value makes every constant zero. The two solutions agree throughout $V$. This is Dirichlet uniqueness for the Poisson equation. For classical solutions merely continuous up to the boundary and twice continuously differentiable in the interior, the maximum principle for harmonic functions gives the same conclusion without requiring boundary derivatives.

Take the Fourier transform in $x$ of the Laplace equation. Writing $U(s,y)$ for the transformed solution gives $U_{yy}-s^{2}U=0$. Its boundary values givewith the multiplier at $s=0$ understood as $1-y$. The particular data simplify this to $U(s,y)=2\sinh (s(1-y))$ for $|s|<1$, and zero outside. Thus, with $a=1-y$, the Fourier solution of the strip Dirichlet problem isThe displayed closed expression applies in the open strip. The integral defines its continuous boundary values, including the removable point $(x,a)=(0,0)$. It gives $u(x,0)=g(x)$ and $u(x,1)=0$. Differentiation under this finite integral verifies $u_{xx}+u_{yy}=0$ directly, since each integrand has opposite second derivatives in $x$ and $y$.

The boundary conditions alone do not specify a unique solution without a growth restriction: adding $C{e}^{\pi x}\sin (\pi y)$ preserves both boundary values and harmonicity. The constructed solution is bounded, and it is unique among bounded solutions continuous on the closed strip. To see this, apply the maximum principle for harmonic functions to a bounded real or imaginary component $w$ of the difference. For $0<\delta <\pi$, the positive harmonic function $B(x,y)=\cosh (\delta x)\cos (\delta (y-1/2))$ has a positive minimum factor in $y\in [0,1]$. For any $\epsilon >0$, the functions $\pm w-\epsilon B$ are nonpositive on the horizontal sides of a sufficiently wide rectangle and on its vertical sides, because $w$ is bounded and $B$ grows there. The maximum principle gives $|w|\le \epsilon B$ at every fixed point. Letting $\epsilon \downarrow 0$ proves uniqueness in this natural Fourier solution class.

A harmonic function on $\Omega$ is a twice continuously differentiable function satisfying $\Delta u=0$ there. For complex-valued functions this means that both its real part and its imaginary part are harmonic functions. The mean value property for harmonic functions characterizes harmonicity: a continuous function is harmonic exactly when its value at each point equals its average over every sphere, or equivalently every ball, whose closed ball lies in $\Omega$. In the forward direction, the derivative of the spherical average vanishes by the divergence theorem. Conversely, radial smoothing leaves a continuous function with the mean value property unchanged locally; it is therefore smooth. Expanding its small-ball average gives $u(x)+r^2\Delta u(x)/(2(d+3))+o(r^2)$ in dimension $d+1$, and hence $\Delta u=0$.

Write $M=\Vert u{\Vert }_{h^1}$ and $v_{d+1}$ for the volume of the unit ball in ${\mathbb{R}}^{d+1}$. We first obtain a pointwise estimate directly from the mean value property for harmonic functions, before using any boundary representation. For the ball of radius $r=t/2$ centered at $(x,t)$,Thus every upward shift of $u$ is a bounded harmonic function, with a continuous bounded boundary value.

The required map is the Poisson integralThe Poisson kernel for the upper half-space is positive, has integral one, and is harmonic as a function of $(x,t)$. Its normalization follows by polar coordinates and $r=\tan \theta$, which reduces the radial integral to ${\int }_0^{\pi /2}{\sin }^{d-1}\theta d\theta$; direct differentiation gives ${\Delta }_{x,t}{P}_t=0$. Differentiation under the integral on compact subsets proves that $P\mu$ is a harmonic function. Tonelli theorem givesHere the total variation norm of a measure is the norm on finite signed or complex Borel measures.

The Poisson kernel for the upper half-space is an approximate identity: for $\psi \in C_0({\mathbb{R}}^d)$, $P_t\ast \psi \to \psi$ uniformly. Indeed, uniform continuity controls translations smaller than a fixed radius, and the kernel mass outside that radius tends to zero. ConsequentlyThe dual characterization of the total variation norm of a measure, as the supremum over $\psi \in C_0$ of supremum norm at most one, now gives the reverse inequality $\Vert \mu {\Vert }_{\mathrm{TV}}\le \Vert P\mu {\Vert }_{h^1}$. Thus $P$ is a linear isometry and is injective.

For surjectivity, fix $s>0$ and put $u_s(x)=u(x,s)$. Bounded harmonic uniqueness in the upper half-space givesTo justify this uniqueness, subtract the Poisson integral of the bounded continuous boundary value from the upward-shifted harmonic function. Their difference is bounded and harmonic and has zero continuous boundary values. Its odd reflection across the boundary is an entire bounded harmonic function; the reflection is harmonic across the flat boundary by harmonic odd reflection across a hyperplane. The Liouville theorem for harmonic functions makes it constant, and oddness makes it zero. The initial pointwise estimate supplies the boundedness needed here. To see the reflection locally, take a ball centered on the hyperplane and extend the boundary data on its upper hemisphere oddly to its lower hemisphere. The harmonic solution on this ball is odd and zero on its equatorial disk. The maximum principle for harmonic functions identifies it with the given difference on the upper half-ball, proving that the reflected function is harmonic across the boundary.

Choose $s_j\downarrow 0$. The measures $u_{s_j}dx$ have total variation norm of a measure at most $M$. By the Banach-Alaoglu theorem, and separability of $C_0({\mathbb{R}}^d)$, a subsequence converges in the weak-star topology to a finite measure $\mu$ with $\Vert \mu {\Vert }_{\mathrm{TV}}\le M$. For fixed $(x,t)$ and $s_j<t$,The kernels in this display belong to $C_0$ and converge uniformly to $P_t(x-\cdot )$. The uniform measure norm bound therefore allows passage to the limit, giving $u(x,t)=P\mu (x,t)$. This proves the Poisson representation of harmonic h1 by finite measures, including onto-ness.

Finally $P_t(x)\le c_d{t}^{-d}$, so the Poisson representation of harmonic h1 by finite measures improves the preliminary pointwise constant:The last occurrence of $h^1({\mathbb{R}}^{d+1})$ in the printed question is understood as the harmonic Hardy space of the upper half-space defined earlier in that question.

Write $\mu =\cos \theta$. Separation of the axisymmetric Laplace equation in spherical coordinates, with regularity at the poles, gives the Legendre polynomial expansionThe discarded interior terms are singular at the origin; the discarded exterior terms fail the prescribed decay at infinity. Continuity at $r=1$ gives $a_{\ell }=b_{\ell }=c_{\ell }$.

Take the radial derivative jump to mean the exterior derivative minus the interior derivative. ThenSince $1-{\mu }^2=\frac{2}{3}(P_0(\mu )-P_2(\mu ))$, comparison with the prescribed jump gives $c_0=-2A/3$, $c_2=2A/15$, and all other coefficients zero. ThusThese expressions satisfy the Laplace equation, continuity, regularity and decay, and their jump is $A{\sin }^2\theta$. If the opposite convention for the word jump is chosen, both expressions change sign. Although only the axisymmetric expansion is needed, no additional nonaxisymmetric solution can be added: zero jump and continuity glue such an addition into an everywhere regular harmonic function decaying at infinity, which vanishes by the maximum principle for harmonic functions.

Seek a power series $y(x)={\sum }_{k\ge 0}{a}_k{x}^k$ in the Legendre differential equation. Equating coefficients of $x^k$ givessoThe even and odd coefficients form separate chains. Choose a nonzero seed of the same parity as $n$ and set the other seed to zero. In the chosen chain, none of the factors vanishes before $k=n$, and the factor $k-n$ vanishes at $k=n$. Therefore this chain terminates with a nonzero $a_n$: it gives a polynomial of degree exactly $n$.

To impose the normalization, its value at $x=1$ must be nonzero. Suppose a nonzero polynomial solution were $y=(x-1{)}^{m}q(x)$ with $m\ge 1$ and $q(1)\ne 0$. The lowest-order coefficient in $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }$ would be $-2m^{2}q(1)(x-1{)}^{m-1}$, whereas $n(n+1)y$ has order $m$. This cannot satisfy the equation. Thus divide by $y(1)$ to obtain the Legendre polynomial $P_n$ with $P_n(1)=1$. Substitution or the coefficient recurrence givesFor an axisymmetric solution of the Laplace equation regular at the two polar axes, the separated modes and their superposition in spherical polar coordinates areThis is the usual separation of variables expansion; the singular angular solutions are excluded by regularity at the axes.

Here ${\sin }^2\theta =\frac{2}{3}(P_0(\cos \theta )-P_2(\cos \theta ))$, so only degrees $0$ and $2$ are required. The degree-zero radial function $A_0+B_0/r$ takes the same value $2/3$ at $r=a,b$; since $a<b$, this forces $B_0=0$, $A_0=2/3$. Write the degree-two coefficient as $-(2/3)T(r)$, where $T=A{r}^2+B{r}^{-3}$ and $T(a)=T(b)=1$. Multiplying these boundary equations by $a^3,b^3$ and subtracting givesConsequentlyEach term is harmonic in the shell, and the radial factor is $1$ at both boundaries, so this satisfies both Dirichlet boundary conditions. The maximum principle for harmonic functions, applied to the difference of two solutions, proves uniqueness among solutions continuous on the closed shell.