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For the usual Dirichlet problem on a bounded regular volume, suppose ${\varphi }_1,{\varphi }_2$ both satisfy the stated data. Their difference $u={\varphi }_1-{\varphi }_2$ is a harmonic function, with ${\nabla }^{2}u=0$ in $V$ and $u=0$ on $\partial V$. In Green's first identity, choose both functions equal to $u$:The nonnegative continuous integrand forces $\nabla u=0$. Thus $u$ is constant on each connected component, and its zero boundary values force each constant to vanish. This proves uniqueness for the Dirichlet problem.

If the boundary value is a constant $c$, the constant function $c$ is itself a harmonic function with those data. Uniqueness therefore yieldsBoundedness, or an appropriate condition at infinity replacing it, is essential. On the upper half-space both $u=0$ and $u=z$ are harmonic functions with zero data on its boundary plane. They show why unrestricted uniqueness must not be asserted for an unbounded volume.

The continuous harmonic function has $m\le \varphi (\mathbf{x})\le M$ on the compact sphere. Integrate these inequalities over its area $4\pi r^2$ and use the mean value property for harmonic functions:The centre value is an average with positive weights, so it lies between the smallest and largest boundary-sphere values.

The construction is well defined. Since $D$ is contained in a ball, its Brownian exit time is finite almost surely: at successive integer times there is a fixed positive probability that the next independent unit-time Brownian motion increment has length greater than the ball's diameter, forcing an exit. The survival probability is therefore bounded by a geometric sequence. Path continuity gives $B_{{\tau }_D}\in \partial D$. Thus the bounded Dirichlet boundary data give a bounded Borel function $u$ on $D$.

For interior harmonicity, fix a ball with closure in $D$, centred at $x$, and let $\sigma$ be its Brownian exit time. The Strong Markov property, followed by the tower property of conditional expectation, givesThe orthogonal invariance of Brownian motion makes $B_{\sigma }$ uniform on the boundary sphere. Hence $u$ has the spherical mean value property for harmonic functions for every such ball. A locally bounded Borel function with this property is smooth and harmonic: integrating the spherical averages against any smooth radial mollifier gives $u=u\ast \rho$ locally, which first proves smoothness; the mean value property for harmonic functions then implies $\Delta u=0$.

For the boundary limit at $\xi$, choose the harmonic barrier for the Dirichlet problem $h_{\xi }$ from (i). The Itô formula, localized inside $D$, makes $h_{\xi }(B_{t\wedge {\tau }_D})$ a bounded martingale. Compact localization and the bounded convergence theorem, first to the exit and then as $t\to \infty$, yieldFor $\epsilon >0$, if $\partial D\setminus B(\xi ,\epsilon )$ is nonempty, compactness and barrier positivity giveIf that boundary subset is empty, the probability is already zero. ThereforeFirst let $x\to \xi$ and then $\epsilon \downarrow 0$. The continuity of the Dirichlet boundary data proves $u(x)\to f(\xi )$.

Finally, the difference of two continuous solutions is a harmonic function vanishing on the boundary. The maximum principle for harmonic functions on the bounded domain gives that it is zero. Thus the Kakutani solution of the Dirichlet problem exists, attains every prescribed boundary value, and is unique.

A standard form of the Kakutani solution of the Dirichlet problem uses a bounded domain $D\subset {\mathbb{R}}^d$, continuous Dirichlet boundary data $f\in C(\partial D)$, and regularity of every boundary point for the Dirichlet problem. The boundedness of $D$ makes $\partial D$ a compact set, so $f$ is a bounded function and a uniformly continuous function. A bounded domain with $C^2$ boundary is a sufficient geometric case; one must not omit boundary regularity for an arbitrary bounded domain.

One standard analytic characterization of a regular boundary point on a bounded domain is the existence of a positive harmonic barrier: for each $\xi \in \partial D$ there is a harmonic function $h_{\xi }\in C(\overline{D})$ with $h_{\xi }(\xi )=0$ and $h_{\xi }(x)>0$ for $x\in \overline{D}\setminus \{\xi \}$. This is the harmonic form of a barrier for the Dirichlet problem; the barrier characterization of regularity is a standard fact of potential theory.

Let $B$ be $d$-dimensional Brownian motion started at $x\in D$ and let ${\tau }_D=\inf \{t\ge 0:B_t\notin D\}$ be its Brownian exit time. Then the unique solution in $C(\overline{D})\cap C^2(D)$ isEquivalently, $u(x)={\int }_{\partial D}f(\xi ){\omega }_D(x,d\xi )$, where ${\omega }_D$ is harmonic measure, the exit probability distribution of Brownian motion. The harmonic function equation is $\Delta u=0$, with the Laplace operator convention $\Delta ={\sum }_j{\partial }_j^2$.

One convention for the Papkovich–Neuber representation uses a harmonic vector potential $\mathbf{\Phi }$ and a harmonic scalar potential $\chi$:These are harmonic functions away from any singular force point. Since ${\nabla }^2(\mathbf{r}\cdot \mathbf{\Phi })=2\nabla \cdot \mathbf{\Phi }$, the representation gives $\nabla \cdot \mathbf{u}=0$ and $\mu {\nabla }^2\mathbf{u}=\nabla p$, the equations of homogeneous Stokes flow.

Place the point force at the origin. A velocity linear in $\mathbf{F}$, decaying as $r^{-1}$ and having the rotational symmetry of a point force is obtained from $\mathbf{\Phi }=c\mathbf{F}/r$, $\chi =0$. The vector components are harmonic functions for $r>0$; scalar dipole potentials would instead generate higher-order decaying singularities. The force normalization fixes $c=-1/(4\pi )$. Indeed, substitution gives the Stokeslet:To verify its strength, the Newtonian fluid stress tensor is ${\sigma }_{ij}=-3(\mathbf{F}\cdot \mathbf{r})r_i{r}_j/(4\pi r^5)$. Its outward traction integrated over any sphere surrounding the origin is $-\mathbf{F}$, because $\int \mathbf{nn}d\Omega =(4\pi /3)I$. Thus the localized force applied to the fluid is $\mathbf{F}$, as required. Translation of the origin gives the same Stokeslet centered at any prescribed force point.

In the bounded-domain setting of the weak maximum principle, let $u_1,u_2$ be two classical solutions with the same forcing and boundary values. Their difference $z=u_1-u_2$ satisfies $Lz=0$ and $z=0$ on the boundary. Apply the weak maximum principle first to $z$ and then to $-z$:Thus $u_1=u_2$, proving uniqueness of the Dirichlet problem. The same proof works on an unbounded domain if the difference and its negative satisfy the required condition at infinity. As literally printed without either restriction, uniqueness need not hold: on $\{x_n>0\}$ the two harmonic functions $0$ and $x_n$ have the same zero boundary values.

Apply the weak weak maximum principle for elliptic operators to the harmonic function $\varphi$ and to $-\varphi$. Their boundary maxima are both zero, giving $\varphi \le 0$ and $\varphi \ge 0$ on the closure. Consequently $\varphi \equiv 0$.

An immersion $\varphi :U\to {\mathbb{R}}^3$ has differential of rank $2$, equivalently linearly independent ${\varphi }_u,{\varphi }_v$. It has isothermal coordinates when its first fundamental form has $E=G>0$, $F=0$. A minimal surface has zero mean curvature.

For an isothermal immersion, differentiation of $E=G$ and $F=0$ givesand similarly its scalar product with ${\varphi }_v$ is zero. Its normal component is $e+g$. The given mean curvature formula reduces to $H=(e+g)/(2E)$, soThus an isothermal immersion is minimal exactly when each coordinate is a harmonic function.

An explicit example is the Enneper surface, parametrized on ${\mathbb{R}}^2$ byEvery coordinate has zero Laplacian. Direct differentiation gives ${\varphi }_u\cdot {\varphi }_v=0$ and $|{\varphi }_u{|}^2=|{\varphi }_v{|}^2=(1+u^2+v^2{)}^2>0$. Thus it is an isothermal immersion and is minimal everywhere.

It is not an open subset of any of the three listed surfaces. The parameter points $(\sqrt{3},0)$ and $(-\sqrt{3},0)$ both map to $(0,0,3)$, but their tangent planes differ: ${\varphi }_u=(-2,0,\pm 2\sqrt{3})$ while ${\varphi }_v=(0,4,0)$. A plane, catenoid, or helicoid is an embedded smooth surface with a unique tangent plane at each point. If our whole image lay in any one of them, both rank-$2$ tangent planes at this point would have to equal its tangent plane, a contradiction.

The identity divergence of a curl is zero follows by the symmetry of second partial derivatives and antisymmetry of the Levi-Civita symbol. ThereforeIf $\mathbf{B}=\nabla \times \mathbf{A}$ and $\mathbf{A}$ is in Coulomb gauge, the curl of the curl identity gives $\mathbf{J}=-{\nabla }^2\mathbf{A}$. Apply the supplied scalar Poisson equation representation to each component of the decaying vector potential. In the units of this problem the Coulomb-gauge vector potential of a localized steady current isHere one needs $\mathbf{A}=O(|\mathbf{x}{|}^{-1})$, $\nabla \mathbf{A}=O(|\mathbf{x}{|}^{-2})$ and enough localization of $\mathbf{J}$ for the stated representation and differentiations. A smooth compactly supported current density is a sufficient convenient case. The local differential equations and gauge alone do not impose these conditions: $\mathbf{A}=(-y,x,0)$, $\mathbf{B}=(0,0,2)$, $\mathbf{J}=0$ satisfy them but do not satisfy the displayed integral formula. The formula selects the decaying solution, excluding nondecaying homogeneous harmonic functions.

With $\mathbf{r}=\mathbf{x}-\mathbf{y}$, the gradient of the kernel is ${\nabla }_x(1/|\mathbf{r}|)=-\mathbf{r}/|\mathbf{r}{|}^3$. Since $\mathbf{J}(\mathbf{y})$ is independent of $\mathbf{x}$,Thus taking the curl gives the Biot-Savart law in the problem's normalization:Finally, change variables $\mathbf{z}=\mathbf{x}-\mathbf{y}$ in the integral for $\mathbf{A}$. This keeps the singular kernel independent of the differentiation variable:For smooth localized $\mathbf{J}$, differentiation under the integral sign is legitimate: $1/|\mathbf{z}|$ is locally integrable in three dimensions, and on a bounded neighbourhood of $\mathbf{x}$ the translated sources and their derivatives have common compact support. The integral indeed satisfies Coulomb gauge.

With the positive logarithmic fundamental solution used here, a Dirichlet Green function for an interior pole ${\mathbf{r}}_0$ satisfiesEquivalently, $G-G_0$ is harmonic near the pole, $G$ is harmonic away from it, and it has the specified logarithmic singularity. Regularity up to the remaining boundary is understood; on an unbounded domain impose the appropriate condition at infinity as well.

Apply Green second identity with $f=\psi$ and $g=G$. Since $\psi$ is a harmonic function, the volume integral is $\int \psi {\delta }_{{\mathbf{r}}_0}dA=\psi ({\mathbf{r}}_0)$. The term $G{\partial }_n\psi$ vanishes on the boundary by the Dirichlet boundary condition. ThereforeOne may equivalently excise a small disc around the pole and take its radius to zero. The sign is positive because this question uses ${\nabla }^{2}G=+\delta$, not the alternative negative-Laplacian convention.

A bounded-domain hypothesis is needed for the requested global construction and uniqueness. The printed question does not include it. For example, the upper half-plane satisfies the stated exterior cone condition, but both $u=0$ and $u(x_1,x_2)=x_2$ are harmonic functions with zero Dirichlet boundary condition. Thus part (iv), in its printed unrestricted solution class, is false. In this question's solutions we add that $\Omega$ is bounded; the exterior cone condition and all other data remain as printed.

The half-plane also rules out the printed global separated barrier. In its angular interval $(0,\pi )$, varying $r$ on a fixed ray forces $\nu =2$ if $r^{\nu -2}(h^{\prime \prime }+{\nu }^{2}h)\ge \delta >0$ for every $r>0$. We would then have $h^{\prime \prime }+4h\ge \delta$ and $h<0$. Put $a=\pi /4$ and $b=3\pi /4$. Twice integration by parts givescontradicting the positive integrand. Thus a source qualification is necessary for part (i) as well as part (iv).

Fix $x_0$ and its exterior cone of half-angle $\alpha$. Choose $0<{\alpha }_0<\min (\alpha ,\pi /2)$ and setOn the complement of the cone, unwrap the angle in polar coordinates as $\vartheta \in (\alpha ,2\pi -\alpha )$, measured from the cone axis. Then $\psi =\vartheta -\pi$ is the angle from the opposite axis and $|\psi |\le \pi -\alpha$ on $\overline{\Omega }\setminus \{x_0\}$. Define the power barrier for an exterior cone byThis has the requested separated form $r^{\nu }h(\vartheta )$. In the printed signed-angle convention, the same function is $-r^{\nu }\cos (\beta (\pi -|\theta |))$ outside the cone. It is smooth across the negative axis: near that axis the angular expression is the even function $\cos (\beta \psi )$.

Since $\beta (\pi -\alpha )<\pi /2$, put $m=\cos (\beta (\pi -\alpha ))>0$. The cosine is at least $m$, so $g\le -m{r}^{\nu }<0$ away from $x_0$, and $g$ is continuous at $x_0$. The Laplacian in polar coordinates givesLet $D={\sup }_{x\in \Omega }|x-x_0|<\infty$. Since $\nu <2$, the barrier for the Dirichlet problem satisfiesBoth the uniform lower bound and the later global comparison use boundedness. In particular, $-g$ is positive at every other boundary point, and is bounded away from zero on boundary sets staying a positive distance from $x_0$.

For the corrected bounded-domain problem, the interior theorem for the Perron method for the Dirichlet problem gives a classical solution of the Poisson equation, and the power barrier for an exterior cone argument gives its continuous extension with the required Dirichlet boundary condition. Hence existence holds in $C^2(\Omega )\cap C^0(\overline{\Omega })$.

If $u_1,u_2$ are two such classical solutions, then $w=u_1-u_2$ is a harmonic function with zero boundary trace. The weak maximum principle for elliptic operators, applied to $w$ and $-w$ on the bounded domain, gives $w\le 0$ and $-w\le 0$. ConsequentlyThere is exactly one solution after adding boundedness of $\Omega$. For the printed unbounded-domain version, the half-plane counterexample in part (i) disproves uniqueness; an additional condition at infinity or a suitable restricted solution class would instead have to be specified.

The real coordinate functions are real and imaginary parts of the holomorphic coordinates $z_j$. Part (a), together with the real coefficients of the Laplace-Beltrami operator, therefore shows that every $x_l$ is a harmonic function. For the given divergence-of-gradient convention, applying the operator to $x_l$ givesThese are the harmonic coordinate identities. Expanding the operator on an arbitrary smooth function now givesThe bracketed coefficients vanish. Hence there is no first-derivative term in these coordinates:The reason is that the real components of holomorphic coordinates are harmonic coordinates for a Kähler metric; a general real coordinate change need not retain this simplification.

Choose a constant positive Hermitian form on $V$. Its underlying flat Kähler metric is translation-invariant and descends through the Euclidean lattice to the compact complex torus $X$. The Dolbeault Hodge decomposition gives a unique Dolbeault Laplacian-harmonic representative of every Dolbeault cohomology class. By the Kähler Laplacian identity, these are also Hodge Laplacian-harmonic.

Choose a constant coframe on $V$. A lifted harmonic form is a sum of constant wedge basis forms with smooth periodic coefficients. For the flat metric, the Hodge Laplacian acts coefficientwise by the scalar constant-coefficient Laplacian. Each coefficient is a harmonic function on a compact flat real torus, hence is constant: integration by parts gives $\int |Da{|}^2=0$. Conversely every constant-coefficient form is harmonic. Thus the harmonic representatives are exactly the translation-invariant differential forms of type (p, q).

The constant $(1,0)$ covectors form $V^{\ast }={\mathrm{Hom}}_{\mathbb{C}}(V,\mathbb{C})$; the constant $(0,1)$ covectors form the space of antilinear maps $\overline{V^{\ast }}={\mathrm{Hom}}_{\overline{\mathbb{C}}}(V,\mathbb{C})$. Sending a constant wedge form to its cohomology class givesThe map is intrinsic and does not depend on the auxiliary flat metric: that metric proves every class has exactly one constant representative, while the inclusion of constant forms is defined directly by the quotient. The Dolbeault cohomology of a complex torus consists of constant forms, and in particularFor indices outside $0,\dots ,g$, both the exterior-power expression and the cohomology group vanish.

The domain is the union of the horizontal and vertical strips. The four connected components of its boundary carry the constants $a,b,c,d$ in the northeast, northwest, southwest and southeast, respectively.

First justify finiteness of $T$. A small open ball around $(2,2)$ lies outside the domain. By part (b), planar Brownian motion visits that ball almost surely in finite time. A continuous path from $z\in D$ to that ball must meet $\partial D$ first. Hence $T<\infty$ almost surely, and $B_T\in \partial D$ by continuity.

The harmonic function $u$ is smooth inside $D$. The Itô formula, localized on relatively compact subdomains, makes $u(B_{t\wedge T})$ a local martingale. Its boundary definition uses the continuous extension of $u$. Since $u$ is bounded, this stopped process is a true martingale with uniform integrability: pass through the compact exhaustion using bounded convergence theorem. In particular ${\mathbb{E}}_{z}u(B_{t\wedge T})=u(z)$.

Now $B_{t\wedge T}\to B_T$ almost surely as $t\to \infty$. The dominated convergence theorem, boundedness of $u$, and its continuous boundary values give the bounded harmonic representation with almost sure Brownian exit:Almost sure exit and boundedness of the harmonic function are both used. The domain's unboundedness causes no loss of expectation here because the stopped function values are uniformly bounded.

Interpret $f$ as a conformal bijection onto $D^{\prime }$. If it is merely an injective map into a larger target, the correct exit domain is $f(D)$ instead. Let $\zeta$ be the exit time from $D$, write $f=u+iv$, and identify planar Brownian motion with $B^1+i{B}^2$. The Cauchy-Riemann equations give $u_x=v_y$, $u_y=-v_x$, and both $u,v$ are harmonic functions. The Itô formula, localized on compact subsets of $D$, yieldsThus both coordinates are continuous local martingales, withSince $f^{\prime }$ never vanishes, $A$ is strictly increasing before $\zeta$. Let ${\eta }_s=\inf \{t:A_t>s\}$, defined for $s<A_{\zeta -}$. The optional time-change theorem gives continuous local martingale coordinates for ${\beta }_s=f(B_{{\eta }_s})$, with bracket matrix $sI$. The Lévy characterization of multidimensional Brownian motion makes $\beta$ a standard planar Brownian motion started at $f(z)$, up to its lifetime. Therefore conformal invariance of planar Brownian motion takes the formTo identify the lifetime, exhaust $D$ by relatively compact open sets $D_n$ with nested closures. The images $f(D_n)$ exhaust $D^{\prime }$ because $f$ is a homeomorphism onto $D^{\prime }$. Each stopped transformed path exits $f(D_n)$ at precisely the clock value of its exit from $D_n$. Increasing these exit times gives the exit lifetime of Brownian motion from $D^{\prime }$, including a possibly infinite lifetime. This identifies $A_{\zeta -}$ with that lifetime and needs no global extension of $f$ to the boundary.

For a harmonic function $u$ on $\mathbb{C}\cong {\mathbb{R}}^2$, the Itô formula givesStop on leaving increasing discs and at increasing deterministic times. On each such interval the gradient is bounded, so the stochastic integral is a martingale. This proves that $u(B)$ is a continuous local martingale.

If $|u|\le K$, this is a bounded local martingale and hence a true martingale. Start Brownian motion at an arbitrary $z$. Its Gaussian heat kernel givesFor a unit vector $e$, ${\partial }_e{p}_t(v)=-(e\cdot v)p_t(v)/t$, so $\int |{\partial }_e{p}_t(v)|dv=\sqrt{2/\pi }/\sqrt{t}$. Integrating the directional derivative along the segment from $z$ to $w$ yieldsLetting $t\to \infty$ gives the harmonic Liouville theorem:This Gaussian heat-kernel proof of the harmonic Liouville theorem uses the bounded martingale identity established above.

Take the convention that the coordinates of planar Brownian motion are independent standard real Brownian motions, so its infinitesimal generator is $\frac{1}{2}\Delta$. Define the forward conformal Brownian clockSince a conformal bijection has nonzero derivative, $A$ is a strictly increasing continuous map from $[0,T)$ onto $[0,\tilde{T})$. The time used inside the original path is its inverse:Thus the interval direction for $\tau$ is the inverse-clock direction.

Write $\varphi =u+iv$. The Cauchy-Riemann equations and the Itô formula make $u(B_s),v(B_s)$ continuous local martingales, withIndeed both components are harmonic functions, and their gradients are orthogonal with the same squared norm. After the time change of a continuous process by $\tau$, their quadratic variations are $t$ and their quadratic covariation is zero. The Lévy characterization of multidimensional Brownian motion therefore identifies $\tilde{B}$ as planar Brownian motion started at $z^{\prime }$, up to its lifetime.

It remains to identify that lifetime as the exit time, rather than merely produce a local Brownian path. Boundedness of $D$ gives $T<\infty$ almost surely. For every compact subset $C\subset D^{\prime }$, its inverse image under $\varphi$ is compactly contained in $D$. As $s\uparrow T$, continuity gives $B_s\to B_T\notin D$, so $\varphi (B_s)$ eventually leaves $C$. Consequently the transformed path leaves every compact subset of $D^{\prime }$ at its lifetime. If $\tilde{T}<\infty$, its Brownian extension has a finite limit, and that limit is outside $D^{\prime }$; if $\tilde{T}=\infty$, the path never exits. In both cases its maximal lifetime is precisely the exit time from $D^{\prime }$.

The killed paths, together with their lifetimes, have the same law: