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The function $u(x)=ϵ/|x|$ is harmonic on $\{|x|>ϵ\}$, equals one on the target sphere, and tends to zero at infinity. Optional stopping at the first hit of radius $ϵ$ and the first exit from a ball of radius $R$ givesLetting $R\to \infty$ and taking $|x|=1$ yields ${\mathbb{P}}_x({\tau }_{ϵ}<\infty )=ϵ$.

Use the Papkovich–Neuber representation in the formwhere $\mathbf{\Phi }$ and $\chi$ are harmonic functions. A torque is a axial vector, so rotational covariance and decay select the harmonic vector fieldHere $\mathbf{x}\cdot \mathbf{\Phi }=0$ and $\nabla \cdot \mathbf{\Phi }=0$, soThis rotlet equals the rotating sphere in Stokes flowwhen $\mathbf{\Omega }=\mathbf{G}/(8\pi \mu a^3)$. It satisfies the no-slip boundary condition on $r=a$, decays at infinity, and its Newtonian fluid stress tensor transmits the applied couple $\mathbf{G}$.

Apply separation of variables to the Laplace equation in polar coordinates by writing $\varphi (r,\theta )=R(r)\Theta (\theta )$. Division by $R\Theta$ givesThe requirement that $\Theta$ be a real $2\pi$-periodic function restricts the separation constants to $n^2$, with angular factors $\cos (n\theta )$ and $\sin (n\theta )$. For $n\ge 1$, the radial ordinary differential equation is an Euler equation with solutions $r^n$ and $r^{-n}$. For the zero mode, $\Theta$ is constant andso $R=a_0+c_0\log r$. By linearity, superposition givesThis is the separated expansion of a harmonic function in a circular region.

Write $s=\log r$, so the two circular boundaries are $s=0$ and $s=2$. The zero angular mode must interpolate between $-1$ and $1$, giving $s-1=\log r-1$.

For the $n$th cosine mode, the radial factor has equal value $A_n$ at both boundaries. The unique harmonic function with those data isHence the solution of the annular Dirichlet problem isAt $r=1$ and $r=e^2$, the hyperbolic cosine quotient equals one, so the boundary conditions are satisfied term by term.

For each $x$, the bounded local martingale $u(x+B_t)$ is a true martingale. Taking expectations givesfor every $t>0$. Thus $u$ equals its convolution with every heat kernel. The convolution is smooth, so the originally Borel function $u$ is smooth. Differentiating the heat semigroup identity at $t=0$ gives $\Delta u=0$, so $u$ is a bounded harmonic function on the plane. The harmonic Liouville theorem now implies that $u$ is constant.

Work in the frame translating with the bubble. The bubble is the fixed circle $r=a$, while the velocity tends to $-U{\mathbf{e}}_x$ at infinity. The potential flow around a circular cylinder has velocity potentialIt is harmonic for $r>a$ and givesThus $u_r=0$ on $r=a$, satisfying the kinematic boundary condition, and $\mathbf{u}\to -U{\mathbf{e}}_x$ as $r\to \infty$. The laboratory-frame flow is obtained by adding $U{\mathbf{e}}_x$.

The displacement must have small relative amplitude and small slope,Write the potentials as perturbations of the base flow,with $|\nabla {\phi }_{\pm }|/U$ and $|\nabla {\phi }_0|/U$ small. A linearization evaluates the conditions at the unperturbed lines $y=\pm h$ and discards products of perturbations. At $y=h$ this giveswhereas at $y=-h$ it givesThe perturbation potentials remain harmonic functions, and their gradients decay at the corresponding far fields.

The interior Schauder estimate iswhere $C$ depends only on $n$, $\alpha$, the ellipticity constant, and the stated coefficient bounds.

It is enough first to prove the estimate for the Hölder seminorm of $D^{2}u$. The Interpolation inequality in Holder spaces controls $Du$ and $D^{2}u$ by a small multiple of $[D^{2}u{]}_{\alpha }$ plus $\Vert u{\Vert }_0$; the small term is later absorbed. Freezing $a^{ij}$ at the centre of each ball and moving coefficient differences and lower-order terms to the right reduces the local estimate toOnce this is known, the Simon absorption lemma gives the desired estimate on $B_{1/2}$.

For completeness, prove the frozen-coefficient estimate by contradiction. If it failed, choose solutions $u_k$, points $x_k$, and scales ${\rho }_k\downarrow 0$ at which the scale-invariant Hölder quotient is almost maximal. Let $q_k$ be the quadratic Taylor polynomial of $u_k$ at $x_k$ and defineThen $v_k(0)=D{v}_k(0)=D^2{v}_k(0)=0$, the Hessians have uniformly bounded local $C^{0,\alpha }$ seminorms, and the normalization makes their oscillation nonzero on a fixed ball. The localized equation iswhere the coefficient matrices converge locally uniformly to one constant positive-definite matrix and the given normalized error $g_k$ tends locally uniformly to zero.

The Arzela-Ascoli theorem produces a locally $C^2$ convergent subsequence with limit $v$ satisfying a constant-coefficient elliptic equation on ${\mathbb{R}}^n$. A linear rotation and scaling turn it into a harmonic function. The normalization gives $D^{2}v(0)=0$ but a nonconstant Hessian, while maximality of the scaled quotient gives growth at most $C(1+|y{|}^{2+\alpha })$. Applying the Liouville theorem to derivatives shows that every second derivative is constant because $0<\alpha <1$. This contradicts the normalized Hessian oscillation. The frozen estimate follows, and interpolation plus absorption completes the proof.

Let $\tau =T_r\wedge T_R$. The function $z\mapsto \log |z|$ is a harmonic function on the annulus $r<|z|<R$, so Itô formula shows thatis a bounded martingale. By the optional sampling theorem for a supermartingale applied to this martingale,where $p={\mathbb{P}}_x(T_r<T_R)$. Solving gives the planar Brownian annulus hitting probability

WriteThis is the harmonic measure of $A$ viewed from $z$, and is a bounded harmonic function outside the closed target disc. The disc is contained in $B(0,r_2)$. In the half-plane cut out by a line $L$ through the origin, the probability of reaching $B(0,r_2)$ before $L$, from a point of modulus $R$, is $O(r_2/R)$ uniformly in the direction. One sees this by mapping the half-plane outside the disc with $z\mapsto \log (z/r_2)$ to a half-strip and solving the corresponding Dirichlet problem by a sine series.

Part (b) consequently shows that the angular oscillation of $H_A$ on a large circle tends to zero, uniformly in the Borel set $A$. The same half-strip estimate in the annulus $R-|a|\le |z|\le R+|a|$ shows that shifting the center of that circle by the fixed vector $a$ changes the average of $H_A$ by $o(1)$, uniformly in $A$. Hence, for every $\epsilon >0$, all sufficiently large $R$ satisfyfor every Borel subset $A$ of the target disc.

The function $z\mapsto \log |z|$ is a harmonic function on the annulus $r_1<|z|<r_2$. The Itô formula therefore makes $\log |B_{t\wedge \tau }|$ a bounded martingale. The optional sampling theorem for a supermartingale giveswhere $p=\mathbb{P}(|B_{\tau }|=r_1)$. Solving this linear equation yields the planar Brownian annulus hitting probability

Take $U=B(0,e^{-t})$ in the Domain Markov property of the Gaussian free field. The field $h^{\mathbb{D}\setminus U}$ is a harmonic function throughout $U$. If $s>t$, the circle $\partial B(0,e^{-s})$ lies inside $U$, so the mean value property for harmonic functions gives

For a harmonic function $u$, letThe divergence theorem givesSince $M(r)\to u(x)$ as $r\downarrow 0$, $M(r)=u(x)$. Integrating the spherical averages in the radial variable gives the corresponding ball average, proving the mean value property for harmonic functions. If $u$ attains its maximum at an interior point, the average of the nonnegative function $\max u-u$ on every sufficiently small centred sphere is zero. Continuity makes $u$ constant on those spheres, and connectedness propagates that value through the domain. Thus the weak maximum principle for elliptic operators gives

For the derivative estimate, choose $r>0$ smaller than half the distance from ${\Omega }^{\prime }$ to $\partial \Omega$, and let ${\rho }_r$ be a smooth radial mollifier supported in $B_r(0)$. Writing its convolution in polar coordinates and using the spherical mean value property shows that $u\ast {\rho }_r=u$ on ${\Omega }^{\prime }$. Hence, for every multi-index $\alpha$,so Holder inequality gives

The function $z\mapsto \log |z|$ is harmonic on the annulus $\epsilon <|z|<R$. Hence $\log |B_{t\wedge {\tau }_{\epsilon }\wedge {\tau }_R}|$ is a bounded martingale by Itô formula. Optional stopping givesSolving yields

The Papkovich–Neuber representation writes a homogeneous Stokes flow in terms of a harmonic vector field $\mathbf{\Phi }$ and a harmonic scalar $\chi$ asFor translation, rotational symmetry and decay at infinity restrict the trial harmonic fields to the fundamental harmonic $1/r$ and its directional derivatives contracted with $\mathbf{U}$. For rotation, the only decaying isotropic axial-vector field with the required boundary value is proportional to $\mathbf{\Omega }\times \mathbf{x}/r^3$. Matching the no-slip boundary condition $\mathbf{u}=\mathbf{U}+\mathbf{\Omega }\times \mathbf{x}$ at $r=a$ gives the superposition of the translating sphere in Stokes flow and the rotating sphere in Stokes flow:Each term decays at infinity, and direct substitution at $r=a$ gives the prescribed rigid velocity.

When $\mathbf{U}=0$, the pressure is constant and may be set to zero. Differentiating the rotational velocity and using the Newtonian fluid stress tensor gives

Let $(B_t)$ be Brownian motion in ${\mathbb{R}}^3$ started at a nonzero point. The function $h(x)=|x{|}^{-1}$ is a positive harmonic function on ${\mathbb{R}}^3\setminus \{0\}$ because its Laplacian vanishes there. Stopping on annuli and applying Itô formula shows that $h(B_t)$ is a local martingale. Letting the inner boundary shrink to zero also shows that three-dimensional Brownian motion does not hit the origin.

A positive local martingale is a supermartingale, so $h(B_t)$ is $L^1$-bounded by $h(B_0)$. The upcrossing proof from part (c), which applies verbatim to a positive supermartingale, therefore gives a finite almost-sure limit

For a unidirectional, $z$-independent velocityincompressibility holds automatically. The axial Stokes flow equation isNo external pressure gradient is imposed, so $p_z=0$ andThus the axial velocity is a harmonic function in the disk.

The Poisson integral convolves boundary data with the Poisson kernel to produce a harmonic function in a half-space. In the upper half-plane,

A prescribed point is almost surely never visited by planar Brownian motion started elsewhere. The harmonic function $\log |z-w|$ gives an annular hitting probability that tends to zero as the inner radius decreases to zero. This does not contradict recurrence of planar Brownian motion, which concerns neighbourhoods.

A potential dipole in three dimensions is the potential flow $\mathbf{u}=\nabla \varphi$ with $\varphi =\mathbf{d}\cdot \mathbf{x}/r^3$. Away from the origin its potential is a harmonic function, so it is also a constant-pressure Stokes flow. Its velocity decays as $r^{-3}$ and its vorticity is zero. It is distinct from a stresslet, whose leading velocity decays as $r^{-2}$.