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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 Arzelà-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 multivariable chain rule first givesTaking a second partial derivative in each variable, with the partial derivatives of $f$ evaluated at $(x(u,v),y(u,v))$, givesThe mixed partial derivatives cancel even without using their equality. Hence the Laplacians satisfyAway from the origin the factor $u^2+v^2$ is positive, so $g$ is a harmonic function if and only if $f$ is a harmonic function on the corresponding local coordinate patch. This is the Laplacian in planar parabolic coordinates.

False. Take $V=\{x^2+y^2+z^2<1\}$, the open unit ball, and $u(x,y,z)=x$. This is a nonzero harmonic function. On the unit sphere, the outward unit normal vector is $\mathbf{n}=(x,y,z)$, soThus the prescribed Robin boundary condition holds. Unlike the preceding case, Green's identities now give ${\int }_V|\nabla u{|}^2={\int }_{\partial V}{u}^2$, which permits nonzero solutions of the differential equation.

False. The constant function $u=1$ is a harmonic function and has zero normal derivative on every smooth boundary, but it is nonzero in $V$.

In fact, the first of Green's identities givesHence $\nabla u=0$; path connectedness makes $u$ constant throughout $V$. Thus this homogeneous Neumann boundary condition determines a harmonic function only up to an additive constant.

Let $w$ be the difference of two solutions. It is a harmonic function in the open rectangle, continuous on its closure, withInterpret these Neumann boundary conditions classically on the open edges, with the corresponding first derivatives extending there. We use the strong maximum principle for harmonic functions together with even reflection across a homogeneous Neumann boundary. This avoids imposing differentiability at the two bottom corners, where the original data fail a corner compatibility condition for mixed boundary data.

By continuity on the compact rectangle, $w$ attains a maximum $M$. Suppose $M>0$. It cannot occur on the bottom edge, where $w=0$. If it occurs in the interior, the strong maximum principle for harmonic functions makes $w$ constant, contradicting its bottom value. If it occurs on an open vertical or top edge, the zero normal derivative allows an even reflection across that edge. The reflected harmonic function then has an interior maximum, giving the same contradiction.

The only remaining possibilities are the two top corners. Reflect evenly across both meeting edges. This gives a bounded harmonic function on a punctured neighbourhood of the corner. The removable singularity for a bounded harmonic function fills the centre with its continuous value. The corner is now an interior maximum, again impossible. Therefore $w\le 0$. Applying the identical argument to $-w$ gives $w\ge 0$.

Thus $\boxed{w=0}$, and the solution is unique among the continuous classical solutions with the stated edge conditions.

A temperature on the full unit disk must be single valued, so its angular factor and angular derivative are $2\pi$-periodic. For positive separation constant $\lambda =k^2$, this requires $k=n$ to be a positive integer. For $\lambda =0$, periodicity removes $D\theta$. For $\lambda <0$, no nonzero linear combination of $e^{k\theta }$ and $e^{-k\theta }$ is $2\pi$-periodic, so that case disappears.

The origin belongs to the disk, so a classical harmonic function must remain bounded and extend smoothly there. This excludes $r^{-n}$ and $\log r$. The remaining individual separated solutions are constants and $r^n(A_n\cos (n\theta )+B_n\sin (n\theta ))$. By superposition, the regular harmonic expansion on a disk isThe admissible Fourier coefficients are fixed by the boundary conditions. Smoothness at the origin can also be seen from $r^n\cos (n\theta )$ and $r^n\sin (n\theta )$ being the real and imaginary parts of $(x+iy{)}^n$, hence harmonic polynomials.