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Let $q$ be a real harmonic function in a semistrip, $g_0(x)=q(x,0)$, $g_l(x)=q(x,l)$ and $h(y)=q(0,y)$. Let $G_j,N_j$ be the negative-exponential Half-range Fourier transforms of $g_j$ and $q_y(x,j)$, and put $H(k)={\int }_0^l{e}^{ky}h(y)dy$, $W(k)={\int }_0^l{e}^{ky}{q}_x(0,y)dy$. The closed differential form $e^{-ikz}{q}_{z}dz$ has zero boundary integral. Integration by parts of tangential derivatives cancels all corner values and gives the displayed global relation, for $\mathrm{Im}k\le 0$.

// Target: analysis.bigb

For $0<x<L$, $0<y<H$, choose $k=m\pi /H>0$ and seek a harmonic function with zero values on the horizontal edges, boundary value $\sin (ky)$ at $x=L$, and $\varphi +\alpha {\partial }_n\varphi =0$ at $x=0$, with constant $\alpha \ge 0$. Separation of variables gives $\varphi =[\alpha k\cosh (kx)+\sinh (kx)]\sin (ky)/[\alpha k\cosh (kL)+\sinh (kL)]$. The outward derivative at the left edge is $-{\partial }_x$, giving $F(0)-\alpha F^{\prime }(0)=0$. The denominator is positive and Poisson uniqueness with a nonnegative Robin normal coefficient proves uniqueness in the regular solution class.

For a positive harmonic function on the unit disc, choose a global harmonic conjugate and apply the Carathéodory derivative bound for the right half-plane to $u+iv$. This gives $|\nabla u(0)|\le 2u(0)$ by the Cauchy-Riemann equations. A disk automorphism sending zero to $z$ has derivative modulus $1-|z{|}^2$, so the chain rule gives the displayed bound. Composing a boundary-pole Poisson kernel with a disk automorphism attains equality, proving the coefficient is optimal at every point.

For $g(1)=1$ and finite displayed energy, the angular mode $g(r)\cos \theta$ has Dirichlet energy equal to $\pi$ times that integral. The harmonic function $r\cos \theta$ minimizes the energy with the same boundary data by the Dirichlet principle, giving the bound. Finite energy forces $g(r)\to 0$ as $r\to 0$: Cauchy–Schwarz bounds changes of $g^2$ by twice the product of the two energy tails, so $g^2$ has a limit, and a nonzero limit would make $\int g^2/r$ divergent. Consequently completing the square gives the exact identity ${\int }_0^1[r{g}^{\prime 2}+g^2/r]=1+{\int }_0^{1}r(g^{\prime }-g/r{)}^2$. Equality occurs exactly for $g(r)=r$. This identity also covers finite-energy functions whose polar expression is not classically smooth at the centre.

In the unit half-cylinder $z\ge 0$, specifying only the curved-side boundary values does not determine a bounded harmonic function. The displayed nonzero function, where $j_{0,1}$ is a positive zero of the Bessel function of the first kind $J_0$, is regular at the axis, bounded, harmonic, and zero on $r=1$. It may be added to any particular solution without changing the side data, and even decays at infinity. Boundary information at the base $z=0$ is still required for uniqueness.

For a Gaussian free field with boundary heights $-\lambda$ and $+\lambda$ on the two sides of the starting point, its distinguished zero-height interface has chordal ${\mathrm{SLE}}_4$ law. Conditional on an initial curve segment, the field is a zero-boundary field in the slit domain plus $m_t(z)=\lambda -(2\lambda /\pi )\arg (g_t(z)-W_t)$. Its mean bracket equals minus the variation of its conditional Green covariance. Subtracting the initial harmonic function gives the corresponding coupling to a zero-boundary field.

A solid harmonic extends a spherical harmonic to a harmonic function in space with either the regular radial power $r^n$ or the irregular radial power $r^{-n-1}$. In three dimensions these are the two separated radial solutions of the Laplace equation at angular degree $n$.

In three dimensions let $v\in H_0^1(B_R)$ solve $-\Delta v=f$, where $f\in L^{6/5}(B_R)$ is supported in $B_a$ and $0<a<R$. Write $M=\int f$ and $\bar{v}(s)=(4\pi {)}^{-1}{\int }_{S^2}v(s\omega )d\omega$. Outside $B_a$, the weak solution is a harmonic function. The weak formulation against radial test functions gives $-4\pi s^2{\bar{v}}^{\prime }(s)=M$ for $a<s<R$. The zero Sobolev trace gives $\bar{v}(R)=0$; integration proves the displayed formula, without assuming the source is a radial function.

Cauchy-Schwarz inequality on spheres gives $\Vert v{\Vert }_2^2\ge M^2(4\pi {)}^{-1}{\int }_a^R(1-s/R{)}^{2}ds$. For fixed nonzero $M$ and fixed $a$, this grows linearly in $R$. Such solutions on expanding balls cannot remain bounded in the full $H^1({\mathbb{R}}^3)$ norm after zero extension of H01.

The map $z\mapsto z^2$ is a bijective conformal map from the open first quadrant to the complex upper half-plane. In polar coordinates it doubles arguments from $(0,\pi /2)$ to $(0,\pi )$, and its derivative $2z$ never vanishes inside the quadrant. Its inverse is the branch of the square root with argument in $(0,\pi /2)$. By conformal invariance of harmonicity, a harmonic function $f$ on the half-plane pulls back to $g(x,y)=f(x^2-y^2,2xy)$. The two boundary rays map to the positive and negative real axes, respectively.

For subsonic motion $|V|<c$, an antiplane shear displacement $w(x_1-Vt,x_2)$ satisfies ${\beta }^2{w}_{xx}+w_{x_2{x}_2}=0$. The displayed rescaling makes it a harmonic function of $(x,y)$, so locally $w=\mathrm{Re}H(x+iy)$ for a holomorphic function $H$. Set $F=\beta H$. The Cauchy-Riemann equations then give $\mu F^{\prime }=\beta {\sigma }_{13}-i{\sigma }_{23}$. Keeping this potential normalization explicit avoids losing a factor of $\beta$ in the dynamic stress field. At sonic or supersonic speeds the elliptic harmonic reduction does not apply.

A locally integrable real function is superharmonic in the distributional sense if its Laplacian is nonpositive on every nonnegative test function. Equivalently its negative is a subharmonic function. The canonical lower semicontinuous representative satisfies the reverse ball mean inequality. A function both subharmonic and superharmonic has zero distributional Laplacian and a smooth harmonic function representative by the Weyl lemma.

In the absence of heat advection, an insulating sphere of radius $a$ in a uniform remote temperature gradient $\mathbf{H}$ has the displayed exterior harmonic function as its temperature field. The reflected dipole term makes ${\partial }_{r}T=0$ at $r=a$. The surface temperature is $T_0+(3/2)a\mathbf{H}\cdot \mathbf{n}$, so the tangential gradient is amplified by $3/2$. If the sphere has thermal conductivity ratio $K$ relative to its surroundings, temperature and heat-flux continuity instead give surface gradient factor $3/(2+K)$.

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.

For a harmonic function on the planar annulus $a<r<b$, the radial amplitude of a sine or cosine Fourier mode of positive integer index $n$ is a linear combination of $r^n$ and $r^{-n}$. If its values at $a,b$ are $A_n,B_n$, respectively, the displayed hyperbolic-sine formula gives that combination. It follows either by solving the two boundary equations or by putting $s=\log (r/a)$, which changes the radial equation to $R_{ss}-n^{2}R=0$. The zero angular mode is instead affine in $\log r$:Superposing the modes gives the harmonic function with suitable prescribed Dirichlet boundary conditions. For a fixed interior radius, high angular modes from either boundary are exponentially suppressed. Piecewise smooth boundary data are attained at continuity points, while jumps have the usual mean-value Fourier limit.

In two-dimensional vortex dynamics, solve the Poisson equation for the streamfunction with the prescribed boundary conditions. On the plane its fundamental solution is $(2\pi {)}^{-1}\log |\mathbf{x}|$, so differentiating the displayed inverse gives the planar Biot-Savart kernel. The harmonic function ${\psi }_h$ represents boundary and global-flow data. An additive constant changes no velocity; circulation around holes and a uniform flow are additional degrees of freedom if their boundary or far-field values are not prescribed. The same inverse acts independently on each level in nonrotating layerwise-two-dimensional vortex dynamics.

A spherical squirmer with tangential surface slip velocity $(I-nn)(A+Bn)$, where $B$ is a symmetric second-rank tensor and a traceless second-rank tensor, translates with $U=-2A/3$ and does not rotate in an unbounded quiescent fluid. With $S=x\cdot Bx$ its exterior Stokes flow isThis follows from the Unscaled Papkovich–Neuber representation using harmonic functions as potentials $\Phi =-a^{2}Bx/(2r^3)$ and $\chi =a^{3}A\cdot x/(3r^3)+a^{4}S/(2r^5)$. Matching the surface coefficients proves the formula. The two modes generate a potential dipole and a stresslet, respectively.

For planar Brownian motion started at $(x,y)$ with $y>0$, the probability that its first hit of the real axis lies in $(0,\infty )$ isThis is the bounded harmonic function in the upper half-plane with boundary values zero on the negative half-axis and one on the positive half-axis.

For a smooth harmonic function on a manifold without boundary, assume its nonzero superlevel sets are compact. On a regular positive superlevel set $\{U>\epsilon \}$, integration by parts gives $\int |\nabla U{|}^2=-\epsilon {\int }_{\partial \{U>\epsilon \}}|\nabla U|\le 0$. Applying the same argument to $-U$ proves $U=0$. This avoids assuming an unstated decay rate for the boundary flux at infinity.

Use a velocity potential to reconstruct an irrotational flow from one scalar field. A velocity potential $\varphi (\mathbf{x},t)$ satisfiesIt has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$ and is unchanged as a description of the velocity field when any function of time alone is added. For twice continuously differentiable $\varphi$, the identity curl of a gradient proves zero vorticity. In incompressible flow, taking the divergence also gives ${\nabla }^2\varphi =0$, the Laplace equation. This turns a three-component velocity field into a single harmonic function with appropriate boundary conditions. ^[ref]

For uniform flow $\mathbf{u}=(U,0,0)$, take $\varphi =Ux$. With $U=2\mathrm{m}{\mathrm{s}}^{-1}$, the change from $x=0$ to $x=3\mathrm{m}$ is $\Delta \varphi =6{\mathrm{m}}^2{\mathrm{s}}^{-1}$. By the fundamental theorem for line integrals, the line integral of velocity between these points equals that change along any path in the domain; a closed curve gives zero circulation.

For the potential line vortex, choose a local angle branch in polar coordinates and set $\varphi =K\theta$. Then $u_r={\partial }_r\varphi =0$ and $u_{\theta }=r^{-1}{\partial }_{\theta }\varphi =K/r$. One counterclockwise turn changes the continued velocity potential by $2\pi K$, although the velocity field returns to its original value. With $K=1{\mathrm{m}}^2{\mathrm{s}}^{-1}$, this change is $2\pi {\mathrm{m}}^2{\mathrm{s}}^{-1}$, exactly the circulation. A full annulus therefore admits local angle branches and a multivalued velocity potential, but no global single-valued velocity potential when $K\ne 0$. Cutting the annulus along a radial line allows a single-valued angle branch on the remaining domain. The obstruction is path independence, not a failure of the local gradient formula.

An element $h\in H_{\mathrm{loc}}^1(U)$ is weakly harmonic on an open set $U$ when ${\int }_U\nabla h\cdot \nabla \varphi =0$ for every test function $\varphi \in C_c^{\infty }(U)$. This is the statement $\Delta h=0$ in distributions. It requires no regularity of the domain boundary. The Weyl lemma, a form of elliptic regularity, identifies such an element with a smooth harmonic function.