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Complex differentiability gives $u_x=v_y$, $u_y=-v_x$. Differentiating once more, for continuously differentiable first derivatives with the needed second derivatives, yields $u_{xx}+u_{yy}=v_{yx}-v_{xy}=0$ and $v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0$. Thus both components satisfy Laplace's equation. The result is local on the domain of holomorphy; it makes no assertion across a branch cut.

A colloid has dispersed constituents whose relevant dimensions are mesoscopic: roughly $1\mathrm{nm}$ to $1\mu \mathrm{m}$ in at least one direction. Interfaces and Brownian motion are consequently important. The size range is a useful convention, not a sharp boundary separating all physical behaviour. A sol of fine silicon dioxide particles in water illustrates solid in liquid; an oil-in-water emulsion illustrates liquid in another immiscible liquid; and a foam illustrates gas dispersed in a liquid, with thin intervening films supplying the small structural dimension. A colloid need not be a thermodynamic equilibrium state.^[ref]

Unprotected particles attract through Van der Waals forces. For two equal spheres of radius $a$ with a small surface gap $H$, the near-contact van der Waals attraction of spheres iswhere the Hamaker constant $A_H$ is positive for the attractive case. The large number of interacting constituent molecules can give a well much deeper than $k_{B}T$, even though individual molecular attractions are weak. Brownian motion brings particles together, and a sufficiently deep attractive well makes separation unlikely. This gives particle aggregation; solid particles can remain distinct within a cluster rather than merging into one continuous solid. Emulsion droplets can additionally coalesce if the separating film ruptures.

Charge stabilization of colloids supplies a repulsive barrier. Like-charged surfaces carry compensating counterions and electric double layers; bringing the layers together costs free energy. DLVO theory combines this repulsion with Van der Waals forces. A barrier large compared with $k_{B}T$ can make aggregation slow, even when sticking would lower the final free energy. Adding salt generally shortens the Debye–Hückel screening length and can lower the barrier; surface charge regulation and specific ion interactions also matter. A negatively charged particle and a positively charged particle need not repel.

Steric stabilization instead coats each particle with adsorbed or grafted polymer chains. In a good solvent, overlap and compression of these layers raise the osmotic pressure and reduce chain entropy, opposing close contact. Electrosteric stabilization combines the two mechanisms. Dense, securely bound layers matter: incomplete adsorption can allow polymer bridging, while nonadsorbing dissolved polymers can create an attractive depletion force. Poor solvent quality can collapse a protective layer. Reducing the density mismatch slows sedimentation but does not remove the interparticle attraction, so uniform suspension and resistance to aggregation are different requirements.

The left panel is an original qualitative potential with arbitrary coefficients: it illustrates a barrier and its reduction by shorter-ranged repulsion, without fitting an experimental DLVO theory curve or extrapolating the continuum attraction to molecular contact. The right panel plots the derived quasi-steady concentration profile in dimensionless variables.

In diffusion-controlled growth, attachment is rapid and delivery through the surrounding medium is the bottleneck. The growing surface depletes the neighboring solute, so a substantial concentration gradient develops. In reaction-controlled growth, arrival is fast compared with incorporation; many arrivals fail to attach, and the surface concentration stays near the reservoir value. The analogous distinction for particle aggregation is between near-unit sticking probability in diffusion-limited cluster aggregation and barrier-suppressed sticking in reaction-limited cluster aggregation. Here the subsequent calculation concerns accretion onto one growing particle, rather than a collision kernel for two mobile clusters.

The printed time-dependent equation is the diffusion equation, not the usual Poisson equation. For constant diffusion coefficient $D$, no bulk sources and negligible advection, Fick's laws give the number flux $\mathbf{j}=-D\nabla c$. Local conservation of mass, expressed here as conservation of the accreting constituent, then givesAn arbitrary time-independent exterior profile does not solve the full moving-boundary problem exactly. The required elementary result uses a quasi-steady approximation: diffusion establishes the exterior profile much faster than the particle radius changes. At each instantaneous radius $a$, set $\partial c/\partial t\simeq 0$. Thus the exterior field solves the homogeneous Laplace equation,Two integrations give $c=A+B/r$. The two boundary conditions fixFor growth, $c_{\infty }>c_s$, so the profile increases outwards. Its radial flux is negative, and the positive inward arrival rate isA material density is needed to turn an arrival rate into a radial velocity. Introduce the constant constituent number density $c_p$ inside the particle, using the same units as $c_s,c_{\infty }$. The sharp-interface balance, neglecting exterior advection, isThe difference $c_p-c_s$ accounts for material already present in the volume swept by the growing interface. In the usual dilute-feed approximation $c_s\ll c_p$, it can be replaced by $c_p$. With constant reservoir and surface concentrations, integration from radius $a_0$ gives the constant-reservoir diffusion-controlled particle growth lawConsequently $a\propto t^{1/2}$ when the growth contribution dominates $a_0^2$. For a finite seed at the literal limit $t\to 0$, the increment $a-a_0$ is instead initially proportional to $t$. Here “early time” means before appreciable reservoir depletion or interactions with other particles, while already beyond the rapid initial establishment of the diffusion field. Indeed, the ratio of diffusion-adjustment time $a^2/D$ to radial-growth time $a/\dot{a}$ is $(c_{\infty }-c_s)/(c_p-c_s)$; small supersaturation relative to the particle density makes the quasi-steady approximation self-consistent. No initial concentration field was specified, so an exact transient solution is not uniquely determined.

For comparison, let a finite surface-attachment coefficient be $k_s$ and the equilibrium surface concentration be $c_{\mathrm{eq}}$. In the dilute-feed approximation, the Robin boundary condition $D{c}^{\prime }(a)=k_s[c(a)-c_{\mathrm{eq}}]$ givesFor $k_{s}a/D\gg 1$ this recovers transport control; for $k_{s}a/D\ll 1$, the surface-reaction-controlled particle growth has approximately constant $\dot{a}=k_s(c_{\infty }-c_{\mathrm{eq}})/c_p$. Thus reaction control gives $a-a_0\propto t$, in contrast to the diffusion-controlled square-radius law. No numerical $c_p$ or $k_s$ is supplied or needed for these scaling results.

Subtract the particular plane Poiseuille flow profile from the full rectangular duct flow. The remainder is a harmonic function:The data are even in $x$ and $y$. The boundary value problem is unique: if two solutions differed by $v$, Green's first identity and the zero Dirichlet boundary condition would giveThus $v$ is constant, and its zero boundary values force $v=0$. Reflection of a solution in either coordinate axis therefore gives the same solution. We may seek even separated eigenfunctions.

Use separation of variables withEach summand solves the Laplace equation, vanishes at $y=\pm b$, and has the required evenness. The hyperbolic cosine ratio equals one at either side wall. The functions $\cos (k_{n}y)$ have orthogonality relationThey are the even Dirichlet eigenfunctions on $[-b,b]$, so their Fourier series represents the even boundary data. Projecting that data onto the $n$th eigenfunction yields the Fourier coefficientFor completeness, two integrations by parts give the quadratic arch expansion in half-integer cosine modes: the displayed integral is $4(-1{)}^n/k_n^3$, agreeing with the identity supplied in the paper.

Adding the particular solution givesThe coefficients are absolutely summable and the hyperbolic cosine ratio is at most one throughout the closed rectangle, giving uniform convergence and the prescribed boundary values. On compact subsets of the interior, the hyperbolic cosine ratio supplies exponential decay in $n$, justifying differentiation of the series and the Laplace equation. The side-wall correction slows the plane Poiseuille flow enough to satisfy all four no-slip boundary conditions.

A regular solid harmonic is $r^l{Y}_l^m(\theta ,\varphi )$: it is homogeneous of degree $l$ and solves the Laplace equation. Thus $\mathbf{r}\cdot \nabla F=lF$ and ${\nabla }^{2}F=0$. These identities turn angular spherical harmonic expansions of displacement, density and Newtonian gravitational potential into radial equations.

For two dilute spherical species with absorbing contact and negligible interaction barriers, relative diffusion has coefficient $D=D_1+D_2$ and contact radius $R=a_1+a_2$. The steady concentration outside one target is $c(r)=c_{\infty }(1-R/r)$, solving the radial Laplace equation with zero contact concentration. Its inward diffusive flux is $4\pi DR{c}_{\infty }$, proving the displayed bimolecular rate coefficient. For identical particles, converting this encounter coefficient into a rate of loss of cluster number additionally needs the factor that prevents counting the same pair twice.

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 $n$.

For a graph $z=x+i\nabla f$ in ${\mathbb{C}}^3$, symmetry of the Hessian matrix makes the graph Lagrangian. Pulling back $\Omega =d{z}^1\wedge d{z}^2\wedge d{z}^3$ gives $\det (I+i{D}^{2}f)$. Its imaginary part is $\mathrm{tr}{D}^{2}f-\det D^{2}f$, while its real part is $1-{\sigma }_2(D^{2}f)$. Vanishing of the imaginary part and positivity of the real part give the displayed equation and oriented branch. The equation is nonlinear even though its linearization near a flat graph is the Laplace equation.

// Target: string-theory.bigb

For viscous fluid inside a circular fluid free boundary between parallel plates of separation $h$, the depth-averaged Hele-Shaw flow satisfies $\bar{u}=-K\nabla p$, $K=h^2/(12\mu )$. With volume injection rate $Q$, ${\bar{u}}_r=Q/(2\pi hr)$ and $R^2=R_0^2+Qt/(\pi h)$. Neglecting surface tension and exterior viscosity, pressure is constant at the interface. Perturb its radius by $a_n\cos n\theta$ and solve the interior Laplace equation with a regular pressure term proportional to $r^n\cos n\theta$. The linearized pressure boundary condition gives $\delta p(R)=a_{n}Q/(2\pi hKR)$, hence velocity perturbation $-n{a}_{n}Q/(2\pi h{R}^2)$. The displaced base velocity adds another $-a_{n}Q/(2\pi h{R}^2)$, proving the displayed growth law. Injection stabilizes the circular blob; suction destabilizes it. An exact radial solution can therefore exist while giving a fragile experimental prediction. These signs concern viscous fluid inside the blob, rather than the opposite geometry of a low-viscosity injected bubble.

For an axisymmetric flow without azimuthal velocity, the spherical Stokes streamfunction gives $u_r={\Psi }_{\theta }/(r^2\sin \theta )$ and $u_{\theta }=-{\Psi }_r/(r\sin \theta )$. These formulas enforce incompressibility. The only possibly nonzero component of vorticity isThus the displayed equation is precisely the condition for irrotational flow, away from coordinate poles, with limits interpreted through regularity of the physical velocity field. It differs from the scalar spherical Laplace equation: $\Psi$ is a Stokes streamfunction, not a velocity potential.

For an inviscid flow of constant mass density $\rho$ and depth $h$, impose excess pressure $p_1\cos (k(x-ct))$ with $k>0$ on a free surface. Neglect surface tension. A travelling solution of the linearized free-surface boundary conditions isThe velocity potential satisfies the Laplace equation and the bed no-penetration boundary condition. Its surface time derivative is $-A{c}^{2}k\coth (kh)\cos (k(x-ct))$, so the dynamic condition gives the displayed signed response. The nonnegative amplitude is $|A|$. It is finite except atthe squared phase speed of a free surface gravity wave at that wavenumber. Equality produces resonance. This describes a travelling forced component; free surface gravity waves must be added to meet general initial conditions. At exact resonance, a solution of the initial value problem generally grows with time rather than having a bounded periodic amplitude.

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.