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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.

TODO: biological interpretations and organism-specific propagation are user-deferred. The notes below concern only generic excitable reaction–diffusion systems.

A front solution joins different asymptotic states in space. In a reaction–diffusion system, local nonlinear dynamics select possible states while diffusion couples neighboring points. A travelling wave with profile $u(x,t)=U(\xi )$, $\xi =x-ct$, converts a partial differential equation into a boundary value problem for the shape and speed. A pulse instead returns to the same resting state on both sides and can contain leading and trailing fronts.

A useful nondimensional cubic fast-slow reaction-diffusion model, in a convention of the FitzHugh-Nagumo model, iswhere $D>0$, $\epsilon \ll 1$, and $\beta >0$. Here $u$ is a fast activation variable, $w$ a slow negative-feedback variable and $I$ a generic drive; no physiological variable is assigned. Without spatial coupling, nullclines are $w=f(u)+I$ and $w=u/\beta$. At $I=0$, the resting state $(0,0)$ has Jacobian matrixIt has linear stability. With appropriate parameter choices a sufficiently large perturbation makes a fast excursion to the other outer branch of the cubic, followed by slow feedback and return. Thus a stable rest state can be excitable without being linearly unstable. Other parameters or applied drive can yield multiple equilibria or oscillations; these behaviours should not be asserted for every parameter set.

The fast-subsystem front can be obtained explicitly by freezing $w=I=0$. For the bistable cubic reaction-diffusion equation, $u=0,1$ are stable and $u=\alpha$ is unstable. The decreasing logistic front of a bistable cubic equation isTo check it, $U^{\prime }=-U(1-U)/\sqrt{2D}$ and substitution into $D{U}^{\prime \prime }+c{U}^{\prime }+f(U)=0$ cancels every term. For $\alpha <1/2$, the high state advances to the right; for $\alpha >1/2$, the direction reverses. At $\alpha =1/2$ the front is stationary. More generally, multiplying the travelling-wave equation by $U^{\prime }$ and integrating yieldsThis links propagation direction to the bias between the two stable states and recovers Maxwell balance for a scalar reaction-diffusion front at zero speed. It is a bistable selected-speed problem, rather than a universal Fisher-type minimum-speed formula.

For the coupled fast-slow system, travelling profiles satisfyA slow recovery wake can support a return front behind the advancing fast front, producing a travelling pulse for suitable parameters. Its width reflects the slow recovery time and propagation speed, while each sharp edge has the faster diffusion–reaction scale. The frozen-recovery speed above is a controlled subsystem result, not an exact speed for every coupled pulse. A definite pulse speed requires parameters and connecting boundary condition; none are specified by this brief-note question.

Use SI units, uniform absolute permittivity $\epsilon$ and a neutral bulk reservoir. Local equilibrium of ideal mobile ions gives $n_i=n_{i,\infty }{e}^{-z_{i}e\psi /(k_{B}T)}$. Combining their charge with the Poisson equation yieldsThe reference is $\psi =0$ in the bulk. Boundary data specify fixed charge, fixed potential or a surface reaction/charge regulation law. These are physically different problems. The mean-field theory assumes point ions, thermal equilibrium and a continuum solvent; ion-size packing, dielectric structure and strong electrostatic correlations can invalidate it.

For a symmetric monovalent electrolyte let $n_{\infty }$ count each sign and define $u=e\psi /(k_{B}T)$, ${\kappa }_D^2=2e^2{n}_{\infty }/(\epsilon k_{B}T)$. Outside fixed sources,For $|u|\ll 1$, the Debye-Hückel approximation is ${\nabla }^{2}u={\kappa }_D^{2}u$. A planar wall bounding electrolyte in $z>0$ has $\psi ={\psi }_0{e}^{-{\kappa }_{D}z}$; assuming zero field on the other side, its charge is $\sigma =\epsilon {\kappa }_D{\psi }_0$. A sphere of radius $a$ and total charge $Q$ instead givesThe factors arise from geometry and boundary conditions; planar screening cannot be substituted unchanged for a small sphere. The weak-potential condition must hold where the potential is largest, particularly at the surface.

The nonlinear planar Poisson-Boltzmann profile is also exactly integrable. Multiply $u^{\prime \prime }={\kappa }_D^2\sinh u$ by $u^{\prime }$ and use $u,u^{\prime }\to 0$ at infinity:The sign selects a profile decaying to the bulk for either sign of surface potential. Integration givesFor weak $u_0$ this reduces to the linear solution. Strong surface potentials have a bounded far-field amplitude $4\gamma$, while the near-wall ion density still grows strongly. There is no need to permit the negative ion densities produced by an invalid extrapolation of a linearized density formula.

The same first integral yields the Grahame equation and its diffuse-layer differential capacitance per unit area:These describe the diffuse layer. A compact interfacial layer adds a voltage drop and can place its capacitance in series with $C_D$; a measured electrode capacitance need not equal the ideal diffuse contribution.^[ref]

For comparison, a planar counterion-only double layer next to a negative plane has positive counterions and no salt reservoir. Its profile is algebraic:Integrating $en(z)$ over $z>0$ gives $|\sigma |$, checking charge neutrality. This problem cannot be obtained by putting a finite reservoir density into the symmetric-electrolyte answer without changing its boundary conditions.

Applications include screening around synthetic colloids, diffuse-layer electrode capacitance and forces between charged plates. In planar geometry, the bulk-subtracted normal stress is constant:At the midpoint of identical plates, symmetry sets $E=0$, so the remaining excess ionic osmotic pressure determines their repulsion. A comparison with force or capacitance measurements should vary salt and surface potential independently, test charge regulation and adsorption, and distinguish compact-layer effects from failures of diffuse screening. Nonlinear Poisson-Boltzmann theory retains finite-potential Boltzmann populations, but still omits correlations and finite-ion-volume saturation.

Consider generic synthetic filaments and films. buckling occurs when compression makes a transverse deformation lower the energy. For a straight filament of bending rigidity $B=EI$ under compressive force $P$,Pinned ends admit $y=A\sin (\pi s/L)$, giving ${\delta }^{2}F=A^{2}L[B(\pi /L{)}^4-P(\pi /L{)}^2]/4$. The first instability is therefore $P_c={\pi }^{2}B/L^2$. True clamping raises this to $4{\pi }^{2}B/L^2$; a clamped-free rod has ${\pi }^{2}B/(4L^2)$. boundary conditions are part of the threshold, not a minor correction. Above threshold geometric nonlinearities and end constraints determine the buckled amplitude.

For buckling of a synthetic film, let the bending modulus be $\kappa$ and the isotropic in-plane tension be $\sigma$. The small-slope quadratic energy isA mode of wave number $q$ has stiffness $\kappa q^4+\sigma q^2$. tension stabilizes it; compression corresponds to $\sigma <0$ and makes modes with $q^2<|\sigma |/\kappa$ unstable. A finite sheet has a smallest allowed $q$, giving a finite threshold. A supporting foundation adds positive stiffness and can select a nonzero wrinkle wavelength.

Approaching either threshold softens a fluctuation mode: its thermal variance is proportional to $k_{B}T$ divided by modal stiffness. The apparent quadratic-theory divergence is rounded by nonlinear geometry, finite size and imperfections. Experimentally vary end shortening or load, measure the transverse profile, and control attachment stiffness and initial curvature. Load control and displacement control need not trace the same postbuckling response. Biological applications are outside this generic mechanics note.

A dead-load traction is prescribed force per unit reference area whose vector is independent of the deformation. It is imposed through $T_{R}N=t_R$ on the reference boundary. It does not rotate or rescale with the current surface as a follower load would.

Elastic incompressibility imposes $\det F=1$ and adds a scalar Lagrange multiplier to the constitutive law. With the pressure sign convention used here,The pressure field is determined by equilibrium and boundary conditions, rather than by the strain energy density alone.

For the diagonal homogeneous deformation, $F=\mathrm{diag}(v_1,v_2,v_3)$ and $v_1{v}_2{v}_3=1$. Isotropy makes $DW(F)$ diagonal here, with entries ${\Phi }_i$, soThe six face conditions are exactly $T_R{e}_i=\tau e_i$. Thus ${\Phi }_i-p/v_i=\tau$ for each $i$, orNecessity follows from the boundary conditions. Conversely these equations give a constant pressure and constant stress satisfying those conditions, while $\mathrm{Div}{T}_R=0$ because the stress is constant and there is no body force. This proves sufficiency as well.

For the particular energy, ${\Phi }_i=\mu (v_i-v_i^{-2})$. Put $S=v_1+v_2+v_3$ and $Q=v_1{v}_2+v_1{v}_3+v_2{v}_3$. Elastic incompressibility gives $v_i^{-1}=v_j{v}_k$, henceEquality for every pair is therefore equivalent to $(\mu S-\tau )(v_i-v_j)=0$. If all stretches are equal, their positive product one gives $v_1=v_2=v_3=1$, with $p=-\tau$. Otherwise $S=\tau /\mu$. Conversely any positive triple with this sum and product one solves the equations, with $p=-\mu Q$. Thus the complete solutions within the stated homogeneous diagonal ansatz are the unit triple and the stated sum-product family.

By the arithmetic-geometric mean inequality, $S\ge 3$, with equality only at the unit triple. Consequently $\tau \le 3\mu$ admits only that triple, including equality, when the two descriptions coincide.

To prove infinitely many for $s=\tau /\mu >3$, choose $v_1=t$ near one and setAt $t=1$ the discriminant is $(s-1{)}^2-4>0$; it stays positive on an interval about one, and both roots remain positive. Every such $t$ gives a different admissible triple. HenceThe degeneracy has an energy explanation: under the product constraint the strain energy density is $\mu S^2/2$, and the total potential under dead loading is $\mu S^2/2-\tau S$. It depends on the three stretches only through their sum. Its minimum over attainable $S\ge 3$ moves from $S=3$ to the whole level set $S=\tau /\mu$ when the load exceeds $3\mu$.

For a radial function $f(r)$ in three dimensions, the Laplacian is $f^{\prime \prime }+2f^{\prime }/r$, while the imposed radial drift contributes $-\alpha f^{\prime }/r$. Thus the radial generator of an SDE diffusion isThe hitting probability for the inner sphere is harmonic for this diffusion generator in the shell, with value one on the inner sphere and zero on the outer sphere. At $\alpha =1$, the boundary value problem is thereforeIntegrating $(r{p}^{\prime }{)}^{\prime }=0$ gives $p(r)=A\log r+B$. The two boundary conditions determineBoth arguments of the logarithms are dimensionless. The result decreases from one to zero across the shell and lies strictly between zero and one for $R_1<r_0<R_2$.

For a direct probability interpretation, stop at the first contact $T$ with either sphere. The stopped process $p(r(t\wedge T))$ is a bounded martingale by Itô's formula. The shell has positive inner radius, so its drift is bounded and its diffusion is nondegenerate; exit occurs almost surely. Optional stopping and dominated convergence therefore giveThe logarithm appears because the inward drift cancels one of the two geometric $f^{\prime }/r$ terms in the three-dimensional radial Laplacian, leaving the same radial diffusion generator as a planar Brownian motion with diffusion coefficient one.

Two-dimensional incompressibility permits $u={\psi }_y$, $v=-{\psi }_x$. Taking the curl of the Stokes momentum equation eliminates pressure and gives ${\nabla }^2(v_x-u_y)=0$. Since $v_x-u_y=-{\nabla }^2\psi$, the stream function obeys $\boxed{{\nabla }^4\psi =0}$.

Scale radius by $a$, velocity by $U$ and stream function by $Ua$. In polar coordinates $u_r=r^{-1}{\psi }_{\theta }$, $u_{\theta }=-{\psi }_r$. No penetration makes $\psi$ constant on the cylinder; choose that constant zero. No slip also requires ${\psi }_r=0$ at $r=1$. The uniform stream has $\psi =r\sin \theta$, specifying the far-field condition.

The boundary forcing has angular mode $\sin \theta$, so its corresponding mode must have $\psi =f(r)\sin \theta$. Define $L_1=d^2/d{r}^2+r^{-1}d/dr-r^{-2}$. Then ${\nabla }^4\psi =(L_1^{2}f)\sin \theta$, while $L_1{r}^s=(s^2-1)r^{s-2}$. Thus the indicial equation of $L_1^2$ is $(s^2-1)[(s-2{)}^2-1]=(s-3)(s-1{)}^2(s+1)=0$, givingThe repeated exponent one supplies the logarithmic term. The condition $f^{\prime }\to 1$ first forces $A=B=0$ and $C=1$. Then $f(1)=0$ forces $D=-1$, but $f^{\prime }(1)=C-D=2\ne 0$. No additional Fourier mode can repair the missing boundary condition in this angular mode. Therefore the unbounded-domain Stokes problem has no solution with both no slip and a nonzero uniform far stream.

This is Stokes' paradox. It identifies a nonuniform approximation, rather than the absence of physical flow. For any small but nonzero Reynolds number, convective inertia becomes important sufficiently far away. An inner Stokes region must be matched to an outer Oseen or Navier-Stokes region that retains that inertia. Alternatively, finite external boundaries introduce a different well-posed physical problem; they cannot be sent to infinity while keeping the naive Stokes solution unchanged.

Take the tangential length and velocity scales as order one. In the boundary layer, continuity reads $u_x+({\delta }_2/{\delta }_1)V_Y=0$, so a nontrivial balance requires ${\delta }_2\sim {\delta }_1$. Tangential inertia is order one, and the normal viscous term has size ${\mathrm{Re}}^{-1}{\delta }_1^{-2}$. Balancing them givesup to constants absorbable in the scaled variables. Streamwise diffusion is then smaller by ${\mathrm{Re}}^{-1}$. The normal momentum equation has a pressure term of size ${\delta }_1^{-1}{p}_Y$ and inertial/viscous terms of size ${\delta }_1$, hence $p_Y=0$ at leading order. The remaining Prandtl boundary-layer equations areThe boundary conditions are no slip and no penetration at the plate, and matching to the outer tangential velocity:The outer inviscid tangential momentum equation supplies $p_x=-U_s{U}_s^{\prime }$; pressure is shared between the boundary layer and outer flow because it has no leading normal variation.

Use autonomous third-order reduction using the dependent variable. On a monotone branch where $f$ can serve as the independent variable, the chain rule givesTherefore the reduced second-order equation isThe wall and matching conditions becomeFor the usual positive outer flow this is the limit $f\to +\infty$. The reduced equation is singular at $G=0$: if $f^{\prime \prime }(0)=b\ne 0$, then $f\sim b{\eta }^2/2$ and $G\sim b\eta$, so $G$ has square-root behavior as a function of $f$ near the wall. A finite value for $G_f(0)$ should not be imposed as an extra boundary condition.

Linearize about a flat quiescent surface. The product ${\varphi }_x{\eta }_x$ and the kinetic-energy term are quadratic, and the boundary conditions can be evaluated at $z=0$ to first order. Thus ${\eta }_t={\varphi }_z$ and ${\varphi }_t+g\eta =0$ there. Fourier transformation in $x$ changes ${\partial }_x^2$ into multiplication by $-k^2$, givingThe transform uses $e^{-ikx}$ in its forward definition; the inverse transform therefore uses $e^{ikx}/(2\pi )$. The zero-wavenumber potential has an irrelevant spatially constant gauge freedom.

Adding a constant to ${\psi }_y$ changes neither its time nor space derivatives, so the action is invariant. The associated Noether density and flux areThe wave equation gives ${\partial }_t{j}_y^0+{\partial }_x{j}_y^x=0$. ThereforeThis is total physical transverse momentum. It is conserved for a whole isolated infinite string with vanishing boundary flux, periodic boundary conditions, or endpoints exerting no net transverse force. For a finite subsystem momentum can flow through its ends, and fixed supports can exchange momentum with the string. Thus the local conservation law always holds in this model, while constancy of the integrated charge requires the stated boundary condition.

With endpoints held at zero displacement but free to rotate, variation permits arbitrary endpoint $\delta {\psi }_x$ and gives the natural zero-moment condition ${\psi }_{xx}=0$ there. These are simply supported endpoints. The normal modes areThe first mode becomes unstable when $F+B(\pi /l{)}^2<0$, soEquivalently the compressive load exceeds $|F{|}_{\mathrm{crit}}={\pi }^{2}B/l^2$. At equality the linear mode is neutral and nonlinear terms are needed to determine its post-buckling amplitude. The quadratic energy of $A\sin (\pi x/l)$ is $A^{2}l[F(\pi /l{)}^2+B(\pi /l{)}^4]/4$, giving the same sign test.

The endpoint displacement condition alone does not impose clamping. The quoted threshold uses the natural free-slope boundary condition of the stated variational problem. If rotations were also fixed, the admissible modes and threshold would change; one must not import a clamped-end formula into this result.

For uniform unconstrained boundary conditions, the positive gradient coefficient favors a spatially constant minimizer. Put $u=|\mathbf{m}{|}^2\ge 0$. The remaining free-energy difference is $a(T-T_c)u+b{u}^2$. Its minimum isBelow $T_c$ the amplitude is fixed but its direction is arbitrary; the set of uniform minima is a two-sphere. The free-energy minimum is $f_0-a^2(T_c-T{)}^2/(4b)$ below $T_c$ and $f_0$ above. The order parameter vanishes continuously, so this is a continuous, second-order transition within Landau theory.

The free energy has internal $O(3)$ spin-rotation/reflection symmetry; a chosen nonzero vector leaves an $O(2)$ subgroup invariant. Restricting to proper rotations, this is $SO(3)\to SO(2)$ spontaneous symmetry breaking. There are two transverse orientation directions and a massive amplitude direction below the transition. Spatially varying directions cost gradient energy, and imposed boundaries or defects can prevent a uniform minimizer. In strictly two-dimensional equilibrium with short-range interactions, the fluctuation caveat of part (d) still applies: minimizing this polynomial functional at mean-field level does not overturn the absence of finite-temperature spontaneous magnetization.

Differentiation gives $f^{\prime }(m)=2m(b+2d{m}^2)$. For $b>0$ the origin is the global minimum; for $b<0$, $m=\sqrt{-b/(2d)}$ has $f^{\prime \prime }=-4b>0$ and $f=-b^2/(4d)<f(0)$. Therefore, close enough to the transition for the quartic expansion to apply,The order parameter approaches zero continuously with mean-field critical exponent $\beta =1/2$: the transition is continuous. In the aligned phase a particular direction is chosen. The vector-spin rotation symmetry is spontaneously broken from $SO(3)$ to $SO(2)$, or from $O(3)$ to $O(2)$ including reflections. The manifold of vector order parameter directions is a sphere, so this is not just a two-state sign choice.^[ref]

The plot uses illustrative $d=1/2$ and temperature measured in units of $T_c$. A unit-spin average must satisfy $m\le 1$; if the quartic model is used literally far below the transition, its constrained minimum is $\min \{1,\sqrt{(T_c-T)/(2d{T}_c)}\}$. In practice the near-transition expansion is not a controlled prediction of deep low-temperature saturation. Also, a finite zero-field statistical ensemble that averages all directions has $⟨\mathbf{m}⟩=0$ even below $T_c$; the curve describes the magnitude in a selected ordered state, obtained by appropriate boundary conditions or the thermodynamic limit before removing a small magnetic field.

Substitute $\omega ={\alpha }_3/\epsilon +\cdots$ into the cubic denominator. Its leading terms areThe nonzero large root obeys $1+i{\alpha }_3=0$, soThe sum of the three roots is exactly $i/\epsilon$. Combining this with the two slow-root expansions gives the refinement ${\omega }_3=i/\epsilon +i\epsilon +O({\epsilon }^3)$.

For physical radiative sign $\epsilon >0$, the fast pole lies high in the upper half-plane. Its homogeneous mode is $e^{-i{\omega }_{3}t}\simeq e^{t/\epsilon }$, a runaway. It prevents a bounded retarded response with the real-axis Fourier prescription even though the two slow poles are damped. If $\epsilon <0$, the fast pole decays but the slow pair grow. No nonzero real sign puts every pole in the lower half-plane.

There are two distinct boundary choices. A strictly causal initial-value Green function exists and includes the growing mode for $\epsilon >0$; it must be represented by a Laplace/Bromwich contour above that pole rather than the stated ordinary Fourier integral. Alternatively the real-axis inverse gives a bounded stationary no-runaway response and preacceleration: for $\tau <0$, upper closure encloses the fast pole and yields a nonzero response before the force. Its leading contribution isThis follows from the upper-pole residue $\mathrm{Res}G\simeq -i\epsilon$ and the factor $i$ from upper contour closure. Eliminating runaway growth by a future boundary condition therefore sacrifices strict retardation on the short scale $\epsilon$.

The singular root also shows why a perturbative expansion around only $\omega =\pm 1$ misses the problem. For a low-frequency effective description, order reduction of a radiation-reaction oscillator uses $\stackrel{...}{x}\simeq \dot{f}-\dot{x}$ in the small correction and gives $\ddot{x}+\epsilon \dot{x}+x=f+\epsilon \dot{f}$. For positive $\epsilon$ this approximate second-order equation has a causal damped response and discards the fast mode outside its intended regime; it does not make the original third-order equation simultaneously bounded and retarded.

The half-line square-root potential well is confining for $a>0$, which is assumed here. The hard wall imposes $\psi (0)=0$ and the smooth turning point is $x_{\mathrm{t}}=(E/a{)}^2$. With $u=a\sqrt{x}/E$, the WKB action becomesThe hard-wall and soft-turning-point WKB quantization condition therefore givesThe coefficient $a$ here denotes the potential strength, not the integration limit in the generic quantization formula.

The WKB approximation is reliable in the interior where $|k^{\prime }|/k^2\ll 1$, but fails near the soft turning point where $k\to 0$; a connection formula supplies its phase. The hard wall also requires its boundary condition, and $V^{\prime }=a/(2\sqrt{x})$ makes the local slow-variation condition fail sufficiently close to zero. Higher-energy states have a larger action relative to $\hslash$, with smaller relative endpoint regions, so the leading WKB spectrum is generally more reliable at large $n$ than for the ground state.

For $b>0$, the exponential variational trial for a square-root half-line well satisfies the wall boundary condition and has norm ${\int }_0^{\infty }{x}^2{e}^{-2bx}dx=1/(4b^3)$. integration by parts has no boundary term and givesThe supplied gamma integral givesSetting $E^{\prime }(b)=0$ yields the unique minimum,At the minimum $⟨V⟩=4⟨T⟩$, so the variational ground-state estimate isThe energy diverges as $b\to 0$ or $b\to \infty$, confirming that the stationary point minimizes the trial energy.

Keep the same measure normalization as in the free generating functional and defineHere $W$ is the source-dependent partition function, despite the common alternative convention in which that letter denotes its logarithm. A functional derivative of $e^{\int Jx}$ brings down one factor $x(\tau )$ at the differentiated point. Expanding the interaction exponential and replacing each such factor by $\delta /\delta J(\tau )$ therefore givesCompleting the square in the free Gaussian integral gives $W_0[J]=W_0[0]\exp (\frac{1}{2}\int J{G}^{(0)}J)$ because $G^{(0)}=D^{-1}$ with periodic boundary conditions. Restoring $\hslash$ replaces the upper limit by $\hslash \beta$ and the interaction coefficients by $\lambda /(4!\hslash )$ and $\mu /(6!\hslash )$, exactly as in the source.

Choose a coarse-grained magnetization $m(\mathbf{x})$ as the order parameter: it vanishes in the paramagnetism phase and develops a spontaneous nonzero value in the ferromagnetism phase. In zero applied magnetic field, spin reversal sends $m\mapsto -m$, so the local Landau free energy contains only even powers. A minimal spatially inhomogeneous free-energy functional isThe positive gradient coefficient penalizes rapid spatial variations. Near an ordinary transition take $a(T)=a_0(T-T_0)$, $a_0>0$, and positive $b$; retain the sextic term when $b$ can approach or cross zero. The source $h$ is conjugate to the magnetization and explicitly breaks spin reversal.

Determine equilibrium by minimizing this free-energy functional, imposing the appropriate boundary conditions. Its Euler-Lagrange equation isFor a uniform system at $h=0$, compare the values of all stable stationary solutions, rather than accepting every solution of this equation. The selected global minimum gives the equilibrium magnetization; derivatives of its free energy give entropy, heat capacity and magnetic susceptibility. At an ordinary transition the two ordered minima emerge continuously from zero. Spatial dependence also permits domain walls and an estimate of the correlation length.

This is a mean-field approximation: it selects a minimizing profile and neglects statistical fluctuations around it. The symmetry constrains the expansion, but its numerical coefficients are phenomenological. Close enough to a transition, thermal fluctuations near a critical point can invalidate the resulting mean-field critical exponents.^[ref]

For $H=0$ and $a<0$, the free energy is a symmetric double well. The central stationary point is a maximum; the two stable minima areThese are the magnetisations of the two symmetry-broken phases, selected by an infinitesimal field or a boundary condition. An exactly symmetric finite-system ensemble can have $⟨M⟩=0$ by averaging both wells; that does not make the central maximum a stable homogeneous phase.