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Let $\mathbf{e}=\mathbf{X}/R$, $s=\mathbf{e}\cdot \mathbf{Be}$, and $A_{\parallel }=\mathbf{A}\cdot \mathbf{e}$. Applied forces and couples below are forces and torques on the solid; by force balance their values are also the strengths exerted by that solid on the fluid. This fixes the signs in Faxén's first law and Faxén's rotational law. To hold the passive sphere fixed, set its translation and angular velocity to zero in those laws. The incident two-mode tensorial squirmer flow givesOnly the stresslet part has vorticity:Consequently, for fixed $\mathbf{A},\mathbf{B}$ as $a/R\to 0$,The leading force is the radial stresslet force. Its next correction, $O(\mu a^4|\mathbf{A}|/R^3)$, comes from the incident potential dipole. The next $\mathbf{B}$ correction, $O(\mu a^5\Vert \mathbf{B}\Vert /R^4)$, contains both the swimmer's $r^{-4}$ field and the Laplacian term in Faxén's first law. If the relevant coefficient vanishes, these next terms must be retained instead of calling the vanished term a nonzero leading approximation. Further exchanges in the method of reflections for Stokes flow are smaller still.

The passive sphere reflects primarily a Stokeslet with strength ${\mathbf{F}}_h$. At the swimmer, its velocity is $(\mathbf{I}+\mathbf{ee}){\mathbf{F}}_h/(8\pi \mu R)$. The unchanged surface slip velocity adds the free swimming velocity to the incident-flow contribution in Faxén's first law. ThusThe leading change is $O(a^3\Vert \mathbf{B}\Vert /R^3)$; its next correction is the displayed $O(a^4|\mathbf{A}|/R^4)$ term. Finite-radius Faxén's first law corrections, the passive sphere's reflected stresslet, and its rotlet contribute at order $a^5\Vert \mathbf{B}\Vert /R^5$ to translation.

For a Stokeslet at the origin,The curl of the first term is $\mathbf{F}\times \mathbf{x}/r^3$; that of the second is also $\mathbf{F}\times \mathbf{x}/r^3$, before multiplying by $1/(8\pi \mu )$. Hence ${\mathbf{\omega }}_F=\mathbf{F}\times \mathbf{x}/(4\pi \mu r^3)$. At the swimmer use displacement $-\mathbf{X}$. The leading $\mathbf{B}$ force is parallel to $\mathbf{X}$, so its Stokeslet has zero vorticity there. This eliminates the putative $O(a^3\Vert \mathbf{B}\Vert /R^4)$ rotation. The potential dipole itself has no vorticity, but the holding force it induces is not radial. Applying Faxén's rotational law to its reflected Stokeslet yieldsThe passive sphere's rotlet and reflected stresslet can affect that last order, so they cannot restore the larger missing rotation. When $\mathbf{A}\times \mathbf{X}=0$, the boxed coefficient is zero and the higher-order terms determine any rotation.

Take $h=2k_c$ and $D=d/dz$. Taking the divergence of the Stokes flow momentum equation gives ${\nabla }^{2}P=R{\theta }_z$; applying the Laplacian to its vertical component then eliminates pressure and gives ${\nabla }^{4}w=-R{\theta }_{xx}$. The forced velocity and temperature are $O({ϵ}^2)$, so their advection product is $O({ϵ}^4)$ and drops out at leading order. The steady heat equation reduces to $w_0+{\nabla }^2{\theta }_0=0$.

Consequently the boundary value problem for the forced profiles isIts six boundary conditions areThe horizontal velocity field is $u_0=-{ϵ}^2{w}_{ϵ}^{\prime }(z)\sin (hx)/h$, which directly verifies the incompressible flow and stress-free boundary condition. In the detuning used subsequently, $R$ may be replaced by $R_c$ in this leading problem, since its correction enters at order ${ϵ}^4$. The vertical profiles depend linearly on $\gamma$.

For reference, the unforced free-slip convection neutral curve is $R_0(k)=(k^2+{\pi }^2{)}^3/k^2$, with $k_c=\pi /\sqrt{2}$ and $R_c=27{\pi }^4/4$. At $h=2k_c$ the neutral curve has $R_0(h)=2R_c$, so the forced second horizontal harmonic is noncritical at $R_c$, and the stated linear problem has no critical-mode resonance.

Take the layer depth as the length unit, its thermal diffusion time as the time unit, and its imposed temperature difference as the temperature unit. Then $\mathbf{u}=(u,0,w)$ is the velocity field, $P$ the pressure after the conductive hydrostatic pressure has been subtracted, $\theta$ the departure from the conductive temperature $T_0=1-z$, ${\mathbf{e}}_z$ the upward unit vector, and $R=g\alpha \Delta T{d}^3/(\nu {\kappa }_T)$ the Rayleigh number. Here $d$ is the dimensional depth, $\nu$ the kinematic viscosity, ${\kappa }_T$ the thermal diffusivity and $\alpha$ the thermal expansion coefficient. The gradient, divergence and Laplacian act on the dimensionless coordinates. The term $w$ in the heat equation is advection of the background gradient, since $-\mathbf{u}\cdot \nabla T_0=w$. Mass density variations enter only the buoyancy term under the Boussinesq approximation; the incompressible flow constraint is $\nabla \cdot \mathbf{u}=0$.

For fixed, impermeable, stress-free plates, the no-penetration boundary condition, stress-free boundary condition and perfectly conducting thermal boundary condition areFor a nonzero horizontal Fourier mode, incompressibility gives $u=i{w}_z/k$, so the velocity conditions are equivalently $w=w_{zz}=0$ on both plates. A uniform horizontal velocity is not fixed by these boundary conditions; take its mean to be zero when discussing the stationary forced pattern.

The printed small-Prandtl-number label does not generally justify these equations. In thermal-time units the full momentum equation contains ${\mathrm{Pr}}^{-1}({\partial }_t\mathbf{u}+\mathbf{u}\cdot \nabla \mathbf{u})$, where the Prandtl number is $\mathrm{Pr}=\nu /{\kappa }_T$. Dropping that term gives infinite-Prandtl-number convection or another explicitly justified Stokes flow approximation. It is not the general $\mathrm{Pr}\to 0$ limit. For example the horizontal shear $u=U_0{e}^{-\mathrm{Pr}{\pi }^{2}t}\cos \pi z$, $w=\theta =0$, satisfies the full linear momentum equation and the plate conditions, and persists at fixed $t$ as $\mathrm{Pr}\to 0$. The printed inertia-free equation would instead require this shear to vanish. The following parts solve the equations actually displayed; applying them at small Prandtl number needs an additional slaving or creeping-flow ordering.

For Darcy-Bénard convection, the dimensionless Darcy law is $\mathbf{u}=-\nabla p+R\theta {\mathbf{e}}_z$, with incompressibility and a heat equation about $T_0=1-z$. Here $R$ is a Darcy thermal-time Rayleigh number, with permeability replacing the squared-depth dependence in the fluid-layer Rayleigh number. The stated stream function convention gives $\mathbf{u}=(-{\psi }_z,0,{\psi }_x)$, so $w={\psi }_x$. The vertical-plane curl of Darcy law gives ${\nabla }^2\psi =R{\theta }_x$; omitting the quadratic perturbation advection from the heat equation gives ${\theta }_t={\psi }_x+{\nabla }^2\theta$.

On the impermeable connected square boundary, zero normal velocity makes $\psi$ constant along the boundary; choose that constant as zero. The prescribed conductive boundary temperature fixes $\theta =0$ on all four sides. Thus the perfect-conductor wording is interpreted through the supplied homogeneous perturbation data: the side temperatures retain the background vertical profile, rather than imposing a different isothermal side state.

At marginal stability put $q=\sqrt{R}$ and $P=\psi +iq\theta$. The two coupled equations implyThe gauge transformation for conducting-square Darcy onset $P=e^{-iq(x-1/2)/2}F$ removes the first derivative:The square's Dirichlet Laplacian eigenfunctions are $\sin (m\pi x)\sin (n\pi z)$, with eigenvalues of $-{\nabla }^2$ equal to ${\pi }^2(m^2+n^2)$, $m,n\ge 1$. The first marginal value is thereforeIt is genuinely the first instability threshold, not just an isolated neutral value. Eliminating $\psi$ gives the temperature linear operator $L_R={\Delta }_D+R{\partial }_x{\Delta }_D^{-1}{\partial }_x$, where ${\Delta }_D$ is the Dirichlet realization of an elliptic operator for the Laplacian. Integration by parts shows the second term is self-adjoint and nonnegative, since its quadratic form is $-⟨{\theta }_x,{\Delta }_D^{-1}{\theta }_x⟩\ge 0$. Thus the spectrum is real, starts negative at $R=0$, and its leading eigenvalue cannot cross into growth without a zero eigenvalue. The smallest zero value is the one found above.

Let $S=\sin \pi x\sin \pi z$, $\xi =x-1/2$ and $a=q_c/2=\sqrt{2}\pi$. Multiplying the lowest real $F=S$ by independent real and imaginary constants gives two real physical eigenfunctions:The subscripts indicate even and odd parity about $x=1/2$. Because $S$ is even in $\xi$, the first has even $\psi$ and odd $\theta$, while the second reverses those parities. They are linearly independent and satisfy all four Dirichlet boundary conditions. The lowest complex $F$ eigenfunction is simple over the complex numbers, but its arbitrary complex multiplier supplies two independent real pairs; this explains the physical degeneracy.

The conclusion holds in dimension two. Shift and change sign to obtain a nonnegative harmonic function $v$. Its circular mean satisfiesHere the angular second derivative in the polar-coordinate Laplacian integrates to zero. Thus $m(r)=a\log r+b$; nonnegativity for every $r>0$ forces $a=0$. For $|x|=r$, the mean value property for harmonic functions and nonnegativity giveThe removable singularity for a bounded harmonic function fills in the origin, and the harmonic Liouville theorem makes the extension constant.

The conclusion fails for $n\ge 3$: $\boxed{u(x)=|x{|}^{2-n}}$ is positive and nonconstant, and its radial Laplacian $u^{\prime \prime }+(n-1)u^{\prime }/r$ vanishes away from zero. For $n=1$, $\boxed{u(x)=|x|}$ is a nonconstant function bounded below whose second derivative vanishes on both components of the punctured line.

Let $u=v+\psi$, so $v$ has zero Dirichlet boundary condition. Using the supplied form of the minimal surface equation for a graph, setFor $\Vert \psi {\Vert }_{C^{2,\alpha }}\le \beta \le 1$ and $\Vert v\Vert ,\Vert w\Vert \le 1$, the difference bound impliesandThe shifted arguments stay in the permitted radius-two ball, and $\Vert f_{\psi }{\Vert }_{C^{0,\alpha }}\le C{\beta }^2+C_n\beta$. The Laplacian with zero Dirichlet boundary condition has trivial null space by the maximum principle for harmonic functions. Choose $\beta >0$ small enough that $2C\beta \le \delta$ and $C{\beta }^2+C_n\beta \le {\epsilon }_0$, for the constants in part (c) with $L=\Delta$. Its small-data existence for a nonlinear elliptic Dirichlet problem supplies $v$, and $\boxed{u=v+\psi }$ solves the minimal surface equation for a graph with the required boundary values.

Put $r^2=y_1^2+y_2^2$ and write the Riemannian metric as a conformal rescaling of a Riemannian metric:The base metric is flat, so its Ricci scalar is zero and its Laplacian is $\Delta ={\partial }_1^2+{\partial }_2^2$. The formula for scalar curvature under conformal rescaling simplifies in dimension two toDifferentiate explicitly:Since $e^{-2\omega }=1+r^2$, the required Ricci scalar is

The same polytrope of index one interior equation is ${\nabla }^2\rho +(2\pi G/K)\rho =0$. A positive separated solution with zero density on all six cube faces is the cubic polytropic interiorIts Laplacian is $-3{\pi }^2\rho /L^2$, so the required side length isWith $\Phi =C-2K\rho$ and $P=K{\rho }^2$, it satisfies the interior hydrostatic equilibrium and Poisson equation for Newtonian gravity. Integrating each sine factor yieldsThis formal interior solution is not an isolated physical cubic star. The cubic polytrope fails isolated gravitational matching: at a vertex, the product of sines has $\nabla \rho =0$, so the interior potential predicts zero gravitational acceleration. At the vertex $(0,0,0)$, however, the field generated by its own positive mass isand each component is strictly positive. There can be no continuous matching to the isolated external field without additional forces or mass sources. Thus solving the interior density equation and imposing zero face values is insufficient. Fluid stars also have no rigid structure to maintain sharp cubic faces, and observed stellar shapes are approximately spherical or rotationally flattened rather than cubic.

The zero streamfunction gives zero velocity. The temperature ${\theta }_0=-z$ obeys both prescribed temperature values, has no horizontal gradient, and has zero Laplacian. Thus it satisfies both equations with ${\psi }_0=0$:This is the conductive state of Darcy-Bénard convection.

Put $u_j={\dot{p}}_j+{\partial }_i{W}_{ij}$. The noises required by a joint trajectory are $f_j=u_j+\Gamma {\mu }_j$ and $N_{ij}=W_{ij}+M{\partial }_i{\mu }_j$. Their independent Gaussian white noise weights give the joint path probability of an order parameter and its flux:For the physically reversed path, $p$ is time-even and the transport flux $W$ is time-odd. Therefore $u\mapsto -u$, $W\mapsto -W$, andThe common factor includes the two noise normalizations and the midpoint Jacobian determinant. Its calculation must include both relaxation channels. On a finite spatial grid, let $\mathsf{D}$ and $\mathsf{G}$ represent the divergence and gradient, with $\Delta =\mathsf{DG}$ the discrete Laplacian, and let ${\mathcal{H}}_n=\partial \mu /\partial p$ be the Hessian matrix of $F$ at the midpoint of time step $n$. Differentiating the two required noises with respect to the next configuration and the interval flux gives the Jacobian matrixUsing the Schur complement, its Jacobian determinant, apart from the path-independent power of $\Delta t$, isReversal visits the same midpoint configurations in reverse order, so the product of these factors is unchanged. The joint time-reversal invariance of a path Jacobian therefore involves the complete relaxation operator $\Gamma I-M\Delta$.

Taking the action difference givesRepeated component indices are summed. The second squared norm sums over both flux indices. This is mixed conserved and nonconserved order-parameter dynamics; reversing the order parameter history while leaving the flux unreversed would give the wrong probability ratio.

Applying the Laplacian to $\mathbf{u}$ and using ${\nabla }^2\mathbf{\Phi }=0$ givesForone therefore hasTogether with part (i), these are the Stokes equations, so the fields form the Unscaled Papkovich–Neuber representation.

Suppose $u_1,u_2$ are solutions and put $w=u_1-u_2$. Then $\Delta w=0$ and $w$ satisfies the homogeneous Robin boundary conditionFor $0<\lambda <1$, Green's first identity givesThe left side is nonnegative, so both sides vanish and $w=0$. For $\lambda =0$, the Dirichlet boundary condition $w=0$ on $S$ and the same identity again imply $w=0$. The solution is therefore unique for every $0\le \lambda <1$.

For a regular spherically symmetric solution, the Laplacian equation isRegularity at the origin gives $u^{\prime }=2r$, hence $u=r^2+C$. Applying the boundary condition at $r=b$ yieldsAt $\lambda =1$ the Neumann boundary condition would require $u^{\prime }(b)=2b=0$, which is impossible for $b>0$. Equivalently, its compatibility condition would require ${\int }_{V}6dV=0$.

Computing the Laplacian givesandThus Laplace's equation holds exactly when

For $z=x+iy$,soOne suitable analytic function is therefore

The magnetic field is $B=\nabla \times A$. The magnetostatic Ampère-Maxwell equation is $\nabla \times B={\mu }_{0}J$, so the curl of the curl identity givesA gauge transformation $A\mapsto A+\nabla \chi$ does not change $B$. We may therefore choose the Coulomb gauge $\nabla \cdot A=0$ by solving the Poisson equation ${\nabla }^2\chi =-\nabla \cdot A$. It follows component by component in a Cartesian coordinate system thatThe free-space Green function of the Laplacian then givesFor a localized steady current density, conservation of electric charge gives ${\nabla }^{\prime }\cdot J=0$; integration by parts confirms directly that this integral has zero divergence and hence obeys the Coulomb gauge.

Let the displayed integrand be $F$. Since it has no explicit dependence on $u$ or $v$, the Euler-Lagrange equations for two fields areIts derivatives areThusUsing the Laplacian, gradient, and divergence, these combine into the static Navier-Cauchy equation

On a relatively compact ball, mollification commutes with the Laplacian, so $u_{\epsilon }=u\ast {\eta }_{\epsilon }$ is a smooth harmonic function. Repeated interior derivative estimates, together with the Caccioppoli inequality, bound every derivative of $u_{\epsilon }$ on a smaller ball by the local $L^2$ norm of $u$, uniformly as $\epsilon \downarrow 0$. The Arzelà-Ascoli theorem and a diagonal argument give a smooth local limit, while mollification gives $u_{\epsilon }\to u$ in $L_{\mathrm{loc}}^2$. Thus the limit equals $u$ almost everywhere. After choosing this smooth representative, $u$ is harmonic pointwise. This is the Weyl lemma for an $H^1$ weak solution.

For $\psi (x)=e^{-|x{|}^2/2}$ in two dimensions,Thus the Laplacian satisfies $\Delta \psi =(|x{|}^2-2)\psi$, and the isotropic harmonic-oscillator operator gives

For Maxwell theory the group is Abelian, so write $J=e^{\chi }$. The reduced equation loses its commutators and becomesSince the Euclidean Laplacian isthe anti-self-dual Maxwell equations in this gauge are equivalent to

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.

If two solutions exist, their difference $w$ satisfies ${\nabla }^{2}w-w=0$ inside $B$ and $w=0$ on the boundary. Suppose $w$ has a positive value. Continuity on the closed ball gives a positive maximum at an interior point, where the Hessian matrix is negative semidefinite and hence ${\nabla }^{2}w\le 0$. But the equation gives ${\nabla }^{2}w=w>0$, a contradiction. A negative minimum similarly contradicts ${\nabla }^{2}w=w<0$.

Thus $\boxed{w=0}$, proving uniqueness by the maximum principle for the modified Helmholtz equation. Here a classical solution is understood to be $C^2$ inside and continuous up to the boundary; mere first differentiability would not define the displayed Laplacian classically. With enough boundary regularity, the alternative Green's first identity proof is $0={\int }_{B}w({\nabla }^{2}w-w)=-{\int }_B(|\nabla w{|}^2+w^2)$.