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Choose the homogeneous dilation $\boxed{r(R)=tR}$, which has the required endpoint values. For a radial field, put $J=r^{\prime }{r}^2/R^2$. The radial energy density, after omitting its constant angular factor, is $R^2[r^{\prime 2}+2r^2/R^2+(J-1{)}^2]$. Its Euler equation simplifies toThe chosen dilation has $r^{\prime \prime }=0$, $J=t^3$ constant, and satisfies it. More directly, the full first Piola-Kirchhoff response for the given energy isAt $F=tI$ this is $[2t+2(t^3-1)t^2]I$, constant throughout the body. Its divergence vanishes, so the deformation satisfies the full force-free equilibrium equations as well.

Its energy isThis grows like $(4\pi /3)t^6$, whereas the same-boundary cavitating competitor has energy asymptotic to $8\pi t^2$. Therefore for sufficiently large $t$, the homogeneous equilibrium is not a global energy minimum in the finite-energy class admitting the cavitating deformation.

The admissible class matters: the competitor from the preceding part is not continuous at the center. This comparison is in that Sobolev class, not a claim about minimization restricted to smooth cavity-free deformations.

The surface element relation is Nanson's formula. The force on a corresponding pair of current and reference surface elements must agree:Therefore the first Piola-Kirchhoff stress tensor isIt measures force per reference area, with a current force direction and a reference normal direction.

In components, differentiate $T_{iA}=S_{ij}(y(x))(\mathrm{cof}F{)}_{jA}$. The Piola identity says ${\partial }_A(\mathrm{cof}Dy{)}_{jA}=0$. One proof writes the cofactor matrix as $\frac{1}{2}{\epsilon }_{jkl}{\epsilon }_{ABC}{y}_{k,B}{y}_{l,C}$; differentiating in $A$ produces symmetric second derivatives contracted against antisymmetric epsilon symbols, hence zero. The chain rule consequently givessince $F_{kA}(\mathrm{cof}F{)}_{jA}=J{\delta }_{kj}$. Thus, with the coordinate composition made explicit,The distinction between reference divergence and current divergence is essential to this identity.

For this homogeneous motion, suppose the matrix $\mathbf{A}=\ddot{\mathbf{F}}{\mathbf{F}}^{-1}$ is symmetric. Put $\mathbf{y}=\mathbf{x}-\mathbf{c}(t)$. The acceleration potential of a homogeneous motion iswhere $g(t)$ is arbitrary. When taking the spatial gradient, $t$, $\mathbf{c}$, $\ddot{\mathbf{c}}$, and $\mathbf{A}$ are fixed. Symmetry givesThis verifies the potential directly and also proves sufficiency of the symmetric matrix condition within the static-family setting. Its dimension is ${\mathsf{L}}^2{\mathsf{T}}^{-2}$, so the pressure correction $-\rho \xi$ has the dimension of stress. Combining with part (b)(ii), the corresponding dynamic Cauchy stress tensor iswhere $\tilde{C}(t)=C(t)-\rho g(t)$. Substituting its divergence into the Cauchy momentum equation reproduces $\rho [\mathbf{Ay}+\ddot{\mathbf{c}}]$, as required.

Take the divergence of the Ampère-Maxwell equation and use that the divergence of a curl is zero:The last step uses Gauss's law. Thus the charge continuity equation is exactlyFor a direct justification that this four-current is a four-vector, take one locally comoving population of electric charges, with rest charge density ${\rho }_0$ and four-velocity $U^{\mu }=\gamma (c,\mathbf{v})$. Invariance of electric charge and length contraction give $\rho =\gamma {\rho }_0$ and $\mathbf{J}=\rho \mathbf{v}$, hence $J^{\mu }={\rho }_0{U}^{\mu }$. Summing populations preserves the four-vector rule. In particular, $c{\rho }^{\prime }=\gamma (c\rho -\beta J_x)$ and $J_x^{\prime }=\gamma (J_x-\beta c\rho )$. The Lorentz invariance of its four-divergence then explains why charge conservation holds in every inertial frame. A single zero-divergence condition alone would not determine a field's transformation law; the physical charge density and current density provide the additional content.

Balance streamwise convection of order one with transverse viscous diffusion $R{e}^{-1}{\partial }_y^{2}u$. This requires a thickness $R{e}^{-1/2}$. Continuity then makes the normal velocity of that same small order. SetThe normal momentum equation makes the leading pressure independent of $Y$. It equals the outer pressure, whose inviscid tangential momentum balance gives $-p_x=U_s{U}_s^{\prime }$. The leading tangential equation and continuity are consequentlyThe divergence-free field $(u,V)$ has a streamfunction with $u={\Psi }_Y$, $V=-{\Psi }_x$. Substituting yieldsThe stationary impermeable wall has $u=V=0$, so ${\Psi }_Y(x,0)=0$ and $\Psi (x,0)$ is constant, chosen to be zero. Matching to the outer slip velocity gives ${\Psi }_Y(x,\infty )=U_s(x)$.

Let $\mathrm{Re}=\rho UL/\mu$ and use the given anisotropic scales. Continuity makes the transverse velocity scale $\delta U$ consistent with the gap height $\delta L$. The streamwise viscous stress divergence has scale $\mu U/({\delta }^2{L}^2)$, while the streamwise pressure gradient has scale $\Pi /L$. Their balance selectsThe streamwise equation then has the dimensionless formThe normal equation becomesWhen $\delta \ll 1$ and ${\delta }^2\mathrm{Re}\ll 1$, neglecting the indicated smaller terms givesThe bottom conditions express no slip on a plate moving at speed $U$ and no penetration. The top conditions express no slip and no penetration on the stationary wall. The leading pressure is a function of the streamwise coordinate alone.

Linearize about rest, with the equilibrium forces already absorbed in the reduced pressure. The quadratic convective acceleration drops out. Since $\mathbf{k}\times (u\mathbf{i}+v\mathbf{j}+w\mathbf{k})=-v\mathbf{i}+u\mathbf{j}$, the component equations areThe scan prints $v_w$ in the continuity equation; the derivative required by incompressibility is $v_y$.

To eliminate pressure, put $\zeta =v_x-u_y$ and ${\nabla }_h^2={\partial }_x^2+{\partial }_y^2$. Taking the vertical component of the curl of momentum givesTaking horizontal divergence of momentum and substituting continuity givesDifferentiate this in $z$ and use $P_z/\rho =-w_t$:One more time derivative, followed by the vorticity relation, yields the Poincaré equation for inertial wavesThis derivation retains all three velocity components, including the transverse component coupled by rotation.

For fields ${\phi }_A$ and a continuous infinitesimal internal symmetry $\delta {\phi }_A=ϵ\Delta {\phi }_A$, suppose the density changes by a divergence, $\delta \mathcal{L}=ϵ{\partial }_{\mu }{K}^{\mu }$. Varying the action and using the Euler-Lagrange equations gives the Noether currentThe conserved charge is $Q=\int j^0{d}^{3}x$ when the flux through the boundary vanishes; more generally $dQ/dt=-\oint \mathbf{j}\cdot d\mathbf{S}$. If the Lagrangian is exactly invariant, $K^{\mu }=0$. Treat a complex field and its conjugate as independent variables during variation, imposing conjugacy on physical solutions. Time and space translations give energy and momentum through the corresponding spacetime version of the theorem. Continuous global symmetry gives an ordinary charge; local gauge redundancy requires additional care and is not an independent conserved charge for every arbitrary function.

Treat $\varphi$ and ${\varphi }^{\ast }$ as independent fields, keep $A_{\mu }(x)$ as the prescribed background, and use the source's interaction with no added $e^2{A}^2|\varphi {|}^2$ term. The momentum derivatives areVarying ${\varphi }^{\ast }$ and $\varphi$ gives respectivelyNo Lorenz gauge condition has been assumed. The background divergence term must be retained unless such a condition is separately imposed.

The canonical momenta conjugate to the two scalar fields areso $\dot{\varphi }={\pi }^{\ast }-ie{A}_0\varphi$ and ${\dot{\varphi }}^{\ast }=\pi +ie{A}_0{\varphi }^{\ast }$. Define the physical three-vector potential by $A_{\mu }=(A_0,-\mathbf{A})$. The Legendre transform in the proper canonical variables isEquivalently the first product is $\pi {\pi }^{\ast }+ie{A}_0({\varphi }^{\ast }{\pi }^{\ast }-\varphi \pi )+e^2{A}_0^2|\varphi {|}^2$. The last quadratic term arises from replacing velocities by canonical momenta even though the original Lagrangian has no quadratic interaction. It is not permission to modify the specified Lagrangian. An explicitly time-dependent background generally makes the scalar-field energy time dependent.

For $\Delta \varphi =-i\varphi$ and $\Delta {\varphi }^{\ast }=i{\varphi }^{\ast }$, the interaction is invariant under the constant phase because every product has net zero phase. Applying the Noether formula to the two independent fields givesThe sign corresponds to the printed transformation $\varphi \to e^{-iϵ}\varphi$. Reversing the phase parameter would reverse the entire current, with no change in conservation.

The derivative products in the divergence of the first term cancel pairwise:The two field equations from part (c) implyMultiplication by $i$ exactly cancels the remaining background terms, proving ${\partial }_{\mu }{J}^{\mu }=0$ without requiring $A_{\mu }$ to be constant or divergence free. The global phase symmetry survives even though this truncated interaction is not exactly invariant under arbitrary local gauge transformations.

Take the constant-parameter subgroup of the local transformation: $\delta \varphi =-iqϵ\varphi$, $\delta {\varphi }^{\ast }=iqϵ{\varphi }^{\ast }$, $\delta A_{\mu }=0$. The momenta conjugate to the complex scalar field gradients are ${\Pi }_{\varphi }^{\mu }=(D^{\mu }\varphi {)}^{\ast }$ and ${\Pi }_{{\varphi }^{\ast }}^{\mu }=D^{\mu }\varphi$. Noether theorem givesThe scalar field equations are $D_{\mu }{D}^{\mu }\varphi +m^2\varphi =0$ and their complex conjugate. Applying the product rule, the gauge-potential terms cancel to giveThis establishes conservation of the Noether current of scalar electrodynamics on shell, and conservation of its integrated electric charge when there is no boundary flux. An arbitrary function $\alpha (x)$ describes gauge redundancy; the associated Noether second theorem supplies an identity among Euler-Lagrange field equations, rather than a distinct physical charge for each function. Indeed its on-shell current can be written $j^{\mu }[\alpha ]=\alpha J^{\mu }-F^{\mu \nu }{\partial }_{\nu }\alpha =-{\partial }_{\nu }(\alpha F^{\mu \nu })$. The antisymmetric tensor makes its divergence vanish. Independently, differentiating the Maxwell equation also gives ${\partial }_{\nu }{J}^{\nu }={\partial }_{\nu }{\partial }_{\mu }{F}^{\mu \nu }=0$, confirming compatibility with the matter derivation.^[ref]

For a translation-invariant potential, the on-shell energy-momentum tensor obeys ${\partial }_{\mu }{\theta }^{\mu \nu }=0$. The divergence of the dilatation current is thereforeUse the equation of motion $\square \varphi =-V^{\prime }(\varphi )$ to obtain the useful general identityFor $D=4$, $m=0$ and only ${\lambda }_4$ nonzero, every term vanishes. More explicitly, $\varphi \square \varphi =-{\lambda }_4{\varphi }^4/6$ while ${\theta }^{\mu }{}_{\mu }=-(\partial \varphi {)}^2+{\lambda }_4{\varphi }^4/6$; adding the first kinetic term cancels both remaining contributions. Thus ${\partial }_{\mu }{j}^{\mu }=0$, and $\int d^{3}x{j}^0$ is conserved when the boundary flux vanishes. Outside the parameter conditions of solution, the boxed formula displays the classical sources of scale breaking; the same current need not be conserved.

The term $\alpha {\nabla }^{2}q$ is cell diffusion arising from random movement. Chemotaxis up the concentration gradient gives the cell flux $\beta q\nabla c$, whose negative divergence is the transport contribution $-\beta \nabla \cdot (q\nabla c)$. The reaction term $q(1-q)$ gives growth governed by the logistic differential equation with dimensionless carrying capacity one.

The chemoattractant itself diffuses, is produced by cells at rate $\kappa \eta q$, and decays at rate $\kappa c$. The homogeneous Neumann boundary conditions suppress the normal diffusive flux of each quantity. They also suppress the normal chemotactic cell flux because the normal derivative of $c$ vanishes. Thus the rectangular boundary is impermeable, and the spatially integrated cell count changes through cell proliferation rather than transport across that boundary.

The planar divergence is $\nabla \cdot F=-(\Omega +x^2)<0$ for $\Omega >0$. Since the plane is simply connected, the Bendixson-Dulac theorem excludes any nonconstant periodic orbit: the flux across a closed trajectory would be zero because the vector field is tangent, whereas Green's theorem makes it the strictly negative integral of the divergence over the enclosed region.

Alternatively define $E=y^2/2-x^2/2+x^4/4$. Direct differentiation givesA nonconstant periodic solution would require the integral of this strictly dissipative expression over a period to vanish. That forces $y\equiv 0$ and hence an equilibrium, a contradiction. This energy argument identifies the damping mechanism behind the geometric criterion.

The Lorenz vector field has constant divergence $\nabla \cdot F=-\sigma -1-b$. Liouville formula for a fundamental matrix for the derivative of the flow therefore givesThis concerns transported three-dimensional phase volume, not the Euclidean distance between each pair of trajectories.

A quasiperiodic orbit with the usual smooth two-frequency torus closure would lie on an invariant embedded two-torus. That torus bounds a solid torus of positive finite volume in ${\mathbb{R}}^3$. Since the vector field is tangent to its boundary, its outward flux is zero, contradicting the divergence theorem, whose volume integral is $-(\sigma +1+b)V<0$. This negative-divergence obstruction to invariant tori excludes such quasiperiodic tori. The argument uses global negative divergence and the enclosed volume; saying only that an attracting torus has zero volume would not exclude it.

At an equilibrium the sum of eigenvalues is $-(\sigma +1+b)<0$, so they cannot all have positive real parts: no equilibrium is a fully repelling source. For a periodic orbit of period $T$, one Floquet multiplier equals one in the flow direction. The product of the other two is $e^{-(\sigma +1+b)T}<1$, so they cannot both have modulus greater than one: no periodic orbit is fully repelling either. Saddle equilibria and saddle periodic orbits remain possible; contraction of volume does not imply contraction in every direction.

The Poincaré-Bendixson theorem states that the omega-limit set of a forward orbit confined to a compact region of the plane, containing no equilibrium points, is a periodic orbit.

Put $x=R\cos \theta$, $y=R\sin \theta$. Direct calculation givesHere $1/2\le H\le 17/16$ and $|\cos \theta {\sin }^3\theta |\le 3\sqrt{3}/16$. Thus on the inner circle $R=\sqrt{3}/2$ the radial velocity is positive, and on the outer circle $R=3/2$ it is negative. In their closed annulus,so there is no equilibrium. The theorem gives a periodic orbit in this trapping annulus.

It is asymptotically stable: the vector-field divergence isthroughout the annulus. Its nontrivial Floquet multiplier is $\exp {\int }_0^T\mathrm{div}Fdt<1$. This proves an attracting periodic orbit exists. There can be only one there: two such nested orbits would enclose an annular region with zero outward flux, contradicting the strictly negative divergence via the divergence theorem.

The four dimensional equations express mass conservation, momentum balance with pressure, gravity and viscous stress, heat advection and conduction, and a linear thermal-expansion equation of state. For heating from below, suitable plate conditions are $\mathbf{u}=0$ and $T=T_0+\Delta T$ at $z=0$, and $\mathbf{u}=0$, $T=T_0$ at $z=d$. No slip and thermal insulation on the sidewall provide one suitable lateral completion; prescribed lateral temperatures would be another specified thermal setup.

Let $\kappa =k/({\rho }_0{c}_p)$ be thermal diffusivity and $\nu =\mu /{\rho }_0$ kinematic viscosity. Use $d$, $d^2/\kappa$, $\kappa /d$ and $\Delta T$ as length, time, velocity and temperature scales. Subtract the static reference pressure $p_H$ with $\nabla p_H=-{\rho }_{0}g\mathbf{k}$ and scale its remainder by $\mu \kappa /d^2$. ThenDensity becomes ${\rho }_0(1-BT)$. Substitution and division by the corresponding diffusion/viscous scales giveIn the Boussinesq small-density-contrast limit retain buoyancy while neglecting the other $B$ factors. At infinite Prandtl number, after any fast initial adjustment, the equations areThe velocity is instantaneously constrained by a viscous elliptic balance; the temperature still evolves.

Take the divergence of the vector Poisson equation. Derivatives commute, and divergence of a curl is zero, so $\Delta (\nabla \cdot \mathbf{A})=0$. Its boundary value is zero, hence the maximum principle gives $\nabla \cdot \mathbf{A}=0$ throughout $D$. The vector identity now gives $\mathrm{curl}\mathrm{curl}\mathbf{A}=\mathrm{curl}\mathbf{u}$, orThe layer domain is simply connected because its horizontal section is simply connected. For a curl-free field on such a domain, define $\chi (x)$ as its line integral from a fixed base point to $x$. Contractibility of loops and Stokes theorem make the integral path independent, and differentiation gives its gradient equal to the field. Thus $\mathrm{curl}\mathbf{A}-\mathbf{u}=\nabla \chi$.

Incompressibility then gives $\Delta \chi =0$, while the prescribed normal curl gives ${\partial }_n\chi =0$. Green's identity yieldsHence $\chi$ is constant on the connected domain and $\boxed{\mathrm{curl}\mathbf{A}=\mathbf{u}}$. The extra condition $\mathbf{A}\cdot \mathbf{n}=0$ belongs to the stated potential boundary problem; the reconstruction argument above uses the divergence and normal-curl conditions. On a multiply connected domain a curl-free field need not be a globally single-valued gradient.

Finally $\mathrm{curl}(T\mathbf{k})=(T_y,-T_x,0)$. Taking the curl of the infinite-Prandtl momentum balance and using $\mathrm{curl}\mathbf{u}=-\Delta \mathbf{A}$ givesThe derivation is an instance of velocity reconstruction from a mixed-boundary vector Poisson problem.

Define ${\chi }_1$ to be one in phase 1 and zero in phase 2, including its dependence on both transverse coordinates even though the averaged fields depend only on $x,t$. With no interphase mass transfer, its distributional phase-mass equation is ${\partial }_t(\rho {\chi }_1)+\nabla \cdot (\rho \mathbf{v}{\chi }_1)=0$. At a pipe cross section $\Sigma$ of area $A$, setThus ${\rho }_1$ is an intrinsic phase density and $u_1$ is the density-weighted mean axial velocity. This weighting is necessary when density varies within the phase; it becomes the ordinary intrinsic mean for an incompressible phase of constant density.

Integrate over the cross section. The transverse divergence contributes no boundary mass flux at the impermeable pipe wall, and the material interface has no mass leakage. The section area is constant, soFor phase 2 use ${\chi }_2=1-{\chi }_1$ and fraction $1-\alpha$, with the analogous intrinsic density and density-weighted velocity:The intrinsic quantities are defined where their phase fraction is nonzero. These are the intrinsic phase averages in a two-fluid pipe model.

The product rule and fluctuation incompressibility give the pointwise identityAveraging therefore givesThe fluctuation term is consequently the divergence of a Reynolds stress. The eddy viscosity model represents this momentum transport through a coefficient multiplying the mean strain-rate tensor:for its shear or deviatoric part. For example, $-\rho \overline{u^{\prime }{v}^{\prime }}\simeq {\mu }_T{U}_y$ in a thin parallel shear flow. The dynamic viscosity ${\mu }_T$ is an effective flow-dependent transport coefficient, rather than a molecular material property. Large eddies carry linear momentum between regions of different mean velocity.

There is a necessary qualification to the printed tensor equation (4). Taking its trace literally gives $0=-\rho \overline{u_i^{\prime }{u}_i^{\prime }}$, which would exclude nonzero turbulence. A consistent deviatoric eddy-viscosity closure in $d$ fluctuating dimensions isThe isotropic term is absorbed into the effective pressure $P_{\mathrm{eff}}=P+2\rho e_t/d$. Here $d=2$ for a strictly two-dimensional fluctuation model. The subsequent equations use $P$ for this effective pressure, or equivalently use (4) as a shear-stress closure. This preserves the intended thin-layer model without asserting an impossible full covariance identity.

For turbulent motions of scale $l$ and velocity scale $u_l$, the estimate ${\mu }_T\sim \rho u_{l}l$ gives ${\mu }_T/\mu \sim u_{l}l/\nu$. It is much greater than one on energetic large scales and in the inertial range well above the Kolmogorov length scale. At the molecular dissipation scales it is no longer large, and molecular transport also dominates very near a smooth solid wall.

Let $\delta \ll L$ be a representative mixing-layer thickness, ${ϵ}_m=\delta /L$, and let $U_s$ be a representative streamwise velocity. Use the scalesThe transverse velocity scale follows from the mean continuity equation. For turbulent transverse transport to balance streamwise inertia,With this ordering, the streamwise eddy-stress divergence isThus transverse diffusion of streamwise linear momentum is retained, while streamwise diffusion and the $V_x$ contribution are smaller by ${ϵ}_m^2$. Molecular kinematic viscosity is neglected where ${\nu }_T\gg \nu$.

In the transverse momentum equation, inertia and the retained eddy stresses have scale ${ϵ}_m{U}_s^2/L$, whereas an order-one transverse pressure variation would give $P_y/\rho \sim U_s^2/({ϵ}_{m}L)$. Hence the leading dimensionless transverse balance is $P_y=0$. The resulting boundary layer equations for the planar turbulent mixing layer areHere ${\nu }_T={\mu }_T/\rho$. The effective pressure convention explained in part (a)(iii) is understood. Since the outer streams have uniform horizontal velocities, their inviscid momentum equations give $P_x=0$; choosing the far-field gauge pressure as zero gives $P=0$ throughout this approximation.