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Rotational symmetry of the vector field means $\mathbf{F}(r\mathbf{n})=h(r)\mathbf{n}$: rotations fixing $\mathbf{n}$ force the field to be parallel to $\mathbf{n}$, and rotations between directions make its magnitude depend only on $r$. Its flux integral over the radius-$r$ sphere is $4\pi r^{2}h(r)$. Applying the flux identity from part (i) to the ball, with the spherically symmetric source, givesTherefore the origin-regular spherical Poisson flux law isOnly source values at $s\le |\mathbf{x}|$ occur. Changing the exterior source can change the scalar potential by an interior constant, but cannot change this radial gradient. Regularity on all of ${\mathbb{R}}^3$, inherited from the Poisson equation setting, matters: on a punctured domain alone an additional $C\mathbf{x}/r^3$ would have zero divergence away from the origin and would not be fixed by the local source. It is excluded here by the origin regularity needed for the ball flux identity.

The divergence formula reduces the problem to the ordinary differential equation $r{f}^{\prime }+nf=0$ on $r>0$. Multiplication by $r^{n-1}$ makes it an exact derivative:Since the interval $r>0$ is connected, $r^{n}f(r)$ is a single constant $C$. Conversely, substituting $f=C{r}^{-n}$ into the divergence formula gives zero. ThusThis proves both existence and uniqueness up to the constant within the specified radial class. For $n=1$, the two punctured half-lines still share this same constant because the stipulated coefficient is one function of $r=|x|$.

In cylindrical coordinates $r=\sqrt{x^2+y^2}$, the bounded solid enclosed by the two conical sheets and the circular cylinder isIts meridional section consists of two triangular regions; revolving them about the $z$-axis gives the solid. The two conical sheets meet at the origin and meet the circular cylinder in its circles at $z=\pm 1$.

Since $\mathbf{F}=(x,y,z)$, its divergence is $3$. Integrate directly with the cylindrical coordinates volume element $rdzdrd\theta$:The same calculation gives $\mathrm{Vol}(V)=4\pi /3$.

Write $D_t={\partial }_t+\mathbf{u}\cdot \nabla$ and $\mathbf{L}=(\nabla \times \mathbf{B})\times \mathbf{B}/{\mu }_0$. The continuity equation converts a material specific-energy balance into a conservative energy density balance. Dot the ideal magnetohydrodynamic momentum equation with $\mathbf{u}$, and use the time independence of the Newtonian gravitational potential:For the energy density $U=p/(\gamma -1)$ associated with internal energy, the adiabatic pressure equation givesThe ideal magnetohydrodynamic induction equation and the cross-product divergence identity give the magnetic energy balanceIndeed $(\mathbf{u}\times \mathbf{B})\cdot (\nabla \times \mathbf{B})=-{\mu }_0\mathbf{u}\cdot \mathbf{L}$. The magnetic work cancels the kinetic magnetic work. The remaining pressure terms are $-\nabla \cdot (p\mathbf{u})$. Adding all three balances proves ideal magnetohydrodynamic energy conservation:whereThe last term is the Poynting vector with the ideal electric field $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$. A time-dependent imposed potential would instead supply the source $\rho {\partial }_t\Phi$.

Use axisymmetry and $\nabla \cdot \mathbf{B}=0$. Since the purely azimuthal velocity has zero divergence, the ideal magnetohydrodynamic induction equation becomes ${\partial }_t\mathbf{B}=(\mathbf{B}\cdot \nabla )\mathbf{u}-(\mathbf{u}\cdot \nabla )\mathbf{B}$. Differentiation of the cylindrical unit vectors contributes to its azimuthal component, givingThe azimuthal component of the magnetic tension force isThere is no azimuthal pressure or gravitational force and no azimuthal advective acceleration for this motion. Because the poloidal magnetic field is divergence-free,These coupled induction and tension equations describe a torsional Alfvén wave.

The magnetic field satisfies $\nabla \cdot \mathbf{B}=0$. Expanding the divergence of its Maxwell stress tensor givesFor the Newtonian gravitational field $\mathbf{g}=-\nabla \Phi$, Poisson equation for Newtonian gravity gives ${\partial }_j{g}_j=-4\pi G\rho$, and the gradient representation gives ${\partial }_j{g}_i={\partial }_i{g}_j$. ConsequentlyThe negative of the Newtonian gravitational stress tensor, in this force-stress convention, therefore supplies $\rho g_i$. Adding the pressure stress supplies $-{\partial }_{i}p$, soEach summand is symmetric. Absence of external gravitational sources is needed to represent the full gravitational force by this self-gravitating stress.

Apply the continuity equation and integrate by parts, with the stated vanishing boundary terms and finite moments. The second mass moment tensor satisfiesand differentiation again givesInsert the stress-divergence equation from part (a). The two force integrals becomeThe first term is $4K_{ij}$. Hence the magnetized-fluid tensor virial theorem isHere $\mathcal{T}$ is the volume-integrated stress, whose sign differs from some gravitational potential-energy tensor conventions. The derivation also requires the advective mass-moment surface terms to vanish; this is automatic for an isolated sufficiently decaying configuration.

Integrate Poisson equation for Newtonian gravity through a narrow slab around the disk. The horizontal derivative contributions vanish as its thickness tends to zero, leaving the normal-derivative jumpThe even function symmetry of $\Phi$ makes the derivatives opposite, soIn the current-free simply connected upper half-space, Ampère's circuital law gives $\nabla \times \mathbf{B}=0$ and permits a magnetic scalar potential. Rescale it so that $\sqrt{4\pi G/{\mu }_0}\mathbf{B}=-\nabla {\Phi }_M$. The divergence-free condition makes ${\Phi }_M$ satisfy Laplace's equation, withCompare with the gravitational jump condition. Subject to the same isolated-field boundary condition at infinity, ${\Phi }_M$ is the harmonic potential of the effective surface densityThis gravity-equivalent magnetic surface density can have either sign; it is a mathematical representation of the exterior magnetic field, not physical negative mass. An imposed nondecaying field would require additional boundary data and would not be fixed by the disk surface density alone.

Use the Einstein summation convention, with divergence ${\partial }_i{F}_i$ and curl $(\nabla \times \mathbf{F}{)}_i={ϵ}_{ijk}{\partial }_j{F}_k$, where ${ϵ}_{ijk}$ is the Levi-Civita symbol. For the first vector field,The derivative matrix is symmetric, so its contraction with the Levi-Civita symbol vanishes. For the second vector field,For the third vector field, $H_i={ϵ}_{ikl}{a}_k{x}_l/r$. Its divergence is zero because the Levi-Civita symbol contracts with symmetric products. For its curl, either contract two Levi-Civita symbols or use the vector triple product:ThusAll formulas hold on $r>0$, the domain where the radial unit vector is defined.

In cylindrical coordinates, $\rho =(x^2+y^2{)}^{1/2}$ and $\mathbf{F}=(f(\rho )x/\rho ,f(\rho )y/\rho ,0)$. The chain rule gives ${\partial }_x\rho =x/\rho$ and ${\partial }_y\rho =y/\rho$. ThereforeThis is cylindrically radial divergence, valid for $\rho >0$.

The bounded domain is outside the unit cylinder, below the inverse-square surface and above the lower plane:The upper plane touches only the rim $\rho =1,z=1$; it is not an extra positive-area top face. Revolving the shaded meridional region below around the $z$ axis gives the three-dimensional domain.

With the Jacobian determinant $dV=\rho d\rho d\varphi dz$ of cylindrical coordinates,Zero divergence gives $(\rho f{)}^{\prime }=0$, hence the most general solution on the positive radial interval isAlthough this vector field is singular on the axis if $A\ne 0$, the entire closure of $V$ lies at $\rho \ge 1$, so it is smooth on a neighbourhood of $V$ and the divergence theorem applies.

Its volume integral of divergence is zero. Compute the outward surface integral directly. On the inner cylinder, $\mathbf{n}=-{\mathbf{e}}_{\rho }$, $dS=d\varphi dz$, and the flux is $-2\pi A(1-1/4)=-3\pi A/2$. On the curved roof $z=g(\rho )={\rho }^{-2}$, the outward vector surface element of a graph isTherefore its flux isThe horizontal bottom has zero flux and the top rim has zero surface area. Summing giveswhich verifies the divergence theorem with every boundary contribution accounted for.

The product of the magnetic field with the velocity in this question is the three-dimensional cross product. Fix ${ϵ}_{123}=1$ and write $K^i{}_j={ϵ}^i{}_{kj}{B}^k$, so $K\mathbf{v}=\mathbf{B}\times \mathbf{v}$. In coordinates $y^0=t$, $y^i=x^i$ on ${\mathbb{R}}^3\times \mathbb{R}$, prescribe an affine connection by the following Christoffel symbols:with all other Christoffel symbols zero. These are smooth globally in the given Cartesian coordinates. The lower-index symmetry makes this a torsion-free connection. It is the geometrization of the Lorentz force by an affine connection; no condition on the curl of $\mathbf{E}$ or the divergence of $\mathbf{B}$ is required for this construction.

An affinely parametrized geodesic, with primes denoting differentiation with respect to $\lambda$, satisfiesOn the branch $t^{\prime }=c\ne 0$, use $t$ itself as an affine parameter. Dividing the spatial equation by $c^2$ givesConversely, each solution of this equation makes $\lambda =t$, $y(t)=(t,\mathbf{x}(t))$ an affinely parametrized geodesic of the constructed affine connection. Therefore its image is also an unparametrized geodesic. A general change of parameter adds a term proportional to the tangent in the geodesic equation, leaving the curve unchanged. The branch $t^{\prime }=0$ is not a trajectory with time as parameter.

For the metric realization, assume $\mathbf{E}=-\nabla U$ and $\nabla \cdot \mathbf{B}=0$. Introduce the differential formThen $dF=(\nabla \cdot \mathbf{B})d{x}^1\wedge d{x}^2\wedge d{x}^3=0$. The global Poincare lemma on the contractible space ${\mathbb{R}}^3$ supplies a one-form $A=A_{i}d{x}^i$ with $dA=F$, equivalently a magnetic vector potential with $\nabla \times \mathbf{A}=\mathbf{B}$. An explicit radial-gauge potential for a divergence-free magnetic field isThis is the radial homotopy formula for a closed two-form and is smooth even at the origin. For a constant magnetic field it gives $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{x}$.

Consider the Eisenhart-Duval lift with the Lorentzian metricThe independent one-forms $d{x}^1,d{x}^2,d{x}^3,dt,du-2A_{i}d{x}^i-Udt$ exhibit three positive directions and a two-dimensional block with one positive and one negative direction. Thus the metric is nondegenerate, of signature $(4,1)$. Its coefficients do not depend on $u$, and $g_{uu}=0$, so $\xi ={\partial }_u$ is a null Killing vector field. Indeed $g_{Au}$ are constant, so ${\Gamma }^A{}_{Bu}=0$ for the Levi-Civita connection and $\xi$ is parallel.

The geodesic Lagrangian for an affine parameter $\lambda$ isThe cyclic coordinate $u$ gives the conserved quantity $p_u=\partial L/\partial u^{\prime }=t^{\prime }$. Work at a nonzero value of $p_u$ and rescale the affine parameter to set $t^{\prime }=1$. This is the essential step in the null Kaluza-Klein reduction of a stationary force: one fixes the momentum along the null isometry and projects its geodesics, rather than dividing by $g_{uu}$.

The spatial Euler-Lagrange equations, before setting $t^{\prime }=1$, areSince $t^{\prime \prime }=0$ and $F_{ij}{v}^j=-(\mathbf{B}\times \mathbf{v}{)}_i$, their reduction isEquivalently, after taking $t$ as the affine parameter, the term $\dot{u}$ is a total derivative and the reduced Lagrangian is $L_{\mathrm{red}}=\frac{1}{2}|\dot{\mathbf{x}}{|}^2-2\mathbf{A}\cdot \dot{\mathbf{x}}-U$. Its Euler-Lagrange equations give the same sign and factor two.

There is also an explicit converse using null geodesics. For any physical trajectory, setThis makes its five-dimensional tangent null. The conserved momentum of the cyclic coordinate $t$ becomesThe quantity in parentheses is conserved, because its derivative is $\dot{\mathbf{x}}\cdot (2\mathbf{B}\times \dot{\mathbf{x}})=0$. Hence the $t$ equation, as well as the spatial and $u$ Euler-Lagrange equations, is satisfied. Every trajectory admits a null geodesic lift with $p_u=1$, and every such lift projects to the required trajectory.

Finally, a time-independent gauge transformation $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$ is absorbed by $u\mapsto u+2\chi$. The one-form $du-2A_{i}d{x}^i-Udt$ and the five-dimensional metric are unchanged. These gauge transformations of an Eisenhart-Duval lift alter the reduced Lagrangian only by the total derivative $-2d\chi /dt$.

Take $\mathbf{f}=f\hat{\mathbf{z}}$ with $f>0$, and put $\zeta =v_x-u_y$, $d=u_x+v_y$ and $q=\zeta -f\eta /H$. Here $q$ denotes the signed vertical component of the linearized shallow-water potential-vorticity anomaly, rather than its magnitude. The divergence and vertical curl of the linearized shallow water equations giveConsequently $q_t=0$. Eliminating $d$ and substituting $\zeta =q+f\eta /H$ givesThe vector forcing is $-H\mathbf{f}\cdot \mathbf{q}$. This is potential-vorticity conservation in its linear, f-plane form: the conserved anomaly forces a stationary balanced part, while the homogeneous equation supports inertial-gravity waves.

For the initial strip, differentiating the discontinuous velocity in the sense of distributions givesThese are two oppositely signed vortex sheets. In the final geostrophic balance, $v_f=0$ and $u_f=-g{\eta }_f^{\prime }/f$. The height therefore solvesThis is the Rossby deformation radius. Decay at infinity fixes the Green function to $G(y)=-\lambda e^{-|y|/\lambda }/2$. Thus the geostrophic adjustment of a finite-width current has the particularly useful representationExpanding the exponentials inside the strip gives ${\eta }_f/H=-(U/c)e^{-a/\lambda }\sinh (y/\lambda )$; above the strip it gives $-(U/c)e^{-y/\lambda }\sinh (a/\lambda )$, and below it gives $(U/c)e^{y/\lambda }\sinh (a/\lambda )$. The height is continuous, odd and exponentially localized. Its derivative has the jumps required by the two Dirac delta functions.

Differentiating the height, rather than assuming a uniform final current, gives the complete velocity:The one-sided velocity jump is $U$ at each edge, with opposite orientations. The value exactly on an idealized vortex sheet is immaterial. Inside the strip the current remains in the original direction; outside it a return current develops.

Let $r=a/\lambda$ and $s=y/\lambda$. For $r\ll 1$, $u_f/U=1-r+O(r^2)$ over the narrow strip, while the outside return flow is approximately $-r{e}^{-|s|}$. The height varies almost linearly across the strip, $c{\eta }_f/(HU)\simeq -s$, and has extrema of magnitude $(1-e^{-2r})/2\simeq r$ at its edges. For $r\gg 1$, the central current is exponentially small: $u_f(0)/U=e^{-r}$. Each edge supports a layer of width $\lambda$, with $u_f/U\simeq \frac{1}{2}{e}^{-(a-|y|)/\lambda }$ on its inner side and $-\frac{1}{2}{e}^{-(|y|-a)/\lambda }$ outside. The corresponding height extrema approach $\pm HU/(2c)$, with almost zero height deep inside and far outside. Both requested profiles are drawn from the exact functions below; the dashed jumps represent one-sided velocity limits.

For balanced shallow-water energy as a signed potential-vorticity pairing, divide out the constant density and work per unit distance in $x$. The kinetic energy plus surface potential energy isMultiplication of the stationary height equation by ${\eta }_f$ and integration by parts, with vanishing end terms, yieldsThe same expression holds with $dA$ for a finite two-dimensional domain or a finite periodic length in $x$. The infinite strip's total energy is infinite, so both energies and their ratio are understood per unit $x$-length.

There is a sign defect in the printed energy expression: $|\mathbf{q}|$ must be replaced by the signed component $\hat{\mathbf{f}}\cdot \mathbf{q}$. Indeed, $|q_0|=|U|[\delta (y-a)+\delta (y+a)]$, so pairing it with the odd ${\eta }_f$ gives zero, although the balanced state has positive energy. The signed pairing givesSince $E_i=H{U}^{2}a$, the energy retention in finite-width geostrophic adjustment isFor $r\ll 1$ this ratio is $1-r+\frac{2}{3}{r}^2+O(r^3)$: nearly all energy remains in the narrow balanced current. For $r\gg 1$ it is approximately $1/(2r)$: only the edge regions retain balanced energy. Conservation of energy still holds for the inviscid evolution. The missing balanced energy is carried away by inertia-gravity waves; “final state” means the local balanced limit after those waves leave, rather than a loss of the total energy over the entire infinite domain.

On a sphere of radius $R$ rotating at angular speed $\Omega$, the Coriolis parameter is $f=2\Omega \sin \varphi$. Near latitude ${\varphi }_0$, write $y=R(\varphi -{\varphi }_0)$, use locally eastward and northward Cartesian coordinates, and expandThe beta plane retains the first northward variation of planetary vorticity while neglecting higher latitude dependence and metric curvature. A midlatitude local calculation assumes $L/R\ll 1$ and $|\beta |L/|f_0|\ll 1$; near the equator the distinct equatorial beta plane has $f_0=0$, so the midlatitude low-frequency reduction used below does not apply.

For a homogeneous shallow layer, the hydrostatic approximation gives $p=p_{\mathrm{atm}}+\rho g(\eta -z)$ and hence $p_x/\rho =g{\eta }_x$, $p_y/\rho =g{\eta }_y$, independently of depth. The approximation follows from the small aspect ratio $H/L$ and neglect of vertical acceleration. Differentiate the linear horizontal momentum equations in $z$. Their depth derivatives obeyStarting from rest gives $u_z=v_z=0$ initially, and the unique solution remains zero. This establishes depth independence of hydrostatic shallow-water flow; an arbitrary pre-existing shear would not be removed merely by taking the hydrostatic approximation. Integrating the continuity equation between the rigid bottom and the moving surface gives ${\eta }_t+H(u_x+v_y)=0$ at linear order. With the depth-integrated shallow-water transports $U=Hu$, $V=Hv$, one obtains

It is useful to make the coefficient approximation in the height reduction explicit. Put ${\mathcal{F}}_y={\partial }_t^2+f(y{)}^2$. Differentiating the two momentum equations in time and eliminating the other transport givesSince ${\partial }_y({\mathcal{F}}_{y}V)={\mathcal{F}}_y{V}_y+2f\beta V$, taking their horizontal divergence and using the continuity equation givesApplying ${\mathcal{F}}_y$ proves the exact variable-Coriolis shallow-water height equationAt the reference latitude, or after the usual local freezing of undifferentiated $f$ factors to $f_0$, this gives the height relation written with ${\mathcal{F}}_0={\partial }_t^2+f_0^2$. With $f=f_0+\beta y$ over a finite region, that constant-coefficient version is a local approximation, not an exact identity. Keeping ${\mathcal{F}}_y$ as above avoids silently commuting a variable Coriolis parameter through a spatial derivative.

For the slow Rossby wave branch, take $|\omega |/|f_0|\ll 1$, approximate ${\mathcal{F}}_0$ by $f_0^2$, and discard the two time-derivative terms on the right compared with $-f_0^2{\eta }_x$. This is the regular midlatitude long-time ordering, with nondegenerate zonal variation. If $a=f_0/c$, the result isThe printed low-frequency equation has the opposite right-hand sign. The minus sign follows directly from the preceding eliminated equation and is also the sign needed for the printed isofrequency-circle centre. An independent check uses geostrophic balance: $\psi =g\eta /f_0$, $(u,v)=(-{\psi }_y,{\psi }_x)$, and the shallow-water quasi-geostrophic potential vorticity is ${\nabla }_h^2\psi -a^2\psi +\beta y$. Linear potential-vorticity conservation gives $({\nabla }_h^2\psi -a^2\psi {)}_t+\beta {\psi }_x=0$.

For $\eta =A{e}^{i(kx+ly-\omega t)}$, the shallow-water Rossby-wave dispersion relation isFor fixed $l$ and $\beta >0$, it is odd in $k$, zero at $k=0$, negative for $k>0$, and tends to zero from below as $k\to +\infty$. Its minimum occurs at $k=\sqrt{l^2+a^2}$ with ${\omega }_{\min }=-\beta /[2\sqrt{l^2+a^2}]$. Thus both long and short waves have small frequency. The zonal phase velocity is westward, $\omega /k<0$, whereas the zonal group velocity isIt changes sign at the frequency minimum.

At fixed nonzero frequency, completing the square yields the Rossby-wave isofrequency circleThe radius is real only if $|\omega |\le |\beta |/(2|a|)$. For $\beta >0$, the branch with $k>0$ has $\omega <0$ and a circle centred on the positive $k$-axis. If instead the printed plus-sign wave equation is taken literally with this same Fourier convention, its dispersion is $\omega =+\beta k/(k^2+l^2+a^2)$ and its circle is centred at $(+\beta /(2\omega ),0)$. These two conventions cannot be mixed.

For reflection of a Rossby wave at a meridional wall, the stationary wall preserves frequency, and its translation invariance in $y$ preserves the tangential wavenumber. Thus ${\omega }_r={\omega }_i$, $l_r=l_i$. Both $k$ values solveTheir sum and product giveAn incident wave in $x>0$ travels toward the wall in group velocity, not necessarily in phase velocity. For $\beta >0$, $0<k_i<\sqrt{l_i^2+a^2}$ has $c_{gx,i}<0$, while its partner $k_r>\sqrt{l_i^2+a^2}$ has $c_{gx,r}>0$. The double-root case has zero normal group velocity and does not describe a wave packet incident on the wall.

At leading quasi-geostrophic approximation, the impermeability condition is $u=-g{\eta }_y/f_0=0$. For $l_i\ne 0$, the boundary condition givesThe reflected height therefore has equal amplitude and a phase change of $\pi$. Equality here concerns the height or quasi-geostrophic streamfunction amplitudes in the reduced model. At the degenerate $l_i=0$, geostrophic no-normal-flow is automatically satisfied and alone does not determine their ratio; the usual homogeneous wall-streamfunction condition supplies $A_r=-A_i$ if imposed. Retaining the small ageostrophic transport changes the boundary condition towhich tends to $-1$ in the regular low-frequency ordering but is not generally of unit modulus. In particular, $l_i=0$ gives $A_r/A_i=-k_i/k_r$ in that more complete boundary relation. Thus the printed equal-amplitude assertion requires the nondegenerate leading quasi-geostrophic approximation, or an explicit homogeneous wall condition; it is not a general exact shallow-water reflection law.

The product rule for the divergence givesSubtracting cancels the two gradient products. ThereforeTo obtain the requested order of terms, apply this identity with $u,v$ interchanged and use the divergence theorem on the spherical shell. On the outer sphere the shell's outward unit normal is $\hat{\mathbf{r}}$; on the inner sphere it is $-\hat{\mathbf{r}}$. Write ${\partial }_r=\hat{\mathbf{r}}\cdot \nabla$ on both spheres. Then the resulting Green second identity isThe minus sign records the inward radial direction of the shell's normal on its inner boundary. Thus the normal derivatives in the displayed difference are outward from each individual ball; using shell-outward normals instead would put a plus sign between the two boundary integrals.

Put $r^2=x^2+y^2$. On the specified domain, $r>0$, and direct differentiation gives the divergence and vorticityFor example, the third component of the curl is $A(r^2-2x^2)/r^4+A(r^2-2y^2)/r^4=0$; the other two components vanish because $U$ depends only on $z$ and the transverse velocity is independent of $z$. Hence the flow is both incompressible and irrotational exactly whenThe exclusion of the axis matters: these are ordinary derivatives on the domain, with no assertion of regularity on the removed axis.

Take the curl of the linear momentum equation. For constant $\mathbf{\Omega }$ and zero divergence, $\nabla \times (\mathbf{\Omega }\times \mathbf{u})=-(\mathbf{\Omega }\cdot \nabla )\mathbf{u}$. With $\mathbf{\Omega }=\Omega {\mathbf{e}}_z$, this givesTake another curl and use part (a): $-{\partial }_t{\nabla }^2\mathbf{u}=2\Omega {\partial }_z\mathbf{\omega }$. Differentiate in time and substitute the previous equation:ConsequentlyThe two curl operations remove the pressure gradient while retaining the rotational restoring force.