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Use signed velocities in the shock frame; the Mach number depends on their squares. For a genuine nonradiative perfect gas shock wave, the Rankine-Hugoniot conditions for a perfect gas give a common mass flux $j={\rho }_L{u}_L={\rho }_R{u}_R$, a common momentum flux $p+\rho u^2$, and a common specific stagnation enthalpy $\gamma p/[(\gamma -1)\rho ]+u^2/2$. Here $e$ in the given equation of state is specific internal energy, despite the source's informal “energy density” wording.

Put $r={\rho }_R/{\rho }_L$, $P=p_R/p_L$ and $M^2={\rho }_L{u}_L^2/(\gamma p_L)$. Mass and momentum give $u_R=u_L/r$ and $P=1+\gamma M^2(1-1/r)$. Substituting these into the energy relation and clearing denominators givesThe $r=1$ factor is the no-jump solution. On the nontrivial shock wave branch,In particular $r-1=2(M^2-1)/[(\gamma -1)M^2+2]$. With $\gamma >1$, compression is therefore equivalent to $M_L^2>1$. The other Mach number isPositive pressure makes its denominator positive, and $M_R^2-1$ has the opposite sign to $M_L^2-1$. Thus a nontrivial physical shock wave has a supersonic side and a subsonic side; the entropy-admissible flow goes from the former to the latter.

The shock compression ratio also yields $u_L{u}_R=(\gamma -1)u_L^2/(\gamma +1)+2\gamma p_L/[(\gamma +1){\rho }_L]$. Rearranging produces the first velocity identity in the question. The pressure relation gives the second identity in the equivalent formThe numerator here contains mass density, as the original PDF shows; the converted TeX incorrectly replaces that mass density by pressure.

For reflection, the once-shocked gas is the common left state in both applications. Before rebound, its velocity in the shock frame is $a=u_s-U_{+}<0$ and the gas ahead has velocity $-U_{+}$; after rebound, it is $b=u_s+U_{-}>0$ and the gas next to the wall has velocity $U_{-}$. In both cases the right velocity minus the left velocity equals $-u_s$. Applying the first identity therefore gives the same quadratic equation,with roots $a$ and $b$. Their opposite signs distinguish the two roots, so $\boxed{(u_s-U_{+})(u_s+U_{-})=-\gamma p_s/{\rho }_s}$.

Write $q_0=p_0/p_s$, $q_1=p_1/p_s$. The pressure identity applied to each shock wave givesSquaring the root-product relation and eliminating the velocities yieldsand hence the normal shock reflection at a rigid wall relationTaking $p_0/p_s\to 0$ gives $\boxed{p_1/p_s\to (3\gamma -1)/(\gamma -1)}$; for a monatomic gas this limiting amplification is six.

Linearizing the ideal magnetohydrodynamic induction equation about a static magnetic field and using $u^{\prime }={\partial }_t\xi$ gives the fluid Lagrangian displacement-generated magnetic field perturbation $b=\nabla \times (\xi \times B)$. The two solenoidal vector field conditions reduce the curl identity toIts divergence vanishes because it is a curl. Directly, the two products of first derivatives in ${\partial }_i{b}_i$ cancel after exchanging their indices, while the remaining terms differentiate $\nabla \cdot \xi$ or $\nabla \cdot B$, both zero. Similarly the integrated linear continuity equation gives ${\rho }^{\prime }=-\nabla \cdot (\rho \xi )$, so $\boxed{{\rho }^{\prime }=-\xi \cdot \nabla \rho }$.

Let $\Pi =p+B^2/(2{\mu }_0)$ be the magnetohydrodynamic total pressure. Magnetostatic equilibrium and the momentum equation in the linearized ideal magnetohydrodynamic equations areThere is no perturbed gravitational potential here because the field is fixed. Multiply by ${\xi }_i^{\ast }$ and integrate. The pressure term vanishes by the assumed boundary cancellation and $\nabla \cdot {\xi }^{\ast }=0$. Thuswhere $\mathcal{M}=-\int {\xi }_i^{\ast }(B_j{\partial }_j{b}_i+b_j{\partial }_j{B}_i)dV$.

Integrate the first magnetic term along $B$, using $\nabla \cdot B=0$, then insert the expression for $b$. This gives the positive term $\int |(B\cdot \nabla )\xi {|}^{2}dV$ and two mixed derivative terms. For clarity, the troublesome term transforms asThe first term on the right cancels the other mixed term. Combining the remaining derivative with the term containing ${\partial }_j{B}_k{\partial }_k{B}_i$ givesDifferentiating the equilibrium relation now combines the gradient of mass density and magnetostatic equilibrium terms:We obtain the incompressible magnetic displacement energy identityThe two Hessian matrices are real symmetric, making their Hermitian form evaluations real. The last term is a squared norm, and $I>0$ for a nonzero displacement in fluid with positive mass density. Thus $\boxed{{\sigma }^2\text{ is real}}$. It may be negative, in which case one temporal branch has exponential growth; real squared frequency is not itself a proof of stability.

For the atmosphere with a horizontal magnetic field, $\Phi =gz$ has zero Hessian matrix and the equilibrium equation isConsequently the interchange stability of an incompressible magnetized atmosphere is governed by the numeratorFor ${\xi }_z=\sin ky$, the magnetic tension term is zero: these fluid Lagrangian displacements interchange neighboring magnetic field lines without varying along them. A mass density increasing with height gives negative energy and instability; decreasing mass density gives positive energy for these vertical interchanges. If ${\rho }_z\le 0$ everywhere, the displayed form is nonnegative for every admitted incompressible displacement. If the gradient changes sign, suitably localized divergence-free interchange trials detect an unstable inverted layer; For example, take ${\xi }_z=f(z)e^{iky}$ and ${\xi }_y=i{f}^{\prime }(z)e^{iky}/k$, with $f$ supported in that layer. Their divergence is zero, and neither component varies along the magnetic field.

For ${\xi }_z=\sin kx$, magnetic field lines are bent. Averaging over a full horizontal wavelength gives the trial Rayleigh quotientunder the stipulated admissible boundary/integrability conditions. It is a trial quotient, not a claim that this displacement is an exact eigenfunction for every profile. mass density decreasing upward is stable here too. For inverted mass density, magnetic tension can stabilize sufficiently short waves along the magnetic field, but does not remove the transverse interchange instability. Locally the two contributions have the familiar scale $-g{\rho }_z/\rho +k^2{v}_A^2$, with the Alfvén speed $v_A=B/\sqrt{{\mu }_0\rho }$.

A spatially uniform pressure is a useful leading approximation when the sound speed makes the acoustic crossing time much shorter than the radiative cooling and thermal conduction times. The motions must remain at low Mach number, and gravity and other stresses must be negligible. Sound then redistributes pressure before appreciable thermal evolution occurs. Small pressure corrections can still drive the slow motion; imposing exactly zero pressure gradient in the full momentum equation at every order would be a stronger assumption.

The equation of state gives the specific internal energy of an ideal gas as $e=p/[(\gamma -1)\rho ]$. With $D/Dt={\partial }_t+v{\partial }_x$, the continuity equation is $D\rho /Dt=-\rho v_x$. Differentiating $e$ therefore yieldsInsert this into the internal energy equation, whose compressional term is $-p{v}_x$, and use spatially uniform pressure, so $Dp/Dt=p_t$. The result isFor temporally constant pressure put $A=\mu p/\mathcal{R}$, so the mass density is $\rho =A/T$. The continuity equation now gives $v_x=T^{-1}DT/Dt$. Multiplication of the last equation by $(\gamma -1)T/(\gamma p)$, with $\lambda ={\lambda }_0{T}^{\alpha }$, givesThis is the isobaric conductive cooling in mass coordinates transformation before changing coordinates.

There is a boundary assumption implicit in the proposed Lagrangian coordinate. For a mass integral from a fixed origin, integration of the continuity equation givesThus $m$ is material when the base mass flux vanishes, as appropriate for the stationary cold reservoir below. If the origin permits mass flux, replace $m$ by $m-{\int }^t\rho (0,s)v(0,s)ds$; this corrected label has zero material derivative. In a material mass coordinate, $D/Dt={\partial }_t{|}_m$ and ${\partial }_x=(A/T){\partial }_m$. In particularThe conduction coefficient therefore becomes $C{\lambda }_0$, with the same $C$ as the cooling coefficient. With $\tau =Ct$, the equation is

In the hot gas there is no volumetric radiative cooling. The initial temperature is uniform and the cold boundary introduces no finite mass length. The remaining nonlinear diffusion equation is invariant under $m\mapsto bm$, $\tau \mapsto b^2\tau$ with the same temperature scale. Its natural similarity solution is thereforeThe far-field condition also recovers the initial temperature at every fixed $m>0$ as $\tau \downarrow 0$. Substitution gives the ordinary differential equationFor $\alpha =1$, this reduces to $f^{\prime \prime }+\xi f^{\prime }/2=0$. Integrating first for $f^{\prime }$ gives $f^{\prime }=K\exp (-{\xi }^2/4)$. The two boundary conditions fix $K=1/\sqrt{\pi }$, hence the error-function isobaric cooling front isHere the error function has normalization $2/\sqrt{\pi }$, as printed in the original PDF; the converted TeX's $2/\pi$ is incorrect. The following original sketch shows the temperature profiles. The cooling layer broadens as $\sqrt{\tau }$, and at fixed positive mass label the temperature falls as time increases, while the remote hot gas remains at $T_0$.

For $\alpha =1$, the magnitude of the heat flux into the cold reservoir isAlthough the thermal conductivity tends to zero there, the physical temperature gradient diverges: their product has this finite limit. The stationary reservoir has no incoming advective energy flux and radiates the conducted energy immediately. Since $T_m(0,\tau )=T_0/\sqrt{\pi {\lambda }_0\tau }$, the radiated power per unit area isThus the requested exponent is $\boxed{k=1/2}$. The initial divergence is the usual idealization of an instantaneous temperature discontinuity in a diffusion equation.

Set $a=2\pi G{\rho }_0$. The axisymmetric Poisson equation for Newtonian gravity integrates to $(R{\Phi }_R{)}^{\prime }=4\pi G{\rho }_{0}R$. Regularity on the axis fixes the integration constant, so ${\Phi }_R=aR$. Radial force balance in the rotating fluid givesWith the stated central pressure, $p_c={\rho }_0{a}^2/(4k^2)$, this is a perfect square:The free surface lies at the first zero of pressure. For the source's positive rotation parameter, $R_0=\sqrt{2\pi G{\rho }_0}/k$; allowing either rotation direction requires $|k|$ in the denominator. The inward effective gravity is ${\Phi }_R-R{\Omega }^2=R(a-k^2{R}^2)$, so $\boxed{g_{\mathrm{eff}}(R_0)=0}$. In particular $p_R(R_0)=0$. This is the critically rotating self-gravitating cylinder.

For a normal mode proportional to $e^{i\omega t+im\varphi }$, the background material derivative becomes $i\sigma$, where $\sigma =\omega +m\Omega (R)$. Linearizing the centrifugal term gives $-2\Omega u_{\varphi }$ in the radial equation. In the azimuthal equation the background velocity gradient and cylindrical geometric term together give $(2\Omega +R{\Omega }^{\prime })u_R=3\Omega u_R$. Since the interior mass density is constant, write $W=p^{\prime }/{\rho }_0+{\Phi }^{\prime }$. The two momentum equations and incompressible flow condition are consequentlyFor $m\ne 0$, the last two equations express the other amplitudes in terms of $u=u_R$:Differentiate the second expression using $R{\Omega }^{\prime }=\Omega$ and $R{\sigma }^{\prime }=m\Omega$, then substitute into the radial momentum equation. Before any division by $\sigma$, the result isAway from corotation this gives the required ordinary differential equationAt corotation the original coupled equations, or the undivided equation, must be used. A regular axisymmetric radial velocity would vanish by the incompressible flow condition; the requested dipolar calculation has $m=1$.

For $m=1$, use the amplitude normalization $u=A(\omega +kR)$, which remains useful when $\omega$ tends to zero. Substitution into the undivided equation gives zero identically. For nonzero $\omega$ this is the source's $1+kR/\omega$ after choosing $A=1/\omega$. The other amplitudes areThese formulas also satisfy the three original equations directly. By the regularity assumption in the question, this is the interior velocity family to which the free surface conditions must be applied.

The gravitational potential of a displaced cylindrical interface supplies an essential part of those conditions. For surface fluid displacement $\eta e^{im\varphi }$, the bulk mass density perturbation vanishes but the displaced mass density step contributes ${\rho }_0\eta \delta (R-R_0)$. The interior and exterior gravitational potential amplitudes solve Laplace's equation, are proportional to $R^m$ and $R^{-m}$ for $m\ge 1$, and are continuous at the surface. Integrating the Poisson equation for Newtonian gravity across the mass density step givesSince the two radial derivatives at the surface are $-m{\Phi }_s^{\prime }/R_0$ and $m{\Phi }_s^{\prime }/R_0$, respectively,It would be incorrect to set the entire potential perturbation to zero merely because the bulk mass density is unchanged.

Let ${\Omega }_s=k{R}_0$, so $a={\Omega }_s^2$. Zero fluid Lagrangian perturbation of surface pressure gives $p_s^{\prime }+\eta p_R(R_0)=0$, hence $p_s^{\prime }=0$. The surface momentum potential and kinematic condition for $m=1$ are thereforeIf $\omega +{\Omega }_s\ne 0$, the kinematic condition fixes $\eta =-iA$. The same value follows if the radial fluid Lagrangian displacement $u_R/[i(\omega +kR)]=-iA$ is required to extend continuously to the free surface, including a possible surface-corotation limit. Comparing the gravitational condition with the interior value of $W_s$ then givesfor a nontrivial perturbation. These dipolar perturbations of a critically rotating cylinder describe a displacement of the whole cylinder's axis. Translating an isolated self-gravitating system gives another equilibrium, so there is no external restoring force and the translation is neutral. A uniform drift of the centre also follows from invariance under a Galilean transformation, explaining the double zero of the frequency equation. The singular $1/\omega$ in the originally chosen velocity normalization does not obstruct the finite-amplitude limiting solution.

There is an admissibility qualification to the frequency conclusion as printed. Regular interior fluid Eulerian perturbation of velocity alone does not justify cancelling the surface factor $\omega +{\Omega }_s$. Eliminating $\eta$ without that cancellation instead givesThe additional surface-corotation case $\omega =-{\Omega }_s$ formally satisfies every displayed velocity equation and the usual linearized surface conditions withHere $u_R(R_0)=p^{\prime }(R_0)=0$, and the velocity is finite on the axis. However, its interior radial fluid Lagrangian displacement tends to $-iA$, whereas the undisplaced surface has $\eta =0$. Thus a regular single-frequency fluid Lagrangian displacement continuous up to the surface excludes this degenerate branch. The intended ${\omega }^2=0$ conclusion holds with that regular-displacement condition, or with surface corotation excluded; the printed velocity regularity assumption by itself leaves this formal exception.

Work at one radius and write $\mathcal{R}=k/({\mu }_m{m}_p)$. Use the surface density to fix the mass density normalization and introduce the height scaleChoose the dimensionless variablesThe pressure and temperature scales make the hydrostatic and ideal gas equations parameter-free. The flux scale fixes the heating coefficient at $9/4$, and the choice of $h$ makes the diffusion coefficient unity. Direct substitution gives the universal ordinary differential equationsFor example, the dimensionless coefficient in the temperature equation is $3\alpha \kappa {\Sigma }^2{\mathcal{R}}^4/(16\sigma {\Omega }^5{h}^6)=1$.

Let the upper free surface be at $\zeta ={\zeta }_s$, so its physical semithickness is $H=h{\zeta }_s$. Midplane symmetry and the radiative-zero disk surface conditions giveThe last condition is precisely $\Sigma ={\int }_{-H}^H\rho dz$. The unknown surface position and the two positive midplane values $P(0),\Theta (0)$ are determined as part of the normalized boundary value problem. Symmetry makes $P,D,\Theta$ even and $Q$ odd; $P^{\prime }(0)={\Theta }^{\prime }(0)=0$ already follow from the equations and are not additional independent boundary conditions.

The zero conditions do not mean zero surface flux. Integrating the heating equation gives $Q({\zeta }_s)=(9/4){\int }_0^{{\zeta }_s}Pd\zeta >0$. Near a regular surface with finite outgoing flux, dividing the hydrostatic equation by the temperature equation yields $dP/d\Theta \simeq {\zeta }_s{\Theta }^3/Q_s$. Thus $P\propto {\Theta }^4$, $D\propto {\Theta }^3$ and $\Theta \propto {\zeta }_s-\zeta$, so mass density also vanishes. The apparent singular quotient in the diffusion equation has a finite regular limit. All boundary conditions and equations are independent of the physical parameters: this is homologous vertical structure of a constant-opacity alpha disk.

The first equation is vertical hydrostatic equilibrium: the upward pressure gradient balances the central mass's vertical gravitational force. For $|z|\ll r$, the point-mass acceleration is $-GMz/(r^2+z^2{)}^{3/2}\simeq -{\Omega }_K^{2}z$, where Keplerian rotation gives ${\Omega }_K^2=GM/r^3$. The disc's self-gravity and vertical inertia are neglected.

The second equation is local thermal balance. The divergence of the vertical radiative flux equals viscous dissipation. In the convention of this alpha disk, $\nu =\alpha p/(\rho \Omega )$, and the dominant Keplerian shear has $rd\Omega /dr=-3\Omega /2$. ThereforeThe radiative flux increases from zero at the symmetric midplane to a positive outgoing value at the upper surface. Radial heat transport, thermal advection and rapid storage of thermal energy are excluded from this local balance.

The third equation is radiative diffusion, equivalentlyIt applies to optically thick material, with $\kappa$ the opacity per unit mass and $\sigma$ the Stefan-Boltzmann constant. Here constant electron-scattering opacity represents Thomson scattering; the formula uses a diffusive, nearly thermal radiation field. Positive upward flux requires temperature to decrease upwards.

The final equation is the ideal gas pressure law, with mean molecular weight ${\mu }_m$ in proton-mass units. Writing $\mathcal{R}=k/({\mu }_m{m}_p)$ gives $p=\mathcal{R}\rho T$. The pressure is gas pressure because radiation pressure is assumed negligible. The four equations express vertical support, local heating balance, diffusive cooling and the gas equation of state.

Estimate characteristic bulk terms. Put $ϵ=H/r\ll 1$ and assume smooth radial variation on scale $r$, secular evolution, and $\alpha \lesssim 1$. Vertical support gives $c_s^2\sim p/\rho \sim {\Omega }^2{H}^2$, while the alpha disk prescription and slow accretion implyThese velocity estimates follow from the viscous radial evolution scale and mass conservation. They are estimates for a slowly evolving thin disc, not a claim that geometric thinness excludes independently forced rapid vertical motion.

For vertical balance, the exact central force expands asAt $z\sim H$, the neglected correction has magnitude $\rho {\Omega }^{2}H{ϵ}^2$, compared with the retained force $\rho {\Omega }^{2}H$. Radial pressure support similarly changes the rotation from its Keplerian value byVertical meridional inertia scales as $u_r{u}_z/r$ or $u_z^2/H$, both $O({\alpha }^2{ϵ}^4{\Omega }^{2}H)$. Meridional viscous accelerations can scale as $\nu u_z/H^2$ or $\nu u_r/(rH)$, both $O({\alpha }^2{ϵ}^2{\Omega }^{2}H)$. Thus the gravity correction and possible viscous correction giveSome omitted terms are smaller than this leading order.

For the heating equation, the retained dissipation is $q_0^{+}\sim \rho \nu {\Omega }^2\sim \alpha p\Omega$. Radial pressure support and the finite-height correction change the dominant shear by a relative $O({ϵ}^2)$. Smooth vertical variation of the rotation gives ${\partial }_z(r\Omega )=O(ϵ\Omega )$, whose additional squared shear contributes $O({ϵ}^2{q}_0^{+})$. Meridional shear of order $u_r/H$ gives an additional relative $O({\alpha }^2{ϵ}^2)$ contribution.

Thermal advection and compressional work are of order $\rho u_r{c}_s^2/r$ and $p{u}_r/r$, respectively. Relative to $\alpha p\Omega$ they are $O({ϵ}^2)$. Thermal storage on the viscous evolution timescale $t_{\nu }\sim r^2/\nu$ has the same order, since $p/t_{\nu }\sim \alpha p\Omega {ϵ}^2$. For radiative diffusion with the same local diffusivity in different directions, smooth gradients give $F_r/F_z\sim H/r$, soConsequentlyThis is the thin-accretion-disk truncation error for the secular, radially smooth approximation. Neglect of disc self-gravity and radiation pressure, optical thickness, the opacity model and the idealized zero surface boundary have their own validity conditions; their errors do not become small solely because $H/r$ does. Near a sharp radial boundary layer or during thermal evolution on the heating timescale, the above estimates need not apply.

For $k_z=v_z=0$, define the complex vertical vorticity amplitudeTake the curl of the amplitude equations, including ${\dot{k}}_x=2A{k}_y$. Incompressibility cancels the pressure and rotation terms, givingEquivalently, the uniform basic absolute vorticity has no gradient for the perturbation to advect. The general two-dimensional viscous shearing wave solution iswhere $Z_0$ is an arbitrary complex constant and $k_x(t_0)$ is an arbitrary real initial radial wavenumber. The perturbation pressure follows from part (ii), or explicitly $\Pi =(4A{k}_y^2-2\Omega k^2)Z/k^4$.

For $A\ne 0$, shift the time origin to the swing $k_x=0$ and put $T=k_x/k_y=2At$, $T_0=2A{t}_0$ and $\mathrm{Re}=A/(\nu k_y^2)$. ThenThe spatially averaged perturbation kinetic-energy mass density is $E=\rho ⟨|{\mathbf{u}}^{\prime }{|}^2⟩/2=\rho |\mathbf{v}{|}^2/4$, and $|\mathbf{v}{|}^2=|Z{|}^2/k^2$. ConsequentlyThe proportionality constant depends on the initial data; the exponential profile is not a claim of infinite physical growth from the remote past. For $A>0$, leading waves with $T<0$ can gain velocity amplitude as their wavevector magnitude decreases, before viscosity and later winding damp them. This is the Orr mechanism, and it is independent of uniform rotation for these two-dimensional disturbances.

If $A=0$, the wavevector is constant and $\mathbf{v}(t)=\mathbf{v}(t_0)e^{-\nu k^2(t-t_0)}$, so there is no shear amplification. The formulas with the physical-time integral remain valid, although $T=2At$ is no longer a useful coordinate. The high-positive-Reynolds-number case in the next part takes $A>0$.

Let the perturbation phase be $\vartheta =\mathbf{k}(t)\cdot \mathbf{r}$ and write $\psi ={\psi }_0+\mathrm{Re}[\Pi (t)e^{i\vartheta }]$. Incompressible flow imposes $\mathbf{k}\cdot \mathbf{v}=0$. Applying the background advective derivative to the phase givesCancel this position-dependent phase by choosingThe perturbation acting on the basic flow gives $({\mathbf{u}}^{\prime }\cdot \nabla ){\mathbf{u}}_0=-2A{u}_x^{\prime }{\mathbf{e}}_y$. Its self-advection vanishes exactly: the physical velocity ${\mathbf{u}}^{\prime }$ is perpendicular to $\mathbf{k}$, while every spatial derivative of ${\mathbf{u}}^{\prime }$ is proportional to a component of $\mathbf{k}$. Thus ${\mathbf{u}}^{\prime }\cdot \nabla {\mathbf{u}}^{\prime }=0$, including the products between the complex mode and its conjugate. Substituting into the full Navier-Stokes equation therefore gives an exact incompressible shearing wave, with no small-amplitude approximation:In components these equations areDifferentiating the transversality constraint closes the pressure:The coefficient $4A$ includes both the shear term in the momentum equation and $\dot{\mathbf{k}}\cdot \mathbf{v}$; treating the wavevector as fixed would miss one contribution. Initial data with $\mathbf{k}(t_0)\cdot \mathbf{v}(t_0)=0$ remain transverse under this closed evolution. These shearing waves are exact in an unbounded local shear flow or with compatible shearing-periodic boundaries; arbitrary rigid boundary conditions need not admit a single such wave.

The variable $\psi$ is a rotating-flow pressure potential including the conservative body-force and centrifugal potentials. For constant mass density,up to an additive function of time, where $\Phi$ is any imposed conservative potential. The last term is the centrifugal potential; its negative gradient is the centrifugal acceleration. The Coriolis force remains explicit in the equation because it is velocity-dependent and cannot be absorbed into a scalar position potential.

The further constraint is incompressible flow:For the specified shear, ${\mathbf{u}}_0\cdot \nabla {\mathbf{u}}_0=0$ and ${\nabla }^2{\mathbf{u}}_0=0$. Its Coriolis acceleration is $4\Omega Ax{\mathbf{e}}_x$, so the corresponding total basic pressure potential can be chosen as ${\psi }_0=-2\Omega A{x}^2$. This supplies the force balance required for the basic state used below.