Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 64 (Astrophysical Fluid Dynamics)

Let ${\mathbf{x}}^{\prime }=\mathbf{x}-\mathbf{v}t$, $t^{\prime }=t$ for a constant relative velocity $\mathbf{v}$. The Galilean transformation gives ${\nabla }^{\prime }=\nabla$ and ${\partial }_{t^{\prime }}={\partial }_t+\mathbf{v}\cdot \nabla$. In the magnetic Galilean limit of the pre-Maxwell equations, takeThe Ampère's law without displacement current and the solenoidal constraint are immediately unchanged. For Faraday's law, the constant-velocity vector identity is $\nabla \times (\mathbf{v}\times \mathbf{B})=\mathbf{v}\nabla \cdot \mathbf{B}-(\mathbf{v}\cdot \nabla )\mathbf{B}$. ThusThese are transformations of the magnetic nonrelativistic approximation. The exact charge-current transformation also contains the convective term $-{\rho }_e\mathbf{v}$ in ${\mathbf{J}}^{\prime }$; ignoring it requires the quasineutral magnetic limit. They are not exact transformations of the full Maxwell equations.

Since ${\mathbf{u}}^{\prime }=\mathbf{u}-\mathbf{v}$, the combination ${\mathbf{E}}^{\prime }+{\mathbf{u}}^{\prime }\times {\mathbf{B}}^{\prime }$ equals $\mathbf{E}+\mathbf{u}\times \mathbf{B}$. For an ideally conducting fluid the rest-frame electric field vanishes, so the ideal moving-conductor Ohm law is $\mathbf{E}+\mathbf{u}\times \mathbf{B}=0$. Substitution in Faraday's law givesThe ideal magnetohydrodynamic induction equation requires nonrelativistic fluid motion, negligible displacement current, and negligible nonideal terms in the electric-field relation. In particular, magnetic diffusion must be slow compared with advection: the magnetic Reynolds number $\mathrm{Rm}=UL/{\eta }_m\gg 1$, where ${\eta }_m=({\mu }_0\sigma {)}^{-1}$ is the magnetic diffusivity. The single-fluid moving-conductor Ohm law must apply on the scales considered, with corrections to that relation negligible; viscosity is not itself excluded by this induction equation.

For a smooth ideal evolution, magnetic flux freezing states that the magnetic flux through every material surface is conserved and that magnetic field lines are carried with the fluid. Their connectivity is preserved: ideal smooth advection does not reconnect them. These implications are stated here without proof, as requested.

Put $D_t={\partial }_t+\mathbf{u}\cdot \nabla$. The continuity equation and the expanded ideal magnetohydrodynamic induction equation areso $D_t(B_i/\rho )=(B_j/\rho ){\partial }_j{u}_i$. Because $\mathbf{B}=\nabla \times \mathbf{A}$, the induction equation permits the magnetic vector potential equation ${\partial }_t\mathbf{A}=\mathbf{u}\times \mathbf{B}-\nabla \phi$, with an arbitrary scalar electric potential $\phi$ carrying the gauge choice. Using $(\mathbf{u}\times \nabla \times \mathbf{A}{)}_i=u_j{\partial }_i{A}_j-u_j{\partial }_j{A}_i$ yieldsContracting with $B_i/\rho$ and differentiating that factor givesThe last two terms cancel by exchanging their dummy indices. Thus the requested material derivative isThis is material transport of magnetic helicity density.

For a material volume, mass conservation gives $d{\int }_{V(t)}\rho hdV/dt={\int }_{V(t)}\rho D_{t}hdV$. Apply this with $h=\mathbf{A}\cdot \mathbf{B}/\rho$, and use $\nabla \cdot \mathbf{B}=0$ and the divergence theorem:If the magnetic field lines do not penetrate the boundary, $\mathbf{B}\cdot \mathbf{n}=0$, so $\boxed{d{H}_m/dt=0}$.

For uniqueness under a regular, single-valued gauge transformation $\mathbf{A}\mapsto \mathbf{A}+\nabla \Lambda$,Consequently this magnetic helicity is both conserved and gauge independent. As usual, vector potentials are understood to be globally regular potentials for the same field, rather than unrelated harmonic-circulation choices on a multiply connected restricted domain.

For $ab\ne 0$, a magnetic field line stays at fixed $R$ and satisfies $d\varphi /dz=B_{\varphi }/(R{B}_z)=a/b$. It is therefore a helix with axial pitch $2\pi b/a$; the sign of $a/b$ fixes its handedness. The axial magnetic flux is $\pi b{R}_0^2$, and the toroidal magnetic field supplies the twist. If $a=0$ the field is untwisted; if $b=0$ it consists of azimuthal circles rather than helical lines carrying axial flux.

A convenient globally regular magnetic vector potential has $A_R=0$ andIndeed, the cylindrical curl gives $B_z=R^{-1}d(R{A}_{\varphi })/dR$ and $B_{\varphi }=-d{A}_z/dR$, with $B_R=0$. The tangential potential components are continuous at $R_0$, so their derivatives introduce no unwanted delta-function magnetic field at the tube boundary. The $C$ term is a constant axial potential.

The most general regular potential is this one plus $\nabla \Lambda (R,\varphi ,z)$ for an arbitrary single-valued gauge function. A putative extra $C_{\varphi }/R$ term in the inner $A_{\varphi }$ is excluded by regularity at the axis; the exterior $1/R$ circulation is already fixed by the enclosed magnetic flux. Since the potential is globally defined across the axis and exterior, any remaining difference is a curl-free vector field, hence a gradient. The constant $C$ can itself be absorbed into a gauge function proportional to $z$.

For the displayed gauge family, inside the tubeand henceFor example, requiring $A_z=0$ outside sets $C=a{R}_0^2/2$ and gives $\boxed{H_m/L=\pi ab{R}_0^4/2}$. The equally regular choice $C=0$ gives zero. More generally, for a length-$L$ segment the gauge transformation contributesIn particular, $\Lambda =\Delta Cz$ shifts $H_m/L$ by $\pi b{R}_0^2\Delta C$. This gauge dependence of helicity in an open magnetic flux tube is possible because field lines penetrate the end caps when $b\ne 0$. There is no contradiction with part (b), whose no-penetration condition fails at those caps. A gauge or end-cap convention is needed to give a unique value per unit length. For $b=0$ the ordinary helicity is zero and this end-cap ambiguity disappears.

For the isothermal equation of state, $p=c_s^2\rho$, with constant isothermal sound speed $c_s$. A nonzero steady plane-parallel flow has constant mass flux $\rho u=j$. Choose the positive flow direction for convenience. The continuity equation then gives ${\rho }^{\prime }/\rho =-u^{\prime }/u$, and the steady Euler momentum equation becomesIntegrating and absorbing the reference velocity into the constant yields, for the Mach number $\mathcal{M}=u/c_s$,For the opposite flow direction use $\mathcal{M}=|u|/c_s$; the same relation holds.

A smooth sonic point has $u=c_s$, so its momentum equation requires ${\Phi }^{\prime }=0$. Moreover $f(\mathcal{M})={\mathcal{M}}^2/2-\ln \mathcal{M}$ satisfies $f^{\prime }(1)=0$, $f^{\prime \prime }(1)=2$ and has its minimum $f(1)=1/2$. Thus, if the flow reaches $\mathcal{M}=1$ at $x_{\ast }$,A sonic crossing is therefore possible at a maximum of the potential, withAt a nondegenerate maximum, the expansions give $(\mathcal{M}-1{)}^2=-{\Phi }^{\prime \prime }(x_{\ast })(x-x_{\ast }{)}^2/(2c_s^2)+\cdots$. Choosing either smooth sign gives a crossing slope ${\mathcal{M}}^{\prime }(x_{\ast })=\pm \sqrt{-{\Phi }^{\prime \prime }(x_{\ast })/(2c_s^2)}$, joining a subsonic flow to a supersonic flow, or the reverse. This establishes the plane-parallel isothermal sonic transition at a potential maximum. For a globally defined crossing, the same maximum must also be high enough for the Bernoulli relation to remain real over the intended domain.

Decompose $\mathbf{u}={\mathbf{u}}_p+u_{\varphi }{\mathbf{e}}_{\varphi }$ and $\mathbf{B}={\mathbf{B}}_p+B_{\varphi }{\mathbf{e}}_{\varphi }$. In a steady axisymmetric vector field, $\nabla \times \mathbf{E}=0$ and $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$. The radial and vertical components of its curl areTherefore $R{E}_{\varphi }=C$ is a constant. Regularity at the symmetry axis sets $C=0$; equivalently, one imposes no toroidal loop electromotive force, $\oint \mathbf{E}\cdot d\mathbf{l}=2\pi C=0$. Under this usual condition,which implieswhere ${\mathbf{B}}_p\ne 0$. This is the regular-axis alignment of steady poloidal ideal flow. It allows nonzero toroidal components as well.

The axis or zero-loop-voltage condition is necessary for the unqualified conclusion. On an annular domain, $\mathbf{B}=(B_0/R){\mathbf{e}}_R$ and $\mathbf{u}=U_0{\mathbf{e}}_z$ give $\mathbf{E}=-(U_0{B}_0/R){\mathbf{e}}_{\varphi }$ with zero curl, yet the poloidal vectors are perpendicular. This can satisfy the remaining steady equations as well: choose constant mass density ${\rho }_0$, $\Phi =\pi G{\rho }_0{R}^2$ and $p=p_0-\pi G{\rho }_0^2{R}^2>0$ on the annulus. The field has zero curl, radial pressure balances gravity, the flow advects no equilibrium quantity, and ${\nabla }^2\Phi =4\pi G{\rho }_0$. It carries a nonzero loop voltage. Steady axisymmetry alone on an annulus therefore does not exclude such cross-field flow.

Use the aligned poloidal flow from part (i), and define the magnetohydrodynamic mass loading $\eta =\rho \alpha$, so $\rho {\mathbf{u}}_p=\eta {\mathbf{B}}_p$. Axisymmetric mass conservation and $\nabla \cdot {\mathbf{B}}_p=0$ give ${\mathbf{B}}_p\cdot \nabla \eta =0$. Introduce the poloidal magnetic flux function $\Psi$ by ${\mathbf{B}}_p=R^{-1}\nabla \Psi \times {\mathbf{e}}_{\varphi }$, and write $u_{\varphi }=R\Omega$. ThenThe steady induction equation says the curl of this expression is zero. Hence $\nabla {\Omega }_F\times \nabla \Psi =0$, wherethe field-line angular velocity ${\Omega }_F$ is constant on each connected magnetic surface and along its field lines.

To control the difference between matter rotation and field-line rotation, use the azimuthal component of the ideal magnetohydrodynamic momentum equation. The inertial and magnetic terms are respectively $\rho R^{-1}{\mathbf{u}}_p\cdot \nabla (R{u}_{\varphi })$ and $({\mu }_{0}R{)}^{-1}{\mathbf{B}}_p\cdot \nabla (R{B}_{\varphi })$; pressure and gravity have no azimuthal derivative. Multiplying by $R$ and using constant mass loading givesThus the magnetohydrodynamic angular-momentum invariant $L=R^2\Omega -R{B}_{\varphi }/({\mu }_0\eta )$ is constant along the line when $\eta \ne 0$. Let $v_{Ap}=|{\mathbf{B}}_p|/\sqrt{{\mu }_0\rho }$ and define $M_{Ap}^2=|{\mathbf{u}}_p{|}^2/v_{Ap}^2={\mu }_0{\eta }^2/\rho$. Eliminating $B_{\varphi }$ between the two invariants givesFor a highly sub-Alfvénic poloidal flow, $M_{Ap}^2\ll 1$, this is $\Omega \simeq {\Omega }_F$ along a nonsingular field-line segment. This proves sub-Alfvénic isorotation in a steady axisymmetric wind, with the correction explicitly displayed. If the poloidal flow is exactly zero, the induction equation directly gives ${\mathbf{B}}_p\cdot \nabla \Omega =0$, namely Ferraro's law of isorotation. The approximate claim presumes regular invariants: small $M_{Ap}$ alone is not a uniform relative bound if $L/R^2$ is allowed to become arbitrarily large.

Matter corotating with the footpoint has fixed angular velocity ${\Omega }_0=(GM/R_0^3{)}^{1/2}$. Its effective potential in that rotating frame is the sum of Newtonian gravitational potential and centrifugal potential,Along an outward-inclined straight magnetic field line, write $R=R_0+s\sin i$ and $z=s\cos i$, where $s$ is distance from the footpoint. The exact potential along it isIts first derivative at $s=0$ is zero by the Keplerian rotation balance. At the footpoint the meridional Hessian is $\mathrm{diag}(-3{\Omega }_0^2,{\Omega }_0^2)$ with zero mixed derivative. HenceWhen $0\le i<30^{\circ }$, the potential initially rises and a cold parcel must overcome a barrier. When $i>30^{\circ }$, its quadratic curvature is negative, so the footpoint is a local maximum along the field and a small outward displacement is accelerated further outward:This is the thirty-degree magnetocentrifugal launching criterion. It describes the lack of a local cold-launch barrier, not a nonzero force on a parcel exactly at rest at $s=0$. At exactly $30^{\circ }$ the quadratic term vanishes; for the specified straight line the next term is $-7{\Omega }_0^2{s}^3/(16R_0)$, as in marginal straight-line launch at thirty degrees. Global wind escape additionally depends on conditions farther along the line.

Work in a gas region with $\rho ,p>0$ and physical adiabatic exponent $\gamma >1$. With $\xi =x/t$, each time derivative is $-\xi /t$ times differentiation in $\xi$, and each spatial derivative is $1/t$ times it. The continuity equation, Euler momentum equation and the entropy advection equation for pressure reduce toTheir coefficient determinant is $(u-\xi )[(u-\xi {)}^2-c^2]$, where the adiabatic sound speed is $c=\sqrt{\gamma p/\rho }$. If the determinant is nonzero, all derivatives vanish and the solution is uniform.

The apparent third characteristic $u=\xi$ does not give a smooth fan: on such an interval $u^{\prime }=1$, whereas continuity requires $\rho u^{\prime }=0$, impossible for $\rho >0$. It cannot support an isolated smooth entropy variation either. Subtracting $\gamma$ times the logarithmic density equation from the logarithmic pressure equation gives $(u-\xi )(\ln (p/{\rho }^{\gamma }){)}^{\prime }=0$. A nonzero continuous entropy derivative would force $u=\xi$ on an interval and give the same contradiction. Thus the specific entropy is constant throughout each smooth branch.

Every nonconstant branch therefore lies in one acoustic characteristic family. Let $\sigma =\pm 1$ denote it, with $\xi =u+\sigma c$, so $u-\xi =-\sigma c$. The constant entropy means $p=K{\rho }^{\gamma }$, $K>0$, and $c^{\prime }/c=(\gamma -1){\rho }^{\prime }/(2\rho )$. Continuity gives $u^{\prime }=\sigma c{\rho }^{\prime }/\rho =2\sigma c^{\prime }/(\gamma -1)$. Differentiating $\xi =u+\sigma c$ then provesIn particular, $u-2\sigma c/(\gamma -1)=J$ is a constant Riemann invariant. All nonconstant smooth local solutions have the explicit formon an interval where $c>0$. Along the left-going family ($\sigma =-1$) sound speed decreases with $\xi$; along the right-going family ($\sigma =+1$) it increases. These are the two self-similar ideal-gas rarefaction branches. Together with arbitrary uniform positive states they exhaust the smooth local solutions. A global solution can join them piecewise at fan edges, where derivatives generally jump, and can also have discontinuities; those do not constitute additional smooth branches.

The initial-value problem is a Riemann problem. Both the one-dimensional Euler equations and the two constant initial states are unchanged under the simultaneous scaling $(x,t)\mapsto (\lambda x,\lambda t)$ for $\lambda >0$. There is no imposed length or time scale. Rescaling a physically admissible entropy solution therefore gives another solution with exactly the same initial data. Uniqueness of the physical Riemann solution makes the two coincide, soThe smooth pieces are consequently those in part (a), while shocks and contacts lie on rays $x=st$. Across shocks one uses the conservative mass, momentum and total-energy equations and the entropy admissibility condition, rather than extending the smooth entropy-advection equation through the discontinuity. Scale invariance by itself, without selecting the unique admissible solution, would not justify the uniqueness step.

Let $c_L=(\gamma p_L/{\rho }_L{)}^{1/2}$ and $c_R=(\gamma p_R/{\rho }_R{)}^{1/2}$, with positive pressures and densities. A nontrivial single rarefaction wave has constant specific entropy and constant Riemann invariant, so for one family label $\sigma =\pm 1$ the necessary matching conditions areThe final inequality says that the fan opens rather than compresses. Its edge speeds are ${\xi }_L=u_L+\sigma c_L$ and ${\xi }_R=u_R+\sigma c_R$, and their difference is $\sigma (c_R-c_L)(\gamma +1)/(\gamma -1)>0$. The profiles in part (a) then interpolate the two end states and supply a sufficient construction. For the left family, $c_R<c_L$, hence ${\rho }_R<{\rho }_L$ and $p_R<p_L$; for the right family all three inequalities reverse. In both cases $u_R>u_L$. These are the single-rarefaction matching conditions for an ideal-gas Riemann problem.

For a concrete nontrivial example, take $\gamma =5/3$ andThe adiabatic sound speeds are $c_L=1$ and $c_R=1/2$, with common $K=3/5$ and left-family invariant $u+3c=3$. The fan edges are ${\xi }_L=-1$ and ${\xi }_R=1$. The complete self-similar solution is uniform at these stated left and right values outside the fan, while inside $-1<\xi <1$,It matches both states continuously, has $u^{\prime }=3/4=2/(\gamma +1)$, and has constant specific entropy. For fixed $x\ne 0$, its $t\downarrow 0$ limit is the required initial step.

A single rarefaction wave enforces two matching equalities: equal specific entropy and one common Riemann invariant. The inequality $u_R>u_L$ does not enforce either. If the entropies differ, one smooth isentropic fan cannot connect them; even if they agree, arbitrary end velocities and pressures need not satisfy either acoustic-family invariant.

The general positive-gas Riemann problem has a left acoustic wave, a middle contact discontinuity, and a right acoustic wave. Across the contact the two intermediate states have the same velocity $u_{\ast }$ and pressure $p_{\ast }$, but can have different mass densities and entropies. Each acoustic wave is a rarefaction wave or a shock wave, depending on the required intermediate pressure relative to its adjacent initial state. Shocks satisfy the conservative jump conditions and increase entropy. A sufficiently large pressure contrast can produce a shock on one side even when the initial velocities obey $u_R>u_L$.

For sufficiently rapid separation, the two rarefactions instead exhaust the gas and create vacuum between them. Their vacuum-edge velocities are $u_L+2c_L/(\gamma -1)$ and $u_R-2c_R/(\gamma -1)$, so a finite vacuum gap occurs whenEquality is the limiting case in which the vacuum edges meet. Thus the additional features can be two acoustic waves, a contact carrying an entropy jump, shocks, or a vacuum region, rather than one fan joining the two original uniform states.

The equilibrium quantities $\rho ,p,\mathbf{B},\Phi$ are real, and the gravitational Hessian ${\Phi }_{,ij}={\partial }_i{\partial }_j\Phi$ is symmetric. The crucial tensor property isThis follows directly from the displayed expression for $V$: the products ${\delta }_{ij}{\delta }_{kl}$ and ${\delta }_{il}{\delta }_{jk}$ are unchanged by interchanging the pairs $(i,j)$ and $(k,l)$; $B_j{B}_l{\delta }_{ik}$ is also unchanged; and the two terms involving $B_i{B}_j{\delta }_{kl}$ and $B_k{B}_l{\delta }_{ij}$ exchange with each other. All coefficients are real.

Multiply $F\mathbf{\xi }$ by $\rho {\mathbf{\eta }}^{\ast }$ and integrate by parts over all space. For admissible displacement fields, the boundary contribution at infinity vanishes: the fluid pressure and density are confined to the body, and the magnetic stress decays in the surrounding conducting medium. The conducting exterior is included in the integration. For a piecewise description, the usual displacement and traction matching conditions cancel the two opposite interface contributions; alternatively, take a smooth transition and then its limit. ThusThe first term is a Hermitian pairing because ${\Phi }_{,ij}={\Phi }_{,ji}$, and the second is Hermitian because $V_{ijkl}=V_{klij}$. Taking the complex conjugate of the same expression with $\mathbf{\eta }$ and $\mathbf{\xi }$ interchanged and relabelling $(i,j)\leftrightarrow (k,l)$ reproduces it. HenceThis establishes the required self-adjoint operator identity in the mass-weighted inner product, on the physical displacement domain with the stated decay and interface conditions. The negligible-density exterior can be treated as a small-positive-density conducting limit when defining this weighted pairing; its magnetic contribution is retained even as its inertia becomes negligible.

Self-adjointness gives $⟨\mathbf{\xi },F\mathbf{\xi }⟩=⟨F\mathbf{\xi },\mathbf{\xi }⟩=⟨\mathbf{\xi },F\mathbf{\xi }{⟩}^{\ast }$, soThe integrated expression from part (a) also writes it explicitly asThis is the quadratic change in potential energy associated with the perturbation, including pressure, magnetic and imposed-gravity restoring effects. Its sign need not be positive term by term.

The perturbation kinetic energy is $T=\frac{1}{2}⟨\dot{\mathbf{\xi }},\dot{\mathbf{\xi }}⟩$, and the total quadratic perturbation energy isSince the equilibrium and $F$ are time independent, differentiating and using the self-adjoint operator identity givesThus $\boxed{dE/dt=0}$ along the linearized motion. This conserved kinetic-plus-restoring energy is the basis of the magnetohydrodynamic energy principle.

First suppose $W[\mathbf{\xi }]>0$ for every nonzero admissible displacement. The conserved energy from part (b) then bounds both $T$ and $W$ by $E$. In particular,where $\Vert \mathbf{\xi }{\Vert }_{\rho }^2=⟨\mathbf{\xi },\mathbf{\xi }⟩$. The positive quadratic form $\Vert \dot{\mathbf{\xi }}{\Vert }_{\rho }^2+2W[\mathbf{\xi }]$ defines the perturbation energy norm and is conserved. Small perturbations in this norm remain small, proving energetic linear stability. Also $\Vert \mathbf{\xi }(t){\Vert }_{\rho }\le \Vert \mathbf{\xi }(0){\Vert }_{\rho }+\sqrt{2E}t$, excluding exponential displacement growth. For any admissible normal mode $\mathbf{\xi }=\mathbf{\zeta }{e}^{-i\omega t}$, multiplication of $-{\omega }^2\mathbf{\zeta }=F\mathbf{\zeta }$ by ${\mathbf{\zeta }}^{\ast }$ gives directlyso all such frequencies are real. No variational principle has been assumed.

To prove instability, suppose some admissible ${\mathbf{\xi }}_0$ has $W_0=W[{\mathbf{\xi }}_0]<0$, and put $N_0=\Vert {\mathbf{\xi }}_0{\Vert }_{\rho }^2>0$. Choose the initial velocityIts kinetic energy exactly cancels $W_0$, so $E=0$ for the entire subsequent motion. Write $N(t)=\Vert \mathbf{\xi }(t){\Vert }_{\rho }^2$. Since $F$ is self-adjoint,Zero energy gives $⟨\mathbf{\xi },F\mathbf{\xi }⟩=\Vert \dot{\mathbf{\xi }}{\Vert }_{\rho }^2$, and hence $N^{\prime \prime }=4\Vert \dot{\mathbf{\xi }}{\Vert }_{\rho }^2\ge 0$. Initially $N^{\prime }(0)=2\alpha N_0>0$, so $N$ stays positive and its logarithm is defined. The Cauchy-Schwarz inequality now givesTherefore $(\ln N{)}^{\prime }(t)\ge (\ln N{)}^{\prime }(0)=2\alpha$, and integration provesThis is an exponentially growing perturbation. Scaling the initial displacement and velocity down leaves $\alpha$ unchanged, so arbitrarily small initial perturbations display the same linear growth mechanism. It proves that a negative-energy direction gives exponential growth in a self-adjoint wave equation, without assuming a discrete spectrum or quoting a trial-function variational theorem.

Positive $W$ on nonzero displacements gives energetic stability; a negative direction gives instability. For mathematical precision, a uniform bound on the mass-weighted displacement norm follows if $W\ge \frac{1}{2}{\omega }_{\min }^2\Vert \mathbf{\xi }{\Vert }_{\rho }^2$ for some ${\omega }_{\min }>0$, giving $\Vert \mathbf{\xi }{\Vert }_{\rho }^2\le 2E/{\omega }_{\min }^2$. Strict positivity alone in an infinite-dimensional space need not supply such a uniform spectral gap; the standard magnetohydrodynamic energy principle asserts energetic stability and absence of growing modes. Cases with zero directions are marginal and require separate consideration.