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For an outgoing simple wave into water initially at rest at depth $H$, the incoming Riemann invariant is $u-2\sqrt{gh}=-2\sqrt{gH}$. Thus $\sqrt{gh}=\sqrt{gH}+u/2$ and the outgoing characteristic speed is $v=\sqrt{gH}+3u/2$. A boundary characteristic emitted at time $s$ is $x=v(s)(t-s)$. Its envelope satisfies $t=s+v/v^{\prime }$ and $x=v^2/v^{\prime }$. At a zero-velocity onset with positive acceleration, the leading-edge intersection is the displayed distance. Later characteristic curves can form a shock earlier elsewhere in a sufficiently strong waveform.

In a right-going simple wave, the state emitted at boundary time $s$ travels along $X=c_b(s)(t-s)$. The mapping from emission time to position first becomes singular when ${\dot{c}}_b(t-s)-c_b=0$, giving the displayed shock time and position $X=c_b^2/{\dot{c}}_b$. A smooth increase of traction produces steepening only if it increases the propagation speed. This describes the first gradient singularity; continuation as a shock wave needs admissible jump conditions.

Use material coordinates $X$ and write the placement as $y=X+u$. The strain energy density $W$ in the printed convention is per unit reference mass. It is convenient to use the conventional first Piola-Kirchhoff stress tensor $P=s^{\mathsf{T}}$, soAssume the undeformed reference state is stress-free. With $H=A-I$ and infinitesimal strain tensor $e=(H+H^{\mathsf{T}})/2$, material isotropy and material frame indifference give the quadratic expansionConsequently the Lamé parameters from hyperelastic energy Hessian areIndeed, differentiating the quadratic expression twice givesthe elastic stiffness tensor of linear elasticity. A prestressed reference state requires the corresponding initial-stress corrections; the displayed formula concerns the natural state.

For a plane elastic wave, $u=u(X_1,t)$, put $r=e_1+u_{,X_1}$ and $\Psi (r)=W([r,e_2,e_3])$. Balance of momentum without body force becomesThis is a material-coordinate equation: the reference density cancels between inertia and nominal stress. If $C{a}_{\alpha }=c_{\alpha }^2{a}_{\alpha }$, the first-order system has characteristic speeds $\pm c_{\alpha }$ and right eigenvectors $(a_{\alpha },\mp c_{\alpha }{a}_{\alpha })$. Along a characteristic curve with $d{X}_1/dt=s=\pm c_{\alpha }$, its differential compatibility relation isThis follows by multiplying the system by the left eigenvector $(-s{a}_{\alpha }^{\mathsf{T}},a_{\alpha }^{\mathsf{T}})$. It does not assert the existence of globally integrable Riemann invariants for an arbitrary three-component constitutive law.

There is a necessary qualification to the finite-amplitude transverse request. Set $u=(0,w,0)$ and $q=w_{,X_1}$, so $r=(1,q,0)$. The longitudinal momentum equation requiresThus arbitrary nonuniform pure transverse profiles are possible only if the compatibility condition for pure nonlinear transverse waves holds:The third component vanishes by the reflection symmetry implied by isotropy and objectivity. The shear equation then closes:

General compressible hyperelastic material assumptions alone do not imply the required compatibility. For a concrete counterexample, take the Saint Venant–Kirchhoff hyperelastic energyFor $A=[(a,q,0{)}^{\mathsf{T}},e_2,e_3]$, this becomesAt $a=1$, $P_{11}=(\lambda +2\mu )q^2/2$. A varying shear therefore produces a longitudinal force $(\lambda +2\mu )q{q}_{,X_1}$ and cannot preserve $u_1=0$. Infinitesimal transverse waves do exist, because this coupling is second order, but a general pure finite-amplitude transverse wave does not.

The compatible nonlinear shear energy class is nonempty and can be genuinely nonlinear. Define $J=\det A$, $S=\mathrm{tr}(A^{\mathsf{T}}A)-2J$, and, for $h\ge 0$, takeThis is objective and isotropic. Near $A=I$, $S-1=2\mathrm{tr}(e^2)-(\mathrm{tr}e{)}^2+O(|H{|}^3)$, so it has the specified Lamé parameters. On the plane deformation above, $S-1=(a-1{)}^2+q^2$ and $J=a$. Hence $P_{11}=0$ at $a=1$, whileFor positive $\lambda ,\mu$, the relevant incremental plane moduli are positive. This supplies an explicit stable pure transverse family and, when $h>0$, amplitude-dependent wave speed.

For any compatible family, letThe equations $q_t=v_{,X_1}$ and $v_t=c(q{)}^2{q}_{,X_1}$ give the Riemann invariantsConsider the initially undisturbed half-space $X_1>0$. A right-going simple wave has the incoming invariant $v+H=0$, so $v=-H(q)$. Denote the prescribed boundary nominal stress by $T(t)=P_{21}(0,t)=\tau (q_b(t))$. The physical boundary traction, whose outward normal is $-e_1$, is $-T(t)e_2$. Invert the monotone law $\tau$ to obtain $q_b$. The wave state emitted at boundary time $s$ is transported alongAhead of the leading front $X_1=c(0)t$ the state is zero. Before characteristic crossing, these equations determine the unique emission time $s$ for each disturbed point. The displacement itself is also determined, rather than just its derivatives:Differentiating the implicit emission relation verifies $w_{,X_1}=q_b(s)$ and $w_t=-H(q_b(s))$, so this is the complete pre-shock simple wave solution.

Write $c_b(s)=c(q_b(s))$. The boundary-generated characteristic curves lose invertibility whenThus the characteristic crossing in a boundary-generated elastic simple wave criterion isHere ${\dot{c}}_b=c^{\prime }(q_b)T^{\prime }/({\rho }_0{c}_b^2)$. A smoothly increasing boundary stress therefore steepens into a shock wave only when later states travel faster: $c^{\prime }(q_b)T^{\prime }>0$. Increasing stress alone is insufficient. For the compatible example with $h>0$ and $q_b>0$, $c^{\prime }(q_b)=3h{q}_b/({\rho }_0{c}_b)>0$, so a nonzero increasing ramp produces finite-time crossing. For $h=0$ all rays are parallel and no such shock forms. This criterion locates the first gradient singularity; subsequent discontinuous evolution additionally requires the jump conditions and an admissibility condition.

For a general vector simple wave, replace the pure-shear line by an integral curveSubstitution gives $a_t+c_{\alpha }(a)a_{,X_1}=0$, with the same emission construction and crossing criterion. Boundary tractions must lie on this one-family curve. Generic finite-amplitude quasi-transverse waves may contain longitudinal displacement; arbitrary vector boundary forcing need not generate a single simple wave.

On a region with smooth positive area, set $q=G(Z)Q(\theta ,\zeta )$, $\zeta =h(Z)$. Substitution into the simple wave equation givesRemove the last term by $G^{\prime }/G=-A^{\prime }/(2A)$, and choose $h^{\prime }=G$. Starting at a regular section $Z_0$, one convenient normalized area transformation of a nonlinear acoustic simple wave isIn the notation $g(\zeta )$, set $g(\zeta )=G(h^{-1}(\zeta ))$. The remaining equation is $Q_{\zeta }-Q{Q}_{\theta }=0$. With initial value $f$ at $Z_0$, the method of characteristics givesThe map from the initial label loses invertibility when $1-f^{\prime }(\xi )\zeta =0$. For smooth data with bounded derivative and $M=\sup f^{\prime }>0$, the first forward characteristic crossing is at ${\zeta }_s=1/M$; if $M\le 0$, no forward shock occurs.

For spherical spreading, write physical radius as $R>0$ and specify data at an emitting radius $R_0>0$. Since $A(R)=R^2$,The shock distance for a spherical simple wave launched at finite radius isIf the axial distance is measured from the emitting surface, $Z=R-R_0$, then $A(Z)=(R_0+Z{)}^2$ andNo finite shock distance exists in this model when $M\le 0$.

The printed combination $A=Z^2$ with general bounded data at $Z=0$ is singular and does not define the intended outgoing spherical initial-value problem. Along a spherical characteristic, $d\theta /dR=-q$ and $dq/dR=-q/R$, so $Rq$ is constant. A bounded trace at $R=0$ forces that constant to be zero for every characteristic curve emanating from the origin; arbitrary nonzero bounded $f$ cannot be launched there. For example, the ostensibly admissible data $f\equiv 1$ would demand both $Rq=0$ and a nonzero trace, a contradiction. The logarithmic transformed distance is also divergent at the origin. Thus an unqualified numerical shock formula from those literal data would be unjustified. The finite-radius formulas above give the physical interpretation once the launching radius, or the shifted area law, is supplied.

Before surface-wave breaking or reflected waves return, the shallow water equations are $h_t+(hu{)}_x=0$ and $u_t+u{u}_x+g{h}_x=0$. Set $c=\sqrt{gh}$ and $c_0=\sqrt{gH}$. Their Riemann invariants are $u\pm 2c$ along $dx/dt=u\pm c$. An outgoing simple wave entering resting water has $u-2c=-2c_0$, so the imposed boundary velocity givesThis formula requires $c_0+u/2>0$ and an appropriate inflow/outflow characteristic curve count. In particular $U<2c_0/3$ ensures that only the outgoing family enters the domain throughout the cycle; stronger forcing can change the boundary problem.

A characteristic curve emitted at time $s$ carries constant $u=U\sin \omega s$ and $h=h(0,s)$ and followsIncreasing boundary velocity produces faster following characteristic curves, steepening the advancing side of a crest; decreasing boundary velocity spreads the curves and produces a rarefaction wave. A trough has reduced depth and smaller propagation speed. Once the compressive curves intersect, the single-valued simple wave must be replaced by an admissible hydraulic bore, with mass conservation and momentum conservation across it and loss of mechanical energy. In a very weak or dispersive wavetrain an undular bore can replace the ideal discontinuity.

For a bore of speed $s_b$ entering undisturbed depth $H$, the Rankine-Hugoniot condition gives $s_b[h]=[hu]$ and $s_b[hu]=[h{u}^2+g{h}^2/2]$. If $h_2>H$ is its depth just behind, elimination of the behind-bore velocity yields $s_b^2=g{h}_2(h_2+H)/(2H)$ and $u_2=s_b(h_2-H)/h_2$. Thus the newly formed front travels faster than $c_0$, carries a depth jump, and dissipates energy; the behind-bore state must satisfy these balances rather than continuing intersecting smooth curves. The final sketch is schematic and does not claim a solved post-breaking boundary-value problem.

There is a numerical defect in the printed leading-edge claim. The first leading edge is the $s=0$ curve $x=c_{0}t$. The envelope condition ${\partial }_{s}x=0$ givesAt the leading edge,This follows directly from the specified boundary acceleration and the exact Riemann invariant. The printed coefficient $4/9$ is incompatible with these data: at that distance the leading-edge characteristic curves have not yet intersected. For the first compression interval, $0<\omega s<\pi /2$, one has $v^{\prime \prime }<0$. Consequently $d(s+v/v^{\prime })/ds=2-v{v}^{\prime \prime }/v^{\prime 2}>0$, so the first envelope occurs at $s=0$. This boundary-driven shallow-water wave steepening calculation also establishes that no later characteristic in the initial compression forms a bore earlier.

For a one-dimensional isentropic perfect gas, continuity and momentum are $D\rho =-\rho u_x$ and $Du=-c^2{\rho }_x/\rho$, where $D={\partial }_t+u{\partial }_x$ and $c^2=\partial P/\partial \rho$. Since $P=K{\rho }^{\gamma }$, differentiation givesThus applying $D+c{\partial }_x$ to $u+2(c-c_0)/(\gamma -1)$ cancels both terms:The complementary invariant $u-2(c-c_0)/(\gamma -1)$ is transported along $dx/dt=u-c$. In the right-running piston simple wave, those characteristics carry the undisturbed value zero, giving $c=c_0+(\gamma -1)u/2$.

A right-running characteristic emitted by the piston at time $\tau$ has $u=f\tau$ and speed $u+c=c_0+(\gamma +1)f\tau /2$, constant along that characteristic. It starts at $x=f{\tau }^2/2$, soNeighboring characteristics first intersect when $\partial x/\partial \tau =0$. This derivative is $\frac{\gamma +1}{2}ft-c_0-\gamma f\tau$, so the envelope time isFor $\gamma >1$, it is smallest at $\tau =0$. Consequently the first gradient catastrophe and ensuing shock form atThis uses the smooth isentropic simple-wave solution up to its first characteristic crossing, not beyond the newly formed shock.

For an isentropic perfect gas, $p=K{\rho }^{\gamma }$ and $c^2=dp/d\rho$. The one-dimensional Euler equations give $D\rho =-\rho u_x$ and $Du=-c^2{\rho }_x/\rho$, where $D={\partial }_t+u{\partial }_x$. Since $dc/d\rho =(\gamma -1)c/(2\rho )$, these becomeAdding the appropriate $\pm c{\partial }_x$ terms showsThese are the Riemann invariants. A simple wave is a solution in which one invariant is spatially and temporally constant, so all local state variables depend on the other one. For a right-moving wave connected to rest, the minus invariant is zero and $c=c_0+(\gamma -1)u/2$. The plus characteristic then has speed $u+c=c_0+(\gamma +1)u/2$, givingFor the supplied damped equation put $b=(\gamma +1)/2$. Along $\dot{x}=c_0+bu$, $\dot{u}=-\alpha u$, so an initial label $x_0$ givesThis solves the initial-value problem while the label map remains invertible. Its derivative is $1+(b/\alpha )(1-e^{-\alpha t})v^{\prime }(x_0)$. Nonnegative slopes cannot cause crossing; for negative slopes, the condition $\alpha >b{\sup }_{v^{\prime }<0}|v^{\prime }|$ keeps that derivative strictly positive. Hence the stated damping condition prevents shock formation at every finite time. Equality also avoids a finite-time crossing, although the derivative may tend to zero as time tends to infinity. The strict inequality in the question is sufficient rather than necessary.

Write $u=u_x$, $w=u_y$ and $B=B_y$. The isothermal equation of state replaces the adiabatic thermal equation by $p=c_s^2\rho$. Since the fields depend only on $x,t$, the solenoidal constraint gives ${\partial }_x{B}_x=0$ and the $x$ component of the ideal magnetohydrodynamic induction equation gives ${\partial }_t{B}_x=0$. Hence $B_x$ is constant in both space and time.

The continuity, two momentum and transverse induction equations areHere $B$ is the transverse magnetic component and the constant $B_x$ is the longitudinal component. The longitudinal magnetic force is the gradient of transverse magnetic pressure; the transverse force is magnetic tension. In primitive variables the isothermal coplanar magnetohydrodynamic characteristic matrix isThe imposed $z$ components stay zero because their corresponding equations have zero right-hand sides under this reduction.

Let $c=v-u$ be a characteristic speed relative to the fluid and define the Alfvén velocity ${\mathbf{v}}_a=\mathbf{B}/\sqrt{{\mu }_0\rho }$. To derive the characteristic polynomial, an eigenvector with components $(\delta \rho ,\delta u,\delta w,\delta B)$ obeysEliminating $\delta u$ and $\delta w$ on a nondegenerate branch givesTheir consistency yields $(c^2-c_s^2)(c^2-v_{ax}^2)=v_{ay}^2{c}^2$. Equivalently, direct expansion of $\det (\mathbf{A}-vI)$ gives the same polynomial without dividing by $c$ or a possibly vanishing field component:ThusThe discriminant is $(c_s^2-v_a^2{)}^2+4c_s^2{v}_{ay}^2\ge 0$, and both squared speeds are nonnegative. These are the fast magnetosonic wave and slow magnetosonic wave families, transported with the mean flow. The equations are hyperbolic; on a generic state the four distinct real eigenvalues give four characteristic directions. Special field alignments can make speeds coincide or the slow speed vanish. There is no separate entropy characteristic because the thermal model is isothermal, and the independent out-of-plane Alfvén wave polarization is excluded by $u_z=B_z=0$.

To construct a simple wave in magnetohydrodynamics, take $\mathbf{U}=\mathbf{U}(a)$ along a state-space curve whose tangent is a right eigenvector of one selected branch:Substitution into the partial differential equations then gives the scalar transport equationSuch curves exist locally by integrating the eigenvector field away from degeneracies. For example, parameterizing by $a=\rho$ where its eigenvector density component is nonzero gives the explicit state-curve equationswith signed $c$ equal to one of the four relative wave speeds. These formulas are used only where their denominators are nonzero.

A particularly transparent nonlinear example is an isothermal perpendicular magnetosonic simple wave. Set the allowed constant $B_x=0$, choose $B=K\rho$ and keep $w$ constant. For the fast branch,These relations satisfy the eigenvector equations and therefore all four original equations wheneverThe amplitude-dependent characteristic speed has derivativeThis already proves the existence of genuinely nonlinear waves in the stated system.

More generally, the method of characteristics applied to initial data $a(x,0)=a_0(x)$ givesIf the initial speed decreases towards increasing $x$ somewhere, the denominator reaches zero in finite time. When the negative minimum is attained, the first steepening time isFaster trailing states catch slower leading states, giving shock formation by characteristic intersection. For the right-moving perpendicular fast example, a negative density gradient provides such a compression. Beyond $t_{\ast }$ a smooth simple wave is no longer sufficient and an appropriate shock solution is required. Rarefactive profiles instead spread. A linearly degenerate characteristic field has $v^{\prime }=0$ along its state curve and avoids this mechanism; the conclusion is that compressive nonlinear magnetosonic simple waves steepen, not that every possible MHD waveform must do so.

For smooth isentropic flow with $p=K{\rho }^{\gamma }$, $\gamma >1$, the continuity and momentum equations givewhere $c^2=dp/d\rho$. Adding and subtracting these equations showsThese are the Riemann invariants on the two characteristic families.

Before any shock, the incoming $C_{-}$ characteristics originate in the initially uniform gas, so their invariant is zero. Thus $c=c_0+(\gamma -1)u/2$ throughout the disturbed simple wave. A $C_{+}$ characteristic leaving the piston at time $\tau$ has constant $u=\dot{X}(\tau )$ and speed $u+c=c_0+(\gamma +1)\dot{X}(\tau )/2$. ThereforeThis parametrization applies for $0\le \tau \le t$ in $X(t)\le x\le c_{0}t$ as long as the characteristics remain ordered. The gas for $x\ge c_{0}t$ is still at rest. The values $\tau =t$ and $\tau =0$ give the piston and leading disturbance respectively.

Characteristic crossing first occurs when the derivative of the displayed $x(\tau ,t)$ with respect to $\tau$ vanishes. With $U(\tau )=\dot{X}(\tau )$ this givesFor the printed $X=2c_0{t}^3/(3T^2)$, $U=2c_0{\tau }^2/T^2$, soIts minimum over $\tau >0$ occurs at ${\tau }_{\ast }=T/\sqrt{3\gamma +1}$ and satisfies $t_s>{\tau }_{\ast }$. Hence the first shock forms atThe characteristic formula is restricted to earlier times; crossing afterwards must be replaced by a shock satisfying the conservation laws.

Write $u=u_x$ and $B=B_z$. In this geometry the Lorentz force density is purely longitudinal and equals $-{\partial }_x(B^2/2{\mu }_0){\mathbf{e}}_x$. The ideal magnetohydrodynamic induction equation becomes $B_t+{\partial }_x(uB)=0$. Together with the continuity equation and adiabatic pressure evolution, the four primitive-variable equations areThey have the required quasilinear form ${\mathbf{U}}_t+A{\mathbf{U}}_x=0$ withHere the notation $u_x$ in derivatives means ${\partial }_{x}u$, whereas $u$ denotes the longitudinal velocity. Spatial dependence is one-dimensional; time dependence is allowed in the wave problem.

Put $c_f^2=\gamma p/\rho +B^2/({\mu }_0\rho )$, and assume $\rho >0$ and $c_f>0$. Expansion of the characteristic polynomial givesTwo independent eigenvectors for ${\lambda }_0$ can be chosen as $(1,0,0,0{)}^T$ and $(0,-B/{\mu }_0,0,1{)}^T$. For the two propagating branches, useful eigenvectors areThe repeated branch advects an entropy disturbance and a disturbance with unchanged total gas-plus-magnetic pressure. The $\pm$ branches are perpendicular fast magnetosonic waves, with speed $c_f$ relative to the fluid.

On a uniform background, a Fourier amplitude $\hat{\mathbf{U}}{e}^{i(kx-\omega t)}$ obeys $(kA-\omega I)\hat{\mathbf{U}}=0$. Thus the eigenvalues are the phase velocities $\omega /k$ for $k\ne 0$, and the four dispersion branches are $\omega =ku$ twice and $\omega =k(u\pm c_f)$. The reduced geometry retains neither an independent transverse velocity nor an additional propagating Alfvén wave.

To construct each finite-amplitude simple wave in magnetohydrodynamics, trace a state-space curve with tangent $r_{\pm }$, using $\rho$ as its parameter. Integrating the pressure and magnetic components givesThe constants $K$ and $b$ are the entropy and transverse magnetic flux freezing invariants along this state curve. HenceSince $A{\mathbf{U}}_{\rho }={\lambda }_{\pm }{\mathbf{U}}_{\rho }$, substituting this curve into the original system reduces it to ${\rho }_t+{\lambda }_{\pm }(\rho ){\rho }_x=0$. By the chain rule,Thus the characteristic speed itself satisfies the Inviscid Burgers equation. For initial speed $L(a)$, its smooth solution is ${\lambda }_{\pm }=L(a)$ along $x=a+tL(a)$, until the characteristic curves intersect. This derivation supplies genuine state curves for simple waves, not merely a linear dispersion relation.

Take the primitive state vector $\mathbf{U}=(\rho ,p,u_x,u_y,u_z,B_x,B_y,B_z{)}^T$, with $\rho >0$. In the absence of gravity, the ideal magnetohydrodynamic equations are first order: expanding the material derivatives, pressure gradient and Lorentz force makes every term linear in a spatial derivative of $\mathbf{U}$, with coefficients depending on $\mathbf{U}$. Dividing momentum balance by $\rho$ therefore gives a quasilinear partial differential equation system ${\partial }_t\mathbf{U}+A_i(\mathbf{U}){\partial }_i\mathbf{U}=0$ with eight-by-eight matrices. The solenoidal magnetic-field constraint is an additional constraint on initial data; the ideal magnetohydrodynamic induction equation preserves it because the divergence of a curl vanishes.

For a one-dimensional simple wave in magnetohydrodynamics, write $\mathbf{U}=\mathbf{U}(q(x,t))$ with a nonzero state-space tangent. Substitution gives ${\mathbf{U}}_q{q}_t+A_x{\mathbf{U}}_q{q}_x=0$. A nonconstant profile requires the tangent to be a right eigenvector of $A_x$:The chain rule then gives the wave-speed equation:This is the Inviscid Burgers equation, including the special case of constant $v$. Its characteristic curves are $x=a+v_0(a)t$, with $v=v_0(a)$. As long as the mapping remains invertible,Thus a region with $v_0^{\prime }<0$ steepens: faster characteristics catch slower ones, and the first gradient catastrophe occurs at $t_{\ast }=-1/{\min }_a{v}_0^{\prime }(a)$ when the minimum is negative. A genuinely nonlinear compressive simple wave therefore forms a shock wave. Rarefactive profiles can spread instead; steepening is not inevitable for every initial profile.

To continue past this time, use weak solutions of the conservative mass, momentum, fluid total-energy equation and ideal magnetohydrodynamic induction equation. The Rankine-Hugoniot conditions fix the jumps and shock velocity, while physical entropy production selects admissible magnetohydrodynamic shocks. Diffusive shock wave layers are replaced by moving discontinuities, so their microscopic structure need not be explicitly resolved. In particular, the smooth adiabatic pressure equation must be replaced by total-energy conservation when imposing the ideal magnetohydrodynamic shock conditions.

For the Alfvén waves, impose the one-dimensional solenoidal magnetic-field constraint, so $B_x$ is constant and $\delta B_x=0$. Put ${\mathbf{B}}_{\perp }=(B_y,B_z)$ and $c=v-u_x$. The transverse components of the linearized ideal magnetohydrodynamic equations giveThe longitudinal momentum equation additionally gives $\delta p+{\mathbf{B}}_{\perp }\cdot \delta {\mathbf{B}}_{\perp }/{\mu }_0=0$. The Alfvén eigenvectors have $\delta \rho =\delta p=\delta u_x=0$, hence their transverse polarization is perpendicular to ${\mathbf{B}}_{\perp }$. For $s=\pm 1$, their speed and explicit right Alfvén characteristic eigenvectors areFor ${\mathbf{B}}_{\perp }\ne 0$ one may choose $(h_y,h_z)=(B_z,-B_y)$, up to a nonzero scalar factor. If ${\mathbf{B}}_{\perp }=0$, either transverse polarization is allowed and this characteristic speed is degenerate. These expressions describe the two propagating Alfvén branches for $B_x\ne 0$; if $B_x=0$, they coalesce with advected, nonpropagating transverse disturbances.