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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.

For positive depth and reduced gravity, the characteristic speeds are $u,u\pm c$. Along the material characteristic, $D{g}^{\prime }=-2g^{\prime }{w}_e/h$. Along $D_{\pm }={\partial }_t+(u\pm c){\partial }_x$, the left-eigenvector compatibility equations areIf $w_e=0$ and $g^{\prime }$ is spatially constant, these reduce to the Riemann invariants $u\pm 4c$. Variable buoyancy or entrainment prevents that simplification.

A channel with width $b(z)=\alpha z^2$ has wetted area $A=\alpha h^3/3$. Its inviscid shallow water equations have characteristic speeds $u\pm c$ and the displayed Riemann invariants. Releasing a reservoir of depth $H$ at rest into a dry channel gives a left-family rarefaction wave: $c=(6c_0-x/t)/7$, $u=6(c_0+x/t)/7$, $h/H=[(6-x/(c_{0}t))/7{]}^2$ between $x=-c_{0}t$ and $x=6c_{0}t$, where $c_0=\sqrt{gH/3}$. This width law differs from a geometrically parabolic trough, whose width is proportional to $\sqrt{z}$.

In a prismatic triangular channel with constant reduced gravity, the wave speed is $c=\sqrt{g^{\prime }h/2}$ and the Riemann invariants are $u\pm 4c$. A dam release into a deep ambient produces a simple-wave rarefaction with $u+4c=4c_0$. The imposed gravity-current front condition $u_f=F\sqrt{g^{\prime }{h}_f}$ gives the displayed uniform nose-adjacent state. The fan has $c=(4c_0-x/t)/5$ and $u=4(c_0+x/t)/5$, extending from $x/t=-c_0$ to $u_f-c_f$; a uniform region then reaches the front $x/t=u_f$. The nose closure supplies physics absent from the one-layer interior equations.

With $A$ invertible, write the system as ${\mathbf{u}}_x+M{\mathbf{u}}_y=\mathbf{d}$, $M=A^{-1}B$. Real distinct eigenvalues of $M$ give strict hyperbolicity; a real complete eigenbasis gives the usual diagonalizable hyperbolic form. The characteristic curves satisfy $dy/dx={\lambda }_i$. Left eigenvectors form characteristic dependent combinations $w_i={\ell }_i\mathbf{u}$; for constant coefficients they obey separate transport equations. Characteristic coordinates are functions constant on those curves, such as $y-{\lambda }_{i}x$ in the constant case. For variable coefficients extra derivatives of the eigenvectors can couple the combinations; local characteristic directions still follow the real matrix pencil. If $A$ is singular one first chooses a noncharacteristic coordinate direction with an invertible coefficient combination.

For the specified constants,Its speeds are $2$ and $-1$, with left eigenvectors $(2,1)$ and $(1,1)$. HenceThe transported Riemann invariants are $\boxed{R_{+}=2u+v-4x,R_{-}=u+v-x}$, constant on $y-2x=\text{constant}$ and $y+x=\text{constant}$, respectively. Initial data give $w_{+}(0,s)=s-\sin (s/2)$ and $w_{-}(0,s)=-e^s$, soInvert these two combinations:Both characteristic feet must lie on the supplied positive initial half-line. ThusOutside this sector at least one invariant needs unsupplied initial data. The closed expressions give one possible extension there but are not uniquely forced by the question's initial half-line.

Write the system as $U_x+A(U)U_y=0$ with $U=(u,v{)}^T$ andIts two characteristic speeds are ${\lambda }_{\pm }=v\pm \sqrt{u}$, so it is strictly hyperbolic precisely when $u>0$. At $u=0$ they coincide and the matrix is not diagonalizable; for $u<0$ the speeds are nonreal.

For $u>0$ define $R_{\pm }=v\pm 2\sqrt{u}$ and $D_{\pm }={\partial }_x+(v\pm \sqrt{u}){\partial }_y$. Substitution from the two differential equations givesThereforeThese are the two Riemann invariants; their constancy applies within smooth regions, not across a shock.

Take a dense lower layer and write $g^{\prime }=g({\rho }_1-{\rho }_0)/{\rho }_0>0$. In the Boussinesq approximation the fractional density difference is small; use the reference density in inertia and retain the difference in buoyancy. The one-layer model additionally treats the ambient as much deeper, with negligible distributed horizontal inertia away from the head. The shallow water approximation requires depth small relative to the horizontal variation length, small vertical accelerations, nearly hydrostatic pressure, and an approximately depth-uniform horizontal velocity. Neglect mixing, viscosity, bed friction and rotation. A slowly varying rectangular channel allows cross-sectional averages without large lateral separation.

A slice of width $b(x)$ contains volume $bhdx$. Its outgoing volume flux is $bhu$, so volume conservation gives $(bh{)}_t+(bhu{)}_x=0$. The integrated hydrostatic excess pressure is ${\rho }_0{g}^{\prime }b{h}^2/2$. Pressure on the sloping side walls contributes ${\rho }_0{g}^{\prime }{h}^2{b}^{\prime }/2$ to the longitudinal force. The conservative momentum equation is thereforeUse the volume equation to simplify it. The resulting shallow water equations areOmitting the wall force would incorrectly introduce a width term into the material acceleration equation.

For $(h,u)$ the principal matrix is $\left(\begin{array}{cc}u& h\\ g^{\prime }& u\end{array}\right)$. Its eigenvalues are $u\pm c$, where $c=\sqrt{g^{\prime }h}$. For $h>0$ they are real and distinct, so the system is strictly hyperbolic. It degenerates at a dry front. From the volume equation, $c_t+u{c}_x+(c/2)u_x=-(cu/2)b^{\prime }/b$. Combine this with the momentum equation to obtain the variable-width shallow-water characteristics:In a constant-width channel $u\pm 2c$ are Riemann invariants along the corresponding characteristic curves; with varying width they satisfy compatibility equations with a source term.

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.

Write $g^{\prime }=g({\rho }_f-{\rho }_a)/{\rho }_f>0$. The Saint-Venant dry-front condition describes a liquid advancing over a dry bed, or into an ambient whose inertia is negligible, as when ${\rho }_a/{\rho }_f\ll 1$. The edge has $h_f=0$; its velocity is obtained by continuing the appropriate shallow water Riemann invariant to zero depth. There is no independent universal finite-depth head Froude number. In the triangular geometry a release of a resting layer gives $u_f=2\sqrt{2g^{\prime }{h}_0}$, whereas a rectangular release gives $2\sqrt{g^{\prime }{h}_0}$.

The Benjamin deep-ambient front condition includes the inertia of an ambient that must turn around a finite-depth gravity current head. In the ideal deep-ambient Boussinesq approximation, ${\rho }_a/{\rho }_f$ is close to one andHere $h_f$ is the current depth just behind its head, $u_f$ its speed relative to the far ambient, and $F$ the ideal head Froude number. The ambient is much deeper than $h_f$; a finite upper-layer depth changes the closure. The lower density must exceed the ambient density for a bottom current. A non-Boussinesq head requires an appropriate density-ratio correction rather than using $F=\sqrt{2}$ at every ratio. The Saint-Venant and Benjamin models therefore describe different front physics, even though both use the same long-wave equations in the interior.

Constant density in each layer and negligible ambient acceleration far from the front giveThis primitive momentum conservation equation remains valid with slowly varying $B$. In conservative form it iswhere the right side is the longitudinal component of the hydrostatic pressure force on the sloping sidewalls. Omitting that term while retaining variable $B$ would give the wrong acceleration.

Together with volume conservation, the principal coefficient matrix in $(h,u)$ isIts real eigenvalues and characteristic curves areThese are downstream and upstream gravity waves relative to the moving fluid. The factor $1/2$ is geometric: the hydraulic depth is $A/b(h)=h/2$. Both waves move downstream in a sufficiently fast current, whereas one can carry information upstream in a slower current.

For $B=1$, differentiate $c^2=g^{\prime }h/2$ and combine the two equations. The Riemann invariants areFor variable $B$ they instead obeywhich also checks that the invariant claim depends on a prismatic channel.

Put $c_0=\sqrt{g^{\prime }{h}_0/2}$. The undisturbed reservoir has $u=0$ and Riemann invariants $u\pm 4c=\pm 4c_0$. The right-going characteristic curves supply the invariant $u+4c=4c_0$ throughout the release. The spreading disturbance is the other family, whose speed is $u-c$. In its rarefaction wave let $\xi =x/t=u-c$. ThenAt the reservoir edge $u=0,c=c_0$, so the back of the rarefaction wave travels at $-c_0$. At the dry edge $c=0$ the Saint-Venant dry-front condition givesThus the fan occupies $-c_{0}t\le x\le 4c_{0}t$, while $x<-c_{0}t$ remains undisturbed. Its full length grows at $5c_0$; there is no finite-depth uniform shelf between the fan and the dry front. The optional interior depth is $h=h_0[(4c_0-x/t)/(5c_0){]}^2$.

The fan's minus-family characteristic curves are rays $x=\xi t$. For the plus family in the fan, $dx/dt=u+c=(8c_0+3x/t)/5$; integration gives $x=4c_{0}t+C{t}^{3/5}$, with constants selected at entry from the reservoir. These curved paths transport the constant plus Riemann invariant.

Put the broken dam at $x=0$ and let $c_0=\sqrt{gH/3}$. A dry bed is assumed downstream. The left-family rarefaction wave joins the undisturbed reservoir $h=H,u=0$ to the dry front. The other Riemann invariant is fixed at $u+6c=6c_0$. Within the fan its characteristic speed equals $\xi =x/t=u-c$, soThus the dry-bed dam break with cubic wetted area has the fieldsVelocity is not a meaningful fluid quantity in the dry region; its wet-side limiting value is $u_f=6c_0=2\sqrt{3gH}$. The upstream head travels at $-c_0$, while the front travels at the stated constant speed in this inviscid model.

Bottom drag slows the flow and changes the dry-tip structure, where the depth is small and bed stress per unit fluid mass becomes especially important. It can replace the inviscid zero-depth high-speed tail by a drag-controlled nose. An exact altered profile or front speed requires a specified drag law; the inviscid speed should not be extrapolated to a thin real irrigation-channel front without that qualification.

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.

Put $G=g({\rho }_p-{\rho }_0)/{\rho }_0$ and $b=G\varphi >0$, the reduced gravity. Use the Boussinesq approximation, hydrostatic pressure, uniform vertical velocity and concentration profiles, negligible fluid entrainment and drag, and a flat bed. Integrating volume conservation, momentum conservation and the particle budget over the layer givesThe hydrostatic pressure resultant is ${\rho }_{0}b{h}^2/2$; the last right-hand side is the deposition flux. With $D={\partial }_t+u{\partial }_x$, these areThe quasilinear coefficient matrix in variables $(h,u,b)$ isIts eigenvalues are $u,u\pm c$, where $c=\sqrt{bh}$. Therefore the characteristic curves and their compatibility ordinary differential equations areFor the last equation the left eigenvectors are $(\pm c/h,1,\pm h/(2c))$, and multiplying the quasilinear equations by them gives the displayed source terms. This is the sedimenting rectangular-channel characteristic compatibility relation; it supplies one differential relation on each wave family, rather than three independently closed scalar equations there. The contact eigenvector has $du=0$ and $d(b{h}^2)=0$. For reference, an equivalent form isOnly for constant $b$ without settling do $u\pm 2c$ become genuine Riemann invariants. With settling, the spatial buoyancy gradient cannot be silently omitted.

Without settling, $b=b_0=G{\varphi }_0$ and $c_0=\sqrt{b_{0}H}$. The left-running rarefaction wave has the unchanged Riemann invariant $u+2c=2c_0$ and similarity coordinate $\xi =x/t=u-c$. HenceMatching to the Froude number condition $u_f=F{c}_f$ givesThe full outer profiles are $(h,u)=(H,0)$ for $x/t<-c_0$; the displayed fan for $-c_0<x/t<{\xi }_s$; and $(h,u)=(h_f,u_f)$ for ${\xi }_s<x/t<u_f$, whereAhead of $x_f$ there is only ambient fluid, so the particle-laden layer has zero depth. The shelf width is $(u_f-{\xi }_s)t=c_{f}t>0$ for every finite positive $F$. It disappears only in the formal $F\to \infty$ dry-front limit, where $u_f\to 2c_0$. For $F<1$, the fan/shelf junction lies to the left of the initial barrier; the profiles remain consistent. This is the finite-Froude lock-release rarefaction, valid while the reservoir can be treated as semi-infinite.

The undisturbed reservoir remains at $h=H$, $u=0$, while the particle budget gives $\varphi ={\varphi }_0{e}^{-V_{s}t/H}$. The edge of the rarefaction wave advancing into that reservoir travels with the local left-running characteristic speed, $-\sqrt{b_{0}H}{e}^{-V_{s}t/(2H)}$. ThusThis is the disturbance head into the initially resting reservoir, also called the leading edge of the rarefaction wave; it reduces to $-c_{0}t$ as $V_s\to 0$. The rightmost fan/shelf junction is the rarefaction tail, not this reservoir edge. Its trajectory in the full sedimenting system cannot be obtained by retaining the constant-buoyancy Riemann invariants.

For the requested near-front approximation, hold $h_f=4H/(F+2{)}^2$ fixed and take the shelf's concentration initially to be ${\varphi }_0$. The locally uniform shelf approximation gives $b_f=b_0{e}^{-V_{s}t/h_f}$, soThis tends to $u_{f}t$ at small settling and predicts a finite limiting front location within the fixed-depth closure. It is not an exact solution of the spatially varying variable-buoyancy shallow water equations: different material histories and resulting buoyancy gradients eventually invalidate a perfectly uniform shelf. No exact finite-time stoppage is claimed.

Interpret ${\rho }_2$ as the ambient density: the PDF's phrase assigning it a “depth” is a typographical error. Take $0<{\rho }_1-{\rho }_2\ll {\rho }_2$, and use a common reference density in inertia while retaining the density difference in buoyancy. This is the Boussinesq approximation, with reduced gravity $g^{\prime }=g({\rho }_1-{\rho }_2)/{\rho }_{\mathrm{ref}}>0$. For the shallow water approximation, the lower-layer depth must be small compared with longitudinal variation scales, vertical accelerations must be negligible and pressure hydrostatic. The deep ambient is approximated as a hydrostatic pressure reservoir away from the nose; assume a uniform velocity over each lower-layer cross-section and neglect drag and entrainment.

Write the prismatic triangular area as $A(h)=k{h}^2$ with fixed $k>0$, so the width at height $z$ is $2kz$. The excess hydrostatic pressure force divided by reference density isThe prismatic-channel shallow water equations are thereforeFor positive depth, divide the first equation by $2kh$ and use it in the second to obtainThe coefficient matrix in variables $(h,u)$ is $\left(\begin{array}{cc}u& h/2\\ g^{\prime }& u\end{array}\right)$. Its two distinct real characteristic speeds areThus the system is strictly hyperbolic for $h>0$ and $g^{\prime }>0$, though it degenerates at zero depth. Since $dc/dh=c/(2h)$, direct substitution in the two equations givesHence $R_{\pm }=u\pm 4c$ are the Riemann invariants conserved along their respective characteristic curves. The factor four, rather than the rectangular-channel factor two, comes from the triangular relation $A\propto h^2$.

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.