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Momentum conservation states that total momentum stays constant when the net force exerted from outside the chosen collection is zero. Use a closed collection, meaning the same fixed-mass point masses are followed throughout, and an inertial frame. ThenTo see why, sum Newton's second law over the members. If ${\mathbf{F}}_{ij}$ is the force on member $i$ from member $j$, Newton's third law assumes ${\mathbf{F}}_{ij}=-{\mathbf{F}}_{ji}$. Each pair cancels in the total:The internal forces act on different bodies and can change their individual momenta; their contributions cancel only in the collection's sum. An isolated system has no external interaction, but balanced external forces also suffice for momentum conservation. A closed system merely excluding matter exchange need not have conserved momentum. ^[ref]

The usefulness is that the detailed internal forces during a collision need not be known. Both an elastic collision and a perfectly inelastic collision obey the same isolated momentum balance; kinetic energy conservation is an additional condition for the elastic collision. If the external impulse is small but nonzero, the total momentum change is correspondingly small. For a variable-mass system exchanging matter, the entering or leaving momentum must also be included.

In a translation-invariant field theory, integrating over the spacetime zero mode enforces momentum conservation at interactions and scattering amplitudes.

In a sectionally uniform flow, a volume source $Q_x$ per downstream length brings axial velocity $u_a$. Momentum conservation gives the displayed source correction, with $A$ the wetted area. The head derivative is $[h+H+u^2/(2g){]}_x=Q_x(u_a-u)/(gA)$. Removing fluid at the local velocity preserves the head in this approximation. Adding fluid at rest does not: its acceleration reduces that head. The volume source alone cannot specify the momentum exchange without a velocity assumption.

In the pair-interaction model of Newtonian mechanics, two bodies exert equal and opposite forces on each other:Both vectors are expressed in the same inertial frame at the same time. The subscripts name the agent and the recipient: $A$ acts on $B$, and $B$ acts on $A$. These are forces on different bodies, so they cannot cancel in the net force on either body. Newton's second law still uses all the forces on the particular body whose acceleration is sought. Identifying this third-law interaction pair does not require the bodies to be isolated: additional external forces can act at the same time.

The distinction makes Newton's third law useful both for drawing a free-body diagram and for deriving momentum conservation. A body-by-body calculation retains one member of each third-law interaction pair; a calculation for a collection containing both bodies cancels that pair in the total momentum balance. Equal interaction forces consequently coexist with nonzero, generally unequal accelerations.

Isaac Newton stated the equality and opposition of mutual actions in Law III of his 1687 Philosophiae Naturalis Principia Mathematica. The explanation distinguishes equal opposite changes of momentum from changes of velocity, which are inversely proportional to the bodies' masses when other influences are absent. Corollary III then explains why mutual interactions preserve the signed total momentum. The equations here are modern notation for that distinction. ^[ref]

The instantaneous body-pair equation has a scope: use it for the mechanical interactions assumed by this Newtonian mechanics model. In electromagnetism, bodies can exchange momentum with an electromagnetic field; their instantaneous mutual forces need not sum to zero. Momentum conservation then accounts for electromagnetic field momentum as well as matter, and for momentum crossing the chosen boundary. This complication occurs in classical electromagnetism, without invoking quantum mechanics.^[ref]

A steady well-mixed dilute particle-laden gravity current above an absorbing porous bed loses carrier fluid at speed $W$ and particles at total downward speed $W+V$, where $V$ is their relative settling velocity magnitude. Volume conservation and the particle budget give $Q=Q_0-Wx$ and $\varphi =(Q/Q_0{)}^{V/W}$ for $W>0$. The momentum conservation equation is $(Q^2/h+g_0^{\prime }\varphi h^2/2{)}^{\prime }=-Wu$ if the withdrawn fluid carries its local horizontal velocity. Consequently $u{u}^{\prime }+g_0^{\prime }\varphi h^{\prime }+g_0^{\prime }h{\varphi }^{\prime }/2=0$. At $W=0$, $\varphi =e^{-Vx/Q_0}$ and the momentum flux is constant.

Work in the parent's rest frame. Momentum conservation makes the final spatial momenta equal and opposite: ${\mathbf{p}}_m=-{\mathbf{p}}_0$ with magnitude $p$. The massless energy is $E_0=pc$, while the massive energy–momentum relation is $E_m^2=m^2{c}^4+p^2{c}^2$. Conservation of energy gives $E_m+pc=M{c}^2$. Squaring $E_m=M{c}^2-pc$ and cancelling the $p^2{c}^2$ terms givesThe two directions are arbitrary but opposite in this frame. These are total energies, including the massive daughter's rest energy, as required for relativistic two-body decay.

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.

At $V=0$, the particle budget gives $\varphi =1$, so $g^{\prime }=g_0^{\prime }$ is constant. The primitive momentum conservation equation integrates toIntroduce $q=Q/Q_0=1-Wx/Q_0$, $y=h/h_0$ and $\epsilon ={\mathrm{Fr}}_0^2=Q_0^2/(g_0^{\prime }{h}_0^3)\ll 1$. The exact relation on the subcritical flow branch isExpanding uniformly for $0\le q\le 1$ givesThe depth increases slightly because horizontal kinetic energy decreases as fluid is removed. In fact the depth equation has $h^{\prime }\ge 0$ on this branch, andThus the small-Froude number approximation becomes better downstream, including as the flux vanishes. At leading order $u=(Q_0-Wx)/h_0$.

Use $G=g\beta (T-T_0(z))$ for the plume reduced gravity, and take $z$ upwards. For a symmetric line plume define fluxes per unit source length,The integrals span the plume; its edges can move horizontally with $z$. The fluid entrainment rate at one moving edge is the inward normal transport relative to that edge. For an edge $x=b(z)$ it is $E=w{b}^{\prime }-u$; symmetry supplies equal transport through the other edge. The Batchelor entrainment hypothesis closes its mean value as $E=\alpha w_c$, where $w_c$ is a representative vertical velocity and $\alpha$ is the entrainment coefficient. The integrated volume conservation equation is therefore $q^{\prime }=2\alpha w_c$.

Using volume conservation, the vertical acceleration equation has conservative form ${\partial }_x(uw)+{\partial }_z(w^2)=G$. Incoming ambient has zero vertical momentum; the integrated momentum conservation equation is consequently $m^{\prime }=\int Gdx$. For the temperature anomaly, $T_0$ depends on height andAmbient fluid enters with $G=0$, giving the buoyancy flux balance. The integral model isHere $N$ is the buoyancy frequency when $N^2>0$; the same algebra permits unstable temperature gradients with $N^2<0$.

These are mean integral balances. The advective equations printed in the question omit explicit turbulent stresses and heat fluxes; a turbulent plume interpretation requires the usual entrainment closure for that unresolved transverse transport. Assuming vanishing edge stresses and heat fluxes, and negligible additional axial turbulent-flux contributions, gives the displayed bulk model. It would be inconsistent to treat a discontinuous top-hat profile as a smooth exact pointwise solution of the inviscid temperature-advection equation.

A simple fluid reduces to a linear viscoelastic fluid when its deformation and rotation over the relevant memory interval are small enough to linearize the constitutive history about an isotropic relaxed state. Small amplitude is essential; small frequency is not. For example, arbitrarily high-frequency oscillation can still be linear if its strain amplitude is sufficiently small. The relaxation modulus $G(t)$ is the shear-stress response to a unit small step of shear strain.

Choose the harmonic convention $e^{i\omega t}$. If the imposed shear rate is $\mathrm{Re}(\hat{\dot{\gamma }}{e}^{i\omega t})$, the corresponding shear stress is $\mathrm{Re}({\eta }^{\ast }(\omega )\hat{\dot{\gamma }}{e}^{i\omega t})$. Substitution into the constitutive convolution gives the complex viscosity:For a nondecaying elastic part this formula is understood through the limit from a positive Laplace damping parameter. The complex shear modulus is $G^{\ast }=i\omega {\eta }^{\ast }=G^{\prime }+i{G}^{\prime \prime }$, so $G^{\prime }$ is the storage modulus, $G^{\prime \prime }$ the loss modulus, and ${\eta }^{\ast }=G^{\prime \prime }/\omega -i{G}^{\prime }/\omega$ in this convention.

Creep is the progressive deformation produced by a held stress. For a small step $S_{0}H(t)$ of shear stress, define the creep compliance $J$ by $\gamma (t)=S_{0}J(t)$ from a relaxed state. Writing hats for Laplace transforms, the same constitutive law gives $\hat{S}=p\hat{G}\hat{\gamma }$. Since $\hat{S}=S_0/p$, this yieldsHere $H$ is the Heaviside step function. Inversion determines the full creep response, including any instantaneous elastic strain jump. Equivalently, the causal relaxation/creep convolution includes that jump through ${\int }_{0^{-}}^{t}G(t-s)dJ(s)=1$ for $t>0$.

For the Maxwell fluid, $\hat{G}(p)=G_0/(p+1/\tau )$. Thus the two material responses areThe creep response is an elastic jump followed by constant-rate viscous deformation. The zero-frequency shear viscosity is $\eta =G_0\tau$.

For the wall problem, take $v=u(y,t)e_x$ and denote the shear component of stress by $S$. The incompressibility constraint is satisfied, and the advective part of the material derivative vanishes because no field depends on $x$. With no imposed longitudinal pressure gradient, momentum conservation and the linear Maxwell law giveThe initial conditions are $u=S=0$ in $y>0$; the no-slip boundary condition is $u(0,t)=UH(t)$, with decay as $y\to \infty$. Transforming these equations givesChoose the analytic square-root branch in $\mathrm{Re}p>0$ with $\mathrm{Re}q>0$, so that the spatially increasing solution is excluded. The Maxwell start-up shear layer consequently hasThe Bromwich inversion formula now givesThe vertical Bromwich contour is directed upward and lies to the right of all singularities, including the origin and the branch points at $0$ and $-1/\tau$. A cut joining these branch points is a possible continuation into the left half-plane; inversion only requires the stated right-half-plane branch.

The Newtonian limit holds $\eta =G_0\tau$ fixed while $\tau \to 0$, so $G_0\to \infty$. Then $q(p)\to \sqrt{\rho p/\eta }$. With $\nu =\eta /\rho$, the inverse Laplace transform gives the complementary error function profileThis is diffusive momentum penetration in a Newtonian fluid, with thickness of order $\sqrt{\nu t}$.

The elastic limit instead holds $G_0$ fixed while $\tau \to \infty$. Write $c=\sqrt{G_0/\rho }$. On the chosen branch, $q(p)\to p/c$, so the Laplace transform time-shift rule yieldsAn elastic shear front travels at speed $c$: the material ahead is at rest, while that behind moves at the wall speed. Indeed the limiting displacement is $d(y,t)=U(t-y/c)H(t-y/c)$, whose shear strain is $d_y=-(U/c)H(t-y/c)$. A nonzero pointwise velocity gradient at the front is interpreted distributionally, or by smoothing the ideal wall start.

The use of linear viscoelasticity is conditional, rather than automatically justified for arbitrary $U$. The material derivative simplifying to a time derivative does not remove the nonlinear constitutive effects of finite shear strain. In this Maxwell problem the rapid elastic response supplies the natural strain scale $U/c=U\sqrt{\rho /G_0}$, which must be small for linearization through start-up. Away from the front, the strain accumulated over an effective memory interval must likewise be small. In the late diffusive regime its characteristic size is $\tau U/\sqrt{\nu t}$, which decreases with time. Large absolute wall displacement $Ut$ alone is not a disqualification for a fluid with finite memory; it is relative deformation during that memory that matters. In the elastic limit, where memory does not decay, the explicit strain behind the front confirms the condition $U/c\ll 1$. Without such a small-strain restriction, the computed field is a formal linear response and need not describe an arbitrary simple fluid.

Let $s=m_0/T$ be the constant fuel mass rate, and let $q(t)=M+m_0-st$ be the remaining rocket mass. Take positive motion opposite to the exhaust direction. Over a short interval $dt$, momentum conservation including the mass $sdt$ expelled at velocity $v-u$ givesConsequently, while the rocket slides forward,This is the constant-burn rocket with ground friction. Assuming forward sliding begins immediately and persists, integration from rest givesand henceA necessary physical assumption is missing from the printed request. With a common static friction and kinetic friction coefficient $\mu >0$, immediate motion requires $su\ge \mu g(M+m_0)$; equality allows motion as soon as fuel loss lowers the normal force. Then the acceleration is nonnegative initially and increases thereafter, so forward sliding is consistent. Without this assumption the displayed expression need not be the final speed and can even be negative.

For completeness, the physical rest/sliding solution with these equal coefficients is as follows. If $su\le \mu gM$, the rocket never starts before burnout and its final speed is zero. If $\mu gM<su<\mu g(M+m_0)$, static friction initially balances the thrust and release occurs atIntegrating only the sliding interval yieldsThis speed is positive because, with $y=q_s/M>1$, it equals $u[\log y-1+1/y]$, whose derivative with respect to $y$ is $(y-1)/y^2>0$. The $\mu =0$ case recovers the frictionless rocket equation. If static and kinetic coefficients differ, the release condition uses the static coefficient and the subsequent integral uses the kinetic coefficient; the prompt supplies no second coefficient.

Use mass conservation and conservative momentum conservation, with viscous stress ${\sigma }_{ij}$ defined so that its divergence is the viscous force:Differentiate the first equation in time and eliminate the momentum time derivative using the second. This givesSet ${\rho }^{\prime }=\rho -{\rho }_0$ and $p^{\prime }=p-p_0$, where the reference values and $c_0$ are constant. Subtract $c_0^2{\nabla }^2{\rho }^{\prime }$ to obtain the exact Lighthill acoustic analogyThe Lighthill stress tensor includes nonlinear momentum transport, departure from the reference linear pressure-density relation, and viscous stress. For an inviscid fluid the last term is absent. The derivation is an exact rearrangement of the conservation equations; treating the tensor as a known localized source is a subsequent approximation.

Use the explicit printed width law $b(z)=\alpha z^2$. Its wetted area and hydrostatic pressure integral areAssuming hydrostatic pressure, a horizontal prismatic bed and cross-sectionally uniform velocity, volume conservation and momentum conservation give the prismatic-channel shallow water equationsEquivalently,The coefficient matrix in variables $(h,u)$ is $\left(\begin{array}{cc}u& h/3\\ g& u\end{array}\right)$. Its eigenvalues and the corresponding gravity-wave speed areFor $h>0$ these eigenvalues are distinct and real, establishing strict hyperbolicity in the hyperbolic partial differential equation sense. Since $dc/dh=c/(2h)$, substitution in the two equations showsFor example $({\partial }_t+(u+c){\partial }_x)(u+6c)=0$. This provides both the slopes and the conserved characteristic quantities. At a dry bed $h=0$, strict hyperbolicity degenerates and the front is interpreted as the limit from positive depth.

The descriptive word “parabolic” alone would normally suggest $b\propto \sqrt{z}$ and different coefficients. The explicit $z^2$ width in the PDF determines the calculation here; it must not be silently replaced by that other geometry.

Write $q=bu$, $\rho ={\rho }_{\infty }$ and $r=\rho -{\rho }_s>0$. The first balance is depth-integrated momentum conservation. Its left side is the downstream increase of horizontal momentum flux. The wind contributes the applied stress $S$; the remaining term is the resultant hydrostatic force associated with the sloping base of the light layer. Because $\overline{\rho }-\rho <0$, that term opposes the wind when $b^{\prime }>0$.

For a strip of unit alongshore width and length $dx$, let $E=q^{\prime }$ be the entrained ocean-volume flux per unit area. Ocean entering the layer carries temperature $T_{\infty }$ and concentration $C_{\infty }$ but no horizontal momentum in this model. Take the reference liquid enthalpy at $T_{\infty }$ to be zero and setThe latent term is negative because ice has lower enthalpy than liquid at the same temperature. The incoming entrained water has zero relative enthalpy, and the outgoing minus incoming horizontal enthalpy flux is $[eq{]}^{\prime }dx$. Atmospheric heat loss therefore gives $[eq{]}^{\prime }=-f$. Similarly, solute conservation givesThese are the second and third balances. Their reference-flux form has already subtracted the heat and solute brought in by entrainment; those fluxes must not be subtracted again.

There is a dimensional normalization to make explicit. With the printed specific $c_p,L$, the physical heat-flux balance is $\rho [eq{]}^{\prime }=-F_{\mathrm{phys}}$. Thus $f=F_{\mathrm{phys}}/\rho$. The displayed exam equation uses its symbol $F$ for $f$, or implicitly works with density-normalized heat flux. The derivation below uses $f$ to keep these two meanings distinct.

With no incoming coastal jet or ice source, the conditions at $x=0$ areThe regular emerging layer also has $T\to T_{\infty }$, $\varphi \to 0$, $\overline{C}\to C_{\infty }$ and $b\to 0$. Prescribed nonzero inlet fluxes would replace these conditions and destroy the coast-origin power law. A zero-thickness coast does not justify independently prescribing an arbitrary nonzero velocity there.

Integration of the solute and enthalpy balances yields $\overline{C}=C_{\infty }$ wherever $q>0$, and $eq=-fx$. Local phase equilibrium now givesFor $\varphi \ll 1$, putTo leading order, $\varphi q=Kx$. The potential-energy closure implies $rg\varphi b{b}^{\prime }=S/4$, and substituting it in the momentum equation givesDividing the ice flux by this last relation gives $\varphi =(K/J)u$. If $b\propto x^p$, $u\propto x^q$, momentum gives $p+2q=1$, while the potential-energy balance gives $2p+q=1$. Thus $p=q=1/3$. Substitution fixes the coefficients:This is the dilute frazil polynya similarity solution. In particular, all three variables grow as $x^{1/3}$, $q$ grows as $x^{2/3}$, and the solution stops being dilute when $\varphi$ is no longer small. Density differences have been retained in buoyancy, but not in inertia or the normalized heat capacity.

To account for upward settling, one must distinguish retained suspension from crystals that leave it. A consistent depth-averaged extension has an upward exported ice-volume flux $R$ and an entrainment flux $E$. If all phases retain the same mean horizontal velocity within the layer, its balances becomeThe last term removes the momentum carried out by the ice; its density may be restored to ${\rho }_s$ outside the Boussinesq approximation. Consequentlyso one may no longer set $\overline{C}=C_{\infty }$. The local equilibrium relations are unchanged, but use the new bulk concentration. The frazil export enthalpy balance also shows why a simple replacement $-f\mapsto -f-LR$ has the wrong sign: exported ice carries negative relative enthalpy.

A slip speed $w$ requires a vertical particle-flux closure, for example $R=w\varphi$ for a well-mixed top boundary. A nonuniform suspension instead requires its surface concentration and buoyancy moment. The potential-energy closure must then include the gravitational energy released by rising light crystals, with scale $rgw\varphi b$, and any energy and buoyancy moment carried through the boundaries. For example, a well-mixed closure that transfers a fraction $\eta$ of this released power to entrainment and neglects additional boundary potential-energy transport would replace the original work balance byThe allocation $\eta$ and exported potential-energy flux are extra closure information, not supplied constants of the original model. The original work balance cannot be retained unchanged while treating settling as an additional loss of suspended ice. Phase-dependent horizontal speeds would also replace $q\varphi$ and $q(1-\varphi )$ by their separate phase fluxes.

It is important to compare total production, not merely the retained ice flux. LetIntegrated solute conservation gives $C(q-I)=C_{\infty }(q+R_{\ast })$, henceIntegrating the heat balance, and neglecting products of the small concentration departure with the exported flux, givesto dilute order, provided the total processed fraction $P/q$, not only the retained fraction $\varphi$, remains small. Thus upward settling reduces retained suspension, but does not increase the leading-order total production at fixed atmospheric heat loss. Beyond that order, efficient removal can reduce the subsequent sensible cooling of crystals and enhance mixing of ocean water; the usual expectation is a larger latent-heat fraction and hence greater production. Its amount, or an unconditional larger/smaller claim, needs the settling, mixing and surface-export closures just identified. At fixed $f$, energy conservation bounds the production by the available heat loss; it must not be inferred simply by observing fewer suspended crystals.

Use Euclidean worldsheet conventions with ${ϵ}^{12}=1$. Including a target metric, a Kalb–Ramond field $B_{\mu \nu }$ and the dilaton $\Phi$ givesThe imaginary coefficient of the antisymmetric term comes from Euclidean continuation; the Lorentzian action has the corresponding real coupling. For constant ${\Phi }_0$, the Gauss-Bonnet theorem gives $S_{\Phi }={\Phi }_0\chi (\Sigma )$, with Euler characteristic $\chi =2-2g$ on a connected closed oriented surface. Thus dilaton Euler-characteristic weighting givesAdding one handle multiplies a contribution by $g_s^2$, identifying $g_s$ as the string coupling. With canonically normalized external closed-string states, $n$ vertices add $g_s^n$, giving the string genus expansion $g_s^{2g-2+n}$.

Before gauge fixing, the Polyakov path integral for inserted string vertex operators is schematicallyThe quotient removes descriptions related by worldsheet diffeomorphisms and Weyl transformations. Gauge fixing has a nontrivial Jacobian, the Faddeev-Popov determinant. Anticommuting vector ghosts $c^a$ and symmetric trace-free antighosts $b_{ab}$ exponentiate this determinant as a local worldsheet ghost action. These worldsheet ghost fields implement the gauge quotient; they are not additional spacetime particles. Their zero modes must be treated separately rather than included in an invertible determinant.

Three kinds of geometry enter the scattering calculation. The worldsheet moduli describe genuinely different conformal shapes that remain after local gauge fixing. Each such shape must be integrated over. The mapping class group consists of orientation-preserving diffeomorphisms modulo those continuously deformable to the identity; it identifies different markings of the same conformal surface, so only one representative of each class is counted. The worldsheet conformal Killing group is the continuous residual conformal symmetry of a chosen surface. It moves insertion positions without changing the geometry and must be divided out, usually by fixing enough vertex positions.

For a torus, write $z\sim z+1\sim z+\tau$, with $\mathrm{Im}\tau >0$. Its shape modulus is $\tau$, its mapping class group acts through the modular group, and its connected worldsheet conformal Killing group acts by translations. Thus the shape integral runs over the standard fundamental domain of the modular group, while one insertion position can be fixed by translation. On the sphere, three positions can instead be fixed by Möbius transformations; for genus at least two the continuous conformal Killing group is trivial.

Now use the half-normalized worldsheet diffeomorphism ghost operator in the question. A vector field $v^a$ can preserve the representative metric after a compensating Weyl transformation precisely when $P_{1}v=0$. The trace of the metric variation sets $\delta \omega =\frac{1}{2}{\nabla }_a{v}^a$, leaving the Conformal Killing equationThe kernel of this operator gives the infinitesimal worldsheet conformal Killing group.

For a surviving infinitesimal worldsheet modulus, choose its metric variation $t_{ab}$ orthogonal to all gauge variations. Orthogonality to Weyl transformations requires $h^{ab}{t}_{ab}=0$. On the closed surface, the formal adjoint of the worldsheet conformal Killing operator follows by integration by parts:Orthogonality to every diffeomorphism variation is therefore the adjoint-kernel condition, givingIn a local conformal coordinate these imply ${\partial }_{\bar{z}}{t}_{zz}=0$ and its conjugate: the moduli variations are represented by holomorphic quadratic differentials. Multiplying the definition of $P_1$ by two changes its adjoint by two but changes neither kernel.

For four tachyon vertex operators on the torus, there is one complex shape modulus, four complex insertion positions and one complex translation to remove. The worldsheet integration count after conformal gauge fixing is thereforeFix the fourth insertion at $z_4=0$. A schematic gauge-fixed expression isHere the antighost pair supplies the modulus measure, and the ghost pair at the fixed vertex absorbs the translation zero modes. The three unfixed positions and $\tau$ are the four complex integrations. The punctured Riemann surface moduli dimension gives the same result, $3g-3+n=4$. Target-spacetime zero-mode integrals producing momentum conservation are separate from this requested count of worldsheet integrations.

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.

Close the unresolved head of the gravity current byHere the speed is measured relative to the resting deep ambient, and $h_f,b_f$ are the depth and reduced gravity immediately behind the head. The Froude number $F$ expresses conversion of the head's hydrostatic pressure head into motion and depends on the ambient-depth ratio and head physics.

A front condition can be motivated by a steady control volume in the front frame: balance mass conservation, momentum conservation and the upstream/downstream Bernoulli equation, with a stagnation region and the appropriate ambient return flow. Assumptions include small density contrast, approximately hydrostatic far fields, negligible imposed ambient current, specified far-field depths, and a prescription for any mixing or energy loss. Viscous/turbulent head structure and finite-depth return flow change $F$; it is not fixed by the outer one-layer equations alone. A deep-ambient ideal head often motivates a constant order-one value, but no numerical choice is needed here.

The shallow-water approximation fails locally in the head because its vertical and horizontal dimensions are comparable and vertical accelerations are appreciable. The Froude number closure is a boundary condition for the outer shallow water equations, not a claim that they resolve the front itself.

Use the Burgers reduction of weakly nonlinear rightgoing acoustics, with $\theta =t-x$, $X=Mx$, ${\partial }_t={\partial }_{\theta }$ and ${\partial }_x=-{\partial }_{\theta }+M{\partial }_X$. The equation of state expands asAt order one, mass conservation and momentum conservation give $({\rho }_0-u_0{)}_{\theta }=0$ and $(u_0-p_0{)}_{\theta }=0$. For the pure rightgoing acoustic wave with no independent leading mean density or velocity shift, ${\rho }_0=u_0=p_0$. This is the background choice implicit in the requested traveling-wave expansion; additional mean components would modify its phase convention. The source's duplicated phrase about expanding $p$ should refer to expanding $\rho$ as well as $u$.

At order $M$, the mass conservation equation becomeswhereas the momentum conservation equation becomesIn the latter equation, the two nonlinear products ${\rho }_0{u}_{0,\theta }$ and $-u_0{u}_{0,\theta }$ cancel. Substitute the equation of state relation into the former equation and add the two equations to eliminate the first corrections. The result isThe same calculation gives $u_1-p_1=-(\gamma +1)p_0^2/4+C(X)$, so the first corrections may be chosen without an explicit secular factor of $\theta$. This is why the retarded-time description can remain uniform for $|\theta |=O(M^{-1})$, provided the leading profile and its relevant derivatives remain bounded over that range and the slow propagation interval.

This Inviscid Burgers equation can develop a shock, so the source's uniformity assertion needs a pre-shock qualification. If $p_0(\theta ,0)=F(\theta )$ and $a=(\gamma +1)/2$, its method of characteristics givesThe smooth construction ends when this denominator vanishes; if $\max F^{\prime }>0$, the first such distance is $X_s=[a\max F^{\prime }{]}^{-1}$.