Codex @codex  0

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.

The radius law integrates to $r(z)=r_0{e}^{-\beta z}$, so the cross-sectional area is $A(z)=\pi r_0^2{e}^{-2\beta z}$. Conserving gas volume in a horizontal slice yieldsor the kinematic bubble transport in an exponentially narrowing vessel equationThe geometrical term increases the concentration as a given gas flux enters a smaller area. On concentration characteristic curves,Put $c=2\beta V_s$. A characteristic starting at $(z_0,0)$ with ${\varphi }_i=\varphi (z_0,0)$ hasThe first formula follows by integrating $d\varphi /[\varphi (1-\varphi )]=cdt$; inserting it into $dz/dt$ gives the second. For initially uniform ${\varphi }_0$, the uninterrupted bubbly region remains spatially uniform, with this time-dependent concentration. These characteristic curves carry concentration information; they are not individual bubble trajectories, whose speed is $V_s(1-\varphi )$.

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.

Assume $U>0$, $k=\alpha >0$, constant $N$, negligible rotation, dissipation and reflection from infinity, small $k{\eta }_0$ and $m{\eta }_0$, and no wave overturning. The terrain is stationary, so laboratory frequency is zero and intrinsic frequency is $\hat{\omega }=-kU$. Propagation requires $kU<N$. PutThe kinematic boundary condition at the linearized bed is $w(x,0)=U{\eta }_x=-kU{\eta }_0\sin (kx)$. The upward radiation condition selects the positive $m$ branch because $c_{gz}=-\hat{\omega }m/(k^2+m^2)>0$. ThusThe internal-wave polarization and every other perturbation follow from this vertical velocity and the linear equations.

In the frame moving with the fluid, $x^{\prime }=x-Ut$, the phase is $k{x}^{\prime }+mz+kUt$, andThe intrinsic energy ray slopes upward and upstream, $dz/d{x}^{\prime }=-k/m$, along a constant-phase line of an internal gravity wave. At fixed intrinsic frequency the spatial wave equation is hyperbolic, with the two characteristic curve line slopes $dz/d{x}^{\prime }=\pm k/m$; the radiation condition selects the upward ray of the appropriate horizontal propagation branch.

In the terrain frame, phase lines remain fixed and the pattern phase velocity is zero. A transported packet hasIts energy path slopes upward and downstream, $dz/dx=m/k$, and need not coincide with a stationary phase line. One should not add $(U,0)$ to the chosen normal intrinsic phase velocity and call that the normal ground-frame phase velocity: tangential motion along a phase line is arbitrary; zero laboratory frequency makes the entire crest pattern stationary. The sketch distinguishes stationary crests, intrinsic characteristics and transported energy.

If $kU>N$, $m$ is imaginary and the bounded field is an evanescent wave; at $kU=N$ the vertical group velocity vanishes and an ordinary upward radiation interpretation degenerates.

Put $a={\sigma }_{13}$, $b={\sigma }_{23}$, $w=v_3$, and write $f_a=\partial f/\partial a$, $f_b=\partial f/\partial b$. In antiplane perfect plasticity, force balance and the fixed yield surface giveDifferentiate the second relation in $x_2$. Along the proposed characteristic curve,Similarly, differentiating the yield relation in $x_1$ givesThe associated flow rule and ${\dot{ϵ}}_{i3}=w_{,i}/2$ give $w_{,1}=2\dot{\lambda }{f}_a$, $w_{,2}=2\dot{\lambda }{f}_b$. ConsequentlyThus both shear stresses and the antiplane velocity are constant on each characteristic. Assume a regular yield gradient $(f_a,f_b)\ne 0$. Since the stress is constant on a characteristic, its tangent $(-f_b,f_a)$ has fixed direction, so the curves are straight lines, apart from arbitrary reparameterization by $\alpha$.

In a centred antiplane plastic fan those lines emanate from the crack tip. Henceand a radial line at angle $\varphi$ from the $x_1$ axis must haveThe crack faces have normal in the $x_2$ direction, so traction freedom requires $b=0$ there. The fan joins the adjacent constant crack-face stress region along the characteristic carrying that stress. Its boundary line therefore satisfieswith the direction interpreted directly when the denominator vanishes. The original PDF reverses this ratio. Its displayed ratio is $d{x}_1/d{x}_2$, whereas the tangent of an angle from the $x_1$ axis is $d{x}_2/d{x}_1$. The distinction cannot be removed while retaining both its stated characteristic direction and its angle convention.

For the elliptic yield surface, with $A,B>0$,On the forward fan $-\pi /2<\varphi <\pi /2$, choose the loading branch $b>0$. The radial tangent condition gives $a/A^2=-(b/B^2)\tan \varphi$. Substituting in the yield equation and choosing the positive square root gives exactlyFor the endpoints, use the equivalent nonsingular expressionsThey give $(a,b)=(-A,0)$ on the upper boundary ray and $(A,0)$ on the lower one. The correct boundaries are $\varphi =\pm \pi /2$, the vertical line through the tip. At either of these stress states $f_b=0$ and $f_a\ne 0$, so the characteristic tangent is vertical. The printed boundary ratio would instead give a horizontal line, providing a concrete counterexample to that intermediate assertion.

Extend the upper and lower rear sectors by the constant stresses $(-A,0)$ and $(A,0)$ respectively; these satisfy the traction-free crack faces and match the fan stresses continuously. The reversed loading has all stresses negated. To check equilibrium within the forward fan, differentiation givesand hence $a_{,1}+b_{,2}=(-\sin \varphi a_{,\varphi }+\cos \varphi b_{,\varphi })/r=0$. The yield equation is satisfied identically as well.

Finally $\nabla w=w^{\prime }(\varphi )e_{\varphi }/r$, while $(f_a,f_b)=2e_{\varphi }/H$. The associated flow rule reduces tofor this stress branch. Any such angular velocity field is constant along the rays and has constant values in the adjacent rear sectors. Its magnitude and angular variation require velocity or loading data beyond the specified traction-free faces; the local stress fan alone does not determine them.

A material derivative measures the rate experienced by a moving fluid particle. If its position is $\mathbf{X}(t)$ with $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$, the chain rule for any scalar field $F$ givesHence the operator is $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$, and the material derivative differs from the time derivative at a fixed spatial point.

In the steady temperature problem ${\partial }_t\theta =0$ and the transverse velocity vanishes. Thus at each fixed $y\in (-1,1)$,Integrating in $x$ and applying the inlet value givesA particle's travel time from the inlet is $x/(1-y^2)$, so the formula is also the exponential decay along its characteristic curve. For fixed $x>0$ the temperature tends to zero near either wall, because the travel time becomes large. No value at $y=\pm 1$ is imposed by an inlet condition there, since the wall velocity is zero and the stated channel is open in $y$.

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 $A\propto h^2$.

Assume the shallow-water approximation: the horizontal scale is long compared with the depth, the hydrostatic approximation applies, and the cross-section-averaged velocity is approximately uniform. Neglect viscosity and drag for this part. Volume conservation gives the constant discharge $Q=bhu$. The steady horizontal momentum equation isso its integral is the Bernoulli equation head $\mathcal{H}=H+h+u^2/(2g)$. The shallow-water specific energy measured above the bed is thereforeThe Froude number compares the flow velocity with the long-wave speed $\sqrt{gh}$:At fixed $Q,b$, the shallow-water specific energy tends to infinity as $h$ tends to zero or infinity. Its unique minimum occurs at the critical depthThe two depth branches above that minimum are subcritical flow, $F<1$ with $h>h_c$, and supercritical flow, $F>1$ with $h<h_c$. The long-wave characteristic speeds relative to the bed are $u\pm \sqrt{gh}$. Thus a subcritical flow can receive information from downstream, whereas both characteristic curves travel downstream in a positive supercritical flow.

A hydraulic control is a critical section where one characteristic speed vanishes and a smooth flow can pass between the two branches. Regularity there supplies a condition relating $Q$, the depth and the geometry; together with reservoir or inlet data, it can determine the discharge. Merely finding $F=1$ is not enough: a finite depth gradient also requires the geometric forcing to vanish in the critical momentum equation.

For the state vector $q=(h,u,g^{\prime }{)}^T$, the primitive equations are $q_t+\mathsf{A}{q}_x=S$, withThe characteristic polynomial isConsequently the three characteristic speeds areFor $h>0$ and $g^{\prime }>0$, the hyperbolic system has three distinct real eigenvalues. The long-wave speed relative to the current is $c$, while $u$ transports a buoyancy/contact disturbance.

A left eigenvector for the material characteristic is $(0,0,1)$, givingFor $u\pm c$, left eigenvectors can be chosen asMultiplying $q_t+\mathsf{A}{q}_x=S$ by these rows gives the characteristic compatibility for an entraining triangular-channel current. With all differentials evaluated along the indicated characteristic,The source follows directly from ${\ell }_{\pm }S=-2u{w}_e/h\pm 2c{w}_e/h\mp 2g^{\prime }{w}_e/(3c)$ and $g^{\prime }/c=2c/h$.

Since $4dc=(2c/h)dh+(h/c)d{g}^{\prime }$, an equivalent form isCombining it with the material buoyancy equation also givesOnly when entrainment vanishes and $g^{\prime }$ is spatially constant do $u\pm 4c$ become constant Riemann invariants on these characteristic curves. Dropping the buoyancy differential in the general problem would lose one of the required transport couplings. The characteristic formulas with $1/c$ apply to the nondegenerate region $g^{\prime }h>0$; dry or neutrally buoyant limits require their conservative form instead.

Hindered settling is the reduction of the particle settling velocity by displaced-fluid backflow and interactions with other particles. In the specified model, the downward speed is $V_s$ at vanishing concentration and falls to zero at the maximum packing fraction ${\varphi }_{\max }$. The characteristic speed of a concentration disturbance will differ from this individual-particle speed.

Take $z$ positive upwards. For a locally uniform horizontal suspension without diffusion or bulk fluid motion, the upward particle flux isParticle volume conservation gives the scalar conservation lawThis is the quadratic hindered-settling flux. Along its characteristic curves,For a smooth initial profile ${\varphi }_0(\zeta )$, the map is $z=\zeta +c({\varphi }_0(\zeta ))t$. Its Jacobian is $1+(2V_s/{\varphi }_{\max }){\varphi }_0^{\prime }(\zeta )t$. A shock develops where this first reaches zero: the compressive condition is ${\varphi }_0^{\prime }<0$, andwhen the minimum is negative and no earlier boundary discontinuity is imposed. A discontinuous compressive initial state can instead contain a shock from $t=0$.

Let $z=s(t)$ separate states ${\varphi }_L$ below the shock and ${\varphi }_R$ above it. Integrating the conservation law across a moving thin interval gives the Rankine-Hugoniot conditionEquivalently, the flux relative to the shock agrees on both sides. For this quadratic flux,Since $j^{\prime \prime }=2V_s/{\varphi }_{\max }>0$, a physical compression shock has ${\varphi }_L>{\varphi }_R$ and satisfies $c({\varphi }_L)>\dot{s}>c({\varphi }_R)$: characteristic curves enter it from both sides. These conditions select the shock over a nonphysical expansive discontinuity.

Start a characteristic curve at $(0,\eta )$ with $u=-\eta$. The Characteristic curve equations give $u=-\eta$, $y=\eta +s$, $x=-\eta s$, so $x={\eta }^2-\eta y$. Therefore the two branches areEach is $C^1$ on $D=\{y^2+4x>0\}$. The plus branch matches $u(0,y)=-y$ for $y>0$, whereas the minus branch matches it for $y<0$. Differentiating $u^2+yu-x=0$ gives $u_x=1/(2u+y)$ and $u_y=-u/(2u+y)$, verifying $u_y+u{u}_x=0$ wherever $2u+y\ne 0$.

On the initial line, the normal Characteristic curve speed is $u=-y$, nonzero away from the origin. Thus the local existence theorem gives precisely these separate local branches near positive and negative initial points. At $y^2+4x=0$ the branches merge and their derivatives become singular. There is no $C^1$ solution realizing the entire initial line in a neighborhood of the origin: the data force $u_y(0,0)=-1$, while the equation and $u(0,0)=0$ force $u_y(0,0)=0$. The failure at this Characteristic curve point is a genuine limitation of the printed global initial-data request.