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Let $w$ be downward Darcy flux, measured per unit total horizontal area. Neglect the resistance of the displaced phase, capillary pressure, and horizontal subsurface flow. Taking atmospheric pressure as zero, the upper end of the saturated column has pressure $\rho gh$ and its advancing lower end has pressure zero. Across a column of depth $l$, Darcy law therefore givesThe pressure head from the overlying layer supplies $h/l$, and gravity supplies one. Because the new liquid fills a pore volume $\varphi dl$ per unit area, the vertical imbibition under a draining liquid layer obeys $\varphi \dot{l}=w$. Volume conservation gives $h+\varphi l=h_0$, so, until the surface layer is exhausted,The initial condition is understood as a limiting solution of this singular equation. For $0<\varphi <1$, separation of variables givesAt small penetration, expansion of the logarithm yields $t\sim \varphi \nu l^2/(2gk{h}_0)$, henceA sufficient dimensional early-time condition is $t\ll \varphi \nu h_0/(gk)$, for which $l\ll h_0$. The printed $t\ll 1$ needs a chosen time unit or nondimensionalization. In the limiting case $\varphi =1$, $l^2=2gk{h}_{0}t/\nu$ holds exactly until $h=0$. The divergent initial Darcy flux is an idealization of sudden contact with a dry continuum; it cannot describe penetration below the pore scale.

For the spreading surface layer, measure $z$ upwards from the substrate. A thin, slow gravity current has hydrostatic pressure $p=\rho g(h-z)$ and a leading horizontal viscous balance $\nu u_{zz}=g{h}_x$. A no-slip boundary condition at $z=0$ and a stress-free boundary condition at $z=h$ giveThis lubrication gravity-current flux drives horizontal spreading down the surface-pressure gradient. Surface volume conservation is $h_t+q_x=-w$, while conservation in the pores is $\varphi l_t=w$. Combining these balances gives the deep-substrate drainage of a gravity current modelDrainage transfers liquid from the surface layer into the substrate. Indeed $(h+\varphi l{)}_t+q_x=0$.

For definiteness take a one-sided current on $0<x<x_N(t)$, supplied at $x=0$. The imposed volume is per unit span, so $V_0$ has dimensions of area. The boundary conditions and volume conservation constraint areThe source starts from zero injected volume. Initially dry points acquire $l>0$ only when the advancing current reaches them; the formula $h/l$ is not applied outside that region. At the advancing nose these are limiting conditions for a singular drainage equation. For a source feeding two identical sides, each side has half the volume and inlet flux; only the corresponding numerical prefactors change. The integral and inlet-flux conditions are equivalent once the local balance and nose conditions hold.

The physical approximations are lubrication theory, negligible inertia, slowly varying surface height, small pore-scale Reynolds number, and negligible capillary entry pressure. Treating the substrate as vertically draining columns also requires its lateral flow to be negligible and its depth to exceed the penetration depth. The model is valid while surface liquid remains available above each draining column.

For a similarity solution, let $h$ and $l$ scale as $t^a$, and let the current length scale as $t^b$. Since $h/l$ then has no time dependence, drainage requires $a-1=0$. Balancing $h_t$ with the horizontal spreading term gives $a-1=4a-2b$, so $a=1$ and $b=2$. These exponents also give injected volume $t^{a+b}=t^3$, explaining the particular supply law.

Define scaleswhich satisfy $CD=V_0/{\tau }^3$ and $D^2=(g/\nu )C^3$. The cubic-input similarity for a draining gravity current isSubstitution into the two conservation equations giveswhereThus only $K$ and $\varphi$ remain in the dimensionless problem. Its constraints areIntegrating the sum of the two ordinary differential equations verifies that the inlet flux and integral normalization agree. The dimensionless endpoint ${\xi }_N={\xi }_N(K,\varphi )$ is the undetermined multiplicative constant in the length law; solving for it is not required.

For an axisymmetric thin elastic plate, define ${\nabla }_r^2=r^{-1}{\partial }_r(r{\partial }_r)$. Uniform excess pressure and bending stiffness $B$ give $B{\nabla }_r^{4}h=p$. The roof is attached to undeformed rock at the front, giving clamped boundary conditions $h(R)=h_r(R)=0$, and regularity excludes singular displacement or curvature at $r=0$. A regular particular solution is $p{r}^4/(64B)$; adding a constant and a multiple of $r^2$ and enforcing the two edge conditions gives the axisymmetric clamped-plate deflectionIntegrating the magma intrusion volume,therefore yieldsHere the axisymmetric phrase requires the radial plate biharmonic operator; a Cartesian one-dimensional beam equation would not give the same pressure coefficient. Hydrostatic contributions are excluded as instructed.

For the early front, let $U_p=\dot{R}$ be the leading propagation speed. The fluid and fracture radii differ only within a local region of length $l_p\ll R$. In lubrication theory, the elastic pressure gradient near the front is of order $B{h}_f/l_p^5$, and the volume flux through the gap is of order $B{h}_f^4/(\mu l_p^5)$. In a frame following the fluid front that flux is of order $U_p{h}_f$. The viscous peeling of an elastic plate law is thereforeMatch the peeling region to the interior curvature: $h_f/l_p^2\sim {\kappa }_i$. The stated vapour-gap relation $h_f\simeq p_T{l}_p^4/(24B)$, with $p_T>0$ interpreted as the relevant pressure magnitude, then givesSubstitution gives the vapour-tip peeling of a magma intrusion speedWith constant supplied volume flux $Q$ and negligible initial intrusion volume, $V=Qt$. Integrating $R^{14}\dot{R}\sim B^{3/2}{Q}^{7/2}{t}^{7/2}/(\mu p_T^{1/2})$ gives $R^{15}\sim B^{3/2}{Q}^{7/2}{t}^{9/2}/(\mu p_T^{1/2})$, orOrder-one coefficients require the full local peeling profile. This early regime is a front-controlled asymptotic regime, not an assertion that a continuum peeling zone remains thin at arbitrarily small $t$.

When fracture toughness instead imposes fixed ${\kappa }_f>0$, equating it to the outer edge curvature gives the toughness-controlled magma intrusion lawIts pressure can be written in either useful formThe pressure decreases as the intrusion grows on this advancing branch.

If $V_{r0}$ is the reservoir volume before any magma was transferred to the initially negligible intrusion, volume conservation and negligible conduit storage give $V_r=V_{r0}-V$. With the supplied pressure-volume law and conduit conductance, the reservoir-fed magma intrusion obeysIn this law $E$ is an effective reservoir pressure-volume stiffness, with dimensions of pressure per volume. If the stated initial reservoir volume refers instead to the start of the late regime, when the intrusion already has volume $V_i$, replace $V_{r0}$ by the total $V_T=V_{r0}+V_i$ throughout. An independent value of $V_i$ would then be needed.

For an advancing solution, inflow ceases when the reservoir and intrusion pressures equalize. Its final volume is the larger positive root in $0<V<V_{r0}$ ofEquivalently, $y=\sqrt{V_{\infty }}$ solves $y^3-V_{r0}y+\mathcal{A}/E=0$. The larger root is essential: $\sqrt{V}(V_{r0}-V)$ rises from zero to a maximum $2V_{r0}^{3/2}/(3\sqrt{3})$ at $V=V_{r0}/3$ and then decreases. A nonempty interval of positive inflow exists only ifThere are then two positive equilibria $V_{-}<V_{r0}/3<V_{+}$. At equilibrium the derivative of the volume ordinary differential equation is $\beta E(V_{r0}-3V)/(2V)$, so $V_{-}$ is unstable and $V_{+}$ has asymptotic stability. An initial advancing intrusion with $V_{-}<V_i<V_{+}$ approaches $V_{\infty }=V_{+}$. The equality case has a double equilibrium and no positive-inflow interval; above the threshold there is no advancing equilibrium branch of this model.

The late formula $p\propto V^{-1/2}$ is singular at $V=0$ and cannot nucleate the intrusion: the earlier propagation regime must supply its finite seed. Nor does this fixed-edge-curvature propagation law prescribe how an irreversible fracture closes under negative inflow. The final-size answer is conditional on reaching the advancing late branch. In particular the reservoir is generally not emptied, and $\beta$ controls the approach rate rather than the equilibrium size.

Use the shallow-water approximation, a hydrostatic vertically mixed layer, a deep motionless ambient, and no ambient fluid entrainment. Particle loss changes reduced gravity but changes total layer volume only at the discarded dilute-particle order. Likewise thermal expansion changes the equation of state while the leading Boussinesq approximation retains volume conservation. DefineDepth integration of mass conservation, horizontal momentum balance, particle transport, and the heating law givesHere the integrated excess hydrostatic pressure is $b{h}^2/2$. There is no particle source from the bed. Bed drag and the particle contribution to inertia are neglected at this order. The equivalent variable-buoyancy shallow water equations areThe term $h{b}_x/2$ is necessary when heating or particle loss makes the reduced gravity vary horizontally.

For $v=(h,u,\varphi ,\theta {)}^T$, the quasilinear system $v_t+A(v)v_x=S(v)$ hasIts characteristic polynomial factors asThus the characteristic curves have slopesOn the two repeated contact characteristic curves, the particle and heating laws are $D\varphi =-W_s\varphi /h$ and $D\theta =\beta \varphi$. Writing $c_g=\sqrt{bh}$ and $D_{\pm }={\partial }_t+(u\pm c_g){\partial }_x$, the gravity-wave compatibility relations areFor $h>0$ and $b>0$, gravity-wave right eigenvectors can be taken as $(1,\pm c_g/h,0,0{)}^T$. The contact eigenspace has dimension two: $\delta u=0$ and $b\delta h+(h/2)\delta b=0$, with independent $\delta \varphi ,\delta \theta$. Therefore $A$ has a full set of real eigenvectors.

The system is hyperbolic throughout strict static stability, but not strictly hyperbolic because the contact speed is repeated. At $b=0$ the eigenvalues coalesce and this diagonalizability is lost; when $b<0$ the gravity-wave speeds become complex, reflecting the loss of a statically stable lower layer.

The shallow-water approximation requires a small depth-to-horizontal-scale ratio, slow streamwise variation of the channel and interface, small vertical acceleration, and rapid enough cross-sectional mixing for $u$ and $\varphi$ to be representative averages. Near the channel apex, or during the initial collapse at a sharp front, these conditions need not hold. The high Reynolds number permits neglect of leading viscous resistance; weak stirring mixes the particles without appreciable ambient fluid entrainment.

Put $\lambda =\beta /H$. At elevation $z$, the channel width is $2\lambda xz$, so the occupied area and interface width areVolume conservation and suspended-particle conservation giveThe excess hydrostatic pressure is ${\rho }_{0}d\varphi (h-z)$. Its streamwise gradient averaged over the triangular section gives acceleration $-d\varphi h_x-(dh/3){\varphi }_x$: the weighted mean of $h-z$ is $h/3$. Thus the triangular-channel shallow water equations areThe quasilinear coefficient matrix in $(h,u,\varphi )$ isIts three distinct real eigenvalues, for $h,d\varphi >0$, establish strict hyperbolicity:On the material characteristic curve, $d\varphi /dt=-2W_s\varphi /h$. On each of the other two characteristic curves, multiplying by the left eigenvector $(\pm 2c/h,1,\pm dh/(3c))$ gives the compatibility ordinary differential equationThe derivatives are along the indicated characteristic, with $g^{\prime }=d\varphi$. At vanishing concentration, the gravity-wave speeds coalesce and strict hyperbolicity is lost.

The shallow-water approximation requires $h/L\ll 1$, a nearly hydrostatic pressure, negligible vertical acceleration, and approximately depth-independent horizontal velocity, concentration, and density. The large Reynolds number allows viscous stress to be neglected away from thin boundary layers, while the deep ambient is taken to remain stationary.

For unit channel width, volume conservation, chemical conservation, and mass conservation are respectivelyandEntrained ambient fluid contains no chemical and enters with density ${\rho }_0$; detrained fluid carries the local concentration and density. The ambient has no horizontal momentum, whereas detrained fluid carries horizontal momentum $\rho u$ per unit volume. The depth-integrated momentum conservation law is thereforeso $M=-\rho u{w}_d$.

The final state is displacement ventilation: fresh air of density ${\rho }_0$ forms a cool lower layer, radiator-heated air forms a well-mixed upper layer, and the wall plume crosses their interface at height $h$. Steady volume conservation requires the plume volume flux there to equal the imposed ventilation flux,

Below the interface the ambient is uniform, so $B_0=F_0/{\rho }_0$ is constant. Put $J=M/{\rho }_0$. The triangular-profile wall line plume equations reduce toThe pure plume conditions $V,J\to 0$ at the radiator select the similarity solutionConsequently the displacement-ventilation interface height isThe upper-layer reduced gravity follows from its steady buoyancy balance as $g_u^{\prime }=B_0/A=F_0/({\rho }_{0}A)$. The two-layer solution applies when the calculated interface satisfies $0<h<H$.

In a channel whose transverse width $b(z)$ does not depend on downstream position, the wetted area is $A(h)={\int }_0^{h}b(z)dz$ and the integrated hydrostatic pressure force divided by density is $P(h)=g{\int }_0^h(h-z)b(z)dz$. With cross-sectionally uniform velocity, volume conservation and momentum conservation give the displayed shallow water equations. Since $P^{\prime }(h)=gA(h)$, the two characteristic speeds are $u\pm c$ with $c^2=gA/A^{\prime }$. For $b(z)=C{z}^p$, $c^2=gh/(p+1)$ and the Riemann invariants are $u\pm 2(p+1)c$.

For fixed channel width $b(z)=\beta z$, cross-sectional area is $A=\beta h^2/2$. With reduced gravity $G\varphi$, uniform cross-sectional fields and downward settling velocity magnitude $w$, volume conservation, momentum and particle deposition flux giveThe hydrostatic pressure force contains ${\int }_0^h(h-z)\beta zdz=\beta h^3/6$. No streamwise widening term occurs in a prismatic channel. Deposition uses the projected boundary width $\beta h$ under the vertical-settling model.

A dilute, vertically mixed particle-laden gravity current with no fluid entrainment, bed erosion or carrier-fluid leakage obeys volume conservation $q=uh=\text{constant}$. With downward settling velocity magnitude $v_s$, an absorbing bed gives $qd{g}^{\prime }/dx=-v_s{g}^{\prime }$. The variable-buoyancy shallow water equations conserve the kinematic momentum flux including hydrostatic pressure,These equations describe a steady interior supplied continuously from upstream, not the moving nose of a finite-volume release. The depth follows one of the depth branches of a steady depositing gravity current.

In a channel with width $2\lambda xz$ at height $z$, a well-mixed particle-laden gravity current of depth $h$ occupies area $\lambda x{h}^2$. Its volume conservation, momentum and particle deposition flux balances givewhere the reduced gravity is $g^{\prime }=d\varphi$. The $h/3$ coefficient is the cross-section-weighted mean of $h-z$ in the hydrostatic pressure force. The characteristic curves have speeds $u$ and $u\pm \sqrt{g^{\prime }h/2}$.

A U-tube differential manometer contains a connected column of liquid with one gas connection above each liquid meniscus. Write $p_L,p_R$ for the left and right pressures, and $z_L,z_R$ for the meniscus heights on an upward vertical coordinate. Assume constant mass density $\rho$ and downward gravitational acceleration of magnitude $g$; neglect the gas column's mass density and differences in capillary pressure. The signed vertical length difference $\Delta h=z_R-z_L$ then givesLarger left pressure means a lower left liquid meniscus: $p_L>p_R$ implies $z_L<z_R$. Equal pressures give equal meniscus heights.

The balance follows directly from hydrostatic pressure. At a common height $z_0$ below both surfaces, the two points in the connected homogeneous liquid have equal pressure:Subtracting cancels the common bottom depth and leaves $p_L-p_R=\rho g(z_R-z_L)$. The same-height rule follows from hydrostatic equilibrium, $\nabla p=-\rho g{\mathbf{e}}_z$: a horizontal displacement changes no pressure, and for constant $\rho$ the quantity $p+\rho gz$ is constant throughout the connected liquid. Equal height alone is insufficient for unrelated vessels or for fluid flow with significant acceleration.

For an original constructed example, choose water of stipulated mass density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, and vertical length difference $\Delta h=0.060\mathrm{m}$. ThenThe SI pressure unit is the pascal: the product has units $\mathrm{kg}{\mathrm{m}}^{-3}\times \mathrm{m}{\mathrm{s}}^{-2}\times \mathrm{m}=\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}=\mathrm{Pa}$. Swapping the two connections gives $-588.6\mathrm{Pa}$ and reverses which liquid meniscus is lower. In equal-bore arms, volume conservation makes each surface move $30\mathrm{mm}$ from its initial equal-height position; using only one surface's displacement would halve the inferred pressure. With unequal arm areas $A_L,A_R$, a left fall $a$ and right rise $b$ instead obey $A_{L}a=A_{R}b$, while the vertical length difference is still $a+b$.

If the right port is open to the local atmosphere, $p_L-p_R$ is the left gauge pressure. Only with that reference and known atmospheric pressure can one obtain absolute pressure fromFor an explicitly stipulated atmospheric pressure of $100000\mathrm{Pa}$, the example gives $100588.6\mathrm{Pa}$ absolute pressure. If the right port is connected to another vessel or a pressure tapping, the same reading is a differential pressure and supplies no absolute pressure without knowing that reference.^[ref]

For cross-sectional area $S(x,h)$ and cross-sectionally uniform velocity, volume conservation gives $S_h(h_t+u{h}_x)+S{u}_x+u{S}_x=0$, while hydrostatic inviscid momentum gives $u_t+u{u}_x+g{h}_x=0$. Their characteristic speeds are $u\pm c$. With $D_{\pm }={\partial }_t+(u\pm c){\partial }_x$, the robust compatibility equation is $(g/c)D_{\pm }h\pm D_{\pm }u=-uc{S}_x/S$. Replacing $D_{\pm }h$ by $D_{\pm }c/c_h$ requires $c_x=0$ at fixed height; otherwise an explicit geometry term remains.

For a rectangular channel with stationary width $b(x)$ and constant reduced gravity $g^{\prime }$, volume conservation and momentum conservation give $h_t+u{h}_x+h{u}_x=-hu{b}^{\prime }/b$ and $u_t+u{u}_x+g^{\prime }{h}_x=0$. With $c=\sqrt{g^{\prime }h}$ the characteristic curves have speeds $u\pm c$. The combinations $u\pm 2c$ obey the displayed compatibility equations; they are Riemann invariants only when the width source term vanishes.

A liquid layer of depth $h$ above a dry deep porous substrate drives a Darcy flux $w=(gk/\nu )(1+h/l)$ through its saturated depth $l$, assuming negligible capillary pressure and negligible resistance from the displaced phase. The pressure head contributes $h/l$ and gravity contributes one. Volume conservation gives $\varphi \dot{l}=w$. For a finite column, $h+\varphi l=h_0$ and initiallyThe singular initial flux is an idealization; the continuum model requires penetration beyond the pore scale.