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For a material volume, mass conservation is $d{\int }_{V(t)}\rho dV/dt=0$. The transport theorem and arbitrariness of that volume give $\boxed{{\rho }_t+\nabla \cdot (\rho u)=0}$. Balance of linear momentum includes the pressure traction and body force:Apply transport, the divergence theorem and the mass equation to obtainFor small perturbations of the uniform resting barotropic state, write $\rho ={\rho }_0+{\rho }^{\prime }$, $p=p_0+c_0^2{\rho }^{\prime }+\cdots$. Linearized mass and momentum areTaking the curl shows that initially vanishing vorticity remains zero, so the acoustic velocity has a potential $u=\nabla \varphi$. Integrating momentum, using the additive freedom of a velocity potential to remove the spatially constant integration term, gives ${\rho }^{\prime }=-({\rho }_0/c_0^2){\varphi }_t$. Continuity now yields $\boxed{{\varphi }_{tt}=c_0^2{\nabla }^2\varphi }$.

For spherical symmetry, set $\varphi =e^{-i\omega t}F(r)$ and $k=\omega /c_0$. The radial equation is $(r^2{F}^{\prime }{)}^{\prime }+k^2{r}^{2}F=0$, or $(rF{)}^{\prime \prime }+k^2(rF)=0$. HenceRigid boundaries require radial velocity $F^{\prime }=0$ at both radii. A form automatically satisfying the condition at $a$ isAt $b$, the condition $b(rF{)}^{\prime }-(rF)=0$ becomesTherefore the nonzero oscillatory eigenfrequencies obeyThe undivided sine-cosine form avoids artificial issues from dividing by a cosine. The formal $k=0$ root is the spatially constant zero-frequency potential additive freedom of a velocity potential, not a propagating acoustic oscillation. This is the radial acoustic spectrum between rigid concentric spheres.

For a material volume $V(t)$ moving with velocity $\mathbf{u}$, the Reynolds transport theorem isIt applies componentwise to vector quantities. Cauchy's stress theorem represents traction on a surface with outward unit normal $\mathbf{n}$ as $t_i={\sigma }_{ij}{n}_j$. Conservation of mass gives $d{\int }_{V(t)}\rho dV/dt=0$ and hence ${\partial }_t\rho +\nabla \cdot (\rho \mathbf{u})=0$. Constant density reduces this to $\boxed{\nabla \cdot \mathbf{u}=0}$.

With no body force, conservation of momentum and the divergence theorem giveApplying transport and then mass conservation cancels the terms proportional to $u_i$, leaving $\rho ({\partial }_t{u}_i+u_j{\partial }_j{u}_i)={\partial }_j{\sigma }_{ij}$. For the stated Newtonian stress, constant viscosity gives ${\partial }_j{\sigma }_{ij}=-{\partial }_{i}p+\mu \Delta u_i+\mu {\partial }_i(\nabla \cdot \mathbf{u})$. The last term vanishes. Since the material volume was arbitrary,A body force, if present, would have to be added to the momentum balance; it is absent from the specified model.

Integrate each steady one-dimensional conservation law across a thin region enclosing the shock. Mass conservation gives constant mass flux $j=\rho u$. Momentum conservation includes both advected momentum and pressure force, so $p+\rho u^2$ is constant. The total energy flux is $u(\rho e+p)=j(e+p/\rho )$, where $e=c_{v}T+u^2/2$. For a genuine shock $j\ne 0$, division by the mass flux and the equation of state $c_{v}T=p/[(\gamma -1)\rho ]$ give the conserved stagnation enthalpyThus the Rankine-Hugoniot conditions for a perfect gas areThe velocities here are measured in the stationary shock frame; a moving-shock problem must first be transformed to that frame.

The material derivative $\boxed{D/Dt={\partial }_t+\mathbf{u}\cdot \nabla }$ measures change while following a moving gas parcel. The four equations respectively express mass conservation, momentum conservation, internal-energy balance, and the ideal gas law. The first accounts for density change under expansion or compression. The second is Newton's law with pressure as the only force. In the third the specific internal energy is $c_{v}T$ and pressure work $-p\nabla \cdot \mathbf{u}$ heats compression and cools expansion. The fourth relates thermodynamic pressure, density and temperature; $R$ is the gas constant per unit mass.

Let $J$ be population flux. For any fixed control volume $V\subseteq \Omega$, mass conservation with local production givesAssume local, isotropic random dispersal, no imposed advection, and constant diffusion coefficient $D$. Fick's first law is $J=-D\nabla u$. The divergence theorem and arbitrariness of $V$ yieldThe coefficient has units length squared per time; $f$ has units population density per time. Smoothness sufficient to pass from the integral balance to the local equation is understood. This is a reaction-diffusion equation.

The four dimensional equations express mass conservation, momentum balance with pressure, gravity and viscous stress, heat advection and conduction, and a linear thermal-expansion equation of state. For heating from below, suitable plate conditions are $\mathbf{u}=0$ and $T=T_0+\Delta T$ at $z=0$, and $\mathbf{u}=0$, $T=T_0$ at $z=d$. No slip and thermal insulation on the sidewall provide one suitable lateral completion; prescribed lateral temperatures would be another specified thermal setup.

Let $\kappa =k/({\rho }_0{c}_p)$ be thermal diffusivity and $\nu =\mu /{\rho }_0$ kinematic viscosity. Use $d$, $d^2/\kappa$, $\kappa /d$ and $\Delta T$ as length, time, velocity and temperature scales. Subtract the static reference pressure $p_H$ with $\nabla p_H=-{\rho }_{0}g\mathbf{k}$ and scale its remainder by $\mu \kappa /d^2$. ThenDensity becomes ${\rho }_0(1-BT)$. Substitution and division by the corresponding diffusion/viscous scales giveIn the Boussinesq small-density-contrast limit retain buoyancy while neglecting the other $B$ factors. At infinite Prandtl number, after any fast initial adjustment, the equations areThe velocity is instantaneously constrained by a viscous elliptic balance; the temperature still evolves.

Take the divergence of the vector Poisson equation. Derivatives commute, and divergence of a curl is zero, so $\Delta (\nabla \cdot \mathbf{A})=0$. Its boundary value is zero, hence the maximum principle gives $\nabla \cdot \mathbf{A}=0$ throughout $D$. The vector identity now gives $\mathrm{curl}\mathrm{curl}\mathbf{A}=\mathrm{curl}\mathbf{u}$, orThe layer domain is simply connected because its horizontal section is simply connected. For a curl-free field on such a domain, define $\chi (x)$ as its line integral from a fixed base point to $x$. Contractibility of loops and Stokes theorem make the integral path independent, and differentiation gives its gradient equal to the field. Thus $\mathrm{curl}\mathbf{A}-\mathbf{u}=\nabla \chi$.

Incompressibility then gives $\Delta \chi =0$, while the prescribed normal curl gives ${\partial }_n\chi =0$. Green's identity yieldsHence $\chi$ is constant on the connected domain and $\boxed{\mathrm{curl}\mathbf{A}=\mathbf{u}}$. The extra condition $\mathbf{A}\cdot \mathbf{n}=0$ belongs to the stated potential boundary problem; the reconstruction argument above uses the divergence and normal-curl conditions. On a multiply connected domain a curl-free field need not be a globally single-valued gradient.

Finally $\mathrm{curl}(T\mathbf{k})=(T_y,-T_x,0)$. Taking the curl of the infinite-Prandtl momentum balance and using $\mathrm{curl}\mathbf{u}=-\Delta \mathbf{A}$ givesThe derivation is an instance of velocity reconstruction from a mixed-boundary vector Poisson problem.

Let ${\chi }_1(\mathbf{x},t)$ equal one in phase one and zero in phase two; its interface value is immaterial for area integrals. In the absence of phase change or interfacial mass transfer, the distributional phase mass conservation is ${\partial }_t({\chi }_1\rho )+\nabla \cdot ({\chi }_1\rho \mathbf{u})=0$. For a cross-section $S$ of area $A$, defineHere the symbol $u_x$ in the integral denotes the axial velocity component, not a derivative. Integrating through an arbitrary axial slice and using zero normal flux at the rigid pipe wall givesThe velocity is mass weighted, so the averaged mass flux factors exactly; an unweighted phase mean would not in general have this property. This is the cross-section averaged phase mass balance.

For phase two use ${\chi }_2=1-{\chi }_1$, volume fraction $1-\alpha$, andThen $\boxed{{\partial }_t[{\rho }_2(1-\alpha )]+{\partial }_x[{\rho }_2(1-\alpha )u_2]=0}$. Mean quantities in a phase that is absent are unnecessary; the products giving conserved masses and fluxes remain the meaningful variables.

Use an asymptotic expansion at fixed $z$, evaluating the boundary conditions at the instantaneous $z=h(x,t)$. At leading order, the downslope equation gives ${\hat{u}}_{0,zz}=0$. The wall condition and the surface condition ${\bar{u}}_z(h)+{\hat{u}}_{0,z}(h)=0$ implyHere $u^{(0)}$ denotes the total leading velocity, to distinguish it from the dimensional scale $U_0$. The continuity equation and wall impermeability then giveThe transverse momentum equation gives ${\hat{p}}_{0,z}=0$. The normal stress condition therefore yieldsThe leading volume flux per unit width is $q^{(0)}={\int }_0^h{u}^{(0)}dz=2h^3/3$. Consequently mass conservation gives $h_t=-2h^2{h}_x+O(\epsilon )$. This relation eliminates $h_t$ inside the first inertial correction:Set $G=2h_x\cot \alpha -3S{h}_{xxx}$. The $O(\epsilon )$ downslope equation and boundary conditions areTwo integrations giveThe pressure correction ${\hat{p}}_1$ is not needed for this calculation because its downslope pressure gradient enters only at $O({\epsilon }^2)$.

Integrating the total velocity across the film givesIndeed the three inertial polynomial terms integrate to $\mathrm{Re}{h}^6{h}_x(1/30-1/6+2/3)=8\mathrm{Re}{h}^6{h}_x/15$. Integrating the continuity equation and using the kinematic boundary condition gives $h_t+q_x=0$. Hence the Benney equation isNormal gravity smooths height variations on the upper side of the slope, while the inertial correction can amplify them. Surface tension suppresses sufficiently short wavelengths.

The hydrostatic pressure gives $p_x=\rho g{h}_x$, independent of depth. Thus Darcy's law gives $u=-K{h}_x$ and the horizontal volume flux per unit width isThe coefficient $K$ is the hydraulic conductivity, with units of velocity. For a column of unit horizontal area, the stored water volume is $\varphi h$, and rainfall supplies volume flux $R$. The local mass conservation equation is $\varphi h_t+q_x=R$. Therefore the Boussinesq equation for an unconfined aquifer isThis assumes rainfall is entirely converted into recharge and neglects unsaturated storage and losses before the water reaches the water table.

Let $J_X={\lim }_{x\downarrow 0}X(x)X^{\prime }(x)>0$. The spatial ordinary differential equation gives $(X{X}^{\prime }{)}^{\prime }=-\lambda X$, so its leading outlet behavior isAgain this is a square-root drainage boundary. For the separable aquifer drawdown, the dimensional drainage flux isEquivalently, with ${\tau }_d=\varphi /(K\lambda )$, it is $Q_0(t^{\ast })[1+(t-t^{\ast })/{\tau }_d{]}^{-2}$ for an exact separated profile. Integrating the spatial ordinary differential equation gives $J_X=\lambda {\int }_0^{L}Xdx$. This verifies the global mass conservation checkThus the late drainage flux follows an inverse-square power law, while stored water decays inversely with time.

The aquifer drainage hydrograph starts at zero, rises linearly in the early rainfall similarity solution, and approaches the total recharge $RL$ if rainfall lasts long enough for a finite-length steady state. Indeed, steady mass conservation gives $q_x=R$ and $q(L)=0$, hence $Q_0=RL$. At $t^{\ast }$, the outlet flux stays continuous for a regular inherited flux profile, then recedes as water drains; the eventual decay is proportional to $(t-t^{\ast }+\text{constant}{)}^{-2}$. An arbitrary transition requires the full time-dependent partial differential equation, so matching an early rainfall formula directly to a separated drought formula is only a schematic approximation.

The curve illustrates these regimes and a possible $t^{\ast }$. Its numerical shape and the relative duration of the regimes are schematic; the early rainfall and late drought formulas apply in their respective asymptotic ranges.

Let $V(t)$ have a boundary moving with velocity ${\mathbf{v}}_b$, and let $\mathbf{n}$ be its outward unit normal vector. For a smooth scalar field $q$, the Reynolds transport theorem isIn the last expression, extend ${\mathbf{v}}_b$ smoothly into $V(t)$ and use the divergence theorem. For a material volume, its boundary moves with the fluid velocity $\mathbf{u}$, so ${\mathbf{v}}_b=\mathbf{u}$. Equivalently, the Reynolds transport theorem for a material volume readswhere $D/Dt$ is the material derivative.

For an arbitrary moving region, mass conservation includes the relative mass flow rate across its boundary:Substitution of the Reynolds transport theorem and the divergence theorem givesSince the region is arbitrary, a continuous integrand must vanish: if it had a positive or negative value at a point, a sufficiently small surrounding region would have an integral of the same sign. This gives the continuity equation. With constant positive mass density $\rho$, it reduces toThus the flow is incompressible. For a material volume, the same argument starts directly from conservation of its total mass, with zero relative boundary flux.

Take a unit-width control volume spanning the stationary hydraulic jump. Mass conservation makes the discharge $q=hu$ constant. The horizontal momentum flux is $\rho h{u}^2$, and the integrated hydrostatic pressure force on a vertical cross-section is ${\int }_0^h\rho g(h-z)dz=\rho g{h}^2/2$. Momentum conservation therefore givesThese are the Rankine-Hugoniot conditions for the shallow water equations in the stationary shock frame. With depths $h_1\ne h_2$ on the two sides, eliminating the velocity gives

Peristaltic pumping transports fluid by a travelling deformation or travelling external pressure acting on a tube. With wave coordinate $X=x-ct$, mass conservation makes the relative flux $Q=\alpha (u-c)$ constant. The laboratory flux is $\alpha u=c\alpha +Q$, so a negative relative flux does not rule out forward mean pumping.

For a planar magnetohydrodynamic shock with normal velocity and tangential magnetic field, mass conservation gives $\rho u=m$ and the ideal magnetohydrodynamic induction equation gives $uB$ constant across the front. Hence $B/\rho$ is unchanged, expressing magnetic flux freezing. Normal momentum flux is $\rho u^2+p+B^2/(2{\mu }_0)$, and energy flux per unit mass flux is $u^2/2+\gamma p/[(\gamma -1)\rho ]+B^2/({\mu }_0\rho )$, apart from a smooth common gravitational potential. The amplified tangential field contributes both magnetic pressure and magnetic energy transport.

For constant phase mass densities ${\rho }_s,{\rho }_l$, a fixed solid framework has mixture density $\bar{\rho }={\rho }_s\phi +{\rho }_l(1-\phi )$. Mass conservation giveswhere $\mathbf{u}$ is Darcy flux. Equal phase densities give a divergence-free flux. Unequal densities require expansion or contraction flow during freezing or melting.

For $\mathbf{r}=a[\mathbf{q}+b(t)(p(q_1),0,0)]$, put $J=1+b{p}^{\prime }$. Before shell crossing, mass conservation gives $\rho =\bar{\rho }/J$. The dust background equation $3\ddot{a}/a=-4\pi G\bar{\rho }$ and the growing-mode equation $\ddot{b}+2H\dot{b}=4\pi G\bar{\rho }b$ imply ${\nabla }_r\cdot \ddot{\mathbf{r}}=-4\pi G\bar{\rho }+4\pi G\bar{\rho }b{p}^{\prime }/J=-4\pi G\rho$. Thus planar Zeldovich approximation trajectories satisfy the full nonlinear dust force-divergence equation until the map loses invertibility. All three physical coordinates are included in the divergence; the two unperturbed directions still contribute background acceleration.

Take depth $z$ positive downward, vertical fluid displacement $\xi$ positive upward, and Lagrangian pressure perturbation $\Pi$. In a plane-parallel atmosphere with $p^{\prime }=\rho g$, the Cowling approximation and adiabatic relation giveIndeed the Eulerian perturbations are $p_1=\Pi +\rho g\xi$ and ${\rho }_1=\Pi /c^2+{\rho }^{\prime }\xi$. Horizontal momentum gives ${\mathbf{\xi }}_h=i\mathbf{k}{p}_1/(\rho {\omega }^2)$; combine this with mass conservation and vertical momentum to eliminate the horizontal displacement. Reversing the sign convention for vertical displacement reverses both displacement-pressure coupling signs.

A potential flow has a velocity field represented by a scalar velocity potential, $\mathbf{v}=\nabla \varphi$. Its vorticity vanishes wherever $\varphi$ is sufficiently smooth. In a constant-density incompressible flow, mass conservation additionally gives Laplace's equation ${\nabla }^2\varphi =0$.

This reduces a vector-flow problem to a scalar boundary value problem. Linear combinations of elementary source and sink potentials can construct flows around simple geometries. The representation can be local even when a globally single-valued potential fails: a circulation-carrying flow around an excluded core has zero vorticity in the fluid region but can have nonzero circulation around the hole. Neglect of viscous effects also limits predictions near solid surfaces and in a wake. A useful potential-flow approximation therefore needs both its domain and its treatment of boundaries stated explicitly.

With well-mixed gas-phase metallicity $Z$, zero mass return, star formation rate $\psi$, pristine galactic gas inflow $\nu \psi$, and constant stellar yield $y$, mass conservation and metal accounting give $\dot{M}=-(1-\nu )\psi$ and $M\dot{Z}=(y-\nu Z)\psi$. Thus, for $0<\nu <1$,The initial gas-phase metallicity $Z_0=0$ gives the usual pristine-infall expression. Neglecting returned bulk mass while retaining a nonzero metal yield is a trace-metal idealization.

For the planar porous medium equation $n_t=[D_0/(3n_0^2)]\Delta (n^3)$, the mass-$Q$ point-source similarity solution is $n=n_0{\lambda }^{-2}\sqrt{[1-r^2/(r_0^2{\lambda }^2){]}_{+}}$, with $\lambda =(6D_{0}t/r_0^2{)}^{1/6}$ and $r_0^2=3Q/(2\pi n_0)$. Its front grows as $t^{1/6}$ and its central density falls as $t^{-1/3}$. The mass conservation integral uses the planar area element $2\pi rdr$. Although the density slope diverges at the front, its diffusion flux vanishes there, giving a compactly supported weak solution with initial measure $Q{\delta }^{(2)}$.