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Choose a length $\ell$, velocity $U_s$, time $\ell /U_s$, and pressure $\rho U_s^2$, giving Reynolds number $Re=U_s\ell /\nu$. Write perturbation velocity as $(u,v,w)$ and let $\mathcal{L}={\partial }_t+U{\partial }_x-R{e}^{-1}{\nabla }^2$. Linearizing the incompressible Navier-Stokes equations givesTaking the divergence gives ${\nabla }^{2}p=-2U^{\prime }{v}_x$. Applying ${\nabla }^2$ to the wall-normal momentum equation and usingthen cancels the pressure derivatives and yields $\mathcal{L}{\nabla }^{2}v-U^{\prime \prime }{v}_x=0$. Define wall-normal vorticity $\eta =u_z-w_x$. Applying ${\partial }_z$ to the streamwise equation minus ${\partial }_x$ to the spanwise equation gives $\mathcal{L}\eta =-U^{\prime }{v}_z$.

For the assumed normal modes, set $D=d/dy$ and ${\kappa }^2={\alpha }^2+{\beta }^2$. The two equations become the Orr-Sommerfeld equation and Squire equation:At rigid no-slip walls the conditions are $\hat{v}=D\hat{v}=\hat{\eta }=0$ for nonzero horizontal wavenumber.

Take $x$ down the plane and $y$ normally away from it, with the solid at $y=0$ and the free surface at $y=h$. A steady unidirectional falling film flow has velocity $\mathbf{u}=u(y){\mathbf{e}}_x$. The streamwise Navier-Stokes equation reduces towhere $\nu$ is the kinematic viscosity. The lower no-slip boundary condition and upper stress-free boundary condition areIntegrating gives the velocity profileThe volume flux per unit width is

If the upper surface is replaced by a stationary solid plane, the second boundary condition becomes $u(h)=0$. The resulting plane-channel profile and flux areThe free surface exerts no tangential shear stress, whereas the stationary upper wall enforces no slip and exerts a retarding shear. This additional drag accounts for the reduced flux.

Let $u(r,z)$ and $w(r,z)$ be the radial and vertical velocity components. For steady incompressible axisymmetric flow without swirl, mass conservation and the relevant cylindrical components of the Navier-Stokes equation are

The vertical scale is $w\sim V$. Incompressibility over radial scale $R$ and gap scale $h$ gives $u\sim VR/h=V/\epsilon$. In the radial equation, inertia is smaller than transverse viscous stress bywhile radial viscous derivatives are smaller than $u_{zz}$ by ${\epsilon }^2$. The pressure scale is $\mu V{R}^2/h^3$. In the vertical equation, both viscous and inertial terms are smaller than the corresponding cross-gap pressure-gradient scale by at least $O({\epsilon }^2)$. Thus the leading lubrication theory equations are

Write $d_{+}=\sigma +h$ and $d_{-}=h-\sigma$ for the lower and upper layer thicknesses. A steady unidirectional velocity $\mathbf{u}=u(y){\mathbf{e}}_x$ with no pressure gradient satisfies the Navier-Stokes equations through $u^{\prime \prime }=0$ in each layer. Continuity of velocity and tangential stress at the interface gives one constant shear stressThe two-layer Couette flow is therefore

Let $x$ point down the plane and $z$ normal to it, with velocity $(u,w)$. The two-dimensional incompressible Navier-Stokes equation isThe lubrication theory limit has film thickness much smaller than its streamwise length. Transverse viscous derivatives dominate, inertia is negligible at late times, and the leading equations areAt the wall, $u=w=0$ by the no-slip boundary condition. At the free surface $z=h(x,t)$, take $p=p_{\mathrm{atm}}$ and $u_z=0$, neglecting surface tension. The normal balance gives the hydrostatic pressureIf the current has thickness scale $H$ and length scale $L$, thenThus the streamwise pressure gradient is asymptotically smaller than the downslope body force for a fixed nonzero inclination. The streamwise equation becomesUsing $u(0)=0$ and $u_z(h)=0$ givesThe volume flux isIntegrating incompressibility across the film and using the kinematic boundary condition gives $h_t+q_x=0$, hence

Choose $x$ tangential to the solid boundary and locally along the outer inviscid streamlines, and let $y$ be the inward normal coordinate. Write $u$ and $v$ for the corresponding tangential and normal velocity field components, $U(x)$ for the outer tangential velocity, and $\nu$ for the kinematic viscosity.

For a layer of streamwise scale $L$ and thickness $\delta \ll L$, incompressible flow givesso $u=O(U)$ and $v=O(U\delta /L)$. The normal Navier-Stokes equation then gives $p_y=0$ to leading order, while the outer Euler equations giveIn the tangential Navier-Stokes equation, streamwise viscous diffusion is smaller than normal diffusion by $O((\delta /L{)}^2)$. Retaining the leading inertial and normal-diffusion terms produces the Prandtl boundary-layer equationIt is supplemented by $u=v=0$ at a stationary wall and $u\to U(x)$ on matching to the outer flow.

For the parallel flow $\mathbf{u}=(u(y,t),0)$ with no pressure gradient, the Navier-Stokes equations reduce to the diffusion equationSubstitution ofgivesThe boundary velocities require $f(L_0)=1$ and $f(-L_0)=-1$. The problem is antisymmetric under $y\mapsto -y$, so $f$ is odd. Withthe unique odd solution is

For the steady, fully developed ansatz $\mathbf{u}=w(r){\mathbf{e}}_z$, both ${\partial }_t\mathbf{u}$ and $(\mathbf{u}\cdot \nabla )\mathbf{u}$ vanish. The axial Navier-Stokes equation reduces toTwo integrations giveRegularity at the axis forces $C_1=0$, and the no-slip boundary condition $w(R)=0$ gives the Hagen-Poiseuille flowFor flow in the positive $z$ direction, $\partial p/\partial z<0$.

Take the curl of the Navier-Stokes equation. The pressure term disappears, curl commutes with the Laplacian, and the incompressible vector identitygivesTherefore the material derivative form is

Write the perturbation as $\eta e^{ikx+st}$ and use the same factor for all velocity and pressure amplitudes. At the upper interface, the linearized kinematic boundary condition, zero tangential traction, and normal-stress balance areThe first term in the normal stress is the linearization of the attractive disjoining pressure $V/h^3$, while the second is the stabilizing capillary pressure. Symmetry about $y=0$ makes $u$ even and $v$ odd, and supplies the lower-interface conditions.

For a two-dimensional Fourier mode, the Papkovich–Neuber representation is equivalently expressed by the odd biharmonic stream function for planar Stokes flowThe tangential-stress condition at $y=h_0$, with $K=k{h}_0$, giveswhereas the kinematic condition gives $iB\sinh K=s\eta$. The associated normal traction isEquating this with the linearized interfacial traction yields the dispersion relationThis is the Van der Waals rupture instability of a viscous sheet.

For $\Gamma =0$, the growth rate is positive, starts from $3V/(4\mu h_0^3)$ as $K\to 0$, and decreases to zero like $3V/(2\mu h_0^{3}K)$ as $K\to \infty$. For $\Gamma =1$, it has the same long-wave limit, vanishes at $K=1$, and is negative for $K>1$: surface tension damps wavelengths shorter than the cutoff. Long waves feel the attractive interaction but require coherent flow over a large distance; at short wavelengths viscous resistance suppresses the clean-film instability, while capillarity adds direct decay. A finite film size, fluid inertia, surrounding-fluid stresses, gravity, surface viscosity, and failure of the continuum disjoining pressure law can shift the observable most unstable wavelength.

During a growing thin spot, interfacial flow stretches the surface and dilutes its surfactant, thereby increasing the local surface tension above ${\gamma }_0$. Adjacent less-stretched regions retain more surfactant and lower tension. The resulting surface-tension gradient pulls toward the thin spot and opposes the outward flow that drives thinning. This is surfactant stabilization of film rupture; in the strong limit the surfaces behave almost as immobile boundaries.

For strong surfactant and $\Gamma \gg 1$, instability requires $K=O({\Gamma }^{-1/2})$. The Taylor expansionreduces the supplied relation toIt is maximal atThus the characteristic rupture time is $s_{\max }^{-1}=4\mu {\gamma }_0{h}_0^5/(3V^2)$. Surfactant greatly extends the life of a soap bubble while its film is moderately thick, but the $h_0^{-5}$ growth-rate dependence predicts rapid final rupture after drainage has made the film sufficiently thin.

The radial momentum equation makes $p$ independent of $r$. The axial Navier-Stokes equation reduces toThe no-slip boundary conditions are

The Navier-Stokes equation for constant mass density $\rho$, dynamic viscosity $\mu$, and body force per unit mass $\mathbf{f}$ iswhere the Newtonian fluid stress tensor isTaking the dot product with $\mathbf{u}$ givesIncompressibility impliesand converts the left side into a local time derivative plus the divergence of kinetic-energy flux. The divergence theorem therefore gives the kinetic-energy balance for an incompressible Newtonian fluidThe four terms are respectively outward advective transport of kinetic energy, mechanical power supplied by surface traction, power supplied by the body force, and irreversible viscous dissipation into heat.

Let $H$ be a representative positive gap thickness and setthe larger-order horizontal speed after including the squeeze-induced flux. The lubrication-limit scaling for a moving thin gap requiresEquivalently, the inertial condition containsThe gap must also remain open, $h(x)>0$. These conditions make streamwise derivatives small relative to transverse derivatives, keep the pressure uniform across the gap to leading order, and reduce the Navier-Stokes equation to Stokes flow locally.

The thermodynamic order parameter is the conserved scalar composition difference $\varphi$ of the binary fluid mixture. The momentum density $\rho \mathbf{v}$ is an additional conserved hydrodynamic mode; it is not a symmetry-breaking order parameter, but it must be retained because momentum relaxes only through spatial transport.

Composition conservation gives a continuity equation,The current $\mathbf{J}=-M\nabla \mu$ is the leading isotropic, dissipative constitutive law: it drives material down gradients of the chemical potential. For the Landau-Ginzburg theorythe functional derivative isThese statements give the advective Cahn--Hilliard equation.

Constant mass density and incompressible flow require $\nabla \cdot \mathbf{v}=0$. Momentum conservation gives the Navier-Stokes equation: material acceleration equals the sum of the Newtonian viscous force $\eta {\nabla }^2\mathbf{v}$, the pressure force $-\nabla P$, and the Korteweg force $-\varphi \nabla \mu$. The pressure is the Lagrange multiplier enforcing incompressibility. Together these equations are Model H dynamics; isothermality removes the need for a separate energy equation.

Write the density as a hydrostatic reference profile $\hat{\rho }(z)$ plus a small perturbation ${\rho }^{\prime }$, and define buoyancy and buoyancy frequency byThe Boussinesq approximation to the Navier-Stokes equation, together with mass conservation, iswhere $D/Dt$ is the material derivative, $\nu$ is kinematic viscosity, and $\kappa$ is mass diffusivity. Linearizing about rest and eliminating pressure and buoyancy givesFor a plane wave proportional to $e^{i(kx+mz-\omega t)}$, with $K^2=k^2+m^2$, the viscous-diffusive dispersion relation isIn the inviscid limit this becomesFor weak diffusion the two oscillatory roots areso a freely evolving Fourier mode decays.

A single plane wave is also an exact solution of the nonlinear equations. Every field depends only on its phase $\Theta =\mathbf{k}\cdot \mathbf{x}-\omega t$, while incompressible flow gives $\mathbf{u}\cdot \mathbf{k}=0$. Therefore $\mathbf{u}\cdot \nabla$ annihilates both $\mathbf{u}(\Theta )$ and $b(\Theta )$, and all nonlinear advection terms vanish.

For a boundary-forced wave with real $\omega$, nonzero $\nu$ or $\kappa$ instead makes the bulk vertical wavenumber complex, attenuating the propagating beam. Because diffusion raises the spatial order of the equations, additional short vertical-wavenumber roots form viscous and scalar boundary layers; they allow a no-slip velocity condition and a scalar no-flux condition to accompany impermeability. These layers and bulk attenuation become essential near critical internal-wave reflection, where the inviscid reflected wavelength collapses.

The horizontal velocities $u$ and $v$ point east and north, $\eta$ is the displacement of the free surface from its mean level, $H_0$ is the undisturbed depth, $g$ is gravitational acceleration, and $f$ is the constant Coriolis parameter on an f-plane. The three linearized shallow water equations are horizontal momentum balance and mass conservation:They follow from the rotating Navier-Stokes equation by assuming an inviscid homogeneous layer, hydrostatic pressure, horizontal scales much larger than $H_0$, depth-independent horizontal velocity, a flat impermeable bottom, constant $f$, and small surface displacement and velocity so that nonlinear products are neglected.

In a steady state, geostrophic balance givesThe relative vorticity is $\zeta =v_x-u_y$. Expanding the shallow-water potential vorticity $(f+\zeta )/(H_0+\eta )$ to first order givesThus one convenient normalization of its disturbance isTaking the curl of momentum and using continuity shows $q_t=0$.

For a smooth incompressible Navier-Stokes equation trajectory with zero velocity at a fixed flat wall, the normal momentum equation determines the normal pressure gradient. In a perturbation model with buoyancy term ${\mathrm{Ri}}_B\theta \hat{\mathbf{y}}$, it gives ${\partial }_{n}p=\nu \mathbf{n}\cdot \Delta \mathbf{u}-{\mathrm{Ri}}_B\theta \mathbf{n}\cdot \hat{\mathbf{y}}$. An extra homogeneous pressure Neumann boundary condition is valid only when the right side vanishes; it does not follow from the no-slip boundary condition.

A unidirectional flow has velocity parallel to a single fixed direction, for example $\mathbf{u}=u(y){\mathbf{e}}_x$. If the speed is independent of the coordinate along that direction, the convective acceleration $(\mathbf{u}\cdot \nabla )\mathbf{u}$ vanishes. The steady Navier-Stokes equation then reduces to a linear balance of the pressure gradient and viscosity, as in Couette flow and Poiseuille flow.