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A stationary viscous wall imposes the no-slip boundary condition, $u=v=0$, whereas the outer inviscid flow has $u=Ax$. At large Reynolds number, the transition is confined to a thin viscous boundary layer, where normal velocity gradients amplify the otherwise small viscous term.

Take a horizontal scale $L$, an outer speed $U=AL$, and a boundary layer thickness $\delta \ll L$. The continuity equation gives $v=O(U\delta /L)$. Write $ϵ=\delta /L$ and ${\mathrm{Re}}_L=UL/\nu$. The tangential Navier-Stokes equation has scalesBalancing convective acceleration with normal viscous diffusion gives $ϵ=O({\mathrm{Re}}_L^{-1/2})$. The leading tangential equation and incompressible flow constraint areHere and below $\nu$ is the kinematic viscosity, while $u$ is the horizontal velocity.

For the normal equation, first remove the hydrostatic pressure by writing $P=p+\rho gy$. Normal inertia and normal viscous terms are $O(\rho U^2\delta /L^2)$, so the change of $P$ across the layer is only $O(\rho U^2{ϵ}^2)$. Matching to the outer pressure therefore makes the leading $O(\rho U^2)$ excess pressure independent of $y$:The outer expression similarly has $(p_{\infty }{)}_y=-\rho g-\rho A^{2}y$. Thus the printed conditionis a leading-order boundary layer matching statement, with the smaller normal dynamic terms discarded on both sides. It is not an exact identity at the next order. In particular, retaining the outer $-\rho A^{2}y$ while dropping normal inertia and normal viscous terms inconsistently mixes orders. The tangential driving term is unambiguous: $-(p_{\infty }{)}_x/\rho =A^{2}x$.

For an inviscid flow, the force per unit volume is $\mathbf{F}=-\nabla p-\rho g\hat{\mathbf{z}}$. The gravitational acceleration has zero curl. Using the curl product rule,The supplied two-dimensional vorticity equation therefore becomesThe out-of-plane vorticity has no stretching term in this two-dimensional setting. The source is baroclinic vorticity generation: differently directed mass density and pressure gradients generate rotation even from zero initial vorticity.

For an explicit sign check, set ${\omega }_y={\partial }_{z}u-{\partial }_{x}w$, with $z$ upward. A thin smoothed interface has $\rho \simeq \bar{\rho }(z-\eta (x,t))$, with ${\bar{\rho }}^{\prime }>0$. If $P(x,t)$ denotes the total pressure evaluated on the interface, then within the thin layer $p_x+{\eta }_x{p}_z\simeq P_x$ andIntegrating through that layer and taking its thickness to zero gives the vortex sheet strength $\gamma =u_2-u_1$ and its linear evolution:For a growing Fourier mode, the potentials from the preceding calculation with $\alpha$ replaced by $k$ giveThus the right-hand descending slope of a light-fluid crest produces positive ${\omega }_y$, while the left-hand ascending slope produces negative ${\omega }_y$. Their induced velocity raises the light-fluid crest and lowers the heavy-fluid trough, increasing the slope and generating more vorticity. This is the positive feedback underlying Rayleigh-Taylor instability.

One can close this mechanism quantitatively. For $\eta =H(t)\cos (kx)$, the kinematic boundary conditions give $\gamma =2\dot{H}\coth (kh)\sin (kx)$. Combining with the sheet-generation equation givesFor $kh\ll 1$, the lids confine motion throughout both layers and horizontal redistribution is slow, yielding $\sigma \sim \sqrt{\mathcal{A}gh}k$. For $kh\gg 1$, motion is localized within depth $k^{-1}$ on each side. Less fluid must accelerate at shorter wavelength, giving $\sigma \sim \sqrt{\mathcal{A}gk}$. Since $k_n=n\pi /b$, the ideal growth rate increases with $n$ with no short-wave cutoff.

The sharp-interface result requires $k\delta \ll 1$ for the smoothed layer. When $k\delta$ becomes order one, it no longer applies. A smooth adverse mass density gradient has a local overturning time of order $[g{\bar{\rho }}^{\prime }/\bar{\rho }{]}^{-1/2}$ rather than the unbounded sharp-interface rate. Finite molecular diffusion also relaxes small-scale mass density perturbations on a time of order $(D{k}^2{)}^{-1}$, and viscosity, if included, damps small-scale velocity. Weak diffusion motivates the thin-layer picture; it does not justify extrapolating the discontinuous dispersion relation to arbitrary $n$.

For inviscid flow that is also incompressible flow and irrotational flow with velocity potential $\mathbf{u}=\nabla \varphi$, the Euler equations integrate to the Unsteady Bernoulli equationThe same $F(t)$ applies throughout a connected irrotational fluid region; an additive time-dependent potential can absorb it. With gravity absent, fluid at rest at infinity and $\varphi \to 0$, it is $p_{\infty }(t)/\rho$.

Take the cylinder centre at laboratory position $(X(t),0)$ with $\dot{X}=U(t)$, and define $r,\theta$ relative to this moving centre. The circulation-free potential flow satisfies Laplace's equation, decays at infinity, and has surface condition ${\varphi }_r(a,\theta )=U\cos \theta$. The $\cos \theta$ harmonic solution is consequentlyThe velocities are $u_r=U{a}^2\cos \theta /r^2$ and $u_{\theta }=U{a}^2\sin \theta /r^2$, hence $|\mathbf{u}{|}^2=U^2{a}^4/r^4$.

The time derivative must be taken at fixed laboratory coordinates, not fixed $r,\theta$. At a fixed fluid point $\dot{r}=-U\cos \theta$, $\dot{\theta }=U\sin \theta /r$, soSubstituting into the Unsteady Bernoulli equation gives the entire pressure fieldAt the cylinder surface this is $p_{\infty }+\rho a\dot{U}\cos \theta +\rho U^2(\cos 2\theta -1/2)$. Pressure acts against the outward radial normal of the cylinder, so the force per unit length isAll steady terms cancel by trigonometric orthogonality. ThereforeThis identifies the added mass of a circular cylinder as the mass of displaced fluid per unit length. The solution assumes the ambient fluid is initially at rest and has no separately prescribed circulation; these select the usual pure translating-cylinder flow.

The Lighthill elongated-body theory is a reactive, nearly inviscid flow description of a long slender swimmer, appropriate when the longitudinal Reynolds number is large but its lateral displacement and slope are small. To leading order, crossflow past each section is the two-dimensional potential flow generated by lateral motion of that section. Longitudinal variation is slow on the cross-sectional scale, and axial speed $U$ is prescribed at this order. With $x$ increasing from nose to tail, fluid passes sections in increasing $x$, and the linearized relative lateral velocity isThe section's added mass per unit length $m(x)$ is defined by its crossflow kinetic energy $\frac{1}{2}m(x)w^2$. Equivalently, if $\Phi$ is the velocity potential for unit lateral section motion, then $m=\rho \int |{\nabla }_{\perp }\Phi {|}^{2}dA$ over the exterior crossflow plane. It has units of mass per length and depends on cross-sectional shape and density. For a circular section of radius $a$, $\Phi =-a^2\cos \phi /r$ gives $m=\rho \pi a^2$. It is not the body's mass per unit length.

The transverse fluid impulse attached to a section is $m(x)w$. A control segment has both changing impulse and an impulse flux $Umw$ through its ends. Taking their sum per unit length gives the reaction on the fluid:The force on the fish is $-F_z$. In particular, $m$ is inside the convective derivative; dropping its spatial derivative or the endpoint fluxes is not justified merely because the fish is slender.

The recoil correction in elongated-body theory adds unknown whole-body translation and yaw to a prescribed active bending shape:They are determined by total lateral linear momentum and yaw angular momentum balance, rather than by imposing that the deforming swimmer remains on its original axis. Body inertia and hydrodynamic added inertia have the same sign. If the translation variable is required to be the literal centre of mass, the active deformation must first have its mass-weighted mean removed: the reference displacement $Z_r$ is otherwise not the centre of mass displacement during the stroke. The corresponding deformation contribution to angular momentum must also be retained when it is significant.

For an inviscid flow that is an incompressible flow and an irrotational flow, write $\mathbf{u}=\nabla \varphi$. The Euler equations for an inviscid fluid, without gravity, integrate spatially to the Unsteady Bernoulli equationA time-dependent addition to $\varphi$ can absorb $C(t)$.

In the ideal uniform tube-plug model, let $u(t)$ be the axial speed toward the outlet. Incompressibility and fixed cross-section make the speed the same at both ends. With $\varphi =u(t)x$, the kinetic terms cancel when the Unsteady Bernoulli equation is subtracted between entrance and exit, givingConservation of volume in the spherical balloon gives $4\pi R^2\dot{R}=-\pi a^{2}u$, or $u=-4R^2\dot{R}/a^2$. Differentiating and eliminating $u$ yieldsThis is the constant-tension balloon discharge equation; entrance losses and a reservoir-to-tube kinetic-head correction are excluded by this ideal plug model.

Multiplication by $2R\dot{R}$ turns the equation intoThe printed time formula additionally assumes the water is initially at rest, $\dot{R}(0)=0$, although only $R_0$ is explicitly specified in the question. Under that assumption $C=K{R}_0^2$. On the shrinking branch,Set $r=R_0\sin \vartheta$ and $\theta =\arcsin (R/R_0)$. The integral becomes $R_0^2{K}^{-1/2}{\int }_{\theta }^{\pi /2}{\sin }^2\vartheta d\vartheta$, hencePutting $R=0$, or $\theta =0$, gives the ideal emptying timeFor an independently prescribed initial rate $v_0=\dot{R}(0)$, the first integral instead has $C=K{R}_0^2+R_0^4{v}_0^2$, so radius alone does not determine that emptying time. This explicitly identifies the extra initial datum needed for the displayed answer.

For steady inviscid flow at atmospheric pressure, Bernoulli's equation along a freely falling streamline gives $u^2/2+gz=u_0^2/2+gH$. The downward speed is thereforeFor the incompressible jet, mass conservation fixes its volume flux: $\pi R^{2}u=\pi a^2{u}_0$. ConsequentlyFar from impact, the inviscid falling jet and radial impact film is thin and nearly horizontal. Let $v$ be its radial speed and $h$ its thickness. The same volume flux now crosses a cylindrical surface of area $2\pi rh$, giving $v=a^2{u}_0/(2rh)$. At the free surface, atmospheric pressure and Bernoulli's equation give $v^2=u_0^2+2g(H-h)$. Eliminate $v$ to obtainThis description assumes a steady incompressible jet, negligible surface tension, and negligible dissipation between the tube and the thin far-field film. It is a free-surface relation, not an assertion that pressure is atmospheric throughout the film or that the impact region itself is thin.

For an inviscid flow of constant density $\rho$, driven by a conservative body force $-\nabla \Phi$, the Euler equation isThe vector identity $(u\cdot \nabla )u=\nabla (|u{|}^2/2)-u\times (\nabla \times u)$ and irrotationality give the Unsteady Bernoulli equation. Specifically, wherever a single-valued potential $u=\nabla \varphi$ is available,On a connected potential-flow region this meansThe integration function is spatially constant but need not be constant in time; it can be removed by changing the potential's time-dependent gauge. A simply connected irrotational region provides the requisite global potential. For a barotropic fluid the pressure term is replaced by $\int dp/\rho (p)$; here the liquid density is constant.

For the tube, let $U(t)$ be its approximately uniform axial velocity. The large reservoir's surface velocity is negligible and gravity is neglected as stipulated. Entrance-region inertia contributes an effective length of order $a$, negligible against $L$. The potential difference between the outlet and the reservoir is consequently $LU$ to leading order. Subtract Bernoulli at the reservoir surface, where pressure is $P+p_a$, from its value at the outlet, where pressure is $p_a$ and speed is $U$. This givesThe inertial term cannot be dropped during startup. With $U_{\infty }=\sqrt{2P/\rho }$, separate variables:Hence the inviscid startup in a pressure-driven tube givesInitially $U\sim Pt/(\rho L)$, the acceleration of the stationary liquid column. At long times $U\to \sqrt{2P/\rho }$, the inviscid pressure-to-kinetic-energy limit. The result applies while the tube remains full and reservoir dimensions are large enough that surface motion and the neglected entrance inertia remain small.

Assume a steady, incompressible, inviscid flow with a nearly uniform velocity $u(x)$ across each water section. The channel varies slowly enough for vertical accelerations to be negligible, pressure to be hydrostatic and the water surface to be nearly horizontal across the width. Atmospheric pressure is uniform; neglect surface tension, wall friction and energy losses. The hydraulic equations apply away from jumps and short nonhydrostatic regions.

The Euler equation for horizontal momentum, with hydrostatic pressure, givesIntegrating gives a constant Bernoulli equation head $\mathcal{B}=H+h+u^2/(2g)$. Conservation of mass gives the constant volume discharge $Q=bhu$. Thus the shallow-water specific energy, measured above the local bed, isDefine the Froude number by $F=u/\sqrt{gh}$ for the positive flow direction. ThenAt fixed discharge and width, energy tends to infinity at both zero and infinite depth. It has a unique minimum atIndeed $E_{hh}=3Q^2/(g{b}^2{h}^4)>0$. Above the minimum there are two depth branches: the shallower one is supercritical flow and the deeper one is subcritical flow.

The local long-wave propagation speeds from the shallow water equations are $u\pm \sqrt{gh}$. In subcritical positive flow, one wave family travels upstream and one downstream. In supercritical flow both travel downstream. At critical flow the upstream-directed speed is zero, so the flow can provide hydraulic control separating upstream adjustment from downstream conditions. This statement concerns long waves in the slowly varying hydraulic approximation, rather than all short-wave disturbances.

The rigid-rotation velocity field is $\mathbf{v}=(-\Omega y,\Omega x,0)$, with spatially uniform $\Omega (t)$. Its divergence is zero, since ${\partial }_x(-\Omega y)={\partial }_y(\Omega x)=0$. It therefore satisfies the incompressible continuity equation. On the cylindrical wall, $\mathbf{v}\cdot (x,y,0)=0$, so it has zero normal velocity; on either end cap its vertical component is zero. These are the impermeability boundary conditions for an inviscid flow. Tangential slip is permitted; a no-slip condition is not part of this inviscid model.

Take $\alpha >0$; a real disturbance with negative wavenumber is recovered by complex conjugation. To obtain the pressure equation for a shear-flow normal mode, apply $i\alpha$ to the streamwise momentum equation and differentiate the normal momentum equation. The mass conservation equation cancels the viscous divergence and the terms proportional to $U-c$, leaving $2i\alpha U^{\prime }v={\alpha }^{2}p-p^{\prime \prime }$. Thus the fluid pressure equation and wall value areThe wall value follows from streamwise momentum with no-slip boundary condition. Normal momentum also gives $p^{\prime }(0)=v^{\prime \prime }(0)/R$.

In the lower viscous boundary layer, mass conservation gives $v=O(\alpha \delta )$ if $u=O(1)$. Since $U\sim y$, convection and shear have size $\alpha \delta$ when $c=O(\delta )$; matching these with transverse viscous diffusion gives $R^{-1}{\delta }^{-2}\sim \alpha \delta$. The pressure gradient balance then gives $p=O(\delta )$. Thus the lower scalings areThe neglected streamwise viscous term is smaller by $(\alpha \delta {)}^2$. In these variables the leading equations areDifferentiating the second and using the first yields the Airy reduction $\boxed{u_{YYY}=i(Y-C)u_Y}$.

Put $q=e^{i\pi /6}$, so $q^3=i$. The bounded matching branch of the Airy function gives $u_Y=A\mathrm{Ai}(q(Y-C))$; the other independent Airy solution grows along the ray and is excluded. Integrating with the two wall conditions giveswhile the wall momentum equation gives the constant fluid pressureDefine the convergent Airy displacement integralsThe Airy function equation and its decaying derivative imply $I_1-C{I}_0=iq{\mathrm{Ai}}^{\prime }(-qC)$. Consequently, as $Y\to \infty$,Both differentiated lower momentum and the constant of integration are satisfied; retaining that constant is essential for the final matching.

For the middle region introduce a perturbation stream function $\Phi$ with $u={\Phi }^{\prime }$ and $v=-i\alpha \Phi$. The leading inviscid flow streamwise equation isFor a shear flow, this is the integrated form of the Rayleigh equation for inviscid shear flow. The leading matched solution for fixed $y=O(1)$ is $\Phi \sim BU$, so $u\sim B{U}^{\prime }$ and $v\sim -i\alpha BU$. In the overlap with the lower viscous boundary layer, the first constant correction iswhich agrees with the integrated lower solution. For a regular smooth profile, the normal momentum equation gives $p^{\prime }\sim -{\alpha }^2(U-c)\Phi$. Thus fluid pressure variation across a fixed-width middle region is $O({\alpha }^{2}B)$; in the final distinguished scaling this is smaller than its $O(\delta )$ value, so $p\sim \delta P_0$ throughout that region. Since $U\to 1$, the middle velocity tends to $\boxed{v\sim -i\alpha B}$, which does not decay and requires another region.

The fluid pressure equation becomes $p^{\prime \prime }-{\alpha }^{2}p=0$ when $U\sim 1$. Therefore the upper scale is $\Delta =1/\alpha$, with $z=\alpha y$. The inviscid decaying upper solution isMatching velocity as $z\downarrow 0$ gives $P_{\infty }\sim \alpha B$; matching fluid pressure to the middle region gives $\delta P_0\sim \alpha B$. This is three-layer long-wave shear-flow matching, expressed as a matched asymptotic expansion.

Under $\alpha =k{R}^{-1/4}$ with fixed positive $k$, $\delta =k^{-1/3}{R}^{-1/4}$, and $\alpha /\delta =k^{4/3}$. Eliminating the nonzero amplitude $A$ gives the Airy wall-layer matching determinant and dispersion relationEquivalently, with $\mathcal{T}(C)={\int }_{-qC}^{\infty e^{i\pi /6}}\mathrm{Ai}(\zeta )d\zeta =q{I}_0(C)$,The determinant form does not divide by a possibly vanishing $I_0$. The wavespeed is $c=\delta C$; temporal growth has sign $\mathrm{Im}C$ for positive $\alpha$.

There is a genuine conflict in the printed assumptions: fixed-$k$ $\alpha =k{R}^{-1/4}$ cannot satisfy $R^{-1/7}\ll \alpha$ as $R\to \infty$. For order-one lower matching amplitudes, constant leading middle fluid pressure requires ${\alpha }^2\ll \delta$, equivalently $\alpha \ll R^{-1/7}$; this is the reverse inequality and is satisfied by the final distinguished scaling. Thinness also requires $\alpha R\gg 1$. The dispersion relation above is therefore derived under the final distinguished limit, for which the complete consistent window is $R^{-1}\ll \alpha \ll R^{-1/7}$. Both printed limits cannot be imposed simultaneously.

Use the standard ideal-fluid assumptions of incompressible flow, inviscid flow, gravity $-g{\mathbf{e}}_y$, and no surface tension. Irrotational flow gives ${\mathbf{u}}_j=\nabla {\phi }_j$ in each layer, and incompressibility gives Laplace equationThe rigid-wall boundary conditions are ${\phi }_{1y}=0$ at $y=h_1$ and ${\phi }_{2y}=0$ at $y=-h_2$. At the moving interface, each fluid has the same normal velocity as the interface:The dynamic condition is continuity of pressure, $p_1=p_2$ at $y=\eta$. Bernoulli equation in each fluid reads ${\phi }_{jt}+\frac{1}{2}|\nabla {\phi }_j{|}^2+p_j/{\rho }_j+gy=C_j(t)$. Choosing the potential gauges to subtract a common reference pressure, the interface condition becomesTangential velocities need not agree across an inviscid interface.

For a material volume, conservation of mass and the Reynolds transport theorem give ${\rho }_t+\nabla \cdot (\rho u)=0$. Constant material density, or equivalently incompressibility with the stated constant initial density, reduces this to $\nabla \cdot u=0$.

In an fluid undergoing inviscid flow the stress is the isotropic pressure stress $-pI$. Momentum conservation for a material volume then givesTransport and the divergence theorem localize this toFor a steady flow, $u_t=0$. For an irrotational flow the vorticity is zero, so the supplied identity gives $(u\cdot \nabla )u=\nabla (|u{|}^2/2)$. Uniform downward gravity is $g=-\nabla (gz)$, and constant density permits pulling $1/\rho$ inside the pressure gradient. ThereforeThe Bernoulli invariant is constant throughout each connected fluid region. Without irrotationality the corresponding steady statement generally gives a constant only along each streamline.

For a material fluid volume, mass conservation yields ${\rho }_t+\nabla \cdot (\rho \mathbf{u})=0$. Constant mass density reduces this to $\nabla \cdot \mathbf{u}=0$. For inviscid flow, the Cauchy stress tensor is $-pI$; its net surface force is the volume integral of $-\nabla p$ by the divergence theorem. Newton's second law for each material volume therefore gives $\rho D\mathbf{u}/Dt=-\nabla p$. Since $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$, these are exactly the two stated equations.

In two dimensions they readDifferentiate the second momentum equation in $x$ and subtract the $y$ derivative of the first. The pressure terms cancel. For $\omega =v_x-u_y$, collecting terms givesIncompressibility removes the last term, givingThus each fluid parcel retains its scalar vorticity as it moves. In particular an initially irrotational flow remains irrotational in this smooth two-dimensional inviscid evolution. There is no vortex stretching in this two-dimensional setting.

Following the moving fluid. The Eulerian velocity field $\mathbf{u}(\mathbf{x},t)$ assigns a velocity to each position. A fluid particle satisfies $\dot{\mathbf{x}}=\mathbf{u}$, so the chain rule gives the material derivative, also called the convective derivative,Even in steady flow a particle can accelerate because it moves between regions with different velocity. For example, flow around a curved obstacle changes direction despite ${\partial }_t\mathbf{u}=0$. Mass conservation isFor a constant-density incompressible fluid, $\nabla \cdot \mathbf{u}=0$.

Euler equations for an inviscid fluid and Bernoulli's equation. Neglect viscosity and take the body force per unit mass to be $-\nabla \Phi$. Newton's second law for a fluid element givesFor steady flow, dot with displacement along a streamline. The tangential component of $\mathbf{u}\cdot \nabla \mathbf{u}$ is the derivative of $u^2/2$ along that line. For constant density, integration givesIn uniform gravity, $\Phi =gz$. Thus pressure, kinetic energy per unit mass and gravitational potential energy per unit mass exchange along a streamline. For a barotropic compressible fluid, replace $p/\rho$ by $\int dp/\rho (p)$. Different streamlines need not have the same Bernoulli constant.

Vorticity and potential flow. Vorticity is $\mathbf{\omega }=\nabla \times \mathbf{u}$. A small fluid element rotates locally with angular velocity $\mathbf{\omega }/2$ after its deformational part is separated. The vector identitygives $\nabla B=\mathbf{u}\times \mathbf{\omega }$ for steady constant-density inviscid flow. Thus an irrotational region has one common Bernoulli constant on each connected region. In a simply connected irrotational domain, introduce a velocity potential $\mathbf{u}=\nabla \varphi$. Incompressibility then gives Laplace's equation, ${\nabla }^2\varphi =0$. For unsteady irrotational flow,An impermeable stationary wall fixes ${\partial }_n\varphi =0$, but ideal potential flow does not enforce no slip. In a multiply connected domain, irrotational flow may still have nonzero circulation around an excluded vortex core, so a globally single-valued potential requires care.

For example, uniform two-dimensional flow past a circular cylinder has $\varphi =U(r+a^2/r)\cos \theta$ outside $r=a$. Its normal velocity vanishes at the cylinder, and its surface tangential velocity is $-2U\sin \theta$. Bernoulli's equation predicts a symmetric pressure distribution and zero drag. This is D'Alembert's paradox: the assumptions omit the viscous boundary layers, separation and wake responsible for real drag.

For an incompressible fluid of constant kinematic viscosity $\nu$, the curl of the motion equation givesThe first term stretches and tilts vortex lines; the second diffuses vorticity. In two-dimensional incompressible inviscid flow the stretching term is absent and vorticity is carried with the fluid. In a three-dimensional inviscid flow it can be amplified by stretching. Density gradients not parallel to pressure gradients provide an additional source in a general compressible fluid.

Viscous flow. A Newtonian fluid's shear stress is proportional to rate of strain. For constant viscosity and incompressibility the Navier-Stokes equation iswith dynamic viscosity $\eta$ and kinematic viscosity $\nu =\eta /\rho$. At an ordinary solid wall, the no-slip boundary condition sets fluid velocity equal to the wall velocity. Viscosity transfers momentum between neighbouring layers and converts mechanical energy into heat; unlike the pressure force, its shear contribution depends on velocity gradients.

For plane Couette flow between plates $y=0,h$, with the upper plate moving at $U$, the steady solution is $u_x=Uy/h$, with shear stress $\eta U/h$. For pressure-driven flow through a pipe of radius $a$, a constant pressure gradient gives the parabolic Poiseuille profileThe Reynolds number $\mathrm{Re}=\rho UL/\eta$ compares convective inertia with viscous force. Small Reynolds number often permits the Stokes-flow approximation; large Reynolds number can produce thin boundary layers and, under suitable conditions, turbulence. It does not justify neglecting the wall conditions. Bernoulli's inviscid constant is generally not conserved along a viscous streamline, and pipe flow instead shows a pressure loss corresponding to dissipation.

Flow visualization and the idealization. A streamline is everywhere tangent to the instantaneous velocity field at one fixed time. A streakline is the locus, at one observation time, of particles that previously passed through a specified injection point. A marked particle's own trajectory is a pathline. These coincide geometrically in steady flow, but need not do so in an unsteady draining flow.

The intended ideal liquid is approximately incompressible and inviscid: viscosity and associated shear stresses are negligible on the scales and times being considered, the Mach number is small, and the density is effectively constant. This does not require zero vorticity; the initially rotating state is already a counterexample. It also does not justify ignoring viscosity in an arbitrarily thin boundary layer or at a real drain.

Spiralling flow and angular momentum. In an axisymmetric inviscid flow, pressure and gravity exert no torque about the vertical axis. A parcel therefore conserves axial specific angular momentum,The late, shallow remnant in the intended draining picture is supplied by parcels initially near the outer rim, so $r_0\simeq R$. The slow-drain angular-momentum amplification then givesThe assumption about the parcels' origin is necessary: torque conservation by itself does not assign $r_0=R$ to every parcel at every time.

A continuous dye feed near the outer free surface produces inward spirals that wind increasingly rapidly near the drain. The figure uses a late quasisteady model, $u_r=-D/r$ and $u_{\varphi }=C/r$ with positive constants $D,C$, giving $\varphi ={\varphi }_0-(C/D)\ln (r/r_0)$. Its streaklines and streamlines approximately coincide over an observation interval short compared with the overall draining time. Strictly, individually marked particles at the initial surface trace pathlines, not a continuously injected streakline.

The free-surface depression. Let $z=H(r)$ be the surface. Slow radial draining makes radial inertial corrections small compared with the azimuthal centripetal acceleration. The radial Euler equation and vertical hydrostatic equilibrium giveThus $g{H}^{\prime }(r)=u_{\varphi }^2/r=C^2/r^3$, with $C={\omega }_0{R}^2$. Integrating from $r$ to the rim givesFor $r\ll R$ this becomesThe surface rises outward because the inward pressure gradient force must sustain the circular motion. The dimensions are metres, and the $1/r^2$ profile is not valid inside the finite drain or vortex core.

The terrestrial estimate. Use the vertical component of Earth's rotation as the ideal initial angular speed, ${\omega }_0\simeq {\Omega }_E\sin \lambda$. At Cambridge, taking $\lambda \simeq 52.2^{\circ }$ and ${\Omega }_E=7.292\times 10^{-5}{\mathrm{s}}^{-1}$, this gives ${\omega }_0\simeq 5.76\times 10^{-5}{\mathrm{s}}^{-1}$. Hence, for $R=1\mathrm{m}$, $r=0.01\mathrm{m}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$,Taking the whole terrestrial angular speed rather than its local vertical projection would give about $2.7\mu \mathrm{m}$. The Cambridge estimate uses the projection appropriate to the vertical vortex. This very small head assumes terrestrial rotation supplies the only initial swirl; imposed residual rotation would change it quadratically.

Curl of the azimuthal velocity. For axisymmetric $u_{\varphi }=r\omega (r)$ with no vertical dependence, the rotational part of the vorticity is vertical:The ideal quasisteady radial sink has no curl outside its core. The source's symbol $\Omega$ denotes this vertical vorticity component, not the fluid's angular velocity. In general its magnitude is the absolute value of the bracket.

The two limiting states. For initial uniform rotation, $\omega ={\omega }_0$, whereas the late approximation has $\omega =C/r^2$. ThereforeThe late flow can have large angular velocity and nonzero circulation while being locally irrotational. The drain core is excluded from the domain of the $1/r$ velocity field.

What is actually conserved. Inviscid does not mean that each parcel's vorticity magnitude is fixed in three dimensions. For an incompressible barotropic liquid with conservative body forces, the vorticity equation isVertical compression of a material column changes its vorticity through vortex stretching, even while its volume is conserved by horizontal expansion. In the columnar shallow-layer limit, vorticity dilution in a shallow draining layer givesThus a remnant much shallower than its original column can have much smaller vertical vorticity. At finite depth there is generally a residual value; the exactly irrotational late profile is a leading limiting approximation, not an exact parcelwise conversion of nonzero vorticity to zero in finite time.

Kelvin's circulation theorem conserves circulation around a material loop, not around a fixed spatial circle. A loop whose original radius is near $R$ has circulation about $2\pi {\omega }_0{R}^2$, consistent with the late $u_{\varphi }=C/r$ field. Fluid also leaves through the drain, so the remaining tank contents are not a fixed material system. If the calculation is expressed explicitly in the Earth-fixed rotating frame, the corresponding shallow-water potential vorticity is $(\zeta +f)/H$, with $f=2{\Omega }_E\sin \lambda$; the conserved quantity involves absolute vorticity.

For steady inviscid flow of constant density without a body-force potential, Bernoulli's theorem givesThe constant can differ between streamlines, because this flow has nonzero vorticity. Dotting the steady Euler equation with $\mathbf{u}$ gives $\mathbf{u}\cdot \nabla (p/\rho +|\mathbf{u}{|}^2/2)=0$, which proves precisely the along-streamline form.

Heinrich Blasius developed a similarity solution for the thin boundary layer along a flat plate. Even when viscosity is small, a fluid must satisfy the no-slip boundary condition at the surface, so an inviscid flow approximation cannot describe the whole region. Balancing downstream advection against transverse diffusion of momentum gives the boundary layer scale $\delta (x)\sim \sqrt{\nu x/U}$, where $\nu$ is kinematic viscosity, $U$ the external speed, and $x$ the downstream distance from the plate's leading edge. The similarity solution reduces the developing boundary layer profile to a function of one scaled coordinate.^[ref]

The pressure coefficient compares a local static pressure with a specified reference static pressure, using a fixed reference dynamic pressure:The reference mass density, speed, pressure and location must accompany a reported pressure coefficient. Keep the same $q_{\infty }$ throughout a surface distribution; replacing it by local $\rho U^2/2$ gives a different quantity and fails at a stagnation point. The definition itself requires neither Bernoulli's principle nor inviscid flow, so measured pressure coefficients remain useful when those idealizations fail.

Positive $C_p$ means $p>p_{\infty }$; zero means equality; negative $C_p$ means $p<p_{\infty }$. This sign does not specify the sign of absolute pressure. For stipulated $p_{\infty }=101325\mathrm{Pa}$ absolute and $q_{\infty }=240\mathrm{Pa}$, $C_p=-3$ gives $p=100605\mathrm{Pa}$ absolute: a $720\mathrm{Pa}$ deficit, not a negative absolute pressure. If $p_{\infty }$ equals atmospheric pressure, the numerator is gauge pressure; a wind tunnel's reference static pressure need not equal room atmospheric pressure. These are reference choices, not different physical pressures.

Under the same-height, constant-mass density $\rho ={\rho }_{\infty }$, steady flow and inviscid flow hypotheses of the Bernoulli equation, and with the same Bernoulli invariant as the reference flow,The first equation immediately gives the second after subtracting $p_{\infty }$ and dividing by $q_{\infty }$. Hence a lossless stagnation point has $C_p=1$, $U=U_{\infty }$ gives $C_p=0$, and $U=2U_{\infty }$ gives $C_p=-3$. For a real stationary wall satisfying the no-slip boundary condition, the velocity is zero right at the wall; inserting that wall velocity in an inviscid flow formula would wrongly predict $C_p=1$ everywhere. A thin boundary layer approximation instead relates the wall pressure to the outer-flow pressure and uses the outer velocity where the Bernoulli equation applies.