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Assume all rings have the same circulation $\Gamma$ and orientation, and that there is no imposed background flow. Vortex lines describe the direction of vorticity, not the direction of velocity. For an incompressible fluid, the velocity induced by a thin vortex filament has the same Biot-Savart integral form as the magnetic field of a current filament:A closely spaced coaxial array of vortex rings is therefore analogous to a solenoid. The velocity field passes along the common axis through the rings and returns outside a finite array. Its orientation reverses if every circulation is reversed. In a long array, far from its ends and vortex cores, the velocity field of a dense coaxial vortex-ring array is approximatelyThe exterior field is exactly zero only in the ideal continuous infinitely long limit, with the background velocity set to zero. In a finite array the streamlines spread out at one end, run back outside, and converge into the other. A large number of closely spaced rings alone does not guarantee that the total array length greatly exceeds $r$; if it does not, the end and return flows remain substantial. Different unspecified ring circulations would require superposing their actual signed contributions.

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.

A streamline is a curve whose tangent is parallel to the instantaneous fluid velocity field. In steady flow one may parametrize it by $d\mathbf{r}/ds=\mathbf{u}(\mathbf{r})$. Dotting the steady Euler equations with $\mathbf{u}$ eliminates the cross-product term, since $\mathbf{u}\cdot ((\nabla \times \mathbf{u})\times \mathbf{u})=0$. Hencewhich is Bernoulli's principle along each streamline.

For the supplied field, $xu+yv=0$, so along a streamlineWriting $r^2=x^2+y^2$, the tangential speed is $v_{\theta }=-br+c/r$ and the radial speed is zero. Both $r$ and $|\mathbf{u}{|}^2=(-br+c/r{)}^2$ are constant around each nonstagnant circular streamline. The Bernoulli quantity is constant there, so the pressure is constant on every such circle.

If a circle is wholly stagnant, the same conclusion follows directly from the steady Euler equations: on it $\mathbf{u}=0$ and $\nabla |\mathbf{u}{|}^2=0$, hence $\nabla p=0$. Equivalently the radial momentum equation gives $\nabla p=\rho (-br+c/r{)}^2{\mathbf{e}}_r/r$ and no angular pressure variation, including at zero tangential speed. Thus $p$ is constant on $x^2+y^2=a^2$ for every nonzero $a$, with no exception for a stagnation circle.

For an isotropic Newtonian fluid in incompressible flow, the constitutive equation isHere $p$ is pressure, ${\delta }_{ij}$ is the Kronecker delta, $\mu$ is the constant dynamic viscosity, and $E$ is the strain-rate tensor. The more general compressible constitutive equation can include a term proportional to $(\nabla \cdot \mathbf{u}){\delta }_{ij}$; that term vanishes here.

Differentiate the Cauchy stress tensor, retaining the constant dynamic viscosity outside the derivatives:The last equality uses incompressibility. Substitution into the Cauchy momentum equation gives the Navier-Stokes equationThe pressure gradient and the Laplacian of the velocity field therefore give the two contact force densities in this model.

For the unidirectional flow $\mathbf{u}=(0,0,w)$, incompressibility gives ${\partial }_{z}w=0$. Hence $w=w(x,y,t)$, and the convective acceleration $w{\partial }_{z}w$ vanishes. The transverse components of the Navier-Stokes equation give $p_x=p_y=0$, while the axial component givesA steady-flow assumption is implicit in the question. Neither unidirectional flow nor a time-independent pressure gradient alone eliminates ${\partial }_{t}w$: a decaying transient can be superposed on the final pipe profile. The native PDF does not explicitly say steady flow in this part. For the intended steady flow, $w_t=0$, and thereforeTaking the pipe walls to be stationary, the no-slip boundary condition isThe no-penetration boundary condition is already satisfied by the assumed velocity field. This is the Poisson equation defining rectangular duct flow.

For the proposed velocity field, $u_x+w_z=1-1=0$, so incompressibility holds. Its convective acceleration isand both components have zero Laplacian. The Navier-Stokes equation therefore requires $p_x=-x$ and $p_z=-z$. Hencewhere $p_{\ast }$ is an arbitrary constant reference pressure. The same pressure gradient balances the same convective acceleration in the Euler equations for an inviscid fluid. Since the viscous terms vanish identically, this is an exact solution for every positive Reynolds number, as well as for the inviscid limit.

At $z=0$, the normal velocity is zero, and the prescribed dimensionless tangential traction isThis stagnation point flow therefore has zero surface shear stress, although its strain-rate tensor and its viscous normal stresses are nonzero.

Write $H(x^{\prime })=h^{\prime }(x^{\prime })$ and $p_x^{\prime }=d{p}^{\prime }/d{x}^{\prime }$ below, so the subscript $x$ in this shorthand denotes differentiation with respect to $x^{\prime }$. The vertical pressure gradient is zero, and integrating $u_{z^{\prime }{z}^{\prime }}^{\prime }=p_x^{\prime }$ twice with the wall boundary conditions gives the local Couette-Poiseuille flowThe first two terms are Couette flow due to the moving plate; the last is plane Poiseuille flow due to the local pressure gradient. Its dimensionless volume flux per unit width isUse the Leibniz integral rule and incompressibility to differentiate this integral:Thus the flux is independent of position, andThe normal velocity can then be recovered from incompressibility:It is zero on the lower wall by construction and zero on the upper wall by the constant-volume flux relation. This supplies the remaining component of the leading velocity field.

The endpoint pressure conditions are matching conditions, not an assertion that the lubrication approximation remains valid at arbitrarily large dimensional $x$. On an outer length scale $R$, a viscous pressure variation is of size $\mu U/R$, whose ratio to the inner scale isEven an outer inertial pressure scale $\rho U^2$ has ratioConsequently the gap pressure is much larger than either outer variation. Matching to fluid with a common ambient pressure at both ends, and taking that value as zero gauge pressure, givesThere is an overlap region $1\ll |x^{\prime }|\ll {\delta }^{-1}$ before the wall slope becomes order one. The infinite endpoints represent matching through that region. An externally imposed pressure difference comparable with $P$ would change these conditions and the resulting volume flux; equal outer pressure is the physical assumption being used here.

A pathline records the motion of one identified fluid element: keep its starting position $\mathbf{a}$ and starting time $t_0$ fixed, and let physical time $t$ advance. In a velocity field $\mathbf{u}(\mathbf{x},t)$ its Lagrangian trajectory $\mathbf{X}(t;t_0,\mathbf{a})$ solves the initial value problemA streamline instead freezes the time argument of the velocity field; a streakline fixes the observation time and varies the release time of many fluid elements. A pathline is useful for answering where a particular parcel goes and how long it takes, rather than merely which way the fluid is moving now. ^[ref]

For a concrete analytical model, take spatial coordinates $x,y$ in metres, time $t\ge 0$ in seconds, and the spatially uniform but time-dependent velocity fieldIts numerical components in these units are $(1,t)$. A fluid element starting at the origin at $t_0=0$ has the pathlineAt $t=1\mathrm{s}$ it is at $(1,0.5)\mathrm{m}$; at $t=2\mathrm{s}$ it is at $(2,2)\mathrm{m}$. The parabola records earlier, shallower velocity directions as well as the final direction. Differentiating gives $(\dot{x},\dot{y})=(U,at)$ and acceleration $(0,a)$, so this curved pathline needs no spatial variation of the velocity field. More generally the chain rule along a pathline gives $df(\mathbf{X}(t),t)/dt={\partial }_{t}f+\mathbf{u}\cdot \nabla f$, the material derivative of a transported scalar field.

A motivating tracer demonstration appears in S. J. Kline's film notes: water passes through a contracting channel, and hydrogen bubbles mark a small element at a wire at time $t_0$. The illustrated element is shown again $4.7\mathrm{s}$ later, with its accumulated pathline superposed. Following that same marked element across successive film frames records a history; the entire bubble pattern in one frame has a different meaning. ^[ref]

For a modern version of the readout, use sparse small markers in a transparent water channel, illuminate the viewed plane, and keep an optical camera fixed with a ruler in that plane. Record timestamps and track the same isolated marker between frames; successive positions estimate its pathline and $[\mathbf{X}(t+\Delta t)-\mathbf{X}(t)]/\Delta t$ estimates its velocity. A long exposure records a pathline segment but loses the timing within it, so its total length divided by exposure time gives only mean speed. These are proposed measurement steps based on Jean Hertzberg's teaching guidance, rather than additional measurements from Kline's film. Markers leaving the viewed plane, overlapping tracks and perspective can break particle identification; repeat with a thinner viewed slice and a lower marker concentration. ^[ref]

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.

Potential-vorticity inversion recovers a balanced velocity field from potential vorticity and specified boundary and global data. In two-dimensional vortex dynamics this is a Poisson equation; shallow-water quasi-geostrophic inversion and stratified quasi-geostrophic inversion use different elliptic boundary value problems. Advance the materially conserved scalar, then invert again to recover the evolving flow. Boundary buoyancy can induce flow with zero interior anomaly. The balance relation and boundary conditions are essential: unrestricted gravity waves are additional degrees of freedom. McIntyre's potential-vorticity account explains the general invertibility principle.

For a twice continuously differentiable interaction potential with $\Vert D^{2}W\Vert \le L$ and a probability measure with finite first absolute moment, this subtraction produces a finite potential even if $W\ast \mu$ is infinite. Taylor's formula and linear growth of the gradient giveDominated differentiation under the integral sign gives $\nabla V_{\mathrm{ren}}=(\nabla W)\ast \mu$ and $D^2{V}_{\mathrm{ren}}=(D^{2}W)\ast \mu$. If the unrenormalized convolution is finite, the two potentials differ only by a constant, so they generate the same velocity field when that field depends on their gradient.

For an incompressible flow of constant mass density $\rho$ and kinematic viscosity $\nu$, Reynolds averaging of the Navier-Stokes equation, without a body force, givesHere $\mathbf{u}=\mathbf{U}+{\mathbf{u}}^{\prime }$ and $\overline{{\mathbf{u}}^{\prime }}=0$, with averaging assumed to commute with the displayed derivatives. The mean acceleration ${\partial }_t{U}_i$ vanishes only when the averaged velocity field is stationary; taking an average does not by itself remove it.

To derive the extra term, average the conservative momentum flux: $\overline{u_i{u}_j}=U_i{U}_j+\overline{u_i^{\prime }{u}_j^{\prime }}$. Mean incompressibility turns the divergence of the first product into $U_j{\partial }_j{U}_i$. The second product is the Reynolds stress divided by mass density, using the positive-covariance convention. It describes transport of mean momentum by fluctuating velocities. A turbulence closure is needed because that covariance is an additional unknown, rather than a specified molecular transport coefficient.

For a rigid frame with constant angular velocity $\mathbf{\Omega }$, add the Coriolis force and centrifugal force to the inviscid magnetohydrodynamic momentum equation. A constant magnetic permeability and solenoidal magnetic-field constraint split the Lorentz force density into magnetic pressure and magnetic tension. This nonrelativistic conducting-fluid approximation neglects the displacement current, net electric-charge force and viscosity. Differential rotation belongs to the fluid's velocity field, not to a spatially varying rigid-frame rotation vector. A time-dependent frame requires the additional Euler force.

Scalar transport follows a material quantity through advection and diffusion equation terms, for example ${\Theta }_t+\mathbf{U}\cdot \nabla \Theta =\kappa \Delta \Theta$. A passive scalar does not affect the velocity field; an active scalar feeds back through forces such as buoyancy. With incompressible flow bounded by impermeable walls and homogeneous Neumann boundary conditions, the spatial mean is conserved.

The molecular Schmidt number is the ratio of kinematic viscosity to scalar diffusivity. A large value allows a passive scalar to develop structure below the viscous smoothing scale of the velocity field. It is distinct from the turbulent Schmidt number, which compares effective turbulent transport coefficients, and from the Schmidt number used for quantum entanglement.

The screw pitch of a rigid-body twist is its signed axial translation per unit rotation angle about the instantaneous motion axis. For angular velocity $\mathbf{\omega }\ne 0$ and linear twist component $\mathbf{v}$ in the same reference frame,The units are length per radian. This mechanics convention differs from using the word pitch for an axial distance per complete revolution.^[ref]

To derive the formula, write the instantaneous velocity field as $\mathbf{u}(\mathbf{x})=\mathbf{v}+\mathbf{\omega }\times \mathbf{x}$. At a point $\mathbf{q}$ on the instantaneous axis, the velocity is parallel to $\mathbf{\omega }$, so $\mathbf{u}(\mathbf{q})=h\mathbf{\omega }$. Taking its scalar product with $\mathbf{\omega }$ removes the cross-product term and gives the displayed result. A suitable closest axis point iswhich verifies $\mathbf{u}(\mathbf{q})=h\mathbf{\omega }$ directly.

Although $\mathbf{v}$ depends on the chosen origin, $h$ does not. Translating that origin by $\mathbf{a}$ replaces $\mathbf{v}$ by $\mathbf{v}+\mathbf{\omega }\times \mathbf{a}$; the added term is perpendicular to $\mathbf{\omega }$. A common rotation of the coordinate axes preserves the scalar product and the norm as well. Multiplying the entire rigid-body twist by a nonzero scalar changes the motion rate but preserves $h$.

For an original mechanism with constant axial advance $2\mathrm{mm}$ per full turn,At five revolutions per second, $\dot{\theta }=10\pi rad/s$ and the axial speed is $h\dot{\theta }=10mm/s$. The advance per revolution is $2\pi h$, which corresponds to screw lead. In a multiple-start screw thread, the spacing between neighbouring threads need not equal that lead, so a thread's engineering pitch must not silently replace the twist parameter.

When $h=0$ and $\mathbf{\omega }\ne 0$, the instantaneous motion is pure rotation about an axis, even if points away from that axis have nonzero linear velocity. When $\mathbf{\omega }=0$ and $\mathbf{v}\ne 0$, the motion is pure rigid-body translation and the displayed quotient is undefined; an infinite-pitch convention can describe that limiting case. A finite nonzero constant pitch produces a screw motion, with off-axis point paths following helices.

A separated vortex ring is a ring-shaped recirculating flow structure that remains detached from the body producing it while occupying a relatively stable downstream position in the body's frame. It differs from a ring attached to the body and from a ring that is shed and then travels away. The stable pattern does not require the same parcel of fluid to remain permanently in one place.

The term was used for the flow behind a dandelion seed's pappus. Flow visualization showed a detached recirculation ring downstream of the porous bristles. Air passing through the structure, rather than simply being excluded by a solid disk, was important to the observed pattern.^[ref]

To test the role of structure, the investigators compared microfabricated disks with different porosities. A disk with a pappus-like porosity reproduced the characteristic flow behavior, supporting a mechanism involving permeability and geometry rather than a special biological material alone.^[ref]

For an invented geometric example, a screen with total projected area $10\mathrm{m}{\mathrm{m}}^2$ and solid area $1\mathrm{m}{\mathrm{m}}^2$ has open-area fraction $0.9$. That fraction alone does not fully specify its aerodynamic behavior: the distribution and shape of openings, fluid properties and flow speed also matter. This is not the measured porosity of the paper's seeds.

The example shows why an apparently empty part of a structure can control its drag. A vortex ring is a property of the surrounding velocity field, so visualizing that field can reveal mechanisms invisible in a static outline of the object.

A shear flow has a velocity field varying across adjacent fluid layers, giving a nonzero shear rate. A parallel example is $\mathbf{u}=U(z){\mathbf{e}}_x$, with vertical shear $U^{\prime }$. Linear shear flow has constant shear, while a shear layer concentrates the velocity change in a narrow region.

A stagnation point of a fluid flow is a point at which the velocity field vanishes. For a two-dimensional streamfunction, it is a critical point satisfying both first spatial derivatives of the streamfunction equal to zero.