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Conservation of volume flux through a circular arc of radius $r$ givesso the outer flow is the sink flow in a sectorThis is a potential flow and already satisfies the no-penetration boundary condition on the radial walls. It has nonzero tangential speed $Q/r$ there, however, and therefore violates the no-slip boundary condition. Thin vorticity-bearing boundary layers are needed to reduce that speed to zero at each wall.

The axial velocity $u(r)$ decreases toward the wall, so $\dot{\gamma }=-du/dr\ge 0$. For a power-law fluid, the stress magnitude isHenceIntegrating inward from the no-slip boundary condition $u(R)=0$ givesFor $n=1$ this reduces to the parabolic Hagen--Poiseuille profile.

Both candidate fields are divergence-free. For solid-body rotation,each component is linear in $\mathbf{x}$, so ${\nabla }^2\mathbf{u}=0$. Also,is harmonic for $r>0$, because ${\nabla }^2(1/r)=0$ there and constant-coefficient derivatives commute with the Laplacian. Thus both solve the Stokes equations with $p=0$.

The rotationally symmetric solution is consequentlyThe no-slip boundary condition givesThereforeandThis is the Rotational Stokes flow between concentric spheres.

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, $\partial p/\partial z<0$.

Use the Papkovich–Neuber representation in the formwhere $\mathbf{\Phi }$ and $\chi$ are harmonic functions. A torque is a axial vector, so rotational covariance and decay select the harmonic vector fieldHere $\mathbf{x}\cdot \mathbf{\Phi }=0$ and $\nabla \cdot \mathbf{\Phi }=0$, soThis rotlet equals the rotating sphere in Stokes flowwhen $\mathbf{\Omega }=\mathbf{G}/(8\pi \mu a^3)$. It satisfies the no-slip boundary condition on $r=a$, decays at infinity, and its Newtonian fluid stress tensor transmits the applied couple $\mathbf{G}$.

The Reynolds number isWhen it is large, viscosity is negligible in most of the flow, but the inviscid outer solution cannot generally satisfy the no-slip boundary condition at a rigid wall. A thin boundary layer resolves this mismatch: tangential gradients remain of outer scale $\mathcal{L}$, normal gradients have the much shorter scale $\delta$, and the small viscous coefficient is offset by $u_{yy}=O(\mathcal{U}/{\delta }^2)$. Matching a viscous inner solution to an inviscid outer solution retains the leading viscous effects without solving the full equations uniformly everywhere.

The Papkovich–Neuber representation writes a homogeneous incompressible Stokes flow in terms of a harmonic vector field $\mathbf{\Phi }$ and harmonic scalar $\chi$:with ${\nabla }^2\mathbf{\Phi }=0$ and ${\nabla }^2\chi =0$. The representation satisfies incompressibility because ${\nabla }^2(\mathbf{x}\cdot \mathbf{\Phi })=2\nabla \cdot \mathbf{\Phi }$, and substitution then verifies the Stokes equation.

For a sphere translating with constant vector velocity $\mathbf{U}$, rotational covariance and decay at infinity suggest a harmonic vector monopole and scalar dipole:Substitution gives the translating sphere in Stokes flowAt $r=a$ the radial tensor terms cancel and $\mathbf{u}=\mathbf{U}$, while $\mathbf{u}\to 0$ as $r\to \infty$, so the no-slip boundary condition and far-field condition hold. The resulting traction integrates to the Stokes drag law $6\pi \mu a\mathbf{U}$ in magnitude.

A rigid body obeys the no-slip boundary condition. Its outer potential flow generally has an $O(U)$ tangential slip velocity at the surface, so the boundary layer must make an $O(U)$ velocity correction across thickness $\delta$. Its strain is therefore $O(U/\delta )$, and in two dimensions its dissipation per unit axial length iswhich exceeds the $O(\mu U^2)$ outer dissipation by $a/\delta \sim {\mathrm{Re}}^{1/2}$.

The dominant drag consequently depends on the detailed boundary-layer velocity field and possibly on boundary-layer separation. It cannot be obtained by integrating the outer irrotational strain alone.

The leading streamwise momentum equation in lubrication theory iswith $p=p(x)$. The no-slip boundary condition isTwice integrating in $y$ gives the local Couette-Poiseuille flow in a thin gapIts volume flux per unit span is

Let $z$ measure height above the sponge. Under the long-wave approximation, hydrostatic pressure gives $p_x=\rho g{H}_x$, and the horizontal balance for viscous fluid flow isThe no-slip boundary condition $u(0)=0$ and stress-free boundary condition $u_z(H)=0$ giveIntegrating this lubrication theory profile gives the horizontal volume flux per unit span

Local mass conservation includes the downward loss found in part i:Consequently the required nonlinear partial differential equation isIf the imposed inlet flux is $Q_0{t}^{\alpha }$ and the moving front is $x=x_N(t)$, sufficient boundary conditions areFor a current released onto a dry substrate one also takes the initial condition $H(x,0)=0$ away from the source. The front position is part of this moving-boundary problem.

Let $d(z,t)$ be the local water-film thickness. To first order in the interface amplitudes,The lubrication theory flux down the vertical surface, with a no-slip boundary condition at the ice and a stress-free boundary condition at the water-air interface, isFor the unperturbed film $p_z=0$, so

The linearized Young–Laplace equation gives the capillary-pressure perturbationBecause the prescribed volume flux is uniform, its first-order perturbation must vanish. Linearization of the flux law givesThus, with $\Gamma =\gamma /(3\rho g)$,and henceThe complex amplitude ratio records the phase shift caused by surface tension in the long-wave approximation.

Let $z$ measure distance across the narrow gap and let $\theta$ be polar angle about the tube axis. In lubrication theory, radial velocity and radial pressure variation are negligible. Axisymmetric incompressible flow is obtained frombecause ${\partial }_{\theta }(u_{\theta }\sin \theta )=0$. The tangential Stokes equation then separates:and hence, after choosing an irrelevant pressure constant,

In the sphere frame, $u_z=-u_{\theta }\sin \theta =-\varphi$. The no-slip boundary condition gives $\varphi (0)=0$ on the sphere and $\varphi (\delta )=U$ on the membrane translating backward relative to it. The sphere-frame volume flux inherited from the narrow remote tube is $-\pi r_0^{2}U$, soUnder the asymptotic condition $r_0^2\ll R\delta$, the right-hand side is negligible at leading order. Solving the quadratic profile subject to the two wall values and zero leading-order integral givesChanging the chosen positive tube direction reverses both signs but leaves the drag magnitude unchanged.

The pressure scale is $p\sim \mu UR/{\delta }^2$, whereas the viscous shear scale is $\tau \sim \mu U/\delta$. After multiplication by comparable areas, pressure drag exceeds shear drag by $R/\delta \gg 1$, an instance of lubrication pressure dominates shear stress. Put $c=\cos \Delta \theta$. The axial pressure force isAs $\Delta \theta \to 0$, the bracket tends to $-2$, and thereforeThe resulting confined-sphere drag coefficient isThus it exceeds the free Stokes drag law coefficient by $2R^2/{\delta }^2$. The Stokes–Einstein relation then gives

The Papkovich–Neuber representation writes a homogeneous Stokes flow in terms of a harmonic vector field $\mathbf{\Phi }$ and a harmonic scalar $\chi$ asFor translation, rotational symmetry and decay at infinity restrict the trial harmonic fields to the fundamental harmonic $1/r$ and its directional derivatives contracted with $\mathbf{U}$. For rotation, the only decaying isotropic axial-vector field with the required boundary value is proportional to $\mathbf{\Omega }\times \mathbf{x}/r^3$. Matching the no-slip boundary condition $\mathbf{u}=\mathbf{U}+\mathbf{\Omega }\times \mathbf{x}$ at $r=a$ gives the superposition of the translating sphere in Stokes flow and the rotating sphere in Stokes flow:Each term decays at infinity, and direct substitution at $r=a$ gives the prescribed rigid velocity.

When $\mathbf{U}=0$, the pressure is constant and may be set to zero. Differentiating the rotational velocity and using the Newtonian fluid stress tensor gives

For a Newtonian fluid with velocity $u(z)\hat{\mathbf{x}}$, $\tau =\eta {\partial }_{z}u$ is the $x$ component of traction on a plane whose normal points along $+z$. The opposite face has the opposite traction. The coefficient $\eta$ is dynamic viscosity, with units $\mathrm{Pa}\mathrm{s}$. Faster layers transfer positive $x$ momentum toward slower neighbours; the internal shear stress resists relative sliding.

Take a thin fluid element with horizontal area $A$ and thickness $dz$. The difference between its upper and lower viscous tractions is $A\eta u_{zz}dz$. The pressure gradient supplies $-A{p}_{x}dz$. Incompressibility is satisfied because $u$ has no $x$ dependence, and the convective acceleration vanishes: $(\mathbf{u}\cdot \nabla )\mathbf{u}=u{\partial }_x\mathbf{u}=0$. Equating net force to mass times acceleration therefore gives the relevant component of the Navier-Stokes equation:For steady flow, integrate $\eta u^{\prime \prime }=p^{\prime }$ twice and impose the no-slip boundary condition at each plate, $u(0)=0$ and $u(a)=u_0$. The Couette-Poiseuille flow profile isThe linear term is Couette flow and the parabolic term is plane Poiseuille flow. A negative $p^{\prime }$ drives flow in $+x$. Integrating the profile across the gap gives the volumetric flow per unit width,It has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$; multiplying by the plate width gives the actual volume per time.

With zero pressure gradient, substituting $u=\mathrm{Re}[A{e}^{i(kz-\omega t)}]$ gives $-i\rho \omega =-\eta k^2$. Select the root that decays into fluid occupying $z>0$:The other sign applies to decay in the opposite half-space. With a real wall velocity amplitude $U_0$, the established periodic solution of Stokes' second problem isThe oscillation's amplitude decays exponentially and its phase lags with distance from the plate. This is diffusion of tangential momentum, not a freely propagating undamped elastic wave. Initial transients must have decayed before this purely periodic expression applies.

The fluid-displacement amplitude and the wall-displacement amplitude have the same ratio as their velocity amplitudes, since both are divided by the common $\omega$. Thus for the specified amplitudes,Using the standard approximate water density $\rho \simeq 10^3\mathrm{kg}{\mathrm{m}}^{-3}$, explicitly an approximation rather than another value printed on the paper, gives $\delta =0.564\mathrm{mm}$ and $z\simeq 2.21\mathrm{mm}$ from either face of a sufficiently large plate. The viscous penetration depth varies as ${\rho }^{-1/2}$ if a more precise water density is used. The ideal parallel-flow solution ignores finite plate edges and tank walls.

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.

Meaning and microscopic origin. For a Newtonian fluid, the dynamic viscosity $\eta$ measures tangential momentum transfer between fluid layers moving at different speeds. For the Newtonian shear considered here, ${\tau }_{xz}=\eta d{v}_z/dx$, with units $\mathrm{Pa}\mathrm{s}$ for $\eta$. Molecular motion and collisions transport momentum across layers in a gas; intermolecular interactions contribute strongly in liquids. The resulting viscous forces oppose relative shear motion and convert organized kinetic energy to heat. The constitutive law is Newtonian and isotropic; incompressibility alone does not establish it.

Vertical force balance. Choose $z$ downward, as in the original diagram, and $x$ outward normal to the plate. In a uniform steady film, $\mathbf{v}=v_z(x)\hat{\mathbf{z}}$ has zero advective acceleration and no along-plate pressure gradient. On a slice of thickness $dx$, the net shear force per unit face area is $\tau (x+dx)-\tau (x)$, while its weight per unit area is $\rho gdx$. ThereforeHere $\tau =\eta d{v}_z/dx$ denotes the downward traction on a face with positive $x$ normal. The opposite face has the opposite traction orientation, accounting for the sign in the force difference.

Boundary conditions. At the fixed solid wall, the no-slip boundary condition is $v_z(0)=0$, with no normal penetration. At the free surface $x=d$, neglect air shear, so $\tau (d)=0$ and hence $v_z^{\prime }(d)=0$. The flat free surface has normal pressure $p_{\mathrm{atm}}$, neglecting surface-curvature effects. These are a velocity condition at the wall and a stress condition at the free surface, not zero velocity at both boundaries.

The parabolic speed profile. Integrating ${\tau }^{\prime }=-\rho g$ from the shear-free surface gives $\tau (x)=\rho g(d-x)$. Thusand the wall condition yieldsIt is zero at the wall and largest at the free surface, where its gradient vanishes. The wall shear $\rho gd$ supports the downslope weight of the film per unit plate area.

Volume flux per unit width. Integrate through the depth:This is volume flow per unit length in $y$, with units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$.

Inclining the plate. Keep $x$ normal outward from the plate and use $s$ downslope. The gravitational components are $g\sin \theta$ downslope and $-g\cos \theta$ normal to the plate. For a uniform film of normal thickness $d$, the steady slice equations areThe free-surface pressure condition givesUniform $d$ makes ${\partial }_{s}p=0$. The same no-slip and zero-surface-shear conditions then give the uniform gravity-driven Newtonian film,The component normal to the plate changes the hydrostatic pressure, while the tangential component drives the flow. The thickness is measured normal to the tilted plate.

The inclined-film volume flux. Integrating the new profile givesIt recovers the vertical result at $\theta =\pi /2$ and vanishes on a horizontal plate at fixed uniform depth. A varying depth, imposed pressure gradient or significant air shear would require additional terms; the solution here uses the uniform film specified by the simple draining-plate model.

The Magnus effect is the transverse force produced when a spinning body moves relative to a fluid. Rotation changes the flow and circulation around the body; the resulting asymmetric pressure and wake deflect the fluid, with the opposite transverse force on the body. Examples include the curved flight of spinning balls and the lift of rotating-cylinder sails.

A useful quantitative model is steady two-dimensional incompressible flow past a cylinder of radius $a$, with fluid velocity $\mathbf{U}=U\hat{\mathbf{x}}$ far away. Let circulation $\Gamma$ be positive anticlockwise as viewed from $+z$. The potential-flow velocity at the cylinder surface isThe Bernoulli equation outside the boundary layer givesIntegrating the pressure force around the cylinder, per unit span, yieldsThus the Kutta–Joukowski theorem givesThe sign here refers to the fluid velocity relative to the cylinder. For a body moving through otherwise stationary fluid, that relative fluid velocity points opposite to the body's translational velocity.

Real viscosity and the no-slip boundary condition transmit the surface rotation to the fluid and help set the circulation and separation pattern. Pure inviscid flow with only an impermeability condition does not determine $\Gamma$ from the cylinder's angular speed. Therefore the circulation must not be assigned a universal spin-rate formula without additional assumptions. The cylinder result illustrates the lift mechanism; quantitative forces on a finite spinning ball also depend on its geometry, Reynolds number, rotation rate, surface roughness and wake.

Let $z$ increase down the pipe, $a=0.015\mathrm{m}$ be its radius and $r$ the distance from its axis. For steady, fully developed laminar flow of a Newtonian fluid, the axial Navier-Stokes equations reduce toWrite $K=-dp/dz+\rho g\sin 20^{\circ }$. Integrating twice, regularity at $r=0$ excludes a logarithmic term, and the no-slip boundary condition $u(a)=0$ gives the Hagen-Poiseuille flow profileThe volumetric flow rate follows by integrating over the circular cross-section:For the intended gravity-driven case with no imposed axial pressure gradient, $dp/dz=0$. Substitution givesandThe corresponding mean speed is about $0.1047\mathrm{m}{\mathrm{s}}^{-1}$ and the diameter-based Reynolds number is $\rho \overline{u}(2a)/\eta \simeq 3.49$, consistent with laminar flow. The original PDF does not specify inlet and outlet pressures. If they impose an additional pressure gradient, the general result with $K$ above applies; gravity alone does not determine that extra contribution.

Take $\mathbf{u}=(u(y),0)$ and no imposed streamwise pressure gradient. The normal momentum balance gives $p=p_{\ast }-\rho gy\cos \alpha$. The streamwise equation is $\mu u^{\prime \prime }+\rho g\sin \alpha =0$, since the convective acceleration vanishes. With no slip at the lower wall and signed upper-wall traction $\tau$ in the positive $x$ direction, the conditions areIntegration givesThe required upper-wall velocity is its endpoint value $u(h)=\rho g{h}^2\sin \alpha /(2\mu )+\tau h/\mu$. The unfactored formula remains meaningful when $\sin \alpha =0$.

The fluid exerts shear $\boxed{\mu u^{\prime }(0)=\rho gh\sin \alpha +\tau }$ per unit length on the lower plane in the positive $x$ direction. The force of that plane on the fluid has the opposite sign. The total wall force on the fluid balances its downslope weight, providing an independent sign check.

No slip on the two fixed radial walls gives $g(\pm \alpha )=0$. The outward flux through a circular arc is ${\int }_{-\alpha }^{\alpha }{u}_{r}rd\theta ={\int }_{-\alpha }^{\alpha }gd\theta$, so source strength $Q$ requires that integral to equal $Q$. Set $g=QG$ and $ϵ=Q/\nu$. ThenAt leading order $G_0^{(3)}+4G_0^{\prime }=0$, whose general solution is a constant plus sine and cosine of $2\theta$. The symmetric source-driven branch has no sine term; for generic $\alpha$ this also follows directly by subtracting its two wall conditions. ThusIndeed its integral is $C(\sin 2\alpha -2\alpha \cos 2\alpha )=1$. At isolated angles where an odd zero-flux homogeneous mode satisfies both wall conditions, symmetry selects the stated branch.