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The initially irrotational inviscid flow remains irrotational, so $u=\nabla \varphi$; incompressibility gives ${\nabla }^2\varphi =0$. At the plates, normal velocity matches their motion:With $\varphi =r^{2}f(\theta )$, Laplace's equation gives $f^{\prime \prime }+4f=0$. The symmetric solution isHenceA streamfunction isso streamlines are $r^2\sin 2\theta =\text{constant}$ and fluid is expelled radially as the plates close. The outward flux through $r=R$ is

In the fluid region $y>\eta (x)$, incompressibility and potential flow giveNo penetration through the exact surface gives

The condition $hk\ll 1$ says the hill slope is small; $|\nabla \varphi |\ll U$ says the disturbance velocity is small compared with the background wind. Dropping the product ${\varphi }_x{\eta }_x$ and Taylor-shifting the boundary from $y=\eta$ to $y=0$ therefore givesThe decaying solution is

Bernoulli's equation isTo first order on the surface,ThusEquivalently, crest minus trough is the negative of this. For $k{U}^2/g\ll 1$, hydrostatic elevation dominates; for $k{U}^2/g\gg 1$, the Bernoulli pressure drop caused by faster crest flow dominates.

The steady Stokes equations are linear and contain no inertial term. If every imposed boundary velocity and body force is reversed, then the velocity and pressure departure from any uniform reference pressure reverse sign. Thus every fluid particle retraces its path in the reversed experiment. This is the principle of kinematic reversibility.

Reflect the configuration in the horizontal plane through the cylinder axis. The wall and cylinder geometry are unchanged. Downward translation becomes upward translation, a possible wall-normal migration velocity is unchanged, and the axial angular velocity changes sign because it is an axial vector.

On the other hand, reversing the original Stokes flow makes all three velocities change sign: translation becomes upward, wall-normal migration reverses, and rotation reverses. The reflected and reversed problems therefore have the same forcing and geometry except that their wall-normal velocities have opposite signs. Uniqueness of Stokes flow forces that velocity to equal its own negative, so it is zero. The rotational velocities already agree under the two operations, so this argument places no restriction on rotation.

For real physical fields of period $T=2\pi /(kc)$, defineEquivalently, for complex amplitudes,HereTheir product is purely imaginary, soPressure and normal velocity are in temporal quadrature: energy is stored and returned by the evanescent field, but none is transported away from the membrane on average.

With $P^{\mu }=m{U}^{\mu }$, $U^{\mu }=d{x}^{\mu }/d\tau$, and $F^{\mu \nu }$ the field tensor, separating temporal and spatial components yields the Lorentz force and power equations. The invariants are $E\cdot B$ and $E^2-c^2{B}^2$. For perpendicular fields with $E<cB$, boost with $v=E\times B/B^2$ (here $-E\hat{y}/B$); then $E\prime =0$ and $B\prime =\sqrt{B^2-E^2/c^2}\hat{z}$. Perpendicular motion is circular with $\omega =|q|B\prime /(\gamma m)$ and radius $p/(|q|B\prime )$. Transforming back adds the frame velocity, giving the nonrelativistic $E\times B/B^2$ drift, independent of $m,q$; a parallel velocity would additionally produce a helix.

$Y(s)={\int }_0^{\infty }{e}^{-st}y(t)dt$, with transforms $sY-y(0)$ and $s^{2}Y-sy(0)-\dot{y}(0)$; $e^{\lambda t}$ transforms to $(s-\lambda {)}^{-1}$ and $\delta$ to $1$. Algebra gives $Y=GH$ with $H=F+(s+2\gamma )y(0)+\dot{y}(0)$, so $y=g\ast h$. Writing $a=\sqrt{|{\gamma }^2-{\omega }^2|}$ gives $g=e^{-\gamma t}\sinh (at)/a$ for $\gamma >\omega$, $e^{-\gamma t}\sin (at)/a$ for $\gamma <\omega$, and $t{e}^{-\gamma t}$ at equality. If $y(0)=0$, $h=f+\dot{y}(0)\delta$, so the initial velocity contributes $\dot{y}(0)g(t)$.

Boundary-layer theory assumes steady incompressible high-Reynolds-number flow, $\delta /x\ll 1$, weak streamwise viscous diffusion, nearly constant pressure across the layer, and outer pressure gradient $-{\rho }^{-1}{p}_x=U{U}^{\prime }$.

For $\varphi =-A{r}^k\cos (k\theta )/k$,At $\theta =\pm \pi /k$ the normal velocity vanishes; along the upper wall $U(x)=A{x}^{k-1}$. Boundary-layer scaling givesSet $m=k-1$. The Falkner-Skan equation becomesso $\alpha =k-1$, $\beta =-k/2$, $\gamma =1-k$, with
$f(0)=f^{\prime }(0)=0$ and $f^{\prime }(\infty )=1$. Finallyrequires $x\gg (\nu /A{)}^{1/k}$; thus $a=1/k$ and the tip region violates the approximation.

Let $H=k{\beta }_{0}h$, ${\beta }_0=\sqrt{1-V^2/c_0^2}$, and
${\beta }_1=\sqrt{V^2/c_1^2-1}$. Match pressure and normal velocity between an evanescent lower potential and an outgoing upper wave. The transmitted amplitude hasThereforeFor $H\gg 1$ this isshowing tunnelling suppression. For $H\ll 1$ it tends to
${\rho }_1{k}^2{V}^3{ϵ}^2/(2{\beta }_1)$, as if the upper medium met the wall.

The axial equation is $\mu r^{-1}(r{u}^{\prime }{)}^{\prime }=-G$. Regularity at zero, no slip $u_a(a)=0$, continuity of velocity and shear at $c$ giveThe annulus is controlled only by its own viscosity and the outer no-slip condition, explaining the absence of ${\mu }_c$. For the annular control volume, kinetic-energy rates vanish; pressure work plus interfacial traction work equals ${\int }_{V}2{\mu }_a{e}_{ij}{e}_{ij}dV$. Substituting $u_a^{\prime }=-Gr/(2{\mu }_a)$ evaluates both sides to the same boundary expression, verifying the mechanical-energy equation.

Apply the normal-shock relations to $M_{1n}=M_1\sin {\theta }_1$. Tangential velocity continuity and mass conservation giveAs $M_1\to \infty$, maximize ${\theta }_1-{\theta }_2$; with $k=(\gamma -1)/(\gamma +1)$ the optimum has ${\tan }^2{\theta }_1=1/k$, and the maximum deflection is $\arcsin (1/\gamma )$.

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]