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Take the difference $w$ of two solutions with the same body force, initial displacement and initial velocity. It satisfies the homogeneous linear elasticity equations. With $e=e(w)$ the infinitesimal strain tensor, define the bulk energyThe usual stability assumptions make the elastic quadratic form nonnegative and $\rho >0$. Multiplying momentum balance by $w_t$, using integration by parts and using symmetry of the stress tensor givesAll interior terms cancel: the strain term is exactly the time derivative of elastic energy.

For the restoring boundary law, $t_w=kw$. The prescribed support displacement cancels in the difference. Since $k$ is a time-independent symmetric matrix,Hence the elastic energy uniqueness with restoring boundary springs identity isThe boundary contribution is nonnegative because $k$ is negative semidefinite. The initial difference and its velocity vanish, so the conserved energy is zero. Its kinetic term implies $w_t=0$ everywhere; the zero initial displacement then gives $w=0$. This proves uniqueness without needing to exclude rigid displacements separately.

Physically, $-k$ is a nonnegative boundary spring-stiffness matrix, and the prescribed vector $U$ is the moving support position. The traction restores the displacement towards that support; null directions of $k$ have no spring force. The interpretation and conserved-energy proof use a fixed spring matrix, as indicated by the given boundary law.

On the mixed part $S_1$, write $t=\sigma n$. The condition isThus the tangential traction and normal displacement are prescribed. The normal traction is a constraint reaction; tangential displacement is free. Only the tangential component of $F$ matters. On $S_2$, the entire displacement is prescribed.

For the difference of two such solutions, $w\cdot n=0$ and $(I-n{n}^{\mathsf{T}})t_w=0$ on $S_1$. Consequently $w_t$ is tangential and $t_w$ normal, so $w_t\cdot t_w=0$. On $S_2$, $w_t=0$. The boundary power vanishes everywhere, giving $d{E}_{\mathrm{bulk}}/dt=0$. The same zero-initial-energy argument gives $w=0$. Therefore the mixed boundary problem also has at most one solution.

The constitutive convolution is the response of a linear viscoelastic fluid close to its relaxed state. Deformations and rotations during the material's memory time must be small, so the stress is linear in the deformation history and evaluating that history at a fixed position is consistent with linearization about rest. For an isotropic incompressible flow, the deviatoric stress is proportional to the history of the rate-of-strain tensor; an arbitrary isotropic pressure is separate. Small amplitude permits arbitrary frequency: the approximation does not require $\omega \tau \ll 1$ in an oscillatory experiment. Large accumulated deformation, substantial advection of memory, and nonlinear normal-stress differences are outside this formula's scope.

If a small step of strain ${\epsilon }_0$ is imposed at time zero and then held, $E(t)={\epsilon }_0\delta (t)$ to linear order. The convolution consequently gives ${\sigma }^{\prime }(t)=2G(t){\epsilon }_0$ for $t>0$. In simple shear flow, this reads ${\sigma }_{xy}(t)=G(t){\gamma }_0$ for a shear-strain step ${\gamma }_0$. Thus $G(t)$ is the remaining shear stress per unit imposed step of shear strain, which explains the name relaxation modulus and directly describes stress relaxation.

Use the convention $E(t)=\mathrm{Re}[\hat{E}{e}^{i\omega t}]$. Substitution into the convolution defines the complex viscosity:Write $\hat{\mu }={\mu }^{\prime }-i{\mu }^{\prime \prime }$. Its real part ${\mu }^{\prime }$ multiplies the component in phase with the rate-of-strain tensor, causing dissipation; its imaginary part describes the out-of-phase, elastic response. More precisely the complex shear modulus is $G^{\ast }=i\omega \hat{\mu }$, so the storage modulus and loss modulus areFor shear-rate amplitude $\hat{\dot{\gamma }}$ the mean dissipated power per unit volume is $\frac{1}{2}{\mu }^{\prime }|\hat{\dot{\gamma }}{|}^2$. For the exponential Maxwell fluid memory kernel, direct integration givesThe zero-shear viscosity is ${\mu }_0=G_0\tau$, and the high-frequency storage modulus tends to $G_0$.

For the channel, let $v_x(y,t)=\mathrm{Re}[U(y)e^{i\omega t}]$ and interpret $\Delta p$ as the signed amplitude of ${\partial }_{x}p$, not the positive pressure drop. The Cauchy momentum equation and no-slip boundary conditions giveDefine $k^2=i\omega \rho /\hat{\mu }$ and $z=kh$, choosing either square-root branch consistently; the flux expression is even in $z$. The symmetric solution isIts integral across the channel gives the oscillatory channel flux of a linear Maxwell fluid:This includes inertia while using the linear constitutive response. The pressure-gradient amplitude must be small enough that the resulting deformation stays in that response regime.

As $\omega \to 0$, $\tanh z/z=1-z^2/3+2z^4/15+\cdots$. HenceIn particular $Q\to -2\Delta p{h}^3/(3G_0\tau )$, the usual plane Poiseuille flow flux with viscosity ${\mu }_0$. The fluid has enough time to relax and behaves as a Newtonian fluid at leading order.

For fixed positive $h,\tau ,\rho ,G_0$, let $c=\sqrt{G_0/\rho }$. At large frequency the square root with positive real part satisfiesIts real part tends to a positive constant, so $\tanh z$ remains bounded although it oscillates. Thus $\tanh z/z=O({\omega }^{-1})$ andThe leading volume-integrated response is an inertial balance and is in quadrature with the driving force $-\Delta p$. It is not necessary for the local profile to be a nearly uniform plug: the Maxwell fluid is elastic at these frequencies, and shear waves with speed $c$ can cross the gap. The oscillatory part of $U$ largely cancels on integration. Small attenuation allows resonance-like enhancements at finite frequencies; the leading flux limit still holds for fixed nonzero relaxation and gap parameters. A Newtonian thin viscous-layer argument would not justify the Maxwell profile here.

Finally take the different ordering $\omega \tau \gg 1$ but $\rho {\omega }^2{h}^2/G_0\ll 1$. Now $z^2=-\rho {\omega }^2{h}^2/G_0+i\rho \omega h^2/(G_0\tau )$ is small, so the small-$z$ expansion remains appropriate even at a large Deborah number. The result isHere inertia is negligible throughout the narrow gap. Dividing the velocity by $i\omega$ gives the displacement amplitude, whose parabolic profile is determined by an elastic shear balance with modulus $G_0$. The velocity therefore grows like frequency for a fixed displacement response. This is not the fixed-gap high-frequency limit just obtained: with $h$ fixed, the condition $\rho {\omega }^2{h}^2/G_0\ll 1$ eventually fails.

The usual elastic secondary flow around a rotating sphere is inward towards the equator, then poleward beside the sphere and outward near both poles. It is superposed on the primary azimuthal rotating sphere in Stokes flow. Tensile normal stresses along curved azimuthal streamlines act as hoop stresses, pulling liquid towards the rotation axis. Incompressibility supplies the axial return flow. Unlike the primary rotation, this meridional circulation does not reverse when the rotation is reversed.

The direction assumes the usual polymeric sign $c={\psi }_1+2{\psi }_2>0$ and negligible inertia. A general second-order fluid has circulation proportional to $c$: a negative value reverses the arrows and $c=0$ eliminates the leading elastic circulation. Inertial secondary circulation is a different contribution, so slow rotation alone does not determine which contribution dominates. The sketch shows the elastic contribution for $c>0$:

For a quantitative check of the sketch, take sphere radius $a$, angular velocity $\Omega$ and $K=\Omega a^3$. The primary azimuthal velocity is $v_{\varphi }=K\sin \theta /r^2$, with $\theta$ measured from the rotation axis. Its only nonzero strain components in the spherical orthonormal basis are $E_{r\varphi }=E_{\varphi r}=-3K\sin \theta /(2r^3)$. The leading elastic secondary circulation around a rotating sphere has the stream functionwith $v_r=H(3{\cos }^2\theta -1)/r^2$ and $v_{\theta }=-H^{\prime }\sin \theta \cos \theta /r$. The bracket is $(1-a/r{)}^2(1+2a/r)>0$ outside the sphere. Thus $v_r<0$ at the equator and $v_r>0$ near the poles when $c>0$, and both components vanish on the sphere. The diagram uses these velocities, not independently chosen arrows.

Two other manifestations of the same elastic normal-stress differences are rod climbing, in which inward hoop stresses raise a free surface around a rotating rod, and die swell, in which relaxation of the axial stress and stored polymer deformation allows an extrudate to expand sideways after leaving the die. These are not explained by a scalar shear viscosity alone. A Newtonian extrudate can also swell through velocity-profile rearrangement, so the elastic contribution should not be identified with every instance of swelling.

The second-order fluid is a slow, weak-flow expansion of the objective constitutive history of an isotropic simple fluid. Material frame indifference requires objective tensor rates. For an incompressible flow, the leading isotropic stress is arbitrary pressure, the linear deviatoric term is $2\mu E$, and at the next order the independent contributions can be represented by $E^2$ and an objective rate of $E$. Isotropic scalar contributions are absorbed into the pressure. Their coefficients can be expressed through the two low-rate normal-stress differences, givingHere $\stackrel{△}{E}$ is the lower-convected derivative, not a division by $E$. This local expansion is appropriate when deformation and temporal change over the relevant viscoelastic relaxation time are small: both the Weissenberg number and the appropriate Deborah number must be small. It captures leading normal stresses in slow flows, but not arbitrary finite-strain history, rapid transients or strongly nonlinear shear thinning. In particular its restrictions are stronger than small-amplitude linear oscillatory response at arbitrary frequency.

For steady simple shear flow $\mathbf{v}=(\dot{\gamma }y,0,0)$, use component-first $L_{ij}={\partial }_j{v}_i$, so the paper's $\nabla v$ is $L^T$. The relevant tensors areSubstitution givesThus the first and second normal-stress differences are $N_1={\psi }_1{\dot{\gamma }}^2$ and $N_2={\psi }_2{\dot{\gamma }}^2$, and the shear stress is $\mu \dot{\gamma }$.

For the velocity-preservation results, the definitions imply $\stackrel{△}{E}=\stackrel{\circ }{E}+2E^2$. Consequently the non-Newtonian stress can be writtenwhere $\stackrel{\circ }{E}$ is the Jaumann derivative. Let ${\mathbf{v}}_N$ solve the steady incompressible Stokes flow problem with the prescribed boundary velocities. The supplied identity says that $\nabla \cdot {\stackrel{\circ }{E}}_N$ is curl-free. If ${\psi }_1+2{\psi }_2=0$, then $\nabla \cdot S_N=\nabla \Phi$ for a scalar $\Phi$ on the ordinary simply connected flow domain. The momentum equation at ${\mathbf{v}}_N$ isHence the Newtonian velocity itself is a solution of the second-order constitutive model, with altered pressure $p=p_N+\Phi$. For nontrivial topology one also needs the curl-free force to admit a single-valued pressure; this is implicit in the usual boundary-value problem.

On the regular weakly non-Newtonian branch, Uniqueness of Stokes flow fixes the velocity correction. At first order it solves homogeneous Stokes flow, with zero boundary velocity, after the gradient force has been absorbed into pressure. Multiplying its equation by the correction and integrating gives $2\mu \int E_1:E_{1}dV=0$, hence the correction vanishes. Higher perturbative corrections vanish in turn because the previous velocities remain ${\mathbf{v}}_N$. The velocity is $\mathbf{v}={\mathbf{v}}_N$ in this second-order-fluid solution; only the pressure and stresses change. This is Newtonian velocity preservation in a second-order fluid. It does not assert global uniqueness of distant nonlinear solution branches or exact equality for a real fluid whose higher-order constitutive terms have been omitted.

For a planar incompressible velocity $(u(x,y),v(x,y),0)$, the rate-of-strain tensor has the formAll quantities are independent of $z$, so $\nabla \cdot E_N^2=\nabla (a^2+b^2)$. The extra quadratic term is therefore also a pressure gradient. Together with the supplied identity for the Jaumann derivative, the entire non-Newtonian force is a gradient for arbitrary ${\psi }_1,{\psi }_2$. The same boundary-value and weak-branch argument proves $\mathbf{v}={\mathbf{v}}_N$ for every such planar second-order-fluid flow with prescribed boundary velocities, without requiring ${\psi }_1+2{\psi }_2=0$. Specifying stresses or a free surface instead of velocities would not give the same conclusion automatically, because the extra stresses affect those boundary conditions.

Take $\omega >0$, with time convention $e^{-i\omega t}$, and use the Fourier transform pair $\hat{h}(k)=\int h(x)e^{-ikx}dx$, $h(x)=(2\pi {)}^{-1}\int \hat{h}(k)e^{ikx}dk$. For an acoustic velocity potential, $\mathbf{u}=\nabla \varphi$ and $p=-{\rho }_0{\varphi }_t$. Its harmonic amplitude solves the Helmholtz equation. The outgoing Fourier component has the formThe outgoing acoustic square-root branch is fixed by decay for evanescent components and upward radiation for propagating components:A precise causal prescription is to replace $\omega$ by $\omega +i\sigma$, $\sigma >0$, select $\mathrm{Re}\gamma >0$ for real $k$, invert on the real $k$ axis, and take $\sigma \downarrow 0$. It corresponds to switching on the forcing from the remote past. The limiting inversion contour $C$ passes below the positive outgoing surface-wave pole and above the negative one. The branch points are approached with the same causal prescription; analytic continuation during contour deformation must remain on this sheet. A principal real square root with positive imaginary part inside the acoustic interval would produce incoming sound and the wrong solution.

The elastic membrane's normal velocity equals the fluid velocity at the boundary, so $-\gamma A=-i\omega \hat{\eta }$. ConsequentlyThe elastic membrane balance therefore givesWriting $D=(T{k}^2-m{\omega }^2)\gamma -{\rho }_0{\omega }^2$, the one-sided fluid-loaded membrane radiation formulas areThe sign of the fluid term describes positive added mass of an evanescent fluid layer, not negative inertia.

For $0<\theta <\pi$, the acoustic saddle point is $k_s=k_0\cos \theta$, with ${\gamma }_s=-i{k}_0\sin \theta$. Applying the stated method of steepest descent result givesSince $p=i{\rho }_0\omega \varphi$ in amplitude notation, the angular acoustic directivity of the pressure, apart from angle-independent factors, isThe intensity directivity is proportional to ${\mathcal{D}}^2$. This cylindrical radiating field has amplitude of order $r^{-1/2}$. The formula is for fixed interior angles; grazing-angle limits require a separate uniform approximation.

For real $k>k_0$, $\gamma$ is positive, $D\to -{\rho }_0{\omega }^2$ as $k\downarrow k_0$, and $D\to +\infty$ as $k\to \infty$. A root exists. A root also requires $T{k}^2-m{\omega }^2>0$. In that range,so the root is unique. Evenness supplies the second root, $-k_{\ast }$. There are no real roots inside $|k|<k_0$: the fluid-loading term is real and nonzero while the other term is purely imaginary. Thus there are exactly two real physical-sheet roots, $\pm k_{\ast }$, for nonzero forcing frequency, positive elastic-sheet tension and positive fluid mass density. The zero-frequency static limit is not a two-distinct-root statement.

These roots are counterpropagating evanescent acoustic surface waves coupled to the membrane:Their normal profiles decay as $e^{-{\gamma }_{\ast }y}$, and their phase speed is below both $c_0$ and the vacuum membrane speed. They matter for the elastic membrane motion and for observations near its surface, but are not automatically part of the leading far-field radiation. A residue is picked up only when the causal contour deformation crosses its pole; at fixed $y/r>0$ its magnitude is exponentially small, $e^{-{\gamma }_{\ast }r\sin \theta }$.

For example, in the right half-plane put $k=k_0\cos \zeta$, $\gamma =-i{k}_0\sin \zeta$ and $k_{\ast }=k_0\cosh a$. The pole is at $\zeta =ia$ and the saddle at $\zeta =\theta$. The steepest-descent path has $\mathrm{Re}\cos (\zeta -\theta )=1$. At the pole height its left branch has real coordinate $\theta -\arccos (\mathrm{sech}a)$; comparing it with the original contour just to the right of the pole shows that its residue is crossed forThe analogous sector lies near $\theta =\pi$ for the negative pole. This is why the existence of real dispersion zeros alone does not make their residues relevant in every observation direction.

Let the perturbation velocity and pressure be $(u,w,p)(z)e^{ik(x-ct)}$, with real $k>0$. Linearizing the Euler equations about the inviscid parallel shear flow givesUse $u={\varphi }^{\prime }$, $w=-ik\varphi$ to satisfy incompressibility. The first momentum equation gives $p/{\rho }_0=-(U-c){\varphi }^{\prime }+U^{\prime }\varphi$; differentiate it and use the second equation. The Rayleigh equation for inviscid shear flow follows:A growing temporal normal mode has $c_i=\mathrm{Im}c>0$ and growth rate $k{c}_i$.

Rayleigh's inflection-point theorem states that such an instability of a smooth parallel profile requires $U^{\prime \prime }$ to change sign somewhere. To prove it, divide by $U-c$, multiply by ${\varphi }^{\ast }$, integrate over $z$ and integrate the second derivative by parts. Decay removes the endpoint terms, givingTaking the imaginary part yieldsThe weighting is nonnegative. A one-signed $U^{\prime \prime }$ that is not identically zero cannot satisfy this identity for a nontrivial mode; otherwise the mode vanishes on an interval and uniqueness of the differential equation forces it to vanish throughout. If $U^{\prime \prime }\equiv 0$, the real part of the identity is already a strictly positive integral for a nontrivial decaying mode. This proves the necessary sign-change condition.

Fjørtoft's theorem strengthens this: if $U_s$ is the velocity at an inflection point, instability requires $U^{\prime \prime }(U-U_s)<0$ somewhere. This is a necessary condition, not a sufficient criterion. For $U=\tanh z$, $U_s=0$ andThe possibility of instability is therefore consistent with Fjørtoft's theorem.

For the piecewise-linear approximation, $U^{\prime \prime }=0$ within each of the three regions. The decaying solutions outside and the two independent interior solutions can be writtenAt each corner, the normal velocity is continuous, so $\varphi$ is continuous. Pressure matching at a piecewise-linear shear interface additionally gives $[(U-c){\varphi }^{\prime }-U^{\prime }\varphi ]=0$, hence $(U-c)[{\varphi }^{\prime }]=[U^{\prime }]\varphi$. At $z=1$ the slope jumps from $1$ to $0$; at $z=-1$ it jumps from $0$ to $1$. Substituting the decaying outer solutions givesThus the inviscid instability of a piecewise-linear mixing layer has dispersion relationAt $k=1/2$, $c=\pm i/e$, so the root with positive imaginary part is an explicitly growing mode, with growth rate $1/(2e)$. More generally the unstable interval is $0<k<k_c$, where the unique $k_c>1/2$ solves $2k_c-1=e^{-2k_c}$. This direct calculation establishes instability of the approximation; passing a necessary inflection test alone would not have established it.

Neglecting gravity, the equation of motion is $m\ddot{\mathbf{x}}=a\dot{\mathbf{x}}\times \mathbf{x}/r^3$, where $r=|\mathbf{x}|>0$. Dotting with the velocity gives $d(m|\dot{\mathbf{x}}{|}^2/2)/dt=0$, because a cross product is perpendicular to its factors. Thus the speed is constant. The vector triple product givesHence the generalized angular momentum in a radial monopole field isDotting with $\mathbf{x}$ gives $\mathbf{L}\cdot \mathbf{x}=-ar$, or $\hat{\mathbf{L}}\cdot \hat{\mathbf{x}}=-a/|\mathbf{L}|$. For $\mathbf{L}\ne 0$, the direction of $\mathbf{x}$ therefore makes the fixed angle $\boxed{\cos \beta =-a/|\mathbf{L}|}$ with the axis $\mathbf{L}$, describing a right circular conical surface with vertex at the North Pole. Orthogonality gives $|\mathbf{L}{|}^2=m^2|\mathbf{x}\times \dot{\mathbf{x}}{|}^2+a^2$, so this cosine is in $[-1,1]$. Radial motion gives a degenerate cone; when $a=0$ and the ordinary angular momentum is nonzero the cone is the usual orbital plane. If $\mathbf{L}=0$, the norm identity forces $a=0$ and $\mathbf{x}\times \dot{\mathbf{x}}=0$, giving a radial straight line.

Apply the rotating-vector derivative rule first to the position vector and then to its velocity. Since $\mathbf{\omega }$ is constant,For example, the derivative of ${\mathbf{v}}^{\prime }$ contributes ${\mathbf{a}}^{\prime }+\mathbf{\omega }\times {\mathbf{v}}^{\prime }$, while that of $\mathbf{\omega }\times \mathbf{x}$ contributes the other two terms. This is the equation of motion in a rotating frame. The mutual central forces are unchanged in form because rotations preserve lengths and differences of position vectors. Newton's second law for particle $i$ becomesSet $\mathbf{\omega }=-e\mathbf{B}/(2m)$. The velocity-dependent Lorentz force is $e{\mathbf{v}}_i^{\prime }\times \mathbf{B}=-e\mathbf{B}\times {\mathbf{v}}_i^{\prime }$, and it cancels the Coriolis force $-2m\mathbf{\omega }\times {\mathbf{v}}_i^{\prime }=e\mathbf{B}\times {\mathbf{v}}_i^{\prime }$. The remaining terms combine to $m\mathbf{\omega }\times (\mathbf{\omega }\times {\mathbf{x}}_i)$ and are quadratic in the magnetic field. Neglecting them givesidentical to the zero-field equations to first order. The Larmor theorem requires the common charge-to-mass ratio; a mixture with different ratios cannot use one rotating frame to cancel every magnetic term.

For distinct positions and positive masses, Newton's law of universal gravitation givesThe proposed homothetic gravitational motion has ${\ddot{\mathbf{x}}}_i=-(2/9)t^{-4/3}{\mathbf{a}}_i$ and pair distances $t^{2/3}|{\mathbf{a}}_i-{\mathbf{a}}_j|$, so cancellation of $t^{-4/3}$ gives the time-independent equationsThese define a Newtonian central configuration with this choice of scale. Fordifferentiation in ${\mathbf{a}}_i$ gives ${\nabla }_{{\mathbf{a}}_i}\Phi =(2/9)m_i{\mathbf{a}}_i-G{\sum }_{j\ne i}{m}_i{m}_j({\mathbf{a}}_i-{\mathbf{a}}_j)/|{\mathbf{a}}_i-{\mathbf{a}}_j{|}^3=0$, exactly the displayed equations. The double sum in the paper counts each pair twice, so it agrees with this expression.

Multiply the configuration equations by $m_i$ and sum. Pair contributions cancel, giving ${\sum }_i{m}_i{\mathbf{a}}_i=0$. Thus the total linear momentum is $\mathbf{P}=(2/3)t^{-1/3}{\sum }_i{m}_i{\mathbf{a}}_i=0$. Each position is parallel to its velocity, so every $m_i{\mathbf{x}}_i\times {\dot{\mathbf{x}}}_i=0$ and the total angular momentum is zero. About any other fixed point $\mathbf{b}$, it is ${\mathbf{L}}_{\mathbf{b}}={\mathbf{L}}_O-\mathbf{b}\times \mathbf{P}=\boxed{0}$.

For an inviscid fluid of constant density, with irrotational velocity $\mathbf{u}=\nabla \Phi$ and conservative body-force potential $\Psi$, the Unsteady Bernoulli equation isIt holds throughout a connected irrotational fluid region, up to the time-dependent gauge of $\Phi$. Indeed the Euler equations and $\nabla \times \mathbf{u}=0$ imply that the gradient of the left side is zero. In the bubble problem there is no body-force contribution compatible with the assumed spherical symmetry.

Spherical incompressibility gives ${\partial }_r(r^2{u}_r)=0$. The moving interface imposes $u_r(R,t)=\dot{R}$, soThis potential tends to zero at infinity, where the velocity tends to zero and $C(t)=p_{\infty }/\rho$. Differentiate $\Phi$ at fixed spatial $r$, not along the moving interface: ${\Phi }_t=-(2R{\dot{R}}^2+R^2\ddot{R})/r$. At $r=R$, pressure continuity without surface tension gives $p=p_{\infty }-\Delta p$. Bernoulli therefore yields $-2{\dot{R}}^2-R\ddot{R}+{\dot{R}}^2/2-\Delta p/\rho =0$, or the Rayleigh equation for an inviscid spherical bubble

Integrating the kinetic energy over the exterior fluid givesDifferentiate and use the radius equation:With $R(0)=R_0$ and $\dot{R}(0)=0$, the pressure-work energy balance for a spherical bubble integrates toorThe collapsing branch takes the negative square root for $\dot{R}$. The energy gain is exactly the pressure difference times the lost bubble volume, providing a check on the sign and numerical factor.

The divergence is ${\partial }_x(kty)+{\partial }_{y}0+{\partial }_{z}0=0$, so this is an incompressible flow. With the planar convention $(u,v)=({\psi }_y,-{\psi }_x)$, one stream function isAt a fixed nonzero $kt$, the streamlines have $y=\mathrm{constant}$ and $z=\mathrm{constant}$, since the instantaneous velocity is purely axial. If $kt=0$, every point is instantaneously stationary and there is no distinguished streamline direction.

Follow a marked particle initially at $(0,y_0,0)$. Its trajectory satisfies $\dot{y}=\dot{z}=0$ and $\dot{x}=kt{y}_0$, soThese points form a straight segment on $x=(k{t}^2/2)y$, $z=0$, not an instantaneous horizontal streamline. Its length isThe directed angle of its vector from the origin to the marked endpoint isFor $k>0$ and $t>0$, this is $\arctan [2/(k{t}^2)]$; it equals $\pi /2$ at $t=0$. If an unoriented acute angle to the axis is intended, use $\arctan [2/(|k|t^2)]$, with $\pi /2$ when $kt=0$. This distinction handles a negative value of the given constant as well. If $a=0$, the marked set is a single point: the length formula gives zero and no line direction is defined.

Use the standard constant-density incompressible flow model, with conservative body force. Write $\mathbf{u}=(u,v,0)$ independent of $z$, and $\omega =v_x-u_y$. Differentiate the $y$-momentum Euler equations with respect to $x$ and subtract the $y$-derivative of the $x$-momentum equation. Mixed pressure derivatives and the curl of a conservative body force cancel, givingThe last term vanishes by incompressibility, so the material derivative of vorticity is zero. Thus $D\omega /Dt=0$. Incompressibility is important: inviscidness alone does not suffice; in a compressible barotropic planar flow the equation is $D\omega /Dt=-\omega \nabla \cdot \mathbf{u}$, conserving $\omega /\rho$ instead.

Take vortex strength $\kappa$ to denote circulation, positive counterclockwise. A point vortex at ${\mathbf{x}}_0$ generatesIt has zero vorticity away from its core, with circulation $\kappa$ around the core. Each vortex moves with the other's regular induced velocity, not its own singular field. With $\mathbf{d}={\mathbf{x}}_1-{\mathbf{x}}_2$,Multiplication by the two strengths proves $\dot{\mathbf{q}}={\kappa }_1{\dot{\mathbf{x}}}_1+{\kappa }_2{\dot{\mathbf{x}}}_2=0$. Also $\dot{\mathbf{d}}=({\kappa }_1+{\kappa }_2)J\mathbf{d}/(2\pi |\mathbf{d}{|}^2)$, so $d|\mathbf{d}{|}^2/dt=2\mathbf{d}\cdot \dot{\mathbf{d}}=0$.

For equal strengths $\kappa$ and the specified initial positions, $\mathbf{q}=0$ and $|\mathbf{d}|=2a$. The motion of two point vortices is rotation with angular velocity ${\Omega }_v=\kappa /(4\pi a^2)$:Negative circulation reverses the rotation. If vortex strength is defined instead as the coefficient of $1/r$ velocity, absorb the factor $2\pi$ consistently into its definition.

Choose the poloidal magnetic field convention ${\mathbf{B}}_p=\nabla \times (A\hat{\mathbf{y}})=(-A_z,0,A_x)$. The term $\alpha B$ regenerates this poloidal flux function from the toroidal magnetic field through the alpha effect. It represents a component of the mean-field electromotive force produced by correlated velocity and magnetic fluctuations. The shear $G=\pi V/d$ stretches the poloidal vertical component $B_{p,z}=A_x$ into toroidal field, giving $G{A}_x$: this is the Omega effect. Both $\eta {\nabla }^2$ terms diffuse and damp the field. In a turbulent mean-field model, $\eta$ can denote an effective diffusivity including turbulent transport. The alpha-Omega dynamo approximation retains the shear-dominated toroidal source, and the kinematic dynamo assumption prescribes the velocity and coefficients without their Lorentz force feedback.

For a horizontal wavenumber $k$ and vertical wavenumber $\pi /d$, set $K^2=k^2+{\pi }^2/d^2$ and seek amplitudes proportional to $e^{st}$. The two modal equations areA nonzero mode requires the dispersion relationFor real nonzero $\alpha Gk$, the root with positive real square-root part isThe other root has strictly negative real part. For $k=0$ there is no oscillatory feedback and both roots have real part $-\eta {\pi }^2/d^2$; the possible triangular-system prefactor cannot yield exponential growth.

Let $h=|k|d/\pi >0$ and $D=\alpha V{d}^2/({\pi }^2{\eta }^2)$. The growing-branch exponent satisfiesConsequently the plane-layer alpha-Omega dynamo threshold at fixed $h$ isTo find its least value, differentiate $D_c(h)=2(h^{-1}+2h+h^3)$:The unique positive solution is $h^2=1/3$. The threshold diverges at both ends $h\downarrow 0$ and $h\to \infty$, so this is its global minimum. HenceWith the usual positive-$D$ convention, the onset value is $D_c=32/(3\sqrt{3})$ and exponentially growing waves exist for $D>D_c$. Exactly at the threshold the real part is zero, so it is a neutral dynamo wave: the quoted minimum is the onset threshold, or infimum for strict growth, not a growing solution at equality. If signed $D$ is allowed, the growth condition concerns $|D|$; there is no smallest signed negative value among all growing cases.