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A smooth wire has instantaneous shape $z=f(r)$ in cylindrical coordinates and rotates at constant angular velocity $\dot{\theta }=\omega$. The tangent vector at fixed time is ${\mathbf{e}}_r+f^{\prime }(r){\mathbf{e}}_z$. Its dot product with the constraint force vanishes. Projecting Newton's second law along this tangent vector gives the equation in the title.

Multiplying that equation by $\dot{r}$ proves conservation ofThis is a conserved quantity in the rotating description; the motor can exchange ordinary mechanical energy with the bead. A relative equilibrium at $a>0$ satisfies $U^{\prime }(a)=g{f}^{\prime }(a)-{\omega }^{2}a=0$. Its linearized equation is $(1+f^{\prime }(a{)}^2)\ddot{ϵ}=-U^{\prime \prime }(a)ϵ$. A strict local minimum with $U^{\prime \prime }(a)>0$ gives Lyapunov stability, while $U^{\prime \prime }(a)<0$ gives an unstable equilibrium. If $U^{\prime \prime }(a)=0$, higher orders are needed.

Average the equations in part b and make exactly the prescribed scale-separation approximation: the retained perturbations vary on lengths much greater than $\ell$, and perturbations at the tangle scale are omitted. Products of two retained perturbations can then be neglected. Writing $\mathbf{v}=\delta \mathbf{u}$, $m=\delta M$ and $\pi =\delta p$, the linearized equations areNo unmodelled small-scale response has been inserted into this closure.

Introduce the fluid displacement $\mathbf{\xi }$ by $\mathbf{v}={\partial }_t\mathbf{\xi }$ and choose $\nabla \cdot \mathbf{\xi }=0$. Integrating the magnetic second-moment tensor equation giveswhere $s$ is a time-independent integration tensor set by the initial stress. For a pure oscillatory normal mode, $s=0$. Its restoring force is therefore $v_A^2{\nabla }^2\mathbf{\xi }$: the isotropic magnetic tangle has the same transverse response as an incompressible elastic medium with shear modulus ${\rho }_0{v}_A^2$.

For a plane wave proportional to $e^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, with $\omega \ne 0$, incompressibility gives $\mathbf{k}\cdot \hat{\mathbf{v}}=0$. The tensor equation and magnetohydrodynamic momentum equation becomeDotting the second equation with $\mathbf{k}$ yields $\hat{\pi }=0$ for $k\ne 0$, since $k_i{k}_j{\hat{m}}_{ij}=0$. Substitution then givesThere are two independent transverse polarizations. The phase speed is $v_A$ in every direction, and the two signs of $\omega$ give group velocities $\pm v_A\hat{\mathbf{k}}$. These magnetoelastic waves in an isotropic magnetic tangle exist because a shear deformation distorts the second moment of the tangled magnetic field, producing a restoring magnetic tension even though its mean is zero.

For comparison, a uniform magnetic field with Alfvén velocity ${\mathbf{B}}_0$ gives the incompressible Alfvén wave relation ${\omega }^2=(\mathbf{k}\cdot {\mathbf{B}}_0{)}^2$. Its frequency depends on the direction of $\mathbf{k}$, and energy propagates along ${\mathbf{B}}_0$. Here the averaged restoring response is isotropic: there is no preferred mean-field direction. Both systems resist transverse deformation through magnetic tension, rather than through compressive sound.

The propagating branch is not the entire perturbation space. For example, $\mathbf{v}=0$, $m_{ij}=f(\mathbf{r}){\delta }_{ij}$ and $\pi =f(\mathbf{r})$ solve the linearized equations without propagation, for any small, slowly varying $f$. Such a static stress perturbation in magnetoelastic wave equations balances a pressure gradient. Thus the requested wave conclusion applies to the oscillatory branch; it must not exclude stationary initial stress perturbations.

The equilibrium points satisfy $y=\beta$ and $x(x^2-1)=0$, so they areThe Jacobian matrix and its eigenvalue equation areAt $x=\pm 1$ the eigenvalues are $\pm \sqrt{2}$, so both points are saddle equilibria. Their stable and unstable tangent directions have slopes $-\sqrt{2}$ and $\sqrt{2}$, respectively.

At $x=0$ the eigenvalues are $\pm i$, giving a centre for the linearized equation. Purely imaginary eigenvalues alone do not classify the nonlinear system. The first integral established in part (ii) proves that $(0,\beta )$ is in fact a centre equilibrium. It is clockwise: immediately to its right, with $0<x<1$ and $y=\beta$, the vector field points downwards. Thus the final nonlinear classification is

The delayed positive term models recruitment depending on the population a time $T$ earlier; the quadratic negative term represents instantaneous density-dependent losses. Thus this delay differential equation delays recruitment rather than the strength of crowding acting on the present population.

With $x=1+y$, the linearized equation is $\dot{y}=\alpha y(t-T)-2\alpha y(t)$. Its characteristic equation of a delay differential equation is $\lambda +2\alpha =\alpha e^{-\lambda T}$. A root with nonnegative real part would satisfy $|\lambda +2\alpha |\ge 2\alpha$, whereas $|\alpha e^{-\lambda T}|\le \alpha$. This is impossible. Consequently the equilibrium is locally exponentially asymptotically stable for every $\alpha >0$ and $T>0$. There is no upper bound on $\alpha T$ in this model.

Put $x=1+y$ and retain linear terms. The delay differential equation is $\dot{y}(t)=-\alpha y(t-T)$, with characteristic equation of a delay differential equation $\lambda +\alpha e^{-\lambda T}=0$. Write $z=\lambda T$ and $q=\alpha T$. If a root has $\mathrm{Re}z\ge 0$, then $|z|=q{e}^{-\mathrm{Re}z}\le q$. Writing $z=a+ib$ gives $a=-q{e}^{-a}\cos b$. For $q<\pi /2$, $|b|<\pi /2$ and the right side is negative, contradicting $a\ge 0$. All characteristic roots therefore lie in the left half-plane.

An imaginary root $z=ib$ requires $\cos b=0$ and $b=q\sin b$. The first crossing is $q=\pi /2$, $z=\pm i\pi /2$. Implicit differentiation gives $dz/dq=-e^{-z}/(1+z)$, whose real part at the positive crossing is $(\pi /2)/(1+{\pi }^2/4)>0$. The later imaginary crossings occur at $q=\pi /2+2\pi k$ and also enter the right half-plane. Thus the equilibrium is locally exponentially asymptotically stable whenand unstable when $\alpha T>\pi /2$. At equality the linearized equation has neutral oscillatory modes; it is the Hopf bifurcation threshold. If stability is meant for the nonlinear equation itself, the borderline can also be resolved. In dimensionless time $s=t/T$, let $q_0=\pi /2$. A small center oscillation has the form $y=R\cos \theta +R^2(\cos 2\theta /5+\sin 2\theta /10)+O(R^3)$. Substitution at quadratic order gives these second-harmonic coefficients. At cubic order the fundamental forcing is $q_0{R}^3(\cos \theta /20+3\sin \theta /20)$. Projection onto the center mode divides the complex forcing by the characteristic derivative $1+i{q}_0$, yielding the Hopf normal form amplitude equationIts cubic coefficient is negative, so the equilibrium at equality is locally asymptotically stable, with algebraic rather than exponential decay. Thus the usual strictly decaying linear condition is $\alpha T<\pi /2$, while nonlinear local asymptotic stability includes $\alpha T=\pi /2$. The purely imaginary linear modes alone would not justify that borderline conclusion.

The razor-thin disk approximation uses vertically integrated pressure and surface density of a disk. To close the printed equations, interpret the barotropic equation of state as $P=P(\Sigma )$, or supply an equivalent fixed vertical conversion from the printed $\rho$ to $\Sigma$. Without that conversion, the literal $P=P(\rho )$ does not close this two-dimensional system. Set $c_s^2=(dP/d\Sigma {)}_0>0$, and write $\Sigma ={\Sigma }_0+\sigma$, $\mathbf{u}={\mathbf{u}}_0+{\mathbf{u}}^{\prime }$, $h^{\prime }=P^{\prime }/{\Sigma }_0=c_s^2\sigma /{\Sigma }_0$. Axisymmetric flow here means ${\partial }_y=0$, so background advection vanishes. The background shear enters through ${\mathbf{u}}^{\prime }\cdot \nabla {\mathbf{u}}_0=-3\Omega u_x^{\prime }{\mathbf{e}}_y/2$. The complete linearized equations areThese are the compressible Keplerian shearing sheet equations for an inertial-acoustic wave. No perturbed self-gravity is present in the given force law, so no gravitational potential perturbation is to be added. This closure is the barotropic closure of a razor-thin disk.

If $g{f}^{\prime \prime }(a)<{\omega }^2$, putThe linearized equation becomes $\ddot{ϵ}-{\lambda }^2ϵ=0$, soArbitrarily small initial perturbations can have $C_1\ne 0$ and grow exponentially. Hence the strict unstable equilibrium condition isFor the particular wire, this occurs precisely at $a=2j\pi b$, $j=1,2,\dots$; the alternating radii $a=(2j-1)\pi b$ satisfy the stable equilibrium condition.

If $g{f}^{\prime \prime }(a)>{\omega }^2$, defineThe linearized equation is $\ddot{ϵ}+{\nu }^2ϵ=0$, with bounded simple harmonic motionThus the strict stable equilibrium condition isThis is also a sufficient condition for local nonlinear Lyapunov stability. Indeed the equation for the bead on a uniformly rotating wire has the conserved quantityIts derivative is $\dot{r}$ times the governing equation moved to one side, so it vanishes. At $a$, the last two terms have zero first derivative and positive second derivative $g{f}^{\prime \prime }(a)-{\omega }^2$. Their strict local minimum traps sufficiently small perturbations. The motion is bounded rather than asymptotically returning to rest.

For the cosine mode in the question, the Laplacian acts by multiplication by $-s$, whereThese modes obey the Neumann boundary conditions. In the chemotactic term, put $q=1+u$, $c=\eta +v$. Then $\nabla c=\nabla v$, and $q\nabla c=\nabla v+u\nabla v$; the second product is a quadratic polynomial in perturbation amplitude. Thus the linearized equations areSubstituting the modal time dependence givesSeparating this matrix into its transport part and the reaction Jacobian matrix gives the required form, with the same $J$ as above.