Let $\tau$ be the time at which a right-running characteristic curve leaves the piston. The boundary velocity and the characteristic it emits areWiththese relations becomeBecause $d\varphi /d\theta =1-ϵ\cos \theta >0$ and $\varphi (\theta +2\pi )=\varphi (\theta )+2\pi$, $U$ is single-valued and $2\pi$-periodic behind the wavefront. Ahead of the first characteristic, where $\varphi <0$, the fluid remains undisturbed and $U=0$. The implicit relation is Kepler equation; unless $ϵ=0$, its inverse is not a pure sinusoid.

Series reversion givesExpanding the cosine then givesand hence

Forthe adiabatic sound speed isSince $\int (c/\rho )d\rho =-B/\rho =-c$, choose invariants that vanish in the equilibrium state $(u,c)=(0,c_0)$:ConsequentlyIn a right-moving simple wave with $R_{-}=0$, the $R_{+}$ characteristic speed is therefore $u+c=c_0$, independent of amplitude. The transport equation becomes $({\partial }_t+c_0{\partial }_x)R_{+}=0$, so the whole velocity profile translates rigidly at the equilibrium sound speed $c_0$ without nonlinear distortion.

A real transmitted normal wavenumber exists only ifso the critical angle for total internal reflection isFor ${\theta }_I>{\theta }_I^{\mathrm{cr}}$, putFor incident amplitude $E_0\hat{\mathbf{y}}$, the transverse-electric Fresnel reflection coefficient and transmission coefficient areA boundary-matched solution iswith each magnetic component obtained from ${\mathbf{B}}_X={\mathbf{k}}_X\times {\mathbf{E}}_X/\omega$, now using the complex transmitted wavevector $(i\kappa ,0,k_z)$. Since $|r|=1$, the wave undergoes total internal reflection; the transmitted field is an evanescent wave that propagates parallel to the interface and decays exponentially into $x>0$.

The condition ${\mathbf{E}}_I\cdot \hat{\mathbf{x}}=0$, together with transversality ${\mathbf{k}}_I\cdot {\mathbf{E}}_I=0$, selects polarization along $\hat{\mathbf{y}}$ away from the normal-incidence degeneracy. Phase matching at a planar wave interface givesThe two dispersion relations give $|{\mathbf{k}}_T{|}^2=({\epsilon }_{+}/{\epsilon }_{-})k_I^2$. Choosing propagation into $x>0$ therefore yieldsEquivalently, this is Snell law for electromagnetic waves with refractive index proportional to $\sqrt{\epsilon }$ because the two media have equal magnetic permeability.

For each $X\in \{I,R,T\}$, substitution of a complex plane electromagnetic wave in a linear medium into the source-free Maxwell equations in a linear medium requireswhere ${\epsilon }_I={\epsilon }_R={\epsilon }_{-}$ and ${\epsilon }_T={\epsilon }_{+}$. The accompanying magnetic field isThe incident and transmitted waves have $k_{Ix}>0$ and $k_{Tx}>0$, while the reflected wave has $k_{Rx}<0$. Matching the fields at the stationary interface additionally imposes the phase matching at a planar wave interface conditions ${\omega }_I={\omega }_R={\omega }_T$ and equality of all tangential wavevector components.

For a time-independent field, Faraday's law gives $\nabla \times \mathbf{E}=0$. Integrating around a thin rectangular loop crossing $x=0$ and shrinking its height gives continuity of the tangential component:A thin Gaussian pillbox and $\nabla \cdot \mathbf{D}=0$ give continuity of the normal electric displacement field:These are the electrostatic cases of the dielectric-interface boundary conditions.

For $f(w)=w^M=r^M{e}^{iM\phi }$, the angular momentum operator $L_z=-i\hslash {\partial }_{\phi }$ givesThe probability density is proportional towhose maximum occurs atIn a large disk of area $A=\pi R^2$, the states that fit inside satisfy $M\lesssim qB{R}^2/(2\hslash )$. Thus the degeneracy of each Landau level is asymptoticallywhere ${\Phi }_B=BA$ is the magnetic flux. There is one state per magnetic flux quantum $h/q$, up to boundary corrections.

Let $\mathbf{\Pi }=\mathbf{p}-q\mathbf{A}$ be the kinetic momentum. The quantum Hamiltonian isAssume $qB>0$ and zero longitudinal momentum, as in the displayed states. Sincewe have $[a,a^{†}]=1$ andFor $w=x+iy$,which annihilates everywith $f$ holomorphic. Hence all these states haveThe Landau levels are $E_n=\hslash {\omega }_B(n+1/2)$ for $n=0,1,\dots$; the arbitrary holomorphic function exhibits the degeneracy of the lowest level. For $qB<0$, the normalizable lowest-level states are anti-holomorphic instead.

These potentials give the uniform magnetic field $\mathbf{B}=B\hat{\mathbf{z}}$ and zero electric field. With the signed frequency ${\omega }_c=qB/m$, the Lorentz force law becomesThusand explicitlyThe transverse trajectory is a circle about the fixed guiding centre $(x+\dot{y}/{\omega }_c,y-\dot{x}/{\omega }_c)$. The positive cyclotron frequency is

For $0<\mu <1$, all orbits converge to one fixed point, and for $\mu =1$ every orbit is eventually fixed or period two, so no iterate has a horseshoe. For the symmetric tent map with slope magnitude $\mu >1$, the topological entropy is $\log \mu >0$. The interval-map positive-entropy horseshoe theorem says that a continuous interval map has positive topological entropy exactly when some iterate has a horseshoe. HenceFor $\mu \ge \sqrt{2}$ this already follows from part (iii); for $1<\mu <\sqrt{2}$, a higher iterate supplies the horseshoe.

The map $F$ itself has a horseshoe only for $\mu =2$. Indeed, its Lipschitz constant is $\mu$. If two intervals $J_0,J_1$ form a horseshoe and their convex hull has length $L$, thenwhich forces $\mu \ge 2$. At $\mu =2$, take $J=(-1,1)$, $K_0=(-1,0)$ and $K_1=(0,1)$; both $K_i$ map onto $J$.

The Lipschitz constant of $F^2$ is ${\mu }^2$, so the same length argument makes $\mu \ge \sqrt{2}$ necessary for an $F^2$ horseshoe. It is sufficient. PutFor $\mu \ge \sqrt{2}$, one has $a<1/\mu \le b\le 1$. Withthe two relevant monotone branches satisfyThus $K_0,K_1\subset J$ are disjoint and $F^2(K_0)=F^2(K_1)=J$. Therefore

For $0<\mu <1$, one iterate sends the whole interval into $[1-\mu ,1]$, whereThis is a contraction mapping with unique fixed pointEvery orbit converges to $x_{\ast }$, alternating around it because the multiplier is $-\mu$.

For $\mu =1$, the first iterate lies in $[0,1]$ and $F(x)=1-x$ there. Hence $F^2$ is the identity on $[0,1]$: $1/2$ is fixed, while every other point in that interval belongs to the period-two orbit $\{x,1-x\}$.

The graph of $F$ is the symmetric tent with verticesIf $0<\mu <1$, then $F\ge 0$ andso $F^2$ is an upward-pointing V with its minimum at $(0,1-\mu )$ and equal endpoint values $1-\mu +{\mu }^2$.

If $1<\mu <2$, the zeros of $F$ are $\pm 1/\mu$ andThus its five vertices are

An interval map $F:I\to I$ has a horseshoe when there are an open interval $J\subseteq I$ and two disjoint open intervals $K_0,K_1\subseteq J$ such that

A finite linear combination of independent Gaussian random variables is Gaussian, soSince $v_n\to v:={\sum }_{k\ge 1}{a}_k^2<\infty$, the characteristic functions converge pointwise:The Lévy continuity theorem yields convergence in distribution toOn the given common probability space, the same variance calculation actually proves convergence in $L^2$ as well.

For every finite Rademacher sum, independence and the vanishing of odd moments giveThe same calculation applied to $S_n-S_m$ shows that $(S_n)$ is Cauchy in $L^4$, because the tail of $\sum a_k^2$ tends to zero. Its $L^4$ limit agrees almost surely with its $L^2$ limit $S$, since both modes imply convergence in probability. Passing to the limit givesand therefore

For $n>m$, independence and $\mathbb{E}{X}_k=0$ giveThus $(S_n)$ is a Cauchy sequence in the complete L2 space and converges in $L^2$ to some random variable $S$. The Cauchy-Schwarz inequality gives $\Vert Z{\Vert }_1\le \Vert Z{\Vert }_2$ on a probability space, so the convergence also holds in $L^1$. Finally, convergence in $L^2$ implies convergence in probability, which implies convergence in distribution.

For the tori in part (c), the distance from the origin is as large as $R_n+1\to \infty$, so condition (1) fails. Their surface areas areso condition (2) fails as well.

False with the $\mathrm{limsup}$ printed in the paper. Let the even-indexed surfaces be one fixed unit sphere and the odd-indexed surfaces one fixed ring torus. Thenbut infinitely many surfaces are tori. The two fixed surfaces lie in one ball and have uniformly bounded area. Their injectivity radii have a positive common lower bound, and compactness gives a common bound on $|\nabla K|$, hence on every radial derivative $|{\partial }_{r}K|$. Thus this alternating sequence satisfies all four additional conditions as well.

The intended statement becomes true if $\mathrm{limsup}$ is replaced byHere is the proof. Put ${\delta }_n=\max (0,-{\inf }_{S_n}{K}_n)$, so ${\delta }_n\to 0$. Choose $p_n$ maximizing distance from the centre of the fixed containing ball. The supporting-sphere curvature bound givesWrite $K_0=K_n(p_n)$ and $j(r,\theta )=\sqrt{G(r,\theta )}$ in geodesic polar coordinates. Choose a fixed sufficiently small $c>0$ and set ${\rho }_n=c{K}_0^{-1/2}$. Conditions (3) and (4) allow $c$ to be chosen so that ${\rho }_n\le {ϵ}_0$ andThe Jacobi equation $j_{rr}=-K_{n}j$, with $j(0)=0$ and $j_r(0)=1$, and the Sturm comparison theorem then give $j(r,\theta )\ge c_{1}r$ on this ball for a constant $c_1>0$ independent of $n$. ConsequentlySince $K_n\ge -{\delta }_n$ elsewhere and $\mathrm{Area}(S_n)\le M$,for all sufficiently large $n$. Gauss-Bonnet forces $\chi (S_n)>0$, and the classification theorem for surfaces makes each such $S_n$ a sphere.

The preceding estimate makes $(x_k)$ equicontinuous, andmakes it uniformly bounded. The Arzelà-Ascoli theorem therefore supplies a subsequence converging uniformly to a continuous function $x:[0,1]\to {\mathbb{R}}^n$. On every mesh interval,including an endpoint by using the adjacent interval. Consequently $y_k\to x$ uniformly along the same subsequence.

All $x_k(t)$ and $y_k(t)$ lie in one compact ball. The restriction of the continuous vector field $F$ to that ball is uniformly continuous, so $F\circ y_k\to F\circ x$ uniformly. Passing to the limit in the defining equation givesThe fundamental theorem of calculus now makes $x$ differentiable and yields $x^{\prime }(t)=F(x(t))$. This proves the required special case of the Peano existence theorem; uniqueness need not hold because $F$ was assumed continuous but not locally Lipschitz continuous.