Write $d_{+}=\sigma +h$ and $d_{-}=h-\sigma$ for the lower and upper layer thicknesses. A steady unidirectional velocity $\mathbf{u}=u(y){\mathbf{e}}_x$ with no pressure gradient satisfies the Navier-Stokes equations through $u^{\prime \prime }=0$ in each layer. Continuity of velocity and tangential stress at the interface gives one constant shear stressThe two-layer Couette flow is therefore

The simple zeros $r_{-}$ and $r_{+}$ are Killing horizons: respectively the black-hole event horizon and the cosmological horizon of the static region.

The function $f$ tends to $-\infty$ both as $r\to 0^{+}$ and as $r\to \infty$. Its only positive stationary point isand it is a maximum withTherefore $f$ has two positive simple roots exactly whenAt $r_c=1/(\sqrt{3}H)$,Since $f$ is positive precisely between its two positive roots, this proves

For a nonradial null geodesic,It tends to $-\infty$ as $r\to 0^{+}$, rises to one maximum, and tends to $-L^2{H}^2$ as $r\to \infty$. A circular null geodesic requires $V^{\prime }(r)=0$, soSince $V^{\prime \prime }(3m)<0$, this photon sphere is unstable. Its energy satisfies $E^2=L^2(1/(27m^2)-H^2)$.

This is the Schwarzschild-de Sitter spacetime. Its geodesic Lagrangian may be taken asTime-translation and rotational symmetry give three independent integrals of motion, which may be chosen asThe constant norm of the tangent is $g_{ab}{\dot{x}}^a{\dot{x}}^b=-\kappa$, where $\kappa =1$ for a timelike geodesic parametrized by proper time and $\kappa =0$ for a null geodesic. Substitution givesand hence

In the interaction picture, first-order time-dependent perturbation theory gives the transition amplitudeTaking its squared modulus and using $|e^{i\theta }-1{|}^2=4{\sin }^2(\theta /2)$ gives

As $|x|\to \infty$, the operator $A$ tends to $-4{\partial }_x^3$. Evolve the normalized scattering eigenfunction byThe added scalar term keeps the coefficient of $e^{-ikx}$ at $-\infty$ equal to one. At $+\infty$,ConsequentlyIt follows that

Differentiatingwith respect to time givesIf $p_t=p^3{p}_{xxx}$, thenThus $L_t=[L,A]$ is an Isospectral Lax equation, and $(L,A)$ is a Lax pair for the Harry Dym equation.

Suppose a normalizable solution satisfies $L\psi =-{\kappa }^2\psi$. Multiplication by $\overline{\psi }$, integration over $\mathbb{R}$, and integration by parts giveThe boundary term vanishes for a normalizable eigenfunction. The left side is negative unless $\psi =0$, so no nontrivial normalizable solution exists.

Normalize the solution by its incoming amplitude at $+\infty$. The transmission coefficient and reflection coefficient areFor two solutions of the same equation in Sturm-Liouville form, the modified Wronskianis independent of $x$. Take ${\psi }_1={\varphi }_k$ and ${\psi }_2=\overline{{\varphi }_k}$. At $-\infty$ this gives $W=-2ik$, while at $+\infty$ it givesTherefore $|a{|}^2-|b{|}^2=1$, and division by $|a{|}^2$ yields conservation of scattering flux:

LetSince $d(e^{-t^2})=-2t{e}^{-t^2}dt$, repeated integration by parts givesThe final remainder estimate follows by one further integration by parts, or by bounding the remaining positive integral by its leading boundary term. Therefore the requested first three terms are

Because $K\subseteq K_n$, one of the two directed terms in the given Hausdorff distance vanishes, andSuppose this does not tend to zero. Then some $\epsilon >0$, a strictly increasing sequence $(n_j)$ and points $x_j\in K_{n_j}$ satisfyAll $x_j$ lie in the compact set $K_1$, so after passing to a subsequence, $x_j\to x$. For every fixed $m$, eventually $n_j\ge m$, so $x_j\in K_m$ and the closedness of $K_m$ gives $x\in K_m$. Hence $x\in K$, contradicting $|x_j-x|\to 0$. Therefore $\rho (K_n,K)\to 0$.

Every $K_n$ is contained in the bounded set $K_1$, so the intersection $K={\bigcap }_{n\ge 1}{K}_n$ is bounded. It is closed because an arbitrary intersection of closed sets is closed.

It remains to prove that $K$ is nonempty. Choose $x_n\in K_n$. The set $K_1$ is a compact set, so some subsequence $x_{n_j}$ converges to a point $x\in K_1$. For each fixed $m$, all sufficiently late terms satisfy $n_j\ge m$ and hence $x_{n_j}\in K_{n_j}\subseteq K_m$. Since $K_m$ is closed, $x\in K_m$. This holds for every $m$, so $x\in K$. Thus $K\in \mathcal{K}$.

The sum is nonempty because $A$ and $B$ are nonempty. It is a bounded set: if $|a|\le M_A$ and $|b|\le M_B$, then $|a+b|\le M_A+M_B$.

To prove that it is a closed set, let $x_j=a_j+b_j\in A+B$ be a convergent sequence with $x_j\to x$. By the Heine-Borel theorem, $A$ is a compact set, so $(a_j)$ has a convergent subsequence $a_{j_r}\to a\in A$. ThenSince $B$ is closed, $x-a\in B$, and hence $x=a+(x-a)\in A+B$. Therefore $A+B$ is nonempty, closed and bounded, so $A+B\in \mathcal{K}$.

For each period setand defineIndependence of the $Z_i$ shows that this strictly positive density has expectation one. Under $\mathbb{Q}$, the conditional mean of each gross return isConsequentlyso the discounted price process is a $\mathbb{Q}$-martingale. This equivalent martingale measure rules out arbitrage.