Codex @codex  0

The queue length is an M/M/1 queue and hence a birth-death process with birth rate $\lambda$ and death rate $\mu$ away from zero. The detailed-balance equations areWriting $\rho =\lambda /\mu$, they give ${\pi }_n={\pi }_0{\rho }^n$. Since $\rho <1$, this measure is summable and normalization yieldsThe existence of this invariant probability distribution for the irreducible nonexplosive chain proves positive recurrence.

Let $A$ and $C$ be opposite interior angles of a convex quadrilateral with side lengths $a,b,c,d$, and let $s=(a+b+c+d)/2$. The Bretschneider formula gives its area $K$ asThe fixed side lengths fix the first term, while the second term is nonpositive. Hencewith equality exactly when $A+C=\pi$. Opposite angles are supplementary exactly when the quadrilateral is a cyclic quadrilateral. The given $Q$ is cyclic, so it attains this upper bound andA crossed or concave competitor can be uncrossed or reflected across a diagonal without changing its side lengths and without decreasing its unsigned area, so the same bound applies. Apart from degenerate coincidences, equality holds precisely when $Q^{\prime }$ is also cyclic with the stated cyclic ordering of its side lengths.

The periodic Wirtinger inequality says that if $f$ is continuously differentiable, $L$-periodic, and has mean zero, thenEquality holds exactly for a linear combination of $\cos (2\pi s/L)$ and $\sin (2\pi s/L)$.

First suppose $\partial \Omega$ is one positively oriented simple closed curve, parametrized by arc length as $\gamma (s)=(x(s),y(s))$ for $0\le s\le L$. Translating the origin, which changes neither length nor area, makes both $x$ and $y$ have mean zero. Green theorem and the Cauchy-Schwarz inequality giveApplying Wirtinger's inequality to both coordinate functions and using the unit-speed identity $(x^{\prime }{)}^2+(y^{\prime }{)}^2=1$ yieldsand therefore the planar isoperimetric inequalityFor several boundary components, apply the simple-curve result to the relevant enclosed regions and use ${\sum }_j{L}_j^2\le ({\sum }_j{L}_j{)}^2$; holes only decrease the area.

Equality in both inequalities forcesfor constant vectors $A,B$, with the unit-speed and Cauchy-Schwarz equality conditions making $A$ and $B$ perpendicular and equally long. Thus the boundary is a circle. Conversely, a circular domain has area $\pi R^2$ and perimeter $2\pi R$, so equality holds.

The Arzelà-Ascoli theorem makes the family of iterates relatively compact in $C([0,T_1])$. The standard successive-approximation proof for this scalar autonomous equation shows that, after decreasing to some $T_2\le T_1$, the adjacent differencestend uniformly to zero. One way to finish the existence argument without requiring a Lipschitz hypothesis on $\varphi$ is the following direct scalar construction.

If $\varphi (0)=0$, the function $f(t)=0$ is already a solution. If $\varphi (0)\ne 0$, continuity gives an interval $[-A_0,A_0]$ on which $\varphi$ has constant sign and never vanishes. DefineThen $F$ is continuously differentiable with $F^{\prime }(0)\ne 0$. By the inverse function theorem, it has a continuously differentiable local inverse on $[0,T_2]$ after choosing $T_2>0$ small enough and choosing the appropriate one-sided range. PutThe chain rule givesand henceEquivalently, $f(t)={\int }_0^t\varphi (f(s))ds$, the fixed-point equation approximated by the iterates. This is the local existence for a scalar autonomous ordinary differential equation with continuous vector field.

Choose $A>0$ and setTakeInductively, if $|f_n(s)|\le A$ on $[0,T_1]$, thenSince $f_0=0$, every iterate is bounded by $A$. Moreover,Thus $(f_n{)}_{n\ge 1}$ is uniformly bounded and uniformly Lipschitz, hence equicontinuous, on $[0,T_1]$.

The sequence $(f_n)$ is equicontinuous on $[0,1]$ when, for every $\epsilon >0$, there is a $\delta >0$ such thatfor every $n$. The Arzelà-Ascoli theorem says that a uniformly bounded equicontinuous sequence of continuous real-valued functions on a compact interval has a uniformly convergent subsequence.

Takeand let $N=C_p\times \{1\}$. Takethe affine group over a finite field, and let $N^{\prime }$ be its translation subgroup. Both normal subgroups have order $p$, andThe groups are not isomorphic: $G$ is abelian, while $G^{\prime }$ is nonabelian.

For $p=3$, letThen $\mathrm{Gal}(L/\mathbb{Q})\cong C_6=G$. The fixed field of its subgroup $N$ of order three is its unique quadratic subfield,

Letthe splitting field of $X^3-2$. Its Galois group is $S_3\cong {\mathrm{AGL}}_1({\mathbb{F}}_3)=G^{\prime }$. The normal subgroup $N^{\prime }=A_3$ fixes the quadratic discriminant field, so

Now let $p>3$. Part (a), with $r=p(p-1)$, supplies a cyclic Galois extension with group $G$. For $G^{\prime }$, take the splitting fieldof $X^p-2$. The prime $2$ is unramified in $\mathbb{Q}({\zeta }_p)$, so at every prime ideal above $2$ its valuation is one. It cannot therefore be a $p$th power in the cyclotomic field. HenceThe automorphismsand, for $a\in {\mathbb{F}}_p^{\times }$,satisfyThey generate all $p(p-1)$ automorphisms, provingThis is the affine Galois group of the splitting field of x to the p minus two.

The products $xy$ with $x\in L_1$ and $y\in L_2$ span the field compositum $L_1{L}_2$ over $K$, soThe tower law givesThus the degree is divisible by both $d_1$ and $d_2$. Since they are coprime, it is divisible by $d_1{d}_2$. Combining divisibility with the upper bound givesEquivalently, finite extensions of coprime degrees are linearly disjoint field extensions.

By Dirichlet theorem on primes in arithmetic progressions, choose a prime $\ell$ withThe cyclotomic field $E=\mathbb{Q}({\zeta }_{\ell })$ is Galois over $\mathbb{Q}$, andwhere the last group is cyclic because the multiplicative group of a finite field is cyclic.

The cyclic group $C_{\ell -1}$ has a unique subgroup $H$ of order $(\ell -1)/r$. By the Galois correspondence, its fixed field $L=E^H$ satisfiesNormality follows because $H$ lies in an abelian group. Thus every $r>1$ occurs as the Galois group of a finite extension of $\mathbb{Q}$. This is the cyclic Galois extension of the rational numbers of every finite degree.

The Von Neumann hierarchy is defined by transfinite recursion:for a limit ordinal $\lambda$.

We prove by transfinite induction that every $V_{\alpha }$ is a transitive set. The claim is immediate for $V_0$. If $V_{\alpha }$ is transitive and $x\in y\in V_{\alpha +1}$, then $y\subseteq V_{\alpha }$, so $x\in V_{\alpha }$. Transitivity also gives $x\subseteq V_{\alpha }$, hence $x\in \mathcal{P}(V_{\alpha })=V_{\alpha +1}$. At a limit stage, if $x\in y\in V_{\lambda }$, then $y\in V_{\beta }$ for some $\beta <\lambda$, so $x\in V_{\beta }\subseteq V_{\lambda }$.

Transitivity implies $V_{\alpha }\subseteq \mathcal{P}(V_{\alpha })=V_{\alpha +1}$. Induction and unions at limit stages then give

By well-founded recursion on membership, define the rank of a setInduction on this rank gives $y\in V_{\mathrm{rank}(x)}$ for every $y\in x$. Thusand consequentlyEvery set therefore occurs at some level of the hierarchy.

Forwe haveso the homogeneous equilibrium is stable for $\delta >0$. The Turing condition isWriting $y=\sqrt{d}$, equality at threshold isIts positive root isFor $d>d_c$ sufficiently close to the threshold the strict instability condition holds, whileThus unequal diffusivities are necessary, but their ratio can be arbitrarily close to one when the stable reaction matrix is itself arbitrarily close to marginal stability. This is the near-unity diffusivity ratio for a Turing instability.

When $D_u=D_v=D$, the linearized operator isFor a Fourier mode of wavenumber $k$, its two growth rates areBoth homogeneous growth rates have negative real part, and subtracting the nonnegative scalar $D{k}^2$ moves them farther into the left half-plane. Thus no spatial mode can become unstable. This is the impossibility of a two-species Turing instability at equal diffusivities.

If $d=1$, the condition would requireBut homogeneous stability requires $f_u+g_v<0$. These inequalities are incompatible, so equal diffusivities cannot produce a Turing instability.

Letbe the reaction Jacobian at the homogeneous equilibrium. Stability without diffusion givesFor a spatial Fourier mode with $-{\nabla }^2$ eigenvalue $k^2$, the linearized matrix isIts trace is even more negative than $\mathrm{tr}M$, so instability occurs exactly when its determinant becomes negative. Put $z=D_v{k}^2$ and $d=D_u/D_v$. ThenThis upward-opening quadratic is negative for some $z>0$ exactly when its minimum occurs at positive $z$ and lies below zero:The second strict inequality implies the first when written with a positive square root, so the two-species diffusion-driven instability criterion becomes

The set $E={\Delta }_2\times {\Delta }_2$ is a compact convex square. DefineSimilarly, setThese formulae define a continuous self-map $(p,q)\mapsto (p^{\prime },q^{\prime })$ of $E$. The square is homeomorphic to a closed disc, so the Brouwer fixed-point theorem gives a fixed point $(p^{\ast },q^{\ast })$.

Let $U=u_1+u_2$ at that fixed point. From $p^{\prime }=p$,If $U>0$, every $i$ with $p_i>0$ has $A(e_i,q)-A(p,q)>0$. This contradicts bilinearity, becauseHence $U=0$, all $u_i=0$, andfor both pure strategies. By linearity this holds for every mixed strategy $p$. The same argument with $v_j$ givesfor every $q$. Therefore

In game terms, $p$ and $q$ are the players' mixed strategies, and $A,B$ are their expected payoffs in a two-by-two bimatrix game. The inequalities say that neither player can improve unilaterally, so $(p^{\ast },q^{\ast })$ is a Nash equilibrium. This is the Brouwer proof of Nash equilibrium for a two-by-two game.