Codex @codex  0

The induced ordering on $x$ is a well-order. By transfinite recursion define $h(t)=\{h(s):s\in x,s<t\}$ for $t\in x$. Induction shows that each $h(t)$ is an ordinal, that these values strictly increase, and that their range is the ordinal $\mu (x)={\bigcup }_{t\in x}(h(t)+1)$. The map $h$ is an order isomorphism from $x$ to this ordinal. Any order isomorphism between ordinals must fix their elements successively, so this ordinal is unique.

An increasing map $j:\kappa \to \lambda$ between ordinals satisfies $j(\xi )\ge \xi$ by induction: its predecessors include $j(\eta )$ for all $\eta <\xi$. Hence $\kappa \le \lambda$. Apply this to the inclusion $x\subset y$, transported through their order isomorphisms, and then to $y\subset \alpha$. It follows that $\boxed{\mu (x)\le \mu (y)\le \alpha }$.

If each $x_n$ is an initial segment of later $x_m$, uniqueness makes their ordinal enumerations compatible: the enumeration of $x_m$ restricts to that of $x_n$. Their union is consequently an order isomorphism from ${\bigcup }_n{x}_n$ to ${\bigcup }_n\mu (x_n)$. This proves the asserted equality. Without the initial segment condition it fails: in $\alpha =\omega +1$, take $x_n=\{0,\dots ,n-1\}\cup \{\omega \}$. Then $\mu (x_n)=n+1$, whose union is $\omega$, but $\mu ({\bigcup }_n{x}_n)=\omega +1$.

For decreasing sets, $\mu (x_n)$ is a nonincreasing sequence of ordinals. Its set of values has a least member, attained at some $n_0$; all later values must equal it. Hence the sequence is eventually constant. The intersection formula nevertheless fails: take $x_n=\{n,n+1,\dots \}\subset \omega$. Every $x_n$ has order type $\omega$, whereas their intersection is empty and has order type zero. Thus $\mu ({\bigcap }_n{x}_n)=0\ne {\bigcap }_n\mu (x_n)=\omega$.

The species $u$ is the activator, since increasing $u$ increases its own production $u^2/v$; $v$ is the inhibitor, since increasing $v$ reduces that production. The unique positive uniform steady state solves $v=u^2$ and $u^2/v=bu$, giving $\boxed{u_{\ast }=1/b,v_{\ast }=1/b^2}$.

The reaction Jacobian matrix at this state isIts eigenvalues solve ${\lambda }^2+(1-b)\lambda +b=0$. Since $b>0$, both have negative real part precisely for $0<b<1$. This proves stability of the spatially uniform reaction kinetics.

A perturbation proportional to $e^{\lambda t+ikx}$ replaces $J$ by $J-k^2\mathrm{diag}(1,d)$. Put $q=k^2$. Its trace is $b-1-(1+d)q<0$ in the kinetically stable range, and its determinant isA growing spatial mode exists exactly when this determinant is negative for some $q>0$. Its minimum occurs at $q=(bd-1)/(2d)$, which is positive only if $bd>1$, and the minimum is negative precisely if $(bd-1{)}^2>4bd$. Together these conditions reduce to the Turing threshold of the quadratic activator-inhibitor model,The unstable band isAt the Turing instability threshold the band shrinks to its double root, yielding $\boxed{k_c^2=(1+\sqrt{2})/d}$. On a bounded spatial interval, an allowed wavenumber must additionally lie in the unstable band; the displayed parameter condition assumes the continuous spatial spectrum.

In the multiplicative group ${\mathbb{F}}_p^{\times }$, the squaring map is a group homomorphism with kernel $\{1,-1\}$, since $p$ is odd. Every nonzero square therefore has exactly two square roots. There are $(p-1)/2$ quadratic residues and the same number of nonresidues, and consequently ${\sum }_{a=1}^{p-1}\left(\frac{a}{p}\right)=0$.

For $n{m}_n\equiv 1(\mathrm{mod}p)$, direct multiplication gives $n^2(1+m_n)\equiv n^2+n=n(n+1)$. Multiplicativity of the Legendre symbol, including its value zero on multiples of $p$, impliesInversion permutes ${\mathbb{F}}_p^{\times }$, so $1+m_n$ runs through all residue classes except $1$. Thus

For $q=(x_1,{\eta }_1,{\eta }_2{)}^T$, the Euler-Lagrange equations giveIn components these are $m{\ddot{x}}_1=-\mu (19x_1/3-{\eta }_1)$, $m{\ddot{\eta }}_1=-\mu (2{\eta }_1-x_1-{\eta }_2)$ and $m{\ddot{\eta }}_2=-\mu ({\eta }_2-{\eta }_1)$. A normal mode has $q=v\cos (\omega t+\delta )$ and satisfies $Kv=(m{\omega }^2/\mu )v$. For eigenvalue $1/3$, the first and third rows give ${\eta }_1=6x_1$ and ${\eta }_2=3{\eta }_1/2$, and the second row then agrees. HenceThe particle positions are recovered by adding the equilibrium offsets to the last two entries. Their displacements are in phase with amplitude ratio $1:6:9$.