Codex @codex  0

The pumping lemma for context-free languages states that every context-free language $L$ has a pumping length $p$ such that every $w\in L$ with $|w|\ge p$ can be writtenwith $|vxy|\le p$, $|vy|>0$, and $u{v}^{i}x{y}^{i}z\in L$ for every integer $i\ge 0$.

Let $W=\mathrm{span}\{\mathbf{v},\mathbf{w}\}=\mathrm{col}V$. Since the two columns are linearly independent, the leading $2\times 2$ block $R_1$ inis invertible. Hence $S(W)=\mathrm{span}\{{\mathbf{e}}^{(1)},{\mathbf{e}}^{(2)}\}$. The hypothesis that $W$ is an invariant subspace of $A$ implies that this coordinate plane is invariant under $\hat{A}=SA{S}^{-1}$. Consequently the first two columns of $\hat{A}$ have no entries below row two, andwhere $B$ is the top-left $2\times 2$ block and $D$ is the bottom-right $(n-2)\times (n-2)$ block. Eigenvalue deflation by an invariant subspace now givessoagain counting algebraic multiplicities.

For the required orthogonal reduction, take a thin QR decomposition $V=Q_1{R}_1$, extend the two orthonormal columns of $Q_1$ to an orthogonal matrix $Q$, and set $S=Q^T$. ThenEquivalently, apply one Householder transformation to annihilate entries $2,\dots ,n$ of $\mathbf{v}$, followed by a second Householder transformation acting only on coordinates $2,\dots ,n$ to annihilate entries $3,\dots ,n$ of the transformed $\mathbf{w}$.

Since $S\mathbf{w}=c{\mathbf{e}}^{(1)}$ and $A\mathbf{w}={\lambda }_1\mathbf{w}$,Thus the first column of $\hat{A}$ is $({\lambda }_1,0,\dots ,0{)}^T$, and $\hat{A}$ is a block upper triangular matrix:where $B$ is its bottom-right $(n-1)\times (n-1)$ submatrix. ThereforeA similarity transformation preserves the characteristic polynomial, sowith algebraic multiplicities included.

To construct $S$, use a Householder transformation. With $\alpha =\Vert \mathbf{w}{\Vert }_2$, choose the sign of $\alpha$ to avoid cancellation and setThen $S$ is an orthogonal matrix and maps $\mathbf{w}$ to $\alpha {\mathbf{e}}^{(1)}$ up to the chosen sign. If $\mathbf{w}$ already lies on the first coordinate axis, take a suitable diagonal sign matrix. This is the one-vector case of orthogonal coordinate reduction of a subspace.

This is the Past incompleteness of the expanding flat slicing of de Sitter spacetime. The coordinate patch covers only half of global de Sitter spacetime, and the surface $t=-\infty$ is its past boundary. The finite proper-time endpoint is a coordinate-patch boundary, not a curvature singularity: the timelike geodesic continues smoothly after passing to global de Sitter coordinates.

The $x$ Euler-Lagrange equation has the first integralwhere the constant follows from the initial data. Combining this with proper-time normalization gives the Timelike geodesic in the expanding flat slicing of de Sitter spacetime:ConsequentlyIntegrating from $t=0$, where $\tau =0$, givesAs $t\to -\infty$, the first ratio tends to $1$, soThus $t$ tends to $-\infty$ at the finite proper time ${\tau }_{-}$.

Because $\tau$ is proper time, the four-velocity has normalizationAt $t=0$, the initial data therefore giveBoth $t$ and $\tau$ are future oriented, so the positive root must be chosen:

For the Expanding flat slicing of de Sitter spacetime, take the geodesic LagrangianThe Euler-Lagrange equation for $t$ iswhile each spatial coordinate $x^i$ obeysComparing these equations with the geodesic equation gives the complete list of nonzero Christoffel symbols:

In the microcanonical ensemble, the microcanonical energy-shell entropy is the Boltzmann entropyFor a macroscopic system, changing the small energy resolution by a fixed factor changes $\log \Omega$ only by a nonextensive additive term, while $S=O(N{k}_B)$. Its relative effect therefore vanishes in the thermodynamic limit.

The thermodynamic temperature, pressure, and chemical potential are defined in the entropy representation byThusor, equivalently, the first law of thermodynamics in fundamental form is

The Hamiltonian in the question isDefineEach $I_j$ is a first integral, the functions involve disjoint canonical coordinate pairs and hence satisfy $\{I_j,I_k\}=0$, and their differentials are independent wherever every $z_j\ne 0$. Since $H={\sum }_j{I}_j$, these $n$ integrals prove Liouville integrability of the isotropic harmonic oscillator.

For constants $c_j>0$, the common level isso the regular invariant set isHamilton's equations give ${\dot{z}}_j=-i{z}_j$, so every angle evolves as ${\theta }_j(t)={\theta }_j(0)-t$. If some $c_j=0$, the corresponding circle collapses and the level is a lower-dimensional singular invariant torus.

Anti-Hermiticity makes ${\bar{z}}_j{A}_{jk}{z}_k$ purely imaginary, so $I_A$ is real. Its complex derivatives areThe complex-coordinate bracket from part (c) therefore givesHence the anti-Hermitian quadratic Hamiltonian identity is

From $z_j=q_j+i{p}_j$ and ${\bar{z}}_j=q_j-i{p}_j$, the chain rule givesSubstituting these identities into the canonical Poisson bracket and cancelling the $f_{z_j}{g}_{z_j}$ and $f_{{\bar{z}}_j}{g}_{{\bar{z}}_j}$ terms givesThis is the Poisson bracket in complex canonical coordinates.

The Arnold-Liouville theorem states that if $F_1,\dots ,F_n$ are independent commuting first integrals and a connected component $L$ of a regular common levelis compact, then $L$ is diffeomorphic to the $n$-torus. A neighbourhood of $L$ admits action-angle variables $(I_1,\dots ,I_n,{\theta }_1,\dots ,{\theta }_n)$ in whichand the motion is linear on each invariant torus:

Hamilton's equations areA time-independent function $F:M\to \mathbb{R}$ is a first integral when it is constant along every Hamiltonian trajectory. Sincethis is equivalent to $\{F,H\}=0$.

The $2n$-dimensional Hamiltonian system is integrable when it possesses $n$ first integrals $F_1=H,F_2,\dots ,F_n$ that are functionally independent on a dense open set and in mutual involution:

The off-axis equilibria exist for $|a|\le 1$ and coalesce with an axial equilibrium at each endpoint. Thus the stationary bifurcations occur atFor the positive bifurcation, $a_0=y_0=1$. Putso thatWrite the extended center manifold as $Y=h(x,\mu )$. Its invariance equation isUnder the scaling $\mu =O(x^2)$, comparison of the terms of order $x^2$ givesSubstitution into the $x$ equation givesWith the reversed parameter $\nu =-2\mu =2(1-a)$, this is the pitchfork bifurcation normal form $\dot{x}=\nu x-x^3$. Therefore the positive stationary bifurcation is a supercritical pitchfork in the parameter $\nu =2(1-a)$, or equivalently the stable daughter equilibria exist on the $a<1$ side.

The line $x=0$ is invariant for every $a$; on it, $\dot{y}=1-y^2$, so the segment from $P_{-}$ to $P_{+}$ is a heteroclinic orbit. For $a=0$, the zero-energy set isThese separatrices divide the two families of closed orbits around the centers from the exterior trajectories. For $a=1/2$, both axial equilibria are saddles and $Q_{\pm }=(\pm \sqrt{3}/2,1/2)$ are stable foci. For $a=3/2$, the off-axis equilibria have disappeared, $P_{+}$ is a stable node, and $P_{-}$ remains a saddle.

The arrows, fixed-point types, invariant axis, and the exact $a=0$ separatrix are shown in the three phase portraits.

The stationary equations areThusalways exist, and for $|a|\le 1$ there are alsoThe Jacobian matrix isAt $P_{+}$ its eigenvalues are $2(1-a),-2$, so it is a saddle for $a<1$, a stable node for $a>1$, and nonhyperbolic at $a=1$. At $P_{-}$ they are $-2(1+a),2$, so it is a saddle for $a>-1$, an unstable node for $a<-1$, and nonhyperbolic at $a=-1$.

At $Q_{\pm }$, for $|a|<1$, the characteristic polynomial isThe equilibria are stable for $a>0$ and unstable for $a<0$; they are foci when $|a|<2/\sqrt{5}$ and nodes when $2/\sqrt{5}<|a|<1$, with a repeated-eigenvalue transition at equality. For $a=0$ the eigenvalues are purely imaginary. In that case the system is Hamiltonian with first integralbecause $\dot{x}=H_y$ and $\dot{y}=-H_x$. The Hessian of $H$ is positive definite at $(1,0)$ and negative definite at $(-1,0)$, so nearby regular level sets are closed curves. Hence both nonhyperbolic equilibria are genuine nonlinear centers.

Finally,For $a\ne 0$ this has a strict constant sign throughout the simply connected plane, so the Bendixson-Dulac criterion proves that there are no periodic orbits. These results comprise the phase portrait of x dot equals two x times y minus a.