Codex @codex  0

The Fourier coefficients obey ${\dot{u}}_n+(1+n^2)u_n=f_n$, soThe coefficient decay of $u_0,f$ and the additional exponential damping justify all derivatives for $t>0$ and recovery of the initial data. Thus this is a smooth periodic solution. Every transient mode decays, giving $\boxed{u(t,\cdot )\to u_f}$ uniformly, indeed with all spatial derivatives. The elliptic solution in part (i) is the attracting steady state of this forced heat equation.

Expand the quadratic functional and use periodic integration by parts:The first integral is zero because $-u_f^{\prime \prime }+u_f=f$. Thus the difference is $\frac{1}{2}\int ({\phi }^{\prime 2}+{\phi }^2)>0$ for every nonzero $\phi$. $u_f$ is the unique strict minimizer.

Write the Fourier series $f(x)={\sum }_{n\in \mathbb{Z}}{f}_n{e}^{inx}$, with $f_n=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }f(x)e^{-inx}dx$. Smoothness makes its coefficients decay faster than every inverse power. Settingtherefore gives a smooth periodic function, and termwise differentiation proves $-u_f^{\prime \prime }+u_f=f$. Conversely every solution must have these coefficients, proving uniqueness. Equivalently, a homogeneous solution has $\int (|u^{\prime }{|}^2+|u{|}^2)=0$ by integration by parts, so vanishes.

For this alternating arrival and service queue, set $\theta =\lambda (\mu +\beta )/[\mu (\lambda +\alpha )]$. Direct multiplication gives $(\theta ,\lambda /\mu )B=\theta (\theta ,\lambda /\mu )$, so the initial row is a left eigenvector and $({\pi }_{iC},{\pi }_{iW})={\theta }^{i-1}({\pi }_{1C},{\pi }_{1W})$. ConsequentlyThese nonnegative probabilities can be normalized exactly when $\theta <1$, equivalently $\mu \alpha >\lambda \beta$. Summing the geometric series givesWhen $\theta \ge 1$ no positive multiple of the invariant sequence is summable, so no stationary distribution exists. This rules out positive recurrence; absence of a stationary probability alone should not be used to distinguish null recurrence from transience.

The stationary balance equations at level zero are $(\lambda +\alpha ){\pi }_{0C}=\beta {\pi }_{0W}$ and $\beta {\pi }_{0W}=\alpha {\pi }_{0C}+\mu {\pi }_{1W}$. At every $i\ge 1$ they areThe zero-level equations give ${\pi }_{0C}=\beta {\pi }_{0W}/(\lambda +\alpha )$ and ${\pi }_{1W}=\beta \lambda {\pi }_{0W}/[\mu (\lambda +\alpha )]$. Substituting the latter in the level-one $C$ equation givesSolve the $W$ equation for ${\pi }_{i+1,W}$, then substitute it into the next $C$ equation. This yields the required row-vector recurrence with

Uniform integrability means ${\sup }_n\mathbb{E}[|X_n|1_{\{|X_n|>K\}}]\to 0$ as $K\to \infty$. In particular the first absolute moments are uniformly bounded. By the Portmanteau theorem, $\mathbb{E}|X|\le {\mathrm{liminf}}_n\mathbb{E}|X_n|<\infty$. Let $T_K(x)=\max (-K,\min (x,K))$, a bounded continuous truncation. Convergence in distribution gives $\mathbb{E}{T}_K(X_n)\to \mathbb{E}{T}_K(X)$. ThereforeThe first term tends to zero by uniform integrability, the second by integrability of $X$. Hence $\boxed{\mathbb{E}{X}_n\to \mathbb{E}X}$.

Convergence in probability means $\mathbb{P}(|X_n-X|>ϵ)\to 0$ for every $ϵ>0$. Convergence in distribution means $F_{X_n}(t)\to F_X(t)$ at every continuity point $t$ of $F_X$. The inclusions obtained by separating the event $|X_n-X|>ϵ$ giveTake lower and upper limits as $n\to \infty$, then $ϵ\downarrow 0$ at a continuity point. Both bounds tend to $F_X(t)$, proving that convergence in probability implies convergence in distribution.

There is no compact boundaryless minimal surface in ${\mathbb{R}}^3$. For an immersion $X$ of a minimal surface, the coordinate functions are harmonic: ${\Delta }_{S}X=0$. Consequently ${\Delta }_S|X{|}^2=2|{\nabla }_{S}X{|}^2=4$, because the immersion has a two-dimensional orthonormal tangent frame. Integrating over a compact surface without boundary gives zero on the left by the divergence theorem and $4\mathrm{Area}(S)>0$ on the right, a contradiction. If surfaces with boundary are allowed, a closed planar disk is a compact minimal example; the boundaryless convention matters.

A minimal surface has $H=0$, so $k_2=-k_1$ and $\boxed{K=-k_1^2\le 0}$. For a nonflat example take the catenoid parametrization $X(u,v)=(\cosh v\cos u,\cosh v\sin u,v)$. Its first fundamental form has $E=G={\cosh }^{2}v$ and $F=0$. Choosing the normal appropriately, its second fundamental form has coefficients $e=-1$, $f=0$, $g=1$. Thus its principal curvatures are $\pm {\mathrm{sech}}^{2}v$, proving $H=0$ and $\boxed{K=-{\mathrm{sech}}^{4}v<0}$.

For an oriented surface with unit normal $N$, the Gauss map sends $p$ to $N(p)\in S^2$. The shape operator is $-dN$ on the tangent plane; its eigenvalues are the principal curvatures $k_1,k_2$. The Gaussian curvature is $K=k_1{k}_2$ and the mean curvature is $H=(k_1+k_2)/2$. Reversing orientation reverses both principal curvatures and $H$, but preserves $K$. The Theorema Egregium says that Gaussian curvature is determined by the first fundamental form, so is invariant under local isometries.

A separable algebraic element has a minimal polynomial with distinct roots in an algebraic closure. A separable field extension is an algebraic extension all of whose elements are separable. In characteristic $p$, the extension ${\mathbb{F}}_p(t)/{\mathbb{F}}_p(t^p)$ is inseparable: the minimal polynomial of $t$ is $T^p-t^p$, irreducible over the smaller field but having the single root $t$ with multiplicity $p$.

If $L$ is a finite field of order $q$, every element of $L$ is a root of $T^q-T\in K[T]$, because $K$ contains its prime field. This polynomial has derivative $-1$, so it has no repeated roots. Every element's minimal polynomial divides it and is therefore separable. Hence $L/K$ is separable.

An algebraic element $\alpha$ over $K$ is a root of a nonzero polynomial in $K[T]$. Its degree is the degree of its minimal polynomial, equivalently $[K(\alpha ):K]$. Since $\alpha$ satisfies $T^2-{\alpha }^2$ over $K({\alpha }^2)$, the tower law gives $[K(\alpha ):K({\alpha }^2)]\le 2$ and makes that degree divide the odd integer $[K(\alpha ):K]$. It must therefore be one, proving $\boxed{K(\alpha )=K({\alpha }^2)}$.

For the addition assertion, let polynomial relations of degrees $r,s$ express ${\alpha }^r$ and ${\beta }^s$ in terms of lower powers. The finite-dimensional $K$-space spanned by ${\alpha }^i{\beta }^j$ for $0\le i<r$, $0\le j<s$ contains one and is closed under multiplication by $\alpha +\beta$. Thus $1,\alpha +\beta ,\dots ,(\alpha +\beta {)}^{rs}$ are linearly dependent. Their dependence is a nonzero polynomial relation over $K$, so $\alpha +\beta$ is algebraic.

Normalize $a(t_0)=1$. In the matter era, horizon crossing $c{t}_H=2\pi a(t_H)/k$ with $a=(t/t_0{)}^{2/3}$ gives $t_H/t_0=(k_0/k{)}^3$. In the radiation era, match $a=(t_{\mathrm{eq}}/t_0{)}^{2/3}(t/t_{\mathrm{eq}}{)}^{1/2}$ to obtainThe boundary mode therefore has $k_{\mathrm{eq}}=k_0(1+z_{\mathrm{eq}}{)}^{1/2}$. For $k<k_{\mathrm{eq}}$, the growing mode multiplies the primordial amplitude by $(t_0/t_H{)}^{2/3}=(k/k_0{)}^2$. For $k>k_{\mathrm{eq}}$, growth is approximately frozen until equality, then multiplied by $(t_0/t_{\mathrm{eq}}{)}^{2/3}=1+z_{\mathrm{eq}}=(k_{\mathrm{eq}}/k_0{)}^2$. Squaring these factors gives the matter power spectrumThe large-$k$ estimate omits the logarithmic radiation-era transfer correction, consistently with the stipulated approximation.

In a flat matter-dominated universe, $a\propto t^{2/3}$, $H=2/(3t)$ and $4\pi G{\bar{\rho }}_c=2/(3t^2)$. Substitution of ${\delta }_k\propto t^{\beta }$ gives $\beta (\beta -1)+(4/3)\beta -2/3=0$, whose roots are $2/3$ and $-1$. HenceDuring radiation domination, $a\propto t^{1/2}$ while ${\bar{\rho }}_c/{\bar{\rho }}_{\mathrm{total}}\propto a\ll 1$. The matter self-gravity term is suppressed relative to $H^2$, leaving approximately ${\ddot{\delta }}_k+t^{-1}{\dot{\delta }}_k=0$, with solutions constant and $\log t$. Thus there is no growing power-law mode comparable to the later $t^{2/3}$ growth. The elementary estimate in part (ii) neglects this slow logarithmic growth of matter perturbations during radiation domination.

The second-order equation is equivalent to the planar system after setting $y=\dot{x}-ax+b{x}^3$. Any nonconstant periodic solution has a maximum $M$ and minimum $m$. At an extremum $\ddot{x}=x^3-x$, so $M>1$ and $m<-1$ are impossible. Equality at either endpoint would give one of the equilibrium points, and uniqueness would force a constant solution. Therefore a nonconstant periodic orbit lies in $-1<x<1$.

If $b/a<1$, throughout this striphas the sign of $a$ whenever $x\ne 0$. A nonconstant orbit spends a positive time away from $x=0$, so its integral cannot vanish, contradicting part (b). Thus there are no nonconstant periodic solutions. The constant equilibrium solutions of course remain; as usual the question's periodic-orbit assertion excludes equilibria.

Take $H(x,y)=y^2/2+x^2/2-x^4/4$. Along a trajectory,Integrating over a periodic orbit returns $H$ to its starting value, and henceThis is the energy obstruction for a cubic planar oscillator.

The equilibrium points are $(0,0)$ and $(\pm 1,\pm (b-a))$, with correlated signs. The Jacobian matrix is $\left(\begin{array}{cc}a-3b{x}^2& 1\\ 3x^2-1& 0\end{array}\right)$. At the origin its eigenvalues satisfy ${\lambda }^2-a\lambda +1=0$. It is a focus for $0<|a|<2$, a repeated-eigenvalue node for $|a|=2$, and a node for $|a|>2$; stability is determined by the sign of $a$, with stability for $a<0$. Its Poincaré index is $+1$. At either other equilibrium the determinant is $-2$, so it is a saddle point with Poincaré index $-1$.

The matrix recurrence gives $p_{n+1}{q}_n-p_n{q}_{n+1}=(-1{)}^n$ and $q_{n+1}=a_{n+1}{q}_n+q_{n-1}$. Successive continued fraction convergents bracket $x$, henceThe infinite continued fraction is irrational: if $x=p/q$ were rational, a distinct convergent would be at distance at least $1/(q{q}_n)$, contradicting the first bound once $q_{n+1}>q$. If $x^2$ were rational, this irrational $x$ would have degree two. The Liouville approximation theorem would then give $|x-p_n/q_n|\ge C/q_n^2$. Comparing with the displayed bound would force $a_{n+1}\le 1/C$ for every $n$, contrary to the unbounded coefficients. Thus $x^2$ is irrational, by the unbounded continued-fraction coefficients exclude quadratic irrationality argument.

The Liouville approximation theorem says that for a real algebraic number $\xi$ of degree $d\ge 2$, there is a constant $C(\xi )>0$ such thatfor all integers $p$ and $q\ge 1$ with $p/q\ne \xi$. In particular a quadratic irrational cannot have arbitrarily accurate rational approximations of order $o(q^{-2})$.

Insert the Fourier mode $u_m^n=g^n{e}^{im\theta }$ into the leapfrog advection scheme. The amplification roots satisfyFor $0<\mu <1$, both roots have modulus one and remain uniformly separated by at least $2\sqrt{1-{\mu }^2}$. The two-level Fourier amplification matrices are therefore uniformly diagonalizable, and their powers stay bounded. The Parseval identity gives stability in the discrete two-norm. For $\mu >1$, some frequencies have $|\mu \sin \theta |>1$ and one root has modulus greater than one, giving instability.

For two-level stability at the leapfrog Courant boundary, at $\mu =1$ and $\theta =\pi /2$, the repeated root is $i$ and the recurrence has solutions $a_n=(A+Bn)i^n$. For example $a_0=0,a_1=1$ gives $a_n=n{i}^{n-1}$. Thus arbitrary two-level perturbations grow without a uniform power bound. The precise answer for unrestricted starting data isThe often-used root-modulus CFL condition is $\mu \le 1$, but by itself it misses this boundary Jordan growth. A startup that selects the exact translating branch at $\mu =1$ removes that growth for the selected data; it does not make the unrestricted two-level update power-bounded.