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No: pointwise convergence need not be uniform convergence. The continuous function $f(x)=x^2$ maps $(0,1)$ into itself and is strictly below the identity. Its iteration of a map givesFor every finite $n$, these values tend to $1$ as $x\uparrow 1$, so ${\sup }_{0<x<1}{f}^n(x)=1$. With $\epsilon =1/2$, choosing $x>2^{-1/2^n}$ gives $f^n(x)>1/2$. Thus no single iterate works for every starting point, although each fixed starting point tends to zero.

Fix $x\in (0,1)$ and put $x_0=x$, $x_{n+1}=f(x_n)$. Then $0<x_{n+1}<x_n\le x_0<1$. The bounded monotone sequence theorem gives a limit $\ell \in [0,x_0]$. If $\ell >0$, it lies in the domain, where continuity allows passage to the limit in the recurrence: $\ell =f(\ell )$. That is a fixed point, contrary to $f(\ell )<\ell$. Hence $\boxed{f^n(x)\to 0\text{ for every }x\in (0,1)}$. Continuity is used at a possible positive limiting point, not at the excluded endpoint zero.

Yes. The arbitrary variable-lag condition forces the Cauchy sequence property. Suppose otherwise. Then some ${\epsilon }_0>0$ has the property that, for every $N$, there exist $p,q\ge N$ with $|a_p-a_q|\ge {\epsilon }_0$. In particular, for each $n$, choose such $p,q\ge n$. By the triangle inequality, at least one satisfies $|a_p-a_n|\ge {\epsilon }_0/2$ or $|a_q-a_n|\ge {\epsilon }_0/2$. That index must be strictly greater than $n$, since the difference at $n$ itself is zero.

It follows that the following positive integer is defined for every $n$:For this particular function, $|a_{n+f(n)}-a_n|\ge {\epsilon }_0/2$ for every $n$, contradicting the assumed limit zero. Therefore $(a_n)$ is a Cauchy sequence, and the completeness of the real numbers proves it converges to a finite real limit. This is convergence from arbitrary variable-lag increments; the quantifier over every function is much stronger than the fixed-lag condition in the previous part.

No. Take $\boxed{a_n=\log n}$. For each fixed positive integer $k$,by continuity of the logarithm at $1$. Yet $a_n\to +\infty$, so $(a_n)$ is not a convergent sequence of real numbers. Vanishing differences at every fixed lag do not control the accumulation of many small increments.

Take $\boxed{a_n=(-1{)}^n}$. Then $|a_n/n|=1/n\to 0$, but $a_{n+1}-a_n=-2(-1{)}^n$ alternates between $2$ and $-2$. Thus convergence of the scaled sequence does not imply convergence of its successive differences.

Let $d_j=a_{j+1}-a_j$ and suppose $d_j\to L$. Telescoping givesTo justify the Cesàro mean limit directly, choose $J$ so that $|d_j-L|<\epsilon$ for $j\ge J$. The finite initial sum contributes at most $n^{-1}{\sum }_{j<J}|d_j-L|$, which tends to zero. The remaining terms contribute at most $\epsilon$, and replacing $(n-1)L/n$ by $L$ contributes $|L|/n$. Consequently $\boxed{a_n/n\to L}$.

The autonomous Hamiltonian is constant and positive on every trajectory, since $q\ne 0$. Hamilton's equations give $\dot{q}=2p$, $\dot{p}=2/q^3$, and henceThus $u=pq-2Ht$ is conserved. Put $s=u+2Ht$. Multiplying the energy equation by $q^2$ gives $H{q}^2=s^2+1$. The full classical path is thereforeThe sign $\eta$ labels the two disconnected half-lines and is fixed by initial data. Differentiating verifies both Hamilton equations. The inverse-square potential prevents the path from crossing zero.

The Poisson bracket in canonical coordinates isUsing Hamilton's equations ${\dot{q}}_i=H_{p_i}$ and ${\dot{p}}_i=-H_{q_i}$ in the chain rule yieldsThe partial time derivative holds phase-space coordinates fixed; the total derivative follows the actual trajectory.

Set $H=({\dot{x}}^2+x^2)/2$. ThenOn an unperturbed circle write $x=r\cos t$, $\dot{x}=-r\sin t$. The supplied integrals giveEquivalently, the leading averaged amplitude equation isWith $s=r^2$, its nonzero stationary amplitudes solve $\beta s^2-2\alpha s+8=0$. If ${\alpha }^2>8\beta$, both roots are positive and simple:The averaged drift is positive below $r_{-}$, negative between $r_{-}$ and $r_{+}$, and positive above $r_{+}$. Thus the inner orbit is attracting and the outer orbit repelling, with leading displacement amplitudes $r_{-}$ and $r_{+}$ and period $2\pi +O(ϵ)$. The simple zeros of the Poincaré return map persist for sufficiently small $ϵ$.

If ${\alpha }^2<8\beta$, the leading quadratic is strictly positive for all $s\ge 0$; the energy increases on every nonzero unperturbed orbit, so the energy balance method yields no periodic orbits in any fixed compact range of nonzero amplitudes for sufficiently small $ϵ$. This perturbative calculation does not itself establish a uniform global exclusion at amplitudes growing as $ϵ$ tends to zero. At equality, the leading drift has a double zero at $r^2=\alpha /\beta$. It is positive on both sides, with a semistable candidate at that amplitude. First-order averaging alone cannot decide whether the exact system has zero, one, or two nearby periodic orbits at equality: higher-order drift determines the unfolding of this degenerate zero. The original PDF has ${\alpha }^2=8\beta$; the extra exponent on $\beta$ in the converted TeX is an OCR error.

For a nearly Hamiltonian system, the unperturbed Hamiltonian is conserved and its level curves typically contain a family of periodic orbits. A weak perturbation changes the energy slowly. The energy balance method integrates $\dot{H}$ over one unperturbed period to obtain the leading energy change $\Delta H(H)$. A zero of this change selects a possible persistent periodic orbit. A simple zero persists for sufficiently small perturbation by a Poincaré return map and the implicit function theorem. A negative slope of the averaged energy drift gives attraction; a positive slope gives repulsion. A multiple zero requires higher-order analysis, and a zero-free leading drift rules out periodic motion in the corresponding perturbative energy range.

Let $K_N=\{k\in {\mathbb{Z}}^2:|k_1|,|k_2|\le N\}$, use $u_N={\sum }_{k\in K_N}{u}_k{e}^{i\pi k\cdot x}$ and impose $u_0=0$. The coefficient has $a_0=3$ and $a_{\pm e_1}=a_{\pm e_2}=1/2$, with all other coefficients zero. Fourier differentiation and convolution giveConsequently the explicit Galerkin system for each $k\in K_N\setminus \{0\}$ iswhere coefficients outside $K_N$ and $u_0$ are zero. The forcing is $f_{e_1}=f_{e_2}=1/(2i)$, $f_{-e_1}=f_{-e_2}=-1/(2i)$ and zero elsewhere. The $k=0$ equation is $0=0$ and is replaced by normalization. For a real solution $u_{-k}=\overline{u_k}$, so the system can equivalently be written in real sine/cosine coordinates.

Since $a\ge 1$, the negative of this operator has quadratic form $\int a|\nabla u_N{|}^2$, strictly positive on nonzero zero-mean trigonometric polynomials. Thus the finite linear system is invertible. This construction multiplies the coefficient and derivative before projecting; it retains the explicit nearest-neighbour mode coupling rather than dividing the forcing by $a$ pointwise.

For period two in each direction, use Fourier coefficientsIntegration of the periodic Poisson equation gives the necessary compatibility condition $f_0=0$. With it, the normalization fixes $u_0=0$ and every other Fourier mode gives $-{\pi }^2|k{|}^2{u}_k=f_k$. The Fourier-Galerkin approximation is thereforeOne may evaluate the coefficients by a discrete Fourier transform for a pseudospectral implementation, with the corresponding quadrature/aliasing error distinguished from truncation.

Convergence with spectral accuracy means faster decay than every fixed algebraic power of $N$. A smooth periodic extension gives this rate; an analytic periodic extension to a complex strip gives the stronger exponential rate $O(e^{-cN})$ on a slightly narrower strip. Division by $|k{|}^2$ preserves that Fourier decay, so the normalized solution has the same type of spectral convergence.

Two qualifications are required by the literal source assumptions. The mean-zero condition on $f$ was not stated: $f\equiv 1$ meets the printed analyticity and endpoint equalities but has no periodic solution. Also endpoint equality of $f$ alone does not make its periodic extension analytic. For example $f(x,y)=x^2-1/3$ is analytic, has matching endpoint values and zero mean, but its nonzero Fourier coefficients are $f_{(k,0)}=2(-1{)}^k/({\pi }^2{k}^2)$. Then $u_{(k,0)}=-2(-1{)}^k/({\pi }^4{k}^4)$, with only algebraic decay. Thus the standard spectral-speed conclusion is valid under the intended analytic-periodic torus assumption and compatibility, not for every function satisfying only the displayed endpoint values.

Use reduced variables $t=T/T_c$, $v=V/V_c$ and $P=p/p_c$, so $P=8t/(3v-1)-3/v^2$. Write $t=1-\tau$, $v=1+s$. Near the critical point,Set $s=\sqrt{\tau }z$. The leading departure from the coexistence pressure is ${\tau }^{3/2}(6z-3z^3/2)$. Equal pressure at the two outer endpoints together with the Maxwell equal-area condition selects the symmetric pair $z_l=-2$, $z_g=2$, with coexistence pressure $1-4\tau +O({\tau }^2)$. To see why, a horizontal line through the odd cubic has zero signed area between its outer equal-height roots only at zero height: the signed area is strictly decreasing with line height, and symmetry makes it zero there. ThusThis proves the coexistence exponent one half.

Along the critical isotherm $T=T_c$ itself, the expansion instead givesThe cubic scaling therefore has exponent three, with a negative coefficient. A fixed noncritical isotherm does not pass through the critical point, so that qualification is essential.

At a fixed subcritical temperature, coexistence requires equal pressure and equal chemical potential of the liquid and gas. The Maxwell equal-area construction chooses $p_0$ and the two outer intersection volumes $V_l,V_g$ so thatIndeed $d\mu =(V/N)dp$, and integration by parts gives $N({\mu }_g-{\mu }_l)=p_0(V_g-V_l)-{\int }_{V_l}^{V_g}p(V)dV$, so the equal-area condition is exactly equality of chemical potentials.

Let $V_{s,l}<V_{s,g}$ be the two spinodal volumes where $p^{\prime }(V)=0$. A homogeneous phase is mechanically unstable on $V_{s,l}<V<V_{s,g}$, since $p^{\prime }>0$ gives negative compressibility. The negative-slope branches $V_l<V<V_{s,l}$ and $V_{s,g}<V<V_g$ are metastable homogeneous liquid and gas respectively: they resist small perturbations but a two-phase state has lower free energy. The single-phase stable branches are $V<V_l$ and $V>V_g$. Throughout the coexistence interval $[V_l,V_g]$, the equilibrium state is a mixture at pressure $p_0$ with its phase fractions chosen to produce the total volume. The preceding figure labels both the formal homogeneous branches and the equilibrium coexistence segment.

In the Van der Waals equation, $b>0$ accounts for excluded volume per particle, while $a>0$ accounts for attractive interactions that reduce the pressure relative to a hard-core ideal gas. Writing $v=V/N$ gives $p=k_{B}T/(v-b)-a/v^2$. At the critical point the first two volume derivatives vanish:Dividing gives $v_c=3b$ and thenAbove $T_c$ the isotherms decrease monotonically; at $T_c$ there is a horizontal inflection; below $T_c$ the formal curve develops a local minimum and maximum. The physically equilibrated isotherm replaces part of that loop by a constant-pressure coexistence segment.

The Gibbs free energy is $G=U-TS+pV$. For an extensive thermodynamic system, scaling the amount of matter at fixed pressure, temperature and composition scales $G$ by the same factor, hence $G(T,p,N)=N\mu (T,p)$. Euler's relation $U=TS-pV+\mu N$ gives the same conclusion. This uses the ordinary thermodynamic-limit extensivity assumption, rather than a universal claim about finite-size or long-range-interacting systems.

The fundamental thermodynamic identity gives $dG=-SdT+Vdp+\mu dN$. Comparing with $d(N\mu )$ at fixed $N$ yieldsThus the chemical potential changes with pressure at a rate equal to the volume per particle.

Convexity and the quadratic growth bound imply $|F_u(v,x)|\le C(1+|v|)$ uniformly in $x$. Indeed bound the derivative between the two secant slopes with increments $\pm (1+|v|)$ and use the upper and lower bounds on $F$. This makes $F_u(u,x)$ square integrable and justifies differentiating the energy in every $H_0^1$ direction by domination. At the minimizer,Thus the Euler-Lagrange equation is the elliptic boundary probleminitially in the weak sense. The zero trace is part of the variational space, not an additional unconstrained endpoint variation.

For $u\in H_0^1(\Omega )$, both $u$ and its weak gradient are square integrable. The assumed bounds give $|F(u,x)|\le K(1+|u{|}^2)$, so the potential term is integrable and $E(u)$ is finite. Alsowhich is bounded below and coercive in $H_0^1$ by the Poincaré inequality.

For distinct $u,v$ and $0<t<1$, convexity of $F$ and expansion of the squared gradient giveThe last norm is nonzero: a zero weak gradient makes $u-v$ constant, and its zero trace forces that constant to vanish. Thus $E$ is strictly convex.

Weak lower semicontinuity means that $u_j\rightharpoonup u$ in $H_0^1$ implies $E(u)\le {\mathrm{liminf}}_{j}E(u_j)$. A minimizing sequence is bounded by coercivity. Reflexivity of this Hilbert space gives a weakly convergent subsequence with limit still in $H_0^1$. The assumed weak lower semicontinuity makes that limit a minimizer, and strict convexity makes it unique. Hence $\boxed{E\text{ has exactly one minimizer in }{H}_0^1(\Omega )}$.