Energy Growth of Infinite Harmonic Chain under Microscopic Random Influence
aa r X i v : . [ m a t h - ph ] M a y Energy Growth of Infinite Harmonic Chain underMicroscopic Random Influence
A.A. Lykov ∗ May 5, 2020
Abstract
Infinite harmonic chains of point particles with finite range translation invariantinteraction have considered. It is assumed that the only one particle influenced bythe white noise. We studied microscopic and macroscopic behavior of the system’senergies (potential, kinetic, total) when time goes to infinity. We proved that underquite general condition on interaction potential the energies grow linearly with time onmacroscopic scale, and grow as ln( t ) on microscopic scale. Moreover it is turned outthat the system exhibit some equipartition properties in this non equilibrium settings. We consider infinite number of point particles of unit masses on R with formal Hamiltonian H ( q, p ) = X k ∈ Z p k X k,j a ( k − j ) q k q j , p k , q k ∈ R where q k = x k − kδ denotes displacement of the particle with index k from the point kδ forsome δ > p k is the impulse of the particle k . We will assume that function a ( n ) satisfiesthe following three natural conditions:1. symmetry: a ( n ) = a ( − n ),2. finite range interaction: a ( n ) has finite support, i.e. there is number r > a ( n ) = 0 if | n | > r ,3. for every λ ∈ R there is the bound ω ( λ ) = X n ∈ Z a ( n ) e inλ > . Next we will see that this condition guarantees that the energy H ( q, p ) is non-negativefor all p, q ∈ l ( Z ). ∗ Mechanics and Mathematics Faculty, Lomonosov Moscow State University, Leninskie Gory 1, Moscow,119991, Russia H C ( q, p ) = X k ∈ Z p k ω X k ( q k +1 − q k ) + ω X k q k for some non-negative constants ω , ω >
0. It is easy to see that ω ( λ ) = ω + 2 ω (1 − cos λ )and the conditions 1–3 obviously hold.We will always assume that conditions 1–3 are fulfilled.Suppose that the particle with index 0 is influenced by the white noise. Then the equa-tions of motion are: ¨ q k = − X j a ( k − j ) q j + σδ k, ˙ w t , k ∈ Z , (1)where δ k, = ( , k = 0 , , k = 0 ,σ > w t is a standard Brownian motion.In this article we study the solution q ( t ) , p ( t ) of latter equations with initial data in l ( Z ).Mainly we are interested in the energies behavior as t → ∞ at the microscopic (local) andmacroscopic (global) scales. By energies we mean three quantities: kinetic, potential andfull energy.The first and more simple objects to study are global energies. By global we mean theenergy of the whole system. We will prove that the expectation values of the energies growlinearly with time t up to terms ¯¯ o ( t ). Moreover, the mean values of the kinetic and potentialenergies grow with the same speed σ /
4. It shows that our system, in some sense, is transientand goes to infinity. Despite this fact, we see the equipartition property at this pure non-equilibrium case. Recall that accordingly to the equipartition theorem ([1], p. 136–138) forsystems with quadratic Hamiltonian at the equilibrium, the expected values of the kineticand potential energies coincide.Next, we investigate the local energies. It turns out that on microscopic level the equipar-tition property holds too. We prove that under some additional assumptions the n -th particlemean kinetic and potential energies asymptotically equal d n ln t as t → ∞ for some constants d n > d > d . In general d n depends on theindex n , but if ω ( λ ) has no critical points on interval [0 , π ] except 0 and π then d n does notdepend on n . Thus in the latter case we have the equipartition property of the energy byparticles in some asymptotic sense. The related effect was mentioned and proved in the book[6] for some class of finite linear Hamiltonian systems with random initial conditions. Theorder of growth ln t is not the same as in a finite case. Indeed, in the finite case for linearHamiltonian system when only few degrees of freedom are influenced by the white noise thelocal energies grow like t ([3], p. 68–73). Therein was also proved equipartition property ofthe local energies. The slower order of growth ln t in infinite case is natural since a part ofthe energy goes to infinity. 2nfinite harmonic chain is a quite standard object in mathematical physics. It is used forstudy various physical phenomenons: convergence to equilibrium, heat transport, hydrody-namics, etc. Most of the works dealing with systems are not far away from an equilibrium.In the present article we have considered, in some sense, the opposite situation — our systemgoes to infinity, i.e. away from equilibrium. The effect of the total energy’s growth in thecase when only few degrees of freedom are influenced by a white noise was proved in [2]for finite linear Hamiltonian systems. Seminal physical papers [4, 5] (and reference therein)are closely related to ours. In these articles authors have considered system with a discreteLaplacian at the right hand side of (1). In [5] instead of white noise there is a periodic forcesin ωt acting on the fixed particle with index 0 and additional pinning term. The authorsstudied behavior of energies and solution as time t goes to infinity. Denote p k ( t ) = ˙ q k ( t ). We say that sequences q k ( t ) , p k ( t ) , k ∈ Z of stochastic processes solvethe equation (1) if they satisfy the following system of stochastic differential equations (inItˆo sense) dq k = p k dt,dp k = − X j a ( k − j ) q j dt + σδ k, dw t , k ∈ Z . Denote the phase space of our system by L = { ψ = ( q, p ) : q ∈ l ( Z ) , p ∈ l ( Z ) } . It isevident that L is the Hilbert space. Lemma 2.1.
For all ψ ∈ L there is a unique solution ψ ( t ) of (1) with initial condition ψ such that P ( ψ ( t ) ∈ L ) = 1 for all t > . By unique we mean that if ψ ′ ( t ) is another solution of (1) such that ψ ′ (0) = ψ and ψ ′ ( t ) ∈ L almost sure for all t then ψ ( t ) and ψ ′ ( t ) are stochastically equivalent, i.e. P ( ψ ( t ) = ψ ′ ( t )) = 1 for all t > T ( t ) = 12 X k p k ( t ) , U ( t ) = 12 X k,j a ( k − j ) q k ( t ) q j ( t ) , where ψ ( t ) = ( q ( t ) , p ( t )) T is a solution of (1). Then evidently H ( t ) = H ( q ( t ) , p ( t )) = T ( t ) + U ( t ). Theorem 2.1 (Global energy behavior) . For all initial condition ψ (0) ∈ L the followingequalities hold E H ( t ) = σ t + H (0) , (2) E T ( t ) = σ t + ¯¯ o ( t ) , (3) E U ( t ) = σ t + ¯¯ o ( t ) , (4)3 s t → ∞ . Moreover the full energy has the following representation H ( t ) = σ t + H (0) + ξ ( t ) + ξ ( t ) , (5) where a Gaussian random process ξ ( t ) depends on initial conditions and can be expressedas follows ξ ( t ) = Z t f ( s ) dw s , and the random process ξ ( t ) has a representation via a multiple Itˆo integral ξ ( t ) = Z t (cid:18)Z s h ( s − s ) dw s (cid:19) dw s , where h ( t ) = σ π Z π cos( tω ( λ )) dλ,f ( t ) = σ π Z π − Q ( λ ) ω ( λ ) sin( tω ( λ )) + P ( λ ) cos( tω ( λ )) dλ,Q ( λ ) = X n q n (0) e inλ , P ( λ ) = X n p n (0) e inλ . If the initial conditions are zero, then as a consequence of presentation (5), we can writethe variance of the full energy: D H ( t ) = Z t ( t − s ) h ( s ) ds. (6)Later on we will prove that the following formula holds: D H ( t ) = ¯¯ o ( t ) , as t → ∞ . (7)We see that the mean energies of the whole system grows linearly with time. Next wewill see that locally the energies grow like logarithm of time t . Additionally we should notethat the mean kinetic and potential energies grow with the same rate: σ /
4. It turns outthat locally this picture is the same, i.e. the leading asymptotic term in the correspondingmean energies coincides. To formulate the corresponding result we need more assumptions.We will suppose that
A1) ω ( λ ) is strictly greater than zero: ω ( λ ) > λ ∈ R . A2) each critical point of ω ( λ ) is non-degenerate, it means that if ω ′ ( λ ) = 0 for some λ then ω ′′ ( λ ) = 0. 4ince a ( n ) is symmetric and has finite support, we can write ω ( λ ) = a (0) + 2 r X n =1 a ( n ) cos( nλ ) . Thus points 0 and π are critical and ω ( λ ) has a finite number of critical points on interval(0 , π ). Let us denote their by λ , . . . , λ m and put by definition λ = 0 , λ m +1 = π . A3)
The third assumption is ω ( λ j ) = ω ( λ i ) for all i = j .Let us introduce local energies T n ( t ) = p n ( t )2 , U n ( t ) = 12 X j a ( n − j ) q n ( t ) q j ( t ) , H n = T n + U n , kinetic, potential and full energy respectively. Theorem 2.2 (Local energy behavior) . Suppose assumptions A1, A2, A3 are fulfilled. Thenfor all n ∈ Z the following equalities hold: E T n ( t ) = d n ln( t ) + O (1) , (8) E U n ( t ) = d n ln( t ) + O (1) , (9) where we denote d n = σ π m +1 X k =0 cos nλ k | ω ′′ ( λ k ) | χ k , χ k = ( , k = 1 , . . . , m , k ∈ { , m + 1 } . (10)We emphasize that inf n d n >
0. Indeed, since λ = 0 , λ m +1 = π , we obtain the bound: d n > σ π (cid:18) | ω ′′ (0) | + 1 | ω ′′ ( π ) | (cid:19) > . First we prove the existence and uniqueness lemma 2.1. Let us rewrite the system (1) in amatrix form: ˙ ψ = Aψ + σg ˙ w t , (11)where a matrix A is given by the formula (in q, p decomposition of the phase space): A = (cid:18) I − V (cid:19) , (12)5 is the unit matrix, V i,j = a ( i − j ) and a vector g = (0 , e ) T , e is a vector of a standardbasis (it has all zero entities except of zero component which is equal to one). We agreevectors are the column vectors. Since a ( n ) has a compact support, V acts on l ( Z ) andhence A defines a bounded linear operator on L .Uniqueness easily follows from the linearity of system (11). Indeed, let ψ ( t ) and ψ ′ ( t ) aretwo solutions of (11) with the same initial condition. Then δ ( t ) = ψ ( t ) − ψ ′ ( t ) is a solutionof homogeneous equation: ˙ δ = Aδ and δ (0) = 0 and moreover δ ( t ) ∈ L almost surely for all t >
0. Thus the same arguments asin the classical theory of ODE in the Banach spaces (see [7]) show us that δ ( t ) = 0 almostsurely for all t >
0. So uniqueness has been proved.A solution of (11) can be found via the classical formula for the solution of inhomogeneousODE [7]: ψ ( t ) = e tA ψ (0) + σ Z t e ( t − s ) A g dw s = ψ ( t ) + ψ ( t ) , (13)where ψ ( t ) = e tA ψ (0) and ψ ( t ) = σ R t e ( t − s ) A g dw s . As we mentioned above A is abounded operator on L , and so e tA is correctly defined bounded operator on L . Therefore ψ ( t ) ∈ L for all t >
0. To prove the same statement with probability one for ψ ( t ) we needthe following lemma. Lemma 3.1.
For all t > the following formula holds: e tA = (cid:18) C ( t ) S ( t ) V S ( t ) C ( t ) (cid:19) , where C ( t ) = ∞ X n =0 ( − n t n (2 n )! V n , S ( t ) = ∞ X n =0 ( − n t n +1 (2 n + 1)! V n . Proof.
Straightforward calculation or see at [7]. Note that formal series expansion forcos tV / equals to C ( t ). The same is true for the pair V − / sin tV / and S ( t ).Denote ψ ( t ) = ( q (1) ( t ) , p (1) ( t )) T . From lemma 3.1 we have q (1) k ( t ) = σ Z t S k, ( t − s ) dw s , p (1) k ( t ) = σ Z t C k, ( t − s ) dw s . (14)We want to prove that q (1) ( t ) ∈ l ( Z ) and p (1) ( t ) ∈ l ( Z ) almost surely. The proof will bebased on the following lemma. Lemma 3.2.
For all t > the following inequalities hold: | C i,j ( t ) | v ρ t ρ (2 ρ )! e √ vt , | S i,j ( t ) | v ρ t ρ +1 (2 ρ + 1)! e √ vt , (15) where v = || V || l ( Z ) , ρ = ⌈| i − j | /r ⌉ and the number r is a radius of interaction, which isdefined in assumption 2 on the function a . By ⌈ x ⌉ we denote a ceiling function of x , i.e. theleast integer greater than or equal to x . roof. Since matrix V is translation invariant, it suffices to prove the assertion for i = k > j = 0. For all n > V n ) k, = X i ,...,i n − ∈ Z V i ,i V i ,i . . . V i n − ,i n , i = k, i n = 0 . We see that if ( V n ) k, = 0 then | i j − i j − | r for all j and thus we get : | k | = | i − i + i − i + . . . + i n − − i n | rn. And we obtain the inequality: n > k/r . Therefore the following is true: C k, ( t ) = ∞ X n > kr ( − n t n (2 n )! V n . Now we derive a bound for C k, ( t ): | C k, ( t ) | ∞ X n = ρ t n (2 n )! v n = ∞ X n = ρ ( √ vt ) n (2 n )! ∞ X n =2 ρ ( √ vt ) n n ! v ρ t ρ (2 ρ )! e √ vt . It is easy to see that S ( t ) = R t C ( s ) ds and so the second inequality in (15) follows.Everything has done to prove the existence and uniqueness lemma 2.1. Indeed, from (14)we obtain : E ( q (1) k ( t )) = σ Z t S k, ( t − s ) ds = σ Z t S k, ( s ) ds. Using inequalities (15) we have the bound: E ( q (1) k ( t )) σ Z t (cid:18) v ρ s ρ +1 (2 ρ + 1)! e √ vs (cid:19) ds σ v ρ t ρ +2 (2 ρ + 2)! e √ vt , where ρ = ⌈| k | /r ⌉ . Hence we conclude: X k ∈ Z E ( q (1) k ( t )) < ∞ and due to monotone convergence theorem we get: X k ∈ Z ( q (1) k ( t )) < ∞ with probability one. Thus q (1) ( t ) ∈ l ( Z ) almost sure for all t >
0. The similar argumentsshow that p (1) ( t ) ∈ l ( Z ) almost sure for all t >
0. Thus lemma 2.1 is proved.7 .2 Solution via the Fourier transform
Consider the Fourier transform of a solution: Q t ( λ ) = X n q n ( t ) e inλ . Simple calculation gives us: d dt Q t = − ω ( λ ) Q t + σ ˙ w t . Thus Q t ( λ ) for every λ satisfies the equation of harmonic oscillator with frequency ω ( λ )influenced by white noise. Its solution is unique and can easily be found using standardtools (see [12, 13]) : Q t ( λ ) = Q (0) t ( λ ) + Q (1) t ( λ ) , where we denote: Q (0) t = Q ( λ ) cos( tω ( λ )) + P ( λ ) sin( tω ( λ )) ω ( λ ) ,Q (1) t = σω ( λ ) Z t sin(( t − s ) ω ( λ )) dw s , and P ( λ ) = ˙ Q ( λ ). Note that Q (0) t is a solution of the homogeneous equation (with σ = 0)with initial data Q (0)0 = Q , ˙ Q (0)0 = P and Q (1) t is a solution of the inhomogeneous equationwith zero initial conditions.Using the inverse transformation we obtain: q n ( t ) = 12 π Z π e − inλ Q t ( λ ) dλ. In formula (13) we denote ψ k ( t ) = ( q ( k ) ( t ) , p ( k ) ( t )) T , k = 0 ,
1. It now follows that: q ( k ) n ( t ) = 12 π Z π e − inλ Q ( k ) t ( λ ) dλ, k = 0 , . (16)Thus we almost have proved the following lemma. Lemma 3.3.
The following formulas hold: q (0) n ( t ) = 12 π Z π e − inλ (cid:18) Q ( λ ) cos( tω ( λ )) + P ( λ ) sin( tω ( λ )) ω ( λ ) (cid:19) dλ, (17) p (0) n ( t ) = 12 π Z π e − inλ ( − Q ( λ ) ω ( λ ) sin( tω ( λ )) + P ( λ ) cos( tω ( λ ))) dλ, (18) q (1) n ( t ) = Z t x n ( t − s ) dw s , x n ( t ) = σ π Z π e − inλ sin( tω ( λ )) ω ( λ ) dλ, (19) p (1) n ( t ) = Z t y n ( t − s ) dw s , y n ( t ) = σ π Z π e − inλ cos( tω ( λ )) dλ. (20)8 roof. Formula (17) was derived above at (16). (18) obtained from (17) by differentiating.(19) follows from (16) after switching the order of integration. We can change the order ofintegration between Itˆo integral and Lebesgue one because the integrand is a deterministicsmooth function and due to “integration by parts” formula Z t f ( s ) dw s = f ( t ) w t − Z t f ′ ( s ) w s ds which is true for any smooth function f . The last formula (20) deduced from (19) and theequality dq (1) n = p (1) n dt .Let us prove here that U ( q ) = P k,j a ( k − j ) q k q j > q ∈ l ( Z ). Using operator V defined above in (12) we can write: U ( q ) = ( q, V q ) . Denote b f ( λ ) = P k f ( k ) e ikλ the Fourier transform of the sequence f ( k ). Thus, due toParseval’s theorem we obtain U ( q ) = 12 π Z π b q ( λ ) [ ( V q )( λ ) dλ = 12 π Z π | b q ( λ ) | ω ( λ ) dλ > . In the last equality we have used an obvious relation [ ( V q )( λ ) = ω ( λ ) b q ( λ ). To prove formula (2), we need find the expression for differential dH . By definition usingItˆo formula we have: dp k = 2 p k dp k + ( dp k ) = 2 p k (cid:16) − X j a ( k − j ) q j dt + σδ k, dw t (cid:17) + σ δ k, dt. Denote X t = H ( t ) = H ( ψ ( t )). Whence for the energy we obtain: dH = dX t = 12 X k dp k + 12 X k,j a ( k − j ) d ( q k q j )= − X k,j a ( k − j ) p k q j dt + 12 X k,j a ( k − j )( p k q j + p j q k ) dt + σp dw t + σ dt = σ dt + σp dw t . This is equivalent to the equality: H ( t ) = H (0) + σ t + σ Z t p ( s ) dw s = H ( ψ (0)) + σ t + σ Z t p (0)0 ( s ) + p (1)0 ( s ) dw s .
9n the last equality we have used (13). Substituting (18) and (20) into the last expressionwe get (5).Formulas (2) and (6), for the expected value and variance of the energy H ( t ) immediatelyfollows from (5). Now we prove equality (7). From (6) we have: D H ( t ) = t Z (1 − s ) h ( ts ) ds. Lemma 3.4 gives us: lim T →∞ T Z T h ( s ) ds = 0 . Therefore due to Bochner’s theorem (see [11], p. 182, th. 5.5.1) the following limit holds: Z (1 − s ) h ( ts ) ds → , as t → ∞ . So (7) has proved.Note that our derivation of (5) has some disadvantages in a place where we sum upthe infinite number of Itˆo differentials. It is not hard to legitimize this procedure using acorresponding limit or one can use a more direct approach based on lemma 3.3.Now let us prove (3). From presentation (13) and lemma 3.3 we have: E T ( t ) = 12 X n ( p (0) n ( t )) + 12 X n E ( p (1) n ( t )) = O (1) + 12 X n Z t y n ( s ) ds. The last equality is due to Itˆo isometry. Next we study the sum. Note that y n ( t ) is a Fouriercoefficient of the function cos( tω ( λ )). Hence using the Parseval’s theorem we obtain: X n Z t y n ( s ) ds = Z t X n y n ( s ) ds = σ Z t π Z π cos ( sω ( λ )) dλds == σ t Z t π Z π cos(2 sω ( λ )) dλds. In the last equality we have used a school formula cos x = [1 + cos(2 x )] /
2. Lemma 3.4 givesus: 12 Z t π Z π cos(2 sω ( λ )) dλds = O ( t − ε )for some ε >
0. Thus we have proved the formula for the mean kinetic energy (3). Equality(4) immediately follows from (2) and (3) because of relation H = T + U . This completesthe proof of Theorem 2.1. 10 .4 Local energy asymptotics Now we prove Theorem 2.2. Let us begin with kinetic energy: T n ( t ) = p n ( t )2 . Remark that according to formulas (17),(18) and Riemann – Lebesgue lemma we have thelimit: lim t →∞ ψ ( t ) = 0 . (21)Hence we obtain: E T n ( t ) = 12 E (cid:0) ( p (0) n ( t )) + ( p (1) n ( t )) + 2 p (0) n ( t ) p (1) n ( t ) (cid:1) = 12 (cid:0) ( p (0) n ( t )) + E ( p (1) n ( t )) (cid:1) = 12 E ( p (1) n ( t )) + ¯¯ o (1) , as t → ∞ . The application of (20) and Itˆo isometry yields the equality: E ( p (1) n ( t )) = Z t y n ( t − s ) ds = Z t y n ( s ) ds. (22)Since ω ( λ ) is even function, for y n ( t ) we get: y n ( t ) = σ π Z π e − inλ cos( tω ( λ )) dλ = σ π Z π cos( nλ ) cos( tω ( λ )) dλ. Lemma 3.5 gives us: y n ( t ) = σ √ t m +1 X k =0 θ k s π | ω ′′ ( λ k ) | cos( nλ k ) cos (cid:16) tω ( λ k ) + π s ( λ k ) (cid:17) + O (cid:16) t (cid:17) = 1 √ t m +1 X k =0 b k u k ( t ) + O (cid:16) t (cid:17) , where we denote: b k = σθ k s π | ω ′′ ( λ k ) | cos( nλ k ) , u k ( t ) = cos (cid:16) tω ( λ k ) + π s ( λ k ) (cid:17) For the square we get: y n ( t ) = 1 t m +1 X k =0 b k u k ( t ) + 1 t X k = j b k b j u k ( t ) u j ( t ) + O (cid:18) t √ t (cid:19) . y n ( t ) to (22): E ( p (1) n ( t )) = Z y n ( s ) ds + Z t y n ( s ) ds = O (1) + Z t y n ( s ) ds = O (1) + m +1 X k =0 b k Z t u k ( s ) s ds + X k = j b k b j Z t u k ( s ) u j ( s ) s ds. (23)Using the formula cos x = [1 + cos(2 x )] / Z t u k ( s ) s ds = Z t cos ( sω ( λ k ) + π s ( λ k )) s ds = ln t Z t cos(2 sω ( λ k ) + π s ( λ k )) s ds = 12 ln t + O (1) . The last equality follows from the fact that integrals:ci(1) = Z + ∞ cos xx dx < ∞ , si(1) = − Z + ∞ sin xx dx < ∞ (24)are converged (see [8] p. 656, 3.721) and that ω ( λ k ) > k = 0 , . . . , m + 1 due toassumption A1). To estimate the remainder term in (23) we use the formula cos( a ) cos( b ) = (cos( a + b ) + cos( a − b )): Z t u k ( s ) u j ( s ) s ds = 12 Z t cos( s ( ω ( λ k ) + ω ( λ j )) + π ( s ( λ k ) + s ( λ j ))) s ds + 12 Z t cos( s ( ω ( λ k ) − ω ( λ j )) + π ( s ( λ k ) − s ( λ j ))) s ds = O (1) . The last equality is derived from (24) using assumption A3). Substituting it to (23) weobtain: E ( p (1) n ( t )) = O (1) + ln( t )2 m +1 X k =0 b k . Thus the equality (8) has proved.Now prove equality (9) of theorem 2.2. The idea of the proof is the same as for thekinetic energy. From (21) we obtain: E U n ( t ) = 12 X j a ( n − j ) E ( q (1) n ( t ) q (1) j ( t )) + ¯¯ o (1) , as t → ∞ . (25)Equality (19) and Itˆo isometry give us: E ( q (1) n ( t ) q (1) j ( t )) = Z t x n ( t − s ) x j ( t − s ) ds = Z t x n ( s ) x j ( s ) ds = Z t x n ( s ) x j ( s ) ds + O (1) . (26)12ince sin( tω ( λ )) ω ( λ ) is an even function and due to lemma 3.5, we have for all n ∈ Z : x n ( t ) = σ π Z π cos( nλ ) sin( tω ( λ )) ω ( λ ) dλ = σ √ t m +1 X k =0 θ k s π | ω ′′ ( λ k ) | cos( nλ k ) ω ( λ k ) sin( tω ( λ k ) + π s ( λ k )) + O (cid:18) t (cid:19) = 1 √ t m +1 X k =0 e ( n ) k v k ( t ) + O (cid:18) t (cid:19) , where we denote: e ( n ) k = σθ k s π | ω ′′ ( λ k ) | cos( nλ k ) ω ( λ k ) , v k ( t ) = sin( tω ( λ k ) + π s ( λ k )) . Substitute the last expression to (26): E ( q (1) n ( t ) q (1) j ( t )) = m +1 X k =0 e ( n ) k e ( j ) k Z t v k ( s ) s ds + X k = k e ( n ) k e ( j ) k Z t v k ( s ) v k ( s ) s ds + O (1) . The same arguments as in the case of the kinetic energy give us equalities: Z t v k ( s ) s ds = 12 ln t + O (1) , Z t v k ( s ) v k ( s ) s = O (1)if k = k . Therefore we have E ( q (1) n ( t ) q (1) j ( t )) = 12 ln t m +1 X k =0 e ( n ) k e ( j ) k + O (1) . Put this expression to formula (25). Then we obtain: E U n ( t ) = D n ln t + O (1) , D n = 14 X j a ( n − j ) m +1 X k =0 e ( n ) k e ( j ) k . Now we prove that D n = d n where d n defined in (10). At the first step we change thesummation order: D n = 14 m +1 X k =0 e ( n ) k X j a ( n − j ) e ( j ) k . For the internal sum we get: X j a ( n − j ) e ( j ) k = σθ k s π | ω ′′ ( λ k ) | ω ( λ k ) X j a ( n − j ) cos( jλ k ) . X j a ( n − j ) cos( jλ k ) = X j a ( n − j ) e ijλ k + e − ijλ k nλ k ) ω ( λ k ) . Whence we have: X j a ( n − j ) e ( j ) k = ω ( λ k ) e ( n ) k . Thus we obtain: D n = 14 m +1 X k =0 ω ( λ k )( e ( n ) k ) = d n . This completes the proof of Theorem 2.2.
Lemma 3.4.
There are positive constants b, ε such that for all sufficiently large t the fol-lowing inequality holds: (cid:12)(cid:12)(cid:12)(cid:12)Z π e itω ( λ ) dλ (cid:12)(cid:12)(cid:12)(cid:12) bt − ε . (27) Proof.
Recall that ω ( λ ) = a (0) + 2 r X n =1 a ( n ) cos( nλ ) . If ω ( λ ) is strictly greater than zero (i.e. assumption A1 holds) then lemma immediatelyfollows from the stationary phase method. Indeed in that case ω ( λ ) is an analytic functionand by the stationary phase method the asymptotic of the integral at (27) is determinedby stationary points of ω ( λ ) (see [9, 10]). Since ω ( λ ) is an analytic then ω ( λ ) has a finitenumber of critical points on [0 , π ] and each of this point has finite multiplicity. Hence theinequality (27) follows from the corresponding asymptotic formulas of the stationary phasemethod.Now suppose that ω ( λ ) has zeros on [0 , π ]. Denote f ( λ ) = ω ( λ ). Since f is an analytic, ω ( λ ) has a finite number of zeros on [0 , π ]. Consider some zero z ∈ [0 , π ] of ω ( λ ) and studyintegral (27) over a small neighborhood of z . From analyticity of f follows that there is anumber n > f ( z ) = 0 , f ′ ( z ) = 0 , . . . f ( n − ( z ) = 0 , f ( n ) ( z ) = 0 . Since f is non-negative, n = 2 m is an even number. This is a well-known fact (see [9, 10]) thatin the given case there is a C ∞ -smooth one-to-one function ϕ ( y ) mapping some neighborhoodof zero, say [ − δ, δ ], to a small vicinity of z , denote it by [ z − δ ′ , z + δ ′ ], such that f ( ϕ ( y )) = y n , ϕ (0) = z. Therefore for the integral we have Z z + δ ′ z − δ ′ e itω ( λ ) dλ = Z δ − δ exp( it | y | n/ ) ϕ ′ ( y ) dy = Z δ exp( ity m )( ϕ ′ ( y ) + ϕ ′ ( − y )) dy = O (cid:0) t − /m (cid:1) . z of f there is a neighborhood of z such that theintegral over this neighborhood satisfies inequality (27) . The same statement is evidentlytrue for the critical points of f . Hence, splitting integral (27) into the integrals over suchneighborhoods and remaining part without zeros and critical points, we get the proof of(27). Lemma 3.5.
Consider the integral E f ( t ) = Z π g ( λ ) e itω ( λ ) dλ, t > for some π -periodic real-valued C ∞ ( R ) -smooth even function g . Then under the assump-tions A1) and A2) the following formula holds: E f ( t ) = 1 √ t m +1 X k =0 θ k s π | ω ′′ ( λ k ) | g ( λ k ) e itω ( λ k )+ iπ s ( λ k ) + O (cid:18) t (cid:19) (28) where θ k = ( , k = 1 , . . . , m, , k ∈ { , m + 1 } , s ( λ ) = sgn( ω ′′ ( λ )) , and λ , . . . , λ m +1 are critical points of the function ω ( λ ) introduced in assumption A2.Proof. We will use the stationary phase method (see [9, 10]). Note that ω ( λ ) = ω (2 π − λ )for all λ . Hence the only critical points of ω ( λ ) on the interval [0 , π ) are λ , . . . , λ m +1 and µ , . . . , µ m where µ j = 2 π − λ j , j = 1 , . . . , m . Recall that λ = 0 and we want to shift theinterval of integration from the boundary stationary point. Since functions g and ω are 2 π periodic, we can write E f ( t ) = Z π − δ − δ g ( λ ) e itω ( λ ) dλ, where we choose small number δ in such a way all critical points λ , . . . , λ m +1 and µ , . . . , µ m lie strongly inside the interval ( − δ, π − δ ). By stationary phase method we have the asymp-totic formula: E f ( t ) ∼ √ t m +1 X k =0 s π | ω ′′ ( λ k ) | g ( λ k ) exp n itω ( λ k ) + iπ s ( λ k ) o + 1 √ t m X k =1 s π | ω ′′ ( µ k ) | g ( µ k ) exp n itω ( µ k ) + iπ s ( µ k ) o . Since ω ( µ k ) = ω ( λ k ) , ω ′′ ( µ k ) = ω ′′ ( λ k ) , g ( µ k ) = g ( λ k ), we obtain the leading term in (28).The term O ( t − ) comes from contribution of the boundary points.15 eferences [1] K. Huang (1987) Statistical Mechanics . John Wiley and Sons, second edition.[2] A. Lykov, V. Malyshev and S. Muzychka (2013) Linear Hamiltonian systems undermicroscopic random influence.
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