Long Time Behavior of Infinite Harmonic Chain with l 2 Initial Conditions
aa r X i v : . [ m a t h - ph ] M a y Long Time Behavior of Infinite Harmonic Chain with l Initial Conditions
A.A. Lykov ∗ M.V. Melikian ∗ May 5, 2020
Abstract
We consider infinite harmonic chain with l -initial conditions and deterministicdynamics (no probability at all). Main results concern the question when the solutionwill be uniformly bounded in time and space in the l ∞ -norm. Consider a countable system of point particles with unit masses on R with coordinates { x k } k ∈ Z and velocities { v k } k ∈ Z . We define formal energy (hamiltonian) by the followingformula: H = X k ∈ Z v k ω X k ∈ Z ( x k ( t ) − ka ) + ω X k ∈ Z ( x k ( t ) − x k − ( t ) − a ) , with parameters a > , ω > , ω >
0. Particle dynamics is defined by the infinite systemof ODE: ¨ x k ( t ) = − ∂H∂x k = − ω ( x k ( t ) − ka ) + ω ( x k +1 ( t ) − x k ( t ) − a ) − ω ( x k ( t ) − x k − ( t ) − a ) , k ∈ Z (1)with initial conditions x k (0) , v k (0). The equilibrium state (minimum of the energy) is x k = ka, v k = 0 , k ∈ Z . This means that if the initial conditions are in the equilibrium state then the system willnot evolve, i.e. x k ( t ) = ka, v k ( t ) = 0 for all t >
0. Let us introduce deviation variables: q k ( t ) = x k − ka, p k ( t ) = ˙ q k ( t ) = v k ( t ) . ∗ Mechanics and Mathematics Faculty, Lomonosov Moscow State University, Leninskie Gory 1, Moscow,119991, Russia q (0) = { q k (0) } k ∈ Z ∈ l ( Z ) , p (0) = { p k (0) } k ∈ Z ∈ l ( Z ). In thepresent article we study the long time behavior of q k ( t ) depending on initial conditions andparameters a, ω , ω . Namely, we are interested in the uniform boundedness (in k and t ),the order of growth (in t ) and exact asymptotic behavior of q k ( t ).It is easy to see that q k ( t ) satisfies the following system of ODE:¨ q k = − ω q k + ω ( q k +1 − q k ) − ω ( q k − q k − ) , k ∈ Z . (2)The system of coupled harmonic oscillators (2) and its generalizations is a classical object inmathematical physics. The existence of solution and its ergodic properties were studied in[7]. There has been an extensive research of convergence to equilibrium for infinite harmonicchain coupled with a heat bath [1, 6, 8, 9]. The property of uniform boundedness (by time t and index k ) is crucial in some applications. For instance, uniform boundedness in finiteharmonic chain allows to derive Euler equation and Chaplygin gas without any stochastics(see [2]). Uniform boundedness of a one-side non-symmetrical harmonic chain play importantrole in some traffic flow models [3]. We should note some physical papers [10, 11, 12]. Themost closely related works to ours are [4, 5], where the author studied weighted l norms ofinfinite harmonic chains, whereas our main interest is a max-norm.The paper is organized as follows. Section 2 contains definitions and formulation of themain results, remainder sections contain detailed proofs. Lemma 2.1. If q (0) , p (0) ∈ l ( Z ) then there exists unique solution q ( t ) , p ( t ) of (2) whichbelongs to l ( Z ) , i.e. q ( t ) , p ( t ) ∈ l ( Z ) for all t > .Proof. This assertion is well known (see [7, 13, 14]), and easily follows from the boundednessof the operator W on l ( Z ):( W q ) k = − ω q k + ω ( q k +1 − q k ) − ω ( q k − q k − ) . The first question of our interest is uniform boundedness (in k and time t >
0) of | q k ( t ) | .Define the max-norm of q k ( t ): M ( t ) = sup k | q k ( t ) | . We shall say that the system has the property of uniform boundedness if:sup t > M ( t ) < ∞ . Theorem 2.1.
The following assertions hold: . If ω > , then: sup t > M ( t ) < ∞ .
2. If ω = 0 then we have the following results.(a) For all t > the following inequality holds: M ( t ) √ ω || p (0) || √ t + || q (0) || . (3) (b) Suppose that X k =0 | p k (0) | ln | k | < ∞ . (4) Then there is a constant c > such that for all t > : M ( t ) √ ω π | P | ln( t ) + || q (0) || + c, P = X k p k (0) . (c) For all δ > / there exists at least one initial condition q (0) = 0 , p (0) ∈ l ( Z ) such that lim t →∞ q ( t ) √ t ln δ t = Γ( δ ) √ ω > where Γ is the gamma function. From the case 2.a we see that if ω = 0 and initial velocities of the particles are all zero,then | q k ( t ) | are uniformly bounded.The assertions 2.c is an attempt to answer the question on the accuracy in the basicinequality (3) from 2.a with respect to the rate of growth in t . Next we will formulate theorems concerning asymptotic behavior of q k ( t ) in several cases.Define Fourier transform of the sequence u = { u k } ∈ l ( Z ): b u ( λ ) = X k u k e ikλ , λ ∈ R . Note that b u ( · ) ∈ L ([0 , π ]), i.e. Z π | b u ( λ ) | dλ = 2 π X k | u k | < ∞ . Further on we will use the Fourier transform of the initial conditions: Q ( λ ) = d q (0)( λ ) , P ( λ ) = d p (0)( λ ) . For complex valued functions f, g on R and constant c ∈ C we will write f ( x ) ≍ c + g ( x ) / √ x , if f ( x ) = c + g ( x ) / √ x + ¯¯ o (1 / √ x ) as x → ∞ .3 heorem 2.2 ( ω > ) . Suppose that ω > and Q, P are of class C n ( R ) for some n > .Then1. For any fixed t > we have q k ( t ) = O ( k − n ) .2. For any fixed k ∈ Z and t → ∞ we have the following asymptotic formula: q k ( t ) ≍ √ t (cid:0) C cos( ω ( t )) + S sin( ω ( t ))+ ( − k C cos( ω ( t )) + ( − k S sin( ω ( t ))) , where C = 1 ω r ω π Q (0) , S = 1 ω ω r ω π P (0) C = 1 ω r ω ′ π Q ( π ) , S = 1 ω ω ′ r ω ′ π P ( π ) ,ω ( t ) = tω + π , ω ( t ) = tω ′ − π , ω ′ = q ω + 4 ω .
3. Let t = β | k | , β > and k → ∞ . Put γ ( β ) = β ω − − βω . (a) If γ ( β ) > then q k ( t ) ≍ p | k | (cid:16) F + k [ Q ] − i F − k h P ( λ ) ω ( λ ) i(cid:17) where we introduce the following functionals for a complex valued function g ( λ ) defined on the real line: F ± k [ g ] = c + ( g ( µ + ) e iω + ( k ) ± g ( − µ + ) e − iω + ( k ) )+ c − ( g ( µ − ) e iω − ( k ) ± g ( − µ − ) e − iω − ( k ) ) ,ω ± ( k ) = k ( µ ± + βω ( µ ± )) ± π k ) ,c ± = 12 r βω ( µ ± )2 π ∆ ,µ ± = − arccos 1 β ω (1 ± ∆) , ∆ = q ( β ω − − β ω ,ω ( λ ) = q ω + 2 ω (1 − cos λ ) . (b) If γ ( β ) = 0 and n > then q k ( t ) = O ( k − ) .(c) if γ ( β ) < then q k ( t ) = O ( k − n ) for n defined above. z ∈ l ( Z ) for b z ∈ C n ( R ) is X k | k | n | z k | < ∞ . Thus if the following series converge for some n > X k | k | n | q k (0) | < ∞ , and X k | k | n | p k (0) | < ∞ , then Theorem 2.2 holds. Theorem 2.3 ( ω = 0 ) . Suppose that ω = 0 and Q, P ∈ C n ( R ) , n > then1. For any fixed t > we have q k ( t ) = O ( k − n ) .2. For any fixed k ∈ Z and t → ∞ one has: q k ( t ) ≍ P (0)2 ω + ( − k √ t (cid:16) C cos (cid:16) ω t − π (cid:17) + S sin (cid:16) ω t − π (cid:17)(cid:17) , where C = 1 √ πω Q ( π ) , S = 12 ω √ πω P ( π ) . If ω = 0 then it makes sense to consider the displacement variables: z k ( t ) = x k +1 ( t ) − x k ( t ) − a, u k ( t ) = ˙ z k = v k +1 ( t ) − v k ( t ) , k ∈ Z . Suppose that { z k (0) } k ∈ Z ∈ l ( Z ) and { u k (0) } k ∈ Z ∈ l ( Z ). It is easy to see that z k ( t ) solves(2) with ω = 0. Consequently all formulated assertions for q k ( t ) in the case ω = 0 hold forvariables z k ( t ). It is interesting to note that the quantity P from item 2.b of theorem 2.1 interms of variables z, u equals P = X k u k (0) = lim n →∞ ( v n (0) − v − n (0)) . So if X k =0 | u k | ln | k | < ∞ and initial velocities of the “right” particles and “left” particles are equal, i.e. lim n →∞ ( v n (0) − v − n (0)) = 0, then from the case 2.b of theorem 2.1 follows uniform boundedness of displace-ments z k ( t ). 5 Proofs
Let us introduce the energy (hamiltonian): H = H ( q, p ) = X k p k ω X k q k + ω X k ( q k − q k − ) . One can easily check that the energy is conserved under the dynamics (2). It means that H ( q ( t ) , p ( t )) = H ( q (0) , p (0)) for all t > q ( t ) , p ( t ) solves (2). If ω > k | q k ( t ) | p H ( q ( t ) , p ( t )) ω the uniform boundedness of | q k ( t ) | follows, i.e.sup t > sup k ∈ Z | q k ( t ) | < ∞ and so item (1) of Theorem 2.1 is proved.Let us analyze the Fourier transform of the solution (2): d q ( t )( λ ) = X k q k ( t ) e ikλ . The inverse transformation is given by the formula: q k ( t ) = 12 π Z π d q ( t )( λ ) e − ikλ dλ, k ∈ Z . (5) Lemma 3.1.
The solution of (2) can be expressed as q k ( t ) = Q k ( t ) + P k ( t ) , (6) where Q k ( t ) = 12 π Z π Q ( λ ) cos( tω ( λ )) e − ikλ dλ,P k ( t ) = 12 π Z π P ( λ ) sin( tω ( λ )) ω ( λ ) e − ikλ dλ,Q ( λ ) = d q (0)( λ ) , P ( λ ) = d p (0)( λ ) ,ω ( λ ) = q ω + 2 ω (1 − cos( λ )) . Proof.
Using (2) we obtain : d dt d q ( t )( λ ) = − ω d q ( t )( λ ) + ω X k q k +1 ( t ) e ikλ + ω X k q k − ( t ) e ikλ − ω d q ( t )( λ )6 − ω ( λ ) d q ( t )( λ ) . Thus d q ( t )( λ ) for fixed λ is coordinate of the harmonic oscillator with the frequency ω ( λ ) andso the solution of equation for d q ( t )( λ ) is d q ( t )( λ ) = d q (0)( λ ) cos( tω ( λ )) + d p (0)( λ ) sin( tω ( λ )) ω ( λ )) . From this equality and the formula of the inverse transformation (5) lemma follows.
The case ω > ω = 0. We will use therepresentation (6) and some upper bounds for Q k , P k .At first we will prove the part 2.a of Theorem 2.1. From the Cauchy – Bunyakovsky –Schwarz inequality we obtain the inequalities: | Q k ( t ) | s π Z π | Q ( λ ) | dλ s π Z π cos ( tω ( λ )) dλ s π Z π | Q ( λ ) | dλ = || q (0) || , | P k ( t ) | s π Z π | P ( λ ) | dλ s π Z π sin ( tω ( λ )) ω ( λ )) dλ = || p (0) || s π Z π sin ( tω ( λ )) ω ( λ )) dλ. In the case ω = 0 one has ω ( λ ) = p ω (1 − cos( λ )) = 2 ω sin( λ/
2) and I = Z π sin ( tω ( λ )) ω ( λ ) dλ = 2 Z π sin (2 ω t sin( u ))(2 ω sin( u )) du = 4 Z π/ sin (2 ω t sin( u ))(2 ω sin( u )) du. Substituting x = sin u in the last integral we get: I = 1 ω Z sin (2 ω tx ) x √ − x dx = 1 ω (cid:18)Z / √ . . . dx + Z / √ . . . dx (cid:19) . (7)The integrals in the latter formula will be estimated separately. Z / √ sin (2 ω tx ) x √ − x dx √ Z / √ sin (2 ω tx ) x dx ω t √ Z √ ω t sin ( x ) x dx ω t √ Z ∞ sin ( x ) x dx = 2 ω t √ π πω t √ . One can find the value of the last integral in [15], p. 713, 3.821 (9). For the second integralin (7) we have Z / √ sin (2 ω tx ) x √ − x dx ω t Z / √ | sin(2 ω tx ) | x √ − x dx ω t √ Z / √ √ − x dx = 2 ω t √ π πω t √ . Thus we obtain the inequality for P k : | P k ( t ) | s πω (cid:18) πω t √ πω t √ (cid:19) || p (0) || r tω || p (0) || . This proves the case 2.a of Theorem 2.1.Next we will check assertion 2.b. Condition (4) implies p (0) ∈ l ( Z ) and consequently P ( λ ) = d p (0)( λ ) is a bounded continuous function on R . We have the following representationof P k ( t ): P k ( t ) = 12 π Z π P ( λ ) − P (0) ω ( λ ) sin( tω ( λ )) e − ikλ dλ + P (0)2 π Z π sin( tω ( λ )) ω ( λ ) e − ikλ dλ. (8)We estimate the first integral using condition (4): I k = (cid:12)(cid:12)(cid:12)(cid:12)Z π P ( λ ) − P (0) ω ( λ ) sin( tω ( λ )) e − ikλ dλ (cid:12)(cid:12)(cid:12)(cid:12) Z π (cid:12)(cid:12)(cid:12) P ( λ ) − P (0) ω ( λ ) (cid:12)(cid:12)(cid:12) dλ X j | p j (0) | Z π (cid:12)(cid:12)(cid:12) e ijλ − ω sin( λ/ (cid:12)(cid:12)(cid:12) dλ. On the other hand Z π (cid:12)(cid:12)(cid:12) e ijλ − λ/ (cid:12)(cid:12)(cid:12) dλ = Z π (cid:12)(cid:12)(cid:12) sin( jλ/ λ/ (cid:12)(cid:12)(cid:12) dλ = 2 Z π (cid:12)(cid:12)(cid:12) sin( ju )sin u (cid:12)(cid:12)(cid:12) dλ = 4 Z π/ (cid:12)(cid:12)(cid:12) sin( ju )sin u (cid:12)(cid:12)(cid:12) dλ (1) π Z π/ (cid:12)(cid:12)(cid:12) sin( ju ) u (cid:12)(cid:12)(cid:12) dλ (2) π (ln | j | + c ) .
8n the inequality (1) we have used the fact that sin x > (2 /π ) x for x ∈ [0; π/
2] and (2) followsfrom Lemma 3.2 below. Thus we have obtained I k πω X j =0 | p j (0) | (ln | j | + c ) < ∞ . Further we will estimate the second integral in (8): J k = (cid:12)(cid:12)(cid:12)(cid:12)Z π sin( tω ( λ )) ω ( λ ) e − ikλ dλ (cid:12)(cid:12)(cid:12)(cid:12) Z π | sin(2 ω t sin u ) | ω sin u du = 2 ω Z | sin(2 ω tx ) | x √ − x dx ω (cid:18)Z / √ . . . dx + Z / √ . . . dx (cid:19) ω ( √ t + c )for some constant c > t > k . In the latter inequality we againhave applied lemma 3.2.Finally from (8) and due to the bounds for I k , J k we obtain: | P k ( t ) | √ ω π | P (0) | ln t + c for some constant c > t > k . This completes the proof of the part2.b. Lemma 3.2.
For all b > and all t > there is a constant c > such that for any t > t the following inequality holds: Z b | sin( tx ) | x dx ln t + c. Proof.
Substituting tx = y we get: Z b | sin( tx ) | x dx = Z tb | sin y | y dy = Z t b | sin y | y dy + Z tbt b | sin y | y dy c + Z tbt b y dy = ln − ln t + c. This proves the assertion.Finally we will prove part 2.c of Theorem 2.1. We will construct the required initialcondition in two steps. At first step for any 0 < α < / α ) such that the corresponding solution satisfies lim t →∞ q ( α )0 ( t ) /t α >
0. Atthe next step we will integrate these initial conditions with an appropriate weight and provethat the resulting function gives us the answer.9irstly we prove the assertion if ω = 1 /
2. Consider initial conditions q ( α ) (0) , p ( α ) (0) withthe following Fourier transforms: Q ( α ) ( λ ) = 0 , P ( α ) ( λ ) = a α ( | ω ( λ ) | ) α = a α | sin λ/ | α where 0 < α < / a α > || P ( α ) ( λ ) || L ([0 , π ]) = Z π | P ( α ) ( λ ) | dλ = 1 . Exact formula for a α will be given below. It is obvious that P ( α ) ( λ ) ∈ L ([0 , π ]). So thecorresponding initial conditions q ( α ) k (0) = 0 , p ( α ) k (0) = 12 π Z π P ( α ) ( λ ) e − ikλ dλ lie in l ( Z ). From (6) we have q ( t ) = q ( α )0 ( t ) = P ( t ) = a α π Z π sin( t sin λ/ | sin λ/ | α +1 dλ. Lemma 3.3.
For all t > the following equality holds: q ( α )0 ( t ) = ϕ ( α ) t α + R ( α, t ) , ϕ ( α ) = 2 a α Γ(1 − α ) πα cos πα , where for the remainder term R we have | R ( α, t ) | a α (cid:16) t (cid:17) , where Γ is the gamma function.Proof. From the definition we have q ( α )0 ( t ) = a α π Z π sin( t sin λ )sin α +1 λ dλ = a α π (cid:16) I h , π i + I h π , π i + I h π , π i(cid:17) , where I [ a, b ] = Z ba sin( t sin λ )sin α +1 λ dλ. The second integral can easily be estimated: (cid:12)(cid:12)(cid:12) I h π , π i(cid:12)(cid:12)(cid:12) Z π/ π/ dλ (sin λ ) α +1 π √ α +1 π − / < . The third integral I [3 π/ , π ] evidently equals to the first one I [0 , π/
4] (to see this it issufficient to make the substitution x = π − λ ). Let us evaluate the first integral. Substituting x = sin λ we have: I h , π i = Z / √ sin( tx ) x α +1 √ − x dx (9)10 Z / √ sin( tx ) x α +1 dx + Z / √ (cid:16) sin( tx ) x α +1 √ − x − sin( tx ) x α +1 (cid:17) dx. Due to the mean-value theorem for all 0 s / (cid:12)(cid:12)(cid:12) √ − s − (cid:12)(cid:12)(cid:12) s max ≤ θ ≤ s √ − θ ) s (cid:0)p − / (cid:1) = s √ . Consequently for the second integral in (9) we have the estimates: (cid:12)(cid:12)(cid:12)(cid:12)Z / √ (cid:16) sin( tx ) x α +1 √ − x − sin( tx ) x α +1 (cid:17) dx (cid:12)(cid:12)(cid:12)(cid:12) Z / √ | sin( tx ) | x α +1 (cid:12)(cid:12)(cid:12) √ − x − (cid:12)(cid:12)(cid:12) dx Z / √ √ x x α +1 dx = √ − α √ − α √ − α < . The first integral in (9) can be expressed as: Z / √ sin( tx ) x α +1 dx = t α Z t/ √ sin uu α +1 du = t α Z + ∞ sin uu α +1 du − t α Z + ∞ t/ √ sin uu α +1 du. In the latter formula the first integral is Bohmer integral (generalized Fresnel integral) whichvalue can be found in [15], p. 648, 3.712: Z ∞ sin uu α +1 du = 1 α (1 − α ) Z ∞ cos y / (1 − α ) dy = 1 α (1 − α ) Γ(1 − α ) sin( π (1 − α ))1 / (1 − α ) = Γ(1 − α ) α cos πα . Integrating by parts we estimate the remainder term: (cid:12)(cid:12)(cid:12)(cid:12)Z + ∞ t/ √ sin uu α +1 du (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) cos( t/ √ t/ √ α +1 − (1 + α ) Z ∞ t/ √ cos uu α +2 du (cid:12)(cid:12)(cid:12)(cid:12) t α +1 + (1 + α ) Z ∞ t/ √ u α +2 du t α +1 . These inequalities complete the proof.For any 0 < ε < / w ε ( α ) = 1 ϕ ( α ) 1(1 / − α ) / − ε where ϕ ( α ) is defined in Lemma 3.3. 11 emma 3.4. w ε ( α ) is absolutely integrable w.r.t. α on [0 , / : Z / w ε ( α ) dα < ∞ . Proof.
Now we need the exact expression of a α :1 a α = Z π λ ) α dλ = 4 Z π/ α x dx = 2B (cid:16) − α , (cid:17) = 2 Γ( − α )Γ( )Γ(1 − α )where B is the beta function ([15], p. 610, 3.621). Hence a α = s Γ(1 − α )2 √ π Γ( − α ) . It is well-known that Γ( z ) = 1 /z + O (1) as z →
0. Thus w ε ( α ) has the only one singularpoint on [0 , /
2] at α = 1 / w ε ( α ) ∼ c (1 / − α ) − ǫ as α → / c . So w ε ( α ) is absolutely integrable on [0 , / Q ( ε ) ( λ ) = 0 , ˜ P ( ε ) ( λ ) = Z / w ε ( α ) P ( α ) ( λ ) dα. The latter integral we understand in the following sense:˜ P ( ε ) ( λ ) = lim δ → Z / − δ w ε ( α ) P ( α ) ( λ ) dα. (10)Since L -norm of P ( α ) ( · ) equals to one and w ε ( α ) is absolutely integrable, the limit in (10)exists and, moreover, one has the inequality: || ˜ P ( ε ) ( · ) || L ([0 , π ]) Z / w ε ( α ) dα. Thus the corresponding to ˜ Q ( ε ) ( λ ) , ˜ P ( ε ) ( λ ) initial conditions:˜ q ( ε ) k (0) = 0 , ˜ p ( ε ) k (0) = 12 π Z π ˜ P ( ε ) ( λ ) e − ikλ dλ (11)12ie in l ( Z ). Denote ˜ q ( ε ) k ( t ) , ˜ p ( ε ) k ( t ) , k ∈ Z the solution of (2) with initial condition (11). Dueto the lemmas 3.1 and 3.3 and Fubini – Tonelli theorem we have:˜ q ( ε )0 ( t ) = 12 π Z π sin( t sin λ )sin λ ˜ P ( ε ) ( λ ) dλ = Z / w ε ( α ) 12 π Z π sin( t sin λ )sin λ P ( α ) ( λ ) dλdα = Z / w ε ( α ) q ( α )0 ( t ) dα = Z / ϕ ( α ) 1(1 / − α ) / − ε (cid:0) ϕ ( α ) t α + R ( α, t ) (cid:1) dα = Z / t α (1 / − α ) / − ε dα + Z / R ( α, t ) ϕ ( α ) 1(1 / − α ) / − ε dα. The remainder term in the latter formula can be easily estimated: (cid:12)(cid:12)(cid:12)(cid:12)Z / R ( α, t ) ϕ ( α ) 1(1 / − α ) / − ε dα (cid:12)(cid:12)(cid:12)(cid:12) c + c t for some nonnegative constants c , c . Let us find the value of the first integral. Put δ = ε + 1 / Z / t α (1 / − α ) − δ dα = √ t Z / t − u u δ − du = √ t Z / e − u ln t u δ − du = √ t (ln t ) − δ Z (ln t ) / e − y y δ − dy. Thus we have proved Theorem 2.1 item 2.c for the case ω = .Now suppose that ω is an arbitrary positive number. Consider solution with initialcondition ˜ q ( ε ) k (0) , ˜ p ( ε ) k (0) defined in (11). Denote it by ˜ q ω , ( ε ) k ( t ). From (6) it is easy to seethat ˜ q ω , ( ε ) k ( t ) = 12 ω ˜ q / , ( ε ) k (2 ω t ) = 12 ω ˜ q ( ε ) k (2 ω t ) . Hence we obtain limiting equalities:lim t →∞ ˜ q ω , ( ε ) k ( t ) √ t ln δ t = 12 ω lim t →∞ ˜ q ( ε ) k (2 ω t ) √ t ln δ t = 1 √ ω Γ( δ ) . This completes the proof of Theorem 2.1.
We will use Lemma 3.1. The first part of Theorem 2.2 can be easily derived by integratingby parts n times the following integral: Z π f ( λ ) e − ikλ dλ where f ( λ ) is a corresponding C n smooth 2 π -periodic function.Next we will need the following lemma. 13 emma 3.5. Consider the integral: E [ g ]( t ) = 12 π Z π g ( λ ) exp( itω ( λ )) dλ, ω ( λ ) = q ω + 2 ω (1 − cos λ ) where g ( λ ) ∈ C n ( R ) , n > is a complex valued π -periodic function. Then, as t → ∞ ,E [ g ]( t ) ≍ √ t (cid:16) c g (0) exp (cid:16) i (cid:16) tω + π (cid:17)(cid:17) + c g ( π ) exp (cid:16) i (cid:16) tω ′ − π (cid:17)(cid:17)(cid:17) , where c = 1 ω r ω π , c = 1 ω r ω ′ π and ω ′ is defined in Theorem 2.2.Proof. We will apply stationary phase method (see [16, 17]). Let us find critical points ofthe phase function, i.e. zeros of the ddλ ω ( λ ): ω ′ ( λ ) = ddλ ω ( λ ) = ω sin λω ( λ ) = 0 . So the critical points lying on the interval [0 , π ] are only 0 , π, π . Zero and 2 π are theboundary points, but as g ( λ ) is a 2 π -periodic function we can replace the interval [0 , π ]from the definition of E [ g ]( t ) by the interval [ − π/ , π/ E [ g ]( t ) = 12 π Z π/ − π/ g ( λ ) exp( itω ( λ )) dλ. Thus the critical points of ω ( λ ) lying on [ − π/ , π/
2] are only 0 , π and we can apply sta-tionary phase method for the internal critical points. Let us find the second derivative ofthe ω ( λ ): ω ′′ ( λ ) = ω cos λω ( λ ) − ω sin λω ( λ ) . Consequently ω ′′ (0) = ω /ω , ω ′′ ( π ) = ω /ω ′ . To prove the lemma it remains to apply theformula (2) from [16], p. 51 (or [17], p. 163).Further we will use the equality (6). For Q k we have: Q k ( t ) = 12 π Z π Q ( λ ) cos( tω ( λ )) e − ikλ dλ = 12 ( E [ Qe − ikλ ]( t ) + E [ Qe − ikλ ]( t )) ,P k ( t ) = 12 π Z π P ( λ ) sin( tω ( λ )) ω ( λ ) e − ikλ dλ = 12 i (cid:16) E [ g ]( t ) − E [ g ]( t ) (cid:17) ,g = P e − ikλ ω ( λ )where z denotes the complex conjugate number of z ∈ C . Applying Lemma 3.5 to theseexpressions one can easily obtain the part 2 of Theorem 2.2.Let us prove part 3 of Theorem 2.2. We need the following lemma.14 emma 3.6. Consider the integral F [ g ]( β, k ) = 12 π Z π g ( λ ) e ik ( λ + βω ( λ )) dλ where g satisfies the conditions of Lemma 3.5, β > , k ∈ Z . Define the constant: γ ( β ) = β ω − − βω . The following assertions hold:1. if γ ( β ) > then as k → ∞ : F [ g ]( β, k ) ≍ p | k | ( c + g ( µ + ) e iω + ( k ) + c − g ( µ − ) e iω − ( k ) ) (12) where ω ± ( k ) = k ( µ ± + βω ( µ ± )) ± π k ) ,c ± = r βω ( µ ± )2 π ∆ , µ ± = − arccos 1 β ω (1 ± ∆) , ∆ = q ( β ω − − β ω ;
2. if γ ( β ) = 0 and n > then F [ g ]( β, k ) = O ( k − ) ;3. if γ ( β ) < then F [ g ]( β, k ) = O ( k − n ) .Proof. We will again use the stationary phase method. Consider the phase function: h ( λ ) = λ + βω ( λ ) . Let us find the critical points: h ′ ( λ ) = 1 + βω sin λ p ω + 2 ω (1 − cos λ ) = 0 . (13)From this equation we have: ω + 2 ω (1 − cos λ ) = β ω (1 − cos λ ) . Making the substitution x = cos λ rewrite the last equation: β ω x − ω x + ( ω + 2 ω − β ω ) = 0 . (14)The discriminant of this equation is: D = 4 ω (1 − β ( ω + 2 ω − β ω )) = 4 ω (cid:0) ( β ω − − β ω (cid:1)
15 4 ω ( β ω − − βω )( β ω − βω ) . Suppose that γ ( β ) >
0. In that case the roots of the equation (14) are: x ± = 1 β ω (cid:16) ± q ( β ω − − β ω (cid:17) = 1 β ω (1 ± ∆) . From the condition γ ( β ) > β ω − > | x ± | < β ω (1 + | β ω − | ) = 1 . Consequently the phase function has only two critical points λ ± on [0 , π ]: cos( λ ± ) = x ± with the additional condition sin( λ ± ) < λ ± = 2 π − arccos 1 β ω (1 ± ∆) = 2 π + µ ± . Obviously λ + = λ − iff γ ( β ) >
0. Moreover, λ ± are internal points, i.e. λ ± ∈ (0 , π ).To apply the stationary phase method we should find the signs of h ′′ ( λ ± ). Rewrite thederivative of the phase function: h ′ ( λ ) = 1 + βω sin λω ( λ ) . So at λ ± we have: ω ( λ ± ) = − βω sin λ ± . Further we obtain h ′′ ( λ ) = βω cos λω ( λ ) − βω sin λω ( λ ) . Thus h ′′ ( λ ± ) = − cos λ ± sin λ ± + 1 βω λ ± = − λ ± (cid:18) cos λ ± − β ω (cid:19) = ± (cid:16) − λ ± (cid:17) β ω q ( β ω − − β ω = ± βω ( λ ± ) ∆ . Since ω ( λ ) > h ′′ ( λ ± )) = ± γ ( β ) >
0. Hence applying(2) from [16], p. 51 we get: F [ g ]( β, k ) ≍ p | k | (cid:0) c + g ( λ + ) e i ˜ ω + ( k ) + c − g ( λ − ) e i ˜ ω − ( k ) (cid:1) where c ± are defined in (12) (we have used the fact that ω ( λ ) = ω ( − λ )). Note that:˜ ω + ( k ) = k ( λ ± + βω ( λ ± )) ± π k ) = 2 πk + ω ± ( k ) ,g ( λ ± ) = g ( µ ± ) 16here ω ± ( k ) is defined in (12). Since k ∈ Z we have e i ˜ ω ± ( k ) = e iω ± ( k ) . This completes theprove in the case γ ( β ) > γ ( β ) = 0 then λ + = λ − and h ′′ ( λ + ) = 0, i.e. λ + is a degenerate critical point and so F [ g ]( β, k ) = O ( k − ) due to [16], p. 52 (or [17], p. 163).Consider the case γ ( β ) <
0. We will prove that h ( λ ) has no critical points on [0 , π ].Rewrite the discriminant using γ : D = 4 ω γ ( γ + 2 βω ) . If γ < γ + 2 βω > D < γ + 2 βω
0. In that case∆ = D ω = (( γ + 2 βω ) − βω ( γ + 2 βω )) > ( γ + 2 βω ) . For the roots x ± we have the following estimates: | x ± | > β ω | − ∆ | > β ω | − | γ + 2 βω || = 1 β ω | γ + 2 βω | = 1 β ω ( β ω + βω ) > . Thus we have proved that h ( λ ) has no critical points on [0 , π ]. Consequently, F [ g ]( β, k ) = O ( k − n ). This completes the proof.Let us prove the remainder part of Theorem 2.2. From Lemma 3.1 we have Q k ( t ) = 12 π Z π Q ( λ ) cos( tω ( λ )) e − ikλ dλ = 14 π Z π Q ( λ ) e itω ( λ ) − ikλ dλ + 14 π Z π Q ( λ ) e − itω ( λ ) − ikλ dλ. Put t = βk and rewrite Q k ( t ): Q k ( t ) = 14 π Z π Q ( λ ) e itω ( λ ) − ikλ dλ + 12 F [ Q ]( β, − k )= 12 ( F [ Q ∗ ]( β, k ) + F [ Q ]( β, − k )) , where we have used the following notation: f ∗ ( λ ) = f (2 π − λ ) = f ( − λ ) . Note that ω ± ( − k ) = − ω ± ( k )for all k ∈ Z . So in the case γ ( β ) >
0, as k → ∞ , t = βk, β >
0, due to Lemma 3.6 we get: Q k ( t ) ≍ p | k | (cid:0) c + ( Q ( µ + ) e iω + ( k ) + Q ( − µ + ) e − iω + ( k ) )17 c − ( Q ( µ − ) e iω − ( k ) + Q ( − µ − ) e − iω − ( k ) ) (cid:1) = 1 p | k | F + k [ Q ] , where F ± k are defined in Theorem 2.2. Similarly one can obtain the expression for P k ( t ): P k ( t ) = 12 i (cid:0) F [ g ∗ ]( β, k ) − F [ g ]( β, − k ) (cid:1) , g = P ( λ ) ω ( λ ) . If γ ( β ) > P k ( t ) ≍ − i p | k | F − [ g ] . This completes the proof of Theorem 2.2.
We will use Lemma 3.1. The fact that Q k ( t ) = O ( k − n ) as t is fixed easily follows from theintegrating by parts the corresponding integral n times. Let us consider the term P k ( t ): P k ( t ) = 12 π Z π P ( λ ) sin( tω ( λ )) ω ( λ ) e − ikλ dλ. In the case ω = 0 we have ω ( λ ) = 2 ω sin λ . Thus the integrand contains two points where the denominator equals to zero. Sincesin( tω ( λ )) ω ( λ ) = ∞ X k =0 t k +1 ω k ( λ )(2 k + 1)! , [sin( tω ( λ ))] /ω ( λ ) is C ∞ ( R ) smooth function w.r.t. λ with period 2 π . Thus P k ( t ) = O ( k − n )as t is fixed. Hence the part 1 of Theorem 2.3 is proved.Next we will prove the remainder part of Theorem 2.3. We need the following lemma. Lemma 3.7.
For all fixed k ∈ Z the folowing limit holds lim t →∞ P k ( t ) = P (0)2 ω . (15) Proof.
We need another expression for P k : P k ( t ) = 12 π Z π P ( λ ) sin( tω ( λ )) ω ( λ ) e − ikλ dλ = 12 π Z π P ( λ ) − P (0) ω ( λ ) sin( tω ( λ )) e − ikλ dλ P (0) 12 π Z π sin( tω ( λ )) ω ( λ ) e − ikλ dλ = ˜ P k ( t ) + P (0) I k ( t ) , where ˜ P k ( t ) = 12 π Z π P ( λ ) − P (0) ω ( λ ) sin( tω ( λ )) e − ikλ dλ,I k ( t ) = 12 π Z π sin( tω ( λ )) ω ( λ ) e − ikλ dλ. Since the integrand [ P ( λ ) − P (0)] /ω ( λ ) in ˜ P k ( t ) is absolutely integrable, we have due toRiemann – Lebesgue theorem lim t →∞ ˜ P k ( t ) = 0 . Using lemma 3.8 we obtain (15).
Lemma 3.8.
The following equalities hold: I k ( t ) = Z t J k (2 ω s ) ds, lim t →∞ I k ( t ) = 12 ω (16) where J k ( t ) is the Bessel function of the first kind.Proof. From the definition we have: I k ( t ) = 12 π Z π sin( tω ( λ )) ω ( λ ) e − ikλ dλ = 1 π Z π sin(2 ω t sin λ )2 ω sin λ e − ikλ dλ. Thus ddt I k ( t ) = 1 π Z π cos(2 ω t sin λ ) e − ikλ dλ = 1 π Z π cos(2 ω t sin λ ) cos(2 kλ ) dλ − iπ Z π cos(2 ω t sin λ ) sin(2 kλ ) dλ. The second term equals zero. To see this one should make the substitution x = π − λ .Whence ddt I k ( t ) = 12 π Z π cos(2 ω t sin λ + 2 kλ ) dλ + 12 π Z π cos(2 ω t sin λ − kλ ) dλ = 12 ( J − k (2 ω t ) + J k (2 ω t )) = J k (2 ω t ) . It proves the first equality in (16). The second one follows from [15], p. 1036, 6.511 (1).19 emma 3.9.
Consider the integral C [ g ]( t ) = 12 π Z π g ( λ ) cos( tω ( λ )) dλ, where g ( λ ) ∈ C n ( R ) , n > is a complex valued π -periodic function. Then C [ g ]( t ) = 1 √ t g ( π ) √ πω cos (cid:16) ω t − π (cid:17) + b cos(2 ω t − π/ t √ t + O ( t − ) , for some complex constant b ∈ C .Proof. We apply the stationary phase method. The phase function ω ( λ ) = 2 ω sin( λ/
2) hasthe only one critical point λ = π on the interval [0 , π ]. The contribution to the asymptoticsof C [ g ]( t ) as t → ∞ is determined by the boundary points 0 , π and the critical point π .The contribution from the boundary points has the order of t − . Indeed the leading term ofthe asymptotic expansion corresponding to the point 0 is: − (cid:16) g (0) e iω (0) itω ′ (0) + g (0) e − iω (0) − itω ′ (0) (cid:17) = 0 . Analogously one has for the point 2 π . Hence the contribution to the asymptotics of C [ g ]( t )up to the order t − determined by the stationary point. From [16] we have the formula C [ g ]( t ) = 12 π cos(2 ω t − π/ √ t (cid:16) g ( π ) r πω + b t + O ( t − ) (cid:17) for some constant b . This completes the proof of Lemma 3.9.Let us continue the proof of Theorem 2.3. From Lemmas 3.1 and 3.9 we have: Q k ( t ) ≍ √ t ( − k Q ( π ) √ πω cos (cid:16) ω t − π (cid:17) as t → ∞ . Using (15) we obtain Z + ∞ t dds P k ( s ) ds = lim T →∞ Z Tt dds P k ( s ) ds = lim T →∞ P k ( T ) − P k ( t ) = P (0)2 ω − P k ( t ) . Whence due to Lemma 3.9 we get P k ( t ) = P (0)2 ω − Z + ∞ t dds P k ( s ) ds = P (0)2 ω − Z + ∞ t C [ P ( λ ) e − ikλ ]( s ) ds = P (0)2 ω − + ∞ Z t (cid:16) √ s ( − k P ( π ) √ πω cos (cid:16) ω s − π (cid:17) + b cos(2 ω s − π/ s √ s (cid:17) ds + O ( t − ) . (17)20pplying the second mean-value theorem we have that for all T > t there is a τ ∈ [ t, T ]such that Z Tt cos(2 ω s − π/ s √ s ds = 1 t √ t Z τt cos (cid:16) ω s − π (cid:17) dτ + 1 T √ T Z Tτ cos (cid:16) ω s − π (cid:17) ds. It is clear that sup −∞
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