Behavior for large time of an infinite chain of harmonic oscillators with defects
aa r X i v : . [ m a t h - ph ] S e p Behavior for large time of an infinite chainof harmonic oscillators with defects
T.V. Dudnikova
Keldysh Institute of Applied Mathematics of Russian Academy of Sciences
Miusskaya sq. 4, Moscow 125047, RussiaE-mail: [email protected]
Abstract
An infinite irregular harmonic chain of particles is considered. We assume that someparticles (“defects”) in the chain have masses and force constants of interaction differentfrom the masses and the interaction constants of the other particles. We study the Cauchyproblem for this model. The main goal is to study the long-time behavior and derive thedispersive bounds for the solutions in the energy weighted norms.
Key words and phases: infinite chain of harmonic oscillators with defects, Cauchyproblem, Fourier–Laplace transform, Puiseux expansion, dispersive estimatesAMS Subject Classification 2010: 35L15, 35B40, 35Q70, 70F45 Introduction
We consider a Hamiltonian infinite system of particles having harmonic nearest-neighbor inter-actions with the Hamiltonian functional of a form H ( u, v ) = 12 X n ∈ Z (cid:16) | v ( n ) | m n + γ n | u ( n + 1) − u ( n ) | + µ n | u ( n ) | (cid:17) , (1.1)where m n , γ n > , µ n ≥ . Then, the displacement of the n -th particle from its equilibriumposition obeys the following equations: ˙ u ( n, t ) = δ H δv = v ( n, t ) m n ˙ v ( n, t ) = − δ H δu = γ n ∇ L u ( n, t ) − γ n − ∇ L u ( n − , t ) − µ n u ( n, t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n ∈ Z , t > . (1.2)Here u ( n, t ) ∈ R , ∇ L denotes the derivative on Z = { , ± , ± , . . . } , ∇ L u ( n ) = u ( n + 1) − u ( n ) , n ∈ Z . We denote by γ n the force constant of interaction between the nearest neighbors, by v ( n, t ) = m n ˙ u ( n, t ) the moment of the n -th particle, by ˙ u ( n, t ) its velocity. We fix some N ≥ andassume that particles located at points n ≥ N +1 and n ≤ − have the same mass m n = m + > and m n = m − > , respectively. Furthermore, particles are affected by the same externalharmonic forces with constants µ n = µ + ≥ for n ≥ N + 1 and µ n = µ − ≥ for n ≤ − .The force constants of interaction are of a form γ n = γ − for n ≤ − , γ n = γ + for n ≥ N .At the same time, the particles (so-called “defects”) located at the points n = 0 , , . . . , N haveconstants m n , µ n , γ n , generally speaking, different from m ± , µ ± , γ ± .Therefore, the system (1.2) becomes m − ¨ u ( n, t ) = ( γ − ∆ L − µ − ) u ( n, t ) , n ≤ − , t > , (1.3) m n ¨ u ( n, t ) = γ n ∇ L u ( n, t ) − γ n − ∇ L u ( n − , t ) − µ n u ( n, t ) , n = 0 , , . . . , N, (1.4) m + ¨ u ( n, t ) = ( γ + ∆ L − µ + ) u ( n, t ) , n ≥ N + 1 , t > . (1.5)Here ∆ L denotes the second derivative on Z : ∆ L u ( n ) = u ( n + 1) − u ( n ) + u ( n −
1) = ∇ L u ( n ) − ∇ L u ( n − , n ∈ Z . For system (1.2), we study the Cauchy problem with the initial data u ( n,
0) = u ( n ) , v ( n,
0) = m n ˙ u ( n,
0) = v ( n ) , n ∈ Z . (1.6)Write Y ( t ) = ( u ( · , t ) , v ( · , t )) , Y ( · ) ≡ ( Y ( · ) , Y ( · )) = ( u ( · ) , v ( · )) . We assume that the initialdata Y belong to the Hilbert space H α , α ∈ R , defined below. Definition 1.1 ℓ α ≡ ℓ α ( Z ) , α ∈ R , is the Hilbert space of sequences u ( n ) , n ∈ Z , withnorm k u k α = P n ∈ Z h n i α | u ( n ) | < ∞ , h n i := (1 + n ) / . H α = ℓ α ⊗ ℓ α is the Hilbert space of pairs Y = ( u, v ) of sequences equipped with norm k Y k α = k u k α + k v k α < ∞ . m n , γ n , µ n (see conditions C and C in Sec. 2) under which for any initial data Y ∈ H α with α > / , the solution Y ( t ) ofsystem (1.3)–(1.6) obeys the following bound k Y ( t ) k − α ≤ C (1 + | t | ) − β/ k Y k α , t ∈ R , α > / , (1.7)where β = 1 if condition C holds, and β = 3 if condition C holds. The last bound is usefulfor applications to scattering problems. In particular, we prove that there exists a bounded‘wave’ operator Ω : H α → H − α such that Y ( t ) = Ω( ˜ Y ( t )) + δ ( t ) , where k δ ( t ) k − α ≤ C (1 + | t | ) − / k Y k α . Here ˜ Y ( t ) ≡ for n = 0 , , . . . , N and ˜ Y ( t ) is a solution of Eqn (1.3) for n ≤ − and ofEqn (1.5) for n ≥ N + 1 with initial data Y , see Theorem 2.11.Finally, we note that instead of system (1.3)–(1.5), it is possible to consider a more generalmodel with additional friction terms − β n ˙ u ( n, t ) in Eqn (1.4): m n ¨ u ( n, t ) = γ n ∇ L u ( n, t ) − γ n − ∇ L u ( n − , t ) − µ n u ( n, t ) − β n ˙ u ( n, t ) , n = 0 , . . . , N, where β n ≥ . If β n > for some n , then conditions on the constants could be weakened.This model was studied in [3] in the case when N = 0 and m ± = m n = 1 for all n . In thispaper, for simplicity, we study only the Hamiltonian model (1.3)–(1.5) without friction terms.Furthermore, we consider here the harmonic chain with nearest neighbor interaction. However,the results can be generalized to a more general case of interaction between the particles of thechain. 3 Main Results
At first, introduce two initial-boundary value problems with zero boundary condition m ± ¨ z ± ( n, t ) = ( γ ± ∆ L − µ ± ) z ± ( n, t ) , ± n ≥ , t > ,z ± (0 , t ) = 0 , t > ,z ± ( n,
0) = u ( n ) , m ± ˙ z ± ( n,
0) = v ( n ) , ± n ≥ . (2.1)Denote by H α, ± = ℓ α, ± ⊗ ℓ α, ± the Hilbert space of pairs Y = ( u, v ) of sequences equipped withnorm k Y k α, ± = k u k α, ± + k v k α, ± < ∞ , where ℓ α, ± ≡ ℓ α ( Z ± ) , α ∈ R , is the Hilbert space ofsequences u ( n ) , n ∈ Z ± , with norm k u k α, ± = P ± n ≥ h n i α | u ( n ) | < ∞ . The results concerningthe solutions of problem (2.1) are stated in [3, 4]. In particular, the following result is provedin [3]. Lemma 2.1
Assume that α ∈ R . Then (i) for any initial data Y ∈ H α, ± , there exists aunique solution Z ± ( t ) = ( z ± ( · , t ) , m ± ˙ z ± ( · , t )) ∈ C ( R , H α, ± ) to problem (2.1);(ii) the operator W ± ( t ) : Y Z ± ( t ) is continuous on H α, ± . Furthermore, the following boundholds, k W ± ( t ) Y k α, ± ≤ C h t i σ k Y k α, ± (2.2) with some constants C = C ( α ) , σ = σ ( α ) < ∞ . The proof of Lemma 2.1 is based on the following formula for the solutions of problem (2.1): Z i ± ( n, t ) = X ± k ≥ G ijt, ± ( n, k ) Y j ( k ) , ± n ≥ , i = 0 , , (2.3)where Z ± ( n, t ) ≡ z ± ( n, t ) , Z ± ( n, t ) ≡ m ± ˙ z ± ( n, t ) , Y ( n ) ≡ u ( n ) , Y ( n ) ≡ v ( n ) , the Greenfunction G t, ± ( n, k ) = ( G ijt, ± ( n, k )) i,j =0 is a matrix-valued function of a form G t, ± ( n, k ) := G t, ± ( n − k ) − G t, ± ( n + k ) , G t, ± ( n ) ≡ π Z T e − inθ ˆ G t, ± ( θ ) dθ, (2.4) T ≡ R / (2 π Z ) denotes torus, ˆ G t, ± ( θ ) = (cid:16) ˆ G ijt, ± ( θ ) (cid:17) i,j =0 , = (cid:18) cos( φ ± ( θ ) t ) sin( φ ± ( θ ) t ) / ( m ± φ ± ( θ )) − m ± φ ± ( θ ) sin( φ ± ( θ ) t ) cos( φ ± ( θ ) t ) (cid:19) φ ± ( θ ) = p ν ± (2 − θ ) + κ ± , (2.5)where, by definition, ν ± = γ ± /m ± > , κ ± = µ ± /m ± ≥ . (2.6)We see that z ± (0 , t ) ≡ for any t , since G ijt, ± ( − n ) = G ijt, ± ( n ) . For the solutions of prob-lem (2.1), the following bound is true. Theorem 2.2 (see [1, Theorem 2.2]) Let Y ∈ H α, ± and α > / . Then k W ± ( t ) Y k − α, ± ≤ C h t i − / k Y k α, ± , t ∈ R . (2.7)4ntroduce the boundary-initial value problem in Z N := { n ∈ Z : n ≥ N + 1 } : m + ¨ z N ( n, t ) = ( γ + ∆ L − µ + ) z N ( n, t ) , n ∈ Z N , t > , (2.8) z N ( N, t ) = 0 , t > , (2.9) z N ( n,
0) = u ( n ) , m + ˙ z N ( n,
0) = v ( n ) , n ∈ Z N . (2.10)Denote by W N ( t ) the solving operator of this problem, W N ( t ) : Y Z N ( t ) = ( z N ( · , t ) , m + ˙ z N ( · , t )) . (2.11)Then, ( W N ( t ) Y ) ( n ) = ( W + ( t ) ˜ Y )( n − N ) for n ∈ Z N , where ˜ Y ( k ) := Y ( k + N ) for k ≥ , W + ( t ) is introduced in Lemma 2.1. Write H α,N the Hilbert space of pairs Y = ( u , v ) withfinite norm k Y k α,N < ∞ , where k Y k α,N = k u k α,N + k v k α,N , k u k α,N = P n ∈ Z N h n i α | u ( n ) | .Hence, for the solutions z N ( n, t ) the following bound holds (cf (2.7)) k W N ( t ) Y k − α,N ≤ C h t i − / k Y k α,N , t ∈ R , α > / , (2.12)for any Y ∈ H α,N .The following theorem can be proved in a similar way as [2, Theorem 2.2]. Theorem 2.3 (i) Let Y ∈ H α , α ∈ R . Then the Cauchy problem (1.3)–(1.6) has a uniquesolution Y ( t ) ∈ C ( R , H α ) .(ii) The operator U ( t ) : Y → Y ( t ) is continuous on H α . Moreover, there exist constants C, B < ∞ such that k U ( t ) Y k α ≤ Ce B | t | k Y k α .(iii) Let Y ∈ H . Then the following identity holds: H ( Y ( t )) = H ( Y ) , t ≥ , (2.13) where H ( Y ) is the Hamiltonian defined in (1.1). Below we assume that α > / .We represent the solution of problem (1.3)–(1.6) in the following form u ( n, t ) = z ( n, t ) + r ( n, t ) , n ∈ Z , t > , (2.14)where, by definition, z ( n, t ) = z − ( n, t ) if n ≤ − , if n = 0 , . . . , N,z N ( n, t ) if n ≥ N + 1 . (2.15)Therefore, r ( n, t ) is a solution of the following problem m − ¨ r ( n, t ) = ( γ − ∆ L − µ − ) r ( n, t ) , n ≤ − , t > , (2.16) m n ¨ r ( n, t ) = γ n ∇ L r ( n, t ) − γ n − ∇ L r ( n − , t ) − µ n r ( n, t )+ δ n γ − z − ( − , t ) + δ nN γ + z N ( N + 1 , t ) , n = 0 , . . . , N, t > , (2.17) m + ¨ r ( n, t ) = ( γ + ∆ L − µ + ) r ( n, t ) , n ≥ N + 1 , t > , (2.18) r ( n,
0) = 0 , ˙ r ( n,
0) = 0 for n ≤ − and n ≥ N + 1 , (2.19) r ( n,
0) = u ( n ) , m n ˙ r ( n,
0) = v ( n ) for n = 0 , . . . N. (2.20)Here δ ij denotes the Kronecker symbol. 5 .2 The problem in the Fourier–Laplace transform To construct the solutions of problem (2.16)–(2.20) we use the Fourier–Laplace transform.
Definition 2.4
Let | r ( t ) | ≤ Ce Bt . The Fourier–Laplace transform of r ( t ) is given by theformula ˜ r ( ω ) = Z + ∞ e iωt r ( t ) dt, ℑ ω > B. The Gronwall inequality implies standard a priori estimates for the solutions r ( n, t ) , n ∈ Z .In particular, there exist constants C, B < ∞ such that X n ∈ Z ( | r ( n, t ) | + | ˙ r ( n, t ) | ) ≤ Ce Bt as t → + ∞ . At first, we study the solutions r ( n, t ) for n / ∈ { , . . . , N } . The Fourier–Laplace transform of r ( n, t ) with respect to t -variable, r ( n, t ) → ˜ r ( n, ω ) , exists at least for ℑ ω > B and satisfiesthe following equation ( − ν ± ∆ L + κ ± − ω )˜ r ( n, ω ) = 0 for n ≤ − and n ≥ N + 1 , ℑ ω > B, (2.21)where ν ± and κ ± are defined in (2.6). Now we construct the solution of (2.21). Note thatthe Fourier transform of the lattice operator − ν ± ∆ L + κ ± is the operator of multiplicationby the function φ ± ( θ ) = ν ± (2 − θ ) + κ ± . Thus, − ν ± ∆ L + κ ± is a self-adjoint operatorand its spectrum is absolutely continuous and coincides with the range of φ ± ( θ ) , i.e., with thesegment [ κ ± , a ± ] , a ± := κ ± + 4 ν ± .We introduce a critical set Λ : Λ = Λ + ∪ Λ − , Λ ± := [ − a ± , − κ ± ] ∪ [ κ ± , a ± ] , a ± = q κ ± + 4 ν ± , (2.22)and denote Λ = Λ − ∪ Λ , where Λ ± := {− a ± , − κ ± , κ ± , a ± } . Lemma 2.5 (see [6, Lemma 2.1]) For given ω ∈ C \ Λ ± , the equation ν ± (2 − θ ) + κ ± = ω (2.23) has the unique solution θ ± ( ω ) in the domain { θ ∈ C : ℑ θ > , ℜ θ ∈ ( − π, π ] } . Moreover, θ + ( ω ) ( θ − ( ω ) ) is an analytic function in C \ Λ + (in C \ Λ − , respectively). Since we seek the solution r ( · , t ) ∈ ℓ α with some α , ˜ r ( n, ω ) has a form ˜ r ( n, ω ) = (cid:26) e − iθ − ( ω ) n ˜ r (0 , ω ) if n ≤ − ,e iθ + ( ω )( n − N ) ˜ r ( N, ω ) if n ≥ N + 1 . Write ˜Γ ± n ( ω ) = e ± iθ ± ( ω ) n for ± n ≥ . Applying the inverse Fourier–Laplace transform withrespect to ω -variable, we write the solution of (2.16), (2.19) in the form r ( n, t ) = Z t Γ − n ( t − s ) r (0 , s ) ds for n ≤ − , t > , (2.24)6nd the solution of (2.18), (2.19) in the form r ( n, t ) = Z t Γ + n − N ( t − s ) r ( N, s ) ds for n ≥ N + 1 , t > , (2.25)where Γ ± n ( t ) := 12 π + ∞ + ic Z −∞ + ic e − iωt ˜Γ ± n ( ω ) dω with some c > , ± n ≥ , t > . (2.26) Theorem 2.6 (see [2, Theorem 3.3]) For any α > / , the following bounds hold, k Γ + n ( t ) k − α, + ≤ C h t i − / , k Γ − n ( t ) k − α, − ≤ C h t i − / , t > . (2.27) In particular, | Γ +1 ( t ) | ≤ C (1 + t ) − / , | Γ −− ( t ) | ≤ C (1 + t ) − / , t > . (2.28)To study the solution r ( n, t ) of Eqn (2.17) for n = 0 , . . . , N we first consider the solutionsof the corresponding homogeneous equation m n ¨ r ( n, t ) = γ n ∇ L r ( n, t ) − γ n − ∇ L r ( n − , t ) − µ n r ( n, t ) , t > , (2.29)where n = 0 , . . . , N , γ − ≡ γ − , γ N ≡ γ + , with the initial data (2.20). Applying the Fourier–Laplace transform to the solutions of problem (2.29), (2.20), we obtain ˜ r ( ω ) = ˜ N ( ω ) ( v − iω u ) for ℑ ω > B, (2.30)where ˜ r ( ω ) denotes a column ˜ r ( ω ) = (˜ r (0 , ω ) , . . . , ˜ r ( N, ω )) T and u := ( m u (0) , . . . , m N u ( N )) T , v := ( v (0) , . . . , v ( N )) T , (2.31) ˜ N ( ω ) = ( ˜ D ( ω )) − . If N ≥ , then ˜ D ( ω ) = ( ˜ D kn ( ω )) Nk,n =0 , ω ∈ C + , is a tridiagonal symmetricmatrix with entries of a form ˜ D ( ω ) = µ − m ω + γ − (1 − e iθ − ( ω ) ) + γ , ˜ D nn ( ω ) = µ n − m n ω + γ n + γ n − , n = 1 , . . . , N − , ˜ D NN ( ω ) = µ N − m N ω + γ + (1 − e iθ + ( ω ) ) + γ N − ˜ D nn +1 ( ω ) = ˜ D n +1 n ( ω ) = − γ n , n = 0 , . . . , N − , ˜ D kn ( ω ) = 0 for | k − n | ≥ . (2.32)If N = 0 , then ˜ D ( ω ) = µ − m ω + X ± γ ± (1 − e iθ ± ( ω ) ) , ω ∈ C + . (2.33)7 .3 Conditions on the constants The properties of the function ˜ D ( ω ) , ω ∈ C , play a key role in the proof of the bound (1.7).This function is studied in Appendices A and B. In particular, we check that det ˜ D ( ω ) = 0 forany ω ∈ C ± = { ω ∈ C : ±ℑ ω > } . Also, we prove that det ˜ D ( ω ± i = 0 for any ω ∈ Λ \ Λ ,where ˜ D ( ω ± i
0) := lim ε → +0 ˜ D ( ω ± iε ) . However, for some constants m ± , γ ± , µ ± , m n , γ n , µ n ,det ˜ D ( ω ) = 0 at some point ω ∈ ( R \ Λ) ∪ Λ . Therefore, to obtain the bound (1.7) we haveto find and eliminate such values of the constants. We divide all values of the constants intothree groups. The first group ( condition C ) includes all values for which det ˜ D ( ω ) = 0 forany ω ∈ R . The second group includes values under which det ˜ D ( ω ) = 0 for ω ∈ R \ Λ ,there exists a point ω ∈ Λ such that det ˜ D ( ω ) = 0 and det ˜ D (0) = 0 if ∈ Λ (wecall these restrictions by condition C ). The remaining values of the constants we call the resonance cases . For example, the case of the homogeneous chain without external forces (i.e.,when m ± = m n , µ ± = µ n = 0 for n = 0 , . . . , N and γ ± = γ n for n = 0 , . . . , N − ) is aresonance case, since in this case ∈ Λ and det ˜ D (0) = 0 . Thus, we impose the followingconditions C or C on the system. Condition C : det ˜ D ( ω ) = 0 for ω ∈ ( R \ Λ) ∪ Λ . Condition C : The following three restrictions hold.1) det ˜ D ( ω ) = 0 for ω ∈ R \ Λ .2) There exists some ω ∈ Λ \ { } such that det ˜ D ( ω ) = 0 .3) If µ − = µ + = 0 , then det ˜ D (0) = 0 .Our main objective is to derive conditions C and C in the terms of the restrictions onthe constants. For example, the third restriction in condition C is equivalent to the conditionthat if µ ± = 0 , then µ n = 0 for some n ∈ { , . . . , N } .To state conditions C and C in the case N = 0 , we introduce the functions K ( ω ) and K ± ( ω ) by the rule K ( ω ) := ¯ κ − q κ − ω q a − ω , ω ∈ R : | ω | ≤ κ + ; (2.34)where ¯ κ := (cid:0) ( κ − + κ ) / (cid:1) / , κ ± = ( µ ± /m ± ) / , ν ± = ( γ ± /m ± ) / , a ± = ( κ ± + 4 ν ± ) / , K ± ( ω ) := ¯ κ + 12 q ω − κ ± q ω − a ± , ω ∈ R : | ω | ≥ a ± . (2.35) Theorem 2.7
Let N = 0 . Then condition C is the following restrictions. µ = 0 if µ ± = 0; µ < κ − ( m − m + ) + 4 ν − (cid:0) m − m − + m + (cid:1) + m + K + ( a − ) if a − ≥ a + ; µ < κ ( m − m − ) + 4 ν (cid:0) m − m − + m + (cid:1) + m − K − ( a + ) if a + ≥ a − ; µ > κ − ( m − m − ) + m + K ( κ − ) if κ − = 0; µ > κ ( m − m − ) + m − K − ( κ + ) or µ < κ − ( m − m + ) + 4 ν − ( m − m − + m + ) + m + K ( a − ) if a − < κ + . Condition C is the following restrictions: the inequalities (4.3)–(4.6) (see Appendix A) holdand one of the following conditions is fulfilled. i) a − > a + , µ = κ − ( m − m + ) + 4 ν − (cid:0) m − m − + m + (cid:1) + m + K + ( a − ) (ii) a − < a + , µ = κ ( m − m − ) + 4 ν (cid:0) m − m − + m + (cid:1) + m − K − ( a + ) (iii) a + = a − , ( κ − , κ + ) = (0 , , µ = m a − − γ − − γ + (iv) κ − = 0 , µ = κ − ( m − m + ) + m + K ( κ − ) .In particular, if κ − = κ + = 0 , then µ = κ − m = µ − m /m − (v) a − ≤ κ + , µ = κ ( m − m + ) + m − K − ( κ + ) (vi) a − < κ + , µ = κ − ( m − m + ) + 4 ν − (cid:0) m − m − + m + (cid:1) + m + K ( a − ) . Theorem 2.7 is proved in Appendix A. Note that condition C excludes the case when m ± = m , γ − = γ + , µ − = µ + . However, the case when m ± = m , γ − = γ + , µ − = µ + = 0 is included in condition C . Remark 2.8
Let us consider a particular case of the chain, which we call ( P1 ) case, when N = 0 and the oscillators of the chain are identical, excluding defects, i.e., m − = m + =: m, γ − = γ + =: γ, µ − = µ + =: µ. (2.36) Set ν = p γ/m , ν = p γ/m , κ = p µ/m , κ = p µ /m , a ≡ a ± = p ( µ + 4 γ ) /m , a = p ( µ + 4 γ ) /m . In this case, conditions C and C are simplified as follows C κ > κ and a < a . C One of the following restrictions holds.(i) κ = κ = 0 and ν = ν (and hence, a = a );(ii) κ = κ = 0 and a < a ;(iii) κ > κ and a = a .In particular, it follows from conditions C and C that κ = 0 , κ ≥ κ and ν ≤ ν .In this particular case, condition C and C can be rewritten in the terms m , m, γ, µ , µ asfollows C µ m − µm > and µ m − µm + 4 γ (cid:18) m − m (cid:19) < . C One of the following restrictions hold.(i) µ = µ = 0 and m = m ;(ii) µ = 0 , µ = 0 , µ m = µm and m > m ;(iii) µ = 0 , µ m > µm and µ m − µm + 4 γ (cid:18) m − m (cid:19) = 0 .Condition C implies that m > m and µ > µ ≥ . Condition C implies that m ≥ m and µ ≥ µ , µ > .
9e see that conditions C and C are tedious even for N = 0 . Therefore, in the case N ≥ , we assume, in addition, that the oscillators in the chain are identical except thedefects, i.e., (2.36) holds. In this case, we derive conditions C and C as restrictions on theconstants m, γ, µ, m n , γ n , µ n in Theorem 2.9.Write a = p ( µ + 4 γ ) /m , κ = p µ/m . Let ˜ D ( κ ) ( ˜ D ( a ) ) denote the matrix ˜ D ( ω ) with ω = κ ( ω = a , resp.), where in the entries we put e iθ ( κ ) := 1 ( e iθ ( a ) := − , resp.). Denote by α n ( ω ) , n = 0 , , . . . , N , the principal (corner) minors of the matrix ˜ D ( ω ) , see formula (5.20)below. Theorem 2.9
Let N ≥ and (2.36) hold. Then conditions C and C are the followingrestrictions. C
1) If µ = 0 , then µ n = 0 for some n ∈ { , . . . , N } .2) The matrix ˜ D ( a ) is negative–definite, ˜ D ( a ) < .3) If µ = 0 , then the matrix ˜ D ( κ ) is positive–definite, ˜ D ( κ ) > . C If µ = 0 , then µ n = 0 for some n ∈ { , . . . , N } . Moreover, the principal minors of thematrix ˜ D ( a ) have the following property: α n ( a )( − n < for n = 0 , , . . . , N − , and α N ( a ) ≡ det ˜ D ( a ) = 0 .If µ = 0 , then either (i) α n ( a )( − n < for n = 0 , , . . . , N − , α N ( a ) = 0 , and ˜ D ( κ ) > or (ii) α n ( κ ) > for n = 0 , , . . . , N − , α N ( κ ) = 0 , and ˜ D ( a ) < . Theorem 2.9 is proved in Appendix B. The proof of this theorem is based on the propertiesof the tridiagonal matrix ˜ D ( ω ) . Remark 2.10
It follows from conditions C and C that ˜ D nn ( a ) ≤ and ˜ D nn ( κ ) ≥ for all n . In addition, in the case N ≥ , it follows that the masses of the first and last defects mustbe greater than half the mass of the “non-defective” particles, m > m/ and m N > m/ , andalso, µ > µ/ − γ and µ N > µ/ − γ N − . For details, see Appendix B. Define an operator W ( t ) , t ∈ R , on the space H α by the rule ( W ( t ) Y )( n ) = ( W − ( t ) Y ) ( n ) for n ≤ − , for n = 0 , . . . N, ( W N ( t ) Y ) ( n ) for n ≥ N + 1 , (2.37)where the operators W − ( t ) and W N ( t ) are introduced in Lemma 2.1 and in (2.11), respectively.The main result is the following theorem. Theorem 2.11
Let Y ∈ H α , α > / , and conditions C or C hold. Then the followingassertions are fulfilled.(i) There exists a bounded operator Ω : H α → H − α such that U ( t ) Y = Ω( W ( t ) Y ) + δ ( t ) , where k δ ( t ) k − α ≤ C h t i − β/ k Y k α , (2.38) β = 3 if condition C holds and β = 1 if condition C holds. ii) k Ω( W ( t ) Y ) k − α ≤ C h t i − β/ k Y k α . Hence, k Y ( t ) k − α ≤ C h t i − β/ k Y k α , (2.39) where β = 3 if condition C holds and β = 1 if condition C holds. Theorem 2.11 can be proved using the technique from [3]. We outline the main steps of theproof.
Step (1) : We first apply the Fourier–Laplace transform to the solutions of problem (2.29),(2.20) and obtain Eqn (2.30). Then, we study the behavior of ˜ N ( ω ) for different values of ω ∈ C : for ω ∈ C ± , ω ∈ Λ \ Λ , ω ∈ R \ Λ and for ω ∈ Λ , see Appendices A and B. Thisallows us to prove the bound for the matrix N ( t ) = ( N nk ( t )) Nn,k =0 , where N nk ( t ) = 12 π + ∞ + ic Z −∞ + ic e − iωt ˜ N nk ( ω ) dω with some c > , t > , (2.40) ˜ N nk ( ω ) are entries of the matrix ˜ N ( ω ) = ( ˜ D ( ω )) − , n, k = 0 , . . . , N . Theorem 2.12
Let condition C or C hold. Then |N ( j ) nk ( t ) | ≤ C (1 + t ) − β/ , t ≥ , j = 0 , , , (2.41) where N ( j ) nk ( t ) ≡ d j dt j N nk ( t ) , β = 3 if condition C holds and β = 1 if condition C holds. We prove this theorem in Appendix A for N = 0 and in Appendix B for N ≥ . Step (2) : Applying the inverse Fourier–Laplace transform to Eqn (2.30), we write the solu-tion r ( t ) = ( r (0 , t ) , . . . , r ( N, t )) of problem (2.29), (2.20) in a form r ( t ) = ˙ N ( t ) u + N ( t ) v , t > , where u and v are defined in (2.31). For t = 0 we put N nk (0) = 0 and ˙ N nk (0) = δ nk /m k for n, k = 0 , . . . , N . Using the variation constants formula, we obtain the following representationfor the solutions r ( t ) = ( r (0 , t ) , . . . , r ( N, t )) of Eqs (2.17) with the initial data (2.20): r ( t ) = ˙ N ( t ) u + N ( t ) v + Z t N ( t − τ ) F ( τ ) dτ. (2.42)where F ( τ ) = ( F n ( τ )) Nn =0 is a column with entries F ( τ ) = γ − z − ( − , τ ) , F n ( τ ) = 0 for n = 1 , . . . , N − , F N ( τ ) = γ + z N ( N + 1 , τ ) . Applying the bounds (2.7) and (2.12) to F ( τ ) ,using (2.42) and the bound (2.41), we obtain the bound for r ( t ) : sup n =0 , ,...,N ( | r ( n, t ) | + | ˙ r ( n, t ) | ) ≤ C (1 + t ) − β/ k Y k α , t ≥ , α > / , (2.43)where β is introduced in Theorem 2.12. Step (3) : Using formulas (2.24), (2.25), (2.27) and (2.43), we obtain the following estimate k ( r ( · , t ) , ˙ r ( · , t )) k − α ≤ C (1 + t ) − β/ k Y k α , t ≥ . (2.44)Finally, the bound (2.39) follows from the decomposition (2.14), Theorem 2.2 and the bound(2.44). The construction of the operator Ω and the proof of Theorem 2.11 are given in Sec. 3.11 emark 2.13 Let conditions C and C be not hold. Then ˜ N nk ( ω ) have either the simplepole at zero or poles at points ω = ± ω ∗ , where a point ω ∗ ∈ R \ Λ such that det ˜ D ( ω ∗ ) = 0 .Let the initial data Y and Y ( n ) ≡ for n
6∈ { , . . . , N } . Therefore, z ( n, t ) ≡ for any n and there exist solutions r ( t ) of problem (2.29), (2.20) which don’t satisfy the bound (2.43).Hence, there are solutions r ( t ) of (2.42) which don’t satisfy the bound (2.43). Thus, there existsolutions u ( · , t ) of problem (1.3)–(1.6) which do not satisfy the bound (2.39). For details, seeSections 4.6 and 5.6 below. Denote by W ′− ( t ) the operator adjoint to W − ( t ) , t ∈ R , h Y, W ′− ( t )Ψ i − = h W − ( t ) Y, Ψ i − , Y ∈ H α, − , Ψ = (Ψ , Ψ ) ∈ [ S ( Z − )] , and by W ′ N ( t ) the operator adjoint to W N ( t ) , t ∈ R , h Y, W ′ N ( t )Ψ i N = h W N ( t ) Y, Ψ i N , Y ∈ H α,N , Ψ = (Ψ , Ψ ) ∈ [ S ( Z N )] , Here S ( Z − ) denotes the class of rapidly decreasing sequences in Z − , h· , ·i − stands for theinner product in H , − or for its different extensions. Similarly, h· , ·i N stands for the innerproduct in H ,N . Below we also use the notation h· , ·i for the inner product in H or for itsdifferent extensions. Write g − ( n, t ) = ( W ′− ( t ) Y )( n ) with Y ( n ) = ( δ ( − n , , n ≤ , (3.1) g N ( n, t ) = ( W ′ N ( t ) Y )( n ) with Y ( n ) = ( δ ( N +1) n , , n ≥ N. (3.2)Let G j − ,n ( k ) , G jN,n ( k ) , j = 0 , , denote the vector valued functions G j − ,n ( k ) = Z + ∞ N ( j ) n ( s ) g − ( k, − s ) ds, k ≤ , n = 0 , . . . , N, (3.3) G jN,n ( k ) = Z + ∞ N ( j ) nN ( s ) g N ( k, − s ) ds, k ≥ N, n = 0 , . . . , N, (3.4)where N nk ( s ) are defined in (2.40). Note that g − (0 , t ) = g N ( N, t ) = 0 and G j − ,n (0) = G jN,n ( N ) = 0 for any n ∈ { , . . . , N } . Introduce G jn ( k ) = γ − G j − ,n ( k ) for k ≤ − , for k = 0 , . . . Nγ + G jN,n ( k ) for k ≥ N + 1 , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n = 0 , . . . , N, j = 0 , . (3.5)By (2.37) and (3.5), we have h W ( t ) Y ( · ) , G jn ( · ) i = γ − h W − ( t ) Y , G j − ,n i − + γ + h W N ( t ) Y , G jN,n i N , t ≥ . (3.6)Set r (0) ( n, t ) = r ( n, t ) , r (1) ( n, t ) = ˙ r ( n, t ) .12 emma 3.1 Let Y ∈ H α , α > / , condition C or C hold, and r ( n, t ) be a solution ofproblem (2.42). Then for t > , j = 0 , , n = 0 , , . . . , N , r ( j ) ( n, t ) = h W ( t ) Y ( · ) , G jn ( · ) i + δ jn ( t ) , sup n =0 , ,...,N | δ jn ( t ) | ≤ C h t i − β/ k Y k α , (3.7) where the operator W ( t ) is defined in (2.37), β is introduced in Theorem 2.12. Proof
Using (2.42) and the bound (2.41), we obtain for t > r ( j ) ( n, t ) = Z t (cid:16) N ( j ) n ( τ ) γ − z − ( − , t − τ ) + N ( j ) nN ( τ ) γ + z N ( N + 1 , t − τ ) (cid:17) dτ + O ( h t i − β/ ) . (3.8)We estimate the fist term in the r.h.s. of (3.8). The second term is estimated by a similar way.The bounds (2.7) and (2.41) give (cid:12)(cid:12)(cid:12) Z + ∞ t N ( j ) n ( τ ) z − ( − , t − τ ) dτ (cid:12)(cid:12)(cid:12) ≤ C Z + ∞ t h τ i − β/ h t − τ i − / k Y k α dτ ≤ C h t i − β/ k Y k α . Using (3.1) and the equality W − ( t − τ ) = W − ( t ) W − ( − τ ) , we have z − ( − , t − τ ) = D ( W − ( t − τ ) Y ) ( k ) , (cid:18) δ − k (cid:19) E − = h W − ( t ) Y ( · ) , g − ( · , − τ ) i − . Hence, r ( j ) ( n, t ) = Z + ∞ (cid:16) N ( j ) n ( τ ) γ − z − ( − , t − τ ) + N ( j ) nN ( τ ) γ + z N ( N + 1 , t − τ ) (cid:17) dτ + O ( h t i − β/ )= γ − h ( W − ( t ) Y ) ( · ) , G j − ,n ( · ) i − + γ + h ( W N ( t ) Y ) ( · ) , G jN,n ( · ) i N + O ( h t i − β/ ) . Together with (3.6), this implies the representation (3.7).Now we estimate the first term in the r.h.s. of (3.7).
Lemma 3.2
Let Y ∈ H α , α > / , condition C or C hold. Then for n = 0 , , . . . , N , |h W ( t ) Y ( · ) , G jn ( · ) i| ≤ C h t i − β/ k Y k α , t > , j = 0 , . (3.9) Proof
At first, we prove the following estimate k (cid:0) W ′− ( t ) G j − ,n (cid:1) ( · ) k − α, − + k (cid:0) W ′ N ( t ) G jN,n (cid:1) ( · ) k − α,N ≤ C h t i − β/ (3.10)for any n = 0 , , . . . , N . Indeed, applying (3.1), we have W ′− ( t ) g − ( k, − s ) = g − ( k, t − s ) .Hence, by (3.3), (2.7) and (2.41), we obtain k (cid:0) W ′− ( t ) G j − ,n (cid:1) ( · ) k − α, − ≤ Z + ∞ |N ( j ) n ( s ) |k g − ( · , t − s ) k − α, − ds ≤ C Z + ∞ h s i − β/ h t − s i − / ds ≤ C h t i − β/ . k (cid:0) W ′ N ( t ) G jN,n (cid:1) ( · ) k − α,N . Since (cid:12)(cid:12) h W − ( t ) Y , γ − G j − ,n i − (cid:12)(cid:12) ≤ γ − k Y k α, − k W ′− ( t ) G j − ,n k − α, − , (cid:12)(cid:12) h W N ( t ) Y , γ + G jN,n i N (cid:12)(cid:12) ≤ γ + k Y k α,N k W ′ N ( t ) G jN,n k − α,N , the bound (3.9) follows from (3.6) and (3.10).Introduce a vector-valued function Γ j ( n, k ) , j = 0 , , by the rule Γ j ( n, k ) = γ − R + ∞ Γ − n ( s ) (cid:16) W ′− ( − s ) G j − , (cid:17) ( k ) ds, if n ≤ − , k ≤ − ,γ + R + ∞ Γ − n ( s ) (cid:16) W ′ N ( − s ) G jN, (cid:17) ( k ) ds, if n ≤ − , k ≥ N + 1 ,γ − R + ∞ Γ + n − N ( s ) (cid:16) W ′− ( − s ) G j − ,N (cid:17) ( k ) ds, if n ≥ N + 1 , k ≤ − ,γ + R + ∞ Γ + n − N ( s ) (cid:16) W ′ N ( − s ) G jN,N (cid:17) ( k ) ds, if n ≥ N + 1 , k ≥ N + 1 , G jn ( k ) , if n = 0 , . . . , N, k ∈ Z , otherwise . (3.11)Now we study the large time behavior of r ( n, t ) for n = 0 , . . . , N . Lemma 3.3
Assume that Y ∈ H α , α > / , and condition C or C hold. Then thesolution r ( n, t ) with n
6∈ { , . . . , N } of problem (2.16), (2.18), (2.19) admits the followingrepresentation r ( j ) ( n, t ) = h (cid:0) W ( t ) Y (cid:1) ( · ) , Γ j ( n, · ) i + δ j ( n, t ) , j = 0 , , t > , (3.12) where k δ j ( · , t ) k − α ≤ C h t i − β/ k Y k α . Proof
We consider the case n ≤ − only. For n ≥ N + 1 , the proof is similar. By (2.24) and(3.7), r ( j ) ( n, t ) = Z t Γ − n ( t − s ) h W ( s ) Y , G j i ds + δ ′ j ( n, t ) for n ≤ − , (3.13)where k δ ′ j ( · , t ) k − α, − ≤ C h t i − β/ k Y k α . Indeed, by (3.7) and (2.27), k δ ′ j ( · , t ) k − α, − = (cid:13)(cid:13)(cid:13) Z t Γ − n ( t − s ) δ j ( s ) ds (cid:13)(cid:13)(cid:13) − α, − ≤ Z t (cid:13)(cid:13) Γ − n ( t − s ) (cid:13)(cid:13) − α, − | δ j ( s ) | ds ≤ C Z t (1 + t − s ) − / (1 + s ) − β/ ds k Y k α ≤ C h t i − β/ k Y k α . The first term in the r.h.s. of (3.13) has a form Z t Γ − n ( s ) h W ( t − s ) Y , G j i ds = Z + ∞ Γ − n ( s ) h W ( t − s ) Y , G j i ds + δ ′′ j ( n, t ) , n ≤ − , (3.14)where, by definition, δ ′′ j ( n, t ) = − Z + ∞ t Γ − n ( s ) h W ( t − s ) Y , G j i ds . The bounds (2.27) and (3.9)yield (cid:13)(cid:13) δ ′′ j ( · , t ) (cid:13)(cid:13) − α, − ≤ Z + ∞ t (cid:13)(cid:13) Γ − n ( s ) (cid:13)(cid:13) − α, − (cid:12)(cid:12)(cid:12) h W ( t − s ) Y , G j i (cid:12)(cid:12)(cid:12) ds ≤ C h t i − β/ k Y k α . (3.15)14inally, applying (3.6) we obtain Z + ∞ Γ − n ( s ) h W ( t − s ) Y , G j i ds = Z + ∞ Γ − n ( s ) h W − ( t − s ) Y , γ − G j − , i − ds + Z + ∞ Γ − n ( s ) h W N ( t − s ) Y , γ + G jN, i N ds = Z + ∞ Γ − n ( s ) h W − ( t ) Y , γ − W ′− ( − s ) G j − , i − ds + Z + ∞ Γ − n ( s ) h W N ( t ) Y , γ + W ′ N ( − s ) G jN, i N ds = h W − ( t ) Y , Γ j ( n, · ) i − + h W N ( t ) Y , Γ j ( n, · ) i N = h W ( t ) Y , Γ j ( n, · ) i . Hence, the bounds (3.13)–(3.15) imply (3.12) with δ j ( n, t ) = δ ′ j ( n, t ) + δ ′′ j ( n, t ) for n ≤ − . Remark 3.4
We put Γ j ( n, k ) = G jn ( k ) for n = 0 , , . . . , N . Hence, the representation (3.7)is a particular case of (3.12). Now we estimate the first term in the r.h.s. of (3.12).
Lemma 3.5
Assume that Y ∈ H α , α > / , and condition C or C hold. Then kh (cid:0) W ( t ) Y (cid:1) ( k ) , Γ j ( · , k ) ik − α ≤ C h t i − β/ k Y k α , t ≥ , j = 0 , . (3.16) Proof
For any n , we have (cid:12)(cid:12)(cid:12) h (cid:0) W ( t ) Y (cid:1) ( · ) , Γ j ( n, · ) i (cid:12)(cid:12)(cid:12) ≤ k Y k α (cid:16) k W ′− ( t ) Γ j ( n, · ) k − α, − + k W ′ N ( t ) Γ j ( n, · ) k − α,N (cid:17) . (3.17)Using (3.11), we estimate the first term in the r.h.s. of (3.17): k W ′− ( t ) Γ j ( n, · ) k − α, − ≤ γ − Z + ∞ | Γ − n ( s ) | k W ′− ( t − s ) G j − , k − α, − ds for n ≤ − ,γ − Z + ∞ | Γ + n − N ( s ) | k W ′− ( t − s ) G j − ,N k − α, − ds for n ≥ N + 1 ,γ − k W ′− ( t ) G j − ,n k − α, − for n = 0 , , . . . , N. Hence, applying (2.27) and (3.10), we obtain sX n ∈ Z h n i − α k W ′− ( t ) Γ j ( n, · ) k − α, − ≤ C N γ − h t i − β/ . (3.18)The similar bound is valid for kk W ′ N ( t ) Γ j ( n, · ) k − α,N k − α . The bound (3.18) implies (3.16).Introduce an operator Ω : H α → H − α , α > / , by the rule Ω : Y → Y ( n ) + (cid:16) h Y ( · ) , Γ ( n, · ) i , m n h Y ( · ) , Γ ( n, · ) i (cid:17) , n ∈ Z . (3.19)It follows from (3.16) that the operator Ω is bounded, k Ω Y k − α ≤ C k Y k α for any α > / . Proof of Theorem 2.11
Theorem 2.2 and Lemmas 3.1–3.5 imply assertions of Theorem 2.11.Indeed, the representation (2.38) follows from (2.14), (2.37), (3.7) and (3.12). The bounds (2.7),(2.12), (3.9) and (3.16) imply the bound (2.39).15
Appendix A: Case N = 0 If N = 0 , then there is the unique “defect” in the chain, which is the particle located at originwith mass m different, generally speaking, from masses of the other particles and with theconstant of the external force µ unequal to µ ± , in general. C and C in the particular cases Now we simplify conditions C and C for N = 0 and for some particular cases of the chain.In Remark 2.8, we consider a particular case ( P1 ). Now we study another two cases. Particular case (P2) : Assume that the oscillators in the chain have identical masses equalto unity, i.e., m ± = m = 1 , and let, for simplicity, κ − ≤ κ + . This case was considered in [3].In this case, conditions C and C are of the following form C µ < K + ( a − ) if a − ≥ a + ; µ < K − ( a + ) if a + ≥ a − ; µ > K ( κ − ) if κ − = 0 ; µ > K − ( κ + ) or µ < K ( a − ) if a − ≤ κ + ; µ = 0 if µ − = µ + = 0 .For example, µ ∈ (cid:16) , ν − , ν + ) p | ν − − ν | (cid:17) if κ ± = 0 and ν − = ν + . Note that firsttwo restrictions of condition C exclude the case when m − = m + , γ − = γ + and µ − = µ + ,since in this case K + ( a − ) = K − ( a + ) = κ − and K ( κ − ) = κ − if κ − = 0 . However, the case κ − = κ + = 0 and ν − = ν + is included in condition C . C One of the following restrictions is fulfilled.(i) a − > a + , µ = K + ( a − ) .(ii) a − < a + , µ = K − ( a + ) .(iii) a + = a − , ( κ − , κ + ) = (0 , , µ = ¯ κ .(iv) κ − = 0 , µ = K ( κ − ) (if K ( κ − ) ≥ ).(v) a − ≤ κ + , µ = K − ( κ + ) .(vi) a − < κ + , µ = K ( a − ) (if K ( a − ) ≥ ). Remark 4.1
Condition C (i)–(iv) includes the following particular cases: • κ − = κ + , ν − = ν + , µ = κ − + 2 max( ν − , ν + ) p | ν − − ν | (see cases (i) and (ii)); • κ − = κ + = 0 , µ = K ( κ − ) = ¯ κ (see case (iv)); • κ − = κ + = 0 , ν − = ν + , µ = µ + (see cases (iii) and (iv)). Particular case (P3) : Assume that external forces don’t act on the oscillators in the chainexcept the defect, i.e., µ ± = 0 , µ > . In this case, κ ± = 0 , a ± = 2 ν ± and conditions C and C are of a form C µ > ; µ < ν − (cid:0) m − m − + m + (cid:1) + 2 m + ν − p ν − − ν if ν − ≥ ν + ; µ < ν (cid:0) m − m − + m + (cid:1) + 2 m − ν + p ν − ν − if ν + ≥ ν − .16 µ > and one of the following restrictions is fulfilled.(i) µ = 4 ν − (cid:0) m − m − + m + (cid:1) + 2 m + ν − p ν − − ν if ν − ≥ ν + .(ii) µ = 4 ν (cid:0) m − m − + m + (cid:1) + 2 m − ν + p ν − ν − if ν − ≤ ν + .Note that if ν − = ν + , then conditions C and C imply that the mass of the defect satisfiesthe following restriction m > ( m − + m + ) / . ˜ D ( ω ± i for ω ∈ Λ \ Λ If N = 0 , then ˜ D ( ω ) is of the form (2.33). We have ˜ D ( ω ) = 0 for ω ∈ C ± by Lemma 5.1from Appendix B. Now we study ˜ D ( ω ± i
0) = lim ε → +0 ˜ D ( ω ± iε ) for ω ∈ Λ \ Λ . Lemma 4.2 ˜ D ( ω ± i = 0 for ω ∈ Λ \ Λ . Proof
Let ω ∈ Λ − \ Λ − and ω / ∈ Λ + . Then ℜ θ − ( ω + i ∈ ( − π, ∪ (0 , π ) and ℑ θ − ( ω + i
0) = 0 .Moreover, sign(sin θ − ( ω + i ω . Therefore, ℑ ˜ D ( ω + i
0) = − γ − sin θ − ( ω + i − m − (cid:26) sign( ω ) p ω − κ − p a − − ω = 0 , κ − = 0 ,ω p a − − ω = 0 , κ − = 0 . Hence, ˜ D ( ω + i = 0 for such values of ω . Similarly, we can check that for ω ∈ Λ + \ Λ and ω / ∈ Λ − , ℑ ˜ D ( ω + i = 0 . For ω ∈ (Λ − ∩ Λ + ) \ ∪ ± Λ ± , we have ℑ ˜ D ( ω + i
0) = − γ − sin θ − ( ω + i − γ + sin θ + ( ω + i − sign( ω ) P ± m ± p ω − κ ± p a ± − ω , if κ ± = 0sign( ω ) (cid:16) m − | ω | p ν − − ω + m + p ω − κ p a − ω (cid:17) , if κ − = 0 , κ + = 0 ω P ± m ± p ν ± − ω , if κ − = κ + = 0 Therefore, ℑ ˜ D ( ω + i = 0 for such values of ω . Since ˜ D ( ω − i
0) = ˜ D ( ω + i for ω ∈ Λ \ Λ , ˜ D ( ω − i = 0 for ω ∈ Λ \ Λ . ˜ D ( ω ) for ω ∈ R \ Λ To prove Theorem 2.7, we first prove the following lemma, using the technique of [3].
Lemma 4.3 ˜ D ( ω ) = 0 for ω ∈ R \ Λ iff the following conditions hold.(i) ˜ D (max( a − , a + )) ≤ .(ii) If min( κ − , κ + ) > , then ˜ D (min( κ − , κ + )) ≥ .(iii) If a ∓ < κ ± , then either ˜ D ( κ ± ) ≥ or ˜ D ( a ∓ ) ≤ . roof To prove this lemma we apply the following formulas from [3].Let κ + = 0 , ω ∈ R and | ω | < κ + . Then, ℜ θ + ( ω ) = 0 and e iθ + ( ω ) = e −ℑ θ + ( ω ) < .Moreover, for ω ∈ R : | ω | < κ + , ν (1 − e iθ + ( ω ) ) = 12 (cid:18) ω − κ + q κ − ω q a − ω (cid:19) = 12 ( ω + κ − ) − K ( ω ) , (4.1)where K ( ω ) is defined in (2.34).If ω ∈ R and | ω | > a ± , then e iθ ± ( ω ) = − e −ℑ θ ± ( ω ) and ν ± (1 − e iθ ± ( ω ) ) = 12 (cid:18) ω − κ ± − q ω − κ ± q ω − a ± (cid:19) = 12 ( ω + κ ∓ ) − K ± ( ω ) , (4.2)where K ± ( ω ) is defined in (2.35). Using formulas (4.1) and (4.2), we rewrite conditions (i)–(iii)from Lemma 4.3. Condition (i) . Let max( a − , a + ) = a − . By (4.2), we have ˜ D ( a − ) = µ − m a − + γ − (1 − e iθ − ( a − ) ) + γ + (1 − e iθ + ( a − ) )= µ − m a − + 2 γ − + m + (cid:18)
12 ( a − + κ − ) − K + ( a − ) (cid:19) . Note that ℜ ˜ D ( ω ) < ℜ ˜ D ( ω ) for | ω | > | ω | ≥ max( a − , a + ) , and ℜ ˜ D ( ω ) → −∞ as | ω | → ∞ .Therefore, ˜ D ( ω ) = 0 for | ω | > a − iff ˜ D ( a − ) ≤ . In turn, ˜ D ( a − ) ≤ if and only if µ ≤ κ − ( m − m + ) + 4 ν − ( m − m − + m + m + K + ( a − ) , where a − ≥ a + . (4.3)Similarly, if max( a − , a + ) = a + , then ˜ D ( a + ) ≤ iff µ ≤ κ ( m − m − ) + 4 ν ( m − m − + m + m − K − ( a + ) , where a + ≥ a − . (4.4)If a − = a + , then (4.3) and (4.4) become µ ≤ m ( µ − + 4 γ − ) /m − + 2 γ − + 2 γ + . Condition (ii) . Let min( κ − , κ + ) = κ − . By (4.1), we have ˜ D ( κ − ) = µ − m κ − + γ + (1 − e iθ + ( κ − ) ) = µ − m κ − + m + ( κ − − K ( κ − )) . Note that ℜ ˜ D ( ω ) > ℜ ˜ D ( ω ) for | ω | < | ω | < min( κ − , κ + ) , and ℜ ˜ D (0) > . Then, ˜ D ( ω ) =0 for | ω | < κ − iff ˜ D ( κ − ) ≥ . In turn, ˜ D ( κ − ) ≥ iff µ ≥ κ − ( m − m + ) + m + K ( κ − ) , where κ − = 0 . (4.5) Condition (iii) . Let a − < κ + . Note that ℜ ˜ D ( ω ) < ℜ ˜ D ( ω ) for | ω | > | ω | . Therefore, ˜ D ( ω ) = 0 for a − < | ω | < κ + iff either ˜ D ( κ + ) ≥ or ˜ D ( a − ) ≤ . then ˜ D ( κ + ) ≥ or ˜ D ( a − ) ≤ . Using (4.2), we have ˜ D ( κ + ) = µ − m κ + γ − (1 − e iθ − ( κ + ) ) = µ − m κ + m − (cid:0) κ − K − ( κ + ) (cid:1) . ˜ D ( κ + ) ≥ ⇐⇒ µ ≥ κ ( m − m − ) + m − K − ( κ + ) , where a − < κ + . (4.6)Secondly, using (4.1), we have ˜ D ( a − ) = µ − m a − + 2 γ − + γ + (1 − e iθ + ( a − ) )= µ − m ( κ − + 4 ν − ) + 2 m − ν − + m + (cid:0) κ − + 2 ν − − K ( a − ) (cid:1) . Therefore, ˜ D ( a − ) ≤ ⇐⇒ µ ≤ κ − ( m − m + )+4 ν − (cid:16) m − m − + m + (cid:17) + m + K ( a − ) . (4.7) Remark 4.4
The formulas (4.3)–(4.7) are the part of the restrictions on the constants inconditions C and C . If restrictions (4.3)–(4.7) are not fulfilled, then there is a point ω ∗ ∈ R \ { Λ ∪ } such that ˜ D ( ω ∗ ) = 0 . Furthermore, ˜ D ′ ( ω ∗ ) = 0 , since ˜ D ′ ( ω ∗ ) = − | ω ∗ | m + X ± γ ± e −ℑ θ ± ( ω ∗ ) | κ ± − ω ∗ | − / | a ± − ω ∗ | − / ! . Hence, ˜ N ( ω ) has simple poles at points ω = ± ω ∗ . Then, the bound (2.41) does not hold. ˜ D ( ω ) near points in Λ Now we study the asymptotic behavior of ˜ D ( ω ) near singular points in Λ using the followingformulas e iθ ± ( ω ) = 1 + iν ± q ω − κ ± − ν ± ( ω − κ ± ) + . . . , ω → κ ± , (4.8)where ω ∈ C + , ℑ ( p ω − κ ± ) > . Here sign( ℜ p ω − κ ± ) = sign( ℜ ω ) for ω ∈ C + . Thischoice of the branch of the complex root p ω − κ ± follows from the condition ℑ θ ± ( ω ) > .Similarly, e iθ ± ( ω ) = − iν ± q a ± − ω + 12 ν ± ( a − ω ) − i ν ± ( a ± − ω ) / + . . . (4.9)for ω → a ± , ω ∈ C + . Here the branch of the complex root p a ± − ω is chosen so that sign( ℜ p a ± − ω ) = sign( ℜ ω ) , by the condition ℑ θ ± ( ω ) > . If κ ± = 0 , then e iθ ± ( ω ) = ( iων ± − ω ν ± − iω ν ± + . . . for ω → − i p ν ± − ω /ν ± + . . . for ω → ν ± (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ω ∈ C + . (4.10) Lemma 4.5
Let ω ∈ Λ and ω = 0 . If condition C holds, then ˜ N ( ω ) = C + iC ( ω − ω ) / + . . . as ω → ± ω , ω ∈ C ± , (4.11) with some C , C = 0 . If condition C holds, then either (4.11) is true or ˜ N ( ω ) = iC ( ω − ω ) − / + C + . . . as ω → ± ω , ω ∈ C ± , (4.12)19 ith some C = 0 .Let ω = 0 ∈ Λ and conditions C or C hold. Then ˜ N ( ω ) = C + iC ω + . . . as ω → , ω ∈ C ± , (4.13) with some C , C = 0 . Proof (1)
Let ω = ± κ − . We consider the following cases: (1.1) κ − = κ + = 0 ; (1.2) κ − = κ + = 0 ; (1.3) κ − = 0 , κ + > ; (1.4) < κ − < κ + . (1.1) If κ − = κ + = 0 , then we apply the representation (4.10) to e iθ ± ( ω ) and obtain ˜ D ( ω ) = µ − iω ( √ m − γ − + √ m + γ + ) + . . . as ω → . In particular, ˜ D (0) = µ = 0 by condition C or C . Hence, (4.13) is true with C = µ − .Note that if µ ± = µ = 0 (this case is excluded by conditions C and C ), then ˜ N ( ω ) has asimple pole at zero, ˜ N ( ω ) = iω ( √ m − γ − + √ m + γ + ) + m − + m + − m √ m − γ − + √ m + γ + ) + . . . , ω → . (4.14) (1.2) If κ − = κ + = 0 , then we apply the representation (4.8) to e iθ ± ( ω ) and obtain ˜ D ( ω ) = ˜ D ( ± κ − ) − i ( √ m − γ − + √ m + γ + )( ω − κ − ) / + . . . as ω → ± κ − , ω ∈ C + , where ˜ D ( ± κ − ) = µ − m κ − = 0 iff µ = m κ − (or µ m = µ − m − ). Hence, if condition C holds,then (4.11) is true with ω = ± κ − , C = ( ˜ D ( ± κ − )) − . If condition C (iv) holds, then (4.12)is true. (1.3) If κ − = 0 and κ + > , then the function e iθ + ( ω ) is an analytic function in a smallneighborhood of origin. Applying the representation (4.10) to e iθ − ( ω ) , we obtain ˜ D ( ω ) = ˜ D (0) + c ω + c ω + . . . , ω → , with ˜ D (0) = µ + γ + (cid:16) − e −ℑ θ + (0) (cid:17) and some constants c , c . ˜ D (0) > , since e −ℑ θ + (0) = 4 κ + ν + + s κ ν ! − < if κ + > . (4.15)Hence, (4.13) holds with C = ( ˜ D (0)) − > . (1.4) If < κ − < κ + , then e iθ + ( ω ) is an analytic function in a small neighborhood of thepoints ω = ± κ − . Applying the representation (4.8) to e iθ − ( ω ) , we obtain ˜ D ( ω ) = ˜ D ( ± κ − ) − i √ m − γ − ( ω − κ − ) / + O ( | ω ∓ κ − | ) as ω → ± κ − ,ω ∈ C + , and (4.1) gives ˜ D ( ± κ − ) = µ − m κ − + m + ( κ − − K ( κ − )) = µ − κ − ( m − m + ) − m + K ( κ − ) . ˜ D ( ± κ − ) = 0 ⇐⇒ µ = κ − ( m − m + ) + m + K ( κ − ) . Therefore, if condition C holds, then (4.11) is true with ω = ± κ − . If condition C (iv) holds, then (4.12) is true. (2) Let ω = ± κ + . There are five cases: (2.1) κ + > a − ; (2.2) κ + = a − ; (2.3) κ − <κ + < a − ; (2.4) κ − = κ + = 0 (see case (1.1) ); (2.5) κ − = κ + = 0 (see case (1.2) ). (2.1) Let κ + > a − . Then the function e iθ − ( ω ) = − e −ℑ θ − ( ω ) is analytic in a small neighbor-hood of the points ω = ± κ + . Applying the representation (4.8) to e iθ + ( ω ) , we obtain ˜ D ( ω ) = ˜ D ( ± κ + ) − i √ m + γ + ( ω − κ ) / + O ( | ω ∓ κ + | ) as ω → ± κ + , ω ∈ C + . Using (4.2), we obtain ˜ D ( ± κ + ) = µ − m κ + γ − (1 + e −ℑ θ − ( κ + ) ) = µ − κ ( m − m − ) − m − K − ( κ + ) . Hence, ˜ D ( ± κ + ) = 0 ⇐⇒ µ = κ ( m − m − ) + m − K − ( κ + ) . Therefore, condition C implies (4.11) with ω = ± κ + , C = ( ˜ D ( ± κ + )) − . If condition C (v) holds, then (4.12) is true. (2.2) If κ + = a − , then we apply (4.8) and (4.9) to e iθ + ( ω ) and e iθ − ( ω ) , respectively, andobtain ˜ D ( ω ) = ˜ D ( ± κ + ) − ( κ − ω ) / ( √ m + γ + + i √ m − γ − ) + O ( | ω ∓ κ + | ) , ω → ± κ + ,ω ∈ C + . Here ˜ D ( ± κ + ) = µ − m κ + 2 γ − . Therefore, condition C implies (4.11) with ω = ± κ + . If condition C (v) holds, then (4.12). (2.3) If κ + ∈ ( κ − , a − ) , then we apply (4.8) to e iθ + ( ω ) and obtain ˜ D ( ω ) = ˜ D ( κ + ± i − i √ m + γ + ( ω − κ ) / + O ( | ω − κ + | ) , ω → κ + ± i , where ˜ D ( κ + ± i
0) = µ − m κ + γ − (cid:0) − e ± iθ − ( κ + + i (cid:1) . Therefore, ℑ ˜ D ( κ + ± i
0) = ∓ γ − sin θ − ( κ + + i ∓ m − (cid:26) p κ − κ − p a − − κ , if κ − > κ + p a − − κ , if κ − = 0 Then, ℑ ˜ D ( κ + ± i = 0 , and (4.11) follows. (3) Let ω = ± a − . There are the following cases: (3.1) a − > a + ; (3.2) a − = a + ; (3.3) a − ∈ ( κ + , a + ) ; (3.4) a − = κ + (see the case (2.2) ); (3.5) a − < κ + . (3.1) If a − > a + , then the function e iθ + ( ω ) = − e −ℑ θ + ( ω ) is analytic in a small neighborhoodof the points ω = ± a − . We apply (4.9) to e iθ − ( ω ) and obtain ˜ D ( ω ) = ˜ D ( ± a − ) − i √ m − γ − ( a − − ω ) / + O ( | ω ∓ a − | ) , ω → ± a − , ω ∈ C + , where (see (4.2)) ˜ D ( a − ) = µ − m a − + 2 γ − + γ + (1 − e iθ + ( a − ) )= µ − m ( κ − + 4 ν − ) + 2 m − ν − + m + (cid:0) κ − + 2 ν − − K + ( a − ) (cid:1) . ˜ D ( ± a − ) = 0 ⇐⇒ µ = κ − ( m − m + ) + 4 ν − (cid:18) m − m − + m + (cid:19) + m + K + ( a − ) . If condition C holds, then (4.11) is true with ω = ± a − , C = ( ˜ D ( ± a − )) − . If condition C (i) holds, then (4.12) is true. (3.2) In the case when a − = a + , we apply (4.9) to e iθ ± ( ω ) and obtain ˜ D ( ω ) = ˜ D ( ± a − ) − i ( √ m − γ − + √ m + γ + )( a − − ω ) / + O ( | ω ∓ a − | ) , ω → ± a − , where ˜ D ( ± a − ) = µ − m a − + 2 γ − + 2 γ + . Hence, ˜ D ( ± a − ) = 0 ⇐⇒ µ = m a − − γ − − γ + . Condition C implies (4.11) with ω = ± a − . If condition C (iii) holds, then (4.12) is truewith ω = ± a − . (3.3) If a − ∈ ( κ + , a + ) , then we apply (4.9) to e iθ − ( ω ) and obtain ˜ D ( ω ) = ˜ D ( a − ± i − i √ m − γ − ( a − − ω ) / + O ( | ω − a − | ) , ω → a − ± i , where ˜ D ( a − ± i
0) = µ − m a − + 2 γ − + γ + (cid:0) − e ± iθ + ( a − + i (cid:1) with ℑ θ + ( a − + i
0) = 0 . ℑ ˜ D ( a − ± i = 0 , since ℑ ˜ D ( a − ± i
0) = ∓ m + (cid:26) p a − − κ p a − a − , if κ + > , a − p ν − ν − , if κ + = κ − = 0 . Hence, ˜ D ( a − + i = 0 and (4.11) follows. Similarly, for ω → − a − ± i . (3.5) If a − < κ + , then e iθ + ( ω ) is an analytic function in a small neighborhood of the points ω = ± a − . Applying the representation (4.9) to e iθ − ( ω ) , we obtain ˜ D ( ω ) = ˜ D ( ± a − ) − i √ m − γ − ( a − − ω ) / + O ( | ω ∓ a − | ) , ω → ± a − , ω ∈ C + , where (see (4.1)) ˜ D ( a − ) = µ − m a − + 2 γ − + γ + (1 − e iθ + ( a − ) )= µ − κ − ( m − m + ) − ν − (cid:18) m − m − + m + (cid:19) − m + K ( a − ) . Hence, ˜ D ( ± a − ) = 0 ⇐⇒ µ = κ − ( m − m + ) + 4 ν − (cid:18) m − m − + m + (cid:19) + m + K ( a − ) . Condition C implies (4.11) with ω = ± a − . If condition C (vi) holds, then (4.12) is truewith ω = ± a − . (4) Let ω = ± a + . There are the following cases: (4.1) a + < a − ; (4.2) a + = a − (seecase (3.2) ); (4.3) a + > a − . 22 If a + < a − , then a + ∈ Λ − \ Λ − . Similarly to case (3.3) , we obtain ˜ D ( ω ) = ˜ D ( a + ± i − i √ m − γ − ( a − ω ) / + O ( | ω − a + | ) , ω → a + ± i . ˜ D ( a + ± i
0) = µ − m a + 2 γ + + γ − (cid:0) − e ± iθ − ( a + + i (cid:1) with ℑ θ − ( a + + i
0) = 0 . Since ℑ ˜ D ( a + ± i
0) = ∓ m − (cid:26) p a − κ − p a − − a , if κ − > a + p ν − − ν , if κ − = 0 , ℑ ˜ D ( a + ± i = 0 . Hence, ˜ D ( a + + i = 0 and (4.11) follows. Similarly, for ω → − a + ± i . (4.3) If a + > a − , then e iθ − ( ω ) is an analytic function in a small neighborhood of the points ω = ± a + . Using (4.9), we have ˜ D ( ω ) = ˜ D ( ± a + ) − i √ m − γ − ( a − ω ) / + O ( | ω ∓ a + | ) , ω → ± a + . where by (4.2), ˜ D ( a + ) = µ − m a + 2 γ + + γ − (1 + e −ℑ θ − ( a + ) )= µ − κ ( m − m − ) − ν (cid:18) m − m − + m + (cid:19) − m − K − ( a + ) . Hence, ˜ D ( ± a + ) = 0 ⇐⇒ µ = κ ( m − m − ) + 4 ν (cid:18) m − m − + m + (cid:19) + m − K − ( a + ) . Condition C implies (4.11) with ω = ± a + . If condition C (ii) holds, then (4.12) is truewith ω = ± a + . Lemma 4.5 is proved. By Lemma 5.1, we can vary the contour of integration in (2.40) as follows: N ( t ) = − π Z | ω | = R e − iωt ˜ N ( ω ) dω, t > . (4.16)where the number R is chosen enough large, R > max { a − , a + } . By Lemmas 5.1 and 4.2, werewrite N ( t ) in the form N ( t ) = − π Z Λ ε e − iωt ˜ N ( ω ) dω, t > , where ε ∈ (0 , δ ) , the contour Λ ε surrounds the segments of Λ and belongs to ε -neighborhoodof Λ (the contour Λ ε is oriented anticlockwise). Passing to a limit as ε → , we obtain N ( t ) = 12 π Z Λ e − iωt (cid:16) ˜ N ( ω + i − ˜ N ( ω − i (cid:17) dω = X ± X j =1 π Z Λ e − iωt P ± j ( ω ) dω + o ( t − K ) , t → + ∞ , with any K > . P ± j ( ω ) := ζ ± j ( ω )( ˜ N ( ω + i − ˜ N ( ω − i , ζ ± j ( ω ) are smooth functions such that P ± ,j ζ ± j ( ω ) = 1 , ω ∈ R , supp ζ ± ⊂ O ( ± κ ± ) , supp ζ ± ⊂ O ( ± a ± ) ( O ( b ) denotes a neighbor-hood of the point ω = b ). In the case κ ± = 0 , instead of ζ ± ( P ± ) we introduce the function ζ (respectively, P ) with supp ζ ⊂ O (0) . Then Lemma 4.5 implies the bound (2.41) with k = 0 . Here we use the following estimate (cid:12)(cid:12)(cid:12) Z R ζ ( ω ) e − iωt ( a − ω ) j/ dω (cid:12)(cid:12)(cid:12) ≤ C (1 + t ) − − j/ при t → + ∞ , where j = ± ( j = − , if condition C holds, j = 1 , if condition C holds), ζ ( ω ) is asmooth function, and ζ ( ω ) = 1 for | ω − a | ≤ δ with some δ > . The bound (2.41) with k = 1 , can be proved by a similar way. N = 0 The cases of the constants when conditions C and C are not fulfilled we call by resonancecases . In these cases, N ( t ) = C + C sin( at + b ) + O ( t − / ) as t → ∞ with some constants C , C , a, b . In the particular case (P2) , the resonance cases were studied in [3]. Let us considerthe particular case (P1) . Then, the resonance cases are the following three cases:case ( R : κ = κ = 0 ;case ( R : ( κ , κ ) = (0 , and a > a ;case ( R : κ < κ .Resonance cases can be rewritten in the terms of restrictions on m , m, γ, µ , µ as follows • µ = µ = 0 • ( µ , µ ) = (0 , and m < m • µ < µ We construct the solutions u ( · , t ) which do not satisfy the bound (1.7) and (2.39). In thecase (R1), ˜ N ( ω ) has a simple pole at zero, ˜ N ( ω ) = i ω √ γ m + m − m γ m + . . . as ω → . Hence, N ( t ) = (2 √ γ m ) − + O ( t − / ) as t → ∞ . Suppose that Y ( n ) ≡ for n = 0 . Then, Y ∈ H α for any α and z ( · , t ) ≡ . Using (2.14) and (2.42), we obtain u (0 , t ) = r (0 , t ) = v (0)2 √ γ m + O ( t − / ) as t → ∞ . Hence, if v (0) = 0 , then the constructed solution u ( n, t ) does not satisfy the bound (2.39).In the cases (R2) and (R3), there exists a point ω ∈ R \ Λ such that ˜ N ( ω ) has simplepoles at the points ω = ± ω with ω ∈ R \ Λ . Hence, N ( t ) = C sin( ω t )+ O ( t − / ) as t → ∞ .Note that ℑ θ ( ω ) > , ℜ θ ( ω ) = ± π . Therefore, the function of the form u ( n, t ) = e iθ ( ω ) | n | sin( ω t ) , t ≥ is a solution of the system. However, this solution does not satisfy the bound (2.39).24 Appendix B: Case N ≥ ˜ D ( ω ) for ω ∈ C ± We prove the following lemma for any N ≥ without the additional condition (2.36). Lemma 5.1 (i) ˜ N ( ω ) is meromorphic-valued matrix for ω ∈ C \ Λ .(ii) | ˜ N ( ω ) | = O ( | ω | − ) as | ω | → ∞ .(iii) det ˜ D ( ω ) = 0 for all ω ∈ C ± . Proof
The first assertion of the lemma follows from the formula (2.32) and the analyticity of ˜ D ( ω ) for ω ∈ C \ Λ . By (2.23) and the condition ℑ θ ± ( ω ) > , we have (cid:12)(cid:12) e iθ ± ( ω ) (cid:12)(cid:12) ≤ C | ω | − as | ω | → ∞ . (5.1)Then the second assertion follows from (2.32) and (5.1). The third assertion follows from theenergy bound (2.13). Indeed, let us assume that det ˜ D ( ω ) = 0 for some ω ∈ C + . Then,there exists a nonzero vector ξ = ( ξ , . . . , ξ N ) such that ˜ D ( ω ) ξ = 0 . (5.2)Introduce a function u ∗ ( n, t ) , n ∈ Z , t ≥ , as u ∗ ( n, t ) = e − iω t e − iθ − ( ω ) n ξ , n ≤ ,e − iω t ξ , n = 1 ,. . . . . .e − iω t ξ N − , n = N − ,e − iω t e iθ + ( ω )( n − N ) ξ N , n ≥ N. Using (2.23) and (5.2), it is easy to check that Y ∗ ( t ) = ( u ∗ ( · , t ) , v ∗ ( · , t )) , where v ∗ ( n, t ) = m n ˙ u ∗ ( n, t ) , is a solution of the problem (1.2) with the initial data Y ∗ ( n ) = e − iθ − ( ω ) n ξ (1 , − im − ω ) , n ≤ ,ξ (1 , − im ω ) , n = 1 ,. . . . . .ξ N − (1 , − im N ω ) , n = N − ,e iθ + ( ω )( n − N ) ξ N (1 , − im + ω ) , n ≥ N. Therefore, the Hamiltonian (see (1.1)) is H ( Y ∗ ( t )) = e t ℑ ω H ( Y ∗ ( · )) for any t > , where H ( Y ∗ ) > . Since ℑ ω > and Y ∗ ∈ H , this exponential growth contradicts the energy estimate (2.13).Hence, det ˜ D ( ω ) = 0 for any ω ∈ C + . Since θ ± ( ω ) = − θ ± (¯ ω ) for ω ∈ C \ Λ , then ˜ D ( ω ) =˜ D (¯ ω ) . Therefore, det ˜ D ( ω ) = 0 for any ω ∈ C − .25 .2 ˜ D ( ω ) for ω ∈ Λ \ Λ For simplicity of the further calculations, we impose condition (2.36). Then, the system (1.3)–(1.5) is of a form (cid:26) m ¨ u ( n, t ) = ( γ ∆ L − µ ) u ( n, t ) for n ≤ − , n ≥ N + 1 ,m n ¨ u ( n, t ) = γ n ∇ L u ( n, t ) − γ n − ∇ L u ( n − , t ) − µ n u ( n, t ) for n = 0 , , . . . , N, with γ N = γ − ≡ γ . Set ν := ν ± , κ := κ ± , a := a ± , where ν = γ/m , κ = µ/m , a = ( µ + 4 γ ) /m , θ ( ω ) := θ ± ( ω ) , Λ = [ − a, − κ ] ∪ [ κ, a ] , Λ = {± κ, ± a } . Lemma 5.2 det ˜ D ( ω ± i = 0 for ω ∈ Λ \ Λ . Proof
For N = 0 , we prove this result in Appendix A. Now we assume that N ≥ . For ω ∈ Λ \ Λ , ℜ θ ( ω + i ∈ ( − π, ∪ (0 , π ) and ℑ θ ( ω + i
0) = 0 . Moreover, sign(sin θ ( ω + i ω . Write y := ℑ ˜ D ( ω + i
0) = ℑ ˜ D NN ( ω + i
0) = − γ sin θ ( ω + i = 0; r := ℜ ˜ D ( ω + i , r N := ℜ ˜ D NN ( ω + i ,d n ≡ d n ( ω ) := ˜ D nn ( ω ) ∈ R , n = 1 , . . . , N − . Assume the contrary, that det ˜ D ( ω + i
0) = 0 . Then, by (5.19),det ˜ D ( ω + i
0) = α N ( ω + i
0) = ( r N + iy ) α N − − γ N − α N − , where α − = 0 , α − = 1 and α n ≡ α n ( ω + i for n = 0 , . . . , N , α n ( ω ) for ω ∈ C aredefined in (5.20). Hence, (cid:26) r N ℜ α N − − y ℑ α N − − γ N − ℜ α N − = 0 y ℜ α N − + r N ℑ α N − − γ N − ℑ α N − = 0 (5.3)At first, we prove that equalities in (5.3) is impossible in the case N = 1 . Indeed, in this case, α N − ≡ α = r + iy , α N − ≡ α − = 1 , and the system (5.3) becomes r r − y − γ = 0 (5.4) yr + yr = 0 (5.5)By (5.5), r = − r , since y = 0 . Then, using (5.4), we have − r − y − γ = 0 what isimpossible. Therefore, det ˜ D ( ω + i = 0 for any ω ∈ Λ \ Λ in the case N = 1 .Now we prove that equalities in (5.3) is impossible if N ≥ . For N ≥ , introduce thefollowing determinants ∆ jk ≡ ∆ jk ( ω ) by the rule ∆ jk ( ω ) = det d j − γ j . . . − γ j d j +1 − γ j +1 . . . − γ j +1 . . . . . . ... . . . . . . d k − − γ k − . . . − γ k − d k , j ≤ kk = 1 , . . . , N − N ≥ . (5.6)26he matrices in (5.6) are real-valued symmetric and tridiagonal (i.e., normal Jakobi matrices).Moreover, for fixed j , the minors { ˜ α p ≡ ∆ jk | k = j + p , p = 0 , , . . . } satisfy the recurrence relation(cf (5.19)) ˜ α p = d j + p ˜ α p − − γ j + p − ˜ α p − , p = 0 , , . . . , (5.7)with the initial conditions ˜ α − = 0 and ˜ α − = 1 . In turn, for fixed k , the minors { ˜ β p ≡ ∆ jk | j = p , p = k, k − , . . . } satisfy the three-term recurrence (cf (5.22)) ˜ β p = d p ˜ β p +1 − γ p ˜ β p +2 , p = k, k − , . . . , (5.8)with the initial conditions ˜ β k +1 = 1 , ˜ β k +2 = 0 . We return to the proof of Lemma 5.2 in thecase N ≥ . Applying notation (5.6), we have α = ( r + iy ) d − γ , α k = ( r + iy )∆ k − γ ∆ k , k = 2 , . . . , N − , N ≥ . Since y = 0 and ∆ jk ∈ R for ω ∈ R , the system (5.3) becomes r N ℜ α N − − y ∆ N − − γ N − ℜ α N − = 0 (5.9) ℜ α N − + r N ∆ N − − γ N − ∆ N − = 0 (5.10)where ℜ α k = r ∆ k − γ ∆ k , ∆ ≡ and ∆ ≡ .Assume that ∆ N − = 0 . Then, ∆ N − · ∆ N − > . This inequality follows from two factsfor determinants ∆ jk : (1) if ∆ N − = 0 , then ∆ N − · ∆ N − = 0 by the properties of theminors of Jakobi matrices or by the relations (5.7) and (5.8); (2) nonzero major minors of sameorder (as ∆ N − and ∆ N − ) have the same sign (see, e.g., [5, p.31, p.83 in Russian edition]).However, this inequality contradicts to (5.10), because by (5.10) and ∆ N − = 0 , one obtains − γ ∆ N − − γ N − ∆ N − = 0 and then, sign ∆ N − = − sign ∆ N − .If ∆ N − = 0 , then we express r N from Eqn (5.10) and substitute in (5.9). For N = 2 , if ∆ N − ≡ ∆ = d = 0 , then r = γ ∆ − ℜ α ∆ = γ − ( x d − γ ) d and Eqn (5.10) becomes − ( x d − γ ) − y d − γ γ d = 0 , what is impossible since γ γ = 0 . For N ≥ , we use the following equality ∆ N − · ∆ N − − ∆ N − · ∆ N − = γ · . . . · γ N − , where N ≥ , ∆ ≡ , which can be proved by induction. Therefore, Eqn (5.9) writes − ( ℜ α N − ) − y (cid:0) ∆ N − (cid:1) − γ γ · . . . · γ N − ∆ N − = 0 what is impossible. Thus, det ˜ D ( ω + i = 0 for any ω ∈ Λ \ Λ . Since ˜ D ( ω − i
0) = ˜ D ( ω + i for ω ∈ Λ \ Λ , det ˜ D ( ω − i = 0 for ω ∈ Λ \ Λ .27 .3 ˜ D ( ω ) for ω ∈ R \ Λ For ω ∈ R \ Λ , the matrix ˜ D ( ω ) is symmetric and real-valued. As before, we write d n ( ω ) =˜ D nn ( ω ) . We consider separately two cases of values of ω : | ω | > a and | ω | < κ (if κ = 0 ).For | ω | > a , the following result holds. Lemma 5.3
Write R a := { ω ∈ R : | ω | > a } . The following assertions are equivalent. (A1) det ˜ D ( ω ) = 0 for any ω ∈ R a . (A2) ( − N det ˜ D ( ω ) < for any ω ∈ R a . (A3) For every n = 0 , , . . . , N , α n ( ω )( − n < for any ω ∈ R a , i.e., the symmetric matrix ˜ D ( ω ) is negative definite for ω ∈ R a . (A4) For every n = 0 , , . . . , N , β n ( ω )( − N + n < for any ω ∈ R a . (A5) For every n = 0 , , . . . , N , c n ( ω ) < for any ω ∈ R a . (A6) α n ( a )( − n < for n = 0 , , . . . , N − , and α N ( a )( − N ≤ . (A7) c n ( a ) < for n = 1 , . . . , N − , and c N ( a ) ≤ . Proof
For ω ∈ R a , ˜ D ( ω ) is a real-valued symmetric (Jakobi) matrix with diagonal terms ofa form d ( ω ) = µ − m ω + γ (1 + e −ℑ θ ( ω ) ) + γ ,d n ( ω ) = µ n − m n ω + γ n + γ n − , n = 0 , . . . N − ,d N ( ω ) = µ N − m N ω + γ (1 + e −ℑ θ ( ω ) ) + γ N − . Note that for ω ∈ R a , d n ( ω ) < d n ( ω ) for | ω | > | ω | and d n ( ω ) → −∞ as ω → ±∞ . (5.11)It follows from (5.11) and (5.25) that all c n ( ω ) are even, strictly decrease for ω > a and c n ( ω ) → −∞ for ω → ±∞ . Furthermore, (5.11) (or (5.24)) implies that ( − N det ˜ D ( ω ) → −∞ as ω → ±∞ . (5.12)Hence, assertion (A1) is equivalent to (A2). Furthermore, (A7) = ⇒ (A5). By formula (5.24),(A5) implies (A2). Evidently, (A3) = ⇒ (A2). Assertions (A3), (A4) and (A5) are equivalentby Remark 5.7, and (A6) ⇐⇒ (A7). Therefore, (A6) = ⇒ (A2).It remains to prove that (A2) implies (A3) and (A6). Assume, for simplicity, that N = 1 .Let assertion (A2) hold. Then, α ( a ) ≡ d ( a ) ≤ . Indeed, if d ( a ) > , then, by (5.11),there is a point ω > a such that d ( ω ) = 0 . Hence, det ˜ D ( ω ) = − γ < . This contradictsto (A2). Similarly, we can check that (A2) implies that d ( a ) ≤ . Hence, d ( ω ) < and d ( ω ) < for all ω ∈ R a , by (5.11), and (A3) holds. Moreover, α ′ ( ω ) ≡ (det ˜ D ( ω )) ′ > for ω > a . Therefore, α ( a ) ≥ . If α ( a ) = 0 , then α ( a ) = − γ < what is impossible. Hence, α ( a ) > and (A6) is true. In the case N ≥ , the proof of implications (A2) = ⇒ (A3) and(A2) = ⇒ (A6) is similar and is based on two facts: α n ( ω )( − n → −∞ as ω → ±∞ , (5.13)28nd 2) if for some point ω , α n ( ω ) = 0 , then α n − ( ω ) α n +1 ( ω ) < by virtue of (5.19).Now we prove these implications for N = 2 , . Assume that α ( a ) ≡ d ( a ) > . It followsfrom (5.11) that there exists a point ω > a such that α ( ω ) = 0 and α ( ω ) < for all ω > ω . (5.14)Then, α ( ω ) = − γ < . Therefore, by (5.13), there is a point ω > ω such that α ( ω ) = 0 . (5.15)Hence, α ( ω ) = − γ α ( ω ) > , (5.16)by (5.14). If N = 2 , then (5.16) contradicts (A2). Hence, α ( a ) ≤ and α ( ω ) < for any | ω | > a, (5.17)by (5.11). Moreover, from reasonings above we see that α ( ω ) > for all ω ∈ R a and (A3) istrue. If α ( a ) = 0 , then α ( a ) = − γ < and there is a point ω > a such that α ( ω ) = 0 that is impossible by (A3). Hence, α ( a ) < . Also, α ( a ) = 0 , since if α ( a ) = 0 , then α ( a ) = − α α ( a ) > , what contradicts (A3). Therefore, (A6) is true. If N = 3 , then by(5.13) and (5.16), there is a point ω > ω such that α ( ω ) = 0 . Hence, α ( ω ) α ( ω ) < by (5.19). We can choose a point ω in (5.15) such that α ( ω ) > for any ω > ω . Hence, α ( ω ) < . But this inequality contradicts (A2) with N = 3 . Hence, (5.17) is valid. Moreover,using reasonings above we obtain that α ( ω ) < and α ( ω ) > for any ω ∈ R a and (A3) istrue.Now we check (A6). Since c n ( ω ) < for any n and ω ∈ R a , then ( − n α n ( ω ) < forany ω ∈ R a by virtue to Remark 5.7 (2). Hence, ( − n α n ( a ) ≤ for any n. (5.18)It remains to prove that α n ( a ) = 0 for n = N . We check this fact by induction. Indeed, if α ( a ) = 0 , then α ( a ) < , what is impossible by (5.18). Hence, α ( a ) < . Assume that ( − k α k ( a ) < for k ≤ n − . If α n ( a ) = 0 , then ( − n − α n − ( a )( − n +1 α n +1 ( a ) < .Hence, ( − n +1 α n +1 ( a ) > what contradicts (5.18).Let κ = 0 . At first, we prove the following auxiliary lemma. Lemma 5.4 If κ = 0 , then c n (0) > and α n (0) > for any n = 0 , , . . . , N . Proof
At first note that K := 1 − e − iθ (0) = 4 κν + r κ ν ! − < . Then, if N = 0 , then by (2.33), we have ˜ D (0) ≡ c (0) ≡ α (0) = µ + 2 γK > . If N ≥ , then d (0) = µ + γK + γ > γ ,d n (0) = µ n + γ n + γ n − ≥ γ n + γ n − for n = 1 , . . . , N − ,d N (0) = µ N + γK + γ N − ≥ γK + γ N − . c n (0) > γ n for n = 0 , , . . . , N − , and c N (0) > γK > . Since α (0) = c (0) and α n (0) = α n − (0) c n (0) , then α n (0) > γ · . . . · γ n for n = 0 , . . . , N − and α N (0) > γ · . . . · γ N − γK > . By Sylvester’s criterion, the symmetric matrix ˜ D (0) is positivedefinite. Lemma 5.5
Write R κ := { ω ∈ R : | ω | < κ } . Then the following assertions are equivalent.(A1) det ˜ D ( ω ) = 0 for any ω ∈ R κ .(A2) det ˜ D ( ω ) > for any ω ∈ R κ .(A3) For every n = 0 , , . . . , N , α n ( ω ) > for any ω ∈ R κ , i.e., the matrix ˜ D ( ω ) ispositive definite for ω ∈ R κ .(A4) For every n = 0 , , . . . , N , β n ( ω ) > for any ω ∈ R κ .(A5) For every n = 0 , , . . . , N , c n ( ω ) > for any ω ∈ R κ .(A6) α n ( κ ) > for n = 0 , , . . . , N − , and α N ( κ ) ≥ .(A7) c n ( κ ) > for n = 1 , . . . , N − , and c N ( κ ) ≥ . Proof
For ω ∈ R κ , all functions d n ( ω ) ∈ R and d n ( ω ) < d n ( ω ) for | ω | < | ω | < κ .Hence, all c n ( ω ) are even, strictly decrease for ω ∈ [0 , κ ) . Furthermore, ˜ D (0) is positivedefinite by Lemma 5.4. Hence, (A1) ⇐⇒ (A2); (A3) = ⇒ (A2); (A3) ⇐⇒ (A4) ⇐⇒ (A5);(A7) = ⇒ (A5); (A6) = ⇒ (A2); (A6) ⇐⇒ (A7). The proof of implications (A2) = ⇒ (A3)and (A2) = ⇒ (A6) can be proved by a similar way as Lemma 5.3. Remark 5.6 (i) If det ˜ D ( ω ) = 0 for any ω ∈ R : | ω | > a , then d n ( ω ) < for | ω | ≥ a .Furthermore, ( − n ( α n ( ω )) ′ < for ω > a .(ii) If det ˜ D ( ω ) = 0 for any ω ∈ R : | ω | < κ , then d n ( ω ) > for | ω | ≤ κ .(iii) If assertions (A2) of Lemmas 5.3 and 5.5 are not fulfilled, then there exists a point ω ∗ ∈ R \ { Λ ∪ } such that det ˜ D ( ω ∗ ) = 0 and the entries of the inverse matrix ˜ N ( ω ) havepoles at points ω = ± ω ∗ . Then, bound (2.41) does not hold.(iv) By Lemmas 5.3 and 5.5, α ( a ) < and α ( κ ) > . Then, m > m/ . Similarly,since β N ( a ) < and β N ( κ ) > , then m N > m/ . ˜ D ( ω ) and ˜ N ( ω ) Let N ≥ . If ω ∈ R \ Λ , then ˜ D ( ω ) ∈ R . Furthermore, ˜ D ( ω ) is a symmetric tridiagonalmatrix with non-diagonal entries ˜ D ii +1 ( ω ) = − γ i = 0 . If ˜ D ( ω ) ∈ R , then ˜ D ( ω ) is a well-known normal Jakobi matrix. Tridiagonal matrices are widely studied in the literature, see e.g.,[5, 7]. We mark some interesting facts on the matrix ˜ D ( ω ) . Write d i ≡ d i ( ω ) = ˜ D ii ( ω ) ∈ R .To find the determinant of ˜ D ( ω ) , we can use the three-term recurrence relation α i = d i α i − − γ i − α i − , i = 0 , . . . , N, (5.19)with initial conditions α − = 0 and α − = 1 . Here α i , i = 0 , . . . , N , is the i -th leading(principal corner) minor of the matrix ˜ D ( ω ) , i.e., α i = det d − γ . . . − γ d − γ . . . − γ d . . . ... . . . . . . . . . − γ i − . . . − γ i − d i (5.20)30imilarly, introduce the sequence { β i } Ni =0 of the determinants by the rule β i = det d i − γ i . . . − γ i d i +1 − γ i +1 . . . − γ i +1 d i +2 . . . ... . . . . . . . . . − γ N − . . . − γ N − d N (5.21)Then β i satisfies the three-term recurrence (cid:26) β i = d i β i +1 − γ i β i +2 , i = 0 , . . . , N,β N +1 = 1 , β N +2 = 0 . (5.22)In particular, det ˜ D ( ω ) = α N ( ω ) = β ( ω ) . The entries of the symmetric matrix ˜ N ( ω ) = (cid:16) ˜ D ( ω ) (cid:17) − are of a form (see [7, Theorem 1]) ˜ N ij ( ω ) = γ j γ j +1 . . . γ i − α j − ( ω ) β i +1 ( ω ) α N ( ω ) , j = 0 , , , . . . , i − ,α i − ( ω ) β i +1 ( ω ) α N ( ω ) j = i (5.23) i = N, N − , . . . , . Remark 5.7 (1) Note that if α n = 0 for some n = 1 , , . . . , then α n − α n +1 < . The lastinequality can be proved using induction, recurrence (5.19) and the fact that all γ n > .(2) The following relation holds, α k ( ω ) = c ( ω ) · . . . · c k ( ω ) . In particular, det ˜ D ( ω ) ≡ α N ( ω ) = c ( ω ) · . . . · c N ( ω ) , (5.24) where c n ≡ c n ( ω ) = d ( ω ) , n = 0 ,d n ( ω ) − γ n − c n − , n = 0 . (5.25) (3) c n = α n /α n − , n = 0 , , . . . , N .(4) The following relation holds, β k ( ω ) = e k ( ω ) · . . . · e N ( ω ) . In particular, det ˜ D ( ω ) ≡ β ( ω ) = e ( ω ) · . . . · e N ( ω ) , where e n ≡ e n ( ω ) = d N ( ω ) , n = N,d n ( ω ) − γ n e n +1 , n = N, and e n = β n /β n +1 , n = 0 , , . . . , N . .5 Asymptotics of N ( t ) for large times To prove Theorem 2.12, it remains to study the behavior of ˜ N ( ω ) near singular points in Λ .Let µ = 0 . Then κ = 0 and ∈ Λ . We apply the representation (4.10) to e iθ ( ω ) andobtain that ˜ D nn ( ω ) ≡ d n ( ω ) = d n (0) − iω √ mγ − ω ( m n − m/
2) + . . . as ω → , for n = 0 , n = N, where d (0) = µ + γ , d N (0) = µ N + γ N − . Using (5.19) and (5.22), we find the asymptoticsof determinants α k ( ω ) and β k ( ω ) as α k ( ω ) = α k (0) − iA k (0) √ mγ ω + . . . , k = 0 , . . . , N − ,β k ( ω ) = β k (0) − iB k (0) √ mγ ω + . . . , k = 1 , . . . , N, (cid:12)(cid:12)(cid:12)(cid:12) ω → , (5.26)where α k (0) > for k = 0 , . . . , N − , β k (0) > for k = 1 , . . . , N , A ≡ , B N ≡ , A k ( ω ) = ∆ k , B k ( ω ) = ∆ kN − , k = 1 , . . . , N − , where ∆ jk are defined in (5.6).If µ = . . . = µ N = 0 , then det ˜ D (0) = 0 and α N ( ω ) ≡ det ˜ D ( ω ) = − iC √ mγω + . . . as ω → , with C = β (0) + α N − (0) = 2 β (0) = 2 γ · . . . · γ N − . Furthermore, α k (0) = γ · . . . · γ k , k = 0 , . . . , N − ,β k (0) = γ k − · . . . · γ N − , k = 1 , . . . , N. Therefore, by (5.23) and (5.26), the entries ˜ N nk ( ω ) have a simple pole at zero, ˜ N nk ( ω ) = i √ γm ω + C nk + . . . as ω → , n, k = 0 , . . . , N, (5.27)where C nk are some constants. If there is nonzero µ n for some n ∈ { , . . . , N } , thendet ˜ D (0) ≥ µ n γ · . . . · γ N − > . Therefore, (4.13) follows by conditions C and C .Let κ = 0 . Then we apply the representation (4.8) to e iθ ( ω ) and obtain d n ( ω ) = d n ( κ ) − i √ mγ ( ω − κ ) / + . . . as ω → κ, ω ∈ C + , n = 0; N, where d ( κ ) = µ − m κ + γ , d N ( κ ) = µ N − m N κ + γ N − . Therefore, the asymptotics ofdeterminants α k ( ω ) and β k ( ω ) is α k ( ω ) = α k ( κ ) − iA k ( κ ) √ mγ ( ω − κ ) / + . . . , k = 0 , . . . , N − ,β k ( ω ) = β k ( κ ) − iB k ( κ ) √ mγ ( ω − κ ) / + . . . , k = 1 , . . . , N. (cid:12)(cid:12)(cid:12)(cid:12) ω → κ. If det ˜ D ( κ ) = 0 , then condition C holds and we obtain the representation (4.11). If det ˜ D ( κ ) =0 , then condition C holds and we obtain the representation (4.12).For ω → a , we apply the representation (4.9) to e iθ ( ω ) and obtain d n ( ω ) = d n ( a ) − i √ mγ ( a − ω ) / + . . . as ω → a, ω ∈ C + , n = 0; N, where d ( a ) = µ − m a + 2 γ + γ , d N ( a ) = µ N − m N a + 2 γ + γ N − . If det ˜ D ( a ) = 0 , thencondition C holds and we obtain the representation (4.11), if det ˜ D ( a ) = 0 , then condition C holds and we obtain the representation (4.12). Therefore, Theorem 2.12 in the case N ≥ canbe proved by a similar way as in the case N = 0 , see Appendix A.32 .6 Resonance cases: N ≥ In the case of N = 0 , the resonance cases are considered in Section 4.6. Now we considerthe case N ≥ and construct the solutions u ( n, t ) which do not satisfy the bound (2.39). Ifconditions C and C are not satisfied, then there are two possible cases: (1) µ = µ = . . . = µ N = 0 ; (2) There is a point ω ∈ R \ Λ such that det ˜ D ( ω ) = 0 .In the case (1) , ˜ N nk ( ω ) have a simple pole at zero and, by (5.27), N nk ( t ) = (2 √ γ m ) − + O ( t − / ) as t → ∞ , n, k = 0 , , . . . , N. Suppose that the initial data Y ( n ) ≡ for n = { , . . . , N } . Then, Y ∈ H α for any α and z ( n, t ) ≡ for any n , where z ( n, t ) is defined in (2.15). Using (2.14) and (2.42), we obtain u (0 , t ) = r (0 , t ) = 12 √ γ m N X k =0 v ( k ) + O ( t − / ) as t → ∞ . Hence, if v (0) + . . . + v ( N ) = 0 , then the solution u ( · , t ) does not satisfy the bound (2.39).In the case (2) , there is a nonzero vector ξ = ( ξ , . . . , ξ N ) ∈ R N +1 such that det ˜ D ( ω ) ξ = 0 .Note that for ω ∈ R \ Λ , ℑ θ ( ω ) > , ℜ θ ( ω ) = ± π . Therefore, the function of the form u ( n, t ) = v ( n ) sin( ω t ) , where v ( n ) = e − iθ ( ω ) n ξ if n ≤ ξ n if n = 1 , . . . , N − e iθ ( ω )( n − N ) ξ N if n ≥ N, is a solution of the system with the initial data Y = (0 , v ) , where v ( n ) = ω m n v ( n ) with m n = m for n ≤ − and n ≥ N + 1 . Note that Y ∈ H α for any α . However, k u ( · , t ) k α = k v k α | sin( ω t ) | ≥ C | sin( ω t ) | with some constant C > . Therefore, the bounds (1.7) and(2.39) are not fulfilled for this solution. References [1] T.V. Dudnikova, Long-time asymptotics of solutions to a Hamiltonian system on a lattice,
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