aa r X i v : . [ m a t h . O C ] J un On Zero Controllability of EvolutionEquations
B. ShklyarNovember 21, 2018
Abstract
The exact controllability to the origin for linear evolution controlequation is considered.The problem is investigated by its transforma-tion to infinite linear moment problem.Conditions for the existence of solution for infinite linear momentproblem has been obtained. The obtained results are applied to thezero controllability for control evolution equations.
Introduction
Let X be a separable complex Hilbert space.Given sequences { c n , n = 1 , . . . , } and { x n ∈ X, n = 1 , . . . , } find nec-essary and sufficient conditions for the existence of an element g ∈ X suchthat c n = ( x n , g ) , n = 1 , . . . , . The problem formulated above is called the linear moment problem. Ithas a long history and many applications in geometry, physics, mechanics.The goal of this paper is to establish necessary and sufficient condi-tions of exact null-controllability for linear evolution control equations withunbounded input operator by transformation of exact null-controllabilityproblem (controllability to the origin) to linear infinite moment problem.It is well-known, that if the sequence { x n , n = 1 , , ..., } forms a Rieszbasic in the closure of its linear span, the linear moment problem has asolution if and only if P ∞ n =1 | c n | < ∞ and vice-versa [3], [7], [16], [17].This well-known fact is one of main tools for the controllability analysisof various partial hyperbolic control equations and functional differentialcontrol systems of neutral type. 1owever the sequence { x n , n = 1 , , ..., } doesn’t need to be a Riesz basicfor the solvability of linear moment problem. This case appears under the in-vestigation of the controllability of parabolic control equations or hereditaryfunctional differential control systems. In this paper we consider the zerocontrollability of control evolution equations for the case when the sequence { x n , n = 1 , , ..., } of the moment problem obtained by the transformationof the source control problem doesn’t form a Riesz basic in its closed linearspan. Let
X, U be complex Hilbert spaces, and let A be infinitesimal generator ofstrongly continuous C -semigroups S ( t ) in X [8],[10]. Consider the abstractevolution control equation [8], [10]˙ x ( t ) = Ax ( t ) + Bu ( t ) , x (0) = x , ≤ t < + ∞ , (1.1)where x ( t ) , x ∈ X, u ( t ) , u ∈ U, B : U → X is a linear possibly un-bounded operator, W ⊂ X ⊂ V are Hilbert spaces with continuous denseinjections, where W = D ( A ) equipped with graphic norm, V = W ∗ , theoperator B is a bounded operator from U to V (see more details in [14], [4],[11], [15]).It is well-known that [4], [11],[14], [15]), etc. : • for each t ≥ S ( t ) has an unique continuous extension S ( t ) on the space V and the family of operators S ( t ) : V → V is the semi-group in the class C with respect to the norm of V and the correspondinginfinitesimal generator A of the semigroup S ( t ) is the closed dense extensionof the operator A on the space V with domain D ( A ) = X ; • the sets of eigenvalues and of generalized eigenvectors of operators A , A ∗ and A, A ∗ are the same; • for each µ / ∈ σ ( A ) the resolvent operator R A ( µ ) has a unique contin-uous extension to the resolvent operator R A ( µ ) : V → X ; • a mild solution x ( t, x , u ( · )) of equation (1.1) with initial condition x (0) = x is obtained by the following representation formula x ( t, x , u ( · )) = S ( t ) x + t Z S ( t − τ ) Bu ( τ ) dτ , (1.2)where the integral in (2.3) is understood in the Bochner’s sense [8]. Toassure x ( t, x , u ( · )) ∈ X, ∀ x ∈ X, u ( · ) ∈ L loc2 [0 , + ∞ ) , t ≥ , we assume2hat t R S ( t − τ ) Bu ( τ ) dτ ∈ X for any u ( · ) ∈ L loc2 [0 , + ∞ ) , t ≥ Definition 1.1
Equation (1.1) is said to be exact null-controllable on [0 , t ] by controls vanishing after time moment t if for each x ∈ X there exists acontrol u ( · ) ∈ L ([0 , t ] , U ) , u ( t ) = 0 a.e. on [ t , + ∞ ) such that x ( t , x , u ( · )) = 0 . (1.3) The assumptions on A are listed below.1. The operators A has purely point spectrum σ with no finite limitpoints. Eigenvalues of A have finite multiplicities.2. There exists T ≥ x ( t ) = Ax ( t ) are expanded in a series of generalized eigenvectors ofthe operator A converging uniformly for any t ∈ [ T , T ] , T < T < T . For the sake of simplicity we consider the following:1. The operator A has all the eigenvalues with multiplicity 1.2. U = R (one input case). It means that the possibly unbounded opera-tor B : U → R is defined by an element b ∈ V , i.e. equation (1.1) canbe written in the form˙ x ( t ) = Ax ( t ) + bu ( t ) , x (0) = x , b ∈ V, ≤ t < + ∞ . (2.1)The operator defined by b ∈ V is bounded if and only if b ∈ X. Let the eigenvalues λ j ∈ σ, j = 1 , , . . . of the operator A be enu-merated in the order of non-decreasing of their absolute values, and let ϕ j , ψ j , j = 1 , , . . . , be eigenvectors of the operator A and the adjoint oper-ator A ∗ respectively. It is well-known, that( ϕ k , ψ j ) = δ kj , j, k = 1 , . . . , (2.2)3here δ kj , j, k = 1 , . . . is the Kroneker delta.Denote: x j ( t ) = (cid:0) x ( t, x , u ( · )) , ψ j (cid:1) , x j = (cid:0) x , ψ j (cid:1) , b j = (cid:0) b, ψ j (cid:1) , j = 1 , , .... (2.3)All scalar products in (2.3) are correctly defined, because ψ j ∈ W, b ∈ V = W ∗ . Theorem 2.1
For equation (1.1) to be exact null-controllable on [0 , t ] ,t > T, by controls vanishing after time moment t − T , it is necessary andsufficient that the following infinite moment problem x j = − Z t − T e − λ j τ b j u ( τ ) dτ , j = 1 , , ... (2.4) with respect to u ( · ) ∈ L [0 , t − T ] is solvable for any x ∈ X . Proof. Necessity.
Multiplying (1.1) by ψ j , j = 1 , , ..., and using (2.3)we obtain˙ x j ( t ) = (cid:0) Ax ( t ) , ψ j (cid:1) + b j u ( t ) = (cid:0) x ( t ) , A ∗ ψ j (cid:1) + b j u ( t ) == λ j x j ( t ) + b j u ( t ) , j = 1 , , ..., . (2.5)Here x j ( t ) , ˙ x j ( t ) and b j , j = 1 , , ..., are well-defined because ψ j ∈ W, ˙ x ( t ) , Ax ( t ) , b ∈ V = W ∗ . From (2.5) it follows that x j ( t ) = e λ j t (cid:18) x j + Z t e − λ j t b j u ( τ ) dτ (cid:19) , j = 1 , , ..., . (2.6)In accordance with the definition of exact null-controllability there exists u ( · ) ∈ L ([0 , t − T ] , U ) , u ( t ) = 0 a.e. on [ t − T, + ∞ ) such that (1.3)holds. Using u ( t ) and t in (2.6), we obtain by (1.3) and (2.5), that x j ( t ) = e λ j t (cid:18) x j + Z t − T e − λ j t b j u ( τ ) dτ (cid:19) = 0 , j = 1 , , ..., . (2.7)Hence we have (2.4) to be true. This proves the necessity. Sufficiency.
Let the control u ( · ) ∈ L ([0 , t − T ] , U ) , u ( t ) = 0 a.e. on[ t − T, + ∞ ) satisfies (2.4). It follows from (2.4) and (2.7) that x j ( t − T ) = (cid:0) x ( t − T ) , ψ j (cid:1) = 0 , j = 1 , , .... (2.8)4enote z ( t ) = x ( t + t − T ) , t ≥ T. Obviously, z ( t ) is a mild solutionof the equation ˙ z ( t ) = Az ( t ) with initial condition z (0) = x ( t − T ) . Byassumption 3 (see the list of assumptions) z ( t ) is expanded in a series z ( t ) = ∞ X j =1 e λ j t (cid:0) x ( t − T ) , ψ j (cid:1) , t ≥ T, (2.9)so by (2.8) and (2.9) we obtain z ( t ) = x ( t + t − T, x , u ( · )) ≡ , t ≥ T ⇔ x ( t, x , u ( · )) ≡ , t ≥ t . This proves the sufficiency.
The solvability of moment problem (2.4) for each x ∈ X essentially dependson the properties of eigenvalues λ j , j = 1 , , ..., . If the sequence of exponents (cid:8) e − λ n t b n , n = 1 , , ..., (cid:9) forms a Riesz basicin L [0 , t − T ] , then the moment problem c j = − Z t − T e − λ j τ b j u ( τ ) dτ , j = 1 , , ... (2.10)is solvable if and only if ∞ X j =1 | c j | < ∞ (2.11)There are very large number of papers and books devoted to conditions forsequence of exponents to be a Riesz basic. All these conditions can be usedfor sufficient conditions of zero controllability of equation (1.1). They arevery useful for the investigation of the zero controllability of hyperbolic par-tial control equations and functional differential control systems of neutraltype [13].However moment problem (2.10) may also be solvable when the sequence (cid:8) e − λ n t b n , n = 1 , , ..., (cid:9) doesn’t form a Riesz basic in L [0 , t − T ] . Below wewill try to find more extended controllability conditions which are applicablefor the case when the sequence (cid:8) e − λ n t b n , n = 1 , , ..., (cid:9) doesn’t form a Rieszbasic in L [0 , t − T ] . Definition 2.1
The sequence { x j ∈ X, j = 1 , , ..., } is said to be minimal,if there no element of the sequence belonging to the closure of the linear spanof others. By other words, x j / ∈ span { x k ∈ X, k = 1 , , ..., k = j } . Theorem
Let x j ∈ X, j = 1 , , ..., . The linear moment problem c j = ( x j , g ) , j = 1 , , ... has a solution g ∈ X for each square summable sequence { c j , j = 1 , , ... } ifand only if there exists a positive constant γ such that all the inequalities γ n X k =1 | c k | ≤ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) n X j =1 c j x j (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) , n = 1 , , ..., . (2.12) are valid. Let { x j ∈ X, j = 1 , , ..., } a sequence of elements of X , and let G n = { ( x i , x j ) , i, j = 1 , , ..., n } be the Gram matrix of n first elements { x , ..., x n } of above sequence. De-note by γ min n the minimal eigenvalue of the n × n -matrix G n . Each mini-mal sequence { x j ∈ X, j = 1 , , ..., } is linear independent, hence any first n elements { x , ..., x n } , n = 1 , , ..., of this sequence are linear indepen-dent, so γ min n > , ∀ n = 1 , , ..., . It is easily to show that the sequence (cid:8) γ min n , n = 1 , , ..., (cid:9) decreases , so there exists lim n →∞ γ min n ≥ . Definition 2.2
The sequence { x j ∈ X, j = 1 , , ..., } is said to be stronglyminimal, if γ min = lim n →∞ γ min n > . It is well-known that for Hermitian n × n -matrix G n = { ( x j , x k ) , j, k = 1 , , ..., n } γ min n n X k =1 | c k | ≤ n X j =1 n X k =1 c j ( x j , x k ) c k , n = 1 , , ..., . (2.13)From the well-known formula P mj =1 P mk =1 c j ( x j , x k ) c k = (cid:13)(cid:13)(cid:13)P mj =1 c j x j (cid:13)(cid:13)(cid:13) , (2.12)and the inequality γ min n ≥ γ min > γ min n X k =1 | c k | ≤ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) n X j =1 c j x j (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) (2.14)Hence the above theorem can be reformulated as follows6 heorem 2.2 The linear moment problem c j = ( x j , g ) , j = 1 , , ... (2.15) has a solution g ∈ X for any sequence { c n , n = 1 , , ... } , ∞ P j =1 c j < ∞ if andonly if the sequence { x n , n = 1 , , .., } is strongly minimal. Theorem 3.1
For equation (1.1) to be exact null-controllable on [0 , t ] ,t > T, by controls vanishing after time moment t − T , it is necessary, thatthe sequence n e − λ j τ b j , t ∈ [0 , t − T ] , j = 1 , , ..., o (3.1) is minimal, and sufficient , that: • the sequence (cid:8) e − λ j τ b j , t ∈ [0 , t − T ] , j = 1 , , ... (cid:9) is strongly minimal; • ∞ X j =1 (cid:12)(cid:12)(cid:0) x , ψ j (cid:1)(cid:12)(cid:12) < + ∞ , ∀ x ∈ X. (3.2) Proof. Necessity.
If the problem (2.4) has a solution for any x ∈ X, then it has a solution for any eigenvector ϕ k , k = 1 , , ..., of the operator A, so for each k = 1 , , ..., there exists a function u k ( · ) ∈ L [0 , t − T ] suchthat (cid:0) ϕ k , ψ j (cid:1) = − Z t − T e − λ j τ b j u k ( τ ) dτ , j = 1 , , ..., . (3.3)The sequence { ϕ k , k = 1 , , ..., } of eigenvectors of the operator A is biorthog-onal to the sequence { ψ k , k = 1 , , ..., } of eigenvectors of the operator A ∗ . Hence it follows from (3.3) and (2.2) that δ jk = (cid:0) ϕ k , ψ j (cid:1) = − Z t − T e − λ j τ b j u k ( τ ) dτ , j = 1 , , ..., . i.e. the sequence {− u k ( t ) , t ∈ [0 , t − T ] , k = 1 , , ..., } is biorthogonal tothe sequence (cid:8) e − λ j t b j , t ∈ [0 , t − T ] , j = 1 , , ..., (cid:9) . It proves the necessity.
Sufficiency.
The sufficiency follows immediately from (3.2) and Theo-rem 2.2.It proves the theorem. 7 .1 The case of the strongly minimal sequence of eigenvec-tors of the operator A . Obviously the sequence of eigenvectors of the operator A being consideredis a minimal sequence.Below we consider the operator A having the strongly minimal sequenceof eigenvectors. Theorem 3.2
Let the sequence (cid:8) ϕ j , j = 1 , , ... (cid:9) of eigenvectors of the op-erator A be strongly minimal.For equation (1.1) to be exact null-controllable on [0 , t ] , t > T, bycontrols vanishing after time moment t − T , it is necessary, that the se-quence (cid:8) e − λ j τ b j , t ∈ [0 , t − T ] , j = 1 , , ... (cid:9) is minimal, and sufficient, that Re λ j ≥ β for some β ∈ R and the sequence (cid:8) e − λ j t b j , t ∈ [0 , t − T ] , j = 1 , , ... (cid:9) is strongly minimal. Proof.
The necessity follows from Theorem 3.1.
Sufficiency.
By Assumption 3 of the list of assumptions the series ∞ X j =1 (cid:0) x , ψ j (cid:1) e λjt ϕ j , ∀ t > T (3.4)converges. Since the sequence (cid:8) ϕ j , j = 1 , , ... (cid:9) of eigenvectors of the oper-ator A is strongly minimal, then on account of property (2.10 there exists anumber α such that α n X j =1 (cid:12)(cid:12)(cid:0) x , ψ j (cid:1)(cid:12)(cid:12) e λ j t ≤ n X j =1 n X k =1 (cid:0) x , ψ j (cid:1) e λjt (cid:0) ϕ j , ϕ k (cid:1) ( x , ψ k ) e λ k t , (3.5) ∀ x ∈ X, ∀ n ∈ N , ∀ t > T. It follows from (3.4) and (3.5) that ∞ X j =1 (cid:12)(cid:12)(cid:0) x , ψ j (cid:1)(cid:12)(cid:12) e λ j t < + ∞ , ∀ x ∈ X, ∀ t > T. (3.6)As Re λ j ≥ β for some β ∈ R , we have by (3.6) that (3.2) holds.In accordance with Theorem 3.1 condition (3.2) and the strong mini-mality of the sequence (3.1) imply the exact null-controllability of equation(1.1). It proves the theorem. 8 .2 The case when the eigenvectors of the operator A forma Riesz basic One of the important problems of the operator theory is the case when thegeneralized eigenvectors of the operator A being considered form a Rieszbasic in X. The problem of expansion into a Riesz basic of eigenvectors ofthe operator A is widely investigated in the literature (see, for example, [1],[6], [7], [12] and references therein). Obviously the sequence of these vectorsis strongly minimal. In this case one can set T = 0 , so the Theorems 3.1,3.2 and Lemma 3.1 can be proven with T = 0 . Theorem 3.3
Let the sequence of operator A forms a Riesz basic in X. For equation (1.1) to be exact null-controllable on [0 , t ] , t > T, bycontrols vanishing after time moment t − T , it is necessary and sufficient,that the sequence sequence (cid:8) e − λ j t b j , t ∈ [0 , t − T ] , j = 1 , , ... (cid:9) is stronglyminimal . Proof.
Let { c j , j = 1 , , ..., } be any complex sequence satisfying thecondition P ∞ j =1 | c j | < ∞ . Since the sequence (cid:8) ϕ j , j = 1 , , ..., (cid:9) of eigenvectors of the operator A forms the Riesz basic, there exists a vector x ∈ X such that c j = (cid:0) x , ψ j (cid:1) , j = 1 , , ..., so in virtue of Theorem 2.1 the exact null controllability being consideredin the paper is equivalent to the solvability of the linear moment problem c j = Z t − T e − λ j τ b j u ( τ ) dτ , j = 1 , , ..., (3.7)for any complex sequence { c j , j = 1 , , ..., } satisfying the condition P ∞ j =1 | c j | < ∞ . By above mentioned results of [2] and [3] the linear moment problem (3.7)is solvable for any complex sequence { c j , j = 1 , , ..., } satisfying the condi-tion P ∞ j =1 | c j | < ∞ if and only if the sequence (cid:8) e − λ j t b j , t ∈ [0 , t − T ] , j = 1 , , ... (cid:9) is strongly minimal . It proves the theorem.Obviously, the condition b j = 0 , j = 1 , , ..., is the necessary conditionfor the solvability of the moment problem (2.1). Lemma 3.1
If the sequence n e − λ j t , t ∈ [0 , t − T ] , j = 1 , , ... o (3.8)9 s strongly minimal and inf n ∈ N | b n | = β > holds, then the sequence (cid:8) e − λ j t b j , t ∈ [0 , t − T ] , j = 1 , , ... (cid:9) is also stronglyminimal. Proof.
Let the sequence (cid:8) e − λ j t , t ∈ [0 , t − T ] , j = 1 , , ... (cid:9) be stronglyminimal. From (2.12) it follows that α n X k =1 | c k | | b j | ≤ Z t − T (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X j =1 c j e − λ j t b j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) dt (3.10)for some positive α and for every finite sequence { c , c , ..., c n } . By (3.9)and (3.10) we have γ n X k =1 | c k | ≤ Z t − T (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X j =1 c j e − λ j t b j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) dt, n = 1 , , ..., γ = αβ > . (3.11)where γ = αβ > . It proves the lemma.
Example of strongly minimal sequence.
Below we will prove thatthe sequence n e n π t , n = 1 , , ..., t ∈ [0 , t ] o is strongly minimal for any t > . Let t = 2 t . The series P ∞ n =1 1 n π converges and ( n + 1) − n ≥ n e n π t , n = 1 , , ..., t ∈ [0 , t ] o is minimal [5]. In virtue ofTheorem 1.5 of [5] for each ε > K ε suchthat the biorthogonal sequence { w n ( t ) , n = 1 , , ..., t ∈ [0 , t ] } satisfies thecondition k w n ( · ) k < K ε e εn π , n = 1 , , ..., . (3.12)The positive constant ε can be chosen such that t − ε > . By the Minkowsky inequality and (3.12) one can show that P pn =1 P pm =1 c n e − n π t (cid:16)R t w n ( t ) w m ( t ) dt (cid:17) e − m π t c m = R t (cid:16)P pn =1 c n e − n π t w n ( t ) dt (cid:17) dt ≤≤ R t P pn =1 | c n | P pn =1 (cid:12)(cid:12)(cid:12) e − n π t w n ( t ) (cid:12)(cid:12)(cid:12) dt = P pn =1 | c n | P pn =1 e − n π t R t | w n ( t ) | dt ≤≤ P pn =1 | c n | P pn =1 e − n π t k w n ( · ) k ≤ K ε P pn =1 | c n | P pn =1 e − n π ( t − ε ) . The series P ∞ n =1 e − n π ( t − ε ) converges for any t , ε, t > ε, so P pn =1 e − n π ( t − ε ) ≤ M , where M is a positive constant.Hence 10 X n =1 p X m =1 c n e − n π t (cid:18)Z t w n ( t ) w m ( t ) dt (cid:19) e − m π t c m ≤ K ε M p X n =1 | c n | (3.13)for every finite sequence { c , c , ..., c p } . Obviously the sequence (cid:26) h n ( t ) = (cid:26) e − n π t w n ( t − t ) , t ∈ [ t , t ] , , t ∈ [0 , t ) , n = 1 , , ..., (cid:27) is the biorthog-onal sequence to the sequence n e n π t , n = 1 , , ..., t ∈ [0 , t ] o , and (cid:16)R t h n ( t ) h m ( t ) dt (cid:17) = e − n π t (cid:16)R t t w n ( t − t ) w m ( t − t ) dt (cid:17) e − m π t = e − n π t (cid:16)R t w n ( t ) w m ( t ) dt (cid:17) e − m π t , so it follows from (3.13) that p X n =1 p X m =1 c n (cid:18)Z t h n ( t ) h m ( t ) dt (cid:19) c m ≤ K ε M p X n =1 | c n | . Hence [9] p X n =1 p X m =1 c n (cid:18)Z t e n π τ e m π τ (cid:19) c m dτ ≥ γ p X n =1 | c n | , p = 1 , , ..., (3.14)for every finite sequence { c , c , ..., c p } , where γ = K ε M > . It proves thatthe sequence n e n π t , t ∈ [0 , t ] , n = 1 , , ... o is strongly minimal for any t > As was said at the end of the previous section the condition lim n →∞ λ min n > { y j ∈ X, j = 1 , , ... } can be approximated in the some sense by strongly minimal sequence { x j ∈ X, j = 1 , , ... } , Theorem 4.1
If the sequence { x j ∈ X, j = 1 , , ... } is strongly minimal, letthe sequence { y j ∈ X, j = 1 , , ... } be such that the sequence { P n y j − x j , j = 1 , , ... } is linear independent and (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) n X j =1 c j ( y j − x j ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ q (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) n X j =1 c j x j , (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) , n = 1 , , ... , (4.1) where { c j , j = 1 , , ... } is any sequence of complex numbers, q is a constant, < q < , then the sequence { y j ∈ X, j = 1 , , ... } also is strongly minimal. Proof . Let { c k , k = 1 , , ... } be an arbitrary sequence of complex num-ber. Denote: x = n X k =1 c k x k , x = n X k =1 c k ( x k − y k ) . (4.2)From (4.2) it follows, that x = x + n X k =1 c k y k , n = 1 , , .... (4.3)By (4.1) we obtain that (cid:13)(cid:13) x (cid:13)(cid:13) ≤ q (cid:13)(cid:13) x (cid:13)(cid:13) . (4.4)Hence using (4.4) in (4.3) we obtain (cid:13)(cid:13) x (cid:13)(cid:13) ≤ − q (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) n X k =1 c k y k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) , n = 1 , , .... (4.5)Since the sequence { x j ∈ X, j = 1 , , ... } is strongly minimal and x = P nk =1 c k x k , we have n X k =1 | c k | ≤ α (cid:13)(cid:13) x (cid:13)(cid:13) , n = 1 , , ..., (4.6)for some α > . By (4.6) and (4.5) we obtain α P nk =1 | c k | ≤ − q k P nk =1 c k y k k , n =1 , , ..., so 12 (1 − q ) n X k =1 | c k | ! ≤ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) n X k =1 c k y k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) , n = 1 , , ..., . (4.7)Using in (4.7) the formula (2.14) we obtain γ n X k =1 | c k | ! ≤ n X k =1 n X l =1 c k ( y k , y l ) c l , γ = α (1 − q ) > µ [ n ]min be a minimal eigenvalue of the Gram matrix G n = { ( y k , y l ) , k, l = 1 , .... } for the sequence { y j , j = 1 , , ..., n } . From (4.8), it follows that lim n →∞ µ [ n ]min ≥ γ > . This proves the theorem.
Let X = l be the Hilbert space of square summable sequences. Considerthe evolution system (cid:26) ˙ x k ( t ) = λ k x k ( t ) + u ( t ) , k = 1 , , ..., < t < t ,x k (0) = x k , n = 1 , , ..., k = 1 , , ..., (4.9)where u ( t ) , < t < t is a scalar control function, { x k ( t ) , k = 1 , , ..., } , { x k , k = 1 , , ..., } ∈ l , the complex numbers λ k ,k = 1 , , ..., belong to the strip { z ∈ C : | Re z | ≤ γ } , i.e. | Re λ k | ≤ γ, k =1 , , ..., . Definition 4.1
Equation (4.9) is said to be exact null-controllable on [0 , t ] by controls vanishing after time moment t , if for each x ( · ) = { x k , k = 1 , , ..., } ∈ l there exists a control u ( · ) ∈ L [0 , t ] , u ( t ) = 0 a.e. on [ t , + ∞ ) such that x k ( t ) ≡ , k = 1 , , ..., ∀ t ≥ t . Control problem (4.9) can be written in the form of (1.1), where x ( t ) = { x k ( t ) , k = 1 , , ..., } ∈ l , u ( · ) ∈ L [0 , t ]; the self-adjoint operator A : l → l is defined for x = { x k , k = 1 , , ..., } ∈ l by Ax = { λ k x k , k = 1 , , ..., } (4.10)with domain D ( A ) = { x ∈ l : Ax ∈ l } , and the unbounded operator B isdefined by Bu = bu, u ∈ R , (4.11)13here b = { , , ..., , ... } / ∈ l .One can show that all the assumptions imposed on equation (1.1) arefulfilled for equation (4.9) with T = 0.Obviously, the numbers λ k , k = 1 , , ..., are eigenvalues of the operator A defined above; the sequences e k = , ..., , , , ..., | {z } k -th place are correspondingeigenvectors, forming the Riesz basic of l , so b j = 1 , j = 1 , , ..., . Together with system (4.9) consider the other evolution system (cid:26) ˙ x k ( t ) = µ k x k ( t ) + u ( t ) , n = 1 , , ..., < t < t ,x k (0) = x k , k = 1 , , ..., n = 1 , , ..., (4.12)where µ k = λ k + O (cid:18) k (cid:19) , k = 1 , , ..., . (4.13) Proposition 1
If system (4.9) is exact null-controllable on [0 , t ] by con-trols vanishing after time moment t , then the same is valid for system (4.12). Proof.
From the Caushy-Schvartz inequality it follows that R t (cid:12)(cid:12)P nk =1 c k (cid:0) e − µ k t − e − λ k t (cid:1)(cid:12)(cid:12) dt ≤ P nk =1 | c k | R t P nk =1 (cid:12)(cid:12) e − µ k t − e − λ k t (cid:12)(cid:12) dt == P nk =1 | c k | R t P nk =1 e − λ k t (cid:12)(cid:12)(cid:12) e O ( k ) t − (cid:12)(cid:12)(cid:12) dt ≤ P nk =1 | c k | R t e γt P nk =1 (cid:12)(cid:12)(cid:12) e O ( k ) t − (cid:12)(cid:12)(cid:12) dt. The series P ∞ k =1 (cid:12)(cid:12)(cid:12) e O ( k ) t − (cid:12)(cid:12)(cid:12) converges for any t ≥
0. Denote M ( t ) = Z t e γt ∞ X k =1 (cid:12)(cid:12)(cid:12) e O ( k ) t − (cid:12)(cid:12)(cid:12) dt. (4.14)Hence Z t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 c k (cid:16) e − µ k t − e − λ k t (cid:17)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) dt ≤ M ( t ) n X k =1 | c k | . (4.15)By Theorem 3.2 we have the sequence (cid:8) e − λ j t , t ∈ [0 , t ] , j = 1 , , ... (cid:9) to bestrongly minimal, so n X k =1 | c k | ≤ α Z t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 c k e − λ k t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) dt for some α > . (4.16)14oining (4.15) and (4.16) we obtain Z t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 c k (cid:16) e − µ k t − e − λ k t (cid:17)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) dt ≤ q Z t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 c k e − λ k t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) dt, (4.17)where q = M ( t ) α . Since from (4.14) it follows that lim t →∞ M ( t ) = 0 , one can choose thenumber t such that 0 < q < . Hence conditions (4.17) are the same as(4.1) for x k = e − λ k t , y k = e − µ k t , k = 1 , , ..., t ∈ [0 , t ] ; q = M ( t ) α . As it was said abov by Theorem 3.2 we have the sequence (cid:8) e − λ j t , t ∈ [0 , t ] , j = 1 , , ... (cid:9) to be strongly minimal .In accordance with Theorem 4.1 the sequence (cid:8) y k = e − µ k t , k = 1 , , ..., t ∈ [0 , t ] (cid:9) is also strongly minimal, provided that t is chosen such that M ( t ) α < . In accordance with Theorem 3.1 the strong minimality of the sequence (cid:8) y k = e − µ k t , k = 1 , , ..., t ∈ [0 , t ] (cid:9) provides the zero controllability of equa-tion (4.11) on [0 , t ] by controls vanishing after time moment t , M ( t ) α < , for any t ≥ t . References [1] N. Ahiezer, I. Glazman,
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