Lack of controllability of thermal systems with memory
aa r X i v : . [ c s . S Y ] A p r Lack of controllability of thermal systems withmemory ∗ Andrei Halanay † L. Pandolfi ‡ September 10, 2018
Abstract
Heat equations with memory of Gurtin-Pipkin type (i.e. Eq. (1)with α = 0) have controllability properties which strongly resemblethose of the wave equation. Instead, recent counterexamples showthat when α > L (Ω) which cannot be controlled to zero. The proof of thisfact consists in the construction of two quite special examples of sys-tems with memory which cannot be controlled to zero. Here we provethat lack of controllability holds in general, for every smooth memorykernel M ( t ). AMS classification:
The following integro-differential equation is often used to model thermalsystems with memory, see [6, 25]: ∗ The research of the first author is partially supported by Romanian CNCS Grant PN-II-ID-PCE-2011-3-0211. The research of the second author fits the plans of INDAM-CNRand of the project “Groupement de Recherche en Contrˆole des EDP entre la France etl’Italie (CONEDP)”. † Department of Mathematics and Informatics, University Politehnica of Bucharest, 313Splaiul Independentei, 060042 Bucharest, Romania ([email protected]) ‡ Dipartimento di Scienze Matematiche, Politecnico di Torino, Corso Duca degli Abruzzi24—10129 Torino, Italy (luciano.pandolfi@polito.it) t = α ∆ w + Z t M ( t − s )∆ w ( s ) d s , w (0) = ξ . (1)Here w = w ( x, t ) and x ∈ Ω, a bounded region with smooth boundary (werequire of class C , and Ω locally on one side of ∂ Ω).The time t = 0 is the time after which a boundary control f is applied tothe system, w ( x, t ) = f ( x, t ) x ∈ Γ = ∂ Ω , t > . Note that we implicitly assume that the system is at rest for negative times, w ( t ) = 0 if t < α is nonnegative. If α is zero then we get a model proposedby Gurtin and Pipkin in [11]. The controllability, when α = 0, has beenstudied in several paper, see references below. So, here we explicitly assume α > ξ = 0 canbe hitted at time T >
0, as it is the case for the memoryless heat equation,i.e. the special case of (CGM) obtained when M ( t ) ≡ Theorem 1
Let M ( t ) ∈ C (0 , + ∞ ) . For every f ∈ L (0 , T ; L (Γ)) and forevery initial condition ξ ∈ L (Ω) there exists a unique solution w ( · , T ) = w f,ξ ( · , T ) ∈ L (0 , T ; L (Ω)) . The solution is not continuous in time, see the examples in [18, p. 217], unless f ( t ) is smooth. So, pointwise computation of w ( · , t ) in L (Ω) is meaninglessin general. However, let A be the operator in L (Ω):dom A = H (Ω) ∩ H (Ω) , Aφ = ∆ φ . (2)Then we have: Corollary 2
Let M ( t ) ∈ C (0 , + ∞ ) . For every function f ∈ L (0 , T ; L (Γ)) and for every initial condition ξ ∈ L (Ω) , the function t A − w f,ξ ( · , t ) iscontinuous from [0 , + ∞ ) to L (Ω) . Definition 3
We say that the initial condition ξ is controllable to at time T if there exists f ∈ L (0 , T ; L (Γ)) such that A − w f,ξ ( · ; T ) = 0 ∈ L (Ω) .We say that (CGM) is null controllable at time T if for every ξ ∈ L (Ω) there exists f ∈ L (0 , T ; L (Γ)) such that A − w f,ξ ( · ; T ) = 0 ∈ L (Ω) . In the memoryless case, M ( t ) ≡
0, the system is null controllable at any time
T >
0. When M ( t ) = 0 but M ( t ) = 0 for 0 ≤ t ≤ T then Eq. (1)for t ≤ T coincide with the memoryless heat equation w t = α ∆ w and anyinitial condition can be controlled to 0 at any time T < T . Keeping this factin mind, our main result is: Theorem 4
Let α > and let M ( t ) ∈ C (0 , T ) , not identically zero. Let T be any time such that R ( T ) = 0 , where R ( t ) is the resolvent kernel of M ( t ) .There exist initial data ξ which cannot be controlled to at time T . Under smoothness assumption on the kernel M ( t ), when α = 0 and M (0) >
0, Eq. (1) can be seen as a perturbation of the wave equation and its proper-ties resemble those of the wave equation. In particular, the solutions belongto C (0 , + ∞ ; L (Ω)) for every f ∈ L (0 , T ; L (Γ)) and every initial condition ξ ∈ L (Ω). Furthermore, there exists T such that the reachable set (cid:8) w f, ( · , T ) , f ∈ L (0 , T ; L (Γ)) (cid:9) is equal to L (Ω). Several different techniques have been used in the proof,but the basic idea is always to compare with the wave equation, see [1, 15,19, 21, 23]. Furthermore, the infimum of the control times is the same asthat for the (memoryless) wave equation (see [3, 8, 15, 22, 24]).Instead, when α > t > T , where T is a preassigned time, see [14]. So, it is a natural conjecture that thecontrollability properties of system (1) with α > T > L (Ω) and this supports3he conjecture that every initial condition ξ ∈ L (Ω) can be controlled tohit the target ξ ( x ) ≡ T , of course without remainingequal to zero in the future, due to the negative results in [14]. Hence, thefollowing negative fact was a surprise: there exist kernels M ( t ) which areeven C ∞ , and such that for every T > every smooth kernel M ( t ) not identically zero, there exist initial conditionswhich cannot be controlled to zero, as stated in Theorem 4. The number α has to be positive and so, changing the time scale, i.e. replac-ing w ( x, t ) with w ( x, rt ), we can assume α = 1 . We present a transformation which simplifies the computations in this paper.We consider a Volterra integral equation on t ≥ y ( t ) + Z t M ( t − s ) y ( s ) d s = f ( t ) . It is known (see [9, Ch. 2]) that it is uniquely solvable for every squareintegrable f ( t ), and that the solution is given by y ( t ) = f ( t ) − Z t R ( t − s ) f ( s ) d s . The function R ( t ), the resolvent kernel of M ( t ), solves R ( t ) = M ( t ) − Z t M ( t − s ) R ( s ) d s . We apply formally this transformation, “solving” Eq. (1) with respect to the“unknown” ∆ w . We get w t = ∆ w + Z t R ( t − s ) w s ( s ) d s . w t = ∆ w + aw ( t ) + Z t L ( t − s ) w ( s ) d s − R ( t ) ξ , w (0) = ξ . (3)Here, a = R (0) = M (0) , L ( t ) = R ′ ( t ) . By, definition, a solution of Eq. (1) is a solution of the Volterra integro-differential equation (3) (solutions can be defined in several different butequivalent ways).Let us consider the operator A defined in (2). It is a selfadjoint operatorwith compact resolvent, which generates a holomorphic semigroup e At .Let D be the “Dirichlet operator”, u = Df ⇐⇒ u solves (cid:26) ∆ u = 0 in Ω u = f on Γ = ∂ Ω . A known fact (see [16, p. 180]) is the following:
Theorem 5
Let f ∈ L (0 , T ; L (Γ)) and ξ ∈ L (Ω) . The solution to theheat equation θ t = ∆ θ + g , θ ( x,
0) = ξ ( x ) , θ ( x, t ) = f ( x, t ) x ∈ ∂ Ω is given by θ ( · , t ) = θ f,ξ,g ( · , t ) = e At ξ + Z t e A ( t − s ) g ( s ) d s − A Z t e A ( t − s ) Df ( s ) d s . (4) The solution is unique in L (0 , + ∞ ; L (Ω)) and A − θ ( · , t ) ∈ C (0 , + ∞ ; L (Ω)) .Furthermore, if ξ = 0 then θ ( · , t ) ∈ L (0 , + ∞ , H / (Ω)) . We apply formula (4) to (3) with g ( t ) = aw ( t ) + Z t L ( t − s ) w ( s ) d s − R ( t ) ξ and we find the following Volterra integral equation for w ( x, t ): w ( x, t ) − Z t e A ( t − s ) (cid:20) aw ( s ) + Z s L ( s − r ) w ( r ) d r (cid:21) d s = (cid:26) e At ξ − Z t e A ( t − s ) R ( s ) ξ d s (cid:27) − A Z t e A ( t − s ) Df ( s ) d s (5)5heorem 1 and Corollary 2 follow from this formula, thanks to the prop-erties of the solutions of the (memoryless) heat equation stated in Theorem 5.See [17] for the theory of Volterra integral and integro-differential equa-tions in Banach spaces, and [5] for further information on the semigroupapproach to boundary value problems for parabolic equations. The previous results allows us to project the system on the eigenvectors ofthe operator A . Let { φ n } be an orthonormal basis of L (Ω), whose elementsare eigenvectors of the operator A in (2). So we have:∆ φ n = − λ n φ n , φ n ( x ) = 0 on Γ = ∂ Ω . Note that λ n > w n ( t ) = Z Ω w ( x, t ) φ n ( x ) d x ξ n = Z Ω ξ ( x ) φ n ( x ) d x . Then w n ( t ) solves w ′ n ( t ) = ( a − λ n ) w n + Z t L ( t − s ) w n ( s ) d s − R ( t ) ξ n − g n ( t )where g n ( t ) = Z Γ ( γ φ n ) f ( x, t ) d Γ( γ denotes normal derivative on Γ) and w ( x, t ) = X φ n ( x ) w n ( t ) . (6)We introduce µ n = λ n − a (we have µ n > n ) so that w n ( t ) − Z t e − µ n ( t − τ ) Z τ L ( τ − s ) w n ( s ) d s d τ = (cid:18) e − µ n t − Z t e − µ n ( t − s ) R ( s ) d s (cid:19) ξ n − Z t e − µ n ( t − s ) g n ( s ) d s . (7)6et T >
0. We define a transformation L in L (0 , T ; L (Ω)), as follows: L (cid:16)X φ n ( x ) h n ( t ) (cid:17) = X φ n ( x ) ( L n h n ) ( t )where ( L n h ) ( t ) = Z t e − µ n ( t − s ) Z s L ( s − r ) h ( r ) d r d s . Then we have( I − L ) w = X φ n ( x ) (cid:26)(cid:18) e − µ n t − Z t e − µ n ( t − s ) R ( s ) d s (cid:19) ξ n − Z t e − µ n ( t − s ) g n ( s ) d s (cid:27) . (8)We prove: Lemma 6
The transformation L in L (0 , T ; L (Ω)) is linear and continu-ous. The transformation ( I − L ) is invertible and its inverse is continuous. Proof.
Linearity is clear. We prove the continuity of L and of its inverse,using the fact that { φ n } is an orthonormal basis of L (Ω). This implies that (cid:13)(cid:13)(cid:13)(cid:16)X h n ( t ) φ n ( x ) (cid:17)(cid:13)(cid:13)(cid:13) L (0 ,T ; L (Ω)) = X Z T | h n ( t ) | d t . Then we have: Z T | ( L n h ) ( t ) | d t = Z T (cid:12)(cid:12)(cid:12)(cid:12)Z t e − µ n ( t − s ) Z s L ( s − r ) h ( r ) d r d s (cid:12)(cid:12)(cid:12)(cid:12) d t ≤ T (cid:18)Z T e − µ n s d s (cid:19) (cid:18)Z T L ( s ) d s (cid:19) (cid:18)Z T h ( r ) d r (cid:19) d s ≤ C Z T | h ( s ) | d s . We can chose the constant C independent of n thanks to the fact that µ n > n . So, we have (cid:13)(cid:13)(cid:13) L (cid:16)X h n ( t ) φ n ( x ) (cid:17)(cid:13)(cid:13)(cid:13) L (0 ,T ; L (Ω)) = Z T Z Ω (cid:12)(cid:12)(cid:12)X ( L n h n ) ( t ) φ n ( x ) (cid:12)(cid:12)(cid:12) d x d t = Z T X | ( L n h n ) ( t ) | d t ≤ C X Z T | h n ( s ) | d s = C (cid:13)(cid:13)(cid:13)(cid:16)X h n ( t ) φ n ( x ) (cid:17)(cid:13)(cid:13)(cid:13) L (0 ,T ; L (Ω)) . L and so also of I − L . Inorder to prove that this last transformation has a bounded inverse, we exibitexplicitly its inverse.To compute the inverse, we must solve, for every k ( x, t ) = P φ n ( x ) k n ( t ),( I − L ) (cid:16)X φ n ( x ) f n ( t ) (cid:17) = k ( x, t ) = X φ n ( x ) k n ( t )i.e. X φ n (cid:26) f n ( t ) − Z t f n ( τ ) Z t − τ L ( t − τ − s ) e − µ n s d s d τ (cid:27) = X φ n ( x ) k n ( t ) . We introduce H n ( t ), the resolvent kernels of Z n ( t ) = − Z t L ( t − s ) e − µ n s d s . (9)Then we must choose f n ( t ) = k n ( t ) − Z t H n ( t − s ) k n ( s ) d s and so( I − L ) − X φ n ( x ) k n ( t ) = X φ n ( x ) (cid:26) k n ( t ) − Z t H n ( t − s ) k n ( s ) d s (cid:27) . Continuity of this transformation is seen as above, using the fact that µ n > n , so that | Z n ( t ) | ≤ M/µ n (for large n ) where M = M T . So,Gronwall inequality applied to | H n ( t ) | ≤ | Z n ( t ) | + Z t | Z n ( s ) | · | H n ( s ) | d s gives | H n ( t ) | ≤ Mµ n , M = M T . Continuity now follows as above. 8sing (8) we find that w ( x, t ) = ( I − L ) − X φ n ( x ) (cid:26)(cid:18) e − µ n t − Z t e − µ n ( t − s ) R ( s ) d s (cid:19) ξ n − Z t e − µ n ( t − s ) g n ( s ) d s (cid:27) = X φ n ( x ) (cid:26) − (cid:20)Z t e − µ n ( t − s ) g n ( s ) d s + Z t H n ( t − τ ) Z τ e − µ n ( τ − s ) g n ( s ) d s d τ (cid:21) + (cid:20) e − µ n t − Z t e − µ n ( t − s ) R ( s ) d s − Z t H n ( t − τ ) (cid:18) e − µ n τ − Z τ e − µ n ( τ − s ) R ( s ) d s (cid:19) d τ (cid:21) ξ n (cid:27) (10)Now we recall the definition of controllability at time T and we can state: Theorem 7
Equation (1) is controllable to at time T if for every sequence { ξ n } ∈ l there exists a function f ∈ L (0 , T ; L (Γ)) which solves the follow-ing moment problem: (cid:20)Z T e − µ n ( T − s ) (cid:18)Z Γ ( γ φ n ) f ( x, s ) d Γ (cid:19) d s − Z T H n ( T − τ ) Z τ e − µ n ( τ − s ) (cid:18)Z Γ ( γ φ n ) f ( x, s ) d Γ (cid:19) d s d τ (cid:21) = (cid:20) e − µ n T − Z T e − µ n ( T − s ) R ( s ) d s − Z T H n ( T − τ ) (cid:18) e − µ n τ − Z τ e − µ n ( τ − s ) R ( s ) d s (cid:19) d τ (cid:21) ξ n (11)The proof of Theorem 4 is then reduced to the proof that this momentproblem is not solvable. Let N be such that n ≥ N = ⇒ µ n > .
9e shall consider the moment problem in Theorem 7 only for the indices n ≥ N and we shall prove that it can’t be solved.We first examine the right hand side of (11). We recall that H n ( t ) is theresolvent kernel of Z n ( t ) in (9) so that the following equality holds: H n ( t ) = − Z t L ( t − s ) e − µ n s d s + Z t (cid:20)Z t − τ L ( t − τ − s ) e − µ n s d s (cid:21) H n ( τ ) d τ The function L ( t ) is bounded on [0 , T ] for every T > µ n >
0, so, usingGronwall inequality, there exists C (which depends on T but not on n ) suchthat | H n ( t ) | ≤ C µ n (a fact already used in the proof of Lemma 6).We fix T such that R ( T ) = 0. On every compact interval, using bound-edness of M ′ ( t ) hence of R ′ ( t ), we have: Z T R ( s ) e − µ n ( T − s ) d s = 1 µ n (cid:18) R ( T ) − e − µ n T R (0) − Z T e − µ n ( T − s ) R ′ ( s ) d s (cid:19) , (cid:12)(cid:12)(cid:12)(cid:12)Z T e − µ n ( T − s ) R ′ ( s ) d s (cid:12)(cid:12)(cid:12)(cid:12) ≤ const µ n , (cid:12)(cid:12)(cid:12)(cid:12)Z T H n ( T − τ ) (cid:18) e − µ n τ + Z τ e µ n ( τ − s ) R ( s ) d s (cid:19) d τ (cid:12)(cid:12)(cid:12)(cid:12) ≤ const µ n . Let d n = (cid:20) e − µ n T − Z T e − µ n ( T − s ) R ( s ) d s − Z T H n ( T − τ ) (cid:18) e − µ n τ − Z τ e − µ n ( τ − s ) R ( s ) d s (cid:19) d τ (cid:21) ξ n = (cid:20) e − µ n T − µ n (cid:18) R ( T ) − e − µ n T R (0) − Z T e − µ n ( T − s ) R ′ ( s ) d s (cid:19) − Z T H n ( T − τ ) (cid:18) e − µ n τ − Z τ e − µ n ( τ − s ) R ( s ) d s (cid:19) d τ (cid:21) ξ n . Using the existence of C such that µ n e − µ n T < Cµ n R ( T ) = 0, give µ n d n = (cid:18) − R ( T ) + M n µ n (cid:19) ξ n where { M n } is a bounded sequence. Hence, we get: Lemma 8
Let R ( T ) = 0 . There exists N > N with the following property:for every { c n } ∈ l ([ N, + ∞ )) the equation in l ([ N, + ∞ )) µ n d n = (cid:18) − R ( T ) + M n µ n (cid:19) ξ n = c n admits a solution { ξ n } ∈ l ([ N, + ∞ )) . We go back to the moment problem (11) for n ≥ N . If equation (1) iscontrollable to 0 at time T , then the moment problem (cid:20)Z T e − µ n ( T − s ) (cid:18)Z Γ µ n ( γ φ n ) f ( x, s ) d Γ (cid:19) d s − Z T H n ( T − τ ) Z τ e − µ n ( τ − s ) (cid:18)Z Γ µ n ( γ φ n ) f ( x, s ) d Γ (cid:19) d s d τ (cid:21) = c n is solvable for every sequence { c n } ∈ l = l ( N, + ∞ ). We exchange the orderof integration and we rewrite this equalities as Z T Z Γ f ( x, T − s ) E n ( x, s ) d Γ d s = c n , n ≥ N (12)where E n ( x, s ) = µ n ( γ φ n ) (cid:18) e − µ n s − Z s H n ( s − τ ) e − µ n τ d τ (cid:19) ∈ L (0 , T ; L (Γ)) . We recall from [2, Theorem I.2.1] that the moment problem (12) is solvablefor every l -sequence { c n } ( n ≥ N ) if and only if the sequence { E n ( x, t ) } admits a bounded biorthogonal sequence { χ n ( x, t ) } in L (0 , T ; L (Γ)); i.e.if and only if there exists a bounded sequence { χ n ( x, t ) } in L (0 , T ; L (Γ))such that Z T Z Γ E n ( x, t ) χ k ( x, t ) d Γ d t = δ n,k = (cid:26) n = k n = k .
11e are going to prove that this sequence does not exist, relaying on knownproperties of the (memoryless) heat equation. We proceed in two steps: thefirst step computes “explicitly” H n ( t ). The second step, using this expressionof H n ( t ), shows that a bounded sequence { χ n ( x, t ) } does not exist, i.e. themoment problem is not solvable. We proceed with the proof.
Step 1: a formula for H n ( t ) . Here we find a formula for H n ( t ), forevery fixed index n . So, for clarity, the fixed index n is not indicated in thecomputations and H n ( t ) (any fixed n ) is denoted H ( t ). Analogously, µ n , with n fixed, is indicated as µ . Furthermore, we use ⋆ to denote the convolution, f ⋆ g = ( f ⋆ g )( t ) = Z t f ( t − s ) g ( s ) d s . We shall use the commutativity and the associativity of the convolution: f ⋆ g = g ⋆ f , f ⋆ ( g ⋆ h ) = f ⋆ ( g ⋆ h ) . The convolution of a function with itself is denoted as follows: f ⋆ = f , f ⋆ = f ⋆ f , f ⋆k = f ⋆ f ⋆ ( k − . Let e k ( t ) = t k k ! e − µ t so that e ⋆ e k = e k +1 . By definition, H ( t ) is the resolvent kernel of Z ( t ) = − Z t L ( t − s ) e − µ s d s = − L ⋆ e . We shall use:
Lemma 9
Let G ( t ) be any (integrable) function and ˜ G = G ⋆ e k . Then, Z ⋆ ˜ G = e k +1 ⋆ ( − L ⋆ G )In fact:
Z ⋆ ˜ G = ( − L ⋆ e ) ⋆ ( G ⋆ e k ) = ( e ⋆ e k ) ⋆ ( − L ⋆ G ) = e k +1 ⋆ ( − L ⋆ G ) . Z ⋆k = ( − k L ⋆k ⋆ e k − . The known formula of the resolvent ([9, p. 36]) gives H ( t ) = + ∞ X k =1 ( − k − Z ⋆k = − + ∞ X k =1 L ⋆k ⋆ e k − == − Z t + ∞ X k =1 L ⋆k ( t − s ) s k − ( k − ! e − µ s d s . (13)The series converges uniformly since the following holds: | L ( t ) | < M ≤ t ≤ T = ⇒ | L ⋆k ( t ) | ≤ T k M k k ! 0 ≤ t ≤ T .
Step 2: the bounded biorthogonal sequence does not exist.
Wereintroduce dependence on the index n . So e k ( t ) = t k k ! e − µ n t . We go back to the moment problem (12). We prove that it is not solvableas follows: we prove that if the sequence { E n ( x, t ) } admits a biorthogonalsequence { χ k ( x, t ) } , then this sequence cannot be bounded. So, let δ n,k = µ n Z Γ ( γ φ n ( x )) (cid:20)Z T χ k ( x, t ) (cid:18) e − µ n t − Z t H n ( t − τ ) e − µ n τ d τ (cid:19) d t (cid:21) d Γ . (14)We have, using (13): Z t H n ( t − τ ) e − µ n τ d τ = e ⋆ H n = − e ⋆ + ∞ X k =1 L ( ⋆k ) ⋆ e k − ! = − + ∞ X k =1 L ( ⋆k ) ⋆ e k = − Z t " + ∞ X k =1 L ( ⋆k ) ( t − s ) s k k ! e − µ n s d s = Z t G ( t, s ) e − µ n s d s . G ( t, s ) does not depend on n and equality (14) can be writtenas δ n,k = Z Γ Z T ( γ φ n ( x )) (cid:16) µ n e − µ n r (cid:17) (cid:20) χ k ( x, r ) − Z Tr G ( s, r ) χ k ( x, s ) d s (cid:21) d r d Γ= Z T (cid:16) µ n e − µ n r (cid:17) (cid:26)Z Γ ( γ φ n ( x )) (cid:20) χ k ( x, r ) − Z Tr G ( s, r ) χ k ( x, s ) d s (cid:21) d Γ (cid:27) d r . Hence, the sequence { ˜Ψ k ( r ) } ,˜Ψ k ( r ) = Z Γ ( γ φ n ( x )) (cid:20) χ k ( x, r ) − Z Tr G ( s, r ) χ k ( x, s ) d s (cid:21) d Γ , n ≥ N is biorthogonal to { µ n e − µ n r } in L (0 , T ).Now we invoke the following result from [13]: Lemma 10
There exist two positive numbers m and M such that the fol-lowing holds for every index n : < m ≤ Z Γ (cid:12)(cid:12)(cid:12)(cid:12) γ φ n ( x ) λ n (cid:12)(cid:12)(cid:12)(cid:12) d Γ ≤ M .
Consequently, the sequence { Ψ k ( t ) } , Ψ k ( t ) = 1 λ n ˜Ψ k ( t )is a bounded biorthogonal sequence of { µ n λ n e − µ n t } in L (0 , T ).We proved in [12] that for every T > { µ n λ n e − µ n t } doesnot admit any bounded biorthogonal sequence in L (0 , T ) and this completesthe proof of Theorem 4. For completeness, we sketch the proof of this last statement (see [12] foradditional details):
Lemma 11
Any sequence { Ψ n ( t ) } which is biorthogonal to { µ n λ n e µ n t } in L (0 , T ) is unbounded. Proof.
Let e n be the function e − µ n t in L (0 , ∞ ) and denote by e Tn its restric-tion to (0 , T ). E ( ∞ ) = cl span { e n } ⊆ L (0 , ∞ ) , E ( T ) = cl span { e Tn } ⊆ L (0 , T ) . ( ∞ ) is a proper subspace of L (0 , ∞ ) (M¨untz Theorem, see [26]) . Let P T : L (0 , ∞ ) → L (0 , T ) be the operator P T f = f | (0 ,T ) . The operator P T isan isomorphism between E ( ∞ ) and E ( T ) (see [26, formula (9.a) p. 55]).Suppose that { ˜ ψ n } is any biorthogonal to { e Tn } in L (0 , T ). We provethat the sequence { ˜ ψ n } is exponentially unbounded. Let ψ n be the orthogonal projection of ˜ ψ n on E ( T ). Then, { ψ n } isbiorthogonal to { e Tn } too and k ψ n k L (0 ,T ) ≤ k ˜ ψ n k L (0 ,T ) . We have ( ( · , · ) is the inner product in the indicated spaces) δ jn = ( ψ j , e Tn ) L (0 ,T ) = ( ψ j , e Tn ) E ( T ) = ( ψ j , P T e n ) E ( T ) = ( P ∗ T ψ j , e n ) E ( ∞ ) . Hence { P ∗ T ψ n } is biorthogonal to { e n } and furthermore ϕ n = P ∗ T ψ n ∈ E ( ∞ )since P T ∈ L ( E ( ∞ ) , E ( T )). Hence, { ϕ n } is the biorthogonal sequence of { e n } whose L -norm is minimal.Using [7, Lemma 3.1] we have: || ϕ n || L (0 , ∞ ) = 2 n e [ π + O (1)] n , n → ∞ . (15)Since P ∗ T ∈ L ( E ( T ) , E ( ∞ )) is boundedly invertible, there exist positivenumbers m and M such that for every n we have m k ψ n k L (0 ,T ) ≤ k P ∗ T ψ n k L (0 , + ∞ ) ≤ M k ψ n k L (0 ,T ) since P ∗ T ψ n = ϕ n . It follows that k ˜ ψ n k L (0 ,T ) ≥ k ψ n k L (0 ,T ) ≥ M k ϕ n k L (0 , ∞ ) ∀ n . (16)So, any biorthogonal sequence of { e − µ n t } in L (0 , + ∞ ) is exponentially un-bounded and from (16) we see that any biorthogonal sequence of { e − µ n t } n ≥ N T in L (0 , T ) is exponentially unbounded too.Let’s go back to the sequence { Ψ n ( t ) } . This sequence cannot be bounded.Otherwise, the sequence { µ n λ n Ψ k ( t ) } is a biorthogonal sequence to { e − µ n t } such that k µ n λ n Ψ k ( t ) k L (0 ,T ) ≤ Cµ n λ n ≤ n /d , d = dim Ω , using known estimates on the eigenvalues of the laplacian (see [20, p. 192]).15 emark 12 Instead of a time T in which R ( T ) = 0 we might have used atime T at which R ( k ) ( T ) = 0 and R ( m ) ( T ) = 0 for m < k , but this does notchange the content of Theorem 4 in an essential way. References [1] S. Avdonin, B.P. Belinskiy, On controllability of an homogeneous stringwith memory, J. Mathematical Analysis Appl., 398 (2013) 254–269.[2] S. A. Avdonin, S.A. Ivanov,
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