Lack of controllability of thermal systems with memory

Heat equations with memory of Gurtin-Pipkin type have controllability properties which strongly resemble those of the wave equation. Instead, recent counterexamples show that when the laplacian appears also out of the memory term, the control properties do not parallel those of the (memoryless) heat equation, in the sense that there are $L^2$-initial conditions which cannot be controlled to zero. The proof of this fact (presented in previous papers) consists in the construction of two quite special examples of systems with memory which cannot be controlled to zero. Here we prove that lack of controllability holds in general, for every systems with smooth memory kernel.

Here w = w(x, t) and x ∈ Ω, a bounded region with smooth boundary (we require of class C 1 , and Ω locally on one side of ∂Ω).
The time t = 0 is the time after which a boundary control f is applied to the system, w(x, t) = f (x, t) x ∈ Γ = ∂Ω , t > 0 .
Note that we implicitly assume that the system is at rest for negative times, The number α is nonnegative. If α is zero then we get a model proposed by Gurtin and Pipkin in [11]. The controllability, when α = 0, has been studied in several paper, see references below. So, here we explicitly assume α > 0 and we call Eq. (1) the (CGM) model (after Colemann and Gurtin).
It appears that (CGM) has been rarely studied from the control point of view. Our goal in this paper is to understand whether the point ξ 0 = 0 can be 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) ≡ 0.
Thanks to this result, the following definition makes sense: Definition 3 We say that the initial condition ξ is controllable to 0 at time We say that (CGM) is null controllable at time T if for every ξ ∈ L 2 (Ω) there exists f ∈ L 2 (0, T ; L 2 (Γ)) such that A −1 w f,ξ (·; T ) = 0 ∈ L 2 (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 0 then Eq. (1) for t ≤ T 0 coincide with the memoryless heat equation w t = α∆w and any initial condition can be controlled to 0 at any time T < T 0 . Keeping this fact in mind, our main result is: There exist initial data ξ which cannot be controlled to 0 at time T .

Comments and references
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 properties resemble those of the wave equation. In particular, the solutions belong to C(0, +∞; L 2 (Ω)) for every f ∈ L 2 loc (0, T ; L 2 (Γ)) and every initial condition ξ ∈ L 2 (Ω). Furthermore, there exists T such that the reachable set w f,0 (·, T ) , f ∈ L 2 (0, T ; L 2 (Γ)) is equal to L 2 (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 as that for the (memoryless) wave equation (see [3,8,15,22,24]).
Instead, when α > 0 the properties of Eq. (1) strongly resemble those of the standard, memoryless, heat equation in spite that it is not possible to control an initial condition to be identically zero for every t > T , where T is a preassigned time, see [14]. So, it is a natural conjecture that the controllability properties of system (1) with α > 0 should be similar to those of the (memoryless) heat equation. Along this line of thought, it was proved in [4] that, for a suitable class of completely monotonic kernels, the reachable states at every time T > 0 are dense in L 2 (Ω) and this supports the conjecture that every initial condition ξ ∈ L 2 (Ω) can be controlled to hit the target ξ 0 (x) ≡ 0 at a certain time T , of course without remaining equal to zero in the future, due to the negative results in [14]. Hence, the following negative fact was a surprise: there exist kernels M(t) which are even C ∞ , and such that for every T > 0 there exist initial data which cannot be controlled to hit 0, see [10,12]. The proofs in these papers exibits two particular counterexamples.
The goal of this paper is the proof that in the presence of memory, i.e. for every smooth kernel M(t) not identically zero, there exist initial conditions which cannot be controlled to zero, as stated in Theorem 4.

Preliminaries
The number α has to be positive and so, changing the time scale, i.e. replacing w(x, t) with w(x, rt), we can assume We present a transformation which simplifies the computations in this paper. We consider a Volterra integral equation on t ≥ 0 It is known (see [9,Ch. 2]) that it is uniquely solvable for every square integrable f (t), and that the solution is given by The function R(t), the resolvent kernel of M(t), solves We apply formally this transformation, "solving" Eq. (1) with respect to the "unknown" ∆w. We get Integrating by parts we get By, definition, a solution of Eq. (1) is a solution of the Volterra integrodifferential equation (3) (solutions can be defined in several different but equivalent ways).
Let us consider the operator A defined in (2). It is a selfadjoint operator with compact resolvent, which generates a holomorphic semigroup e At .
See [17] for the theory of Volterra integral and integro-differential equations in Banach spaces, and [5] for further information on the semigroup approach to boundary value problems for parabolic equations.

Projection of the system on the eigenspaces
The previous results allows us to project the system on the eigenvectors of the operator A. Let {φ n } be an orthonormal basis of L 2 (Ω), whose elements are eigenvectors of the operator A in (2). So we have: Note that λ 2 n > 0. Let Then w n (t) solves We introduce µ 2 n = λ 2 n − a (we have µ n > 0 for large n) so that Let T > 0. We define a transformation L in L 2 (0, T ; L 2 (Ω)), as follows: Then we have We prove:

Lemma 6
The transformation L in L 2 (0, T ; L 2 (Ω)) is linear and continuous. 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 2 (Ω). This implies that Then we have: We can chose the constant C independent of n thanks to the fact that µ 2 n > 0 for large n. So, we have .
This proves continuity of the transformation L and so also of I − L. In order to prove that this last transformation has a bounded inverse, we exibit explicitly its inverse.
To compute the inverse, we must solve, for every k(x, t) = φ n (x)k n (t), We introduce H n (t), the resolvent kernels of Then we must choose Continuity of this transformation is seen as above, using the fact that µ 2 n > 0 for large n, so that |Z n (t)| ≤ M/µ 2 n (for large n) where M = M T . So, Gronwall inequality applied to Continuity now follows as above.
Using (8) we find that Now we recall the definition of controllability at time T and we can state: (1) is controllable to 0 at time T if for every sequence {ξ n } ∈ l 2 there exists a function f ∈ L 2 (0, T ; L 2 (Γ)) which solves the following moment problem: The proof of Theorem 4 is then reduced to the proof that this moment problem is not solvable.

The proof of Theorem 4
Let N 0 be such that n ≥ N 0 =⇒ µ 2 n > 0 .
We shall consider the moment problem in Theorem 7 only for the indices n ≥ N 0 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 the resolvent kernel of Z n (t) in (9) so that the following equality holds: The function L(t) is bounded on [0, T ] for every T > 0 and µ 2 n > 0, so, using Gronwall inequality, there exists C (which depends on T but not on n) such that |H n (t)| ≤ C 1 µ 2 n (a fact already used in the proof of Lemma 6).
We fix T such that R(T ) = 0. On every compact interval, using boundedness of M ′ (t) hence of R ′ (t), we have: Using the existence of C such that We go back to the moment problem (11) for n ≥ N. If equation (1) is controllable to 0 at time T , then the moment problem is solvable for every sequence {c n } ∈ l 2 = l 2 (N, +∞). We exchange the order of integration and we rewrite this equalities as where We recall from [2, Theorem I.2.1] that the moment problem (12) We are going to prove that this sequence does not exist, relaying on known properties of the (memoryless) heat equation. We proceed in two steps: the first step computes "explicitly" H n (t). The second step, using this expression of H n (t), shows that a bounded sequence {χ n (x, t)} does not exist, i.e. the moment 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), for every fixed index n. So, for clarity, the fixed index n is not indicated in the computations and H n (t) (any fixed n) is denoted H(t). Analogously, µ 2 n , with n fixed, is indicated as µ 2 . Furthermore, we use ⋆ to denote the convolution, We shall use the commutativity and the associativity of the convolution: The convolution of a function with itself is denoted as follows: Let e k (t) = t k k! e −µ 2 t so that e 0 ⋆ e k = e k+1 .
By definition, H(t) is the resolvent kernel of We shall use: The previous lemma shows that The known formula of the resolvent ([9, p. 36]) gives The series converges uniformly since the following holds: Step 2: the bounded biorthogonal sequence does not exist. We reintroduce dependence on the index n. So We go back to the moment problem (12). We prove that it is not solvable as follows: we prove that if the sequence {E n (x, t)} admits a biorthogonal sequence {χ k (x, t)}, then this sequence cannot be bounded. So, let We have, using (13): Note that G(t, s) does not depend on n and equality (14) can be written as Hence, the sequence {Ψ k (r)}, Ψ k (r) = Γ (γ 1 φ n (x)) χ k (x, r) − T r G(s, r)χ k (x, s) d s d Γ , n ≥ N is biorthogonal to {µ 2 n e −µ 2 n r } in L 2 (0, T ). Now we invoke the following result from [13]: Lemma 10 There exist two positive numbers m and M such that the following holds for every index n: Consequently, the sequence is a bounded biorthogonal sequence of {µ 2 n λ n e −µ 2 n t } in L 2 (0, T ). We proved in [12] that for every T > 0 the sequence {µ 2 n λ n e −µ 2 n t } does not admit any bounded biorthogonal sequence in L 2 (0, T ) and this completes the proof of Theorem 4.
For completeness, we sketch the proof of this last statement (see [12] for additional details): Lemma 11 Any sequence {Ψ n (t)} which is biorthogonal to {µ 2 n λ n e µ 2 n t } in L 2 (0, T ) is unbounded.

Remark 12
Instead of a time T in which R(T ) = 0 we might have used a time T at which R (k) (T ) = 0 and R (m) (T ) = 0 for m < k, but this does not change the content of Theorem 4 in an essential way.