Approximate controllability of the FitzHugh-Nagumo equation in one dimension

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Approximate controllability of the FitzHugh-Nagumo equation in one dimension Shirshendu Chowdhury ∗ , Mrinmay Biswas, Rajib Dutta Indian Institute of Science Education and Research, Kolkata, West Bengal, India Received 26 July 2018; revised 30 September 2019; accepted 1 October 2019

Abstract The FitzHugh-Nagumo (FHN) equation is a simplified model of a nerve axon. We explore the controllability of this model using a localized interior control only for the first equation. The Linearized system is not null controllable using a localized interior control since the spectrum of the linearized system has an accumulation point though it is approximate controllable. We show that the solution of the FHN equation fails to be globally approximate controllable in a given time. But it is possible to move from any steady state to any other steady state arbitrarily close after some appropriate time by a localized interior control, provided that both steady states are in the same connected component of the set of steady states. Finally we make some additional remarks and comments and we mention some open questions for our system. For the sake of completeness, we give the details of the existence, uniqueness and uniform bound of the solution in Appendix. © 2019 Elsevier Inc. All rights reserved. MSC: 93B05; 93B52; 93C20; 93D30 Keywords: FHN equation; Rogers-McCulloch model; Approximate controllability; Riesz basis; Pole shifting; Lyapunov functional

* Corresponding author.

E-mail addresses: [email protected] (S. Chowdhury), [email protected] (M. Biswas), [email protected] (R. Dutta). https://doi.org/10.1016/j.jde.2019.10.001 0022-0396/© 2019 Elsevier Inc. All rights reserved.

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1. Introduction The FitzHugh-Nagumo equation is a simpler version of the Hodgkin-Huxley model [27] from where several interesting problems in ordinary and partial differential equations originate. Let us begin by giving a brief description of the physical setting of the problem. The function of the nervous system is to transmit information. In order to do so, it initiates and transmits electrical impulses through individual nerve cell or neuron. Hodgkin and Huxley modelled the ionic and electrical activities at the time of transmission of an impulse along an axon by a system of differential equations. The Hodgkin-Huxley model can be described by the following non-linear system (see Hastings [26])  ut − uxx + I (u, w) = 0, (1.1) wt − P (u)w − Q(u) = 0, where u = u(t, x), a function of time t and distance x along the axon from a reference point on the axon, determines the electrical potential across the axonal membrane and w = w(t, x), a three-dimensional vector valued function of both x and t , represents the permeability of the membrane to the main ionic component of the transmembrane current. Both the three by three matrix-valued function P and the vector-valued function Q are non-linear functions of u, whereas, I (u, w) is linear in w but non-linear in u. See Hastings [26] and references therein for more details. Due to the appearance of complicated non-linear structure of functions I, P and Q in (1.1), the mathematical analysis of the Hodgkin-Huxley equations becomes very difficult. Simpler formulations have been studied, see FitzHugh [20], with I = IF (u, w) := −[a1 u(u − a)(u − 1) − a2 w], P (u) := γ , Q(u) := δu, in (1.1), which is known as the FHN model, and also with I = IR (u, w) := −[a1 u(u − a)(u − 1) − a2 uw], P (u) := γ , Q(u) := δu, in (1.1), known as the Rogers-McCulloch (RM) model, see Rogers-McCulloch [42], where a1 , a2 , a, δ, γ > 0 are parameters and w is a scalar valued function. Both FHN and RM models manage to reproduce many of the salient qualitative features of the original system and, at the same time, is more tractable. In this article, we consider both FHN and RM models. In FitzHugh [20], FitzHugh studied the system of ordinary differential equations in two unknowns, that is actually the system (1.1), where the unknowns are independent of the space variable and I = IF . FitzHugh’s work was crucially used by Nagumo-Arimoto-Yoshizawa [36], to describe the phenomenon of producing an active pulse transmission line, simulating the nerve axon, in tunnel diodes. This work relied heavily on FitzHugh’s work. In this paper we study the following control system corresponding to the FHN equation with the Dirichlet boundary conditions ⎧ ut − uxx = −u(u − a)(u − 1) − w + f χO , in (0, T ) × (0, 1), ⎪ ⎪ ⎪ ⎨w = δu − γ w, in (0, T ) × (0, 1), t ⎪ u(0, x) = u0 (x), and w(0, x) = w0 (x), x ∈ (0, 1) ⎪ ⎪ ⎩ u(t, 0) = u(t, 1) = 0, t ∈ (0, T ),

(1.2)

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where a ∈ (0, 1), δ, γ > 0, f ∈ L2 ((0, T ) × (0, 1)) is a control acting on the given set O := (b, c) ⊂ (0, 1) with 0 < b < c < 1 and χO denotes the characteristic function of the set O. The variable u stands for the electrical potential across the axonal membrane, w represents a recovery variable, corresponding to the permeability of the membrane for the main ionic component of the trans-membrane current and f stands for the medicine actuator (the control variable localized at O). See Hastings [26], Hodgkin-Huxley [27] for more details. This model is known to describe cardiac excitation phenomena, see Aliev-Panilov [1], Rogers-McCullough [42], and labyrinth pattern formation in an activator inhibitor system, cf. [23]. For more details in this direction, we also refer [45] and the references therein. As this model is relatively simple and it describes appropriately the excitability and bi-stability phenomena, this system attracts our attention from analytical point of view. We are interested in the role played by the actuators and we will answer several controllability questions for this model on the way. As in Chaves-Silva-Zhang-Zuazua [10], we observed that it is technically difficult to answer the null controllability question because of the presence of the t memory term δ 0 e−γ (t−s) u(s)ds in its equivalent formulation

ut − uxx = −u(u − a)(u − 1) − e

−γ t

t w0 − δ

e−γ (t−s) u(s)ds + χO f,

in (0, T ) × (0, 1).

0

(1.3) The linearized system around zero of the above system is given by

ut − uxx = −au − e

−γ t

t w0 − δ

e−γ (t−s) u(s)ds + χO f

in (0, T ) × (0, 1).

(1.4)

0

Equation (1.4) is basically the heat equation with a parabolic memory kernel. Due to the presence of this memory kernel, the resolvent of the linearized operator is not compact. More precisely, one can prove that the linearized operator has a branch of eigenvalues accumulating near a real number. See Lemma 3.7 in Section 3. This accumulating nature of the spectrum does not allow the system to go to the zero state in given time T > 0. In this case, it has been shown, [24], [25], [37], [47], that the null controllability may fail. Nevertheless, there is a possibility that the approximate controllability property still holds for the same equation, at least for some special cases, see [2], [5], [16], [47]. In Chaves-Silva-Zhang-Zuazua [10], it was shown that (1.4) is controllable by a control with moving (with respect to time) supports covering the whole interval (0, 1). Since the original system (1.2) is constituted by the coupling of a nonlinear (cubic nonlinearity) heat equation with an ordinary differential equation (ODE), the presence of the ODE component creates difficulty to establish the null controllability result of the system if the control is confined only on a proper subset O of (0, 1). That is why to get an observability estimate, support of the control must need to move to cover the interval (0, 1) as shown in Kunisch-Souza [31]. In fact, in [31], the local exact controllability to the trajectories for the one-dimensional FHN was established using distributed controls with a moving support, whereas, we consider approximate controllability of the full nonlinear system (1.2) (the FHN with the Dirichlet boundary conditions) using a control confined only on a fixed proper subset O of (0, 1).

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We now present an ODE set-up for the FHN model in an appropriate Hilbert space. In order to do so, let us consider the Hilbert space Z := L2 (0, 1) × L2 (0, 1) over the complex field C endowed with the inner product

  1  u1 u2 := u1 (x)u2 (x) + w1 (x)w2 (x) dx. , w1 w2 Z 0

System (1.2) can be written as an ODE in the Hilbert space Z as 

U (t) = AU(t) + F(U)(t) + B(f )(t), U(0) = U0

t ∈ (0, T ),

(1.5)

where U := (u, w)T ∈ L2 (0, T ; Z), f ∈ L2 (0, T ; L2 (0, 1)), the unbounded linear operator (A, D(A)) in Z is defined by   D(A) := U = (u, w)T ∈ Z | u ∈ H01 (0, 1) ∩ H 2 (0, 1) , and  A :=

d2 dx 2

− aI δI

−I −γ I

 ,

the operators F : D(A) → Z and B : L2 (0, 1) → Z are defined by

 (a + 1)u2 − u3 F(U ) := , 0



χO f B(f ) := 0

 .

(1.6)

It is well known (see Brandão et al. [5]) that for f ∈ L2 ((0, T ) × (0, 1)) and u0 ∈ L2 (0, 1) and w0 ∈ L2 (0, 1), there exists a unique solution u ∈ L2 (0, T ; H01 (0, 1)) ∩ C([0, T ]; L2 (0, 1)) ∩ L4 (0, T ; L4 (0, 1)) and w ∈ C 1 ([0, T ]; L2 (0, 1)) of the system (1.2). See Theorem 3.18 in this paper. Before we state main results, let us introduce the basic definition of approximate controllability. Definition 1.1 (Approximate controllability). Let T > 0. The system (1.5) (or (1.2)) is approximate controllable in time T if for given any U0 , UT ∈ Z and δ > 0, there exists a control f ∈ L2 ((0, T ) × (0, 1)) such that the solution of the system (1.5) satisfies U(T ) − UT Z ≤ δ.

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Our first result regarding approximate controllability is the following: Theorem 1.2. Let T > 0. Then, the system (1.2) is not approximate controllable in Z with respect to the set of controls f ∈ L2 ((0, T ) × (0, 1)). Next, we consider the following boundary control system for the FHN ⎧ ⎪ ⎨ut − uxx = −u(u − a)(u − 1) − w, in (0, T ) × (0, 1), wt = δu − γ w, in (0, T ) × (0, 1), ⎪ ⎩ u(0, x) = u0 (x) and w(0, x) = w0 (x), x ∈ (0, 1),

(1.7)

with the Dirichlet boundary control u(t, 0) = f1 (t), u(t, 1) = f2 (t),

t ∈ (0, T ),

(1.8)

or with the Neumann boundary control ux (t, 0) = f1 (t), ux (t, 1) = f2 (t),

t ∈ (0, T ),

(1.9)

where a, δ and γ are as above and fi ∈ L2 (0, T ) as controls for i = 1, 2. In this case we have the following negative result similar to the Theorem 1.2. Theorem 1.3. Let T > 0. Then, the system (1.7)–(1.8) (or (1.9)) is not L2 (0, 1) ×L2 (0, 1)-approximate controllable with respect to the set of controls fi ∈ L2 (0, T ), i = 1, 2. Idea of proofs of Theorem 1.2, 1.3 is in the spirit of [21], which involves some kind of weighted energy estimates of the system. To establish our next positive result we introduce the notion of steady state for the control system (1.2). Definition 1.4 (Steady state). A function U = (u, w)T ∈ C 2 [0, 1] × C 2 [0, 1] is called a steady state of the control system (1.2) if U satisfies ⎧ ⎪ ⎨uxx − u(u − a)(u − 1) − w = 0, δu − γ w = 0, in (0, 1), ⎪ ⎩ u(0) = 0 = u(1).

in (0, b) ∪ (c, 1),

Let us denote the set of all steady states by S endowed with the C 2 topology. We ask the following controllability question: Can we go from any given steady state to another staying in the same connected component of S? The main positive result in this direction is the following Theorem 1.5. Let U0 , U1 ∈ S be in the same connected component of S. Then, for given any δ > 0, there exists Tδ > 0 such that, for every T ≥ Tδ , there exists a control f ∈ L2 ((0, T ) × (0, 1)) such that the solution U of (1.5) satisfies U(T ) − U1 Z ≤ δ.

(1.10)

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Moreover, the following holds |u(T ) − u1 |H 1 (0,1) + |w(T ) − w1 |L2 (0,1) ≤ δ. 0

In order to prove the above result we follow the idea of steady-state controllability results for semi-linear heat and wave equations by using the quasi-static deformation method introduced in Coron-Trélat [14], [15] (see also Schmidt-Trélat [44]). We have adapted the strategy of the quasi-static deformation method and applied to a system of nonlinear equations whose spectrum of the linearized operator possesses an accumulation point, whereas, in Coron-Trélat [14], [15], the quasi-static deformation method was applied to a single equation whose linearized operator has no accumulation point. We agree that the unstable part of the spectrum of our system is of finite dimensional as in the models discussed in Coron-Trélat [14], [15]. The idea of the proof of Theorem 1.5 is as follows. Linearizing the system (1.2) along a path of steady states joining U0 and U1 , we have a one parameter dependent system. The spectral analysis of the corresponding parameter dependent linearized operator shows that it has two branches of eigenvalues. One branch of eigenvalues tends to −∞ and other branch accumulates near a negative real number uniformly with respect to the parameter. Thus, it may have finitely many eigenvalues with positive real part. Therefore, using one parameter dependent Riesz basis of eigenvectors of the linearized operator which depends smoothly on the parameter, we decompose the system into two parts; one is infinite dimensional stable part and another is finite dimensional unstable part. Then, we stabilize the finite dimensional unstable part of the nonautonomous linear control system using the method of pole shifting. This method yields a proper choice of a control function f ∈ L2 ((0, T ) × (0, 1)) in the feedback form. Then, choosing a suitable Lyapunov functional, we show that the original fully nonlinear system (FHN) is approximately controllable. The main novelty of our work lies in showing that for a given time T > 0, if the target lies outside a large enough ball, it is not possible to reach arbitrarily near the target state in time T . Finally, using the quasi-static deformation method, we show that if the initial and target state lie on the same connected component of S, we obtain a control by which the solution of the system can reach arbitrarily close to the target state at some time T . But we do not know whether the system is approximate controllable if both the initial and target state do not lie in the same connected component of S and yet stay in a ball of small radius. (not exceeding the bound given in (2.10) in Section 2). Since, there are other physical systems whose behaviour of the spectrum of the linearized operator, is similar to the spectrum of the linearized FHN, for example, (i) linearized compressible Navier-Stokes equations linearized around (1, 0) state (see [11]), (ii) structurally damped wave equation (see [43]), (iii) incompressible viscoelastic linearized Jeffrey fluids (see [18]), we emphasize that our result is important and expect our approach may help to conclude such kind of approximate controllability results for the above systems, which is our future direction of research. Let us now give additional references on stabilization and optimal control results for the FHN equation. Optimal control problems related to the FHN system using an interior control are studied in [5]. Local boundary feedback stabilization for the FHN with the Dirichlet and Neumann boundary condition in two dimensions and three dimensions has been studied in [7]. Local exponential stabilizability of the non-linear closed loop system around a stationary solution using an interior control with the homogeneous Neumann boundary condition has been studied in [6], [8].

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Local feedback stabilization to nonstationary solutions of both the FHN and Rogers-McCulloch using interior controls with the homogeneous Neumann boundary condition has been established in [9]. This paper is organized as follows. In Section 2, we study the global approximate controllability results and provide the proof of Theorem 1.2 and Theorem 1.3. Section 3 is devoted to approximate controllability to steady states, using the property of the spectrum of the parameter dependent linearized operator. Thereafter, we give the proof of Theorem 1.5. In Section 4, we mention some additional remarks on approximate controllability of the FHN with the Neumann boundary conditions and the Rogers-McCulloch model both with the Dirichlet and Neumann boundary conditions. We also discuss some open questions on the connected components of the steady states of the FHN equation, local approximate controllability of the FHN equation, controls with non-negativity constraints, approximate controllability for some specific steady states and approximate controllability of the FHN equation for arbitrary states as sources and targets. For the sake of completeness, we give the details of the existence, uniqueness and uniform bound of solution of the FHN equation in our setting in Appendix. Acknowledgment. The authors deeply acknowledge Prof. Jean-Michel Coron for sending us some useful references. We deeply acknowledge Prof. Emmanuel Trélat for his detailed email with reference regarding some questions in his paper [15]. The authors would like to express sincere thanks to the reviewers for their helpful comments, valuable questions and suggestions to improve the first version of the paper. Shirshendu Chowdhury and Rajib Dutta acknowledge financial support from INSPIRE Fellowship (Reference No. DST/INSPIRE/04/2014/002199 and DST/INSPIRE/04/2015/002388 respectively) and Mrinmay Biswas acknowledges supports from CSIR Fellowship Grant No. SPM-07/921(0214)/2014-EMR-I. 2. Lack of global approximate controllability In this section we will prove Theorem 1.2 and Theorem 1.3. Let us define the function g˜ 1 : [0, b] ⊂ [0, 1] → R by g˜ 1 (x) := (b − x)N ,

x ∈ [0, b]

with N > 5.

We deduce the following estimate. Lemma 2.1. Let T > 0. Let (u, w)T be the solution of the system (1.2). Then we have, for t ∈ (0, T ), 1 d 2 dt

b

 g˜ 1 (x) u2 (t, x) + w 2 (t, x) dx ≤ L(a, b, δ, γ , N),

(2.1)

0

where L(a, b, δ, γ , N ) is given by  L(a, b, δ, γ , N) :=

3 N 2 (N − 1)2 N−3 1 + b 16 (N − 3) N +1



  3(1 − δ)4 729 N+1 (a + 1)4 + b . 256 64γ 2 (2.2)

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Proof. Let us denote Eg˜1 (t, x) by Eg˜1 (t, x) :=

 1 d  g˜ 1 (x) u2 (t, x) + w 2 (t, x) , 2 dt

(t, x) ∈ (0, T ) × [0, 1].

(2.3)

Then, using (1.2) and Young’s inequality we have Eg˜1 = g˜ 1 {uut + wwt } ,   = g˜ 1 u (uxx − u(u − a)(u − 1) − w + f χO ) + w (δu − γ w) ,   = g˜ 1 uuxx − u4 + (a + 1)u3 − au2 − (1 − δ)uw + uf χO − γ w 2 ,   = g˜ 1 (uux )x − u2x − u4 + (a + 1)u3 − au2 − (1 − δ)uw − γ w 2 + g˜ 1 uf χO ,   (1 − δ)2 2 4 3 u + g˜ 1 uf χO , ≤ g˜ 1 (uux )x − u + (a + 1)u + 4γ = g˜ 1 (uux )x − g˜ 1 u4 + (a + 1)g˜ 1 u3 +  = (g˜ 1 uux )x −

g˜ 1 u2 2

+

 + x

(1 − δ)2 g˜ 1 u2 + g˜ 1 uf χO , 4γ

g˜ 1 u2 − g˜ 1 u4 + (a + 1)g˜ 1 u3 2

(1 − δ)2 g˜ 1 u2 + g˜ 1 uf χO . 4γ

Integrating with respect to x over the interval [0, b] and using the boundary conditions u(t, 0) = g˜ 1 (b) = 0 and the fact that g˜ 1 is decreasing, we note that b Ig˜1 (t) :=

Eg˜1 (t, x)dx,

t ∈ (0, T ),

0

satisfies, for t ∈ (0, T ), b  Ig˜1 (t) ≤

 2 g˜ 1 u2 (1 − δ) − g˜ 1 (x)u4 + (a + 1)g˜ 1 (x)u3 + g˜ 1 (x)u2 dx. 2 4γ

0

Let us denote, for (t, x) ∈ (0, T ) × [0, 1], L1 (t, x) :=

g˜ 1 u2 1 = N (N − 1)(b − x)N−2 u2 , 2 2

L2 (t, x) := (a + 1)g˜ 1 u3 = (a + 1)(b − x)N u3 ,

(2.4)

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and L3 (t, x) :=

(1 − δ)2 (1 − δ)2 g˜ 1 (x)u2 = (b − x)N u2 . 4γ 4γ

Applying Young’s inequality we have 1 L1 ≤ (b − x)N u4 + 3 1 L2 ≤ (b − x)N u4 + 3 1 L3 ≤ (b − x)N u4 + 3

3 2 N (N − 1)2 (b − x)N−4 , 16 729 (a + 1)4 (b − x)N , 256 3(1 − δ)4 (b − x)N . 64γ 2

(2.5)

Using (2.5) in (2.4) we have b  Ig˜1 (t) ≤

  729 3(1 − δ)4 3 2 N N (N − 1)2 (b − x)N−4 + (a + 1)4 + (b − x) dx 16 256 64γ 2

0

= L(a, b, δ, γ , N). This completes the proof.

(2.6)

2

Using Lemma 2.1, we now prove Theorem 1.2. Proof of Theorem 1.2. Let UT = (uT , wT )T ∈ Z. Let U(t) = (u(t, ·), w(t, ·))T be the solution of the system (1.2) and T > 0. Then, we obtain ⎛ 1 ⎞ 12  UT − U(T )Z = ⎝ |uT (x) − u(T , x)|2 + |wT (x) − w(T , x)|2 dx ⎠ , 0

⎛

⎞1 2 b/2 ⎜ ⎟ 2 2 ≥ ⎝ |uT (x) − u(T , x)| + |wT (x) − w(T , x)| dx ⎠ , 0

⎞1 2 b/2 ⎜ 2⎟ ≥ ⎝ |uT (x) − u(T , x)| ⎠ , ⎛

0

≥ uT L2 (0,b/2) − u(T , ·)L2 (0,b/2) . In virtue of (2.6) for any T > 0 we have

(2.7)

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b (b − x)

N

b  

u (T , x) + w (T , x) dx ≤ (b − x)N u20 (x) + w02 (x) dx 2

2

0

(2.8)

0

+ 2T L(a, b, δ, γ , N). Using the inequality b

−N

−N

≤ (b − x)

−N b ≤ , 2

for 0 ≤ x ≤ b/2,

we obtain that b/2 

≤ (b − x)−N (b − x)N u2 (T , x) + w 2 (T , x) dx,

u(T , ·)2L2 (0,b/2)

0

−N b/2 

b ≤ (b − x)N u2 (T , x) + w 2 (T , x) dx, 2 0

−N b 

b (b − x)N u2 (T , x) + w 2 (T , x) dx, ≤ 2

(2.9)

0

−N b 

b ≤ (b − x)N u20 (x) + w02 (x) dx, 2 0

+ 2T

−N b L(a, b, δ, γ , N). 2

Let T > 0 be fixed. Let 0 > 0 be given. Let UT = (uT , wT )T ∈ Z satisfying the condition 1

− N  b 2 

2 b uT L2 (0,b/2) > (b − x)N u20 (x) + w02 (x) dx + 2T L(a, b, δ, γ , N) + 0 . 2 0

(2.10) Equation (2.9) and (2.10) together imply uT L2 (0,b/2) > u(T , ·)L2 (0,b/2) + 0 .

(2.11)

Using (2.11) in (2.7) we obtain UT − U(T , ·)Z > 0 . This inequality implies the system (1.2) is not approximate controllable. Thus we have the result. 2

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We now prove Theorem 1.3. Proof of Theorem 1.3. Taking the multiplier g˜ 2 (x) := x N (b − x)N , N > 5, for all (t, x) ∈ (0, T ) × [0, b] we get 1 d 2 dt

b

 g˜ 2 (x) u2 (t, x) + w 2 (t, x) dx ≤ L1 (a, b, δ, γ , N),

0

where L1 (a, b, δ, γ , N ) is given by 

3 N 2 (N − 1)2 N−3 1 L1 (a, b, δ, γ , N) := + b 16 (N − 3) N +1



  3(1 − δ)4 729 4 N+1 (a + 1) + b . 256 64γ 2

Using similar calculations as in Theorem 1.2, we observe that for b/4 ≤ x ≤ b/2, 3−N b−2N b−2N −N −N ≤ x (b − x) ≤ . 2−3N 2−3N Using the above inequality we obtain

u(T , ·)2L2 (b/4,b/2) ≤

b−2N 2−3N

⎞ ⎛ b  

⎝ x N (b − x)N u20 (x) + w02 (x) dx + 2T L1 (a, b, δ, γ , N)⎠ . 0

Hence, for 0 > 0 and for any target UT = (uT , wT )T ∈ Z satisfying

uT L2 (b/4,b/2) >

b−N

 b

1 2 

2 2 x (b − x) u0 (x) + w0 (x) dx + 2T L1 (a, b, δ, γ , N) N

2−3N/2

N

0

+ 0 , we have UT − U(T , ·)Z > 0 . This implies the system (1.7)–(1.8) (or (1.9)) is not approximate controllable. 2 Remark 2.2. We point out the following two observations. (i) From (2.9) one can observe that for a given T > 0 and U0 = (u0 , w0 )T ∈ Z, if the target UT = (uT , wT )T lies outside of a large ball i.e. in particular the first component uT of UT satisfies (2.10) with 0 = 0, then, it is not possible to reach arbitrarily close the state UT in time T .

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(ii) For given U0 , UT ∈ Z satisfying 1

− N  b 2 

2 b N 2 2 uT L2 (0,b/2) > (b − x) u0 (x) + w0 (x) dx , 2 0

the time T of approximate controllability cannot be taken arbitrarily small. Now, the question is: what is next best possible result one can expect? Remark 2.3. Following Coron [14], [15] we expect some (partial) approximate controllability result. More precisely, we have the following query, given any two states U0 , U1 ∈ Z and given any δ > 0, whether there exist T = T (δ, U0 , U1 ) > 0 and f ∈ L2 (0, T ; L2 (0, 1)) such that the solution U of (1.5) satisfies U(T ) − U1 Z ≤ δ. We don’t have the complete answer yet to the question of approximate controllability. We have positive results only when U0 , U1 are two steady states lying in the same connected components of the set of steady states in C 2 -topology (see Section 3). For further open questions and unknown results in this regard, we refer Section 4. 3. Quasi-static deformation 3.1. Characterization of steady state In order to discuss the controllability results, we need to characterize the set of steady states S of the given system (1.2). To compute a steady state U = (u, w)T , we look for u ∈ C 2 ([0, 1]) such that  uxx − u(u − a)(u − 1) − γδ u = 0, in (0, b) ∪ (c, 1), (3.1) u(0) = u(1) = 0, and the corresponding w satisfies the relation w=

δ u. γ

More precisely, to compute the first component u of a steady state, we seek u ∈ C 2 ([0, b] ∪ [c, 1]) satisfying (3.1) and then we extend u on [0, 1] as a C 2 [0, 1] function. A simple way to extend u on [b, c] is using a six degree polynomial where the coefficients of the polynomial can be computed from the known information of u at b and c. It is easy to observe that the extension of u on [b, c] may not be unique. But, for any two u1 , u2 ∈ C 2 [0, 1] such that u1 = u2 on [0, b] ∪ [c, 1] the corresponding U1 = (u1 , w1 ) and U2 = (u2 , w2 ) belong to the same connected component of S. To discuss the connected components of S, we introduce the following notation. For a given U = (u, w)T ∈ S, we can find a PU = (u, ¯ w) ¯ T ∈ S such that u = u¯ in [0, b] ∪ [c, 1] and u¯ in

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[b, c] is given by a suitable six degree polynomial as before and w¯ = γδ u. ¯ Note that U and PU belong to the same connected component of S. The following lemma gives a characterization of the connected components of S. Lemma 3.1. [Characterization of steady states] Let U0 = (u0 , w0 ), U1 = (u1 , w1 ) ∈ S. Then, U0 and U1 belong to same connected component of S if and   only if for any two real number (α, β) ∈     τ u0 (0), u0 (1) + (1 − τ ) u1 (0), u1 (1) : τ ∈ [0, 1] , the maximal solution of the system (3.1) with u (0) = α, u (1) = β, denoted by U(α,β) , is defined in [0, b] ∪ [c, 1]. For a proof of the above lemma see Appendix. 3.2. Reduction of the problem Let  > 0, τ ∈ [0, 1]. Let τ → U(τ ) be a C 1 path in S, connecting U0 and U1 . Let us define τ → f (τ ) by f (τ, x) = u(τ, x)(u(τ, x) − a)(u(τ, x) − 1) + w(τ, x) − uxx (τ, x), for x ∈ [0, 1] and τ ∈ [0, 1]. For t ∈ [0, 1 ] and x ∈ [0, 1], let us define ⎧ ⎪ ⎨z(t, x) = u(t, x) − u(t, x), v(t, x) = w(t, x) − w(t, x), ⎪ ⎩ g(t, x) = f (t, x) − f (t, x).

(3.2)

The time reparametrization introduced above will give us the advantage to perform a pole shifting procedure on the linear finite dimensional system representing the unstable part of the equation. An appropriate choice of feedback control, suggested by the pole shifting of the unstable part, will make the whole system stable. Using (1.2), (3.1) and (3.2) we see that z and v satisfy the following equation, with T = 1/, ⎧ ⎪ zt = zxx + (2(a + 1)u − 3u2 − a)z + z2 (a + 1 − 3u) − z3 − v ⎪ ⎪ ⎪ ⎪ ⎪ +χO g − uτ , in (0, T ) × (0, 1), ⎨ vt = δz − γ v − wτ , in (0, T ) × (0, 1), ⎪ ⎪ ⎪ z(0, x) = 0 = v(0, x), x ∈ (0, 1), ⎪ ⎪ ⎪ ⎩z(t, 0) = z(t, 1) = 0, t ∈ (0, T ).

(3.3)

We introduce an  one-parameter family of unbounded linear operators for 0 ≤ τ ≤ 1

A1 (τ ), D(A1 (τ )) in Z defined by   D(A1 (τ )) := W = (z, v)T ∈ Z | z ∈ H01 (0, 1) ∩ H 2 (0, 1) ,  2  d − p(τ, ·)I −I A1 (τ ) := dx 2 δI −γ I

(3.4)

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with p(τ, ·) := a − 2(a + 1)u(τ, ·) + 3u2 (τ, ·). We note that the domain D(A1 (τ )) of A1 (τ ) are same for all τ ∈ [0, 1]. Hence, onward we

T denote D(A1 (τ )) by D for all τ ∈ [0, 1]. Let W(t, x) = (z(t, x), v(t, x) . Therefore, we can write (3.3) as 

Wt (t) = A1 (t)W(t) + B(g)(t) + R(, t, ·), W(0) = 0,

(3.5)

where

 z2 (a + 1 − 3u) − z3 − uτ (t) R(, t, ·) = , −wτ (t)

(3.6)

and B is given in (1.6). Thus, in order to prove Theorem 1.5, it is enough to show that for a given neighbourhood V0 of (0, 0) in L2 (0, 1) ×L2 (0, 1) and  > 0 small enough, there exists a control g ∈ L2 (0, T ; L2 (0, 1)) such that the solution (z, v)T of (3.3) corresponding to this g, we have (z(T ), v(T ))T ∈ V0 with T = 1 . To achieve this we shall construct an appropriate control function and a Lyapunov function which stabilizes the system (3.3). Note that here “stabilize” is not in the usual sense (see [13], Section 7.1, Chapter 7), since t ∈ [0, 1/] instead of t ∈ [0, ∞). Let us prove the following estimate for R. Lemma 3.2. Let L > 0. Then, there exists C1 > 0 such that

 R(, t, ·)∞ ≤ C1  + |z(t, ·)|2∞ ,

t ∈ [0, 1/],

whenever |z(t, ·)|∞ ≤ L. Proof. We note that using |z(t, ·)|∞ ≤ L,



 R(, t, ·)∞ ≤  |uτ |∞ + |wτ |∞ + |z(t, ·)|2∞ a + 1 + 3|u|∞ + |z|∞ ,



 ≤  |uτ |∞ + |wτ |∞ + |z(t, ·)|2∞ a + 1 + L + 3|u|∞ .   As both u, w ∈ C 1 [0, 1]; C 2 [0, 1] and |z(t, ·)|∞ ≤ L, we have

 R(, t, ·)∞ ≤ C1  + |z(t, ·)|2∞ , for some C1 > 0. Hence, we have the lemma. 2

(3.7)

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Remark 3.3. In Lemma 3.2 we get an estimate for R(, t, ·)∞ provided |z(t, ·)|∞ ≤ L for some L > 0, whereas in Theorem 3.21 we actually establish that there exists L > 0 such that the solution of (3.35) satisfies |z(t, ·)|∞ ≤ L for all t ≥ 0. So the L in Lemma 3.2 is same as the L in Theorem 3.21. 3.3. Spectral analysis of the operator A1 (τ ) The proof of the Theorem 1.5 is based on the spectral analysis of the operator defined in (3.4). Let τ ∈ [0, 1]. We are interested to compute the eigenvalues of the operator A1(τ ). In order to do that we consider the following eigenvalue problem  ϕ  (x) − p(τ, x)ϕ(x) = λϕ(x), ϕ(0) = ϕ(1) = 0.

in (0, 1),

(3.8)

From the Sturm-Liouville Theorem [see [12], Theorem 2.1, Theorem 4.2], we conclude the following result. Theorem 3.4. The eigenvalues and the corresponding eigenvectors of (3.8) satisfy the following properties: (i) For each τ , the eigenvalue problem (3.8) has an infinite sequence of real eigenvalues which satisfy λ1 (τ ) > λ2 (τ ) > · · · λk (τ ) > λk+1 (τ ) · · · · · ·

with

lim λk (τ ) = −∞.

k→∞

(ii) For each k ∈ N and τ ∈ [0, 1] the eigenvalue λk (τ ) is simple and the corresponding eigenfunctions ϕk (τ, ·) can be chosen so that the set {ϕk (τ, ·)}k∈N forms an orthonormal basis of L2 (0, 1). (iii) For each k ∈ N, the map τ → λk (τ ) is C 1 ([0, 1]) and the map τ → ϕk (τ, ·) is C 1 ([0, 1]; L2 (0, 1)). (see [29], [41]). (iv) Using (i) and (iii), an application of Dini’s theorem gives that lim λk (τ ) = −∞

k→∞

uniformly on [0, 1].

 The adjoint of the one-parameter family of unbounded linear operator A1 (τ ), D(A1 (τ )) in 

Z for 0 ≤ τ ≤ 1, is denoted by defined by A∗1 (τ ), D(A∗1 (τ )) . The following lemma is easy to observe. Lemma 3.5. Let τ ∈ [0, 1]. Then,   D(A∗1 (τ )) = D := W = (z, v)T ∈ Z | z ∈ H01 (0, 1) ∩ H 2 (0, 1) ,

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 A∗1 (τ ) =

d2 dx 2



− p(τ, ·)I −I

δI −γ I

(3.9)

,

with p(τ, ·) := a − 2(a + 1)u(τ, ·) + 3u2 (τ, ·). We use Theorem 3.4 to compute the spectrum of the operator A1 (τ ). To do this computation we split the space Z into two dimensional invariant subspaces of A1 (τ ). Let τ ∈ [0, 1]. Let us consider a basis {k (τ )}k≥1 in Z as follows

 ϕk (τ, ·) 2k (τ ) := , 0



0 2k−1 (τ ) := ϕk (τ, ·)

 for all k ≥ 1.

Let us consider two dimensional subspaces Vk (τ ) := span {2k (τ ), 2k−1 (τ )} for all k ≥ 1 of Z. It can be checked that Z = ⊕k≥1 Vk (τ ). Lemma 3.6. For all k ≥ 1, the subspace Vk (τ ) is invariant under the actions of the operators   ∗  A1 (τ ) and A∗1 (τ ). Moreover, the operators Ak1 (τ ) := A1 (τ )V (τ ) and Ak∗ 1 (τ ) := A1 (τ ) Vk (τ ) k have the following matrix representations

Ak1 (τ ) =

λk (τ ) δ

 −1 , −γ

Ak∗ 1 (τ ) =

λk (τ ) −1

δ −γ



respectively, with respect to the basis {2k (τ ), 2k−1 (τ )} of Vk (τ ). Proof. We will prove the result for A1 (τ ) and the proof for A∗1 (τ ) is similar. We note that 

 

  d2 ϕk (τ, ·) − p(τ, ·)I −I ϕ (τ, ·) − p(τ, ·)ϕ (τ, ·) k k 2 A1 (τ )2k (τ ) = , = dx 0 δI −γ I δϕk (τ, ·) 

λk (τ )ϕk (τ, ·) = λk (τ )2k (τ ) + δ2k−1 (τ ) ∈ Vk (τ ). = δϕk (τ, ·) d2 dx 2

Also, we have  A1 (τ )2k−1 (τ ) =

d2 dx 2

− p(τ, ·)I δI

−I −γ I





 −ϕk (τ, ·) 0 = , −γ ϕk (τ, ·) ϕk (τ, ·)

= −2k (τ ) − γ 2k−1 (τ ) ∈ Vk (τ ). Therefore, the matrix representation of Ak1 (τ ) with respect to the basis {2k (τ ), 2k−1 (τ )} of Vk (τ ) is 

λk (τ ) −1 . δ −γ This completes the proof.

2

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17

k Lemma 3.7. For each τ ∈ [0, 1] and k ∈ IN , let μ± k (τ ) denote the eigenvalues of A1 (τ ). Then, μ± k (τ ) satisfy the following properties. + − (i) For at most finitely many k’s μ± k (τ ) are complex and satisfy μk (τ ) = μk (τ ) and

|Im(μ+ k (τ ))| ≤

√

δ.

(ii) There exists a k1 ∈ IN such that for all k ≥ k1 , μ± k (τ ) are real for all τ ∈ [0, 1] and satisfy lim μ+ k (τ ) = −∞,

lim μ− k (τ ) = −γ

k→∞

k→∞

uniformly τ ∈ [0, 1].

Hence, the eigenvalues of A1 (τ ) and A∗1 (τ ) are given by μ± k (τ ), k ∈ IN and the eigenvectors of (τ ) are given by A1 (τ ) and A∗1 (τ ) corresponding to μ± k ⎛

ξk± (τ, x) = ⎝

ϕk (τ,  x) δ γ +μ± k (τ )

ϕk (τ, x)

⎞ ⎠

⎛

and

ηk± (τ, x) = ⎝

ϕk (τ,  x)

−1 γ +μ± k (τ )

ϕk (τ, x)

⎞ ⎠.

(3.10)

Moreover, the spectra σ (A1 (τ )) of A1 (τ ) and σ (A∗1 (τ )) of A∗1 (τ ) are given by ∗ σ (A1 (τ )) = {μ± k (τ ) : k ∈ IN} ∪ {−γ } = σ (A1 (τ )).

(3.11)

Proof. First we compute the eigenvalues of Ak1 (τ ). The eigenvalues of Ak1 (τ ) are given by the roots of the equation μ2 − (λk (τ ) − γ )μ + δ − γ λk (τ ) = 0.

(3.12)

The roots are given by  1 λk (τ ) − γ ∓ 2  1 = λk (τ ) − γ ∓ 2

μ± k (τ ) =

 (λk (τ ) − γ )2 − 4(δ − γ λk (τ )) ,  (λk

(τ ) + γ )2

(3.13)

− 4δ .

This shows that the roots are complex if and only if (λk (τ ) + γ )2 < 4δ.

(3.14)

Using (iv) of Theorem 3.4, we see that the above inequality is true for at most finitely many k’s. Hence, it turns out that, for at most finitely many k’s the eigenvalues μ± k (τ ) are complex and their imaginary part satisfy    1  √ + 2  |Im(μk (τ ))| ≤ − 4δ − (λk (τ ) + γ )  ≤ δ. 2 This completes the proof of (i).

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Next we prove (ii). First, from (i) and (iv) of Theorem 3.4 we see that there exists a k0 ∈ N large enough such that for all k ≥ k0 and for all τ ∈ [0, 1], (λk (τ ) + γ )2 ≥ 4δ. Therefore, for all k ≥ k0 and for all τ ∈ [0, 1], the eigenvalues μ± k (τ ) are real. From the explicit (τ ) and from (iv) of Theorem 3.4 we conclude that expressions of μ± k  1 λk − γ − k→∞ 2

lim μ+ k = lim

k→∞

 (λk + γ )2 − 4δ = −∞

uniformly on [0, 1] and lim μ− k→∞ k

  1 2 = lim λk − γ + (λk + γ ) − 4δ , k→∞ 2   2 |λk | γ + δ ! = −γ , = lim − k→∞ |λk | + γ + (− |λk | + γ )2 − 4δ

uniformly on [0, 1]. This completes the proof of (ii). Using the fact that μ± k (τ ) satisfy (3.12) and the pair (λk (τ ), ϕk (τ, ·)) satisfy (3.8), we observe  A1 (τ )ξk± (τ, x) =

d2 dx 2

− p(τ, x)I δI ⎛

⎝ = μ± k (τ )



−I −γ I

⎛ ⎝

ϕk (τ,  x) δ γ +μ± k (τ )

ϕk (τ, x)

ϕk (τ,  x)



⎞

δ γ +μ± k (τ )

ϕk (τ, x)

⎞ ⎠,

⎠ = μ± (τ )ξ ± (τ, x). k

k

Hence, ξk± (τ, ·) are eigenfunctions of A1 (τ ) corresponding to the eigenvalues μ± k (τ ). Similarly,

we see that ηk± (τ, ·) are eigenfunctions of A∗1 (τ ) corresponding to the eigenvalues μ± k (τ ). The above properties of the eigenvalues and eigenfunctions show that A1 (τ ) is a Rieszspectral operator (see [17]). Thus the spectrum of A1 (τ ) satisfies (3.11). This completes the proof. 2 Remark 3.8. Using the property that λ1 (τ ) > λ2 (τ ) > · · · λk (τ ) > λk+1 (τ ) · · · , an explicit analy− + + − − sis shows that μ+ j = l. The equality j (τ ) = μl (τ ), μj (τ ) = μl (τ ) and μ√ j (τ ) = μl (τ ) for any√ + − μj (τ ) = μj (τ ) can occur only when λj (τ ) = −γ +2 δ or λj (τ ) = −γ −2 δ for some j . This − ±  yields that μ+ j (τ ) = μj (τ ) for atmost two j s. Hence, corresponding to the eigenvalue μj (τ ) for that j , we need to consider a generalized eigenfunction. We also note that the corresponding solution behaves like t exp(−μ± j (τ )t). This does not change the analysis in an essential way; simply we would have to modify the sums of exponentials appearing in the analysis below by the inclusion of such a term. Now onward, we assume that the spectrum of A1 (τ ) has only simple eigenvalues.

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3.4. Decomposing the system into stable and unstable part From Lemma 3.7 of previous section we see that only a finite number of eigenvalues may have non-negative real part. By property (ii) of Lemma 3.7, there exists n ∈ N and η > 0 such that all μ± k (τ ) are real for all τ ∈ [0, 1] and μ± k (t) < −η

for all t ∈ [0, 1/],

for all

(3.15)

k > n.

It can be easily check that for j, l ∈ N  # " 1− ξj± (τ, ·), ηl± (τ, ·) = Z 0,

δ 2, |γ +μ± j (τ )|

for

j = l,

for

j = l,

" # " # and for j, l ∈ N ξj+ (τ, ·), ηl− (τ, ·) = 0 = ξj− (τ, ·), ηl+ (τ, ·) . Let us denote for j ∈ N, Z

cj± (τ ) =

Z

1 1−

δ 2 |γ +μ± j (τ )|

,

and ηˆ j± (τ, x) = cj± (τ )ηj± (τ, x).

(3.16)

So {ξk± (τ )}k∈N and {ηˆ k± (τ )}k∈N form a bi-orthonormal Riesz basis of Z. Let W(t, ·) ∈ D be a solution of (3.5). Therefore, we can expand W(t, ·) as series W(t, ·) =

∞ $

Wj+ (t)ξj+ (t, ·) +

j =1

∞ $

Wl− (t)ξl− (t, ·),

(3.17)

l=1

" # where Wj± (t) = W(t, ·), ηˆ j± (t, ·) . We note that the Wj± (t) depend on  and should be deZ

+ noted, for example, by Wj, (t). For simplicity of notation we omit the index , and we shall also omit the index  for other functions later on. For any τ ∈ [0, 1], let us define (τ ) be the projection onto the 2n dimensional subspace of Z spanned by {ξj± (τ, ·) : 1 ≤ j ≤ n} and be defined by  : [0, 1] → L(Z), for τ ∈ [0, 1],

(τ )H =

n " $

n " # # $ H, ηˆ j+ (τ, ·) ξj+ (τ, ·) + H, ηˆ l− (τ, ·) ξl− (τ, ·) Z

j =1

Z

l=1

for

H ∈ Z,

(3.18)

where L(Z) denotes the set of bounded linear operators on Z. Next lemma shows that the operators (τ ) and A1 (τ ) commute. Lemma 3.9. For each τ ∈ [0, 1], the operators (τ ) and A1 (τ ) commute. Proof. We see from (3.18) that for all H ∈ Z, A1 (τ )(τ )H =

n " n " # # $ $ H, ηˆ j+ (τ, ·) A1 (τ )ξj+ (τ, ·) + H, ηˆ l− (τ, ·) A1 (τ )ξl− (τ, ·), j =1

Z

l=1

Z

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=

n " n " # # $ $ + − − (τ )ξ (τ, ·) + (τ, ·) μ− H, ηˆ j+ (τ, ·) μ+ H, η ˆ j j l l (τ )ξl (τ, ·), Z

j =1

=

l=1

n " n " # # $ $ + + − − (τ ) η ˆ (τ, ·) ξ (τ, ·) + H, μ+ H, μ− j j j l (τ )ηˆ l (τ, ·) ξl (τ, ·), Z

j =1

=

l=1

Z

n " n " # # $ $ H, A∗1 (τ )ηˆ j+ (τ, ·) ξj+ (τ, ·) + H, A∗1 (τ )ηˆ l− (τ, ·) ξl− (τ, ·), Z

j =1

=

Z

l=1

Z

n " n " # # $ $ A1 (τ )H, ηˆ j+ (τ, ·) ξj+ (τ, ·) + A1 (τ )H, ηˆ l− (τ, ·) ξl− (τ, ·), Z

j =1

l=1

Z

= (τ )A1 (τ )H. Hence, the operators (τ ) and A1 (τ ) commute.

2

We note that the family of operators (τ ) is continuously differentiable with respect to τ . In fact, we have the following.   Lemma 3.10. Let  be defined in (3.18). Then, we have (·) ∈ C 1 [0, 1]; L(Z) and for all τ ∈ [0, 1] and H ∈ Z there holds  (τ )H =

n " $

H,

j =1

+

l=1

n " $ j =1

n " # # $ ∂ ηˆ − (τ, ·) ξj+ (τ, ·) + H, l (τ, ·) ξl− (τ, ·) Z Z ∂τ ∂τ

∂ ηˆ j+

# H, ηˆ j+ (τ, ·)

∂ξj+ Z

∂τ

(τ, ·) +

n " # ∂ξ − $ l (τ, ·). H, ηˆ l− (τ, ·) Z ∂τ

(3.19)

l=1

Proof. By property (iii) in Theorem 3.4 we note that τ → ηˆ j± (τ, ·) and τ → ξj± (τ, ·) are C 1 . Hence, we have the result. 2 Let us consider the 2n dimensional subspace Z2n (τ ) of Z defined by Z2n (τ ) := span{ξj± (τ, ·) : 1 ≤ j ≤ n}. We note that the projected component (τ )W(t, ·) of W(t, ·) belongs to the subspace Z2n (τ ). Now we will consider the system of equations satisfied by the components Wj± (t) of (τ )W(t, ·). Lemma 3.11. Let n be as in (3.15). Let  > 0. For all t ∈ [0, 1/], let W(t, ·) be the solution of (3.5) corresponding to a g ∈ L2 ((0, T ) × (0, 1)) which is of the form g(t, ·) =

n $ i=1

gi (t)ϕi (τ, ·),

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with gi ∈ L2 (0, T ) and let (t)W(t, ·) =

n $

Wj+ (t)ξj+ (t, ·) +

j =1

n $

Wl− (t)ξl− (t, ·),

(3.20)

l=1

" # where Wj± (t) = W(t, ·), ηˆ j+ (t, ·) for j = 1, 2, . . . , n. Then, Wj± (t) satisfy Z

dWj± dt

± (t) = μ± j (t)Wj (t) +

n $

gi (t)cj± (t)bij (t) + rj± (, t),

j = 1, 2, ..., n, (3.21)

i=1

where

 bij (t) =

i, j = 1, 2, ..., n,

for

ϕi (t, x)ϕj (t, x)dx O

and " # rj± (, t) = R 1 (, t, ·), ηˆ j± (t, ·)

Z

for j = 1, 2, ..., n,

and R 1 (, t, ·) is given by R 1 (, t, ·) = (t)R(, t, ·) + 

n " $

W(t, ·),

∂ ηˆ j+ ∂τ

j =1

+

n " $ l=1

# (t, ·) ξj+ (t, ·) Z

# ∂ ηˆ − W(t, ·), l (t, ·) ξl− (t, ·). Z ∂τ

(3.22)

Proof. Differentiating (3.20) with respect to t and using (3.19), (3.5) and Lemma 3.9, we have n dW + $ j j =1

dt

(t)ξj+ (t, ·) +

n $ dW − l

l=1

+

n $

Wl− (t)

l=1

dt

(t)ξl− (t, ·) + 

n $

Wj+ (t)

j =1

∂ξj+ ∂τ

(t, ·)

∂ξl− (t, ·), ∂τ

=  (t)W(t, ·) + (t)Wt (t, ·), n " n " # # $ $ ∂ ηˆ j+ ∂ ηˆ − (t, ·) ξj+ (t, ·) +  W(t, ·), W(t, ·), l (t, ·) ξl− (t, ·) = Z Z ∂τ ∂τ j =1

l=1

+

n $ j =1

Wj+ (t)

∂ξj+ ∂τ

(t, ·) + 

n $ l=1

Wl− (t)

∂ξl− (t, ·) ∂τ

+ A1 (t)(t)W(t) + (t)B(g)(t) + (t)R(, t, ·).

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22

So, n dW + $ j j =1

=

dt

n $

(t) ξj+ (t, ·) +

n $ dW − l

dt

l=1

+ + μ+ j (t)Wj (t) ξj (t, ·) +

(t) ξl− (t, ·)

n $

j =1

− − μ− l (t)Wl (t) ξl (t, ·)

l=1 n " $

+ (t)B(g)(t) + (t)R(, t, ·) + 

W(t, ·),

j =1

+

∂ ηˆ j+ ∂τ

# (t, ·) ξj+ (t, ·) Z

n " # $ ∂ ηˆ − W(t, ·), l (t, ·) ξl− (t, ·), Z ∂τ l=1

=

n $

+ + μ+ j (t)Wj (t)ξj (t, ·) +

j =1

n $

− − μ− l (t)Wl (t)ξl (t, ·)

l=1

+ (t)B(g)(t) + R 1 (, t, ·). Let us define  g(t, x)χO . G(t, x) := B(g)(t) = 0

Then, we have

(τ )G(t, x) = (τ ) =

n $ i=1

=

n $

g(t, ·)χO 0

gi (t)(τ ) 

i=1

i=1

n

$

n $

+

n $ m=1

, ηˆ k+ (τ, x) ξk+ (τ, ·) Z

 ϕi (τ, ·)χO − − , ηˆ m (τ, x) ξm (τ, ·) , 0 Z 

⎛

ck+ (τ ) ⎝

k=1

 ϕi (τ, ·)χO , 0

ϕi (τ, ·)χO , 0

0

m=1



gj (t)



k=1

gi (t)

n $

 n

$ ϕi (τ, ·)χ

+

=

= (τ )

O

gi (t)

i=1

n $



⎞



ϕi (τ, x)ϕk (τ, x)dx ⎠ ξk+ (τ, ·)

O

⎛

− cm (τ ) ⎝



O

⎞



ϕi (τ, x)ϕm (τ, x)dx ⎠ ξm− (τ, ·)

,

(3.23)

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=

n n $ $

gi (t)ck+ (τ )bik (τ )ξk+ (τ, ·) +

i=1 k=1

n n $ $

23

− gi (t)cm (τ )bim (τ )ξm− (τ, ·),

i=1 m=1

where  bik (τ ) =

ϕi (τ, x)ϕk (τ, x)dx, O

for i, k = 1, 2, ..., n. We introduce the operator B1 ∈ L(C n , Z2n (τ )) defined by B1 (τ )g =

n n $ $

gi ck+ (τ )bik (τ )ξk+ (τ, ·) +

i=1 k=1

n n $ $

− gi c m (τ )bim (τ )ξm− (τ, ·),

(3.24)

i=1 m=1

for g = (g1 , g2 , · · · , gn ) ∈ C n . Therefore, we note that B1 (τ ) can be represented by the 2n × n matrix  B1 (τ ) =

C + (τ )B2 (τ ) C − (τ )B2 (τ )

 (3.25)

,

where   C ± (τ ) = diag c1± (τ ), · · · , cn± (τ )

⎛

and

b11 (τ ) B2 (τ ) = ⎝ . bn1 (τ )

··· ··· ···

⎞ b1n (τ ) . ⎠. bnn (τ )

Substituting (3.24) in the equation (3.23) we have n dW + $ j j =1

=

dt

n $

(t)ξj+ (t, ·) +

n $ dW − l

l=1

dt

+ + μ+ j (t)Wj (t)ξj (t, ·) +

n $

j =1

=

n $ j =1

+

(t)xil− (t, ·) − − 1 μ− l (t)Wl (t)ξl (t, ·) + B1 (t)(g)(t) + R (, t, ·),

l=1 + + μ+ j (t)Wj (t)ξj (t, ·) +

n $

 n $

k=1

i=1

n $

 gi (t)ck+ (t)bik (t)

− − 1 μ− l (t)Wl (t)ξl (t, ·) + R (, t, ·)

l=1

ξk+ (t, ·) +

 n n $ $ m=1

 − gi (t)cm (t)bim (t)

ξm− (t, ·).

i=1

Projecting the above system on each ηˆ j± (t, ·) we obtain (3.21). Hence, we have the result. 2 Let us estimate the finite dimensional projected part of the non-linear term.

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24

Lemma 3.12. Let R 1 (, t, ·) be given in previous lemma. Then, there exists C3 > 0 such that

 R 1 (, t, ·)∞ ≤ C3  + |v(t, ·)|22 + |z(t, ·)|2∞ ,

(3.26)

whenever |z(t, ·)|L∞ (0,1) ≤ L for all t ∈ [0, 1/]. Proof. Using Cauchy-Schwarz inequality we note that % & n  $ ∂ ηˆ j+  (τ, ·)  W(t, ·),  ∂τ j =1

Z

 % & n    $ ∂ ηˆ l−   +  +  ξj (τ, ·) ∞  W(t, ·),   L (0,1) ∂τ l=1



Z

    −  ξl (τ, ·)L∞ (0,1) 

 ≤ CW(t, ·)Z ≤ C |z(t, ·)|L2 (0,1) + |v(t, ·)|L2 (0,1) ≤ C |z(t, ·)|L∞ (0,1) + |v(t, ·)|L2 (0,1) .

Therefore, using Lemma 3.2 we have

  R 1 (, t, ·)∞ ≤ C R(, t, ·)L∞ (0,1) +  |z(t, ·)|L∞ (0,1) + |v(t, ·)|L2 (0,1) , 

≤ C  +  2 + |z(t, ·)|2L∞ (0,1) + |v(t, ·)|2L2 (0,1) ,

 ≤ C3  + |z(t, ·)|2L∞ (0,1) + |v(t, ·)|2L2 (0,1) . Hence, we have the result. 2 Let us denote ⎞ W1+ (t) ⎜ . ⎟ ⎜ + ⎟ ⎜ W (t) ⎟ n ⎟ X1 (t) = ⎜ ⎜ W − (t) ⎟ , ⎟ ⎜ 1 ⎝ . ⎠ Wn− (t) ⎛

⎞ r1+ (, t) ⎟ ⎜ ⎟ ⎜ +. ⎜ r (, t) ⎟ n ⎟ R1 (, t) = ⎜ ⎜ r − (, t) ⎟ , ⎟ ⎜ 1 ⎠ ⎝ . − rn (, t) ⎛

A1 (τ ) =

M + (τ ) On

 On , M − (τ )

where ± M ± (τ ) = diag(μ± 1 (τ ), · · · , μn (τ )),

and On is the zero square matrix of order n. Therefore, system (3.21) can be written as X1 (t) = A1 (t)X1 (t) + B1 (t)g(t) + R1 (, t).

(3.27)

In order to study the controllability of the system (3.27), let us recall the Popov-BelevitchHautus criterion. Let us denote the k × k identity matrix by Ik . Theorem 3.13. [Popov-Belevitch-Hautus] For every τ ∈ [0, 1], the system (A1 (τ ), B1 (τ )) is controllable if and only if the matrix [λI2n − A1 (τ ), B1 (τ )] has rank 2n for all λ ∈ C.

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25

Proof of Theorem 3.13 can be found in [33] Theorem 3.5, page 62. Lemma 3.14. The matrix B2 (τ ) defined in (3.25) is invertible. Proof. First, we will show that the family {ϕj }nj=1 is linearly independent in L2 (O). In order to do this if for some aj ∈ R for j = 1, · · · , n, n $

aj ϕj (τ, x) = 0 for all x ∈ O,

(3.28)

j =1

then, we need to show that aj = 0 for all j = 1, · · · , n. Let us consider the function φ˜ 1 (τ, x) :=

n $

aj ϕj (τ, x),

x ∈ (0, 1).

j =1

Now using (3.8) we obtain n  

d2

$ ˜ − p(τ, ·) − λ (τ ) φ (τ, ·) = a (τ ) − λ (τ ) ϕj (τ, ·), λ 1 1 j j 1 dx 2 j =1

=

n $



aj λj (τ ) − λ1 (τ ) ϕj (τ, ·) := φ˜ 2 (τ, ·).

j =2

As φ˜ 1 (τ, x) = 0 for all x ∈ O, we also note that φ˜ 2 (τ, x) = 0 for all x ∈ O. Again applying d2 − p(τ, ·) − λ2 (τ ), dx 2 on φ˜ 2 (τ, ·) we obtain n   

d2

$ ˜ − p(τ, ·) − λ (τ ) φ (τ, ·) = aj λj (τ ) − λ1 (τ ) λj (τ ) − λ2 (τ ) ϕj (τ, ·), 2 2 2 dx j =2

=

n $

 

aj λj (τ ) − λ1 (τ ) λj (τ ) − λ2 (τ ) ϕj (τ, ·),

j =3

:= φ˜ 3 (τ, ·). Repeating this procedure after finite number of steps we obtain 

d2 − p(τ, ·) − λ (τ ) φ˜ n−1 (τ, ·) n−1 dx 2  

= an λn (τ ) − λ1 (τ ) · · · λn (τ ) − λn−1 (τ ) ϕn (τ, ·) := φ˜ n (τ, ·),

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26

and φ˜ n (τ, x) = 0 for all x ∈ O. Now we see that φ˜ n (τ, ·) satisfies 

d2 ˜ φ (τ, x) − p(τ, ·)φ˜ n (τ, x) = λn (τ )φ˜ n (τ, x) dx 2 n

in O,

φ˜ n (τ, 0) = 0 = φ˜ n (τ, 1).

We will show φ˜ n (τ, x) = 0 for all x ∈ [0, 1]. Suppose that, on the contrary, it is not true. Then, by the Theorem 2.1 of Chapter 8 (page 212) of [12], we note that φ˜ n (τ, ·) can have at most n − 1 zeros in (0, 1) which is in contradiction with the fact φ˜ n (τ, x) = 0 for all x ∈ O. Hence, φ˜ n (τ, x) = 0 for all x ∈ [0, 1]. As λn (τ ) = λj (τ ) for j = 1, .., n −1. Therefore, an = 0. Following the same analysis we can show aj = 0 for j = 1, .., n. Hence, the family {ϕj (τ )}nj=1 is linearly independent in L2 (O). This yields B2 (τ ) is invertible because it is a Gram matrix in Rn×n . 2 Now we will verify the Popov-Belevitch-Hautus condition given in the Theorem 3.13. Theorem 3.15. Let τ ∈ [0, 1]. Then, the matrix [λI2n − A1 (τ ), B1 (τ )] has rank 2n for all λ ∈ C. Proof. The matrix [λI2n − A1 (τ ), B1 (τ )] can be written as

[λI2n − A1 (τ ), B1 (τ )] =

λIn − M + (τ ) On On λIn − M − (τ )

 C + (τ )B2 (τ ) . C − (τ )B2 (τ )

± If λ ∈ C \ {μ± 1 (τ ), · · · , μn (τ )}, then, we note that λI2n − A1 (τ ) has rank 2n and so has [λI2n − + ± A1 (τ ), B1 (τ )]. Now, consider the case λ ∈ {μ± 1 (τ ), · · · , μn (τ )}. For simplicity, let λ = μ1 (τ ). Therefore, we have

[μ+ 1 (τ )I2n − A1 (τ ), B1 (τ )] =



+ μ+ 1 (τ )In − M (τ ) On

 On C + (τ )B2 (τ ) . − − μ+ 1 (τ )In − M (τ ) C (τ )B2 (τ )

Since we have assumed that all the eigenvalues are simple (see Remark 3.8), therefore, the 2n order square matrix [μ+ 1 (τ )I2n − A1 (τ )] is actually a diagonal matrix with first diagonal element is zero and rest diagonal elements are non zero. So it has rank (2n − 1). Let us denote

T bC (τ ) := c1+ (τ )b11 (τ ), · · · , cn+ (τ )bn1 (τ ), c1− (τ )b11 (τ ), · · · , cn− (τ )bn1 (τ ) , the 2n × 1 column vector. Then, the 2n × (2n + 1) order sub matrix [μ+ 1 (τ )I2n − A1 (τ ), bC (τ )] has all zero entries in the first column and all other columns are nonzero and it can be written as ⎛ ⎞ 0 ··· 0 0 ··· 0 c1+ (τ )b11 (τ ) ⎜ . ··· ⎟ . . ··· . . ⎜ ⎟ + (τ )b (τ ) ⎟ ⎜ 0 · · · μ+ (τ ) − μ+ (τ ) 0 · · · 0 c n1 n n 1 ⎜ ⎟. − ⎜0 · · · 0 μ+ 0 c1− (τ )b11 (τ ) ⎟ 1 (τ ) − μ1 (τ ) · · · ⎜ ⎟ ⎝ . ··· ⎠ . . ··· . . + − − 0 ··· 0 0 · · · μ1 (τ ) − μn (τ ) cn (τ )bn1 (τ )

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27

Since c1+ (τ )b11 (τ ) = 0, therefore, the matrix [μ+ 1 (τ )I2n − A1 (τ ), bC (τ )] has rank 2n and so has + [μ1 (τ )I2n − A1 (τ ), B1 (τ )]. Hence, we have the result. 2 Remark 3.16. It is known in the literature that, for a finite dimensional linear autonomous control system, the Popov-Belevitch-Hautus condition, which is equivalent to the controllability of the system, implies the stabilizability of the same system. This implication is no longer true for non autonomous linear systems but for the system with slowly-varying time, where the importance of the role played by the parameter  is essential. As a consequence we have the smoothly varying pole shifting property (see [30], Ch 9.6). Corollary 3.17. Let τ ∈ [0, 1]. Then, there exist k1+ (τ ), ..., kn+ (τ ), k1− (τ ), ..., kn− (τ ) such that the matrix A1 (τ ) + B1 (τ )K1 (τ ) admits −1 as an eigenvalue of order 2n, where K1 (τ ) is a n × 2n matrix given by ⎛

k1+ (τ ) · · · K1 (τ ) = ⎝ . ··· 0 ···

0 k1− (τ ) · · · . . ··· 0 ··· kn+ (τ )

⎞ 0 . ⎠. kn− (τ )

Moreover, there exists a C 1 map τ → P (τ ) on [0, 1], where P (τ ) is an 2n × 2n symmetric positive definite matrix such that the following identity

 ∗ P (τ ) A1 (τ ) + B1 (τ )K1 (τ ) + A1 (τ ) + B1 (τ )K1 (τ ) P (τ ) = −I2n ,

(3.29)

holds for all τ ∈ [0, 1]. The matrix K1 (τ ), obtained in Corollary 3.17, helps us to construct the feedback control function g(t) of the form g(t) = K1 (t)X1 (t), on [0, 1/] and stabilizes the finite dimensional linear control system X1 (t) = A1 (t)X1 (t) + B1 (t)g(t). We will show later that this feedback control actually stabilizes the whole infinite dimensional system (3.5), provided  > 0 is chosen small enough, indeed this will guide us to construct appropriate Lyapunov functional for that system to stabilize. 3.5. Existence, uniqueness and regularity Let T > 0. In this subsection we will consider the existence, uniqueness and regularity of the solution the system (3.5) which we write as  Wt (t) = A1 (t)W(t) + BK1 (t)(t)W(t) + F(W(t)), W(0) = 0,

in (0, T ),

(3.30)

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28

where

F(W(t)) =

z2 (a + 1 − 3u) − z3 − uτ −wτ

 ,

W(t) = (z(t, ·), v(t, ·))T and the control g(t) is replaced by the feedback control g(t) = K1 (t)(τ )W(t). Regularity of the solution will be used to construct appropriate Lyapunov functional in the next section. We note that BK1 (τ )(τ )W(t) = B1 (τ )K1 (τ )(τ )W(t) and B1 (τ )K1 (τ ) is equal to ⎛

k1+ (τ )c1+ (τ )b11 (τ ) ⎜ . ⎜ + ⎜ k (τ )c+ (τ )bn1 (τ ) n ⎜ 1+ ⎜ k (τ )c− (τ )b11 (τ ) 1 ⎜ 1 ⎝ . k1+ (τ )cn− (τ )bn1 (τ )

·· kn+ (τ )c1+ (τ )b1n (τ ) ·· . + + ·· kn (τ )cn (τ )bnn (τ ) ·· kn+ (τ )c1− (τ )b1n (τ ) ·· . ·· kn+ (τ )cn− (τ )bnn (τ )

k1− (τ )c1+ (τ )b11 (τ ) . − + k1 (τ )cn (τ )bn1 (τ ) k1− (τ )c1− (τ )b11 (τ ) . k1− (τ )cn− (τ )bn1 (τ )

⎞ ·· kn− (τ )c1+ (τ )b1n (τ ) ⎟ ·· . ⎟ − + ·· kn (τ )cn (τ )bnn (τ ) ⎟ ⎟. ·· kn− (τ )c1− (τ )b1n (τ ) ⎟ ⎟ ⎠ ·· − − ·· kn (τ )cn (τ )bnn (τ )

Hence, we have by the definition of B BK1 (τ )(τ )W(t) n $ n  $ ki+ (τ )Wi+ + ki− (τ )Wi− cj+ (τ )bij (τ )ξj+ (τ, ·) = i=1 j =1

+

n $ n $

 ki+ (τ )Wi+ + ki− (τ )Wi− cj− (τ )bij (τ )ξj− (τ, ·),

i=1 j =1

=

n n $

  $ ki+ (τ )Wi+ + ki− (τ )Wi− bij (τ ) cj+ (τ )ξj+ (τ, ·) + cj− (τ )ξj− (τ, ·) , i=1 j =1

⎛ n $ n  $ ki+ (τ )Wi+ + ki− (τ )Wi− bij (τ ) ⎝ = i=1 j =1

' =

n 'n i=1 j =1

cj+ (τ )ϕj (τ ) + cj− (τ )ϕj (τ )

cj+ (τ )δ

γ +μ+ j (τ )

(τ )ϕj (τ ) +

cj− (τ )δ

ϕj (τ ) γ +μ− j (τ )

   ki+ (τ )Wi+ + ki− (τ )Wi− bij (τ ) cj+ (τ ) + cj− (τ ) ϕj (τ ) . 0

Using the explicit expression of Wj± , given in (3.47), we simplify as follows: n $ n

  $ ki+ (τ )Wi+ + ki− (τ )Wi− bij (τ ) cj+ (τ ) + cj− (τ ) ϕj (τ ) i=1 j =1

⎞ ⎠,

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=

n $ n $



 ki+ (τ )

ci+ (τ )zi (t) −

i=1 j =1

 + ki− (τ ) 

=

n $ n

$

 −

n n $ $

γ + μ+ i (τ )



ci− (τ )

γ

vi (t) + μ− i (τ )

ki+ (τ )ci+ (τ ) γ + μ+ i (τ )

+

ki− (τ )ci− (τ )



γ + μ− i (τ )

 · bij (τ ) cj+ (τ ) + cj− (τ ) ϕj (τ ),



 vi (t) · bij (τ ) cj+ (τ ) + cj− (τ ) ϕj (τ ),



 ki+ (τ )ci+ (τ ) + ki− (τ )ci− (τ ) zi (t)bij (τ ) cj+ (τ ) + cj− (τ ) ϕj (τ )

i=1 j =1

−

vi (t)

 ki+ (τ )ci+ (τ ) + ki− (τ )ci− (τ ) zi (t)

i=1 j =1

=



ci+ (τ )

ci− (τ )zi (t) −

29

 n n $ $ ki+ (τ )ci+ (τ ) γ

i=1 j =1

+ μ+ i (τ )

+

ki− (τ )ci− (τ ) γ



+ μ− i (τ )

 vi (t)bij (τ ) cj+ (τ ) + cj− (τ ) ϕj (τ ),

:=d1 (τ )z(t) + d2 (τ )v(t), where d1 (τ ), d2 (τ ) is defined for z, v ∈ L2 (0, 1) by d1 (τ )z =

n $ n

  $ ki+ (τ )ci+ + ki− (τ )ci− zi (τ )bij (τ ) cj+ (τ ) + cj− (τ ) ϕj (τ ),

(3.31)

i=1 j =1

with z =

∞ $

zi (τ )ϕi (τ ) and

i=1

d2 (τ )v = −

 n $ n $ ki+ (τ )ci+ (τ ) γ + μ+ i (τ )

i=1 j =1

+

ki− (τ )ci− (τ ) γ + μ− i (τ )



 vi (τ )bij (τ ) cj+ (τ ) + cj− (τ ) ϕj (τ ), (3.32)

with v =

∞ $

2 vi (τ )ϕi (τ ). As {ϕi (τ )}∞ j =1 is an orthonormal sequence in L (0, 1), we get

i=1

2    d1 (τ )z 2

L (0,1)

=

n $ j =1

≤C

 n 2 $

cj+ (τ ) + cj− (τ )

n $



ki+ (τ )ci+ (τ ) + ki− (τ )ci− (τ )

2 zi (τ )bij (τ )

,

i=1

|zi (τ )|2 ≤ C|z|2L2 (0,1) .

i=1

Hence, d1 (τ ) ∈ L(L2 (0, 1)). Similarly we obtain d2 (τ ) ∈ L(L2 (0, 1)). Thus the system (3.30) can be written, as

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⎧ ⎪ zt = zxx + 2(a + 1)u(t) − 3u2 (t) − a z + z2 a + 1 − 3u(t) − z3 ⎪ ⎪ ⎪ ⎪ ⎪ −v + d1 (t)z + d2 (t)v − uτ (t), in (0, T ) × (0, 1), ⎨ vt = δz − γ v − wτ (t), in (0, T ) × (0, 1), ⎪ ⎪ ⎪ ⎪ z(t, 0) = z(t, 1) = 0, t ∈ (0, T ), ⎪ ⎪ ⎩ z(0, x) = 0 = v(0, x), x ∈ (0, 1).

(3.33)

An equivalent formulation of (3.33) can be obtained by solving for v(t, ·) in (3.33) which yields t v(t, x) =



e−γ (t−s) δz(s, x) − wτ (s, x) ds.

(3.34)

0

Hence, the system (3.33) can be written as

 ⎧ 2 ⎪ = z + 2(a + 1)u(t) − 3u (t) − a + d (t) z z t xx 1 ⎪ ⎪ 



⎪  t ⎪ t 2 3 −γ (t−s) ⎪ ⎪ z(s)ds − 0 e−γ (t−s) z(s)ds +z a + 1 − 3u(t) − z + δ d2 (t) 0 e ⎨

 t t −uτ (t) −  d2 (t) 0 e−γ (t−s) wτ (s)ds − 0 e−γ (t−s) wτ (s)ds in (0, T ) × (0, 1), ⎪ ⎪ ⎪ ⎪ ⎪ z(t, 0) = z(t, 1) = 0, t ∈ (0, T ), ⎪ ⎪ ⎩ z(0, x) = 0, x ∈ (0, 1). (3.35) Let us denote Q = (0, T ) × (0, 1) and consider the Hilbert space   H 1,2 (Q) = z ∈ L2 (0, T ; H 2 (0, 1)) : z(t, 0) = 0 = z(t, 1), zt ∈ L2 (Q) . Theorem 3.18. The system (3.33) (or (3.30)) has a unique solution W(t, ·) = (z(t, ·), v(t, ·))T with  z ∈ L2 (0, T ; H 2 (0, 1)) ∩ C([0, T ]; H01 (0, 1)), zt ∈ L2 (Q), (3.36) v ∈ C([0, T ]; H 2 (0, 1)), vt ∈ L2 (0, T ; H 2 (0, 1)). Theorem 3.18 has been proved in the same spirit of Theorem 1 in [5] and is given in Appendix. We have already shown in Theorem 3.18 that W ∈ L2 (0, T ; D). We also have v ∈ C([0, T ]; H 2 (0, 1) ∩ H01 (0, 1)), vt ∈ L2 (0, T ; H 2 (0, 1) ∩ H01 (0, 1)) (see also [5]). Let us denote   Λ = u ∈ H 3 (0, 1) : u(0) = u(1) = 0 = uxx (0) = uxx (1) . We define D((A1 (τ ))2 ) domain of (A1 (τ ))2 as   D((A1 (τ ))2 ) := W ∈ D(A1 (τ )) : A1 (τ )W ∈ D(A1 (τ )) . We note that D((A1 (τ ))2 ) are same for all τ ∈ [0, 1]. Let us denote D2 := D((A1 (τ ))2 ) for all 0 ≤ τ ≤ 1. We have the following regularity result.

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Theorem 3.19. Let W(t) be the solution of (3.30). Then, W ∈ L2 (0, T ; D2 ) and Wt ∈ L2 (0, T ; D). Proof. As uτ (τ, ·), wτ (τ, ·) ∈ C 2 and W ∈ L2 (0, T ; D), so one can observe that F(W(·)) ∈ L2 (0, T ; D). Also from (3.31)-(3.32), we note that B(τ )K1 (τ )(τ )W ∈ L2 (0, T ; D). Let us consider W(t) = A(t)W(t). We note that W(t) satisfies ⎧  ⎪ ⎨Wt (t) = A1 (t)W(t) + A1 (t)W(t) + A1 (t)BK1 (t)(τ )W(t) +A1 (t)F(W(t)) in (0, T ), ⎪ ⎩ W(0) = 0.

(3.37)

We note that A1 (τ )W ∈ L2 (0, T ; Z). Again, by parabolic regularity theory (see [3]) we have W ∈ L2 (0, T ; D). Hence W ∈ L2 (0, T ; D2 ). We note from the definition of W(t) that A(t)Wt (t) = Wt (t) − A (t)W(t). We observe from (3.37) that Wt ∈ L2 (0, T ; Z). Hence, Wt ∈ L2 (0, T ; D). This completes the proof.

2

As a corollary of the above theorem we obtain that Corollary 3.20. Let W(t) = (z(t, ·), v(t, ·))T be the solution of (3.30). Then z ∈ L2 (0, T ; H 4 (0, 1)) ∩ C([0, T ]; Λ).

(3.38)

By the Theorem 3.18 we conclude that the time T may be taken as any arbitrary positive number, i.e. the solution exists for all time t ∈ [0, ∞). But the bounds in (3.36) may depend on T . Now we will find uniform L∞ (0, 1) bound of the solution. Let us denote Q∞ := (0, ∞) × (0, 1). Then, we have the following result. Theorem 3.21. Let z ∈ H 1,2 (Q∞ ) be a solution of (3.35) and v be defined in (3.34). Then, there exists L > 0 such that |z(t)|L2 (0,1) , |v(t)|L2 (0,1) ≤ L,

(3.39)

|z(t)|L∞ (0,1) ≤ L,

(3.40)

and

for all t ≥ 0. Moreover, we have |v(t)|L∞ (0,1) ≤ L˜ for all t ≥ 0, for some L˜ > 0. Proof of Theorem 3.21 has been given in Appendix.

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3.6. Construction of Lyapunov functional In this section we will construct a Lyapunov functional so that the original system (3.5) should be stabilizable. Let c > 0 (will be chosen later). For any t ∈ [0, 1/] and W ∈ D let # 1 " T V (t, W) = cX1 (t)P (t)X1 (t) − Re W, A1 (t)W , Z 2

(3.41)

where we denote Wj± (t) = W, ηˆ j± (t)Z for j ∈ N and ⎞ W1+ (t) ⎜ . ⎟ ⎜ + ⎟ ⎜ Wn (t) ⎟ 2n ⎟ X1 (t) = ⎜ ⎜ W − (t) ⎟ ∈ C . ⎟ ⎜ 1 ⎝ . ⎠ Wn− (t) ⎛

3.6.1. Notation ) ( Let  be a set and let Ξ = (, t) : 0 <  ≤ 1, 0 ≤ t ≤ 1/ . Let F1 , F2 be two real valued functions defined on Ξ × . The notation F1  F2 means that F2 ≥ 0 and that there exists a constant C > 0 such that ∀ (, t) ∈ Ξ, ∀ θ ∈ , F1 (, t, θ) ≤ CF2 (, t, θ). We say that F1 ∼ F2 if both F1  F2 and F2  F1 . Let us denote the Hermitian norm in C 2n by  · 2 . As τ → P (τ ) is C 1 ([0, 1], R2n×2n ) map and P (τ ) is real symmetric positive definite. We can choose C11 , C12 > 0 independent of τ ∈ [0, 1] such that T

C11 X22 ≤ X P (τ )X ≤ C12 X22 ,

∀ X ∈ C 2n .

T

Hence, cX1 (t)P (t)X1 (t) ∼ X1 (t)22 .

 z Lemma 3.22. Let V be defined in (3.41). Then for all t ∈ [0, 1/] and for W = ∈ D, we v have V (t, W) ∼ |z|2H 1 (0,1) + |v|2L2 (0,1) ,

(3.42)

V (t, W))  X1 22 + A1 (t)W2Z .

(3.43)

0

and

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Proof. Using integration by Parts we note that − W, A1 (t)WZ = − =−



 z z − p(t, ·)z − v , xx , v δz − γ v Z 1

 zzxx − p(t, ·)|z|2 − zv + δzv − γ |v|2 dx,

0

=

1

 |zx |2 + p(t, ·)|z|2 + zv − δzv + γ |v|2 dx.

0

Hence, we have "

− Re W, A1 (t)W

# Z

1  = |zx |2 + p(t, ·)|z|2 + Re(zv − δzv) + γ |v|2 dx.

(3.44)

0

Using Cauchy-Schwarz inequality from (3.44), we obtain 1 " #  −Re W, A1 (t)W ≤ |zx |2 + p(t, ·)|z|2 + (1 + δ)|z||v| + γ |v|2 dx, Z

0

1  1 |zx |2 + p(t, ·)|z|2 + (1 + δ)(|z|2 + |v|2 ) + γ |v|2 dx. ≤ 2 0

Using the fact that p(τ, ·) is bounded (uniformly with respect to τ ) we get that # 1 " − Re W, A1 (t)W  |z|2H 1 (0,1) + |v|22 . Z 2 0 Therefore, we have # 1 " T V (t, W) = cX1 (t)P (t)X1 (t) − Re W, A1 (t)W  |z|2H 1 (0,1) + |v|2L2 (0,1) . Z 2 0 Again using Cauchy-Schwarz inequality from (3.44), we have for some c1 > 0 1 " #  −Re W, A1 (t)W = |zx |2 + p(t, ·)|z|2 + Re(zv − δzv) + γ |v|2 dx, Z

0

≥

1 0

 |zx |2 + p(t, ·)|z|2 − (1 + δ)|z||v| + γ |v|2 dx,

(3.45)

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1

≥

 γ 2 (1 + δ)2 2 2 |z| + γ |v| dx, |zx | + p(t, ·)|z| − |v| − 2 2γ 2

2

0

≥

1

|zx |2 +

 γ 2 |v| − c1 |z|2 dx. 2

0

Hence, using the series expansion of W defined in (3.17) we see that # 1 2 γ c1 1 " |z|H 1 (0,1) + |v|2L2 (0,1) ≤ |z|22 − Re W, A1 (t)W Z 2 4 2 2 0  ∞ ∞   *2 *

 2 1 $ c1 $   + 2 * + * ≤ Re μ+ (t) W (t) (t, ·) zj (t) − ξ   * * j j j Z 2 2 j =1

(3.46)

j =1

+

∞ $



  * * W − (t)2 *ξ − (t, ·)*2 . Re μ− (t) l l l Z

l=1

Let us find the relation of z(t) and v(t) with Wj± (t). We recall that for j = 1, 2, · · · , n, %  ⎛ " # z ⎝

Wj± (t) = W, ηˆ j± (t) = cj± (t) , v Z ⎛

⎛

= cj± (t) ⎝zj (t) − ⎝

⎞&

ϕj (t,  ·)

−1 γ +μ± j (t)

ϕj (t, ·) ⎞

1 γ + μ± j (t)

⎠ ⎞

, Z

(3.47)

⎠ vj (t)⎠ ,

where     zj (t) := z, ϕj (t, ·) L2 (0,1) , vj (t) := v, ϕj (t, ·) L2 (0,1) , are the coordinates of z, v respectively in the basis given by ϕj (t, ·). Hence, solving for zj (t) and using the explicit expression of cj± (t) given in (3.16) we have + − 1 (γ + μj (t))) 1 (γ + μj (t))) + W W − (t), (t) − cj+ μ+ (t) − μ− (t) j cj− μ+ (t) − μ− (t) j j j j j * * * * * + * * − * + := ej (t) *ξj (, ·)* Wj (t) + fj (t) *ξj (, ·)* Wj− (t),

zj (t) =

Z

Z

where 

δ ej (t) = 1 − 2 |γ + μ+ j (t)|



(γ + μ+ j (t)) − μ+ j (t) − μj (t)

+ 1+

1 δ2 2 |γ +μ+ j (t)|

,

(3.48)

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and 

δ fj (t) = 1 − 2 |γ + μ− j (t)|



(γ + μ− j (t)) − μ+ j (t) − μj (t)

+

1 1+

.

δ2 2 |γ +μ− j (t)|

Therefore, we obtain 2 * *2  2 * *2        * * − − zj (t)2 ≤ 2 ej (t)2 W + (t) * fj (t)2 W − (t) ) * ξ (, ·) + ξ (, ·) * * * * . j j j j Z

Z

Using Lemma 3.7 (ii), we note that ej (t) → 1 and fj (t) → 0 as j → ∞ uniformly for t ∈ [0, 1 ]. Using (3.15) and the property (ii) of the Lemma 3.7 we note that there exists C4 > 0 such that for j > n, c1 |ej (t)|2 ≤ −

C4 + μ (t) 2 j

c1 |fj (t)|2 ≤ −

C4 − μ (t). 2 j

(3.49)

Substituting (3.48) in (3.46) and then, using (3.49), we have 1 2 γ + |v|2 2 |z| 1 2 H0 (0,1) 4 L (0,1) ∞

2 * *2  2 * *2  $      * * + − fj (t)2 W − (t) * ej (t)2 W + (t) * ξ ξ ≤ c1 (, ·) + (, ·) * * * * j j j j Z

j =1

−

Z

∞ 2 * *2  1$ +   * * Re μj (t) Wj+ (t) *ξj+ (, ·)* Z 2 j =1

− ≤ c1

n

$

∞  2 * *2 1$ − Re μl (t) Wl− (t) *ξl− (, ·)*Z , 2 l=1

2 * *2  *2    2  − 2 *  * − * * +   ej (t)2 W + (t) * *ξj (, ·)* + fj (t) Wj (t) *ξj (, ·)* j Z

j =1

+ c1

∞

2 * *2  2 * *2  $      * * + − fj (t)2 W − (t) * ej (t)2 W + (t) * ξ ξ (, ·) + (, ·) * * * * j j j j j =n+1

−

Z

Z

n 2 * *2  1$ +   * * Re μj (t) Wj+ (t) *ξj+ (, ·)* Z 2 j =1

−

n  2 * *2 1$ − Re μl (t) Wl− (t) *ξl− (, ·)*Z 2 l=1

−

∞  * *2

 1 $  + 2 * + * Re μ+ j (t) Wj (t) *ξj (, ·)*Z 2 j =n+1

Z

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−

∞

  * * 1 $ W − (t)2 *ξ − (, ·)*2 , Re μ− (t) l l l Z 2 l=n+1

≤ c1

n

$

2 * *2  *2    2  − 2 *  * * * − +   ej (t)2 W + (t) * *ξj (, ·)* + fj (t) Wj (t) *ξj (, ·)* j Z

j =1

−

Z

n 2 * *2  1$ +   * * Re μj (t) Wj+ (t) *ξj+ (, ·)* Z 2 j =1

−

n  2 * *2 1$ − Re μl (t) Wl− (t) *ξl− (, ·)*Z 2 l=1

−

∞  * *2

 (C4 + 1) $  + 2 * + * Re μ+ j (t) Wj (t) *ξj (, ·)*Z 2 j =n+1

−

∞

  * * (C4 + 1) $ W − (t)2 *ξ − (, ·)*2 , Re μ− (t) l l l Z 2 l=n+1

≤ c1

n

$

2 * *2  *2    2  − 2 *  * − * * +   ej (t)2 W + (t) * *ξj (, ·)* + fj (t) Wj (t) *ξj (, ·)* j Z

j =1

+

Z

n 2 * *2  C4 $ +   * * Re μj (t) Wj+ (t) *ξj+ (, ·)* Z 2 j =1

+

n  2 * *2 C4 $ − Re μl (t) Wl− (t) *ξl− (, ·)*Z 2 l=1

−

∞ 2 * *2  (C4 + 1) $ +   * * Re μj (t) Wj+ (t) *ξj+ (, ·)* Z 2 j =1

−

∞  2 * *2 (C4 + 1) $ − Re μl (t) Wl− (t) *ξl− (, ·)*Z 2

(3.50)

l=1

*2

 * * ± * Using a uniform bound of |ej (t)|2 , |fj (t)|2 , Re μ± (t) , ξ (, ·) * * for 1 ≤ j ≤ n we obj j Z tain # 1 2 γ 1 " |z|H 1 (0,1) + |v|2L2 (0,1)  X1 (t)22 − Re W, A1 (t)W  V (t, W). Z 2 4 2 0

(3.51)

Therefore, we conclude from (3.45) and (3.50) that |z|2H 1 (0,1) + |v|2L2 (0,1) ∼ V (t, W). Hence, we 0

have (3.42). For the second part, using (3.15), from (3.50) we observe that

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1 2 γ |z|H 1 (0,1) + |v|2L2 (0,1) 2 4 0 ∞

*2  *2  $ 2  + 2 * 2  − 2 *  * + * − * *     ≤ c1 ej (t) Wj (t) *ξj (, ·)* + fj (t) Wj (t) *ξj (, ·)* j =1

−

Z

⎛

1⎝ 2

∞ $ j =1

Z

⎞ ∞ 2 * *2 $ 

  * *  +  * + *  − 2 *ξ − (, ·)*2 ⎠ , Re μ+ Re μ− j (t) Wj (t) *ξj (, ·)* + l (t) Wl (t) l Z

Z

l=1

⎛ ⎞ n  2  2 $    +  ⎝ Wj (t) + Wj− (t) ⎠ j =1

⎛ ⎞ ∞ ∞  2 * *2 $  *    * 1 ⎝ $  + 2 2   * * μ− (t) W − (t) *ξ − (, ·)* ⎠ , + μj (t) Wj+ (t) *ξj+ (, ·)* + l l l Z Z 2 j =n+1

⎞ ⎛ n  2  2 $     ⎝ Wj+ (t) + Wj− (t) ⎠ j =1

l=n+1

⎛

⎞ ∞  ∞  *   * $ $ 2 2 2   *   * 1  +    * * μ− (t)2 W − (t)2 *ξ − (, ·)*2 ⎠ , + ⎝ μj (t) Wj+ (t) *ξj+ (, ·)* + l l l Z Z 2 j =1

 X1 (t)22

l=1

+ A1 (t)W2Z .

Hence, we have the result. 2 Remark 3.23. Before going to further analysis we would like to comment that we have considered here the Lyapunov functional V (t, W) which is equivalent to |z|2H 1 (0,1) + |v|2L2 (0,1) . Instead 0

of taking this type of Lyapunov functional, we could have considered another V (t, W) which is equivalent to |z|2L2 (0,1) + |v|2L2 (0,1) . Because of the presence of accumulating nature of a part of the spectrum of A(τ ) about a real number, that may lead us to conclude only that the solution (z, v)T is bounded in a neighbourhood of (0, 0)T with respect to time, but not the decay of (z, v)T to (0, 0)T . Hence, we would not have been able to show that the solution would reach near (0, 0)T after a large time. Let W(t) = (z(t, ·), v(t, ·))T be the solution of (3.30) for t ∈ [0, 1/]. Let us set # 1 " T V1 (t) := V (t, W(t)) = cX1 (t)P (t)X1 (t) − Re W(t), A1 (t)W(t) . Z 2 Now we compute

dV1 dt (t)

and derive the differential inequality satisfied by V1 (t).

Lemma 3.24. Let W(t) = (z(t, ·), v(t, ·))T be the solution of (3.30) for t ∈ [0, 1/] and V1 (t) be defined above. Then, V1 (t) satisfies the differential inequality

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dV1 (t) + X1 (t)22 + A1 (t)W(t)2Z dt 3 !  2( V1 (t) + V12 (t)) + 2V1 (t)

(3.52)

+  A1 (t)W(t)Z + V1 (t) A1 (t)W(t)Z . Proof. Using Theorem 3.19 and differentiating V1 (t) with respect to t we have 

 dV1 T (t) = c (X1 )T (t)P (t)X1 (t) + X1 (t)P (t)X1 (t) dt # 1 " T +  cX1 (t)P  (t)X1 (t) − Re Wt (t), A1 (t)W(t) Z 2 # # 1 "  " − Re W(t), A1 (t)Wt (t) − Re W(t), A1 (t)W(t) . Z Z 2 2

(3.53)

Using (3.27) and (3.29) we have 

(X1 )T (t)P (t)X1 (t) + X1 (t)P (t)X1 (t) T

T

= −X1 (t)22 + R1T (t)P (t)X1 (t) + X1 (t)P (t)R1 (t).

(3.54)

Using (3.30) we obtain " # Re Wt (t), A1 (t)W(t)

Z

" # = A1 (t)W(t)2Z + Re BK1 (t)X1 (t), A1 (t)W(t) Z " # + Re R(, t, ·), A1 (t)W(t) .

(3.55)

Z

As A∗1 (t) = A1 (t) + D where

0 D= −(δ + 1)I

(δ + 1)I 0

 ,

so using Theorem 3.19, " # " # Re W(t), A1 (t)Wt (t) = Re A∗1 (t)W(t), Wt (t) , Z Z " # " # = Re A1 (t)W(t), Wt (t) + Re DW(t), Wt (t) , Z Z " # = A1 (t)W(t)2Z + Re A1 (t)W(t), BK1 (t)X1 (t) Z " # " # + Re A1 (t)W(t), R(, t, ·) + Re DW(t), A1 (t)W(t) Z Z " # " # + Re DW(t), BK1 (t)X1 (t) + Re DW(t), R(, t, ·) . Z

Z

(3.56)

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Also we note that A1 (t) =



−p  (t, ·) 0

   0 2(a + 1) − 6u(t, ·) uτ (t, ·) = 0 0

 0 . 0

Therefore, we get " # Re W(t), A1 (t)W(t) Z "

 # = Re z(t, ·), 2(a + 1) − 6u(t, ·) uτ (t, ·)z(t, ·)

(3.57) L2 (0,1)

.

Using (3.54)-(3.57), we have from (3.53) dV1 (t) = − cX1 (t)22 − A1 (t)W(t)2Z dt " # − Re A1 (t)W(t), BK1 (t)X1 (t) Z " # − Re A1 (t)W(t), R(, t, ·) Z

 T T + c R1 (t)P (t)X1 (t) + X1 (t)P (t)R1 (t) # 1 " − Re DW(t), A1 (t)W(t) Z 2 # " 1 − Re DW(t), BK1 (t)X1 (t) Z 2 # 1 " T − Re DW(t), R(, t, ·) +  cX1 (t)P  (t)X1 (t) Z 2  #

 " . − Re z(t, ·), 2(a + 1) − 6u(t, ·) uτ (t, ·)z(t, ·) 2 L (0,1;C) 2

(3.58)

Using Cauchy-Schwarz inequality we note that  " #    Re A1 (t)W(t), BK1 (t)X1 (t)  Z

≤ A1 (t)W(t)Z BK1 (t)X1 (t)Z , 1 ≤ A1 (t)W(t)2Z + 2BK1 (t)X1 (t)2Z , 8 1 ≤ A1 (t)W(t)2Z + M1 X1 (t)22 , 8

(3.59)

for some M1 > 0. Using the definition of R in (3.6) we have

2  " # z (a + 1 − u) − z3 Re A1 (t)W(t), R(, t, ·) = Re A1 (t)W(t), 0 Z Z 

uτ (t) −  Re A1 (t)W(t), . wτ (t) Z

(3.60)

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Using (3.39)-(3.40) we note that 

2   3   Re A1 (t)W(t), z (a + 1 − u) − z    0 Z * 2 * * z (a + 1 − u) − z3 * * , ≤ A1 (t)W(t)Z * * * 0 Z

 2 3  A1 (t)W(t)Z |z(t, ·)|L4 (0,1) + |z(t, ·)|L6 (0,1) ,

  A1 (t)W(t)Z |z(t, ·)|2L6 (0,1) + |z(t, ·)|3L6 (0,1) ,

  A1 (t)W(t)Z |z(t, ·)|2L6 (0,1) 1 + |z(t, ·)|L∞ (0,1) ,

(3.61)

 A1 (t)W(t)Z |z(t, ·)|2H 1 (0,1) , 0

 A1 (t)W(t)Z V1 (t), and



   Re A1 (t)W(t), uτ (t)  wτ (t)

Z

    A1 (t)W(t)Z . 

(3.62)

Using (3.61)-(3.62), we have from (3.60) that  " #    Re A1 (t)W(t), R(, t, ·)   A1 (t)W(t)Z V1 (t) + A1 (t)W(t)Z . Z

We note from (3.13) that μ± j (τ ) + γ Hence, cj± (τ ) and we get

1 γ +μ± j (τ )

X1 (t)22

=

n $

 1 = λj (τ ) + γ ∓ 2

(λj

− 4δ ,

δ > 0.

|Wj+ (t)|2 + |Wj− (t)|2 ,

n $

⎛

⎞

|cj+ (t)|2 ⎝|zj (t)|2 +

j =1

+2

n $

⎛

n $ j =1

1 |γ

2 + μ+ j (t)|

|cj− (t)|2 ⎝|zj (t)|2 +

j =1

≤C

 (τ ) + γ )2

are uniformly bounded for 1 ≤ j ≤ n. Thus, using (3.47) and (3.42)

j =1

≤2

(3.63)

|vj (t)|2 ⎠

1 |γ

2 + μ− j (t)|

⎞ |vj (t)|2 ⎠ ,



 |zj (t)|2 + |vj (t)|2 ≤ C |z(t)|2L2 (0,1) + |v(t)|2L2 (0,1) ,

 ≤ C |z(t)|2H 1 (0,1) + |v(t)|2L2 (0,1)  V1 (t).

0

(3.64)

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Using Lemma 3.12, Equation (3.64) and (3.39)-(3.40), we obtain that   T   c(R1T (t)P (t)X1 (t) + X1 (t)P (t)R1 (t))  cX1 (t)2 R 1 (, t, ·)L∞ (0,1) ,

  cX1 (t)2  + |v(t, ·)|2L2 (0,1) + |z(t, ·)|2L∞ (0,1) ,

 !  c V1 (t)  + V1 (t) .

(3.65)

Let us calculate DW(t)Z as follows

1 2 DW(t)Z = (1 + δ) |z(t, ·)|2L2 (0,1) + |v(t, ·)|2L2 (0,1) ,

1 ! 2  |z(t, ·)|2H 1 (0,1) + |v(t, ·)|2L2 (0,1)  V1 (t).

(3.66)

0

Using the fact that μ± j (t) are real for all j > n (see (3.15)), (3.66) and Theorem 3.19, we have for some M2 > 0, " # Re DW(t), A1 (t)W(t) Z " # ∗ = Re (A1 (t) − A1 (t))W(t), A1 (t)W(t) , Z # " # " ∗ = Re A1 (t)W(t), A1 (t)W(t) − A1 (t)W(t), A1 (t)W(t) , Z Z # " = Re W(t), A21 (t)W(t) − A1 (t)W(t)2Z , Z

=

∞ $

∞ 2

  $ 2  + 2  + 2 2  − Re (μ+ (t)) (t) ξ (t, ·) + Re μ− W Wl (t) ξl− (t, ·)2Z  j  j Z j l (t))

j =1

l=1

− A1 (t)W(t)2Z , =

n $

n 2

  $ 2  + − 2  + 2 2  − Re (μ+ (t)) (t) ξ (t, ·) + Re (μ (t)) W Wl (t) ξl− (t, ·)2Z   Z j j j l

j =1

l=1

+

∞ ∞  2  2  $ $   + 2 + 2 W − (t)2 ξ − (t, ·)2 μ+ μ− Z j (t) Wj (t) ξj (t, ·)Z + l (t) l l j =n+1

l=n+1

− A1 (t)W(t)2Z , ≥

n $

n 2

  $ 2  + − 2  + 2 2  − Re (μ+ (t)) (t) ξ (t, ·) + Re (μ (t)) W Wl (t) ξl− (t, ·)2Z   Z j j j l

j =1

− A1 (t)W(t)2Z ,

l=1

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⎛ ⎞ n  n 2 $ $  − 2   + W (t) ⎠ − A1 (t)W(t)2 , ≥ −M2 ⎝ Wj (t) + Z l j =1

≥ −M2 X(t)22

l=1

− A1 (t)W(t)2Z .

Hence, # 1 " M2 1 − Re DW(t), A1 (t)W(t) ≤ X(t)22 + A1 (t)W(t)2Z . Z 2 2 2 Using Cauchy-Schwarz inequality we obtain, for some M3 > 0  #  1 "  Re DW(t), BK1 (t)X1 (t)  ≤ 1 DW(t)Z BK1 (t)X1 (t)Z , 2 Z 2 !  V1 (t)X1 (t)2 ,

(3.67)

(3.68)

 X1 (t)22 + A1 (t)W(t)Z X1 (t)2 , 1 ≤ A1 (t)W(t)2Z + M3 X1 (t)22 4 and  #  1 "  Re DW(t), R(, t, ·)  ≤ 1 DW(t)Z R(, t, ·)Z , 2 Z 2

 ! !  V1 (t)R(, t, ·)L∞ (0,1)  V1 (t)  + V1 (t) . Using Corollary 3.17, we have that P  (t) is bounded on [0, 1]. Hence,   T    cX1 (t)P  (t)X1 (t)  c X1 (t)22  c V1 (t).

(3.69)

(3.70)

As both u and uτ are bounded, applying Poincaré’s inequality we obtain that    #

   "  ·) u (t, ·)z(t, ·) Re z(t, ·), 2(a + 1) − 6u(t, τ  L2 (0,1;C)  2   |z(t, ·)|2L2 (0,1)   |z(t, ·)|2H 1 (0,1)   V1 (t).

(3.71)

0

Using (3.59)-(3.71) in (3.58) we thus obtain

 dV1 1 1 (t)  − c − M1 − M2 − M3 X1 (t)22 − A1 (t)W(t)2Z dt 2 8 + A1 (t)W(t)Z V1 (t) + A1 (t)W(t)Z

 ! + 2c V1 (t)  + V1 (t) + 2c V1 (t). We choose c > 0 large enough such that c4 := c − M1 − 12 M2 − M3 > 0. Hence we have (3.52). This finishes the proof. 2

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Lemma 3.25. Let  > 0 be given. Let U0 , U1 ∈ Z. Let U be the solution of (1.5). Then, we have U (1/) − U1 2Z ≤ γ1 

(3.72)

where γ1 > 0 is independent of . Proof. Let us recall the differential inequality satisfied by V1 that 3 ! dV1 (t) + X1 (t)22 + A1 (t)W(t)2Z  2( V1 (t) + V12 (t)) + 2V1 (t) + A1 (t)WZ dt +V1 (t)A1 (t)WZ .

Now using Young’s inequality we have for θ > 0 ! 1 2 V1 (t) ≤ θ V1 (t) +  2 , θ 3 1 2V12 (t) ≤ θ V1 (t) + V1 (t)2 , θ 2V1 (t) ≤ V12 (t) +  2 , θ A1 (t)WZ ≤ A1 (t)W2Z + 2 θ V1 (t)A1 (t)WZ ≤ A1 (t)W2Z + 2

1 2  , θ 1 2 V (t). θ 1

Choosing θ > 0 small enough we have V1 (t) + V1 (t)  V12 (t) +  2 .

(3.73)

We can write the above inequality as V1 (t) ≤ CV12 (t) + C 2 . Integrating in [0, t], using V1 (0) = 0 we have V1 (t)

v2 0

dv ≤ Ct, + 2



1 −1 V1 (t) tan ≤ Ct,   V1 (t) ≤  tan(Ct). As t ∈ [0, 1 ] we have V1 (t) ≤  tan(C) for all t ∈ [0, 1 ]. In particular we obtain V1 (1/) ≤  tan(C).

(3.74)

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Now, using Lemma 3.22, we get |z(1/)|2H 1 (0,1) + |v(1/)|2L2 (0,1) ≤ C1 . 0

2

Using (3.2), we have the result.

Let us now prove our main theorem. Proof of Theorem 1.5. Using Lemma 3.25 and choosing  = δ/γ1 and Tδ = γ1 /δ we conclude the result. 2 Remark 3.26. Note that the time Tδ in the Theorem 1.5 also depends on U0 , U1 ∈ S. Aposteriori Tδ is chosen so that the first component u1 of U1 satisfies 1

− N  b 2 

2 b N 2 2 u1 L2 (0,b/2) ≤ (b − x) u0 (x) + w0 (x) dx + 2Tδ L(a, b, δ, γ , N) . 2 0

Otherwise, it would contradict the comment in Remark 2.2(i). 4. Further comments, remarks and open questions This section is devoted to discuss some additional facts concerning the controllability of the FHN model. Some of them lead to open problems that, in our opinion, are of considerable interest. 4.1. Approximate controllability of the FHN with the Neumann boundary data In order to study approximate controllability of the FHN with the Neumann boundary condition, let us consider the following system: ⎧ ut − uxx = −u(u − a)(u − 1) − w + f χO , in (0, T ) × (0, 1), ⎪ ⎪ ⎪ ⎨w = δu − γ w, in (0, T ) × (0, 1), t ⎪ u(0, x) = u0 (x) and w(0, x) = w0 (x), x ∈ (0, 1), ⎪ ⎪ ⎩ ux (t, 0) = ux (t, 1) = 0, t ∈ (0, T ).

(4.1)

Theorem 1.2 can be extended to the system (4.1) without any major changes. The steady states of (4.1) are given by the solutions of the system 

uxx − u(u − a)(u − 1) − γδ u = 0, ux (0) = ux (1) = 0,

and the corresponding w satisfies the relation w=

δ u. γ

in (0, b) ∪ (c, 1),

(4.2)

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Similar result to Lemma 3.1 holds with the parameters (α, β) belonging to the set       τ u0 (0), u0 (1) + (1 − τ ) u1 (0), u1 (1) : τ ∈ [0, 1] . Here, one needs to consider the one-parameter family of unbounded linear operators (AN 1 (τ ), N D(A1 (τ ))) in Z for 0 ≤ τ ≤ 1, defined by   T 2 D(AN (τ )) := W = (z, v) ∈ Z | z ∈ H (0, 1) , 1   2 d −I N 2 − p(τ, ·)I dx . A1 (τ ) := δI −γ I

(4.3)

Analogous results to the Lemma 3.22, for the operator AN 1 (τ ) hold. But only thing we notice here is that for W = (z, v)T ∈ D we have V (t, W) ∼ |z|2H 1 (0,1) + |v|2L2 (0,1) ,

(4.4)

that is |z|H 1 (0,1) is replaced with |z|H 1 (0,1) . Hence, Theorem 1.5 holds true as in the Neumann 0 boundary case as well. 4.2. Approximate controllability of the Rogers-McCulloch model Now we will discuss the approximate controllability of the Rogers-McCulloch model: ⎧ ⎪ ut − uxx = −u(u − a)(u − 1) − ρuw + f χO , in (0, T ) × (0, 1), ⎪ ⎪ ⎪ ⎪ ⎪ ⎨wt = δu − γ w, in (0, T ) × (0, 1), u(0, x) = u0 (x) and w(0, x) = w0 (x), x ∈ (0, 1), ⎪ ⎪ ⎪ u(t, 0) = u(t, 1) = 0, t ∈ (0, T ), ⎪ ⎪ ⎪ ⎩or u (t, 0) = u (t, 1) = 0, t ∈ (0, T ), x x

(4.5)

where δ, γ , ρ > 0, a ∈ (0, 1). √ Theorem 1.2 can be extended to the system (4.5) provided ρ < 1 γ . In this model if 1

− N  b 2 

2 b N 2 2 uT L2 (0,b/2) > (b − x) u0 (x) + w0 (x) dx + 2T LRM (a, b, δ, γ , N, ρ) , 2 0

where LRM (a, b, δ, γ , N, ρ) is given by LRM (a, b, δ, γ , N, ρ)  3γ N 2 (N − 1)2 N−3 = b 2 16(4γ − ρ ) (N − 3)

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1 + N +1



3γ 729γ 3 (a + 1)4 + 2 3 4(4γ − ρ ) 16(4γ − ρ 2 )



δ2 −a (4γ − ρ 2 )

2 

, b

N+1

,

then, it is not possible to reach arbitrarily close to the state UT in time T . For the steady state of the Rogers-McCulloch model we consider the system ⎧ ⎪ ⎨uxx − u(u − a)(u − 1) −

ρδ 2 γ u

= 0,

in (0, b) ∪ (c, 1),

u(0) = u(1) = 0, ⎪ ⎩ or ux (0) = ux (1) = 0

(4.6)

and the corresponding w satisfies the relation w=

δ u. γ

For the Rogers-McCulloch model the perturbed functions z and v (as in (3.2)-(3.3)) satisfy the following equations, with T = 1/, ⎧

 

2 2 a + 1 − 3u − z3 ⎪ = z + 2(a + 1)u − 3u − a − ρw z + z z ⎪ t xx ⎪ ⎪ ⎪ ⎪ ⎪ −ρzv − ρuv + χO g − uτ , in (0, T ) × (0, 1), ⎪ ⎪ ⎨ vt = δz − γ v − wτ , in (0, T ) × (0, 1), ⎪ z(0, x) = 0 = v(0, x), x ∈ (0, 1), ⎪ ⎪ ⎪ ⎪ ⎪ z(t, 0) = z(t, 1) = 0, t ∈ (0, T ), ⎪ ⎪ ⎪ ⎩or z (t, 0) = z (t, 1) = 0, t ∈ (0, T ). x x

(4.7)

In this case we need to consider the one-parameter family of unbounded linear operators

RD ARD 1 (τ ), D(A1 (τ )) in Z for 0 ≤ τ ≤ 1, defined by   T 1 2 D(ARD 1 (τ )) := W = (z, v) ∈ Z | z ∈ H0 (0, 1) ∩ H (0, 1) ,  2  d − p (τ, ·)I −ρuI RD R A1 (τ ) := dx 2 δI −γ I

(4.8)

 RN with the Dirichlet boundary conditions or ARN 1 (τ ), D(A1 (τ )) in Z for 0 ≤ τ ≤ 1, defined by   T 2 D(ARN (τ )) := W = (z, v) ∈ Z | z ∈ H (0, 1) , 1  2  d − p (τ, ·)I −ρuI RN R 2 A1 (τ ) := dx δI −γ I with the Neumann boundary conditions and with

(4.9)

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pR (τ, ·) := a − 2(a + 1)u(τ, ·) + 3u2 (τ, ·) + ρw(τ, ·). Therefore, we can write (4.7) as 

Wt (t) = ARD 1 (t)W(t) + B(g)(t) + R(, t, ·), W(0) = 0,

(4.10)

where R (, t, ·) = R

   z2 a + 1 − 3u − z3 − ρzv − uτ (t) −wτ (t)

.

(4.11)

To understand whether the result similar to Theorem 1.5 holds for the Rogers-McCulloch model, RN we need to analyse the spectral behaviour of the operator ARD 1 (τ ) (or A1 (τ )) and carryout the same analysis as we have done for the FHN. In case of FHN, to show A1 (τ ) is a Riesz operator we have used the spectral analysis of the system  ϕ  (x) − p(τ, x)ϕ(x) = λϕ(x), ϕ(0) = ϕ(1) = 0.

in (0, 1),

Following the same analysis to investigate similar results (as in FHN equation) for the RogersMcCulloch system we need to compute the eigenvalues of the operator ARD 1 (τ ). In order to do that we must consider the following weighted eigenvalue problem ϕ  (x) − pR (τ, x)ϕ(x) = λρuϕ(x),

in (0, 1)

ϕ(0) = ϕ(1) = 0, where u ∈ C 1 ([0, 1]; C 2 ([0, 1])). But according to our knowledge spectral behaviour, which is essential to carry our analysis for this system, is known only when u > 0. See [[46], Section 5 and 6; [4], Chapter 10]. RN Furthermore, we observe form (4.8) (or (4.9)) that the operator ARD 1 (τ ) (or A1 (τ )) is a bounded perturbation of a Riesz spectral operator by a Nilpotent type operator. According to our knowledge, there is no such result whether bounded perturbation of a Riesz spectral operator is again a Riesz spectral operator. Hence, we are unable to carryout the similar analysis to conclude the result similar to the Theorem 1.5 for the Rogers-McCulloch model. Therefore, we remark here that if we would have the result that bounded perturbation of a Riesz spectral operator was a Riesz spectral operator, then going by the similar analysis, we could have the result similar to Theorem 1.5 for the Rogers-McCulloch model. 4.3. Connected components and approximate controllability of the FHN It is not known to us yet how many connected components of steady states for the FHN exist in our case depending on δ and γ . In the case of the heat equation, some sufficient conditions on nonlinearity give connectedness. See the Proposition 3.1 of [15]. We would like to mention that this result is not applicable to our system (FHN). Moreover, it was proved in [15] that, if the

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steady states y0 and y1 belong to distinct connected components of the set S of steady-states, then it is impossible, either to move from y0 to y1 , or the reverse. Here, in the case of the FHN equation, the question is completely open. Phase plane analysis may help to conclude some information about connectedness of the steady states. For example in [40], using phase plane analysis the authors have shown that the steady state y 0 ≡ 0 and y 1 ≡ 1 belong to the distinct component of the set of steady states. In our case we have only one constant steady state namely, u0 ≡ 0. We are not able to apply directly the argument of [40]. Some modified strategy may be required and this could be a future direction of research for our model (FHN). 4.4. Local approximate controllability of the FHN Note that given T > 0, and U0 = (u0 , w0 )T , U1 = (u1 , w1 )T ∈ Z, if u1 satisfies ⎞ 12 ⎛

− N b 

2 b ⎝ (b − x)N u20 (x) + w02 (x) dx + 2T L⎠ , u1 L2 (0,b/2) > 2 0

then, we have shown in Section 2 that the system is not approximately controllable. But, if u1 satisfies ⎞ 12 ⎛

− N b 

2 b ⎝ (b − x)N u20 (x) + w02 (x) dx + 2T L⎠ , u1 L2 (0,b/2) ≤ 2 0

then, the result is not known to us. The known way to show local approximate controllability of non-linear model is to first prove the approximate controllability of the linear model with non zero forcing term. Then, using fixed point argument we expect to replace the forcing term with the desirable non-linearity. But in this process we need an estimate of the control term obtained in terms of the given data say U0 and UT . It seems difficult in our case to get such an estimate because our system is not null controllable and does not satisfy the observability inequality. Another way is due to Zuazua [48], where he showed approximate controllability results for the globally Lipschitz non-linearity. Unfortunately in our case the nonlinearity is not globally Lipschitz. So this method does not conclude any positive result in our case. Another possible way to deduce local approximate controllability could be the use of the inverse mapping argument. But to use such argument for the FHN we face at least the following two difficulties: i) It is not evident which is a proper functional framework to use the inverse mapping theorem. This theorem (for example in [28]) has been applied for several control problems in the literature. To mention a few we refer [19], [22], [32]. But in all cases, it is connected with exact controllability questions. ii) We do not know how to deduce the density property of the range of the non-linear mapping from the density property of the range of its derivative at zero.

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4.5. Controls with non-negativity constraints In order to get controls with non-negative constraints the strategies available in the literature are explained in [34], [38], [39]. In those papers the authors use the two crucial facts for the given system. i) Local exact controllability of the system [[38], Lemma 2.1; [39], Lemma 2.5]. ii) Estimate of the controls in terms of initial conditions [[38], Equation (6), Lemma 2.1; [39], Equation (19), Lemma 2.5]. But unfortunately the above information is not available for our system (FHN). Due to this fact we are unable to follow the strategies of [38], [39] to conclude the non-negativity of the controls. One needs to develop some other strategies. This could be a future direction for research. 4.6. Approximate controllability for some specific steady states The steady-state equation for our system (FHN) has similar structure as the steady states for the reaction-diffusion equation given in [40]. For example (population dynamics), ⎧ ⎪ ⎨yt − yxx = −y(y − a)(y − 1), in (0, T ) × (0, 1), (4.12) y(0, x) = y0 (x), x ∈ (0, 1), ⎪ ⎩ y(t, 0) = u(t), y(t, 1) = v(t), t ∈ (0, T ), with 0 < a < 1. The above equation has following three important steady states namely, y 0 ≡ 0,

y a ≡ a,

y 1 ≡ 1.

Steady state y 0 describes the extinction, y 1 describes the invasion of a population and y a describes an intermediate state below invasion. In practical application it is common to target extinction or invasion. An intermediate state is also desirable to reach. In our case, that is, for the FHN system the component u denotes the potential accross the axonal membrane and w represents the permeability of the principal ionic component of the trans-membrane current. We recall that for FHN, a steady state u satisfies  uxx − u(u − a)(u − 1) − γδ u = 0 in (0, b) ∪ (c, 1), (4.13) u(0) = u(1) = 0. We observe u0 ≡ 0 is the only constant steady state of (4.13). In the spirit of [40] we feel that it is one of the desirable steady states for FHN. To reach the steady states y 0 and y 1 in [40], the authors follow a strategy based on the Theorem 1 in [40] (or Theorem B, [35]) and the phase plane analysis. Also to reach the steady state y a , the authors use the Theorem 1 in [40] (or Theorem B, [35]), local exact controllability (estimates of the control in terms of initial conditions) of the given system in [40] and stair case method (mentioned in [38], [39]). In our system, non-availability of the Theorem 1 in [40] (or Theorem B [35]), local exact controllability and estimate of the control in terms of initial data prevent us to carry similar analysis.

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4.7. Approximate controllability of the FHN for arbitrary states Given U0 = (u0 , w0 )T , U1 = (u1 , w1 )T ∈ Z, and given any δ > 0, then, we have shown in Section 3 that there exists T = T (δ, U0 , U1 ) > 0 and f ∈ L2 (0, T ; L2 (0, 1)) such that the solution U of (1.5) satisfies U(T ) − U1 Z ≤ δ, provided U0 , U1 are two steady states lying in the same connected components of the set of steady states in C 2 -topology. But, we don’t know whether the same result holds if U0 , U1 are any two arbitrary source and target respectively, say, • one of them is steady state of the system and other one is non-steady state, • both of them are steady states but lying in different connected components of steady states, • none of them are steady states of the system. Appendix A Here we provide proof of Theorem 3.18, Theorem 3.21 and Lemma 3.1. A.1. Proof of Theorem 3.18 We will prove the Theorem 3.18. First we will show that the system (3.35) possesses a solution z ∈ H 1,2 (Q) by using Leray-Schauder’s principle. Let us write the system (3.35) as ⎧ ⎪ ⎨zt = zxx + F (z) + G(z) + H, in (0, T ) × (0, 1), (A.1) z(t, 0) = z(t, 1) = 0, t ∈ (0, T ), ⎪ ⎩ z(0, x) = 0, x ∈ (0, 1), where

 

F (z) = 2(a + 1)u(t) − 3u2 (t) − a + d1 (t) z + z2 a + 1 − 3u(t) − z3 ,

G(z) = δ d2 (t) − I

 t

e−γ (t−s) z(s, ·)ds,

(A.2) (A.3)

0

  H (t) = − uτ (t) −  d2 (t) − I

t

e−γ (t−s) wτ (s)ds.

(A.4)

0

To apply Leray-Schauder’s principle we need to consider the following auxiliary equation for λ ∈ [0, 1] ⎧

 ⎪ ⎪ = z + λ F (z) + G(z) + H , in (0, T ) × (0, 1), z t xx ⎨ (A.5) z(t, 0) = z(t, 1) = 0 t ∈ (0, T ), ⎪ ⎪ ⎩z(0, x) = 0, x ∈ (0, 1).

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We consider the mapping  : L6 (Q) × [0, 1] → L6 (Q), with z = (˜z, λ) if and only if z is the unique solution to ⎧

 ⎪ ⎪ ⎨zt = zxx + λ F (˜z) + G(˜z) + H , z(t, 0) = z(t, 1) = 0, t ∈ (0, T ), ⎪ ⎪ ⎩z(0, x) = 0, x ∈ (0, 1).

in

(0, T ) × (0, 1), (A.6)

Before stating the results we set few notations. Since τ ∈ [0, 1] → u(τ, ·), w(τ, ·) ∈ C 2 ([0, 1]), and τ ∈ [0, 1] → d1 (τ ), d2 (τ ) ∈ L(L2 (0, 1)), all are C 1 so there exists M > 0 such that             sup u(τ ) ∞ + uτ (τ ) ∞ + wτ (τ ) τ ∈[0,1]

L (0,1)

    sup d1 (τ )

τ ∈[0,1]

L (0,1)

L(L2 (0,1))

    + d2 (τ )

L∞ (0,1)

    + ux (τ )

 L∞ (0,1)

≤ M,

(A.7)

 L(L2 (0,1))

≤ M.

(A.8)

Using these bounds we prove the following lemmas. In the subsequent discussion we will use C as the generic constant which may depend on M, , δ, γ , a. Lemma A.1. The mapping  : L6 (Q) × [0, 1] → L6 (Q) defined above is well-defined, continuous and compact. Proof. We note that             |F (˜z)| ≤  2(a + 1) u(t) − 3 u2 (t) − a  |˜z| + d1 (t)˜z +  a + 1 − 3u(t)  |˜z|2 + |˜z|3 . Using the bounds (A.7)-(A.8)we obtain T 1 |F (˜z)|2 dxdt ≤ C 0 0

T 1

 |˜z|2 + |˜z|4 + |d1 (t)˜z|2 + |˜z|6 dxdt,

0 0

≤C

T 1

 |˜z|2 + |˜z|4 + |d1 (t)|2L(L2 (0,1)) |˜z|2 + |˜z|6 dxdt,

0 0

≤C

T 1 0 0

 |˜z|2 + |˜z|4 + |˜z|2 + |˜z|6 dxdt.

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Now applying Hölder’s inequality we get

2  1 |F (˜z)|2L2 (Q) ≤ C T 3 |˜z|2L6 (Q) + T 3 |˜z|4L6 (Q) + +|˜z|6L6 (Q) .

(A.9)

An application of Hölder’s inequality shows that ⎛ |G(˜z)| ≤ δ ⎝

⎞ 16 ⎛ t ⎞ 56  6γ (t−s) |(d2 (s) − I ) z˜ (s)|6 ds ⎠ ⎝ e− 5 ds ⎠ .

t 0

0

This implies ⎛

1 |G(˜z)| dx ≤ δ 6

6⎝

0

⎛ ≤δ

6⎝

1  t

⎞

5(1 − e− |(d2 (s) − I )˜z(s)| dsdx ⎠ 6γ 6

0 0

1 T

⎞ |(d2 (s) − I )˜z(s, x)| dsdx ⎠ 6



5 6γ

6γ t 5

)

5 ,

5 ,

0 0

 ≤ C |d2 (·)˜z|6L6 (Q) + |˜z|6L6 (Q) . Now using the fact that d2 (τ )˜z lies in the finite dimensional subspace of L2 (0, 1) and (A.7)-(A.8), we get T 1 |d2 (·)˜z|6L6 (Q)

=

T |d2 ()˜z(t, x)| dxdt =

|d2 (t)˜z(t)|6L6 (0,1) dt,

6

0 0

0

T ≤C

|d2 (t)|6L(L2 (0,1)) |˜z(t)|6L2 (0,1) dt ≤ C |˜z|6L6 (Q) . 0

Hence, we have |G(˜z(t))|6L6 (0,1) ≤ C|˜z|6L6 (Q) .

(A.10)

Therefore, for z˜ ∈ L6 (Q), F (˜z) ∈ L2 (Q) and G(˜z) ∈ L∞ (0, T ; L6 (0, 1)). Also we note that the mapping z˜ → F (˜z) and z˜ → G(˜z) are continuous. Hence, by existence and uniqueness theory of heat equation with the Dirichlet boundary condition the map (˜z, λ) → z = (˜z, λ) is well defined and continuous. Compactness of the mapping  will follow from parabolic regularity theory, that is, if (˜z, λ) ∈ L6 (Q) × [0, 1] and z = (˜z, λ) then z ∈ L2 (0, T ; H 2 (0, 1)) ∩ C([0, T ]; H01 (0, 1)),

zt ∈ L2 (Q),

hence, z ∈ H 1,2 (Q). Now by the estimate (which will be proved in Lemma A.2) one notes that if (˜z, λ) lies in a bounded set of L6 (Q) × [0, 1], then the associated z lies in a bounded set of

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H 1,2 (Q). As H 1,2 (Q) continuously embedded in H 1 (Q) and by Rellich-Kondrachov theorem H 1 (Q) is compactly embedded in L6 (Q), hence H 1,2 (Q) is compactly embedded in L6 (Q). Therefore, we conclude that the mapping  : L6 (Q) × [0, 1] → L6 (Q) is compact. This completes the proof. 2 Next we prove the following lemma. Lemma A.2. Let us consider the set   = z ∈ L6 (Q) : z = (z, λ)

 for some λ ∈ [0, 1] .

Then the set  is uniformly bounded in L6 (Q). Proof. Let λ ∈ [0, 1], z ∈ L6 (Q) and z = (z, λ). We will show that there exists C > 0 (independent of λ, z) such that |z|L2 (0,T ;H 2 (0,1)) and |zt |L2 (Q) are bonded by C. This gives us the desired result. Let us write the equation (A.5) as

 ⎧ ⎪ = z + λF (z) + d (t) − I λv − λuτ (t), z ⎪ t xx 2 ⎪ ⎪ ⎨ vt = δz − γ v − wτ (t), in (0, T ) × (0, 1), ⎪ z(t, 0) = z(t, 1) = 0, t ∈ (0, T ), ⎪ ⎪ ⎪ ⎩ z(0, x) = 0, x ∈ (0, 1). Multiplying first equation by z and second equation by have in (0, T )

λv δ

in (0, T ) × (0, 1), (A.11)

and integrating and then adding, we

1 d λ d λγ |z(t)|2L2 (0,1) + |v(t)|2L2 (0,1) + |zx (t)|2L2 (0,1) + |v(t)|2L2 (0,1) 2 dt 2δ dt δ 1 1 1 1 λ = λ F (z)zdx − λ uτ (t)zdx + λ d2 (t)vzdx − wτ (t)vdx. δ 0

0

0

(A.12)

0

Using Hölder’s inequality and (A.7)-(A.8), we estimate the first term of the right hand side of (A.12) as follows 1 F (z)zdx ≤ C 0



|z|2L2 (0,1)

+ |z|3L3 (0,1)



1 + |d1 (t)z|L2 (0,1) |z|L2 (0,1) −



≤ C |z|2L2 (0,1) + |z|3L3 (0,1) − |z|4L4 (0,1) ,

 ≤ C |z|2L4 (0,1) + |z|3L4 (0,1) − |z|4L4 (0,1) , 1 1 ≤ |z|4L4 (0,1) + C − |z|4L4 (0,1) = C − |z|4L4 (0,1) . 2 2

z4 dx, 0

(A.13)

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Similarly, we estimate the other terms of the right hand side of (A.12) as 1 −

4 1 uτ (t, x)zdx ≤ C|z(t)|L1 (0,1) ≤ C|z(t)|L4 (0,1) ≤ |z(t)|4L4 (0,1) + 2C 3 , 8

(A.14)

0

1 d2 (t)vz dx ≤|z(t)|L2 (0,1) |d2 (t)v(t)|L2 (0,1) 0

≤C|z(t)|L2 (0,1) |v(t)|L2 (0,1) , ≤

C2δ γ |v(t)|2L2 (0,1) + |z(t)|2L2 (0,1) , 4δ γ

≤

γ 1 2C 4 δ 2 , |v(t)|2L2 (0,1) + |z(t)|4L4 (0,1) + 4δ 8 γ2

(A.15)

and  − δ

1 wτ (t)vdx ≤

C C γ C2 |v(t)|L1 (0,1) ≤ |v(t)|L2 (0,1) ≤ |v(t)|2L2 (0,1) + . δ δ 4δ δγ

(A.16)

0

Combining (A.13)-(A.16), we have from (A.12) 1 d λ d |z(t)|2L2 (0,1) + |v(t)|2L2 (0,1) 2 dt 2δ dt λγ λ + |zx (t)|2L2 (0,1) + |v(t)|2L2 (0,1) + |z(t)|4L4 (0,1) ≤ λC, 2δ 4

(A.17)

for some positive constant C which does not depend on λ. Now, integrating over (0, t) we have

1 λ |z(t)|2L2 (0,1) + |v(t)|2L2 (0,1) + 2 2δ

t |zx (s)|2L2 (0,1) ds 0

λγ + δ

t |v(s)|2L2 (0,1) ds

λ + 4

t |z(s)|4L4 (0,1) ds ≤ λCT .

(A.18)

|z|L∞ (0,T ;L2 (0,1)) , |z|L2 (0,T ;H 1 (0,1)) , |z|L4 (Q) , |v|L∞ (0,T ;L2 (0,1)) ≤ CT .

(A.19)

0

0

As λ ∈ [0, 1] and δ, γ > 0, we have

0

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Again multiplying the first equation of (A.11) by zt and integrating over (0, 1) we have in (0, T )

|zt (t)|2L2 (0,1)

1 d + |zx (t)|2L2 (0,1) = λ 2 dt

1

1 F (z)zt dx − λ

0

uτ (t)zt dx 0

(A.20)

1 +λ

(d2 (t) − I )vzt dx. 0

Again as before we estimate each term of the right hand side of (A.20). Using Hölder’s inequality and (A.7)-(A.8), we estimate the term involving F as follows 1 F (z)zt dx 0

=

1

1  2(a + 1)u(t) − 3u2 (t) − a zzt dx + d1 (t)zzt dx

0

0

1 1  2 + a + 1 − 3u(t) z zt dx − z3 zt dx, 0

0

1 ≤C

|z||zt | dx + |d1 (t)|L(L2 (0,1)) |z|L2 (0,1) |zt |L2 (0,1) 0

1 +C

z2 |zt | dx −

(A.21)

1 d 4 , |z| 4 4 dt L (0,1)

0

 1 d ≤ C |z|L2 (0,1) |zt |L2 (0,1) + |z|2L4 (0,1) |zt |L2 (0,1) − , |z|4 4 4 dt L (0,1)

 1 1 1 d 4 ≤ |zt |2L2 (0,1) + C |z|2L2 (0,1) + |z|4L4 (0,1) + |zt |L2 (0,1) − , |z| 4 8 8 4 dt L (0,1)

 1 d 1 ≤ |zt |2L2 (0,1) + C |z|2L4 (0,1) + |z|4L4 (0,1) − , |z|4 4 4 4 dt L (0,1) 1 1 d 4 ≤ |zt |2L2 (0,1) + C|z|4L4 (0,1) + C − . |z| 4 4 4 dt L (0,1) Similarly, we estimate the last two terms of (A.20) as 1 − 0

1 uτ (t)zt dx ≤ C|zt (t)|L1 (0,1) ≤ C|zt (t)|L2 (0,1) ≤ |zt (t)|2L2 (0,1) + C 2 , 4

(A.22)

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and 1

 d2 (t) − I v zt dx ≤ C |zt (t)|L2 (0,1) |v(t)|L2 (0,1) , (A.23)

0

1 ≤ |zt (t)|2L2 (0,1) + C 2 |v(t)|2L2 (0,1) . 4 Combining (A.21)-(A.23), and using (A.19) and as λ ∈ [0, 1], we have from (A.20) |zt (t)|2L2 (0,1) +

1 d 3λ |zx (t)|2L2 (0,1) ≤ |zt |2L2 (0,1) + Cλ|z|4L4 (0,1) + Cλ 2 dt 4 λ d 4 − + Cλ|v(t)|2L2 (0,1) . |z| 4 4 dt L (0,1)

This implies 1 1 d λ d 4 ≤ C|z|4L4 (0,1) + C, |zt (t)|2L2 (0,1) + |zx (t)|2L2 (0,1) + |z| 4 4 2 dt 4 dt L (0,1)

(A.24)

where the positive constant C does not depend on λ. Integrating over (0, t) we have 1 4

t

1 λ |zt (s)|2L2 (0,1) ds + |zx (t)|2L2 (0,1) + |z(t)|4L4 (0,1) , 2 4

0

1 ≤ |zx (0)|2L2 (0,1) + C 2

T |z(s)|4L4 (0,1) ds + C,

(A.25)

0

1 ≤ |zx (0)|2L2 (0,1) + C. 2 Hence, we have |zt |L2 (Q) , |z|L∞ (0,T ;H 1 (0,1)) , λ|z|L∞ (0,T ;L4 (0,1)) ≤ C. 0

(A.26)

We can write first equation of (A.11) as zxx = zt − λF (z) − (d2 (τ ) − I )λv − λH. In view of (A.19) we note that z is uniformly bounded in L2 (0, T ; L6 (0, 1)) and hence F (z) is uniformly bounded in L2 (Q). Now again using (A.19) and (A.26) we have zxx is uniformly bounded in L2 (Q). Therefore, |z|L2 (0,T ;H 2 (0,1)) ≤ CT . This completes the proof.

2

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Lemma A.3. Let (zi , vi ) ∈ H 1,2 (Q) × C([0, 1]; H 2 (0, 1)) for i = 1, 2 be two solutions of (3.35). Then, z1 = z2 and v1 = v2 . Proof. Let us take z := z1 − z2 and v := v1 − v2 . Then z, v satisfy ⎧ zt = zxx + (F (z1 ) − F (z2 )) + (d2 (t) − I ), v ⎪ ⎪ ⎪ ⎨v = δz − γ , v in (0, T ) × (0, 1), t ⎪ z(t, 0) = z(t, 1) = 0, t ∈ (0, T ), ⎪ ⎪ ⎩ z(0, x) = 0, x ∈ (0, 1).

in

(0, T ) × (0, 1), (A.27)

Multiplying first equation of (A.27) by z and the second equation by adding we have in (0, T )

v δ

and integrating and then

1 d 1 d γ |z(t)|2L2 (0,1) + |v(t)|2L2 (0,1) + |zx (t)|2L2 (0,1) + |v(t)|2L2 (0,1) 2 dt 2δ dt δ 1 1    = F (z1 ) − F (z2 ) zdx + d2 (t)vzdx. 0

(A.28)

0

We note that 1

 F (z1 − F (z2 ) z dx

0

=

1

2(a + 1)u(t) − 3u2 (t) − a

  z1 − z2 zdx

0

1 +

1

   d1 (t) z1 − z2 zdx + a + 1 − 3u(t) z12 − z22 zdx

0

−

0

1

 z13 − z23 zdx,

(A.29)

0

1 ≤C

1  z dx + |d1 (t)z|L2 (0,1) |z|L2 (0,1) + C |z1 | + |z2 | z2 dx 2

0

0

1 +2

 z12 + z22 z2 dx,

0

≤ C|z|2L2 (0,1)

1 1   2 +C |z1 | + |z2 | z dx + 2 z12 + z22 z2 dx, 0

0

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and 1 d2 (t)vzdx ≤ |z(t)|L2 (0,1) |d2 (t)v(t)|L2 (0,1) (A.30)

0

≤ C|z(t)|L2 (0,1) |v(t)|L2 (0,1) ≤

γ |v(t)|2L2 (0,1) + C|z(t)|2L2 (0,1) . 2δ

Using (A.29)-(A.30) we have from (A.28) that 1 d 1 d γ |z(t)|2L2 (0,1) + |v(t)|2L2 (0,1) + |zx (t)|2L2 (0,1) + |v(t)|2L2 (0,1) ≤ σ (t)|z(t)|2L2 (0,1) , 2 dt 2δ dt 2δ where



 σ (t) := C + C |z1 (t)|L∞ (0,1) + |z2 (t)|L∞ (0,1) + 2 |z1 (t)|2L∞ (0,1) + |z2 (t)|2L∞ (0,1) . Using Gronwall’s inequality, we observe that as z(0, x) = 0, z(t, x) = 0 for all (t, x) ∈ Q. Hence, z1 = z2 and v1 = v2 . Thus, (3.35) has exactly one solution in H 1,2 (Q). 2 Therefore, proof of the Theorem 3.18 follows from Lemma A.1-A.3. A.2. Proof of Theorem 3.21 We will prove the Theorem 3.21 using the following Lemma. Lemma A.4. Let h ∈ C([0, ∞); [0, ∞)) be an unbounded differentiable function. Then, for each M > h(0) there exists tM > 0 such that h(tM ) > M,

h(tM ) = max h(t) t∈[0,tM ]

and



h (tM ) > 0.

(A.31)

Proof. Consider the set   AM = t ∈ [0, ∞)| h(t) > M . Note that as h is unbounded and nonnegative, so AM = φ. We will show that there exists tM ∈   AM such that h (tM ) > 0. Suppose not. Then, for all t ∈ AM , h (t) ≤ 0. As AM = φ there exists t > 0 such that h(t) > M. Since AM is open, so there exists maximal interval (a, t] ⊂ AM . Therefore, by the continuity of h one notes that h(a) = M and as M > h(0) we have a > 0. As   for all t ∈ AM , h (t) ≤ 0, so in particular h (t) ≤ 0 for all t ∈ (a, t] and hence, for all t ∈ (a, t] we have M < h(t) ≤ h(t) ≤ h(a) = M. 

This is a contradiction. Therefore, there exists tM ∈ AM such that h (tM ) > 0.

2

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Proof of Theorem 3.21. We note that t |H (t)| ≤ |uτ (t)| + 

e−γ (t−s) |(d2 (τ ) − I )wτ (s)|ds,

0

t ≤ |uτ |L∞ (0,1) + |(d2 (·) − I )wτ |L∞ (0,1)

e−γ (t−s) ds,

(A.32)

0

≤ |uτ |L∞ (0,1) + |(d2 (·) − I )wτ |L∞ (0,1)

1 − e−γ t ≤ M. γ

Multiplying (A.1) with z and integrating over the interval (0, 1) we have

1 d |z(t)|2L2 (0,1) + |zx (t)|2L2 (0,1) = 2 dt

1

1 F (z)zdx +

0

1 G(z)zdx +

0

H zdx.

(A.33)

0

Let us compute 1

1 F (z)zdx =

0

(2(a + 1)u(t) − 3u2 (t) − a)z2 dx 0

1 +

1 d1 (t)zzdx +

0

1 (a + 1 − 3u(t))z dx − 3

0

0

1 ≤C

z4 dx,

1 z dx + |d1 (t)z|L2 (0,1) |z|L2 (0,1) + C

0

1 |z| dx −

2

3

0

≤ C|z|2L2 (0,1) +

1 2

1

z4 dx,

(A.34)

0

1 |z|4 dx −

0

z4 dx, 0

1 = C|z|2L2 (0,1) − |z|4L4 (0,1) , 2 and 1 H (t, x)zdx ≤ C|z(t)|L1 (0,1) ≤ C|z(t)|L2 (0,1) . 0

Using (A.34)-(A.35) from (A.33) we have

(A.35)

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1 d 1 |z(t)|2L2 (0,1) + |zx (t)|2L2 (0,1) ≤ C|z(t)|2L2 (0,1) − |z(t)|4L2 (0,1) + C|z(t)|L2 (0,1) 2 dt 2 t + C|z(t)|L2 (0,1) e−γ (t−s) |z(s)|L2 (0,1) ds.

(A.36)

0

First, we will show that there exists C > 0 such that |z(t)|L2 (0,1) ≤ C, for all t ≥ 0. Suppose (3.39) is not true. Since z ∈ C 1 (0, ∞; L2 (0, 1)), so t → |z(t)|L2 (0,1) is of class C 1 . So there exists (by Lemma A.4) sequence tk ↑ ∞ such that |z(tk )|2L2 (0,1) ↑ ∞ as k → ∞ and |z(tk )|L2 (0,1) = max |z(s)|L2 (0,1) , s∈[0,tk ]

d |z(tk )|2L2 (0,1) > 0, dt

for all k large. Now from (A.36) we have at t = tk and for large k 1 d |z(tk )|2L2 (0,1) + |zx (tk )|2L2 (0,1) 2 dt 1 ≤ C|z(tk )|2L2 (0,1) − |z(tk )|4L2 (0,1) 2 tk + C|z(tk )|L2 (0,1) e−γ (tk −s) |z(s)|L2 (0,1) ds + C|z(tk )|L2 (0,1) , 0

1 ≤ C|z(tk )|2L2 (0,1) − |z(tk )|4L2 (0,1) 2 tk + C|z(tk )|L2 (0,1) e−γ (tk −s) |z(s)|L2 (0,1) ds + C|z(tk )|L2 (0,1) , 0

1 ≤ C|z(tk )|2L2 (0,1) − |z(tk )|4L2 (0,1) 2 tk + C|z(tk )|2L2 (0,1) e−γ (tk −s) ds + C|z(tk )|L2 (0,1) , 0

1 ≤ C|z(tk )|2L2 (0,1) − |z(tk )|4L2 (0,1) 2 1 − e−γ tk + C|z(tk )|2L2 (0,1) + C|z(tk )|L2 (0,1) , γ 1 ≤ C|z(tk )|2L2 (0,1) − |z(tk )|4L2 (0,1) + C|z(tk )|L2 (0,1) < 0. 2

(A.37)

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This gives a contradiction. Hence, there exists C > 0 such that |z(t)|L2 (0,1) ≤ C for t ≥ 0. Now using the expression of v(x, t) in (3.34) and Hölder’s inequality, we have t |v(t, x)| ≤δ

e

−γ (t−s)

t |z(s, x)|ds + 

0

e−γ (t−s) |wτ (s, x))|ds,

0

⎞ 12 ⎛ t ⎞ 12 t  t −γ (t−s) 2 ⎠ ⎝ −γ (t−s) ⎠ ⎝ e |z(s, x)| ds e ds +  C e−γ (t−s) ds, ≤δ ⎛

0

0

0

⎞ 12 ⎛ t ⎞ 12 t   ≤δ ⎝ e−γ (t−s) |z(s, x)|2 ds ⎠ ⎝ e−γ (t−s) ds ⎠ + C. γ ⎛

0

0

This implies, 1

1 |v(t, x)|2 dx ≤2δ 2

0

⎞⎛ t ⎞ ⎛ t   ⎝ e−γ (t−s) |z(s, x)|2 ds ⎠ ⎝ e−γ (t−s) ds ⎠ dx + C.

0

0

0

Hence, we obtain

|v(t)|2L2 (0,1)

2δ 2 ≤ γ

t e 0

≤

2δ 2

−γ (t−s)

1 |z(s, x)|2 dxds + C, 0

t

γ

(A.38)

e−γ (t−s) |z(s)|2L2 (0,1) ds + C ≤ C.

0

This proves (3.39). In order to show (3.40), let us multiply (A.1) with zxx and integrating over the interval (0, 1) we have 1 d |zx (t)|2L2 (0,1) + |zxx (t)|2L2 (0,1) = − 2 dt

1

1 F (z)zxx dx −

0

1 G(z)zxx dx −

0

Using (3.39) we get 1 −

F (z)zxx dx 0

=−

1 0

1  2(a + 1)u(t) − 3u (t) − a zzxx dx + d1 (t)zzxx dx 2

0

H zxx dx. 0

(A.39)

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−

1

1  2 a + 1 − 3u(t) z zxx dx + z3 zxx dx,

0

1 ≤ C|z|L2 (0,1) |zxx |L2 (0,1) −

0

 − 3ux (t) z2 zx dx + 2

0

1 (a + 1 − 3u(t))zzx2 dx 0

1 −3

z2 zx2 dx, 0

1

 ≤ C|z|L2 (0,1) |zxx |L2 (0,1) + |zzx |L2 (0,1) + C|zzx |L2 (0,1) |zx |L2 (0,1) − 3 z2 zx2 dx, 0 1 2

1

1 2

≤ C|zxx |L2 (0,1) + C|zzx |L2 (0,1) + C|zzx |L2 (0,1) |zxx |L2 (0,1) |z|L2 (0,1) − 3

z2 zx2 dx, 0

1 1 ≤ C|zxx |L2 (0,1) + |zzx |2L2 (0,1) + C + C|zzx |L2 (0,1) |zxx |L2 2 (0,1) − 3 2

1 z2 zx2 dx, 0

≤ C|zxx |L2 (0,1) + |zzx |2L2 (0,1) + C − 3|zzx |2L2 (0,1) , ≤ C|zxx |L2 (0,1) + C − 2|zzx |2L2 (0,1) .

(A.40)

Using (3.39) we have 1 −

G(z)zxx dx 0

1  t = −δ

e−γ (t−s) (d2 (t) − I )z(s, x)zxx (t, x)dsdx,

0 0

t ≤δ

e−γ (t−s) |d2 (t) − I |L(L2 (0,1)) |z(s)|L2 (0,1) |zxx (t)|L2 (0,1) ds,

0

t ≤C

e−γ (t−s) |z(s)|L2 (0,1) |zxx (t)|L2 (0,1) ds,

0

t = C|zxx (t)|L2 (0,1) 0

e−γ (t−s) |z(s)|L2 (0,1) ds ≤ C|zxx (t)|L2 (0,1) ,

(A.41)

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and 1 −

H (t, x)zxx dx ≤ C|zxx (t)|L1 (0,1) ≤ C|zxx (t)|L2 (0,1) .

(A.42)

0

Combining (A.40)-(A.42) we have from (A.39) 1 d |zx (t)|2L2 (0,1) + |zxx |2L2 (0,1) ≤ C|zxx |L2 (0,1) + C − 2|zzx |2L2 (0,1) . 2 dt Hence, we obtain 1 d |zx (t)|2L2 (0,1) + |zxx (t)|2L2 (0,1) ≤ C|zxx (t)|L2 (0,1) + C. 2 dt

(A.43)

We will show that there exists C > 0 such that |zx (t)|L2 (0,1) ≤ C for all t ≥ 0. Suppose it is   not true. Since zx ∈ C 1 0, ∞; L2 (0, 1) , so t → |zx (t)|L2 (0,1) is of class C 1 . So there exists (by Lemma A.4) a sequence tk ↑ ∞ such that |zx (tk )|2L2 (0,1) ↑ ∞ as k → ∞ and |zx (tk )|L2 (0,1) = max |zx (s)|L2 (0,1) , s∈[0,tk ]

d |zx (tk )|2L2 (0,1) > 0, dt

for all k large. But using (3.39), we note that 1 |zx (tk )|2L2 (0,1)

=

1 zx zx dx = −

0

zxx zdx ≤ |zxx (tk )|L2 (0,1) |z(tk )|L2 (0,1) ≤ C|zxx (tk )|L2 (0,1) . 0

Therefore, |zxx (tk )|L2 (0,1) ↑ ∞ as k → ∞. Hence from (A.43) we have d |zx (tk )|2L2 (0,1) < 0, dt for large k. This is a contradiction. Hence we have |zx (t)|L2 (0,1) ≤ C for some C > 0 and for all t ≥ 0. Sobolev embedding yields us (3.40). On the other hand we have from (3.34), t |v(t)|L∞ (0,1) ≤

  e−γ (t−s) δ|z(t)|L∞ (0,1) + |wτ |L∞ (0,1) ds,

0

  ≤ δ|z(t)|L∞ (0,1) + |wτ |L∞ (0,1)

t

e−γ (t−s) ds,

0

 1 1 ˜ ≤ (δL + C) 1 − e−γ t ≤ (δL + C) := L. γ γ This completes the proof.

2

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A.3. Proof of Lemma 3.1 Let us consider the set       P := τ u0 (0), u0 (1) + (1 − τ ) u1 (0), u1 (1) : τ ∈ [0, 1] ⊂ R2 . Let us first assume that for any pair of real number (α, β) ∈ P , the maximal solution of the system (3.1) with u (0) = α, u (1) = β, which is denoted by U(α,β) , is defined in [0, b] ∪ [c, 1]. Let us consider the map H : P → S defined by H (α, β) = U(α,β) . We note that the solution U(α,β) , which is defined in [0, b] ∪ [c, 1], depends continuously on (α, β) in C 2 topology i.e. for any sequence (αn , βn ) → (α, β) then U(αn ,βn ) → U(α,β) in C 2 topology as n → ∞. Therefore, the map H is continuous. As the set P is path-connected, so is H (P ) ⊂ S. Hence, U0 , U1 ∈ S belong to the same path component of S. On the other hand, let U0 , U1 ∈ S belong to the same path component of S. Let F : [0, 1] → S be a path in S such that F (0) = U0 and F (1) = U1 . Therefore, we note that  

δ F (0)(x) = u0 (x), w0 (x) = u0 (x), u0 (x) , γ

and 



δ F (1)(x) = U1 = u1 (x), w1 (x) = u1 (x), u1 (x) . γ This yields 

d δ F (0)(x) = u0 (x), u0 (x) dx γ

and



d δ F (1)(x) = u1 (x), u1 (x) . dx γ

(A.44)

Hence, at x = 0,

  d δ  = u0 (0), u0 (0) F (0) x=0 dx γ

and

  d δ  = u1 (0), u1 (0) . F (1) x=0 dx γ

   d Let (α, β) ∈ P . Without loss of generality, let us assume u0 (0) < α < u1 (0). As dx F (t) : x=0      d d F (0) and dx F (1) , there exists t0 ∈ [0, 1] such that t ∈ [0, 1] is a path joining dx x=0

x=0

  d δ  = α, α . F (t0 ) x=0 dx γ So, the solution F (t0 )(·) exists in [0, b] ∪ [c, 1] in particular in [0, b]. Similarly, from (A.44) (evaluating at x = 1), we get

  d δ  = u0 (1), u0 (1) F (0) x=1 dx γ

and

  d δ  = u1 (1), u1 (1) . F (1) x=1 dx γ

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   d Again as before, without loss of generality, let us assume u0 (1) < β < u1 (1). As dx F (t) : x=1      d d F (0) and dx F (1) , there exists t1 ∈ [0, 1] such that t ∈ [0, 1] is a path joining dx x=1

x=1



γ  d  = β, β . F (t1 ) x=1 dx δ Therefore, the solution F (t1 )(·) exists in [0, b] ∪ [c, 1] in particular in [c, 1]. Combining all these from above, let us define W = (u, ˜ w) ˜ : [0, 1] → R2 by  W (x) =

F (t0 )(x) F (t1 )(x)

x ∈ [0, b], x ∈ [c, 1],

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