Electronic Journal of Qualitative Theory of Differential Equations 2007, No. 19, 1-21; http://www.math.u-szeged.hu/ejqtde/

Null Controllability of Some Impulsive Evolution Equation in a Hilbert Space

R. BOUKHAMLA^{(1)} & S. MAZOUZI^{(2)}

(1) Centre Universitaire de Souk-Ahras, Souk-Ahras 41000, Algeria.

(2) Département de Mathématiques, Université d’Annaba, BP 12, ANNABA 23000, Algeria.

Abstract

We shall establish a necessary and sufficient condition under which we have the null controllability of some first order impulsive evolution equation in a Hilbert space.

MSC(2000) : 34A37, 93B05, 93C15.

Keywords: Null-controllability, impulsive conditions, mild solutions, evolution equation.

1 Introduction

The problem of exact controllability of linear systems represented by infinite conservative systems has been extensively studied by several authors A. Haraux [8], R.Triggiani [16], Z.H. Guan, T.H. Qian, and X.Yu [7], see also the references [1, 2, 6, 10,15]. In the sequel, we shall be concerned with the problem of null controllability of some first order evolution equation subject to impulsive conditions and so we shall derive a necessary and sufficient condition under which null controllability occurs. Actually, we shall establish an equivalence between the null-controllability and some ”observability” inequality in somehow more general framework than that proposed by A Haraux [8]. Regarding the literature on the impulsive differential equations we refer the reader to the works of D.D. Bainov and P.S. Simeonov [3, 4] and the references [5, 9,11, 12, 13]. We are going to study the following problem

y′ (t) + Ay (t) = Bu (t) , t ∈ (0, T) \ {t_{k}}_{k∈σ_{1}^{m}} ,  (1)
y (0) = y^{0},
Δy (t_{k}) = I_{k}y (t_{k}) + D_{k}v_{k}, k ∈ σ_{1}^{m},  (1_{k})

where the final time T is a positive number, y^{0} is an initial condition in a Hilbert space H endowed with an inner product ⟨., .⟩_{H}, y (t) : [0, T] → H is a vector function, σ_{1}^{m} is a subset of ℕ given by σ_{1}^{m} = {1, 2, ..., m}, and finally, {t_{k}}_{k∈σ_{1}^{m}} is an increasing sequence of numbers in the open interval (0, T) , and Δy (t_{k}) denotes the jump of y (t) at t = t_{k}, i.e.,

Δy (t_{k}) = y (t_{k}^{+}) − y (t_{k}^{−})

where y (t_{k}^{+}) and y (t_{k}^{−}) represent the right and left limits of y (t) at t = t_{k} respectively. On the other hand, the operators A, B, I_{k}, D_{k} : H → H are given linear bounded operators. Moreover, we set the following assumptions:

(H1) A^{∗} = −A,

(H2) I_{k}^{∗} = −I_{k}, for every k ∈ σ_{1}^{m}, and for each k ∈ σ_{1}^{m}, the operator I_{k} = I_{k} + I is invertible,

(H3) B^{∗} = B ≥ 0 and there is d_{0} > 0 such that

(Bu, u)_{H} ≤ d_{0} ∥u∥_{H}^{2} , for all u ∈ H,

(H4) D_{k}^{∗} = D_{k} ≥ 0, for every k ∈ σ_{1}^{m}, and for each k ∈ σ_{1}^{m} there is d_{k} > 0 such that

(D_{k}u, u)_{H} ≤ d_{k} ∥u∥_{H}^{2} , for all u ∈ H.

In the sequel we shall designate by h the function

h (t) = (u (t) , {v_{k}}_{k∈σ_{1}^{m}}) ,

where u (t) ∈ L^{2} ((0, T) \ {t_{k}}_{k∈σ_{1}^{m}}; H) and

{v_{k}}_{k∈σ_{1}^{m}} ∈ l^{2} (σ_{1}^{m}; H) ≑ {{v_{k}}_{k∈σ_{1}^{m}} , v_{k} ∈ H} .

We point out that the space K_{m} = L^{2} ((0, T) \ {t_{k}}_{k∈σ_{1}^{m}}; H) × l^{2} (σ_{1}^{m}; H) is a Hilbert space with respect to the inner product

(h, h̃)_{K_{m}} = ∫_{0}^{T} (u (t) , ũ (t))_{H} dt + ∑_{k=1}^{m} (v_{k}, ṽ_{k})_{H} ,

defined for all h = (u (t) , \{v_{k}\}_{k=1}^{m}) and h̃ = (ũ (t) , \left\{{ \atop v_{k}}\right\}_{k=1}^{m}) ∈ K_{m}. We shall denote by B the control operator given by

B = (B, {D_{k}}_{k∈σ_{1}^{m}}) ∈ L (L^{2} ((0, T) \ {t_{k}}_{k∈σ_{1}^{m}}; H) × l^{2} (σ_{1}^{m}; H)) ,

so that

Bh (t) = (Bu (t) , {D_{k}v_{k}}_{k∈σ_{1}^{m}}) .

We have for every h = (u(t), {v_{k}}_{k=1}^{m}) ∈ K_{m}

(Bh, h)_{K_{m}} = ∫_{0}^{T} (Bu (t) , u (t))_{H} dt + ∑_{k=1}^{m} (D_{k}v_{k}, v_{k})_{H}

= ∫_{0}^{T} (u (t) , Bu (t))_{H} dt + ∑_{k=1}^{m} (v_{k}, D_{k}v_{k})_{H}
= (h, Bh)_{K_{m}} ,

which shows that B^{∗} = B, that is, B is self-adjoint. On the other hand, we have

(Bh, h)_{K_{m}} = ∫_{0}^{T} (Bu (t) , u (t))_{H} dt + ∑_{k=1}^{m} (D_{k}v_{k}, v_{k})_{H}

≤ d_{0} ∫_{0}^{T} ∥u (t)∥_{H}^{2} dt + ∑_{k=1}^{m} d_{k} ∥v_{k}∥_{H}^{2}
≤ δ ∥h∥_{K_{m}}^{2} ,

where δ = max {d_{0}, d_{1}, ..., d_{m}} . Thus, the operator is B bounded in K_{m}. Next, we consider the homogeneous system associated with (1) :

φ′ (t) + Aφ (t) = 0, t ∈ (0, T) \ {t_{k}}_{k∈σ_{1}^{m}} ,  (2)
φ (0) = φ^{0},
Δφ (t_{k}) = I_{k}φ (t_{k}) , k ∈ σ_{1}^{m}.  (2_{k})

We point out that on each interval [t_{k}, t_{k+1}), for k = 0, ..., m, the solution φ is left continuous at each time t_{k}. Consider the corresponding homogeneous backward problem :

−φ̃′(t) + Aφ̃(t) = 0, , t ∈ (0, T) \ {t_{k}}_{k∈σ_{1}^{m}} ,  (3)
φ̃(T) = φ^{0},
Δφ̃(t_{m−(k−1)}) = −Ĩ_{m−(k−1)}φ̃(t_{m−(k−1)}^{+}), k ∈ σ_{1}^{m},  (3_{k})

where

A = A^{∗} = −A, Ĩ_{m−(k−1)} = I_{m−(k−1)}^{∗} = −I_{m−(k−1)}, k ∈ σ_{1}^{m}.

We observe that the problem (3) on the interval [t_{m}, T] is equivalent to the classical backward problem

\begin{matrix}−φ̃′(t) + Aφ̃(t) = 0, t ∈ [t_{m}, T] , \\ φ̃(T) = φ^{0}. \end{matrix}

We introduce the following space : PC ([0, T] ; H) = {y, y : [0, T] → H such that y(t) is continuous at t ≠ t_{k}, and has discontinuities of first kind at t = t_{k}, for every k ∈ σ_{1}^{m}}. Evidently, PC ([0, T] ; H) is a Banach space with respect to the norm

∥y∥_{PC} = sup_{t∈(0,T)} ∥y(t)∥ .

On the other hand, we define the subspaces PLC, (respectively, PRC)= {y, y ∈ PC such that y(t) is left ( respectively, right) continuous at t = t_{k}, for every k ∈ σ_{1}^{m}}.

Remark 1 1) The space PLC, (respectively, PRC) can be identified to a subspace of K_{m}. That is, to each y ∈ PLC, (respectively, ỹ ∈ PRC) is assigned the function h (respectively, h̃) defined by

h (t) = (y (t) , {y (t_{k})}_{k∈σ_{1}^{m}}) ,

and

h̃ (t) = (ỹ(t) , {ỹ(t_{k})}_{k∈σ_{1}^{m}}) .

The mapping y ↦ h (t) (respectively, ỹ ↦ h̃) is a linear injection.

2) Let ỹ ∈ PRC, the function y can be written as :

ỹ(t) = \{ \begin{matrix}ỹ_{[0]}(t) & if & t ∈ [t_{0}, t_{1}) \\ ỹ_{[k]}(t) & if & t ∈ [t_{k}, t_{k+1}) \\ ỹ_{[m]}(t) & if & t ∈ [t_{m}, T] . \end{matrix}

Next, let τ_{k} = t_{k} − t_{k−1}, we define the operator T : D(T ) = PRC ⊂ K_{m} → K_{m} by

(T ỹ)(t) = \{ \begin{matrix}ỹ_{[0]}((T − t)\frac{τ_{1}}{τ_{m+1}} + t_{0}) & if & t ∈ [t_{m}, T] , \\ ỹ_{[k]}((t_{m−(k−1)} − t)\frac{τ_{k+1}}{τ_{m−(k−1)}} + t_{k}) & if & t ∈ [t_{m−k}, t_{m−(k−1)}) , & k ∈ σ_{1}^{m−1}, \\ ỹ_{[m]}((t_{1} − t)\frac{τ_{m+1}}{τ_{1}} + t_{m}) & if & t ∈ (0, t_{1}] . \end{matrix}

(4)

We note that the range of T is exactly PLC. The function (T ỹ)(t) can be written as follows:

(T ỹ)(t) = \{ \begin{matrix}y_{[0]}(t) & if & t ∈ [t_{0}, t_{1}] , \\ y_{[k]}(t) & if & t ∈ (t_{k}, t_{k+1}] , & k ∈ σ_{1}^{m−1}, \\ y_{[m]}(t) & if & t ∈ (t_{m}, T] . \end{matrix}

Let X (t) be the resolvent solution of the operator system

X′(t) + AX(t) = 0, 0 = t_{0} < t < t_{m+1} = T, t ≠ t_{k}, k = 1, 2, ..., m, X(0) = I, X(t_{k} + 0) − X(t_{k} − 0) = I_{k}X(t_{k}), k = 1, 2, ..., m,

where I : H → H is the identity operator. We shall suppose that the operator I_{k} = I_{k} + I has a bounded inverse.

Definition 1 A function y ∈ PC ([0, T] ; H)is a mild solution to the impulsive problem (1) if the impulsive conditions are satisfied and

y(t) =  G(t, 0^{+})y^{0} + ∫_{0}^{t} G(t, s)Bu (s) ds
+ ∑_{0<t_{k}≤t} G(t, t_{k})(D_{k}v_{k}), for every t ∈ (0, T),

where the evolution operator G(t, s) is given by

G(t, s) = X(t)X^{−1}(s).

It is not hard to check that the operator G(t, t_{k}) satisfies the operator system

G′(t, t_{k}) + AG(t, t_{k}) = 0, t ∈ [t_{k}, t_{k+1}) , k ∈ σ_{0}^{m},
G(t_{k}, t_{k}) = I,
G(t_{k+1}^{+}, t_{k}) − G(t_{k+1}^{−}, t_{k}) = I_{k+1}G(t_{k+1}^{−}, t_{k}).

It is well known that (1) has a unique solution y such that

y ∈ PLC ([0, T] ; H) ∩ C^{1} ([0, T] \ {t_{k}}_{k∈σ_{1}^{m}}; H) .

Now, we define the concept of mild solution for the backward impulsive system (3) associated with system (2).

Definition 2 We say that φ̃ ∈ PRC ([0, T] ; H) is a mild solution for the backward impulsive system (3) if T φ̃ is a mild solution for the homogeneous impulsive system (2).

Let us introduce the notion of the null controllability of the initial state as follows:

Definition 3 We say that the initial state y^{0} ∈ H is null controllable at time T, if there is a control function h ∈ K_{m} for which the solution y of system (1) satisfies y (T) = 0.

2 Main Results

First we begin by the following lemma.

Lemma 1 Assume that ξ (t) , ζ (t) ∈ L^{1} ([0, T] ; H) and {ξ_{k}}_{k=1}^{m} , {ζ_{k}}_{k=1}^{m} ∈ l^{1}(σ_{1}^{m}, H). Then, for every vector functions

γ (t) ∈ PLC ([0, T] ; H) ∩ C^{1} ([0, T] \ {t_{k}}_{k∈σ_{1}^{m}}; H)

and

η (t) ∈ PRC ([0, T] ; H) ∩ C^{1} ([0, T] \ {t_{k}}_{k∈σ_{1}^{m}}; H)

satisfying the problem

\frac{d}{dt}⟨γ (t) , η (t)⟩ = ⟨ξ (t) , ζ (t)⟩, t ≠ t_{k}, for k ∈ σ_{1}^{m},
Δ⟨γ (t_{k}) , η (t_{k})⟩ = ⟨Δγ (t_{k}) , η (t_{k})⟩ + ⟨γ (t_{k}) , Δη (t_{k})⟩ = ⟨ξ_{k}, ζ_{k}⟩, k ∈ σ_{1}^{m},

we have the following identity

⟨γ (t) , η (t)⟩|_{0}^{T} = ⟨γ (T) , η (T)⟩ − ⟨γ (0) , η (0)⟩  (5)
= ∫_{0}^{T} ⟨ξ (t) , ζ (t)⟩dt + ∑_{k=1}^{m}⟨ξ_{k}, ζ_{k}⟩.

Proof. It is straightforward. □ We also need the following Lemmas.

Lemma 2 [14] If B ∈ L(K_{m}) is self-adjoint and nonnegative, then

∥Bh∥ ≤ ∥B∥^{1/2} (Bh, h)_{K_{m}}^{1/2}, h ∈ K_{m}.

Lemma 3 If τ_{k+1} = τ_{m−(k−1)}, k ∈ σ_{0}^{m−1}, then for the mild solution φ̃ of (3), the identity holds :

∫_{0}^{T} |Bφ̃|_{H}^{2} dt + ∑_{k=1}^{m} |D_{k}φ̃(t_{k}^{+})|_{H}^{2} = ∫_{0}^{T} |Bφ|_{H}^{2} dt + ∑_{k=1}^{m} |D_{k}φ (t_{m−(k−1)})|_{H}^{2} .

(6)

Proof. For each k ∈ σ_{0}^{m}, using the change of variable t → (t_{m−(k−1)} − t)\frac{τ_{k+1}}{τ_{m−(k−1)}} + t_{k} we have

∫_{t_{m−k}}^{t_{m−(k−1)}}(Bφ_{[m−k]}(t), Bφ_{[m−k]}(t))dt
= ∫_{t_{m−k}}^{t_{m−(k−1)}}(Bφ̃_{[k]}((t_{m−(k−1)} − t)\frac{τ_{k+1}}{τ_{m−(k−1)}} + t_{k}), Bφ̃_{[k]}((t_{m−(k−1)} − t)\frac{τ_{k+1}}{τ_{m−(k−1)}} + t_{k}))dt
= \frac{−τ_{m−(k−1)}}{τ_{k+1}} ∫_{t_{k+1}}^{t_{k}}(Bφ̃_{[k]}(s), Bφ̃_{[k]}(s))ds
= ∫_{t_{k}}^{t_{k+1}}(Bφ̃_{[k]}(s), Bφ̃_{[k]}(s))ds.

Summing up with respect to k, we get

∑_{k=0}^{m} ∫_{t_{m−k}}^{t_{m−(k−1)}}(Bφ_{[m−k]}((t)), Bφ_{[m−k]}(t))dt = ∑_{k=0}^{m} ∫_{t_{k}}^{t_{k+1}}(Bφ̃_{[k]}(t), Bφ̃_{[k]}(t))dt.

Thus, we obtain

∫_{0}^{T} |Bφ̃|_{H}^{2} dt = ∫_{0}^{T} |Bφ|_{H}^{2} dt.

On the other hand, by virtue of the definition of the function φ̃ we get

φ (t_{m−k}) = φ̃(t_{k+1}) , k ∈ σ_{0}^{m−1}.

Also, we have

φ (t_{m−(k−1)}) = φ̃(t_{k}) , k ∈ σ_{1}^{m},

and

φ̃(t_{m−k}) = φ (t_{k+1}) , k ∈ σ_{0}^{m−1}.

This implies that

∑_{k=1}^{m} |D_{k}φ̃(t_{k})|_{H}^{2} = ∑_{k=0}^{m−1} ⟨D_{m−k}φ̃(t_{m−k}) , D_{m−k}φ̃(t_{m−k})⟩_{H}
= ∑_{k=0}^{m−1} ⟨D_{m−k}φ (t_{k+1}) , D_{m−k}φ (t_{k+1})⟩_{H}
= ∑_{l=1}^{m} ⟨D_{l}φ (t_{m−(l−1)}) , D_{l}φ (t_{m−(l−1)})⟩_{H}
= ∑_{k=1}^{m} ⟨D_{k}φ (t_{m−(k−1)}) , D_{k}φ (t_{m−(k−1)})⟩_{H}
= ∑_{k=1}^{m} |D_{k}φ (t_{m−(k−1)})|_{H}^{2} ,

which gives (6). □ Corollary 1 If τ_{k+1} = τ_{m−(k−1)}, for k ∈ σ_{0}^{m−1}, and B, D_{k} are nonnegative in H, then the following holds:

∫_{0}^{T} ⟨Bφ̃(t), φ̃(t)⟩dt + ∑_{k=1}^{k=m}⟨D_{k}φ̃(t_{k}), φ̃(t_{k})⟩

= ∫_{0}^{T} ⟨Bφ(t), φ(t)⟩dt + ∑_{k=1}^{k=m}⟨D_{k}φ(t_{m−(k−1)}), φ(t_{m−(k−1)})⟩.

Proof. This follows immediately from Lemma 3 if we substitute B by B^{\frac{1}{2}} ,

and D_{k} by Dk^{\frac{1}{2}} . □

Now, we state and establish the following Theorem.

Theorem 1 Let y^{0} ∈ H be a given initial state for the system (1), then y^{0} is null controllable at time T if and only if there is a positive constant C such that

|⟨y^{0}, φ̃^{0}⟩_{H} | ≤ C {∫_{0}^{T} |Bφ|_{H}^{2} dt + ∑_{k=1}^{m} |D_{k}φ (t_{m−(k−1)})|_{H}^{2}}^{1/2} , ∀φ̃^{0} ∈ H,

(7)

where φ ∈ PLC ([0, T] ; H) is the unique mild solution to (2) with φ (T) = φ̃^{0}.

Proof. It suffices to prove this Theorem for the special case τ_{k+1} = τ_{m−(k−1)},

for k ∈ σ_{0}^{m−1}, because the norm ∥|.|∥ ≑ \{∑_{k=0}^{m} \frac{τ_{m−(k−1)}}{τ_{k+1}} ∫_{t_{k}}^{t_{k+1}} |.|_{H}^{2} dt\}^{1/2} is

equivalent to the usual norm of L^{2} ([0, T] ; H) .

We shall proceed in several steps.

Step 1: Let y and φ̃ be strong solutions to (1) and (3), respectively. Then, for t ≠ t_{k}, k ∈ σ_{1}^{m}, we have

\frac{d}{dt}⟨y (t) , φ̃(t)⟩ = ⟨y (t) , φ̃′ (t)⟩ + ⟨y′ (t) , φ̃(t)⟩  (8)
= ⟨y (t) , −Aφ̃(t)⟩ + ⟨−Ay (t) + Bu (t) , φ̃(t)⟩
= ⟨y (t) , −Aφ̃(t)⟩ + ⟨−Ay (t) , φ̃(t)⟩ + ⟨Bu (t) , φ̃(t)⟩
= ⟨Bu (t) , φ̃(t)⟩.

Multiplying equation (3_{k}) in (3) from the left by y (t_{m−(k−1)}) the solution of (1), and multiplying equation (1_{k}) in (1) from the right by φ̃(t_{k}) the solution of (3), and finally adding memberwise we get

Δ⟨y(t), φ̃(t)⟩_{|t=t_{k}} = ⟨y(t_{k}), Δφ̃(t_{k})⟩ + ⟨Δy(t_{k}), φ̃(t_{k})⟩  (9)
= ⟨y(t_{k}), I_{k}φ̃(t_{k})⟩ + ⟨I_{k}y(t_{k}) + D_{k}v_{k}, φ̃(t_{k})⟩
= ⟨y(t_{k}), I_{k}φ̃(t_{k})⟩ + ⟨I_{k}y(t_{k}), φ̃(t_{k})⟩ + ⟨D_{k}v_{k}, φ̃(t_{k})⟩
= ⟨D_{k}v_{k}, φ̃(t_{k})⟩.

Setting γ(t) = y(t), η(t) = φ̃(t), ξ(t) = Bu(t), ζ(t) = φ̃(t), ξ_{k} = D_{k}v_{k}, ζ_{k} = φ̃(t_{k}), then equations (5), (8) and (9) give

⟨y(T), φ̃(T)⟩ − ⟨y(0), φ̃(0)⟩ = ∫_{0}^{T} ⟨Bu(t), φ̃(t)⟩dt + ∑_{k=1}^{k=m}⟨D_{k}v_{k}, φ̃(t_{k})⟩. (10)

Since B is bounded, self-adjoint and B ≥0, then by density the latter identity is still valid for mild solutions y of (1). Identity (10) can be written as follows

⟨y(T), φ̃(T)⟩ − ⟨y(0), φ̃(0)⟩ = ∫_{0}^{T} ⟨u(t), Bφ̃(t)⟩dt + ∑_{k=1}^{k=m}⟨v_{k}, D_{k}φ̃(t_{k})⟩. (11)

Next, if there is a certain h(t) ∈ K_{m} such that the mild solution of (1) with y(0) = y^{0} satisfies y(T) = 0, then

−⟨y(0), φ̃(0)⟩ = ∫_{0}^{T} ⟨u(t), Bφ̃(t)⟩dt + ∑_{k=1}^{k=m}⟨v_{k}, D_{k}φ̃(t_{k})⟩,

and so by Cauchy-Schwarz Inequality we obtain

|⟨y(0), φ̃(0)⟩_{H}| ≤ {∫_{0}^{T} ∥u(t)∥_{H}^{2} dt + ∑_{k=1}^{k=m} ∥v_{k}∥_{H}^{2}}^{1/2}  (12)

× {∫_{0}^{T} ∥Bφ̃(t)∥_{H}^{2} dt + ∑_{k=1}^{k=m} ∥D_{k}φ̃((t_{k})∥_{H}^{2}}^{1/2} .

Using Lemma 3, and equation (12) we have

|⟨y(0), φ̃(0)⟩_{H}| ≤ {∫_{0}^{T} ∥u(t)∥_{H}^{2} dt + ∑_{k=1}^{k=m} ∥v_{k}∥_{H}^{2}}^{1/2}

× {∫_{0}^{T} ∥Bφ(t)∥_{H}^{2} dt + ∑_{k=1}^{k=m} ∥D_{k}φ(t_{m−(k−1)})∥_{H}^{2}}^{1/2} .

Setting

C = ∥h(t)∥_{K_{m}} = {∫_{0}^{T} ∥u(t)∥_{H}^{2} dt + ∑_{k=1}^{k=m} ∥v_{k}∥_{H}^{2}}^{1/2}

we find that

|(⟨y(0), φ̃(0)⟩_{H}| ≤ C {∫_{0}^{T} ∥Bφ(t)∥_{H}^{2} dt + ∑_{k=1}^{k=m} ∥D_{k}φ(t_{m−(k−1)})∥_{H}^{2}}^{1/2} .

This shows the necessary condition of the Theorem.

Step 2: To prove the sufficiency we need the following result when B ≥ α > 0.

Claim 1 Assume that there is α > 0 such that

{∫_{0}^{T} ∥Bu(t)∥_{H}^{2} dt + ∑_{k=1}^{k=m} ∥D_{k}v_{k}∥_{H}^{2}} ≥ α {∫_{0}^{T} ∥u(t)∥_{H}^{2} dt + ∑_{k=1}^{k=m} ∥v_{k}∥_{H}^{2}}

then, for every y^{0} ∈ H there is φ^{0} ∈ H such that the mild solution of (1) with

h(t) = (φ̃(t), φ̃(t_{1}), .., φ̃(t_{k}).., φ̃(t_{m})) ∈ K_{m} and y(0) = y^{0}

satisfies y(T) = 0.

To prove this Claim, we consider for every z ∈ H the solution φ of (2) satisfying φ(T) = z and the unique mild solution y to the problem

y′(t) + Ay(t) = Bφ̃(t), t ∈ (0, T) \{t_{k}}_{k∈σ_{1}^{m}} ,
Δy(t_{k}) = I_{k}y(t_{k}) + D_{k}φ̃(t_{k}),
y(T) = 0.

Next, we introduce a bounded linear operator Λ : H → H defined by

Λz = −y(0).

According to formula (11) and the Corollary 1 we have

|⟨Λz, z⟩| = |−⟨y(0), φ̃(0)⟩| = |∫_{0}^{T} ⟨Bφ̃(t), φ̃(t)⟩dt + ∑_{k=1}^{k=m}⟨D_{k}φ̃(t_{k}), φ̃(t_{k})⟩|

= |∫_{0}^{T} ⟨Bφ(t), φ(t)⟩dt + ∑_{k=1}^{k=m}⟨D_{k}φ(t_{m−(k−1)}), φ(t_{m−(k−1)})⟩|

≤ ς {∫_{0}^{T} ∥φ(t)∥^{2} dt + ∑_{k=1}^{k=m} ∥φ(t_{k})∥^{2}} ,

where

ς = sup_{k∈σ_{0}^{m}} {d_{k}} < ∞.

We have

∫_{0}^{T} ∥φ(t)∥^{2} dt = ∫_{0}^{t_{1}} ∥φ(t)∥^{2} dt + ∫_{t_{1}}^{t_{2}} ∥φ(t)∥^{2} dt + ... + ∫_{t_{m}}^{T} ∥φ(t)∥^{2} dt.

Since there is no impulse in the interval [t_{k}, t_{k+1}) we have

∥φ(t)∥ = ∥φ(t_{k}^{+})∥ , for every t ∈ [t_{k}, t_{k+1}) , k ∈ σ_{0}^{m},
∥φ(t_{k+1}^{−})∥ = ∥φ(t_{k−1}^{+})∥ , k ∈ σ_{0}^{m}.  (13)

Therefore, there are τ_{k+1} = t_{k+1} − t_{k} > 0, k ∈ σ_{0}^{m} such that

∫_{t_{k}}^{t_{k+1}} ∥φ(t)∥^{2} dt ≤ ρ_{k} ∥φ(t_{k}^{+})∥^{2} = τ_{k+1} ∥I_{k}φ(t_{k}^{−}) + φ(t_{k}^{−})∥^{2} , k ∈ σ_{1}^{m}. (14)

On the other hand, the continuity of I_{k} implies that

∥φ(t_{k}^{+})∥^{2} = ∥(I_{k} + I)φ(t_{k}^{−})∥^{2} ≤ (1 + L(I_{k}))^{2} ∥φ(t_{k}^{−})∥^{2} , k ∈ σ_{1}^{m}. (15)

It follows from (14) and (15) that

∫_{t_{k}}^{t_{k+1}} ∥φ(t)∥^{2} dt ≤ τ_{k+1}(1 + L(I_{k}))^{2} ∥φ(t_{k}^{−})∥^{2} , k ∈ σ_{1}^{m}.  (16)

Since m is finite, and due to (13),(16), then there is a constant 0 < μ < ∞ such that ⟨Λz, z⟩ ≤ μ ∥z∥^{2} , and thus, Λ is bounded.

Now, as B is nonnegative in K_{m}, we have

∥Bξ (t)∥ ≥ α {(ξ (t) , ξ (t))_{K_{m}}}^{1/2}

for all ξ ∈ K_{m}; thus, by virtue of Lemma 2, we have

{∫_{0}^{T} (Bu(t), u(t))_{H}dt + ∑_{k=1}^{k=m}(D_{k}v_{k}, v_{k})_{H}}  (17)

≥ α ∥B∥ {∫_{0}^{T} ∥u(t)∥_{H}^{2} dt + ∑_{k=1}^{k=m} ∥v_{k}∥_{H}^{2}} .

It follows from (11), (17) and Corollary 1 that

⟨Λz, z⟩ = −⟨y(0), φ̃(0)⟩
= ∫_{0}^{T} ⟨Bφ(t), φ(t)⟩dt + ∑_{k=1}^{k=m}⟨D_{k}φ(t_{m−(k−1)}), φ(t_{m−(k−1)})⟩

≥ α ∥B∥ {∫_{0}^{T} ∥φ(t)∥^{2} dt + ∑_{k=1}^{k=m} ∥φ(t_{k})∥^{2}}

≥ α ∥B∥ ∫_{0}^{t_{1}} ∥φ(t)∥^{2} dt = ∥B∥ αt_{1} ∥z∥^{2} = θ ∥z∥^{2} ,

because there is no impulse before time t_{1}. Therefore, Λ is coercive on H. To show that there is a bijection from H onto H, it suffices to prove that Λ + I is a bijection from H onto H. Clearly, Λ + I is injective since

⟨Λz + z, z⟩ = ⟨Λz, z⟩ + ⟨z, z⟩ ≥ (θ + 1) ∥z∥^{2} .

On the other hand, let y^{0} ∈ H, as the form a(f, g) + ⟨f, g⟩ = ⟨Λf, g⟩ + ⟨f, g⟩ is symmetric and coercive, then, by virtue of Lax-Milgram Theorem, there is an element f ∈ H such that

a(f, g) + ⟨f, g⟩ = ⟨y^{0}, g⟩, for all g ∈ H.

This implies that Λ(H) = H. Thus, for every y^{0} ∈ H , there is a unique z ∈ H such that Λ(z) = −y^{0}, which completes the proof of Claim 1.

Step 3: Assume that B, D_{k} ≥ 0, then B ≥ 0,

B̃^{2} = B, D̃_{k}^{2} = D_{k}.

We define for ε > 0,

β^{ε} ≑ B̃^{2} + εI,
δ_{k}^{ε} ≑ D̃_{k}^{2} + εI,

and

B^{ε} ≑ (β^{ε}; δ_{1}^{ε}, .., δ_{m}^{ε}) = (B̃^{2} + εI; D̃_{1}^{2} + εI, .., D̃_{m}^{2} + εI).

According to Claim 1, there is φ̃^{0,ε} ∈ H such that the mild solution y_{ε} of (1) with y_{ε}(0) = y^{0} satisfies y_{ε}(T) = 0; where B(h) has been replaced by

B^{ε}(φ̃(t), φ̃(t_{1}), .., φ̃(t_{k}).., φ̃(t_{m})) ∈ K_{m}.

We obtain from (11) and Corollary 1

−⟨y(0), φ̃_{ε}(0)⟩ = ∫_{0}^{T} ⟨β_{ε}^{ε}φ̃(t), φ̃_{ε}(t)⟩dt + ∑_{k=1}^{k=m}⟨δ_{k}^{ε}φ̃_{ε}(t_{k}), φ̃_{ε}(t_{k})⟩,  (18)

and (7) gives

−⟨y(0), φ̃_{ε}(0)⟩ ≤ C {∫_{0}^{T} ⟨B̃^{2}φ_{ε}(t), φ_{ε}(t)⟩dt + ∑_{k=1}^{k=m}⟨D̃_{k}^{2}φ_{ε}(t_{m−(k−1)}), φ_{ε}(t_{m−(k−1)})⟩}^{1/2} .

(19)

Whence,

−⟨y(0), φ̃_{ε}(0)⟩ ≤ C {∫_{0}^{T} ⟨β^{ε}φ_{ε}(t), φ_{ε}(t)⟩dt + ∑_{k=1}^{k=m}⟨δ_{k}^{ε}φ_{ε}(t_{m−(k−1)}), φ_{ε}(t_{m−(k−1)})⟩}^{1/2} .

(20)

It follows at once from (18), (19) and (20) that

ε {∫_{0}^{T} ∥φ_{ε}(t)∥^{2} dt + ∑_{k=1}^{k=m} ∥φ_{ε}(t_{k})∥^{2}}

+ ∫_{0}^{T} ⟨B̃φ_{ε}(t), B̃φ_{ε}(t)⟩dt + ∑_{k=1}^{k=m}⟨D̃_{k}φ_{ε}(t_{m−(k−1)}), D̃_{k}φ_{ε}(t_{m−(k−1)})⟩

= ∫_{0}^{T} (β^{ε}φ_{ε}(t), φ_{ε}(t))dt + ∑_{k=1}^{k=m}⟨δ_{k}^{ε}φ_{ε}(t_{m−(k−1)}), φ_{ε}(t_{m−(k−1)})) ≤ C^{2}.

(21)

Step 4: According to the estimate (20) the family

b_{ε} = B^{ε}(φ̃_{ε}(t); φ̃_{ε}(t_{1})..., φ̃_{ε}(t_{m})
= (B̃_{ε}^{2}φ̃(t); D̃_{1}^{2}φ̃_{ε}(t_{1})..., D̃_{m}φ̃_{ε}(t_{m})) + ε(φ̃_{ε}(t); φ̃_{ε}(t_{1})..., φ̃_{ε}(t_{m}))

is contained in a bounded subset K_{m}. Thus, both of the families

\sqrt{ε}(φ̃_{ε}(t); φ̃_{ε}(t_{1})..., φ̃_{ε}(t_{m})) and (Bφ̃_{ε}(t); D_{1}φ̃_{ε}(t_{1})..., D_{m}φ̃_{ε}(t_{m}))

are bounded in K_{m}. Therefore, we may extract a subfamily, say

(Bφ̃_{ε}(t); D_{1}φ̃_{ε}(t_{1})..., D_{m}φ̃_{ε}(t_{m})) ⇀ h, weakly in K_{m}.

Then clearly

(B̃^{2}φ̃_{ε}(t); D̃_{1}^{2}φ̃_{ε}(t_{1})..., D̃_{m}^{2}φ̃_{ε}(t_{m}))+ε(φ̃_{ε}(t); φ̃_{ε}(t_{1})..., φ̃_{ε}(t_{m})) ⇀ Bh, weakly in K_{m}.

Step 5: Taking the limit as ε → 0, we see that the solution y of (1) with initial condition y(0) = y^{0}, h being as in step 4 satisfies y(T) = 0. This completes the proof of Theorem 1. □

As an immediate application of the foregoing Theorem we give the following example.

Example. One dimensional impulsive Schrödinger equation : We consider the problem

\frac{∂y (t, x)}{∂t} + i\frac{∂^{2}y}{∂x^{2}} (t, x) = χ_{ω_{0}}u (t, x) , t ∈ (0, T) \ \{t_{k}\}_{k∈σ_{1}^{m}} , x ∈ Ω = (0, 2π),
y(t, 0) = y(t, 2π) = 0,  (22)
y (0, x) = y^{0},
Δy (t_{k}, x) = iα_{k}y (t_{k}, x) + χ_{ω_{k}}v_{k}(x), k ∈ σ_{1}^{m},

where

t_{k+1} − t_{k} > 2π, ω_{k} = (a_{1}^{k}, a_{2}^{k}) ⊂ Ω, k ∈ σ_{0}^{m}, {α_{k}}_{k∈σ_{1}^{m}} ⊂ ℝ^{+}.

Let

H = L^{2}(Ω, ℂ), Aw(x) = i\frac{∂^{2}w}{∂x^{2}}(x), D(A) = \{w ∈ H, \frac{∂^{2}w}{∂x^{2}} ∈ H, w(0) = w(π) = 0\} ,

and I_{k}w(x) = iα_{k}w(x) and the control operator is given by B = χ_{ω_{0}}, D_{k} = χ_{ω_{k}}, then the system (22) becomes an abstract formulation of (1). As a consequence of Theorem 1, the initial state y^{0} ∈ L^{2}(Ω, ℂ) = H of the solution of (22) is null-controllable at t = T, if and only if, there is C > 0 such that

|∫_{Ω} y^{0}(x)φ̃^{0}(x)dx|  (23)

≤ C \{∫_{0}^{T} ∫_{ω_{0}} |φ|^{2} (t, x)dxdt + ∑_{k=1}^{m} ∫_{ω_{k}} |φ|^{2} (t_{m−(k−1)}, x)\}^{\frac{1}{2}} , ∀φ̃^{0} ∈ L^{2}(Ω, ℂ),

where φ̃^{0}(x) = φ(T, x) and φ is the mild solution of

\frac{∂φ(t, x)}{∂t} + i\frac{∂^{2}φ(t, x)}{∂x^{2}} = 0, t ∈ (0, T) \ \{t_{k}\}_{k∈σ_{1}^{m}} , x ∈ Ω,
φ(t, 0) = φ(t, 2π) = 0,
φ (0, x) = φ^{0}(x), x ∈ Ω,
Δφ (t_{k}, x) = iα_{k}φ (t_{k}, x) , x ∈ Ω, k ∈ σ_{1}^{m}.

Here φ is given by

φ(t) = \{ \begin{matrix}φ_{[0]}(t) & , if & t ∈ [t_{0}, t_{1}) \\ φ_{[k]}(t) & , if & t ∈ [t_{k}, t_{k+1)}) \\ φ_{[m]}(t) & , if & t ∈ [t_{m}, T] , \end{matrix}

where φ_{[k]}(t) is a solution of the classical Schrödinger equation

\frac{∂φ_{[k]} (t, x)}{∂t} + i\frac{∂^{2}φ_{[k]}}{∂x^{2}} (t, x) = χ_{ω_{0}}u (t, x) , t ∈ (t_{0}, t_{1}) , x ∈ Ω = (0, 2π),
φ_{[k]}(t, 0) = φ_{[k]}(t, 2π) = 0,
φ_{[0]} (t_{0}, x) = φ^{0}(x), x ∈ Ω,

and

\frac{∂φ_{[k]} (t, x)}{∂t} + i\frac{∂^{2}φ_{[k]}}{∂x^{2}} (t, x) = χ_{ω_{0}}u (t, x) , t ∈ (t_{k}, t_{k+1}) , x ∈ Ω = (0, 2π),
φ_{[k]}(t, 0) = φ_{[k]}(t, 2π) = 0,
φ_{[k]} (t_{k}, x) = (1 + iα_{k})φ_{[k−1]} (t_{k}, x) , x ∈ Ω, k ∈ σ_{1}^{m}.

Then a standard application of a variant of Ingham’s Inequality [8] shows that

∫_{t_{k}}^{t_{k+1}} ∫_{w_{0}} |φ_{[k]}| (t, x)dtdx ≥ c(τ_{k}, w_{0}) ∫_{Ω} |φ_{[k]}| (t_{k}^{+}, x)dx,

for some positive constants c(τ_{k}, w_{0}) > 0. Summing up we get

∑_{k=0}^{m} ∫_{t_{k}}^{t_{k+1}} ∫_{w_{0}} |φ_{[k]}| (t, x)dtdx = ∫_{0}^{T} ∫_{ω_{0}} |φ|^{2} (t, x)dxdt

≥ c_{1} ∑_{k=1}^{m} ∫_{Ω} |φ_{[k]}| (t_{k}^{+}, x)dx,

where c_{1} = min_{k∈σ_{0}^{m}}c(τ_{k}, w_{0}) > 0.

On the other hand, there is a positive constant c_{2} > 0 such that

∑_{k=1}^{m} ∫_{ω_{k}} |φ|^{2} (t_{m−(k−1)}, x) ≥ c_{2} ∑_{k=1}^{m} ∫_{Ω} |φ_{[k]}|^{2} (t_{k}^{+}, x)dx.

It follows that

∫_{0}^{T} ∫_{ω_{0}} |φ|^{2} (t, x)dxdt

+ ∑_{k=1}^{m} ∫_{ω_{k}} |φ|^{2} (t_{m−(k−1)}, x)
≥ (c_{1} + c_{2}) ∑_{k=1}^{m} ∫_{Ω} |φ_{[k]}|^{2} (t_{k}^{+}, x)dx
≥ (c_{1} + c_{2}) ∫_{Ω} |φ_{[m]}|^{2} (t_{m}^{+}, x)dx
= (c_{1} + c_{2}) ∫_{Ω} |φ|^{2} (T, x)dx.

Now, since φ̃^{0}(x) = φ̃(0, x) = φ(T, x), then,

∫_{0}^{T} ∫_{ω_{0}} |φ|^{2} (t, x)dxdt + ∑_{k=1}^{m} ∫_{ω_{k}} |φ|^{2} (t_{m−(k−1)}, x) ≥ m(c_{1} + c_{2}) ∫_{Ω} |φ̃^{0}|^{2} (x)dx,

from which we get

∫_{Ω} |φ̃^{0}|^{2} (x)dx ≤ \frac{1}{m(c_{1} + c_{2})} (∫_{0}^{T} ∫_{ω_{0}} |φ|^{2} (t, x)dxdt + ∑_{k=1}^{m} ∫_{ω_{k}} |φ|^{2} (t_{m−(k−1)}, x)) .

We conclude by Cauchy-Schwarz inequality that

|∫_{Ω} y^{0}(x)φ̃^{0}(x)dx| ≤ {∫_{Ω} |y^{0}|^{2} (x)dx ∫_{Ω} |φ̃^{0}|^{2} (x)dx}^{1/2}
≤ \{\frac{∫_{Ω} |y^{0}|^{2} (x)dx}{m(c_{1} + c_{2})}\}^{1/2} (∫_{0}^{T} ∫_{ω_{0}} |φ|^{2} (t, x)dxdt

+ ∑_{k=1}^{m} ∫_{ω_{k}} |φ|^{2} (t_{m−(k−1)}, x)dx)^{1/2} ,

EJQTDE, 2007 No. 19, p. 17 which establishes the necessary and sufficient condition of null controllability stated in Theorem 1.

We conclude our paper by a special case when our initial state is an eigensolution of the following linear operator Γ : H → H defined by

Γ(ψ) = ∫_{0}^{T} X^{−1}(s)B^{2}X(s)ψds + ∑_{k=1}^{k=m} X^{−1}(t_{k})D_{k}^{2}X(t_{k})ψ.

We have the following result of null-controllability.

Proposition 1 Let λ > 0 be an eigenvalue of Γ with eigenvector ψ ∈ H. Then, the solution y to the problem

\{ \begin{matrix}y′(t) + Ay(t) = −\frac{1}{λ}B^{2}(X(t)ψ), & t ∈ (0, T) \\{t_{k}\}_{k∈σ_{1}^{m}} , \\ Δy(t_{k}) = I_{k}y(t_{k}) − \frac{1}{λ}D_{k}^{2}(X(t_{k})ψ), & k ∈ σ_{1}^{m} \\ y(0) = ψ, \end{matrix}  (24)

satisfies

y(T) = 0.

Proof.

Write system (24) into the form

\{ \begin{matrix}y′(t) + Ay(t) = −\frac{1}{λ}B^{2}(X(t)ψ), & t ∈ (0, T) \\{t_{k}\}_{k∈σ_{1}^{m}} , \\ y(t_{k}^{+}) = I_{k}y(t_{k}) − \frac{1}{λ}D_{k}^{2}(X(t_{k})ψ), & k ∈ σ_{1}^{m} \\ y(0) = ψ. \end{matrix}

Therefore, this impulsive problem has a solution which can be represented explicitly as follows

y(t) = X(t)ψ+∫_{0}^{t} G(t, s) [−\frac{1}{λ}B^{2}(X(s)ψ] ds+ ∑_{0<t_{k}≤t} G(t, t_{k}) [−\frac{1}{λ}D_{k}^{2}X(t_{k})ψ] ,

where the evolution operator G(t, s) is given by

G(t, s) = X(t)X^{−1}(s).

On the other hand, the system (24) yields

y(T) = X(T)ψ + ∫_{0}^{T} G(T, s) \{−\frac{1}{λ}B^{2}(X(s)ψ\} ds

+ ∑_{0<t_{k}≤T} G(T, t_{k}) \{−\frac{1}{λ}D_{k}^{2}X(t_{k})ψ\}

= X(T) [ψ + ∫_{0}^{T} X^{−1}(T)G(T, s) \{−\frac{1}{λ}B^{2}(X(s)ψ\} ds

−\frac{1}{λ} ∑_{0<t_{k}≤T} X^{−1}(T)G(T, t_{k}) \{D_{k}^{2}X(t_{k})ψ\}]

= X(T) [ψ + ∫_{0}^{T} X^{−1}(s) \{−\frac{1}{λ}B^{2}(X(s)ψ\} ds

−\frac{1}{λ} ∑_{0<t_{k}≤T} X^{−1}(t_{k}) \{D_{k}^{2}X(t_{k})ψ\}]

= X(T)) [ψ − \frac{1}{λ}Γ (ψ)] = 0.

This shows that the initial state ψ is null-controllable at time T with control

h (t) = (u (t) , \{v_{k}\}_{k∈σ_{1}^{m}}) = (−\frac{1}{λ}X (t) ψ, \{−\frac{1}{λ}X(t_{k})ψ\}_{k∈σ_{1}^{m}}) ,

which completes the proof of the Proposition. □

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(Received October 17, 2006)