Exact controllability of nonlinear diffusion equations arising in ...

Nonlinear Analysis: Real World Applications 9 (2008) 2029–2054www.elsevier.com/locate/na

Exact controllability of nonlinear diffusion equations arising inreactor dynamics

K. Sakthivela, K. Balachandrana, S.S. Sritharanb,∗aDepartment of Mathematics, Bharathiar University, Coimbatore 641 046, IndiabDepartment of Mathematics, University of Wyoming, Laramie WY 82071, USA

Received 25 June 2007; accepted 27 June 2007

Abstract

This paper studies the problems of local exact controllability of nonlinear and global exact null controllability of linear parabolicintegro-differential equations, respectively, with mixed and Neumann boundary data with distributed controls acting on a subdomain� of � ⊂ Rn. The proof of the linear problem relies on a Carleman-type estimate and observability inequality for the adjointequations and that the nonlinear one, on the fixed point technique.� 2007 Elsevier Ltd. All rights reserved.

MSC: 93B05; 93C20; 45K05; 35K50

Keywords: Carleman estimate; Parabolic integro-differential equation; Controllability; Observability

1. Introduction

In the problem of partial integro-differential equation modeled by including the integral term with a well-known basicpartial differential equation in various fields of physics and engineering, it is essential to take into account the effect ofpast history while describing the system as a function at a given time. Consider for example, a physical situation whichgives rise to a parabolic partial integro-differential equation of the form⎧⎪⎪⎪⎨⎪⎪⎪⎩

�y

�t− �y +

∫ t

0a(t, r)g(y(r, x)) dr = f (t, x) ∈ (0, T ) × �,

y(0, x) = y0(x), x ∈ �,

y(t, x) = 0 (t, x) ∈ (0, T ) × ��,

(1.1)

where � ⊂ Rn is a connected bounded domain with smooth boundary ��, is a feedback heat control in the interiorof some heat conducting medium, where the control mechanism possesses some inertia or a similar control situationfor a reaction–diffusion problem. In the analysis of space time dependent nuclear reactor dynamics, if the effect of alinear temperature feedback is taken into consideration and the reactor model is considered as an infinite rod, then the

∗ Corresponding author. Tel.: +1 307 766 4221.E-mail addresses: [email protected] (K. Sakthivel), [email protected] (K. Balachandran), [email protected] (S.S. Sritharan).

1468-1218/$ - see front matter � 2007 Elsevier Ltd. All rights reserved.doi:10.1016/j.nonrwa.2007.06.013

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2030 K. Sakthivel et al. / Nonlinear Analysis: Real World Applications 9 (2008) 2029–2054

one group neutron flux y(t, x) and the temperature v(t, x) in the reactor are given by the following coupled equations(see for instance [15,22] and also [18]):⎧⎪⎪⎨⎪⎪⎩

�y

�t− (a(x)yx)x = (c1v + c2 − 1)�f y(x) (t > 0, −∞ < x < ∞),

�c�v

�t= c3�gy,

(1.2)

where a is the diffusion coefficient and �f , �g, �, c, ci = (i =1, 2, 3) are physical quantities. By integrating the secondequation in (1.2) in the interval (0, t) and substituting it into the first equation, we obtain the following nonlinearintegro-differential diffusion equation:

�y

�t− (a(x)yx)x = �y

∫ t

0y(r, x) dr + by (t > 0, −∞ < x < ∞), (1.3)

where �, b are the constants associated with the initial temperature and various physical parameters. However, in theactual reactor systems, the temperature is a function of position x, which may be one, two or three dimensional. Thus itis more realistic to consider the heat equation for y in a higher dimensional spatial domain (see, for example, [2,20–22]).In this paper we consider a more general system of integro-differential equations of the form⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

�y

�t− Ly = g(t, x) + f

(t, x, y,

∫ t

0K(t, x, r, y(r, x)) dr

)(t, x) ∈ (0, T ) × �,

y(0, x) = y0(x), x ∈ �,

d(x, t)�y

��+ y = 0 (t, x) ∈ (0, T ) × ��,

(1.4)

where � is a connected bounded domain in Rn with smooth boundary �� and

Ly =n∑

i,j=1

aij (x)�2y

�xi�xj

+n∑

i=1

ai(x)�y

�xi

− a0(x)y.

The existence, uniqueness and asymptotic behavior of solutions of the system of the form (1.4) has been studiedin [20]. This problem governs many physical systems occurring in diffusion problems and includes (1.2) as a specialcase. With reference to Pachpatte [20], other problems in nuclear reactor dynamics give rise to equations of the form(1.1) but with the integral term replaced by∫ t

0k(t, r, x)y2(r, x) dr .

A general reference to solvability theory of the type of nonlinear integro-differential equations considered here isVrabie [28]. Moreover one can refer the introduction of Yanik and Fairweather [29] for a rich set of integro-differentialmodels and good number of references.

Controllability of dynamical systems represented by partial differential equations has been studied by several authorsby using different techniques (see, [3,6] and the references cited therein). One such technique is duality arguments.By using this, the exact controllability of a linear system can be reduced to the observability estimate of its dualsystem. In the same way, the exact controllability of a semilinear system can be reduced to an estimate, provided theobservability constant depends on the coefficients of the “linearized” systems. Thus, one of the main problems in thetheory of exact controllability is how to construct the observability estimates for the linear system. Broadly, for derivingthe observability estimates, we have the following three important methods: multiplier techniques [16,19], Carlemanestimates [9,10,25] and microlocal analysis [7].

However, of all these methods so far the most effective method in establishing such a result is the method of Carleman-type estimates. There have been great concerns in the Carleman inequalities after the publication of the basic paperby Carleman in 1939. In particular, after the appearance of the fundamental results by Hörmander [11], this theory isone of the most developing areas of linear partial differential equations and also for the detailed results of Carlemaninequalities for parabolic equations one can refer [12]. Using this technique the exact controllability of the heat equation

K. Sakthivel et al. / Nonlinear Analysis: Real World Applications 9 (2008) 2029–2054 2031

and the local controllability of the Navier–Stokes equation are obtained during the last decade [4,5,10,13,14]. Both theresults and techniques are derived from a Carleman inequality which is the key result of the whole theory. As a matterof fact, in both the cases the proof of Carleman inequality is essentially due to Imanuvilov [9,12].

This paper is organized as follows: in Section 3 we state a Carleman estimate as well as controllability results for thelinearized version of the general parabolic problem (3.1), (3.2) and at the end mainly we discuss the nonlinear problemby Schauder’s fixed point theorem through the results quoted for the linear case in the same section. Before that inSection 2 we establish a Carleman-type estimate in Theorem 2.1 for the backward adjoint problem of (2.1) stated in(2.2) to guarantee the existence (i.e., to settle the convolution term properly so as to get the same upper bound) of thisestimate for the partial integro-differential equations and also in the usual way we derive an observability estimate asa corollary of this theorem. Based on these two estimates and additionally by using classical Pontryagin’s maximumprinciple we prove the global exact null controllability of the linear problem (2.1) at the end of Section 2.

2. Linear parabolic equations with memory effects

In this section we shall discuss the controllability and observability of the linearized version of the problem (1.1) withNeumann boundary data and the results of the corresponding nonlinear model will be carried out in a forthcoming paper.Before getting into the discussion on controllability, we establish a Carleman estimate for the dual problem of (2.1) andit is essentially an a priori estimate of solutions by their restrictions on a subdomain together with the multiplicationof suitable weight functions and in fact major part of this paper is devoted to establishing such an estimate. The proofof this estimate is in many ways identical to, the proofs of different cases given by Fursikov and Imanuvilov [9] andsome simple techniques adopted in Barbu [4], Barbu et al. [5] and then Sakthivel et al. [23].

2.1. Formulation of the problem

Consider the linear partial integro-differential equation⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

�z(t, x)

�t− �z(t, x) +

∫ t

0a(t, r)z(r, x) dr = 1�u(t, x) + g(t, x), ∀ (t, x) ∈ Q,

�z(t, x)

��= 0, ∀ (t, x) ∈ �,

z(0, x) = z0(x), ∀ x ∈ �,

(2.1)

where � ⊂ Rn is a connected bounded domain with boundary �� of class C2 and 1� is the characteristic function ofan open set � of �, more precisely

1� ={1 for x ∈ �,

0 for x ∈ �\�.

The function �(x) is the external normal to �� and 0 < T < ∞ is an arbitrary moment of time. We also used thenotations Q = (0, T ) × � and � = (0, T ) × ��.

Moreover, the kernel a is a smooth given function with supp(a(·, t)) ⊂ (t0, t1), where 0 < t0 < t1 < T and g ∈ L2(Q)

is also given, while u ∈ L2(Q) is a control input and let z0 ∈ H 10 (�) be arbitrary but fixed initial data. System (2.1) is

said to be exactly null controllable if there is u ∈ L2(Q) such that z(T , .) ≡ 0.

In order to establish the controllability of (2.1) it is sufficient to prove a Carleman-type estimate, concerning thefollowing backward dual problem of (2.1), which is the key result in proving the observability of the same:⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

�y(t, x)

�t+ �y(t, x) −

∫ T

t

a(r, t)y(r, x) dr = f (t, x), ∀ (t, x) ∈ Q,

�y(t, x)

��= 0, ∀ (t, x) ∈ �,

y(T , x) = yT (x), ∀ x ∈ �,

(2.2)

where f ∈ L2(Q) and yT ∈ L2(�).


To formulate our results, we give some of the frequently used notations and function spaces

Wpk (�) =

⎧⎪⎨⎪⎩w(x): ‖w‖Wpk

=⎛⎝∑

|�|�k

∫�

|D�w|p dx

⎞⎠1/p

< ∞

⎫⎪⎬⎪⎭ ,

when p = 2, instead of W 2k we shall write Hk(�).

|z|2 =(∫

�|z|2 dx

)1/2

,

W1,22 (Q) =

{w(t, x): w,

�w

�t,�w

�xi

,�2w

�xixj

∈ L2(Q), i, j = 1, 2, . . . , n

},

C1,2(Q) ={y(t, x): y,

�y

�t,

�y

�xi

,�2y

�xixj

∈ C(Q), i, j = 1, 2, . . . , n

},

H 1([0, T ]; L2(�)) ={w ∈ L2(0, T ; L2(�)):

dw

dt∈ L2(0, T ; L2(�))

},

where dw/dt is taken in the sense of distributions. For more details of the basic function spaces and estimations onecan refer [1,17,27].

The following lemma is the most fundamental tool in proving the Carleman estimate of the problem (2.2). The proofof this lemma can be found in [9, Chapter 1, Lemma 1.1].

Lemma 2.1. Let �0�� be an arbitrary fixed subdomain of �. Then there exists a function ∈ C2(�) such that

(x) > 0 ∀x ∈ �, |�� = 0, |∇(x)| > 0 ∀x ∈ �\�0.

Here we introduce functions , � and , � by the formulas

(t, x) = e�(x)

�(t), (t, x) = e−�(x)

�(t)

and

�(t, x) = e�(x) − e2�

�(t), �(t, x) = e−�(x) − e2�

�(t), (2.3)

where

�(t) = t (T − t) and = ‖(x)‖C(�),

the parameter � > 1 and the function is defined in Lemma 2.1.Here we see that the weight functions � and � approached −∞ at t = 0 and t =T , and this helps us to get the desired

observability estimate. Also the additional parameter � is essential in order to obtain the control of the constant whichenables us to handle arbitrarily large coefficients in the coupling terms. Estimates of this type have been proved in[9,14,26].

Now we are ready to formulate the Carleman inequality concerning the problem (2.2).


Theorem 2.1 (Carleman estimate). Let the functions y and f satisfy (2.2) and suppose the kernel a(·, t) has supportin (t0, t1), where 0 < t0 < t1 < T. Then for any � > �0 and s > s0 the following inequality holds:∫

Q

((s)3|y|2 + s|∇y|2)e2s� dx dt

+∫

Q

1

s

(∣∣∣∣�y

�t

∣∣∣∣2 + |�y|2 +∣∣∣∣∫ T

t

a(r, t)y(r, x) dr

∣∣∣∣2)

e2s� dx dt

�c(�, s)

(∫Q

e2s�|f |2 dx dt +∫

Q�

e2s�3|y|2 dx dt

), (2.4)

where Q� = � × (0, T ), the functions , � are as defined in (2.3) and the constant c(�, s) > 0 is independent off and y.

Proof. Let us make the change of variables for the unknown function p(t, x)=es�y(t, x) in Eq. (2.2). Then it becomes⎧⎪⎪⎪⎨⎪⎪⎪⎩�p

�te−s� − ��

�tse−s�p − 2s�∇e−s�∇p + s2�22|∇|2e−s�p

−s�2|∇|2e−s�p + e−s��p − s��e−s�p

− ∫ T

ta(r, t)e−s�(r)p(r, x) dr = f.

We rewrite this equation in terms of two differential operators L1p(t, x), L2p(t, x) as

�p(t, x)

�t+ L1p(t, x) − L2p(t, x) = fs(t, x), ∀(t, x) ∈ Q, (2.5)

�p

��

∣∣∣∣�

= 0, (2.6)

where

L1p = −2s�∇∇p − 2s�2|∇|2p, (2.7)

L2p = −�2s22|∇|2p + s��

�tp − �p − s�2|∇|2p + s��p, (2.8)

fs = es�f +∫ T

t

a(r, t)es(�(t)−�(r))p(r, x) dr. (2.9)

From the definitions of � we see that �(t, x)|t=0 = �(t, x)|t=T = −∞. This implies that

p(t, x)|t=0 = p(t, x)|t=T = 0. (2.10)

In virtue of (2.5), we obtain∥∥∥∥�p

�t+ L1p

∥∥∥∥2

L2(Q)

+ ‖L2p‖2L2(Q)

− 2

⟨�p

�t, L2p

⟩L2(Q)

− 2〈L1p, L2p〉L2(Q) = ‖fs‖2L2(Q)

.

Clearly, this would imply the following inequality:

−〈L1p, L2p〉L2(Q) �⟨�p

�t, L2p

⟩L2(Q)

+ 1

2‖fs‖2

L2(Q). (2.11)

Eqs. (2.7) and (2.8) show that −〈L1p, L2p〉L2(Q) =∑6i=1Ii , where

I1 = −2∫

Q

(s3�33|∇|2p − s2�

��

�tp

)∇ · ∇p dx dt ,


I2 = −2∫

Q

(s�∇ · ∇p)�p dx dt ,

I3 = −2∫

Q

(s2�32|∇|2p − s2�22�p

)∇ · ∇p dx dt ,

I4 = −2∫

Q

(�4s33|∇|4p − s2�2|∇|2��

�tp

)p dx dt ,

I5 = −2∫

Q

(s�2|∇|2p

)�p dx dt ,

I6 = 2∫

Q

(s2�32�|∇|2p − s2�42|∇|4p

)p dx dt .

Now we estimate the integrals one by one. Integrating by parts in I1, we get

I1 = −∫

Q

(�3s33|∇|2 − �s2

��

�t

)∇∇|p|2 dx dt

=∫

Q

(3�4s33|∇|4 + �3s33(∇|∇|2∇ + |∇|2�))|p|2 dx dt

−∫

Q

� ln(�−1(t))

�t(�2s2|∇|2� + �2s22|∇|2 + �s2��)|p|2 dx dt

−∫�

(�3s33|∇|2 − �s2

��

�t

)�

��|p|2 d�

�∫

Q

3�4s33|∇|4|p|2 dx dt − c

∫Q

(�3s33 + �2s23 + �2s22 + �s23)|p|2 dx dt

−∫�

(�3s33|∇|2 − �s2

��

�t

)�

��|p|2 d�, (2.12)

where the constant c depends on and T . From the definitions of and �, we note that

�

�t= e�(x) (2t − T )

�2(t),

��

�t=(

e� − e2� ) (2t − T )

�2(t),

and so∣∣∣∣�

�t

∣∣∣∣ �c12,

∣∣∣∣��

�t

∣∣∣∣= ∣∣∣∣� �

�tln(�−1(t))

∣∣∣∣ �c12,

∣∣∣∣�2�

�t2

∣∣∣∣ �c13, (2.13)

where c1 > 0 is a constant and it is independent of (t, x) ∈ Q and � > 1. Further, is a continuous function withcompact support in �; then there exist constants c2, c3, c4 such that

max1� i,j �n

supx∈�

∣∣∣∣ �2

�xi�xj

∣∣∣∣= c2, max1� i �n

supx∈�

∣∣∣∣ �

�xi

∣∣∣∣= c3 and max1� i �n

supx∈�

∣∣∣∣ �

�xi

∣∣∣∣2 = c4


hold. But, for simplicity, throughout the proof we shall use the generic constant c alone. Next, integrating by partstwice (using Green’s theorem) in I2 with the help of the boundary condition (2.6), we have

I2 =∫

Q

⎛⎝2�2s(∇p)2(∇)2 + 2s�n∑

i,j=1

�2

�xi�xj

�p

�xi

�p

�xj

⎞⎠ dx dt

+∫

Q

s�∇∇(|∇p|2) dx dt − 2∫�

�p

��∇p�s∇ d�

=∫

Q

⎛⎝2s�2(∇p)2(∇)2 + 2�sn∑

i,j=1

�2

�xi�xj

�p

�xi

�p

�xj

⎞⎠ dx dt

−∫

Q

(s�2|∇|2|∇p|2 + s��|∇p|2) dx dt +∫�

s�|∇p|2 �

��d�

� −∫

Q

s�2|∇|2|∇p|2 dx dt − c

∫Q

s�|∇p|2 dx dt +∫�

s�|∇p|2 �

��d�. (2.14)

Integrating by parts in I3 and estimating the same, we get

I3 = −∫

Q

s2�32|∇|2∇∇|p|2 dx dt + 2∫

Q

s2�22�p∇ · ∇p dx dt

= 2∫

Q

s2�42|∇|4|p|2 dx dt +∫

Q

s2�32(∇|∇|2∇ + |∇|2�)|p|2 dx dt

+ 2∫

Q

(s�)3/2�p(s�)1/2∇ · ∇p dx dt −∫�

s2�32|∇|2 �

��|p|2 d�

�2∫

Q

s2�42|∇|4|p|2 dx dt − c

(∫Q

(s2�32 + s3�33)|p|2 dx dt +∫

Q

s�|∇p|2 dx dt

)

−∫�

s2�32|∇|2 �

��|p|2 d�, (2.15)

where the constant c in the estimations I3, I4 depends only on . Next, we estimate the integral I4 using the inequalities(2.13):

I4 = − 2∫

Q

�4s33|∇|4|p|2 dx dt + 2∫

Q

s2�2|∇|2(

��

�t

)|p|2 dx dt

� − 2∫

Q

�4s33|∇|4|p|2 dx dt − c

∫Q

s2�23|p|2 dx dt . (2.16)

Adding the lower bounds of I1 and I4, we have the following estimation:

I1 + I4 �∫

Q

�4s33|∇|4|p|2 dx dt − c

∫Q

(�3s33 + �2s23 + �2s22 + �s23)|p|2 dx dt

−∫�

(�3s33|∇|2 − �s2

��

�t

)�

��|p|2 d�, (2.17)


where the constant c depends on and T. Again, integrating by parts (in fact using Green’s theorem) in I5 with thehelp of the boundary condition (2.6), we arrive at

I5 = 2∫

Q

s�3|∇|2p∇ · ∇p dx dt + 2∫

Q

s�2∇(|∇|2)p∇p dx dt

+ 2∫

Q

s�2|∇|2|∇p|2 dx dt − 2∫�(s�2|∇|2p)

�p

��d�. (2.18)

Here we need to estimate the integrals 2∫Q

s�3|∇|2∇p∇p dx dt and 2∫Q

s�2∇(|∇|2)p∇p dx dt. By the simpleapplication of Cauchy’s inequality with � = 4, one can obtain that

2∫

Q

(s)1/2�2|∇|2p(s)1/2�∇ · ∇p dx dt

� − c

∫Q

s�4|p|2 dx dt − 1

4

∫Q

s�2|∇|2|∇p|2 dx dt (2.19)

and

2∫

Q

(s)1/2��p(s)1/2�∇ · ∇p dx dt

� − c

∫Q

s�2|p|2 dx dt − 1

4

∫Q

s�2|∇|2|∇p|2 dx dt , (2.20)

where the constant c in both the above inequalities depends on only. Substituting the inequalities (2.19)–(2.20) into(2.18) we get

I5 � 3

2

∫Q

s�2|∇|2|∇p|2 dx dt − c

∫Q

(s�4 + s�2)|p|2 dx dt . (2.21)

We note that the addition of the inequality (2.21) with the estimate of I2 yields

I2 + I5 � 1

2

∫Q

s�2|∇|2|∇p|2 dx dt

− c

(∫Q

(s�4 + s�2)|p|2 dx dt +∫

Q

s�|∇p|2 dx dt

)+∫�

s�|∇p|2 �

��d�, (2.22)

where the positive constant c depends only on . Clearly from the integral I6, we have

I6 = 2∫

Q

(s2�32�|∇|2 − s2�42|∇|4)|p|2 dx dt

� − 2∫

Q

s2�42|∇|4|p|2 dx dt − c

∫Q

s2�32|p|2 dx dt . (2.23)

Adding the inequality (2.23) with the estimation of the integral I3, we get

I3 + I6 � − c

(∫Q

(s2�32 + s3�33)|p|2 dx dt +∫

Q

s�|∇p|2 dx dt

)

−∫�

s2�32|∇|2 �

��|p|2 d�. (2.24)


Now, we estimate the integrals involving the term 〈�p/�t, L2p〉L2(Q). From Eq. (2.8), we observe that〈�p/�t, L2p〉L2(Q) = J1 + J2, where

J1 =∫

Q

(s��

�tp − s2�22|∇|2p − s�2|∇|2p + s��p

)�p

�tdx dt ,

J2 = −∫

Q

�p

�t�p dx dt .

Estimating the integrals J1 and J2 simultaneously,

J1 + J2 = 1

2

∫Q

(s��

�t− s2�22|∇|2 − s�2|∇|2 + s��

)�|p|2�t

dx dt

+∫

Q

∇p∇(

�p

�t

)dx dt −

∫�

(�p

��

)�p

�td�

= −∫

Q

(s

2

�2�

�t2− s2�2

�

�t|∇|2 − s

2�2 �

�t|∇|2 + s

2��

�

�t

)|p|2 dx dt

+ 1

2

∫Q

�|∇p|2�t

dx dt

�c

∫Q

(s3 + s2�23 + s�22 + s�2)|p|2 dx dt , (2.25)

where the positive constant c is independent of (t, x) ∈ Q and recall the boundary conditions (2.6), (2.10) and theinequalities (2.13).

Before going to frame the main inequality we need to eliminate the surface integrals occurring in (2.17), (2.22) and(2.24). In this regard we use the weight functions defined in (2.3) with negative order exponents. Then we repeat the proofwith a similar computation as explained in the above inequalities by defining the transformation p(t, x) = es�y(t, x),where the weight functions and � are as defined in (2.3).

By replacing the functions , � and the parameter �, respectively, by , � and −�, one can see that p also satisfiessystem (2.5)–(2.6). Therefore we have

�p

�t+ L1p(t, x) − L2p(t, x) = fs(t, x), ∀(t, x) ∈ Q, (2.26)

�p

��

∣∣∣∣�

= 0, (2.27)

where

L1p = 2s�∇∇p − 2s�2|∇|2p,

L2p = −�2s22|∇|2p + s

��

�tp − �p − s�2|∇|2p − s��p,

fs = es�f +∫ T

t

a(r, t)es(�(t)−�(r))p(r, x) dr .

Corresponding to the inequality written in (2.11), we write the following one:

−〈L1p, L2p〉L2(Q) �⟨�p

�t, L2p

⟩L2(Q)

+ 1

2‖fs‖2

L2(Q). (2.28)


We repeat the same calculations starting from the integrals I1 to I6 and those in J1, J2 by replacing , � and theparameter �, respectively, by the functions , � and −�, and after renaming the integrals as Ii , i = 1, 2, . . . , 6 andJj , j = 1, 2, we arrive at the following inequalities:

I1 �∫

Q

3�4s33|∇|4|p|2 dx dt − c

∫Q

(�3s33 + �2s2

3 + �2s22 + �s2

3)|p|2 dx dt

+∫�

(�3s3

3|∇|2 − �s2��

�t

)�

��|p|2 d�. (2.29)

From Lemma 2.1 we note that = 0 on �� and hence the definitions of , and �, � give (remember that � =(0, T ) × ��) = , � = �, p = p on �. Clearly we can rewrite the surface integral in (2.29) as∫

�

(�3s3

3|∇|2 − �s2��

�t

)�

��|p|2 d� =

∫�

(�3s33|∇|2 − �s2

��

�t

)�

��|p|2 d�.

We note that

I2 � −∫

Q

s�2|∇|2|∇p|2 dx dt − c

∫Q

s�|∇ p|2 dx dt −∫�

s�|∇p|2 �

��d�. (2.30)

Repeating the calculation for I3, I4, we have

I3 �2∫

Q

s2�42|∇|4|p|2 dx dt

− c

∫Q

((s2�32 + s3�3

3)|p|2 + s�|∇p|2) dx dt +

∫�

s2�32|∇|2 �

��|p|2 d� (2.31)

and

I4 � − 2∫

Q

�4s33|∇|4|p|2 dx dt − c

∫Q

s2�23|p|2 dx dt . (2.32)

The inequalities (2.29), (2.32) show that

I1 + I4 �∫

Q

�4s33|∇|4|p|2 dx dt − c

∫Q

(�3s33 + �2s2

3 + �2s22 + �s2

3)|p|2 dx dt

+∫�

(�3s33|∇|2 − �s2

��

�t

)�

��|p|2 d�. (2.33)

Computation corresponding to the integral I5 leads to the following inequality:

I5 � 3

2

∫Q

s�2|∇|2|∇p|2 dx dt − c

∫Q

(s�4 + s�2)|p|2 dx dt . (2.34)

Here we note from the estimations of I2, I5 that,

I2 + I5 � 1

2

∫Q

s�2|∇|2|∇p|2 dx dt − c

(∫Q

(s�4 + s�2)|p|2 dx dt +∫

Q

s�|∇p|2 dx dt

)

−∫�

s�|∇p|2 �

��d�. (2.35)

Further

I6 � − 2∫

Q

s2�42|∇|4|p|2 dx dt − c

∫Q

s2�32|p|2 dx dt . (2.36)


Adding the lower bounds of I3, I6, we get

I3 + I6 � − c

(∫Q

(s2�32 + s3�3

3)|p|2 dx dt +

∫Q

s�|∇p|2 dx dt

)

+∫�

s2�32|∇|2 �

��|p|2 d�. (2.37)

Finally, estimating the integrals J1 and J2, we get

J1 + J2 �c

∫Q

(s3 + s2�2

3 + s�22 + s�

2)|p|2 dx dt .

But the definitions of , and �, � clearly show that

�, �� and p�p in Q. (2.38)

Then the previous inequality can be written as

J1 + J2 �c

∫Q

(s3 + s2�23 + s�22 + s�2)|p|2 dx dt . (2.39)

In order to complete the estimations involving the terms of (2.11), we need to obtain the upper bounds for the L2

integrals of fs . From Eq. (2.9), we see that

‖fs‖2L2(Q)

�2

(∫Q

e2s�|f |2 dx dt +∫

Q

∣∣∣∣∫ T

t

a(r, t)es(�(t)−�(r))p(r, x) dr

∣∣∣∣2 dx dt

). (2.40)

Now we are ready to frame the required inequality. First, we add the inequalities (2.11), (2.28) and after we use theestimates (2.17), (2.22), (2.24)–(2.25) and (2.33), (2.35), (2.37) through (2.39)–(2.40). Then for s > 1, � > 1, we havethe following key inequality:∫

Q

s3�43|∇|4|p|2 dx dt +∫

Q

s�2|∇|2|∇p|2 dx dt

�c

∫Q

(s2�43 + s3�33)|p|2 dx dt + c

∫Q

s�|∇p|2 dx dt +∫

Q

e2s�|f |2 dx dt

+∫

Q

∣∣∣∣∫ T

t


∣∣∣∣2 dx dt , (2.41)

where the constant c is independent of (t, x) ∈ Q, � > 1 which may depend on , T . Here we have used the obviousinequalities ()−1 �(T /2)2, ()−2 �(T /2)4 and the inequalities (2.38) and the fact that p = es(�−�)p, gives ∇p =∇pes(�−�) + ses(�−�)∇ (� − �)p = es(�−�)(∇p) − 2s�∇p so that for some positive constant c depending on , wehave

|∇p|�c(|∇p| + s�|∇||p|)�c(|∇p| + s�|p|) in Q.

From Lemma 2.1, we see that |∇| has a lower bound on �\�0 and therefore on Q\Q�0 (note that does notdepend on t), then there exists a constant � > 0 such that |∇|�� in Q\Q�0 , so the left-hand side terms of (2.41) havethe following lower bounds:∫

Q

s3�43|∇|4|p|2 dx dt ��4

(∫Q

s3�43|p|2 dx dt −∫

Q�0

s3�43|p|2 dx dt

)


and similarly∫Q

s�2|∇|2|∇p|2 dx dt ��2

(∫Q

s�2|∇p|2 dx dt −∫

Q�0

s�2|∇p|2 dx dt

).

Making use of these estimates in (2.41), we have

�4∫

Q

s3�43|p|2 dx dt + �2∫

Q

s�2|∇p|2 dx dt

�c

(∫Q�0

s3�43|p|2 dx dt +∫

Q�0

s�2|∇p|2 dx dt

+∫

Q

s2�43|p|2 dx dt +∫

Q

(s3�33|p|2 + s�|∇p|2) dx dt

)

+∫

Q

e2s�|f |2 dx dt +∫

Q

∣∣∣∣∫ T

t


∣∣∣∣2 dx dt . (2.42)

Now we note that for any s sufficiently large such that s > s0 = 2c/�4, the third integral on the right-hand side goeswith the similar left-hand side integral and for any � such that � > �0 =max{4c/�4, 2c/�2}, the fourth integral can alsobe absorbed. Thus we have∫

Q

s3�43|p|2 dx dt +∫

Q

s�2|∇p|2 dx dt

�c

(∫Q�0

s3�43|p|2 dx dt +∫

Q�0

s�2|∇p|2 dx dt

+∫

Q

e2s�|f |2 dx dt +∫

Q

∣∣∣∣∫ T

t


∣∣∣∣2 dx dt

). (2.43)

Thanks to the assumption on the kernel, we also estimate the last integral as∫Q

∣∣∣∣∫ T

t


∣∣∣∣2 dx dt

�∫

Q

e2s�(∫ t1

t0

|a(r, t)|2−3(r)e−2s�(r) dr

)(∫ t1

t0

3(r)|p(r, x)|2 dr

)dx dt

�c‖a‖2L∞

∫Q

e2s�(∫ t1

t0

3(r)|p(r, x)|2 dr

)dx dt

�c

∫Q

3|p|2 dx dt . (2.44)

It is easy to see that for any sufficiently large � > �0 and s > s0, we have∫Q

s3�43|p|2 dx dt +∫

Q

s�2|∇p|2 dx dt

�c

(∫Q�0

s3�43|p|2 dx dt +∫

Q�0

s�2|∇p|2 dx dt +∫

Q

e2s�|f |2 dx dt

).


Next we come back to the original variable y by substituting p = es�y into the above inequality and after somecalculation, we get∫

Q

(s|∇y|2 + s33|y|2)e2s� dx dt

�c(�)

(∫Q�0

e2s�(s33|y|2 + s|∇y|2) dx dt +∫

Q

e2s�|f |2 dx dt

)(2.45)

for � > �0 and s > s0, where the constant c is somehow greater than the constant defined in the preceding estimate.We now express the integral of e2s�s|∇y|2 over Q�0 in the right-hand side by the integral of e2s�(s)3|y|2 over asome how larger domain Q� (recall that �0��). To this end, we multiply the first equation in (2.2) by e2s�s�y andintegrate over Q, where � is some function in C∞

0 (�) such that � ≡ 1 if x ∈ �0 and � ≡ 0 if x ∈ �\� to get

−∫

Q

e2s��sy�y dx dt � �

2

∫Q

e2s�|f |2 dx dt + 1

�

∫Q�

e2s�s22|y|2 dx dt

+∫

Q

e2s��sy�y

�tdx dt + �

2

∫Q�

e2s�∣∣∣∣∫ T

t

a(r, t)y(r, x) dr

∣∣∣∣2 dx dt . (2.46)

Integrating by parts with respect to t on the third integral and estimating the same, we get (taking �(0, x)=�(T , x)=−∞into account)∫

Q

e2s��sy�y

�tdx dt = −

∫Q

s2e2s� ��

�t�|y|2 dx dt − 1

2

∫Q

se2s� �

�t�|y|2 dx dt

�c

∫Q�

(s)3e2s�|y|2 dx dt , (2.47)

where the positive constant c is independent of (t, x) ∈ Q. Integration by parts (using Green’s theorem) with respectto x in the left-hand side integral of (2.46), we have

−∫

Q

e2s��sy�y dx dt =∫

Q

s∇(�e2s�)y(∇y) dx dt +∫

Q

s�e2s�|∇y|2 dx dt

−∫�

se2s��y

(�y

��

)d�. (2.48)

Here it is easy to see that ∇(�e2s�) = 2s�e2s��2∇ + e2s�∇� + �e2s��∇ and so we have the following threeinequalities:

2∫

Q

((s)3/2�(s)1/2�e2s�∇)y∇y dx dt

� − 6c4

∫Q�

e2s�(s)3�2|y|2 dx dt − 1

6

∫Q

e2s��s|∇y|2 dx dt

and ∫Q

(s∇�e2s�)y∇y dx dt

� − 3

2supx∈�

|∇�|∫

Q�

e2s�s|y|2 dx dt − 1

6

∫Q

e2s�∇�s|∇y|2 dx dt .

Further,∫Q

(s��e2s�∇)y∇y dx dt � − 3

2c4

∫Q�

e2s�s�2|y|2 dx dt − 1

6

∫Q

e2s��s|∇y|2 dx dt .


Then applying the above three inequalities to the relation (2.48), we get

−∫

Q

s�e2s�y�y dx dt � − c(�)

∫Q�

e2s�(s)3|y|2 dx dt + 1

2

∫Q

e2s��s|∇y|2 dx dt , (2.49)

for � > �0, s > s0. Here we used the readymade inequality −2 �(T /2)4 and the inequalities (2.13). Further thecomputation similar to (2.44) shows that

∫Q�

e2s�∣∣∣∣∫ T

t

a(r, t)y(r, x) dr

∣∣∣∣2 dx dt

�∫

Q�

e2s�(∫ t1

t0

e−2s�(r)|a(r, t)|2−3(r) dr

)(∫ t1

t0

e2s�(r)3(r)|y(r, x)|2 dr

)dx dt

�c‖a‖2L∞

∫Q�

e2s�(∫ t1

t0

e2s�(r)3(r)|y(r, x)|2 dr

)dx dt

�c‖a‖2L∞

∫Q�

e2s�3|y|2 dx dt . (2.50)

Consequently, with the help of the estimates (2.47)–(2.50), we obtain∫Q�0

e2s�s|∇y|2 dx dt �c(�, s)

(∫Q

e2s�|f |2 dx dt +∫

Q�

e2s�3|y|2 dx dt

). (2.51)

Eventually making use of the estimation (2.51), the inequality (2.45) can now be written as∫Q

e2s�s|∇y|2 dx dt +∫

Q

e2s�s33|y|2 dx dt

�c(�, s)

(∫Q

e2s�|f |2 dx dt +∫

Q�

e2s�3|y|2 dx dt

), (2.52)

for � > �0, s > s0. To complete the proof of this theorem, we have to estimate the integral

∫Q

e2s�(s)−1

(∣∣∣∣�y

�t

∣∣∣∣2 + |�y|2 +∣∣∣∣∫ T

t

a(r, t)y(r, x) dr

∣∣∣∣2)

dx dt .

By squaring the main equation of (2.2), multiplying it with e2s�(s)−1 and integrating on Q, we find that

∫Q

e2s�(s)−1(

�y

�t+ �y +

∫ T

t

a(r, t)y(r, x) dr

)2

dx dt �c

∫Q

e2s�|f |2 dx dt . (2.53)

Now it remains to obtain the estimation for the cross terms appearing in the left-hand side. First, we note that∫Q

e2s�(s)−1(

�y

�t

)�y dx

= −∫

Q

∇(

e2s�(s)−1 �y

�t

)∇y dx dt +

∫�

e2s�(s)−1(

�y

�t

)�y

��d�. (2.54)


Integrating by parts with respect to t (here we remember that � blows up to −∞ when t = 0 and t = T )

−∫

Q

∇(

e2s�(s)−1 �y

�t

)∇y dx dt

= −1

2

∫Q

e2s�(s)−1 �|∇y|2�t

dx dt −∫

Q

∇(e2s�(s)−1)�y

�t∇y dx dt

=∫

Q

e2s�(

−1 ��

�t− s−1

2−2 �

�t

)|∇y|2 dx dt −

∫Q

e2s�(2� − �(s)−1)�y

�t∇ · ∇y dx dt ,

then through the inequalities (2.13), one can easily see that∫Q

e2s�(

−1 ��

�t− s−1

2−2 �

�t

)|∇y|2 dx dt � − c

∫Q

e2s�s|∇y|2 dx dt ,

where the constant c depends only on T . By Cauchy’s inequality with � = 2, we get

−∫

Q

e2s�(2� − �(s)−1)�y

�t∇ · ∇y dx dt

� − 1

4

∫Q

e2s�(s)−1∣∣∣∣�y

�t

∣∣∣∣2 dx dt − c

∫Q

e2s�s�2|∇y|2 dx dt .

Therefore, by the substitution of the preceding estimations into (2.54), we immediately obtain

2∫

Q

e2s�(s)−1 �y

�t�y dx dt

� − 1

2

∫Q

e2s�(s)−1∣∣∣∣�y

�t

∣∣∣∣2 dx dt − c

∫Q

e2s�s�2|∇y|2 dx dt . (2.55)

Next we obtain the lower estimates for the product of the terms �y/�t and �y with the integral term. We notice, bythe definition of , that (t, x)�c > 0 for all (t, x) ∈ Q and e�� c < ∞ for all � > 0, � ∈ R and taking this intoaccount, we have

2∫

Q

e2s�(s)−1 �y

�t

(∫ T

t

a(r, t)y(r, x) dr

)dx dt

= 2∫

Q

e2s�s−1−2 �

�ty

(∫ T

t

a(r, t)y(r, x) dr

)dx dt

− 4∫

Q

e2s� ��

�t−1y

(∫ T

t

a(r, t)y(r, x) d�

)dx dt + 2

∫Q

e2s�(s)−1a(t, t)y2 dx dt

− 2∫

Q

e2s�(s)−1y

(∫ T

t

�a

�t(r, t)y(r, x) dr

)dx dt

= M1 + M2 + M3 + M4.

Choosing a parameter � such that � ∈ (0, 1), we obtain that for any s�s0,

M1 + M2 � − c

�

∫Q

e2s�s33|y|2 dx dt − �∫

Q

e2s�(s)−1∣∣∣∣∫ T

t

a(r, t)y(r, x) dr

∣∣∣∣2 dx dt


and

M4 � − c

∫Q

e2s�s33|y|2 dx dt −∫

Q

e2s�∣∣∣∣∫ T

t

�a


∣∣∣∣2 dx dt .

Since the estimation similar to (2.50), further yields∫Q

e2s�∣∣∣∣∫ T

t

�a


∣∣∣∣2 dx dt

�∫

Q

e2s�

(∫ t1

t0

e−2s�(r)

∣∣∣∣�a

�t(r, t)

∣∣∣∣2−3(r) dr

)(∫ t1

t0

e2s�(r)3(r)|y(r, x)|2 dr

)dx dt

�c

∥∥∥∥�a

�t

∥∥∥∥2

L∞

∫Q

e2s�3|y|2 dx dt .

Moreover we note that∫Q

e2s�(s)−1�y

(∫ T

t

a(r, t)y(r, x) dr

)dx dt

� − �∫

Q

e2s�(s)−1|�y|2 dx dt − c

�

∫Q

e2s�∣∣∣∣∫ T

t

a(r, t)y(r, x) dr

∣∣∣∣2 dx dt

and the last integral can further be bounded by

c

�‖a‖2

L∞

∫Q

e2s�s33|y|2 dx dt for any s�s0.

With the help of the preceding estimates for the cross terms, we can rewrite (2.53) as follows:∫Q

e2s�s−1−1

(∣∣∣∣�y

�t

∣∣∣∣2 + |�y|2 +∣∣∣∣∫ T

t

a(r, t)y(r, x) dr

∣∣∣∣2)

dx dt

�c(�)

(∫Q

e2s�|f |2 dx dt +∫

Q

e2s�s33|y|2 dx dt +∫

Q

e2s�s|∇y|2 dx dt

). (2.56)

Evidently, in view of the inequalities (2.52) and (2.56), one can conclude this theorem. �

Now we reduce the following observability estimate from Theorem 2.1 for the adjoint problem of (2.1) and it will bethe main ingredient of the proof of the controllability of the linear problem in the next section. This inequality enablesus to estimate the solutions in the entire domain by observing them in small subdomain only.

Corallary 1. Suppose all the assumptions of Theorem 2.1 are satisfied. Then there exist �0 > 0, s0 > 0 independent ofy and f such that for � > �0, s > s0, the following observability estimate holds:∫

�|y(0, x)|2 dx�c(�, s)

(∫Q

|f |2 dx dt +∫

Q�

e2s�3|y|2 dx dt

). (2.57)

Proof. First, we multiply Eq. (2.2) by y and integrate on �. Integrating by parts with respect to x and using Young’sinequality, we get

−1

2

d

dt|y|2

L2(�)+ |∇y|2

L2(�)� 1

2(2|y|2

L2(�)+ |f |2

L2(�)) + 1

2

∣∣∣∣∫ T

t

a(r, t)y(r, x) dr

∣∣∣∣2L2(�)

. (2.58)


It follows that

− d

dt(exp(2t)|y|2

L2(�))� exp(2t)

(|f |2

L2(�)+∣∣∣∣∫ T

t

a(r, t)y(r, x) dr

∣∣∣∣2L2(�)

)

for a.e. t ∈ (0, T ). Clearly one can have the following inequality:

|y(0, x)|2L2(�)

�e2T

(|y|2

L2(�)+ |f |2

L2(Q)+∣∣∣∣∫ T

t

a(r, t)y(r, x) dr

∣∣∣∣2L2(Q)

)

and so

|y(0, x)|2L2(�)

�c

(�(t)

∫�

e2s�s33|y|2 dx

+∫

Q

|f |2 dx +∫

Q

∣∣∣∣∫ T

t

a(r, t)y(r, x) dr

∣∣∣∣2 dx

)for s > s0,

where

�(t) = sup{e−2s�(x,t) : x ∈ �}�e��s/t (T −t), �� = 2e2� ,

and note that −3 �(T /2)6.Now we fix t1 and t2 such that 0 < t1 < t2 < T and integrating on the interval (t1, t2), we have

|y(0, x)|2L2(�)

∫ t2

t1

e−��s/t (T −t) dt �c

∫ t2

t1

(∫�

e2s�s33|y|2 dx +∫

Q

|f |2 dx

+∫

Q

∣∣∣∣∫ T

t

a(r, t)y(r, x) dr

∣∣∣∣2 dx

)dt ,

since inf t∈(t1,t2){e−��s/t (T −t)}�c > 0 for any s > s0 and therefore

|y(0, x)|2L2(�)

�c

(∫Q

e2s�s33|y|2 dx dt +∫

Q

|f |2 dx dt

+∫

Q

∣∣∣∣∫ T

t

a(r, t)y(r, x) dr

∣∣∣∣2 dx dt

). (2.59)

As a result, applying the estimate (2.50) for the last integral in (2.59) and combining the resultant estimate withTheorem 2.1, we arrive at the proof of this corollary. �

2.2. Controllability of linear equations

In this section, we obtain a solution to the global controllability of problem (2.1) as the limit of an approximationprocess, constructed with the aid of a family of appropriate optimal control problems for system (2.1). To derive theneeded estimates for the solutions of the optimal control problems we shall use Pontryagin’s maximum principle andthe Carleman estimate (2.4) derived for the backward adjoint equation of (2.1). Throughout this sequel we use thefunctions � and as defined in (2.3).

Theorem 2.2. Let � be a C2-open and bounded domain of Rn and suppose the kernel a(·, t) has support in (t0, t1),where 0 < t0 < t1 < T . Then there are � > �0 for �0 > 0, and s > s0 as defined in Theorem 2.1 such that for anyg ∈ L2(Q), z0 ∈ H 1

0 (�) there exist

(u, z) ∈ L2(Q) × H 1([0, T ]; L2(�)) ∩ L2(0, T ; H 10 (�) ∩ H 2(�))


which satisfies Eqs. (2.1) and the terminal condition

z(T , x) ≡ 0 a.e. x ∈ � (2.60)

and the control u satisfies the inequality∫Q

e−2s�−3|u|2 dx dt �c(�, s)

(∫�

|z0|2 dx +∫

Q

e−2s�−3|g|2 dx dt

). (2.61)

Proof. We start with the optimal control problem subject to (2.1) with

J (z, u) =∫

Q

e−2s�−3|u|2 dx dt + 1

�

∫�

|z(T , x)|2 dx → inf (2.62)

and this problem has a unique solution (z�, u�) for every � > 0. Next we shall show that this (z�, u�) converges (on asubsequence of {�}) to (z, u) and this will be proved to be a solution of the control problem (2.1). To this end we need toobtain the estimates for (z�, u�). This is achieved by Pontryagin’s maximum principle and it states that (z�, u�) satisfiesthe following necessary condition for optimality. Here the maximal condition on the control u� is described as

u� = 1�e2s�3y� a.e. in Q,

where y� ∈ H 1([0, T ]; L2(�)) ∩ L2(0, T ; H 10 (�) ∩ H 2(�)) is a solution of the following backward adjoint equations

of (2.1):⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

�y�

�t+ �y� −

∫ T

t

a(r, t)y�(r, x) dr = 0, ∀ (t, x) ∈ Q,

�y�

��= 0, ∀ (t, x) ∈ �,

y�(T , ·) = −1

�z�(T , ·), ∀ x ∈ �.

(2.63)

Now we multiply (2.63) by z� and Eq. (2.1) by y� (after replacing z by z� and control u by u�) and add those twoequations by integrating over Q. Then integrating by parts with respect to x, one can arrive at∫

Q�

e2s�3y2� dx dt + 1

�

∫�

z2� (T , x) dx = −

∫�

z0(x)y�(0, x) dx −∫

Q

gy� dx dt . (2.64)

Using Hölder’s inequality and Corollary 2.1, we estimate the integral∫� z0(x)y�(0, x) dx as∣∣∣∣∫

�z0(x)y�(0, x) dx

∣∣∣∣ �(∫

�|y�(0, x)|2 dx

)1/2(∫�

|z0(x)|2 dx

)1/2

�c(�, s)

(∫Q�

e2s�3|y�|2 dx dt

)1/2

|z0|L2(�)

and hence for every � > 0 we have∣∣∣∣∫�

z0(x)y�(0, x) dx

∣∣∣∣�c(�, s)

(∫Q�

e2s�3|y�|2 dx dt + 1

�

∫�

|z�(T , x)|2 dx

)1/2

|z0|L2(�). (2.65)


Again, using Hölder’s inequality and Theorem 2.1, we arrive at∣∣∣∣∫Q

gy� dx dt

∣∣∣∣ �∫

Q

|gy�| dx dt

�(∫

Q

e2s�3|y�|2 dx dt

)1/2(∫Q


)1/2

�c(�, s)

(∫Q�

e2s�3|y�|2 dx dt + 1

�

∫�

|z�(T , x)|2 dx

)1/2

‖e−s�−3/2g‖L2(Q). (2.66)

Substitution of the estimations (2.65)–(2.66) into (2.64) yields the following inequality:∫Q�

e2s�3|y�|2 dx dt + 1

�

∫�

|z�(T , x)|2 dx�c(�, s)(|z0|L2(�) + ‖e−s�−3/2g‖L2(Q))2

�c(�, s)(|z0|2L2(�)+ ‖e−s�−3/2g‖2

L2(Q)), (2.67)

where the constant c(�, s) is somehow greater than that defined in the previous inequality. So, from the definition ofthe control u� we have

J (z�, u) =∫

Q

e−2s�−3|u�|2 dx dt + 1

�

∫�

|z�(T , x)|2 dx

�c(�, s)

(∫�

|z0|2 dx +∫

Q


)(2.68)

for every � > 0 and the positive constant c(�, s) is independent of �. Clearly from the inequality (2.67), we see that∫�

|z�(T , x)|2 dx�c� (2.69)

for some positive constant c independent of �.From (2.68) we note that u� is bounded in L2(Q) and from (2.1), z� is bounded in L2(0, T ; H 1

0 (�) ∩ H 2(�)) ∩H 1([0, T ]; L2(�)). Hence we infer that there exist (u, z) ∈ L2(Q) × L2(0, T ; H 1

0 (�) ∩ H 2(�)) such that, on asubsequence of � (also denoted by {�}), as � → 0:

u� → u weakly in L2(Q),

z� → z weakly in L2(0, T ; H 10 (�) ∩ H 2(�)) (hence strongly in C([0, T ]; L2(�))).

Also one can obtain by squaring Eq. (2.1) (after replacing z and u, respectively, by z� and u�) and using standardtechnique (see [24])

‖z�(t)‖2H 1

0 (�)+∫

Q

|�z�|2 dx dt +∫

Q

∣∣∣∣�z�

�t

∣∣∣∣2 dx dt +∫

Q

∣∣∣∣∫ t

0a(t, r)z�(r, x) dr

∣∣∣∣2 dx dt

�c(‖z0‖2

H 10 (�)

+ |z0|2L2(�)+ ‖u�‖2

L2(Q�)+ ‖g‖2

L2(Q)

)�c(�, s)(‖z0‖2

H 10 (�)

+ ‖e−s�−3/2g‖2L2(Q)

), (2.70)

where we have used the inequality (2.68) (recall that es�k �c < ∞ for all � > 0, k ∈ R) and the Sobolev imbeddingtheorem.

On addition to (2.70), by the above two finite limits of (u�, z�), we have the following obvious convergences as� → 0:

z� → z strongly in C([0, T ]; L2(�)),


�z� → �z weakly in L2(Q),

�z�

�t→ �z

�tweakly in H 1([0, T ]; L2(�)),

z�(T , x) → z(T , x) weakly in L2(�).

So, letting � → 0 in (2.1) where u, z are replaced, respectively, by u�, z� we obtain that the pair (u, z) satisfy Eqs.(2.1). Consequently, by Fatou’s Lemma and the inequality (2.69), we have∫

�|z(T , x)|2 dx� lim

�→0inf∫�

|z�(T , x)|2 dx = 0.

This would imply that

z(T , x) ≡ 0 a.e. x ∈ �.

Moreover, letting � → 0 in (2.68) we obtain the estimate (2.61) for the control u. This concludes the proof. �

3. Nonlinear diffusion equations

In this section, we study the solvability of problems of exact controllability for nonlinear parabolic integro-differentialequation (described in Section 1) with mixed boundary conditions with the nonlinear term f (t, ·, ·, ·) increasing as|y| → +∞ not faster than a associated linear function.

3.1. Formulation of the problem

Consider the nonlinear parabolic integro-differential control system of the form

G(y) = �y

�t−

n∑i,j=1

�

�xi

(aij (t, x)

�y

�xj

)+

n∑i=1

bi(t, x)�y

�xi

+ c(t, x)y

+ f

(t, x, y(t, x),

∫ t

0K(t, s, x, y(s, x)) ds

)= u + g in Q, (3.1)

u ∈ U(�),

(�y

��+ l(t, x)y

)∣∣∣∣�

= 0, y(0, x) = v0(x), y(T , x) = v1(x) in �, (3.2)

where � ⊂ Rn is a connected bounded domain with boundary �� of class C2, �(x), the external normal to �� and0 < T < ∞ is an arbitrary moment of time. The functions g, v0 are given and u(t, x) is a control in the space

U(�) = {u(t, x) ∈ L2(Q)|supp u ⊂ [0, T ] × �}.Here � is an arbitrarily fixed subdomain of �. For simplicity, let us follow the notation:

z =∫ t

0K(t, s, x, y(s, x)) ds.

In order to prove the required results, we need to assume the following conditions:

(i) The functions aij (t, x) ∈ C1,2(Q), aij (t, x) = aji(t, x), bi(t, x) ∈ C0,1(Q), and c(t, x) ∈ L∞(Q), wherei, j = 1, 2, . . . , n.

(ii) The function a(t, x) satisfies the uniform ellipticity condition, more precisely, there exists � > 0 such that

a(t, x, �, �) =n∑

i,j=1

aij (t, x)�i�j ��|�|2 ∀� ∈ Rn, (t, x) ∈ Q.


(iii) The function l(t, x) > 0 and l(t, x) ∈ C1,1(�).(iv) The functions K(t, s, x, y), f (t, x, y, z) satisfy K(t, s, x, 0) = 0 ∀(t, s, x) ∈ Q, 0�s� t �T and f (t, x, 0, 0) =

0 ∀(t, x) ∈ Q, respectively, and also satisfy the Lipschitz conditions

|K(t, s, x, y1) − K(t, s, x, y2)|�M1|y1 − y2| ∀(t, s, x, y) ∈ Q × R

and

|f (t, x, y1, z1) − f (t, x, y2, z2)|�M2(|y1 − y2| + |z1 − z2|) ∀(t, x, y, z) ∈ Q × R × R.

For brevity, let us take

� =n∑

i,j=1

‖aij (t, x)‖C1,2(Q) +n∑

i=1

‖bi(t, x)‖C0,1(Q) + ‖c(t, x)‖L∞(Q),

and

� =n∑

i,j=1

‖aij (t, x)‖C1,2(Q).

Consider the linear boundary value problem

Lz = �z

�t−

n∑i,j=1

�

�xi

(aij (t, x)

�z

�xj

)+

n∑i=1

bi(t, x)�z

�xi

+ c(t, x)z = g in Q, (3.3)

(�z

��+ l(t, x)z

)∣∣∣∣�

= 0, z(0, x) = z0(x). (3.4)

Then we state the following estimate for the linear boundary value problem (3.3)–(3.4) without proof, which is vital tothe criterion of controllability.

Theorem 3.1 (Carleman estimate). Assume that (i)–(iii) be fulfilled and functions , �, and � be defined in (2.3).Then there exists a number �0 > 0 such that for any arbitrary ��0 there exists s0 such that for each s�s0 the solutionsof the problem (3.3)–(3.4) satisfy the following estimate:∫

Q

(s)−1

⎛⎝∣∣∣∣�z(t, x)

�t

∣∣∣∣2 +n∑

i,j=1

∣∣∣∣�2z(t, x)

�xi�xj

∣∣∣∣2⎞⎠ (e2s� + e2s�) dx dt

+∫

Q

(s

n∑i=1

∣∣∣∣�z(t, x)

�xi

∣∣∣∣2 + (s)3|z(t, x)|2)

(e2s� + e2s�) dx dt

�c(�)

(∫Q

(e2s� + e2s�)|g(t, x)|2 dx dt +∫

Q�

(e2s� + e2s�)(s)3|z(t, x)|2 dx dt

), (3.5)

where Q� = � × (0, T ) and constant c(�) > 0 depends continuously on � and constant �0 depends continuouslyon �. �

For the proof of Theorem 3.1, the interested reader can refer [8, Lemma 1.2] (or) [9, Chapter 1, Lemma 1.2].To establish our results, we have to introduce the following function spaces:

Y (Q) ={y(t, x) : y ∈ L2((0, T ); W 2

2 (�)),�y

�t∈ L2(Q)

},

X(Q) = {y(t, x) : es�y ∈ L2(Q)}, Z(Q) ={y(t, x) : es�y

(T − t)3/2∈ L2(Q)

},


E(Q) ={y(t, x) ∈ Z(Q) : |∇y|es�

√(T − t)

,√

(T − t)

(∣∣∣∣�y

�t

∣∣∣∣+ |�y|)

es� ∈ L2(Q)

}equipped with the norms

‖y‖2Y (Q) = ‖y‖2

L2(0,T ;W 22 (�))

+∥∥∥∥�y

�t

∥∥∥∥2

L2(Q)

,

‖y‖X(Q) = ‖es�y‖L2(Q), ‖y‖Z(Q) =∥∥∥∥ es�y

(T − t)3/2

∥∥∥∥L2(Q)

,

‖y‖E(Q) =(

‖y‖2Z(Q) +

∥∥∥∥ |∇y|es�√

(T − t)

∥∥∥∥2

L2(Q)

+∥∥∥∥√(T − t)

(∣∣∣∣�y

�t

∣∣∣∣+ |�y|)

es�∥∥∥∥2

L2(Q)

)1/2

.

Since the function �(t, x, �) is defined as follows:

�(t, x, �) = e2� − e�

(T − t)l(t), (3.6)

where ��0 and = supx∈C(�)

|(x)|. The function (x) and the parameter � are as defined in Lemma 2.1 andTheorem 2.1. We also assume that l(t) is a fixed function which satisfies the following conditions:

l(t) ∈ C1[0, T ], l(t) = t, ∀t ∈(

3T

4, T

], l(t) > 0, ∀t ∈ [0, T ]. (3.7)

3.2. Controllability results

Consider the problem of exact controllability for linear parabolic equations

Ly = �y

�t−

n∑i,j=1

�

�xi

(aij (t, x)

�y

�xj

)+

n∑i=1

bi(t, x)�y

�xi

+ c(t, x)y = u + g in Q, (3.8)

u ∈ U(�),

(�y

��+ l(t, x)y

)∣∣∣∣�

= 0, y(0, x) = v0(x), y(T , x) = v1(x) in �. (3.9)

Then the solution of the problem (3.8)–(3.9) satisfies the following estimate. The proofs are established with the helpof the Carleman estimate stated in (3.5).

Theorem 3.2. Let ��0 and v0 ∈ W 12 (�), v1 ≡ 0, and let the assumptions (i)–(iii) be fulfilled. Then there exists

a constant s0 depending on � such that if g ∈ X(Q) with s�s0, the problem (3.8)–(3.9) has a solution (y, u) ∈(Y (Q) ∩ Z(Q)) × (U(�) ∩ X(Q)) which satisfies the following estimate:

‖(y, u)‖(Y (Q)∩Z(Q))×(U(�)∩X(Q)) �c(�, s)(‖v0‖W 1

2 (�) + ‖g‖X(Q)

). (3.10)

The following theorem illustrates some regularity of the solution of the problem (3.8)–(3.9) via Theorem 3.1.

Theorem 3.3. Let ��0 and v0 ∈ W 12 (�), v1 ≡ 0, and let the assumptions (i)–(iii) be fulfilled. Then there exists a

constant s0 depending on � such that if g ∈ X(Q), with s�s0, then the problem (3.8)–(3.9) has a solution (y, u) ∈E(Q) × (U(�) ∩ X(Q)) which satisfies the following estimate:

‖(y, u)‖E(Q)×(U(�)∩X(Q)) �c(�, s)(‖v0‖W 1

2 (�) + ‖g‖X(Q)

). (3.11)


The proof of Theorems 3.2 and 3.3, respectively, can be found in [9, Chapter 1, Theorems 2.1 and 2.2]. Now weprove the main result of this section.

Theorem 3.4. Let v0 ∈ W 12 (�), v1 ≡ 0 and let the assumptions (i)–(iv) be fulfilled. Then there exists a constant

�0 > 0 such that for ��0, there exists a constant s0 with s�s0 such that if g ∈ X(Q), then the problem (3.1)–(3.2)has a solution (y, u) ∈ Y (Q) × U(�).

Proof. Let us consider the following family of problems of exact controllability:

G�(y) = Ly + f�(t, x, y, z) − f�(t, x, 0, 0) = u + g in Q, u ∈ U(�), (3.12)(�y

��+ l(t, x)y

)∣∣∣∣�

= 0, y(0, x) = v0(x), y(T , x) = 0 in �, (3.13)

where the symbol f� stands for the averaging of f with smooth kernel �. More precisely,

f�(t, x, y, z) = 1

�2

∫R

∫R

�

(1

�(|� − y|, |� − z|)

)f (t, x, �, �) d� d�,

�(p, q)�0 ∀p, q ∈ R, and �(p, q) = �(|p|, |q|); here the support of � belongs to the unit ball, that is,

supp(�) ⊂ {(p, q)||p|�1, |q|�1},∫

R

∫R

� dp dq = 1

and the operator L is as defined in (3.8). We note that

(f�(t, x, y, z) − f�(t, x, 0, 0))|y=z=0 = 0 ∀(t, x) ∈ Q (3.14)

and

|f�(t, x, �1, �1) − f�(t, x, �2, �2)|

=∣∣∣∣ 1

�2

∫R

∫R

�

(1

�(|� − �1|, |� − �1|)

)f (t, x, �, �) d� d�

− 1

�2

∫R

∫R

�

(1

�(|� − �2|, |� − �2|)

)f (t, x, �, �) d� d�

∣∣∣∣� 1

�2

∣∣∣∣∫R

∫R

�

(1

�(|�|, |�|)

)f (t, x, � + �1, � + �1) d� d�

−∫

R

∫R

�

(1

�(|�|, |�|)

)f (t, x, � + �2, � + �2) d� d�

∣∣∣∣� 1

�2

∫R

∫R

�

(1

�(|�|, |�|)

) ∣∣f (t, x, � + �1, � + �1) − f (t, x, � + �2, � + �2)∣∣ d� d�

� M2

�2

∫R

∫R

�

(1

�(|�|, |�|)

)d� d�(|�1 − �2| + |�1 − �2|) ∀(t, x) ∈ Q. (3.15)

From Eqs. (3.14) and (3.15), we obtain

f�(t, x, �, �) − f�(t, x, 0, 0) = f�(t, x, �, �)(� + �)

and

|f�(t, x, �, �)|�M2 ∀(t, x, �, �) ∈ Q × R × R, (3.16)


where M2 is a constant defined in assumption (iv). It follows from (3.16) that for the linear parabolic operator L�(y)p=Lp + f�(t, x, y, z)p, the parameter �(y) defined as

�(y) =n∑

i,j=1

‖aij (t, x)‖C1,2(Q) +n∑

i=1

‖bi(t, x)‖C0,1(Q) + ‖c(t, x) + f�(t, x, y, z)‖L∞(Q),

for every y ∈ L2(Q) satisfies the inequality

�(y)�c6, (3.17)

where c6 is a constant independent of y and �.Therefore by Theorem 3.3 there exists a constant �0 > 0 satisfying ��0 and there exists s0 with s�s0(�) such

that for any linear operator L�(y), the problem of exact controllability (3.8), (3.9) has a solution in the space E(Q) ×(U(�) ∩ X(Q)) for all the initial data (v0, g) ∈ W 1

2 (�) × X(Q). Further this solution satisfies (3.11) with c(�, s)

independent of y ∈ L2(Q) and � ∈ (0, 1). Also it should be noted that the functions , , �, � and �, respectively,defined in (2.3) and (3.6) do not depend on y since the parameter L�(y) has no influence on the terms standing in themain part.

Let us introduce the mappings (�): y → p� and (�)1 : y → (p�, u�) as follows: for y ∈ L2(Q), the pair (p�, u�) is

defined as a solution of the extremal problem

J (p, u) =∫

Q

e2s�(t,x,�)

(T − t)3p2 dx dt +

∫Q

e2s�(t,x,�)u2 dx dt → inf ,

L�(y)p = u + g in Q, u ∈ U(�), (3.18)(�p

��+ l(t, x)p

)∣∣∣∣�

= 0, p(0, x) = v0(x), p(T , x) = 0 in �. (3.19)

Clearly by Theorem 3.2 for y ∈ L2(Q), there exists a unique solution (p�, u�) ∈ (Y (Q)∩Z(Q))× (U(�)∩X(Q))

of the problem (3.18)–(3.19). So, the mappings (�) and (�)1 are well defined on the whole space L2(Q).

Next we use the Schauder fixed point theorem with L2(Q) topology to show that (�) has a fixed point. Denote byBr a ball in L2(Q) with center at zero and having radius r . By Theorem 3.2 and (3.17) for all sufficiently large r , weobtain �(Br) ⊂ Br where r is independent of �.

Further if S is a bounded set in L2(Q), then, by Theorem 3.2, the set �(S) is bounded in Y (Q). Since imbeddingY (Q) ⊂ L2(Q) is compact and hence the mapping � is a compact mapping. To attain this end, we only need to provethat (�) ∈ C(L2(Q), L2(Q)) is a continuous mapping.

Suppose on the contrary assume that (�) is not a continuous mapping. Let � > 0 be fixed. Then there exists a functiony ∈ L2(Q) and sequence {(yi, pi , ui )} such that

yi → y in L2(Q),

(�)(yi) = pi → p weakly in Y (Q) ∩ Z(Q),

ui → u weakly in U(�) ∩ X(Q), (3.20)

and

(�)1 (y) = (p, u) �= (p, u), p ∈ Z(Q), (3.21)

the triple (yi, pi , ui ) satisfying (3.19) and

J (p, u) < �0 < J(pi, ui) ∀i ∈ Z+. (3.22)


By Eq. (3.20) and (3.21) and the assumption (iv), we have

p

{f�

(t, x, yi,

∫ t

0K(t, s, x, yi(s)) ds

)− f�

(t, x, y,

∫ t

0K(t, s, x, y(s)) ds

)}→ 0

in Z(Q) as i → ∞. (3.23)

By (3.23) and Theorem 3.2, there exists a subsequence {(�i , qi)}∞i=1 ⊂ (Y (Q) ∩ Z(Q)) × (U(�) ∩ X(Q)) such that

L�i + f�

(t, x, yi,

∫ t


)�i = p

{f

(t, x, yi,

∫ t


)

− f

(t, x, y,

∫ t

0K(t, s, x, y(s)) ds

)}+ qi in Q, (3.24)(

��i

��+ l(t, x)�i

)∣∣∣∣�

= 0, �i (0, x) = 0, �i (T , x) = 0 in �, (3.25)

‖�i‖Y (Q)∩Z(Q) + ‖qi‖X(Q) → 0 as i → +∞. (3.26)

We set

pi = p − �i , ui = u − qi . (3.27)

By Eqs. (3.21), (3.24) and (3.25), the following holds:

Lpi + f�

(t, x, yi,

∫ t


)pi = g + ui in Q, ui ∈ U(�), (3.28)

(�pi

��+ l(t, x)pi

)∣∣∣∣�

= 0, pi(0, x) = v0(x), pi(T , x) = 0 in �. (3.29)

But, from (3.26), clearly we have

J (pi , ui) → J (p, u). (3.30)

Eqs. (3.22), (3.28)–(3.30) show that the pair (pi , ui ) is an admissible solution of the extremal problem (3.18)–(3.19).So, by the definition of the mapping �

1, we have the following inequality:

J (pi , ui)�J (pi , ui). (3.31)

Combining (3.30) and (3.31), we arrive at a contradiction to (3.21). This proves the continuity of �. Then, bySchauder’s fixed point theorem, there exists a fixed point y� of the mapping �(y�) such that �(y�) = y� and

‖ �1(y�)‖Y (Q)×U(�) �c7, (3.32)

where c7 is a constant independent of �. Thus the pair �1(y�)=(y�, u�) is a solution of the exact controllability problem

Ly� + f�

(t, x, y�,

∫ t

0K(t, s, x, y�(s)) ds

)− f�(t, x, 0, 0) = u� + g in Q, (3.33)

u� ∈ U(�),

(�y�

��+ l(t, x)y�

)∣∣∣∣�

= 0, y�(0, x) = v0(x), y�(T , x) = 0 in �. (3.34)

Then, by (3.32), there exists a subsequence (again denoted by {�}) (y�, u�) such that

(y�, u�) → (y, u) weakly in Y (Q) × L2(Q)

and

u� → u weakly in L2((0, T ) × �).


Hence passing to the weak limit (remembering the assumption (iv)) in (3.33)–(3.34), we arrive at the solution of theproblem (3.1)–(3.2) satisfying y(T , x) = 0. This concludes the proof of main theorem. �

Acknowledgments

The work of the first and second authors is supported by the NBHM, Department of Atomic Energy, India (Grantno: 48/3/2006/R&D-II/8336). The work of the third author is supported by the Army Research Office, Probability andStatistics Program.

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