EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 12, No. 3, 2019, 870-892
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

Simultaneous null controllability for two stroke
nonlinear systems: Application to the sentinel of
detection in population dynamics model with

incomplete data

Cédric Kpèbbèwèrè Somé1, Somdouda Sawadogo2,∗

1 Département de Mathématiques, Sciences Exactes et Appliquées,
Université Joseph Ki-Zerbo, Ouagadougou, Burkina Faso
2 Département de Mathématiques, Institut Des Sciences, Ouagadougou, Burkina Faso

Abstract. This paper deals with the simultaneous null controllability for some nonlinear two
stroke systems. We shall solve this problem by transforming the simultaneous null controllability
of uncoupled initial systems into a null controllability of a coupled system via a change of vari-
ables. This last problem is solved thanks to a global Carleman inequality, appropriates estimates
adapted to the system and via some fixed point theorems. The obtained results are used to build
a simultaneous sentinel of detection in a population dynamics model with incomplete data.

2010 Mathematics Subject Classifications: 49J20, 93B05, 92D25, 35Q92,35Q93

Key Words and Phrases: Population dynamics, Null controllability, Carleman inequality,
Simultaneous sentinel

1. Introduction

Let Ω be a bounded open subset of RN , N ∈ {1, 2, 3} with boundary Γ of class
C2. Let ω ⊂ Ω be an open nonempty subset. For a time T > 0 and the common life
expectancy A > 0 of species, we set U = (0, T ) × (0, A), Q = U × Ω, Qω = U × ω, QT =
(0,T) × Ω, QA = (0,A) × Ω,Σ = U × Γ, ΣT = (0, T ) × Γ and we consider the following

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v12i3.3459

Email addresses: cedrickpebsom@yahoo.fr (C. K. Somé), sawasom@yahoo.fr (S. Sawadogo)

http://www.ejpam.com 870 c© 2019 EJPAM All rights reserved.



C. K. Somé, S. Sawadogo / Eur. J. Pure Appl. Math, 12 (3) (2019), 870-892 871

nonlinear two stroke systems :

−∂q1

∂t
− ∂q1

∂a
−∆q1 + µ1q1 = β1F

(∫ A
0 β1q1da

)
q1(t, 0, x)

+ h+ wχω in Q,

−∂q2

∂t
− ∂q2

∂a
−∆q2 + µ2q2 = β2G

(∫ A
0 β2q2da

)
q2(t, 0, x)

+ h+ wχω in Q,
q1(T, a, x) = q2(T, a, x) = 0 in QA,
q1(t, A, x) = q2(t, A, x) = 0 in QT ,

q1 = q2 = 0 on Σ,

(1)

for some functions F,G defined on R. We assume that

(H0) the functions F,G belong to L∞(R) and F (0) = G(0) = 0.

The simultaneous null controllability problem can be stated as follows : Given h ∈
L2(Q) find w ∈ L2(Qω) such that the solution of (1) satisfies

q1(0, a, x) = q2(0, a, x) = 0 a.e (a, x) in QA. (2)

The null controllability problem for one two stroke system with one and only one control
is well understood: it has been studied by several authors using different methods. We
refer to B. Ainseba and M. Langlais [2], B. Ainseba and S. Anita [3]. We also refer to S.
Sawadogo [9], O. Traoré [12], Y. Simporé and O. Traoré [10] and their bibliography for
other related controllability problems. As far as we know, there is no results on simulta-
neous null controllability for nonlinear two stroke systems. In this paper we focus on the
previous problem in order to applicate it to build a simultaneous sentinel of detection in
population dynamics problem with incomplete data.

The remainder of this paper is organized as follows : In order to well pose our problem,
in section 2 we make some assumptions, transform the system (1) into an equivalent
cascade problem and we state the main result of this paper. In section 3, we state first
some Carleman’s inequalities that we had established in [11]. Afterwards, we study the
controllability for a linear intermediate problem and for another nonlinear. The section 4
is devoted to the proof of the main result and in the last section we use the result obtained
in section 4 to build a simultaneous sentinel.

2. Assumptions and main result

For the sequel, the following assumptions hold:

(H1)

 (µi, ∇µi) ∈ (L∞(Q))N+1 for all i ∈ {1; 2}, N ∈ {1, 2, 3},
µi ≥ 0 in Q for all i ∈ {1; 2},
µ1 6= µ2 in Qω.



C. K. Somé, S. Sawadogo / Eur. J. Pure Appl. Math, 12 (3) (2019), 870-892 872

(H2)

{
βi ∈ C2(Q) for all i ∈ {1; 2},
βi ≥ 0 in Q for all i ∈ {1; 2}.

(H3) There exists positive constants non null a0 and a1 with a0 < a1 < A such that
for each i ∈ {1; 2}, βi(t, a, x) = 0 a.e (t, a, x) ∈ (0, T )× ([0, a0] ∪ [a1, A])× Ω.

Under the assumptions (H0) − (H3), for all h ∈ L2(Q), w ∈ L2(Qω) the system (1) ad-
mits an unique solution (q1, q2) in L2(U,H1

0 (Ω))2 such that ∂qi
∂t + ∂qi

∂a ∈ L2(U ;H−1(Ω))
where H−1(Ω) is the dual of the Hilbert space H1

0 (Ω). Moreover (q1, q2) belong to
C((0, T );L2(QA)) ∩ C((0, A);L2(QT )) ∩ L2(U,H1

0 (Ω))2 (see Lemma 0 in [5]).

Remark 1. Assume that (H1) holds and set

p1 = q1 + q2 ; p2 = q1 − q2. (3)

Thus, the condition (2) is equivalent to p1(0, a, x) = p2(0, a, x) = 0 a.e (a, x) in QA. The
following changes are required :

µ̂1 = 1
2(µ1 + µ2) , µ̂2 = 1

2(µ1 − µ2) , f = 2h , k = 2w,

β̂1(p1, p2) = 1
2

[
β1F

(
1
2

∫ A
0 β1(p1 + p2)da

)
+ β2G

(
1
2

∫ A
0 β2(p1 − p2)da

)]
,

β̂2(p1, p2) = 1
2

[
β1F

(
1
2

∫ A
0 β1(p1 + p2)da

)
− β2G

(
1
2

∫ A
0 β2(p1 − p2)da

)]
.

Then, the null controllability problem (1)-(2) is equivalent to the problem : for any
µ̂1, µ̂2 ∈ L∞(Q) and for f ∈ L2(Q) find a control

k ∈ L2(Qω) (4)

such that the pair p = (p1, p2) solution of the system

−∂p1

∂t
− ∂p1

∂a
−∆p1 + µ̂1p1 + µ̂2p2 = β̂1(p)p1(t, 0, x)

+ β̂2(p)p2(t, 0, x)
+ f + kχω in Q ,

−∂p2

∂t
− ∂p2

∂a
−∆p2 + µ̂1p2 + µ̂2p1 = β̂2(p)p1(t, 0, x)

+ β̂1(p)p2(t, 0, x) in Q ,
p1 = p2 = 0 on

∑
,

p1(T, a, x) = p2(T, a, x) = 0 in QA,
p1(t, A, x) = p2(t, A, x) = 0 in QT ,

(5)

satisfies
p1(0, a, x) = p2(0, a, x) = 0 in QA. (6)

Notice that system (5) admits an unique solution (p1, p2) in
(
C((0, T );L2(QA))∩C((0, A);L2(QT ))∩

L2(U,H1
0 (Ω))

)2
for each control k verifying (4). The main goal of this paper is to prove

the following result :



C. K. Somé, S. Sawadogo / Eur. J. Pure Appl. Math, 12 (3) (2019), 870-892 873

Theorem 1. Let Ω be an open subset of RN with boundary Γ of class C2 and ω be
a non empty subset of Ω. Assume that the hypothesis (H0) − (H3) hold. There exists
a positive real function θ ( θ is defined by (13)) such that for any function f ∈ L2 (Q)
with θf ∈ L2 (Q), there exists an unique control k̃, of minimal norm in L2(Qω) such that(
k̃, p̃1, p̃2

)
is solution of the simultaneous null controllability problem (5)-(6). Moreover,

the control k̃ is given by
k̃ = η̃1χω (7)

and verifies
‖k̃‖L2(Qω) ≤ C

(
‖θf‖L2(Q) + ‖f‖L2(Q)

)
(8)

where η̃ = (η̃1, η̃2) satisfies

∂η̃1

∂t
+
∂η̃1

∂a
−∆η̃1 + µ̂1η̃1 + µ̂2η̃2 = 0 in Q,

∂η̃2

∂t
+
∂η̃2

∂a
−∆η̃2 + µ̂1η̃2 + µ̂2η̃1 = 0 in Q,

η̃1 = η̃2 = 0 on Σ,

η̃1(t, 0, x) =

∫ A
0

(
β̂1(p)η̃1 + β̂2(p)η̃2

)
da in QT ,

η̃2(t, 0, x) =

∫ A
0

(
β̂2(p)η̃1 + β̂1(p)η̃2

)
da in QT .

(9)

with p̃ = (p̃1, p̃2).

3. Null controllability result for some coupled models

Before tackling the controllability problem, we will state the following results.

3.1. Global Carleman’s inequality and observability inequality result

For any positive parameters λ and τ , we define the positive functions:

α(t, a, x) = τ
e

4
3λ‖ψ‖∞ − eλψ(x)

at (T − t)
and ϕ(t, a, x) =

eλψ(x)

at (T − t)
, ∀ (t, a, x) ∈ Q.

Remark 2. As a reminder (see [4]) the function ψ ∈ C2(Ω) is such that :

∀x ∈ Ω ;ψ(x) > 0 ; ∀x ∈ Γ, ψ(x) = 0 and ∀x ∈ Ω \ ω0 , ∇ψ(x) 6= 0

where ω0 is an open set such that ω0 ⊂ ω ⊂ Ω. In the sequel :

• C represent different positive constants,

• we will use the following notations :

V =

{
ρ ∈ C∞

(
Q
)

such that ρ|Σ = 0

}
; W = V × V ,



C. K. Somé, S. Sawadogo / Eur. J. Pure Appl. Math, 12 (3) (2019), 870-892 874

Lρ = −∂ρ
∂t
− ∂ρ

∂a
−∆ρ ; L∗ρ =

∂ρ

∂t
+
∂ρ

∂a
−∆ρ

M(ρ1, ρ2) = L∗ρ1 + µ̂1ρ1 + µ̂2ρ2 ; N(ρ1, ρ2) = L∗ρ2 + µ̂1ρ2 + µ̂2ρ1.

‖ µ̂1, µ̂2 ‖2∞=‖ µ̂1 ‖2∞ + ‖ µ̂2 ‖2∞ and dQ = dtdadx

Theorem 2. [11] There exists λ0 > 0 , τ0 > 0 and a positive constant C such that for
all λ ≥ λ0 , τ ≥ τ0 and for all s ≥ −3 , the inequality∫

Q

(
1

λ

∣∣∣∣∂ρ∂t +
∂ρ

∂a

∣∣∣∣2 +
1

λ
|∆ρ|2 + λτ2ϕ2 |∇ρ|2 + λ4τ4ϕ4 |ρ|2

)
ϕ2s−1e−2αdQ

≤ C

(
τ

∫
Q

∣∣∣∣∂ρ∂t +
∂ρ

∂a
±∆ρ

∣∣∣∣2 ϕ2se−2αdQ + λ4τ4

∫ T

0

∫ A

0

∫
ω
|ρ|2 ϕ2s+3e−2αdQ

)
(10)

holds for any function ρ ∈ V such that the member on the right hand side of the inequality
(10) is finite.

Lemma 1. [11] Let C be the constant given by the theorem 2. Assume that for λ ≥
λ0 , τ ≥ 1 and s ≥ −3, there exists a constant b0 > 0 and a set ωb such that

ωb ⊂ ω and |µ̂2| ≥ b0 in (0 ; T )× (0 ; A)× ωb. (11)

Then, for all r ∈ [0 ; 2[, there exists a constant C = C(A, T, ‖ µ̂1 , µ̂2 ‖∞, b0, r) such that
for all ρ = (ρ1, ρ2) ∈ W, we have :∫ T

0

∫ A

0

∫
ω′

(
|ρ1|2 + |ρ2|2

)
e−2αdQ ≤ C

(∫
Q

[
|M(ρ)|2 + |N(ρ)|2

]
ϕ2se−2αdQ

+

∫
Qω

|ρ1|2 e−rαdQ

)
(12)

with ω′ ⊂ ωb.
Setting

θ = eα and δ = θ
r
2−1, (13)

we have the following result

Lemma 2. [11] Under the hypothesis of the lemma 1, for all ρ = (ρ1, ρ2) ∈ W, there
exists a positive constant C = C(A, T, ‖ aµ , bµ ‖∞, c0, r) such that∫

Q

1

θ2

(
|ρ1|2 + |ρ2|2

)
dQ ≤ C

(∫
Q

(
|M(ρ)|2 + |N(ρ)|2

)
dQ +

∫
Qω

δ2 |ρ1|2 dQ

)
. (14)

At last, we deduct the following result.

Proposition 1. [11] Under the hypothesis of the lemma 2 , there exists a positive constant
C such that for all ρ = (ρ1 , ρ2) ∈ W, we have∫ T

0

∫
Ω

(
|ρ1(t, 0, x)|2 + |ρ2(t, 0, x)|2

)
dxdt+

∫ A

0

∫
Ω

(
|ρ1(0, a, x)|2 + |ρ2(0, a, x)|2

)
dxda

≤ C

(∫
Q

(
|M(ρ)|2 + |N(ρ)|2

)
dQ +

∫
Qω

δ2 |ρ1|2 dQ

)
(15)



C. K. Somé, S. Sawadogo / Eur. J. Pure Appl. Math, 12 (3) (2019), 870-892 875

3.2. Study of the linear case :

In this paragraph, we study the following problem : For given functions µ̃1, µ̃2, b1, b2 ∈
L2(QT ), β̃1, β̃2 ∈ C2(Q) and f ∈ L2(Q) find v ∈ L2(Qω) such that the solution (z1, z2) of
: 

−∂z1

∂t
− ∂z1

∂a
−∆z1 + µ̃1z1 + µ̃2z2 = G1(t, a, x)z1(t, 0, x) + f + vχω

+G2(t, a, x)z2(t, 0, x) in Q

−∂z2

∂t
− ∂z2

∂a
−∆z2 + µ̃1z2 + µ̃2z1 = G2(t, a, x)z1(t, 0, x)

+G1(t, a, x)z2(t, 0, x) in Q
zi = 0 on Σ, i = 1, 2

zi(T, a, x) = 0 in QA, i = 1, 2
zi(t, A, x) = 0 in QT , i = 1, 2

(16)

verifies
zi(0, a, x) = 0 in QA, i = 1, 2. (17)

where,

G1(t, a, x) = β̃1(t, a, x)b1(t, x) + β̃2(t, a, x)b2(t, x)

G2(t, a, x) = β̃1(t, a, x)b1(t, x)− β̃2(t, a, x)b2(t, x)

and for all i ∈ {1, 2}, µ̃i verifies (H1), β̃i satisfies (H2)− (H3).
We can state the following result:

Theorem 3. Suppose that assumptions (H1) − (H3) hold and b1, b2 ∈ L2(QT ). For any
function f ∈ L2(Q) such that θf ∈ L2(Q), there exists a control ṽ in L2(Qω) such that
(ṽ, z̃1, z̃2) is solution of simultaneous null controllability problem (16)-(17). Moreover,
(ṽ, z̃1, z̃2) verifies

ṽ = ũ1χω (18)

‖z̃1‖L2(U ;H1(Ω) ≤ C
(
‖θf‖L2(Q) + ‖f‖L2(Q)

)
(19)

‖z̃2‖L2(U ;H1(Ω) ≤ C
(
‖θf‖L2(Q) + ‖f‖L2(Q)

)
(20)

where ũ = (ũ1, ũ2) satisfies

∂ũ1

∂t
+
∂ũ1

∂a
−∆ũ1 + µ̃1ũ1 + µ̃2ũ2 = 0 in Q,

∂ũ2

∂t
+
∂ũ2

∂a
−∆ũ2 + µ̃1ũ2 + µ̃2ũ1 = 0 in Q,

ũ1(0, a, x) = ũ2(0, a, x) = 0 in QA,
ũ1 = ũ2 = 0 on Σ,

ũ1(t, 0, x) = Υ1(ũ) in QT ,
ũ2(t, 0, x) = Υ2(ũ) in QT .

(21)



C. K. Somé, S. Sawadogo / Eur. J. Pure Appl. Math, 12 (3) (2019), 870-892 876

where

Υ1(ũ) = b1

∫ A

0
β̃1(ũ1 + ũ2)da+ b2

∫ A

0
β̃2(ũ1 − ũ2)da

Υ2(ũ) = b1

∫ A

0
β̃1(ũ1 + ũ2)da+ b2

∫ A

0
β̃2(ũ2 − ũ1)da

Proof. We will do it in two steps as follows :

Step 1: There exists a control vε that leads to extinction each distribution z1ε , z2ε . For
any ε > 0, we consider the functional defined on L2(Qω) by

Jε(v) =
1

2
‖v‖2L2(Qω) +

1

2ε

∫
QA

(
z2

1(0, a, x) + z2
2(0, a, x)

)
dQA, (22)

where z = (z1, z2) is solution of (16). It is clear that Jε is continuous, convex and coer-
cive on L2(Qω). Hence, the minimization problem of Jε admits at least one solution vε
associated to (z1ε , z2ε) solution of (16). From the maximum principle (see [10]), we get

vε = η1εχω in Q (23)

where ηε = (η1ε , η2ε) verifies the system



∂η1ε

∂t
+
∂η1ε

∂a
−∆η1ε + µ̃1η1ε + µ̃2η2ε = 0 in Q,

∂η2ε

∂t
+
∂η2ε

∂a
−∆η2ε + µ̃1η2ε + µ̃2η1ε = 0 in Q,

η1ε = η2ε = 0 on Σ,
η1ε(0, a, x) = −1

εz1ε(0, a, x) in QA,

η2ε(0, a, x) = −1
εz2ε(0, a, x) in QA,

η1ε(t, 0, x) = Υ1(ηε) in QT
η2ε(t, 0, x) = Υ2(ηε) in QT ,

(24)

herein zε = (z1ε , z2ε) is the solution of (16) associated to vε.
Let us multiply the first (with v = vε and z1 = z1ε) and the second (with z2 = z2ε)
equalities of (16) by η1ε and η2ε respectively, and integrate each equality by parts over Q.
Using (24) we deduct that∫

Q
(−f)η1εdQ = ‖vε‖2L2(Qω) + 1

ε‖z1ε(0, ·, ·)‖2L2(Q) + 1
ε‖z2ε(0, ·, ·)‖2L2(Q). (25)

Elsewhere, Young’s inequality gives:

∫
Q
|fη1ε | dQ ≤ 2C‖θf‖2L2(Q) +

1

2C

∫
Q

1

θ2
η2

1εdQ for

any C > 0. Thus,∫
Q

(−f)η1ε ≤ 2C‖θf‖2L2(Q) +
1

2C

∫
Q

1

θ2

(
η2

1ε + η2
2ε

)
dQ.



C. K. Somé, S. Sawadogo / Eur. J. Pure Appl. Math, 12 (3) (2019), 870-892 877

The lemma 2 allows, choosing C, the constant defined therein, to deduct that∫
Q

(−f)η1εdQ ≤ 2C‖θf‖2L2(Q) +
1

2
‖vε‖2L2(G). (26)

From (25) and (26) one obtains :

‖vε‖L2(G) ≤ 2
√
C‖θf‖L2(Q) (27)

‖z1ε(0, ·, ·)‖L2(Q) ≤
√

2εC‖θf‖L2(Q) (28)

‖z2ε(0, ·, ·)‖L2(Q) ≤
√

2εC‖θf‖L2(Q) (29)

We can extract subsequences denoted again (vε)ε and (zε)ε such that vε ⇀ ṽ weakly in
L2(Qω) and ziε ⇀ z̃i , i = 1, 2 weakly in L2(U,H1

0 (Ω)). Note that (z̃1, z̃2) is the unique
couple solution of (16)-(17) associated to ṽ. In the same ways, it follows that (η1ε , η2ε)
converge weakly to (η̃1, η̃2) and that (η̃1, η̃2) satisfies (21). From (23) and (27) we obtain
that ṽ = η̃1χω in Q.

Step 2 : Now we prove the inequalities (19)and (20).
Let set ẑiε = e−λ0tziε, i = 1, 2 where (z1ε, z2ε) verifies (16)-(17) and λ0 is a positive real
constant. Then ẑ1ε, ẑ2ε verify the system

−∂ẑ1ε

∂t
− ∂ẑ1ε

∂a
−∆ẑ1ε + µ̂1ẑ1ε + µ̃2ẑ2ε = Ĝ1(t, a, x)z1ε(t, 0, x) + f̂ + v̂εχω

+ Ĝ2(t, a, x)z2ε(t, 0, x) in Q

−∂ẑ2ε

∂t
− ∂ẑ2ε

∂a
−∆ẑ2ε + µ̂1ẑ2ε + µ̃2ẑ1ε = Ĝ2(t, a, x)z1ε(t, 0, x)

+ Ĝ1(t, a, x)z2ε(t, 0, x) in Q
ẑiε = 0 on Σ, i = 1, 2

ẑiε(T, a, x) = 0 in QA, i = 1, 2
ẑiε(t, A, x) = 0 in QT , i = 1, 2

(30)

where :
Ĝi = e−λ0tGi, f̂ = e−λ0tf, v̂ε = e−λ0tvε and µ̂1 = µ̃1 + λ0.

Multiplying the first and the second equations of (30) by ẑ1ε and ẑ2ε respectively, and
integrating by parts over Q, we have thanks to Young’s inequality :∫

Q
|∇ẑ1ε|2 dQ+ Γ1

∫
Q
|ẑ1ε|2dQ−

‖µ̃2‖∞
2C1

∫
Q
|ẑ2ε|2dQ+

(
1− A

2C2

)∫
QT

ẑ2
1ε(t, 0, x)dQT

+

∫
QA

ẑ2
1ε(0, a, x)dQA − A

2C3

∫
QT

ẑ2
2ε(t, 0, x)dQT ≤

1

2C4

∫
Q

∣∣∣f̂ ∣∣∣ dQ+ 1
2C5

∫
G
v̂2
εdQ (31)

and∫
Q
|∇ẑ2ε|2 dQ+ Γ2

∫
Q
|ẑ2ε|2dQ−

‖µ̃2‖∞
2K1

∫
Q
|ẑ1ε|2dQ+

(
1− A

2K3

)∫
QT

ẑ2
2ε(t, 0, x)dQT



C. K. Somé, S. Sawadogo / Eur. J. Pure Appl. Math, 12 (3) (2019), 870-892 878

+

∫
QA

ẑ2
2ε(0, a, x)dQA −

‖µ̃1‖∞
2K2

‖∞
∫
QT

ẑ2
1ε(t, 0, x)dQT ≤ 0 (32)

where :
Γ1 = λ0 − 2C1‖µ̃2‖∞ − 4A (C2 + C3) ‖β̃1, β̃2‖2∞‖b1, b2‖2QT

− ‖µ̃1‖∞ − 2C5,

Γ2 = λ0 − 2K1‖µ̃1‖∞ − 4A (K2 +K3) ‖β̃1, β̃2‖2∞‖b1, b2‖2QT
− ‖µ̃1‖∞ and the Ci,Ki are

Young’s constants for i = 1, 2, 3, 5.
Summing (31) and (32), one obtains :∫

Q
|∇ẑ1ε|2 dQ+ Π1

∫
Q
|ẑ1ε|2dQ+

∫
Q
|∇ẑ2ε|2 dQ+ Π2

∫
Q
|ẑ2ε|2dQ+

(
1− A

2C2
− A

2K3

)∫
QT

ẑ2
1ε(t, 0, x)dQT +

(
1− A

2C3
− A

2K2

)∫
QT

ẑ2
2ε(t, 0, x)dQT

+

∫
QA

ẑ2
1ε(0, a, x)dQA +

∫
QA

ẑ2
2ε(0, a, x)dQA ≤

1

2C4

∫
Q

∣∣∣f̂ ∣∣∣ dQ+ 1
2C5

∫
G
v̂2
εdQ (33)

with :

Π1 = Γ1 −
‖µ̃2‖∞
2K1

and Π2 = Γ2 −
‖µ̃2‖∞

2C1
.

Choosing λ0 and the Young’s constants such that:
λ0 ≥ max

{
2C1‖µ̃2‖∞ + 4A (C2 + C3) ‖β̃1, β̃2‖2∞‖b1, b2‖2QT

+ ‖µ̃1‖∞ + 2C5 + ‖µ̃2‖∞
2K1

+ 1;

2K1‖µ̃1‖∞ + 4A (K2 +K3) ‖β̃1, β̃2‖2∞‖b1, b2‖2QT
+ ‖µ̃1 + ‖µ̃2‖∞

2C1
+ 1
}

and min
{

1− A
2C2
− A

2K3
, 1− A

2C3
− A

2K2

}
≥ 1, One deducts from (27) and (33) that∫

Q
|∇ẑ1ε|2 dQ+

∫
Q
|ẑ1ε|2dQ ≤ C

(
‖f̂‖2L2(Q) + ‖θf̂‖L2(Q)

)
(34)∫

Q
|∇ẑ2ε|2 dQ+

∫
Q
|ẑ2ε|2dQ ≤ C

(
‖f̂‖2L2(Q) + ‖θf̂‖L2(Q)

)
(35)∫

QT

ẑ2
1ε(t, 0, x)dQT ≤ C

(
‖f̂‖2L2(Q) + ‖θf̂‖L2(Q)

)
(36)∫

QT

ẑ2
2ε(t, 0, x)dQT ≤ C

(
‖f̂‖2L2(Q) + ‖θf̂‖L2(Q)

)
(37)

Consequently, the sequences (ẑ1ε)ε, (ẑ2ε)ε, (ẑ1ε(·, 0, ·))ε and (ẑ2ε(·, 0, ·))ε are bounded re-
spectively in L2(U,H1

0 (Q)) and L2(QT ). That ends this proof, thanks to limit’s results
obtained in the step 1.



C. K. Somé, S. Sawadogo / Eur. J. Pure Appl. Math, 12 (3) (2019), 870-892 879

3.3. Study of the nonlinear case

Let bi(t, x) = Ti

(∫ A
0 βi(t, a, x)zi(t, a, x)da

)
, i = 1; 2 where Ti ∈ L∞(R) , βi i = 1, 2

verify (H2)− (H3). we study here, the null controllability of the following system :

−∂z1

∂t
− ∂z1

∂a
−∆z1 + µ̃1z1 + µ̃2z2 = β1T1(ξ1)z1(t, 0, x) + f + vχω

+ β2T2(ξ2)z2(t, 0, x) in Q

−∂z2

∂t
− ∂z2

∂a
−∆z2 + µ̃1z2 + µ̃2z1 = β1T2(ξ2)z1(t, 0, x)

+ β2T1(ξ1)z2(t, 0, x) in Q
zi = 0 on Σ, i = 1, 2

zi(T, a, x) = 0 in QA, i = 1, 2
zi(t, A, x) = 0 in QT , i = 1, 2

(38)

The system (38) is nonlinear. Let

A =
{
ṽ ∈ L2(Qω) : (z̃1, z̃2) solves (38), verifies (17) and ṽ satifies (27)

}
,

N = L2(QT )× L2(QT ),

and define the multivalued mapping :

Λ : N −→ 2N , (ξ1, ξ2) 7−→ Λ(ξ1, ξ2) by

Λ(ξ1, ξ2) =

{(∫ A

0
β1z̃1da,

∫ A

0
β2z̃2da

)
: (z̃1, z̃2) is associated to ṽ ∈ A

}
.

The null controllability problem of (38) is reduced to find a fixed point of Λ. In order to
use the generalization of the Leray-Schauder’s fixed point theorem, we set

Nρ = {(ξ1, ξ2) ∈ N : ∃ρ ∈ (0, 1), (ξ1, ξ2) ∈ ρΛ(ξ1, ξ2)}.

The following proposition is a direct consequence of the Leray-Schauder’s fixed point
theorem (see [1]).

Proposition 2. Under the assumptions (H1)− (H3), the multivalued mapping Λ admits
at least one fixed point.

Proof. We proceed in four steps :

Step 1: Nρ is bounded in N .

Let (ξ1, ξ2) ∈ Nρ. Then, there exists ρ ∈ (0, 1), z̃1, z̃2 such that ξ1 = ρ
∫ A

0 β1z̃1da and

ξ2 = ρ
∫ A

0 β2z̃2da. Then,
∫
QT
|ξi|2dQT ≤ ‖β1, β2‖2∞

∫
Q z̃

2
i dQ , i = 1; 2. So,

‖ξ1‖L2(QT ) + ‖ξ2‖L2(QT ) ≤ ‖β1, β2‖∞
(
‖z̃1‖L2(Q) + ‖z̃2‖L2(Q)

)
(39)

From the theorem 3, one deducts that there exists a positive constant C such that

‖ξ1‖L2(QT ) + ‖ξ2‖L)2(QT ) ≤ 2C‖β1, β2‖∞
(
‖θf‖L2(Q) + ‖f‖L2(Q)

)
(40)

Hence, Nρ is bounded in N since L2(U ;H1(Ω)) ⊂ L2(Q).



C. K. Somé, S. Sawadogo / Eur. J. Pure Appl. Math, 12 (3) (2019), 870-892 880

Step 2 : For all (ξ1, ξ2) ∈ N , Λ(ξ1, ξ2) is closed and convex subset of N .
Let (ξ1, ξ2) ∈ Λ(ξ1, ξ2). Under the assumptions (H1) − (H3), the system (38) admits a
solution and the corresponding control verifies (27). So, the set Λ(ξ1, ξ2) is non empty.
Elsewhere, like the mapping (ξ1, ξ2) 7−→ (z̃1, z̃2) is affine, then, the set Λ(ξ1, ξ2) is convex.
There rest to prove that this set is closed.
Let (η1n , η2n)n ⊂ Λ(ξ1, ξ2) which converges strongly towards (η1, η2) in N . Then, for
each n ∈ N, there exists a control ṽn ∈ A and a corresponding solution (z̃1n , z̃2n) of

(38) such that ηin =
∫ A

0 βiz̃in , i = 1, 2. From the inequalities (27), (34) and (35) one
deduces that (z̃1n , z̃2n) and ṽn are bounded respectively in (L2(Q))2 and L2(Qω). Thus,
(η1n , η2n) is bounded in N . Hence, we can extract subsequences denoted still (z̃1n , z̃2n), ṽn
and (η1n , η2n) respectively such that (z̃1n , z̃2n), ṽn and (η1n , η2n) converge weakly towards

(z̃1, z̃2), ṽ and (η1, η2) respectively in (L2(Q))2, L2(Qω) and N with ηi =
∫ A

0 βiz̃ida, i =
1; 2. Notice that (z̃1, z̃2) is solution of (38) and ṽ verifies (27). So, (z̃1, z̃2) satisfies (17).
As consequence, (η1, η2) ∈ Λ(ξ1, ξ2).

Step 3 : Λ is a compact multivalued mapping.
Let B be a bounded subset of N , (ξ1, ξ2) ∈ B. Let (ρ1n , ρ2n) ∈ Λ(ξ1, ξ2). Then, for all

n ∈ N, there exists (z̃1n , z̃2n), solution of (38), and ṽn in
(
L2(Q)

)2
and L2(Qω) respec-

tively such that ρin =
∫ A

0 βiz̃inda, i = 1; 2 and ṽn satisfies (27). So, (ṽn)n is bounded
in L2(Qω). Proceeding in the similar ways that the step 2 of the proof of the theo-
rem 3, one deducts from (27), (34)-(37) and the fact that H1(Ω) ⊂ L2(Ω) that (z̃1n , z̃2n)n
is bounded in

(
L2(Q)

)2
, and then, (ρ1n , ρ2n) is bounded in N . Thus, there exists subse-

quences of (z̃1n , z̃2n) and ṽn also denoted by (z̃1n , z̃2n) which converges weakly in
(
L2(Q)

)2
and L2(Qω). Moreover, the subsequences ρin =

∫ A
0 βiz̃inda, i = 1; 2 of (ρin)n verify the

following system :

−∂ρ1n

∂t
−∆ρ1n +

∫ A
0 µ̂1β1z̃1nda+

∫ A
0 β1µ2z̃2nda = K1(ξn) in QT

−∂ρ2n

∂t
−∆ρ2n +

∫ A
0 µ̂1β2z̃2nda+

∫ A
0 β2µ2z̃1nda = K2(ξn) in QT

ρ1n = ρ2n = 0 on ΣT

ρ1n(0, x) = ρ2n(0, x) = 0 in Ω
ρ1n(T, x) = ρ2n(T, x) = 0 in Ω

(41)

where ΣT = (0, T )× Γ, µ̂1 = µ1 + λ0 and for all n ∈ N,

K1n(ξ) = −
∫ A

0

(
∂β1
∂t + ∂β1

∂a + ∆β1 + µ2β2

)
z̃1nda+

∫ A

0
β1(f + ṽnχω)da

+

∫ A

0
β2

1T1(ξ1n)z̃1n(t, 0, x)da+

∫ A

0
β1β2T2(ξ2n)z̃2n(t, 0, x)da

− 2
n∑
i=1

∫ A

0

∂β1
∂xi

.
∂z̃1n
∂xi

da



C. K. Somé, S. Sawadogo / Eur. J. Pure Appl. Math, 12 (3) (2019), 870-892 881

K2n(ξ) = −
∫ A

0

(
∂β2
∂t + ∂β2

∂a + ∆β2 + µ2β1

)
z̃2nda+

∫ A

0
β2

2T1(ξ1n)z̃2n(t, 0, x)da

+

∫ A

0
β1β2T2(ξ2n)z̃1n(t, 0, x)da− 2

n∑
i=1

∫ A

0

∂β2
∂xi

.
∂z̃2n
∂xi

da

Under the assumptions (H1)− (H3) the boundedness of B and of sequences (z̃in)n i = 1; 2,
from (27), (34)-(37), one deducts that there exists positive constants Ci which depend on
‖∇βi‖∞, ‖β1, β2‖2∞, ‖T1, T2‖∞ for i = 1; 2 such that

‖Ki(ξn)‖2L2(QT ) ≤ Ci
(
‖θf‖2L2(Qω) + ‖f‖2L2(Q)

)
(42)

Now, multiplying the first and the second equations of (41) by ρ1n and ρ2n respectively
and proceeding by integrations by parts over QT , one has∫

QT

|∇ρ1n |2dQT + λ0

∫
QT

ρ2
1ndQT =

∫
QT

(
K1(ξn)−

∫ A

0
β1(µ̃1z̃1n + µ̃2z̃2n)da

)
ρ1ndQT

Since z̃1n , z̃2n verify (35)-(36), one deducts that K1(ξn)−
∫ A

0 β1(µ̃1z̃1n + µ̃2z̃2n)da verifies
(42). So, using Young inequality, one has∫

QT

|∇ρ1n |2dQT + (λ0 −
λ1

2
)

∫
QT

ρ2
1ndQT ≤

C1
2λ1

(
‖θf‖2L2(Qω) + ‖f‖2L2(Q)

)
(43)

By analogy we show that∫
QT

|∇ρ2n |2dQT + (λ0 −
λ2

2
)

∫
QT

ρ2
2ndQT ≤

C2
2λ2

(
‖θf‖2L2(Qω) + ‖f‖2L2(Q)

)
(44)

Taking λ0−1 ≥ max(λ12 ,
λ2
2 ), one deducts that (ρ1n)n and (ρ2n)n are bounded in L2((0, T );H1(Ω)).

Let remark that the system (41) is equivalent to the system

−∂ρ1n

∂t
−∆ρ1n + λ0ρ1n = K ′1(ξn) in QT

−∂ρ2n

∂t
−∆ρ2n + λ0ρ2n = K ′2(ξn) in QT

ρ1n = ρ2n = 0 on ΣT

ρ1n(0, x) = ρ2n(0, x) = 0 in Ω
ρ1n(T, x) = ρ2n(T, x) = 0 in Ω

(45)

with K ′1 = K1(ξn) −
∫ A

0 β1(µ1z̃1n + µ2z̃2n)da , K ′2 = K2(ξn) −
∫ A

0 β2(µ1z̃2n + µ2z̃1n)da
and (45) is a system of retrograde heat equations which the source terms are bounded in
L2(QT ) and the distributions are bounded in L2((0, T );H1(Ω)). So, the sequences

(ρ1n
∂t

)
n

and
(ρ2n
∂t

)
n

are bounded in L2((0, T );H−1(Ω)). Thus, we deduct from Aubin-Lions lemma
that there exists subsequences (ρ1nk

)k and (ρ2nk
)k of (ρ1n)n and (ρ2n)n respectively that

converge strongly towards ρ1 and ρ2 respectively in L2(QT ) . Hence, (ρ1n)n and (ρ2n)n



C. K. Somé, S. Sawadogo / Eur. J. Pure Appl. Math, 12 (3) (2019), 870-892 882

converge weakly towards ρ1 and ρ2 respectively in L2(QT ) . Elsewhere, there exists sub-
sequences (z̃ink

)k of z̃in , i = 1, 2 associated to (ρink
)k, i = 1, 2 respectively that converge

weakly towards z̃i, i = 1, 2 respectively in L2(U ;H1(Ω)), say us more precisely in L2(Q),
since, L2(U ;H1(Ω)) ⊂ L2(Q). Thus, we have firsly

ρink
⇀ ρi weakly in L2(QT ) i = 1; 2 (46)

and secondly

ρink
⇀

∫ A

0
βiz̃ida weakly in L2(QT ) i = 1; 2, (47)

then, from the uniqueness of the limit, for all i ∈ {1, 2}, one deducts that

ρi =

∫ A

0
βiz̃ida. (48)

Similarly, we can prove that (ṽn)n converges towards ṽ ∈ L2(Qω). Moreover, (z̃1, z̃2)
verifies (38) and ṽ satisfies (27). From the theorem 3, one deducts that z̃i, i = 1; 2 satisfies
(17).
Step 4 : Λ is upper semi-continuous on N .
Let K be a closed subset of N . Let (k1n , k2n)n ⊂ Λ−1(K) that converges strongly towards
(k1, k2) in N . Then, (k1n , k2n)n is bounded in N . Since Λ−1(K) = {(k1, k2) ∈ K :
Λ(k1, k2) ∩K 6= ∅}, there exists, a sequence (ρ1n , ρ2n)n ∈ K that belongs to Λ(k1n , k2n).
Now, proceeding as in the previous step with K instead of B and with Λ−1(k1n , k2n) instead
of Λ−1(ξ1, ξ2), one deduces that there exists subsequences still denoted by (ρ1n , ρ2n) and
(ṽn) which converge weakly to (ρ1, ρ2) and ṽ respectively in N and L2(Qω), and for all
i ∈ {1, 2}, there exists z̃i ∈ L2(U,H2(Ω)) such that ρin verifies (47). So, for all i ∈ {1, 2},
ρi verifies (48). Let mention that (z̃1, z̃2) solves (38), ṽ verifies (27) and z̃i i = 1, 2
satisfies (17). Consequently,

(ρ1, ρ2) ∈ Λ(k1, k2) (49)

From (43), (44) and Lions-Aubin lemma one deduces that the subsequence (ρ1n , ρ2n) of
the closed set K, converges strongly towards (ρ1, ρ2) in N . Then,

(ρ1, ρ2) ∈ K. (50)

(49) and (50) say that (k1, k2) ∈ Λ−1(K).

4. Proof of the main result

In this section, we study the controllability of the (8)-(9). In view of the above, let’s
set for any ξ = (ξ1, ξ2) ∈ L2(QT )× L2(QT )

T1(ξ) = F (ξ1 + ξ2) ; T2(ξ) = G(ξ1 − ξ2),
G1(ξ) = β1(t, a, x)T1(ξ) + β2(t, a, x)T2(ξ),
G2(ξ) = β1(t, a, x)T1(ξ)− β2(t, a, x)T2(ξ).

(51)



C. K. Somé, S. Sawadogo / Eur. J. Pure Appl. Math, 12 (3) (2019), 870-892 883

Now, we consider the system that follows

−∂p̂1ε

∂t
− ∂p̂1ε

∂a
−∆p̂1ε + µ̃1p̂1ε + µ2p̂2ε = G1(ξ)p̂1ε(t, 0, x) + f̂ + v̂εχω

+G2(ξ)p̂2ε(t, 0, x) in Q,

−∂p̂2ε

∂t
− ∂p̂2ε

∂a
−∆p̂2ε + µ̃1p̂2ε + µ2p̂1ε = G2(ξ)p̂1ε(t, 0, x)

+G1(ξ)p̂2ε(t, 0, x) in Q,
p̂1ε = p̂2ε = 0 on Σ,

p̂1ε(T, a, x) = p̂2ε(T, a, x) = 0 in QA,
p̂1ε(t, A, x) = p̂2ε(t, A, x) = 0 in QT ,

(52)

where : p̂iε = e−λ0tpiε , i = 1; 2, f̂ = e−λ0tf, µ̃1 = µ̃1 + λ0 and v̂ε = e−λ0tvε for any λ0 ≥ 0
with (p1ε , p2ε) a solution of (8) associated to vε.
The controllability of the system (8) -(9) is summarized in the study of the null control-
lability of system (52). We consider the operator Λ̂ from N = L2(QT )× L2(QT ) into 2N

defined by

(ξ1, ξ2) 7−→ Λ̂(ξ1, ξ2) = Λξ2(ξ1)× Λξ1(ξ2) (53)

such that

Λξ2(ξ1) =

{∫ A

0
β1 (p̂1ε(ξ1) + p̂2ε(ξ2)) da

}
Λξ1(ξ2) =

{∫ A

0
β2 (p̂1ε(ξ1)− p̂2ε(ξ2)) da

}
where (p̂1ε(ξ1), p̂2ε(ξ2)) solves (52), verifies (28)-(29) and the associated control v̂ε satisfies
(27).
The controllability of (52) is summarized to the study of the existence of a fixed point of
the mapping Λ̂ [8]. We are going to show that Λ̂ admits a fixed point. To do that, we
have to demonstrate that for each (ξ1, ξ2) ∈ N , Λξ2(ξ1) and Λξ1(ξ2) are bounbed closed

convex sets in L2(QT ) and Λ̂(ξ1, ξ2) is upper semicontinuous. Let set

Y1(ξ)(t, x) =

∫ A

0
β1p̂1ε(ξ1)da+

∫ A

0
β1p̂2ε(ξ2)da (54)

Y2(ξ)(t, x) =

∫ A

0
β2p̂1ε(ξ1)da−

∫ A

0
β2p̂2ε(ξ2)da (55)

Proceeding as in the step 2 of the proof of the Proposition 3, one deducts from (41)-(42)
that Yi(ξ) , i = 1; 2 verify for any positive real λ0 the following system :

−∂Yi(ξ)
∂t

−∆Yi(ξ) + λ0Yi = Ri(ξ) in QT

Yi(ξ) = 0 on ΣT

Yi(ξ)(0, x) = 0 in Ω

(56)



C. K. Somé, S. Sawadogo / Eur. J. Pure Appl. Math, 12 (3) (2019), 870-892 884

where

R1(ξ) = −
∫ A

0

(
∂β1
∂t + ∂β1

∂a + ∆β1 + (µ1 + µ2)β1

)
(p̂1ε(ξ1) + p̂2ε(ξ2))da

+

∫ A

0
β1(G1(ξ)p̂1ε(ξ1)(t, 0, x) +G2(ξ)p̂2ε(ξ2)(t, 0, x) + f̂ + v̂εχω)da

+

∫ A

0
β2 (G2(ξ)p̂1ε(ξ1)(t, 0, x) +G1(ξ)p̂2ε(ξ2)(t, 0, x)) da

− 2

n∑
i=1

∫ A

0

∂β1
∂xi

.
(
∂p̂1ε
∂xi

+ ∂p̂2ε
∂xi

)
da

R2(ξ) =

∫ A

0

(
∂β2
∂t + ∂β2

∂a + ∆β2 + β2(µ1 − µ2)
)

(p̂1ε(ξ1)− p̂2ε(ξ2))da

+

∫ A

0
β2(G1(ξ)p̂1ε(t, 0, x) +G2(ξ)p̂2ε(t, 0, x) + f̂ + v̂εχω)da

−
∫ A

0
β2 (G2(ξ)p̂1ε(t, 0, x) +G1(ξ)p̂2ε(t, 0, x)) da

− 2
n∑
i=1

∫ A

0

∂β2
∂xi

.
(
∂p̂1ε
∂xi
− ∂p̂2ε

∂xi

)
da.

Under the hypothesis (H1) − (H4), taking λ0 as in the proof of the theorem 1, one
deducts from (27), (34)-(37) that there exists a positive reals C1, C2 which depend on
‖β1, β2‖∞, ‖F,G‖∞ and ‖µ1, µ2‖∞ such that

‖R1(ξ)‖2∞ ≤ C1

(
‖θf‖2L2(Qω) + ‖f‖2Q

)
(57)

‖R2(ξ)‖2∞ ≤ C2

(
‖θf‖2L2(Qω) + ‖f‖2Q

)
. (58)

Multiplying respectively the first equation of (56) by Yi(ξ), i = 1; 2 and by integrating
by parts over QT , we show (using Young’s inequality as in the step 2 of the proof of the
Proposition 3) that Yi, i = 1; 2 are bounded in L2(0, T ;H1

0 (Ω)). Thus, for each i ∈ {1; 2},
the system (56) is a retrograde heat equation with the source term and the initial condition

are bounded respectively in L2(QT ) and L2(Q). Moreover, Yi,
∂Yi(ξ)
∂t i = 1, 2 are bounded

respectively in L2(0, T ;H1
0 (Ω)) and L2(0, T ;H−1(Ω)). Consequently, we conclude, thanks

to Lions-Aubin Lemma, that Λξi i = 1, 2 are bounded and compact in L2(QT ) . Thus, Λ̂
is bounded and compact in N .

Now, let K a closed subset of N . Let (ξ1n , ξ2n)n ⊂ Λ̂−1(K) that converges strongly
towards (ξ1, ξ2) in N . Then, ((ξ1n , ξ2n))n is bounded in N . Let remember that
Λ̂−1(K) = {(ξ1, ξ2) ∈ K : Λ(ξ1, ξ2) ∩ K 6= ∅}. So, there exists a sequence (Y1n , Y2n)n ∈
K that belongs to Λ−1

ξ2
(ξ1n) × Λ−1

ξ1
(ξ2n) = Λ̂−1(ξ1n , ξ2n) such that Y1n and Y2n verifies

respectively (54) and (55) with respectively ξ1n and ξ2n instead of ξ1 and ξ2, and moreover,
the pair (p̂1ε(ξ1n), p̂2ε(ξ2n)) satisfies (52) and the associated control v̂ε verifies (27). Using



C. K. Somé, S. Sawadogo / Eur. J. Pure Appl. Math, 12 (3) (2019), 870-892 885

(56) and the estimations (34)-(37), we show (as the step 4 in the section 4) that the sequel
(Yin)n , i = 1, 2 converge strongly to Yi i = 1, 2. Since p̂iε(ξin) , i = 1, 2 and η1ε(ξ1n) are
bounded independently to (ξin) , i = 1; 2, then, for all n, Ri(ξn) i = 1, 2 are bounded in
L2(QT ). Consequently, one can extract a subsequence still denoted by Yin , Ri(ξn) i = 1, 2
such that

Yin −→ Yi in L2(QT ) i = 1, 2 ;

Ri(ξn) −→ Ri(ξ) i = 1, 2 ;∫ A

0
µ̃1β̃ip̂iε(ξin) −→

∫ A

0
µ̃1βip̂iε(ξi)da weakly in L2(QT ) i = 1, 2 ;∫ A

0
µ̃1β1p̂2ε(ξ2n)da −→

∫ A

0
µ̃1β1p̂2ε(ξ2)da weakly in L2(QT ) ;∫ A

0
µ2β2p̂1ε(ξ1n)da −→

∫ A

0
µ2β2p̂1ε(ξ1)da weakly in L2(QT ) ;

So, for each i ∈ {1; 2}, Yi(ξ) is solution of (56), (p̂1ε(ξ1), p̂2ε(ξ2)) solves (52) and the
associated control v̂ε = η1(ξ1) verifies (29). Hence, (Y1, Y2) ∈ Λ−1

ξ2
(ξ1) × Λ−1

ξ1
(ξ2) and so,

(ξ1, ξ2) ∈ Λ̂−1(K). Endly, since ξ1 7−→ p̂1ε and ξ2 7−→ p̂2ε are affine, then Λξ2(ξ1) and

Λξ1(ξ2) are nonempty convex sets in L2(QT ) . Thus, the gragh G
Λ̂

= {〈(ξ1, ξ2), Λ̂(ξ1, ξ2)〉}
of Λ̂ is closed. Then, Λ̂(ξ1, ξ2) = Λξ2(ξ1) × Λξ1(ξ2) is upper semicontinuous, and from

the Kakutani’s fixed point theorem [8], we conclude that Λ̂ admits a fixed point. More
precisely, there exists ξ = (ξ1, ξ2) ∈ N such that

Λ̂(ξ) = ξ =

(∫ A

0
β1(p̂1ε(ξ1) + p̂2ε(ξ2))da ,

∫ A

0
β2(p̂1ε(ξ1)− p̂2ε(ξ2))da

)
where (p̂1ε , p̂2ε) is solution of the system (52) with

G1

(∫ A

0

βp̂εda

)
= β1(t, a, x)F

(∫ A

0

β1(p̂1ε + p̂2ε)da

)
+ β2(t, a, x)G

(∫ A

0

β2(p̂1ε − p̂2ε)da

)

G2

(∫ A

0

β2p̂εda

)
= β1(t, a, x)F

(∫ A

0

β1(p̂1ε
+ p̂2ε)da

)
− β2(t, a, x)G

(∫ A

0

β2(p̂1ε
− p̂2ε)da

)
.

instead of G1(ξ) and G2(ξ) respectively.



C. K. Somé, S. Sawadogo / Eur. J. Pure Appl. Math, 12 (3) (2019), 870-892 886

5. Application to the sentinel of detection

We consider for given positive functions Gi = 1; 2 the following systems :

∂yi
∂t

+
∂yi
∂a
−∆yi + µiyi = 0 in Q,

yi(0, a, x) = y0
i + τiŷ

0
i in QA,

yi(t, 0, x) = Gi

(∫ A

0
βiyida

)
in QT ,

yi =

{
gi + λiĝi on Σi,

0 on Σ \ Σi.

(59)

where Σi = (0, T ) × (0, A) × Γi i = 1; 2, the Γi, i = 1; 2 are such that Γ1 ∪ Γ2 = Γ and
Γ = ∂Ω is the smooth boundary of Ω, the functions µi, βi and the reals T,A are defined
respectively as in section 1. y(t, a, x) is the distribution of individuals of age a at time

t and location x ∈ Ω. The expressions
∫ A

0 βiyida, i = 1; 2 denote the distribution of
newborn individuals at time t and location x. In an ovipare species it represents the total

eggs hatch at time t and the position x and Gi

(∫ A
0 βiyida

)
denote the distribution of

eggs that at time t and the position x. The functions Gi i = 1; 2 are of class C1, globally
lipschitz and their derivate functions verify G′i(0) = 0 and moreover G′i ∈ L∞(R) are
globally lipschitz. The system (59) describes the evolution of the populations under the
inhospitable boundary conditions when the flux of population takes the form −∇y(t, a, x).
As for the initial and boundary conditions of (59), y0

i and gi are given respectively in
L2(QA)τiŷ

0
i , λiĝi i = 1; 2 are unknown where τi, λi i = 1; 2 are reals. As a matter of fact

the terms y0
i + τiŷ

0
i and gi + λiĝi are qualified as incomplete data. Suppose that :

(i) for i=1;2 ĝi ∈ L2(Σi) and ‖ĝi‖L2(Σi) ≤ 1,

(ii) for i=1;2 ŷ0
i ∈ L2(QA) and ‖ŷ0

i ‖L2(QA) ≤ 1,

(iii) for i = 1; 2 the reals τi and λi are unknown and small enough.

It is now assumed that measures yiobs , i = 1; 2 are available on QO = U×O where O ⊂ Ω
is the observation set and O ∩ ω 6= ∅. Assume moreover that

yi = yiobs = m0i , i = 1; 2 on QO. (60)

where m0i, i = 1; 2 are known functions belonging to L2(QO). The aim is to calculate the
pollution terms λ1ĝ1 and λ2ĝ2 independently from the missing terms τ1ŷ

0
1 and τ2ŷ

0
2 with

one and only one sentinel. One of the methods to solve this problem is the least squares
method. The sentinel concept was introduced by J.L. Lions [7] to study the systems with
incomplete data. This concept relies on the following elements : the state y described by
a equation or a partial differential equations system, an observation function yobs defined
on U × O where O is the observation set and a control function v to be determined.
Many papers use the definition of Lions in the theoretical aspect. As to applications,
we quote S. Sawadogo in [9] who studied the detection of incomplete parameters for a



C. K. Somé, S. Sawadogo / Eur. J. Pure Appl. Math, 12 (3) (2019), 870-892 887

linear population dynamic model. In [10] the author made the same study for a nonlinear
population dynamic model. For the sentinel concept we refer to [9, 10] and the references
therein. In this paragraph we study the simultaneous sentinel concept for a coupled
nonlinear population dynamic model. We begin by the following proposition

Proposition 3. For each i = 1; 2, the functions λi 7−→ yi(λi, τi) and τi 7−→ yi(λi, τi) are
differentiable at the point 0.

Proof. Let ŷi(t, a, x) = e−λ0t (yi(λi, τi)− y0i) i = 1; 2 with y0i = yi(λi, 0) and for each
i = 1; 2, yi(λi, τi) and y0i solve (59). Then ŷi i = 1; 2 verify

∂ŷi
∂t

+
∂ŷi
∂a
−∆ŷi + (µi + λ0)ŷi = 0 in Q,

ŷi(0, a, x) = τiŷ
0
i in QA,

ŷi(t, 0, x) = e−λ0t
(
Gi

(∫ A

0
βiyida

)
−Gi

(∫ A

0
βiy0ida

))
in QT ,

ŷi = 0 on Σ.

(61)

System (61) is this one obtained in the proof of the Proposition 9 in [10] with here βi(t, a, x),
Gi respectively in the place of β(a), F and τi = τ, λi = λ i = 1; 2. Let multiply (61) by
ŷi and integrate by parts over Q. Since Gi, i = 1; 2 is globally lipschitz, proceeding as in
[10], we have

‖ŷi(·, 0, ·)‖L2(QT ) ≤ C‖βi‖2∞‖ŷi‖L2(QT ). (62)

One deducts from (62) that

‖∇ŷi‖L2(QT ) + ‖ŷi‖L2(QT ) ≤ Cτ2
i . (63)

According to the expression of ŷi and the relation (61), we get yi converges uniformly to

y0i on Q and

∫ A

0
βiyi(λi, τi)da converges uniformly to

∫ A

0
βiy0ida on QT . Set now zτi = ŷi

τi

and pτi = zτi − zi for i = 1; 2, where zi verifies

∂zi
∂t

+
∂zi
∂a
−∆zi + µizi = 0 in Q,

zi(0, a, x) = ŷ0
i in QA,

zi(t, 0, x) = G′i

(∫ A

0
βiy0ida

)∫ A

0
βizida in QT ,

yi = 0 on Σ.

(64)

we show as in [10] that :
pτi −→ 0, zτi −→ zi i = 1; 2 respectively in L2

(
U ;H1

0 (Ω)
)

as τi → 0.



C. K. Somé, S. Sawadogo / Eur. J. Pure Appl. Math, 12 (3) (2019), 870-892 888

Likewise let ûi(t, a, x) = e−λ0t (yi(λi, τi)− yi0) i = 1; 2 with yi0 = yi(0, τi) and for each
i = 1; 2, yi(λi, τi) and yi0 solve (59). Then ûi, i = 1; 2 verify

∂ûi
∂t

+
∂ûi
∂a
−∆ûi + (µi + λ0)ûi = 0 in Q,

ûi(0, a, x) = 0 in QA,

ûi(t, 0, x) = e−λ0t
(
Gi

(∫ A

0
βiyida

)
−Gi

(∫ A

0
βiy0ida

))
in QT ,

ûi =

{
λiĝi on Σi

0 on Σ \ Σi.

(65)

Multiplying (65) by ûi and by integrating by parts over Q, we have

1

2

∫
QA

û2
i (T, a, x)dQA +

1

2

∫
QT

û2
i (t, A, x)dQT +

∫
Q
|∇ûi|2dQ

+

∫
Q

(µi + λ0)û2
i dQ = τi

∫
Σi

∂ûi
∂σi

ĝidΣi +
1

2

∫
QT

û2
i (t, 0, x)dQT (66)

From (62), taking λ0 = 1 + C‖βi‖2∞, one has

‖ûi‖2L2(Q) + ‖∇ûi‖2(L2(Q))N ≤ λi
∫

Σi

∇ûiĝidΣi (67)

Using Young inequality and according to hypothesis (i), there exists a positive constant
CY such that

‖ûi‖2L2(Q) + ‖∇ûi‖2(L2(Q))N
≤ λ2i

2CY
. (68)

Then ŷi converges uniformly to yi0 on Q and from the regularity of Gi, i = 1; 2 we proove

that

∫ A

0
βiyi(λi, τi)da converges uniformly to

∫ A

0
βiyi0da on QT .

One deducts from the proposition 9 in [10], that the functions λi 7−→ y(λi, τi) i = 1; 2 are
differentiable. Set now zλi = ûi

λi
and pλi = zλi − zi for i = 1; 2, where zi verifies

∂zi
∂t

+
∂zi
∂a
−∆zi + µizi = 0 in Q,

zi(0, a, x) = 0 in QA,

zi(t, 0, x) = G′i

(∫ A

0
βiy0ida

)∫ A

0
βizida in QT ,

yi =

{
ĝi on Σi

0 on Σ \ Σi.

(69)



C. K. Somé, S. Sawadogo / Eur. J. Pure Appl. Math, 12 (3) (2019), 870-892 889

Then pλi solves

∂pλi

∂t
+
∂pλi

∂a
−∆pλi + µipλi = 0 in Q,

pλi
(0, a, x) = 0 in QA,

pλi(t, 0, x) = e−λ0t

[
Gi

(∫ A
0

βiyida

)

−Gi

(∫ A
0

βiy0ida

)]
−G′i

(∫ A
0

βiy0ida

)∫ A
0

βizida in QT ,

pλi
= 0 on Σ.

(70)

We obtain the equality (66) when we multiply (70) by pλi and integrate by parts over Q.
From the fact that the functions Gi i = 1; 2 are globally lipschitz and λi 7−→ yi(λi, τi) con-
verge uniformly, one deduces that the functions λi 7−→ yi(λi, τi) i = 1; 2 are differentiable
(see Proposition 9 in [10]).

In the sequel, we consider for h ∈ L2(QO) and w ∈ L2(Qω), the following functionals :

Si(λi, τi) =

∫
QO

hyi(λi, τi)dQ+

∫
Qω

wyi(λi, τi)dQ i = 1; 2. (71)

We obtain from the Proposition 3 the following result.

Corollary 1. The functionals Si i = 1; 2 are differentiable at the point (0, 0) and

∂Si
∂τi

(0, 0) =

∫
QO

hyτidQ+

∫
Qω

wyτidQ i = 1; 2 (72)

∂Si
∂λi

(0, 0) =

∫
QO

hyλidQ+

∫
Qω

wyλidQ i = 1; 2 (73)

where for each i = 1; 2, yτi solves the system :

∂yτi
∂t

+
∂yτi
∂a
−∆yτi + µiyτi = 0 in Q,

yτi(0, a, x) = ŷ0(a, x) in QA,

yτi(t, 0, x) = G′i

(∫ A

0
βiy0ida

)∫ A

0
βizida in QT ,

yτi = 0 on Σ,

(74)

and yλi solves the system

∂yλi

∂t
+
∂yλi

∂a
−∆yλi + µiyλi = 0 in Q,

yλi
(0, a, x) = 0 in QA,

yλi
(t, 0, x) = G′i

(∫ A
0

βiy0ida

)∫ A
0

βiyλi
da in QT ,

yλi =

{
ĝi on Σi
0 on Σ \ Σi.

(75)



C. K. Somé, S. Sawadogo / Eur. J. Pure Appl. Math, 12 (3) (2019), 870-892 890

Moreover

yλi , yτi ∈ C((0, T );L2(QA)) ∩ C((0, A);L2(QT )) ∩ L2(U,H1
0 (Ω)) i = 1; 2. (76)

Proof. We know that for each pair (λi, τi) ∈ R2, (59) admits an unique solution
y(λi, τi) in C((0, T );L2(QA)) ∩ C((0, A);L2(QT )) ∩ L2(U,H1

0 (Ω))2 (see [5]). We have

Si(λi = 0, τi) =

∫
QO

hy(λi = 0, τi)dQ+

∫
Qω

wy(λi = 0, τi)dQ. So

Si(λi = 0, τi)− Si(0, 0)

τi
=

∫
QO

h
y(λi = 0, τi)− yi(0, 0)

τi
dQ

+

∫
Qω

w
y(λi = 0, τi)− yi(0, 0)

τi
dQ

Passing to the limit as τi → 0 one obtain (72). Likewise, since y(λi = 0, τi) − yi(0, 0)
verifies (61) with λ0 = 0, then from the regularities of the functions Gi 1; 2 and from the

Proposition 3, one shows that yτi = lim
τi→0

y(λi = 0, τi)− yi(0, 0)

τi
solves (74) and verifies

(76) for i = 1; 2. In the same ways setting yλi = lim
λi→0

y(λi, τi = 0)− yi(0, 0)

λi
, we proof

that yλi satisfies (73), (75) and (76).

Remark 3. Si i = 1; 2 is say to be a simultaneous sentinel if there exists a control
w ∈ L2(Qω) such that

∂Si
∂τi

(0, 0) = 0 i = 1; 2 (77)

and

‖w‖L2(Qω
= min

{
‖k‖L2(Qω

: k ∈ L2(Qω) and k verifies (77)
}

(78)

Following [9, 10], we show that the simultaneous sentinel problem is equivalent to the
following null controllability problem : find w ∈ L2(Qω) with minimal norm such that
(q1, q2) satisfies

−∂qi
∂t
− ∂qi
∂a
−∆qi + µiqi = βiG

′
i

(∫ A

0
β1y0ida

)
qi(t, 0, x)

+ hχO + wχω in Q,
qi(T, a, x) = 0 in QA,
qi(t, A, x) = 0 in QT ,

qi = 0 on Σ,

(79)

and

q1(0, a, x) = q2(0, a, x) = 0 in QA (80)

‖w‖L2(Qω) = min
k∈E
{‖k‖} (81)

where E =
{
k ∈ L2(Qω) such that (k, Si) satisfies (71) and (77)

}
.



REFERENCES 891

Remark 4. Setting G′1 = F and G′2 = G the problem (79)-(80) is exactly the problem
(1) that we have solved. Since E is closed and convex subset of L2(Qω), we can obtain w
to be of minimal norm in L2(Qω) by minimizing the norm of k, when k ∈ E.

6. Detection of the pollution term λiĝi i = 1; 2.

We know from the Corollary 1 that for each i = 1; 2 the function

yλi = lim
λi→0

y(λi, 0)− yi(0, 0)

λi
(82)

solve (75). Using the Taylor formula at the neighbourhood of (0; 0) we have :

Si(λi, τi) ≈ Si(0, 0) + λi
∂Si
∂λi

(0, 0) + τi
∂Si
∂τi

(0, 0) , i = 1; 2. (83)

According to (77), one deducts from (71), (73) and from the expression of Si(0, 0) that
(83) is equivalent to∫

Q

(hχO + wχω)yi(λi, τi)dQ =

∫
Q

(hχO + wχω)yi(0, 0)dQ+ λi

∫
Q

(hχO + wχω)yλidQ (84)

Thanks to (60), the equality (84) becomes

λi

∫
Q

(hχO + wχω)yλidQ =

∫
Q

(hχO + wχω)(m0i − yi(0, 0))dQ , i = 1; 2. (85)

Elsewhere, multiplying the first equation of (79) by yλi , i = 1; 2 and by integratings by
parts over Q, we have thanks to (75) and (80)∫

Σi

ĝi
∂qi
∂σ

dΣ =

∫
Q

(hχO + wχω)yλidQ i = 1; 2. (86)

where σ is the external unitary normal vector of Γ. Then (73) becames∫
Σi

λiĝi
∂qi
∂σ

dΣ ≈
∫
Q

(hχO + wχω)(m0i − yi(0, 0))dQ , i = 1; 2. (87)

Since qi, h, w, and yi(0, 0) i = 1; 2 are known, (87) is a integral equation in λiĝi that supply
some informations on the terms λiĝi i = 1; 2.

References

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