Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 167, pp. 1–26.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
LIPSCHITZ STABILITY FOR LINEAR PARABOLIC SYSTEMS
WITH INTERIOR DEGENERACY
IDRISS BOUTAAYAMOU, GENNI FRAGNELLI, LAHCEN MANIAR
Abstract. In this article, we study an inverse problem for linear degenerate
parabolic systems with one force. We establish Lipschitz stability for the source
term from measurements of one component of the solution at a positive time
and on a subset of the space domain, which contains degeneracy points. The
key ingredient is the derivation of a Carleman-type estimate.
1. Introduction
The null controllability and inverse problems of parabolic equations and parabolic coupled systems have attracted much interest in these last years, see [3,
4, 5, 6, 8, 15, 16, 17, 18, 23, 24, 25, 28, 29, 30]. The main result in these papers is the development of suitable Carleman estimates, which are crucial tools
to obtain observability inequalities and Lipschitz stability for term sources, initial
data, potentials and diffusion coefficients. The above systems are considered to
be non degenerate. In other words, the diffusion coefficients are uniformly coercive. On the contrary, the case of degenerate coefficients at the boundary is also
considered in several papers by developing adequate Carleman estimates. The null
controllability and inverse problems of degenerate parabolic equations are studied
in [10, 12, 13, 14, 27, 32], and for the coupled degenerate parabolic systems in
[1, 2, 7, 11, 26]. In these papers, the degeneracy considered is at the boundary of
the spatial domain.
After the pioneering works [19, 20], there has been substantial progress in understanding the null controllability of parabolic equations with interior degeneracy
(see, e.g., [21]). In this scope, the goal of this paper is to study an inverse source
problem of a 2×2 parabolic systems with interior degeneracy and different diffusion
coefficients
ut − (a1 ux )x + b11 u + b12 v = f, (t, x) ∈ Q,
vt − (a2 vx )x + b22 v = 0,
(t, x) ∈ Q,
u(t, 0) = u(t, 1) = v(t, 0) = v(t, 1) = 0,
u(0, x) = u0 (x),
v(0, x) = v0 (x),
t ∈ (0, T ),
x ∈ (0, 1),
2000 Mathematics Subject Classification. 35k65.
Key words and phrases. Parabolic system; interior degeneracy; Carleman estimates;
Lipschitz stability.
c
2014
Texas State University - San Marcos.
Submitted June 28, 2014. Published July 30, 2014.
1
(1.1)
2
I. BOUTAAYAMOU, G. FRAGNELLI, L. MANIAR
EJDE-2014/167
where u0 , v0 ∈ L2 (0, 1), T > 0 fixed, Q := (0, T ) × (0, 1), bij ∈ L∞ (0, 1), i, j = 1, 2,
and every ai , i = 1, 2, degenerates at an interior point xi of the spatial domain
(0, 1) (for the precise assumptions we refer to Section 2). For t0 ∈ (0, T ) given, let
0
QTt0 = (t0 , T ) × (0, 1) and T 0 := T +t
2 . For a given C0 > 0, we denote by S(C0 ) the
space
S(C0 ) := {f ∈ H 1 (0, T ; L2 (0, 1)) : |ft (t, x)| ≤ C0 |f (T 0 , x)|, a.e. (t, x) ∈ Q}.
More precisely, we want to establish Lipschitz stability for the source term f
from measurements of the component u at time T 0 and on a subset ω ⊂ (0, 1),
which contains the degeneracy points.
The main ingredient to obtain Lipschitz stability is Carleman estimates for degenerate equations. For null controllability of a parabolic equation with interior
degeneracy, Carleman estimates were obtained in [21] and in [20]. For inverse
problems, these estimates are not sufficient, and one needs also some additional
estimates on the term u with a special weight and the derivative term ut . We prove
first these for parabolic equations with interior degeneracy similar to the ones obtained in [10, 32] in the case of a boundary degeneracy. This will lead to obtain
our Carleman estimates for system (1.1). At the end having these Carleman estimates in hand, we follow the method developed in [5, 8, 24] to obtain the Lipschitz
stability for the source term f . The main task here is to estimate the source f by
the measurements, on the domain ω, of the first component u of the solutions of
system (1.1).
To prove our Carleman estimates, we use the following Hardy-Poincaré inequality
proved in [20, Proposition 2.1]
Z 1
Z 1
p(x)
2
w (x) dx ≤ CHP
p(x)|wx (x)|2 dx
(1.2)
2
0
0 (x − x0 )
for all functions w such that
Z
w(0) = w(1) = 0
and
1
p(x)|wx (x)|2 dx < ∞.
0
Here p is any continuous function in [0, 1], with p > 0 on [0, 1] \ {x0 }, p(x0 ) = 0,
for some x0 in (0, 1), and such that there exists ϑ ∈ (1, 2) so that the function
x 7→ p(x)/|x − x0 |ϑ is non-increasing on the left of x0 and nondecreasing on the
right of x0 .
This article is organized as follows: in Section 2, we discuss the well-posedness
of the system (1.1). Then, in Section 3, we establish different Carleman estimates
for parabolic equations and parabolic systems (1.1). Finally, in Section 4, we apply
the Carleman estimates to prove the Lipschitz stability result.
2. Assumptions and well-posedness
To study the well-posedness of system (1.1), we consider two situations, namely
the weakly degenerate (WD) and the strongly degenerate (SD) cases. The associated weighted spaces and assumptions on diffusion coefficients are the following:
Case (WD): for i = 1, 2, let
√
Ha1i (0, 1) := u abs. cont. in [0, 1], ai ux ∈ L2 (0, 1), u(0) = u(1) = 0 ,
EJDE-2014/167
LIPSCHITZ STABILITY FOR LINEAR PARABOLIC SYSTEMS
3
where the functions ai satisfy
there exists xi ∈ (0, 1), i = 1, 2 such that ai (xi ) = 0, ai > 0 in
[0, 1] \ {xi }, ai ∈ C 1 ([0, 1] \ {xi });
and there exists Ki ∈ (0, 1) such that (x − xi )a0i ≤ Ki ai , a.e. in
[0, 1].
(2.1)
Case (SD): for i = 1, 2, let
n
Ha1i (0, 1) := u ∈ L2 (0, 1) : u is locally abs. cont. in [0, 1] \ {xi },
o
√
ai ux ∈ L2 (0, 1), u(0) = u(1) = 0 ,
where the functions ai satisfy
there exists xi ∈ (0, 1), i = 1, 2 such that ai (xi ) = 0, ai > 0
in [0, 1] \ {xi }, ai ∈ C 1 ([0, 1] \ {xi }) ∩ W 1,∞ (0, 1); there exists
Ki ∈ [1, 2) such that (x − xi )a0i ≤ Ki ai a.e. in [0, 1], and if
Ki > 4/3, there exists γ ∈ (0, Ki ] such that ai /|x − xi |γ is nonincreasing on the left of xi and nondecreasing on the right of xi .
In both cases, for i = 1, 2, we consider the space
Ha2i (0, 1) := u ∈ Ha1i (0, 1) : ai ux ∈ H 1 (0, 1)
(2.2)
with the norms
√
kuk2Ha1 := kuk2L2 (0,1) + k ai ux k2L2 (0,1) ,
i
kuk2Ha2 := kuk2Ha1 + k(ai ux )x k2L2 (0,1) .
i
i
We recall from [21] that, for i = 1, 2, the operator (Ai , D(Ai )) defined by Ai u :=
(ai ux )x , u ∈ D(Ai ) = Ha2i (0, 1) is closed negative self-adjoint with dense domain
in L2 (0, 1). In the Hilbert space H := L2 (0, 1) × L2 (0, 1), the system (1.1) can be
transformed into the Cauchy problem
X 0 (t) = AX(t) − BX(t) + F (t), t ∈ (0, T ),
u0
X(0) =
,
v0
u(t)
A1 0
f (t)
where X(t) =
,A=
, D(A) = D(A1 ) × D(A2 ), F (t) =
v(t)
0 A2
0
b11 b12
and B =
.
0 b22
Since the operator A is diagonal and B is a bounded perturbation, the following
well-posedness and regularity results hold.
Proposition 2.1. (i) The operator A generates a contraction strongly continuous
semigroup.
(ii) For all (u0 , v0 ) ∈ D(A) and f ∈ H 1 (0, T ; L2 (0, 1)), the problem (1.1) has a
unique solution (u, v) ∈ C [0, T ], D(A) ∩ C 1 (0, T ; H).
(iii) For all f ∈ L2 (Q), u0 , v0 ∈ L2 (0, 1), and ε ∈ (0, T), there exists a unique
mild solution (u, v) ∈ XT := H 1 [ε, T ], H ∩ L2 ε, T ; D(A) of (1.1) satisfying
k(u, v)kXT ≤ CT k(u0 , v0 )k2H + k(F, G)k2H .
Moreover, for f ∈ H 1 (0, T ; L2 (0, 1)) and ε ∈ (0, T ), we have (u, v) ∈ YT :=
C [ε, T ], D(A) ∩ C 1 (ε, T ; H).
4
I. BOUTAAYAMOU, G. FRAGNELLI, L. MANIAR
EJDE-2014/167
3. Carleman estimate
The main topic of this section is to establish a Carleman estimate for a degenerate
parabolic single equation with a boundary observation on the right hand side. Then,
we will deduce the one for the degenerate system (1.1) with distributed observation
of u on the subdomain ω.
Part of these estimates were obtained in [20, 21] under two different assumptions
on the degenerate diffusion coefficient for a null controllability purpose. In the
forthcoming theorems we will prove additional estimates on u and ut , that are
crucial to prove Lipschitz stability results.
Throughout this section, we set ω = (λ, β) and assume, without loss of generality,
˜ ∪ ω := (λ, λ1 +2β1 ) ∪
x1 < x2 . Set also ω 0 := ω̃ ∪ ω := (λ, β1 ) ∪ (λ2 , β) and ω 00 := ω̃
3
2
( λ2 +2β
,
β),
where
λ
and
β
,
for
i
=
1,
2,
satisfy
0
<
λ
<
λ
<
β
<
x
<
x2 <
i
i
1
1
1
3
λ2 < β2 < β < 1.
3.1. Carleman estimate for an equation. In this subsection we shall derive the
Carleman estimate for the solution of the problem
ut − (a(x)ux )x + cu = h,
u(t, 0) = u(t, 1) = 0,
u(0, x) = u0 (x),
(t, x) ∈ Q,
t ∈ (0, T ),
(3.1)
x ∈ (0, 1),
where the diffusion coefficient a satisfies (2.1) or (2.2) in x0 ∈ (0, 1) with K in
place of Ki , i = 1, 2, h ∈ L2 (Q) and c ∈ L∞ (Q). As usual this aim relies on the
introduction of some suitable weight functions. Towards this end, as in [20, 21], we
define the following time and space weight functions
1
, η(t) := T + t0 − 2t,
[(t − t0 )(T − t)]4
Z x y − x
0
dy − c2 , ϕ(t, x) := θ(t)ψ(x),
ψ(x) = c1
x0 a(y)
(1−x0 )2
x20
where t0 ∈ (0, T ) is given, c1 > 0 and c2 > max a(1)(2−K)
, a(0)(2−K)
. For this
choice it is easy to prove that −c1 c2 ≤ ψ(x) < 0 for every x ∈ [0, 1], and that η is
0
0
positive if 0 < t < T +t
and negative if T +t
< t < T.
2
2
Now we are ready to state the Carleman estimate related to (3.1).
θ(t) :=
Theorem 3.1. Assume (2.1) or (2.2) and let T > 0. Then there exist two posi
tive constants C and s0 such that the solution u of (3.1) in H 1 [ε, T ], L2 (0, 1) ∩
L2 ε, T ; Ha2 (0, 1) satisfies, for all s ≥ s0 ,
Z
(x − x0 )2 2
1 sθa(x)u2x + s3 θ3
u + sθ3/2 |ηψ|u2 + u2t e2sϕ dx dt
a(x)
sθ
QT
t0
(3.2)
Z Th
Z
ix=1 ≤C
h2 e2sϕ dx dt + sc1
θa(x − x0 )u2x e2sϕ
dt .
QT
t
x=0
t0
0
Proof. Let u be the solution of (3.1). For s > 0, the function w = esϕ u satisfies
−(awx )x − sϕt w − s2 aϕ2x w + wt + 2saϕx wx + s(aϕx )x w = hesϕ − cw .
|
{z
} |
{z
} | {z }
L+
s w
L−
s w
hs
EJDE-2014/167
LIPSCHITZ STABILITY FOR LINEAR PARABOLIC SYSTEMS
5
Moreover, w(t0 , x) = w(T, x) = 0. This property allows us to apply the Carleman
estimates established in [21] to w with QTt0 in place of (0, T ) × (0, 1)
(x − x0 )2 2
w + sθa(x)wx2 dx dt
a(x)
QT
t0
Z Th
ix=1 dt .
h2 e2sϕ dx dt + sc1
θa(x − x0 )wx2
2
−
2
kL+
s wk + kLs wk +
≤C
Z
QT
t
Z
s3 θ3
x=0
t0
0
(3.3)
−
The operators L+
s and Ls are not exactly the ones of [20, 21]. However, one can
prove that the CarlemanR estimates do not change.
Using the previous estimate we
1 2
ut + sθ3/2 |ηψ|u2 e2sϕ dx dt. In fact, we have
will bound the integral QT sθ
t0
Z
sθ3/2 |ηψ|w2 dx dt
QT
t
0
Z
sθ3/2 w2 dx dt
≤C
QT
t
0
= sC Z
QT
t
0
3
≤ sC
2
3/4 |x − x |2 1/4
a1/3
0
2
w
w2
dx dt
θ3
a
|x − x0 |2/3
Z
2
a1/3
3 |x − x0 |
2
3 1
θ
θ
w
dx
dt
+
s
C
w2 dx dt,
2 QTt
a
|x − x0 |2/3
Z
QT
t
0
θ
0
since |η| ≤ T + t0 and |ψ| ≤ c1 c2 . Now, if K ≤ 4/3, we consider the function
p(x) = |x − x0 |4/3 . Obviously, there exists q ∈ (1, 4/3) such that the function
p(x)
|x−x0 |q is non-increasing on the left of x0 and nondecreasing on the right of x0 .
Then, we can apply the Hardy-Poincaré inequality (1.2), obtaining
Z
0
1
a1/3
w2 dx ≤ max a1/3 (x)
x∈[0,1]
|x − x0 |2/3
Z
1
0
Z
1
w2 dx
|x − x0 |2/3
1
p(x)
w2 dx
2
x∈[0,1]
0 |x − x0 |
Z 1
≤ max a1/3 (x)CHP
p(x)wx2 dx
= max a1/3 (x)
x∈[0,1]
0
= max a1/3 (x)CHP
x∈[0,1]
1
a
0
1/3
= max a
x∈[0,1]
Z
Z
|x − x0 |4/3 2
wx dx
a
1
(x)CHP C1
awx2 dx,
0
where
C1 = max
x4/3 (1 − x )4/3 0
0
,
.
a(0)
a(1)
In the previous inequality, we have used the property that the map x 7→ |x −
x0 |γ /a(x) is non-increasing on the left of x0 and nondecreasing on the right of x0
for all γ > K, see [20, Lemma 2.1]. If K > 4/3, we can consider the function
6
I. BOUTAAYAMOU, G. FRAGNELLI, L. MANIAR
EJDE-2014/167
2/3
2
0)
≤ C1 a(x), where
p(x) = (a(x)|x − x0 |4 )1/3 . Then p(x) = a(x) (x−x
a(x)
C1 := max
n x2 2/3 (1 − x )2 2/3 o
0
0
,
,
a(0)
a(1)
a1/3
p(x)
=
.
(x − x0 )2
|x − x0 |2/3
p(x)
4+γ
Moreover, using hypothesis (2.2), one has that the function |x−x
q , with q :=
3
0|
in (1, 2), is non-increasing on the left of x0 and nondecreasing on the right of x0 .
The Hardy-Poincaré inequality implies
Z 1
Z 1
Z 1
a1/3
p
2
2
w dx =
w dx ≤ CHP
p(wx )2 dx
2
2/3
(x
−
x
)
|x
−
x
|
0
0
0
0
0
Z 1
≤ CHP C1
a(wx )2 dx.
0
Thus, in every case,
Z
QT
t
0
a1/3
w2 dx dt ≤ C
θ
|x − x0 |2/3
Z
QT
t
θawx2 dx dt
(3.4)
0
for a positive constant C. Then, for s large enough, we have
Z
Z
|x − x0 |2 2 sθ3/2 |ηψ|w2 dx dt ≤ C
sθawx2 + s3 θ3
w dx dt,
(3.5)
a
QT
QT
t0
t0
Z
Z Th
Z
ix=1 sθ3/2 |ηψ|w2 dx dt ≤ C
h2 e2sϕ dx dt + sc1
θa(x − x0 )wx2
dt .
QT
t
QT
t
0
x=0
t0
0
(3.6)
On the other hand, we have
√
√
1
1
√ L−
wt + 2c1 sθ(x − x0 )wx + c1 sθw.
s w = √
sθ
sθ
Therefore,
Z
Z
Z
1 2
|x − x0 |2 2
2
wt dx dt ≤ C kL−
wk
+
sθ
aw
dx
dt
+
sθw2 dx dt
s
x
sθ
a
QT
QT
QT
t0
t
t0
Z 0
Z
2
2
2
≤ C kL−
wk
+
sθaw
dx
dt
+
sθw
dx
dt
,
s
x
QT
t
0
QT
t
0
(3.7)
√
since 1/ θ is bounded and
n x2 (1 − x )2 o
|x − x0 |2
0
0
≤ max
,
a(x)
a(0)
a(1)
(see
[20, Lemma 2.1]). Proceeding as in the proof of (3.4), we can estimate
R
2
dx dt thanks to the Hardy-Poincaré inequality (1.2),
sθw
QT
t0
Z
Z
sθw dx dt = s
QT
t
QT
t
0
0
3
≤s
2
3/4 |x − x |2 1/4
a1/3
0
2
2
w
θ
w
dx
dt
2/3
a
|x − x0 |
Z
a1/3
s
|x − x0 |2 2
2
θ
w
dx
dt
+
θ
w dx dt
2 QTt
a
|x − x0 |2/3
2
Z
QT
t
0
θ
0
EJDE-2014/167
LIPSCHITZ STABILITY FOR LINEAR PARABOLIC SYSTEMS
Z
≤C
QT
t
sθawx2 + s3 θ3
7
(x − x0 )2 2 w dx dt.
a
0
Hence, taking s large enough, one has
Z
Z Th
Z
ix=1 1 2
wt dx dt ≤ C
dt .
h2 e2sϕ dx dt + sc1
θa(x − x0 )wx2
sθ
x=0
QT
QT
t0
t
t
0
(3.8)
0
Now from (3.6) and (3.8) we can get the estimate of ut as follows: from the definition
of w, we have wt = ut esϕ + sϕt w. Hence
Z
Z
Z
1 2 2sϕ
1 2
s2 ϕ2t 2
ut e
dx dt ≤ 2
wt dx dt +
w dx dt .
sθ
sθ
sθ
QT
QT
QT
t
t
t
0
0
0
The second term in the above right-hand side is estimated as follows:
Z
Z
s2 ϕ2t 2
w dx dt = 16
sθ3/2 η 2 ψ 2 w2 dx dt
T
sθ
QT
Q
t0
t0
Z
≤ 16(T + t0 )c1 c2
sθ3/2 |ηψ|w2 dx dt.
QT
t
0
Hence using (3.6) and (3.8) we conclude that
Z
Z Th
Z
ix=1 1 2 2sϕ
2 2sϕ
ut e
dx dt ≤ C
h e
dx dt + sc1
θa(x − x0 )wx2
dt . (3.9)
sθ
x=0
QT
QT
t0
t
t
0
0
Thus (3.2) follows by (3.3), (3.4) and (3.9).
3.2. Carleman estimate for systems. By the above Carleman estimate (3.2),
we are able to show the main result of this section, which is the ω-Carleman estimate
for the system (1.1). For x ∈ [0, 1], let us define
Z x
y − xi
1
,
ψ
(x)
=
c
[
dy − di ],
ϕi (t, x) := θ(t)ψi (x), θ(t) :=
i
i
4
[(t − t0 )(T − t)]
xi ai (y)
and, for x ∈ [−1, 1],
Φi (t, x) := θ(t)Ψi (x),
Ψi (x) = e2ρi − eri ζi (x) ,
where
Z
1
ζi (x) =
x
dy
p
ãi (y)
(
,
ρi = ri ζi (−1),
ãi (x) :=
ai (x),
ai (−x),
x ∈ [0, 1],
x ∈ [−1, 0].
Here the functions ai , i = 1, 2, satisfy hypothesis (2.1) or (2.2) and the positive
constants ci , di , and ri are chosen such that
n
o
16A
x22
(1 − x2 )2
15
d2 >
max
,
, d?2 ,
< A < 1 (3.10)
16A − 15
(2 − K2 )a2 (0) (2 − K2 )a2 (1)
16
n
x21
(1 − x1 )2 o
d1 > max
,
,
(2 − K1 )a1 (0) (2 − K1 )a1 (1)
(3.11)
ρ2 > 2 ln(2), e2ρ1 − er1 ζ1 (0) ≥ e2ρ2 − 1,
n e2ρ2 − 1
(2 − K2 )a2 (1)(e2ρ2 − 1)
(2 − K2 )a2 (0)(e2ρ2 − 1) o
max
,
,
d2 − d?2 (2 − K2 )a2 (1)d2 − (1 − x2 )2 (2 − K2 )a2 (0)d2 − x22
(3.12)
4A 2ρ2
r2 ζ2 (0)
≤ c2 <
(e − e
)
3d2
8
I. BOUTAAYAMOU, G. FRAGNELLI, L. MANIAR
c1 ≥ max
EJDE-2014/167
n e2ρ1 − 1
(2 − K1 )a1 (1)(e2ρ1 − 1)
,
,
?
d1 − d1 (2 − K1 )a1 (1)d1 − (1 − x1 )2
(2 − K1 )a1 (0)(e2ρ1 − 1) c2 d2 o
,
,
(2 − K1 )a1 (0)d1 − x21 d1 − d?1
(3.13)
where
min Ψ2 (−x)
A=
,
max Ψ2 (x)
d?i
:= max
nZ
1
xi
y − xi
dy,
ai (y)
Z
0
xi
y − xi o
dy .
ai (y)
Remark 3.2. The interval
h
n e2ρ2 − 1
(2 − K2 )a2 (0)(e2ρ2 − 1) o
(2 − K2 )a2 (1)(e2ρ2 − 1)
,
max
,
,
d2 − d?2 (2 − K2 )a2 (1)d2 − (1 − x2 )2 (2 − K2 )a2 (0)d2 − x22
4A(e2ρ2 − er2 ζ(0) ) i
3d2
is not empty. In fact, from ρ2 > 2 ln 2, A > 15/16 and d2 > 16Ad?2 /(16A − 15), we
have
15
5
4A
d?
d?2
<1−
⇔ ≤
(1 − 2 )
d2
16A
4
3
d2
d?
4A
⇔ 1 + e−ρ2 <
(1 − 2 )
3
d2
2ρ2
4A 2ρ2
4A 2ρ2
e −1
<
(e − eρ2 ) <
(e − er2 ζ2 (0) ).
⇔
d2 − d?2
3d2
3d2
Similarly for
d2 >
n
16A
x22
(1 − x2 )2 o
max
,
16A − 15
(2 − K2 )a2 (0) (2 − K2 )a2 (1)
one has
max
<
n
(2 − K2 )a2 (1)(e2ρ1 − 1)
(2 − K2 )a2 (0)(e2ρ1 − 1) o
,
2
(2 − K2 )a2 (1)d2 − (1 − x2 )
(2 − K2 )a2 (0)d2 − x22
4A 2ρ2
(e − er2 ζ2 (0) ).
3d2
From (3.10)-(3.13), we have the following results.
Lemma 3.3. (i) For (t, x) ∈ [0, T ] × [0, 1],
ϕ1 ≤ ϕ2 ,
−Φ1 ≤ −Φ2 ,
ϕi ≤ −Φi .
(3.14)
4Φ2 (t, −x) + 3ϕ2 (t, x) > 0.
(3.15)
(ii) For (t, x) ∈ [0, T ] × [0, 1],
− Φ2 (t, x) ≤ −Φ2 (t, −x),
Proof. (i)
(1) ϕ1 ≤ ϕ2 : since θ ≥ 0 it is sufficient to prove ψ1 ≤ ψ2 . By the choice
d2
of c1 , we have c1 ≥ dc12−d
Then max{ψ1 (0), ψ1 (1)} ≤ −c2 d2 . Hence,
?.
1
ψ1 (x) ≤ ψ2 (x).
(2) −Φ1 ≤ −Φ2 :
since Ψi is increasing, it is sufficient to prove that min Ψ1 (x) ≥ max Ψ2 (x).
Indeed Ψ1 (0) = e2ρ1 − er1 ζ1 (0) ≥ e2ρ2 − 1 = Ψ2 (1).
EJDE-2014/167
LIPSCHITZ STABILITY FOR LINEAR PARABOLIC SYSTEMS
9
2ρi
(3) ϕi ≤ −Φi : since ci ≥ edi −d−1
? , then max{ψi (0), ψi (1)} ≤ −Ψi (1) and the
i
thesis follows immediately.
(ii)
(1) The first inequality follows from −Ψ2 (x) ≤ −Ψ2 (−x) for all x ∈ [0, 1].
(2) 4Ψ2 (−x) + 3ψ2 (x) > 0 : by definition of A, we have AΨ2 (x) ≤ Ψ2 (−x) and,
obviously, 4Ψ2 (−x) + 3ψ2 (x) ≥ 4AΨ2 (x) + 3ψ2 (x). Thus, to obtain the
thesis, it is sufficient to prove that 4AΨ2 (x)+3ψ2 (x) > 0. This follows easily
observing that, by the assumption 3c2 d2 < 4AΨ2 (0), −3ψ2 (x0 ) < 4AΨ2 (0).
Hence −3ψ2 (x) ≤ −3ψ2 (x0 ) < 4AΨ2 (0) ≤ 4AΨ2 (x) for all x ∈ [0, 1].
We show first an intermediate Carleman estimate with distributed observation
of u and v.
Theorem 3.4. Let T > 0. There exist two positive constants C and s0 such that,
for every (u0 , v0 ) ∈ H and all s ≥ s0 , the solution of (1.1) satisfies
Z
1 (x − x1 )2 2
u + sθ3/2 |ηψ1 |u2 + u2t e2sϕ1 dx dt
sθa1 u2x + s3 θ3
a1
sθ
QT
t0
Z
(x − x2 )2 2
1 +
sθa2 vx2 + s3 θ3
v + sθ3/2 |ηψ2 |v 2 + vt2 e2sϕ2 dx dt
a2
sθ
QT
t0
Z
Z
Z
T
≤C
f 2 e−2sΦ2 (t,−x) dx dt +
s2 θ2 (u2 + v 2 )e−2sΦ2 (t,−x) dx dt .
QT
t
ω0
t0
0
For the proof we shall use the following classical Carleman estimate (see [20]).
Proposition 3.5. Let z be the solution of
zt − (azx )x = h,
x ∈ (A, B), t ∈ (0, T ),
z(t, A) = z(t, B) = 0,
t ∈ (0, T ),
where a ∈ C 1 ([A, B]) is a strictly positive function. Then there exist two positive
constants r and s0 such that for any s ≥ s0 ,
Z TZ B
Z TZ B
rζ 2 −2sΦ
sθe zx e
dx dt +
s3 θ3 e3rζ z 2 e−2sΦ dx dt
(3.16)
t0
A
Z
≤c
t0
T
t0
Z
B
h2 e−2sΦ dx dt − c
A
Z
A
T
σ(t, ·)zx2 (t, ·)e−2sΦ(t,·)
t0
dt
x=0
x=1
(3.17)
for some positive constant c. Here the functions Φ, σ and ζ are defined, for r, s > 0
and (t, x) ∈ [0, T ] × [A, B], by
φ(t, x) := θ(t)Ψ(x), Ψ(x) := e2rζ(A) − erζ(x) > 0,
Z B
1
p
dy, σ(t, x) := rsθ(t)erζ(x) .
ζ(x) :=
a(y)
x
Proof of Theorem 3.4. Consider a cut-off function ξ : [0, 1] → R such that
0 ≤ ξ(x) ≤ 1,
for all x ∈ [0, 1],
ξ(x) = 1,
x ∈ [λ1 , β2 ],
ξ(x) = 0,
x ∈ [0, 1] \ ω.
10
I. BOUTAAYAMOU, G. FRAGNELLI, L. MANIAR
EJDE-2014/167
Define w := ξu and z := ξv, where (u, v) is the solution of (1.1). Hence, w and z
satisfy the system
wt − (a1 wx )x + b11 w = ξf − b12 z − (a1 ξx u)x − ξx a1 ux =: g,
zt − (a2 zx )x + b22 z = −(a2 ξx v)x − ξx a2 vx ,
w(t, 0) = w(t, 1) = z(t, 0) = z(t, 1) = 0,
(t, x) ∈ Q,
(t, x) ∈ Q,
t ∈ (0, T ).
Applying the estimate (3.2) and using w = wx = 0 in a neighborhood of x = 0 and
x = 1, from the definition of ξ, we have
Z
1
(x − x1 )2 2
w + sθ3/2 |ηψ1 |w2 + wt2 e2sϕ1 dx dt
sθa1 wx2 + s3 θ3
a1
sθ
QT
t0
Z
≤C
g 2 e2sϕ1 dx dt
QT
t
0
for all s ≥ s0 . Then using the fact that ξx and ξxx are supported in ω 00 , we can
write
g 2 ≤ C(ξ 2 f 2 + b212 z 2 + (u2 + u2x )χω00 ).
Hence, applying Cacciopoli inequality (5.1) and the previous estimates, we obtain
Z
(x − x1 )2 2
1
sθa1 wx2 + s3 θ3
w + sθ3/2 |ηψ1 |w2 + wt2 e2sϕ1 dx dt
a1
sθ
QT
t0
Z
Z
ξ 2 f 2 e2sϕ1 dx dt +
≤C
b212 z 2 e2sϕ1 dx dt
(3.18)
QT
t
QT
t
0
Z
T
0
Z
((1 + s2 θ2 )u2 + f 2 )e2sϕ1 dx dt .
+
t0
ω0
Arguing as before and applying the estimate (3.2) to the second component z of
the system, we obtain
Z
1 (x − x2 )2 2
sθa2 zx2 + s3 θ3
z + sθ3/2 |ηψ2 |z 2 + zt2 e2sϕ2 dx dt
a2
sθ
QT
t0
Z TZ
≤C
(v 2 + vx2 )e2sϕ2 dx dt.
t0
ω 00
Hence, Cacciopoli inequality (5.1) yields
Z
1 (x − x2 )2 2
z + sθ3/2 |ηψ2 |z 2 + zt2 e2sϕ2 dx dt
sθa2 zx2 + s3 θ3
a2
sθ
QT
t0
Z TZ
≤C
(1 + s2 θ2 )v 2 e2sϕ2 dx dt.
t0
(3.19)
ω0
On the other hand, using the Poincaré inequality and the fact that ϕ1 < ϕ2 , we
have
Z 1
Z 1
b212 z 2 e2sϕ1 dx ≤ kb12 k2∞
(zesϕ2 )2 dx
0
0
= kb12 k2∞
Z
0
1
|x − x |2
1/4 a1/3 (x)
3/4
2
2 2sϕ2
2
z 2 e2sϕ2
z
e
dx
a2 (x)
|x − x2 |2/3
EJDE-2014/167
LIPSCHITZ STABILITY FOR LINEAR PARABOLIC SYSTEMS
≤
kb12 k2∞
4
Z
11
1
0
3kb12 k2∞
+
4
|x − x2 |2 2 2sϕ2
dx
z e
a2 (x)
Z 1
1/3
a2 (x) 2 2sϕ2
dx.
z e
2/3
0 |x − x2 |
Applying the Hardy-Poincaré inequality to w(t, x) := esϕ(t,x) z(t, x) and proceeding
as in the proof of (3.4), one has
Z 1
Z 1
Z 1
1/3
1/3
a2 (x)
a2 (x) 2 2sϕ2
2
dx
=
a(wx )2 dx
z
e
w
dx
≤
C
2/3
2/3
0 |x − x2 |
0 |x − x2 |
0
Z 1
Z 1
|x − x2 |2 2 2sϕ
s2 θ2
≤C
a2 zx2 e2sϕ2 dx,
z e dx + C
a
(x)
2
0
0
2
since ψ2,x (x) = c2 x−x
a2 (x) . Hence, for a universal positive constant C, it results
Z
Z
(x − x2 )2 2 2sϕ2
b212 z 2 e2sϕ1 dx dt ≤ C
(a2 zx2 + s2 θ2
z )e
dx dt.
a2
QT
QT
t
t
0
0
sθ
2 ,
Taking s such that C ≤
we have
Z
Z
(x − x2 )2 2 2sϕ2
1
2 2 2sϕ1
sθa2 zx2 + s3 θ3
z e
dx dt.
b12 z e
dx dt ≤
2 QTt
a2
QT
t
(3.20)
0
0
Combining (3.18), (3.19) and (3.20) we obtain for s large enough
Z
1
(x − x1 )2 2
w + sθ3/2 |ηψ1 |w2 + wt2 e2sϕ1 dx dt
sθa1 wx2 + s3 θ3
a1
sθ
QT
t0
Z
(x − x2 )2 2
1 +
sθa2 zx2 + s3 θ3
z + sθ3/2 |ηψ2 |z 2 + zt2 e2sϕ2 dx dt
a2
sθ
QT
t0
Z
Z
Z
T
≤C
ξ 2 f 2 e2sϕ1 dx dt +
s2 θ2 v 2 e2sϕ2 dx dt
QT
t
0
Z
T
Z
+
t0
t0
ω0
(s2 θ2 u2 + f 2 )e2sϕ1 dx dt .
ω0
By the previous inequality, the definition of w and z, it follows that
Z T Z β1 (x − x1 )2 2
1 sθa1 u2x + s3 θ3
u + sθ3/2 |ηψ1 |u2 + u2t e2sϕ1 dx dt
a1
sθ
t0
λ1
Z T Z β1 2
(x − x2 ) 2
1 +
sθa2 vx2 + s3 θ3
v + sθ3/2 |ηψ2 |v 2 + vt2 e2sϕ2 dx dt
a2
sθ
t0
λ1
Z T Z β1 2
(x − x1 ) 2
1
=
sθa1 wx2 + s3 θ3
w + sθ3/2 |ηψ1 |w2 + wt2 e2sϕ1 dx dt
a1
sθ
t0
λ1
Z T Z β1 2
(x − x2 ) 2
1 +
sθa2 zx2 + s3 θ3
z + sθ3/2 |ηψ2 |z 2 + zt2 e2sϕ2 dx dt
a2
sθ
t0
λ1
(3.21)
Z
2
(x − x1 ) 2
1
≤
sθa1 wx2 + s3 θ3
w + sθ3/2 |ηψ1 |w2 + wt2 e2sϕ1 dx dt
T
a1
sθ
Qt
0
12
I. BOUTAAYAMOU, G. FRAGNELLI, L. MANIAR
Z
+
QT
t
0
≤C
Z
(x − x2 )2 2
1 z + sθ3/2 |ηψ2 |z 2 + zt2 e2sϕ2 dx dt
a2
sθ
Z TZ
s2 θ2 v 2 e2sϕ2 dx dt
ξ 2 f 2 e2sϕ1 dx dt +
sθa2 zx2 + s3 θ3
QT
t
t0
0
Z
T
Z
ω0
(s2 θ2 u2 + f 2 )e2sϕ1 dx dt .
+
t0
EJDE-2014/167
(3.22)
ω0
Now define U = χu and V = χv, where (u, v) is the solution of (1.1) and χ :
[0, 1] → R is a cut-off function defined as
0 ≤ χ(x) ≤ 1,
χ(x) = 1,
χ(x) = 0,
x ∈ [0, 1],
x ∈ [β2 , 1],
x ∈ [0,
λ2 + 2β2
].
3
Then U and V satisfy
Ut − (a1 Ux )x + b11 U = χf − b12 V − (a1 χx u)x − χx a1 ux ,
Vt − (a2 Vx )x + b22 V = −(a2 χx v)x − χx a2 vx ,
U (t, 0) = U (t, 1) = V (t, 0) = V (t, 1) = 0,
(t, x) ∈ Q
(t, x) ∈ Q
t ∈ (0, T ).
Using (3.16) and a technique similar to the one used in (3.7)-(3.8), one has
Z
1
(sθer1 ζ1 Ux2 + s3 θ3 e3r1 ζ1 U 2 + Ut2 )e−2sΦ1 dx dt
T
sθ
Qt
0
Z
Z
≤C
χ2 f 2 e−2sΦ1 dx dt + C̃
V 2 e−2sΦ1 dx dt
QT
t
QT
t
0
Z
0
T
Z
+C
(u2 + u2x )e−2sΦ1 dx dt.
ω
t0
Analogously, one can prove that V satisfies
Z
1
(sθer2 ζ2 Vx2 + s3 θ3 e3r2 ζ2 V 2 + Vt2 )e−2sΦ2 dx dt
sθ
QT
t0
Z TZ
≤C
(v 2 + vx2 )e−2sΦ2 dx dt.
t0
ω
Thus combining the last two inequalities,
Z
1
(sθer1 ζ1 Ux2 + s3 θ3 e3r1 ζ1 U 2 + Ut2 )e−2sΦ1 dx dt
T
sθ
Qt
Z0
1
+
(sθer2 ζ2 Vx2 + s3 θ3 e3r2 ζ2 V 2 + Vt2 )e−2sΦ2 dx dt
T
sθ
Qt
Z0
Z
≤C
χ2 f 2 e−2sΦ1 dx dt + C̃
V 2 e−2sΦ1 dx dt
QT
t
QT
t
0
Z
0
T
Z
+C
t0
ω
(u2 + u2x )e−2sΦ1 dx dt + C
Z
T
t0
Z
ω
(v 2 + vx2 )e−2sΦ2 dx dt.
EJDE-2014/167
LIPSCHITZ STABILITY FOR LINEAR PARABOLIC SYSTEMS
13
Taking s such that C̃ ≤ 12 s3 θ3 e3r2 ζ2 , using −Φ1 ≤ −Φ2 and Cacciopoli inequality
(5.1), we obtain
Z
1
(sθer1 ζ1 Ux2 + s3 θ3 e3r1 ζ1 U 2 + Ut2 )e−2sΦ1 dx dt
T
sθ
Qt
Z0
1
1
+
(sθer2 ζ2 Vx2 + s3 θ3 e3r2 ζ2 V 2 + Vt2 )e−2sΦ2 dx dt
2
sθ
QT
t0
Z TZ
Z
(s2 θ2 u2 + f )e−2sΦ1 dx dt
≤C
χ2 f 2 e−2sΦ1 dx dt + C
QT
t
t0
0
Z
T
Z
+C
t0
ω
s2 θ2 v 2 e−2sΦ2 dx dt.
ω
Then, by Lemma 3.3, one can prove that there exists a positive constant C such
that for every (t, x) ∈ [0, T ] × supp(χ),
C
(x − xi )2 2sϕi (t,x)
e
≤ e3ri ζi e−2sΦi .
ai (x)
2
ai (x)e2sϕi (t,x) ≤ Ceri ζi e−2sΦi ,
(3.23)
Consequently,
Z
1
(x − x1 )2 2
U + sθ3/2 |ηψ1 |U 2 + Ut2 e2sϕ1 dx dt
sθa1 Ux2 + s3 θ3
a1
sθ
QT
t0
Z
1
(x − x2 )2 2
+
V + sθ3/2 |ηψ2 |V 2 + Vt2 e2sϕ2 dx dt
sθa2 Vx2 + s3 θ3
a2
sθ
QT
t0
Z
Z
Z
T
≤C
χ2 f 2 e−2sΦ1 dx dt +
(s2 θ2 u2 + f )e−2sΦ1 dx dt
QT
t
t0
0
Z
T
Z
+
t0
ω
s2 θ2 v 2 e−2sΦ2 dx dt .
ω
By the definitions of U and V , we obtain
Z T Z 1
(x − x1 )2 2
1 sθa1 u2x + s3 θ3
u + sθ3/2 |ηψ1 |u2 + u2t e2sϕ1 dx dt
a1
sθ
t0
β2
Z T Z 1
2
1 (x − x2 ) 2
v + sθ3/2 |ηψ2 |v 2 + vt2 e2sϕ2 dx dt
+
sθa2 vx2 + s3 θ3
a2
sθ
t0
β2
Z T Z 1
2
(x − x1 ) 2
1
=
sθa1 Ux2 + s3 θ3
U + sθ3/2 |ηψ1 |U 2 + Ut2 e2sϕ1 dx dt
a1
sθ
t0
β2
Z T Z 1
2
1
(x − x2 ) 2
V + sθ3/2 |ηψ2 |V 2 + Vt2 e2sϕ2 dx dt
+
sθa2 Vx2 + s3 θ3
a2
sθ
t0
β2
Z
2
(x − x1 ) 2
1
≤
sθa1 Ux2 + s3 θ3
U + sθ3/2 |ηψ1 |U 2 + Ut2 e2sϕ1 dx dt
a1
sθ
QT
t0
Z
2
(x − x2 ) 2
1
+
sθa2 Vx2 + s3 θ3
V + sθ3/2 |ηψ2 |V 2 + Vt2 e2sϕ2 dx dt
a2
sθ
QT
t0
Z
Z
Z
T
≤C
χ2 f 2 e−2sΦ1 dx dt +
(s2 θ2 u2 + f )e−2sΦ1 dx dt
QT
t
0
t0
ω
14
I. BOUTAAYAMOU, G. FRAGNELLI, L. MANIAR
Z
T
Z
s2 θ2 v 2 e−2sΦ2 dx dt .
+
t0
EJDE-2014/167
(3.24)
ω
To complete the proof it is sufficient to prove a similar inequality on the interval
[0, λ1 ]. To this aim define the functions
(
(
u(t, x),
x ∈ [0, 1],
v(t, x),
x ∈ [0, 1],
, Z(t, x) :=
W (t, x) :=
−u(t, −x), x ∈ [−1, 0],
−v(t, −x), x ∈ [−1, 0],
where (u, v) solves (1.1), and
(
f (t, x),
x ∈ [0, 1],
˜
,
f (t, x) :=
−f (t, −x), x ∈ [−1, 0],
(
bij (x),
b̃ij (x) :=
bij (−x),
x ∈ [0, 1],
x ∈ [−1, 0].
Therefore, (W, Z) solves
Wt − (ã1 Wx )x + b̃11 W = f˜ − b̃12 Z,
Zt − (ã2 Zx )x + b̃22 Z = 0,
x ∈ (−1, 1), t ∈ (0, T ),
x ∈ (−1, 1), t ∈ (0, T ),
W (t, −1) = W (t, 1) = Z(t, −1) = Z(t, 1) = 0,
(3.25)
t ∈ (0, T ).
Consider a cut-off function ρ : [−1, 1] → R such that
0 ≤ ρ(x) ≤ 1,
ρ(x) = 1,
x ∈ [−1, 1],
x ∈ [−λ1 , λ1 ],
λ1 + 2β1
λ1 + 2β1
]∪[
, 1].
3
3
The functions p = ρW and q = ρZ, where (W, Z) is the solution of (3.25), satisfy
ρ(x) = 0,
x ∈ [−1, −
pt − (ã1 px )x + b̃11 p = ρf˜ − b̃12 q − (ã1 ρx W )x − ρx ã1 Wx ,
qt − (ã2 qx )x + b̃22 q = −(ã2 ρx Z)x − ρx ã2 Zx ,
x ∈ (−1, 1), t ∈ (0, T ),
x ∈ (−1, 1), t ∈ (0, T ),
p(t, −1) = p(t, 1) = q(t, −1) = q(t, 1) = 0,
t ∈ (0, T ).
Applying (3.16) for the first component p with A = −β1 , B = β1 and proceeding
as in (3.7)-(3.8), we obtain
Z T Z β1
1
(sθer1 ζ1 p2x + s3 θ3 e3r1 ζ1 p2 + p2t )e−2sΦ1 dx dt
sθ
t0
−β1
Z T Z β1
Z T Z β1
≤C
ρ2 f˜2 e−2sΦ1 dx dt + C̃
q 2 e−2sΦ1 dx dt
−β1
t0
Z
T
Z
t0
λ1 +2β1
3
+C
t0
Z T
λ1
Z −λ1
+C
t0
−
λ1 +2β1
3
−β1
(W 2 + Wx2 )e−2sΦ1 dx dt
(W 2 + Wx2 )e−2sΦ1 dx dt.
Using the definition of W , −Φ1 ≤ −Φ2 and e−2sΦ2 (t,x) ≤ e−2sΦ2 (t,−x) , a.e. x ∈
[0, β1 ], it follows that
Z T Z β1
1
(sθer1 ζ1 p2x + s3 θ3 e3r1 ζ1 p2 + p2t )e−2sΦ1 dx dt
sθ
t0
−β1
EJDE-2014/167
LIPSCHITZ STABILITY FOR LINEAR PARABOLIC SYSTEMS
T
Z
β1
Z
≤C
ρ2 f 2 e−2sΦ2 (t,−x) dx dt + C̃
Z
0
t0
Z
T
Z
t0
Z
λ1 +2β1
3
+C
t0
T
β1
15
q 2 e−2sΦ1 dx dt
−β1
(u2 + u2x )e−2sΦ1 (t,−x) dx dt.
λ1
Similarly, for the second component q, we obtain
Z T Z β1
1
(sθer2 ζ2 qx2 + s3 θ3 e3r2 ζ2 q 2 + qt2 )e−2sΦ2 dx dt
sθ
−β1
t0
Z T Z λ1 +2β
1
3
≤C
(v 2 + vx2 )e−2sΦ2 (t,−x) dx dt.
t0
λ1
Thus it follows from the last two inequalities that
Z T Z β1
1
(sθer1 ζ1 p2x + s3 θ3 e3r1 ζ1 p2 + p2t )e−2sΦ1 dx dt
sθ
t0
−β1
Z T Z β1
1
+
(sθer2 ζ2 qx2 + s3 θ3 e3r2 ζ2 q 2 + qt2 )e−2sΦ2 dx dt
sθ
t0
−β1
Z T Z β1
Z T Z β1
2 2 −2sΦ2 (t,−x)
≤C
ρ f e
dx dt + C̃
q 2 e−2sΦ1 dx dt
t0
0
Z
T
t0
Z
λ1 +2β1
3
+C
t0
Z
T
(v 2 + vx2 )e−2sΦ2 (t,−x) dx dt
λ1
Z
λ1 +2β1
3
+C
t0
−β1
(u2 + u2x )e−2sΦ1 (t,−x) dx dt.
λ1
Taking s such that C̃ ≤ 12 s3 θ3 e3r2 ζ2 , using −Φ1 ≤ −Φ2 and Cacciopoli inequality
(5.1), we obtain
Z T Z β1
1
(sθer1 ζ1 p2x + s3 θ3 e3r1 ζ1 p2 + p2t )e−2sΦ1 dx dt
sθ
t0
−β1
Z T Z β1
1
1
+
(sθer2 ζ2 qx2 + s3 θ3 e3r2 ζ2 q 2 + qt2 )e−2sΦ2 dx dt
2
sθ
t0
−β1
Z T Z β1
Z TZ
2 2 −2sΦ2 (t,−x)
≤C
ρ f e
dx dt + C
(s2 θ2 u2 + f )e−2sΦ1 (t,−x) dx dt
t0
Z
0
T Z
+C
t0
t0
ω̃
s2 θ2 v 2 e−2sΦ2 (t,−x) dx dt.
ω̃
(3.26)
Clearly, one can obtain the same estimates (3.23) in [0, T ] × [0, β1 ]. Hence, by
(3.14), (3.26), by the definitions of W , Z, p and q, and proceeding as in (3.5), we
obtain
Z T Z λ1 1 (x − x1 )2 2
u + sθ3/2 |ηψ1 |u2 + u2t e2sϕ1 dx dt
sθa1 u2x + s3 θ3
a1
sθ
t0
0
Z T Z λ1 2
(x − x2 ) 2
1 +
sθa2 vx2 + s3 θ3
v + sθ3/2 |ηψ2 |v 2 + vt2 e2sϕ2 dx dt
a2
sθ
t0
0
16
I. BOUTAAYAMOU, G. FRAGNELLI, L. MANIAR
T
λ1
EJDE-2014/167
(x − x1 )2 2
1
W + sθ3/2 |ηψ1 |W 2 + Wt2 e2sϕ1 dx dt
a1
sθ
0
t0
Z T Z λ1 2
(x − x1 ) 2
1
sθa2 Zx2 + s3 θ3
+
Z + sθ3/2 |ηψ2 |Z 2 + Zt2 e2sϕ2 dx dt
a2
sθ
0
t0
Z T Z λ1 2
(x − x1 ) 2
1 sθa1 p2x + s3 θ3
=
p + sθ3/2 |ηψ1 |p2 + p2t e2sϕ1 dx dt
a1
sθ
0
t0
Z T Z λ1 2
(x − x2 ) 2
1 sθa2 qx2 + s3 θ3
+
q + sθ3/2 |ηψ2 |q 2 + qt2 e2sϕ2 dx dt
a2
sθ
0
t0
Z T Z β1
1
(sθer1 ζ1 p2x + s3 θ3 e3r1 ζ1 p2 + p2t )e−2sΦ1 dx dt
≤C
sθ
−β1
t0
Z T Z β1
1
(sθer2 ζ2 qx2 + s3 θ3 e3r2 ζ2 q 2 + qt2 )e−2sΦ2 dx dt
+C
sθ
t0
−β1
Z T Z β1
Z TZ
≤C
ρ2 f 2 e−2sΦ2 (t,−x) dx dt + C
(s2 θ2 u2 + f )e−2sΦ1 (t,−x) dx dt
Z
Z
=
t0
Z
sθa1 Wx2 + s3 θ3
0
T Z
+C
t0
ω̃
t0
s2 θ2 v 2 e−2sΦ2 (t,−x) dx dt.
(3.27)
ω̃
Finally adding up (3.22), (3.24) and (3.27), the proof is complete.
To estimate the term source with only the first component of solutions of (1.1),
we need to show a Carleman estimate with one force.
Theorem 3.6. Let T > 0. Moreover, assume that
b12 > µ > 0
on ω 0 .
There exist two positive constants C and s0 such that, for every (u0 , v0 ) ∈ H and
for all s ≥ s0 , the solution (u, v) of (1.1) satisfies
Z
1 (x − x1 )2 2
u + sθ3/2 |ηψ1 |u2 + u2t e2sϕ1 dx dt
I(u, v) :=
sθa1 u2x + s3 θ3
a1
sθ
QT
t0
Z
1 (x − x2 )2 2
+
v + sθ3/2 |ηψ2 |v 2 + vt2 e2sϕ2 dx dt
sθa2 vx2 + s3 θ3
a2
sθ
QT
t0
Z
Z TZ
≤C
f 2 e−2sΦ2 (t,−x) dx dt + C
u2 dx dt := I(f, u).
QT
t
t0
0
ω
(3.28)
The above theorem is a consequence of Theorem 3.4 and of the following lemma.
Lemma 3.7. Let ω2 b ω1 such that x1 , x2 ∈
/ ω1 . Moreover, assume that
b12 > µ > 0
on ω1 .
There exists C > 0 such that
Z TZ
Z
2 2 2 −2sΦ2 (t,−x)
s θ v e
dx dt ≤ εJ(v) + C
t0
ω2
2 2sϕ2
f e
QT
t
0
Z
T
Z
dx dt + C
t0
ω
u2 dx dt,
EJDE-2014/167
LIPSCHITZ STABILITY FOR LINEAR PARABOLIC SYSTEMS
17
where ε > 0 is small enough, s is large enough and
Z
(x − x2 )2 2 2sϕ2
J(v) =
sθa2 vx2 + s3 θ3
v e
dx dt.
a2
QT
t
0
Proof. The choice of the weight functions given in Lemma 3.3 will play a crucial role.
We will adapt the technique used in [2]. Let χ ∈ C ∞ (0, 1), such that supp(χ) ⊂ ω1
and χ ≡ 1 on ω2 . Multiplying the first equation of (1.1) by s2 θ2 χe−2sθ(t)Ψ2 (−x) v
and integrating over QTt0 , we obtain
Z
s2 θ2 b12 χe−2sθ(t)Ψ2 (−x) v 2 dx dt
QT
t
Z0
s2 θ2 χe−2sθ(t)Ψ2 (−x) f v dx dt −
=
Z
QT
t0
s2 θ2 χe−2sθ(t)Ψ2 (−x) ut v dx dt
QT
t0
Z
(3.29)
s2 θ2 χe−2sθ(t)Ψ2 (−x) (a1 ux )x v dx dt
+
QT
t
0
Z
s2 θ2 b11 χe−2sθ(t)Ψ2 (−x) uv dx dt.
−
QT
t
0
Integrating by parts and using the second equation in (1.1), we obtain
Z
s2 θ2 χe−2sθ(t)Ψ2 (−x) vut dx dt
QT
t
Z0
−2sθ(t)Ψ2 (−x)
2 2
=
s θ a2 χe
Z
Z0
+
s2 θ2 a2 (χe−2sθ(t)Ψ2 (−x) )x uvx dx dt
ux vx dx dt +
QT
t
QT
t
0
(s θ b22 + 2s θ θ̇Ψ2 (−x) − 2s θθ̇)χe−2sθ(t)Ψ2 (−x) vu dx dt,
2 2
3 2
2
QT
t0
(3.30)
and
Z
s2 θ2 χe−2sθ(t)Ψ2 (−x) (a1 ux )x v dx dt
QT
t0
Z
s2 θ2 a1 χe−2sθ(t)Ψ2 (−x) vx ux dx dt
=−
QT
t
0
Z
(3.31)
s2 θ2 a1 (χe−2sθ(t)Ψ2 (−x) )x vx u dx dt
+
QT
t
0
Z
s2 θ2 (a1 (χe−2sθ(t)Ψ2 (−x) )x )x vu dx dt.
+
QT
t
0
Combining (3.29)-(3.31), we obtain
Z
s2 θ2 b12 χe−2sθ(t)Ψ2 (−x) v 2 dx dt = I1 + I2 + I3 + I4 ,
QT
t
0
where
Z
I1 =
s2 θ2 χe−2sθ(t)Ψ2 (−x) vf dx dt,
QT
t0
Z
s2 θ2 (a1 + a2 )χe−2sθ(t)Ψ2 (−x) ux vx dx dt,
I2 = −
QT
t
0
18
I. BOUTAAYAMOU, G. FRAGNELLI, L. MANIAR
Z
EJDE-2014/167
s2 θ2 (a1 − a2 )(χe−2sθ(t)Ψ2 (−x) )x uvx dx dt
I3 =
QT
t
0
Z
(s2 θ2 χ0 + 2s3 θ3 Ψ2,x (−x)χ)(a1 − a2 )e−2sθ(t)Ψ2 (−x) uvx dx dt,
=
QT
t0
Z
(2s2 θθ̇ − s2 θ2 (b11 + b22 ) − 2s3 θ2 θ̇Ψ2 (−x))χe−2sθ(t)Ψ2 (−x) uv dx dt
I4 =
QT
t
Z0
s2 θ2 (a1 (χe−2sθ(t)Ψ2 (−x) )x )x uv dx dt.
+
QT
t0
For ε > 0, we have
Z
|I1 | =
(f esϕ2 )(s2 θ2 χe−s(2θ(t)Ψ2 (−x)+ϕ2 ) v) dx dt
QT
t
0
≤
Z
1
2
Z
f 2 e2sϕ2 dx dt + C
Z
QT
t0
T
t0
Z
s4 θ4 e−2s(2θ(t)Ψ2 (−x)+ϕ2 ) v 2 dx dt,
ω1
p
1
( sθa2 esϕ2 vx )((sθ)3/2 (a2 )− 2 (a1 + a2 )χe−s(2θ(t)Ψ2 (−x)+ϕ2 ) ux ) dx dt
|I2 | =
QT
t
Z0
≤ε
QT
t0
+
1
ε
sθa2 e2sϕ2 vx2 dx dt
Z
s3 θ3
QT
t
0
|
(a21 + a22 ) 2 −2s(2θ(t)Ψ2 (−x)+ϕ2 ) 2
χ e
ux dx dt .
a2
{z
}
L
The integral L should be estimated by an integral in u2 . For this, we multiply the
(a2 +a2 )
first equation in (1.1) by s3 θ3 a11 a22 χ2 e−2s(2θ(t)Ψ2 (−x)+ϕ2 ) u and we integrate by
parts, obtaining
L = L1 + L2 + L3 + L4 + L5 ,
where
Z
s3 θ3
L1 =
QT
t
0
Z
1
L2 =
2
(a21 + a22 ) 2 −2s(2θ(t)Ψ2 (−x)+ϕ2 )
χ e
uf dx dt,
a1 a2
s3 (3θ2 − 2sθ3 (2Ψ2 (−x) + ψ2 ))θ̇
QT
t
0
×e
L3 =
1
2
Z
(a21 + a22 ) 2
χ
a1 a2
−2s(2θ(t)Ψ2 (−x)+ϕ2 ) 2
u dx dt,
s3 θ3 (a1 (
QT
t
0
Z
L4 = −
s3 θ3
(a21 + a22 ) 2
χ b11 e−2s(2θ(t)Ψ2 (−x)+ϕ2 ) u2 dx dt,
a1 a2
s3 θ3
(a21 + a22 ) 2
χ b12 e−2s(2θ(t)Ψ2 (−x)+ϕ2 ) uv dx dt.
a1 a2
QT
t
0
Z
L5 = −
QT
t
0
(a21 + a22 ) 2 −2s(2θ(t)Ψ2 (−x)+ϕ2 )
χ e
)x )x u2 dx dt,
a1 a2
EJDE-2014/167
LIPSCHITZ STABILITY FOR LINEAR PARABOLIC SYSTEMS
19
Since |θ̇| ≤ Cθ2 and supp(χ) ⊂ ω1 , we obtain
Z
Z TZ
1
|L1 | ≤
f 2 e2sϕ2 dx dt + C
s6 θ6 e−2s(4θ(t)Ψ2 (−x)+3ϕ2 ) u2 dx dt,
2 QTt
t0
ω1
0
Z TZ
|L2 | ≤ C
s4 θ5 e−2s(2θ(t)Ψ2 (−x)+ϕ2 ) u2 dx dt,
t0
T
Z
ω1
Z
s5 θ5 e−2s(2θ(t)Ψ2 (−x)+ϕ2 ) u2 dx dt,
|L3 | ≤ C
t0
Z T
ω1
Z
s3 θ3 e−2s(2θ(t)Ψ2 (−x)+ϕ2 ) u2 dx dt,
|L4 | ≤ C
t0
Z
ω1
s3 θ3
|L5 | ≤ ε
QT
t
0
C
+
ε
Z
T
t0
(x − x2 )2 2sϕ2 2
v dx dt
e
a2
Z
s3 θ3 e−2s(4θ(t)Ψ2 (−x)+3ϕ2 ) u2 dx dt.
ω1
Hence,
Z
f 2 e2sϕ2 dx dt + Cε
|L| ≤ C
Z
QT
t0
T
Z
ω1
t0
Z
+ε
QT
t
0
s6 θ3 e−2s(4θ(t)Ψ2 (−x)+3ϕ2 ) u2 dx dt
(x − x2 )2 2sϕ2 2
s3 θ3
e
v dx dt.
a2
Furthermore,
Z
Z
|I2 | ≤ C
f 2 e2sϕ2 dx dt + Cε
QT
t
T
Z
t0
0
s6 θ6 e−2s(4θ(t)Ψ2 (−x)+3ϕ2 ) u2 dx dt + εJ(v).
ω1
Using the fact that χ0 and χ are supported in ω 0 and x2 6∈ ω 0 , proceeding as before,
one has
Z
|I3 | ≤ C
s3 θ3 (χ0 + χ)e−2sθ(t)Ψ2 (−x) uvx dx dt
QT
t
0
Z
p
1
( sθa2 esϕ2 vx )((sθ)5/2 (a2 )− 2 (χ0 + χ)e−s(2θ(t)Ψ2 (−x)+ϕ2 ) u) dx dt
=C
QT
t
0
Z
≤ε
QT
t
sθa2 vx2 e2sϕ2 dx dt +
0
C
ε
Z
T
t0
Z
(s5 θ5 e−2s(2θ(t)Ψ2 (−x)+ϕ2 ) u2 dx dt,
ω1
and
Z
s4 θ4 (χ00 + χ0 + χ)e−2sθ(t)Ψ2 (−x) uv dx dt
I4 ≤ C
QT
t
0
Z
√
a2
− x2 sϕ2
5/2
e v)((sθ)
(χ00 + χ0 + χ)
√
a2
x − x2
3/2 3/2 x
=C
(s
QT
t
0
θ
× e−s(2θ(t)Ψ2 (−x)+ϕ2 ) u) dx dt
Z
(x − x2 )2 2 2sϕ2
≤ε
s3 θ3
v e
dx dt
a2
QT
t
0
20
I. BOUTAAYAMOU, G. FRAGNELLI, L. MANIAR
+
T
Z
C
ε
Z
t0
EJDE-2014/167
(s5 θ5 e−2s(2θ(t)Ψ2 (−x)+ϕ2 ) u2 dx dt,
ω1
So, thanks to Lemma 3.3, we have
e−2s(2θ(t)Ψ2 (−x)+ϕ2 ) ≤ e−2s(4θ(t)Ψ2 (−x)+3ϕ2 ) ,
sup sr θr (t)e−2s(4θ(t)Ψ2 (−x)+3ϕ2 ) < ∞,
r ∈ R.
(t,x)∈Q
Then, for ε small enough and s large enough, we have
Z
|
s2 θ2 b12 χe−2sθ(t)Ψ2 (−x) v 2 dx dt|
QT
t
0
Z
f 2 e2sϕ2 dx dt + C
≤C
QT
t
Z
T
t0
0
Z
u2 dx dt + εJ(v).
ω
Finally, the definition of χ, (2.1) and the previous inequality give
Z TZ
µ
s2 θ2 e−2sθ(t)Ψ2 (−x) v 2 dx dt
t0
Z
≤|
Z
≤
ω2
T Z
s2 θ2 b12 e−2sθ(t)Ψ2 (−x) v 2 dx dt|
ω2
t0
|s2 θ2 b12 χe−2sθ(t)Ψ2 (−x) v 2 | dx dt
QT
t
0
Z
f 2 e2sϕ2 dx dt + C
≤C
QT
t
0
Z
T
t0
Z
u2 dx dt + εJ(v).
ω
This completes the proof.
4. The Lipschitz stability result
The object of this section is to recover a source term f from the knowledge of
(a1 ux )x (T 0 , ·) and some additional observations of the component u. The main
result of this paper reads as follows.
Theorem 4.1. Let C0 > 0. Then, there exists C = C(T, t0 , x1 , x2 , C0 ) > 0 such
that, for all f ∈ S(C0 ) and u0 ∈ L2 (0, 1),
kf k2L2 (QT ) ≤ C kuk2L2 (ωT ) + kut k2L2 (ωT ) + ku(T 0 , ·)k2L2 (0,1)
t0
t0
t
(4.1)
0
0
2
+ k(a1 ux )x (T , ·)kL2 (0,1) ,
where ωtT0 := (t0 , T ) × ω.
Proof. The functions y = ut and z = vt , where (u, v) is the solution of (1.1), are
solutions of the system
yt − (a1 yx )x + b11 y + b12 z = ft ,
zt − (a2 zx )x + b22 z = 0,
(t, x) ∈ Q,
(t, x) ∈ Q,
y(t, 0) = y(t, 1) = z(t, 0) = z(t, 1) = 0,
t ∈ (0, T ).
EJDE-2014/167
LIPSCHITZ STABILITY FOR LINEAR PARABOLIC SYSTEMS
21
When we apply Carleman estimate (3.28) to (y, z), we obtain
Z
(x − x1 )2 2
1 sθa1 yx2 + s3 θ3
I(y, z) :=
y + sθ3/2 |ηψ1 |y 2 + yt2 e2sϕ1 dx dt
a1
sθ
QT
t0
Z
2
(x − x2 ) 2
1 +
sθa2 zx2 + s3 θ3
z + sθ3/2 |ηψ2 |z 2 + zt2 e2sϕ2 dx dt
a2
sθ
QT
t0
Z
Z TZ
u2t dx dt
≤C
ft2 e−2sΦ2 (t,−x) dx dt + C
QT
t
t0
0
ω
:= I(ft , y).
(4.2)
The terms appearing in (4.1) are well defined,
indeed,
by
Proposition
2.1,
we
have
y ∈ L2 (t0 , T ; D(A1 )) ∩ H 1 t0 , T ; L2 (0, 1) .
As in [8], we divide the proof into three steps.
Step 1. We show first that there exists a constant C > 0 such that
I(f, u) + I(ft , y)
Z 1
0
1
≤C √
f 2 (T 0 , x)e−2sΦ2 (T ,−x) dx + kuk2L2 (ωT ) + kut k2L2 (ωT ) ,
t0
t0
s 0
(4.3)
where, we recall, T 0 = (T + t0 )/2. To obtain (4.3), it remains to prove that
Z 1
Z
0
C
(f 2 + ft2 )e−2sΦ2 (t,−x) dx dt ≤ √
f 2 (T 0 , x)e−2sΦ2 (T ,−x) dx.
T
s
0
Qt
0
Since Φ1,t (T 0 ) = 0 and Φ1,tt (t) ≥ µ0 > 0, then Taylor’s formula provides
µ0
−Φ2 (t, −x) ≤ −Φ2 (T 0 , −x) −
(t − T 0 )2 ,
2
and then
Z T
Z ∞
0
2
1
C
−2sΦ2 (t,−x)
−2sΦ2 (T 0 ,−x)
e
dt ≤ √
e
e−l dl ≤ √ e−2sΦ2 (T ,−x) .
µ
s
s
0
t0
−∞
So,
Z
2
0
f (T , x)e
−2sΦ2 (t,−x)
QT
t
C
dx dt ≤ √
s
0
Z
1
f 2 (T 0 , x)e−2sΦ2 (T
0
,−x)
dx.
0
For f ∈ S(C0 ), one has
|f (t, x)| ≤ |f (T 0 , x)| +
Z
t
|ft (s, x)|ds ≤ C|f (T 0 , x)|.
(4.4)
T0
Thus
Z
QT
t
C
(f 2 + ft2 )(t, x)e−2sΦ2 (t,−x) dx dt ≤ √
s
0
Z
1
f 2 (T 0 , x)e−2sΦ2 (T
0
,−x)
dx.
0
The purpose of the first step is acomplished.
Step 2. Now, let us show that there exists a constant C > 0 such that
Z 1
0
(y(T 0 , x) + b12 v(T 0 , x))2 e2sϕ1 (T ,x) dx ≤ C(I(y, z) + I(u, v)).
0
(4.5)
22
I. BOUTAAYAMOU, G. FRAGNELLI, L. MANIAR
EJDE-2014/167
Since, for a.e. x ∈ (0, 1),
lim (y(t, x) + b12 v(t, x))2 e2sϕ1 (t,x) = 0.
t→t0
Hence
Z
1
(y(T 0 , x) + b12 v(T 0 , x))2 e2sϕ1 (T
0
,x)
dx
0
Z
1
T0
Z
∂
(y + b12 v)2 e2sϕ1 (t,x) dt dx
∂t
=
0
Z
t0
T0
(4.6)
1
Z
(2(y + b12 v)(yt + b12 z) + 2sϕ1,t (y + b12 v)2 )e2sϕ1 (t,x) dx dt.
=
0
t0
Using Young inequality, for s large enough, we obtain
Z T0 Z 1
|
2(y + b12 v)(yt + b12 z)e2sϕ1 dx dt|
t0
0
Z
1 ≤C
sθy 2 + sθz 2 + sθv 2 + yt2 e2sϕ1 dx dt
T
sθ
Qt
(4.7)
0
and, by the Hardy inequality,
Z
sθy 2 e2sϕ1 dx dt
QT
t
0
Z
= s
QT
t
0
≤s
3
2
1/3
3/4 |x − x |2
1/4
a1
1
2 2sϕ1
2 2sϕ1
y
e
y
e
θ
dx
dt
2/3
a1
|x − x1 |
Z
1/3
a1
|x − x1 |2 2 2sϕ1
s
2 2sϕ1
θ
θ
y
e
dx
dt
+
y e
dx dt.
2/3
2 QTt
a1
|x − x1 |
Z
QT
t
0
θ
0
Moreover, by the Hardy-Poincaré inequality applied to yesϕ1 , one has
Z
Z
(x − x1 )2 2 2sϕ1
sθy 2 e2sϕ1 dx dt ≤ C
sθa1 [yx + 2sϕ1,x y]2 + s3 θ3
y e
dx dt
a1
QT
QT
t0
t0
Z
(x − x1 )2 2 2sϕ1
≤C
sθa1 yx2 + s3 θ3
y e
dx dt.
a1
QT
t
0
Similarly, by ϕ1 < ϕ2 , we have
Z
sθ(z 2 + v 2 )e2sϕ1 dx dt
QT
t
0
Z
≤C
QT
t
sθa2 (zx2 + vx2 ) + s3 θ3
0
(x − x2 )2 2
(z + v 2 ) e2sϕ2 dx dt.
a2
On the other hand, since |ϕ1,t | ≤ C|ηψ1 |θ3/2 , we have
Z T0 Z 1
sϕ1,t (y + b12 v)2 e2sϕ
t0
0
Z
≤C
sθ3/2 |ηψ1 |y 2 e2sϕ1 + sθ3/2 |ηψ2 |v 2 e2sϕ2 dx dt.
QT
t
0
Thus, (4.6)-(4.9) yield the estimate (4.5).
(4.8)
(4.9)
EJDE-2014/167
LIPSCHITZ STABILITY FOR LINEAR PARABOLIC SYSTEMS
23
Step 3. Combining (3.28), (4.2), (4.3) and (4.5), we deduce
Z 1
0
(y(T 0 , x) + b12 v(T 0 , x))2 e2sϕ1 (T ,x) dx
0
1 Z 1
0
≤C √
f 2 (T 0 , x)e−2sΦ2 (T ,−x) dx + kuk2L2 (ωT ) + kut k2L2 (ωT ) .
t0
t0
s 0
(4.10)
Since y + b12 v satisfies
y(T 0 , x) + b12 v(T 0 , x) = (a1 ux )x (T 0 , x) − b11 u(T 0 , x) + f (T 0 , x),
it follows that
Z 1
Z 1
0
0
f 2 (T 0 , x)e2sϕ1 (T ,x) dx ≤ C
(y(T 0 , x) + b12 v(T 0 , x))2 e2sϕ1 (T ,x) dx
0
0
+ k(a1 ux )x (T 0 )k2L2 (0,1) + ku(T 0 )k2L2 (0,1) .
(4.11)
Hence, by (4.10) and (4.11) we obtain, for s large enough,
Z 1
f 2 (T 0 , x)dx ≤ C kuk2L2 (ωT ) + kut k2L2 (ωT ) + ku(T 0 )k2L2 (0,1)
t0
t0
0
+ k(aux )x (T 0 )k2L2 (0,1) ,
which, together with (4.4), give the thesis.
5. Appendix
In this section, we show a Cacciopoli’s inequality for inhomogenous degenerate
parabolic equations. This inequality is different from the one shown in [20] for
homogenous case.
Proposition 5.1. Let ω 00 ⊂ ω 0 b ω ⊂ (0, 1) and x0 ∈
/ ω 0 . Let s ≥ s0 > 0, then
there exists a positive constant C = C(s0 , T, inf ω00 a(x), kckL∞ (Q) ) such that every
solution u of (3.1) satisfies
Z TZ
Z TZ
u2x e2sϕ dx dt ≤ C
(s2 θ2 u2 + h2 )e2sϕ dx dt.
(5.1)
t0
ω 00
t0
ω0
Proof. Define a smooth cut-off function ξ ∈ C ∞ ([0, 1]) such that ξ ≡ 1 in ω 00 and
supp(ξ) ⊂ ω 0 . Since u solves (3.1), we have
Z T Z 1
Z T Z 1
d
2 2sϕ 2
0=
ξ e u dx dt =
2sξ 2 ϕt e2sϕ u2 + 2ξ 2 e2sϕ uut dx dt
t0 dt
0
t0
0
Z TZ 1
=
[2sξ 2 ϕt e2sϕ u2 + 2ξ 2 e2sϕ u((aux )x + h − cu)] dx dt
t0
T
Z
0
Z
1
=
t0
Z
0
T
[2ξ 2 (sϕt − c)e2sϕ u2 + 2ξ 2 e2sϕ uh − 2(ξ 2 e2sϕ )x auux − 2ξ 2 e2sϕ au2x ] dx dt
Z
= −2
t0
Z T
ω0
Z
+2
t0
ω0
ξ 2 e2sϕ au2x dx dt
[ξ 2 (sϕt − c)e2sϕ u2 + ξ 2 e2sϕ uh − (ξ 2 e2sϕ )x auux ] dx dt.
24
I. BOUTAAYAMOU, G. FRAGNELLI, L. MANIAR
EJDE-2014/167
Then, integrating by parts and using the Young inequality, we obtain
Z TZ
2
ξ 2 e2sϕ au2x dx dt
ω0
T Z
t0
Z
[ξ 2 (sϕt − c)e2sϕ u2 + ξ 2 e2sϕ uh − (ξ 2 e2sϕ )x auux ] dx dt
=2
Z
≤
t0
ω0
T Z h
ω0
T Z
t0
Z
2ξ 2 (sϕt − c)e2sϕ + ξ 2 e2sϕ +
√
+
aξesϕ
2
ω0
t0
i
√ (ξ 2 e2sϕ ) 2 x
2
2 2sϕ 2
a
u
dx dt
+
ξ
e
h
ξesϕ
u2x dx dt.
Hence, for s large enough,
Z TZ
ξ 2 e2sϕ au2x dx dt
t0
Z
ω0
T Z
h
≤
t0
Z
ω0
T Z
≤C
2ξ 2 (sϕt − c)e2sϕ + ξ 2 e2sϕ +
i
√ (ξ 2 e2sϕ ) 2 x
u2 + ξ 2 e2sϕ h2 dx dt
a
sϕ
ξe
(s2 θ2 u2 + h2 )e2sϕ dx dt.
ω0
t0
Since x0 ∈
/ ω0 ,
Z
inf
a(x)
00
ω
T
t0
Z
ω 00
e2sϕ u2x
Z
T
Z
dx dt ≤
t0
Z
ω0
T Z
ξ 2 e2sϕ au2x dx dt
≤C
t0
(s2 θ2 u2 + h2 )e2sϕ dx dt,
ω0
and the proof is complete.
Acknowledgments. Genni Fragnelli was partially supported by the GNAMPA,
project Equazioni di evoluzione degeneri e singolari: controllo e applicazioni.
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26
I. BOUTAAYAMOU, G. FRAGNELLI, L. MANIAR
EJDE-2014/167
Idriss Boutaayamou
Département de Mathématiques, Faculté des Sciences Semlalia, LMDP, UMMISCO (IRDUPMC), Université Cadi Ayyad, Marrakech 40000, B.P. 2390, Morocco
E-mail address: dsboutaayamou@gmail.com
Genni Fragnelli
Dipartimento di Matematica, Università degli Studi di Bari Aldo Moro, Via E. Orabona
4, 70125 Bari - Italy
E-mail address: genni.fragnelli@uniba.it
Lahcen Maniar
Département de Mathématiques, Faculté des Sciences Semlalia, LMDP, UMMISCO (IRDUPMC), Université Cadi Ayyad, Marrakech 40000, B.P. 2390, Morocco
E-mail address: maniar@uca.ma