Electronic Journal of Differential Equations, Vol. 2010(2010), No. 156, pp.... ISSN: 1072-6691. URL: or

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Electronic Journal of Differential Equations, Vol. 2010(2010), No. 156, pp. 1–8.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
PROPERTIES OF THE FIRST EIGENVALUE OF A MODEL FOR
NON NEWTONIAN FLUIDS
OMAR CHAKRONE, OKACHA DIYER, DRISS SBIBIH
Abstract. We consider a nonlinear Stokes problem on a bounded domain.
We prove the existence of the first eigenvalue which is given by a minimization
formula. Some properties such as strict monotony and the Fredholm alternative are established.
1. Introduction
In studies of semi-linear elliptic equations such as
−∆u = f (x, u) + h(x)
u=0
in Ω,
on ∂Ω
n
where Ω is a bounded domain of R . It is usual to impose conditions on the
asymptotic behavior of the nonlinearity f (x, u) in relation to the spectrum of the
linear part of −∆. In the simplest situations, we consider f (x, u) as a perturbation
of λu. According to that λ being or not an eigenvalue of −∆, the results of such
resonance or non-resonance are then obtained. Among the classical references on
this subject, we can cite [5] (λ < λ1 ), [4] (λ between two consecutive eigenvalues),
[6] (λ = λ1 ). We also cite the Dirichlet problem
− div(|∇u|k−2 ∇u) = λm(x)|u|k−2 u
in Ω,
u = 0 on ∂Ω.
The first eigenvalue λ1 of the Dirichlet problem is simple and isolated. It was proved
that it is the unique positive eigenvalue having a non negative eigenfunction, see
[2].
Now we consider the eigenvalue problem of a non-linear
operator k-Laplacian.
S4
Let Ω ⊂ R2 be a bounded domain with boundary Γ = i=1 Γi , where Γ1 = {0}×] −
1, 1[, Γ2 = {1}×] − 1, 1[ and Γ3 , Γ4 are symmetrical to the X-axis, see Figure 1.
In the interior of this domain, a non-Newtonian liquid is subjected to pressures of
known differences between the two sides Γ1 and Γ2 .
2000 Mathematics Subject Classification. 74S05, 76T10.
Key words and phrases. k-Laplacian; eigenvalue; minimization.
c
2010
Texas State University - San Marcos.
Submitted May 6, 2010. Published October 28, 2010.
1
2
O. CHAKRONE, O. DIYER, D. SBIBIH
EJDE-2010/156
1
Γ3
0.5
Γ1
0.2
0.4
0.6
0.8
Γ2
x
−0.5
Γ4
−1
Figure 1. Geometry of channel Ω
We denote by V the closure of V in the space W 1,k (Ω), where
V = u = (u1 , u2 )t ∈ (C 1 (Ω̄))2 : div u = 0, ui (0, y) = ui (1, y) on [−1, 1]
for i = 1, 2 and u = 0 on Γ3 ∪ Γ4 .
For given α ∈ R, we consider the eigenvalue problem: Find (λ, u, p) ∈ R × V \ {0} ×
L2 (Ω) such that
∂p
= λm(x, y)|u1 |k−2 u1 in Ω,
∂x
∂p
−∆k u2 +
= λm(x, y)|u2 |k−2 u2 in Ω,
∂y
∂u1
∂u2
div u =
+
= 0 in Ω,
∂x
∂y
u1 (0, y) = u1 (1, y) on [−1, 1],
−∆k u1 +
u2 (0, y) = u2 (1, y)
(1.1)
on [−1, 1],
∂u1
∂u1
(0, y) =
(1, y) on [−1, 1],
∂x
∂x
∂u2
∂u2
|∇u2 (0, y)|k−2
(0, y) = |∇u2 (1, y)|k−2
(1, y)
∂x
∂x
p(1, y) − p(0, y) = −α on [−1, 1]
on [−1, 1],
where the weight function m(x, y) ∈ L∞ (Ω) can change the sign and it is positive
in a subset of Ω,
−∆k ui = − div(|∇ui |k−2 ∇ui )
is a k-Laplacian, i = 1, 2 and 1 < k < ∞. In the particular case k = 2; i.e.,
∆k = ∆, and λ = 0, the above problem has been studied by many authors, we
cite for example Amrouche et al. [1]. Here, we give an extension to previous work
in the nonlinear case by applying new methods to characterize the first eigenvalue
for this kind of problem such as minimization and as application is to solving the
problem of Fredholm alternative. This note is organized as follows. In Section 2,
we give the existence and the characterization of the first eigenvalue. In Section
3, we prove the Fredholm alternative and we justify all the given properties. In
Section 4, we give a conclusion.
EJDE-2010/156
PROPERTIES OF THE FIRST EIGENVALUE
3
2. Existence and characterization of the first eigenvalue
Theorem 2.1. There exists one principal eigenvalue λ1 for Problem (1.1). It is
characterized by
Z
1
(ϕ1 )1 (0, y)dy,
λ1 = kβ + (k − 1)α
(2.1)
−1
where ϕ1 is the principal corresponding eigenfunction and
1
β = min
k
Z
Z
k
k
Z
1
|∇u1 | + |∇u2 | − α
u1 (0, y)dy;
−1
Ω
m(x, y)(|u1 |k + |u2 |k ) = 1, u ∈ V ,
Ω
β=
1
k
Z 1
|∇(ϕ1 )1 |k + |∇(ϕ1 )2 |k − α
(ϕ1 )1 (0, y)dy,
Ω
−1
Z
m(x, y)[|(ϕ1 )1 |k + |(ϕ1 )2 |k ] = 1.
Z
Ω
Furthermore, for all α, α0 ∈ R such that αα0 > 0, λ1 (α) is an eigenvalue of Problem
(1.1).
For the sake of simplicity, in what follows, we denote λ1 = λ1 (m) = λ1 (α, m) =
λ1 (α).
Theorem 2.2. (i) λ1 defined by
1
= max{
λ1
Z
k
k
Z
m(x, y)(|u1 | + |u2 | );
Ω
|∇u1 |k + |∇u2 |k = 1, u ∈ V }
(2.2)
Ω
is the first eigenvalue of Problem (1.1) with α = 0 in the sense Σ ⊂ [λ1 , +∞[,
where Σ is the set of the positive
Moreover,
R eigenvalues.
R u is the eigenfunction
k
k
associated
with
λ
if
and
only
if
|∇u
|
+
|∇u
|
−
λ
m(x, y)(|u1 |k + |u2 |k ) =
1
1
2
1
Ω
Ω
R
R
k
k
k
k
inf{ Ω |∇v1 | + |∇v2 | − λ1 Ω m(x, y)(|v1 | + |v2 | ) v ∈ V } = 0.
(ii) λ1 (.) is strictly monotone in L∞ (Ω); i.e., if m1 , m2 are in the set
m ∈ L∞ (Ω); measure{(x, y) ∈ Ω; m(x, y) > 0} =
6 0
such that m1 (x, y) < m2 (x, y) a.e., then λ1 (m1 ) > λ1 (m2 ).
(iii) λ1 (.) is continuous in L∞ (Ω).
4
O. CHAKRONE, O. DIYER, D. SBIBIH
EJDE-2010/156
Theorem 2.3 (Fredholm alternative). Suppose that λ < λ1 , then for f ∈ (C(Ω))2
the problem: Find (u, p) ∈ V × L2 (Ω) such that
∂p
= λm(x, y)|u1 |k−2 u1 + f1
∂x
∂p
−∆k u2 +
= λm(x, y)|u2 |k−2 u2 + f2
∂y
∂u2
∂u1
+
= 0 in Ω,
div u =
∂x
∂y
u1 (0, y) = u1 (1, y) on [−1, 1],
−∆k u1 +
u2 (0, y) = u2 (1, y)
in Ω,
in Ω,
(2.3)
on [−1, 1],
∂u1
∂u1
(0, y) =
(1, y) on [−1, 1],
∂x
∂x
∂u2
∂u2
|∇u2 (0, y)|k−2
(0, y) = |∇u2 (1, y)|k−2
(1, y)
∂x
∂x
p(1, y) − p(0, y) = −α on [−1, 1]
on [−1, 1],
has a solution.
3. Proof of the main theorems
For proving Theorem 2.1, we need the following results.
Proposition 3.1. u = (u1 , u2 )t is a solution of problem: Find (u, p) ∈ V \ {0} ×
L2 (Ω) such that
∂p
= f1 in Ω,
∂x
∂p
−∆k u2 +
= f2 in Ω,
∂y
∂u1
∂u2
div u =
+
= 0 in Ω,
∂x
∂y
u1 (0, y) = u1 (1, y) on [−1, 1],
−∆k u1 +
u2 (0, y) = u2 (1, y)
(3.1)
on [−1, 1],
∂u1
∂u1
(0, y) =
(1, y) on [−1, 1],
∂x
∂x
∂u2
∂u2
|∇u2 (0, y)|k−2
(0, y) = |∇u2 (1, y)|k−2
(1, y)
∂x
∂x
p(1, y) − p(0, y) = −α on [−1, 1]
on [−1, 1],
where f = (f1 , f2 )t ∈ (C(Ω))2 , if and only if u is a solution of problem: Find u ∈ V
such that
Z 1
Z
2 Z
X
|∇ui |k−2 ∇ui ∇vi − α
v1 (0, y)dy = (f1 v1 + f2 v2 )
(3.2)
i=1
for all v ∈ V .
Ω
−1
Ω
EJDE-2010/156
PROPERTIES OF THE FIRST EIGENVALUE
5
Remark 3.2. If we take fi = λm(x, y)|ui |k−2 ui , i = 1, 2. Then (λ, u, p) is a
solution of (1.1) if and only if
Z 1
2 Z
2 Z
X
X
|∇ui |k−2 ∇ui ∇vi − α
v1 (0, y)dy = λ
m(x, y)|ui |k−2 ui vi
i=1
−1
Ω
i=1
Ω
for all v ∈ V . For a proof of this remark see [3].
R
Proof of Theorem 2.1. Since for all v ∈ V , v = 0 on Γ3 ∪ Γ4 , u ∈ V → ( Ω |∇u1 |k +
|∇u2 |k )1/k define a norm in V according to the Poincaré inequality in the space V :
There exists c > 0 such that
Z
Z
c
|u1 |k + |u2 |k ≤
|∇u1 |k + |∇u2 |k .
(3.3)
Ω
Ω
∗
n
n t
Suppose
R by contradiction Rthat for all n ∈ N there exists un = (u1 , u2 ) ∈ V such
that n1 Ω |un1 |k + |un2 |k > Ω |∇un1 |k + |∇un2 |k , then we put
vn = (v1n , v2n )t
R
un
i
where vin = (R |un |k +|u
|v n |k + |v2n |k = 1, so
n |k )1/k , i = 1, 2. Thus
Ω 1
1
2
Ω
Z
1
>
|∇v1n |k + |∇v2n |k .
n
Ω
(3.4)
As (vn )n is bounded in V , we have for a subsequence also denoted (vn )n , vn * v
in V and vn → v in Lk (Ω). Therefore kvkLk (Ω) = 1, so v 6= 0. By passing to the
limit in (3.4), we have
Z
Z
0 ≥ lim inf
|∇v1n |k + |∇v2n |k ≥
|∇v1 |k + |∇v2 |k .
n
R
k
R
Ω
Ω
k
So Ω |∇v1 | = Ω |∇v2 | = 0, hence v = cst, therefore v = 0 because v = 0 on
Γ3 ∪ Γ4 , is a contradiction. By using (3.3) and the Holder’s inequality, we easily
prove that β is well defined. Let (un ) = ((un1 , un2 )) be a suitable minimization of
β, then we have
Z
Z 1
1
k
k
β = lim
|∇un1 | + |∇un2 | − α
un1 (0, y)dy
n→∞ k Ω
−1
and
Z
m(x, y)(|un1 |k + |un2 |k ) = 1.
Ω
R
The sequence (Xn ) := ( k1 Ω |∇un1 |k + |∇un2 |k ) is bounded, if we have not for a
subsequence, also denoted (Xn ), Xn → +∞. Using the Holder’s inequality and the
fact that V ,→ Lk (Γ1 ) we get
Z 1
Z
1
α
un1 (0, y)dy ≤ |α|c(
|∇un1 |k + |∇un2 |k )1/k = |α|cXn1/k
k Ω
−1
R P2
R1
1/k
where c ∈ R. Thus k1 Ω i=1 |∇uni |k − α −1 un1 (0, y)dy ≥ Xn − |α|cXn , this
prove that β = +∞, which is impossible. According to the reflexivity of the space
V and the compact injections V ,→ Lk (Ω) and V ,→ Lk (Γ1 ), there exists a subsequence of (un ) = ((un1 , un2 )), which is also denoted by (un ) = ((un1 , un2 )), such
6
O. CHAKRONE, O. DIYER, D. SBIBIH
EJDE-2010/156
that
un = (un1 , un2 ) * ϕ1 = ((ϕ1 )1 , (ϕ1 )2 )
1
1
1
un = (un1 , un2 ) → ϕ = ((ϕ )1 , (ϕ )2 )
un1 |Γ1 → (ϕ1 )1 |Γ1
Hence
R
Ω
in V,
in Lk (Ω),
in Lk (Γ1 ).
m(x, y)(|(ϕ1 )1 |k + |(ϕ1 )2 |k ) = 1, consequently ϕ1 6= 0 and
Z
Z 1
1
1
k
1
k
β≤
(ϕ1 )1 (0, y)dy
|∇(ϕ )1 | + |∇(ϕ )2 | − α
k Ω
0
Z
Z 1
1
k
k
un1 (0, y)dy,
≤
|∇un1 | + |∇un2 | − α
k Ω
0
so
Z
Z 1
1
|∇(ϕ1 )1 |k + |∇(ϕ1 )2 |k − α
(ϕ1 )1 (0, y)dy.
k Ω
0
On the other hand, for all t > 0, v = (v1 , v2 ) ∈ V , we put wt = (wthm1 , wt2 ) where
β=
(ϕ1 )1 + tv1
wthm1 = R
,
( Ω m(x, y)(|(ϕ1 )1 + tv1 |k + |(ϕ1 )2 + tv2 |k ))1/k
(ϕ1 )2 + tv2
,
wt2 = R
1
( Ω m(x, y)(|(ϕ )1 + tv1 |k + |(ϕ1 )2 + tv2 |k ))1/k
so that
R
Ω
m(x, y)(|wthm1 |k + |wt2 |k ) = 1 and
Z
Z 1
1
β=
|∇(ϕ1 )1 |k + |∇(ϕ1 )2 |k − α
(ϕ1 )1 (0, y)dy
k Ω
0
Z
Z 1
1
|∇wthm1 |k + |∇wt2 |k − α
≤
wthm1 (0, y)dy.
k Ω
0
By developing to order 1 for t → 0 and by applying the same reasoning to (−v),
we obtain
Z 1
2 Z
X
1 k−2
1
|∇(ϕ )i |
∇(ϕ )i ∇vi − α
v1 (0, y)dy
i=1
Ω
0
Z
= (kβ + (k − 1)α
1
2 Z
X
(ϕ1 )1 (0, y)dy) × (
m(x, y)|(ϕ1 )i |k−2 (ϕ1 )i vi ).
0
i=1
Ω
Now we suppose that αα0 > 0. We put ϕ1 = (ϕ1 1 , ϕ1 2 ), where ϕ1 i = ηϕ1i with
0
η k−1 = αα . Then, by replacing in the equation (P1 (α)), we obtain
Z 1
2 Z
2 Z
X
X
k−2
0
1
1
|∇ϕ i |
∇ϕ i ∇vi − α
v1 (0, y)dy = λ1 (α)
m(x, y)|ϕ1 i |k−2 ϕ1 i vi ,
i=1
Ω
0
which completes the proof of Theorem 2.1.
i=1
Ω
Proof of Theorem 2.2. (i) It is easy to prove that for α = 0, λ1 is an eigenvalue of RProblem (1.1) with
if and only
R α = 0 andk u 6= 0k is a eigenfunction
P2
P2 R
k
k
if
|∇u
|
−
λ
(m)
m(x,
y)(|u
|
+
|u
|
)
=
0
=
inf{
i
1
1
2
i=1
i=1 Ω |∇vi | −
Ω
R Ω
k
k
λ1 (m) Ω m(x, y)(|v1 | + |v2 | ); v ∈ V }. The proofs of (ii) and (iii) follow from
(i).
EJDE-2010/156
PROPERTIES OF THE FIRST EIGENVALUE
7
Proof of Theorem 2.3. It is clear that Problem (2.3) is equivalent to the weak formulation: Find u ∈ V such that
Z 1
2 Z
X
k−2
|∇ui |
∇ui ∇vi − α
v1 (0, y)dy
−1
Ω
i=1
=λ
2 Z
X
i=1
k−2
m(x, y)|ui |
ui vi +
Ω
2 Z
X
i=1
(3.5)
fi vi
∀v ∈ V.
Ω
We consider the energy functional defined on V ,
Z 1
2 Z
2 Z
2 Z
X
λX
1X
|∇ui |k − α
m(x, y)|ui |k −
u1 (0, y)dy −
fi ui .
Φ(u) =
k i=1 Ω
k i=1 Ω
−1
i=1 Ω
(3.6)
We verify that u is a solution of Problem (3.5) if and only if u is a critical point of
the function Φ. For the existence, it suffices to prove that there exists u ∈ V such
that
Φ(u) = inf Φ(v).
v∈V
The functional Φ is continuous and convex, it suffices to show that Φ is coercive,
indeed for all u ∈ V , using Theorem 2.2, we obtain
Z
Z
λ1
m(x, y)(|u1 |k + |u2 |k ) ≤
|∇u1 |k + |∇u2 |k .
(3.7)
Ω
Ω
R
R
Since the function u 7→ ( Ω |∇u1 |k )1/k + ( Ω |∇u2 |k )1/k := kukV defines a norm in
V , then we have successively
2 Z
2
X
X
fi ui ≤
kfi kLk0 kui kLk
Ω
i=1
i=1
≤c
2
X
k∇ui kLk = ckukV ,
i=1
where c > 0.
Z 1
Z
α
u1 (0, y)dy ≤ |α|
−1
|u1 |dσ
Z
≤ |α|c0 (
|u1 |k dσ)1/k (Holder’s inequality)
∂Ω
Z
≤ |α|c0 ( |∇u1 |k )1/k (V ,→ Lk (∂Ω) a continuous injection)
∂Ω
Ω
= c00 kukV ,
where c0 and c00 are positive.
2 Z
2 Z
λ̃ X
λX
m(x, y)|ui |k ≤
m(x, y)|ui |k
k i=1 Ω
k i=1 Ω
e Z
λ
≤
|∇u1 |k + |∇u2 |k ,
λ1 k Ω
8
O. CHAKRONE, O. DIYER, D. SBIBIH
EJDE-2010/156
(
0
λ
if λ < 0
According to (3.6), we obtain
if λ ≥ 0.
Z
1
λ̃
Φ(u) ≥ (1 − )
|∇u1 |k + |∇u2 |k − c000 kukV ,
k
λ1 Ω
where c000 > 0. Thus
λ̃
1
Φ(u) ≥ (1 − )kukkV − c000 kukV ,
k
λ1
where c000 > 0. Since λ < λ1 , we deduce that Φ(u) → +∞ when kukV → +∞, so
we have proved the existence.
e :=
where λ
References
[1] C. Amrouche, M. Batchi, J. Batina; Navier-Stokes equations with periodic boundary conditions and pressure loss, Applied Mathematics Letters. 20 (2007), 48-53.
[2] A. Anane; Simplicité et isolation de la premire valeur propre du p-Laplacien avec poids, C.
R. Acad. Sci. Paris, t. 305 (1987), 725-728.
[3] O. Chakrone, O. Diyer, D. Sbibih; A non resonance of non-Newtonian fluid with known
differences pressures between two parallel plates. Submitted.
[4] C. L. Dolph; Nonlinear integral equations of the Hammerstein type, Trans. Amer. Math. Soc.
66 (1949), 289-307.
[5] A. Hammerstein; Nichtlineare Integralgleichungen nebst anwendungen acta, Math. 54 (1930),
117-176.
[6] E. Landesman, A. Lazer; Nonlinear perturbation of linear elliptic boundary value problems
at resonance, J. Amer. Math. Mech. 19 (1970), 609-623.
Omar Chakrone
Université Mohammed I, Faculté des sciences, Laboratoire LANOL, Oujda, Maroc
E-mail address: chakrone@yahoo.fr
Okacha Diyer
Université Mohammed I, Ecole Supérieure de Technologie, Laboratoire MATSI, Oujda,
Maroc
E-mail address: odiyer@yahoo.fr
Driss Sbibih
Université Mohammed I, Ecole Supérieure de Technologie, Laboratoire MATSI, Oujda,
Maroc
E-mail address: sbibih@yahoo.fr
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