Electronic Journal of Differential Equations, Vol. 2013 (2013), No. 225, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
WEAK HETEROCLINIC SOLUTIONS OF ANISOTROPIC
DIFFERENCE EQUATIONS WITH VARIABLE EXPONENT
ABOUDRAMANE GUIRO, BLAISE KONE, STANISLAS OUARO
Abstract. In this article, we prove the existence of heteroclinic solutions for
a family of anisotropic difference equations. The proof of the main result is
based on a minimization method, a change of variables and a discrete Hölder
type inequality.
1. Introduction
In this article we study the existence of heteroclinic solutions for the nonlinear
discrete anisotropic problem
−∆(a(k − 1, ∆u(k − 1))) + g(k, u(k)) = f (k), k ∈ Z∗
(1.1)
u(0) = 0,
lim u(k) = −1,
lim u(k) = 1,
k→−∞
k→+∞
where ∆u(k) = u(k + 1) − u(k) is the forward difference operator.
The study of heteroclinic connections for boundary value problems had a certain
impulse in recent years, motivated by applications in various biological, physical
and chemical models, such has phase-transition, physical processes in which the
variable transits from an unstable equilibrium to a stable one, or front-propagation
in reaction-diffusion equations. Indeed, heteroclinic solutions are often called transitional solutions (see [2, 6] and the references therein).
In this article, we show that the solvability of (1.1) is connected to the behavior of
g(k, s) as k ∈ Z+ and as k ∈ Z− . Problem (1.1) involves variable exponents due to
their use in image restoration (see [3]), in electrorheological and thermorheological
fluids dynamic (see [4, 7, 8]). The paper is organized as follows: In section 2,
we introduce hypotheses on f , g and a, we define the functional spaces and some
of their useful properties and in section 3, we prove the existence of heteroclinic
solutions of (1.1).
2. Auxiliary results
For the rest of this article, we will use the notation:
p− = inf p(k).
p+ = sup p(k),
k∈Z
k∈Z
2000 Mathematics Subject Classification. 47A75, 35B38, 35P30, 34L05, 34L30.
Key words and phrases. Difference equations; heteroclinic solutions; anisotropic problem;
discrete Hölder type inequality.
c
2013
Texas State University - San Marcos.
Submitted December 11, 2012. Published October 11, 2013.
1
2
A. GUIRO, B. KONE, S. OUARO
EJDE-2013/225
We assume that
p(.) : Z → (1, +∞)
and
1 < p− ≤ p(.) < p+ < +∞.
(2.1)
We introduce the spaces:
l1 = {u : Z → R, kukl1 :=
X
|u(k)| < ∞},
k∈Z
l01 = {u : Z → R; u(0) = 0 and kukl01 :=
X
|u(k)| < ∞},
k∈Z
p(.)
l0
= {u : Z → R; u(0) = 0 and ρp(.) (u) :=
X
|u(k)|p(k) < ∞},
k∈Z
p(.)
l0,+
X
+
= {u : Z → R; u(0) = 0 and ρp+ (.) (u) :=
|u(k)|p(k) < ∞},
k∈Z+
p(.)
X
l0,− = {u : Z− → R; u(0) = 0 and ρp− (.) (u) :=
|u(k)|p(k) < ∞},
k∈Z−
1,p(.)
W0,+
= {u : Z+ → R; u(0) = 0 and ρ1,p+ (.) (u) :=
X
|u(k)|p(k)
k∈Z+
+
X
|∆u(k)|p(k) < ∞}
k∈Z+
p(.)
p(.)
= {u : Z+ → R; u ∈ l+ , ∆u(k) ∈ l+
1,p(.)
W0,−
and u(0) = 0}
X
= {u : Z− → R; u(0) = 0 and ρ1,p− (.) (u) :=
|u(k)|p(k)
k∈Z−
+
X
|∆u(k)|p(k) < ∞}
k∈Z−
p(.)
p(.)
= {u : Z− → R; u ∈ l− , ∆u(k) ∈ l−
and u(0) = 0}.
p(.)
On l0,+ we introduce the Luxemburg norm
X u(k)
kukp+ (.) := inf λ > 0 :
|
|p(k) ≤ 1
λ
+
k∈Z
and we deduce that
X u(k)
X ∆u(k)
kuk1,p+ (.) := inf λ > 0;
|
|p(k) +
|
|p(k) ≤ 1
λ
λ
+
+
k∈Z
k∈Z
= kukp+ (.) + k∆ukp+ (.)
1,p(.)
p(.)
is a norm on the space W0,+ . We replace Z+ by Z− to get the norms on l0,− and
1,p(.)
W0,− denoted respectively k · kp− (.) and k · k1,p− (.) .
For the data f , g and a, we assume the following:
a(k, .) : R → R for all k ∈ Z and there exists a mapping A : Z × R → R
such that a(k, ξ) =
∂
A(k, ξ) for all k ∈ Z and A(k, 0) = 0 for all k ∈ Z.
∂ξ
|ξ|p(k) ≤ a(k, ξ)ξ ≤ p(k)A(k, ξ) ∀k ∈ Z and ξ ∈ R.
(2.2)
(2.3)
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WEAK HETEROCLINIC SOLUTIONS
3
There exists a positive constant C1 such that
|a(k, ξ)| ≤ C1 (j(k) + |ξ|p(k)−1 ),
0
for all k ∈ Z and ξ ∈ R where j ∈ lp (.) with
1
p(k)
+
1
p0 (k)
(2.4)
= 1.
f ∈ l1 ,
p(k)−2
g(k, t) = |t − 1|
(2.5)
p(k)−2
(t − 1)χZ+ (k) + |t + 1|
(t + 1)χZ− (k),
(2.6)
where χA (k) = 1 if k ∈ A and χA (k) = 0 if k ∈
/ A.
p(.)
p(.)
p(.)
p(.)
1,p(.)
Remark 2.1. Note that l0,+ ⊂ l0 , l0,− ⊂ l0 , W0,+
1,p(.)
1,p(.)
⊂ W0
and W0,−
⊂
1,p(.)
W0
.
p(.)
p(.)
p(.)
If u ∈ l0,+ (or u ∈ l0,− or u ∈ l0 ) then limk→+∞ u(k) = 0 (or limk→−∞ u(k) = 0
P
p(.)
or lim|k|→+∞ u(k) = 0). Indeed, for instance, if u ∈ l0,+ then k∈Z+ |u(k)|p(k) <
∞. Let
X
X
X
|u(k)|p(k) =
|u(k)|p(k) +
|u(k)|p(k) ,
k∈Z+
k∈S1
k∈S2
where S1 = {k ∈ Z+ ; |u(k)| < 1} and S2 = {k ∈ Z+ ; |u(k)| ≥ 1}. The set S2 is
p(.)
necessarily finite, and |u(k)| < ∞ for any k ∈ S2 since u ∈ l0,+ . We also have that
X
X
+
|u(k)|p ≤
|u(k)|p(k) ,
k∈Z+
k∈S1
+
then k∈S1 |u(k)|p < ∞. As S2 is a finite set then
implies that
X
+
|u(k)|p < ∞.
P
P
k∈S2
+
|u(k)|p < ∞, which
k∈Z+
Thus, limk→+∞ u(k) = 0.
We now give useful properties of the spaces defined above which are similar to
those in [5].
p(.)
Proposition 2.2. Assume that (2.1) is fulfilled. Then l01 ⊂ l0 .
Proposition 2.3. Under conditions (2.1), ρp+ (.) satisfies
p(.)
(a) ρp+ (.) (u + v) ≤ 2p+ (ρp+ (.) (u) + ρp+ (.) (v)), for all u, v ∈ l0,+ .
p(.)
(b) For u ∈ l0,+ , if λ > 1 we have
−
+
ρp+ (.) (u) ≤ λρp+ (.) (u) ≤ λp ρp+ (.) (u) ≤ ρp+ (.) (λu) ≤ λp ρp+ (.) (u)
and if 0 < λ < 1, we have
−
+
λp ρp+ (.) (u) ≤ ρp+ (.) (λu) ≤ λp ρp+ (.) (u) ≤ λρp+ (.) (u) ≤ ρp+ (.) (u).
p(.)
(c) For every fixed u ∈ l0,+ \ {0}, ρp+ (.) (λu) is a continuous convex even function in λ and it increases strictly when λ ∈ [0, ∞).
p(.)
Proposition 2.4. Let u ∈ l0,+ \ {0}, then kukp+ (.) = a if and only if ρp+ (.) ( ua ) = 1.
p(.)
Proposition 2.5. If u ∈ l0,+ and p+ < +∞, then the following properties hold:
(1) kukp+ (.) < 1 (= 1; > 1) if and only if ρp+ (.) (u) < 1 (= 1; > 1);
−
(2) kukp+ (.) > 1 implies kukpp+ (.) ≤ ρp+ (.) (u) ≤ kukp+
p+ (.) ;
4
A. GUIRO, B. KONE, S. OUARO
EJDE-2013/225
+
(3) kukp+ (.) < 1 implies kukpp+ (.) ≤ ρp+ (.) (u) ≤ kukp−
p+ (.) ;
(4) kun kp+ (.) → 0 if and only if ρp+ (.) (un ) → 0 as n → +∞.
1,p(.)
Proposition 2.6. Let u ∈ W0,+
ρ1,p+ (.) (u/a) = 1.
\ {0}. Then kuk1,p+ (.) = a if and only if
1,p(.)
Proposition 2.7. If u ∈ W0,+ and p+ < +∞, then the following properties hold:
(1) kuk1,p+ (.) < 1 (= 1; > 1) if and only if ρ1,p+ (.) (u) < 1 (= 1; > 1);
−
+
+
−
(2) kuk1,p+ (.) > 1 implies kukp1,p+ (.) ≤ ρ1,p+ (.) (u) ≤ kukp1,p+ (.) ;
(3) kuk1,p+ (.) < 1 implies kukp1,p+ (.) ≤ ρ1,p+ (.) (u) ≤ kukp1,p+ (.) ;
(4) kun k1,p+ (.) → 0 if and only if ρ1,p+ (.) (un ) → 0 as n → +∞.
p(.)
Theorem 2.8 (Discrete Hölder type inequality). Let u ∈ l+
1
1
+ q(k)
= 1 for all k ∈ Z+, then
that p(k)
+∞
X
|uv| ≤ (
k=0
q(.)
and v ∈ l+
be such
1
1
+ − )kukp+ (.) kvkq+ (.) .
p−
q
p(.)
Remark 2.9. All the properties above hold for the spaces lp(.) , l−
1,p(.)
and W0,− .
3. Existence of weak heteroclinic solutions
In this section, we study the existence of weak heteroclinic solutions of problem
(1.1).
Definition 3.1. A weak heteroclinic solution of (1.1) is a function u : Z → R such
that
X
X
X
a(k − 1, ∆u(k − 1))∆v(k − 1) +
g(k, u(k))v(k) =
f (k)v(k),
(3.1)
k∈Z
k∈Z
k∈Z
for any v : Z → R, with u(0) = 0, limk→+∞ u(k) = 1 and limk→−∞ u(k) = −1.
Theorem 3.2. Assume that (2.1)–(2.6) hold. Then, there exists at least one weak
heteroclinic solution of (1.1).
To prove Theorem 3.2, we first prove that the problem
−∆(a(k − 1, ∆u(k − 1))) + |u(k)|p(k)−2 u(k) = f (k),
u(0) = 0,
k ∈ Z+
∗
lim u(k) = 0,
(3.2)
k→+∞
admits a weak solution in the following sense.
1,p(.)
Definition 3.3. A weak solution of (3.2) is a function u ∈ W0,+
+∞
X
a(k − 1, ∆u(k − 1))∆v(k − 1) +
k=1
+∞
X
k=1
|u(k)|p(k)−2 u(k)v(k) =
+∞
X
such that
f (k)v(k), (3.3)
k=1
1,p(.)
for any v ∈ W0,+ .
We have the following result.
Theorem 3.4. Assume that (2.1)–(2.5) hold. Then, there exists at least one weak
solution of problem (3.2).
EJDE-2013/225
WEAK HETEROCLINIC SOLUTIONS
5
1,p(.)
The energy functional corresponding to problem (3.2) is J : W0,+
→ R defined
by
J(u) =
+∞
X
A(k − 1, ∆u(k − 1)) +
k=1
+∞
+∞
X
X
1
|u(k)|p(k) −
f (k)u(k).
p(k)
(3.4)
k=1
k=1
We first present some basic properties of J.
1,p(.)
Proposition 3.5. The functional J is well-defined on the space W0,+
class C
1
1,p(.)
(W0,+ , R),
and is of
with the derivative given by
hJ 0 (u), vi =
+∞
X
a(k − 1, ∆u(k − 1))∆v(k − 1)
k=1
+∞
X
+
|u(k)|
p(k)−2
u(k)v(k) −
k=1
+∞
X
(3.5)
f (k)v(k),
k=1
1,p(.)
for all u, v ∈ W0,+ .
Proof. We denote
I(u) =
+∞
X
A(k −1, ∆u(k −1)),
+∞
X
1
|u(k)|p(k) ,
L(u) =
p(k)
Λ(u) =
k=1
k=1
+∞
X
f (k)u(k).
k=1
Using Young inequality, from assumptions (2.2) and (2.4) it follows that
|I(u)| = |
≤
≤
≤
+∞
X
A(k − 1, ∆u(k − 1))|
k=1
+∞
X
|A(k − 1, ∆u(k − 1))|
k=1
+∞
X
k=1
+∞
X
C1 j(k − 1) +
1
|∆u(k − 1)|p(k−1)−1 |∆u(k − 1)|
p(k − 1)
C1 j(k − 1)|∆u(k − 1)| +
k=1
+∞
X
k=1
C1
|∆u(k − 1)|p(k−1) < ∞,
p(k − 1)
+∞
1 X
|u(k)|p(k) < ∞,
|L(u)| ≤
p−
k=1
+∞
+∞
X
X
|Λ(u)| = f (k)u(k) ≤
|f (k)||u(k)| < ∞.
k=1
k=1
Therefore, J is well-defined.
1,p(.)
1,p(.)
Clearly I, L and Λ are in C 1 (W0,+ , R). Let us now choose u, v ∈ W0,+ . We
have
I(u + δv) − I(u)
L(u + δv) − L(u)
hI 0 (u), vi = lim+
, hL0 (u), vi = lim+
,
δ
δ
δ→0
δ→0
Λ(u + δv) − Λ(u)
hΛ0 (u), vi = lim+
.
δ
δ→0
6
A. GUIRO, B. KONE, S. OUARO
Let us denote gδ =
ity,
+∞
X
k=1
EJDE-2013/225
A(k−1,∆u(k−1)+δ∆v(k−1))−A(k−1,∆u(k−1))
.
δ
+∞
+∞
k=1
k=1
Using Young inequal-
1X
1X
|A(k−1, ∆u(k−1)+δ∆v(k−1))|+
|A(k−1, ∆u(k−1))| < +∞.
|gδ | ≤
δ
δ
Thus,
I(u + δv) − I(u)
δ
δ→0
+∞
X A(k − 1, ∆u(k − 1) + δ∆v(k − 1)) − A(k − 1, ∆u(k − 1))
= lim+
δ
δ→0
lim+
k=1
=
=
+∞
X
k=1
+∞
X
lim+
δ→0
A(k − 1, ∆u(k − 1) + δ∆v(k − 1)) − A(k − 1, ∆u(k − 1))
δ
a(k − 1, ∆u(k − 1))∆v(k − 1).
k=1
By the same method, we deduce that
+∞
lim
δ→0+
X |u(k) + δv(k)|p(k) − |u(k)|p(k)
L(u + δv) − L(u)
= lim
δ
p(k)δ
δ→0+
k=1
=
=
+∞
X
k=1
+∞
X
lim
δ→0+
|u(k) + δv(k)|p(k) − |u(k)|p(k)
p(k)δ
|u(k)|p(k)−2 u(k)v(k)
k=1
and
+∞
lim
δ→0+
X f (k)(u(k) + δv(k)) − f (k)u(k)
Λ(u + δv) − Λ(u)
= lim
δ
δ
δ→0+
k=1
=
=
+∞
X
k=1
+∞
X
lim
δ→0+
f (k)(u(k) + δv(k)) − f (k)u(k)
δ
f (k)v(k)
k=1
Lemma 3.6. The functional I is weakly lower semi-continuous.
Proof. From (2.2), I is convex with respect to the second variable. Thus, by [1,
corollary III.8], it is sufficient to show that I is lower semi-continuous. For this, we
1,p(.)
1,p(.)
fix u ∈ W0,+ and > 0. Since I is convex, we deduce that for any v ∈ W0,+ ,
I(v) ≥ I(u) + hI 0 (u), v − ui
≥ I(u) +
+∞
X
k=1
a(k − 1, ∆u(k − 1))(∆v(k − 1) − ∆u(k − 1))
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WEAK HETEROCLINIC SOLUTIONS
≥ I(u) − C(
7
1
1
+ 0− )kgkp0+ (.) k∆(u − v)kp+ (.) ,
−
p
p
with g(k) = j(k) + |∆u(k)|p(k)−1
≥ I(u) − K ku − vkp+ (.) + k∆(u − v)kp+ (.)
≥ I(u) − Kku − vk1,p+ (.)
≥ I(u) − ,
1,p(.)
for all v ∈ W0,+ with ku − vk1,p+ (.) < δ = /K. Hence, we conclude that I is
weakly lower semi-continuous.
Proposition 3.7. The functional J is bounded from below, coercive and weakly
lower semi-continuous.
Proof. By Lemma 3.6, J is weakly lower semi-continuous. We shall only prove the
coerciveness of the energy functional J and its boundedness from below.
J(u) =
≥
+∞
X
k=1
+∞
X
k=1
A(k − 1, ∆u(k − 1)) +
+∞
+∞
X
X
1
|u(k)|p(k) −
f (k)u(k)
p(k)
k=1
k=1
+∞
+∞
X
X
1
1
|∆u(k − 1)|p(k−1) +
|u(k)|p(k) −
|f (k)u(k)|
p(k − 1)
p(k)
k=1
k=1
+∞
+∞
+∞
X
X
1 X
p(s)
p(k)
|∆u(s)|
+
|u(k)|
)−
|f (k)||u(k)|
≥ +(
p s=1
k=1
k=1
1
≥ + ρ1,p+ (.) (u) − c0 kf kp0+ (.) kukp+ (.)
p
1
≥ + ρ1,p+ (.) (u) − Kkuk1,p+ (.) .
p
To prove the coerciveness of J, we may assume that kuk1,p+ (.) > 1 and we get from
the above inequality that
J(u) ≥
−
1
kukp1,p+ (.) − Kkuk1,p+ (.) .
p+
Thus,
J(u) → +∞ as kuk1,p+ (.) → +∞.
As J(u) → +∞ when kuk1,p+ (.) → +∞, then for kuk1,p+ (.) > 1, there exists c ∈ R
such that J(u) ≥ c. For kuk1,p+ (.) ≤ 1, we have
J(u) ≥
1
ρ1,p+ (.) (u) − Kkuk1,p+ (.) ≥ −Kkuk1,p+ (.) ≥ −K > −∞.
p+
Thus J is bounded below.
We can now give the proof of the main result.
Proof of Theorem 3.4. By Proposition 3.7, J has a minimizer which is a weak solution of (3.2)
8
A. GUIRO, B. KONE, S. OUARO
EJDE-2013/225
Now, we consider the problem
−∆(a(k − 1, ∆u(k − 1))) + |u(k)|p(k)−2 u(k) = f (k),
u(0) = 0,
k ∈ Z−
(3.6)
lim u(k) = 0.
k→−∞
A weak solution of problem (3.6) is defined as follows.
1,p(.)
Definition 3.8. A weak solution of (3.6) is a function u ∈ W0,−
0
X
a(k − 1, ∆u(k − 1))∆v(k − 1) +
k=−∞
0
X
|u(k)|p(k)−2 u(k)v(k) =
k=−∞
such that
0
X
f (k)v(k),
k=−∞
(3.7)
for any v ∈
1,p(.)
W0,− .
By mimicking the proof of Theorem 3.4, we prove the following result.
Theorem 3.9. Assume that (2.1)–(2.5) hold. Then, there exists at least one weak
solution of problem (3.6).
Let us now show the existence of weak heteroclinic solutions of problem (1.1).
Proof of Theorem 3.2. We define v1 = u1 +1, where u1 is a weak solution of problem
(3.2) and v2 = u2 − 1, where u2 is a weak solution of problem (3.6). Therefore, we
deduce that
u = v1 χZ+ + v2 χZ−
is an heteroclinic solution of problem (1.1).
Acknowledgments. This work was done while the authors were visiting the Abdus Salam International Centre for Theoretical Physics (ICTP), Trieste, Italy. The
authors want to thank the mathematics section of ICTP for their hospitality for
providing financial support.
References
[1] H. Brezis; Analyse fonctionnelle: thorie et applications. Paris, Masson (1983).
[2] Alberto Cabada, Stepan Tersian; Existence of Heteroclinic solutions for discrete p-laplacian
problems with a parameter, Nonlinear Anal. RWA 12 (2011), 2429-2434.
[3] Y. Chen, S. Levine, M. Rao; Variable exponent, linear growth functionals in image processing,
SIAM J. Appl. Math. 66(4) (2006), 1383-1406.
[4] L. Diening; Theoretical and numerical results for electrorheological fluids, Ph.D. thesis, University of Freiburg, Germany, 2002.
[5] A. Guiro, B. Koné, S. Ouaro; Weak Homoclinic solutions of anisotropic difference equations
with variable exponents. Adv. Differ. Equ. 154 (2012), 1-13.
[6] C. Marcelli, F. Papalini; Heteroclinic connections for fully non-linear non-autonomous
second-order differential equations. J. Differential Equations. 241 (2007), 160-183.
[7] K. R. Rajagopal, M. Ruzicka; Mathematical Modeling of Electrorheological materials, Contin.
Mech. Thermodyn. 13 (2001), 59–78.
[8] M. Ruzicka; Electrorheological fluids: modelling and mathematical theory, Lecture Notes in
Mathematics 1748, Springer-Verlag, Berlin, 2002.
Aboudramane Guiro
Laboratoire d’Analyse Mathématique des Equations (LAME), UFR Sciences et Techniques, Université Polytechnique de Bobo Dioulasso, 01 BP 1091 Bobo-Dioulasso 01,
Bobo Dioulasso, Burkina Faso
E-mail address: abouguiro@yahoo.fr, aboudramane.guiro@univ-ouaga.bf
EJDE-2013/225
WEAK HETEROCLINIC SOLUTIONS
9
Blaise Kone
Laboratoire d’Analyse Mathématique des Equations (LAME), Institut Burkinabé des
Arts et Métiers, Université de Ouagadougou, 03 BP 7021 Ouaga 03, Ouagadougou,
Burkina Faso
E-mail address: leizon71@yahoo.fr, blaise.kone@univ-ouaga.bf
Stanislas Ouaro
Laboratoire d’Analyse Mathématique des Equations (LAME), UFR. Sciences Exactes et
Appliquées, Université de Ouagadougou, 03 BP 7021 Ouaga 03, Ouagadougou, Burkina
Faso
E-mail address: ouaro@yahoo.fr, souaro@univ-ouaga.bf