Existence and Multiplicity Results for Second-Order Nonlinear Differential Equations with Multivalued Boundary Conditions ()
1. Introduction
This paper is devoted to the study of the following problem:
(1)
where
and
are maximal monotone graphs in
and are some multifunctions which describe the boundary conditions,
is a Caratheodory function and
is an increasing homeomorphism such that
.
The tool of investigation for this problem is lower and upper solution’s method. This method provides a precious tool to get existence results for first and second order initial and boundary value problems. The method allows to generate monotone iterative techniques which provide constructive methods (amenable numerical treatment), to obtain solutions. The method was first introduced by Perron [1] . Later Nagumo [2] used upper and lower solutions to study second-order differential equations with Dirichlet boundary conditions. Since then many authors have used that method primarily in the context of single-valued differential equations with linear boundary conditions (Dirichlet, Neumann, Sturm-Liouville or periodic). Very recently, in 2017, combining lower and upper solutions method and theory of topological degree, Goli-Adjé [3] established existence and multiplicity results for the considered problem under Neumann-Steklov boundary value conditions. Shortly before, in 2013, Khattabi-Frigon-Ayyadi [4], by lower and upper solutions method and the fixed point index theory, obtained existence and multiplicity results for the problem under various boundary value conditions (Dirichlet, periodic or Neumann). Other authors have investigated the second-order differential equation with multivalued boundary conditions driven by maximal monotone operators, in this direction, see [5] [6] [7] [8] [9] and references therein. In [6] [7], the problems unify classical problems of Dirichlet, periodic and Neumann and in [5] [8] [9] the problems unify classical problems of Dirichlet, Neumann and Sturm-Louiville. To our knowledge, the lower and upper solutions method for differential inclusions formulation of problems of type (1) was initiated by Bader-Papageorgiou [5] in 2002. Soon after, in 2006, Staicub and Papageorgiou [9] extended the study of that problem to gradient systems with a discontinuous nonlinearity. In 2007, Kyritsi and Papageorgiou [8], in their book (see [8] the problem (5.111), p. 390) investigated the following single-valued version of the problem in Staicub-Papageorgiou [9] :
where
and f are defined as in problem (1) and for all
,
. So, in [5] [8] [9], the authors deal with the homogeneous operator differential p-laplacian
for all
. But they do not establish multiplicity results.
The goal of this paper is to extend the work of Kyritsi-Papageorgiou [8] to a large class of problems. Indeed, we deal with a non-homogeneous operator
-laplacian which contains, for example, some versions of
-Laplacian operators like the case when, for all
,
with
is a continuous map. Moreover, to obtain multiplicity results, we combine lower and upper method used in [5] [8] [9] and the one of Goli-Adjé [3] . So, our aim in this paper is to study existence and multiplicity results concerning solutions of problem (1).
After introducing notations, preliminary results and auxiliary results in Section 2 and Section 3, in Section 4,
and
are assumed respectively an ordered pair of lower and upper solutions of the problem. By combining lower and upper solutions method and theory of topological degree we obtain existence results.
In Section 5,
and
are assumed respectively non-ordered pair of lower and upper solutions of the problem. Also, by combining lower and upper solutions method and topological degree theory, we obtain existence results.
In Section 6, using the aforementioned method in Section 4 and 5, we show multiplicity results at first when the problem admits a pair of lower and strict lower solutions and a pair of upper and strict upper solutions, secondly when the problem admits two lower solutions and a strict upper solution or when the problem admits a strict lower solution and two upper solutions.
In Section 7, we give an example of application and we show also, as in [8], that our method stays true for the periodic problem.
In Section 8, we give a conclusion.
2. Preliminaries
In this section, we introduce our terminology and notations. We also recall some basic definitions and facts from multivalued analysis that we will need in the sequel. Our main sources are the books of Hu-Papageorgiou [10] and Zeidler [11] .
The Sobolev spaces
and
are respectively equipped with the norms:
and
.
The space of continuous function
is endowed with the norm:
.
We denote:
: the weak convergence;
: the strong convergence;
: absolute value of x;
;
: image of operator A;
: Leray-Schauder’s degree;
;
: the family of subsets of space
.
Let X be a reflexive Banach space and
the topological dual of X. A map
is said to be monotone, if for all
and for all
, we have
by
we denote the duality brackets for the pair
. If additionally, the fact that
implies that
, then we say that A is strictly monotone. The map A is said to be maximal monotone, if it is monotone and for all
,
, the fact that
implies that
and
. It is clear from this definition that A is maximal monotone if and only if its graph
is maximal with respect to inclusion among the graphs of monotone maps. If A is maximal monotone, for any
, the set
is non-empty, closed and convex. Moreover,
is demi-closed, i.e., if
, either
in X and
in
, or
in X and
in
, then
. If
is everywhere defined and single-valued, we say that A is demi-continuous, if for every sequence
such that
in X, we have that
in
. If map
is monotone and demi-continuous, then it is also maximal monotone. A map
is said to be coercive, if
is bounded or if
is unbounded and we have that
A maximal monotone and coercive map is surjective. Let
be Banach spaces and
. We say:
a) L is “completely continuous”, if
in Y implies
in Z and
b) L is “compact”, if it is continuous and maps bounded sets into relatively compact sets.
In general, these two notions are distinct. However, if Y is reflexive, then complete continuity implies compactness. Moreover, if Y is reflexive and L is linear, then the two notions are equivalent.
3. Auxiliary Results
Let
such that
. First, let us define what we mean by solution of problem (1).
Definition 1. A function
such that
is said to be a solution of the problem (1), if it verifies (1).
Next, we introduce the notions of upper and lower solutions of problem (1).
Definition 2.
a) A function
such that
is said to be an upper solution of the problem (1), if:
b) A function
such that
is said to be a lower solution of problem (1), if:
Now, let us specify what we mean by strict lower and strict upper solutions of problem (1).
Definition 3. A lower solution
of (1) is said to be strict if all solution u of (1) with
is such that
.
Definition 4. An upper solution
of (1) is said to be strict if all solution u of (1) with
is such that
.
Proposition 5. Let
be a lower solution of (1) such that:
i) For all
, there exists
and
is an open interval such that
and:
for all
;
ii)
;
iii)
;
then
is a strict lower solution of (1).
Proof. Let u be a solution of problem (1) such that
for all
.
Let us assume that u is not strict, then there exists
such that
. Whence
A is closed and bounded. Let
. Then
a) If
, then
and there exist
and
according to (i). We can choose
such that
,
, and
Therefore, for almost
,
Since
is an increasing homeomorphism, we have
and
(2)
Also we have:
which contradicts (2).
b) We suppose that
, then
and it follows that:
(3)
Since
, because of the monotonicity of
, if
, we have:
Then,
So
which contradicts (3).
c) We suppose that
, then
. It follows that:
(4)
Since
, because of the monotonicity of
, if
, we have:
Then,
.
So
which contradicts 4. Then,
does not exist. So,
.
Proposition 6. Let
be an upper solution of (1) such that:
i) For all
, there exist
and
is an open interval such that
and:
for all
;
ii)
;
iii)
;
then
is a strict upper solution of (1).
Proof. Let u be a solution of problem (1) such that
for all
.
Let us assume that u is not strict, then there exists
such that
. Whence
E is closed and bounded. Let
. Then
a) If
there exist
and
according to (i). We can choose
such that
,
and
Therefore, for almost
,
Since
is an increasing homeomorphism, we have
and
(5)
Also we have:
which contradicts (5).
b) We suppose that
, then
and it follows that:
(6)
Since
, because of the monotonicity of
, if
, we have:
Then
So
which contradicts (6).
c) We suppose that
, then
. It follows that:
(7)
Since
, because of the monotonicity of
, if
, we have:
. Then
.
So
which contradicts (7). Then,
does not exist. So,
.
Remark 7. In general, for a given problem, there is not a methodology (single valued and multivalued alike) which allows generating a lower and upper solutions. But, one should try simple functions such as constants, linear, quadratic, exponentials, eigenfunctions of simple operator, etc.
We make the following hypotheses on the data of (1):
: There exists a lower solution
and an upper solution
.
is an increasing continuous map such that:
a)
;
b) there exists
such that:
for all
;
c) there exist
such that for a.e
and for all
:
Remark 8. Suppose that
. Then this function satisfies hypothesis
. This function corresponds to the one-dimensional operator p-Laplacian. Another interesting case which satisfies hypothesis
is when
is defined by
with
continuous,
for all
and
is strictly increasing on
and
. For example, we can have
It is well-known that under the monotonicity condition and the hypotheses (a) and (b),
is an increasing homeomorphism from
onto
. And
is strictly monotone and
as
(See Deimling [12] chap. 3). Our operator
is a slightly more restrictive version of the scalar case of the operator used by Sophia Kirytsi-N. Matzakos [13] and Manasevich-Mawhin [14] where growth condition
is not assumed. Nevertheless, it incorporates the operator p-laplacian and many other classes of operators.
is a fonction such that:
i) for all
,
is measurable;
ii) for a.e
,
is continuous;
iii) for a.e
,
and all
, we have:
where
and
a Borel measurable non-decreasing functions such that:
with
;
iv) for every
, there exists
such that for a.e
and for all
with
we have:
.
Remark 9. Hypothesis
(iii) is known as a Bernstein-Nagumo-Wintner growth condition and produces a uniform a priori bound of the derivatives of the solutions of problem (1). And the hypotheses
(i), (ii) and (iv) are well known as
-Carathéodory conditions.
:
and
:
are maximal monotone maps such that
.
Remark 10. There exist functions
proper, convex and lower semi-continuous which are not identically equal to
such that
. More exactly, there exists some increasing positives functions
and
such that
(the minimum absolute value element in the closed, convex set
). Then
,
. We have
for all
, where
and
,
.
:
is a function that maps bounded sets to bounded sets and there exists
such that
is increasing.
Remark 11. We emphasize that
need not be continuous.
Lemma 1. If
and hypotheses
and
(iii) hold,
and if
for all
then, there exists
(depending only on
) such that:
for all
.
Proof. Set
(See hypothesis
). By hypothesis
(iii), we can find
such that
We claim that
for all
. Let’s suppose that it is not the case. Then, we can find
such that
By the mean value theorem, there exists
such that
. Without any loss of generality, we assume that
. We obtain:
.
Since
, by the intermediate value theorem, there exists
and
with
such that
and
. We have:
Thus:
and then
Setting
, we have:
which contradicts the choice of
.
Now, we introduce the truncation map:
defined by
(8)
where
and the penalty function
defined by:
(9)
We set
. Note that for ae
and all
, we have
. Moreover, for almost all
and all
, we have:
with
. For every
, set
and
the Nemitsky operators corresponding to
and
respectively. We set
for every
.
Proposition 12. If hypothesis
(ii) holds, then:
is continuous.
Proof. Since
and
are Nemitsky operators, it is standard to show that they are continuous. It follows that G is continuous.
We introduce the set
and then we define the non-linear operator:
by
Proposition 13. If the hypotheses
and
hold, then
is maximal monotone.
Proof. Given
, we consider the following nonlinear boundary value problem:
(10)
We show that problem (10) has a unique solution
. To this end, given
, we consider the following two-point boundary value problem:
(11)
We set
. Then
and
. We consider the function y defined by
and rewrite (11) in the terms of this function.
(12)
This is a homogeneous Dirichlet problem for (11). To solve (12), let
be the non-linear operator defined by:
where
denote the duality brackets for the pair
.
Let us show that
is strictly monotone.
Let
. We have
Then
We know that
is monotone. Moreover, it is easy to show that
is strictly monotone. Whence
Therefore,
is strictly monotone.
• Let us show that
is demicontinuous.
Using the extended dominated convergence theorem (see for example Hu-Papageorgiou 10, Theorem A.2.54, p. 907), it follows easily that
is demicontinuous.
Recall that an operator monotone and demicontinuous is maximal monotone. So
is maximal monotone.
Let us show that
is coercive.
For
we have:
Hence, using the hypotheses (b) and (c) on
, we obtain:
with
.
whence:
, for some
.
Therefore,
is coercive.
Recall that an operator maximal monotone which is coercive is surjective.
Moreover, since
is strictly monotone, we infer that there exists a unique
such that
. It follows easily that
and it solves the problem (12). Then
and it is the solution of the problem (11). We can define the solution map
which to each pair
assigns the unique solution of the problem (11). Let
be defined by:
We claim that
is monotone.
Indeed, for
, we have:
where
is the scalar product in
.
From (11), we have
. Because of monotonicity of the operators
and
, we have the monotonicity of
.
We claim that
is continuous.
Indeed, let
and
be real sequences converging respectively to b and e.
Let us set that
Now, we consider the following sequence of problems:
(13)
We claim that
is bounded.
Let us multiply (13) by
and integrate on
, we obtain:
By using green’s formula, we obtain:
with
.
whence:
for some
.
Furthermore, using the Cauchy-Schwartz inequality and then the triangular inequality, we obtain the following inequalities:
Then:
So
for some
.
Therefore, the sequence
is bounded. It follows immediately that the sequences
is bounded. So directly from the problem (11), we get that the sequence
is bounded. By integration, we obtain
. So the sequence
is bounded. Then we have respectively
Due to the compact embedding of
in
, we have:
Since
is an increasing homeomorphism,
exists and is continuous. So, we have:
in
. Whence
(i.e.
). Therefore passing to the limit as
, we have:
(i.e.
is continuous). So,
is continuous.
We claim that
is coercive.
For
, we have:
Then
where
denote the euclidean norm in
.
Since
, by mean value theorem, there exists
such that
.
As
, we have:
for all
. In particular, we have:
Hence
Therefore
is coercive.
We infer that
is maximal monotone (being continuous, monotone) and coercive. Thus
is surjective. Now, we consider
with operator B defined by
for all
. Since
is coercive and B is maximal monotone, we deduce that
is coercive. Also,
is maximal monotone (see Brezis [15] Corollary 2.7, p. 36 or Zeidler [14] Theorem 32.I, p. 897). So
is surjective. We infer that we can find
such that
. Since
, we can find
such that
and
and
and
. So
and
. It follows that
. Therefore
is the solution of the problem (10).
Let
be the operator defined by:
Since
is continuous and monotone, then H is continuous and monotone. Therefore H is maximal monotone. Moreover it is evident to see that H is strictly monotone.
Since in (10) the choice of h is arbitrary, then by the previous arguments, we have:
(14)
We denote by
the duality brackets between the pair
.
Let us show that
surjective implies
is maximal monotone
For this purpose, we suppose that, for some
and some
:
(15)
Because of (14), we can find
such that:
We use this in (16) with
, we obtain:
(16)
Because H is strictly monotone, from (16), we conclude that
and
. So
is maximal monotone. In addition, since
is monotone, we have
. Whence the operator
is maximal monotone, strictly monotone and coercive. Therefore
is well defined, single valued, and maximal monotone (From
into
).
Proposition 14. If hypothesis
holds, then
is completly continuous.
Proof. Suppose that
in
. We have to show that
in
. let us set
for all
. We have
(17)
By integration by part, we obtain:
(18)
Since
, we have
and
for all
. We recall that
, then:
(19)
Moreover, the map
being monotone, we have:
(20)
From (19) and (20), we obtain:
(21)
From (21) and (18), we infer that:
(22)
By hypothesis b) on
, we have:
(23)
It follows from (22) and (23) that:
whence:
for some
.
Therefore the sequence
is bounded. Then we can find a convergente subsequence of
. So
in
. Due to the compact embedding of
in
, we have
in
. Since
is bounded, we have
and
are bounded. It follows immediatly that:
is bounded. Then
imply that
is bounded. Whence, by integration,
is bounded. So we can suppose that
in
. Due to the compact embedding of
in
, we obtain
in
. Since
is an increasing homeomorphism,
exists and is continuous. So, we have
for all
. Then
for all
. It follows that
in
(By Lebesgue dominated convergence theorem). We infer that:
. It follows
But recall that
in
. Thus
in
. This prove that the operator L is completly continuous.
Proposition 15. If the conditions in lemma 1 hold, then a function
is solution of (1) if and only if
and u is a fixed point of
defined by:
with the operator Q defined by:
.
Moreover, for all
,
and K is continuous and completly continuous.
Proof. If u is a solution of (1), then
because of hypothesis
(iii). It follows that
. So,
. We have also
and
. Hence,
.
Furthermore, we have:
Therefore, u is a fixed point of K.
On the other hand, if
and u is a fixed point of K, then we have
,
,
and
a.e on
a.e on
. Hence, u is solution of (1).
Finally, by lemma 1, we have:
,
.
• Let us show that K is continuous.
Let
in
. Then, there exists
such that
and
.
We will show that
in
. That’s mean
in
and
in
. For
, we have:
Since
and N are continuous respectively in
and
, we have:
Also from the monotone convergence theorem, we have:
Since
in
, it follows that
in
. Using the previous arguments and the dominated convergence theorem, we have:
It follows that
Since
is an increasing homeomorphism,
exists and is continuous. Finally, we have:
By integration, we obtain:
in
. Therefore, K is continuous.
• Let us show that K is completely continuous.
Let
be a bounded set of
. We set
. Since
is bounded, there exist
such that:
It follows that:
Therefore, there exist
such that
.
For
and
.
We infer that for all
, there exists
such that
It suffices to take
. Therefore,
being an increasing homeomorphism, for all
, it exists
, such that for all
,
, if
, then
is uniformly equicontinuous and
is bounded on
. By Ascoli-Arzela’s theorem,
is relatively compact in
. Since K is continuous and
is relatively compact in
for every bounded subset
of
, K is completely continuous.
4. Existence Results with Ordered Pair of Lower and Upper Solutions
We consider the operator
defined by:
Evidently,
is bounded (i.e., maps bounded sets to bounded ones) and is continuous.
Theorem 16. Suppose that there exists a lower solution
and an upper solution
such that
,
.
Then the problem (1) admits at least one solution u, such that:
Moreover, if
are strict, then
where
K is the operator associated to the problem (1).
We consider the following auxilary boundary problem:
(24)
A solution of problem (24) is a function
such that
and satisfied (24).
The problem (24) is equivalent to the fixed point problem
and
with
defined by:
We have:
Lemma 2. All solution u of (24) is such that
,
.
Proof. Since
is a lower solution of the problem (1), we have:
(25)
Soustraying (25) from (24), we obtain:
(26)
We multiply (26) by
and then integrate on
. We obtain:
(27)
The integration by parts of the left-hand side in inequality, yields:
(28)
We set
(29)
Also, from the boundary conditions in (24) and (25), we have:
If
, then from the monotony of
(See hypothesis
), we have:
. Whence
.
So, it follows that
(30)
In a similar way, using the boundary conditions
and
with
, if
, we have:
. We infer that
.
It follows that
(31)
Also, since
is an increasing homeomorphism, we have:
(32)
where
.
Using the inequalities (30), (31) and (32) in the first member of (27), we obtain:
(33)
Furthermore:
(34)
Also from the definiton of the penalty map
, if
(By
, we denote the Lebesgue mesure in
), then:
(35)
Finally, by virtue of hypothesis
and since
, we see that:
(36)
Using the inequalities, (34), (35) and (36) in the second member of (27), we infer that:
(37)
We consider (27) and using (33) and (34), we have a contradiction when
. Therefore, for all
,
. In a similar fashion we show that
for all
. Thus
.
Proof. theorem 16: As in the proof of the proposition 15, we can show the complete continuity of the operator
. Moreover,
for all
. Therefore, by Leray-Schauder’s theorem, we can say that the operator
has a fixed point u in the open ball
which is solution of problem (24). It follows, by the lemma 2, that u is also solution of problem (1).
We assume that
is a strict lower solution and
is a strict upper solution of (1). Let
be quite a few such that
for some
.
Because
is completely continuous, we can compute the degree of
. The function H defined by
is compact on
. We assume that there exist
and
such that
, then
. But
, so
which contradict the fact that
. We can apply the homotopic invariance degre property of Leray-Schauder to obtain:
Indeed, let us recall the following set:
By definitions of strict lower and strict upper solutions,
and
cannot be solution of problem (24). Therefore, (24) has not solution on the boundary of
. By using the additivity and excision properties of Leray-Schauder degree, we obtain:
Furthermore, since K is completely continuous operator associate to (1). That equates to
on
, we have
5. Existence Results with Non Ordered Lower and Upper Solutions
Theorem 17. We assume that there exists a lower solution
and an upper solution
of (1) such that
such that
. (38)
Then the problem (1) admits at least one solution u, such that:
for some
(39)
and
(40)
Proof. We set
.
We consider the functions
,
, and for
, the multifunctions
are defined respectively by:
where, for
,
and
are defined as in remark 10,
is
-Caratheodory and
and
are maximal monotone operators. We consider the following modified problem:
(41)
We can verify that
is a lower solution and
is upper solution of the problem (41). Let
be defined by
. We have:
So we can find
such that
and
Therefore,
is an upper solution of (41).
The function
defined by
, verifies:
So we can find
such that
and
Therefore,
is a lower solution of (41). Furthermore,
Let us introduce the sets
and
By using the definition (39), we obtain:
Also we have:
Let us consider:
Then
and
Since all constant function between
and
is into
,
is nonempty.
Let
be the fixed point operator associated to problem (41) given in the proposition 15. We consider
such that
and
. There exists
such that
or
.
Let us consider the case
.
If
, then
and there exists
such that
for all
. Moreover
. Whence,
,
for all
. It follows that u is increasing on
. That contradicts the existence of
.
If
and we obtain the contradiction
.
If
and we obtain the contradiction
.
In the similar fashion, we obtain contradiction with the case
. Therefore
(42)
Let
such that
. It becomes from (42) that
,
, and
. It follows, there exists
such that
and
, that implies
Then,
So,
(43)
We have two cases:
• 1rst case: We assume that there exists
such that
. From (43), we infer that
, that implies that u is a solution of (1), and (39) and (40) are satisfied. Then, there exists
such that
or
,
• 2nd case: We assume that
for all
. Then, as in the proof of theorem 16, we have:
Theorem 18. If there exists a lower solution
and an upper solution
of the problem (1), then the problem (1) admits at least one solution u such that:
(44)
Proof. If
, by theorem 16, the problem (1) admits at least one solution such that:
Moreover, (44) holds.
If
such that
(45)
By the theorem 17, the problem (1) admits at least one solution, such that:
6. Multiplicity Results
Theorem 19. We assume that there exists
a lower solution and
a strict lower solution of problem (1),
an upper solution and
a strict upper solution of problem (1) such that:
Then the problem (1) admits at least three solutions u, v and w such that:
for some
.
Proof. By using the theorem 16 and the fact that
and
are strict, we can say that the problem (1) admits at least one solution u, such that:
(46)
By using the theorem 16, the problem (1) admits at least one solution w such that:
(47)
(46) and (47) yield
and
.
In the following theorem, we show existence of at least two solutions of the problem (1).
Theorem 20. We assume that there exists
and
are two lower solutions of the problem (1),
is a strict upper solution of problem (1) such that:
Then the problem (1) admits at least two solutions u and w such that:
and
such that
. (48)
Proof. By using the theorem 16 and the fact that
is strict, we can say that the problem (1) admits at least one solution u, such that:
(49)
By using the theorem 17, the problem (1) admits at least one solution w such that:
(50)
(49) and (50) yield
.
Theorem 21. We assume that there exists
a strict lower solutions of the problem (1) and two upper solutions
and
of the problem (1) such that:
Then the problem (1) admits at least two solutions u and w such that:
and
such that
. (51)
Proof. The proof is similar to those of theorem 18.
7. Example and Periodic Problem
7.1. Example
Let us consider the following problem:
(52)
where
and
are defined as in problem (1). Here,
,
for all
and by the remark 8, it satisfies hypothesis
. Therefore, theorem 16 and theorem 17 are true for the problem (52). Moreover, by [8] (see example 5.2.25 page 404), this problem unifies classical problems of Dirichlet, Neumann and Sturm-Liouville and go beyong them.
7.2. Periodic Problem
Let us consider the following periodic problem:
(53)
Remark 22. The theorems 16 and 17 stay true for this problem (see 8 remark 5.2.26 page 404 and also [5] section 6 page 23).
8. Conclusion
In this article, by combining lower and upper solutions method, theory of maximal monotone operators and theory of topological degree, we establish existence and multiplicity results for second-order problems with multivalued boundary conditions. We give an example but more examples and applications can be given. In perspective, we will study the same problems under general multivalued boundary conditions. Also, the same problem can be considered for a singular
-Laplacian operator.