Abstract
In this paper, by using the coincidence degree theory and the upper and lower solutions method, we deal with the existence of multiple solutions to threepoint boundary value problems for secondorder differential equation with impulses at resonance. An example is given to show the validity of our results.
Keywords:
resonance; coincidence degree; upper and lower solutions; impulsive; threepoint boundary value problems1 Introduction
The purpose of the present paper is to investigate the following secondorder impulsive differential equations:
together with the boundary conditions:
where , is a continuous differentiable function, is continuous, , for , m is a fixed positive integer, , , denotes the jump of at , . , (, ) represent the right limit (left limit) of and at , respectively.
Impulsive differential equations describe processes which experience a sudden change of their state at certain moments. The theory of impulse differential equations has been a significant development in recent years and played a very important role in modern applied mathematical models of real processes rising in phenomena studied in physics, population dynamics, chemical technology, biotechnology, and economics; see [110] and the references therein.
Recently, several authors (see [6,1114] and the references therein) have studied the existence of nontrivial or positive solutions for secondorder threepoint boundary value problem of the type
Note that the nonlinear term f depends on u and its derivative , then the relative problem becomes more complicated. A general method to deal with this difficulty is to add some conditions to restrict the growth of the term. One condition is the Caratheodory nonlinearity, the other usual condition is Nagumo condition or NagumoWinter condition (see [2,9,12,1520]). When , the linear operator is invertible, this is the socalled nonresonance case. Gupta et al. made use of the LeraySchauder continuation theorem to get the results on the existence of the solution for the problems (1.3) when in [14]. By using the LeraySchauder continuation theorem and in the presence of two pairs of upper and lower solutions, Khan and Webb [12] established the existence of at least three solutions for the problem (1.3) when . The linear operator is noninvertible when , this is the socalled resonance case, and the LeraySchauder continuation theorem cannot be applied. In [11], by using the coincidence degree theory of Mawhin [21] and some linear or nonlinear growth assumptions on f, Feng and Webb obtained the existence of the solution of the problem (1.3) when . By applying the nonlinear alternative of LeraySchauder, Ma [13] have showed the existence of at least one solution for the problem (1.3) when .
Recently, using the coincidence degree theory and the concept of autonomous curvature bound set, Liu and Yu [6] have studied the existence of at least one solution for the problem (1.1)(1.2) when , .
In the present paper, we assume that there exist n ( and ) pairs of upper and lower solutions for problem (1.1)(1.2) and the nonlinear f satisfies a Nagumolike growth condition with respect to . By considering a suitably modified nonlinearity and applying the coincidence degree method of Mawhin [21], the existence of multiple solutions for the problem (1.1)(1.2) is given.
2 Preliminaries
Let
where
Obviously, X is a Banach space with the following norm:
In the following, we recall the concept of strict upper and lower solutions for problem (1.1)(1.2).
Definition 2.1 A function is said to be a strict lower solution of the problem (1.1)(1.2) if
Similarly, a function is said to be a strict upper solution of the problem (1.1)(1.2) if
Remark 2.2 Let be continuous, , and is a solution of the problem (1.1)(1.2), if is a strict lower solution (strict upper solution) for the problem (1.1)(1.2) with (), then () on .
Definition 2.3 Let α be a strict lower solution and β be a strict upper solution for the problem (1.1)(1.2) satisfying on J. We say that has property (H) relative to α and β, if there exists a function such that
3 The key lemmas
Then the problem (1.1)(1.2) can be written as
Lemma 3.1Suppose thatLbe defined in the above. ThenLis a Fredholm operator of index zero. Furthermore
and
Proof Firstly, it is clear that (3.1) holds. Next, we shall prove that (3.2) holds.
The following problem:
has a solution satisfying and if and only if
In fact, if (3.3) has a solution satisfying , , then from (3.3) we have
According to , , we get (3.4).
On the other hand, if (3.4) holds, setting
where c is an arbitrary constant, then is a solution of (3.3) with , . Hence (3.2) holds. □
Take the projector as follows:
Thus, we obtain
Then . Hence . Since , we have , which implies . Hence L is a Fredholm operator of index zero.
Take , . So the generalized inverse of L can be written as
Set , . Then for the function ψ defined by (2.8), let be the solution of the following initial value problem:
and be the solution of the following initial value problem:
Lemma 3.2Suppose that there exists a constant, thenis well defined inand positive on this interval, is also well defined and positive in. Moreover, if, thenfor any; for any.
Proof We only consider the case (in the case , the proof is similar). Assume that there exists a such that . Let . It follows from (3.7) that
Integrating the above equation over we get (let )
However, the left side of the above equation equals ∞ by (2.8). We reach a contradiction. Hence for every . From , by the continuity of solution of differential equations on the initial values, we obtain for . The proof is complete. □
Define the following sets:
Lemma 3.3Let Deg denote the coincidence degree. Let the following conditions hold:
(i) , (Mis given in Lemma 3.2);
Then
Proof Consider the following family of equations:
We will show
If not, then there exist some and such that (3.10) holds. Note that if and only if and either for some , or for some . There are two possibilities.
Case (I). If , , . In this case, , or .
Subcase (1). Suppose . Let . Then , and . When , , we get
If , then . From condition (i), we have
which is a contradiction.
If , then for some and . Thus from (iii), we have
On the other hand, , and , thus
which is a contradiction.
If , it is easy to see that . Since , we have , thus we can obtain . However, from condition (i) we know
which is a contradiction.
If , then . This means that . Since , we have . However, according to the above arguments, we know , which is a contradiction.
Subcase (2). Suppose for , . Obviously, . Since . Thus we get
Let
Without loss of generality, we suppose that for . Then we have three possibilities.
(i) If , then there exists a sufficiently small neighborhood of such that . Thus , and has a local maximum value on , which implies . But in this case, from (3.7), (3.12), and (ii), we have
which is a contradiction.
(ii) If , then there exists a sufficiently small neighborhood of such that . Thus , and has a local maximum value on , which implies . But in this case, from (3.8), (3.12), and (ii), we have
which is a contradiction.
(iii) If , then there exists a sufficiently small neighborhood of such that . Thus . By the same argument as in (3.13), we reach a contradiction.
Case (II). for some . In this case, , or .
Subcase (3). Suppose for some . Let be defined in the above. Obviously, we have for and . It follows from (iii) that
If , then . Form (3.11) and (i), we have
which is a contradiction.
If , together with (iii), we get . But and for , which yields , and we reach a contradiction.
Similar to subcase (2), we can show that is also impossible. Combining the results of case (I) and case (II) we obtain (3.11).
On the other hand, for , it follows from [21] that (3.10) is equivalent to the following family of operator equations:
where E is the identity mapping.
From (3.5) and (3.6), we have
and
By the AscoliArzela theorem, it is easy to show that is bounded and is compact. Thus N is Lcompact on .
Then it is easy to prove that is completely continuous and we claim that
In fact, for , it follows from (3.10) and (3.14) that (3.15) holds. For , if there exists a such that , that is , in this case, , , hence or . However, it follows from (iii) that
Thus we get and , which contradicts for . Therefore (3.15) holds. As follows from [21] and by using the invariance of LeraySchauder degree under homotopy, we obtain
Since KerL is one dimensional and , , we get
From the property of coincidence degree we proved Lemma 3.3. □
Lemma 3.4Assume that
(c1) there exist lower and upper solutions, of the problem (1.1)(1.2), respectively, with;
(c2) is continuous and has property (H) relative toα, β;
(c3) , are continuous for each, and satisfy
Proof Choose large enough for , , , such that
and let () be defined in (3.7) and (3.8), then it follows from Lemma 3.2 that we can choose large enough such that
Define the auxiliary functions F and , as follows:
where
We then generalize F to and . It is easy to see that F, , are continuous and satisfy
Moreover, when , F, , are bounded. It follows from Lemma 3.3 that
where
Next we show
It suffices to show that
In fact, let such that and assume that
Case (1). If there exists , such that the function attains its maximum value , which implies and . But on the other hand,
which implies . We reach a contradiction.
If , then , . On the other hand, since and is a strict upper solution of problem (1.1)(1.2), we see that . Thus . Note that assumes the maximum value at , there exists , such that and . By the same argument as in (3.18) where , we reach a contradiction.
If , then . According to (2.6), we get
if , which is a contradiction. Thus , which implies that also attains its maximum value at . By the same argument as in (3.18) where , we reach a contradiction.
Hence the function cannot have any nonnegative maximum value on the interval , for .
Case (2). If there exists a such that , from case (1), we get for some . Hence , , and consequently we have , which implies , because
By the continuity of f and is a strict upper solution of problem (1.1)(1.2), there exists a sequence with , , as such that
where are from the mean value theorem. As before, we also get
where . As a result, we can obtain
which is a contradiction to (3.19).
Case (3). If for all , then there must be a () such that , and , which implies . Namely, . However, this is impossible because
Thus we have proved that on J. Similarly we can show that on J. It then follows that on J.
We now shall prove that
Assume that (3.20) cannot hold. Thus there are two possibilities:
(a) There exists for some such that .
(b) There exists some such that .
If , then there exists a sufficiently small neighborhood of τ such that . Thus and . However, it follows from (3.7) and (3.21) that
which is a contradiction.
If , then there exists a sufficiently small neighborhood of τ such that . Thus and . However, it follows from (3.8) and (3.21) that
a contradiction.
If , then there exists a sufficiently small neighborhood of τ such that . Thus and . By the similarly argument as in (3.22), we reach a contradiction.
Hence , , . Similarly, we can prove , , .
Likewise, we can show that case (b) is also impossible, and thus (3.20) holds. Combining the above results, we see that if , , then . Hence (3.17) is proved. From the property of coincidence, we know that (3.16) holds. Since in , , we have . The proof is complete. □
4 Main results
We are now in a position to prove our main result on the existence of at least solutions of boundary value problem (1.1)(1.2).
Theorem 4.1Assume that
(H_{1}) there existn (and) pairs of strict lower and upper solutions, of the problem (1.1)(1.2) such that
(H_{2}) , is continuous onand has property (H) relative to, ;
(H_{3}) , are continuous for each, and satisfy
Then BVP (1.1)(1.2) has at leastsolutionssuch that, and
Proof Choose large enough such that, for ,
and let be defined in (3.10), then it follows from Lemma 3.4 that we can take large enough such that, for all ,
Also
Then by Lemma 3.4, we have
From the additive property of coincidence degree, we obtain
Since is arbitrary, we first deal with the case , the above discussion implies that the equation , that is, the problem (1.1)(1.2) has at least one solution in the set , and respectively. That is, there exist at least three different solutions , and such that
For , replacing , , , by , , , respectively, then we can obtain another two different solutions , of the problem (1.1)(1.2) such that
Along this way, we can complete the proof by the induction method. □
We now present an example to illustrate that the assumptions of our theorem can be verified.
Example 1 Consider the following boundary value problem:
Next, it is clear that and are n ( and ) pairs of strict lower and upper solutions of the problem (4.1), respectively. It can be seen that , , , . Thus conditions (H_{1}) and (H_{3}) of Theorem (4.1) hold.
Hence f has property (H). It follows from Theorem 4.1 that the problem (4.1) has at least different solutions such that
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
The authors declare that the study was realized in collaboration with the same responsibility. All authors read and approved the final manuscript.
Acknowledgements
The authors are highly grateful to the referees for careful reading and comments on this paper. The research is supported by Hunan Provincial Natural Science Foundation of China (No. 13JJ3106, and 12JJ2004); it is also supported by the National Natural Science Foundation of China (No. 61074067, 11271372, and 11201138).
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