Multiple Positive Solutions for Nonlinear First-Order Impulsive Dynamic Equations on Time Scales with Parameter
© D.-B.Wang and W.Guan. 2009
Received: 13 February 2009
Accepted: 14 May 2009
Published: 22 June 2009
By using the Leggett-Williams fixed point theorem, the existence of three positive solutions to a class of nonlinear first-order periodic boundary value problems of impulsive dynamic equations on time scales with parameter are obtained. An example is given to illustrate the main results in this paper.
Let be a time scale, that is, is a nonempty closed subset of . Let be fixed and be points in , an interval denoting time scales interval, that is, Other types of intervals are defined similarly. Some definitions concerning time scales can be found in [1–5].
where is a positive parameter, , is right-dense continuous, , and for each and represent the right and left limits of at .
where is a time scale which has at least finitely-many right-dense points, is regressive and right-dense continuous, is given function, . The paper  obtained the existence of one solution to problem (1.2) by using the nonlinear alternative of Leray-Schauder type.
They proved the existence of one solution to the problem (1.3) by applying Schaefer's fixed point theorem and the nonlinear alternative of Leray-Schauder type.
In , Li and Shen studied the problem (1.3). Some existence results to problem (1.3) are established by using a fixed point theorem, which is due to Krasnoselskii and Zabreiko, and the Leggett-Williams fixed point theorem.
In , the first author studied the problem (1.1) when . The existence of positive solutions to the problem (1.1) was obtained by means of the well-known Guo-Krasnoselskii fixed point theorem.
By using the fixed point index, some existence, multiplicity and nonexistence criteria of positive solutions to the problem (1.4) were obtained for suitable .
Motivated by the results mentioned above, in this paper, we shall show that the problem (1.1) has at least three positive solutions for suitable by using the Leggett-Williams fixed point theorem . We note that for the case and problem (1.1) reduces to the problem studied by .
In the remainder of this section, we state the following theorem, which are crucial to our proof.
for all and .
Let be constants,
Theorem 1.1 (see ).
Then has at least three fixed points in satisfying
Throughout the rest of this paper, we always assume that the points of impulse are right-dense for each
where is the restriction of to and
with the norm Then X is a Banach space.
and the periodic boundary condition
Since the method is similar to that of in [27, Lemma 3.1], we omit it here.
It is obvious, so we omit it here.
where It is not difficult to verify that is a cone in
We define an operator by
By [27, Lemmas 3.3 and 3.4], it is easy to see that is completely continuous.
3. Main Result
and for we define
Assume that there exists a number such that the following conditions:
Let it is easy to see that is a nonnegative continuous concave functional on such that
First, we assert that there exists such that is completely continuous.
In fact, by the condition of (H2), there exist and such that
Let if then and we have
Take then the set is a bounded set. According to that is completely continuous, then maps bounded sets into bounded sets and there exists a number such that
If we deduce that is completely continuous. If then from (3.4), we know that for any and hold. Then we have is completely continuous. Take then and are completely continuous.
Second, we assert that and for all
In fact, take so Moreover, for then and we have
Third, we assert that there exist such that if
Indeed, by the condition of (H2), there exist and such that
Then we get
Finally, we assert that if and
To do this, if and then
To sum up, all the hypotheses of Theorem 1.1 are satisfied by taking Hence has at least three fixed points, that is, the problem (1.1) has at least three positive solutions and such that
Using (H3) instead of (H2) in Theorem 3.1, the conclusion of Theorem 3.1 remains true.
Taking then by it is easy to see that So, for all we have Obviously, we have
Therefore, together with Corollary 3.2, it follows that the problem (4.1) has at least three positive solutions for .
The authors express their gratitude to the anonymous referee for his/her valuable suggestions.
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