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Existence Result for Impulsive Differential Equations with Integral Boundary Conditions
We investigate the following differential equations: , with impulsive and integral boundary conditions , , , , where . The expression of Green's function and the existence of positive solution for the system are obtained. Upper and lower bounds for positive solutions are also given. When , , , and take different values, the system can be simplified to some forms which has been studied in the works by Guo and LakshmiKantham (1988), Guo et al. (1995), Boucherif (2009), He et al. (2011), and Atici and Guseinov (2001). Our discussion is based on the fixed point index theory in cones.
The theory of impulsive differential equations in abstract spaces has become a new important branch and has developed rapidly (see [1–4]). As an important aspect, impulsive differential equations with boundary value problems have gained more attention. In recent years, experiments in a variety of different areas (especially in applied mathematics and physics) show that integral boundary conditions can represent the model more accurately. And researchers have obtained many good results in this field.
In this paper, we study the existence of positive solutions for the following system: where , , , , . , are left continuous at , . . And , , are continuous and positive functions.
Throughout the rest of the paper, we assume is a fixed positive number, and is a parameter. , are real-valued measurable functions defined on , and they satisfy the following condition:
(H1) , , almost everywhere, and
This paper aims to obtain the positive solution for (1). In Section 2, we introduce some lemmas and notations. In particular, the expression and some properties of Green's functions are investigated. After the preparatory work, we draw the main results in Section 3.
Theorem 1 (Krasnoselskii's fixed point theorem). Let be a Banach space and . Assume , are open sets in with , and be a completely continuous operator such that either(i), , and , ; or (ii), , and , . Then has a fixed point in .
Definition 2. For two differential functions and , we defined their Wronskian by
Consider the linear nonhomogeneous problem of the form Its corresponding homogeneous equation is
Lemma 3. Suppose that and form a fundamental set of solutions for the homogeneous problem (5). Then the general solution of the nonhomogeneous problem (4) is given by where and are arbitrary constants.
Consider the following boundary value problem with integral boundary conditions: Denote by and the solutions of the homogenous equation (5) satisfying the initial conditions
(H2) Let , denote a function satisfies .
For convenience, we denote , .
Lemma 4. Let be a nonnegative continuous function defined for , and a nonnegative integrable function on . Then for arbitrary nonnegative continuous function defined on , the Volterra integral equation has a unique solution . Moreover, this solution is continuous and satisfied the inequality
Proof. We solve (14) by the method of successive approximations setting If the series converges uniformly with respect to , then its sum will be, obviously, a continuous solution of (14). To prove the uniform convergence of this series, we put Then it is easy to get from (16) that Hence it follows that (14) has a continuous solution and because , , , for this solution the inequality (15) holds. Uniqueness of the solution of (14) can be proved in a usual way. The proof is complete.
Remark 5. Evidently, the statement of Lemma 4 is also valid for the Volterra equation of the form
Lemma 6. For the solution of the BVP (11), the formula holds, where
Proof. By Lemma 3, the general solutions of the nonhomogeneous problem (4) has the form
where and are arbitrary constants. Now we try to choose the constants and so that the function satisfies the boundary conditions of (11).
From (23), we have Consequently, Substituting these values of and into the first boundary condition of (11), we find Similarly from the second boundary condition of (11), we can find Putting these values of and in (23), we get the formula (21), (22).
Lemma 7. Let condition (H1) hold. Then for the Wronskian of solution and , the inequality , holds.
Proof. Using the initial conditions (12), we can deduce from (5) for and the following equations: From (28), by condition (H1) and Lemma 4, it follows that Now from (3), we get , . The proof is complete.
Lemma 8. Under condition (H1) the Green's function of the BVP (11) is positive. That is, for .
Let denote the Banach of all continuous functions equipped with the form , for any . Denote , then is a positive cone in .
Let us set , , and by Lemma 8, obviously, , .
Define a mapping in Banach space by where
Lemma 9. The fixed point of the mapping is a solution of (1).
Proof. Clearly, is continuous in for . For , where We have where We can easy get that which implies that the fixed poind of is a solution of (1). The proof is complete.
Lemma 10. Let , then is a cone.
Proof. (i) For for all and for all , , we have
(ii) If and , we have It implies that . Hence is a cone.
Defined a linear operator by Then we have the following lemma.
Lemma 11. If (H2) is satisfied, then(i) is a bounded linear operator, ;(ii) is invertible;(iii).
for all , .
Using , it is easy to see that .
Let . Then for all . Since , it follows that for each . So .
(ii) We want to show that is invertible, or equivalently is not an eigenvalue of .
Since , it follows from condition (H2) that .
So On the other hand, we suppose 1 is an eigenvalue of , then there exists a such that . Moreover, we can obtain that . So . Thus this assumption is false.
Conversely, 1 is not an eigenvalue of . Equivalently, is invertible.
(iii) We use the theory of Fredholm integral equations to find the expression for .
Obviously, for each , .
By (40), we can get The condition implies that 1 is not an eigenvalue of the kernel . So (43) has a unique continuous solution for every continuous function .
By successive substitutions in (43), we obtain where the resolvent kernel is given by Here , and .
The series on the right in (45) is convergent because .
It can be easily verified that .
So we can get Therefore So Thus . This completes the proof of the lemma.
Remark 12. Since for each , it is easy to prove that .
3. Main Results
Consider the following boundary value problem (BVP) with impulses: Denote a nonlinear operator by It is easy to see that solutions of (49) are solutions of the following equation: According to Lemma 11, is a solution of (51) if and only if it is a solution of It follows from (46) that is a solution of (52) if and only if So, the operator can be written as It satisfies the conditions of Theorem 1 with and the cone .
Let us list some marks and conditions for convenience.
The nonlinearity is continuous and satisfies the following.
(H3) There exist and , with such that for all , .
(H4) There exist and , with such that for all , .
Then, we can get the following theorem.
Theorem 13. Assume (H1), (H2), (H3), and (H4) are satisfied. And then, if satisfies The problem (49) has at least one positive solution.
Proof. First of all, we show that operator is defined by (54) maps into itself. Let .
Then for all that , and Because from the formula (54), we have Hence, inequality (59) is established.
This implies that or equivalently On the other hand, it follows that In fact, we have It follows from (62) that So, we get This show that .
It is easy to see that is the complete continuity.
We now proceed with the construction of the open sets and .
First, let with . Inequality (59) implies By condition (H3) and (58), we obtain So Consequently, .
Let . Then, we have for .
Next, let and set .
For with , we have It follows from (63) that By condition (H4) and (58), we obtain Since we have for all . It follows from the above inequality that Hence for .
It follows from (i) of Theorem 1 that has a fixed point in , and this fixed point is a solution of (49).
This completes the proof.
Next, with and as above, we assume that satisfied the following.
(H5) There exist , with such that for all , .
(H6) There exist , with such that for all , .
Theorem 14. Assume (H1), (H2), (H5), and (H6) are satisfied. And then, if satisfies The problem (49) has at least one positive solution.