Nonlinear boundary value problem for nonlinear second order differential equations with impulses

The paper deals with the impulsive nonlinear boundary value problem u(t) = f(t, u(t), u(t)) for a. e. t ∈ [0, T ], u(tj+) = Jj(u(tj)), u (tj+) = Mj(u (tj)), j = 1, . . . ,m, g1(u(0), u(T )) = 0, g2(u (0), u(T )) = 0, where f ∈ Car([0, T ] × R), g1, g2 ∈ C(R ), Jj , Mj ∈ C(R). An existence theorem is proved for non–ordered lower and upper functions. Proofs are based on the Leray–Schauder degree and on the method of a priori estimates. Mathematics Subject Classification 2000: 34B37, 34B15.


Introduction
The nonlinear impulsive boundary value problem (IBVP) of the second order with nonlinear boundary conditions has been studied by many authors by the lower and upper functions method. For instance, the paper [1] considers such problem provided the nonlinearity in the equation satisfies the Nagumo growth conditions. In [2] the Nagumo conditions are replaced with other ones, which allow more than the quadratic growth of the right-hand side of the differential equation in the third variable. Both these works deal with well-ordered lower and upper functions. Until now there are no existence results available for the above problem such that its lower and upper functions are not well-ordered. The aim of this paper is to fill in this gap. The arguments are based on the ideas of papers [4] and [5], where the periodic nonlinear IBVP in the non-ordered case is investigated.
The paper is organized as follows. The first section contains basic notation and definitions. In the second section the Leray-Schauder Degree Theorem is established (Theorem 9) for the well-ordered case, which is used to prove the main existence result in the third section. As a secondary result the existence theorem with Nagumo conditions is obtained (Theorem 8). The third section contains the existence result (Theorem 21) for non-ordered lower and upper functions, where a Lebesgue bounded right-hand side of the differential equation is considered. T ] → R such that the function u |(t i ,t i+1 ) (and its derivative) is continuous and continuously extendable to [t i , t i+1 ] for i = 0, . . . , m, u(t i ) = lim t→t i − u(t), i = 1, . . . , m + 1 and u(0) = lim t→0+ u(t). Moreover, AC D (or AC 1 D ) stands for the set of functions u ∈ C D (or u ∈ C 1 D ) which are absolutely continuous (or have absolutely continuous first derivatives) on each subinterval (t i , t i+1 ), i = 0, . . . , m. For u ∈ C 1 D and i = 1, . . . , m + 1 we write and Note that the set C 1 D becomes a Banach space when equipped with the norm · D and with the usual algebraic operations. By the symbol R + we denote the set of positive real numbers and R + 0 = R + ∪ {0}.
Let k ∈ N. We say that f : [0, T ] × S → R, S ⊂ R k satisfies the Carathéodory conditions on [0, T ] × S if f has the following properties: For the set of functions satisfying the Carathéodory conditions on [0, T ] × S we write Car([0, T ] × S). For a subset Ω of a Banach space, cl(Ω) stands for the closure of Ω, ∂Ω stands for the boundary of Ω.
We study the following boundary value problem with nonlinear boundary value conditions and impulses: and u (t i ) are understood in the sense of (1) for i = 1, . . . , m.
Definition 2 A function σ k ∈ AC 1 D is called a lower (upper) function of the problem (2) -(4) provided the conditions where k = 1 (k = 2), are satisfied.
Remark 3 If we put for x, y ∈ R, then (4) reduces to the periodic conditions From (14) we see that g 1 is one-to-one in x, which implies that g 1 satisfies (11). Moreover, g 1 fulfils (12) because g 1 is increasing in y. Similarly, since g 2 is increasing in x and decreasing in y, we have that g 2 satisfies (13). We . . , m having the following properties: Next, we consider d 0 , d T ∈ R such that We define an auxiliary impulsive boundary value problem where f ∈ Car(J × R 2 ),J i ,M i ∈ C(R), i = 1, . . . , m, d 0 , d T ∈ R.
is satisfied. We are allowed to use Lemma 5 and get (23). This fact together with (29) and (28) implies that u satisfies (2) and the first condition in (3). From the Mean Value Theorem it follows that for i = 1, . . . , m there exists Due to (24) and (26) we can see that u ∞ ≤ c, which together with (28) implies that u satisfies the second condition in (3).
Step 4. It remains to prove the validity of (4). It is sufficient to prove the inequalities and Let us suppose that the first inequality in (35) is not true. Then In view of (27) and (30) we have u(0) = σ 1 (0), thus it follows from (12) and (23) that which contradicts (7). We prove the second inequality in (35) similarly. Let us suppose that the first inequality in (36) is not valid, i. e. let It follows from (27) and (30) that and by (37) we obtain that 0 > g 2 (u (0), u (T )). Further, by virtue of (7), (30), (35) and (38), we have In view of (23) and (11) we get It follows from (23), (38) and (39) that σ 1 (T ) ≥ u (T ) and u (0) ≥ σ 1 (0). Finally, by (13), we get the inequalities contrary to (7). The second inequality in (36) can be proved by a similar argument. Due to (30), the conditions (35) and (36) imply (4). 2 We can combine Proposition 6 with lemmas on a priori estimates to get the existence of solutions to the problem (2) -(4) when f does not fulfil (24). Here we will use the following lemma from the paper [3]. The existence result is contained in Theorem 8.
Lemma 7 Assume that r > 0 and that Then there exists r * > 0 such that for each function u ∈ AC 1 D satisfying u ∞ ≤ r and holds.
Proof. It is formally the same as the proof of Theorem 3.1 in [3]. We use Proposition 6 instead of Proposition 3.2 in [3]. 2 Now consider an operator F : C 1 D → C 1 D given by the formula The main result of this section is the computation of the Leray-Schauder topological degree of the operator I − F on a certain set Ω which is described by means of lower and upper functions σ 1 , σ 2 . The degree will be denoted by "deg". The degree computation will be used in the next section.

It follows from
Step 3 of the proof of Proposition 6 that each fixed point u ofF satisfies (23) and consequently, Then This completes the proof. 2

Non-ordered lower and upper functions
We consider the following assumptions: σ 1 , σ 2 are lower and upper functions of the problem (2) - (4), for i = 1, . . . , m, g 1 (x, y) is strictly decreasing in x, strictly increasing in y, g 2 (x, y) is strictly increasing in x, strictly decreasing in y, Remark 10 The assumptions (51) allow us to write (4) in the form where h j : (a j , b j ) → R is increasing, −∞ ≤ a j < b j ≤ ∞ for j = 1, 2. In this case, conditions (7) can be replaced by Definition 11 We define an operator K : For the sake of simplicity of notation we will writeÑ (x; A, q) = K(N, A, q)(x) for each x ∈ R, N ∈ C(R), A > 0, q ≥ 0.
Lemma 12 Let N ∈ C(R). Then the condition thus, the constantL does not depend on A, but it does on q.
Proof. Let (55) be valid and A > q ≥ 0. First we see that In general we can putL = K + L + q + 1. 2 Then Proof. Let us assume a > 0, b > a + q and denote N * = sup{|N (x)| : |x| < a}. Then sup{|Ñ (x; A, q)| : |x| < a} ≤ max{a + q, N * } < b. Then there exists d > ρ 1 such that the estimate is valid for every b > 0 and every u ∈ AC 1 D satisfying the conditions whereM i (y; b, 0) is defined in the sense of Definition 11 for i = 1, . . . , m.
Due to Lemma 12,M i (y; b, 0) has the same property as M i independently of b.
Thus, there exists c j−1 = c j−1 (a j + 1) > 0 such that We proceed till j = 1. If ξ u = 0, we can proceed in the estimation in the same way as in Case A. We put d = max{a j : j = 0, . . . , m} + 1. is valid for every a > q and every u ∈ C 1 D satisfying conditions (58), whereJ i (x; a, q) is defined in the sense of Definition 11 for i = 1, . . . , m.
For arbitrary a, b > 0, we consider the conditions Then each function u ∈ B satisfies Proof. Part 1. Let u ∈ B satisfy (73). We consider three cases.
Suppose, on the contrary, that (79) does not hold.
Lemma 20 Each u ∈ B satisfies the condition for some τ u ∈ [0, T ). Now, we are ready to prove the main result of this paper concerning the case of non-ordered lower and upper functions which is contained in the next theorem.
Step 3. We define G by (31), the operator F by and its domain where ( is defined in (25)). It is clear that u is a solution to the problem (89) -(91) if and only if F u = u.
Now, we will prove the implication u ∈ cl(Ω) =⇒ ( u ∞ < c and u ∞ < d) It means u ∈ B (with constants c, d instead of a, b, respectively) and Lemma 19 implies that there exists ξ u ∈ [0, T ] such that |u (ξ u )| < ρ 1 . We apply Lemma 14 and get u ∞ < d. From Lemma 20 and Lemma 15 we get (102).
Analogously we seek an upper function of this problem (for example σ 2 = −σ 1 for suitable constants). We can see the construction of well-ordered upper and lower functions is too difficult (or even impossible).