Electronic Journal of Qualitative Theory of Differential Equations 2005, No. 10, 1-22; http://www.math.u-szeged.hu/ejqtde/

Nonlinear boundary value problem for nonlinear second order differential

equations with impulses

Jan Tomeček

Abstract

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(t_{j}+) = J_{j}(u(t_{j})), u′(t_{j}+) = M_{j}(u′(t_{j})), j = 1, …, m,
g_{1}(u(0), u(T)) = 0, g_{2}(u′(0), u′(T)) = 0,

where f ∈ Car([0, T] × ℝ^{2}), g_{1}, g_{2} ∈ C(ℝ^{2}), J_{j}, M_{j} ∈ 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.

Key words: Ordinary differential equation of the second order, well–ordered lower and upper functions, non–ordered lower and upper functions, nonlinear boundary value conditions, impulses.

1 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.

Let T be a positive real number. For a real valued function u defined a. e. on [0, T], we put

∥u∥_{∞} = sup ess_{t∈[0,T]} |u(t)| and ∥u∥_{1} = ∫_{0}^{T} |u(s)| ds.

For k ∈ ℕ and a given set B ⊂ ℝ^{k}, let C(B) denote the set of real valued functions which are continuous on B. Furthermore, let C^{1}([0, T]) be the set of functions having continuous first derivative on [0, T] and L([0, T]) the set of functions which are Lebesgue integrable on [0, T]. Let m ∈ ℕ and let

0 = t_{0} < t_{1} < … < t_{m} < t_{m+1} = T

be a division of the interval [0, T]. We denote D = {t_{1}, …, t_{m}} and define C_{D} (C_{D}^{1}) as the set of functions u : [0, 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_{D}^{1}) stands for the set of functions u ∈ C_{D} (or u ∈ C_{D}^{1}) which are absolutely continuous (or have absolutely continuous first derivatives) on each subinterval (t_{i}, t_{i+1}), i = 0, …, m. For u ∈ C_{D}^{1} and i = 1, …, m + 1 we write

u′(t_{i}) = u′(t_{i}−) = lim_{t→t_{i}−} u′(t), u′(0+) = lim_{t→0+} u′(t)  (1)

and

∥u∥_{D} = ∥u∥_{∞} + ∥u′∥_{∞}.

Note that the set C_{D}^{1} 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: (i) for each x ∈ S the function f(·, x) is measurable on [0, T]; (ii) for almost each t ∈ [0, T] the function f(t, ·) is continuous on S; (iii) for each compact set K ⊂ S there exists a function m_{K}(t) ∈ L([0, T]) such that |f(t, x)| ≤ m_{K}(t) for a. e. t ∈ [0, T] and all x ∈ K. 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:

u′′(t) = f(t, u(t), u′(t)),  (2)
u(t_{i}+) = J_{i}(u(t_{i})), u′(t_{i}+) = M_{i}(u′(t_{i})), i = 1, …, m,  (3)
g_{1}(u(0), u(T)) = 0, g_{2}(u′(0), u′(T)) = 0,  (4)

where f ∈ Car([0, T] × ℝ^{2}), g_{1}, g_{2} ∈ C(ℝ^{2}), J_{i}, M_{i} ∈ C(ℝ) and u′(t_{i}) are understood in the sense of (1) for i = 1, …, m.

Definition 1 A function u ∈ AC_{D}^{1} which satisfies equation (2) for a. e. t ∈ [0, T] and fulfils conditions (3) and (4) is called a solution to the problem (2) – (4).

Definition 2 A function σ_{k} ∈ AC_{D}^{1} is called a lower (upper) function of the problem (2) – (4) provided the conditions

[σ′′_{k}(t) − f(t, σ_{k}(t), σ′_{k}(t))](−1)^{k} ≤ 0 for a. e. t ∈ [0, T],  (5)
σ_{k}(t_{i}+) = J_{i}(σ_{k}(t_{i})), [σ′_{k}(t_{i}+) − M_{i}(σ′_{k}(t_{i}))](−1)^{k} ≤ 0, i = 1, …, m, (6)
g_{1}(σ_{k}(0), σ_{k}(T)) = 0, g_{2}(σ′_{k}(0), σ′_{k}(T))(−1)^{k} ≤ 0,  (7)

where k = 1 (k = 2), are satisfied.

2 Well–ordered lower and upper functions

Throughout this section we assume:

\begin{matrix}σ_{1} and σ_{2} are lower and upper functions, respectively, \\ of the problem (2) – (4) and σ_{1}(t) ≤ σ_{2}(t) for t ∈ [0, T], \end{matrix} \}  (8)
σ_{1}(t_{i}) ≤ x ≤ σ_{2}(t_{i}) ⟹ J_{i}(σ_{1}(t_{i})) ≤ J_{i}(x) ≤ J_{i}(σ_{2}(t_{i})),  (9)
\begin{matrix}y ≤ σ′_{1}(t_{i}) ⟹ M_{i}(y) ≤ M_{i}(σ′_{1}(t_{i})), \\ y ≥ σ′_{2}(t_{i}) ⟹ M_{i}(y) ≥ M_{i}(σ′_{2}(t_{i})), \end{matrix} \}  (10)

for i = 1, …, m,

\begin{matrix}x > σ_{1}(0) & ⟹ & g_{1}(σ_{1}(0), σ_{1}(T)) ≠ g_{1}(x, σ_{1}(T)), \\ x < σ_{2}(0) & ⟹ & g_{1}(σ_{2}(0), σ_{2}(T)) ≠ g_{1}(x, σ_{2}(T)), \end{matrix} \}  (11)

\begin{matrix}σ_{1}(T) ≤ y & ⟹ & g_{1}(σ_{1}(0), σ_{1}(T)) ≤ g_{1}(σ_{1}(0), y), \\ σ_{2}(T) ≥ y & ⟹ & g_{1}(σ_{2}(0), σ_{2}(T)) ≥ g_{1}(σ_{2}(0), y), \end{matrix} \}  (12)
\begin{matrix}x ≥ σ′_{1}(0) and y ≤ σ′_{1}(T) & ⟹ & g_{2}(σ′_{1}(0), σ′_{1}(T)) ≤ g_{2}(x, y), \\ x ≤ σ′_{2}(0) and y ≥ σ′_{2}(T) & ⟹ & g_{2}(σ′_{2}(0), σ′_{2}(T)) ≥ g_{2}(x, y). \end{matrix} \} (13)

Remark 3 If we put

g_{1}(x, y) = y − x, g_{2}(x, y) = x − y  (14)

for x, y ∈ ℝ, then (4) reduces to the periodic conditions

u(0) = u(T), u′(0) = u′(T).  (15)

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 consider functions f̃ ∈ Car([0, T] × ℝ^{2}), J̃_{i}, M̃_{i} ∈ C(ℝ) for i = 1, …, m having the following properties:

\begin{matrix}f̃(t, x, y) < f(t, σ_{1}(t), σ′_{1}(t)) for a. e. t ∈ [0, T], \\ x < σ_{1}(t), |y − σ′_{1}(t)| ≤ \frac{σ_{1}(t)−x}{σ_{1}(t)−x+1}, \\ f̃(t, x, y) > f(t, σ_{2}(t), σ′_{2}(t)) for a. e. t ∈ [0, T], \\ x > σ_{2}(t), |y − σ′_{2}(t)| ≤ \frac{x−σ_{2}(t)}{x−σ_{2}(t)+1}, \end{matrix} \}  (16)

\begin{matrix}x < σ_{1}(t_{i}) & ⟹ & J̃_{i}(x) < J_{i}(σ_{1}(t_{i})), \\ σ_{1}(t_{i}) ≤ x ≤ σ_{2}(t_{i}) & ⟹ & J̃_{i}(x) = J_{i}(x), \\ x > σ_{2}(t_{i}) & ⟹ & J̃_{i}(x) > J_{i}(σ_{2}(t_{i})), \end{matrix} \}  (17)

\begin{matrix}y ≤ σ′_{1}(t_{i}) & ⟹ & M̃_{i}(y) ≤ M_{i}(σ′_{1}(t_{i})), \\ y ≥ σ′_{2}(t_{i}) & ⟹ & M̃_{i}(y) ≥ M_{i}(σ′_{2}(t_{i})). \end{matrix} \}  (18)

Next, we consider d_{0}, d_{T} ∈ ℝ such that

σ_{1}(0) ≤ d_{0} ≤ σ_{2}(0),  σ_{1}(T) ≤ d_{T} ≤ σ_{2}(T).  (19)

We define an auxiliary impulsive boundary value problem

u′′(t) = f̃(t, u(t), u′(t)),  (20)
u(t_{i}+) = J̃_{i}(u(t_{i})), u′(t_{i}+) = M̃_{i}(u′(t_{i})), i = 1, …, m,  (21)
u(0) = d_{0}, u(T) = d_{T} ,  (22)

where f ∈ Car(J × ℝ^{2}), J̃_{i}, M̃_{i} ∈ C(ℝ), i = 1, …, m, d_{0}, d_{T} ∈ ℝ.

Definition 4 A function u ∈ AC_{D}^{1} which satisfies equation (20) for a. e. t ∈ [0, T] and fulfils conditions (21) and (22) is called a solution to the problem (20) – (22).

Lemma 5 Let (8) – (10), (16) – (19) be true. Then a solution u to the problem (20) – (22) satisfies the inequalities

σ_{1} ≤ u ≤ σ_{2}  on [0, T].  (23)

Proof. Let u be a solution to the problem (20) – (22). Put v(t) = u(t) − σ_{2}(t) for t ∈ [0, T]. Then, by (22), we have v(0) ≤ 0 and v(T) ≤ 0. The rest of the proof is exactly the same as the proof of Lemma 2.1 in [3]. □

Proposition 6 Let (8) – (13) be true and let there exist h ∈ L([0, T]) such that

|f(t, x, y)| ≤ h(t)  for a. e. t ∈ [0, T] and all (x, y) ∈ [σ_{1}(t), σ_{2}(t)]×ℝ. (24)

Then there exists a solution u to the problem (2) – (4) satisfying (23).

Proof. Step 1. We define

△ = min_{i=0,...,m}(t_{i+1} − t_{i}),  (25)
c = ∥h∥_{1} + \frac{∥σ_{1}∥_{∞} + ∥σ_{2}∥_{∞}}{△} + ∥σ′_{1}∥_{∞} + ∥σ′_{2}∥_{∞} + 1,  (26)

α(t, x) = \{ \begin{matrix}σ_{1}(t) & for & x < σ_{1}(t), \\ x & for & σ_{1}(t) ≤ x ≤ σ_{2}(t), \\ σ_{2}(t) & for & σ_{2}(t) < x, \end{matrix}  (27)

for all t ∈ [0, T], x ∈ ℝ,

β(y) = \{ \begin{matrix}y & for & |y| ≤ c, \\ c sgn y & for & |y| > c, \end{matrix}

ω_{i}(t, ϵ) = sup{|f(t, σ_{i}(t), σ′_{i}(t)) − f(t, σ_{i}(t), y)| : |σ′_{i}(t) − y| ≤ ϵ}

for a. e. t ∈ [0, T] and for ϵ ∈ [0, 1], i = 1, 2,

\begin{matrix}J̃_{i}(x) = x + J_{i}(α(t_{i}, x)) − α(t_{i}, x), \\ M̃_{i}(y) = y + M_{i}(β(y)) − β(y), \end{matrix} \}  (28)

and

f̃(t, x, y) = \{ \begin{matrix}f(t, σ_{1}(t), y) − ω_{1}(t, \frac{σ_{1}(t)−x}{σ_{1}(t)−x+1})−\frac{σ_{1}(t)−x}{σ_{1}(t)−x+1} & for & x < σ_{1}(t), \\ f(t, x, y) & for & σ_{1}(t) ≤ x ≤ σ_{2}(t), \\ f(t, σ_{2}(t), y) + ω_{2}(t, \frac{x−σ_{2}(t)}{x−σ_{2}(t)+1})+\frac{x−σ_{2}(t)}{x−σ_{2}(t)+1} & for & σ_{2}(t) < x \end{matrix}

(29) for a. e. t ∈ [0, T] and all x, y ∈ ℝ. It follows from [2], Lemma 4 that f̃ ∈ Car([0, T] × ℝ^{2}). We consider the problem (20), (21) and

\begin{matrix}u(0) & = & α(0, u(0) + g_{1}(u(0), u(T))), \\ u(T) & = & α(T, u(T) + g_{2}(u′(0), u′(T))). \end{matrix} \}  (30)

Step 2. We will prove solvability of the problem (20), (21), (30). We define a function G : [0, T] × [0, T] → ℝ by

G(t, s) = \{ \begin{matrix}\frac{s(t−T)}{T} & for & 0 ≤ s < t ≤ T, \\ \frac{t(s−T)}{T} & for & 0 ≤ t ≤ s ≤ T, \end{matrix}  (31)

and a totally continuous operator F̃ : C_{D}^{1} → C_{D}^{1} by

(F̃u)(t) = \frac{T − t}{T}α(0, u(0) + g_{1}(u(0), u(T))) + \frac{t}{T} α(T, u(T) + g_{2}(u′(0), u′(T)))

+ ∫_{0}^{T} G(t, s)f̃(s, u(s), u′(s)) ds − ∑_{i=1}^{m} \frac{∂G}{∂s}(t, t_{i})(J̃_{i}(u(t_{i})) − u(t_{i}))
+ ∑_{i=1}^{m} G(t, t_{i})(M̃_{i}(u′(t_{i})) − u′(t_{i})),  (32)

where \frac{∂G}{∂s}(t, s) is continuous on [0, T] × [t, T]. Obviously, u is a solution to the problem (20), (21), (30) if and only if u is a fixed point of the operator F̃.

We consider the family of equations

(I − λF̃)u = 0,  λ ∈ [0, 1].  (33)

For R > 0 we define B(R) = {u ∈ C_{D}^{1} : ∥u∥_{D} < R}. Relations (24), (27), (28), (29) imply that there exists R_{0} > 0 such that u ∈ B(R_{0}) for every solution u to the problem (33) and each λ ∈ [0, 1]. Thus, I − λF̃ is a homotopy on cl(B(R)) × [0, 1] for R ≥ R_{0}, λ ∈ [0, 1] and

deg(I − λ̃F̃, B(R)) = deg(I − λF̃, B(R))

for λ, λ̃ ∈ [0, 1]. Since deg(I, B(R)) = 1, we conclude that

deg(I − F̃, B(R)) = 1 for R ≥ R_{0}.  (34)

Thus there exists a fixed point of F̃ in B(R) and the problem (20), (21), (30) is solvable. Step 3. Let u be a solution to (20), (21), (30). The definitions of the functions f̃, J̃_{i}, M̃_{i} for i = 1, …, m and (10) imply that (16), (17), (18) are valid. We put d_{0} = α(0, u(0)+g_{1}(u(0), u(T))) and d_{T} = α(T, u(T)+g_{2}(u′(0), u′(T))). Obviously,

(19) 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 ξ_{i} ∈ (t_{i}, t_{i+1}) such that

|u′(ξ_{i})| < \frac{1}{△}(∥σ_{1}∥_{∞} + ∥σ_{2}∥_{∞}).

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

σ_{1}(0) ≤ u(0) + g_{1}(u(0), u(T)) ≤ σ_{2}(0)  (35)

and

σ_{1}(T) ≤ u(T) + g_{2}(u′(0), u′(T)) ≤ σ_{2}(T).  (36)

Let us suppose that the first inequality in (35) is not true. Then

σ_{1}(0) > u(0) + g_{1}(u(0), u(T)).

In view of (27) and (30) we have u(0) = σ_{1}(0), thus it follows from (12) and (23) that

0 > g_{1}(σ_{1}(0), u(T)) ≥ g_{1}(σ_{1}(0), σ_{1}(T)),

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

σ_{1}(T) > u(T) + g_{2}(u′(0), u′(T)).  (37)

It follows from (27) and (30) that

u(T) = σ_{1}(T)  (38)

and by (37) we obtain that 0 > g_{2}(u′(0), u′(T)). Further, by virtue of (7), (30),

(35) and (38), we have

g_{1}(σ_{1}(0), σ_{1}(T)) = 0 = g_{1}(u(0), u(T)) = g_{1}(u(0), σ_{1}(T)).

In view of (23) and (11) we get

u(0) = σ_{1}(0).  (39)

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

0 > g_{2}(u′(0), u′(T)) ≥ g_{2}(σ′_{1}(0), σ′_{1}(T)),

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). □

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

k ∈ L([0, T]) is nonnegative a. e. on [0, T],  (40)
ω ∈ C([1, ∞)) is positive on [1, ∞) and ∫_{1}^{∞} \frac{ds}{ω(s)} = ∞.  (41)

Then there exists r^{∗} > 0 such that for each function u ∈ AC_{D}^{1} satisfying ∥u∥_{∞} ≤ r and

|u′′(t)| ≤ ω(|u′(t)|)(|u′(t)| + k(t))  (42)

for a. e. t ∈ [0, T] and for |u′(t)| > 1, the estimate

∥u′∥ ≤ r^{∗}  (43)

holds.

Theorem 8 Assume that (8) – (13) hold. Further, let

|f(t, x, y)| ≤ ω(|y|)(|y| + k(t))  (44)

for a. e. t ∈ [0, T] and for each x ∈ [σ_{1}(t), σ_{2}(t)], |y| > 1, where k and ω fulfil (40) and (41). Then the problem (2) – (4) has a solution u satisfying (23).

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]. □

Now consider an operator F : C_{D}^{1} → C_{D}^{1} given by the formula

(F u)(t) = \frac{T − t}{T}(u(0) + g_{1}(u(0), u(T))) + \frac{t}{T} (u(T) + g_{2}(u′(0), u′(T)))

+ ∫_{0}^{T} G(t, s)f(s, u(s), u′(s)) ds − ∑_{i=1}^{m} \frac{∂G}{∂s}(t, t_{i})(J_{i}(u(t_{i})) − u(t_{i}))

+ ∑_{i=1}^{m} G(t, t_{i})(M_{i}(u′(t_{i})) − u′(t_{i})).  (45)

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.

Theorem 9 Let f ∈ Car([0, T] × ℝ^{2}), g_{1}, g_{2} ∈ C(ℝ^{2}), J_{i}, M_{i} ∈ C(ℝ) for i = 1, …, m and let σ_{1}, σ_{2} be lower and upper functions of the problem (2) – (4) such that

σ_{1} < σ_{2} on [0, T] and σ_{1}(t_{i}+) < σ_{2}(t_{i}+) for i = 1, …, m,

σ_{1}(t_{i}) < x < σ_{2}(t_{i}) ⟹ J_{i}(σ_{1}(t_{i})) < J_{i}(x) < J_{i}(σ_{2}(t_{i})),

\begin{aligned}y ≤ σ′_{1}(t_{i}) & ⟹ & M_{i}(y) ≤ M_{i}(σ′_{1}(t_{i})), \\ y ≥ σ′_{2}(t_{i}) & ⟹ & M_{i}(y) ≥ M_{i}(σ′_{2}(t_{i})) \end{aligned}

for i = 1, …, m. Let (11) – (13) be valid and let h ∈ L([0, T]) be such that

|f(t, x, y)| ≤ h(t) for a. e. t ∈ [0, T] and all (x, y) ∈ [σ_{1}(t), σ_{2}(t)] × ℝ. We define the (totally continuous) operator F by (45) and c by (26) and denote

Ω = {u ∈ C_{D}^{1} : ∥u′∥ < c, σ_{1}(t) < u(t) < σ_{2}(t) for t ∈ [0, T],

σ_{1}(t_{i}+) < u(t_{i}+) < σ_{2}(t_{i}+) for i = 1, …, m}.  (46)

Then deg(I − F, Ω) = 1 whenever F u ≠ u on ∂Ω. Proof. We consider J̃_{i}, M̃_{i}, i = 1, …, m defined by (28) and f̃ by (29). Define F̃ by (32). We can see (use Step 4 from the proof of Proposition 6) that

F u = u if and only if F̃u = u on cl(Ω).  (47)

We suppose that F u ≠ u for each u ∈ ∂Ω. Then

F̃u ≠ u on ∂Ω.

It follows from Step 3 of the proof of Proposition 6 that each fixed point u of F̃ satisfies (23) and consequently,

|u′′(t)| = |f(t, u(t), u′(t))| ≤ h(t).

Then

∥u′∥_{∞} ≤ ∥h∥_{1} + \frac{∥σ_{1}∥_{∞} + ∥σ_{2}∥_{∞}}{△} < c.

It means that

u = F̃u ⟹ u ∈ Ω.

Now we choose R > 0 in (34) such that Ω ⊂ B(R). Let Ω_{1} = {u ∈ Ω : (F u)(0) ∈ [σ_{1}(0), σ_{2}(0)]}. If u ∈ Ω is a fixed point of F, then u ∈ Ω_{1}. Hence F and (by (47)) F̃ have no fixed points in cl(Ω) \ Ω_{1}. Moreover,

F = F̃ on cl(Ω_{1}).

Therefore, by the excision property,

deg(I − F, Ω) = deg(I − F, Ω_{1}) = deg(I − F̃, Ω_{1}) = deg(I − F̃, B(R)) = 1. This completes the proof. □

3 Non–ordered lower and upper functions

We consider the following assumptions:

\begin{matrix}σ_{1}, σ_{2} are lower and upper functions of the problem (2) – (4), \\ there exists τ ∈ [0, T] such that σ_{1}(τ) > σ_{2}(τ), \end{matrix} \}  (48)

\begin{matrix}x > σ_{1}(t_{i}) ⟹ J_{i}(x) > J_{i}(σ_{1}(t_{i})), \\ x < σ_{2}(t_{i}) ⟹ J_{i}(x) < J_{i}(σ_{2}(t_{i})), \end{matrix} \}  (49)
\begin{matrix}y ≤ σ′_{1}(t_{i}) ⟹ M_{i}(y) ≤ M_{i}(σ′_{1}(t_{i})), \\ y ≥ σ′_{2}(t_{i}) ⟹ M_{i}(y) ≥ M_{i}(σ′_{2}(t_{i})) \end{matrix} \}  (50)

for i = 1, …, m,

\begin{matrix}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, \end{matrix} \} (51)
lim_{x→±∞} |J_{i}(x)| = ∞,  i = 1, …, m,  (52)
lim_{y→±∞} |M_{i}(y)| = ∞,  i = 1, …, m.  (53)

Remark 10 The assumptions (51) allow us to write (4) in the form

u(T) = h_{1}(u(0)), u′(T) = h_{2}(u′(0)),

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

σ_{k}(T) = h_{1}(σ_{k}(0)), [h_{2}(σ′_{k}(0)) − σ′_{k}(T)](−1)^{k} ≤ 0.

Definition 11 We define an operator K : C(ℝ) × ℝ^{+} × ℝ_{0}^{+} → C(ℝ) by

K(N, A, q)(x) = \{ \begin{matrix}x + q & if & x ≤ −A − 1, \\ N(−A)(A + 1 + x) − (x + q)(x + A) & if & −A − 1 < x < −A, \\ N(x) & if & −A ≤ x ≤ A, \\ N(A)(A + 1 − x) + (x − q)(x − A) & if & A < x < A + 1, \\ x − q & if & x ≥ A + 1 \end{matrix}

(54)

for each N ∈ C(ℝ), A > 0, q ≥ 0. For the sake of simplicity of notation we will write Ñ(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

∀K > 0 ∃L > 0 ∀x ∈ ℝ : |N(x)| < K ⟹ |x| < L  (55)

implies

∀q ≥ 0 ∀K > 0 ∃L̄ > 0 ∀x ∈ ℝ ∀A > q : |Ñ(x; A, q)| < K ⟹ |x| < L̄; (56)

thus, the constant L̄ does not depend on A, but it does on q.

Proof. Let (55) be valid and A > q ≥ 0. First we see that

min{N(A), A − q} ≤ Ñ(x; A, q) ≤ max{N(A), A − q + 1}  (57)

for x ∈ [A, A + 1] and an analogous assertion is valid for x ∈ [−A − 1, −A]. Let K > 0 be arbitrary. We distinguish several cases. If K ≥ A−q+1, then we can set L̄ = K+q. If A−q ≤ K < A−q+1, then L̄ = K+q+1. Let us consider the case for which K < A−q. If min{|N(A)|, |N(−A)|} ≥ K, then in view of (57) and (55) we take L̄ = L, where L is the constant from (55). If max{|N(A)|, |N(−A)|} ≤ K, min{|N(A)|, |N(−A)|} ≤ K or max{|N(A)|, |N(−A)|} ≥ K, we take L̄ = L + 1. In general we can put L̄ = K + L + q + 1. □

Lemma 13 Let N ∈ C(ℝ), A ≥ q > 0. Then

sup{|N(x)| : |x| < a} < b ⟹ sup{|Ñ(x; A, q)| : |x| < a} < b

for a > 0, b > a + q.

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.
□

Lemma 14 Let ρ_{1} > 0, h̄ ∈ L([0, T]), M_{i} ∈ C(ℝ), i = 1, …, m satisfy (53). Then there exists d > ρ_{1} such that the estimate

∥u′∥_{∞} < d  (58)

is valid for every b > 0 and every u ∈ AC_{D}^{1} satisfying the conditions

|u′(ξ_{u})| < ρ_{1}  for some ξ_{u} ∈ [0, T],  (59)
u′(t_{i}+) = M̃_{i}(u′(t_{i}); b, 0),  i = 1, …, m,  (60)
|u′′(t)| < h̄(t)  for a. e. t ∈ [0, T],  (61)

where M̃_{i}(y; b, 0) is defined in the sense of Definition 11 for i = 1, …, m.

Proof. Let us denote

b_{i}(a) = sup_{|y|<a} |M̃_{i}(y; b, 0)| < ∞

for all a > 0, i = 1, …, m. Let ξ_{u} ∈ (t_{j}, t_{j+1}] for some j ∈ {0, …, m}. Then in view of (59), (61) we get

|u′(t)| < ρ_{1} + ∥h̄∥_{1} = a_{j} for t ∈ (t_{j}, t_{j+1}].  (62)

Case A. If j < m, then due to (60) the inequalities |u′(t_{j+1}+)| < b_{j+1}(a_{j}) and |u′(t)| < b_{j+1}(a_{j}) + ∥h̄∥_{1} = a_{j+1} are valid for t ∈ (t_{j+1}, t_{j+2}]. We can proceed in this way till j = m. We get

|u′(t)| < max{a_{j}, …, a_{m}} + 1 for t ∈ (t_{j}, T].

Case B. If j > 0, then we will establish an estimate of |u′| on [0, t_{j}]. If follows from (62) and (60) that

|M̃_{j}(u′(t_{j}); b, 0)| < a_{j} + 1.

The assumption (53) implies that for every K > 0 and for every i = 1, …, m there exists L > 0 such that for y ∈ ℝ

|M_{i}(y)| < K ⟹ |y| < L.

Due to Lemma 12, M̃_{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

|u′(t_{j})| < c_{j−1}

and c_{j−1} is independent of b. Then |u′(t)| < c_{j−1} + ∥h̄∥_{1} = a_{j−1} for t ∈ (t_{j−1}, t_{j}]. 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. □

Lemma 15 Let ρ_{0}, d, q > 0 and J_{i} ∈ C(ℝ), i = 1, …, m satisfy (52). Then there exists c > ρ_{0} + q such that the estimate

∥u∥_{∞} < c  (63)

is valid for every a > q and every u ∈ C_{D}^{1} satisfying conditions (58),

|u(τ_{u})| < ρ_{0}  for some τ_{u} ∈ [0, T],  (64)
u(t_{i}+) = J̃_{i}(u(t_{i}); a, q),  i = 1, …, m,  (65)

where J̃_{i}(x; a, q) is defined in the sense of Definition 11 for i = 1, …, m.

Proof. We argue in the same way as in the proof of Lemma 14. □

Lemma 16 Let J_{i}, M_{i} ∈ C(ℝ), i = 1, …, m, and (49), (50) be valid, let q > 0, σ_{1}, σ_{2} ∈ AC_{D}^{1} satisfy the equalities in (6), let a, b ∈ ℝ be such that

a > ∥σ_{1}∥_{∞} + ∥σ_{2}∥_{∞} + q + 1  and  b > ∥σ′_{1}∥_{∞} + ∥σ′_{2}∥_{∞} + ρ̄ + 1, (66)

where

ρ̄ = ∑_{i=1}^{m} (|M_{i}(σ′_{1}(t_{i}))| + |M_{i}(σ′_{2}(t_{i}))|) ,  (67)

and let J̃_{i}(x; a, q), M̃_{i}(y; b, 0) be defined in the sense of Definition 11. Then the implications

\{ \begin{matrix}x > σ_{1}(t_{i}) & ⟹ & J̃_{i}(x; a, q) > J̃_{i}(σ_{1}(t_{i}); a, q) = J_{i}(σ_{1}(t_{i})), \\ x < σ_{2}(t_{i}) & ⟹ & J̃_{i}(x; a, q) < J̃_{i}(σ_{2}(t_{i}); a, q) = J_{i}(σ_{2}(t_{i})), \end{matrix}  (68)

\{ \begin{matrix}y ≤ σ′_{1}(t_{i}) & ⟹ & M̃_{i}(y; b, 0) ≤ M_{i}(σ′_{1}(t_{i})), \\ y ≥ σ′_{2}(t_{i}) & ⟹ & M̃_{i}(y; b, 0) ≥ M_{i}(σ′_{2}(t_{i})), \end{matrix}  (69)

are valid for i = 1, …, m.

Proof. Obviously, (68) is valid for |x| ≤ a. Let x > a. Then

x > max{|σ_{1}(t_{i})|, |σ_{2}(t_{i})|}

and it is sufficient to prove the first implication in (68). The fact that |σ_{1}(t_{i})| < a implies J̃_{i}(σ_{1}(t_{i}); a, q) = J_{i}(σ_{1}(t_{i})). From the first inequality in (66) we get x − q > a − q > ∥σ_{1}∥_{∞} and thus (49) and (6) yield

J̃_{i}(x; a, q) = J_{i}(a)(a + 1 − x) + (x − q)(x − a) >
J_{i}(σ_{1}(t_{i}))(a + 1 − x) + J_{i}(σ_{1}(t_{i}))(x − a) = J_{i}(σ_{1}(t_{i}))

for x ∈ (a, a + 1). If x ≥ a + 1, then

J̃_{i}(x; a, q) = x − q > σ_{1}(t_{i}+) = J_{i}(σ_{1}(t_{i})).

Similarly, if x < −a, it is sufficient to prove the second implication in (68). The implications in (69) are obvious for |y| ≤ b. Otherwise, due to (66) and (67) we get

M̃_{i}(y; b, 0) = M_{i}(b)(b + 1 − y) + y(y − b) > M_{i}(σ′_{2}(t_{i}))

for y ∈ (b, b + 1) and

M̃_{i}(y; b, 0) = y > M_{i}(σ′_{2}(t_{i}))

for y ≥ b + 1. Analogously, (69) is valid for y < −b. □

Definition 17 We define an operator L : C(ℝ^{2}) × ℝ^{+} × (ℝ \ {0}) → C(ℝ^{2}) by

L(g, A, K)(x, y) = \{ \begin{matrix}g(x, y) & if & (x, y) ∈ [−A, A]^{2}, \\ K(y − x) & if & (x, y) ∈ ℝ^{2} \ (−A − 1, A + 1)^{2} \end{matrix}

and so that L(g, A, K)(x, y) is a linear function of the variable y on rectangles [−A, A] × [A, A + 1] and [−A, A] × [−A − 1, −A], a linear function of the variable x on [−A − 1, −A] × [−A, A] and [A, A + 1] × [−A, A] and a bilinear function on squares [A, A+1]×[A, A+1], [−A−1, −A]×[A, A+1], [−A−1, −A]×[−A−1, −A] and [A, A + 1] × [−A − 1, −A] for each g ∈ C(ℝ^{2}), A > 0 and K ∈ ℝ \ {0}.

Let us consider functions g_{1}, g_{2} ∈ C(ℝ^{2}) satisfying (51) and a, b > 0. We put

g̃_{1}(x, y; a) = L(g_{1}, a, max{|g_{1}(x, y)| : (x, y) ∈ [−a, a]^{2}} + 1)(x, y),  (70)
g̃_{2}(x, y; b) = L(g_{2}, b, − max{|g_{2}(x, y)| : (x, y) ∈ [−b, b]^{2}} − 1)(x, y)  (71)

for each (x, y) ∈ ℝ^{2}.

Remark 18 The functions g̃_{1}(x, y; a) and g̃_{2}(x, y; b) defined by (70) and (71) preserve the properties of g_{1} and g_{2}, respectively, in (51).

For arbitrary a, b > 0, we consider the conditions

g̃_{1}(u(0), u(T); a) = 0,  g̃_{2}(u′(0), u′(T); b) = 0  (72)

and for u ∈ C_{D}^{1}

u(s_{u}) < σ_{1}(s_{u}) and u(t_{u}) > σ_{2}(t_{u}) for some s_{u}, t_{u} ∈ [0, T],  (73)
u ≥ σ_{1} on [0, T] and inf_{t∈[0,T]} |u(t) − σ_{1}(t)| = 0,  (74)
u ≤ σ_{2} on [0, T] and inf_{t∈[0,T]} |u(t) − σ_{2}(t)| = 0.  (75)

Lemma 19 Let σ_{1}, σ_{2} ∈ AC_{D}^{1} be lower and upper functions of the problem (2) – (4), let J_{i}, M_{i} ∈ C(ℝ) satisfy (49), (50), let g_{1}, g_{2} ∈ C(ℝ^{2}) satisfy (51), let q > 0, a, b ∈ ℝ satisfy (66), where ρ̄ is defined in (67). We define J̃_{i}(x; a, q) and M̃_{i}(y; b, 0) in the sense of Definition 11, g̃_{1}(x, y; a), g̃_{2}(x, y; b) by (70), (71), respectively and

B = {u ∈ C_{D}^{1} : u satisfies (72), (60), (65)  (76)

and one of the conditions (73), (74), (75) }.

Then each function u ∈ B satisfies

\{ \begin{matrix}|u′(ξ_{u})| < ρ_{1} for some ξ_{u} ∈ [0, T], where \\ ρ_{1} = \frac{2}{t_{1}} (∥σ_{1}∥_{∞} + ∥σ_{2}∥_{∞})+∥σ′_{1}∥_{∞} + ∥σ′_{2}∥_{∞} + 1. \end{matrix}  (77)

Proof. Part 1. Let u ∈ B satisfy (73). We consider three cases. Case A. If min{σ_{1}(t), σ_{2}(t)} ≤ u(t) ≤ max{σ_{1}(t), σ_{2}(t)} for each t ∈ [0, T], then it follows from the Mean Value Theorem that there exists ξ_{u} ∈ (0, t_{1}) such that

|u′(ξ_{u})| ≤ \frac{2}{t_{1}} (∥σ_{1}∥_{∞} + ∥σ_{2}∥_{∞}).

Case B. Assume that u(s) > σ_{1}(s) for some s ∈ [0, T]. We denote v = u − σ_{1} on [0, T]. According to (73) we have

inf_{t∈[0,T]} v(t) < 0 and sup_{t∈[0,T]} v(t) > 0.  (78)

We will prove that

v′(α) = 0 for some α ∈ [0, T] or v′(τ+) = 0 for some τ ∈ D.  (79)

Suppose, on the contrary, that (79) does not hold. Let v′(0) > 0. In view of (66), (7), (72) and Remark 18 we have

g̃_{2}(σ′_{1}(0), σ′_{1}(T); b) = g_{2}(σ′_{1}(0), σ′_{1}(T)) ≥ 0
= g̃_{2}(u′(0), u′(T); b) > g̃_{2}(σ′_{1}(0), u′(T); b).

Thus the monotony of g̃_{2} yields σ′_{1}(T) < u′(T), i. e. v′(T) > 0. Due to the fact that (79) does not hold and by virtue of (60) and (6) we get

0 < v′(t_{m}+) = u′(t_{m}+) − σ′_{1}(t_{m}+) ≤ M̃_{m}(u′(t_{m}); b, 0) − M_{m}(σ′_{1}(t_{m})).

In view of (69) we get v′(t_{m}) > 0. Continuing by induction we have

v′(t) > 0 for each t ∈ [0, T] and v′(τ+) > 0 for each τ ∈ D.  (80)

If v(0) ≥ 0, then due to (80) we have v > 0 on (0, t_{1}]. The first implication in (68) implies u(t_{1}+) > σ_{1}(t_{1}+). We proceed till t = t_{m+1}. We get v ≥ 0 on [0, T], which contradicts (78). If v(0) < 0 then in view of Remark 18, (7), (72) and (66) we get

g̃_{1}(u(0), σ_{1}(T); a) > g̃_{1}(σ_{1}(0), σ_{1}(T); a)
= g_{1}(σ_{1}(0), σ_{1}(T)) = 0 = g̃_{1}(u(0), u(T); a).

Thus v(T) < 0. Due to (80) we have v < 0 on (t_{m}, T] and the relations (65) and (6) imply J̃_{m}(u(t_{m}); a, q) < J_{m}(σ_{1}(t_{m})). Due to (68) we get u(t_{m}) ≤ σ_{1}(t_{m}). We proceed in the same way till t = t_{0}. The inequality v ≤ 0 on [0, T] contradicts (78). Let v′(0) < 0, then v′(t_{1}) < 0. In view of (60), (69) and (6) we have

u′(t_{1}+) = M̃_{1}(u′(t_{1}); b, 0) ≤ M_{1}(σ′_{1}(t_{1})) ≤ σ′_{1}(t_{1}+)

and (79) implies v′(t_{1}+) < 0. We proceed till t = t_{m+1} again and get

v′(t) < 0 for each t ∈ [0, T] and v′(τ+) < 0 for each τ ∈ D.  (81)

We distinguish two cases

v(0) ≥ 0 and v(0) < 0.

In an analogous way we get a contradiction to (78). Assertion (79) implies that there exists ξ_{u} ∈ [0, T] such that

|u′(ξ_{u})| < ∥σ′_{1}∥_{∞} + 1.

Case C. Assume that u(s) < σ_{2}(s) for some s ∈ [0, T]. We can prove that there exists ξ_{u} ∈ [0, T] such that

|u′(ξ_{u})| < ∥σ′_{2}∥_{∞} + 1

by an argument analogous to Case B. Part 2. Assume that u ∈ B satisfies (74). Then u ≥ σ_{1} on [0, T] and either there exists α_{u} ∈ [0, T] such that u(α_{u}) = σ_{1}(α_{u}) or there exists t_{j} ∈ D such that u(t_{j}+) = σ_{1}(t_{j}+). Case A. Assume that α_{u} ∈ (0, T) \ D is such that u(α_{u}) = σ_{1}(α_{u}). Then u′(α_{u}) = σ′_{1}(α_{u}) and (77) is valid. If α_{u} = 0 then u(0) = σ_{1}(0) and

g̃_{1}(u(0), σ_{1}(T); a) = g̃_{1}(σ_{1}(0), σ_{1}(T); a)
= g_{1}(σ_{1}(0), σ_{1}(T)) = 0 = g̃_{1}(u(0), u(T); a)

according to (66), (7) and (72). The monotony of g̃_{1} implies σ_{1}(T) = u(T) and (74) gives

u′(0) ≥ σ′_{1}(0) and u′(T) ≤ σ′_{1}(T).

Due to (72), Remark 18, (66) and (7), we get

0 = g̃_{2}(u′(0), u′(T); b) ≥ g̃_{2}(σ′_{1}(0), σ′_{1}(T); b) = g_{2}(σ′_{1}(0), σ′_{1}(T)) ≥ 0,

thus σ′_{1}(0) = u′(0) and σ′_{1}(T) = u′(T), i. e. ξ_{u} = 0. If α_{u} = t_{j} ∈ D then u(t_{j}) = σ_{1}(t_{j}),

u(t_{j}+) = J̃_{j}(u(t_{j}); a, q) = J̃_{j}(σ_{1}(t_{j}); a, q) = J_{j}(σ_{1}(t_{j})) = σ_{1}(t_{j}+)

and from (74) we get the inequality u′(t_{j}+) ≥ σ′_{1}(t_{j}+) and u′(t_{j}) ≤ σ′_{1}(t_{j}). From (69) we have M̃_{j}(u′(t_{j}); b, 0) ≤ M_{j}(σ′_{1}(t_{j})), i. e. u′(t_{j}+) ≤ σ′_{1}(t_{j}+). We get σ′_{1}(t_{j}+) = u′(t_{j}+). The estimate (77) is valid for ξ_{u} sufficiently close to t_{j}. Case B. If the second possibility occurs, i. e. u(t_{j}+) = σ_{1}(t_{j}+) for some t_{j} ∈ D, then J̃_{j}(u(t_{j}); a, q) = J_{j}(σ_{1}(t_{j})). We get u(t_{j}) ≤ σ_{1}(t_{j}) and due to (74) we have u(t_{j}) = σ_{1}(t_{j}). Arguing as before, we get (77). Part 3. Assume that u ∈ B satisfies (75). We argue analogously to Part 2. □

We need the following lemma from the paper [4] (Lemma 2.4.), where the periodic problem was considered. The proof for our problem is formally the same.

Lemma 20 Each u ∈ B satisfies the condition

min{σ_{1}(τ_{u}+), σ_{2}(τ_{u}+)} ≤ u(τ_{u}+) ≤ max{σ_{1}(τ_{u}+), σ_{2}(τ_{u}+)}  (82)

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.

Theorem 21 Assume that f ∈ Car([0, T]×ℝ^{2}), J_{i}, M_{i} ∈ C(ℝ) for i = 1, …, m, g_{1}, g_{2} ∈ C(ℝ^{2}), (48) – (53) hold and there exists h ∈ L([0, T]) such that

|f(t, x, y)| ≤ h(t) for a. e. t ∈ [0, T] and each (x, y) ∈ ℝ^{2}.  (83)

Then the problem (2) – (4) has a solution u satisfying one of the conditions (73) – (75).

Proof. Step 1. Let σ_{1}, σ_{2} be lower and upper functions of (2) – (4) and let ρ_{1} be defined by (77). We put h̄ = 2h + 1 a. e. on [0, T]. By Lemma 14 we find d > ρ_{1} satisfying (58). Provided d > ρ_{1} + ρ̄, where ρ̄ is defined in (67), the properties of the constant d remain valid. We put ρ_{0} = ∥σ_{1}∥_{∞} + ∥σ_{2}∥_{∞} + 1 and

q = \frac{T}{m} ∑_{i=1}^{m} max\{ max_{|y|≤d+1} |M_{i}(y)|, d + 1\}.

Lemma 15 guarantees the existence of c > ρ_{0}+q such that (63) is valid. Obviously,

c > ∥σ_{1}∥_{∞} + ∥σ_{2}∥_{∞} + q + 1 and d > ∥σ′_{1}∥_{∞} + ∥σ′_{2}∥_{∞} + ρ̄ + 1.

We define

f̃(t, x, y) = \{ \begin{matrix}f(t, x, y) − h(t) − 1 & if & x ≤ −c − 1, \\ f(t, x, y) + (x + c)(h(t) + 1) & if & −c − 1 < x < −c, \\ f(t, x, y) & if & −c ≤ x ≤ c, \\ f(t, x, y) + (x − c)(h(t) + 1) & if & c < x < c + 1, \\ f(t, x, y) + h(t) + 1 & if & x ≥ c + 1 \end{matrix}  (84)

for a. e. t ∈ [0, T] and each (x, y) ∈ ℝ^{2},

J̃_{i}(x; c, q) = K(J_{i}, c, q)(x), i = 1, …, m,  (85)
M̃_{i}(y; d, 0) = K(M_{i}, d, 0)(y), i = 1, …, m,  (86)
g̃_{1}(x, y; c) = L(g_{1}, c, max{|g_{1}(x, y)| : (x, y) ∈ [−c, c]^{2}} + 1)(x, y), (87)
g̃_{2}(x, y; d) = L(g_{2}, d, − max{|g_{2}(x, y)| : (x, y) ∈ [−d, d]^{2}} − 1)(x, y)  (88)

and consider the problem

u′′ = f̃(t, u, u′),  (89)
u(t_{i}+) = J̃_{i}(u(t_{i}); c, q), u′(t_{i}+) = M̃_{i}(u′(t_{i}); d, 0), i = 1, …, m,  (90)
g̃_{1}(u(0), u(T); c) = 0, g̃_{2}(u′(0), u′(T); d) = 0.  (91)

It follows from definitions (84) – (88) that σ_{1} and σ_{2} are lower and upper functions of the problem (89) – (91). The inequality (83) implies

|f̃(t, x, y)| ≤ h̄(t) for a. e. t ∈ [0, T] and each (x, y) ∈ ℝ^{2}  (92)

and (84) yields

\{ \begin{matrix}f̃(t, x, y) < 0 & for a. e. t ∈ [0, T] and each (x, y) ∈ (−∞, −c − 1] × ℝ, \\ f̃(t, x, y) > 0 & for a. e. t ∈ [0, T] and each (x, y) ∈ [c + 1, ∞) × ℝ. \end{matrix} (93)

Step 2. We construct lower and upper functions σ_{3}, σ_{4} of the problem (89) – (91). We put

A^{∗} = q + ∑_{i=1}^{m} max_{|x|≤c+1} |J̃_{i}(x; c, q)|  (94)

and

\{ \begin{matrix}σ_{4}(0) = A^{∗} + mq, \\ σ_{4}(t) = A^{∗} + (m − i)q + \frac{mq}{T}t, t ∈ (t_{i}, t_{i+1}], \\ σ_{3} = −σ_{4} on [0, T]. \end{matrix}  (95)

It is obvious that σ_{3}, σ_{4} ∈ AC_{D}^{1} and

\{ \begin{matrix}σ_{3}(t) < −A^{∗} < −c − 1, & σ_{4}(t) > A^{∗} > c + 1, \\ σ′_{3}(t) = −\frac{mq}{T} ≤ −d − 1, & σ′_{4}(t) = \frac{mq}{T} ≥ d + 1 \end{matrix}

for t ∈ [0, T]. From these facts we can prove that σ_{3} and σ_{4} are lower and upper functions of the problem (89) – (91). Step 3. We define G by (31), the operator F by

(F u)(t) = \frac{T − t}{T}(u(0) + g̃_{1}(u(0), u(T); c)) + \frac{t}{T} (u(T) + g̃_{2}(u′(0), u′(T); d))

+ ∫_{0}^{T} G(t, s)f̃(s, u(s), u′(s)) ds − ∑_{i=1}^{m} \frac{∂G}{∂s}(t, t_{i})(J̃_{i}(u(t_{i}); c, q) − u(t_{i}))

+ ∑_{i=1}^{m} G(t, t_{i})(M̃_{i}(u′(t_{i}); d, 0) − u′(t_{i}))  (96)

and its domain

Ω_{0} = {u ∈ C_{D}^{1} : ∥u′∥_{∞} < C^{∗}, σ_{3} < u < σ_{4} on [0, T],
σ_{3}(τ+) < u(τ+) < σ_{4}(τ+), for τ ∈ D}  (97)

where

C^{∗} = ∥h̄∥_{1} + \frac{∥σ_{3}∥_{∞} + ∥σ_{4}∥_{∞}}{△} + ∥σ′_{3}∥_{∞} + ∥σ′_{4}∥_{∞} + 1  (98)

(△ is defined in (25)). It is clear that u is a solution to the problem (89) – (91) if and only if F u = u. Step 4. We will prove that for every solution to (89) – (91) the implication

u ∈ cl(Ω_{0}) ⟹ u ∈ Ω_{0}

is true. Let

u ∈ cl(Ω_{0})  (99)

be valid. For each i = 1, …, m there exists ξ_{i} ∈ (t_{i}, t_{i+1}) such that

|u′(ξ_{i})| ≤ \frac{∥σ_{3}∥_{∞} + ∥σ_{4}∥_{∞}}{△} .

Hence we get ∥u′∥_{∞} < C^{∗}. It remains to show that σ_{3} < u < σ_{4} on [0, T] and σ_{3}(τ+) < u(τ+) < σ_{4}(τ+) for τ ∈ D.

Assume the contrary, i. e. let there exists k ∈ {3, 4} such that

u(ξ) = σ_{k}(ξ) for some ξ ∈ [0, T]  (100)

or

u(τ+) = σ_{k}(τ+) for some τ ∈ D.  (101)

Case A. Let (100) be valid for k = 4.

(i) If ξ = 0, then in view of (91) we get u(0) = σ_{4}(0) = σ_{4}(T) = u(T) = A^{∗} + mq and u′(0) = u′(T) = \frac{mq}{T} = σ′_{4}(t) for each t ∈ [0, T]. Then there exists δ > 0 such that

u(t) > c + 1 for each t ∈ [0, δ]

and

u′(t) − u′(0) = ∫_{0}^{t} f̃(s, u(s), u′(s)) ds > 0 for each t ∈ [0, δ].

We have u′(t) > u′(0) = σ′_{4}(t) for every t ∈ (0, δ], thus u > σ_{4} on (0, δ], which contradicts assumption (99). (ii) If ξ ∈ (t_{i}, t_{i+1}) for some i ∈ \{0, …, m\}, then u′(ξ) = σ′_{4}(ξ) = \frac{mq}{T} = σ′_{4}(t) for t ∈ [0, T]. (iii) If ξ = t_{i} ∈ D then u(t_{i}) = σ_{4}(t_{i}) and

u(t_{i}+) = σ_{4}(t_{i}) − q > c + 1 − q > ∥σ_{1}∥_{∞} + ∥σ_{2}∥_{∞}.

From (99) we get u′(t_{i}+) ≤ σ′_{4}(t_{i}+) and u′(t_{i}) ≥ σ′_{4}(t_{i}). This implies u′(t_{i}+) ≥ σ′_{4}(t_{i}+) and

u′(t_{i}+) = σ′_{4}(t_{i}+) = \frac{mq}{T} = σ′_{4}(t) for each t ∈ [0, T].

We get a contradiction as in (i). Case B. Let (101) be valid for k = 4, i. e. u(t_{i}+) = σ_{4}(t_{i}+). Then

J̃_{i}(u(t_{i}); c, q) = σ_{4}(t_{i}+) = σ_{4}(t_{i}) − q > A^{∗} − q

and (94) yields u(t_{i}) > c+1. We get J̃_{i}(u(t_{i}); c, q) = u(t_{i})−q, thus u(t_{i}) = σ_{4}(t_{i}). We get a contradiction as in (iii). We prove (100) and (101) for k = 3 analogously. Step 5. We define

Ω_{1} = {u ∈ Ω_{0} : u(t) > σ_{1}(t) for t ∈ [0, T], u(τ+) > σ_{1}(τ+) for τ ∈ D},
Ω_{2} = {u ∈ Ω_{0} : u(t) < σ_{2}(t) for t ∈ [0, T], u(τ+) < σ_{2}(τ+) for τ ∈ D}

and

Ω̃ = Ω_{0} \ cl(Ω_{1} ∪ Ω_{2}).

We see that

Ω̃ = {u ∈ Ω_{0} : u satisfies (73)}.

Now, we will prove the implication

u ∈ cl(Ω̃) ⟹ (∥u∥_{∞} < c and ∥u′∥_{∞} < d)  (102)

for every solution u to the problem (89) – (91). Let u be a solution to the problem (89) – (91) and u ∈ cl(Ω̃), where

cl(Ω̃) = {u ∈ Ω_{0} : u satisfies one of the conditions (73), (74), (75)}.

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). Step 6. Finally, we will prove the existence result for the problem (2) – (4). We consider the operator F defined by (96). Case A. Let F have a fixed point u on ∂Ω̃, i. e. u ∈ ∂Ω̃. Then (102) implies that u is a solution to the problem (2) – (4). Case B. Let F u ≠ u for each u ∈ ∂Ω̃. Then F u ≠ u for each u ∈ ∂Ω_{0}∪∂Ω_{1}∪∂Ω_{2}. Theorem 9 implies

deg(I − F, Ω(σ_{3}, σ_{4}, C^{∗})) = deg(I − F, Ω_{0}) = 1,
deg(I − F, Ω(σ_{1}, σ_{4}, C^{∗})) = deg(I − F, Ω_{1}) = 1,
deg(I − F, Ω(σ_{3}, σ_{2}, C^{∗})) = deg(I − F, Ω_{2}) = 1.

Using the additivity property of the Leray–Schauder topological degree we obtain

deg(I − F, Ω̃) = deg(I − F, Ω_{0}) − deg(I − F, Ω_{1}) − deg(I − F, Ω_{2}) = −1.

Therefore F has a fixed point u ∈ Ω̃. From (102) we conclude that ∥u∥_{∞} < c and ∥u′∥_{∞} < d. From (84) – (91) we see that u is a solution to the problem (2) – (4). The proof is complete. □

Example 22 Let us consider the interval [0, T] = [0, 2], one impulsive point t_{1} = 1 and the boundary value problem

u′′(t) = e^{t} − 10 arctg(u(t) + 3u′(t)) for a. e. t ∈ [0, 2],
u(1+) = k_{1}u(1), u′(1+) = k_{2}u′(1), k_{1}, k_{2} ∈ (0, 1), k_{1} ≥ k_{2},
u(0) = u(2), u^{′3}(0) = u′(2).

We will seek a lower function of this problem in the form

σ_{1}(t) = \{ \begin{matrix}a + bt & for & t ∈ [0, 1], \\ c + dt & for & t ∈ (1, 2]. \end{matrix}  (103)

First we choose b > 1. From the impulsive and boundary value conditions (6) and (7) respectively we get relations

c + d = k_{1}(a + b), a = c + 2d, k_{2}b ≤ d ≤ b^{3}.

We can put

d = k_{2}b, a = \frac{k_{1} + k_{2}}{1 − k_{1}} b, c = \frac{k_{1} + 2k_{1}k_{2} − k_{2}}{1 − k_{1}} b.

Then a, b, c, d > 0 and all these constants become sufficiently large, when b is taken sufficiently large. The condition (5) can be satisfied for a suitable b. 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).

References

[1] L. H. Erbe and Liu Xinzhi, Existence results for boundary value problems of second order impulsive differential equations, J. Math. Anal. Appl. 149 (1990), 56–69.

[2] I. Rachůnková and J. Tomeček, Impulsive BVPs with nonlinear boundary conditions for the second order differential equations without growth restrictions, J. Math. Anal. Appl., to appear.

[3] I. Rachůnková and M. Tvrdý, Nonmonotone impulse effects in second order periodic boundary value problems, Abstr. Anal. Appl. 7 (2004), 577–590.

[4] I. Rachůnková and M. Tvrdý, Non–ordered lower and upper functions in second order impulsive periodic problems, Dynamics of Contionuous, Discrete and Impulsive Systems, to appear.

[5] I. Rachůnková and M. Tvrdý, Existence results for impulsive second order periodic problems, Nonlinear Anal., Theory Methods Appl., 59 (2004), 133– 146.

(Received October 7, 2004)

Department of Mathematics, Palacký University, Olomouc, Czech Republic E-mail address: jan tomecek@seznam.cz