A new approach to BVPs with state-dependent impulses

  • Irena Rachůnková1Email author and

    Affiliated with

    • Jan Tomeček1

      Affiliated with

      Boundary Value Problems20132013:22

      DOI: 10.1186/1687-2770-2013-22

      Received: 24 October 2012

      Accepted: 15 January 2013

      Published: 11 February 2013

      Abstract

      The paper deals with the second-order Dirichlet boundary value problem with one state-dependent impulse

      z ( t ) = f ( t , z ( t ) ) for a.e.  t [ 0 , T ] , z ( τ + ) z ( τ ) = I ( z ( τ ) ) , τ = γ ( z ( τ ) ) , z ( 0 ) = 0 , z ( T ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equa_HTML.gif

      Proofs of the main results contain a new approach to boundary value problems with state-dependent impulses which is based on a transformation to a fixed point problem of an appropriate operator in the space C 1 ( [ 0 , T ] ) × C 1 ( [ 0 , T ] ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq1_HTML.gif. Sufficient conditions for the existence of solutions to the problem are given here. The presented approach can be extended to more impulses and to other boundary conditions.

      MSC:34B37, 34B15.

      Keywords

      impulsive differential equation state-dependent impulses Dirichlet problem second-order ODE

      1 Introduction

      Differential equations involving impulse effects appear as a natural description of observed evolution phenomena of several real world problems. We refer to the monographs [13].

      Most papers in the literature on impulsive boundary value problems concern the case with fixed moments of impulsive effects. Papers dealing with state-dependent impulses, called also impulses at variable times, focus their attention on initial value problems or periodic problems. Such papers investigate the existence, stability or asymptotic properties of solutions of initial value problems [48] or solvability of autonomous periodic problems [9, 10] and nonautonomous ones [1115]. We can also find papers investigating other boundary value problems with state-dependent impulses through some initial value problems for multi-valued maps [16, 17].

      In this paper we provide a new approach to boundary value problems with state-dependent impulses based on a construction of proper sets and operators and the topological degree arguments. Unlike previous existing results, our approach enables us to find simple existence conditions for data functions and it can be used for other regular (and also singular) problems. We demonstrate it on the second-order Dirichlet boundary value problem with one state-dependent impulse
      z ( t ) = f ( t , z ( t ) ) for a.e.  t [ 0 , T ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ1_HTML.gif
      (1)
      z ( τ + ) z ( τ ) = I ( z ( τ ) ) , τ = γ ( z ( τ ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ2_HTML.gif
      (2)
      z ( 0 ) = 0 , z ( T ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ3_HTML.gif
      (3)
      where we assume
      f Car ( [ 0 , T ] × R ) , I C ( R ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ4_HTML.gif
      (4)
      { there exists  h Car ( [ 0 , T ] × [ 0 , ) )  such that h ( t , )  is nondecreasing for a.e.  t [ 0 , T ]  and | f ( t , x ) | h ( t , | x | )  for a.e.  t [ 0 , T ]  and all  x R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ5_HTML.gif
      (5)
      { there exists  J C ( [ 0 , T ] )  nondecreasing and such that  | I ( x ) | J ( | x | )  for  x R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ6_HTML.gif
      (6)
      K > 0 : 1 K [ 0 T h ( s , K + T J ( K ) ) d s + J ( K ) ] < min { 1 , 1 T } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ7_HTML.gif
      (7)
      and
      { γ C 1 ( [ K 1 , K 1 ] ) , 0 < γ ( x ) < T , | γ ( x ) | < T K 1 for  | x | K 1 , where  K 1 = K + T J ( K ) , K  is from (7). http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ8_HTML.gif
      (8)

      Under assumptions (4)-(8), we prove the solvability of problem (1)-(3). In particular, we transform problem (1)-(3) to a fixed point problem for a proper operator in the space C 1 ( [ 0 , T ] ) × C 1 ( [ 0 , T ] ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq1_HTML.gif. This approach can be also used for other types of boundary conditions and it can be easily extended to more impulses.

      Here, we denote by C ( J ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq2_HTML.gif the set of all continuous functions on the interval J, by C 1 ( J ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq3_HTML.gif the set of all functions having continuous derivatives on the interval J and by L 1 ( J ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq4_HTML.gif the set of all Lebesgue integrable functions on J. For a compact interval J, we consider the linear space of functions from C ( J ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq2_HTML.gif or C 1 ( J ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq3_HTML.gif equipped, respectively, with the norms
      x = max t J | x ( t ) | , x 1 = x + x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equb_HTML.gif
      In this paper we work with the linear space C 1 ( [ 0 , T ] ) × C 1 ( [ 0 , T ] ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq1_HTML.gif, where T > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq5_HTML.gif, equipped with the norm
      ( u , v ) = u 1 + v 1 for  ( u , v ) C 1 ( [ 0 , T ] ) × C 1 ( [ 0 , T ] ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equc_HTML.gif

      It is well-known that the mentioned normed spaces are Banach spaces. Recall that for A R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq6_HTML.gif, a function f : [ a , b ] × A R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq7_HTML.gif satisfies the Carathéodory conditions on [ a , b ] × A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq8_HTML.gif (we write f Car ( [ a , b ] × A ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq9_HTML.gif) if

      • f ( , x ) : [ a , b ] R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq10_HTML.gif is measurable for all x A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq11_HTML.gif,

      • f ( t , ) : A R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq12_HTML.gif is continuous for a.e. t [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq13_HTML.gif,

      • for each compact set K A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq14_HTML.gif, there exists a function m K L 1 ( [ a , b ] ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq15_HTML.gif such that | f ( t , x ) | m K ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq16_HTML.gif for a.e. t [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq13_HTML.gif and each x K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq17_HTML.gif.

      We say that z : [ 0 , T ] R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq18_HTML.gif is a solution of problem (1)-(3), if z is continuous on [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq19_HTML.gif, there exists unique τ ( 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq20_HTML.gif such that γ ( z ( τ ) ) = τ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq21_HTML.gif, z | [ 0 , τ ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq22_HTML.gif and z | [ τ , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq23_HTML.gif have absolutely continuous first derivatives, z satisfies equation (1) for a.e. t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq24_HTML.gif and fulfills conditions (2), (3).

      2 Operators

      In this section we assume that (4)-(8) are fulfilled. We introduce sets and operators corresponding to problem (1)-(3) and prove their properties which are needed for an application of the Leray-Schauder degree theory. Let us consider K of (7) and define the set
      B 1 = { u C 1 ( [ 0 , T ] ) : u < K , u < K / T } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equd_HTML.gif
      Lemma 1 For each u B ¯ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq25_HTML.gif, there exists a unique τ u ( 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq26_HTML.gif such that
      γ ( u ( τ u ) ) = τ u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ9_HTML.gif
      (9)
      Proof Let us take an arbitrary u B ¯ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq25_HTML.gif. Obviously, the constant τ u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq27_HTML.gif is a solution of the equation
      γ ( u ( t ) ) = t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Eque_HTML.gif
      i.e., τ u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq27_HTML.gif is a root of the function
      σ ( t ) = γ ( u ( t ) ) t , t [ 0 , T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equf_HTML.gif
      From (8) it follows σ ( 0 ) = γ ( u ( 0 ) ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq28_HTML.gif, σ ( T ) = γ ( u ( T ) ) T < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq29_HTML.gif. According to (8) and the definition of B 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq30_HTML.gif, we get
      σ ( t ) = γ ( u ( t ) ) u ( t ) 1 | γ ( u ( t ) ) | | u ( t ) | 1 < T K K T 1 = 0 , t ( 0 , T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ10_HTML.gif
      (10)

      Therefore, σ is strictly decreasing on [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq19_HTML.gif and hence it has exactly one root in ( 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq31_HTML.gif. □

      Now, define a functional P : B ¯ 1 ( 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq32_HTML.gif by
      P u = τ u , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equg_HTML.gif

      where τ u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq27_HTML.gif fulfills (9). The next lemma provides an important result about the continuity of P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq33_HTML.gif which is fundamental for our approach.

      Lemma 2 The functional P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq33_HTML.gif is continuous on B ¯ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq34_HTML.gif.

      Proof Let us consider u n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq35_HTML.gif, u B ¯ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq25_HTML.gif for n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq36_HTML.gif such that u n u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq37_HTML.gif in C 1 ( [ 0 , T ] ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq38_HTML.gif. Let us denote
      σ n ( t ) = γ ( u n ( t ) ) t , σ ( t ) = γ ( u ( t ) ) t for  t [ 0 , T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equh_HTML.gif
      By Lemma 1, σ n ( τ n ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq39_HTML.gif and σ ( τ ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq40_HTML.gif, where τ n = P u n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq41_HTML.gif and τ = P u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq42_HTML.gif, respectively. According to (8), we get σ n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq43_HTML.gif, σ C 1 ( [ 0 , T ] ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq44_HTML.gif for n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq45_HTML.gif and
      σ n σ in  C ( [ 0 , T ] ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ11_HTML.gif
      (11)
      We will prove that lim n τ n = τ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq46_HTML.gif. Let us take an arbitrary ϵ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq47_HTML.gif. Since σ ( τ ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq40_HTML.gif and σ ( τ ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq48_HTML.gif (cf. (10)), we can find ξ ( τ ϵ , τ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq49_HTML.gif and η ( τ , τ + ϵ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq50_HTML.gif such that
      σ ( ξ ) > 0 and σ ( η ) < 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equi_HTML.gif
      From (11) it follows the existence of n 0 N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq51_HTML.gif such that
      σ n ( ξ ) > 0 and σ n ( η ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equj_HTML.gif

      for each n n 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq52_HTML.gif. By Lemma 1 and the continuity of σ n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq43_HTML.gif, it follows that τ n ( ξ , η ) ( τ ϵ , τ + ϵ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq53_HTML.gif for n n 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq54_HTML.gif. □

      Further, consider K 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq55_HTML.gif of (8) and define sets B 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq56_HTML.gif and Ω by
      B 2 = { v C 1 ( [ 0 , T ] ) : v < K 1 , v < K 1 T } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equk_HTML.gif
      and
      Ω = B 1 × B 2 C 1 ( [ 0 , T ] ) × C 1 ( [ 0 , T ] ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ12_HTML.gif
      (12)
      Finally, define an operator F : Ω ¯ C 1 ( [ 0 , T ] ) × C 1 ( [ 0 , T ] ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq57_HTML.gif by F ( u , v ) = ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq58_HTML.gif, where
      { x ( t ) = 0 T G ( t , s ) f ˜ ( s , u ( s ) , v ( s ) ) d s + g 1 ( t , τ u ) I ( u ( τ u ) ) , y ( t ) = 0 T G ( t , s ) f ˜ ( s , u ( s ) , v ( s ) ) d s + g 2 ( t , τ u ) I ( u ( τ u ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ13_HTML.gif
      (13)
      for t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq24_HTML.gif, τ u = P u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq59_HTML.gif,
      f ˜ ( t , u ( t ) , v ( t ) ) = { f ( t , u ( t ) ) for a.e.  t [ 0 , τ u ] , f ( t , v ( t ) ) for a.e.  t ( τ u , T ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ14_HTML.gif
      (14)
      g 1 ( t , s ) = t ( s T ) T , g 2 ( t , s ) = s ( t T ) T , s , t [ 0 , T ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ15_HTML.gif
      (15)
      and G is the Green function of the problem u = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq60_HTML.gif, u ( 0 ) = u ( T ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq61_HTML.gif, that is,
      G ( t , s ) = { g 1 ( t , s ) for  0 t s T , g 2 ( t , s ) for  0 s t T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equl_HTML.gif

      Lemma 3 The operatoris compact on Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq62_HTML.gif.

      Proof First, we will prove the continuity of the operator ℱ. Let us choose ( u n , v n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq63_HTML.gif, ( u , v ) Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq64_HTML.gif for n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq65_HTML.gif such that
      ( u n , v n ) ( u , v ) in  C 1 ( [ 0 , T ] ) × C 1 ( [ 0 , T ] ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ16_HTML.gif
      (16)
      Let us denote τ n = P u n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq41_HTML.gif, τ = P u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq42_HTML.gif, ( x n , y n ) = F ( u n , v n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq66_HTML.gif, ( x , y ) = F ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq67_HTML.gif for each n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq36_HTML.gif. We will prove that x n x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq68_HTML.gif in C 1 ( [ 0 , T ] ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq38_HTML.gif. For each t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq24_HTML.gif, we get by (13)-(15)
      x n ( t ) x ( t ) = 0 τ G ( t , s ) [ f ( s , u n ( s ) ) f ( s , u ( s ) ) ] d s + τ T G ( t , s ) [ f ( s , v n ( s ) ) f ( s , v ( s ) ) ] d s + τ τ n G ( t , s ) [ f ( s , u n ( s ) ) f ( s , v n ( s ) ) ] d s + g 1 ( t , τ n ) I ( u n ( τ n ) ) g 1 ( t , τ ) I ( u ( τ ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equm_HTML.gif
      and
      x n ( t ) x ( t ) = 0 τ G t ( t , s ) [ f ( s , u n ( s ) ) f ( s , u ( s ) ) ] d s + τ T G t ( t , s ) [ f ( s , v n ( s ) ) f ( s , v ( s ) ) ] d s + τ τ n G t ( t , s ) [ f ( s , u n ( s ) ) f ( s , v n ( s ) ) ] d s + g 1 t ( t , τ n ) I ( u n ( τ n ) ) g 1 t ( t , τ ) I ( u ( τ ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equn_HTML.gif
      Since
      | G ( t , s ) | T , | G t ( t , s ) | 1 for  t , s [ 0 , T ] , t s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equo_HTML.gif
      we get
      x n x 1 ( T + 1 ) 0 T | f ( s , u n ( s ) ) f ( s , u ( s ) ) | d s + ( T + 1 ) 0 T | f ( s , v n ( s ) ) f ( s , v ( s ) ) | d s + ( T + 1 ) τ τ n | f ( s , u n ( s ) ) f ( s , v n ( s ) ) | d s + max t [ 0 , T ] | g 1 ( t , τ n ) I ( u n ( τ n ) ) g 1 ( t , τ ) I ( u ( τ ) ) | + max t [ 0 , T ] | g 1 t ( t , τ n ) I ( u n ( τ n ) ) g 1 t ( t , τ ) I ( u ( τ ) ) | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equp_HTML.gif
      By (16), there exists a compact set K R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq69_HTML.gif such that u n ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq70_HTML.gif, v n ( t ) K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq71_HTML.gif for each t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq24_HTML.gif and n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq72_HTML.gif. Consequently, by (4), there exists m K L 1 ( [ 0 , T ] ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq73_HTML.gif such that
      | f ( t , u n ( t ) ) | m K ( t ) , | f ( t , v n ( t ) ) | m K ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equq_HTML.gif
      for a.e. t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq74_HTML.gif and all n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq36_HTML.gif. Since
      lim n f ( t , u n ( t ) ) = f ( t , u ( t ) ) , lim n f ( t , v n ( t ) ) = f ( t , v ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equr_HTML.gif
      for a.e. t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq74_HTML.gif, then due to the Lebesgue dominated convergence theorem, it follows that
      0 T | f ( s , u n ( s ) ) f ( s , u ( s ) ) | d s 0 and 0 T | f ( s , v n ( s ) ) f ( s , v ( s ) ) | d s 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equs_HTML.gif
      as n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq75_HTML.gif. Since lim n τ n = τ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq46_HTML.gif, the absolute continuity of the Lebesgue integral yields
      lim n | τ τ n | f ( s , u n ( s ) ) f ( s , v n ( s ) ) | d s | 2 lim n | τ τ n m K ( s ) d s | = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equt_HTML.gif
      Further, we have for g 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq76_HTML.gif the inequality
      | g 1 ( t , τ n ) I ( u n ( τ n ) ) g 1 ( t , τ ) I ( u ( τ ) ) | T | I ( u n ( τ n ) ) I ( u ( τ n ) ) | + | g 1 ( t , τ n ) I ( u ( τ n ) ) g 1 ( t , τ ) I ( u ( τ ) ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equu_HTML.gif
      for each t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq74_HTML.gif and the same is true for g 1 t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq77_HTML.gif. The continuity of g 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq76_HTML.gif, g 1 t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq77_HTML.gif and I imply that
      g 1 ( t , τ n ) I ( u n ( τ n ) ) g 1 ( t , τ ) I ( u ( τ ) ) , g 1 t ( t , τ n ) I ( u n ( τ n ) ) g 1 t ( t , τ ) I ( u ( τ ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equv_HTML.gif

      as n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq75_HTML.gif uniformly w.r.t. t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq24_HTML.gif. Therefore, x n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq78_HTML.gif converges to x in C 1 ( [ 0 , T ] ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq38_HTML.gif. Similar arguments can be applied to the sequence { y n } n = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq79_HTML.gif.

      Now we will prove that F ( Ω ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq80_HTML.gif is relatively compact. The boundedness of Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq62_HTML.gif implies the existence of M > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq81_HTML.gif and m L 1 ( [ 0 , T ] ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq82_HTML.gif such that for all ( u , v ) Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq64_HTML.gif,
      | I ( u ( t ) ) | M for all  t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equw_HTML.gif
      and
      | f ˜ ( t , u ( t ) , v ( t ) ) | m ( t ) for a.e.  t [ 0 , T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equx_HTML.gif
      Therefore, by (13), we get
      | x ( t ) | + | x ( t ) | ( T + 1 ) ( 0 T m ( s ) d s + M ) = : M 1 , t [ 0 , T ] , | y ( t ) | + | y ( t ) | M 1 , t [ 0 , T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equy_HTML.gif
      We have proved that the set F ( Ω ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq80_HTML.gif is bounded in C 1 ( [ 0 , T ] ) × C 1 ( [ 0 , T ] ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq83_HTML.gif. We now show that the set { ( x , y ) : ( x , y ) F ( Ω ¯ ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq84_HTML.gif is equicontinuous on [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq19_HTML.gif. For a.e. t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq85_HTML.gif and all ( x , y ) F ( Ω ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq86_HTML.gif, we have
      | x ( t ) | m ( t ) , | y ( t ) | m ( t ) for a.e.  t [ 0 , T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equz_HTML.gif
      As a result, for each ϵ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq47_HTML.gif, there exists δ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq87_HTML.gif such that for each t 1 , t 2 [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq88_HTML.gif satisfying | t 1 t 2 | < δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq89_HTML.gif, the inequality
      | x ( t 1 ) x ( t 2 ) | + | y ( t 1 ) y ( t 2 ) | 2 | t 2 t 1 m ( t ) d t | < ϵ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equaa_HTML.gif

      holds for all ( x , y ) F ( Ω ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq90_HTML.gif. Consequently, F ( Ω ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq80_HTML.gif is relatively compact in C 1 ( [ 0 , T ] ) × C 1 ( [ 0 , T ] ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq1_HTML.gif by the Arzelà-Ascoli theorem. □

      Lemma 4 Let ( u , v ) Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq64_HTML.gif be a fixed point of ℱ. Then the function
      z ( t ) = { u ( t ) , t [ 0 , τ u ] , v ( t ) , t ( τ u , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ17_HTML.gif
      (17)

      is a solution of problem (1)-(3).

      Proof Let ( u , v ) Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq64_HTML.gif be such that ( u , v ) = F ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq91_HTML.gif, that is,
      { u ( t ) = 0 T G ( t , s ) f ˜ ( s , u ( s ) , v ( s ) ) d s + g 1 ( t , τ u ) I ( u ( τ u ) ) , v ( t ) = 0 T G ( t , s ) f ˜ ( s , u ( s ) , v ( s ) ) d s + g 2 ( t , τ u ) I ( u ( τ u ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ18_HTML.gif
      (18)

      t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq24_HTML.gif, τ u = P u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq59_HTML.gif.

      Let us consider the function z defined in (17). Hence, z ( 0 ) = u ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq92_HTML.gif, z ( T ) = v ( T ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq93_HTML.gif,
      z ( τ u ) = u ( τ u ) = v ( τ u ) = z ( τ u + ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ19_HTML.gif
      (19)
      and by Lemma 1,
      γ ( z ( τ u ) ) = τ u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ20_HTML.gif
      (20)
      In addition, by (17), τ u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq27_HTML.gif is a unique point in ( 0 , τ u ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq94_HTML.gif satisfying (20). Put σ ( t ) = γ ( v ( t ) ) t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq95_HTML.gif, t [ τ u , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq96_HTML.gif. Due to (19) and (20), we get σ ( τ u ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq97_HTML.gif. Further,
      σ ( t ) = γ ( v ( t ) ) v ( t ) 1 | γ ( v ( t ) ) | | v ( t ) | 1 < T K 1 K 1 T 1 = 0 for  t ( τ u , T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equab_HTML.gif

      Therefore, σ is strictly decreasing on [ τ u , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq98_HTML.gif, which yields σ ( t ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq99_HTML.gif for t ( τ u , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq100_HTML.gif. Consequently, τ u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq27_HTML.gif is a unique point in ( 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq31_HTML.gif satisfying (20).

      Further, we get
      u ( t ) = 0 T G t ( t , s ) f ˜ ( s , u ( s ) , v ( s ) ) d s + g 1 t ( t , τ u ) I ( u ( τ u ) ) , v ( t ) = 0 T G t ( t , s ) f ˜ ( s , u ( s ) , v ( s ) ) d s + g 2 t ( t , τ u ) I ( u ( τ u ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equac_HTML.gif
      t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq24_HTML.gif, and by virtue of (14),
      u ( t ) = f ˜ ( t , u ( t ) , v ( t ) ) = f ( t , u ( t ) ) for a.e.  t [ 0 , τ u ) , v ( t ) = f ˜ ( t , u ( t ) , v ( t ) ) = f ( t , v ( t ) ) for a.e.  t ( τ u , T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equad_HTML.gif
      Therefore,
      z ( t ) = f ( t , z ( t ) ) for a.e.  t [ 0 , T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equae_HTML.gif
      Finally,
      z ( τ u + ) = v ( τ u ) = 0 T G t ( τ u + , s ) f ˜ ( s , u ( s ) , v ( s ) ) d s + g 2 t ( τ u , τ u ) I ( u ( τ u ) ) , z ( τ u ) = u ( τ u ) = 0 T G t ( τ u , s ) f ˜ ( s , u ( s ) , v ( s ) ) d s + g 1 t ( τ u , τ u ) I ( u ( τ u ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equaf_HTML.gif
      Since
      G t ( τ u + , s ) = G t ( τ u , s ) for  s [ 0 , T ] , s τ u , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equag_HTML.gif
      we have
      z ( τ u + ) z ( τ u ) = g 2 t ( τ u , τ u ) I ( u ( τ u ) ) g 1 t ( τ u , τ u ) I ( u ( τ u ) ) = τ u T I ( u ( τ u ) ) τ u T T I ( u ( τ u ) ) = I ( u ( τ u ) ) = I ( z ( τ u ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equah_HTML.gif

       □

      3 Main result

      Here, using the Leray-Schauder degree theory, we prove our main result about the solvability of problem (1)-(3). To this end, we will need the following lemma on a priori estimates.

      Lemma 5 Assume (4)-(8). Then for any λ [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq101_HTML.gif and any solution ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq102_HTML.gif of the equation
      ( u , v ) = λ F ( u , v ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ21_HTML.gif
      (21)
      the implication
      ( u , v ) Ω ¯ ( u , v ) Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ22_HTML.gif
      (22)

      holds.

      Proof Let us choose λ ( 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq103_HTML.gif and let ( u , v ) Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq64_HTML.gif satisfy (21), i.e.,
      { u ( t ) = λ ( 0 T G ( t , s ) f ˜ ( s , u ( s ) , v ( s ) ) d s + g 1 ( t , τ u ) I ( u ( τ u ) ) ) , v ( t ) = λ ( 0 T G ( t , s ) f ˜ ( s , u ( s ) , v ( s ) ) d s + g 2 ( t , τ u ) I ( u ( τ u ) ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ23_HTML.gif
      (23)
      for t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq24_HTML.gif. Then
      { u ( t ) = λ ( 0 T G t ( t , s ) f ˜ ( s , u ( s ) , v ( s ) ) d s + g 1 t ( t , τ u ) I ( u ( τ u ) ) ) , v ( t ) = λ ( 0 T G t ( t , s ) f ˜ ( s , u ( s ) , v ( s ) ) d s + g 2 t ( t , τ u ) I ( u ( τ u ) ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ24_HTML.gif
      (24)

      t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq74_HTML.gif. Since ( u , v ) Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq64_HTML.gif, it follows that ( u , v ) B ¯ 1 × B ¯ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq104_HTML.gif and therefore u K / T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq105_HTML.gif, u K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq106_HTML.gif, v K 1 / T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq107_HTML.gif and v K 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq108_HTML.gif. There are two possibilities as follows.

      Case A. Let u < K / T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq109_HTML.gif. Then u < K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq110_HTML.gif and from (15) and (24), it follows
      v ( t ) = u ( t ) + λ ( τ u T τ u T T ) I ( u ( τ u ) ) = u ( t ) + λ I ( u ( τ u ) ) , t ( 0 , T ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equai_HTML.gif
      which implies, due to (6) and (8),
      v u + J ( u ) < K 1 T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equaj_HTML.gif

      Then v < K 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq111_HTML.gif, which yields ( u , v ) Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq112_HTML.gif.

      Case B. Let u = K / T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq113_HTML.gif. From (24), (14), (6), (5) and (8), it follows
      u 0 τ u | f ( s , u ( s ) ) | d s + τ u T | f ( s , v ( s ) ) | d s + J ( K ) 0 τ u h ( s , K ) d s + τ u T h ( s , K 1 ) d s + J ( K ) 0 T h ( s , K 1 ) d s + J ( K ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equak_HTML.gif
      This inequality together with (7) implies
      1 T K [ 0 T h ( s , K 1 ) d s + J ( K ) ] < T min { 1 , 1 T } = min { T , 1 } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equal_HTML.gif

      which is a contradiction.

      For λ = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq114_HTML.gif, the solution of (21) is ( u , v ) = ( 0 , 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq115_HTML.gif, and it clearly belongs to Ω. □

      Theorem 6 Assume (4)-(8). Then the operatorhas a fixed point in Ω.

      Proof According to Lemma 5, the operator I λ F : [ 0 , 1 ] × Ω ¯ ( C 1 ( [ 0 , T ] ) ) 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq116_HTML.gif is a homotopy. Therefore,
      deg ( I F , Ω ) = deg ( I , Ω ) = 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equam_HTML.gif
      and consequently the equation
      ( I F ) ( u , v ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equan_HTML.gif

      has a solution in Ω. This solution is a fixed point of the operator ℱ. □

      Theorem 7 Assume (4)-(8). Then problem (1)-(3) has a solution z such that
      z < K 1 , z < K 1 T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ25_HTML.gif
      (25)

      Proof From Theorem 6 it follows that there exists a fixed point ( u , v ) Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq112_HTML.gif of the operator ℱ. Lemma 4 yields that the function z defined in (17) (with τ u = P u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq117_HTML.gif) is a solution of problem (1)-(3). Estimates (25) follow from (17) and from the definitions of Ω and K 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq55_HTML.gif (cf. (12) and (8)). □

      Remark 8 Let us note that assumption (7) follows from the condition
      lim inf x 1 x [ 0 T h ( s , x + T J ( x ) ) d s + J ( x ) ] < min { 1 , 1 T } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equao_HTML.gif

      4 Examples

      In this section we demonstrate that Theorem 7 can be applied to sublinear, linear and superlinear problems.

      Example 9 (Sublinear problem)

      Let us consider problem (1)-(3) with
      T = 1 , f ( t , x ) = t 2 | x | α sgn x , I ( x ) = | x | β sgn x , α , β ( 0 , 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equap_HTML.gif
      that is, f and I are sublinear in x. Then assumptions (5) and (6) are valid for
      h ( t , x ) = t 2 + x α , t [ 0 , 1 ] , x > 0 , J ( x ) = x β , x > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equaq_HTML.gif
      Since
      lim x 1 x [ 0 1 h ( s , x + J ( x ) ) d s + J ( x ) ] = lim x 1 x [ 1 3 + ( x + x β ) α + x β ] = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equar_HTML.gif
      Remark 8 yields that condition (7) is satisfied for any sufficiently large K. In particular, let us put
      α = β = 1 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equas_HTML.gif
      If we choose K = 10 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq118_HTML.gif, we see that (7) holds. Then by (8), we have
      K 1 = 10 + 10 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equat_HTML.gif
      For instance, if we choose c ( 0 , 1 / ( 2 K 1 2 ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq119_HTML.gif and put
      γ ( x ) = c x 2 + 1 2 , x R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ26_HTML.gif
      (26)
      or if we choose c ( 0 , 1 / 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq120_HTML.gif, n > c K 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq121_HTML.gif and put
      γ ( x ) = c sin x n + 1 2 , x R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ27_HTML.gif
      (27)

      we can check that conditions (8) are satisfied in both cases. Therefore, by Theorem 7, the corresponding problem (1)-(3) has at least one solution.

      Note that (27) shows that γ need not be monotonous.

      Example 10 (Linear problem)

      Let us consider problem (1)-(3) with f and I having the linear behavior in x and put
      T = 1 , f ( t , x ) = a ( t α x ) , I ( x ) = b x , a , b R , α > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equau_HTML.gif
      Then assumptions (5) and (6) are valid for
      h ( t , x ) = | a | ( t α + x ) , t [ 0 , 1 ] , x > 0 , J ( x ) = | b | x , x > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equav_HTML.gif
      Since
      lim x 1 x [ 0 1 h ( s , x + J ( x ) ) d s + J ( x ) ] = lim x 1 x [ | a | ( 1 α + 1 + x ( 1 + | b | ) ) + x | b | ] = | a | ( 1 + | b | ) + | b | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equaw_HTML.gif
      Theorem 7 can be applied, due to Remark 8, under the additional assumption
      | a | < 1 | b | 1 + | b | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ28_HTML.gif
      (28)

      If (28) holds, then for any sufficiently large K, condition (7) is satisfied. By (8), we have K 1 = K ( 1 + | b | ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq122_HTML.gif, and problem (1)-(3) has a solution for any γ satisfying (8). Consequently, if γ is given by (26) or (27), problem (1)-(3) is solvable.

      Example 11 (Superlinear problem) Let us consider problem (1)-(3) with f and I superlinear in x. Put, for example,
      T = 1 , f ( t , x ) = c 1 t 3 + c 2 x 3 , I ( x ) = 1 2 x 2 , c 1 , c 2 R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ29_HTML.gif
      (29)
      Then assumptions (5) and (6) are valid for
      h ( t , x ) = | c 1 | t 3 + | c 2 | x 3 , t [ 0 , 1 ] , x > 0 , J ( x ) = 1 2 x 2 , x > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equax_HTML.gif
      It holds
      1 x [ 0 1 h ( s , x + J ( x ) ) d s + J ( x ) ] = 1 x [ | c 1 | 4 + | c 2 | ( x + 1 2 x 2 ) 3 + 1 2 x 2 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equay_HTML.gif
      By virtue of (7), Theorem 7 can be applied provided there exists K > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq123_HTML.gif such that
      | c 1 | 4 + | c 2 | ( K + 1 2 K 2 ) 3 + 1 2 K 2 < K . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equ30_HTML.gif
      (30)
      Let us search K in the interval ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq124_HTML.gif. Then K 3 < K 2 < K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq125_HTML.gif and it holds
      | c 2 | ( K + 1 2 K 2 ) 3 + 1 2 K 2 < | c 2 | ( 3 2 K ) 3 + 1 2 K 2 < ( 27 8 | c 2 | + 1 2 ) K 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equaz_HTML.gif
      Consequently, each K ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq126_HTML.gif fulfilling the equation
      ( 27 8 | c 2 | + 1 2 ) K 2 K + | c 1 | 4 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_Equba_HTML.gif

      satisfies (30) as well. Put, for example, c 1 = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq127_HTML.gif, c 2 = 4 / 27 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq128_HTML.gif. Then we get that for K = 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq129_HTML.gif inequality (30) holds. Consequently, (8) gives K 1 = 5 / 8 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-22/MediaObjects/13661_2012_Article_282_IEq130_HTML.gif and the corresponding problem (1)-(3) is solvable for any γ satisfying (8). In particular, γ given by (26) or (27) can be considered in this case as well.

      Declarations

      Acknowledgements

      Dedicated to Jean Mawhin on the occasion of his 70th birthday.

      The authors would like to thank the anonymous referees for their valuable comments and suggestions. This work was supported by the grant Matematické modely a struktury, PrF_ 2012_ 017.

      Authors’ Affiliations

      (1)
      Department of Mathematical Analysis and Applications of Mathematics, Faculty of Science, Palacký University

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      Copyright

      © Rachůnková and Tomeček; licensee Springer. 2013

      This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://​creativecommons.​org/​licenses/​by/​2.​0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.