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First-order nonlinear differential equations with state-dependent impulses

Boundary Value Problems20132013:195

https://doi.org/10.1186/1687-2770-2013-195

  • Received: 13 May 2013
  • Accepted: 13 August 2013
  • Published:

Abstract

The paper deals with the state-dependent impulsive problem

z ( t ) = f ( t , z ( t ) ) for a.e.  t [ a , b ] , z ( τ + ) z ( τ ) = J ( τ , z ( τ ) ) , γ ( z ( τ ) ) = τ , ( z ) = c 0 ,

where [ a , b ] R , c 0 R , f fulfils the Carathéodory conditions on [ a , b ] × R , the impulse function is continuous on [ a , b ] × R , the barrier function γ has a continuous first derivative on some subset of and is a linear bounded functional which is defined on the Banach space of left-continuous regulated functions on [ a , b ] equipped with the sup-norm. The functional is represented by means of the Kurzweil-Stieltjes integral and covers all linear boundary conditions for solutions of first-order differential equations subject to state-dependent impulse conditions. Here, sufficient and effective conditions guaranteeing the solvability of the above problem are presented for the first time.

MSC:34B37, 34B15.

Keywords

  • first-order ODE
  • state-dependent impulses
  • transversality conditions
  • general linear boundary conditions
  • existence
  • Kurzweil-Stieltjes integral

1 Introduction

The investigation of impulsive differential equations has a long history; see, e.g., the monographs [13]. Most papers dealing with impulsive differential equations subject to boundary conditions focus their attention on impulses at fixed moments. But this is a very particular case of a more complicated case with state-dependent impulses. Boundary value problems with state-dependent impulses, where difficulties with an operator representation appear (cf. Remark 6.2), are substantially less developed. We refer to the papers [46] and [7] which are devoted to periodic problems, and for problems with other boundary conditions, see [8, 9] or [1012].

Here, in our paper, we present an approach leading to a new existence principle for impulsive boundary value problems. This approach is applicable to each linear boundary condition which is considered with some first-order differential equation subject to state-dependent impulses. The important step is a proof of a transversality (Remark 2.3 and Lemmas 5.1 and 5.2), which makes possible a construction of a continuous operator (Section 6) whose fixed point leads to a solution of our original impulsive problem (Section 7).

Notation

Let M R n , n N , [ a , b ] R .

  • C ( M ) is the set of real functions continuous on M.

  • AC ( M ) is the set of real functions absolutely continuous on M.

  • L 1 [ a , b ] is the set of real functions Lebesgue integrable on [ a , b ] .

  • L [ a , b ] is the set of real functions essentially bounded on [ a , b ] .

  • BV [ a , b ] is the set of real functions with bounded variation on [ a , b ] .

  • G L [ a , b ] is the set of real left-continuous regulated functions on [ a , b ] , that is, z G L [ a , b ] if and only if z : [ a , b ] R , and for each τ 1 ( a , b ] and each τ 2 [ a , b ) ,
    z ( τ 1 ) = z ( τ 1 ) = lim t τ 1 z ( t ) , z ( τ 2 + ) = lim t τ 2 + z ( t ) R .
    (1.1)
  • Car ( [ a , b ] × M ) is the set of functions f : [ a , b ] × M R such that

  1. (i)

    f ( , x ) : [ a , b ] R is measurable for all x M ,

     
  2. (ii)

    f ( t , ) : M R is continuous for a.e. t [ a , b ] ,

     
  3. (iii)
    for each compact set Q M , there exists m Q L 1 [ a , b ] satisfying
    | f ( t , x ) | m Q ( t ) for a.e.  t [ a , b ]  and each  x Q .
     
  • The set L [ a , b ] equipped with the norm
    z = sup ess { | z ( t ) | : t [ a , b ] } for  z L [ a , b ]
    (1.2)

is a Banach space.

  • Since C [ a , b ] G L [ a , b ] L [ a , b ] , we equip the sets C [ a , b ] and G L [ a , b ] with the norm and get also Banach spaces (cf.[13]). Then (1.2) can be written as
    z = sup { | z ( t ) | : t [ a , b ] } for  z G L [ a , b ]
    (1.3)
and
z = max { | z ( t ) | : t [ a , b ] } for  z C [ a , b ] .
(1.4)
  • W 1 , [ a , b ] is the Banach space of functions z : [ a , b ] R such that z AC [ a , b ] and z L [ a , b ] , where the norm 1 , is given by
    z 1 , = z + z for  z W 1 , [ a , b ] .
    (1.5)
  • χ A is the characteristic function of a set A, where A R .

2 Formulation of problem

We investigate the solvability of the nonlinear differential equation
z ( t ) = f ( t , z ( t ) )
(2.1)
subject to the state-dependent impulse condition
z ( τ + ) z ( τ ) = J ( τ , z ( τ ) ) , γ ( z ( τ ) ) = τ ,
(2.2)
and the general linear boundary condition
( z ) = c 0 .
(2.3)
Here we assume that
{ f Car ( [ a , b ] × R ) , J C ( [ a , b ] × R ) , [ a , b ] R , K ( 0 , ) , γ C 1 [ K , K ] , c 0 R ,
(2.4)

and : G L [ a , b ] R is a linear bounded functional.

Definition 2.1 A function z : [ a , b ] R is a solution of problem (2.1), (2.2) if

  • there exists a unique τ ( a , b ) such that γ ( z ( τ ) ) = τ ;

  • the restrictions z | [ a , τ ] and z | ( τ , b ] are absolutely continuous;

  • z ( τ + ) = z ( τ ) + J ( τ , z ( τ ) ) ;

  • z satisfies equation (2.1) for a.e. t [ a , b ] .

Definition 2.2 A graph of a function γ : [ K , K ] R is called a barrier γ.

Remark 2.3 Let be the set of all solutions of problem (2.1), (2.2). According to Definition 2.1, each function z S satisfies a transversality property, which means that the graph of z crosses a barrier γ at a unique point τ ( a , b ) , where the impulse acts on z. After that (for t ( τ , b ] ) the graph of z lies on the right of the barrier γ. This transversality property follows from transversality conditions (cf. (4.5), (4.6)) and it is proved in Section 5.

Assume that z 1 , z 2 S and z 1 z 2 . Then there exists a unique τ i ( a , b ) such that γ ( z i ( τ i ) ) = τ i for i = 1 , 2 and τ 1 τ 2 can occur. Therefore different functions from can have their discontinuities at different points from ( a , b ) . Our aim in this paper is to prove the existence of a solution of problem (2.1), (2.2) satisfying the general linear boundary condition (2.3). To do this, we need a suitable linear space containing . Due to state-dependent impulses, the Banach space of piece-wise continuous functions on [ a , b ] with the sup-norm cannot be used here. Therefore we choose the Banach space G L [ a , b ] . Clearly, by (1.1), S G L [ a , b ] . The operator in the general linear boundary condition (2.3) can be written uniquely in the form
( z ) = k z ( a ) + ( KS ) a b v ( t ) d [ z ( t ) ] ,
(2.5)

where k R , v BV [ a , b ] and a b ( KS ) is the Kurzweil-Stieltjes integral (cf. [14], Theorem 3.8). Representation (2.5) is correct on , because for each z G L [ a , b ] the integral a b ( KS ) v ( t ) d [ z ( t ) ] exists. Its definition and properties can be found in [15] (see Perron-Stieltjes integral based on the work of Kurzweil).

Definition 2.4 A function z : [ a , b ] R is a solution of problem (2.1)-(2.3) if z is a solution of problem (2.1), (2.2) and fulfils (2.3).

3 Green’s function

For further investigation, we will need a linear homogeneous problem corresponding to problem (2.1)-(2.3). Such problem has the form
z ( t ) = 0 ,
(3.1)
( z ) = 0 ,
(3.2)
because the impulse in (2.2) disappears if J 0 . We will also work with the non-homogeneous equation
z ( t ) = q ( t ) ,
(3.3)

where q L 1 [ a , b ] .

Definition 3.1 A solution of problem (3.3), (3.2) is a function z AC [ a , b ] satisfying equation (3.3) for a.e. t [ a , b ] and fulfilling condition (3.2).

Remark 3.2 If x is a solution of problem (3.3), (3.2), then x belongs to AC [ a , b ] , and consequently condition (3.2) can be written in the form (cf. (2.5))
( x ) = k x ( a ) + a b v ( t ) x ( t ) d t = 0 ,
(3.4)

where k R , v BV and the Lebesgue integral a b v ( t ) x ( t ) d t is used.

Definition 3.3 A function G : [ a , b ] × [ a , b ] R is the Green’s function of problem (3.1), (3.2) if
  1. (i)

    for any s ( a , b ) , the restrictions G ( , s ) | [ a , s ) , G ( , s ) | ( s , b ] are solutions of equation (3.1) and G ( s + , s ) G ( s , s ) = 1 , where G ( s , s ) = G ( s , s ) ;

     
  2. (ii)

    G ( t , ) BV [ a , b ] for any t [ a , b ] ;

     
  3. (iii)
    for any q L 1 [ a , b ] , the function
    x ( t ) = a b G ( t , s ) q ( s ) d s
    (3.5)
     

fulfils condition (3.4).

Lemma 3.4 Let be from (2.5) with k R and v BV [ a , b ] .
  1. (i)
    k 0 if and only if there exists the Green’s function G of problem (3.1), (3.2) which has the form
    G ( t , s ) = { v ( s ) k for  a t s b , 1 v ( s ) k for  a s < t b .
    (3.6)
     
  2. (ii)

    k 0 if and only if there exists a unique solution x of problem (3.3), (3.4), which has a form of (3.5) with G from (3.6).

     
Proof Clearly, G given by (3.6) fulfils (i) and (ii) of Definition 3.3 if and only if k 0 . A general solution of equation (3.3) is x ( t ) = c + a t q ( s ) d s , where c R . By (3.4),
( x ) = k c + a b v ( t ) q ( t ) d t = 0 .
The equation
k c = a b v ( t ) q ( t ) d t
has a unique solution c if and only if k 0 . Then a unique solution x of problem (3.3), (3.4) is written as
x ( t ) = 1 k a b v ( s ) q ( s ) d s + a t q ( s ) d s = a t ( 1 v ( s ) k ) q ( s ) d s + t b ( v ( s ) k ) q ( s ) d s , t [ a , b ] .

 □

Lemma 3.5 Let G be the Green’s function of problem (3.1), (3.2), where is from (2.5) and k 0 . Then, for each s [ a , b ) , the function G ( , s ) belongs to G L [ a , b ] and
( G ( , s ) ) = 0 , s [ a , b ) .
(3.7)
Proof Choose s [ a , b ) . By (3.6),
G ( t , s ) = χ ( s , b ] ( t ) v ( s ) k for  t [ a , b ] .
Consequently, the function G ( , s ) belongs to G L [ a , b ] . This yields that the integral a b ( KS ) v ( t ) d [ G ( t , s ) ] exists for each v BV [ a , b ] . Note that since G ( , s ) is not continuous on [ a , b ] , formula (3.4) cannot be used for G ( , s ) in place of x. Instead, we use the properties of the Kurzweil-Stieltjes integral which justify the following computation
a b ( KS ) v ( t ) d [ G ( t , s ) ] = a b ( KS ) v ( t ) d [ χ ( s , b ] ( t ) v ( s ) k ] = a b ( KS ) v ( t ) d [ χ ( s , b ] ( t ) ] ( KS ) a b v ( t ) d [ v ( s ) k ] = v ( s ) .
Hence, by (2.5), we get
( G ( , s ) ) = k G ( a , s ) + ( KS ) a b v ( t ) d [ G ( t , s ) ] = k ( v ( s ) k ) + v ( s ) = 0 .

 □

Example 3.6 Consider a solution x of problem (3.3), (3.2), where has a form of the two-point boundary condition
( x ) = α x ( a ) + β x ( b ) = 0 , α , β R .
(3.8)
We will show that can be expressed in a form of (3.4). If α + β 0 , then k and v can be found from the equality
α x ( a ) + β x ( b ) = k x ( a ) + a b v ( t ) x ( t ) d t .
Assuming that v ( t ) v 0 R , we get
α x ( a ) + β x ( b ) = k x ( a ) + v 0 ( x ( b ) x ( a ) ) ,
and hence k = α + β , v 0 = β . In addition, if α + β 0 , then (cf. (3.6))
G ( t , s ) = { β α + β for  a t s b , 1 β α + β for  a s < t b .
Example 3.7 Consider a solution x of problem (3.3), (3.2), where has a form of the multi-point boundary condition
( x ) = i = 0 n α i x ( t i ) , α i R , i = 0 , 1 , , n , n N .
(3.9)
Here a = t 0 < t 1 < < t n = b . If i = 0 n α i 0 , then k and v of (3.4) can be found from the equality
i = 0 n α i x ( t i ) = k x ( a ) + a b v ( t ) x ( t ) d t .
(3.10)
Assume that v is a piece-wise constant right-continuous function on [ a , b ] , that is,
v ( s ) = v i for  s [ t i , t i + 1 ) , i = 0 , , n 2 , v ( s ) = v n 1 for  s [ t n 1 , b ] ,
where v i R , i = 0 , , n 1 . By (3.10), we get
i = 0 n α i x ( t i ) = k x ( a ) + i = 0 n 1 v i t i t i + 1 x ( t ) d t = k x ( a ) + v 0 ( x ( t 1 ) x ( a ) ) + v 1 ( x ( t 2 ) x ( t 1 ) ) + + v n 1 ( x ( b ) x ( t n 1 ) ) .
Consequently,
v i = j = i + 1 n α j , i = 0 , , n 1 , k = j = 0 n α j .
To summarize, if j = 0 n α j 0 , then
v ( s ) = j = i + 1 n α j for  s [ t i , t i + 1 ) , i = 0 , , n 2 , v ( s ) = α n for  s [ t n 1 , b ] ,
and further (cf. (3.6))
G ( t , s ) = { v ( s ) j = 0 n α j for  a t s b , 1 v ( s ) j = 0 n α j for  a s < t b .
Example 3.8 Consider a solution x of problem (3.3), (3.2), where has a form of the integral condition
( x ) = x ( b ) a b h ( ξ ) x ( ξ ) d ξ ,
where h L 1 [ a , b ] . If a b h ( ξ ) d ξ 1 , then k and v of (3.4) can be found from the equality
x ( b ) a b h ( ξ ) x ( ξ ) d ξ = k x ( a ) + a b v ( t ) x ( t ) d t .
(3.11)
Let us put
v ( s ) = a s h ( ξ ) d ξ + v ( a ) .
Then
a b v ( ξ ) x ( ξ ) d ξ = a b h ( ξ ) x ( ξ ) d ξ + v ( b ) x ( b ) v ( a ) x ( a )
and (3.11) gives v ( a ) = k , a b h ( ξ ) d ξ + k = 1 . Consequently,
k = 1 a b h ( ξ ) d ξ , v ( s ) = 1 s b h ( ξ ) d ξ , s [ a , b ] .
Similarly, if
( x ) = x ( a ) a b h ( ξ ) x ( ξ ) d ξ ,
and a b h ( ξ ) d ξ 1 , we derive
k = 1 a b h ( ξ ) d ξ , v ( s ) = s b h ( ξ ) d ξ , s [ a , b ] .
In both cases, G is written as
G ( t , s ) = { v ( s ) 1 a b h ( ξ ) d ξ for  a t s b , 1 v ( s ) 1 a b h ( ξ ) d ξ for  a s < t b .

4 Assumptions

An existence result for problem (2.1)-(2.3) will be proved in the next sections under the basic assumption (2.4) and the following additional assumptions imposed on f, , and γ.
  1. (i)
    Boundedness of f
    { There exists  h L [ a , b ]  such that | f ( t , x ) | h ( t ) for a.e.  t [ a , b ]  and all  x R .
    (4.1)
     
  2. (ii)
    Boundedness of
    { There exists  J 0 ( 0 , )  such that | J ( t , x ) | J 0 for  t [ a , b ] , x R .
    (4.2)
     
  3. (iii)
    Boundedness of γ
    { There exist  a 1 , b 1 ( a , b )  such that a 1 γ ( x ) b 1 for  x [ K , K ] .
    (4.3)
     
  4. (iv)
    Properties of
     fulfils  (2.5) , where  k R , k 0 , v BV [ a , b ] C [ a 1 , b 1 ] .
    (4.4)
     
  5. (v)
    Transversality conditions
    | γ ( x ) | < 1 h for  x [ K , K ] ,
    (4.5)
     
{ either J ( t , x ) 0 , γ ( x ) 0 for  t [ a 1 , b 1 ] , x [ K , K ] , or J ( t , x ) 0 , γ ( x ) 0 for  t [ a 1 , b 1 ] , x [ K , K ] ,
(4.6)
where h is from (4.1) and a 1 , b 1 are from (4.3).
  1. (vi)
    L -continuity of f
    { For any  ε > 0 , there exists  δ > 0  such that | x y | < δ f ( , x ) f ( , y ) < ε , x , y [ K , K ] .
    (4.7)
     
Remark 4.1
  1. (a)

    Boundedness of f and can be replaced by more general conditions, for example, growth or sign ones, if the method of a priori estimates is used. See, e.g., [16, 17].

     
  2. (b)

    Continuity of v on [ a 1 , b 1 ] is necessary for the construction of a continuous operator in Section 6. Note that then we need t 1 , , t n 1 [ a 1 , b 1 ] in Example 3.7.

     
  3. (c)

    Clearly, if f is continuous on [ a , b ] × [ K , K ] , then f fulfils (4.7).

     
  4. (d)
    Let there exist p N , ψ L [ a , b ] and g i C ( R ) , i = 1 , , p , such that
    | f ( t , x ) f ( t , y ) | ψ ( t ) i = 1 p | g i ( x ) g i ( y ) |
     
for a.e. t [ a , b ] and all x , y [ K , K ] . Then f fulfils (4.7). An example of such a function f is
f ( t , x ) = i = 1 p f i ( t ) g i ( x ) + f 0 ( t ) ,

where f j L [ a , b ] , j = 0 , 1 , , p , g i C [ K , K ] , i = 1 , , p .

5 Transversality

Consider K ( 0 , ) , h L [ a , b ] and define a set by
B = { u W 1 , [ a , b ] : u < K , u < h } .
(5.1)

The following two lemmas for functions from are the modifications of lemmas in [10] and provide the transversality (cf. Remark 2.3) which will be essential for operator constructions in Section 6.

Lemma 5.1 Let γ satisfy (2.4), (4.3) and (4.5). Then, for each u B ¯ , there exists a unique τ ( a , b ) such that
τ = γ ( u ( τ ) ) .
(5.2)

In addition τ [ a 1 , b 1 ] .

Proof Let us take an arbitrary u B ¯ and denote
σ ( t ) = γ ( u ( t ) ) t , t [ a , b ] .
Then, by (2.4) and (5.1), we see that σ AC [ a , b ] and
σ ( t ) = γ ( u ( t ) ) u ( t ) 1 for a.e.  t [ a , b ] .
Since u ( a ) , u ( b ) [ K , K ] , condition (4.3) gives
σ ( a ) = γ ( u ( a ) ) a a 1 a > 0 , σ ( b ) = γ ( u ( b ) ) b b 1 b < 0 .
Consequently, there exists at least one zero of σ in ( a , b ) . Let τ ( a , b ) be a zero of σ. By virtue of (4.5) and (5.1), we get, for t [ a , b ] , t τ ,
sign ( t τ ) σ ( t ) = sign ( t τ ) τ t σ ( s ) d s = sign ( t τ ) τ t ( γ ( u ( s ) ) u ( s ) 1 ) d s sign ( t τ ) τ t ( | γ ( u ( s ) ) | u 1 ) d s < sign ( t τ ) τ t ( 1 h h 1 ) d s = 0 .
That is,
σ > 0 on  [ a , τ ) , σ < 0 on  ( τ , b ] .
(5.3)

Hence τ is a unique zero of σ, and (4.3) yields τ [ a 1 , b 1 ] . □

Due to Lemma 5.1, we can define a functional P : B ¯ [ a 1 , b 1 ] by
P u = τ ,
(5.4)

where τ fulfils (5.2).

Lemma 5.2 Let γ satisfy (2.4), (4.3) and (4.5). Then the functional is continuous.

Proof Let us choose a sequence { u n } n = 1 B ¯ which is convergent in W 1 , [ a , b ] . Then
u n W 1 , [ a , b ] , u n K , u n h , n N ,
(5.5)
and there exists u W 1 , [ a , b ] such that
lim n u n u 1 , = 0 .
(5.6)
So, by virtue of (1.5) and (5.5),
u lim n u u n + lim n u n K , u lim n u u n + lim n u n h .
We see that u B ¯ . For n N , define
σ n ( t ) = γ ( u n ( t ) ) t , σ ( t ) = γ ( u ( t ) ) t , t [ a , b ] .
By Lemma 5.1,
σ n ( τ n ) = 0 , σ ( τ ) = 0 , where  τ n = P u n , τ = P u , n N .
(5.7)
We need to prove that
lim n τ n = τ .
(5.8)
Conditions (2.4), (1.5) and (5.6) yield
lim n σ n = σ in  C [ a , b ] .
(5.9)

Let us take an arbitrary ε > 0 . By (5.3) and (5.9) we can find ξ ( τ ε , τ ) , η ( τ , τ + ε ) and n 0 N such that σ n ( ξ ) > 0 , σ n ( η ) < 0 for each n n 0 . By Lemma 5.1 and the continuity of σ n , we see that τ n ( ξ , η ) ( τ ε , τ + ε ) for n n 0 , and (5.8) follows. □

6 Fixed point problem

In this section we assume that
conditions  (2.4) , (4.1) - (4.7)  are fulfilled ,
(6.1)
and we construct a fixed point problem whose solvability leads to a solution of problem (2.1)-(2.3). To this aim, having the set from (5.1), we define a set Ω by
Ω = B × B W 1 , [ a , b ] × W 1 , [ a , b ] ,
(6.2)
and for u = ( u 1 , u 2 ) Ω , we define a function f u : [ a , b ] R as follows. We set, for a.e. t [ a , b ] ,
f u ( t ) = { f ( t , u 1 ( t ) ) if  t [ a , P u 1 ] , f ( t , u 2 ( t ) ) if  t ( P u 1 , b ] ,
(6.3)
where is defined by (5.4) and the point P u 1 [ a 1 , b 1 ] is uniquely determined due to Lemma 5.1. By (4.1)
f u L [ a , b ] , f u h .
(6.4)
Now, we can define an operator F : Ω ¯ W 1 , [ a , b ] × W 1 , [ a , b ] by F ( u 1 , u 2 ) = ( x 1 , x 2 ) , where
(6.5)
(6.6)
Here the functionals A 1 : Ω ¯ R and A 2 : Ω ¯ R are defined such that the functions x 1 and x 2 are continuous at the point P u 1 . Therefore
{ A 1 u = a b G ( P u 1 , s ) f u ( s ) d s a b G ( P u 1 , s ) f ( s , u 1 ( s ) ) d s , A 2 u = a b G ( P u 1 , s ) f u ( s ) d s a b G ( P u 1 , s ) f ( s , u 2 ( s ) ) d s .
(6.7)
Differentiating (6.5) and using (3.6) and (6.3), we get
x i ( t ) = f ( t , u i ( t ) ) for a.e.  t [ a , b ] , i = 1 , 2 .
(6.8)
This together with (4.1) yields
x i h , i = 1 , 2 .
(6.9)
Since v BV [ a , b ] (cf. (4.4)), we see that (6.4)-(6.6), (3.6), (4.1) and (4.2) give
x i 3 ( 1 + v | k | ) ( b a ) h + | c 0 | | k | + ( 1 + v | k | ) J 0 , i = 1 , 2 .
(6.10)

Due to (6.8)-(6.10), we see that x i W 1 , [ a , b ] , i = 1 , 2 , and the operator is defined well.

Lemma 6.1 Assume that (6.1) holds and that Ω and are given by (6.2) and (6.5), (6.6), respectively. Then the operator is compact on Ω ¯ .

Proof Step 1. We show that is continuous on Ω ¯ . Choose a sequence
{ u [ n ] } n = 1 = { ( u 1 [ n ] , u 2 [ n ] ) } n = 1 Ω ¯
which is convergent in W 1 , [ a , b ] × W 1 , [ a , b ] , that is, (cf. (1.5)) there exists u = ( u 1 , u 2 ) Ω ¯ such that
lim n u 1 [ n ] u 1 1 , = 0 , lim n u 2 [ n ] u 2 1 , = 0 .
(6.11)
Lemma 5.1 and Lemma 5.2 yield
P u 1 , P u 1 [ n ] [ a 1 , b 1 ] , n N , lim n P u 1 [ n ] = P u 1 ,
(6.12)
where is defined by (5.4). Denote
x = ( x 1 , x 2 ) = F ( u 1 , u 2 ) , x [ n ] = ( x 1 [ n ] , x 2 [ n ] ) = F ( u 1 [ n ] , u 2 [ n ] ) , n N .
(6.13)
We will prove that
lim n x 1 [ n ] x 1 1 , = 0 , lim n x 2 [ n ] x 2 1 , = 0 .
(6.14)
By (4.7), (6.8), (6.11) and (6.13),
lim n ( x i [ n ] ) x i = lim n f ( , u i [ n ] ( ) ) f ( , u i ( ) ) = 0 , i = 1 , 2 .
(6.15)
Using (4.1), we get
lim n | τ τ n | f ( s , u 1 [ n ] ( s ) ) f ( s , u 2 [ n ] ( s ) ) | d s | 2 lim n | τ τ n h ( s ) d s | = 0 .
(6.16)
Since
a b ( f u [ n ] ( s ) f u ( s ) ) d s = a τ ( f ( s , u 1 [ n ] ( s ) ) f ( s , u 1 ( s ) ) ) d s + τ b ( f ( s , u 2 [ n ] ( s ) ) f ( s , u 2 ( s ) ) ) d s + τ τ n ( f ( s , u 1 [ n ] ( s ) ) f ( s , u 2 [ n ] ( s ) ) ) d s ,
the Lebesgue dominated convergence theorem and (6.16) give
lim n a b | f u [ n ] ( s ) f u ( s ) | d s = 0 .
(6.17)
Using (6.13) and (6.5), we get
| x 1 [ n ] ( a ) x 1 ( a ) | a b | G ( a , s ) | | f u [ n ] ( s ) f u ( s ) | d s + | v ( P u 1 [ n ] ) k J ( P u 1 [ n ] , u 1 [ n ] ( P u 1 [ n ] ) ) v ( P u 1 ) k J ( P u 1 , u 1 ( P u 1 ) ) | .
The continuity and boundedness of , and v (cf. Lemma 5.2, (2.4), (4.2), (4.4) and (6.12)) imply
lim n | v ( P u 1 [ n ] ) k J ( P u 1 [ n ] , u 1 [ n ] ( P u 1 [ n ] ) ) v ( P u 1 ) k J ( P u 1 , u 1 ( P u 1 ) ) | v | k | lim n | J ( P u 1 [ n ] , u 1 [ n ] ( P u 1 [ n ] ) ) J ( P u 1 , u 1 ( P u 1 ) ) | + J 0 | k | lim n | v ( P u 1 [ n ] ) v ( P u 1 ) | = 0 ,
wherefrom, by the boundedness of G and (6.17),
lim n | x 1 [ n ] ( a ) x 1 ( a ) | = 0 .
(6.18)
Using (6.13) and integrating (6.8), we get
x 1 ( t ) = x 1 ( a ) + a t f ( s , u 1 ( s ) ) d s , x 1 [ n ] ( t ) = x 1 [ n ] ( a ) + a t f ( s , u 1 [ n ] ( s ) ) d s ,
and, due to (6.15) and (6.18), we arrive at
lim n x 1 [ n ] x 1 = 0 .
(6.19)
Similarly, we derive
lim n | x 2 [ n ] ( b ) x 2 ( b ) | = 0 , lim n x 2 [ n ] x 2 = 0 .
(6.20)

Properties (6.15), (6.19) and (6.20) yield (6.14).

Step 2. We show that the set F ( Ω ¯ ) is relatively compact in W 1 , [ a , b ] × W 1 , [ a , b ] . Choose an arbitrary sequence
{ ( x 1 [ n ] , x 2 [ n ] ) } n = 1 F ( Ω ¯ ) W 1 , [ a , b ] × W 1 , [ a , b ] .
We need to prove that there exists a convergent subsequence. Clearly, there exists { ( u 1 [ n ] , u 2 [ n ] ) } n = 1 Ω ¯ such that
F ( u 1 [ n ] , u 2 [ n ] ) = ( x 1 [ n ] , x 2 [ n ] ) , n N .
Choose i { 1 , 2 } . By (5.1) and (6.2), it holds
{ u i [ n ] } n = 1 W 1 , [ a , b ] , u i [ n ] K , | u i [ n ] ( t 1 ) u i [ n ] ( t 2 ) | = | t 1 t 2 ( u i [ n ] ) ( s ) d s | h | t 1 t 2 |
for t 1 , t 2 [ a , b ] , n N . Therefore, the Arzelà-Ascoli theorem yields that there exists a subsequence
{ ( u 1 [ m ] , u 2 [ m ] ) } m = 1 { ( u 1 [ n ] , u 2 [ n ] ) } n = 1
which converges in C [ a , b ] × C [ a , b ] . Consequently, for each ε > 0 , there exists m 0 N such that for each m N ,
m m 0 u i [ m 0 ] u i [ m ] < ε , i = 1 , 2 .
Similarly as in Step 1, we prove (cf. (6.15), (6.19), (6.20))
( x i [ m 0 ] ) ( x i [ m ] ) < ε , x i [ m 0 ] x i [ m ] < ε , i = 1 , 2 ,

which gives by (1.5) that { ( x 1 [ m ] , x 2 [ m ] ) } m = 1 is convergent in W 1 , [ a , b ] × W 1 , [ a , b ] . □

Remark 6.2 If there exists τ 0 [ a 1 , b 1 ] such that γ ( x ) = τ 0 for x [ K , K ] , then problem (2.1)-(2.3) has an impulse at fixed time τ 0 and a standard operator F 0 , acting on the space of piece-wise continuous functions on [ a , b ] and having the form
( F 0 z ) ( t ) = a b G ( t , s ) f ( s , z ( s ) ) d s + c 0 k + G ( t , τ 0 ) J ( τ 0 , z ( τ 0 ) ) , t [ a , b ] ,
(6.21)

can be used instead of the operator from (6.5), (6.6). But this is not possible if γ is not constant on [ K , K ] . The reason is that then an impulse is realized at a state-dependent point τ = γ ( z ( τ ) ) , and F 0 with τ instead of τ 0 should be investigated on the space G L [ a , b ] . But if we write a state-dependent τ instead of a fixed τ 0 in (6.21), F 0 loses its continuity on G L [ a , b ] , which we show in the next example.

Example 6.3 Let a = 0 , b = 2 and be from (2.5) with k R , k 0 and v C [ 0 , 2 ] . Consider the functions
u ( t ) = 1 , u n ( t ) = 1 1 n , t [ 0 , 2 ] , n N .
Clearly, u n u uniformly on [ 0 , 2 ] and hence
lim n u n u = 0 .
For n N , denote x n = F 0 u n and x = F 0 u . Assume that the barrier γ is given by the linear function γ ( x ) = x on and the impulse function J ( t , x ) = 1 for t [ 0 , 2 ] , x R . Then
τ = γ ( u ( τ ) ) = u ( τ ) = 1 , τ n = γ ( u n ( τ n ) ) = u n ( τ n ) = 1 1 n , n N ,
and, according to (6.21), we have for t [ 0 , 2 ]
x n ( t ) = 0 2 G ( t , s ) f ( s , 1 1 n ) d s + c 0 k + G ( t , 1 1 n ) , n N , x ( t ) = 0 2 G ( t , s ) f ( s , 1 ) d s + c 0 k + G ( t , 1 ) .
Consequently,
lim n ( x n ( 1 ) x ( 1 ) ) = lim n 0 2 G ( 1 , s ) ( f ( s , 1 1 n ) f ( s , 1 ) ) d s + lim n ( G ( 1 , 1 1 n ) G ( 1 , 1 ) ) = 1 v ( 1 ) k ( v ( 1 ) k ) = 1

due to (3.6). Hence x n ( 1 ) x ( 1 ) and we have also x n x 0 , and F 0 is not continuous on G L [ 0 , 2 ] .

Lemma 6.1 results in the following theorem.

Theorem 6.4 Assume that (6.1) holds and that the set Ω is given by (6.2), where
K ( 1 + v | k | ) ( 3 ( b a ) h + J 0 ) + | c 0 | | k | .
(6.22)

Further, let the operator be given by (6.5), (6.6). Then has a fixed point in Ω ¯ .

Proof By Lemma 6.1, is compact on Ω ¯ . Due to (5.1), (6.2), (6.5), (6.6), (6.10) and (6.22),
F ( Ω ¯ ) Ω ¯ .

Therefore, the Schauder fixed point theorem yields a fixed point of in Ω ¯ . □

7 Main result

The main result, which is contained in Theorem 7.1, guarantees the solvability of problem (2.1)-(2.3) provided the data functions f, and γ are bounded (cf. (4.1)-(4.3)). As it is mentioned in Remark 4.1, Theorem 7.1 serves as an existence principle which, in combination with the method of a priori estimates, can lead to more general existence results for unbounded f and and concrete boundary conditions.

Theorem 7.1 Assume that (6.1) and (6.22) hold. Then there exists a solution z of problem (2.1)-(2.3) such that
z K .
(7.1)
Proof By Theorem 6.4, there exists u = ( u 1 , u 2 ) Ω ¯ which is a fixed point of the operator defined in (6.5) and (6.6). This means that
(7.2)
(7.3)
where G, , f u , A 1 , A 2 are given by (3.6), (5.4), (6.3), (6.7), respectively. Recall that P u 1 is a unique point in ( a , b ) satisfying
P u 1 = τ 1 [ a 1 , b 1 ] , where  τ 1 = γ ( u 1 ( τ 1 ) ) .
(7.4)
For t [ a , b ] , define a function z by
z ( t ) = { u 1 ( t ) if  t [ a , τ 1 ] , u 2 ( t ) if  t ( τ 1 , b ] .
(7.5)
Differentiating (7.2), (7.3) and using (3.6) and (6.3), we get u i ( t ) = f ( t , u i ( t ) ) for a.e. t [ a , b ] , i = 1 , 2 , and consequently
z ( t ) = f ( t , z ( t ) ) for a.e.  t [ a , b ] .
By virtue of (7.2)-(7.5), we have
z ( τ 1 + ) z ( τ 1 ) = u 2 ( τ 1 ) u 1 ( τ 1 ) = J ( τ 1 , u 1 ( τ 1 ) ) = J ( τ 1 , z ( τ 1 ) ) .
(7.6)
Let us show that τ 1 is a unique solution of the equation
t = γ ( z ( t ) )
(7.7)
in [ a , b ] . According to (7.4) and (7.5), it suffices to prove
t γ ( u 2 ( t ) ) , t ( τ 1 , b ] .
(7.8)
Since ( u 1 , u 2 ) Ω ¯ , we have (cf. (5.1) and (6.2))
u i K , u i h , i = 1 , 2 .
Assume that the first condition in (4.6) is fulfilled. Then J ( τ 1 , x ) 0 , γ ( x ) 0 for x [ K , K ] . Put
σ ( t ) = γ ( u 2 ( t ) ) t , t [ a , b ] .
By (7.6), u 2 ( τ 1 ) u 1 ( τ 1 ) = J ( τ 1 , u 1 ( τ 1 ) ) 0 , and since γ is non-increasing, we have
σ ( τ 1 ) = γ ( u 2 ( τ 1 ) ) τ 1 γ ( u 1 ( τ 1 ) ) τ 1 = 0
due to (7.4). Using (4.5), we derive for t ( τ 1 , b ]
σ ( t ) = τ 1 t ( γ ( u 2 ( s ) ) u 2 ( s ) 1 ) d s τ 1 t ( | γ ( u 2 ( s ) ) | u 2 1 ) d s < τ 1 t ( 1 h h 1 ) d s = 0 .

So, (7.8) is valid. If the second condition in (4.6) is fulfilled, we use the dual arguments.

Finally, let us check that ( z ) = c 0 . By (7.2)-(7.6) and (3.6), we have
z ( t ) = a b G ( t , s ) f ( s , z ( s ) ) d s + c 0 k + G ( t , τ 1 ) J ( τ 1 , z ( τ 1 ) ) .
(7.9)
Put
x ( t ) = a b G ( t , s ) f ( s , z ( s ) ) d s .
(7.10)
Then, according to (iii) of Definition 3.3 and Remark 3.2, we get ( x ) = 0 . Further, using (3.7) from Lemma 3.5, we arrive at ( G ( , τ 1 ) ) = 0 . Consequently, due to (2.5), (7.9) and (7.10), ( z ) results in
( z ) = ( x ) + ( c 0 k ) + ( G ( , τ 1 ) ) J ( τ 1 , z ( τ 1 ) ) = ( c 0 k ) = k c 0 k + ( KS ) a b v ( t ) d [ c 0 k ] = c 0 .

 □

Declarations

Acknowledgements

This research was supported by the grant Matematické modely, PrF_2013_013. The authors thank the referees for suggestions which improved the paper.

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, Palacký University, 17. listopadu 12, Olomouc, 77146, Czech Republic

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