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First-order nonlinear differential equations with state-dependent impulses

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.

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 , nN, [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]×MR such that

  1. (i)

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

  2. (ii)

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

  3. (iii)

    for each compact set QM, there exists m Q L 1 [a,b] satisfying

    |f(t,x)| m Q (t)for a.e. t[a,b] and each xQ.
  • 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 zC[a,b].
(1.4)
  • W 1 , [a,b] is the Banach space of functions z:[a,b]R such that zAC[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 AR.

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 zS 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)=kz(a) + ( KS ) a b v(t)d [ z ( t ) ] ,
(2.5)

where kR, vBV[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 J0. 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 zAC[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)=kx(a)+ a b v(t) x (t)dt=0,
(3.4)

where kR, vBV and the Lebesgue integral a b v(t) x (t)dt 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)ds
    (3.5)

fulfils condition (3.4).

Lemma 3.4 Let be from (2.5) with kR and vBV[a,b].

  1. (i)

    k0 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)

    k0 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 k0. A general solution of equation (3.3) is x(t)=c+ a t q(s)ds, where cR. By (3.4),

(x)=kc+ a b v(t)q(t)dt=0.

The equation

kc= a b v(t)q(t)dt

has a unique solution c if and only if k0. 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 k0. 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 vBV[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 ) ) =kG(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)=αx(a)+βx(b)=0,α,βR.
(3.8)

We will show that can be expressed in a form of (3.4). If α+β0, then k and v can be found from the equality

αx(a)+βx(b)=kx(a)+ a b v(t) x (t)dt.

Assuming that v(t) v 0 R, we get

αx(a)+βx(b)=kx(a)+ v 0 ( x ( b ) x ( a ) ) ,

and hence k=α+β, v 0 =β. In addition, if α+β0, 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,nN.
(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 )=kx(a)+ a b v(t) x (t)dt.
(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,,n1. 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,,n1,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)dξ,

where h L 1 [a,b]. If a b h(ξ)dξ1, then k and v of (3.4) can be found from the equality

x(b) a b h(ξ)x(ξ)dξ=kx(a)+ a b v(t) x (t)dt.
(3.11)

Let us put

v(s)= a s h(ξ)dξ+v(a).

Then

a b v(ξ) x (ξ)dξ= a b h(ξ)x(ξ)dξ+v(b)x(b)v(a)x(a)

and (3.11) gives v(a)=k, a b h(ξ)dξ+k=1. Consequently,

k=1 a b h(ξ)dξ,v(s)=1 s b h(ξ)dξ,s[a,b].

Similarly, if

(x)=x(a) a b h(ξ)x(ξ)dξ,

and a b h(ξ)dξ1, we derive

k=1 a b h(ξ)dξ,v(s)= s b h(ξ)dξ,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 kR,k0,vBV[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 pN, 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 σAC[a,b] and

σ (t)= γ ( u ( t ) ) u (t)1for 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], tτ,

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>0on [a,τ),σ<0on (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

Pu=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 ,nN,
(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 nN, 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=Pu,nN.
(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 =σin C[a,b].
(5.9)

Let us take an arbitrary ε>0. By (5.3) and (5.9) we can find ξ(τε,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)(τε,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 vBV[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 ],nN, 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 ] ) ,nN.
(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 ) ) | ds|2 lim n | τ τ n h(s)ds|=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)|ds=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 ) ) ds, x 1 [ n ] (t)= x 1 [ n ] (a)+ a t f ( s , u 1 [ n ] ( s ) ) ds,

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 ] ) ,nN.

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], nN. 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 mN,

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 ) ) ds+ 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 kR, k0 and vC[0,2]. Consider the functions

u(t)=1, u n (t)=1 1 n ,t[0,2],nN.

Clearly, u n u uniformly on [0,2] and hence

lim n u n u =0.

For nN, 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], xR. 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

tγ ( 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 ) ) ds+ 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 ) ) ds.
(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 .

 □

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Acknowledgements

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

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Rachůnek, L., Rachůnková, I. First-order nonlinear differential equations with state-dependent impulses. Bound Value Probl 2013, 195 (2013). https://doi.org/10.1186/1687-2770-2013-195

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