First-order nonlinear differential equations with state-dependent impulses

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1 Rchůne nd Rchůnová Boundry Vlue Problems 2013, 2013:195 R E S E A R C H Open Access First-order nonliner differentil equtions with stte-dependent impulses Luáš Rchůne nd Iren Rchůnová * * Correspondence: iren.rchunov@upol.cz Deprtment of Mthemtics, Fculty of Science, Plcý University, 17. listopdu 12, Olomouc, 77146, Czech Republic Abstrct The pper dels with the stte-dependent impulsive problem z t)=ft, zt)) for.e. t [, b], zτ +) zτ )=Jτ, zτ )), γ zτ )) = τ, lz)=c 0, where [, b] R, c 0 R, f fulfils the Crthéodory conditions on [, b] R,the impulse function J is continuous on [, b] R, the brrier function γ hs continuous first derivtive on some subset of R nd l is liner bounded functionl which is defined on the Bnch spce of left-continuous regulted functions on [, b] equipped with the sup-norm. The functionl l is represented by mens of the Kurzweil-Stieltjes integrl nd covers ll liner boundry conditions for solutions of first-order differentil equtions subject to stte-dependent impulse conditions. Here, sufficient nd effective conditions gurnteeing the solvbility of the bove problem re presented for the first time. MSC: 34B37; 34B15 Keywords: first-order ODE; stte-dependent impulses; trnsverslity conditions; generl liner boundry conditions; existence; Kurzweil-Stieltjes integrl 1 Introduction The investigtion of impulsive differentil equtions hs long history; see, e.g.,themonogrphs [1 3]. Most ppers deling with impulsive differentil equtions subject to boundry conditions focus their ttention on impulses t fixed moments. But this is very prticulr cse of more complicted cse with stte-dependent impulses. Boundry vlue problems with stte-dependent impulses, where difficulties with n opertor representtionpper cf. Remr 6.2), resubstntillylessdeveloped. Werefertotheppers [4 6] nd [7] which re devoted to periodic problems, nd for problems with other boundry conditions, see [8, 9]or[10 12]. Here, in our pper, we present n pproch leding to new existence principle for impulsive boundry vlue problems. This pproch is pplicble to ech liner boundry condition which is considered with some first-order differentil eqution subject to sttedependent impulses. The importnt step is proof of trnsverslity Remr 2.3 nd Lemms 5.1 nd 5.2), which mes possible construction of continuous opertor Section 6) whose fixed point leds to solution of our originl impulsive problem Section 7) Rchůne nd Rchůnová; licensee Springer. This is n Open Access rticle distributed under the terms of the Cretive Commons Attribution License which permits unrestricted use, distribution, nd reproduction in ny medium, provided the originl wor is properly cited.

2 Rchůne nd Rchůnová Boundry Vlue Problems 2013, 2013:195 Pge 2 of 18 Nottion Let M R n, n N,[, b] R. CM) is the set of rel functions continuous on M. ACM) is the set of rel functions bsolutely continuous on M. L 1 [, b] is the set of rel functions Lebesgue integrble on [, b]. L [, b] is the set of rel functions essentilly bounded on [, b]. BV[, b] is the set of rel functions with bounded vrition on [, b]. G L [, b] is the set of rel left-continuous regulted functions on [, b],thtis, z G L [, b] if nd only if z: [, b] R,ndforechτ 1, b] nd ech τ 2 [, b), zτ 1 )=zτ 1 ) = lim t τ 1 zt), zτ 2+) = lim zt) R. 1.1) t τ 2 + Cr[, b] M) is the set of functions f :[, b] M R such tht i) f, x): [, b] R is mesurble for ll x M, ii) f t, ): M R is continuous for.e. t [, b], iii) for ech compct set Q M,thereexistsm Q L 1 [, b] stisfying f t, x) mq t) for.e. t [, b] nd ech x Q. ThesetL [, b] equipped with the norm z = sup ess { zt) : t [, b] } for z L [, b] 1.2) is Bnch spce. SinceC[, b] G L [, b] L [, b],weequipthesetsc[, b] nd G L [, b] with the norm nd get lso Bnch spces cf. [13]). Then 1.2) cn be written s z = sup { zt) : t [, b] } for z G L [, b] 1.3) nd z = mx { zt) : t [, b] } for z C[, b]. 1.4) W 1, [, b] is the Bnch spce of functions z: [, b] R such tht z AC[, b] nd z L [, b], where the norm 1, is given by z 1, = z + z for z W 1, [, b]. 1.5) χ A is the chrcteristic function of set A,whereA R. 2 Formultion of problem We investigte the solvbility of the nonliner differentil eqution z t)=f t, zt) ) 2.1) subject to the stte-dependent impulse condition zτ+) zτ)=j τ, zτ) ), γ zτ) ) = τ, 2.2)

3 Rchůne nd Rchůnová Boundry Vlue Problems 2013, 2013:195 Pge 3 of 18 nd the generl liner boundry condition lz)=c ) Here we ssume tht f Cr[, b] R), J C[, b] R), [, b] R, K 0, ), γ C 1 [ K, K], c 0 R, 2.4) nd l: G L [, b] R is liner bounded functionl. Definition 2.1 Afunctionz: [, b] R is solution of problem 2.1), 2.2)if thereexistsuniqueτ, b) such tht γ zτ)) = τ ; therestrictionsz [,τ] nd z τ,b] re bsolutely continuous; zτ+) = zτ)+j τ, zτ)); z stisfies eqution 2.1) for.e. t [, b]. Definition 2.2 Agrphoffunctionγ :[ K, K] R is clled brrier γ. Remr 2.3 Let S be the set of ll solutions of problem 2.1), 2.2). According to Definition 2.1, echfunctionz S stisfies trnsverslity property, which mens tht the grph of z crosses brrier γ t unique point τ, b), where the impulse J cts on z. After tht for t τ, b]) the grph of z lies on the right of the brrier γ.thistrnsverslity property follows from trnsverslity conditions cf. 4.5), 4.6)) nd it is proved in Section 5. Assume tht z 1, z 2 S nd z 1 z 2. Then there exists unique τ i, b) suchtht γ z i τ i )) = τ i for i =1,2ndτ 1 τ 2 cn occur. Therefore different functions from S cn hve their discontinuities t different points from, b). Our im in this pper is to prove the existence of solution of problem 2.1), 2.2) stisfying the generl liner boundry condition 2.3). To do this, we need suitble liner spce contining S. Dueto stte-dependent impulses, the Bnch spce of piece-wise continuous functions on [, b] with the sup-norm cnnot be used here. Therefore we choose the Bnch spce G L [, b]. Clerly, by 1.1), S G L [, b]. The opertor l in the generl liner boundry condition 2.3)cnbewrittenuniquelyintheform lz)=z)+ KS) vt)d [ zt) ], 2.5) where R, v BV[, b] nd KS) is the Kurzweil-Stieltjes integrl cf. [14], Theorem 3.8). Representtion 2.5) is correct on S, becuseforechz G L [, b] theintegrl KS) vt)d[zt)] exists. Its definition nd properties cn be found in [15] seeperron- Stieltjes integrl bsed on the wor of Kurzweil). Definition 2.4 Afunctionz: [, b] R is solution of problem 2.1)-2.3)ifzis solution of problem 2.1), 2.2)ndfulfils2.3).

4 Rchůne nd Rchůnová Boundry Vlue Problems 2013, 2013:195 Pge 4 of Green s function For further investigtion, we will need liner homogeneous problem corresponding to problem 2.1)-2.3). Such problem hs the form z t)=0, 3.1) lz)=0, 3.2) becuse the impulse in 2.2) disppersifj 0. We will lso wor with the nonhomogeneous eqution z t)=qt), 3.3) where q L 1 [, b]. Definition 3.1 A solution of problem 3.3), 3.2) is function z AC[, b]stisfyingeqution 3.3) for.e. t [, b] nd fulfilling condition 3.2). Remr 3.2 If x is solution of problem 3.3), 3.2), then x belongs to AC[, b], nd consequently condition 3.2)cnbewrittenintheformcf. 2.5)) lx)=x)+ vt)x t)dt =0, 3.4) where R, v BV nd the Lebesgue integrl vt)x t)dt is used. Definition 3.3 AfunctionG: [, b] [, b] R is the Green s function of problem 3.1), 3.2)if i) for ny s, b),therestrictionsg, s) [,s), G, s) s,b] re solutions of eqution 3.1) nd Gs+, s) Gs, s) =1,whereGs, s) =Gs, s); ii) Gt, ) BV[, b] for ny t [, b]; iii) for ny q L 1 [, b],thefunction xt)= fulfils condition 3.4). Gt, s)qs)ds 3.5) Lemm 3.4 Let l be from 2.5) with R nd v BV[, b]. i) 0if nd only if there exists the Green s function G of problem 3.1), 3.2) which hs the form Gt, s)= vs) 1 vs) for t s b, for s < t b. 3.6) ii) 0if nd only if there exists unique solution x of problem 3.3), 3.4), which hs form of 3.5) with G from 3.6).

5 Rchůne nd Rchůnová Boundry Vlue Problems 2013, 2013:195 Pge 5 of 18 Proof Clerly, G given by 3.6) fulfils i) nd ii) of Definition 3.3 if nd only if 0.Agenerl solution of eqution 3.3)isxt)=c + t qs)ds,wherec R.By3.4), lx)=c + The eqution vt)qt)dt =0. c = vt)qt)dt hs unique solution c if nd only if 0. Then unique solution x of problem 3.3), 3.4) is written s xt)= 1 = t vs)qs)ds + 1 vs) t ) qs)ds + qs)ds t vs) ) qs)ds, t [, b]. Lemm 3.5 Let G be the Green s function of problem 3.1), 3.2),where l isfrom 2.5) nd 0.Then, for ech s [, b), the function G, s) belongs to G L [, b] nd l G, s) ) =0, s [, b). 3.7) Proof Choose s [, b). By 3.6), Gt, s)=χ s,b] t) vs) for t [, b]. Consequently, the function G, s) belongs to G L [, b]. This yields tht the integrl KS) vt)d[gt, s)] exists for ech v BV[, b]. Note tht since G, s) is not continuous on [, b], formul 3.4) cnnot be used for G, s)inplceofx. Insted, we use the properties of the Kurzweil-Stieltjes integrl which justify the following computtion KS) vt)d [ Gt, s) ] [ = KS) vt)d χ s,b] t) vs) ] = KS) vt)d [ χ s,b] t) ] [ ] vs) KS) vt)d = vs). Hence, by 2.5), we get l G, s) ) = G, s)+ KS) vt)d [ Gt, s) ] ) vs) = + vs)=0. Exmple 3.6 Consider solution x of problem 3.3), 3.2), where l hs form of the twopoint boundry condition lx) =αx) +βxb) =0, α, β R. 3.8)

6 Rchůne nd Rchůnová Boundry Vlue Problems 2013, 2013:195 Pge 6 of 18 We will show tht l cn be expressed in form of 3.4). If α + β 0,then nd v cn be found from the equlity αx)+βxb)=x)+ Assuming tht vt) v 0 R,weget vt)x t)dt. αx)+βxb)=x)+v 0 xb) x) ), nd hence = α + β, v 0 = β. In ddition, if α + β 0,thencf. 3.6)) β for t s b, α+β Gt, s)= 1 β for s < t b. α+β Exmple 3.7 Consider solution x of problem 3.3), 3.2), where l hs form of the multipoint boundry condition lx)= n α i xt i ), α i R, i =0,1,...,n, n N. 3.9) i=0 Here = t 0 < t 1 < < t n = b. If n i=0 α i 0,then nd v of 3.4) cnbefoundfromthe equlity n α i xt i )=x)+ i=0 vt)x t)dt. 3.10) Assume tht v is piece-wise constnt right-continuous function on [, b], tht is, vs)=v i for s [t i, t i+1 ), i =0,...,n 2, vs)=v n 1 for s [t n 1, b], where v i R, i =0,...,n 1.By3.10), we get n n 1 ti+1 α i xt i )=x)+ v i x t)dt i=0 i=0 t i = x)+v 0 xt1 ) x) ) + v 1 xt2 ) xt 1 ) ) + + v n 1 xb) xtn 1 ) ). Consequently, n v i = α j, i =0,...,n 1, = j=i+1 n α j. j=0 To summrize, if n j=0 α j 0,then vs)= n α j for s [t i, t i+1 ), i =0,...,n 2, j=i+1 vs)=α n for s [t n 1, b],

7 Rchůne nd Rchůnová Boundry Vlue Problems 2013, 2013:195 Pge 7 of 18 nd further cf. 3.6)) Gt, s)= vs) nj=0 α j for t s b, 1 vs) nj=0 α j for s < t b. Exmple 3.8 Consider solution x of problem 3.3), 3.2), where l hs form of the integrl condition lx)=xb) hξ)xξ)dξ, where h L 1 [, b]. If hξ)dξ 1,then nd v of 3.4)cnbefoundfromtheequlity xb) Let us put Then vs)= s hξ)xξ)dξ = x)+ hξ)dξ + v). vt)x t)dt. 3.11) vξ)x ξ)dξ = hξ)xξ)dξ + vb)xb) v)x) nd 3.11)givesv)=, hξ)dξ + =1.Consequently, =1 Similrly, if hξ)dξ, vs)=1 s hξ)dξ, s [, b]. lx)=x) hξ)xξ)dξ, nd hξ)dξ 1,wederive =1 hξ)dξ, vs)= s hξ)dξ, s [, b]. In both cses, G is written s Gt, s)= 1 vs) 1 hξ)dξ vs) 1 hξ)dξ for t s b, for s < t b.

8 Rchůne nd Rchůnová Boundry Vlue Problems 2013, 2013:195 Pge 8 of Assumptions An existence result for problem 2.1)-2.3) will be proved in the next sections under the bsic ssumption 2.4) nd the following dditionl ssumptions imposed on f, l, J nd γ. i) Boundedness of f There exists h L [, b] such tht 4.1) f t, x) ht) for.e. t [, b] nd ll x R. ii) Boundedness of J There exists J 0 0, ) such tht J t, x) J 0 for t [, b], x R. iii) Boundedness of γ There exist 1, b 1, b) such tht 1 γ x) b 1 for x [ K, K]. 4.2) 4.3) iv) Properties of l l fulfils 2.5), where R, 0,v BV[, b] C[ 1, b 1 ]. 4.4) v) Trnsverslity conditions γ x) 1 < for x [ K, K], 4.5) h either J t, x) 0, γ x) 0 for t [ 1, b 1 ], x [ K, K], 4.6) or J t, x) 0, γ x) 0 for t [ 1, b 1 ], x [ K, K], where h is from 4.1) nd 1, b 1 re from 4.3). vi) L -continuity of f For ny ε >0,thereexistsδ >0such tht x y < δ f, x) f, y) < ε, x, y [ K, K]. 4.7) Remr 4.1 ) Boundedness of f nd J cn be replced by more generl conditions, for exmple, growth orsignones, ifthemethodofprioriestimtes is used. See, e.g.,[16, 17]. b) Continuity of v on [ 1, b 1 ] is necessry for the construction of continuous opertor in Section 6. Note tht then we need t 1,...,t n 1 / [ 1, b 1 ] in Exmple 3.7. c) Clerly, if f is continuous on [, b] [ K, K], then f fulfils 4.7). d) Let there exist p N, ψ L [, b] nd g i CR), i =1,...,p,suchtht p f t, x) f t, y) ψt) gi x) g i y) i=1

9 Rchůne nd Rchůnová Boundry Vlue Problems 2013, 2013:195 Pge 9 of 18 for.e. t [, b] nd ll x, y [ K, K].Thenf fulfils 4.7). An exmple of such function f is f t, x)= p f i t)g i x)+f 0 t), i=1 where f j L [, b], j =0,1,...,p, g i C[ K, K], i =1,...,p. 5 Trnsverslity Consider K 0, ), h L [, b]nddefinesetb by B = { u W 1, [, b]: u < K, u < h }. 5.1) The following two lemms for functions from B re the modifictions of lemms in [10] nd provide the trnsverslity cf. Remr 2.3) which will be essentil for opertor constructions in Section 6. Lemm 5.1 Let γ stisfy 2.4), 4.3) nd 4.5). Then, for ech u B, there exists unique τ, b) such tht τ = γ uτ) ). 5.2) In ddition τ [ 1, b 1 ]. Proof Let us te n rbitrry u B nd denote σ t)=γ ut) ) t, t [, b]. Then, by 2.4)nd5.1), we see tht σ AC[, b]nd σ t)=γ ut) ) u t) 1 for.e.t [, b]. Since u), ub) [ K, K], condition 4.3) gives σ )=γ u) ) 1 >0, σ b)=γ ub) ) b b 1 b <0. Consequently, there exists t lest one zero of σ in, b). Let τ, b) bezeroofσ.by virtue of 4.5)nd5.1), we get, for t [, b], t τ, signt τ)σ t) =signt τ) signt τ) < signt τ) t τ t τ t τ σ s)ds = signt τ) t γ us) ) u 1 ) ds ) 1 h 1 ds =0. h τ γ us) ) u s) 1 ) ds

10 Rchůne nd Rchůnová Boundry Vlue Problems 2013, 2013:195 Pge 10 of 18 Tht is, σ >0 on[, τ), σ <0 onτ, b]. 5.3) Hence τ is unique zero of σ,nd4.3)yieldsτ [ 1, b 1 ]. Due to Lemm 5.1, we cn define functionl P : B [ 1, b 1 ]by Pu = τ, 5.4) where τ fulfils 5.2). Lemm 5.2 Let γ stisfy 2.4), 4.3) nd 4.5). Then the functionl P is continuous. Proof Let us choose sequence {u n } n=1 B which is convergent in W1, [, b]. Then u n W 1, [, b], u n K, u n h, n N, 5.5) nd there exists u W 1, [, b]suchtht lim u n u 1, =0. 5.6) So, by virtue of 1.5)nd5.5), u lim u u n + lim u n K, u lim u u n + lim u n h. We see tht u B.Forn N,define σ n t)=γ u n t) ) t, σ t)=γ ut) ) t, t [, b]. By Lemm 5.1, σ n τ n )=0, σ τ)=0, whereτ n = Pu n, τ = Pu, n N. 5.7) We need to prove tht lim τ n = τ. 5.8) Conditions 2.4), 1.5)nd5.6)yield lim σ n = σ in C[, b]. 5.9) Let us te n rbitrry ε >0.By5.3)nd5.9)wecnfindξ τ ε, τ), η τ, τ + ε)nd n 0 N such tht σ n ξ)>0,σ n η)<0forechn n 0. By Lemm 5.1 nd the continuity of σ n,weseethtτ n ξ, η) τ ε, τ + ε)forn n 0,nd5.8) follows.

11 Rchůne nd Rchůnová Boundry Vlue Problems 2013, 2013:195 Pge 11 of Fixed point problem In this section we ssume tht conditions 2.4), 4.1)-4.7) refulfilled, 6.1) nd we construct fixed point problem whose solvbility leds to solution of problem 2.1)-2.3). To this im, hving the set B from 5.1), we define set by = B B W 1, [, b] W 1, [, b], 6.2) nd for u =u 1, u 2 ), wedefinefunctionf u :[, b] R s follows. We set, for.e. t [, b], f t, u 1 t)) if t [, Pu 1 ], f u t)= 6.3) f t, u 2 t)) if t Pu 1, b], where P is defined by 5.4) ndthepointpu 1 [ 1, b 1 ] is uniquely determined due to Lemm 5.1.By4.1) f u L [, b], f u h. 6.4) Now, we cn define n opertor F : W 1, [, b] W 1, [, b] byfu 1, u 2 )=x 1, x 2 ), where Gt, s)f us)ds + c 0 vpu 1) J Pu 1, u 1 Pu 1 )) if t P u1, x 1 t)= Gt, s)f s, u 1s)) ds + c 0 vpu 1) J Pu 1, u 1 Pu 1 )) + A 1 u if t > P u1, Gt, s)f s, u 2s)) ds + c 0 +1 vpu 1) )J Pu 1, u 1 Pu 1 )) + A 2 u if t P u1, x 2 t)= Gt, s)f us)ds + c 0 +1 vpu 1) )J Pu 1, u 1 Pu 1 )) if t > P u1. 6.5) 6.6) Here the functionls A 1 : R nd A 2 : R re defined such tht the functions x 1 nd x 2 re continuous t the point Pu 1. Therefore A 1 u = GPu 1, s)f u s)ds GPu 1, s)f s, u 1 s)) ds, A 2 u = GPu 1, s)f u s)ds GPu 1, s)f s, u 2 s)) ds. 6.7) Differentiting 6.5)ndusing3.6)nd6.3), we get x i t)=f t, u i t) ) for.e. t [, b], i =1,2. 6.8)

12 Rchůne nd Rchůnová Boundry Vlue Problems 2013, 2013:195 Pge 12 of 18 This together with 4.1)yields x i h, i =1,2. 6.9) Since v BV[, b]cf. 4.4)), we see tht 6.4)-6.6), 3.6), 4.1)nd4.2)give x i 3 1+ v + 1+ v ) b ) h + c 0 ) J 0, i =1, ) Due to 6.8)-6.10), we see tht x i W 1, [, b], i = 1, 2, nd the opertor F is defined well. Lemm 6.1 Assume tht 6.1) holds nd tht nd F re given by 6.2) nd 6.5), 6.6), respectively. Then the opertor F is compct on. Proof Step 1. We show tht F is continuous on.choosesequence { u } n=1 = { u 1, u )} 2 n=1 which is convergent in W 1, [, b] W 1, [, b], tht is, cf. 1.5)) there exists u =u 1, u 2 ) such tht lim u 1 u 1, 1 =0, lim u 2 u 1, 2 = ) Lemm 5.1 nd Lemm 5.2 yield Pu 1, Pu 1 [ 1, b 1 ], n N, lim Pu 1 = Pu 1, 6.12) where P is defined by 5.4). Denote x =x 1, x 2 )=Fu 1, u 2 ), x = x 1, ) x 2 = F u 1, u ) 2, n N. 6.13) We will prove tht lim x 1 x 1, 1 =0, lim x 2 x 1, 2 = ) By 4.7), 6.8), 6.11)nd6.13), lim x ) i x i = lim f, u i ) ) f, u i ) ) =0, i = 1, ) Using 4.1), we get lim τn τ f s, u 1 s) ) f s, u 2 s)) ds 2 lim τn τ hs)ds = )

13 Rchůne nd Rchůnová Boundry Vlue Problems 2013, 2013:195 Pge 13 of 18 Since fu s) f u s) ) ds = τ f s, u 1 s)) f s, u 1 s) )) ds + + τ τn τ f s, u 2 s)) f s, u 2 s) )) ds f s, u 1 s)) f s, u 2 s))) ds, the Lebesgue dominted convergence theorem nd 6.16)give lim fu s) f u s) ds = ) Using 6.13)nd6.5), we get x 1 ) x 1) G, s) f u s) f u s) ds + vpu 1 ) J Pu 1, u 1 Pu 1 )) vpu 1 ) J Pu 1, u 1 Pu 1 ) ). The continuity nd boundedness of P, J nd v cf. Lemm 5.2,2.4), 4.2), 4.4)nd6.12)) imply lim vpu 1 ) J Pu 1, u 1 v lim J Pu 1, u 1 Pu + J 0 lim v Pu ) 1 vpu1 ) =0, )) vpu 1 ) 1 J Pu 1, u 1 Pu 1 ) ) Pu )) 1 J Pu1, u 1 Pu 1 ) ) wherefrom, by the boundedness of G nd 6.17), lim x 1 ) x 1) = ) Using 6.13) nd integrting 6.8), we get x 1 t)=x 1 )+ t f s, u 1 s) ) ds, x 1 t)=x 1 )+ t f s, u 1 s)) ds, nd, due to 6.15)nd6.18), we rrive t lim 1 x 1 = ) x Similrly, we derive lim 2 b) x 2b) =0, lim 2 x 2 = ) x x Properties 6.15), 6.19)nd6.20)yield6.14).

14 Rchůne nd Rchůnová Boundry Vlue Problems 2013, 2013:195 Pge 14 of 18 Step 2. We show tht the set F ) is reltively compct in W 1, [, b] W 1, [, b]. Choose n rbitrry sequence { x 1, )} x 2 n=1 F ) W1, [, b] W 1, [, b]. We need to prove tht there exists convergent subsequence. Clerly, there exists {u 1, u 2 )} n=1 such tht F u 1, u ) 2 = x 1, ) x 2, n N. Choose i {1, 2}.By5.1)nd6.2), it holds { u } i n=1 W1, [, b], u i t 1 ) u i t 2 ) = t2 t 1 u i K, u ) s)ds i h t 1 t 2 for t 1, t 2 [, b], n N. Therefore, the Arzelà-Ascoli theorem yields tht there exists subsequence { u [m] 1, u [m] )} 2 m=1 { u 1, u )} 2 n=1 which converges in C[, b] C[, b]. Consequently, for ech ε >0,thereexistsm 0 N such tht for ech m N, m m 0 u [m 0 ] i u [m] i < ε, i =1,2. Similrly s in Step 1, we prove cf. 6.15), 6.19), 6.20)) x [m ) 0] i x [m]) i < ε, [m x 0 ] i x [m] < ε, i =1,2, i which gives by 1.5)tht{x [m] 1, x [m] 2 )} m=1 is convergent in W1, [, b] W 1, [, b]. Remr 6.2 If there exists τ 0 [ 1, b 1 ]suchthtγx)=τ 0 for x [ K, K], then problem 2.1)-2.3) hs n impulse t fixed time τ 0 nd stndrd opertor F 0,ctingonthespce of piece-wise continuous functions on [, b] nd hving the form F 0 z)t)= Gt, s)f s, zs) ) ds + c 0 + Gt, τ 0)J τ 0, zτ 0 ) ), t [, b], 6.21) cn be used insted of the opertor F from 6.5), 6.6). But this is not possible if γ is not constnt on [ K, K]. The reson is tht then n impulse is relized t stte-dependent point τ = γ zτ)), nd F 0 with τ insted of τ 0 should be investigted on the spce G L [, b]. But if we write stte-dependent τ insted of fixed τ 0 in 6.21), F 0 loses its continuity on G L [, b], which we show in the next exmple.

15 Rchůne nd Rchůnová Boundry Vlue Problems 2013, 2013:195 Pge 15 of 18 Exmple 6.3 Let =0,b =2ndl be from 2.5) with R, 0ndv C[0, 2]. Consider the functions ut)=1, u n t)=1 1, t [0, 2], n N. n Clerly, u n u uniformly on [0, 2] nd hence lim u n u =0. For n N,denotex n = F 0 u n nd x = F 0 u. Assume tht the brrier γ is given by the liner function γ x)=x on R nd the impulse function J t, x)=1fort [0, 2], x R.Then τ = γ uτ) ) = uτ)=1, τ n = γ u n τ n ) ) = u n τ n )=1 1 n, n N, nd, ccording to 6.21), we hve for t [0, 2] 2 x n t)= Gt, s)f s,1 1 ) ds + c 0 n + G t,1 1 ), n N, n xt)= Consequently, Gt, s)f s,1)ds + c 0 + Gt,1). lim xn 1) x1) ) = lim lim =1 v1) G1, s) f s,1 1 ) ) f s,1) ds n G 1, 1 1 ) ) G1, 1) n v1) ) =1 due to 3.6). Hence x n 1) x1) nd we hve lso x n x 0, nd F 0 is not continuous on G L [0, 2]. Lemm 6.1 results in the following theorem. Theorem 6.4 Assume tht 6.1) holds nd tht the set is given by 6.2), where K 1+ v ) 3b ) c 0 ) h + J ) Further, let the opertor F be given by 6.5), 6.6). Then F hs fixed point in. Proof By Lemm 6.1, F is compct on.dueto5.1), 6.2), 6.5), 6.6), 6.10)nd6.22), F ). Therefore, the Schuder fixed point theorem yields fixed point of F in.

16 Rchůne nd Rchůnová Boundry Vlue Problems 2013, 2013:195 Pge 16 of Min result The min result, which is contined in Theorem 7.1, gurntees the solvbility of problem 2.1)-2.3) provided the dt functions f, J nd γ re bounded cf. 4.1)-4.3)). As it is mentioned in Remr 4.1,Theorem7.1 serves s n existence principle which, in combintion with the method of prioriestimtes, cn led to more generl existence results for unbounded f nd J nd concrete boundry conditions. Theorem 7.1 Assume tht 6.1) nd 6.22) hold. Then there exists solution z of problem 2.1)-2.3) such tht z K. 7.1) Proof By Theorem 6.4,thereexistsu =u 1, u 2 ) which is fixed point of the opertor F defined in 6.5)nd6.6). This mens tht Gt, s)f us)ds + c 0 vpu 1) J Pu 1, u 1 Pu 1 )) if t P u1, u 1 t)= Gt, s)f s, u 1s)) ds + c 0 vpu 1) J Pu 1, u 1 Pu 1 )) + A 1 u if t > P u1, Gt, s)f s, u 2s)) ds + c 0 +1 vpu 1) )J Pu 1, u 1 Pu 1 )) + A 2 u if t P u1, u 2 t)= Gt, s)f us)ds + c 0 +1 vpu 1) )J Pu 1, u 1 Pu 1 )) if t > P u1, 7.2) 7.3) where G, P, f u, A 1, A 2 re given by 3.6), 5.4), 6.3), 6.7), respectively. Recll tht Pu 1 is uniquepointin, b)stisfying Pu 1 = τ 1 [ 1, b 1 ], where τ 1 = γ u 1 τ 1 ) ). 7.4) For t [, b], define function z by u 1 t) ift [, τ 1 ], zt)= u 2 t) ift τ 1, b]. 7.5) Differentiting 7.2), 7.3) ndusing3.6) nd6.3), we get u i t) =f t, u it)) for.e. t [, b], i = 1, 2, nd consequently z t)=f t, zt) ) for.e. t [, b]. Byvirtueof 7.2)-7.5), we hve zτ 1 +) zτ 1 )=u 2 τ 1 ) u 1 τ 1 )=J τ 1, u 1 τ 1 ) ) = J τ 1, zτ 1 ) ). 7.6) Let us show tht τ 1 is unique solution of the eqution t = γ zt) ) 7.7)

17 Rchůne nd Rchůnová Boundry Vlue Problems 2013, 2013:195 Pge 17 of 18 in [, b]. According to 7.4) nd7.5), it sufficesto prove t γ u 2 t) ), t τ 1, b]. 7.8) Since u 1, u 2 ),wehvecf. 5.1)nd6.2)) u i K, u i h, i =1,2. Assume tht the first condition in 4.6) is fulfilled. Then J τ 1, x) 0, γ x) 0forx [ K, K]. Put σ t)=γ u 2 t) ) t, t [, b]. By 7.6), u 2 τ 1 ) u 1 τ 1 )=J τ 1, u 1 τ 1 )) 0, nd since γ is non-incresing, we hve 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] t σ t)= γ u 2 s) ) t u 2 s) 1) ds γ u 2 s) ) u 2 1 ) ds τ 1 τ 1 t ) 1 < h 1 ds =0. h τ 1 So, 7.8) is vlid. If the second condition in 4.6) is fulfilled, we use the dul rguments. Finlly, let us chec tht lz)=c 0.By7.2)-7.6)nd3.6), we hve Put zt)= xt)= Gt, s)f s, zs) ) ds + c 0 + Gt, τ 1)J τ 1, zτ 1 ) ). 7.9) Gt, s)f s, zs) ) ds. 7.10) Then, ccording to iii) of Definition 3.3 nd Remr 3.2, we get lx) = 0. Further, using 3.7) from Lemm 3.5, we rrive t lg, τ 1 )) = 0. Consequently, due to 2.5), 7.9) nd 7.10), lz)resultsin lz)=lx)+l = l c0 c0 ) + l G, τ 1 ) ) J τ 1, zτ 1 ) ) ) = c 0 + KS) vt)d [ c0 ] = c 0. Competing interests The uthors declre tht they hve no competing interests. Authors contributions Both uthors contributed eqully to the mnuscript nd red nd pproved the finl mnuscript.

18 Rchůne nd Rchůnová Boundry Vlue Problems 2013, 2013:195 Pge 18 of 18 Acnowledgements This reserch ws supported by the grnt Mtemticé modely, PrF_2013_013. The uthors thn the referees for suggestions which improved the pper. Received: 13 My 2013 Accepted: 13 August 2013 Published: 28 August 2013 References 1. Binov, D, Simeonov, P: Impulsive Differentil Equtions: Periodic Solutions nd Applictions. Longmn, Hrlow 1993) 2. Lshminthm, V, Binov, DD, Simeonov, PS: Theory of Impulsive Differentil Equtions. World Scientific, Singpore 1989) 3. Smoileno, AM, Perestyu, NA: Impulsive Differentil Equtions. World Scientific, Singpore 1995) 4. Bjo, I, Liz, E: Periodic boundry vlue problem for first order differentil equtions with impulses t vrible times. J. Mth. Anl. Appl. 204, ) 5. Belley, J, Virgilio, M: Periodic Duffing dely equtions with stte dependent impulses. J. Mth. Anl. Appl. 306, ) 6. Belley, J, Virgilio, M: Periodic Liénrd-type dely equtions with stte-dependent impulses. Nonliner Anl. TMA 64, ) 7. Frigon, M, O Regn, D: First order impulsive initil nd periodic problems with vrible moments. J. Mth. Anl. Appl. 233, ) 8. Benchohr, M, Gref, JR, Ntouys, SK, Ouhb, A: Upper nd lower solutions method for impulsive differentil inclusions with nonliner boundry conditions nd vrible times. Dyn. Contin. Discrete Impuls. Syst., Ser. A, Mth. Anl. 12, ) 9. Frigon, M, O Regn, D: Second order Sturm-Liouville BVP s with impulses t vrible times. Dyn. Contin. Discrete Impuls. Syst., Ser. A, Mth. Anl. 8, ) 10. Rchůnová, I, Tomeče, J: A new pproch to BVPs with stte-dependent impulses. Bound. Vlue Probl. 2013, ) 11. Rchůnová, I, Tomeče, J: Second order BVPs with stte-dependent impulses vi lower nd upper functions. Cent. Eur. J. Mth. to pper) 12. Rchůnová, I, Tomeče, J: Existence principle for BVPs with stte-dependent impulses. Topol. Methods Nonliner Anl. submitted) 13. Tvrdý, M: Liner integrl equtions in the spce of regulted functions. Mth. Bohem. 123, ) 14. Tvrdý, M: Regulted functions nd the Perron-Stieltjes integrl. Čs. Pěst. Mt. 114, ) 15. Schwbi, Š, Tvrdý, M, Vejvod, O: Differentil nd Integrl Equtions. Acdemi, Prgue 1979) 16. Rchůnová, I, Tomeče, J: Singulr Dirichlet problem for ordinry differentil eqution with impulses. Nonliner Anl. TMA 65, ) 17. Rchůnová, I, Tomeče, J: Impulsive BVPs with nonliner boundry conditions for the second order differentil equtions without growth restrictions. J. Mth. Anl. Appl. 292, ) doi: / Cite this rticle s: Rchůne nd Rchůnová: First-order nonliner differentil equtions with stte-dependent impulses. Boundry Vlue Problems :195.