STRUCTURE OF THE SOLUTION SET TO DIFFERENTIAL INCLUSIONS WITH IMPULSES AT VARIABLE TIMES

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1 Elecronic Journal of Differenial Equaions, Vol ), No. 114, pp ISSN: URL: hp://ejde.mah.xsae.edu or hp://ejde.mah.un.edu fp ejde.mah.xsae.edu STRUCTURE OF THE SOLUTION SET TO DIFFERENTIAL INCLUSIONS WITH IMPULSES AT VARIABLE TIMES AGATA GRUDZKA, SEBASTIAN RUSZKOWSKI Absrac. A opological srucure of he soluion se o differenial inclusions wih impulses a variable imes is invesigaed. In order o do ha an appropriae Banach space is defined. I is shown ha he soluion se is an R δ -se. Resuls are new also in he case of differenial equaions wih impulses a variable imes. 1. Inroducion Impulsive differenial equaions and inclusions have a lo of applicaions in diverse fields. The momens of impulses can be chosen in various ways: randomly, fixed beforehand, deermined by he sae of a sysem. The problems wih fixed ime of impulses were recenly invesigaed [6,7,9,10,12,17] and he resuls considering he srucure of soluion ses were summarised in [11]. The problems wih impulses a variable imes bring much more difficulies and up o now here were only exisence heorems [3,4,5]. Our resuls develop his research area. We need o use sophisicaed assumpions ha would guaranee fixed amoun of impulses and hen we need o deal wih he esimaions more carefully han in case wih fixed imes of jumps. In he resul we show ha he soluion se is an R δ -se. I is worh menioning ha his resul is new also in he case of differenial equaion wih impulses a variable imes ha has no uniqueness of soluions and herefor has a nonrivial soluion se. There are many moivaions o sudy he srucure of soluion ses of differenial equaions and inclusions. One of hem is considering he Poincaré ranslaion operaor and discussing he problem of he exisence of periodic soluions [16,8,14]. We have o have he space of funcions which conains soluions of given problem o sudy he srucure of soluion se. Obviously sandard Banach space wih he supremum norm is insufficien o impulsive problem wih he imes of jumps ha depend on he sae. B-opology on spaces of soluions of impulsive differenial inclusions is inroduced in [1], however, i is only Hausdorff opology. We use his concep o creae Banach space ha have he same opology on common funcions, and is sufficien o he considered problem Mahemaics Subjec Classificaion. 34A37, 34A60, 34K45. Key words and phrases. Soluion se; impulsive differenial inclusions; variable imes; R δ -se; opological srucure. c 2015 Texas Sae Universiy - San Marcos. Submied Ocober 22, 2014 Published April 28,

2 2 A. GRUDZKA, S. RUSZKOWSKI EJDE-2015/114 In Secion 2 we describe he problem, inroduce he suiable Banach space CJ m [0, a]) and recall useful heorems. In Secion 3 we presen he main resuls of he paper. The main idea is included in Theorem 3.1 in which we show ha he soluion se for he problem wih exacly one jump is an R δ -se. In he Theorem 3.2 we are using he resul from previous Theorem proving by inducion analogical saemen for any fixed number of jumps. We also provide he reader wih a ransparen example. We sudy he problem 2. Preliminaries ẏ) F, y)), for [0, a], τ j y)), j = 1,..., m, y0) = y 0, y + ) = y) + I j y)), for = τ j y)), j = 1,..., m, 2.1) where F : [0, a] R N R N, I j : R N R N, j = 1,..., m, are given impulse funcions, τ j C 1 R N, R) wih 0 < τ j y) < a, and y = { = τ k y))}. The hypersurface τ j y) = 0 is called he j-h pulse hypersurface and we denoe i by Σ j. If for each j = 1,..., m, τ j is a differen consan funcion, hen impulses are in he fixed imes. Our goal is o find he srucure of he soluion se of he previous problem, bu o do ha we need a space of funcions wih m jumps. We inroduce considering space as CJ m [0, a]) := C[0, a]) R R N ) m wih following inerpreaion: he elemen ϕ, l j, v j ) m j=1 ), where l j [0, a] we will inerpre as he funcion wih m jumps in he imes j k defined as follows: ϕ), 0 l σ1), ˆϕ) := ϕ) + j v σi), l σj) < l σj+1), i=1 ϕ) + m v σi), l σm) < a, i=1 where σ is a permuaion of {1, 2,..., m} such ha l σi) l σi+1). There is a muual correspondence beween he funcions on inerval [0, a] wih m jumps and he ses {ϕ, l j, v j ) m j=1 ) CJ m[0, a]) : l j < l j+1 }, wih ζ ˇζ, l j, I j ˇζl j ))) m j=1 ), where he funcion ˇζ is ζ wih reduced jumps, l j is j-h ime of jump and he funcion I j is an impulse funcions. The space CJ m [0, a]) wih he norm m ϕ, l j, v j ) m j=1) := sup [0,a] ϕ) + l j + v j ) is a Banach space. We will find ou ha he waned srucure is R δ -ype. To show ha we will use he following well-known heorems: Theorem 2.1 [15]). Le X be an absolue neighbourhood rerac and A X be a compac nonempy subse. Then he following saemens are equivalen: a) A is an R δ -se, b) for every ɛ > 0 he se A is conracible in O ɛ A) = {x X disx, A) < ɛ}, j=1

3 EJDE-2015/114 STRUCTURE OF THE SOLUTION SET 3 c) A is an inersecion of a decreasing sequence {A n } of compac conracible spaces, d) A is an inersecion of a decreasing sequence {A n } of closed conracible spaces, such ha βa n ) 0, where β is he Hausdorff measure of noncompacness. A mulimap F : X E, where E is real Banach space, is called upper hemiconinuous uhc) if for every funcional p E he funcion X x σ F x) p) := sup y F x) p, y R {+ } is upper semiconinuous usc). Theorem 2.2 Convergence heorem [2]). Le E and E be Banach spaces, le he space T, Ω, µ) be a measurable space, and he mulivalued map F : T E E has closed and convex values and for a.e. T he map F, ) : E E is uhc. Le u n : T E) be a sequence of funcions such ha u n u in L p T, E) and le sequence w n ) L p T, E ), 1 p < be such ha w n w in L p T, E ). If for a.e. T and for arbirary ε > 0, here exiss N N such ha w n ) cl conv BF, Bu n ), ε)), ε) for n > N, hen w) F, u)) for a.e. T. We recall Arzela- Ascoli Theorem: Theorem 2.3. If he family F C[a, b], R N ) of coninuous funcions is equiconinuous and uniformly bounded, hen here exiss a subsequence ha converges uniformly. A piecewise absoluely coninuous funcion y : [0, a] R N is a soluion of he problem wih impulses 2.1) if: a) y0) = y 0, b) here exiss a funcion f L 1 [0, a], R N ) such ha f) F, y)) for a.e. [0, a] and y) = y 0 + m j=1 I jy j )) + 0 fs)ds, where j = τ j y j )), c) he funcion y is lef coninuous a = τ j y)) [0, a] and he limi y + ) exiss and y + ) = y) + I j y)) for = τ j y)), j = 1,..., m. 3. Srucure of he soluion se We assume he following condiions on he mulivalued perurbaion F : [0, a] R N R N : F0) F has compac and convex values, F1) F, y) : [0, a] R N has a measurable selecion for every y R N, F2) F is almos uniformly wih respec o H-usc, i.e. for every y R N and ε > 0 here exiss δ > 0 such ha for a.e. [0, a] and for all x R N if y x < δ, hen sup ϕ F,x) dϕ, F, y)) < ε, F3) F has a sublinear growh, i.e., here exiss α L 1 [0, a]) such ha sup ϕ α)1 + y ) for a.e. [0, a] and y R N. ϕ F,y) Moreover, we assume he following hypoheses abou impulse funcions: H1) I j CR N, R N ), j = 1,..., m,

4 4 A. GRUDZKA, S. RUSZKOWSKI EJDE-2015/114 H2) τ j C 1 R N, R), j = 1,..., m, for j = 1,..., m 1 we have: 0 < τ j y) < τ j+1 y) < a, τ j y + I j y)) τ j y) < τ j+1 y + I j y)), τ m y + I m y)) τ m y), and here exiss a consan M 0 such ha τ j y) M for all y R N, j = 1,..., m, H3) here exiss a consan p > 0 such ha for a.e. [0, a] sup τ jy) ϕ 1 p < 0 for all y R N, j = 1,..., m. ϕ F,y) Noe ha if τ j y) = 0 for j = 1,..., m, hen he problem is reduced o a problem wih a fixed impulse ime. Assumpion F2) and compac values of he mulivalued map F implies ha F, ) is usc. Theorem 3.1. Le he assumpions F0) F3) hold, and H1)-H3) hold for m = 1. Then every soluion of he problem 2.1), where m = 1, mees Σ 1 exacly once and he soluion se S of his problem is an R δ -se in he space CJ 1 [0, a]). Proof. To simplify noaion we wrie I and τ insead of I 1 and τ 1. We will proceed in several seps. Sep 1. A Lipschiz selecion and he uniqueness of jump. For each n le {By, r n y))} y R N be an open covering open balls, such ha r n y) 1 n 0) of he space R N, such ha for every x By, r n y)) we have sup dϕ, F, y)) < 1 ϕ F,x) n. 3.1) There exiss locally finie open poin-sar refinemen U n = {U n,s } s S of he cover {By, r n y))} y R N, i.e. for every y R N here exiss x y,n R N such ha sy, U n ) Bx y,n, r n x y,n )). We can choose i in a way ha U n+1 is a refinemen of he cover U n. Le {λ s } s S be a locally Lipschiz pariion of uniy subordinaed o he cover U n i.e. for every s S he funcion λ s : [0, a] R N saisfies he locally Lipschiz condiion. For every y s R N, s S le a funcion q s be a measurable selecion of F, y s ). We define he funcion g n : [0, a] R N R N in he following way g n, y) := s S λ s y) q s ). The se Sy) := {s S λ s y) 0} is finie. If λ s y) > 0, so s Sy), hen y supp λ s U n,s sy, U n ). There exiss x y,n R N such ha sy, U n ) Bx y,n, r n x y,n )). We know ha y s sy, U n ). We obain g n, y) = s S λ s y) q s ) cnv F, sy, U n )) Moreover, G n, y) := cl conv F, sy, U n )). G n, y) cl conv F, Bx y,n, r n x y,n ))),

5 EJDE-2015/114 STRUCTURE OF THE SOLUTION SET 5 so from he inequaliy 3.1) we have We have G n, y) cl O 1/n F, x y,n )). 3.2) F, y) n 1 G n, y). From usc he map F has compac values) we obain ha for every y R N and for every ε > 0 here exiss δ > 0 such ha We have We obain We have F, By, δ)) O ε F, y)). n 1 G n, y) n 1 cl O 1/n F, x y,n )) n 1 cl O 1/n F, By, 2r n x y,n )))) n 1 cl O 1 n +ε F, y)) = F, y). F, y) = n 1 G n, y). G n+1, y) G n, y). Le he Nemiskiĭ subsiuion) operaor P Gn : CJ 1 [0, a]) L 1 J, R N ) be defined by P Gn y) := {φ L 1 [0, a], R N ) φ) G n, y)) Le S n denoe he se of soluions of he problem ẏ) G n, y)), for [0, a], τy)), y0) = y 0, y + ) = y) + Iy)), for = τy)). I is obvious ha he ses S n are nonempy, because he problem ẏ) = g n, y)), for [0, a], τy)), y0) = y 0, y + ) = y) + Iy)), for = τy)), for a.e. [0, a]}. 3.3) 3.4) for every n N has exacly one soluion. Sep 1a. We denoe by j y n he ime of j-h jump for he funcion y n a nd if he funcion y n has less ha j jumps we ake j y n = a. Le y n be an arbirary soluion of he sysem 3.3) for 0 2 y n. For 1 y n we obain he following form of he soluion y n ) = y φ n s)ds, where φ n P Gn y n ). There exiss selecion f n no necessarily measurable) of he mulivalued map F, x y,n )) such ha for a.e. we obain φ n ) f n, x y,n )) 1 n. We have φ n ) 1 n + f n, x y,n )). From he assumpion F 3) we obain f n, x y,n )) α) 1 + x y,n ) ) α) 1 + y) + 1 ). n

6 6 A. GRUDZKA, S. RUSZKOWSKI EJDE-2015/114 So y n ) y 0 + From Gronwall inequaliy, we have sup y n ) [0, 1 yn ] 0 y 0 + αs) 1 + y n s) + 1 ) + 1 ) ds. n n a 0 2αs)ds + a n )e R a 0 αs)ds := K. 3.5) By he coninuiy of I here exiss a consan c > 0 such ha Iy n 1 y n )) c for all n. Nex for 1 y n < 2 y n we obain y n ) y 0 + Iy n 1 y n )) + y 0 + c + Again, from Gronwall inequaliy we obain 0 0 φ n s) ds αs) ) n + y ns) + 1 n ) ds. sup y n ) Ce R 0 αs)ds < Ce R a 0 αs)ds =: K, 3.6) [0, 2 yn ] where C := y 0 + c + a 0 2αs)ds + a n. If he soluion y n does no have jumps, hen of course sup [0,a] y n ) K. Sep 1b. Le ȳ be a fixed funcion wih values in cl B0, K). Le φ n G n, ȳ)), where is such ha H3) is saisfied. We denoe ȳ) =: y. From he assumpions H2) and H3) for some v cl B0, 1) we obain ) τ y) φ n 1 = τ x y,n ) + τ y) τ x y,n ) φ n 1 = τ x y,n ) ϕ + 1 ) n v 1 + τ y) τ x y,n )) φ n where ϕ F, x y,n ). Hence = τ x y,n ) ϕ n τ x y,n ) v + τ y) τ x y,n )) φ n, τ y) φ n 1 p + τ x y,n ) n + τ y) τ x y,n )) φ n p + M n + τ y) τ x y,n ) φ n. The funcion τ is coninuous on he compac se cl B0, K + 1), herefore i is uniformly coninuous. Hence, for y cl B0, K), τ y) τ x y,n ) 0 we have x y,n K + 1 n ). Moreover, he se {φ n} is bounded from he sublinear growh of F ). We have τ y) τ x y,n )) φ n 0. We ake N 0 N such ha for every n N 0 we have p + M n + τ y) τ x y,n ) φ n < p 2. There exiss a consan p > 0 such ha τ y) φ n 1 < p. Sep 1c. Le us fix y n, where n > N 0, he soluion of he problem 3.3). We define he funcion w n : [0, a] R by w n ) := τy n )).

7 EJDE-2015/114 STRUCTURE OF THE SOLUTION SET 7 The funcion w n has value 0 in any ime, in which he funcion y n has a jump. By he condiion H2) wih m = 1, we obain w n 0) = τy 0 ) > 0 and w n a) = τy n a)) a < a a = 0. If w n ) 0 on [0, a], hen here would no be any impulse effec, herefore here would no be any jump ime, so w n would be coninuous, which would conradic wih he earlier inequaliies. Hence every soluion of he problem 3.3) has a leas one jump. Suppose ha 0 < 1 y n < a is he firs ime in which he soluion y n his he hypersurface Σ 1. Then w n 1 y n ) = 0 and w n ) > 0, for [0, 1 y n ). By assumpion H2) wih m = 1 we obain ha w n 1 y n + ) = τyn 1 y n + )) 1 yn = τ y n 1 y n ) + Iy n 1 y n )) ) 1 y n 0. For a.e. 1 y n we have w n) = τ y n )) y n) 1 = τ y n )) φ n ) 1 < p < 0, where φ n P Gn y n ). The funcion w n in [ 1 y n, a] is decreasing, hence y n his he hypersurface Σ 1 exacly once and he ime of his jump we denoe yn. Sep 2. Now we show ha each sequence y n ), where y n S n, has a convergen subsequence o he soluion ỹ of he problem 2.1). There exiss exacly one jump, so from he previous esimaions we have y n ) K. Consequenly, he values of soluions of he problem 2.1) are conained in a ball cl B0, K), which is convex, so in paricular we know ha funcion g n [0,a] cl B0, K) has inegrable Lipschiz consan Λ. For < yn we have y n ) y n ) = φ n s)ds and for yn < < we have 1 n K) y n ) y n ) = φ n s)ds 1 n K) αs) ) n + y ns) ds αs)ds, αs) ) n + y ns) ds αs)ds. 3.7) 3.8) Sep 2.a Le us consider convergence o he ime, where he limi of a convergen subsequence ynk ) of he sequence yn ) wih y n S n ), which exiss due

8 8 A. GRUDZKA, S. RUSZKOWSKI EJDE-2015/114 o he compacness of [0, a]. For every ε > 0 here exiss N 0 such ha for n k > N 0 we have ε < ynk. Noe ha +δ sup [0, ε δ] αs)ds 0 wih δ 0. From inequaliy 3.7) we obain ha for all ξ > 0 and 0 < < ε here exiss δ > 0 such ha for all n k > N 0 and < < +δ ε we have y nk ) y nk ) +2+K) αs)ds < ξ. Therefore, he family {y nk } nk >N 0 is equiconinuous and by he inequaliy 3.5) uniformly bounded. By Arzela-Ascoli Theorem 2.3 possibly going o he subsequences) we can assume ha y nk y ε on [0, ε], where y ε is coninuous funcion. This can be done in such a way ha for any ε 1, ε 2 > 0 such ha ε 1 > ε 2 funcions y ε1, y ε2 fulfil condiion y ε2 [0, ε 1] = y ε1. For ε 0 we obain an exension of he funcion y ε, i.e. he funcion y : [0, ) R N, where y nk converges poinwise o y. Moreover ) φ nk ) α) 1 + x ynk,n k ) + 1 α)2 + n K) ) k We know ha: φ nk ) cl conv F, sy nk ), U nk )) clo 1/nk F inclusion 3.2), y nk ) y ε ) a.e. on [0, ε], φ nk L 1 [0, ε], R N ), )), By nk ), 1 n k ), by by esimaion 3.9) and weak compacness of he closed ball we obain φ nkl φ on [0, ε]. Thus by Theorem 2.2 we obain φ) F, y ε )) for a.e. [0, ε]. By analogy, we conclude ha φ) F, y )) a.e. on [0, ). By weak convergence φ nkl φ on [0, ε], for Ψφ n ) := 0 φ ns)ds we have 0 φs)ds = Ψφ) = lim k Ψφ n k ) = lim k = lim k y n k ) y 0 = y ) y 0. 0 φ nk s)ds For an increasing sequence s n ) convergen o wih n < n we obain y s n ) y s n ) = sn s n φs)ds s n αs)2 + K) + 1)ds. We have convergence of he righ hand side of he inequaliy o 0 wih n, consequenly y s n )) is Cauchy sequence. I is convergen R N is complee) and we denoe is limi y s n ) y 1. We define y ) := y 1 and we obain coninuous exension y on [0, ]. We will show, ha τy )) = 0, which means ha is ime of jump for y. Le ɛ > 0. The funcion αs)2 + K) + 1 is inegrable, so we can choose ɛ < so ha ɛ 2[αs)2 + K) + 1]ds < ɛ 2.

9 EJDE-2015/114 STRUCTURE OF THE SOLUTION SET 9 There exiss K 0 N such ha for k > K 0 we have y ɛ ) y nk ɛ ) < ɛ 2. We can esimae y nk ynk ) y ) y nk ɛ ) + ynk ɛ y nk ɛ ) y ɛ ) + y nk ɛ ) y ɛ ) + ɛ 2 + φ nk s)ds y ɛ ) ynk ɛ φ nk s) ds + ynk ɛ 2αs)2 + K) + 1)ds ɛ. For ynk > we obain ɛ φs)ds ɛ φs) ds ɛ αs)2 + K) + 1 ) ds + ɛ αs)2 + K) + 1 ) ds y ) y nk ynk ) y ) y ε) + y ε) y nk ε) bu i is easy o see ha + y nk ε) y nk ynk ), so ynk ynk y nk ε) y nk ynk ) = φ nk s)ds ε ynk ε ynk = 2 + K) ε αs)2 + K) + 1 ) ds ε φ nk s) ds αs)ds + ynk + ε, y ) y nk ynk ) y ) y ε) + y ε) y nk ε) K) ynk ε αs)ds + ynk + ε y ) y ε) K) ε αs)ds + ε, as k. From he arbirariness of ɛ and ε we obain y ) y nk ynk ) 0. Summarising, we have ha y nk ynk ) y ). By he coninuiy of τ we obain τy )) = lim nk τy nk ynk )) ynk ) = 0 which means ha is he ime of jump for y. Sep 2.b We make a similar reasoning wih he par of segmen [0, a] afer he jump. Form inequaliies 3.6) and 3.8) we conclude ha he family {y nk } is equiconinuous and equibounded on [ + ε, a]. Therefore, from Arzela-Ascoli 2.3 heorem passing o a subsequence if i is needed) we can assume ha y nk y ε, where y ε is a coninuous funcion and we exend i o a coninuous funcion y :, a] R N wih y nk convergen poinwise o y. We know ha: φ nk ) cl conv F, sy nk ), U nk ) ) cl O 1/nk F, By nk ), 1 n k ) )). y nk ) y ε ) a.e. on [ + ε, a], φ nk L 1 [ + ε, a], R N ),

10 10 A. GRUDZKA, S. RUSZKOWSKI EJDE-2015/114 by esimaion 3.9) and he weak compacness of closed ball we obain φ nkl φ on [ + ε, a]. Again, by heorem 2.2 we obain φ) F, y ε )), for a.e. [ + ε, a], so we obain an informaion ha φ) F, y )) a.e. on, a]. By weak convergence φ on [ + ε, a] for Ψφ n ) := a φ ns)ds we have φ nkl a φs)ds = Ψφ) = lim k Ψφ n k ) = lim k a = lim k y n k a) y ) = y a) y ). φ nk s)ds For decreasing sequence s n ) convergen o we obain for n < n : y s n ) y s n ) = sn s n φs)ds sn αs)1 + K)ds. By analogy o Sep 2a, we obain ha y s n )) is he Cauchy sequence, which is convergen in Banach space, so y s n ) y 2 for some y 2 R N. By coninuiy of I we have ha Iy nk ynk )) Iy )), herefore y 2 = y )+ Iy )). We can define a funcion ỹ : [0, a] R N by concaenaion y on [0, ] wih he funcion y on, a]. Obviously, for he funcion ỹ is he soluion of he problem 2.1). For > we have ỹ) = ỹ ) + Iỹ )) + φs)ds = y 0 + φs)ds + Iỹ )) + φs)ds, 0 where φ P Gn ỹ), so ỹ is he soluion of he problem 2.1) for all [0, a], hence ỹ S. The funcion ỹ is he limi of he sequence y nk ) in he space CJ 1 [0, a]). Sep 3. We show, for every n N, he conracibiliy of he se cl S n. Fix n such ha we can define p see Sep 1.) and ake ȳ cl S n. We divide he inerval [0, 1] ino wo halves. Le r [0, 1 2 ]. We consider he problem ẏ) = g n, y)), y) = ȳ), y + ) = y) + Iy)), for [a 2ra ȳ), a], τy)), for [0, a 2ra ȳ)], for = τy)). 3.10) In he previous problem we denoe by g n selecion of he map G n. There exiss exacly one soluion of his problem; we denoe i by y 2 ȳ,r. Then y 2 ȳ,r cl S n, Nex for r 1/2, 1] we consider he problem ẏ) = g n, y)), for [ȳ,r, a], τy)), y) = ȳ), for [0, ȳ,r ], y + ) = y) + Iy)), for = τy)). 3.11) where ȳ,r := ȳ 2r 1 2 ) ȳ. There exiss exacly one soluion of his problem, denoed by yȳ,r, 1 which also belongs o cl S n. Finally we consider he funcion h : [0, 1] cl S n cl S n given by hr, ȳ) := { yȳ,r, 2 r [0, 1 2 ], yȳ,r, 1 r 1 2, 1]. 3.12)

11 EJDE-2015/114 STRUCTURE OF THE SOLUTION SET 11 Now, we show ha he funcion h is coninuous. Due o he coninuous dependence of soluions on iniial condiions [13] we know ha he funcion h is coninuous on [0, 1/2) cl S n and lef coninuous on {1/2} cl S n. Le r k, ȳ k )) k be a sequence convergen o 1 + 2, ȳ). We know ha if ȳ < hrk,ȳ k ) hen τhr k, ȳ k )ȳ)) ȳ > 0 and we have so ȳ τhr k, ȳ k )ȳ)) = hrk,ȳ k ) ȳ τhr k, ȳ k ) )) ) θ)dθ < p hrk,ȳ k ) ȳ), hrk,ȳ k ) < ȳ + τhr k, ȳ k )ȳ)) ȳ)/p, bu for k such ha ȳ hrk,ȳ k ) one sees ha ȳ ȳ k,r k ) + = ȳ ȳk,r k ȳ ȳk,r k ȳ ȳk,r k ȳ ȳk,r k φ n s)ds yȳ 1 k,r ȳk,r k k ) + φ n s) g n s, x 1 ȳ k,r k s))ds ȳ φ n s) + g n s, y 1 ȳ k,r k s)) ds 2 αs)1 + K) + 1 ) ds. n Hence, if k we obain hr k, ȳ k )ȳ) ȳȳ), so For every k we have and τhr k, ȳ k )ȳ)) ȳ 0. ȳk,r k g n s, y 1 ȳ k,r k s))ds) ȳk,r k < hrk,ȳ k ) ȳ + 1 ȳ< hrk,ȳ k ) τhr k, ȳ k )ȳ)) ȳ)/p ȳk,r k ȳ ȳ + 1 ȳ< hrk,ȳ k ) τhr k, ȳ k )ȳ)) ȳ)/p, herefore by he squeeze heorem Le us fix k for a while. By 1 we denoe he funcion { y), for > 0, 1 >0 y)) = 0, for 0. We define funcion ϱȳk,r k : [0, a] R N hrk,ȳ k ) ȳ. 3.13) ϱȳk,r k ) := y 1 ȳ k,r k ) 1 >y 1ȳk,r k Iy 1 ȳ k,r k y 1ȳk,r k ))) y 2 ȳ,1/2 ) 1 >ȳiy 2 ȳ,1/2 ȳ))), which is funcion of differences beween yȳ 1 k,r k, and yȳ,1/2 2 caused by jumps. I is easy o see ha ȳ = y 2. For ȳk,r k ȳ,1/2 ϱȳk,r k ) ȳ ȳ k + ȳ ȳ k + ȳk,r k ȳk,r k g n s, y 1 ȳ k,r k s)) φ n s) ds 3.14) wih deleed changes ȳ we obain φn s) + g n s, y 1 ȳ k,r k s)) ) ds

12 12 A. GRUDZKA, S. RUSZKOWSKI EJDE-2015/114 and for > ȳ we have ϱȳk,r k ) zȳk,r k ȳ) + zȳk,r k ȳ) + = zȳk,r k ȳ) + ȳ ȳ k + =: zȳk,r k ) ȳ ȳ ȳ Iyȳ,1/2 2 ȳ)) ds zȳk,r k ȳ) + ȳk,r k 2 αs)1 + K) + 1 ) ds n g n s, y 1 ȳ k,r k s)) g n s, y 2 ȳ,1/2 s)) ds Λs) y 1 ȳ k,r k s) y 2 ȳ,1/2 s) ds Λs) ˇy 1 ȳ k,r k s) + 1 s>y 1ȳk,r k Iy 1 ȳ k,r k y 1ȳk,r k )) ˇy 2 ȳ,1/2 s) y 1ȳk,r k ȳ Λs) Iyȳ,1/2 2 ȳ) ds + Λs) Iyȳ 1 k,r k y )) Iy 1ȳk,r max{ȳ, y } k ȳ,1/2 2 ȳ)) ds 1ȳk,r k + ȳ Λs) ˇy 1 ȳ k,r k s) ˇy 2 ȳ,1/2 s) ds, hence by he Gronwall inequaliy we obain y 1ȳk,r ϱȳk,r k ) zȳk,r k ȳ) + k Λs) Iyȳ,1/2 2 ȳ)) ds ȳ ) + Λs) Iyȳ 1 k,r k y )) Iy 1ȳk,r max{ȳ, y } k ȳ,1/2 2 ȳ)) ds 1ȳk,r k exp ȳ Λs)ds =: zȳk,r k ). By previous convergences and coninuiy of I, we obain Ihr k, ȳ k ) hrk,ȳ k ))) Ih1/2, ȳ) h1/2,ȳ) )) 0, 3.15) sup ϱȳk,r k ) zȳk,r k a) ) [0,a] Summing up, by 3.16), 3.13) and 3.15), if r k, ȳ k ) converges o 1 + 2, ȳ), hen yȳ 1 k,r k converges o y 2 in norm in he space CJ ȳ, 1 1 [0, a]). 2 The funcion h, as coninuous on [0, 1] cl S n, is a homoopy. By definiion of he funcion h we have h0, ȳ) = ȳ and h1, ȳ) = yȳ,1, 1 so cl S n is a conracible se. Sep 4. We show ha properies needed o heorem 2.1 are fulfilled. The ses cl S n are conracible in he power of Sep 3. If x n N cl S n, hen x cl S n for every n. Therefore, here exiss sequence d n ) R + converges o 0 such ha Bx, d n ) in CJ 1 [0, a])) conains y n y n S n ).

13 EJDE-2015/114 STRUCTURE OF THE SOLUTION SET 13 Hence y n x in he space CJ 1 [0, a]). Moreover, we know ha subsequence y nk converges o a soluion of problem 2.1), where m = 1, so x S. We obain S n N S n n N cl S n S, so S = n N cl S n. We will show ha sup{dz, S) z S n } n 0. Assume ha here exiss ε > 0 and a sequence y n ) such ha y n S n and dy n, S) ε. From he Sep 3 we know ha his sequence has subsequence y nk ) such ha y nk ỹ S, so dy nk, S) 0. I is conrary o he choice of he sequence y n ), hence sup{dz, S) z S n } 0. Therefore sup{dz, S) z cl S n } 0. We obain S n S + B0, p n ), where p n := sup z Sn dz, S) 0 wih n. The compacness of S implies βcl S n ) = βs n ) βs) + p n = p n, so βcl S n ) 0. Summing up, we can use heorem 2.1, which implies ha he se S is an R δ -se. We will use his heorem o prove more general case. Theorem 3.2. Le he assumpions F0) F3), H1) H3) hold. Then every soluion of he problem 2.1) for every j = 1,..., m mees Σ j exacly once and he soluion se S of his problem is an R δ -se in he space CJ m [0, a]). Proof. We show ha we can divide he inerval [0, a] ino m disjoin pars and any of hem will have exacly one jump effec. Then we will be able o use he reasoning of heorem 3.1 on every such par, which will complee he proof. By analogy o Sep 1. in he proof of Theorem 3.1 we define a mulivalued map G n : [0, a] R N R N and consider he problem ẏ) G n, y)), for [0, a], τ j y)), j = 1,..., m, y0) = y 0, y + ) = y) + I j y)), for = τ j y)), j = 1,..., m. 3.17) We denoe by j y n he ime of j-h jump for he funcion y n : [0, a] R N. If he funcion y n has less han j jumps we ake j y n := a. Le y n be an arbirary soluion of he problem 3.17) for 0 2 y n. By analogy o heorem 3.1 in Sep 1a. we show ha here exiss a consan K such ha sup [0, 2 yn ] y n ) K. Nex we will proceed similarly o he proof of heorem 3.1 Sep 1b), we show ha here exiss a consan p > 0 such ha τ jy) φ n 1 < p for all j = 1,..., m, and for enough big n, where ȳ is fixed funcion wih values in cl B0, K), is such ha he assumpion H3) is saisfied, ȳ) = y, φ n G n, ȳ)). We define he funcion w n,j : [0, a] R in he following way: w n,j ) := τ j y n )), j = 1,..., m. Le us fix a soluion y n of problem 3.17). Now we prove by inducion ha he j-h ime of jump is zero of he funcion w n,j.

14 14 A. GRUDZKA, S. RUSZKOWSKI EJDE-2015/114 Basis. By assumpion H2) we have: a) w n,j 0) = τ j y 0 ) > 0, b) w n,j a) = τ j y n a)) a < 0, c) w n,j ) = τ j y n )) < τ j+1 y n )) = w n,j+1 ) for all [0, a]. If here are no impulses hen w n,j ) 0 on [0, a], bu by he definiion of w n,j for every j he funcion w n,j is coninuous, which conradics wih a) and b). Hence here is a leas one jump ime. Le 0 < 1 y n < a be he firs ime of jump for he soluion y n. Then By he assumpion H2) we obain w n,j 1 y n ) = 0, w n,j ) > 0, for [0, 1 y n ). w n,j 1 y n + ) = τj y n 1 y n + )) 1 yn = τ j y n 1 y n ) + I j y n 1 y n ))) 1 y n 0. For a.e. 1 y n we have w n,j) = τ jy n )) y n) 1 = τ jy n )) φ n ) 1 < p < 0. Hence w n,j is decreasing funcion a [ 1 y n, a], so y n for every j mees Σ j exacly once. By c) we know ha he ime 1 y n of he firs jump is zero of he funcion w n,1. Inducive sep. We assume ha he j-h ime of jump is zero of w n,j, j < m. Denoe by j y n he ime for which we have = τ j y n )). We know ha w n,l j y n + ) = τl y n j y n )) j y n > τ j y n j y n )) j y n = 0 for j > 1, l > j. We consider inerval J := j y n, a]. By analogy o Sep 1 in he proof of heorem 3.1 we have w n,i ) 0, J, i j, as long as here is no jump caused by w n,l for l > j. There have o be a leas one jump afer j y n and we denoe i by. By c) here are no jumps before he jump caused by w n,j+1, so = j+1 y n. Since boh he basis and he inducive sep have been proved, i has been proved by mahemaical inducion ha here are exacly m jumps one for each Σ j, j = 1,..., m. Nex we will proceed similarly o he proof of heorem 3.1. Le 1, 2,..., m ) be he limi of he sequence 1 y n, 2 y n,..., m y n )). We can show ha on [0, 1 ] here exiss he subsequence of y nk ) such ha I 1 y nkl ynkl )) I 1 y 1 )) and τ 1 y 1 )) = 1, so τ 2 y 1 ) + )) = τ 2 y 1 ) + I 1 y 1 ))) > τ 1 y 1 )) = 1. By he assumpions H1) and H2) here exiss ε 1 > 0 such ha here is only one jump ime of y on [0, 1 + 2ε 1 ] and we obain ha for all sufficienly big l here is only one jump ime of y nkl on [0, 1 + ε 1 ]. By analogy we find subsequence and ε 2 > 0 such ha here is exacly one jump on [ 1 + ε 1, 2 + ε 2 ], and so on. On every such inerval we proceed wih reasoning from heorem 3.1. The following example shows he case of inclusion wih exacly one jump, bu can easily be rearranged o a mulijump case. Example 3.3. There are wo rus funds wih ineres raes dependen on ime and amoun of money) α, y 1 ) A, y 1 ) and β, y 2 ) B, y 2 ) respecively where A and B are mulivalued) invesmen plans. I is available o ransfer money once in or ou) wihou loosing ineres.

15 EJDE-2015/114 STRUCTURE OF THE SOLUTION SET 15 We decided o sar boh rus funds wih he same amoun of money and ransfer money from worse deposi o beer one afer clarifying which one is beer. The amoun of money ha we wish o ransfer would be proporional o difference in incomes. This siuaion can be represened in he form of following differenial inclusion: wih y 1, y 2 )) F, y 1, y 2 ))), for [0, 1], τy 1, y 2 ))), 3.18) y 1, y 2 )0) = y 0, y 0 ), 3.19) y 1, y 2 ) + ) = y 1, y 2 )) + Iy 1, y 2 ))), for = τy 1, y 2 ))) 3.20) F, y 1, y 2 ) = A, y 1 ) B, y 2 ), where A and B fulfill assumpions F1) and F2) for example coninuous) and have inerval values wih A, y), B, y) [ 6y, 6y] y 1, y 1 ), ρ )y 1 < y 2, y 2, y 2 ), ρ Iy 1, y 2 ) = )y 2 < y 1, ρy 1 y 2 ), ρy 2 y 1 )), 0 y ρ )y ρ )2 y 1, 0, 0), y 1 < 0 or y 2 < 0 and For any α A, y) we have ha τy 1, y 2 ) := arc coy 1 + y 2 )/π. α 31/2 + 2y 2 ) = πp 1))1/2 + 2y 2 ) where 1 > p = 1 3/π > 0. We have analogous inequaliy for β herefore we obain sup τ jy 1, y 2 ) ϕ = ϕ F,y 1,y 2) sup sup α A,y 1) β B,y 2) α + β π1 + y 1 + y 2 ) 2 ) 1 p) 1 + 2y y y 1 + y 2 ) 2 1 p < 1 All assumpions of Theorem 3.1 are saisfied, herefore he soluion se of his problem is an R δ -se. References [1] Akhmed, M.; Principles of Disconinuous Dynamical Sysems, Springer, New York, [2] Aubin, J.-P.; Ekeland, I.; Applied nonlinear analysis, Wiley, New York, [3] Belarbi, A., Benchohra, M., Ouahab, A.; Nonconvex-valued impulsive funcional differenial inclusions wih variable imes, Nonlinear Oscillaions, Vol. 10, No ). [4] Benchohra, M.; Ouahab, A.; Impulsive neural funcional differenial equaions wih variable imes, Nonlinear Anal. 55, ). [5] Benchohra, M.; Ouahab, A.; Impulsive neural funcional differenial inclusions wih variable imes, Elecronic Journal of Differenial Equaions, Vol. 2003, No. 67, pp [6] Benedei, I.; Rubbioni, P.; Exisence of soluions on compac and non-compac inervals for semilinear impulsive differenial inclusions wih delay, Topol. Mehods Nonlinear Anal. 32, ). [7] Cardinali, T.; Rubbioni, P.; On he exisence of mild soluions of semilinear evoluion differenial inclusions, J. Mah. Anal. Appl. 308, ). [8] Cardinali, T.l Servadei, R.; On he exisence of soluions for nonlinear impulsive periodic viable problems, Cen. Eur. J. Mah ), no. 4,

16 16 A. GRUDZKA, S. RUSZKOWSKI EJDE-2015/114 [9] Djebali, S.; Górniewicz, L.; Ouahab, A.; Filippov-Ważewski heorems and srucure of soluion ses for firs order impulsive semilinear funcional differenial inclusions, Topol. Mehods Nonlinear Anal. 32, ). [10] Djebali S., Górniewicz L., Ouahab A. Topological srucure of soluion ses for impulsive differenial inclusions in Fréche spaces, Nonlinear Anal. 74, ). [11] Djebali, S.; Górniewicz, L.; Ouahab, A.; Soluion Ses for Differenial Equaions and Inclusions, De Gruyer Series in Nonlinear Analysis and Applicaions 18, Berlin - New York, 2013). [12] Gabor, G.; Grudzka, A.; Srucure of he soluion se o impulsive funcional differenial inclusions on he half-line, Nonlinear Differenial Equaions and Applicaions ), [13] Hale, J. K.; Theory of funcional diferenial equaions, Springer-Verlag, New York, [14] He, Y.; Xing, Y.; Poincar map and periodic soluions of firs-order impulsive differenial equaions on Moebius sripe, Absr. Appl. Anal. 2013, Ar. ID , 11 pp. [15] Hyman, D. M.; On decreasing sequence of compac absolue reracs, Fund. Mah ), [16] Kryszewski, W.; Plaskacz, S.; Periodic soluions o impulsive diferenial inclusions wih consrains, Nonlinear Anal., 65: , [17] Obukhovskii, V.; Yao, J.-C.; On impulsive funcional differenial inclusions wih Hille-Yosida operaors in Banach spaces, Nonlinear Anal, 736), ). Agaa Grudzka Faculy of Mahemaics and Compuer Science, Nicolaus Copernicus Universiy, Chopina 12/18, Toruń, Poland address: agaa33@ma.uni.orun.pl Sebasian Ruszkowski Faculy of Mahemaics and Compuer Science, Nicolaus Copernicus Universiy, Chopina 12/18, Toruń, Poland address: sebrus@ma.uni.orun.pl