In the present note we are concerned with the question of the existence and uniqueness of solutions of two-point boundary value problem for nonlinear difference equations with parameters in Banach space.The proof of the above problem is based on the method of successive approximation and fixed-point method

A. Puk

I. Kubiaczyk

J. Morchalo

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GLASNIK MATEMATICKIVol. 38(58)(2003), 299 { 309A DISCRETE BOUNDARY VALUE PROBLEM WITHPARAMETERS IN BANACH SPACEI. Kubiaczyk, J. Morcha lo and A. PukA. Mickiewicz University, Poznan University of Technology and State Schoolof Higher Vocational Education in Leszno, PolandAbstract. In the present note we are concerned with the ques-tion of the existence and uniqueness of solutions of two- point boundaryvalue problem for nonlinear dierence equations with parameters in Banachspace.The proof of the above problem is based on the method of successiveapproximation and xed-point method.1. IntroductionIn this paper we present some existence and uniqueness results for discreteboundary value problems where the values of the solution lie in a Banach spaceE.The motivation for the present work comes from many recent investiga-tions. In fact, the continuous versions have been studied by many authors,e.g. see [4, 5, 6, 11, 12] and the references cited therein.Discrete boundaryvalue problems have been discussed widely in the literature, see [1, 3, 7, 10]and their references.In the present note we are concerned with the question of the existenceand uniqueness of solutions of two-point boundary value problem for nonlineardierence equations with parameters in Banach space.The proof of the existence is based on the method of successive approxi-mation and xed-point methods. We use conditions expressed in terms of themeasure of strong or weak noncompactness and weakly{weakly sequentiallycontinuity.2000 Mathematics Subject Classication. 39A10, 47H10.Key words and phrases. Dierence equations, boundary value problems, Banach space.299300 I. KUBIACZYK, J. MORCHA LO AND A. PUKConsider the Banach spaces E and F with the norms kk. Let Z denotesthe set of integers. Given a < b in Z, let [a; b] = fa; a + 1; : : : ; bg. Also thesymbol  denotes the forward dierence operator with stepsize 1.2. Main results2.1. In the rst part of this section we discusses the nonlinear discrete equa-tion of the form(1) x(n) = f(n; x(n); );with the boundary conditions(2) x(0) = x0; x(T + 1) = X;where T is a xed positive integer, f : [0; T ]  Br  C ! E; Br = fx 2E : kx  x0k  rg; C  F some closed, bounded set. For this equation weconsider the following problem (A):(A) There exists parameter  2 C such that equation (1) has a solutionx : [0; T ]! E satisfying conditions (2).Equation (1) is equivalent to the equation(i) x(n+ 1) = x0 +nXs=0f(s; x(s); )for every n 2 [0; T ].Suppose that the function f satises the following conditions:(H1) there exists a function  : [0; T ]! R+ such thatTXs=0(s)  randjf(n; x; )j  (n)for every n 2 [0; T ]; x 2 Br;  2 C,(H2) there exists a function  : [0; T ]! R+ such thatkf(n; u1; 1)  f(n; u2; 2)k  (n)(ku1   u2k+ k1   2k);for every n 2 [0; T ]; u1; u2 2 Br; 1; 2 2 C,(H3) there exists a constant L > 0 such thatTXs=0(f(s; x(s); 1)  f(s; x(s); 2)  L k1   2kfor 1; 2 2 C; x : [0; T ]! Br and q 1 + qL< 1 where q =TPs=0(s),A DISCRETE BOUNDARY VALUE PROBLEM WITH PARAMETERS 301(H4) for every x : [0; T ]! Br there exists  2 C such thatx0 +TXs=0f(s; x(s); ) = X:We are now ready to present the main results.Theorem 2.1. Suppose (H1)   (H4) hold. Then the problem (A) hasexactly one solution.Proof. In order to apply method of successive approximations we denesequences fxm()g1m=0 and fmg1m=0 in the following way:Let x0() : [0; T ] ! Br; x0(n) = x0 for every n 2 [0; T ]. Then by (H4)there exists 0 2 C such thatx0 +TXs=0f(s; x0(s); 0) = X:Next we dene the function x1() as follows:x1(0) = x0andx1(n+ 1) = x0 +nXs=0f(s; x0(s); 0) for n 2 [0; T ]:By (H1) x1(n) 2 Br and x1(T + 1) = X 2 Br. Repeating the procedure, wedene the function x2() by the formulax2(n+ 1) = x0 +nXs=0f(s; x1(s); 1):In this manner we obtain two sequences fxm()g and fmg, whose m-th termare dened by relations(3) xm(n+ 1) = x0 +nXs=0f(s; xm 1(s); m 1); n 2 [0; T ]where m 1 satisfy equality(4) x0 +TXs=0f(s; xm 1(s); m 1) = X:By assumptions we havexm(n) 2 Br for n 2 [0; T ]; xm(T + 1) = X 2 Br:It will be now veried that sequences fxm()g and fmg are convergent.302 I. KUBIACZYK, J. MORCHA LO AND A. PUKFrom (3) and (4) we havexm(T + 1)  xm 1(T + 1) = 0 ==TXs=0ff(s; xm 1(s); m 1)  f(s; xm 1(s); m 2)g++TXs=0ff(s; xm 1(s); m 2)  f(s; xm 2(s); m 2)g:(5)So  TXs=0ff(s; xm 1(s); m 1)  f(s; xm 1(s); m 2)g ==TXs=0ff(s; xm 1(s); m 2(s))  f(s; xm 2(s); m 2g :By (H3) left hand side is greater then L km 1   m 2k and by (H2) righthand side is less thenTXs=0(s) jxm 1(s)  xm 2(s)j ;hencekm 1   m 2k  1L TXs=0(s)!max0sTkxm 1(s)  xm 2(s)k  qLmax0sTkxm 1(s)  xm 2(s)k :(6)From (3),(6) and assumption (H1), we havekxm(n+ 1)  xm 1(n+ 1)k nXs=0(s)km 1   m 2k+nXs=0(s) max0nTkxm 1(n)  xm 2(n)k  q1 +qLmax0nTkxm 1(n)  xm 2(n)k      qm 11 +qLm 1max0nTkx1(n)  x0(n)k:Thus the sequence fxm(i)g forms a Cauchy sequence for each i 2 [0; T ] andthere is x() such thatlimm!1xm(n) = x(n) for each n 2 [0; T ]:A DISCRETE BOUNDARY VALUE PROBLEM WITH PARAMETERS 303Since Br is closed, x(n) is in Br. From this and (6) it is clear thatkm   m 1k  qLhq1 +qLim 1max0nTkx1(n)  x0(n)k:So the sequence fmg Cauchy sequence and converges to some  as m!1.Take the limit as m!1 in (3) and (4) we obtain that x(n) satises problem(A).Now, we prove the uniqueness of the solution of (A).In fact if there exist two functions x(); y() and two parameters ;  sat-isfying the problem (A), thenx(n+ 1) = x0 +nXs=0f(s; x(s); )y(n+ 1) = x0 +nXs=0f(s; y(s); )(7)and(8) x(0) = y(0) = x0; x(T + 1) = X = y(T + 1):From (7) and assumption (H2) we have(9) kx(n)  y(n)k nXs=0(s)[kx(s)   y(s)k+ k  k]:From (8) TXs=0f(s; x(s); )  TXs=0f(s; y(s); ) = 0:Hence analogously to (6) we obtain that(9') k  k  qLmax0sTkx(s)  y(s)k:Applying this inequality to (9) we can show thatmax0nTkx(n)  y(n)k  q1 +qLmax0nTkx(n)  y(n)k;a contradiction. So x(n) = y(n) for n 2 [0; T ] and by (9')  = . Thus theproof is complete.2.2. In this part we will consider problem (A) for the dierence equationwith delay.Let  2 N; ' : [ ; 0]! Br; '(0) = x0; f : [0; T ]BrC ! E. Insteadof the (1) and (2) we take the equation(10) x(n) = f(n; x(n  ); ); n 2 [0; T ];304 I. KUBIACZYK, J. MORCHA LO AND A. PUKwith the boundary conditionsx(n) = '(n) for n 2 [ ; 0]x(T + 1) = X:(11)Analogously to Theorem 1 we can prove the following theorem:Theorem 2.2. Under the assumptions (H1)   (H4) the problem (A)for(10) and (11) has exactly one solution.2.3. Now we consider the boundary value problem which is essentially relatedto the equation with delayed argument: there exist a solution x(n; ) of theproblem(12) x(n) = f(n; x(n); x(n  )); n 2 [0; T ];(13) x(n) = '(n; ) for n 2 [ ; 0]({is nonnegative integer) which satises the condition(14) x(T + 1; ) = X;where (T;X) are given,  will be denite.We shall assume that the functionf : [0; T ]E E ! Esatises the following conditions:(G1) there exists a function  : [0; T ]! R+ such thatkf(n; x; y)k  (n);TXs=0(s)  rfor every n 2 [0; T ]; x; y 2 Br,(G2) there exists a function  : [0; T ]! R+ such thatkf(n; u1; v1)  f(n; u2; v2)k  (n)fku1   u2k+ kv1   v2kg;for every n 2 [0; T ]; uivi 2 Br; i = 1; 2,TXs=0(s) = q <14;(G3) the function' : [ ; 0] C ! Br; '(0; ) = x0;is continuous,(G4) there exists a constant L > 0 such thatXs=0ff(s; x(s); '(s  ; 1))  f(s; x(s); '(s  ; 2))g  Lk1   2kfor 1; 2 2 C,A DISCRETE BOUNDARY VALUE PROBLEM WITH PARAMETERS 305(G5) for every y 2 Br and x : [0; T ]! Br there exists  2 C such thaty +Xs=0f(s; x(s); '(s   ; )) = X:Theorem 2.3. Suppose that the conditions (G1)   (G5) hold. Then theproblem (12)-(14) has solution.Proof. Let x0() : [0; T ] ! Br; x0(n) = x0 for every n 2 [0; T ]. Then,by condition (G5) there exists 0 2 C such thatx0 +Xs=0f(s; x0(s); '(s  ; 0)) +TXs=+1f(s; x0(s); x0(s  )) = X;becausey = x0 +TXs=+1f(s; x0(s); x0(s  )) 2 Br:We build a sequence of x1; x2; : : : according to the following law:x1(n+ 1) =x0 +Xs=0f(s; x0(s); '(s  ; 0))+nXs=+1f(s; x0(s); x0(s  )) = X;for n 2 [0; T ],x1(n) = '(n; 0) for n 2 [ ; 0];xm(n+ 1) = x0 +Xs=0f(s; xm 1(s); '(s  ; m 1))+nXs=+1f(s; xm 1(s); xm 1(s  )); n 2 [0; T ];(15)(16) xm(n) = '(n; m 1) for n 2 [ ; 0]:Evidently, x1(n) 2 Br and x1(T + 1) = X 2 Br,(17) xm(n) 2 Br and xm(T + 1) = X 2 Br:Now, we shall prove that the sequences fxm()g and fmg are convergent.From (15), (16), (17) and assumptions (G1); (G4) we have0 =Xs=0ff(s; xm 1(s); '(s   ; m 1))  f(s; xm 2(s); '(s   ; m 2))g+TXs=+1ff(s; xm 1(s); xm 1(s  ))   f(s; xm 2(s); xm 2(s  ))g;306 I. KUBIACZYK, J. MORCHA LO AND A. PUKLkm 1   m 2k Xs=0(s)kxm 1(s)  xm 2(s)k+TXs=+1(s)fkxm 1(s)  xm 2(s)k+ kxm 1(s  )   xm 2(s  )kg 3TXs=0(s) max0sTkxm 1(s)  xm 2(s)k= 3q max0sTkxm 1(s)  xm 2(s)k:Hence(18) km 1   m 2k  3qLmax[0;T ]kxm 1(s)  xm 2(s)k:By (15) and (18) we getkxm(n+ 1)  xm 1(n+ 1)k ==Xs=0[f(s; xm 1(s); '(s   ; m 1))  f(s; xm 2(s); '(s  ; m 2))]+nXs=+1[f(s; xm 1(s); xm 1(s  ))   f(s; xm 2(s); xm 2(s  ))]TXs=+1kf(s; xm 1(s); xm 1(s  ))   f(s; xm 2(s); xm 2(s  ))k+nXs=+1kf(s; xm 1(s); xm 1(s  ))  f(s; xm 2(s); xm 2(s  ))k 4TXs=+1(s) max0sTkxm 1(s)  xm 2(s)k= q1 max0sTkxm 1(s)  xm 2(s)k;where q1 = 4q. Let Vm = max0sTkxm(s)  xm 1(s)k, thenVm  q1Vm 1 for m = 2; 3; : : :Thus, by induction, the inequalityVm  qm 11 V1is true for m = 2; 3; : : : Since by (G2) q1 < 1, we have that fxm(n)g isuniformly convergent to x(n) for each n 2 [0; T ] as m!1.A DISCRETE BOUNDARY VALUE PROBLEM WITH PARAMETERS 307This implies from (18) thatkm   m 1k  3qLqm 11 V1;so fmg converges to . The relations (15), (16), (17) implies that x(n) isa solution of the problem (12)-(14).2.4. In this part we prove existence theorems for the problem (A). Wewill use the measure of noncompactness and weak noncompactness. We nowgather together some denitions and results which we will be needed in thissection.Definition 2.4 (K. Kuratowski). For any bounded subset A of a metricspace E we denote by (A) the inmum of all " > 0, such that there exists anite covering of A by sets of diameter  ". The number (A) is called themeasure of noncompactness of set A. (For properties of  see [9] and theirreferences.)Definition 2.5. Let B = fx 2 E : kxk  1g and let A be a boundedsubset of E. The (A) measure of weak noncompactness of A is dened by(A) = infft  0 : A  K + tB for some weakly compact K  Eg:(For properties of  see [7].)In particular if  =  or  = , A and B are bounded subset of E, then(1) A  B ) (A)  (B),(2) (A [B) = max((A); (B)),(3) (A+B)  (A) + (B),(4) (e+A) = (A) for e 2 E,(5) (A) = 0, A is compact,(6) (A) = 0, Aw weakly compact.Since [0; T ] = f0; 1; : : : ; Tg is a discrete space, then any mapping of [0; T ]to a topological space E is continuous. We shall denote the set of all suchmappings by C([0; T ]; E). Let eBr = fx : [0; T ]! Brg.Lemma 2.6 (Ambrosetti). Let  be a measure of strong (or weak) non-compactness, V  C([0; T ]; E) be bounded. Then(V ) = (V ([0; T ])) = supf(V (i)) : i 2 [0; T ]g;where V (i) = fv(i) : v 2 V g,V ([0; T ]) =T[i=0V (i) = V (0) [ V (1) [    [ V (T );(V ) denotes the measure of strong (or weak) noncompactness in C([0; T ]; E).Our results will be proved by the following xed point theorem of Darbo[2]:308 I. KUBIACZYK, J. MORCHA LO AND A. PUKTheorem 2.7. Let D be a nonempty, closed, convex and bounded subsetof a Banach space E. Let G : D ! D be a continuous mapping, which iscondensing with respect to the measure of weak noncompactness  i.e.(F (V )) < (V )for (V ) > 0; V  D. Then G has a xed point.Theorem 2.8. Let C be subset of F and f : [0; T ]  Br  C ! E becontinuous and satises H1, and(H 04) for every x 2 eBr their exists (x) 2 C such thatx0 +TXs=0f(s; x(s); (x)) = Xand (x) is continuous with respect x,(H5)(f(n; V; C)  k(V )for each n 2 [0; T ] and for every subset V  Br, with Tk < 1.Then the problem (A) has at last one solution.Proof. By H 04 for each function x 2 eBr there exists (x) such thatx0 +TXs=0f(s; x(s); (x)) = X:We dene the operator G as follows:(Gx)(i + 1) = x0 +iXs=0f(s; x(s); (x)):By our assumptions G : eBr ! eBr is continuous and the xed point of G is asolution of the problem (A).By the properties of  and by Ambrosetti lemma we have((GV )(i+ 1))   x0 +iXs=0f(s; V (s); C)!iXs=0(f(s; V (s); C)) iXs=0k(V (s))  kT(V )so (G(V ))  kT(V ).By the Darbo xed point theorem G has a xed point.Remark 2.9. Using the above method and Darbo theorem of - measureof weak noncompactness (see [3, Thm. 2]) we can prove analogical theoremunder conditions expressed in terms of measure of weak noncompactness andweakly-weakly sequentically continuity.A DISCRETE BOUNDARY VALUE PROBLEM WITH PARAMETERS 309References[1] R.P. Agarwal and D. O'Regan, Singular Discrete Boundary value problems, Appl.Math. Letter 12 (1999), 127{131.[2] G. Darbo, Punti uniti in transformationi a condominio non compacto, Rend. Sem.Mat. Univ. Padova 24 (1955), 84{92.[3] M. Dawidowski, I. Kubiaczyk and J. Morcha lo, A discrete boundary value problem inBanach space, Glas. Math. 36(56) (2001), 233{239.[4] P. Eloe and I. Henderson, Singular nonlinear boundary value problems for higherorder ordinary dierential equations, Nonlinear Analysis 17 (1991), 1{10.[5] J.A. Gatica, V. Oliker and P. Woltman, Singular nonlinear boundary value problemsfor second-order ordinary dierential equations, J. Di. Equal. 79 (1989), 62{78.[6] J. Henderson and K.C. Yin, Signular boundary value problems, Bull. Inst. Math.Acad. Sin. 19 (1991), 229{242.[7] J. Henderson, Singular boundary value problems for dierence equations, DynamicSystems and Appl. 1 (1992), 271{282.[8] J. Kubiaczyk, On a xed point theorem for weakly sequentially continuous mapping,Discussions Mathematicae { Dierential Inclusions 15 (1995), 15{20.[9] J. Kubiaczyk, Kneser type theorems for ordinary dierential equations in Banachspaces, J. Di. Equations 45 (1982), 139{146.[10] A. Lasota, A discrete boundary value problem, Ann. Polon. Math. 20 (1968), 183{190.[11] J. Morcha lo and S. Szua, De Vale Poisson boundary value problem, Glas. Mat. 43(1989), 35{39.[12] J. Wang, A singular nonlinear boundary value problem for a higher order ordinarydierential equation, Nonlinear Analysis 22 (1994), 1051{1056.I. KubiaczykInstitute of MathematicsA. Mickiewicz University, Matejki 48/4960-769 PoznanPolandE-mail : kuba@amu.edu.plJ. Morcha loInstitute of MathematicsPoznan University of TechnologyPiotrowo 3a, 60-965 PoznanPolandE-mail : jmorchal@math.put.poznan.plA. PukState School of Higher Vocational Education in LesznoA. Mickiewicza 5, 61-100 LesznoPolandReceived : 06.09.2002.Revised : 21.01.2003.

A discrete boundary value problem with parameters in Banach space

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GLASNIK MATEMATIČKIVol. 38(58)(2003), 299 – 309A DISCRETE BOUNDARY VALUE PROBLEM WITHPARAMETERS IN BANACH SPACEI. Kubiaczyk, J. Morcha lo and A. PukA. Mickiewicz University, Poznań University of Technology and State Schoolof Higher Vocational Education in Leszno, PolandAbstract. In the present note we are concerned with the ques-tion of the existence and uniqueness of solutions of two- point boundaryvalue problem for nonlinear difference equations with parameters in Banachspace.The proof of the above problem is based on the method of successiveapproximation and fixed-point method.1. IntroductionIn this paper we present some existence and uniqueness results for discreteboundary value problems where the values of the solution lie in a Banach spaceE.The motivation for the present work comes from many recent investiga-tions. In fact, the continuous versions have been studied by many authors,e.g. see [4, 5, 6, 11, 12] and the references cited therein.Discrete boundaryvalue problems have been discussed widely in the literature, see [1, 3, 7, 10]and their references.In the present note we are concerned with the question of the existenceand uniqueness of solutions of two-point boundary value problem for nonlineardifference equations with parameters in Banach space.The proof of the existence is based on the method of successive approxi-mation and fixed-point methods. We use conditions expressed in terms of themeasure of strong or weak noncompactness and weakly–weakly sequentiallycontinuity.2000 Mathematics Subject Classification. 39A10, 47H10.Key words and phrases. Difference equations, boundary value problems, Banach space.299300 I. KUBIACZYK, J. MORCHA LO AND A. PUKConsider the Banach spaces E and F with the norms ‖·‖. Let Z denotesthe set of integers. Given a < b in Z, let [a, b] = {a, a + 1, . . . , b}. Also thesymbol ∆ denotes the forward difference operator with stepsize 1.2. Main results2.1. In the first part of this section we discusses the nonlinear discrete equa-tion of the form(1) ∆x(n) = f(n, x(n), λ),with the boundary conditions(2) x(0) = x0, x(T + 1) = X,where T is a fixed positive integer, f : [0, T ] × Br × C → E, Br = {x ∈E : ‖x− x0‖ ≤ r}, C ⊂ F some closed, bounded set. For this equation weconsider the following problem (A):(A) There exists parameter λ ∈ C such that equation (1) has a solutionx : [0, T ]→ E satisfying conditions (2).Equation (1) is equivalent to the equation(i) x(n+ 1) = x0 +n∑s=0f(s, x(s), λ)for every n ∈ [0, T ].Suppose that the function f satisfies the following conditions:(H1) there exists a function α : [0, T ]→ R+ such thatT∑s=0α(s) ≤ rand|f(n, x, λ)| ≤ α(n)for every n ∈ [0, T ], x ∈ Br, λ ∈ C,(H2) there exists a function β : [0, T ]→ R+ such that‖f(n, u1, λ1)− f(n, u2, λ2)‖ ≤ β(n)(‖u1 − u2‖+ ‖λ1 − λ2‖),for every n ∈ [0, T ], u1, u2 ∈ Br, λ1, λ2 ∈ C,(H3) there exists a constant L > 0 such that∥∥∥∥∥T∑s=0(f(s, x(s), λ1)− f(s, x(s), λ2)∥∥∥∥∥ ≥ L ‖λ1 − λ2‖for λ1, λ2 ∈ C, x : [0, T ]→ Br and q(1 + qL)< 1 where q =T∑s=0β(s),A DISCRETE BOUNDARY VALUE PROBLEM WITH PARAMETERS 301(H4) for every x : [0, T ]→ Br there exists λ ∈ C such thatx0 +T∑s=0f(s, x(s), λ) = X.We are now ready to present the main results.Theorem 2.1. Suppose (H1) − (H4) hold. Then the problem (A) hasexactly one solution.Proof. In order to apply method of successive approximations we definesequences {xm(·)}∞m=0 and {λm}∞m=0 in the following way:Let x0(·) : [0, T ] → Br, x0(n) = x0 for every n ∈ [0, T ]. Then by (H4)there exists λ0 ∈ C such thatx0 +T∑s=0f(s, x0(s), λ0) = X.Next we define the function x1(·) as follows:x1(0) = x0andx1(n+ 1) = x0 +n∑s=0f(s, x0(s), λ0) for n ∈ [0, T ].By (H1) x1(n) ∈ Br and x1(T + 1) = X ∈ Br. Repeating the procedure, wedefine the function x2(·) by the formulax2(n+ 1) = x0 +n∑s=0f(s, x1(s), λ1).In this manner we obtain two sequences {xm(·)} and {λm}, whose m-th termare defined by relations(3) xm(n+ 1) = x0 +n∑s=0f(s, xm−1(s), λm−1), n ∈ [0, T ]where λm−1 satisfy equality(4) x0 +T∑s=0f(s, xm−1(s), λm−1) = X.By assumptions we havexm(n) ∈ Br for n ∈ [0, T ], xm(T + 1) = X ∈ Br.It will be now verified that sequences {xm(·)} and {λm} are convergent.302 I. KUBIACZYK, J. MORCHA LO AND A. PUKFrom (3) and (4) we havexm(T + 1)− xm−1(T + 1) = 0 ==T∑s=0{f(s, xm−1(s), λm−1)− f(s, xm−1(s), λm−2)}++T∑s=0{f(s, xm−1(s), λm−2)− f(s, xm−2(s), λm−2)}.(5)So∥∥∥∥∥−T∑s=0{f(s, xm−1(s), λm−1)− f(s, xm−1(s), λm−2)}∥∥∥∥∥ ==∥∥∥∥∥T∑s=0{f(s, xm−1(s), λm−2(s))− f(s, xm−2(s), λm−2}∥∥∥∥∥ .By (H3) left hand side is greater then L ‖λm−1 − λm−2‖ and by (H2) righthand side is less thenT∑s=0β(s) |xm−1(s)− xm−2(s)| ,hence‖λm−1 − λm−2‖ ≤1L(T∑s=0β(s))max0≤s≤T‖xm−1(s)− xm−2(s)‖ ≤≤ qLmax0≤s≤T‖xm−1(s)− xm−2(s)‖ .(6)From (3),(6) and assumption (H1), we have‖xm(n+ 1)− xm−1(n+ 1)‖ ≤≤n∑s=0β(s)‖λm−1 − λm−2‖+n∑s=0β(s) max0≤n≤T‖xm−1(n)− xm−2(n)‖ ≤≤ q(1 +qL)max0≤n≤T‖xm−1(n)− xm−2(n)‖ ≤ · · · ≤≤ qm−1(1 +qL)m−1max0≤n≤T‖x1(n)− x0(n)‖.Thus the sequence {xm(i)} forms a Cauchy sequence for each i ∈ [0, T ] andthere is x∗(·) such thatlimm→∞xm(n) = x∗(n) for each n ∈ [0, T ].A DISCRETE BOUNDARY VALUE PROBLEM WITH PARAMETERS 303Since Br is closed, x∗(n) is in Br. From this and (6) it is clear that‖λm − λm−1‖ ≤qL[q(1 +qL)]m−1max0≤n≤T‖x1(n)− x0(n)‖.So the sequence {λm} Cauchy sequence and converges to some λ∗ as m→∞.Take the limit as m→∞ in (3) and (4) we obtain that x∗(n) satisfies problem(A).Now, we prove the uniqueness of the solution of (A).In fact if there exist two functions x(·), y(·) and two parameters λ, λ sat-isfying the problem (A), thenx(n+ 1) = x0 +n∑s=0f(s, x(s), λ)y(n+ 1) = x0 +n∑s=0f(s, y(s), λ̄)(7)and(8) x(0) = y(0) = x0, x(T + 1) = X = y(T + 1).From (7) and assumption (H2) we have(9) ‖x(n)− y(n)‖ ≤n∑s=0β(s)[‖x(s) − y(s)‖+ ‖λ− λ‖].From (8) ∥∥∥∥∥T∑s=0f(s, x(s), λ) −T∑s=0f(s, y(s), λ)∥∥∥∥∥ = 0.Hence analogously to (6) we obtain that(9’) ‖λ− λ̄‖ ≤ qLmax0≤s≤T‖x(s)− y(s)‖.Applying this inequality to (9) we can show thatmax0≤n≤T‖x(n)− y(n)‖ ≤ q(1 +qL)max0≤n≤T‖x(n)− y(n)‖,a contradiction. So x(n) = y(n) for n ∈ [0, T ] and by (9’) λ = λ̄. Thus theproof is complete.2.2. In this part we will consider problem (A) for the difference equationwith delay.Let τ ∈ N, ϕ : [−τ, 0]→ Br, ϕ(0) = x0, f : [0, T ]×Br×C → E. Insteadof the (1) and (2) we take the equation(10) ∆x(n) = f(n, x(n− τ), λ), n ∈ [0, T ],304 I. KUBIACZYK, J. MORCHA LO AND A. PUKwith the boundary conditionsx(n) = ϕ(n) for n ∈ [−τ, 0]x(T + 1) = X.(11)Analogously to Theorem 1 we can prove the following theorem:Theorem 2.2. Under the assumptions (H1) − (H4) the problem (A)for(10) and (11) has exactly one solution.2.3. Now we consider the boundary value problem which is essentially relatedto the equation with delayed argument: there exist a solution x(n, λ) of theproblem(12) ∆x(n) = f(n, x(n), x(n− τ)), n ∈ [0, T ],(13) x(n) = ϕ(n, λ) for n ∈ [−τ, 0](τ–is nonnegative integer) which satisfies the condition(14) x(T + 1, λ∗) = X,where (T,X) are given, λ∗ will be definite.We shall assume that the functionf : [0, T ]×E ×E → Esatisfies the following conditions:(G1) there exists a function α : [0, T ]→ R+ such that‖f(n, x, y)‖ ≤ α(n),T∑s=0α(s) ≤ rfor every n ∈ [0, T ], x, y ∈ Br,(G2) there exists a function β : [0, T ]→ R+ such that‖f(n, u1, v1)− f(n, u2, v2)‖ ≤ β(n){‖u1 − u2‖+ ‖v1 − v2‖},for every n ∈ [0, T ], uivi ∈ Br, i = 1, 2,T∑s=0β(s) = q <14,(G3) the functionϕ : [−τ, 0]× C → Br, ϕ(0, λ) = x0,is continuous,(G4) there exists a constant L > 0 such that∥∥∥∥∥τ∑s=0{f(s, x(s), ϕ(s− τ, λ1))− f(s, x(s), ϕ(s− τ, λ2))}∥∥∥∥∥ ≥ L‖λ1 − λ2‖for λ1, λ2 ∈ C,A DISCRETE BOUNDARY VALUE PROBLEM WITH PARAMETERS 305(G5) for every y ∈ Br and x : [0, T ]→ Br there exists λ ∈ C such thaty +τ∑s=0f(s, x(s), ϕ(s − τ, λ)) = X.Theorem 2.3. Suppose that the conditions (G1) − (G5) hold. Then theproblem (12)-(14) has solution.Proof. Let x0(·) : [0, T ] → Br, x0(n) = x0 for every n ∈ [0, T ]. Then,by condition (G5) there exists λ0 ∈ C such thatx0 +τ∑s=0f(s, x0(s), ϕ(s− τ, λ0)) +T∑s=τ+1f(s, x0(s), x0(s− τ)) = X,becausey = x0 +T∑s=τ+1f(s, x0(s), x0(s− τ)) ∈ Br.We build a sequence of x1, x2, . . . according to the following law:x1(n+ 1) =x0 +τ∑s=0f(s, x0(s), ϕ(s− τ, λ0))+n∑s=τ+1f(s, x0(s), x0(s− τ)) = X,for n ∈ [0, T ],x1(n) = ϕ(n, λ0) for n ∈ [−τ, 0],xm(n+ 1) = x0 +τ∑s=0f(s, xm−1(s), ϕ(s− τ, λm−1))+n∑s=τ+1f(s, xm−1(s), xm−1(s− τ)), n ∈ [0, T ],(15)(16) xm(n) = ϕ(n, λm−1) for n ∈ [−τ, 0].Evidently, x1(n) ∈ Br and x1(T + 1) = X ∈ Br,(17) xm(n) ∈ Br and xm(T + 1) = X ∈ Br.Now, we shall prove that the sequences {xm(·)} and {λm} are convergent.From (15), (16), (17) and assumptions (G1), (G4) we have0 =τ∑s=0{f(s, xm−1(s), ϕ(s − τ, λm−1))− f(s, xm−2(s), ϕ(s − τ, λm−2))}+T∑s=τ+1{f(s, xm−1(s), xm−1(s− τ)) − f(s, xm−2(s), xm−2(s− τ))},306 I. KUBIACZYK, J. MORCHA LO AND A. PUKL‖λm−1 − λm−2‖ ≤τ∑s=0β(s)‖xm−1(s)− xm−2(s)‖+T∑s=τ+1β(s){‖xm−1(s)− xm−2(s)‖+ ‖xm−1(s− τ) − xm−2(s− τ)‖}≤ 3T∑s=0β(s) max0≤s≤T‖xm−1(s)− xm−2(s)‖= 3q max0≤s≤T‖xm−1(s)− xm−2(s)‖.Hence(18) ‖λm−1 − λm−2‖ ≤3qLmax[0,T ]‖xm−1(s)− xm−2(s)‖.By (15) and (18) we get‖xm(n+ 1)− xm−1(n+ 1)‖ ==∥∥∥∥∥τ∑s=0[f(s, xm−1(s), ϕ(s − τ, λm−1))− f(s, xm−2(s), ϕ(s− τ, λm−2))]+n∑s=τ+1[f(s, xm−1(s), xm−1(s− τ)) − f(s, xm−2(s), xm−2(s− τ))]∥∥∥∥∥≤T∑s=τ+1‖f(s, xm−1(s), xm−1(s− τ)) − f(s, xm−2(s), xm−2(s− τ))‖+n∑s=τ+1‖f(s, xm−1(s), xm−1(s− τ))− f(s, xm−2(s), xm−2(s− τ))‖≤ 4T∑s=τ+1β(s) max0≤s≤T‖xm−1(s)− xm−2(s)‖= q1 max0≤s≤T‖xm−1(s)− xm−2(s)‖,where q1 = 4q. Let Vm = max0≤s≤T‖xm(s)− xm−1(s)‖, thenVm ≤ q1Vm−1 for m = 2, 3, . . .Thus, by induction, the inequalityVm ≤ qm−11 V1is true for m = 2, 3, . . . Since by (G2) q1 < 1, we have that {xm(n)} isuniformly convergent to x∗(n) for each n ∈ [0, T ] as m→∞.A DISCRETE BOUNDARY VALUE PROBLEM WITH PARAMETERS 307This implies from (18) that‖λm − λm−1‖ ≤3qLqm−11 V1,so {λm} converges to λ∗. The relations (15), (16), (17) implies that x∗(n) isa solution of the problem (12)-(14).2.4. In this part we prove existence theorems for the problem (A). Wewill use the measure of noncompactness and weak noncompactness. We nowgather together some definitions and results which we will be needed in thissection.Definition 2.4 (K. Kuratowski). For any bounded subset A of a metricspace E we denote by α(A) the infimum of all ε > 0, such that there exists afinite covering of A by sets of diameter ≤ ε. The number α(A) is called themeasure of noncompactness of set A. (For properties of α see [9] and theirreferences.)Definition 2.5. Let B = {x ∈ E : ‖x‖ ≤ 1} and let A be a boundedsubset of E. The β(A) measure of weak noncompactness of A is defined byβ(A) = inf{t ≥ 0 : A ⊂ K + tB for some weakly compact K ⊂ E}.(For properties of β see [7].)In particular if µ = α or µ = β, A and B are bounded subset of E, then(1) A ⊂ B ⇒ µ(A) ≤ µ(B),(2) µ(A ∪B) = max(µ(A), µ(B)),(3) µ(A+B) ≤ µ(A) + µ(B),(4) µ(e+A) = µ(A) for e ∈ E,(5) α(A) = 0⇔ A is compact,(6) β(A) = 0⇔ Aw weakly compact.Since [0, T ] = {0, 1, . . . , T} is a discrete space, then any mapping of [0, T ]to a topological space E is continuous. We shall denote the set of all suchmappings by C([0, T ], E). Let B̃r = {x : [0, T ]→ Br}.Lemma 2.6 (Ambrosetti). Let µ be a measure of strong (or weak) non-compactness, V ⊂ C([0, T ], E) be bounded. Thenµ(V ) = µ(V ([0, T ])) = sup{µ(V (i)) : i ∈ [0, T ]},where V (i) = {v(i) : v ∈ V },V ([0, T ]) =T⋃i=0V (i) = V (0) ∪ V (1) ∪ · · · ∪ V (T ),µ(V ) denotes the measure of strong (or weak) noncompactness in C([0, T ], E).Our results will be proved by the following fixed point theorem of Darbo[2]:308 I. KUBIACZYK, J. MORCHA LO AND A. PUKTheorem 2.7. Let D be a nonempty, closed, convex and bounded subsetof a Banach space E. Let G : D → D be a continuous mapping, which iscondensing with respect to the measure of weak noncompactness α i.e.α(F (V )) < α(V )for α(V ) > 0, V ⊂ D. Then G has a fixed point.Theorem 2.8. Let C be subset of F and f : [0, T ] × Br × C → E becontinuous and satisfies H1, and(H ′4) for every x ∈ B̃r their exists λ(x) ∈ C such thatx0 +T∑s=0f(s, x(s), λ(x)) = Xand λ(x) is continuous with respect x,(H5)α(f(n, V, C) ≤ kα(V )for each n ∈ [0, T ] and for every subset V ⊂ Br, with Tk < 1.Then the problem (A) has at last one solution.Proof. By H ′4 for each function x ∈ B̃r there exists λ(x) such thatx0 +T∑s=0f(s, x(s), λ(x)) = X.We define the operator G as follows:(Gx)(i + 1) = x0 +i∑s=0f(s, x(s), λ(x)).By our assumptions G : B̃r → B̃r is continuous and the fixed point of G is asolution of the problem (A).By the properties of α and by Ambrosetti lemma we haveα((GV )(i+ 1)) ≤ α(x0 +i∑s=0f(s, V (s), C))≤≤i∑s=0α(f(s, V (s), C)) ≤i∑s=0kα(V (s)) ≤ kTα(V )so α(G(V )) ≤ kTα(V ).By the Darbo fixed point theorem G has a fixed point.Remark 2.9. Using the above method and Darbo theorem of β- measureof weak noncompactness (see [3, Thm. 2]) we can prove analogical theoremunder conditions expressed in terms of measure of weak noncompactness andweakly-weakly sequentically continuity.A DISCRETE BOUNDARY VALUE PROBLEM WITH PARAMETERS 309References[1] R.P. Agarwal and D. O’Regan, Singular Discrete Boundary value problems, Appl.Math. Letter 12 (1999), 127–131.[2] G. Darbo, Punti uniti in transformationi a condominio non compacto, Rend. Sem.Mat. Univ. Padova 24 (1955), 84–92.[3] M. Dawidowski, I. Kubiaczyk and J. Morcha lo, A discrete boundary value problem inBanach space, Glas. Math. 36(56) (2001), 233–239.[4] P. Eloe and I. Henderson, Singular nonlinear boundary value problems for higherorder ordinary differential equations, Nonlinear Analysis 17 (1991), 1–10.[5] J.A. Gatica, V. Oliker and P. Woltman, Singular nonlinear boundary value problemsfor second-order ordinary differential equations, J. Diff. Equal. 79 (1989), 62–78.[6] J. Henderson and K.C. Yin, Signular boundary value problems, Bull. Inst. Math.Acad. Sin. 19 (1991), 229–242.[7] J. Henderson, Singular boundary value problems for difference equations, DynamicSystems and Appl. 1 (1992), 271–282.[8] J. Kubiaczyk, On a fixed point theorem for weakly sequentially continuous mapping,Discussions Mathematicae – Differential Inclusions 15 (1995), 15–20.[9] J. Kubiaczyk, Kneser type theorems for ordinary differential equations in Banachspaces, J. Diff. Equations 45 (1982), 139–146.[10] A. Lasota, A discrete boundary value problem, Ann. Polon. Math. 20 (1968), 183–190.[11] J. Morcha lo and S. Szufla, De Vale Poisson boundary value problem, Glas. Mat. 43(1989), 35–39.[12] J. Wang, A singular nonlinear boundary value problem for a higher order ordinarydifferential equation, Nonlinear Analysis 22 (1994), 1051–1056.I. KubiaczykInstitute of MathematicsA. Mickiewicz University, Matejki 48/4960-769 PoznańPolandE-mail : kuba@amu.edu.plJ. Morcha loInstitute of MathematicsPoznań University of TechnologyPiotrowo 3a, 60-965 PoznańPolandE-mail : jmorchal@math.put.poznan.plA. PukState School of Higher Vocational Education in LesznoA. Mickiewicza 5, 61-100 LesznoPolandReceived : 06.09.2002.Revised : 21.01.2003.

A discrete boundary value problem with parameters in Banach space

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