# Problems with Mixed Boundary Conditions in Banach Spaces

Authors: Dionicio Pastor Dallos Santos
Journal: Chinese Journal of Mathematics
 (2017)

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## Abstract

Using Leray-Schauder degree or degree forα-condensing maps we obtain the existence of at least one solution for the boundary value problem of the following type: φu′′=ft,u,u′,  u(T)=0=u′(0), where φ:X→X is a homeomorphism with reverse Lipschitz constant such that φ(0)=0, f:0,T×X×X→X is a continuous function, T is a positive real number, and X is a real Banach space.


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## Body

## 1. Introduction

The purpose of this article is to obtain some existence results for the nonlinear boundary value problem of the form(1)φu′′=ft,u,u′,uT=0=u′0,where φ:X→X is a homeomorphism such that φ(0)=0 and φ-1 is Lipschitz constant, f:0,T×X×X→X is a continuous function, T is a positive real number, and X is a real Banach space. Of course, asolution of this problem is a function u:0,T→X of class C1 such that the function t↦φ(u′(t)) is continuously differentiable, satisfying the boundary conditions, and φu′t′=f(t,u(t),u′(t)) for all t∈0,T.The existence of solutions for second-order boundary value problems has been studied by many authors using various methods (see [1–6]).In particular, the authors in [3] have studied the following boundary value problem:(2)u′′=ft,u,u′,au0-bu′0=u0,cu1+du′1=u1,where a,b,c,d, and ad+bc>0. They obtained the existence of solutions of (2) using Darbo fixed point theorem and properties of the measure of noncompactness.Recently, Zhou and Peng [5] have studied the following boundary value problem:(3)-u′′=ft,u,u0=0=u1,where f:0,1×X→X is a continuous function and X is a Banach space. They obtained the existence of solutions of (3), where the main tools used in the study are Sadovskii fixed point theorem and precise computation of measure of noncompactness.Inspired by these results, the main aim of this paper is to study the existence of at least one solution for the boundary value problem (1) using Leray-Schauder degree or degree for α-condensing maps. For this, we reduce the nonlinear boundary value problem to some fixed points problem. Next, we shall essentially consider two types of regularity assumptions for f(t,x,y). In Theorem 10 we suppose that f is completely continuous, which allows us to prove that the associated fixed point operator is completely continuous required by a Leray-Schauder approach. In Theorem 11 we only assume some regularity conditions expressed in terms of the measure of noncompactness, which allows us to apply the methods of topological degree theory for α-condensing maps.The paper is organized as follows. In Section2, we establish the notation, terminology, and various lemmas which will be used throughout this paper. In Section 3, we formulate the fixed point operator equivalent to problem (1). In Section 4, we give main results in this paper. In Section 5, we study the existence of at least one solution for (1) in Hilbert spaces. For these results, we adapt the ideas of [7–9] to the present situation.


## 2. Notations and Preliminary Results

We first introduce some notation. For fixedT, we denote the usual norm in L1=L1(0,T,X) by ·L1. Let C=C(0,T,X) denote the Banach space of continuous functions from 0,T into X, endowed with the uniform norm ·∞ and let C1=C1(0,T,X) denote the Banach space of continuously differentiable functions from 0,T into X, equipped with the usual norm u1=max⁡u∞,u′∞.We introduce the following applications:TheNemytskii operatorNf:C1→C is(4)Nfut=ft,ut,u′t,theintegration operatorH:C→C1 is(5)Hut=∫0tusds,and the continuous linear application K:C→C1 is(6)Kut=-∫tTusds.Throughout this paper, we denote by(X,·) a real Banach space and I=0,T. For A⊆C1, we use the notation(7)At=ut:u∈A,AI=ut:u∈A,t∈I,A′t=u′t:u∈A,A′=u′:u∈A,A′I=u′t:u∈A,t∈I.Definition 1.
LetX be a Banach space and let MX be the family of bounded subsets of X. The Kuratowski measure of noncompactness is the map α:MX→0,∞ defined for(8)αB=inf⁡d>0:Badmitsafinitecoverbysetsofdiameter≤d;hereB∈MX.Properties
(a)
α
(
B
)
=
0 iff B¯ is compact.
(b)
S
⊂
B then α(S)≤α(B).
(c)
α
(
B
¯
)
=
α
(
B
).
(d)
α
(
B
∪
S
)
=
max
⁡
α
(
B
)
,
α
(
S
).
(e)
α
(
λ
B
)
=
λ
α
(
B
), where λ∈R and λB=λb:b∈B.
(f)
α
(
B
+
S
)
≤
α
(
B
)
+
α
(
S
), where B+S=b+s:b∈B,s∈S.
(g)
α
(
c
o
n
v
¯
(
B
)
)
=
α
(
B
).The details ofα and its properties can be found in [10].Definition 2 (see [11]).
Assuming thatD⊂X the mapping A:D→X is said to be a condensing operator if A is continuous and bounded (sends bounded sets into bounded sets), and, for any nonrelatively compact and bounded set S⊂D,(9)αAS<αS.The following lemmas are of great importance in the proof of our main results. The proofs can be found in [11].In the following, we denoteαc and α1 as the noncompactness measure in C and C1, respectively.Lemma 3.
LetS be a bounded subset of real numbers and B a bounded subset of X. Then (10)αSB=supt∈S⁡tαB,where SB=sb:s∈S,b∈B.Lemma 4.
LetA,B be bounded subsets of Banach spaces X and Y, respectively, with (11)x,y=max⁡x,y.Then (12)αA×B=max⁡αA,αB.Lemma 5.
IfH⊂C is bounded and equicontinuous, then we have the following:
(1)αcH=αHI.
(2)αHI=maxIαHt.Lemma 6.
IfH is a bounded set in C1, then
(i)α1(H)≥α(H(I)),
(ii)2α1(H)≥α(H′(I)).Lemma 7.
IfH is a bounded set in C1 and H′ equicontinuous, then (13)α1H=max⁡maxI⁡αHt,maxI⁡αH′t.


## 3. Fixed Point Formulations

Let us consider the operator(14)M1:C1⟶C1,u⟼Kφ-1HNfu.Here φ-1 is understood as the operator φ-1:C→C defined for φ-1(v)(t)=φ-1(v(t)). It is clear that φ-1 is continuous and sends bounded sets into bounded sets.Lemma 8.
u
∈
C
1 is a solution of (1) if and only if u is a fixed point of the operator M1.Proof.
Letu be a solution of (1). This implies that (15)φu′′=ft,u,u′,uT=0=u′0.Integrating of 0 to t and using the fact that u′(0)=0, we deduce that (16)φu′t=HNfut.Applying φ-1 and K to both of its members and using that u(T)=0, we have that(17)ut=Kφ-1HNfut=M1utt∈0,T.Conversely, since, by definition of the mapping M1, (18)Kφ-1HNfuT=0,φ-1HNfu0=0,it is a simple matter to see that if u is such that u=M1(u) then u is a solution to (1).Using the theorem of Arzelà-Ascoli we show that the operatorM1 is completely continuous.Lemma 9.
Iff is completely continuous, then the operator M1:C1→C1 is completely continuous.Proof.
LetΛ⊂C1 be a bounded set. Then, if u∈Λ, there exists a constant ρ>0 such that(19)u1≤ρ.Next, we show that M1(Λ)¯⊂C1 is a compact set. Let (vn)n be a sequence in M1(Λ), and let (un)n be a sequence in Λ such that vn=M1(un). Using (19), we have that there exists a constant W>0 such that, for all n∈N, (20)Nfun∞≤W,which implies that(21)HNfun∞≤WT.Hence the sequence H(Nf(un))n is bounded in C. Moreover, for t,t1∈0,T and for all n∈N, we have that (22)HNfunt-HNfunt1=∫0tfs,uns,un′sds-∫0t1fs,uns,un′sds=∫t1tfs,uns,un′sds≤Wt-t1,which implies that H(Nf(un))n is equicontinuous.
On the other hand, fort∈0,T,(23)Bt=HNfunt:n∈N,where (24)HNfunt=∫0tNfunsds=∫0tfs,uns,un′sds=tlimm→∞∑k=1mfsk,unsk,un′sksk-sk-1t.Recalling that the convex hull of a set A⊆X is given by (25)convA=∑i=1Nαixi:xi∈A,αi∈R,αi≥0,∑i=1Nαi=1,it follows that(26)∫0tfs,uns,un′sds∈tconv¯fs,uns,un′s:s∈0,T,n∈N,which implies that(27)Bt=HNfunt:n∈N⊂tconv¯fs,uns,un′s:s∈0,T,n∈N.Using the fact that f:0,T×X×X→X is completely continuous, we deduce that α(B(t))=0. Hence, B(t) is a relatively compact set in X. Thus, by the Arzelà-Ascoli theorem there is a subsequence of (H(Nf(un)))n, which we call (H(Nf(unj)))j, which is convergent in C. Using the fact that φ-1:C→C is continuous it follows from (28)M1unj′=φ-1HNfunjthat the sequence M1unj′j is convergent in C and hence (vnj)j=(M1(unj))j is convergent in C1. Finally, let (vn)n be a sequence in M1(Λ)¯. Let (zn)n⊆M1(Λ) be such that(29)limn→∞zn-vn1=0.Let (znj)j be a subsequence of (zn)n such that it converges to z. It follows that z∈M1(Λ)¯ and (vnj)j converge to z. This concludes the proof.In order to apply Leray-Schauder degree to the operatorM1, we introduced a family of problems depending on a parameter λ. For λ∈0,1, we consider the family of boundary value problems(30)φu′′=λft,ut,u′tuT=0=u′0.Notice that (30) coincide, for λ=1, with (1). So, for each λ∈[0,1], the operator associated with (30) for Lemma 8 is the operator M(λ,·), where M is defined on [0,1]×C1 by (31)Mλ,u=Kφ-1λHNfu.Using the same arguments as in the proof of Lemma 9 we show that the operator M is completely continuous. Moreover, using the same reasoning as above, system (30) (see Lemma 8) is equivalent to the problem(32)u=Mλ,u.


## 4. Main Results

In this section, we present and prove our main results.Theorem 10.
LetX be a Banach space and φ-1 a homeomorphism with Lipschitz constant k. Suppose that f is completely continuous and that there exist two numbers c0,c1≥0 such that (33)ft,x,y≤c0+c1y,∀t,x,y∈0,T×X×X.Then problem (1) has at least one solution.Proof.
Let(λ,u)∈[0,1]×C1 be such that M(λ,u)=u. Using (32) we have that u is solution of (30), which implies that (34)u′=φ-1λHNfu,u′0=0=uT.Using the fact that φ-1 is a homeomorphism with Lipschitz constant k, we deduce that (35)u′t≤kc0T+kc1∫0tu′sdst∈0,T.By Gronwall’s Inequality, we have (36)u′t≤kc0Te∫0tkc1ds≤kc0Tekc1Tt∈0,T.Hence, u′∞≤kc0Tekc1T≔β. Because u∈C1 is such that u′(T)=0, we have that (37)ut≤∫tTu′sds≤∫0Tu′sds≤βTt∈0,T,and hence (38)u1≤R1,whereR1=max⁡β,βT.Using that M is completely continuous we deduce that, for each λ∈[0,1], the Leray-Schauder degree degLS(I-M(λ,·),Bρ(0),0) is well-defined for any ρ>R1, and by the homotopy invariance we have that (39)degLSI-M1,·,Bρ0,0=degLSI-M0,·,Bρ0,0.Hence, degLS(I-M(1,·),Bρ(0),0)≠0. This, in turn, implies that there exists u∈Bρ(0) such that M1(u)=u, which is a solution for (1).Theorem 11.
LetX be a Banach space and φ-1 a homeomorphism with Lipschitz constant k. Assume that f is continuous and satisfies the following conditions: (1)
There exist two numbersc0,c1≥0 such that (40)ft,x,y≤c0+c1y,∀t,x,y∈0,1×X×X.(2)
For all bounded subsetsA,B in X, (41)αf0,1×A×B≤k1max⁡αA,αB,where0<k1<12k.Then problem (1) has at least one solution.Proof.
Observe thatM1 maps bounded sets into bounded sets. Furthermore, its continuity follows by the continuity of the operators which compose M1. We show that the operator M1 is condensing (α-condensing). In fact, for a bounded set Λ in C1, there exists a constant L1>0 such that (42)Nfu∞≤L1,∀u∈Λ.For t,t1∈0,1 we have that (43)M1u′t-M1u′t1=φ-1HNfut-φ-1HNfunt1≤kHNfut-HNfut1≤k∫t1tfs,us,u′sds≤kL1t-t1,which means M1Λ′ is equicontinuous. Applying Lemma 7 there exists τ∈0,1 or ω∈0,1 with (44)α1M1Λ=αM1Λτor (45)α1M1Λ=αM1Λ′ω.Let us consider the first case. (46)α1M1Λ=αM1Λτ=αKφ-1HNfuτ:u∈Λ.Using the properties of α, we see that (47)α1M1Λ≤α∫τ1φ-1HNfusds:u∈Λ≤1-ταconv¯φ-1HNfus:s∈τ,1,u∈Λ≤αφ-1HNfus:s∈0,1,u∈Λ.Using the fact that φ-1 is a homeomorphism with Lipschitz constant k, we deduce that (48)α1M1Λ≤kαHNfus:s∈0,1,u∈Λ≤kα∫0sft,ut,u′tdt:s∈0,1,u∈Λ≤kα0,1conv¯ft,ut,u′t:t∈0,1,u∈Λ.Applying Lemma 3 and again the properties of α, we obtain that (49)α1M1Λ≤kαft,ut,u′t:t∈0,1,u∈Λ≤kαf0,1×Λ0,1×Λ′0,1.Using (41), we have that(50)α1M1Λ≤kk1max⁡αΛ0,1,αΛ′0,1.This implies, by Lemma 6, that (51)α1M1Λ≤2kk1α1Λ.
Consider the alternative case. Proceeding as before, we obtain(52)α1M1Λ=αM1Λ′ω≤2kk1α1Λ.
Therefore, in either case, we obtain(53)α1M1Λ≤2kk1α1Λ.By (41), we get 0<2kk1<1, and therefore M1 is α-condensing.
Let us consider the function(54)M~:0,1×C1⟶C1,λ,u⟼λM1u.Let (λ,u)∈0,1×C1 be such that u=M~(λ,u). Using the fact that φ-1 is a homeomorphism with Lipschitz constant k and Gronwall’s Inequality, we deduce that there exists a constant r>0 such that u1<r.
Finally, we show the existence of at least one solution of (1) using the homotopy invariance of the degree for α-condensing maps. Let B be bounded in C1. Then (55)α1M~0,1×B=α1M~λ,u:λ∈0,1,u∈B≤2kk1α1B.Then we have that, for each λ∈0,1, the degree degN(I-M~(λ,·),Br(0),0) is well-defined and, by the properties of that degree, that (56)degNI-M~1,·,Br0,0=degNI-M~0,·,Br0,0=1.Then, from the existence property of degree, there exists u∈Br(0) such that u=M~(1,u)=M1(u)=u, which is a solution for (1).Remark 12.
In [3], the nonlinear term f(t,x,y) is bounded; in our result, the nonlinear term f(t,x,y) may possess no more than linear growth.Now we consider an example to illustrate our results.Example 13.
Consider the boundary value problem in(57)X=l∞=x=x1,x2,…,xi,…:supi⁡xi<∞,with norm x=supixi. (58)ui′′=cos⁡uii2+ui′4ett∈0,1,ui1=ui′0=0.Now we can regard this problem as a problem of form (1), where u=(u1,u2,…,ui,…)∈l∞ and f=(f1,f2,…,fi,…) with (59)fit,x,y=git,x,y+hit,x,y=cos⁡xii2+yi4ett,x,y∈0,1×l∞×l∞.It is clear that f:0,1×l∞×l∞→l∞ is a continuous function and that (60)ft,x,y=supifit,x,y≤1+y.On the other hand, for any bounded subsets A,B⊂l∞ we have (61)αg0,1×A×B=0,αh0,1×A×B=αB4,and hence(62)αf0,1×A×B≤14αB≤14max⁡αA,αB.So, by Theorem 11 we get one solution.


## 5. Boundary Value Problems in Hilbert Spaces

Throughout this section, let(X,·,·) denote a real Hilbert space. Assume that φ:X→X satisfies the following conditions:(1)φ-1 is a homeomorphism with Lipschitz constant k.(2) For anyx,y∈X,x≠y, (63)φx-φy,x-y>0.Lemma 14.
Leth∈C([0,T],R+) be such that(64)ft,x,y≤ft,x,y,x+htfor all (t,x,y)∈[0,T]×X×X. If (λ,u)∈[0,1]×C1 is such that M(λ,u)=u, then there exists R>0 such that u1≤R.Proof.
Let(λ,u)∈(0,1]×C1 be such that M(λ,u)=u. Using (32), we have that u is solution of (30), which implies that (65)u′=φ-1λHNfu,u′0=0=uT,where, for all t∈[0,T], we obtain (66)λHNfut≤∫0Tfs,us,u′sds≤∫0Tfs,us,u′s,usds+∫0Thsds.On the other hand, because φ is a homeomorphism such that(67)φy,y≥0for all y∈X, then (68)φu′t,u′t≥0t∈0,T,and hence (69)-∫0Tφu′t,u′tdt≤0.Using the integration-by-parts formula and the boundary conditions, we deduce that (70)∫0Tφu′t′,utdt=-∫0Tφu′t,u′tdt≤0.Since λ∈(0,1] and u is solution of (30) we have that (71)∫0Tft,ut,u′t,utdt≤0.Hence,(72)φu′t≤hL1.It follows that there exists L>0 such that u′∞≤L. Because u∈C1 is such that u(T)=0, we deduce that (73)ut≤∫tTu′sds≤∫0Tu′sds≤LTt∈0,T,and hence (74)u1≤R,whereR=max⁡L,LT.Finally, if u=M(0,u), then u=0, so the proof is complete.Now we show the existence of at least one solution for problem (1) by means of Leray-Schauder degree.Theorem 15.
Letf be completely continuous. Assume that f satisfies the conditions of Lemma 14. Then (1) has at least one solution.Proof.
Letλ∈0,1 and u be a possible fixed point of M(λ,·). Then, using Lemma 14 we deduce that (75)u1≤R,whereR=max⁡L,LT.Therefore, if ρ>R, it follows from the homotopy invariance of Leray-Schauder degree that degLS(I-M(λ,·),Bρ(0),0) is independent of λ∈0,1 so that if we notice that M(0,·)=0, (76)degLSI-M1,·,Bρ0,0=degLSI-M0,·,Bρ0,0=1.Hence there exists u∈Bρ(0) such that it is a solution for (1).Using a proof similar to that of Theorem11, we obtain the following existence result.Theorem 16.
Letf=f(t,x) be continuous. Assume that f satisfies the following conditions: (1)
There existsh∈C([0,1],R+) such that (77)ft,x≤ft,x,x+htfor all(t,x)∈[0,1]×X.(2)
f sends bounded sets into bounded sets.(3)
For any bounded setS in X, (78)αf0,1×S≤k1αS,where0<k1<1k.Then problem (1) has at least one solution.Proof.
LetA be bounded in C1. Applying Lemma 7 there exists τ∈0,1 or ω∈0,1 with (79)α1M0,1×A=αM0,1×Aτor (80)α1M0,1×A=αM0,1×A′ω.Proceeding as Theorem 11, we obtain in either case (81)α1M0,1×A≤kk1α1A,wherekk1<1.Using the homotopy invariance of the degree for α-condensing maps, we obtain (82)degNI-M0,·,Bρ0,0=degNI,Bρ0,0=1,whereρ>L.Then, from the existence property of degree, there exists u∈Bρ(0) such that u=M(1,u), which is a solution for (1).The following corollary is concerned with the existence of one solution for (1).Corollary 17.
Assume thatf=f(t,x) satisfies the following conditions: (1)
Suppose that for anyδ>0 the mapping f:[0,1]×X→X is bounded and uniformly continuous in [0,1]×Bδ(0)¯, where Bδ(0)¯=x∈X:x≤δ.(2)
There existsh∈C([0,1],R+) such that (83)ft,x≤ft,x,x+htfor all(t,x)∈[0,1]×X.(3)
There exists a constantk1 with 0<k1<1/k such that (84)ft,x-ft,y≤k1x-y,∀t,x∈0,1×X.
Then problem (1) has at least one solution.Proof.
LetS be a bounded set in X. Let us consider (85)H=ψx:x∈S,whereψxt=ft,x∀t∈0,1.Cleary, H⊂C, H is bounded and equicontinuous. Thus, by using the conclusion of Lemma 5, we have (86)αcH=αH0,1=αf0,1×S=max0,1⁡αft,x:x∈S.Using (84), we obtain (87)αf0,1×S≤k1αS.By using the arguments of Theorem 16, we can obtain the conclusion of Corollary 17.



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