Boundary Value Problems on Time Scales

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VALUE PROBLEM AT RESONANCE

NICKOLAI KOSMATOV

Received 18 January 2005; Accepted 1 June 2005

We apply a coincidence degree theorem of Mawhin to show the existence of at least one symmetric solution of the nonlinear second-order multipoint boundary value problem u(t)= f(t,u(t),|u(t)|), t∈(0, 1),u(0)=_n

i=1μ_iu(ξ_i), u(1−t)=u(t), t∈[0, 1], where 0< ξ1< ξ2<···< ξn≤1/2,ⁿ_i₌₁μi=1, f : [0, 1]×R²→Rwith f(t,x,y)=f(1−t,x,y), (t,x,y)∈[0, 1]×R², satisfying the Carath´eodory conditions.

Copyright © 2006 Nickolai Kosmatov. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Definitions and technical results

We study symmetric solutions of the multipoint nonlinear boundary value problem u(t)= ft,u(t),u(t), t∈(0, 1), (1.1)

u(0)= n i=1

μiuξi

, (1.2)

u(1−t)=u(t), t∈[0, 1], (1.3) whereξi∈[0, 1] with 0< ξ1< ξ2<···< ξn≤1/2,μi∈Rwith

n i=1

μ_i=1, (1.4)

and the inhomogeneous term satisfies (H0) f : [0, 1]×R²→Rwith

f(t,x,y)= f(1−t,x,y), (t,x,y)∈[0, 1]×R². (1.5)

Hindawi Publishing Corporation Abstract and Applied Analysis

Volume 2006, Article ID 54121, Pages1–11 DOI10.1155/AAA/2006/54121

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If there is aμi>1, we assume, in addition, that n

i=1

μiξi 1−ξi

=0. (1.6)

Due to the condition (1.4) the diﬀerential operator in the left side of (1.1) is not invertible. In the literature, boundary value problems of this type are referred to as problems at resonance. Boundary value problems at resonance have been studied by several authors including the most recent works [1–9,11]. In the recent works [5,6,8,9], the inhomogeneous term is either a continuous function on [0, 1]×R²or the sum of a continuous and a Lebesgue integrable functions. In this note, we merely require measurability off in the first variable, continuity in the rest of variables for a. a. values oft, and, in addition, f being locally bounded by Lebesgue integrable functions for a. a. values oft. The above assumptions constitute the so-called Carath´eodory conditions.

In this section, we provide the necessary background definitions and facts and state the key theorem due to Mawhin [10]. In the second section, we provide additional assumptions on the inhomogeneous term and give the suﬃcient conditions of existence of at least one solution of (1.1)–(1.3). The emphasis in this note is on symmetric solutions at resonance.

Definition 1.1. LetX andZbe normed spaces. A linear mappingL: domL⊂X→Z is called a Fredholm mapping if the following two conditions hold:

(i) kerLhas a finite dimension,

(ii) ImLis closed and has a finite codimension.

If L is a Fredholm mapping, its (Fredholm) index is the integer IndL=dim kerL− codim ImL.

In this paper, we are concerned with a Fredholm mapping of index zero. FromDefini- tion 1.1, it follows that there exist continuous projectorsP:X→X andQ:Z→Zsuch that

ImP=kerL, kerQ=ImL, X=kerL⊕kerP, Z=ImL⊕ImQ, (1.7) and that the mapping

L|domL∩kerP: domL∩kerP−→ImL (1.8) is invertible. We denote the inverse ofL|domL∩kerPbyKP: ImL→domL∩kerP. The gen- eralized inverse ofLdenoted byK_P,Q:Z→domL∩kerPis defined byK_P,Q=K_P(I−Q).

IfLis a Fredholm mapping of index zero, then for every isomorphismJ: ImQ→kerL, the mappingJQ+KP,Q:Z→domLis an isomorphism and, for everyu∈domL,

JQ+KP,Q−1

u=

L+J⁻¹Pu. (1.9)

Definition 1.2. LetL: domL⊂X→Zbe a Fredholm mapping, letEbe a metric space, and letN:E→Zbe a mapping. Say thatNisL-compact onEifQN:E→ZandK_P,QN: E→X are compact onE. In addition, say thatNisL-completely continuous if it isL- compact on every boundedE⊂X.

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When the boundary value problem is shown to be equivalent to the abstract equation Lu=Nu, the existence of a solution will be guaranteed by the following theorem due to Mawhin [10, Theorem IV.13].

Theorem 1.3. LetΩ⊂X be open and bounded,Lbe a Fredholm mapping of index zero, and letNbeL-compact onΩ. Assume that the following conditions are satisfied:

(i)Lu=λNufor every (u,λ)∈((domL\kerL)∩∂Ω)×(0, 1);

(ii)Nu /∈ImLfor everyu∈kerL∩∂Ω;

(iii) deg(QN|kerL∩∂Ω,Ω∩kerL, 0)=0, withQ:Z→Za continuous projector such that kerQ=ImL.

Then the equationLu=Nuhas at least one solution in domL∩Ω.

The following definition introduces the so-called Carath´eodory conditions imposed on a map.

Definition 1.4. Say that the mapf : [0, 1]×Rⁿ→R, (t,z)→f(t,z) satisfies the Carath´eodory conditions with respect toL¹[0, 1] if the following conditions are satisfied:

(i) for eachz_∈_Rⁿ, the mappingt_→f(t,z) is Lebesgue measurable;

(ii) for almost eacht∈[0, 1], the mappingz→f(t,z) is continuous onRⁿ;

(iii) for eachr >0, there existsαr∈L¹([0, 1],R) such that for a.e.t∈[0, 1] and every zsuch that|z| ≤r,|f(t,z)| ≤αr(t).

We introduce the Sobolev space W^2,1(0, 1)=

u: [0, 1]−→R:u,uabsolutely continuous on [0, 1] andu∈L[0, 1]. (1.10) LetX=C¹[0, 1] with the normu =max{u∞,u∞}andZ=L¹[0, 1] with the usual Lebesgue norm denoted by · 1. Consider the mappingL: domL⊂X→Zwith

domL=

u∈W^2,1(0, 1) :usatisfies (1.2) and (1.3) (1.11) by

Lu(t)=u(t), t∈(0, 1). (1.12) Define the mappingN:X→Zby

Nu(t)=ft,u(t),u(t), t∈(0, 1). (1.13) Lemma 1.5. The mappingL: domL_⊂X_→Zis a Fredholm mapping of index zero.

Proof. It is clear that kerL=R.

Letu∈domLand consider the linear equation

u(t)=g(t), (1.14)

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subject to (1.2), (1.3). Theng∈Zis symmetric on the interval [0, 1]. Sinceuis absolutely continuous, it follows from the symmetry condition (1.3) that

u(t)= ^t

0g(s)ds− ¹

0(1−s)g(s)ds. (1.15)

Integrating again, we get u(t)= ^t

0(t−s)g(s)ds−t

1

0(1−s)g(s)ds+c. (1.16) Sinceⁿ_i₌₁μi=1, it follows from (1.2) that we must have

n i=1

μ_i

ξi

0

ξ_i−sg(s)ds−μ_iξ_i

1

0(1−s)g(s)ds

=0. (1.17)

Conversely, if (1.17) holds for someg∈Z, we take the candidate ofu∈domLas given by (1.16) and establish that it is symmetric, absolutely continuous along with its derivative, u(t)=g(t) for a. a.t∈(0, 1) and (1.2) is satisfied. In fact, we have

ImL=

g∈Z:gsatisfies (1.3) and (1.17). (1.18) We recall the condition (1.6) and define the continuous linear mappingQ:Z→Zby

Qg= 2

n

i=1μ_iξ_i1−ξ_i n i=1

μ_iξ_i

1

0(1−s)g(s)ds−μ_i

ξi

0

ξ_i−sg(s)ds

. (1.19) It is easy to see thatQ²g=Qgfor allg∈Z, that is, the mappingQis idempotent. Observe also that (1.17) and (1.19) imply that ImL=kerQ. Takeg∈Zin the formg=(g−Qg) + Qgso thatg−Qg∈ImLandQg∈R. Ifg≡c=0, then, by (1.6),Qg=0, which implies that ImL∩R= {0}. HenceZ=ImL⊕R.

Now, IndL=dim kerL−codim ImL=0 and soL is a Fredholm mapping of index

zero.

The continuous projectorP:X→Xis defined by

Pu(t)=u(0), t∈(0, 1). (1.20)

By takingu∈Xin the formu(t)=u(0) + (u(t)−u(0)), it is clear thatX=kerL⊕kerP.

Note that the projectorsPandQare exact, that is, satisfy the relationships (1.7). Define K_P: ImL_→domL_∩kerPby

KPg(t)= ^t

0(t−s)g(s)ds−t

1

0(1−s)g(s)ds, (1.21) so that

K_Pg(t)= ^t

0g(s)ds− ¹

0(1−s)g(s)ds. (1.22)

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ThenKPg∞≤2g1and(KPg)∞≤2g1, and thus

K_Pg≤2g1. (1.23)

In fact ifg∈ImL, then LKP

g(t)= d² dt²

t

0(t−s)g(s)ds−t

1

0(1−s)g(s)ds

=g(t). (1.24) Also, ifu∈domL∩kerP, then

K_PLu(t)= ^t

0(t−s)u(s)ds−t

1

0(1−s)u(s)ds=u(t)−u(0)−tu(1)−u(0)=u(t) (1.25) (sinceu∈kerPanduis symmetric,u(0)=u(1)=0). Thus, we get that

KP=

L|domL∩kerP

₋1

. (1.26)

For convenience, we introduce a constant

C= 2

n

i=1μ_iξ_i1−ξ_i. (1.27) Now

QNu=C n i=1

μ_iξ_i

1

0(1−s)fs,u(s),u(s)ds−μ_i

ξi

0

ξ_i−sfs,u(s),u(s)ds

, KP,QNu(t)= ^t

0(t−s)Nu(s)ds−t

1

0(1−s)Nu(s)ds

− ^t

0(t−s)(QN)u(s)ds+t

1

0(1−s)(QN)u(s)ds

= ^t

0(t−s)fs,u(s),u(s)ds−t

1

0(1−s)fs,u(s),u(s)ds

−1

2Ct(t−1) n i=1

μiξi

1

0(1−s)fs,u(s),u(s)ds

−μi ξi

0

ξi−sfs,u(s),u(s)ds

.

(1.28) Lemma 1.6. The mappingNisL-completely continuous.

Proof. LetE⊂Xbe bounded and{u_k}⊂E. Define the sequence{v_k}byv_k(t)=K_P,QNu_k(t).

Set

r=supu:u∈E. (1.29)

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Since the function f : [0, 1]×R²→Rsatisfies the Carath´eodory conditions with respect toL¹[0, 1], there exists a Lebesgue integrable functionαr such that for allk∈Nand a.e.

t∈[0, 1],

Nu_k(t)=ft,u_k(t),u_k(t)≤α_r(t). (1.30) Fort∈[0, 1] andk∈N,

vk(t)=KP(I−Q)Nuk(t)

=

t

0(t−s)Nu_k(s)ds−t

1

0(1−s)Nu_k(s)ds

−1

2Ct(t−1) n i=1

μ_iξ_i

1

0(1−s)Nu_k(s)ds−μ_i

ξi

0

ξ_i−sNu_k(s)ds

≤ ^t

0(t−s)Nuk(s)ds+t

1

0(1−s)Nuk(s)ds +1

2Ct(t−1) n i=1

μ_iξ_i

1

0(1−s)Nu_k(s)ds+μ_i

ξi

0

ξ_i−sNu_k(s)ds

≤

1 +C 8

n i=1

μiξi

1 +ξi

αr

1,

(1.31) that is, the sequence{v_k}is uniformly bounded on [0, 1].

Now v_k(t)=

t

0Nu_k(s)ds− ¹

0(1−s)Nu_k(s)ds

−1

2C(2t−1) n i=1

μ_iξ_i

1

0(1−s)Nu_k(s)ds−μ_i

ξi

0

ξ_i−sNu_k(s)ds

≤ ^t

0

Nuk(s)ds+

1

0(1−s)Nuk(s)ds +1

2C|2t−1| n i=1

μiξi

1

0(1−s)Nuk(s)ds+μi ξi

0

ξi−sNuk(s)ds

≤1 2

3 +C n i=1

μ_iξ_i1 +ξ_i α_r₁

(1.32) for allt∈[0, 1] andk∈N, that is, the sequence{v_k}is uniformly bounded on [0, 1] and as such is equicontinuous on [0, 1]. Since{v_k}is uniformly bounded and equicontinuous on [0, 1], by Arzela-Ascoli theorem, it has a subsequence{vkl}that converges to some v∈C[0, 1].

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Consider the sequence{wkl}defined by w_k_l(t)₌ d

dtK_P(I₋Q)Nu_k_l(t)

= ^t

0Nu_k_l(s)ds− ¹

0(1−s)Nu_k_l(s)ds

−1

2C(2t−1) n i=1

μ_iξ_i

1

0(1−s)Nu_k_l(s)ds+μ_i

ξi

0

ξ_i−sNu_k_l(s)ds

.

(1.33)

Employing arguments similar to that for {v_k} one can show that {w_k_l} is uniformly bounded and equicontinuous on [0, 1]. Hence{w_k_l}as a subsequence that converges to somew∈C[0, 1]. In fact,w(t)=v(t),t∈[0, 1] and, thus, there is a subsequence of{vkl} that converges inC¹[0, 1]. Therefore, the image ofEunderK_P,QN is relatively compact.

Since the function f : [0, 1]×R²→Rsatisfies the Carath´eodory conditions with respect toL¹[0, 1], the continuity ofKP,QNonEfollows from the Lebesgue dominated conver- gence theorem.

Similar considerations apply to show thatQN is continuous and thatQN(E) is relatively compact. Now, since the mappingsQN andKP,QN are compact on an arbitrary boundedE⊂X, the mappingN:X→ZisL—completely continuous byDefinition 1.2.

2. Solutions at resonance

Assume that the following conditions on the function f(t,x1,|x2|) are satisfied:

(H1) there exists a constantA >0 such that for eachu∈domL\kerLsatisfying|u(t)|

> Afor allt∈[0, 1], we have n

i=1

μ_iξ_i

1

ξi

0

=0; (2.1) (H2) there exist functionsα,β,γ,ρ∈L¹[0, 1] and a constant∈[0, 1) such that for all

(x1,x2)∈R²and a.e.t∈[0, 1], we have either

ft,x1,x2≤ρ(t) +α(t)x1+β(t)x2+γ(t)x1 (2.2) or

ft,x1,x2≤ρ(t) +α(t)x1+β(t)x2+γ(t)x2; (2.3) (H3) there exists a constantB >0 such that for everyc∈Rwith|c|> B, we have either

c n i=1

μ_iξ_i

1

0(1−s)f(s,c, 0)ds−μ_i

ξi

0

ξ_i−sf(s,c, 0)ds

<0 (2.4) or

c n i=1

μ_iξ_i

1

0(1−s)f(s,c, 0)ds−μ_i

ξi

0

ξ_i−sf(s,c, 0)ds

>0. (2.5)

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Theorem 2.1. If (H0)–(H3) hold, then the boundary value problem (1.1)–(1.3) has at least one solution provided that

α1+β1<2

5. (2.6)

Proof. We construct an open bounded setΩ⊂Xthat satisfies the assumptions ofTheo- rem 1.3. Let

Ω1=

u∈domL\kerL:Lu=λNufor someλ∈(0, 1). (2.7) Foru∈Ω1, we haveu /∈kerL,λ=0 andNu∈ImL. But kerQ=ImLand, thus,

n i=1

μ_iξ_i

1

ξi

0

=0 (2.8) sinceQNu=0. It follows from (H1) that there existst0∈[0, 1] such that|u(t0)| ≤A.

Now,

u(0)= ut0

− ^t⁰

0 u(s)ds≤ut0+

t0

0

u(s)ds≤A+u∞. (2.9)

Also, sinceuis absolutely continuous, and, by symmetry,u(1/2)=0,u(1−t)=u(t), u(t)= − ^1/2

t u(s)ds. (2.10)

Hence

u∞≤1

2u1=1

2Lu1<1

2Nu1. (2.11)

Combining the above inequalities, we get u(0)< A+1

2Nu1. (2.12)

Observe that (I−P)u∈ImKP=domL∩kerPforu∈Ω1. Then, by (1.23) and (1.26), (I−P)u=KPL(I−P)u≤2L(I−P)u₁=2Lu1<2Nu1. (2.13) Using (2.12) and (2.13), we obtain

u =Pu+ (I−P)u≤ Pu+(I−P)u<u(0)+ 2Nu1< A+5 2Nu1,

(2.14) that is, for allu∈Ω1,

u< A+5

2Nu1. (2.15)

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If the second condition of (H2) is satisfied, then u∞,u∞≤ u ≤5

2

ρ1+α1u∞+β1u∞+γ1u_∞

+A, (2.16) and consequently,

u∞≤ 5 2−5α1

ρ1+β1u∞+γ1u_∞+2A 5

(2.17) or

u∞≤ 5β1

2−5α1u∞+ 5γ1

2−5α1u_∞+5ρ1+ 2A 2−5α1

. (2.18)

Also, by (2.16) and (2.17), u∞≤5

2α1u∞+5 2

ρ1+β1u∞+γ1u_∞+2A 5

≤ 5β1

2−5α1u∞+ 5γ1

2−5α1u_∞+5ρ1+ 2A 2−5α1

,

(2.19)

that is,

u∞≤ 5γ1

2−5α1+β1

u_∞+ 5ρ1+ 2A 2−5α1+β1

. (2.20)

But∈[0, 1) andα1+β1<2/5, so there existsM1>0 such thatu∞≤M1for all u∈Ω1. The inequality (2.18) then shows that there existsM2>0 such thatu∞≤M2for allu∈Ω1. Therefore,Ω1is bounded given the second condition of (H2). If, otherwise, the first part of (H2) holds, then with minor adjustments to the arguments above we derive the same conclusion.

Define

Ω2= {u∈kerL:Nu∈ImL}. (2.21) Thenu≡c∈Rand

Nu∈ImL=kerQ (2.22)

imply that n i=1

μiξi

1

0(1−s)f(s,c, 0)ds−μi ξi

0

ξi−sf(s,c, 0)ds

=0. (2.23)

Hence, by (H3),

u =c≤B, (2.24)

that is,Ω2is bounded.

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We take our isomorphism,J, to be the identity mapId: kerL→ImL, that is,Jc=cfor c∈R. Set

Ω3= {u∈kerL:−λJu+ (1−λ)QNu=0,λ∈[0, 1]}. (2.25) For everyc∈Ω3,

λc=(1−λ) n i=1

μiξi

1

0

ξi−sf(s,c, 0)ds

. (2.26)

Ifλ=1, thenc=0 and in the caseλ∈[0, 1), if|c|> B, then by (H3), λc²=(1−λ)c

n i=1

μiξi

1

0

ξi−sf(s,c, 0)ds

<0, (2.27) which, in either case, is a contradiction. If the other part of (H3) is satisfied, then we take

Ω3=

u∈kerL:λJu+ (1−λ)QNu=0,λ∈[0, 1] (2.28) and, again, obtain a contradiction. Thus, in either caseu =c≤Bfor allu∈Ω3, that is, Ω3is bounded.

Let Ωbe open and bounded such that³_i₌₁Ωi⊂Ω. Then the assumptions (i) and (ii) ofTheorem 1.3are fulfilled. ByDefinition 1.2, the mappingN isL-compact onΩ.

Lemma 1.5establishes thatLis Fredholm of index zero. It only remains to verify that the third assumption ofTheorem 1.3applies.

We apply the degree property of invariance under a homotopy. To this end, we define a homotopy

H(u,λ)= ±λJu+ (1−λ)QNu. (2.29)

Ifu∈kerL∩∂Ω, then

degQN|kerL∩∂Ω,Ω∩kerL, 0=degH(·, 0),Ω∩kerL, 0

=degH(·, 1),Ω∩kerL, 0

=deg±J,Ω∩kerL, 0

=0,

(2.30)

so, the third assumption ofTheorem 1.3is fulfilled and the proof is complete.

Acknowledgment

The author wishes to thank the anonymous referee for the comments and suggestions on improving the presentation of the paper.

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Nickolai Kosmatov: Department of Mathematics and Statistics, University of Arkansas at Little Rock, Little Rock, AR 72204-1099, USA

E-mail address:[email protected]

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