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# Existence of solutions for fourth order three-point boundary value problems on a half-line

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## Existence of solutions for fourth order three-pointboundary value problems on a half-line

B1

### and Ravi P. Agarwal

2

1Department of Mathematics, Ege University, Bornova, Izmir, 35100, Turkey

2Department of Mathematics, Texas A&M University-Kingsville, 700 University Blvd., Kingsville, TX 78363-8202, USA

Received 12 June 2015, appeared 12 October 2015 Communicated by Alberto Cabada

Abstract. In this paper, we apply Schauder’s fixed point theorem, the upper and lower solution method, and topological degree theory to establish the existence of unbounded solutions for the following fourth order three-point boundary value problem on a half- line

x0000(t) +q(t)f(t,x(t),x0(t),x00(t),x000(t)) =0, t∈(0,+), x00(0) =A, x(η) =B1, x0(η) =B2, x000(+) =C,

whereη ∈(0,+), but fixed, and f:[0,+R4Rsatisfies Nagumo’s condition.

We present easily verifiable sufficient conditions for the existence of at least one solu- tion, and at least three solutions of this problem. We also give two examples to illustrate the importance of our results.

Keywords: three-point boundary value problem, lower and upper solutions, half-line, Schauder’s fixed point theorem, topological degree theory.

2010 Mathematics Subject Classification: 34B15, 34B40.

### 1Introduction

In this paper, we develop an existence theory for fourth order ordinary differential equations together with boundary conditions on a half-line

x0000(t) +q(t)f(t,x(t),x0(t),x00(t),x000(t)) =0, t ∈(0,+), x00(0) = A, x(η) =B1, x0(η) =B2, lim

t→+x000(t) =x000(+) =C, (1.1) where η ∈ (0,+), but fixed, q: (0,+) → (0,+), f: [0,+R4R are continuous, and A,B1,B2R, C ≥ 0. By using the upper and lower solution method, we present easily verifiable sufficient conditions for the existence of unbounded solutions of (1.1).

BCorresponding author. Email: erbil.cetin@ege.edu.tr

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The upper and lower solution method has been successfully used to provide existence re- sults for two-point and multi-point boundary value problems (in short BVPs) for second-order and higher-order ordinary differential equations, see [6,9,10,14,19,20,26,27] and references therein. All of these works deal with problems on finite intervals. In recent years the study of BVPs on[0,+)has attracted several researchers, for instance see [3,5,11,12,15,16,24,25,28]

and references therein. In these works authors have applied either some fixed point theo- rem or a monotone iterative technique to establish the existence of bounded or unbounded solutions.

Fourth-order differential equations appear in mathematical modeling of physical, biolog- ical, and chemical phenomena such as viscoelastic and inelastic flows, deformation of beams and plate deflection problems [2,7,8,13,18,29]. On a finite interval fourth-order differential equations together with two-point boundary conditions have been studied in [1,17,21,22], and with multi-point conditions in [6,19,23,27]. It seems that [12] is the only paper which considers a particular fourth-order differential equation on [0,+) (with entirely different technique and boundary conditions than ours). Thus, to fill a gap in this paper we present an existence theory of unbounded solutions for the BVP (1.1).

The plan of our paper is as follows: in Section 2, we give some definitions and lemmas which we need to prove the main results. This includes the construction of Green’s function for a related fourth order boundary value problem with nonhomogeneous three-point bound- ary conditions, definitions of upper and lower solutions of (1.1), and Nagumo’s condition. In Section 3, we present two main results. In our first result we use Schauder’s fixed point theo- rem to establish the existence of at least one solution of (1.1) which lies between the assumed pair of upper and lower solutions. In our second result we assume the existence of two pairs of upper and lower solutions and employ the degree theory to prove the existence of at least three solutions of (1.1). We demonstrate the importance of our results through two illustrative examples.

### 2Preliminaries

We begin with constructing Green’s function for the linear boundary value problem x0000(t) +v(t) =0, t ∈(0,+),

x00(0) = A, x(η) =B1, x0(η) = B2, x000(+) =C. (2.1) Lemma 2.1. Let v ∈ C[0,+) and R

0 v(t)dt < +∞. Then the solution x ∈ C3[0,+)∩ C4(0,+)of the problem(2.1)can be expressed as

x(t) =B1+

B2−Aη−2 2

(t−η) + A

2(t2η2) +C

6(t3η3) +

Z

0 G(t,s)v(s)ds, where

G(t,s) =























 s

t2

2 −ηt+η

2

2

, s≤min{η,t}; t3

6 +s

2t

2 −ηts+η

2s 2 − s

3

6, t≤ s≤η;

st2 2 − s2t

2 − η2t 2 + s

3

6 + η

3

3 , η≤s ≤t;

t3 6 −η

2t 2 + η

3

3 , max{η,t} ≤s.

(2.2)

(3)

Proof. Sincev∈ C[0,+)andR

0 v(t)dt<+∞, we can integrate (2.1) from tto +∞, and use x000(+) =C, to get

x000(t) =C+

Z

t

v(s)ds.

Integrating the above equation on[0,t], applying Fubini’s theorem, and using x00(0) = A, we obtain

x00(t) =A+Ct+

Z t

0 sv(s)ds+

Z

t tv(s)ds.

Again integrating the above equation on[0,t], we find x0(t) = x0(0) +At+ C

2t2+

Z t

0

st−s

2

2

v(s)ds+

Z

t

t2

2v(s)ds. (2.3) Since x0(η) =B2, it follows that

x0(0) =

B2−Aη− 2 2

Z η

0

sη− s2 2

v(s)ds−

Z

η

η2

2 v(s)ds.

Hence from (2.3), we have x0(t) =

B2−Aη−

2

2

+At+C 2t2 +

Z t

0

s(t−η)v(s)ds+

Z η

t

t2 2 + s

2

2 −sη

v(s)ds +

Z

η

1

2(t2η2)v(s)ds, if t≤η

(2.4)

and

x0(t) =

B2−Aη−

2

2

+At+C 2t2 +

Z η

0 s(t−η)v(s)ds+

Z t

η

st−s

2

2 −η

2

2

v(s)ds +

Z

t

1

2(t2η2)v(s)ds, if η≤t.

(2.5)

When t ≤ η we integrate (2.4) fromt to η, and when η ≤ t we integrate (2.5) from ηto t, to obtain

x(t) =B1+

B2−Aη−

2

2

(t−η) + A

2(t2η2) +C

6(t3η3)

+

























Z t

0 s t2

2 −ηt+ η

2

2

v(s)ds+

Z η

t

t3 6 +s

2t

2 −ηts+η

2s 2 − s

3

6

v(s)ds +

Z

η

t3 6 − η

2t 2 + η

3

3

v(s)ds, t≤η;

Z η

0 s t2

2 −ηt+ η

2

2

v(s)ds+

Z t

η

st2 2 − s

2t 2 −η

2t 2 + s

3

6 +η

3

3

v(s)ds +

Z

t

t3 6 − η

2t 2 + η

3

3

v(s)ds, η≤t;

(4)

which is the same as x(t) =B1+

B2−Aη−

2

2

(t−η) + A

2(t2η2) + C

6(t3η3) +

Z

0 G(t,s)v(s)ds, ∀t∈ [0,+). This completes the proof of the lemma.

Let X=

x∈ C3[0,+): lim

t→+

x(t) 1+t3, lim

t→+

x0(t) 1+t2, lim

t→+

x00(t)

1+t and lim

t→+x000(t) exist

with the normkxk=max{kxk1,kxk2,kxk3,kxk4}, where kxk1= sup

t∈[0,+)

|x(t)|

1+t3, kxk2= sup

t∈[0,+)

|x0(t)|

1+t2, kxk3= sup

t∈[0,+)

|x00(t)|

1+t , kxk4= sup

t∈[0,+)

|x000(t)|.

Then by standard arguments, it follows that (X,k.k) is a Banach space. In what follows, we shall need the following modified version of the Arzelà–Ascoli lemma [4,24].

Lemma 2.2. Let M⊂X. Then M is relatively compact if the following conditions hold:

(i) M is bounded in X;

(ii) functions in {y : y = x

1+t3, x ∈ M}, {z : z = x0

1+t2, x ∈ M}, {u : u = 1x+00t, x ∈ M}and {w:w= x000(t), x∈ M}are locally equi-continuous on[0,+);

(iii) functions in {y : y = x

1+t3, x ∈ M}, {z : z = x0

1+t2, x ∈ M}, {u : u = 1x+00t, x ∈ M}and {w:w= x000(t), x∈ M}are equi-convergent at+∞.

Definition 2.3. A functionα∈ X∩ C4(0,+)is called a lower solution of (1.1) if

α0000(t) +q(t)f(t,α(t),α0(t),α00(t),α000(t))≥0, t∈ (0,+), (2.6) α00(0)≤ A, α(η)≤ B1, α0(η) =B2, α000(+)≤C. (2.7) Similarly, a functionβ∈ X∩ C3(0,+)is called an upper solution of (1.1) if

β0000(t) +q(t)f(t,β(t),β0(t),β00(t),β000(t))≤0, t∈(0,+), (2.8) β00(0)≥ A, β(η)≥B1, β0(η) =B2, β000(+)≥C. (2.9) Also, we sayα(β)is a strict lower solution (strict upper solution) for problem (1.1) if the above inequalities are strict.

Remark 2.4. If

α00(t)≤β00(t) for allt ∈(0,+), (2.10) then by integrating (2.10) and using the continuity ofα(t)andβ(t), and the fact thatα0(η) = B2= β0(η), it follows thatβ0(t)≤α0(t)for allt ∈[0,η)andα0(t)≤ β0(t)for allt∈[η,+). A further integration then yieldsα(t)≤β(t)for allt ∈[0,+).

(5)

Definition 2.5. Letα,β∈ X∩ C4(0,+)be a pair of lower and upper solutions of (1.1) satisfy- ing α00(t)≤ β00(t), t ∈[0,+). A continuous function f: [0,+R4Ris said to satisfy Nagumo’s condition with respect to the pair of functions α,β, if there exist a nonnegative functionφ∈ C[0,+)and a positive functionh∈ C[0,+)such that

|f(t,y,z,u,w)| ≤φ(t)h(|w|) (2.11) for all (t,y,z,u,w)∈ [0,η)×[α(t),β(t)]×[β0(t),α0(t)]×[α00(t),β00(t)]×Rand (t,y,z,u,w)∈ [η,+)×[α(t),β(t)]×[α0(t),β0(t)]×[α00(t),β00(t)]×R, and

Z

0

s

h(s)ds= +∞. (2.12)

### 3Main results

The following result provides sufficient conditions for the existence of at least one solution of the problem (1.1).

Theorem 3.1. Assume that α,β are lower and upper solutions of (1.1) satisfying α00(t) ≤ β00(t), t ∈ [0,+), and suppose that f: [0,+R4R is continuous satisfying Nagumo’s condition with respect to the pair of functionsα,β. Further, assume that

f(t,α(t),z,u,w)≤ f(t,y,z,u,w)≤ f(t,β(t),z,u,w) (3.1) and

f(t,y,α0(t),u,w)≤ f(t,y,z,u,w)≤ f(t,y,β0(t),u,w) (3.2) for (t,y,z,u,w) ∈ [0,η)×[α(t),β(t)]×[β0(t),α0(t)]×[α00(t),β00(t)]×R and (t,y,z,u,w) ∈ [η,+)×[α(t),β(t)]×[α0(t),β0(t)]×[α00(t),β00(t)]×R.If

Z

0 max{s, 1}q(s)ds<+∞,

Z

0 max{s, 1}φ(s)q(s)ds< + (3.3) and there exists a constantγ>1such that

m= sup

t∈[0,+)

(1+t)γq(t)φ(t)< +∞, (3.4) where φ(t)is the function in Nagumo’s condition of f , then (1.1) has at least one solution x ∈ X∩ C4(0,+)satisfying

α(t)≤x(t)≤β(t), t∈ [0,+),

β0(t)≤x0(t)≤α0(t), t ∈[0,η), α0(t)≤x0(t)≤ β0(t), t∈ [η,+), α00(t)≤x00(t)≤β00(t), |x000(t)|< N, t∈ [0,+);

here, N is a constant depending onα,β, h and C.

Proof. We can choose an

r ≥max (

sup

t∈[0,+)

|α000(t)|, sup

t∈[0,+)

|β000(t)|,C )

(3.5)

(6)

and anN >rsuch that Z N

r

s

h(s)ds>m sup

t∈[0,+)

β00(t)

(1+t)γ − inf

t∈[0,+)

α00(t)

(1+t)γ + γ

γ−1max{kβk3,kαk3}

!

. (3.6) We define the following auxiliary functions

f0(t,y,z,u,w) =





f(t,β,z,u,w), y >β(t);

f(t,y,z,u,w), α(t)≤y≤ β(t); f(t,α,z,u,w), y <α(t),

f1(t,y,z,u,w) =





















t∈[0,η),





f0(t,y,β0,u,w), z< β0(t);

f0(t,y,z,u,w), β0(t)≤z≤α0(t); f0(t,y,α0,u,w), z>α0(t),

t∈[η,+),





f0(t,y,β0,u,w), z> β0(t);

f0(t,y,z,u,w), α0(t)≤z≤ β0(t); f0(t,y,α0,u,w), z<α0(t),

and

f(t,y,z,u,w) =





f1(t,y,z,β00,w) + 1+|β00(t)−u

β00(t)−u|, u>β00(t);

f1(t,y,z,u,w), α00(t)≤ u≤β00(t); f1(t,y,z,α00,w) + 1+|α00(t)−u

α00(t)−u|, u<α00(t),

(3.7)

where

w =





N, w> N;

w, −N≤ w≤ N;

−N, w<−N.

Now we consider the modified problem

x0000(t) +q(t)f(t,x(t),x0(t),x00(t),x000(t)) =0, t∈(0,+),

x00(0) = A, x(η) =B1, x0(η) =B2, x000(+) =C. (3.8) As an application of Schauder’s fixed point theorem first we will prove that (3.8) has at least one solutionx. To show this, forx∈X, we define two operators as follows

(T1x)(t) =

Z

0 G(t,s)q(s)f(s,x(s),x0(s),x00(s),x000(s))ds, t∈ [0,+) and

(Tx)(t) = B1+

B2−Aη−

2

2

(t−η) + A

2(t2η2) + C

6(t3η3) + (T1x)(t), t ∈[0,+).

(3.9)

Now we shall prove that T: X → X is completely continuous. We divide the proof in the following three parts.

(7)

(1) T: X → X is well defined. For each x ∈ X, in view of (2.11), (3.3) and (3.7) as in [5], we have

Z

0 q(s)f(s,x(s),x0(s),x00(s),x000(s))ds

Z

0 q(s)(H0φ(s) +1)ds

Z

0 max{1,s}q(s)(H0φ(s) +1)ds<+,

(3.10)

where H0=max0t≤kxk4h(t). For x∈ X, we find from (3.10) that Z

1 sq(s)(H0φ(s) +1)ds≤

Z

0 max{s, 1}q(s)(H0φ(s) +1))ds<+∞, (3.11) which implies

t→+limtq(t)(H0φ(t) +1) =0. (3.12) Since

Z

t q(s)(H0φ(s) +1)ds≤

Z

t sq(s)(H0φ(s) +1))ds< +∞, t ≥1, (3.13) it also follows that

t→+lim Z

t q(s)(H0φ(s) +1)ds=0. (3.14) Thus by Lebesgue’s dominated convergence theorem, l’Hôpital’s rule, (3.12) and (3.14), we find

t→+lim

(T1x)(t) 1+t3

≤ lim

t→+ Z

0

|G(t,s)|

1+t3 q(s)(H0φ(s) +1)ds

= lim

t→+

"

Z η

0 s(t22ηt+ η22)

1+t3 q(s)(H0φ(s) +1)ds +

Z t

η

(st22s22tη22t+s63 +η33)

1+t3 q(s)(H0φ(s) +1)ds +

Z

t

(t63η22t+η33)

1+t3 q(s)(H0φ(s) +1)ds

#

= lim

t→+ Z t

η

(st− s22η22)

3t2 q(s)(H0φ(s) +1)ds+ lim

t→+

(−η22t +t63 +η33)

3t2 q(t)(H0φ(t) +1) + lim

t→+ Z

t

(t22η22)

3t2 q(s)(H0φ(s) +1)ds− lim

t→+

(t63η22t+ η33)

3t2 q(t)(H0φ(t) +1)

= lim

t→+ Z t

η

s

6tq(s)(H0φ(s) +1)ds+ lim

t→+

(t2t22η22)

6t q(t)(H0φ(t) +1) + lim

t→+ Z

t

t

6tq(s)(H0φ(s) +1)ds− lim

t→+

(t22η22)

6t q(t)(H0φ(t) +1)

= 1 6 lim

t→+ Z

t q(s)(H0φ(s) +1)ds=0,

(8)

which implies that limt→+ (T1x)(t)

1+t3 =0. Therefore, it follows that

t→+lim

(Tx)(t)

1+t3 = lim

t→+

B1+ (B2−Aη− 22)(t−η) + A2(t2η2) + C6(t3η3) 1+t3

+ lim

t→+

(T1x)(t) 1+t3

= C 6.

By Lebesgue’s dominated convergence theorem, l’Hôpital’s rule, (3.12) and (3.14), we have

t→+lim

(T1x)0(t) 1+t2

≤ lim

t→+

"

Z η

0

s(t−η)

1+t2 q(s)(H0φ(s) +1)ds +

Z t

η

(st−s22η22)

1+t2 q(s)(H0φ(s) +1)ds +

Z

t 1

2(t2η2)

1+t2 q(s)(H0φ(s) +1)ds

#

= lim

t→+

Rt

ηsq(s)(H0φ(s) +1)ds+ (t2t22η22)q(t)(H0φ(t) +1) 2t

+ lim

t→+

R

t tq(s)(H0φ(s) +1)ds−12(t2η2)q(t)(H0φ(t) +1) 2t

= 1 2 lim

t→+ Z

t q(s)(H0φ(s) +1)ds=0, which implies that limt→+ (T1x)0(t)

1+t2 =0. Therefore, it follows that

t→+lim

(Tx)0(t)

1+t2 = lim

t→+

(B2−Aη− 22) +At+ C2t2

1+t2 + lim

t→+

(T1x)0(t) 1+t2 = C

2.

Again, using Lebesgue’s dominated convergence theorem, l’Hôpital’s rule, (3.12) and (3.14), we obtain

t→+lim

(T1x)00(t) 1+t

= lim

t→+

1 1+t

Z t

0 sq(s)f(s,x(s),x0(s),x00(s),x000(s))ds +

Z

t tq(s)f(s,x(s),x0(s),x00(s),x000(s))ds

lim

t→+

1 1+t

Z t

0

sq(s)(H0φ(s) +1)ds +

Z

t

tq(s)q(s)(H0φ(s) +1)ds

= lim

t→+ Z

t

q(s)(H0φ(s) +1)ds=0, which implies that limt→+ (T1x)00(t)

1+t =0. Hence, we find

t→+lim

(Tx)00(t)

1+t = lim

t→+

A+Ct

1+t + lim

t→+

(T1x)00(t) 1+t = C.

Now from (3.12), we have

t→+lim Z

t q(s)f(s,x(s),x0(s),x00(s),x000(s))ds

≤ lim

t→+ Z

t q(s)(H0φ(s) +1)ds=0,

(9)

and hence,

t→+lim(Tx)000(t) = lim

t→+C+

Z

t q(s)f(s,x(s),x0(s),x00(s),x000(s))ds=C.

Consequently, it follows that Tx∈ X.

(2) T: X →Xis continuous. For any convergent sequence xn→ xin X, we have

xn(t)→x(t), x0n(t)→x0(t), x00n(t)→x00(t), x000n(t)→x000(t), n→+∞, t∈[0,+). Thus the continuity of f implies that

|f(s,xn(s),x0n(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))| →0, n→+, s∈ [0,+). Since x000n(t)→x000(t), we have sup

nN

kxnk4 <+∞. Let

H1 = max

0tmax{kxk4, supnNkxnk4}h(t). Then we obtain

Z

0 sq(s)|f(s,xn(s),x0n(s),x00n(s),x000(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds

≤2 Z

0 sq(s)(H1φ(s) +1)ds<+∞.

(3.15)

Therefore from Lebesgue’s dominated convergence theorem and (3.15) for η ≤ t it follows that

|Txn(t)−Tx(t)|

1+t3 = |T1xn(t)−T1x(t)|

1+t3

=

Z

0

G(t,s)

1+t3q(s)(f(s,xn(s),xn0(s),x00n(s),xn000(s))− f(s,x(s),x0(s),x00(s),x000(s)))ds

Z η

0

s(t−η)2

2(1+t3)q(s)|f(s,xn(s),x0n(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds +

Z t

η

s(t22st2 +ηt2 +s62 + η32) 1+t3 q(s)

× |(f(s,xn(s),x0n(s),xn00(s),x000(s))−f(s,x(s),x0(s),x00(s),x000(s))|ds +

Z

t

s(t62 + ηt2 +η32) 1+t3 q(s)

× |f(s,xn(s),x0n(s),xn00(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds

Z

0 s t2

1+t3q(s)|f(s,xn(s),x0n(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds

Z

0

sq(s)|f(s,xn(s),x0n(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds

(10)

and for t≤η

|Txn(t)−Tx(t)|

1+t3 = |T1xn(t)−T1x(t)|

1+t3

=

Z

0

G(t,s)

1+t3 q(s)(f(s,xn(s),x0n(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s)))ds

Z t

0

s(t−η)2

2(1+t3)q(s)|f(s,xn(s),x0n(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds +

Z η

t

s(t62 +st2 +ηt+ η22 +s62) 1+t3 q(s)

× |(f(s,xn(s),x0n(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds +

Z

η

s(t62 + η22 + η32) 1+t3 q(s)

× |f(s,xn(s),x0n(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds

Z

0 s

2 3

1+t3q(s)|f(s,xn(s),xn0(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds

2 3

Z

0 sq(s)|f(s,xn(s),x0n(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds.

Combining the above inequalities, we obtain kTxn−Txk1

≤maxn

1,32oZ

0 sq(s)|f(s,xn(s),x0n(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds, which approaches zero asn→. Similarly, from Lebesgue’s dominated convergence theorem and (3.15) forη≤t, we find

|(Txn)0(t)−(Tx)0(t)|

1+t2 = |(T1xn)0(t)−(T1x)0(t)|

1+t2

Z η

0

s(t−η)

1+t2 q(s)|f(s,xn(s),x0n(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds +

Z t

η

(st− s22η22) 1+t2 q(s)

× |(f(s,xn(s),x0n(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds +

Z

t 1

2(t2η2) 1+t2 q(s)

× |f(s,xn(s),x0n(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds

Z

0

st

1+t2q(s)|f(s,xn(s),x0n(s),xn00(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds

1 2

Z

0

sq(s)|f(s,xn(s),x0n(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds

(11)

and for t≤η

|(Txn)0(t)−(Tx)0(t)|

1+t2 = |(T1xn)0(t)−(T1x)0(t)|

1+t2

=

Z

0

G(t,s)

1+t2q(s)(f(s,xn(s),xn0(s),x00n(s),xn000(s))− f(s,x(s),x0(s),x00(s),x000(s)))ds

Z t

0

s(η−t)

1+t2 q(s)|f(s,xn(s),xn0(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds +

Z η

t

(sη−2s22) 1+t2 q(s)

× |(f(s,xn(s),x0n(s),xn00(s),xn000(s))−f(s,x(s),x0(s),x00(s),x000(s))|ds +

Z

η 1

2(η2−t2) 1+t2 q(s)

× |f(s,xn(s),x0n(s),xn00(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds

Z

0

1+t2q(s)|f(s,xn(s),xn0(s),x00n(s),xn000(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds

η Z

0 sq(s)|f(s,xn(s),x0n(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds.

Combining the above inequalities, we obtain kTxn−Txk2

≤max1

2,η Z

0 sq(s)|f(s,xn(s),x0n(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds, which approaches zero as n→∞. We also have

kTxn−Txk3

= sup

t∈[0,+)

|(Txn)00(t)−(Tx)00(t)|

1+t = sup

t∈[0,+)

|(T1xn)0(t)−(T1x)0(t)|

1+t

sup

t∈[0,+)

Z t

0

s 1+tq(s)

× |f(s,xn(s),x0n(s),xn00(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds +

Z

t

t 1+tq(s)

× |f(s,xn(s),x0n(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds

Z

0

sq(s)|f(s,xn(s),x0n(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds,

(12)

which approaches zero asn→∞. Finally, from (3.15) we have k(Txn)−(Tx)k4

= sup

t∈[0,+)

|(Txn)000(t)−(Tx)000(t)|= sup

t∈[0,+)

|(T1xn)000(t)−(T1x)000(t)|

= sup

t∈[0,+)

Z

t q(s)(f(s,xn(s),x0n(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s)))ds

Z

0 q(s)|f(s,xn(s),x0n(s),x00n(s),x000n(s))− f(s,x(s),x0(s),x00(s),x000(s))|ds,

which approaches zero asn→. As a result, we conclude thatkTxn−Txk →0, asn→+; and thereforeT: X→Xis continuous.

(3) T: X → X is compact. For this it is suffices to show that T maps bounded subsets of X into relatively compact sets. We assume thatM1is any bounded subset ofX, then forx ∈ M1, we let H2 =max0t≤kxk4,xM1h(t)<+∞. Now following as above, we get

kTxk1= sup

t∈[0,+)

|Tx(t)|

1+t3

= sup

t∈[0,+)

B1+ (B2−Aη− 22)(t−η) + A2(t2η2) + C6(t3η3) 1+t3

+

Z

0

G(t,s)

1+t3q(s)f(s,x(s),x0(s),x00(s),x000(s))ds

≤ |B1|+

B2−Aη− 2 2

+ |A| 2 + C

6 +max

1,7η2

3 Z

0 sq(s)|f(s,x(s),x0(s),x00(s),x000(s))|ds

≤ |B1|+

B2−Aη−

2

2

+ |A| 2 + C

6 +max

1,7η2 3

Z

0 sq(s)(H2φ(s) +1)ds

< +,

kTxk2 = sup

t∈[0,+)

|(Tx)0(t)|

1+t2

= sup

t∈[0,+)

(B2−Aη− 22) +At+ C2t2 1+t2

+

Z

0

∂G(t,s)

∂t

1+t2q(s)f(s,x(s),x0(s),x00(s),x000(s))ds

B2−Aη−

2

2

+|A|+C 2 +max

1 2,η

Z

0 sq(s)|f(s,x(s),x0(s),x00(s),x000(s))|ds

B2−Aη−

2

2

+|A|+C

2 +max 1

2,η Z

0 sq(s)(H2φ(s) +1)ds

< +∞,

Hivatkozások

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