Positive solutions of boundary value problems with nonlinear nonlocal boundary conditions

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http://dx.doi.org/10.7494/OpMath.2016.36.1.69

POSITIVE SOLUTIONS

OF BOUNDARY VALUE PROBLEMS

WITH NONLINEAR NONLOCAL

BOUNDARY CONDITIONS

Seshadev Padhi, Smita Pati, and D.K. Hota

Communicated by Alexander Domoshnitsky

Abstract.We consider the existence of at least three positive solutions of a nonlinear first order problem with a nonlinear nonlocal boundary condition given by

x′₍_t_{) =}_r₍_t₎_x₍_t_{) +}

m

X

i=1

fi(t, x(t)), t∈[0,1],

λx(0) =x(1) +

n

X

j=1

Λj(τj, x(τj)), τj∈[0,1],

wherer: [0,1]→[0,∞) is continuous; the nonlocal points satisfy 0≤τ1< τ2< . . . < τn≤1,

the nonlinear functionfi andτj are continuous mappings from [0,1]×[0,∞) →[0,∞) for

i= 1,2, . . . , mandj= 1,2, . . . , nrespectively, andλ >0 is a positive parameter.

Keywords: positive solutions, Leggett-Williams fixed point theorem, nonlinear boundary conditions.

Mathematics Subject Classification:34B08, 34B18, 34B15, 34B10.

1. INTRODUCTION

Consider the first order boundary value problem with a nonlinear nonlocal boundary condition

x′

(t) =r(t)x(t) +

m X

i=1

fi(t, x(t)), t∈[0,1], (1.1)

λx(0) =x(1) +

n X

j=1

Λj(τj, x(τj)), τj∈[0,1], (1.2)

c

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wherer: [0,1]→[0,∞) is continuous,fi: [0,1]×[0,∞)→[0,∞), τj : [0,1]×[0,∞)→

[0,∞) are continuous,i= 1,2, . . . , m andj = 1,2, . . . , n, the nonlocal points satisfy 0≤τ1< τ2< . . . < τn≤1, and the scalarλsatisfies

λ >exp

Z1

0

r(η)dη

. (1.3)

The motivation of this present paper has come from a recent paper due to Anderson [1], who used the Leggett-Williams multiple fixed point theorem [8] to establish three positive solutions of the boundary value problem (1.1) and (1.2). For completeness, we state here the statement of the theorem.

Assume that the nonlinear functions Λj satisfy

0≤xψj(t, x)≤Λj(t, x)≤xΨj(t, x), t∈[0,1], x∈[0,∞) (1.4)

for some positive continuous functionsψj,Ψj : [0,1]×[0,∞)→[0,∞). Set

βj = max

[0,1]×[0,c]Ψj(t, x) and αj=[0,1]min×[0,d]ψj(t, x) (1.5)

for some real constantscandd. Then, using the Leggett-Williams multiple fixed point theorem, Anderson proved the following theorem.

Theorem 1.1. Suppose that (1.4)holds and the scalarλsatisfies

λ > 

1 + m X

j=1

βj 

exp 



1

Z

0

r(η)dη 

>1. (1.6)

Further, suppose that there exist constants 0< c1< c2< λc2≤c4 such that

(F1) fi(t, x)≤ M c_m4 fort∈[0,1]andx∈[0, c4];

(F2) fi(t, x)> N c_m2 fort∈[0,1]andx∈[c2, λc2];

(F3) fi(t, x)< M c_m1 fort∈[0,1]andx∈[0, c1]

fori= 1,2, . . . , m, where

M = ₁ 1

R

0

G(1, s)ds "

1−

exp

1

R

0

r(η)dηPn j=1βj

λ−exp

1

R

0

r(η)dη #

(1.7)

and

N = ₁ 1

R

0

G(0, s)ds "

1−

Pn j=1αj

λ−exp

1

R

0

r(η)dη #

. (1.8)

Then the boundary value problem (1.1)and (1.2)has at least three positive solutions x1, x2 and x3 satisfying kx1k =x1(1)< c1, c2 < ψ(x2) =x2(0),kx3k =x3(1)> c1

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We consider a Banach space X =C[0,1] endowed with the sup norm. Define a coneK onX by

K={x∈X;x≥0 andxincreasing}, (1.9) and a nonnegative concave continuous functionalψonK by

ψ(x) = min

t∈[0,1]x(t) =x(0), x∈K. (1.10)

Now, we state the main result of this paper.

Theorem 1.2. Assume that (1.3) holds. Suppose that there exist three constants

0< c1< c2< λc2≤c4 such that

(H1) λ

m X i=1 1 Z 0 exp − s Z 0

r(η)dη

fi(s, x(s))ds+ n X

j=1

Λj(τj, x(τj))≤

λ−exp

1

R

0

r(η)dηc4

exp

1

R

0

r(η)dη

forτj ∈[0,1], j= 1,2, . . . , nandx∈[0, c4];

(H2)

m X i=1 1 Z 0 exp Z1 s

r(η)dη

j=1

Λj(τj, x(τj))> c2

λ−exp

Z1

0

r(η)dη

forτj ∈[0,1],j= 1,2, . . . , n andx∈[c2, λc2];

and

(H3) λ

m X i=1 1 Z 0 exp − s Z 0

r(η)dη

j=1

Λj(τj, x(τj))< λ−exp

1

R

0

r(η)dη c1 exp 1 R 0

r(η)dη

for τj ∈ [0,1], j = 1,2, . . . , n and x ∈ [0, c4] hold. Then the boundary

value problem (1.1) and (1.2) has at least three positive solutions x1, x2 and

x3 satisfying kx1k=x1(1)< c1, c2< ψ(x2) =x2(0) and kx3k = x3(1) > c1 with

ψ(x3) =x3(0)< c4.

We, now show that the conditions of Theorem 1.1 imply the conditions of Theo-rem 1.2. In fact, (1.6) implies (1.3). First, suppose that (1.4) and (F1) hold, that is,

fi(t, x)≤M c_m4 fort∈[0,1] andx∈[0, c4]. Then,

λ m X i=1 1 Z 0 exp  − s Z 0

r(η)dη 

fi(s, x(s))ds+ n X

j=1

Λj(τj, x(τj))

≤λM c4 m m X i=1 1 Z 0 exp  − s Z 0

r(η)dη 

ds+c4

n X

j=1

βj

≤λM c4 1 Z 0 exp  − s Z 0

r(η)dη 

ds+c4

n X

j=1

βj

=c4

 λM 1 Z 0 exp  − s Z 0

r(η)dη 

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Using the value ofM, which is given in Theorem 1.1, the above inequality gives λ m X i=1 1 Z 0 exp  − s Z 0

r(η)dη 

j=1

Λj(τj, x(τj))

=c4

" λ 1 Z 0 exp  − s Z 0

r(η)dη  ds 1 1 R 0

G(1, s)ds

1−

exp

1

R

0

r(η)dηPn j=1βj

λ−exp

1

R

0

r(η)dη ! + n X j=1 βj #

=c4

" λ 1 Z 0 exp  − s Z 0

r(η)dη 

ds

λ−exp

1

R

0

r(η)dη λ 1 R 0 exp 1 R s

r(η)dηds

× 1−

exp

1

R

0

r(η)dηPn j=1βj

λ−exp

1

R

0

r(η)dη ! + n X j=1 βj #

=c4

"

1

R

0

exp−

s R

0

r(η)dηds

1 R 0 exp 1 R s

r(η)dηds 

λ−exp

Z1

0

r(η)dη

−exp

Z1

0

r(η)dη n X j=1 βj  + n X j=1 βj #

=c4

"

1

R

0

exp−

s R

0

r(η)dηdsexp

1

R

0

r(η)dη

1 R 0 exp 1 R s

r(η)dηds

λ−exp

1

R

0

r(η)dη

exp

1

R

0

r(η)dη

− n X j=1 βj ! + n X j=1 βj # =

λ−exp

1

R

0

r(η)dη

exp

1

R

0

r(η)dη c4,

which implies that (H1) holds. In a similar way, we can show that if (1.4) and (F3)

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Finally, suppose that (F2) holds. In addition, suppose that (1.4) holds. Then for

c2≤x≤λc2,t∈[0,1], using the value ofN, given in Theorem 1.1, we obtain

m X i=1 1 Z 0 exp   1 Z s

r(η)dη 

j=1

Λj(τj, x(τj))

>N c2 m m X i=1 1 Z 0 exp 1 Z s

r(η)dη !

ds+c2

n X

j=1

αj

=c2

" N 1 Z 0 exp   1 Z s

r(η)dη 

ds+ n X

j=1

αj #

=c2

" 1 Z 0 exp   1 Z s

r(η)dη  ds 1 1 R 0

G(0, s)ds

1−

Pn j=1αj

α−exp

1

R

0

r(η)dη ! + n X j=1 αj #

=c2

" 1 R 0 exp 1 R s r(η)dη

ds

λ−exp(

1

R

0

r(η)dη)

1 R 0 exp 0 R s r(η)dη

exp

1

R

0

r(η)dη

ds

1−

Pn j=1αj

λ−exp

1

R

0

r(η)dη ! + n X j=1 αj #

=c2



λ−exp





1

Z

0

r(η)dη 

− n X

j=1

αj+ n X j=1 αj  

=c2



λ−exp





1

Z

0

r(η)dη 

 

.

This shows that condition (1.4) and (F2) implies condition (H2). Thus, Theorem 1.2

provides a better condition than the condition given in Theorem 1.1.

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2. PRELIMINARIES

Let X be a Banach space, K be a cone in X, and ψ be a nonnegative continuous functional onK. Further, leta, b, c >0 be constants. Define

Ka ={x∈K;kxk< a}

and

K(ψ, b, c) ={x∈K;ψ(x)≥b,kxk ≤c}.

Theorem 2.1(Leggett-Williams multiple fixed point theorem, [8]). LetX= (X,k·k)

be a Banach space and K ⊂X be a cone, and c4 > 0 be a constant. Suppose that

there exists a concave nonnegative continuous functionψ on K with ψ(x)≤ kxk for x∈Kc₄ and letA:Kc₄ →Kc₄ be a continuous compact map. Assume that there are numbersc1, c2 andc3 with 0< c1< c2< c3≤c4 such that

(i) {x∈K(ψ, c2, c3) :ψ(x)> c2} 6=∅andψ(Ax)> c2 for all x∈K(ψ, c2, c3);

(ii) kAxk< c1 for allx∈Kc1;

(iii) ψ(Ax)> c2 for allx∈K(ψ, c2, c4)withkAxk> c3.

Then A has at least three fixed points x1, x2 and x3 in Kc₄. Furthermore, we have x1∈Kc₁,x2∈ {x∈K(ψ, c2, c4) :ψ(x)> c2}, andx3∈Kc₄\ {K(ψ, c2, c4)∪Kc₁}.

3. PROOF OF THEOREM 1.2

Consider an operatorA:K→X by

Ax(t) =

m X

i=1 1

Z

0

G(t, s)fi(s, x(s))ds+

exp

t R

0

r(η)dηPn

j=1Λj(τj, x(τj))

λ−exp

1

R

0

r(η)dη

, (3.1)

whereG(t, s) is the Green’s Kernel, given by

G(t, s) =

exp

t R

s

r(η)dη

λ−exp

1

R

0

r(η)dη

×

 



λ if 0≤s≤t≤1,

exp(R1

0 r(η)dη) if 0≤t≤s≤1.

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The operatorAin (3.1) can be rewritten as

Ax(t) =

m X

i=1

t Z

0

λexp

t R

s

r(η)dη

λ−exp

t R

s

r(η)dη

fi(s, x(s))ds

+

m X

i=1 1

Z

t

exp

1

R

0

r(η)dηexp

t R

s

r(η)dη

λ−exp

1

R

0

r(η)dη

fi(s, x(s))ds

+ exp

t R

0

r(η)dηPn

λ−exp

1

R

0

r(η)dη

.

(3.3)

Letx∈K. Then the fact thatfi, i= 1,2, . . . , m and Λj, j= 1,2, . . . , nare positive

imply that Ax≥0 for all t∈[0,1]. We claim that fixed points of the operator Aare the solutions of the boundary value problem (1.1) and (1.2). In fact, ifx=Ax, then from (3.3), we have

λx(0)−x(1) =

m X

i=1 1

Z

0

λexp

1

R

s

r(s)ds

λ−exp

1

R

0

r(s)ds

fi(s, x(s))ds

+λ Pn

λ−exp

1

R

0

r(η)dη

−

m X

i=1 1

Z

0

λexp

1

R

s

r(η)dη

λ−exp

1

R

0

r(η)dη

fi(s, x(s))ds

−

exp

1

R

0

r(η)dη

λ−exp

1

R

0

r(η)dη n X

j=1

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Hence, the boundary condition (1.2) is satisfied. Again differentiating (1.9) with re-spect tot, with Ax=x, we obtain

x′ (t) =

m X

i=1

t Z

0

λr(t) exp

t R

s

r(η)dη

λ−exp

1

R

0

r(η)dη

fi(s, x(s))ds

+

m X

i=1

λ

λ−exp

1

R

0

r(η)dη

fi(t, x(t))

+

m X

i=1 1

Z

t

r(t) exp

1

R

0

r(η)dηexp

t R

s

r(η)dη

λ−exp

1

R

0

r(η)dη

fi(s, x(s))ds

−

m X

i=1

exp

1

R

0

r(η)dη

λ−exp

1

R

0

r(η)dη

fi(t, x(t))

+ exp

t R

0

r(η)dηr(t)

λ−exp

1

R

0

r(η)dη n X

j=1

Λj(τj, x(τj))

=r(t)

" m X

i=1

t Z

0

λ

exp

t R

s

r(η)dη

λ−exp

1

R

0

r(η)dη

fi(s, x(s))ds #

+r(t)

" m X

i=1 1

Z

t

exp

1

R

0

r(η)dηexp

1

R

s

r(η)dη

λ−exp

1

R

0

r(η)dη

fi(s, x(s))ds #

+r(t)

" exp _Rt

0

r(η)dη

λ−exp

1

R

0

r(η)dη n X

j=1

Λj(τj, x(τj)) #

+

m X

i=1

fi(t, x(t))

=r(t)x(t) +

m X

i=1

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which shows thatx(t) satisfies (1.1). Moreover,

x′

(t) = (Ax)′

(t) =r(t)x(t) +

m X

i=1

fi(t, x(t)), t∈[0,1]

implies that x is increasing and A : K → K. Further, one may verify that A is completely continuous. We, now show that all conditions of Theorem 2.1 are satisfied. In order to use Theorem 2.1, we use λc2 in place ofc3. Forx∈K, we havex(0) =

ψ(x)≤ kxk = x(1), since x∈ K. Let x∈ Kc4, that is, kxk ≤c4. Then we obtain,

using (H1),

kAxk=Ax(1)

=

m X

i=1 1

Z

0

G(1, s)fi(s, x(s))ds+

exp

1

R

0

r(η)dηPn

λ−exp

1

R

0

r(η)dη

=

exp

1

R

0

r(η)dη

λ−exp

1

R

0

r(η)dη 

λ m X

i=1 1

Z

0

exp−

s Z

0

r(η)dηfi(s, x(s))ds+ n X

j=1

Λj(τj, x(τj)) 



≤c4,

that is,A:Kc₄ →Kc₄. In a similar way, using (H3), one can prove condition (ii) of

Theorem 2.1, thatA:Kc₁→Kc₁.

In order to verify condition (i) of Theorem 2.1, we choosexk(t) =λc2fort∈[0,1].

Since ψ(xk(t)) = mint∈[0,1]xk(t) = λc2 > c2; c2 ≤ ψ(x),kxk = λc2, then the set {x∈ K;c2 ≤ ψ(x),kxk ≤ λc2} 6= ∅. Let x ∈ K(ψ, c2, c3). Then c2 ≤ψ(x) ≤ x≤ kxk=x(1) =λc2 fort∈[0,1], that is,c2≤x(t)≤λc2 fort∈[0,1] holds. Hence

ψ(Ax) =Ax(0)

=

m X

i=1 1

Z

0

G(0, s)fi(s, x(s))ds+ Pn

λ−exp

1

R

0

r(η)dη

=

m X

i=1 1

Z

0

exp

0

R

s

r(η)dηexp

1

R

0

r(η)dη

λ−exp

1

R

0

r(η)dη

j=1

Λj(τj, x(τj))

λ−exp

1

R

0

r(η)dη

= 1

λ−exp

1

R

0

r(η)dη 

 m X

i=1 1

Z

0

exp

1

Z

s

r(η)dηfi(s, x(s))ds+ n X

j=1

Λj(τj, x(τj)) 

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holds. Then, using (H2), the above inequality yields that

ψ(Ax)> c2 for x∈ψ(K, c2, c3), t∈[0,1],

that is, condition (i) of Theorem 2.1 is satisfied.

Finally, suppose that x∈K(ψ, c2, c4) withkAxk> λc2. Then

ψ(Ax) =Ax(0)≥ 1

λAx(1) =

kAxk

λ = λc2

λ =c2

implies that condition (iii) of Theorem 2.1 is satisfied. Consequently, the boundary value problem (1.1) and (1.2) has at least three positive solutions. This completes the proof of the theorem.

Acknowledgments

The authors are thankful to the anonymous referee for their suggestions in improving the results of this paper.

Smita Pati: This work is supported by National Board for Higher Mathematics, Department of Atomic Energy, Govt. of India, vide Post Doctoral Fellowship no. 2/40(49)/2011-R&D-II/3705 dated 25 April 2012.

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Seshadev Padhi ses_2312@yahoo.co.in

Birla Institute of Technology Department of Mathematics Mesra, Ranchi – 835215, India

Smita Pati

spatimath@yahoo.com

Birla Institute of Technology Department of Mathematics Mesra, Ranchi – 835215, India

D.K. Hota

deepakhota.kumar@gmail.com

Indira Gandhi Institute of Technology Department of Mechanical System Design Sarang – 759146, India