Existence of Multiple Positive Solutions for p-Laplacian Fractional Order Boundary Value Problems

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ISSN 2291-8639

Volume 6, Number 1 (2014), 63-81 http://www.etamaths.com

EXISTENCE OF MULTIPLE POSITIVE SOLUTIONS FOR

p-LAPLACIAN FRACTIONAL ORDER BOUNDARY VALUE

PROBLEMS

K. R. PRASAD1 _{AND B. M. B. KRUSHNA}2,∗

Abstract. This paper deals with the existence of at least one and multiple positive solutions forp-Laplacian fractional order two-point boundary value problems,

Dq2

0+

φp

Dq1

0+y(t)

=f(t, y(t)), t∈(0,1),

y(j)(0) = 0, j= 0,1,2,· · ·, n−2, y(q4)_{(1) = 0,} φp

Dq1

0+y(0)

= 0 =Dq3

0+

φp

Dq1

0+y(1)

,

whereq2 ∈(1,2], q1 ∈(n−1, n], n≥2, q3 ∈(0,1], q4 ∈[1, q1−1] is a fixed

integer, φp(y) = |y|p−2y, p > 1, φ−p1 = φq, 1/p+ 1/q = 1, by applying

Krasnosel’skii and five functionals fixed point theorems.

1. _Introduction

The goal of differential equations is to understand the phenomena of nature by developing mathematical models. Fractional calculus is the field of mathematical analysis, which deals with investigation and applications of derivatives and inte-grals of an arbitrary order. Among all, a class of differential equations governed by nonlinear differential operators appears frequently and generated great deal of interest in studying such problems. In this theory, the most applicable operator is the classicalp-Laplacian, given byφp(y) =|y|p−2_{y, p >}₁_._{These types of problems} arise in applied fields such as turbulent flow of a gas in a porous medium, modeling of viscoelastic flows, biophysics, plasma physics and material science.

The existence of positive solutions of boundary value problems (BVPs) as-sociated with integer order differential equations were studied by many authors [11, 1, 9, 2, 19] and extended top-Laplacian boundary value problems [17, 14, 4, 23]. Later, these results are further extended to fractional order boundary value prob-lems [6, 10, 5, 7, 8, 20, 21, 22] by utilizing various fixed point theorems on cones. In recent years, researchers are concentrating on the theory of fractional order bound-ary value problems related withp-Laplacian operator.

2010Mathematics Subject Classification. 26A33, 34B18, 35J05.

Key words and phrases. Fractional derivative, p-Laplacian, boundary value problem, two-point, Green’s function, positive solution.

c

2014 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

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In 2012, Chai [7] investigated the existence and multiplicity of positive solutions for a class of boundary value problem of fractional differential equation with p -Laplacian operator,

D₀₊β (φp(Dα₀₊u(t))) +f(t, u(t)) = 0, 0< t <1,

u(0) = 0, u(1) +σDγ₀₊u(1) = 0, Dα0+u(0) = 0,

where 1< α≤2,0< β, γ ≤1,0≤α−γ−1, σis a positive number,Dα

0+, D β

0+, D γ

0+ are the standard Riemann-Liouville fractional order derivatives.

The purpose of this paper is to establish the existence of at least one and multiple positive solutions forp-Laplacian fractional order boundary value problems,

(1.1) Dq2

0+

φpDq1

0+y(t)

=f(t, y(t)), t∈(0,1),

(1.2)

y(j)(0) = 0, j= 0,1,2,· · ·, n−2, y(q4)_{(1) = 0}_,

φpDq1

0+y(0)

= 0 =Dq3

0+

φpDq1

0+y(1)

,







whereq2∈(1,2], q1∈(n−1, n], n≥2, q3∈(0,1], q4∈[1, q1−1] is a fixed integer,

φp(y) = |y|p−2y, p > 1, φp−1 = φq, 1/p+ 1/q = 1, f : [0,1]×

R

+

→

_R

+ is continuous andDqi

0+, fori = 1,2,3 are the standard Riemann-Liouville fractional order derivatives.

Define the nonnegative extended real numbersf0, f0, f∞ andf∞by

f0= lim

y→0+_t_∈min_[0_,_1] f(t, y)

φp(y)

, f0= lim

y→0+_tmax_∈_[0_,_1] f(t, y)

φp(y) ,

f∞= lim

y→∞t∈min[0,1]

f(t, y)

φp(y)

andf∞= lim

y→∞tmax∈[0,1]

f(t, y)

φp(y) ,

and also assume that they will exist.

The rest of the paper is organized as follows. In Section 2, we construct the Green functions for the homogeneous BVPs corresponding to (1.1)-(1.2) and estimate the bounds for the Green functions. In Section 3, we establish criteria for the existence of at least one positive solution for p-Laplacian fractional order BVP (1.1)-(1.2), by using Krasnosel’skii fixed point theorem. In Section 4, we derive sufficient conditions for the existence of at least three positive solutions for thep-Laplacian fractional order BVP (1.1)-(1.2), by applying five functionals fixed point theorem. We also establish the existence of at least 2k−1 positive solutions for an arbitrary positive integer k. In Section 5, as an application, the results are demonstrated with examples.

2. Green Functions and Bounds

In this section, we construct the Green functions for the homogeneous BVPs and estimate the bounds for the Green functions, which are essential to establish the main results.

LetG(t, s) be the Green’s function for the homogeneous BVP,

(2.1) −Dq1

0+y(t) = 0, t∈(0,1),

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Lemma 2.1. Let h(t)∈C[0,1]. Then the fractional order differential equation,

(2.3) Dq1

0+y(t) +h(t) = 0, t∈(0,1), satisfying (2.2)has a unique solution,

y(t) = Z 1

0

G(t, s)h(s)ds,

where

(2.4) G(t, s) =

G1(t, s), 0≤t≤s≤1,

G2(t, s), 0≤s≤t≤1,

G1(t, s) =

tq1−1₍₁₋_s₎q1−q4−1 Γ(q1)

,

G2(t, s) =

tq1−1₍₁₋_s₎q1−q4−1₋₍_t₋_s₎q1−1 Γ(q1)

.

Proof. Lety(t)∈C[q1]+1_[0_,_{1] be the solution of fractional order differential} equa-tion (2.3) satisfying (2.2). Then

y(t) = −1 Γ(q1)

Z t

0

(t−s)q1−1_h₍_s₎_ds₊_c

1tq1−1+c2tq1−2+· · ·+cntq1−n.

From (2.2),ci = 0, i= 2,3,· · ·, nandc1=R 1 0

(1−s)q1−q4−1

Γ(q1) h(s)ds.Thus, the unique solution of (2.3) with (2.2) is

y(t) = −1 Γ(q1)

Z t

0

(t−s)q1−1_h₍_s₎_ds₊t q1−1

Γ(q1) Z 1

0

(1−s)q1−q4−1_h₍_s₎_ds

= Z 1

0

G(t, s)h(s)ds,

whereG(t, s) is the Green’s function and given in (2.4).

Lemma 2.2. Let z(t)∈C[0,1]. Then the fractional order differential equation,

(2.5) Dq2

0+

φpDq1

0+y(t)

=z(t), t∈(0,1),

satisfying

(2.6) φpDq1

0+y(0)

= 0, Dq3

0+

φpDq1

0+y(1)

= 0,

has a unique solution,

y(t) = Z 1

0

G(t, s)φq

Z 1

0

H(s, τ)z(τ)dτds,

where

(2.7) H(t, s) =

( tq2−1₍₁₋_s₎q2−q3−1

Γ(q2) , 0≤t≤s≤1, tq2−1(1−s)q2−q3−1−(t−s)q2−1

Γ(q2) , 0≤s≤t≤1.

Here H(t, s)is the Green’s function for

−Dq2

0+

φp(x(t))= 0, t∈(0,1),

φp(x(0)) = 0, Dq3

0+

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Proof. An equivalent integral equation for (2.5) is given by

φp

Dq1

0+y(t)

= 1

Γ(q2) Z t

0

(t−τ)q2−1_z₍_τ₎_dτ₊_c

1tq2−1+c2tq2−2.

By (2.6), one can determine c2= 0 andc1=

−1 Γ(q2)

Z 1

0

(1−τ)q2−q3−1_z₍_τ₎_dτ._Then,

φp

Dq1

0+y(t)

= 1

Γ(q2) Z t

0

(t−τ)q2−1_z₍_τ₎_dτ ₋t q2−1

Γ(q2) Z 1

0

(1−τ)q2−q3−1_z₍_τ₎_dτ

=−

Z 1

0

H(t, τ)z(τ)dτ.

Therefore,φ−1

p

φpDq1

0+y(t)

=−φ−1

p

R1

0 H(t, τ)z(τ)dτ

.Consequently,

Dq1

0+y(t) +φq

R1

0 H(t, τ)z(τ)dτ

= 0.

Hence,

y(t) = Z 1

0

G(t, s)φq

Z 1

0

H(s, τ)z(τ)dτds

is the solution of fractional order BVP (2.5), (1.2).

Lemma 2.3. Fort∈I=1 4,

3 4

,the Green’s functionG(t, s)given in (2.4)satisfies the following inequalities

(i)G(t, s)≥0, f or all(t, s)∈[0,1]×[0,1],

(ii)G(t, s)≤G(1, s), f or all(t, s)∈[0,1]×[0,1],

(iii)G(t, s)≥ξG(1, s), f or all(t, s)∈I×[0,1],

whereξ=1₄

q1−1 .

Proof. The Green’s functionG(t, s) of (2.1), (2.2) is given in (2.4). For 0≤t≤s≤1,

G1(t, s) = 1 Γ(q1)

h

tq1−1₍₁₋_s₎q1−q4−1

i

≥0.

For 0≤s≤t≤1,

G2(t, s) = 1 Γ(q1)

h

tq1−1₍₁₋_s₎q1−q4−1₋₍_t₋_s₎q1−1i

≥ 1

Γ(q1) h

tq1−1₍₁₋_s₎q1−q4−1₋₍_t₋_ts₎q1−1

i

=t

q1−1

Γ(q1) h

1 +δs+q4(q4+ 1)

2 s

2₊_{· · ·}₋₁i₍₁₋_s₎q1−1

=t

q1−1

Γ(q1) h

q4s+O(s2) i

(1−s)q1−1_≥₀_.

Hence the Green’s functionG(t, s) is nonnegative. For 0≤t≤s≤1,

∂G1(t, s)

∂t =

1 Γ(q1)

h

(q1−1)tq1−2(1−s)q1−q4−1 i

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Therefore, G1(t, s) is increasing int, which implies G1(t, s) ≤ G1(1, s). Now, for 0≤s≤t≤1,

∂G2(t, s)

∂t

= 1

Γ(q1) h

(q1−1)tq1−2(1−s)q1−q4−1−(q1−1)(t−s)q1−2 i

≥ 1

Γ(q1) h

(q1−1)tq1−2(1−s)q1−q4−1−(q1−1)(t−ts)q1−2 i

=(q1−1)t

q1−2

Γ(q1) h

1−1−(q4−1)s+

(q4−1)(q4−2)

2 s

2₊_{· · ·}i₍₁₋_s₎q1−q4−1

=(q1−1)t

q1−2

Γ(q1) h

(q4−1)s+O(s2) i

(1−s)q1−q4−1_≥₀_.

Therefore,G2(t, s) is increasing int,which impliesG2(t, s)≤G2(1, s). Let 0≤t≤s≤1 andt∈I.Then

G1(t, s) = 1 Γ(q1)

h

tq1−1₍₁₋_s₎q1−q4−1i

=tq1−1 1 Γ(q1)

h

(1−s)q1−q4−1

i

=tq1−1_G

1(1, s)

≥1

4 q1−1

G1(1, s).

Let 0≤s≤t≤1 andt∈I.Then

G2(t, s) = 1 Γ(q1)

h

tq1−1₍₁₋_s₎q1−q4−1₋₍_t₋_s₎q1−1i

≥ 1

Γ(q1) h

tq1−1₍₁₋_s₎q1−q4−1₋₍_t₋_ts₎q1−1i

=tq1−1_G

2(1, s)

≥1

4 q1−1

G2(1, s).

Hence the result.

Lemma 2.4. Fort, s∈[0,1], the Green’s function H(t, s)given in (2.7)satisfies the following inequalities

(i)H(t, s)≥0,

(ii)H(t, s)≤H(s, s).

Proof. The Green’s functionH(t, s) is given in (2.7). For 0≤t≤s≤1,

H(t, s) = 1 Γ(q2)

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For 0≤s≤t≤1,

H(t, s) = 1 Γ(q2)

h

tq2−1₍₁₋_s₎q2−q3−1₋₍_t₋_s₎q2−1

i

≥ 1

Γ(q2) h

tq2−1₍₁₋_s₎q2−q3−1₋₍_t₋_ts₎q2−1i

=t

q2−1

Γ(q2) h

(1−s)−q3₋₁

i

(1−s)q2−1

=t

q2−1

Γ(q2) h

1 +q3s+

q3(q3+ 1)

2 s

2₊

· · ·−1i(1−s)q2−1

=t

q2−1

Γ(q2) h

q3s+O(s2) i

(1−s)q2−1_≥₀_.

For 0≤t≤s≤1,

∂H(t, s)

∂t =

1 Γ(q2)

[(q2−1)tq2−2(1−s)q2−q3−1]≥0.

Therefore,H(t, s) is increasing int,fors∈[0,1], which impliesH(t, s)≤H(s, s).

Now, for 0≤s≤t≤1,

∂H(t, s)

∂t =

1 Γ(q2)

h

(q2−1)tq2−2(1−s)q2−q3−1−(q2−1)(t−s)q2−2 i

≤ (q2−1)

Γ(q2) h

(1−s)q2−q3−1₋₍₁₋_s₎q2−2i

= (q2−1) Γ(q2)

h

(1−s)−q3+1₋₁i₍₁₋_s₎q2−2

= (q2−1) Γ(q2)

h

1−(1−q3)s+

(1−q3)(−q3)

2 s

2₊_{· · ·}₋₁i₍₁₋_s₎q2−2

= (q2−1) Γ(q2)

h

(q3−1)s+O(s2) i

(1−s)q2−2_≤₀_.

Therefore,H(t, s) is decreasing int,fors∈[0,1] which impliesH(t, s)≤H(s, s).

Hence the result.

Lemma 2.5. Letµ∈(1₄,3₄). Then the Green’s functionH(t, s)holds the inequality,

(2.8) min

t∈I H(t, s)≥δ

∗₍_s₎_H₍_{s, s}₎_{, f or}₀_{< s <}₁_,

where

(2.9) δ∗(s) =

( ₍3 4)

q2−1₍₁₋_s₎q2−q3−1₋₍3 4−s)

q2−1

sq2−1₍₁₋_s₎q2−q3−1 , s∈(0, µ],

1

(4s)q2−1, s∈[µ,1).

Proof. For s ∈ (0,1), H(t, s) is increasing in t for t ≤ s and decreasing in t for

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We define

h1(t, s) =

[tq2−1₍₁₋_s₎q2−q3−1₋₍_t₋_s₎q2−1_] Γ(q2)

h2(t, s) =

[tq2−1₍₁₋_s₎q2−q3−1_] Γ(q2)

, and

H(s, s) = 1 Γ(q2)

sq2−1₍₁₋_s₎q2−q3−1_.

Then,

min

t∈I H(t, s) =







h1(3₄, s), s∈(0,1₄], min{h1(3₄, s), h2(1₄, s)}, s∈[1₄,3₄],

h2(1₄, s), s∈[3₄,1),

=

h1(3₄, s), s∈(0, µ],

h2(1₄, s), s∈[µ,1),

= 



 (3

4)

q2−1₍₁₋_s₎q2−q3−1₋₍3 4−s)

q2−1

Γ(q2) , s∈(0, µ],

1 Γ(q2)

(1−s)q2−q3−1

4q2−1 , s∈[µ,1),

≥

( ₍3 4)

q2−1₍₁₋_s₎q2−q3−1₋₍3 4−s)

q2−1

sq₂−1₍₁₋_s₎q₂−q₃−1 H(s, s), s∈(0, µ],

1

(4s)q2−1H(s, s), s∈[µ,1),

=δ∗(s)H(s, s).

LetB={y:y∈C[0,1]} be the real Banach space equipped with the norm

kyk= max

t∈[0,1]

|y(t)|.

Define a coneP ⊂B by

P =ny∈B:y(t)≥0, t∈[0,1] and min

t∈I y(t)≥ξkyk

o

.

Let

K= 1

R1

0 G(1, s)φq

R1

0 H(τ, τ)dτ

ds

and

L= 1

R

s∈IξG(1, s)φq

R

τ∈Iδ∗(τ)H(τ, τ)dτ

ds .

LetT :P →B be the operator defined by

(2.10) T y(t) =

Z 1

0

G(t, s)φq

Z 1

0

H(s, τ)f(τ, y(τ))dτds.

If y ∈ P is a fixed point of T, then y satisfies (2.10) and hence y is a positive solution of thep-Laplacian fractional order BVP (1.1)-(1.2).

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Proof. Lety∈P. Clearly,T y(t)≥0,for allt∈[0,1], and

T y(t) = Z 1

0

G(t, s)φq

Z 1

0

H(s, τ)f(τ, y(τ))dτds

≤

Z 1

0

G(1, s)φq

Z 1

0

so that

kT yk ≤

Z 1

0

G(1, s)φq

Z 1

0

H(s, τ)f(τ, y(τ))dτds.

Next, ify∈P, then by the above inequality we have

min

t∈I T y(t) = mint∈I

Z 1

0

G(t, s)φq

Z 1

0

≥ξ

Z 1

0

G(1, s)φq

Z 1

0

≥ξkT yk.

Hence,T y∈P and soT :P →P.Standard arguments involving the Arzela-Ascoli

theorem shows thatT is completely continuous.

3. Existence of at Least One Positive Solution

In this section, we establish criteria for the existence of at least one positive solution of thep-Laplacian fractional order BVP (1.1)-(1.2) by using Krasnosel’skii fixed point theorem.

To establish the existence of at least one positive solution forp-Laplacian frac-tional order BVP (1.1)-(1.2) by employing the following Krasnosel’skii fixed point theorem.

Theorem 3.1. [15] Let X be a Banach Space, P ⊆ X be a cone and suppose that Ω1,Ω2 are open subsets of X with 0∈Ω1 andΩ1 ⊂Ω2. Suppose further that

T :P∩(Ω2\Ω1)→P is completely continuous operator such that either (i) kT uk≤kuk, u∈P∩∂Ω1 andkT uk≥kuk, u∈P∩∂Ω2, or (ii) kT uk≥kuk, u∈P∩∂Ω1 andkT uk≤kuk, u∈P∩∂Ω2 holds. ThenT has a fixed point inP∩(Ω2\Ω1).

Theorem 3.2. If f0_{= 0}_and_f

∞=∞,then the p-Laplacian fractional order BVP

(1.1)-(1.2)has at least one positive solution that lies inP.

Proof. Let T be the cone preserving, completely continuous operator defined by (2.10). Sincef0_{= 0}_,_{we may choose}_H1_>_{0 so that}

max

t∈[0,1]

f(t, y)

φp(y)

≤η1, for 0< y≤H1,

whereη1>0 satisfies

ηq₁−1

Z 1

0

G(1, s)φq

Z 1

0

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Thus, ify∈P and kyk=H1,then we have

T y(t) = Z 1

0

G(t, s)φq

Z 1

0

H(s, τ)f(τ, y)dτds

≤

Z 1

0

G(1, s)φq

Z 1

0

H(τ, τ)η1φp(y)dτ

ds

= Z 1

0

G(1, s)η₁q−1φq

Z 1

0

H(τ, τ)dτyds

≤η₁q−1

Z 1

0

G(1, s)φq

Z 1

0

H(τ, τ)dτdskyk

=η₁q−11

Kkyk ≤ kyk.

Therefore,

kT yk ≤ kyk.

Now, if we let

Ω1={y∈B:kyk< H1},

then

(3.1) kT yk ≤ kyk, fory∈P∩∂Ω1.

Further, sincef∞=∞,there existsH

2

>0 such that

min

t∈[0,1]

f(t, y)

φp(y) ≥η2, fory≥H 2

,

whereη2>0 is chosen such that

ηq₂−1ξ2

Z

s∈I

G(1, s)φq

Z

τ∈I

δ∗(τ)H(τ, τ)dτds=η₂q−1ξ21 L ≥1.

Let

H2= maxn2H1,1 ξH

2o

,

and define

Ω2={y∈B:kyk< H2}.

Ify∈P∩∂Ω2 andkyk=H2,then

min

t∈I y(t)≥ξkyk ≥H

2

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and so

T y(t) = Z 1

0

G(t, s)φq

Z 1

0

≥

Z

s∈I

ξG(1, s)φq

Z 1

0

≥ξ

Z

s∈I

G(1, s)φq

Z

τ∈I

δ∗(τ)H(τ, τ)η2φp(y)dτ

ds

=ξηq₂−1

Z

s∈I

G(1, s)φq

Z

τ∈I

δ∗(τ)H(τ, τ)dτyds

≥ξηq₂−1

Z

s∈I

G(1, s)φq

Z

τ∈I

δ∗(τ)H(τ, τ)dτξkykds

=ξ2η₂q−1

Z

s∈I

G(1, s)φq

Z

τ∈I

δ∗(τ)H(τ, τ)dτdskyk

=ηq₂−1ξ21 Lkyk ≥ kyk.

Hence,

(3.2) kT yk ≥ kyk, fory∈P∩∂Ω2.

An application of Theorem 3.1 to (3.1) and (3.2) yields a fixed point ofT that lies inP∩(Ω2\Ω1).This fixed point is the solution of thep-Laplacian fractional order

BVP (1.1)-(1.2).

Theorem 3.3. If f0=∞andf∞= 0,then the p-Laplacian fractional order BVP (1.1)-(1.2)has at least one positive solution that lies inP.

Proof. Let T be the cone preserving, completely continuous operator defined by (2.10). Sincef0=∞,we chooseJ1>0 such that

min

t∈[0,1]

f(t, y)

φp(y) ≥η1, for 0< y≤J 1_,

whereη₁>0 satisfies

(η1)q−1ξ2 Z

s∈I

G(1, s)φq

Z

τ∈I

δ∗(τ)H(τ, τ)dτds= (η1)q−1ξ2 1

L ≥1.

In this case, we define

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we havef(τ, y)≥η1φp(y), τ ∈I,and moreovery(t)≥ξkyk, t∈I and so

T y(t) = Z 1

0

G(t, s)φq

Z 1

0

≥

Z

s∈I

ξG(1, s)φq

Z 1

0

≥ξ

Z

s∈I

G(1, s)φq

Z

τ∈I

δ∗(τ)H(τ, τ)η₁φp(y)dτds

=ξ(η1)q−1 Z

s∈I

G(1, s)φq

Z

τ∈I

δ∗(τ)H(τ, τ)dτyds

≥ξ(η₁)q−1 Z

s∈I

G(1, s)φq

Z

τ∈I

δ∗(τ)H(τ, τ)dτξkykds

≥ξ2(η₁)q−1 Z

s∈I

G(1, s)φq

Z

τ∈I

δ∗(τ)H(τ, τ)dτdskyk

= (η₁)q−1ξ21 Lkyk ≥ kyk.

Thus,

(3.3) kT yk ≥ kyk, fory∈P∩∂Ω1.

Now, sincef∞= 0,there existsJ2>0 such that

max

t∈[0,1]

f(t, y)

φp(y)

≤η₂, fory≥J2,

whereη₂>0 satisfies

(η₂)q−1 Z 1

0

G(1, s)φq

Z 1

0

H(τ, τ)dτds= (η₂)q−11

K ≤1.

It follows that

f(t, y)≤η2φp(y), fory≥J 2

.

We establish the result in two subcases.

Case(i): fis bounded. SupposeN >0 is such that maxt∈[0,1]f(t, y)≤N, for all 0<

y <∞.In this case, we may choose

J2= maxn2J1, Nq−11 K

o

,

and let

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Then fory∈P∩∂Ω2 andkyk=J2,we have

T y(t) = Z 1

0

G(t, s)φq

Z 1

0

≤

Z 1

0

G(1, s)φq

Z 1

0

H(s, τ)N dτds

≤

Z 1

0

G(1, s)φq

Z 1

0

H(τ, τ)dτNq−1ds

≤Nq−1

Z 1

0

G(1, s)φq

Z 1

0

H(τ, τ)dτds

=Nq−11 K

=J2=kyk,

and therefore

Case(ii): f is unbounded. LetJ2_>_max_{₂_J1_{, J}2_}_{be such that}_f₍_{t, y}₎_≤_f₍_{t, J}2₎_, for 0 < y ≤ J2_, _{and let Ω}

2 ={y ∈ B : kyk < J2}. Choosing y ∈ P ∩∂Ω2 with

kyk=J2_,_{we have}

T y(t) = Z 1

0

G(t, s)φq

Z 1

0

≤

Z 1

0

G(1, s)φq

Z 1

0

≤

Z 1

0

G(1, s)φq

Z 1

0

H(τ, τ)f(τ, J2)dτds

≤

Z 1

0

G(1, s)φq

Z 1

0

H(τ, τ)η₂φp(J2)dτ

ds

= (η₂)q−1 Z 1

0

G(1, s)φq

Z 1

0

H(τ, τ)dτdsJ2

= (η2)q−1 1

KJ

2

≤J2=kyk.

And so

An application of Theorem 3.1 to (3.3), (3.4) and (3.5) yields a fixed point ofT that lies in P ∩(Ω2\Ω1). This fixed point is the solution of the p-Laplacian fractional

order BVP (1.1)-(1.2).

4. Existence of Multiple Positive Solutions

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2k−1 positive solutions for an arbitrary positive integerk.

Let γ, β, θ be nonnegative continuous convex functionals on P and α, ψ be nonnegative continuous concave functionals on P, then for nonnegative numbers

h0, a0, b0, d0 andc0, convex sets are defined.

P(γ, c0) ={y∈P :γ(y)< c0},

P(γ, α, a0, c0) ={y∈P :a0≤α(y);γ(y)≤c0}, Q(γ, β, d0, c0) ={y∈P :β(y)≤d0;γ(y)≤c0},

P(γ, θ, α, a0, b0, c0) ={y∈P :a0≤α(y);θ(y)≤b0;γ(y)≤c0}, Q(γ, β, ψ, h0, d0, c0) ={y∈P :h0 ≤ψ(y);β(y)≤d0;γ(y)≤c0}.

In obtaining multiple positive solutions for thep-Laplacian fractional order B-VP (1.1)-(1.2), the following so called Five Functionals Fixed Point Theorem is fundamental.

Theorem 4.1. [3] Let P be a cone in the real Banach space B. Suppose α and

ψ are nonnegative continuous concave functionals on P and γ, β, θ are nonnega-tive continuous convex functionals on P, such that for some positive numbers c0

and e0, α(y) ≤ β(y) and k y k≤ e0γ(y), for all y ∈ P(γ, c0₎_. _{Suppose further}

that T : P(γ, c0₎ _→ _P₍_{γ, c}0₎ _{is completely continuous and there exist constants}

h0, d0, a0 and b0≥0 with0< d0 < a0 such that each of the following is satisfied.

(D1) {y∈P(γ, θ, α, a0, b0, c0) :α(y)> a0} 6=∅ and α(T y)> a0 f or y∈P(γ, θ, α, a0, b0, c0),

(D2) {y∈Q(γ, β, ψ, h0, d0, c0) :β(y)> d0} 6=∅ and β(T y)> d0 f or y∈Q(γ, β, ψ, h0, d0, c0),

(D3) α(T y)> a0 provided y∈P(γ, α, a0, c0)with θ(T y)> b0,

(D4) β(T y)< d0 provided y∈Q(γ, β, ψ, h0, d0, c0)with ψ(T y)< h0.

Then T has at least three fixed pointsy1, y2, y3∈P(γ, c0)such thatβ(y1)< d0, a0<

α(y2) andd0 < β(y3)withα(y3)< a0.

Define the nonnegative continuous concave functionalsα, ψand the nonnegative continuous convex functionalsβ, θ, γonP by

α(y) = min

t∈I y(t), ψ(y) = mint∈I1 y(t),

γ(y) = max

t∈[0,1]

y(t), β(y) = max

t∈I1

y(t), θ(y) = max

t∈I y(t),

whereI1=1₃,2₃.For anyy∈P,

(4.1) α(y) = min_t_∈_I y(t)≤max_t_∈_I 1

y(t) =β(y)

and

(4.2) kyk ≤ 1

ξmint∈I y(t)≤

1

ξtmax∈[0,1]y(t) = 1

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Theorem 4.2. Suppose there exist 0< a0 < b0 < b_ξ0 ≤c0 such thatf satisfies the following conditions:

(C1)f(t, y(t))< φpa0K, t∈[0,1]and y∈[ξa0, a0],

(C2)f(t, y(t))> φpb0L, t∈I and y∈hb0,b

0

ξ

i

,

(C3)f(t, y(t))< φpc0K, t∈[0,1]and y∈[0, c0].

Then the p-Laplacian fractional order BVP (1.1)-(1.2) has at least three positive solutions y1, y2 and y3 such that β(y1) < a0, b0 < α(y2)and a0 < β(y3) with

α(y3)< b0.

Proof. We seek three fixed points y1, y2, y3 ∈ P of T defined by (2.10). From Lemma 2.6, (4.1) and (4.2), for each y ∈ P, α(y) ≤ β(y) and kyk ≤ 1

ξγ(y). We

show thatT :P(γ, c0₎_→_P₍_{γ, c}0_{). Let}_y_∈_P₍_{γ, c}0_{). Then 0}_≤_y _≤_c0_{. We may use}

the condition (C3) to obtain

γ(T y) = max

t∈[0,1] Z 1

0

G(t, s)φq

Z 1

0

≤

Z 1

0

G(1, s)φq

Z 1

0

H(τ, τ)f(τ, y(τ))dτds

≤

Z 1

0

G(1, s)φq

Z 1

0

H(τ, τ)φp

c0Kdτds

< c0K

Z 1

0

G(1, s)φq

Z 1

0

H(τ, τ)dτds=c0.

ThereforeT :P(γ, c0₎_→_P₍_{γ, c}0₎_._{Now we verify the conditions (}_D_{1) and (}_D_{2) of}

Theorem 4.1 are satisfied. It is obvious that

b0+b_ξ0

2 ∈

n

y∈Pγ, θ, α, b0,b

0

ξ, c

0_:_α₍_y₎_{> b}0o₆₌_∅

and

ξa0+a0

2 ∈

n

y∈Qγ, β, ψ, ξa0, a0, c0:β(y)< a0o6=∅.

Next, let y∈P(γ, θ, α, b0_,b0 ξ, c

0_{) or}_y_∈_Q₍_{γ, β, ψ, ξa}0_{, a}0_{, c}0_{). Then,} _b0 _≤_y_≤ b0 ξ and ηa0≤y≤a0.Now, we may apply the condition (C2) to get

α(T y) = min

t∈I

Z 1

0

G(t, s)φq

Z 1

0

≥ξ

Z

s∈I

G(1, s)φq

Z

τ∈I

δ∗(τ)H(τ, τ)φp

b0Ldτds

> b0L

Z

s∈I

ξG(1, s)φq

Z

τ∈I

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Clearly, by the condition (C1),we have

β(T y) = max

t∈I1

Z 1

0

G(t, s)φq

Z 1

0

≤

Z 1

0

G(1, s)φq

Z 1

0

H(τ, τ)φpa0Kdτds

< a0K

Z 1

0

G(1, s)φq

Z 1

0

H(τ, τ)dτds

=a0.

To see that (D3) is satisfied, let y∈P(γ, α, b0, c0) withθ(T y)> b0 ξ.Then

α(T y) = min

t∈I

Z 1

0

G(t, s)φq

Z 1

0

≥ξ

Z 1

0

G(1, s)φq

Z 1

0

≥ξ max

t∈[0,1] Z 1

0

G(t, s)φq

Z 1

0

≥ξmax

t∈I

Z 1

0

G(t, s)φq

Z 1

0

=ξθ(T y)

> b0.

Finally, we show that (D4) holds. Lety∈Q(γ, β, a0, c0) with ψ(T y)< ξa0. Then, we have

β(T y) = max

t∈I1

Z 1

0

G(t, s)φq

Z 1

0

≤ max

t∈[0,1] Z 1

0

G(t, s)φq

Z 1

0

≤

Z 1

0

G(1, s)φq

Z 1

0

=1

ξ

h

ξ

Z 1

0

G(1, s)φq

Z 1

0

H(s, τ)f(τ, y(τ))dτi

≤1 ξmint∈I

Z 1

0

G(t, s)φq

Z 1

0

≤1 ξmint∈I1

Z 1

0

G(t, s)φq

Z 1

0

=1

ξψ(T y)< a

0_.

We have proved that all the conditions of Theorem 4.1 are satisfied. Therefore, the

p-Laplacian fractional order BVP (1.1)-(1.2) has at least three positive solutions

y1, y2andy3such thatβ(y1)< a0, b0< α(y2) anda0 < β(y3) withα(y3)< b0.This

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Theorem 4.3. Let k be an arbitrary positive integer. Assume that there exist numbers ar(r = 1,2,· · ·, k) andbs(s = 1,2,· · ·, k−1) with 0 < a1 < b1 < b_ξ1 <

a2< b2 < b_ξ2 <· · ·< ak−1 < bk−1 < bk_ξ−1 < ak such thatf satisfies the following conditions:

(C4) f(t, y(t))< φp

arK, t∈[0,1] andy∈[ξar, ar], r= 1,2,· · ·, k,

(C5) f(t, y(t))> φpbsL, t∈I andy∈hbs,bs ξ

i

, s= 1,2,· · ·, k−1.

Then thep-Laplacian fractional order BVP (1.1)-(1.2)has at least2k−1positive solutions inPak.

Proof. We use induction onk. First, fork= 1, we know from the condition (C4) that T : Pa1 → Pa1, then it follows from the Schauder fixed point theorem that thep-Laplacian fractional order BVP (1.1)-(1.2) has at least one positive solution in Pa₁. Next, we assume that this conclusion holds for k =l. In order to prove that this conclusion holds for k = l+ 1. We suppose that there exist numbers

ar(r = 1,2,· · ·, l+ 1) and bs(s = 1,2,· · ·, l) with 0< a1 < b1 < b_ξ1 < a2 < b2 <

b2

ξ <· · ·< al< bl< bl

ξ < al+1 such thatf satisfies the following conditions:

(4.3) f(t, y(t))< φparK, t∈[0,1] andy∈[ξar, ar], r= 1,2,· · ·, l+ 1,

(4.4) f(t, y(t))> φpbsL, t∈I andy∈hbs,bs ξ

i

, s= 1,2,· · ·, l.

By assumption, the fractional order BVP (1.1)-(1.2) has at least 2l−1 positive solutionsy∗_i(i= 1,2,· · ·,2l−1) inPal.At the same time, it follows from Theorem

4.2, (4.3) and (4.4) that the p-Laplacian fractional order BVP (1.1)-(1.2) has at least three positive solutions y1, y2 and y3 in Pal+1 such that β(y1) < al, bl < α(y2) andal< β(y3) withα(y3)< bl.Obviouslyy2andy3are distinct fromy∗i(i=

1,2,· · ·,2l−1) inPa_l.Therefore, thep-Laplacian fractional order BVP (1.1)-(1.2) has at least 2l+ 1 positive solutions inPal+1,which shows that this conclusion also

holds fork=l+ 1.This completes the proof.

5. _Examples

In this section, as an application, the results of the previous sections are demon-strated with examples.

Example 5.1Consider thep-Laplacian fractional order BVP,

(5.1) D1₀.+7

φpD2₀.+6y(t)

=f(t, y(t)), t∈(0,1),

(5.2) y(0) =y0(0) =y00(1) = 0, φpD2₀.+6y(0)

= 0 =D₀0+.6

φpD2₀.+6y(1)

,

where

f(t, y) = y

2₍₆₅₀₋₆₄₉_e−3y₎

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Then the Green’s functionG(t, s) for the homogeneous BVP,

−D2.6

0+y(t) = 0, t∈(0,1), y(0) =y0(0) =y00(1) = 0,

and is given by

G(t, s) =

( t1.6₍₁₋_s₎−0.6

Γ(2.6) , t≤s,

t1.6(1−s)−0.6−(t−s)1.6

Γ(2.6) , s≤t.

The Green’s functionH(t, s) for the BVP,

−D1.7 0+

φpD₀2+.6y(t)

= 0, t∈(0,1),

φpD2₀.+6y(0)

= 0 =D0₀.+6

φpD₀2+.6y(1)

,

and is given by

H(t, s) =

( t0.7₍₁₋_s₎0.1

Γ(1.7) , t≤s,

t0.7₍₁₋_s₎0.1₋₍_t₋_s₎0.7

Γ(1.7) , s≤t.

Clearly, the Green functionsG(t, s) and H(t, s) are positive. Letp= 2. By direct calculations,ξ= 0.1088,K= 1.0077,L= 27.1209, f0_{= 0 and}_f

∞ =∞.Then, all

the conditions of Theorem 3.2 are satisfied. Thus by Theorem 3.2, thep-Laplacian fractional order BVP (5.1)-(5.2) has at least one positive solution.

Example 5.2Consider thep-Laplacian fractional order BVP,

(5.3) D1₀.+8

φp

D2₀.+7y(t)

=f(t, y(t)), t∈(0,1),

(5.4) y(0) =y0(0) =y00(1) = 0, φpD2₀.+7y(0)

= 0 =D₀0+.7

φpD2₀.+7y(1)

,

where

f(t, y) =

t

100+

15 32y

7_, ₀_≤_y_≤₂_,

y+₁₀₀t +116₂ , y >2.

Then the Green’s functionG(t, s) for the homogeneous BVP,

−D2.7

0+y(t) = 0, t∈(0,1), y(0) =y0(0) =y00(1) = 0,

and is given by

G(t, s) =

( t1.7₍₁₋_s₎−0.7

Γ(2.7) , t≤s,

t1.7(1−s)−0.7−(t−s)1.7

Γ(2.7) , s≤t,

The Green’s functionH(t, s) for the BVP,

−D1.8 0+

φpD₀2+.7y(t)

= 0, t∈(0,1),

φpD2₀.+7y(0)

= 0 =D0₀.+7

φpD₀2+.7y(1)

,

and is given by

H(t, s) =

( t0.8₍₁₋_s₎0.1

Γ(1.8) , t≤s,

t0.8(1−s)0.1−(t−s)0.8

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Clearly, the Green functions G(t, s) and H(t, s) are positive and f is continuous and increasing on [0,∞).Letp= 2.By direct calculations,ξ= 0.0947,K= 1.8901 andL= 29.6238. Choosinga0 = 1, b0 = 2 andc0= 100, then 0< a0< b0 < b0

ξ ≤c

0

andf satisfies

(i)f(t, y)<1.8901 =φp

a0K, t∈[0,1] andy∈[0.0947,1],

(ii)f(t, y)>59.25 =φpb0L, t∈h1

4, 3 4 i

andy∈[2,21.12],

(iii)f(t, y)<189.01 =φpc0K, t∈[0,1] andy∈[0,100].

Then, all the conditions of Theorem 4.2 are satisfied. Thus by Theorem 4.2, the

p-Laplacian fractional order BVP (5.3)-(5.4) has at least three positive solutions

y1, y2 andy3 satisfying

max

t∈[1 3,

2 3]

y1(t)<1,2< min

t∈[1 4,

3 4]

y2(t),

1< max

t∈[1 3,23]

y3(t), min

t∈[1 4,34]

y3(t)<2.

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1

Department of Applied Mathematics, Andhra University, Visakhapatnam, 530 003, India,

2

Department of Mathematics, MVGR College of Engineering, Vizianagaram, 535 005, India,

∗