In this article, we study the existence of positive solutions for the singular fractional boundary value problem −Dαu(t) =Af(t, u(t

Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 130, pp. 1–13.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu

UNIQUENESS AND EXISTENCE OF POSITIVE SOLUTIONS FOR SINGULAR FRACTIONAL DIFFERENTIAL EQUATIONS

NEMAT NYAMORADI, TAHEREH BASHIRI, S. MANSOUR VAEZPOUR, DUMITRU BALEANU

Abstract. In this article, we study the existence of positive solutions for the singular fractional boundary value problem

−D^αu(t) =Af(t, u(t)) +

k

X

i=1

BiI^βigi(t, u(t)), t∈(0,1),

D^δu(0) = 0, D^δu(1) =aD^α−δ−12 (D^δu(t))˛

˛t=ξ

where 1< α≤2, 0< ξ≤1/2,a∈[0,∞), 1< α−δ <2, 0< βi<1,A, Bi, 1≤i≤k, are real constant,D^αis the Riemann-Liouville fractional derivative of orderα. By using the Banach’s fixed point theorem and Leray-Schauder’s alternative, the existence of positive solutions is obtained. At last, an example is given for illustration.

1. Introduction

Fractional calculus is the field of mathematical analysis which deals with the investigation and applications of integrals and derivatives of arbitrary order, the fractional calculus may be considered an old and yet novel topic.

Recently, fractional differential equations have been of great interest. This is because of both the intensive development of the theory of fractional calculus itself and its applications in various sciences, such as physics, mechanics, chemistry, en- gineering, etc. For example, for fractional initial value problems, the existence and multiplicity of solutions were discussed in [2, 4, 10, 11], moreover, fractional deriv- ative arises from many physical processes,such as a charge transport in amorphous semiconductors [18], electrochemistry and material science are also described by differential equations of fractional order [5, 6, 7, 14, 15]. Bai and L¨u [3] considered the boundary value problem of fractional order differential equation

D^α₀+u(t) +f(t, u(t)) = 0, t∈(0,1), u(0) =u(1) = 0,

2000Mathematics Subject Classification. 34A08, 74H20,30E25.

Key words and phrases. Existence of solutions; Banachs fixed point theorem;

Leray-Schauders alternative.

c

2014 Texas State University - San Marcos.

Submitted August 21, 2013. Published June 6, 2014.

1

whereD^α₀+is the standard Riemann-Liouville fractional derivative of order 1< α≤ 2 and

f : [0,1]×[0,∞)→[0,∞) is continuous.

Hussein [9], considered the following nonlinearm-point boundary value problem of fractional type

D₀^α+x(t) +q(t)f(t, x(t)) = 0, a.e. on [0,1], α∈(n−1, n], n≥2, x(0) =x⁰(0) =x⁰⁰(0) =· · ·=x⁽ⁿ⁻²⁾(0) = 0, x(1) =

m−2

X

i=1

ξ_ix(η_i), where 0< η1 <· · ·< η_m−2 <1, ξi >0 with Pm−2

i=1 ξiη_i^α−1 <1, qis a real-valued continuous function andf is a nonlinear Pettis integrable function.

Motivated by the above works, the purpose of this paper is to discuss the fol- lowing singular fractional boundary value problem:

−D^αu(t) =Af(t, u(t)) +

k

X

i=1

BiI^βigi(t, u(t)), t∈(0,1), D^δu(0) = 0, D^δu(1) =aD^α−δ−12 (D^δu(t))

_t=ξ,

(1.1)

where 1 < α ≤ 2, 0 < ξ ≤ ¹₂, a ∈ (0,∞), 1 < α−δ < 2, 0 < βi < 1, A, Bi, 1 ≤i ≤k, are real constant,D^α is the Riemann-Liouville fractional derivative of orderα.

The rest of this paper is organized as follows. In Section 2, we give some pre- liminaries. In Sections 3 and 4, we study the existence and uniqueness of solutions for system (1.1) by Banach’s fixed point theorem and Leray-Schauder’s alternative, respectively. At last, in Section 5, an example is also given to illustrate our theory.

2. Preliminaries

In this section, we present notation and some preliminary lemmas that will be used in the proofs of the main results.

Definition 2.1 ([16, 17]). The Riemann-Liouville fractional integral operator of orderα >0, of functionf ∈L¹(R⁺) is defined as

I₀^α+f(t) = 1 Γ(α)

Z t

0

(t−s)^α−1f(s)ds, where Γ(·) is the Euler gamma function.

Definition 2.2 ([16, 17]). The Riemann-Liouville fractional derivative of order α >0 of a continuous functionf : (0,∞)→Ris defined as

D^α₀+f(t) = 1 Γ(n−α)

d dt

ⁿZ t

0

(t−s)^n−α−1f(s)ds, wheren= [α] + 1.

Lemma 2.3 ([12]). The equality D^γ₀₊I₀^γ₊f(t) = f(t) with γ > 0 holds for f ∈ L¹(0,1).

Lemma 2.4 ([12]). Let α >0. Then the differential equation D^α₀+u= 0

has a unique solution u(t) =c1t^α−1+c2t^α−2+· · ·+cnt^α−n,ci ∈R, i= 1, . . . , n, theren−1< α≤n.

Lemma 2.5 ([12]). Let α >0. Then the following equality holds foru∈L¹(0,1), D₀^α+u∈L¹(0,1);

I₀^α+D^α₀+u(t) =u(t) +c₁t^α−1+c₂t^α−2+· · ·+c_nt^α−n, c_i ∈R,i= 1, . . . , n, theren−1< α≤n.

Lemma 2.6 ([3]). Forλ >−1 andα >0, D^α₀+t^γ = Γ(γ+ 1)

Γ(γ−α+ 1)t^γ−α.

Lemma 2.7 ([13]). Suppose that g ∈L¹(0,1) andα, β be two constant such that 0≤β ≤1< α, then

D₀^β+

Z t

0

(t−s)^α−1g(s)ds= Γ(α) Γ(α−β)

Z t

0

(t−s)^α−β−1g(s)ds.

Now, we consider (1.1). By the substitutionu(t) =I^δy(t) = D^−δy(t), problem (1.1) is turned into

−D^α−δy(t) =Af(t, I^δy(t)) +

k

X

i=1

B_iI^βig_i(t, I^δy(t)), t∈(0,1), y(0) = 0, y(1) =aD^α−δ−12 y(t))

_t=ξ.

(2.1)

Lemma 2.8. For any h ∈ C[0,1]∩L(0,1), the unique solution of the boundary value problem

−D^α−δy(t) =h(t), t∈(0,1), y(0) = 0, y(1) =aD^α−δ−12 y(t))

_t=ξ

(2.2) is

y(t) =−I^α−δh(t) + t^α−δ−1Γ(^α−δ+1₂ ) Γ(^α−δ+1₂ )−aΓ(α−δ)ξ^α−δ−12

I^α−δh(1)−aI^α−δ+12 h(ξ) . Proof. By applying Lemma 2.5, equation (2.2) is equivalent to the integral equation

y(t) =−I^α−δh(t)−c1t^α−δ−1−c2t^α−δ−2, (2.3) for some arbitrary constantsc1, c2∈R.

By the boundary conditiony(0) = 0, we conclude thatc2= 0. Then, we have y(1) =−I^α−δh(1)−c₁,

and it follows from lemma (2.6) that

D^α−δ−12 y(t) =−D^α−δ−12 I^α−δh(t)−c1D^α−δ−12 t^α−δ−1

=−I^α−δ+12 h(t)−c₁ Γ(α−δ)

Γ(^α−δ+1₂ )t^α−δ−12 . Therefore,

D^α−δ−12 y(t)

_t=ξ=− Z ξ

0

(ξ−s)^α−δ−12

Γ(^α−δ+1₂ ) h(s)ds−c1

Γ(α−δ)

Γ(^α−δ+1₂ )t^α−δ−12 .

So, by the boundary conditiony(1) =aD^α−δ−12 y(t))

_t=ξ, we obtain that

c₁= Γ(^α−δ+1₂ )

Γ(^α−δ+1₂ )−aΓ(α−δ)ξ^α−δ−12 h−

Z 1

0

(1−s)^α−δ−1 Γ(α−δ) h(s)ds +a

Z ξ

0

(ξ−s)^α−δ−12

Γ(^α−δ+1₂ ) h(s)dsi . Therefore, the unique solution of equation (2.2) is

y(t) =−I^α−δh(t) + t^α−δ−1Γ(^α−δ+1₂ ) Γ(^α−δ+1₂ )−aΓ(α−δ)ξ^α−δ−12

hZ 1

0

(1−s)^α−δ−1 Γ(α−δ) h(s)ds

−a Z ξ

0

(ξ−s)^α−δ−12

Γ(^α−δ+1₂ ) h(s)dsi .

The proof is complete.

Thus, the solution of the problem (1.1) can be written as u(t) =I^δy(t)

=I^δh

−I^α−δh(t) + t^α−δ−1Γ(^α−δ+1₂ ) Γ(^α−δ+1₂ )−aΓ(α−δ)ξ^α−δ−12

I^α−δh(1)−aI^α−δ+12 h(ξ)i

=−I^αh(t) + Γ(^α−δ+1₂ )

Γ(^α−δ+1₂ )−aΓ(α−δ)ξ^α−δ−12

I^α−δh(1)−aI^α−δ+12 h(ξ)

× Z t

0

(t−s)^δ−1

Γ(δ) s^α−δ−1ds

=−I^αh(t) + Γ(^α−δ+1₂ )

Γ(^α−δ+1₂ )−aΓ(α−δ)ξ^α−δ−12

I^α−δh(1)−aI^α−δ+12 h(ξ)

×nt^α−1 Γ(δ)

Z 1

0

(1−ν)^δ−1ν^α−δ−1dνo ,

where we have used the substitutions=tν in the integral of the last term. Using the relation for the Beta functionB(·,·),

B(α, β) = Z 1

0

(1−u)^α−1u^β−1du= Γ(α)Γ(β) Γ(α+β), one has

u(t) =−I^αh(t) + t^α−1Γ(α−δ)Γ(^α−δ+1₂ ) Γ(α)

Γ(^α−δ+1₂ )−aΓ(α−δ)ξ^α−δ−12

I^α−δh(1)−aI^α−δ+12 h(ξ) . (2.4) The solution of the original nonlinear problem (1.1) can be obtained by replacing hwith the right hand side of the fractional equation of (1.1) in (2.4).

The basic space used in this paper is the real Banach spaceC=C([0,1],R) of all continuous functions from [0,1]→Rendowed with the normkuk= sup_t∈[0,1]|u(t)|.

In relation to problem (1.1), we define an operatorT :C → C as (Tu)(t) =−A

Z t

0

(t−s)^α−1

Γ(α) f(s, u(s))ds−

k

X

i=1

B_i Z t

0

(t−s)^α+βi−1

Γ(α+βi) g_i(s, u(s))ds

+t^α−1A0

h A

Z 1

0

(1−s)^α−δ−1

Γ(α−δ) f(s, u(s))ds +

k

X

i=1

Bi

Z 1

0

(1−s)^{α−δ+βi−1}

Γ(α−δ+β_i) gi(s, u(s))ds

−aA Z ξ

0

(ξ−s)^α−δ−12

Γ(^α−δ+1₂ ) f(s, u(s))ds

−a

k

X

i=1

Bi

Z ξ

0

(ξ−s)^{α−δ−12 +βi}

Γ(^α−δ+1₂ +βi)gi(s, u(s))dsi , where

A0= Γ(α−δ)Γ(^α−δ+1₂ ) Γ(α)

Γ(^α−δ+1₂ )−aΓ(α−δ)ξ^α−δ−12 .

It is clear that the existence of a positive solution for the system (1.1) is equivalent to the existence of a nontrivial fixed point ofT onC.

For convenience of the reader, we set

∆ = sup

t∈[0,1]

n|A|h t^α

Γ(α+ 1)+|A0|t^α−1 1

Γ(α−δ+ 1)+a ξ^α−δ+12 Γ(^α−δ+1₂ + 1)

i

+

k

X

i=1

|Bi|h t^α+βi

Γ(α+βi+ 1)+|A0|t^α−1 1

Γ(α−δ+βi+ 1) +a ξ^{α−δ+12 +βi}

Γ(^α−δ+1₂ +β_i+ 1) io

.

(2.5)

3. Existence results via Banach’s fixed point theorem

In this section, by using Banach’s fixed point theorem, we will establish find a unique solution of problem (1.1). Now, we state our results.

Theorem 3.1. Assume that f, gi : [0,1]×R → R, i = 1, . . . , k, are continuous functions satisfying the condition

(A1) |f(t, u)−f(t, v)| ≤ L1|u−v|, |gi(t, u)−gi(t, v)| ≤ Li+1|u−v|, for i = 1, . . . , k,t∈[0,1],Li >0,(i= 1, . . . , k+ 1),u, v∈R.

Then the boundary-value problem (1.1) has a unique solution if L < _∆¹, where L= max{Li:i= 1, . . . , k+ 1} and∆is given by (2.5).

Proof. Assume thatM = max{Mi:i= 1, . . . , k+ 1}, whereMi are finite numbers given by sup_t∈[0,1]|f(t,0)|=M₁, sup_t∈[0,1]|gi(t,0)|=M_i+1. Selectingr > _1−L∆^∆M , we show thatTB_r⊂B_r, whereB_r={u∈ C:kuk ≤r}. Using that|f(s, u(s))| ≤

|f(s, u(s))−f(s,0)|+|f(s,0)| ≤L₁r+M₁, |gi(s, u(s))| ≤ |gi(s, u(s))−g_i(s,0)|+

|g_i(s,0)| ≤L_i+1r+M_i+1, i= 1, . . . , k, foru∈B_r, and (2.5) we can show that k(Tu)k

≤(Lr+M) sup

t∈[0,1]

n|A|h t^α

Γ(α+ 1)+|A0|t^α−1 1

Γ(α−δ+ 1)+a ξ^α−δ+12 Γ(^α−δ+1₂ + 1)

i

+

k

X

i=1

|Bi|h t^α+βi

Γ(α+β_i+ 1)+|A0|t^α−1 1

Γ(α−δ+β_i+ 1) +a ξ^{α−δ+12 +βi}

Γ(^α−δ+1₂ +βi+ 1) io

≤(Lr+M)∆< r,

which implies thatTBr⊂Br. Now, foru, v ∈ Cwe obtain kTu− Tvk

≤ sup

t∈[0,1]

n|A|

Z t

0

(t−s)^α−1

Γ(α) |f(s, u(s))−f(s, v(s))|ds +

k

X

i=1

|Bi| Z t

0

(t−s)^α+βi−1

Γ(α+β_i) |gi(s, u(s))−gi(s, v(s))|ds +|A0|t^α−1h

|A|

Z 1

0

(1−s)^α−δ−1

Γ(α−δ) |f(s, u(s))−f(s, v(s))|ds +

k

X

i=1

|B_i| Z 1

0

(1−s)^{α−δ+βi−1}

Γ(α−δ+βi) |g_i(s, u(s))−g_i(s, v(s))|ds +|A|a

Z ξ

0

(ξ−s)^α−δ−12

Γ(^α−δ+1₂ ) |f(s, u(s))−f(s, v(s))|ds +a

k

X

i=1

|Bi| Z ξ

0

(ξ−s)^{α−δ−12 +βi−1}

Γ(^α−δ−1₂ +β_i) |gi(s, u(s))−gi(s, v(s))|dsio

≤L sup

t∈[0,1]

n|A|h t^α

Γ(α+ 1)+|A₀|t^α−1 1

Γ(α−δ+ 1)+a ξ^α−δ+12 Γ(^α−δ+1₂ + 1)

i

+

k

X

i=1

|Bi|h t^α+βi

Γ(α+β_i+ 1) +|A0|t^α−1 1

Γ(α−δ+β_i+ 1) +a ξ^{α−δ+12 +βi}

Γ(^α−δ+1₂ +βi+ 1)

ioku−vk

=L∆ku−vk.

By the assumption,L <1/∆. Therefore,T is a contraction. Thus, by the contrac- tion mapping principle (Banach’s fixed point theorem) the proof is complete.

Now we present another variant of existence uniqueness result based on the H¨older inequality.

Theorem 3.2. Suppose that the continuous functions f, gi satisfy the following conditions:

(A2) |f(t, u(t))−f(t, v(t))| ≤m(t)|u−v|,|g_i(t, u(t))−g_i(t, v(t))| ≤n_i(t)|u−v|, for t∈[0,1],u, v∈R,m, ni∈L^1γ([0,1],R⁺), i= 1, . . . , k, andγ∈(0, α−δ−2).

(A3) |A|kmkZ1+Pk

i=1|Bi|knikZi+1<1, where Z₁= 1

Γ(α) 1−γ

α−γ ^1−γ

+ |A₀| Γ(α−δ)

1−γ α−δ−γ

^1−γ

+ a|A0| Γ(^α−δ+1₂ )

1−γ

α−δ+1 2 −γ

1−γ

ξ^{α−δ+12 −γ},

Z_i+1= ( 1

Γ(α+βi)) 1−γ α+βi−γ

1−γ

+ ( |A₀|

Γ(α−δ+βi)) 1−γ α−δ+βi−γ

1−γ

+ ( a|A0|

Γ(^α−δ+1₂ +βi)) 1−γ

α−δ+1

2 +βi−γ 1−γ

ξ^{α−δ+12 +βi−γ}, (i= 1, . . . , k), andkµk= (R1

0 |µ(s)|^1γds)^γ, µ=m, n.

Then, the boundary value problem (1.1)has a unique solution.

Proof. Foru, v∈Randt∈[0,1], by H¨older inequality, we have kTu− Tvk

≤ sup

t∈[0,1]

n|A|

Z t

0

(t−s)^α−1

Γ(α) m(s)|u(s)−v(s)|ds +

k

X

i=1

|Bi| Z t

0

(t−s)^α+βi−1

Γ(α+β_i) ni(s)|u(s)−v(s))|ds +|A0|h

|A|

Z 1

0

(1−s)^α−δ−1

Γ(α−δ) m(s)|u(s)−v(s)|ds +

k

X

i=1

|Bi| Z 1

0

(1−s)^{α−δ+βi−1}

Γ(α−δ+βi) n_i(s)|u(s)−v(s))|ds +a|A|

Z ξ

0

(ξ−s)^α−δ−12

Γ(^α−δ+1₂ ) m(s)|u(s)−v(s)|ds +a

k

X

i=1

|Bi| Z ξ

0

(ξ−s)^{α−δ−12 +βi}

Γ(^α−δ+1₂ +β_i) ni(s)|u(s)−v(s))|dsio

≤ sup

t∈[0,1]

n|A|kmk Γ(α)

1−γ α−γ

^1−γ t^α−γ+

k

X

i=1

( |Bi|knik

Γ(α+βi)) 1−γ α+βi−γ

1−γ

t^α+βi−γ

+|A0|h |A|kmk Γ(α−δ)

1−γ α−δ−γ

1−γ

+

k

X

i=1

( |Bi|knik

Γ(α−δ+β_i)) 1−γ α−δ+β_i−γ

1−γ

+ a|A|kmk Γ(^α−δ+1₂ )

1−γ

α−δ+1 2 −γ

^1−γ

ξ^{α−δ+12 −γ}

+a

k

X

i=1

( |Bi|knik

Γ(^α−δ+1₂ +βi)) 1−γ

α−δ+1

2 +βi−γ 1−γ

ξ^{α−δ+12 +βi−γ}io ku−vk

≤n

A|kmkh 1 Γ(α)

1−γ α−γ

1−γ

+ |A0| Γ(α−δ)

1−γ α−δ−γ

1−γ

+ a|A0| Γ(^α−δ+1₂ )

1−γ

α−δ+1 2 −γ

1−γ

ξ^{α−δ+12 −γ}i

+

k

X

i=1

|B_i|kn_ikh

( 1

Γ(α+βi)) 1−γ α+βi−γ

1−γ

+ ( |A0|

Γ(α−δ+β_i)) 1−γ α−δ+β_i−γ

1−γ

+ ( a|A₀|

Γ(^α−δ+1₂ +βi)) 1−γ

α−δ+1

2 +βi−γ 1−γ

ξ^{α−δ+12 +βi−γ}io ku−vk

= [|A|kmkZ1+

k

X

i=1

|Bi|knikZi+1]ku−vk.

By the condition (A3), it follows that T is a contraction mapping. Hence, by the Banach’s fixed point theorem T has a unique fixed point which is the unique solution of the problem (1.1). Then, the proof is complete.

4. Existence result via Leray-Schauder’s alternative

In this section, by using the Leray-schauder’s alternative, we will find at least one solution to problem (1.1). The proof of the main result in this section is based on the Leray-schauder’s alternative [1, 8] that we recall here for the reader’s convenience.

Lemma 4.1. (Nonlinear alternative for single valued maps [1, 8]). Let E be a Banach space,C a closed, convex subset of E, U an open subset of C and 0∈U.

Suppose thatF: ¯U→C is a completely continuous operator. Then, either (i) F has a fixed point in ¯U, or

(ii) there is au∈∂U (the boundary ofU inC) andλ∈(0,1)withu=λF(u).

We now state our main result in this section.

Theorem 4.2. Suppose that f, gi : [0,1]×R → R, i = 1, . . . , k, are continuous functions. Assume that:

(H1) There exist functionsp, pi∈L¹([0,1],R⁺),i= 1, . . . , k, and nondecreasing functionsψ, ψi:R⁺→R⁺,i= 1, . . . , k, such that

|f(t, x)| ≤p(t)ψ(kxk), |g_i(t, x)| ≤p_i(t)ψ_i(kxk), for all(t, x)∈[0,1]×Randi= 1, . . . , k.

(H2) There exists a constantM >0 such that M

|A|ψ(M)kpk_L1Ω +Pk

i=1|Bi|Ωiψi(M)kpik_L1

>1 where

Ω = 1

Γ(α+ 1) +|A₀| 1

Γ(α−δ+ 1)+a ξ^α−δ+12 Γ(^α−δ+1₂ + 1)

, and

Ωi = 1

Γ(α+βi+ 1)+|A0| 1

Γ(α−δ+βi+ 1) +a ξ^{α−δ+12 +βi} Γ(^α−δ+1₂ +βi+ 1)

, (i= 1, . . . , k).

Then, the boundary-value problem (1.1)has at least one solution on [0,1].

Proof. Consider the operatorT :C → Cwith (Tu)(t)

=−A Z t

0

(t−s)^α−1

Γ(α) f(s, u(s))ds−

k

X

i=1

Bi

Z t

0

(t−s)^α+βi−1

Γ(α+β_i) gi(s, u(s))ds +t^α−1A0

h A

Z 1

0

(1−s)^α−δ−1

Γ(α−δ) f(s, u(s))ds +

k

X

i=1

Bi

Z 1

0

(1−s)^{α−δ+βi−1}

Γ(α−δ+βi) gi(s, u(s))ds

−aA Z ξ

0

(ξ−s)^α−δ−12

Γ(^α−δ+1₂ ) f(s, u(s))ds−a

k

X

i=1

Bi

Z ξ

0

(ξ−s)^{α−δ−12 +βi}

Γ(^α−δ+1₂ +βi) gi(s, u(s))dsi . We show thatT maps bounded sets into bounded sets inC([0,1],R). for a positive number r, let B_r ={u∈C([0,1],R) : kuk ≤r} be a bounded set inC([0,1],R).

Then, we have

|(Tu)(t)|

≤ |A|

Z t

0

(t−s)^α−1

Γ(α) p(s)ψ(kuk)ds+

k

X

i=1

|Bi| Z t

0

(t−s)^α+βi−1

Γ(α+βi) p_i(s)ψ_i(kuk)ds +|A₀|t^α−1h

|A|

Z 1

0

(1−s)^α−δ−1

Γ(α−δ) p(s)ψ(kuk)ds +

k

X

i=1

|Bi| Z 1

0

(1−s)^{α−δ+βi−1}

Γ(α−δ+βi) pi(s)ψi(kuk)ds +|A|a

Z ξ

0

(ξ−s)^α−δ−12

Γ(^α−δ+1₂ ) p(s)ψ(kuk)ds +a

k

X

i=1

|B_i| Z ξ

0

(ξ−s)^{α−δ−12 +βi−1}

Γ(^α−δ−1₂ +βi) p_i(s)ψ_i(kuk)dsi

≤ |A|ψ(r)kpkL¹

h t^α

Γ(α+ 1) +|A0|t^α−1 1

Γ(α−δ+ 1) +a ξ^α−δ+12 Γ(^α−δ+1₂ + 1)

i

+

k

X

i=1

|Bi|ψi(r)kpik_L1

h t^α+βi

Γ(α+βi+ 1)+|A0|t^α−1 1

Γ(α−δ+βi+ 1) +a ξ^{α−δ+12 +βi}

Γ(^α−δ+1₂ +βi+ 1) i

. Consequently,

kTuk

≤ |A|ψ(r)kpk_L1

h 1

Γ(α+ 1)+|A₀| 1

Γ(α−δ+ 1)+a ξ^α−δ+12 Γ(^α−δ+1₂ + 1)

i

+

k

X

i=1

|B_i|ψ_i(r)kp_ik_L1

h 1

Γ(α+βi+ 1) +|A₀| 1

Γ(α−δ+βi+ 1) +a ξ^{α−δ+12 +βi}

Γ(^α−δ+1₂ +β_i+ 1) i

=|A|ψ(r)kpk_L1Ω +

k

X

i=1

|Bi|ψi(r)kpik_L1Ωi,

let K = |A|ψ(r)kpkL¹Ω + Pk

i=1|Bi|ψi(r)kpikL¹Ωi, therefore, we conclude that kTuk ≤K. Thus,T maps bounded sets into bounded sets inC([0,1],R).

Next, we show thatT maps bounded sets into equicontinuous sets ofC([0,1],R).

Lett₁, t₂∈[0,1] witht₁< t₂ andu∈B_r; thus, we have

|(Tu)(t₂)−(Tu)(t₁)|

≤ |A|

Γ(α) Z t₁

0

[(t2−s)^α−1−(t1−s)^α−1]|f(s, u(s))|ds + |A|

Γ(α) Z t₂

t₁

(t2−s)^α−1|f(s, u(s))|ds +

k

X

i=1

|Bi| Γ(α+β_i)

Z t1

0

[(t2−s)^α+βi−1−(t1−s)^α+βi−1]|gi(s, u(s))|ds

+

k

X

i=1

|Bi| Γ(α+βi)

Z t₂

t₁

(t2−s)^α+βi−1|gi(s, u(s))|ds

+ [(t2)^α−1−(t1)^α−1]|A0|h

|A|

Z 1

0

(1−s)^α−δ−1

Γ(α−δ) |f(s, u(s))|ds +

k

X

i=1

|Bi| Z 1

0

(1−s)^{α−δ+βi−1}

Γ(α−δ+β_i) |gi(s, u(s))|ds +a|A|

Z ξ

0

(ξ−s)^α−δ−12

Γ(^α−δ+1₂ ) |f(s, u(s))|ds +a

k

X

i=1

|Bi| Z ξ

0

(ξ−s)^{α−δ−12 +βi}

Γ(^α−δ+1₂ +βi) |gi(s, u(s))|dsi

≤ |A|

Γ(α) Z t₁

0

[(t2−s)^α−1−(t1−s)^α−1]p(s)ψ(r)ds + |A|

Γ(α) Z t₂

t1

(t2−s)^α−1p(s)ψ(r)ds +

k

X

i=1

|B_i| Γ(α+βi)

Z t1

0

[(t₂−s)^α+βi−1−(t₁−s)^α+βi−1]p_i(s)ψ_i(r)ds

+

k

X

i=1

|Bi| Γ(α+βi)

Z t₂

t1

(t2−s)^α+βi−1pi(s)ψi(r)ds

+ [(t2)^α−1−(t1)^α−1]|A0|h

|A|

Z 1

0

(1−s)^α−δ−1

Γ(α−δ) p(s)ψ(r)ds +

k

X

i=1

|B_i| Z 1

0

(1−s)^{α−δ+βi−1}

Γ(α−δ+βi) p_i(s)ψ_i(r)ds

+a|A|

Z ξ

0

(ξ−s)^α−δ−12

Γ(^α−δ+1₂ ) p(s)ψ(r)ds+a

k

X

i=1

|Bi| Z ξ

0

(ξ−s)^{α−δ−12 +βi}

Γ(^α−δ+1₂ +βi) pi(s)ψi(r)dsi Obviously, the right-hand side of the above inequality tends to zero independently of u∈B_r as t₂−t₁ →0. Therefore, T :C([0,1],R)→ C([0,1],R) is completely continuous by application of the Arzel´a-Ascoli theorem.

Now, we can conclude the result by using the Leray-Schauder’s nonlinear alter- native theorem. Consider the equations x=λTx for λ∈(0,1) and assume that ube a solution. Then, using the computations in proving that T is bounded, we have

kuk=kλ(Tu)k

≤ |A|ψ(kuk)kpk_L1

h 1

Γ(α+ 1)+|A0| 1

Γ(α−δ+ 1)+a ξ^α−δ+12 Γ(^α−δ+1₂ + 1)

i

+

k

X

i=1

|Bi|ψi(kuk)kpik_L1

h 1

Γ(α+β_i+ 1) +|A0| 1

Γ(α−δ+β_i+ 1) +a ξ^{α−δ+12 +βi}

Γ(^α−δ+1₂ +βi+ 1) i; therefore,

kuk

|A|ψ(kuk)kpk_L1Ω +Pk

i=1|B_i|Ω_iψ_i(kuk)kp_ik_L1

≤1.

In view of (H2), there existsM such thatkuk 6=M. Let us set U ={x∈C([0,1],R) :kxk< M}.

It is obvious that the operatorT : ¯U →C([0,1],R) is continuous and completely continuous. From the choice of U there is no u ∈ ∂U such that u = λT(u) for some λ∈(0,1). Therefore, by the Leray-Schauder’s nonlinear alternative theorem (Lemma 4.1), we conclude thatT has a fixed point u∈ U which is a solution of¯

the problem (1.1). Thus, the proof is completed.

5. Application

Example 5.1. Consider the singular boundary value problem

−D^3/2u(t) =Af(t, u(t)) +

3

X

i=1

BiI^βigi(t, u(t)), t∈(0,1), D^1/4u(0) = 0, D^1/4u(1) =aD^1/8(D^1/4u(t))

_t=1/2

(5.1)

Here, A =Bi = 1, (i = 1,2,3), β1 = 1/2, β2 = 1/3, β3 = 2/3, a= 2, f(t, u) =

9 125

√1−t(1 +u),gi(t, u) =₂₉³ tan⁻¹u+ cos(e^t). With the given data, we obtain:

A₀= Γ(α−δ)Γ(^α−δ+1₂ ) Γ(α)

Γ(^α−δ+1₂ )−aΓ(α−δ)ξ^α−δ−12 =−1.3365,

∆ = sup

t∈[0,1]

n|A|h t^α

Γ(α+ 1)+|A0|t^α−1 1

Γ(α−δ+ 1)+a ξ^α−δ+12 Γ(^α−δ+1₂ + 1)

i

+

k

X

i=1

|Bi|h t^α+βi

Γ(α+β_i+ 1)+|A0|t^α−1 1

Γ(α−δ+β_i+ 1) +a ξ^{α−δ+12 +βi}

Γ(^α−δ+1₂ +βi+ 1) io

= 8.9046,

andL1= 9/125,L2=L3=L4= 3/29 as|f(t, u)−f(t, v)| ≤ ₁₂₅⁹ |u−v|,|gi(t, u)− g_i(t, v)| ≤ ₂₉³|u−v|. Obviously, L = max{L_i : i = 1, . . . ,4} = 3/29 and L < _∆¹. Hence, all the assumptions of Theorem 3.1 are satisfied. Thus, by the conclusion of Theorem 3.1, problem (5.1) has a unique solution.

Acknowledgments. The authors would like to thank the anonymous referees for their suggestions and helpful comments which improved the presentation of the original manuscript.

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Nemat Nyamoradi

Department of Mathematics, Faculty of Sciences, Razi University, 67149 Kermanshah, Iran

E-mail address:[email protected]

Tahereh Bashiri

Department of Mathematics and Computer Sciences, Amirkabir University of Technol- ogy, Tehran, Iran

E-mail address:[email protected]

S. Mansour Vaezpour

Department of Mathematics and Computer Sciences, Amirkabir University of Technol- ogy, Tehran, Iran

E-mail address:[email protected]

Dumitru Baleanu

Department of Mathematics and Computer Sciences, Faculty of Art and Sciences, Cankaya University, 06530 Ankara, Turkey.

Institute of Space Sciences, P.O.BOX, MG-23, R 76900,Magurele-Bucharest, Romania.

Department of Chemical and Materials Engineering, Faculty of Engineering, King Ab- dulaziz University, P.O. Box 80204, Jeddah 21589, Saudi Arabia

E-mail address:[email protected]