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The existence of solutions for a nonlinear mixed problem of singular fractional differential equations

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Abstract

By using fixed point results on cones, we study the existence of solutions for the singular nonlinear fractional boundary value problem

D α c u ( t ) = f ( t , u ( t ) , u ( t ) , c D β u ( t ) ) , u ( 0 ) = a u ( 1 ) , u ( 0 ) = b c D β u ( 1 ) , u ( 0 ) = u ( 0 ) = u ( n 1 ) ( 0 ) = 0 ,

where n3 is an integer, α(n1,n), 0<β<1, 0<a<1, 0<b<Γ(2β), f is an L q -Caratheodory function, q> 1 α 1 and f(t,x,y,z) may be singular at value 0 in one dimension of its space variables x, y, z. Here, cD stands for the Caputo fractional derivative.

1 Introduction

Fractional differential equations (see, for example, [16] and references therein) started to play an important role in several branches of science and engineering. There are some works about existence of solutions for the nonlinear mixed problems of singular fractional boundary value problem (see, for example, [711] and [12]). Also, there are different methods for solving distinct fractional differential equations (see, for example, [1318] and [19]). By using fixed point results on cones, we focus on the existence of positive solutions for a nonlinear mixed problem of singular fractional boundary value problem. For the convenience of the reader, we present some necessary definitions from fractional calculus theory (see, for example, [20]). The Caputo derivative of fractional order α for a function f:[0,)R is defined by

D α c f(t)= 1 Γ ( n α ) 0 t ( t s ) n α 1 f ( n ) (s)ds ( n 1 < α < n , n = [ α ] + 1 ) .

Let q1. As you know, L q [0,1] denotes the space of functions, whose q th powers of modulus are integrable on [0,1], equipped with the norm x q = ( 0 1 | x ( t ) | q d t ) 1 q . We consider the sup norm

x=sup { | x ( t ) | : t [ 0 , 1 ] }

on the space C[0,1]. Also, AC[0,1] is the set of absolutely continuous functions on [0,1]. Let B be a subset of R 3 . A function f:[0,1]×BR is called an L q -Caratheodory function whenever the real-valued function f(,x,y,z) on [0,1] is measurable for all (x,y,z)B, the function f(t,,,):BR is continuous for almost all t(0,1], and for each compact set UB, there exists a function φ u L q [0,1] such that |f(t,x,y,z)| φ u (t) for almost all t[0,1] and (x,y,z)U. Consider the nonlinear fractional boundary value problem

D α u ( t ) = f ( t , u ( t ) , u ( t ) , c D β u ( t ) ) , u ( 0 ) = a u ( 1 ) , u ( 0 ) = b c D β u ( 1 ) , u ( 0 ) = u ( 0 ) = u ( n - 1 ) ( 0 ) = 0 ,
(*)

where n3 is an integer, α(n1,n), 0<β<1, 0<a<1, 0<b<Γ(2β) and q> 1 α 1 . We say that the function u:[0,1]R is a positive solution for the problem whenever u>0 on [0,1], D α c u is a function in L q [0,1], and u satisfies the boundary conditions almost everywhere on [0,1]. In this paper, we suppose that f is an L q -Caratheodory function on [0,1]×B, where B=(0,)×(0,)×(0,), there exists a positive constant m such that mf(t,x,y,z) for almost all t[0,1] and (x,y,z)B, f satisfies the estimate

f(t,x,y,z)h(x)+r ( | y | ) +k ( | z | ) +γ(t)w ( x , | y | , | z | ) ,

where h,r,kC(0,) are positive and non-increasing, γ L q [0,1] and wC([0,)×[0,)×[0,)) are positive, w is non-decreasing in all its variables, 0 1 h q ( s α )ds<, 0 1 r q ( s α 1 )ds<, 0 1 k q ( s α β )ds<, and lim x w ( x , x , x ) x =0. Since we suppose that problem () is singular, that is, f(t,x,y,z) may be singular at the value 0 of its space variables x, y, z, we use regularization and sequential techniques for the existence of positive solutions of the problem. In this way, for each natural number n define the function f n by

f n (t,x,y,z)=f ( t , χ n + ( x ) , χ n + ( y ) , χ n + ( z ) )

for all t[0,1] and (x,y,z) R 3 , where

χ n + (u)={ u , u 1 n , 1 n , u < 1 n .

It is easy to see that each f n is an L q -Caratheodory function on [0,1]× R 3 , m f n (t,x,y,z),

f n (t,x,y,z)h ( 1 n ) +r ( 1 n ) +k ( 1 n ) +γ(t)w ( 1 + x , 1 + | y | , 1 + | z | )

and

f n (t,x,y,z)h(x)+r ( | y | ) +k ( | z | ) +γ(t)w ( 1 + x , 1 + | y | , 1 + | z | )

for almost all t[0,1] and all (x,y,z)B. In 2012, Agarwal et al. proved the following result.

Lemma 1.1 [7]

Let ρ L q [0,1] and 0 t 1 < t 2 1. Then we have | 0 t ( t s ) α 2 ρ(s)ds| ( t d d ) 1 / p ρ q for all t[0,1] and

| 0 t 2 ( t 2 s ) α 2 ρ ( s ) d s 0 t 1 ( t 1 s ) α 2 ρ ( s ) d s | ( t 1 d + ( t 2 t 1 ) d t 2 d d ) 1 / p ρ q + ( ( t 2 t 1 ) d d ) 1 / p ρ q ,

where d=(α2)p+1.

2 Main results

Now, we are ready to investigate the problem in regular and singular cases. First, we give the following result.

Lemma 2.1 Let yC[0,1]. Then the boundary value problem

D α c u ( t ) = y ( t ) ( t ( 0 , 1 ) ) , u ( 0 ) = a u ( 1 ) , u ( 0 ) = b c D β u ( 1 ) , u ( 0 ) = u ( 0 ) = u ( n 1 ) ( 0 ) = 0

is equivalent to the fractional integral equation u(t)= 0 1 G(t,s)y(s)ds, where

G ( t , s ) = ( t s ) α 1 Γ ( α ) + a Γ ( α β ) ( Γ ( 2 β ) b ) ( 1 s ) α 1 + b Γ ( α ) Γ ( 2 β ) ( a + t a t ) ( 1 s ) α β 1 ( 1 a ) Γ ( α ) Γ ( α β ) ( Γ ( 2 β ) b )

whenever 0st1 and

G(t,s)= a Γ ( α β ) ( Γ ( 2 β ) b ) ( 1 s ) α 1 + b Γ ( α ) Γ ( 2 β ) ( a + t a t ) ( 1 s ) α β 1 ( 1 a ) Γ ( α ) Γ ( α β ) ( Γ ( 2 β ) b )

whenever 0ts1.

Proof From D α c u(t)=y(t) and the boundary conditions, we obtain

u ( t ) = I α y ( t ) + u ( 0 ) + u ( 0 ) t + u ( 0 ) 2 ! t 2 + + u ( n 1 ) ( 0 ) ( n 1 ) ! t n 1 = 1 Γ ( α ) 0 t ( t s ) α 1 y ( s ) d s + u ( 0 ) + u ( 0 ) t .

By properties of the Caputo derivative, we get

D β c u ( t ) = I α β y ( t ) + c D β ( u ( 0 ) + u ( 0 ) t ) = 1 Γ ( α β ) 0 t ( t s ) α β 1 y ( s ) d s + u ( 0 ) t 1 β Γ ( 2 β ) .

Thus, u(1)= 1 Γ ( α ) 0 1 ( 1 s ) α 1 y(s)ds+u(0)+ u (0) and

D β c u(1)= 1 Γ ( α β ) 0 1 ( 1 s ) α β 1 y(s)ds+ u ( 0 ) Γ ( 2 β ) .

By using the boundary conditions u(0)=au(1) and u (0)=b c D β u(1), we get u(0)=a( 1 Γ ( α ) 0 1 ( 1 s ) α 1 y(s)ds+u(0)+ u (0)) and

u (0)=b ( 1 Γ ( α β ) 0 1 ( 1 s ) α β 1 y ( s ) d s + u ( 0 ) Γ ( 2 β ) ) .

Hence, u (0)= b Γ ( 2 β ) Γ ( α β ) ( Γ ( 2 β ) b ) 0 1 ( 1 s ) α β 1 y(s)ds and

u ( 0 ) = a ( 1 a ) Γ ( α ) 0 1 ( 1 s ) α 1 y ( s ) d s + a b Γ ( 2 β ) ( 1 a ) Γ ( α β ) ( Γ ( 2 β ) b ) 0 1 ( 1 s ) α β 1 y ( s ) d s .

Thus,

u ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 y ( s ) d s + u ( 0 ) + u ( 0 ) t = 0 t ( ( t s ) α 1 Γ ( α ) + a Γ ( α β ) ( Γ ( 2 β ) b ) ( 1 s ) α 1 ( 1 a ) Γ ( α ) Γ ( α β ) ( Γ ( 2 β ) b ) + b Γ ( α ) Γ ( 2 β ) ( a + t a t ) ( 1 s ) α β 1 ( 1 a ) Γ ( α ) Γ ( α β ) ( Γ ( 2 β ) b ) ) y ( s ) d s + t 1 ( a Γ ( α β ) ( Γ ( 2 β ) b ) ( 1 s ) α 1 ( 1 a ) Γ ( α ) Γ ( α β ) ( Γ ( 2 β ) b ) + b Γ ( α ) Γ ( 2 β ) ( a + t a t ) ( 1 s ) α β 1 ( 1 a ) Γ ( α ) Γ ( α β ) ( Γ ( 2 β ) b ) ) y ( s ) d s = 0 1 G ( t , s ) y ( s ) d s .

This completes the proof. □

Put k 1 = Γ ( α β ) ( Γ ( 2 β ) b ) + b Γ ( α ) Γ ( 2 β ) ( 1 a ) Γ ( α ) Γ ( α β ) ( Γ ( 2 β ) b ) and k 2 = a b Γ ( 2 β ) ( 1 a ) Γ ( α β ) ( Γ ( 2 β ) b ) . It is easy to check that the Green function G in the last result belongs to C([0,1]×[0,1]), G(t,s)>0 for all (t,s)[0,1)×[0,1),

G(t,s) k 1 ( 1 s ) α β 1 1andG(t,s) k 2 ( 1 s ) α β 1

for all (t,s)[0,1]×[0,1]. Consider the Banach space X= C 1 [0,1] with the norm x =max{x, x } and the cone

P= { x X : x ( t ) 0  and  x ( t ) 0  for all  t [ 0 , 1 ] } .

For each natural number n, define the operator Q n on P by

( Q n x)(t)= 0 1 G(t,s) f n ( s , u ( s ) , u ( s ) , c D β u ( s ) ) ds.

Now, we prove that Q n is a completely continuous operator (see [2]).

Lemma 2.2 The operator Q n is a completely continuous operator.

Proof Let xP. Then, D β c xC[0,1] and D β c x0. Now, define ρ(t)= f n (t,u(t), u (t) , c D β u(t)) for almost all t[0,1]. Then ρ L q [0,1] and ρ(t)m for almost all t[0,1]. By using the properties of fractional integral I α , it is easy to see that Q n xC[0,1], ( Q n x)(t)0 and

( Q n x ) (t)= 1 Γ ( α 1 ) 0 t ( t s ) α 2 ρ(s)ds

for all t[0,1]. This implies that ( Q n x ) C[0,1] and ( Q n x ) 0 on [0,1]. Consequently, Q n maps P into P. In order to prove that Q n is a continuous operator, let x m be a convergent sequence in P and lim m x m =x. Thus, lim m x m ( j ) (t)= x ( j ) (t) uniformly on [0,1] for j=0,1. Since

D β c x ( t ) = 1 Γ ( 1 β ) d d t 0 t ( t s ) β ( x ( s ) x ( 0 ) ) d s = 1 Γ ( 1 β ) 0 t ( t s ) β x ( s ) d s ,

we get | c D β x m (t) c D β x(t)| x m x Γ ( 1 β ) 0 t ( t s ) β ds x m x Γ ( β ) and lim m c D β x m (t) = c D β x(t) uniformly on [0,1]. Also, we have | c D β x m (t)| x m Γ ( β ) on [0,1], and so c D β x m x m Γ ( β ) . Now, put

ρ m (t)= f n ( t , x m ( t ) , x m ( t ) , c D β x m ( t ) ) andρ(t)= f n ( t , x ( t ) , x ( t ) , c D β x ( t ) ) .

Then, it is easy to see that lim m ρ m (t)=ρ(t) for almost all t[0,1], and there exists β L q [0,1] such that 0 ρ m (t)β(t) for almost all t[0,1] and all m1. Since f n is an L q -Caratheodory function, { x m } is bounded in C 1 [0,1], and { c D β x m } is bounded in C[0,1]. Therefore, lim m ( Q n x m )(t)=( Q n x)(t) uniformly on [0,1]. Since { ρ m } is L q -convergent on [0,1],

lim m ( Q n x m ) (t)= 1 Γ ( α 1 ) lim m 0 t ( t s ) α 2 ρ m (s)ds= ( Q n x ) (t)

uniformly on [0,1]. Hence, Q n is a continuous operator. Now, we have to show that for each bounded sequence { x m } in P, the sequence { Q n x m } is relatively compact in C[0,1]. Choose a positive constant k such that x m k and x m k for all m. Note that c D β x m k Γ ( β ) and | 0 t ( t s ) α 2 ρ m (s)ds| ( 0 t ( t s ) ( α 2 ) p d s ) 1 p ( 0 t | ρ m ( s ) | q d s ) 1 q ( t d d ) 1 p ρ m q for all m, where d=(α2)p+1. But we have

0( Q n x m )(t)= 0 1 G(t,s) ρ m (s)ds 0 1 G(t,s)β(s)ds β 1 Γ ( α )

and

0 ( Q n x m ) ( t ) = 1 Γ ( α 1 ) 0 t ( t s ) α 2 ρ m ( s ) d s 1 Γ ( α 1 ) 0 t ( t s ) α 2 β ( s ) d s 1 Γ ( α 1 ) ( 1 ( α 2 ) p + 1 ) 1 p β q

for all t[0,1] and m. This implies that { Q n x m } is bounded in C 1 [0,1]. Also, we have

| ( Q n x m ) ( t 2 ) ( Q n x m ) ( t 1 ) | = 1 Γ ( α 1 ) | 0 t 2 ( t 2 s ) α 2 ρ m ( s ) d s 0 t 1 ( t 1 s ) α 2 ρ m ( s ) d s | ρ m q Γ ( α 1 ) ( ( t 1 d + ( t 2 t 1 ) d t 2 d d ) 1 p + ( ( t 2 t 1 ) d d ) 1 p ) β q Γ ( α 1 ) ( ( t 1 d + ( t 2 t 1 ) d t 2 d d ) 1 p + ( ( t 2 t 1 ) d d ) 1 p )

for all 0 t 1 t 2 1, where d=(α2)p+1. Hence, { ( Q n x m ) } is equicontinuous on [0,1]. Thus, { Q n x m } is relatively compact in C 1 [0,1] by the Arzela-Ascoli theorem. Hence, Q n is a completely continuous operator. □

We need the following result (see [2] and [21]).

Lemma 2.3 [21]

Let Y be a Banach space, P a cone in Y and Ω 1 and Ω 2 bounded open balls in Y centered at the origin with Ω ¯ 1 Ω 2 . Suppose that T:P( Ω ¯ 2 Ω 1 )P is a completely continuous operator such that Txx for all xP Ω 1 and Txx for all xP Ω 2 . Then T has a fixed point in P( Ω ¯ 2 Ω 1 ).

Theorem 2.4 For each natural number n, problem () has a solution u n P such that u n m k 2 α β , u n (t) m t α 1 Γ ( α ) and D β c u n (t) m t α β Γ ( α β + 1 ) for all t[0,1].

Proof Let n1. It is sufficient to show that Q n has a fixed point u n in P with the desired conditions. In this way, note that

( Q n x ) ( t ) = 0 1 G ( t , s ) f n ( s , x ( s ) , x ( s ) , c D β x ( s ) ) d s m 0 1 G ( t , s ) d s m 0 1 k 2 ( 1 s ) α β 1 d s = m k 2 α β ,

and so Q n x Q n (x) m k 2 α β . Put Ω 1 ={xX: x < m k 2 α β }. Then Q n x x for all xP Ω 1 . If v n =h( 1 n )+r( 1 n )+k( 1 n ), then

| ( Q n x ) ( t ) | | 0 1 G ( t , s ) f n ( s , u ( s ) , u ( s ) , c D β u ( s ) ) d s | 0 1 | G ( t , s ) | ( v n + γ ( s ) w ( 1 + | x ( s ) | , 1 + | x ( s ) | , 1 + | c D β x ( s ) | ) ) d s k 1 ( v n + w ( 1 + x , 1 + x , 1 + c D β x ) γ 1 )

and

| ( Q n x ) ( t ) | = | 1 Γ ( α 1 ) 0 t ( t s ) α 2 f n ( s , x ( s ) , x ( s ) , c D β x ( s ) ) d s | 1 Γ ( α 1 ) 0 t ( t s ) α 2 ( v n + γ ( s ) w ( 1 + | x ( s ) | , 1 + | x ( s ) | , 1 + | c D β x ( s ) | ) ) d s 1 Γ ( α 1 ) ( v n t α 1 α 1 + w ( 1 + x , 1 + x , 1 + c D β x ) 0 t ( t s ) α 2 γ ( s ) d s )

for all xP and t[0,1], because w is non-decreasing in all its variables. Since x x , x x , c D β x x Γ ( β ) x Γ ( β ) and

0 t ( t s ) α 2 γ(s)ds ( 1 d ) 1 / p γ q ,

where d=(α2)p+1, we have

Q n ( x ) k 1 ( v n + w ( 1 + x , 1 + x , 1 + x Γ ( β ) ) γ 1 )

and

( Q n x ) 1 Γ ( α 1 ) ( v n α 1 + w ( 1 + x , 1 + x , 1 + x Γ ( β ) ) ( 1 / d ) 1 / p γ q ) .

Hence, Q n x M( v n α 1 +Nw(1+ x ,1+ x ,1+ x Γ ( β ) )), where N=max{ γ 1 , ( 1 / d ) 1 / p γ q } and M=max{ 1 Γ ( α 1 ) , k 1 }. Since

lim v 1 v w(1+v,1+v,1+v)=0,

there exists a positive constant L such that

M ( v n α 1 + N w ( 1 + v , 1 + v , v Γ ( β ) ) ) <v

for all vL. Thus, Q n x < x for all xP with x L. Put Ω 2 ={xX: x <L}. Then Q n x < x for all xP Ω 2 . By using last result, Q n has a fixed point u n in P( Ω ¯ 2 Ω 1 ). But u n =( Q n u n )(t) m k 2 α β and

( Q n x ) ( t ) = 1 Γ ( α 1 ) 0 t ( t s ) α 2 f n ( s , x ( s ) , x ( s ) , c D β x ( s ) ) d s m Γ ( α 1 ) 0 t ( t s ) α 2 d s = m t α 1 Γ ( α )

for all t[0,1] and xP. Since 0 t ( t s ) β s α 1 ds= Γ ( α ) Γ ( 1 β ) Γ ( α β + 1 ) t α β ,

D β c u n ( t ) = 1 Γ ( 1 β ) 0 t ( t s ) β u n ( s ) d s m Γ ( α ) Γ ( 1 β ) 0 t ( t s ) β s α 1 d s = m t α β Γ ( α β + 1 )

for all t[0,1]. This completes the proof. □

Now, we give our last result.

Theorem 2.5 Problem () has a solution u such that u(t) m k 2 α β , u (t) m t α 1 Γ ( α ) and D β c u(t) m t α β Γ ( α β + 1 ) for all t[0,1].

Proof By using Theorem 2.4, one gets that for each natural number n, problem () has a solution u n P with the desired conditions. Thus, h( u n (t))h( m k 2 α β ), r(| u n (t)|)r( m t α 1 Γ ( α ) ) and k( | c D β u n (t)|)k( m t α β Γ ( α β + 1 ) ) for all t[0,1] and n. Also, we have c D β u n u n Γ ( β ) . Suppose that

S(t)=h ( m k 2 α β ) +r ( m t α 1 Γ ( α ) ) +k ( m t α β Γ ( α β + 1 ) ) .

Then

m f n ( t , u n ( t ) , u n ( t ) , c D β u n ( t ) ) S ( t ) + γ ( t ) w ( 1 + u n , 1 + u n , 1 + c D β u n ) S ( t ) + γ ( t ) w ( 1 + u n , 1 + u n , 1 + u n Γ ( β ) )

for almost all t[0,1] and n. Since 0G(t,s) k 1 , we get

0 u n ( t ) = 0 1 G ( t , s ) f n ( s , u n ( s ) , u n ( s ) , c D β u n ( s ) ) d s k 1 ( 0 1 S ( s ) d s + w ( 1 + u n , 1 + u n , 1 + u n Γ ( β ) ) γ 1 )

and

0 u n ( t ) 1 Γ ( α 1 ) ( 0 t ( t s ) α 2 S ( s ) d s + w ( 1 + u n , 1 + u n , 1 + u n Γ ( β ) ) 0 t ( t s ) α 2 γ ( s ) d s ) .

We show that 0 t ( t s ) α 2 S(s)ds is bounded on [0,1]. Let d=(α2)p+1. Note that

0 1 ( t s ) α 2 h ( m k 2 α β ) d s = h ( m k 2 α β ) 0 1 ( t s ) α 2 d s = 1 α 1 h ( m k 2 α β ) = : η 1 < , 0 t ( t s ) α 2 r ( m s α 1 Γ ( α ) ) d s = ( 1 d ) 1 / p ( Γ ( α ) m ) 1 ( α 1 ) q ( 0 ( m Γ ( α ) ) 1 α 1 r q ( s α 1 ) d s ) 1 / q = : η 2 <

and

0 t ( t s ) α 2 k ( m s α β Γ ( α β + 1 ) ) d s = ( 1 d ) 1 / p ( Γ ( α β + 1 ) m ) 1 ( α β ) q ( 0 ( m Γ ( α β + 1 ) ) 1 α β k q ( s α β ) d s ) 1 / q = : η 3 < .

Thus, 0 t ( t s ) α 2 S(s)dsη for all t[0,1], where η= η 1 + η 2 + η 3 . Also, we have

0 1 S ( s ) d s 1 α 1 h ( m k 2 α β ) + ( Γ ( α ) m ) 1 α 1 0 ( m Γ ( α ) ) 1 α 1 r ( s α 1 ) d s + ( Γ ( α β + 1 ) m ) 1 α β 0 ( m Γ ( α β + 1 ) ) 1 α β k ( s α β ) d s < .

Since

u n = k 1 ( 0 1 S ( s ) d s + w ( 1 + u n , 1 + u n , 1 + u n Γ ( β ) ) γ 1 )

and

u n 1 Γ ( α 1 ) ( η + w ( 1 + u n , 1 + u n , 1 + u n Γ ( β ) ) ( 1 d ) 1 / p γ q ) ,

we get u n M(Φ+Kw(1+ u n ,1+ u n ,1+ u n Γ ( β ) )) for all n, where Φ=max{η, 0 1 S(s)ds}, K=max{ γ 1 , ( 1 d ) 1 / p γ q } and also M=max{ k 1 , 1 Γ ( α 1 ) }. On the other hand, there exists a positive constant L such that M(Φ+Kw(1+v,1+v,1+ v Γ ( β ) ))<v for all vL, and so u n <L for all n. Thus, for almost all t[0,1] and all n, we have f n (t, u n (t), u n (t) , c D β u n (t))R(t), where

R(t)=S(t)+γ(t)w ( 1 + L , 1 + L , 1 + L Γ ( β ) ) .

Note that R L q [0,1]. We show that { u n } is equicontinuous on [0,1]. Let ρ n (t)= f n (t, u n (t), u n (t) , c D β u n (t)) and 0 t 1 < t 2 T. Then

| u n ( t 2 ) u n ( t 1 ) | = 1 Γ ( α 1 ) | 0 t 2 ( t 2 s ) α 2 ρ n ( s ) d s 0 t 1 ( t 1 s ) α 2 ρ n ( s ) d s | 1 Γ ( α 1 ) ( 0 t 1 ( ( t 1 s ) α 2 ( t 2 s ) α 2 ) ρ n ( s ) d s + t 1 t 2 ( t 2 s ) α 2 ρ n ( s ) d s ) 1 Γ ( α 1 ) ( 0 t 1 ( ( t 1 s ) α 2 ( t 2 s ) α 2 ) R ( s ) d s + t 1 t 2 ( t 2 s ) α 2 R ( s ) d s ) ,

and so

| u n ( t 2 ) u n ( t 1 ) | R q Γ ( α 1 ) ( ( t 1 d + ( t 2 t 1 ) d t 2 d d ) 1 / p + ( ( t 2 t 1 ) d d ) 1 / p ) .

Hence, { u n } is equicontinuous on [0,1]. Since { u n } is a bounded sequence in C[0,1], by using the Arzela-Ascoli theorem, without loss of generality, we can assume that { u n } is convergent in C[0,1]. Let lim n u n =u. Then, it is easy to see that D β c u n (t)= 1 Γ ( α 1 ) 0 t ( t s ) β u n (s)ds, and D β c u n (t) uniformly converges to 1 Γ ( α 1 ) 0 t ( t s ) β u (s)ds on [0,1]. Thus, D β c u n converges to D β c u in C[0,1]. Hence,

lim n f n ( t , u n ( t ) , u n ( t ) , c D β u n ( t ) ) =f ( t , u ( t ) , u ( t ) , c D β u ( t ) )

for almost all t[0,1]. Since R L q [0,1], by using the dominated convergence theorem on the relation

u n (t)= 0 1 G(t,s) f n ( s , u n ( s ) , u n ( s ) , c D β u n ( s ) ) ds,

we get u(t)= 0 1 G(t,s)f(s,u(s), u (s) , c D β u(s)) for all t[0,1]. This completes the proof. □

2.1 Examples for the problem

Example 2.1 Let ρ 1 , ρ 2 L q [0,1], ρ 1 (t)m>0 for almost all t in [0,1]. Suppose that

f(t,x,y,z)= ρ 1 (t)+ 1 x 1 3 λ + 1 y 1 4 + 1 z 1 4 + | ρ 2 ( t ) | ( x 1 3 + y 1 4 + z 1 4 )

on [0,1]×B, λ= ( a u ( 1 ) ) 1 3 , h(x)= 1 x 1 3 λ whenever x 1 3 λ0 and h(x)=0 whenever x 1 3 λ<0, r(x)= 1 x 1 4 , k(x)= 1 x 1 4 , w(x,y,z)=( x 1 3 + y 1 4 + z 1 4 +1) and γ(t)= ρ 1 (t)+| ρ 2 (t)|. Then Theorem 2.5 guarantees that problem () has a positive solution.

Example 2.2 Consider the nonlinear mixed problem of singular fractional boundary value problem

D 7 3 u ( t ) = t = 1 = 1 u ( t ) 1 3 - ρ = 1 u ( t ) 1 4 = 1 ( c D 1 4 u ( t ) ) 1 4 = 2 ( u ( t ) 1 3 = u ( t ) 1 4 = ( c D 1 4 u ( t ) ) 1 4 = 1 ) )

via boundary value conditions u(0)= 1 4 u(1), u (0)= 1 3 ( c D 1 4 )u(1) and u (0)= u (0)== u ( n 1 ) (0)=0, where ρ= ( ( 1 4 ) u ( 1 ) ) 1 3 . Let f

Then the map f is singular at t=0, and f satisfies the desired conditions, where h(x)= 1 x 1 3 ρ whenever x 1 3 ρ0 and h(x)=0 whenever x 1 3 ρ<0, r(x)= 1 x 1 4 , k(x)= 1 x 1 4 , w(x,y,z)= x 1 3 + y 1 4 + z 1 4 +1, ρ 1 (t)=t+1>1=m, ρ 2 (t)=2 and γ(t)= ρ 1 (t)+| ρ 2 (t)|. Then Theorem 2.5 guarantees that this problem has a positive solution.

3 Conclusions

One of the most interesting branches is obtaining solutions of singular fractional differential via boundary value problems. Having these things in mind, we study the existence of solutions for a singular nonlinear fractional boundary value problem. Two illustrative examples illustrate the applicability of the proposed method. It seems that the obtained results could be extended to more general functional spaces. Finally, note that all calculations in proofs of the results depend on the definition of the fractional derivative.

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Acknowledgements

Research of the second and third authors was supported by Azarbaijan Shahid Madani University. Also, the authors express their gratitude to the referees for their helpful suggestions, which improved the final version of this paper.

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Correspondence to Dumitru Baleanu.

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Keywords

  • boundary value problem
  • fixed point
  • fractional differential equation
  • Green function
  • regularization
  • singular