Baleanu et al. Advances in Difference Equations 2013, 2013:359 http://www.advancesindifferenceequations.com/content/2013/1/359

RESEARCH

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The existence of solutions for a nonlinear mixed problem of singular fractional differential equations Dumitru Baleanu1,2,3* , Hakimeh Mohammadi4 and Shahram Rezapour4 * Correspondence: [email protected] 1 Department of Chemical and Materials Engineering, Faculty of Engineering, King Abdulaziz University, P.O. Box 80204, Jeddah, 21589, Saudi Arabia 2 Department of Mathematics, Cankaya University, Ogretmenler Cad. 14, Balgat, Ankara, 06530, Turkey Full list of author information is available at the end of the article

Abstract By using fixed point results on cones, we study the existence of solutions for the singular nonlinear fractional boundary value problem D u(t) = f (t, u(t), u (t), c Dβ u(t)),

c α

u(0) = au(1),

u (0) = b c Dβ u(1),

u (0) = u (0) = u(n–1) (0) = 0,

where n ≥ 3 is an integer, α ∈ (n – 1, n), 0 < β < 1, 0 < a < 1, 0 < b <  (2 – β ), f is an Lq -Caratheodory function, q > α1–1 and f (t, x, y, z) may be singular at value 0 in one dimension of its space variables x, y, z. Here, c D stands for the Caputo fractional derivative. Keywords: boundary value problem; fixed point; fractional differential equation; Green function; regularization; singular

1 Introduction Fractional differential equations (see, for example, [–] 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, [–] and []). Also, there are different methods for solving distinct fractional differential equations (see, for example, [–] and []). 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, []). The Caputo derivative of fractional order α for a function f : [, ∞) → R is defined by c

Dα f (t) =

 (n – α)



t

(t – s)n–α– f (n) (s) ds



 n –  < α < n, n = [α] +  .



Let q ≥ . As you know, Lq [, ] denotes the space of functions, whose qth powers of mod  ulus are integrable on [, ], equipped with the norm xq = (  |x(t)|q dt) q . We consider the sup norm    x = sup x(t) : t ∈ [, ] ©2013 Baleanu et al.; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Baleanu et al. Advances in Difference Equations 2013, 2013:359 http://www.advancesindifferenceequations.com/content/2013/1/359

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on the space C[, ]. Also, AC[, ] is the set of absolutely continuous functions on [, ]. Let B be a subset of R . A function f : [, ] × B → R is called an Lq -Caratheodory function whenever the real-valued function f (·, x, y, z) on [, ] is measurable for all (x, y, z) ∈ B, the function f (t, ·, ·, ·) : B → R is continuous for almost all t ∈ (, ], and for each compact set U ⊂ B, there exists a function ϕu ∈ Lq [, ] such that |f (t, x, y, z)| ≤ ϕu (t) for almost all t ∈ [, ] and (x, y, z) ∈ U. Consider the nonlinear fractional boundary value problem c

  Dα u(t) = f t, u(t), u (t), c Dβ u(t) ,

u() = au(),

u () = b c Dβ u(),

u () = u () = u(n–) () = ,

(∗)

 where n ≥  is an integer, α ∈ (n – , n),  < β < ,  < a < ,  < b < ( – β) and q > α– . We say that the function u : [, ] → R is a positive solution for the problem whenever u >  on [, ], c Dα u is a function in Lq [, ], and u satisfies the boundary conditions almost everywhere on [, ]. In this paper, we suppose that f is an Lq -Caratheodory function on [, ] × B, where B = (, ∞) × (, ∞) × (, ∞), there exists a positive constant m such that m ≤ f (t, x, y, z) for almost all t ∈ [, ] 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, k ∈ C(, ∞) are positive and non-increasing, γ ∈ Lq [, ] and w ∈ C([, ∞) ×  [, ∞) × [, ∞)) are positive, w is non-decreasing in all its variables,  hq (sα ) ds < ∞,    q α– ) ds < ∞,  k q (sα–β ) ds < ∞, and limx→∞ w(x,x,x) = . Since we suppose that prob r (s x lem (∗) is singular, that is, f (t, x, y, z) may be singular at the value  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 fn by   fn (t, x, y, z) = f t, χn+ (x), χn+ (y), χn+ (z) for all t ∈ [, ] and (x, y, z) ∈ R , where

χn+ (u) =

⎧ ⎨u, u ≥  , n ⎩, u< . n n

It is easy to see that each fn is an Lq -Caratheodory function on [, ] × R , m ≤ fn (t, x, y, z),      +r +k + γ (t)w  + x,  + |y|,  + |z| fn (t, x, y, z) ≤ h n n n and       fn (t, x, y, z) ≤ h(x) + r |y| + k |z| + γ (t)w  + x,  + |y|,  + |z| for almost all t ∈ [, ] and all (x, y, z) ∈ B. In , Agarwal et al. proved the following result.

Baleanu et al. Advances in Difference Equations 2013, 2013:359 http://www.advancesindifferenceequations.com/content/2013/1/359

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Lemma . [] Let ρ ∈ Lq [, ] and  ≤ t < t ≤ . Then we have | d ( td )/p ρq for all t ∈ [, ] and    



t

(t – s) 



≤

α–

t

ρ(s) ds –

(t – s)

α–



td + (t – t )d – td d

/p

 (t

– s)α– ρ(s) ds| ≤

  ρ(s) ds

ρq +

t

(t – t )d d

/p ρq ,

where d = (α – )p + .

2 Main results Now, we are ready to investigate the problem in regular and singular cases. First, we give the following result. Lemma . Let y ∈ C[, ]. Then the boundary value problem c



Dα u(t) = y(t)

 t ∈ (, ) , u () = b c Dβ u(),

u() = au(),

u () = u () = u(n–) () = 

is equivalent to the fractional integral equation u(t) =

G(t, s) =

 

G(t, s)y(s) ds, where

(t – s)α– (α) +

a(α – β)(( – β) – b)( – s)α– + b(α)( – β)(a + t – at)( – s)α–β– ( – a)(α)(α – β)(( – β) – b)

whenever  ≤ s ≤ t ≤  and G(t, s) =

a(α – β)(( – β) – b)( – s)α– + b(α)( – β)(a + t – at)( – s)α–β– ( – a)(α)(α – β)(( – β) – b)

whenever  ≤ t ≤ s ≤ . Proof From c Dα u(t) = y(t) and the boundary conditions, we obtain u(t) = I α y(t) + u() + u ()t + =

 (α)



t

u ()  u(n–) () n– t + ··· + t ! (n – )!

(t – s)α– y(s) ds + u() + u ()t.



By properties of the Caputo derivative, we get c

  Dβ u(t) = I α–β y(t) + c Dβ u() + u ()t  t  u ()t –β . = (t – s)α–β– y(s) ds + (α – β)  ( – β)

Baleanu et al. Advances in Difference Equations 2013, 2013:359 http://www.advancesindifferenceequations.com/content/2013/1/359

Thus, u() =

c

 (α)



 ( – s)

 D u() = (α – β)

α–



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y(s) ds + u() + u () and



β

( – s)α–β– y(s) ds + 

u () . ( – β)

By using the boundary conditions u() = au() and u () = b c Dβ u(), we get u() =   α– a( (α) y(s) ds + u() + u ()) and  ( – s)

 u () = b (α – β) 

Hence, u () =





( – s) 

b(–β) (α–β)((–β)–b)

a u() = ( – a)(α)

α–β–





 ( – s)

u () . y(s) ds + ( – β)

α–β–

y(s) ds and



( – s)α– y(s) ds 

ab( – β) + ( – a)(α – β)(( – β) – b)





( – s)α–β– y(s) ds. 

Thus,  t  (t – s)α– y(s) ds + u() + u ()t u(t) = (α)   t (t – s)α– a(α – β)(( – β) – b)( – s)α– = + (α) ( – a)(α)(α – β)(( – β) – b) 

b(α)( – β)(a + t – at)( – s)α–β– y(s) ds + ( – a)(α)(α – β)(( – β) – b)   a(α – β)(( – β) – b)( – s)α– + ( – a)(α)(α – β)(( – β) – b) t

b(α)( – β)(a + t – at)( – s)α–β– y(s) ds + ( – a)(α)(α – β)(( – β) – b)   = G(t, s)y(s) ds. 



This completes the proof.

ab(–β) Put k = (α–β)((–β)–b)+b(α)(–β) and k = (–a)(α–β)((–β)–b) . It is easy to check that the (–a)(α)(α–β)((–β)–b) Green function G in the last result belongs to C([, ] × [, ]), G(t, s) >  for all (t, s) ∈ [, ) × [, ),

G(t, s) ≤ k ( – s)α–β– ≤  and

G(t, s) ≥ k ( – s)α–β–

for all (t, s) ∈ [, ] × [, ]. Consider the Banach space X = C  [, ] with the norm x∗ = max{x, x } and the cone   P = x ∈ X : x(t) ≥  and x (t) ≥  for all t ∈ [, ] .

Baleanu et al. Advances in Difference Equations 2013, 2013:359 http://www.advancesindifferenceequations.com/content/2013/1/359

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For each natural number n, define the operator Qn on P by  (Qn x)(t) =



  G(t, s)fn s, u(s), u (s), c Dβ u(s) ds.



Now, we prove that Qn is a completely continuous operator (see []). Lemma . The operator Qn is a completely continuous operator. Proof Let x ∈ P. Then, c Dβ x ∈ C[, ] and c Dβ x ≥ . Now, define ρ(t) = fn (t, u(t), u (t), c β D u(t)) for almost all t ∈ [, ]. Then ρ ∈ Lq [, ] and ρ(t) ≥ m for almost all t ∈ [, ]. By using the properties of fractional integral I α , it is easy to see that Qn x ∈ C[, ], (Qn x)(t) ≥  and (Qn x) (t) =

 (α – )



t

(t – s)α– ρ(s) ds 

for all t ∈ [, ]. This implies that (Qn x) ∈ C[, ] and (Qn x) ≥  on [, ]. Consequently, Qn maps P into P. In order to prove that Qn is a continuous operator, let xm be a convergent (j) sequence in P and limm→∞ xm = x. Thus, limm→∞ xm (t) = x(j) (t) uniformly on [, ] for j = , . Since c

    d t (t – s)–β x(s) – x() ds ( – β) dt   t  = (t – s)–β x (s) ds, ( – β) 

Dβ x(t) =

we get |c Dβ xm (t) – c Dβ x(t)| ≤ c

xm –x  (–β)

t

 (t

– s)–β ds ≤

Dβ x(t) uniformly on [, ]. Also, we have |c Dβ xm (t)| ≤

xm  . (β)

xm –x∗ (β) xm on (β)

and limm→∞ c Dβ xm (t) = [, ], and so c Dβ xm  ≤

Now, put

  ρm (t) = fn t, xm (t), xm (t), c Dβ xm (t)

and

  ρ(t) = fn t, x(t), x (t), c Dβ x(t) .

Then, it is easy to see that limm→∞ ρm (t) = ρ(t) for almost all t ∈ [, ], and there exists β ∈ Lq [, ] such that  ≤ ρm (t) ≤ β(t) for almost all t ∈ [, ] and all m ≥ . Since fn is an Lq -Caratheodory function, {xm } is bounded in C  [, ], and {c Dβ xm } is bounded in C[, ]. Therefore, limm→∞ (Qn xm )(t) = (Qn x)(t) uniformly on [, ]. Since {ρm } is Lq -convergent on [, ], lim (Qn xm ) (t) =

m→∞

 lim (α – ) m→∞



t

(t – s)α– ρm (s) ds = (Qn x) (t)



uniformly on [, ]. Hence, Qn is a continuous operator. Now, we have to show that for each bounded sequence {xm } in P, the sequence {Qn xm } is relatively compact in C[, ]. Choose a positive constant k such that xm  ≤ k and xm  ≤ k for all m. Note that c Dβ xm  ≤ t t  t  d  k and |  (t – s)α– ρm (s) ds| ≤ (  (t – s)(α–)p ds) p (  |ρm (s)|q ds) q ≤ ( td ) p ρm q for all (β) m, where d = (α – )p + . But we have 



 ≤ (Qn xm )(t) = 





G(t, s)ρm (s) ds ≤ 

G(t, s)β(s) ds ≤

β (α)

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and  t  (t – s)α– ρm (s) ds  ≤ (Qn xm ) (t) = (α – ) 

p  t    (t – s)α– β(s) ds ≤ βq ≤ (α – )  (α – ) (α – )p +  

for all t ∈ [, ] and m. This implies that {Qn xm } is bounded in C  [, ]. Also, we have   (Qn xm ) (t ) – (Qn xm ) (t )   t   t    α– α–  = (t – s) ρm (s) ds – (t – s) ρm (s) ds  (α – )    d

 ρm q t + (t – t )d – td p (t – t )d p ≤ + (α – ) d d d



 βq t + (t – t )d – td p (t – t )d p ≤ + (α – ) d d for all  ≤ t ≤ t ≤ , where d = (α – )p + . Hence, {(Qn xm ) } is equicontinuous on [, ]. Thus, {Qn xm } is relatively compact in C  [, ] by the Arzela-Ascoli theorem. Hence, Qn is a completely continuous operator.  We need the following result (see [] and []). Lemma . [] Let Y be a Banach space, P a cone in Y and  and  bounded open ¯  ⊂  . Suppose that T : P ∩ ( ¯  \  ) → P is a balls in Y centered at the origin with completely continuous operator such that Tx ≥ x for all x ∈ P ∩ ∂  and Tx ≤ x ¯  \  ). for all x ∈ P ∩ ∂  . Then T has a fixed point in P ∩ ( Theorem . For each natural number n, problem (∗) has a solution un ∈ P such that α– mk mt α–β un ≥ α–β , un (t) ≥ mt and c Dβ un (t) ≥ (α–β+) for all t ∈ [, ]. (α) Proof Let n ≥ . It is sufficient to show that Qn has a fixed point un in P with the desired conditions. In this way, note that 



(Qn x)(t) = 

  G(t, s)fn s, x(s), x (s), c Dβ x(s) ds





≥m 





G(t, s) ds ≥ m

k ( – s)α–β– ds = 

mk , α–β

mk . Put  = {x ∈ X : x∗ < and so Qn x∗ ≥ Qn (x) ≥ α–β   all x ∈ P ∩ ∂  . If vn = h( n ) + r( n ) + k( n ), then

mk }. α–β

Then Qn x∗ ≥ x∗ for

         (Qn x)(t) ≤  G(t, s)fn s, u(s), u (s), c Dβ u(s) ds   



        G(t, s) vn + γ (s)w  + x(s),  + x (s),  + c Dβ x(s) ds



≤ 

       ≤ k vn + w  + x,  + x ,  + c Dβ x γ 

Baleanu et al. Advances in Difference Equations 2013, 2013:359 http://www.advancesindifferenceequations.com/content/2013/1/359

and   (Qn x) (t)    t      α–  c β  = (t – s) fn s, x(s), x (s), D x(s) ds (α – )   t          (t – s)α– vn + γ (s)w  + x(s),  + x (s),  + c Dβ x(s) ds ≤ (α – )  α–

    t    vn t ≤ + w  + x,  + x ,  + c Dβ x (t – s)α– γ (s) ds (α – ) α –   for all x ∈ P and t ∈ [, ], because w is non-decreasing in all its variables. Since x ≤ x∗ , x  ∗ x  ≤ x∗ , c Dβ x ≤ (β) ≤ x and (β) 

t

(t – s)α– γ (s) ds ≤



/p  γ q , d

where d = (α – )p + , we have



  Qn (x) ≤ k vn + w  + x∗ ,  + x∗ ,  + x∗ γ  (β) and   (Qn x)  ≤



 vn x∗ + w  + x∗ ,  + x∗ ,  + (/d)/p γ q . (α – ) α –  (β)

vn ∗ + Nw( + x∗ ,  + x∗ ,  + x )), where N = max{γ  , (/d)/p × Hence, Qn x∗ ≤ M( α– (β)  γ q } and M = max{ (α–) , k }. Since

 w( + v,  + v,  + v) = , v→∞ v lim

there exists a positive constant L such that

v vn + Nw  + v,  + v,
for all v ≥ L. Thus, Qn x∗ < x∗ for all x ∈ P with x∗ ≥ L. Put  = {x ∈ X : x∗ < L}. Then Qn x∗ < x∗ for all x ∈ P ∩ ∂  . By using last result, Qn has a fixed point un in ¯  \  ). But un = (Qn un )(t) ≥ mk and P ∩ ( α–β (Qn x) (t) = ≥

 (α – ) m (α – )



t

  (t – s)α– fn s, x(s), x (s), c Dβ x(s) ds





t

(t – s)α– ds = 

mt α– (α)

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for all t ∈ [, ] and x ∈ P. Since c

 D un (t) = ( – β)



t

β

≥



t

 (t

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– s)–β sα– ds =

(α)(–β) α–β t , (α–β+)

(t – s)–β un (s) ds

m (α)( – β)



t

(t – s)–β sα– ds = 

mt α–β (α – β + )

for all t ∈ [, ]. This completes the proof.



Now, we give our last result. Theorem . Problem (∗) has a solution u such that u(t) ≥ c

D u(t) ≥ β

mt α–β (α–β+)

mk , α–β

u (t) ≥

mt α– (α)

and

for all t ∈ [, ].

Proof By using Theorem ., one gets that for each natural number n, problem (∗) has a α– mk ), r(|un (t)|) ≤ r( mt ) solution un ∈ P with the desired conditions. Thus, h(un (t)) ≤ h( α–β (α) α–β

mt ) for all t ∈ [, ] and n. Also, we have c Dβ un  ≤ and k(|c Dβ un (t)|) ≤ k( (α–β+) pose that

S(t) = h

mk α–β

+r

mt α– (α)

+k

mt α–β . (α – β + )

Then   m ≤ fn t, un (t), un (t), c Dβ un (t)      ≤ S(t) + γ (t)w  + un ,  + un ,  + c Dβ un 

  un ∗   ≤ S(t) + γ (t)w  + un ∗ ,  + un ∗ ,  + (β) for almost all t ∈ [, ] and n. Since  ≤ G(t, s) ≤ k , we get  ≤ un (t)     G(t, s)fn s, un (s), un (s), c Dβ un (s) ds = 



≤ k 





un ∗ γ  S(s) ds + w  + un ∗ ,  + un ∗ ,  + (β)

and  ≤ un (t)

 t  (t – s)α– S(s) ds ≤ (α – ) 

 t

un ∗ α– (t – s) γ (s) ds . + w  + un ∗ ,  + un ∗ ,  + (β) 

un  . (β)

Sup-

Baleanu et al. Advances in Difference Equations 2013, 2013:359 http://www.advancesindifferenceequations.com/content/2013/1/359

We show that

t

 (t

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– s)α– S(s) ds is bounded on [, ]. Let d = (α – )p + . Note that

mk ds α–β 

 

mk mk  =h =: η < ∞, h (t – s)α– ds = α–β α– α–β  α–  t ms ds (t – s)α– r (α)  





(t – s)α– h

 /p

  ( m ) α–

/q   (α)  (α) (α–)q = rq sα– ds =: η < ∞ d m 

and 



t

(t – s)α– k 

msα–β ds (α – β + )



  ( m ) α–β /p

/q   (α–β+)  (α – β + ) (α–β)q q α–β = k s =: η < ∞. ds d m 

Thus,

t

 (t



– s)α– S(s) ds ≤ η for all t ∈ [, ], where η = η + η + η . Also, we have



S(s) ds 



  m  (α) α– ( (α) ) α–  α–  mk  + h ≤ r s ds α– α–β m 

(α – β + ) + m

 

α–β



m ( (α–β+) ) α–β

  k sα–β ds < ∞.



Since 



un  = k 



un ∗ γ  S(s) ds + w  + un ∗ ,  + un ∗ ,  + (β)

and   u  ≤ n



/p

 un ∗  η + w  + un ∗ ,  + un ∗ ,  + γ q , (α – ) (β) d

n ∗ we get un ∗ ≤ M( + Kw( + un ∗ ,  + un ∗ ,  + u )) for all n, where = max{η, (β)   /p   S(s) ds}, K = max{γ  , ( d ) γ q } and also M = max{k , (α–) }. On the other hand, v there exists a positive constant L such that M( + Kw( + v,  + v,  + (β) )) < v for all v ≥ L, and so un ∗ < L for all n. Thus, for almost all t ∈ [, ] and all n, we have fn (t, un (t), un (t), c Dβ un (t)) ≤ R(t), where

R(t) = S(t) + γ (t)w  + L,  + L,  +

L . (β)

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Note that R ∈ Lq [, ]. We show that {un } is equicontinuous on [, ]. Let ρn (t) = fn (t, un (t), un (t), c Dβ un (t)) and  ≤ t < t ≤ T. Then    u (t ) – u (t ) n n   t   t    α– α–  = (t – s) ρn (s) ds – (t – s) ρn (s) ds (α – )     t

 t    (t – s)α– ρn (s) ds (t – s)α– – (t – s)α– ρn (s) ds + ≤ (α – )  t  t

 t    ≤ (t – s)α– R(s) ds , (t – s)α– – (t – s)α– R(s) ds + (α – )  t and so    u (t ) – u (t ) n n d

Rq t + (t – t )d – td /p (t – t )d /p ≤ + . (α – ) d d Hence, {un } is equicontinuous on [, ]. Since {un } is a bounded sequence in C[, ], by using the Arzela-Ascoli theorem, without loss of generality, we can assume that {un } is t  convergent in C[, ]. Let limn→∞ un = u. Then, it is easy to see that c Dβ un (t) = (α–)  (t – t  –β  c β –β  s) un (s) ds, and D un (t) uniformly converges to (α–)  (t – s) u (s) ds on [, ]. Thus, c β D un converges to c Dβ u in C[, ]. Hence,     lim fn t, un (t), un (t), c Dβ un (t) = f t, u(t), u (t), c Dβ u(t)

n→∞

for almost all t ∈ [, ]. Since R ∈ Lq [, ], by using the dominated convergence theorem on the relation 



un (t) = 

we get u(t) =

  G(t, s)fn s, un (s), un (s), c Dβ un (s) ds,

 

G(t, s)f (s, u(s), u (s), c Dβ u(s)) for all t ∈ [, ]. This completes the proof. 

2.1 Examples for the problem Example . Let ρ , ρ ∈ Lq [, ], ρ (t) ≥ m >  for almost all t in [, ]. Suppose that f (t, x, y, z) = ρ (t) +

  

+

x –λ 

on [, ] × B, λ = (au())  , h(x) = λ < , r(x) =

 

x

, k(x) =

 

x

 y

+

 

z 



x  –λ

     + ρ (t) x  + y  + z 

  





whenever x  – λ ≥  and h(x) =  whenever x  – 





, w(x, y, z) = (x  + y  + z  + ) and γ (t) = ρ (t) + |ρ (t)|. Then

Theorem . guarantees that problem (∗) has a positive solution.

Baleanu et al. Advances in Difference Equations 2013, 2013:359 http://www.advancesindifferenceequations.com/content/2013/1/359

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Example . Consider the nonlinear mixed problem of singular fractional boundary value problem c



D  u(t) = t +  +

  

+

u(t) – ρ

 

u (t) 

+

 



( c D  u(t)) 

       +  u(t)  + u (t)  + c D  u(t)  +  

via boundary value conditions u() =  u(), u () =  (c D  )u() and u () = u () = · · · =  u(n–) () = , where ρ = ((  )u())  . Let   f t, u(t), u (t), c D  = t +  +

  

+

u(t) – ρ

 

u (t) 

+

 



( c D  u(t)) 

       +  u(t)  + u (t)  + c D  u(t)  +  . Then the map f is singular at t = , and f satisfies the desired conditions, where h(x) = 



whenever x  – ρ ≥  and h(x) =  whenever x  – ρ < , r(x) =  

 

 

 

x  –ρ  

x

, k(x) =

 

, w(x, y, z) =

x

x + y + z + , ρ (t) = t +  >  = m, ρ (t) =  and γ (t) = ρ (t) + |ρ (t)|. Then Theorem . 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. Competing interests The authors declare that they have no competing interests. Authors’ contributions All authors have equal contributions. All authors read and approved the final manuscript. Author details 1 Department of Chemical and Materials Engineering, Faculty of Engineering, King Abdulaziz University, P.O. Box 80204, Jeddah, 21589, Saudi Arabia. 2 Department of Mathematics, Cankaya University, Ogretmenler Cad. 14, Balgat, Ankara, 06530, Turkey. 3 Institute of Space Sciences, Magurele, Bucharest, Romania. 4 Department of Mathematics, Azarbaijan Shahid Madani University, Azarshahr, Tabriz, Iran. 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. Received: 27 May 2013 Accepted: 28 August 2013 Published: 11 Dec 2013 References 1. Bai, Z, Lu, H: Positive solutions for boundary value problem of nonlinear fractional differential equation. J. Math. Anal. Appl. 311, 495-505 (2005) 2. Baleanu, D, Mohammadi, H, Rezapour, S: Positive solutions of a boundary value problem for nonlinear fractional differential equations. Abstr. Appl. Anal. 2012, Article ID 837437 (2012) 3. Baleanu, D, Mohammadi, H, Rezapour, S: Some existence results on nonlinear fractional differential equations. Philos. Trans. R. Soc. Lond. A 371, 20120144 (2013) 4. Baleanu, D, Mohammadi, H, Rezapour, S: On a nonlinear fractional differential equation on partially ordered metric spaces. Adv. Differ. Equ. 2013, 83 (2013)

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10.1186/1687-1847-2013-359 Cite this article as: Baleanu et al.: The existence of solutions for a nonlinear mixed problem of singular fractional differential equations. Advances in Difference Equations 2013, 2013:359

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