Jiang et al. Boundary Value Problems (2016) 2016:175 DOI 10.1186/s13661-016-0680-x

RESEARCH

Open Access

The existence of positive solutions for p-Laplacian boundary value problems at resonance Weihua Jiang* , Jiqing Qiu and Caixia Yang *

Correspondence: [email protected] College of Sciences, Hebei University of Science and Technology, Shijiazhuang, Hebei 050018, P.R. China

Abstract By using the Leggett-Williams norm-type theorem due to O’Regan and Zima and constructing suitable Banach spaces and operators, we investigate the existence of positive solutions for fractional p-Laplacian boundary value problems at resonance. An example is given to illustrate the main results. MSC: 34B15 Keywords: positive solutions; p-Laplacian operator; boundary value problem; resonance; Fredholm operator

1 Introduction Boundary value problems at resonance have attracted more and more attention. Many authors studied the existence of solutions for these problems by using Mawhin’s continuous theorem [] and its extension obtained by Ge and Ren []; see [–] and the references cited therein. By using Leggett-Williams norm-type theorems due to O’Regan and Zima [], the existence of positive solutions for the boundary value problems at resonance with a linear derivative operator has been investigated (see [–]). To the best of our knowledge, there is no paper to show the existence of a positive solution for boundary value problems with a nonlinear derivative operator (for instance, p-Laplacian operator) at resonance by using Leggett-Williams norm-type theorems. Motivated by the excellent results mentioned above, we will discuss the existence of positive solutions for the p-Laplacian boundary value problem 

β

D+ [ϕp (C Dα+ x)](t) = f (t, (C Dα+ x)(t)), t ∈ (, ), x(i) () = , i = , , , . . . , n – , (C Dα+ x)() = (C Dα+ x)(),

C

(.)

β

where n –  < α ≤ n,  < β < , ϕp (s) = |s|p– s, p > , C D+ is the Caputo fractional derivative (see [, ]).

2 Preliminaries For convenience, we introduce some notations and a theorem. For more details see []. Assume that X, Y are real Banach spaces. A linear mapping L : dom L ⊂ X → Y is a Fredholm operator of index zero (i.e. dim Ker L = codim Im L < +∞ and Im L is closed in © 2016 Jiang et al. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

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Y ) and an operator N : X → Y is nonlinear. P : X → X and Q : Y → Y are projectors with Im P = Ker L and Ker Q = Im L. J : Im Q → Ker L is a isomorphism since dim Im Q = dim Ker L. Denote by LP the restriction of L to Ker P ∩ dom L → Im L and its inverse by KP . So, x is a solution of Lx = Nx if and only if it satisfies x = (P + JQN)x + KP (I – Q)Nx. Let C ⊂ X be a cone, γ : X → C be a retraction,  := P + JQN + KP (I – Q)N and γ :=  ◦γ. Theorem . [] Let  ,  be open bounded subsets of X with  ⊂  and C ∩ ( \  ) = ∅. Assume that L : dom L ⊂ X → Y is a Fredholm operator of index zero and the following conditions are satisfied. (C) QN : X → Y is continuous and bounded and KP (I – Q)N : X → X is compact on every bounded subset of X; (C) Lx = λNx for all x ∈ C ∩ ∂ ∩ dom L and λ ∈ (, ); (C) γ maps subsets of  into bounded subsets of C; (C) dB ([I – (P + JQN)γ ]|Ker L , Ker L ∩  , ) = , where dB stands for the Brouwer degree; (C) there exists u ∈ C \ {} such that x ≤ σ (u )x for x ∈ C(u ) ∩ ∂ , where C(u ) = {x ∈ C : μu x for some μ > } and σ (u ) is such that x + u  ≥ σ (u )x for every x ∈ C; (C) (P + JQN)γ (∂ ) ⊂ C; (C) γ ( \  ) ⊂ C. Then the equation Lx = Nx has at least one solution in the set C ∩ ( \  ). Now, we present some fundamental facts on the fractional calculus theory which can be found in [, ]. Definition . The Riemann-Liouville fractional integral of order α >  of a function y : (, ∞) → R is given by  t  Iα+ y(t) = (t – s)α– y(s) ds, (α)  provided the right-hand side is pointwise defined on (, ∞). Definition . The Caputo fractional derivative of order δ >  of a function y : (, ∞) → R is given by  t  C δ D+ y(t) = (t – s)n–δ– y(n) (s) ds, (n – δ)  provided that the right-hand side is pointwise defined on (, ∞), where n = [δ] + . Lemma . [, ] Assume f ∈ L[, ], q > p ≥ , q > , then C

p

q

q–p

D+ I+ f (t) = I+ f (t),

C

p

p

D+ I+ f (t) = f (t).

Lemma . [, ] Assume p > , then p

p

I+ C D+ f (t) = f (t) + c + c t + · · · + cn– t n– , where n is an integer and n –  < p ≤ n.

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β

Since C D+ [ϕp (C Dα+ ·)] is a nonlinear operator, we cannot solve the problem (.) by Theorem .. Based on this, we prove the following lemma. Lemma . u(t) is a solution of the following problem: 

β

(C D+ u)(t) = f (t, ϕq (u(t))), u() = u(),

t ∈ [, ],

(.)

if and only if x(t) is a solution of (.), where x(t) = Iα+ ϕq (u(t)),

 p

+

 q

= .

Proof Assume that u(t) is a solution of the problem (.) and x(t) = Iα+ ϕq (u(t)). Then u(t) = [ϕp (C Dα+ x)](t) and x(i) () = , i = , , , . . . , n – . Replaces u(t) with [ϕp (C Dα+ x)](t) in (.), we can see that x(t) is a solution of (.). On the other hand, if x(t) is a solution of (.) and u(t) = [ϕp (C Dα+ x)](t), substituting u(t)  for [ϕp (C Dα+ x)](t) in (.), we can see that u(t) satisfies (.). In this paper, we will always suppose that f ∈ [, ] × R → R is continuous, p > , ϕp (s) = s · |s|p– , p + q = , α > ,  < β < .

3 Main result Let X = Y = C[, ] with the norm u = maxt∈[,] |u(t)|. Take a cone   C = u(t) ∈ X | u(t) ≥ , t ∈ [, ] . Define operator L : dom L ⊂ X → Y and N : X → Y as follows: (Lu)(t) =

C

β  D+ u (t),

   (Nu)(t) = f t, ϕq u(t) ,

where   β dom L = u(t) | u(t), C D+ u(t) ∈ X, u() = u() . Then the problem (.) can be written by Lu = Nu,

u ∈ dom L.

Lemma . L is a Fredholm operator of index zero. KP is the inverse of L|dom L∩Ker P , where KP : Im L → dom L ∩ Ker P is given by KP y(t) =

 (β)



t

(t – s)β– y(s) ds – 

 β





( – s)β y(s) ds .



Proof It is easy to see that Ker L = {c | c ∈ R},

   Im L = y ∈ Y  ( – s)β– y(s) ds =  , 

and Im L ⊂ Y is closed.

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Define P : X → X, Q : Y → Y as 





Pu =



( – s)β– y(s) ds.

Qy = β

u(t) dt, 



Obviously, P : X → X, Q : Y → Y are projectors and Im P = Ker L, X = Ker P ⊕ Ker L.

t  It is easy to see that Im L ⊂ Ker Q. Conversely, if y(t) ∈ Ker Q, take u(t) = (β)  (t – β

s)β– y(s) ds. Then u(t) ∈ dom L and Lu = C D+ u(t) = y(t). These imply Ker Q ⊂ Im L. Therefore Im L = Ker Q. For y ∈ Y , y = (y – Qy) + Qy ∈ Im L + Im Q. If y ∈ Im L ∩ Im Q, then y = Qy and y ∈ Im L = Ker Q. This means that y = , i.e. Y = Im L ⊕ Im Q. So, dim Ker L = codim Im L =  < +∞. L is a Fredholm operator of index zero. For y ∈ Im L, it is clear that KP y ∈ dom L ∩ Ker P and LKP y = y. On the other hand, if u ∈ dom L ∩ Ker P, by Lemma ., we get KP Lu(t) = =

 (β)



t

(t – s)β– Lu(s) ds – 

 β







( – s)β Lu(s) ds 

β β β+ β I+ C D+ u(t) – I+ C D+ u() β+ C

= u(t) + c – I+

β

D+ u().



 β+ β Thus,  KP Lu(t) dt =  u(t) dt + c – I+ C D+ u(). It follows from u ∈ Ker P and KP Lu ∈ β+ C β Ker P that c – I+ D+ u() = . So, we have KP Lu = u, u ∈ dom L ∩ Ker P.  Define J : Im Q → Ker L as J(c) = c, c ∈ R. Thus, JQN + KP (I – Q)N : X → X is given by 

 JQN + KP (I – Q)N u(t) =





   G(t, s)f s, ϕq u(s) ds,

(.)



where  G(t, s) =

β( – s)β– ( – β( – s)β– ( –

tβ (β+) tβ (β+)

+ +

 )– (β+)  )– (β+)

β– (–s)β + (t–s) , (β+) (β) β (–s) , (β+)

 ≤ s < t ≤ ,  ≤ t ≤ s < .

Lemma . QN : X → Y is continuous and bounded and KP (I – Q)N :  → X is compact, where  ⊂ X is bounded. Proof Assume that  ⊂ X is bounded. There exists a constant M > , such that |Nu| = |f (t, ϕq (u(t)))| ≤ M, t ∈ [, ], u ∈ . So, |QNu| ≤ M, u ∈ , i.e. QN() is bounded. Based on the definition of Q and the continuity of f we know that QN : X → Y is continuous. For u ∈ , we have   KP (I – Q)Nu(t)   t        β– β  (t – s) (I – Q)Nu(s) ds – ( – s) (I – Q)Nu(s) ds  = (β)  β   t  t       (t – s)β– Nu(s) ds + (t – s)β– QNu(s) ds ≤ (β)  (β) 

Jiang et al. Boundary Value Problems (2016) 2016:175

+ ≤

 β(β)





  ( – s)β Nu(s) ds +



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 β(β)





  ( – s)β QNu(s) ds



M < +∞. (β + )

Thus, |KP (I – Q)N() is bounded. For u ∈ ,  ≤ t < t ≤ , we get   KP (I – Q)Nu(t ) – KP (I – Q)Nu(t )    t    t β– β– (t – s) (I – Q)Nu(s) ds – (t – s) (I – Q)Nu(s) ds =  (β)    t   t      β– β– β– (t – s) – (t – s) (I – Q)Nu(s) ds + = (t – s) (I – Q)Nu(s) ds (β)   t  t  t   M (t – s)β– – (t – s)β– ds + (t – s)β– ds ≤ (β)  t =

 M  β β t – t + (t – t )β . (β + )

It follows from the uniform continuity of t β and t on [, ] that KP (I – Q)N() are equicontinuous on [, ]. By the Arzela-Ascoli theorem, we see that KP (I – Q)N() is compact.  In order to prove our main result, we need the following conditions. (H ) There exists a constant R > , such that f (t, u) < , t ∈ [, ], u > R .

t  (β–) (H ) There exist nonnegative functions a(t), b(t) with maxt∈[,] (β) a(s) ds :=  (t – s)

t  (β–) b(s) ds := B < /, such that A < +∞, maxt∈[,] (β)  (t – s)     f (t, u) ≤ a(t) + b(t)ϕp |u| . (H ) f (t, u) ≥ –( – t)–β ϕp (u)/β, t ∈ [, ], u > . (H ) There exist r > , t ∈ [, ], and M ∈ (, ) such that G(t , s)f (s, u) ≥

 – M ϕp (u), M

s ∈ [, ), M r ≤ u ≤ r.

(H ) G(t, s)f (s, u) ≥ –ϕp (u), s ∈ [, ), t ∈ [, ], u ≥ . Lemma . If the conditions (H ) and (H ) hold, the set    = u(t) | (Lu)(t) = λNu(t), u(t) ∈ C ∩ dom L, λ ∈ (, ) is bounded. β

Proof For u(t) ∈  , we get QNu(t) =  and u(t) = λI+ Nu(t) + u(). By (H ) and QNu(t) = β , there exists t ∈ [, ] such that ϕq (u(t )) ≤ R . This, together with u(t) = λI+ Nu(t) + u(), means  β  β  u() ≤ u(t ) + λI+ Nu(t ) ≤ ϕp (R ) + I+ Nu(t ).

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Thus, we have  β  β  β  u(t) ≤ u() + λI+ Nu(t) ≤ ϕp (R ) + I+ Nu(t ) + I+ Nu(t).

(.)

It follows from (H ) that  t      (t – s)β– f s, ϕq u(s)  ds (β)   t     (t – s)β– f s, ϕq u(s)  ds

u(t) ≤ ϕp (R ) + +

 (β)



 t    (t – s)β– a(s) + b(s)u(s) ds (β)   t   (t – s)β– a(s) + b(s)u(s) ds

≤ ϕp (R ) + +

 (β)



  ≤ ϕp (R ) +  A + Bu . Therefore, u ≤

ϕp (R ) + A < +∞.  – B 

This means that  is bounded.

Theorem . Assume that the conditions (H )-(H ) hold. Then the boundary value problem (.) has at least one positive solution. Proof Set      = u ∈ X | M u < u(t) < r < R, t ∈ [, ] ,

   = u ∈ X | u < R ,

where R > max{ϕp (R ), (β + )A} is large enough such that  ⊃  . Clearly,  and  are open bounded sets of X,  ⊂  and C ∩ ( \  ) = ∅. In view of Lemmas ., ., and ., L is a Fredholm operator of index zero and the conditions (C), (C) of Theorem . are fulfilled. Define γ : X → C as (γ u)(t) = |u(t)|, u(t) ∈ X. Then γ : X → C is a retraction and (C) holds. Let u(t) ∈ Ker L ∩ ∂ , then u(t) ≡ c = ±R, t ∈ [, ]. Define 



H(c, λ) = c – λ|c| – λβ

   ( – s)β– f s, ϕq |c| ds.



If c = R, λ ∈ [, ], by (H ), we get 



H(R, λ) = R – λR – λβ

  ( – s)β– f s, ϕq (R) ds > .



If c = –R, λ ∈ [, ], by (H ), we obtain  H(–R, λ) = –R – λR – λβ



  ( – s)β– f s, ϕq (R) ds < –( + λ)R + λR = –R.



So, we have H(u, λ) = , u ∈ Ker L ∩ ∂ , λ ∈ [, ].

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Therefore,    dB I – (P + JQN)γ |Ker L , Ker L ∩  ,      = dB H(·, )|Ker L , Ker L ∩  ,  = dB H(·, )|Ker L , Ker L ∩  ,  = dB (I|Ker L , Ker L ∩  , ) =  = . Thus, (C) holds. Set u (t) = , t ∈ [, ], then u ∈ C \{}, C(u ) = {u ∈ C | u(t) > , t ∈ [, ]}. Take σ (u ) =  and u ∈ C(u ) ∩ ∂ . Then M r ≤ u(t) ≤ r, t ∈ [, ]. By (H ), we get 





u(t ) =



   G(t , s)f s, ϕq u(s) ds



 – M u(s) ds M

u(s) ds + 







≥



u(s) ds + 



≥ M r + ( – M )r = r = u. Thus, u ≤ σ (u )u, for u ∈ C(u ) ∩ ∂ . So, (C) holds. For u(t) ∈ ∂ , t ∈ [, ], by the condition (H ), we have 



 u(s) ds + β

(P + JQN)γ (u) = 





 u(s) ds –

≥ 





   ( – s)β– f s, ϕq u(s) ds







 u(s) ds = .



So, (P + JQN)γ (∂ ) ⊂ C. Hence, (C) holds. For u(t) ∈  \  , t ∈ [, ], it follows from (H ) that  (γ u)(t) =



 u(s) ds +







   G(t, s)f s, ϕq u(s) ds ≥







 u(s) ds –







 u(s) ds = .



So, (C) is satisfied. By Theorem ., we confirm that the equation Lu = Nu has a positive solution u. Based on Lemma ., the problem (.) has at least one positive solution. 

4 Examples To illustrate our main result, we present an example. Let us consider the following boundary value problem: ⎧    ⎨ C D + [ϕ  (C D + x)](t) =  ( – t)  –  ( – t)  |C D + x(t)|  ,         ⎩ x() = , (C D  x)() = (C D  x)(). +

t ∈ (, ),

(.)

+

On the basis of Lemma ., it is sufficient to examine the issue 





D+ u(t) =  ( – t)  – u() = u(). C

  ( – t)  |u(t)|, 

t ∈ [, ],

(.)

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

Corresponding to the problem (.), we have f (t, u) =  ( – t)  – q = , α =  , β =  . So, ⎧  ⎪ ⎨  ( – s)–  ( – G(t, s) =



⎪ ⎩  ( – s)–  ( – 





t (  )  t (  )

+ +

  ) – (–s) (  (  ) )

 )– (  )

 (–s)   (  )

+ ,

–

(t–s)  (  )

 ( 





– t)  |u|  , p =  ,

,  ≤ s < t ≤ ,  ≤ t ≤ s < .

Take R = , a(t) = , b(t) =  , r = ., t = , and M = .. Clearly, (H ) holds. By simple calculations, we can see that     f (t, u) ≤ a(t) + b(t)ϕp |u| ,  t    A = max < +∞, (t – s)–  ds = t∈[,] (  )  .   t      B = max (t – s)–  · ds = < , t∈[,] (  )   .      f (t, u) ≥ – ( – t)  u  , u > ,  . .  G(t , s)f (s, u) ≥ – u   .  u  , . ≤ u ≤ ., s ∈ [, ), ≥ . 

G(t, s)f (s, u) ≥ –u  ,

u ≥ , s ∈ [, ), t ∈ [, ].

So, the conditions (H )- (H ) hold. By Theorem ., we can conclude that the problem (.) has at least one positive solution.

Competing interests The authors declare that they have no competing interests. Authors’ contributions All results belong to WJ, JQ, and CY. All authors read and approved the final manuscript. Acknowledgements This work is supported by the Natural Science Foundation of Hebei Province (A2017208101). Received: 11 March 2016 Accepted: 13 September 2016 References 1. Mawhin, J: Topological Degree Methods in Nonlinear Boundary Value Problems. NSFCBMS Regional Conference Series in Mathematics. Am. Math. Soc., Providence (1979) 2. Ge, W, Ren, J: An extension of Mawhin’s continuation theorem and its application to boundary value problems with a p-Laplacian. Nonlinear Anal. TMA 58, 477-488 (2004) 3. Ma, R: Existence results of a m-point boundary value problem at resonance. J. Math. Anal. Appl. 294, 147-157 (2004) 4. Xue, C, Ge, W: The existence of solutions for multi-point boundary value problem at resonance. Acta Math. Sin. 48, 281-290 (2005) 5. Du, Z, Lin, X, Ge, W: Some higher-order multi-point boundary value problems at resonance. J. Comput. Appl. Math. 177, 55-65 (2005) 6. Feng, W, Webb, JRL: Solvability of m-point boundary value problems with nonlinear growth. J. Math. Anal. Appl. 212, 467-480 (1997) 7. Lian, H, Pang, H, Ge, W: Solvability for second-order three-point boundary value problem at resonance on a half-line. J. Math. Anal. Appl. 337, 1171-1181 (2008) 8. Zhang, X, Feng, M, Ge, W: Existence result of second-order differential equations with integral boundary conditions at resonance. J. Math. Anal. Appl. 353, 311-319 (2009) 9. Liu, B, Li, J, Liu, L: Existence and uniqueness for an m-point boundary value problem at resonance on infinite intervals. Comput. Math. Appl. 64, 1677-1690 (2012)

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