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Existence criterion for the solutions of fractional order p-Laplacian boundary value problems

  • 1, 2Email author,
  • 3, 4,
  • 5, 6,
  • 5 and
  • 5
Boundary Value Problems20152015:164

  • Received: 19 November 2014
  • Accepted: 25 August 2015
  • Published:


The existence criterion has been extensively studied for different classes in fractional differential equations (FDEs) through different mathematical methods. The class of fractional order boundary value problems (FOBVPs) with p-Laplacian operator is one of the most popular class of the FDEs which have been recently considered by many scientists as regards the existence and uniqueness. In this scientific work our focus is on the existence and uniqueness of the FOBVP with p-Laplacian operator of the form: \(D^{\gamma}(\phi_{p}(D^{\theta}z(t)))+a(t)f(z(t)) =0\), \(3<{\theta}\), \(\gamma\leq{4}\), \(t\in[0,1]\), \(z(0)=z'''(0)\), \(\eta D^{\alpha}z(t)|_{t=1}= z'(0)\), \(\xi z''(1)-z''(0)=0\), \(0<\alpha<1\), \(\phi_{p}(D^{\theta}z(t))|_{t=0}=0 =(\phi_{p}(D^{\theta}z(t)))'|_{t=0}\), \((\phi_{p}(D^{\theta} z(t)))''|_{t=1} = \frac{1}{2}(\phi_{p}(D^{\theta} z(t)))''|_{t=0}\), \((\phi_{p}(D^{\theta}z(t)))'''|_{t=0}=0\), where \(0<\xi, \eta<{1}\) and \(D^{\theta}\), \(D^{\gamma}\), \(D^{\alpha}\) are Caputo’s fractional derivatives of orders θ, γ, α, respectively. For this purpose, we apply Schauder’s fixed point theorem and the results are checked by illustrative examples.


  • FOBVP with p-Laplacian operator
  • fixed point theorems
  • existence and uniqueness

1 Introduction

Fractional calculus has widely been studied by scientists from the era of Leibniz to the present and has drawn the attention of mathematicians, engineers, and physicists in many scientific disciplines based on mathematical modeling, and it was found that the fractional order models are more precise in comparison with integer order models and, therefore, we can see many useful fractional order models in fluid flow, viscoelasticity, signal processing, and many other fields. For instance, see [16].

In fractional calculus, the existence of a positive solution for FOBVP with p-Laplacian operator has extensively attracted the attention of the scientific community. This side of the fractional calculus has a wide range of applications in day life problems and these were investigated for the existence of solutions by different mathematical tools. For instance, Lv [7], has studied existence results for m-point FOBVP with p-Laplacian operator with the help of a monotone iterative technique and produced interesting results which were examined by two examples. Prasad and Krushna [8] studied FOBVPs with p-Laplacian operator with the help of the Krasnosel’skii and five functional fixed point theorems and checked their results by examples. Yuan and Yang [9] studied the existence of a positive solution for a q-FOBVP with a p-Laplacian operator using the upper and lower solution method through Schauder’s fixed point theorem and the results were examined by examples.

Some of the interesting results in the existence of positive solution for FOBVP with p-Laplacian operator which raised our attention toward this project are, for instance, that Zhang et al. [10] has contributed for positive solutions of an FOBVP with p-Laplacian
$$ \begin{aligned} &\bigl(\phi_{p}\bigl(z'\bigr) \bigr)' = k\bigl(\tau,z,z'\bigr), \quad \tau\in[0,T], \\ &z(T) = z(0),\qquad z'(T)= z'(0), \end{aligned} $$
with the help of degree theory, and they also used the upper and lower solution method for this work. Xu and Xu in [11] investigated the p-Laplacian equation for sign changing equations of the form
$$ \begin{aligned} &\bigl(\phi_{p}\bigl(z'(\tau)\bigr) \bigr)' = -f\bigl(\tau,z,z'\bigr), \quad \tau\in(0,1), \\ &z(0) = 0=z(1), \end{aligned} $$
using the method of upper and lower solution method with the help of Leray-Schauder degree theory. Wang in [12] considered three solutions of
$$ \begin{aligned} &\bigl(\phi_{p}\bigl(z'(\tau)\bigr) \bigr)'+m(\tau)f\bigl(\tau,z,z'\bigr) = 0, \quad 0< \tau< \infty, \\ &z(0) = 0,\qquad \lim_{\tau\rightarrow{+\infty}}z=0. \end{aligned} $$
In this paper we consider the FOBVP with p-Laplacian of the form
$$\begin{aligned}& D^{\gamma}\bigl(\phi_{p}\bigl(D^{\theta}z(t) \bigr)\bigr)+a(t)f\bigl(z(t)\bigr) = 0, \quad 3< {\theta}, \gamma \leq{4}, t \in[0,1], \\& z(0)=z'''(0),\qquad \eta D^{\alpha}z(t)|_{t=1} = z'(0),\qquad \xi z''(1)-z''(0)=0,\quad 0< \alpha< 1, \\& \phi_{p}\bigl(D^{\theta}z(t)\bigr)|_{t=0}=0 = \bigl( \phi_{p}\bigl(D^{\theta}z(t)\bigr)\bigr)'|_{t=0}, \\& \bigl(\phi_{p}\bigl(D^{\theta} z(t)\bigr)\bigr)''|_{t=1} = \frac{1}{2}\bigl(\phi_{p}\bigl(D^{\theta} z(t)\bigr) \bigr)''|_{t=0}, \\& \bigl(\phi_{p}\bigl(D^{\theta}z(t)\bigr)\bigr)'''|_{t=0} = 0, \end{aligned}$$
where \(D^{\theta}\), \(D^{\gamma}\), and \(D^{\alpha}\) are Caputo’s fractional derivatives of fractional order θ, γ, α where \(\theta, \gamma\in(3,4]\), and \(\alpha\in(0,1)\). \(\phi_{p}(s)=|s|^{p-2}s\), \(p>1\), \(\phi_{p}^{-1}=\phi_{q}\), \(\frac{1}{p}+\frac{1}{q}=1\). We recall some basic definitions and results. For \(\alpha>0\), choose \(n=[\alpha]+1\) in the case α is not an integer and \(n=\alpha\) in the case α is an integer. We recall the following definitions of a fractional order integral and a fractional order derivative in Caputo’s sense, and some basic results of fractional calculus [2, 13].

Definition 1


For a function \(k:(0,\infty)\rightarrow R\) and \(\gamma>0\), fractional integral of order γis defined by
$$ I^{\gamma}k(\tau)=\frac{1}{\Gamma(\gamma)}\int_{0}^{\tau}( \tau-s)^{\gamma -1} k(s) \, ds, $$
with the condition that the integral converges.

Definition 2


For \(\gamma>0\) the left Caputo fractional derivative of order γ is defined by
$$ \bigl(D^{\gamma}\bigr)h\bigl(z(t)\bigr)=\frac{1}{\Gamma(n-\gamma)}\int ^{t}_{a} (t-x)^{n-\gamma-1}f^{(n)}(x)\, dx, $$
where n is such that \(n-1<\gamma<n\).

Lemma 3


For \(\mu, \beta>{0}\), the following relation holds: \(D^{\mu}t^{\nu}=\frac{\Gamma(\nu+1)}{\Gamma(\nu+1-\mu)}t^{\nu-\mu-1}\), \(\nu>{n}\) and \(D^{\mu}t^{l}=0\), for \(l=0,1,2,\ldots,n-1\).

Lemma 4

For \(\mathcal{H}(t)\in{C(0,1)}\), the solution of the homogeneous FDE \(D_{0^{+}}^{\omega}\mathcal{H}(t)=0 \) is
$$ \mathcal{H}(t)=k_{1}+k_{2}t+k_{3}t^{2}+ \cdots+k_{n}t^{n-1},\quad k_{i}\in {R},i=1,2,3, \ldots,n. $$

Definition 5


A cone P is solid in a real Banach space X if its interior is non-empty.

Definition 6


Let P be a solid cone in a real Banach space X, \(T:P^{0}\rightarrow P^{0}\) be an operator and \(0<\alpha<1\). Then T is called θ-concave operator if \(T(ku)\geq{k^{\alpha}}T(u)\) for any \(0< k<1\) and \(u\in{P^{0}}\).

Lemma 7


Assume that P is a normal solid cone in a real Banach space X, \(0<\alpha<1\), and \(T:P^{0}\rightarrow{P^{0}}\) is a α-concave increasing operator. Then T has only one fixed point in \(P^{0}\).

2 Main results

Lemma 8

For \(z(t)\in C[0,1]\), the FOBVP
$$\begin{aligned}& D^{\theta}z(t)=h(t),\quad 3< {\theta}\leq{4}, \end{aligned}$$
$$\begin{aligned}& z(0)=z'''(0),\qquad \eta D^{\alpha}z(1)=z'(0),\qquad \xi z''(1)-z''(0)=0, \quad 0< \alpha< 1, \end{aligned}$$
has a solution of the form
$$ z(t)=\int_{0}^{1}G(t,s)h(t)\, ds, $$
$$ G(t,s)= \textstyle\begin{cases} \frac{t\eta}{1-\frac{\eta}{\Gamma(2-\alpha)}} [\frac{(1-s)^{\theta -\alpha-1}}{\Gamma(\theta-\alpha)}+\frac{\xi}{(1-\xi)\Gamma(3-\alpha )}\frac{(1-s)^{\theta-3}}{\Gamma(\theta-2)} ] \\ \quad +\frac{t^{2}\xi}{2(1-\xi)}\frac{(1-s)^{\theta-3}}{\Gamma(\theta -2)}+\frac{1}{\Gamma(\theta)}(t-s)^{\theta-1},& 0< {s}\leq{t}< {1}, \\ \frac{t\eta}{1-\frac{\eta}{\Gamma(2-\alpha)}} [\frac{(1-s)^{\theta -\alpha-1}}{\Gamma(\theta-\alpha)}+\frac{\xi}{(1-\xi)\Gamma(3-\alpha )}\frac{(1-s)^{\theta-3}}{\Gamma(\theta-2)} ] \\ \quad {}+\frac {t^{2}\xi}{2(1-\xi)}\frac{(1-s)^{\theta-3}}{\Gamma(\theta-2)}, &0< {t}\leq {s}< {1}. \end{cases} $$


Applying the operator \(I^{\alpha}\) on the differential equation in (4) and using Lemma 4, we obtain
$$ z(t)=I^{\theta} h(t)+c_{1}+c_{2}t+c_{3}t^{2}+c_{4}t^{3}. $$
Using boundary conditions (9) for the values of \(c_{1}\), \(c_{2}\), \(c_{3}\), \(c_{4}\), \(z(0)=0=z'''(0)\) yield to \(c_{1}=0=c_{4}\), and by \(\xi z''(1)-z''(0)=0\), we have \(c_{3}=\frac{\xi}{2(1-\xi)}I^{\theta-2}h(1)\). Using \(\eta D^{\alpha }z(1)=z'(0)\), we obtain \(c_{2}=\frac{\eta}{1-\frac{\eta}{\Gamma(2-\alpha)}} [I^{\theta-\alpha }h(1)+\frac{\xi}{(1-\xi)\Gamma(3-\alpha)}I^{\theta-2}h(1) ]\). Substituting the values of \(c_{1}\), \(c_{2}\), \(c_{3}\), \(c_{4}\), in (12), we get
$$\begin{aligned} z(t) =&I^{\theta} h(t)+\frac{t\eta}{1-\frac{\eta}{\Gamma(2-\alpha )}} \biggl[I^{\theta-\alpha}h(1)+\frac{\xi}{(1-\xi)\Gamma(3-\alpha )}I^{\theta-2}h(1) \biggr] \\ &{}+\frac{t^{2}\xi}{2(1-\xi)}I^{\theta -2}h(1). \end{aligned}$$
The integral form is given as
$$\begin{aligned} z(t) =&\int_{0}^{t} \frac{(t-s)^{\theta-1}h(s)\, ds}{\Gamma(\theta)} +\frac{t^{2}\xi}{2(1-\xi)}\int_{0}^{1} \frac{(1-s)^{\theta-3}h(s)\, ds}{\Gamma (\theta-2)} \\ &{}+\frac{t\eta}{1-\frac{\eta}{\Gamma(2-\alpha)}} \biggl[ \int_{0}^{1} \frac{(1-s)^{\theta-\alpha-1}h(s)\, ds}{\Gamma(\theta-\alpha )} \\ &{}+\frac{\xi}{(1-\xi)\Gamma(3-\alpha)}\int_{0}^{1} \frac {(1-s)^{\theta-3}h(s)\, ds}{\Gamma(\theta-2)} \biggr] =\int_{0}^{1}G(t,s)h(s)\, ds. \end{aligned}$$

Lemma 9

Let \(3<{\theta}, \gamma\leq{4}\). For \(z(t)\in{C [0,1]}\), the FOBVP with p-Laplacian
$$\begin{aligned}& D^{\gamma}\bigl(\phi_{p}\bigl(D^{\theta}z(t) \bigr)\bigr)+a(t)f\bigl(z(t)\bigr) = 0, \quad 3< {\theta}, \gamma \leq{4}, t \in[0,1], \\& z(0)=z'''(0), \qquad \eta D^{\alpha}z(t)|_{t=1} = z'(0), \qquad \xi z''(1)-z''(0)=0,\quad 0< \alpha< 1, \\& \phi_{p}\bigl(D^{\theta}z(t)\bigr)|_{t=0}=0 = \bigl( \phi_{p}\bigl(D^{\theta}z(t)\bigr)\bigr)'|_{t=0}, \\& \bigl(\phi_{p}\bigl(D^{\theta} z(t)\bigr)\bigr)''|_{t=1} = \frac{1}{2}\bigl(\phi_{p}\bigl(D^{\theta} z(t)\bigr) \bigr)''|_{t=0}, \\& \bigl(\phi_{p}\bigl(D^{\theta}z(t)\bigr)\bigr)'''|_{t=0} = 0, \end{aligned}$$
has a solution of the form
$$ z(t)=\int_{0}^{1}G(t,s)\phi_{q}\biggl( \int_{0}^{1}\mathcal{H}(s,x)a(x)f\bigl(z(x)\bigr)\, dx\biggr)\, ds, $$
$$ \mathcal{H}(t,x) = \textstyle\begin{cases} -\frac{(t-x)^{\gamma-1}}{\Gamma\gamma}+\frac {t^{2}}{\Gamma(\gamma-2)}(1-x)^{\gamma-3},& 0< {x}\leq{t}< {1}, \\ \frac{t^{2}}{\Gamma(\gamma-2)}(1-x)^{\gamma-3}, &0< {t}\leq {x}< {1}, \end{cases} $$
and \(G(t,s)\) is given by (11).
Applying the integral \(I^{\gamma}\) on the differential equation in (15) and using Lemma 9, we obtain
$$ \phi_{p}\bigl(D^{\theta}z(t)\bigr)=-I^{\gamma}a(t)f \bigl(z(t)\bigr)+c_{1}+c_{2}t+c_{3}t^{2}+c_{4}t^{3}. $$
The boundary conditions \(\phi_{p}(D^{\theta}z(0))=(\phi_{p}(D^{\theta}z(0)))'=(\phi_{p}(D^{\alpha}z(0)))'''=0\) lead to \(c_{1}=c_{2}=c_{4}=0\). From (18) and \(c_{1}=c_{2}=c_{4}=0\), we deduce
$$ \bigl(\phi_{p}\bigl(D^{\theta}z(t)\bigr)\bigr)''=-I^{\gamma-2}a(t)f \bigl(z(t)\bigr)+2c_{3}, $$
and the boundary condition \((\phi_{p}(D^{\theta}z(1)))''=\frac{1}{2}(\phi_{p}(D^{\theta}z(0)))''\) yield \(c_{3}=\frac{1}{\Gamma(\gamma-2)} \int_{0}^{1}(1-x)^{\gamma-3} a(x)f(z(x))\, dx\). Consequently, (18) takes the form
$$\begin{aligned} \phi_{p}\bigl(D^{\theta}z(t)\bigr) =&-\frac{1}{\Gamma(\gamma)}\int _{0}^{1}(t-x)^{\gamma -1}a(x)f\bigl(z(x)\bigr)\, dx \\ &{}+\frac{t^{2}}{\Gamma(\gamma-2)}\int_{0}^{1}(1-x)^{\gamma-3}a(x)f \bigl(z(x)\bigr)\, dx \\ =&\int_{0}^{1}\mathcal{H}(t,x)a(x)f\bigl(z(x) \bigr)\, dx. \end{aligned}$$
The boundary value problem (15) reduces to the following problem:
$$ \begin{aligned} &D^{\alpha}z(t)=\phi_{q}\biggl(\int _{0}^{1}\mathcal{H}(t,x)a(x)f\bigl(z(x)\bigr)\, dx \biggr), \\ &z(0)=z'''(0), \qquad \eta D^{\alpha}z(1)= y'(0), \qquad \xi z''(1)-z''(0)=0, \quad 0< \alpha< 1, \end{aligned} $$
which, in view of Lemma 8, yields the required result,
$$ z(t)=\int_{0}^{1}G(t,s)\phi_{q}\biggl( \int_{0}^{1}\mathcal{H}(s,x)a(x)f\bigl(z(x)\bigr)\, dx\biggr)\, ds. $$

Lemma 10

Let \(3< \gamma\leq{4}\). The Green’s function \(\mathcal{H}(t,x)\) is a continuous function and satisfies
  1. (A)

    \(\mathcal{H}(t,x)\geq0\), \(\mathcal{H}(t,x)\leq\mathcal{H}(1,x)\), for \(t,x\in(0,1]\),

  2. (B)

    \(\mathcal{H}(t,x) \geq t^{\gamma-1 }\mathcal{H}(1,x)\) for \(t,x \in(0,1]\).

The continuity of \(\mathcal{H}(t,x)\) is ensured by its definition. For (A), considering the case when \(0<{x}\leq{t}\leq1\), we have
$$\begin{aligned} \mathcal{H}(t,x) =&-\frac{(t-x)^{\gamma-1}}{\Gamma(\gamma)}+\frac {t^{2}}{\Gamma(\gamma-2)}(1-x)^{\gamma-3} \\ =&-\frac{t^{\gamma-1}(1-\frac{x}{t})^{\gamma-1}}{\Gamma(\gamma)}+\frac {t^{2}}{\Gamma(\gamma-2)}(1-x)^{\gamma-3} \\ \geq&-\frac{t^{\gamma-1}(1-x)^{\gamma-1}}{\Gamma(\gamma)}+\frac {t^{\gamma-1}}{\Gamma(\gamma-2)}(1-x)^{\gamma-3} \\ \geq& \frac{t^{\gamma-1}(1-x)^{\gamma-1}}{\Gamma(\gamma)\Gamma(\gamma -2)} \bigl[\Gamma(\gamma)-\Gamma(\gamma-2) \bigr]>0. \end{aligned}$$
In the case \(0< t\leq x\leq1\), the \(\mathcal{H}(t,s)>0\) is obvious. Now for \(\mathcal{H}(t,x)\leq\mathcal{H}(1,x)\), for \(t,x\in(0,1]\), we have
$$ \frac{\partial}{\partial t}\mathcal{H}(t,x)= \textstyle\begin{cases} -\frac{(t-x)^{\gamma-2}}{\Gamma(\gamma-1)}+2\frac{t}{\Gamma(\gamma -2)}(1-x)^{\gamma-3},& 0< {x}\leq{t}\leq{1}, \\ \frac{2t}{\Gamma(\gamma-2)}(1-x)^{\gamma-3},& 0< {t}\leq{x}\leq{1}. \end{cases} $$
For \(x, t \in(0,1]\), such that \(0< x\leq t\leq1\), from (24), we deduce
$$\begin{aligned} \frac{\partial}{\partial t}\mathcal{H}(t,x) =&-\frac{(t-x)^{\gamma -2}}{\Gamma(\gamma-1)}+ \frac{2t}{\Gamma(\gamma-2)}(1-x)^{\gamma -3} \\ =&-\frac{t^{\gamma-2}(1-\frac{x}{t})^{\gamma-2}}{\Gamma(\gamma -1)}+\frac{2t}{\Gamma(\gamma-2)}(1-x)^{\gamma-3} \\ \geq& -\frac{t^{\gamma-2}(1-x)^{\gamma-2}}{\Gamma(\gamma-1)}+\frac {2t^{\gamma-2}}{\Gamma(\gamma-2)}(1-x)^{\gamma-3} \\ \geq& \frac{t^{\gamma-2}(1-x)^{\gamma-2}}{\Gamma(\gamma-1)\Gamma(\gamma -2)} \bigl[2\Gamma(\gamma-1)-\Gamma(\gamma-2) \bigr]\geq0, \end{aligned}$$
from (25), we have \(\frac{\partial}{\partial t}\mathcal {H}(t,x)\geq0\). In the case \(0< t\leq x \leq1\), (24) implies that the relation \(\frac{\partial}{\partial t}\mathcal{H}(t,x)\geq0\) is obvious, which implies that \(\mathcal{H}(t,x)\) is an increasing function w.r.t. t. Hence \(\mathcal{H}(t,x)\leq\mathcal{H}(1,x)\).
Now for part (B), using (17) in the case \(0< x\leq t\leq1\), we proceed thus:
$$\begin{aligned} \frac{\mathcal{H}(t,x)}{\mathcal{H}(1,x)} =&\frac{\frac{-(t-x)^{\gamma -1}}{\Gamma(\gamma)}+\frac{t^{2}}{\Gamma(\gamma-2)}(1-x)^{\gamma-3}}{\frac {-(1-x)^{\gamma-1}}{\Gamma(\gamma)}+\frac{1}{\Gamma(\gamma -2)}(1-x)^{\gamma-3}} \\ =& \frac{\frac{-t^{\gamma-1}(1-\frac{x}{t})^{\gamma-1}}{\Gamma(\gamma )}+\frac{t^{2}}{\Gamma(\gamma-2)}(1-x)^{\gamma-3}}{\frac{-(1-x)^{\gamma -1}}{\Gamma(\gamma)}+\frac{1}{\Gamma(\gamma-2)}(1-x)^{\gamma-3}} \\ \geq&\frac{\frac{-t^{\gamma-1}(1-x)^{\gamma-1}}{\Gamma(\gamma)}+\frac {t^{\gamma-1}}{\Gamma(\gamma-2)}(1-x)^{\gamma-3}}{\frac{-(1-x)^{\gamma -1}}{\Gamma(\gamma)}+\frac{1}{\Gamma(\gamma-2)}(1-x)^{\gamma -3}} \\ =&t^{\gamma-1} \biggl[\frac{\frac{-(1-x)^{\gamma-1}}{\Gamma(\gamma )}+\frac{1}{\Gamma(\gamma-2)}(1-x)^{\gamma-3}}{\frac{-(1-x)^{\gamma -1}}{\Gamma(\gamma)}+\frac{1}{\Gamma(\gamma-2)}(1-x)^{\gamma-3}} \biggr]=t^{\gamma-1}. \end{aligned}$$
We can also prove the result for the case of \(0< t\leq x\leq1\), this completes the proof.
Assume that the following hold:

\(0<\int_{0}^{1}\mathcal{H}(1,x)a(x)\, dx<+\infty\).


There exist \(0<\delta<1\) and \(\rho>0\) such that \(f(x)\leq\delta L \phi_{p}(x)\), for \(0\leq x \leq\rho\), where \(0< L\leq(\phi_{p}(\varpi)\delta\int_{0}^{1}\mathcal{H}(1,x)a(x)\, dx)^{-1}\).


There exists \(b>0\), such that \(f(x)\leq M \phi_{p}(x)\), for \(u>b\), \(0< M<( \phi_{p}(\varpi2^{q-1})\int_{0}^{1}\mathcal{H}(1,x) a(x)\, dx)^{-1}\).


\(f(z)\) is non-decreasing with respect to z.


There exists \(0\leq\beta<1\) such that \(f(Kz)\geq(\phi_{p}(K))^{\beta}f(z)\), for any \(0\leq k<1\) and \(0< z<+\infty\).

2.1 Existence and uniqueness of solutions

Theorem 11

Under the assumptions (J1) and (J2), the FOBVP (4) has at least one positive solution.


Define \(K_{1}=\{z\in{C [0,1]}:0\leq z(t) \leq\rho\mbox{ on } [0,1]\}\) a closed convex set [15]. Define an operator \(\mathcal{T}:K_{1}\rightarrow C[0,1]\) by
$$ \mathcal{T}z(t)=\int_{0}^{1}G(t,s) \phi_{q}\biggl(\int_{0}^{1} \mathcal{H}(s,x)a(x)f\bigl(z(x)\bigr)\, dx\biggr)\, ds. $$
By Lemma 9, \(z(t)\) is a solution of the FOBVP with p-Laplacian operator (4) if and only if \(z(t)\) is a fixed point of \(\mathcal{T}\). The compactness of the operator \(\mathcal{T}\) can easily be shown. Now consider
$$\begin{aligned} \int_{0}^{1}G(t,s)\,ds =& \frac{\int_{0}^{t}(t-s)^{\theta-1}\,ds}{\Gamma(\theta)}+ \frac{t\eta}{1-\frac {\eta}{\Gamma(2-\alpha)}} \biggl[\frac{\int_{0}^{1}(1-s)^{\theta-\alpha -1}\,ds}{\Gamma(\theta-\alpha)} \\ &{}+ \frac{\xi\int_{0}^{1}(1-s)^{\theta-3}\,ds}{(1-\xi)\Gamma(3-\alpha)\Gamma (\theta-2)}\biggr]+\frac{t^{2}\xi}{2(1-\xi)}\frac{\int_{0}^{1}(1-s)^{\theta -3}\,ds}{\Gamma(\theta-2)} \\ =&\frac{(t-s)^{\theta}|_{t}^{0}}{\Gamma(\theta+1)}+\frac{t\eta}{1-\frac {\eta}{\Gamma(2-\alpha)}} \biggl[\frac{(1-s)^{\theta-\alpha}|_{1}^{0}}{\Gamma (\theta-\alpha+1)} \\ &{}+\frac{\xi}{(1-\xi)\Gamma(3-\alpha)}\frac {(1-s)^{\theta-2}|_{1}^{0}}{\Gamma(\theta-1)}\biggr]+\frac{t^{2}\xi}{2(1-\xi)} \frac {(1-s)^{\theta-2}|_{1}^{0}}{\Gamma(\theta-1)} \\ =&\frac{t^{\theta}}{\Gamma(\theta+1)}+\frac{t\eta}{1-\frac{\eta}{\Gamma (2-\alpha)}} \biggl[\frac{1}{\Gamma(\theta-\alpha+1)} \\ +&\frac{\xi }{(1-\xi)\Gamma(3-\alpha)}\frac{1}{\Gamma(\theta-1)} \biggr] +\frac{t^{2}\xi }{2(1-\xi)} \frac{1}{\Gamma(\theta-1)} \\ \leq& \frac{1}{\Gamma(\theta+1)}+\frac{ \eta}{1-\frac{\eta}{\Gamma (2-\alpha)}}\biggl[\frac{1}{\Gamma(\theta-\alpha+1)} \\ +&\frac{\xi }{(1-\xi)\Gamma(3-\alpha)}\frac{1}{\Gamma(\theta-1)}\biggr] +\frac{ \xi }{2(1-\xi)} \frac{1}{\Gamma(\theta-1)} \\ =&\varpi, \end{aligned}$$
where \(\varpi=\frac{1}{\Gamma(\theta+1)}+\frac{ \eta}{1-\frac{\eta }{\Gamma(2-\alpha)}} [\frac{1}{\Gamma(\theta-\alpha+1)}+\frac{\xi }{(1-\xi)\Gamma(3-\alpha)}\frac{1}{\Gamma(\theta-1)} ] +\frac{\xi }{2(1-\xi)}\frac{1}{\Gamma(\theta-1)}\). For any \(y\in K_{1}\), using (J2) and Lemma 10, we obtain
$$\begin{aligned} \mathcal{T}\bigl(z(t)\bigr) =&\int_{0}^{1} G(t,s) \phi_{q}\biggl(\int_{0}^{1}\mathcal {H}(s,x)a(x)f\bigl(z(x)\, dx\bigr)\biggr)\, ds \\ \leq&\int_{0}^{1}G(t,s)\phi_{q}\biggl( \int_{0}^{1}\mathcal{H}(1,x)a(x)L\delta\phi _{p}(\rho)\, dx\biggr)\, ds \\ \leq& \varpi\phi_{q}\biggl(\int_{0}^{1} \mathcal{H}(1,x)a(x)L\delta\phi_{p}(\rho )\, dx\biggr) \\ =&\varpi\phi_{q}\biggl(L\delta\int_{0}^{1} \mathcal{H}(1,x)a(x)\, dx\biggr)\rho\leq \rho, \end{aligned}$$
which implies \(\mathcal{T}(K_{1})\subseteq{K_{1}}\). By Schauder’s fixed point theorem \(\mathcal{T}\) has a fixed point in \(K_{1}\). □

Theorem 12

Under the assumptions (J1), (J3) the FOBVP with p-Laplacian operator (4) has at least one positive solution.


Let \(b>0\) as given in (J3). Define \(\mathcal{F}^{*}=\max_{0\leq{x}\leq{b}}f(x)\), then \(\mathcal{F}^{*}\geq f(x)\) for \(0\leq{x}\leq{b}\). In view of (J3), we have
$$ \varpi_{1}2^{q-1}\phi_{q}(M)\phi_{q} \biggl(\int_{0}^{1}\mathcal{H}(1,x)a(x)\, dx \biggr)< 1. $$
Choose \(b^{*}>b\) large enough such that
$$ \varpi_{1}2^{q-1}\bigl(\phi_{q} \bigl(\mathcal{F}^{*}\bigr)+\phi_{q}(M)b^{*}\bigr)\phi_{q} \biggl(\int_{0}^{1}\mathcal{H}(1,x)a(x)\, dx \biggr)< b^{*}. $$
Define \(K_{2}=\{z(t)\in C [0,1]: 0\leq z(t)\leq{b^{*}}\mbox{ on } [0,1]\}\), \(\Omega_{1}=\{t\in[0,1]:0\leq{z(t)\leq{b}}\}\), \(\Omega_{2}=\{ t\in [0,1]:b<{z(t)\leq{b^{*}}}\}\). Then \(\Omega_{1}\cup{\Omega_{2}}=[0,1]\) and \(\Omega_{1}\cap{\Omega_{2}}=\varphi\). For \(z\in{K_{1}}\), (J3) implies \(f(z(t))\leq{M}\phi_{p}(z(t))\leq{M\phi_{p}(b^{*})}\) for \(t\in{\Omega_{2}}\) and
$$\begin{aligned} 0\leq\mathcal{T}\bigl(z(t)\bigr) =&\int_{0}^{1}G(t,s) \phi_{q}\biggl(\int_{0}^{1}\mathcal {H}(s,x)a(x)f\bigl(z(x)\bigr)\,dx\biggr)\,ds \\ \leq&\varpi\phi_{q}\biggl(\int_{S_{1}} \mathcal{H}(1,x)a(x)f\bigl(y(x)\bigr)\,dx+\int_{S_{2}} \mathcal{H}(1,x)a(x)f\bigl(z(x)\bigr)\,dx\biggr) \\ \leq&\varpi\phi_{q}\biggl(\mathcal{F}^{*}\int_{S_{1}} \mathcal{H}(1,x)a(x)\,dx+M\phi _{p}\bigl(b^{*}\bigr)\int _{S_{2}}\mathcal{H}(1,x)a(x)\,dx\biggr) \\ \leq&\varpi\phi_{q}\bigl(\mathcal{F}^{*}+M\phi_{p}\bigl(b^{*} \bigr)\bigr)\phi_{q}\biggl(\int_{0}^{1} \mathcal {H}(1,x)a(x)\,dx\biggr). \end{aligned}$$
From (31) and using the inequality \((a+b)^{r}\leq{2^{r}}(a^{r}+b^{r})\) for any \(a,b,r>0\), with (32) we have
$$ 0\leq\mathcal{T}z(t)\leq\varpi_{1}2^{q-1}\bigl( \phi_{q}\bigl(\mathcal{F}^{*}\bigr)+\phi _{q}(M)b^{*}\bigr) \phi_{q}\biggl(\int_{0}^{1} \mathcal{H}(1,x)a(x)\, dx\biggr)\leq{b^{*}}, $$
thus \(\mathcal{T}(K_{2})\subseteq{K_{2}}\). Hence by Schauder’s fixed point theorem \(\mathcal{T}\) has a fixed point \(z\in{K_{2}}\), therefore, the FOBVP with p-Laplacian operator (4) has at least one positive solution in \(K_{2}\). □

Theorem 13

Assume that (J1), (J4), (J5) hold. Then the FOBVP with p-Laplacian operator (4) has a unique positive solution.


Consider the set \(\Lambda=\{z(t)\in C[0,1]:z(t)\geq0 \mbox{ on }[0,1]\}\), where Λ is a normal solid cone in \(C[0,1]\) with \(\Lambda^{0}=\{z(t)\in C[0,1]:z(t)>0 \mbox{ on }[0,1]\}\). Let \(\mathcal{T}:\Lambda^{0}\rightarrow C[0,1]\) be defined by (27), we prove that \(\mathcal{T}\) is a θ-concave and increasing operator. For \(z_{1}, z_{2}\in\Lambda^{0}\) with \(z_{1}\geq z_{2}\) the assumption (J4) implies that the operator \(\mathcal{T}\) is an increasing operator, i.e., \(\mathcal{T}(z_{1}(t))\geq\mathcal {T}(z_{2}(t))\) on \(t\in[0,1]\). With the help of \(f(kz)\geq\phi_{p}(k^{\alpha})f(z)\), we have
$$\begin{aligned} \mathcal{T}\bigl(kz(t)\bigr) \geq& \int_{0}^{1} G(t,s)\phi_{q}\biggl(\int_{0}^{1}H(t,x) \phi _{p}\bigl(k^{\alpha}\bigr)a(x)f\bigl(z(x)\bigr)\, dx\biggr) \, ds \\ =&k^{\alpha}\int_{0}^{1}G(t,s) \phi_{q}\biggl(\int_{0}^{1}H(t,x)a(x)f \bigl(z(x)\bigr)\, dx\biggr)\, ds \\ =&k^{\alpha}\mathcal{T}\bigl(z(t)\bigr), \end{aligned}$$
which implies that \(\mathcal{T}\) is α-concave operator and with the help of Lemma 7, the operator \(\mathcal{T}\) has a unique fixed point which is the unique positive solution of the FOBVP with p-Laplacian operator (4) in \(\Lambda^{0}\). □

3 Examples

Example 1

Consider the following boundary value problem:
$$ \begin{aligned} &D^{3.5}\bigl(\phi_{p} \bigl(D^{3.5}z(t)\bigr)\bigr)+tz(t)=0, \\ &z(0)=0=z'''(0),\qquad \frac{1}{2}D^{0.5} z(t)|_{t=1}-z'(0)=0,\qquad \frac {1}{2}z''(1)-z''(0)=0, \\ &\phi_{p}\bigl(D^{3.5}z(t)\bigr)|_{t=0}=0=\bigl( \phi_{p}\bigl(D^{3.5}z(t)\bigr)\bigr)'|_{t=0}, \\ &\bigl(\phi_{p}\bigl(D^{3.5}z(t)\bigr)\bigr)''|_{t=1}= \frac{1}{2} \bigl(\phi_{p}\bigl(D^{3.5}z(t)\bigr) \bigr)'',\qquad \bigl(\phi_{p} \bigl(D^{3.5}z(t)\bigr)\bigr)''|_{t=0}=0. \end{aligned} $$
Here we have \(\alpha= \beta=3.5\), \(\alpha=0.5\), \(\xi=\gamma=\frac{1}{2}\), \(a(t)=t\), \(f(z(t))=z(t)\). By simple computation, we obtain \(0< L\leq{1.9092}\), choose \(L=1.5\), \(\delta=1\), \(p=2\), and then the conditions (J1) and (J2) are satisfied. Hence, by Theorem 11, the FOBVP with p-Laplacian operator (35) has at least one positive solution.

Example 2

For the following boundary value problem:
$$ \begin{aligned} &D^{3.5}\bigl(\phi_{p} \bigl(D^{3.5}z(t)\bigr)\bigr)+t\sqrt[3]{z}(t)=0, \\ &z(0)=0=z'''(0),\qquad 0.1D^{0.5} z(t)|_{t=1}-z'(0)=0,\qquad 0.1 z''(1)-z''(0)=0, \\ &\phi_{p}\bigl(D^{3.5}z(t)\bigr)|_{t=0}=0=\bigl( \phi_{p}\bigl(D^{3.5}z(t)\bigr)\bigr)'|_{t=0}, \\ &\bigl(\phi_{p}\bigl(D^{3.5}z(t)\bigr)\bigr)''|_{t=1}= \frac{1}{2} \bigl(\phi_{p}\bigl(D^{3.5}z(t)\bigr) \bigr)'',\qquad \bigl(\phi_{p} \bigl(D^{3.5}z(t)\bigr)\bigr)''|_{t=0}=0, \end{aligned} $$
we have \(\alpha=\beta=3.5\), \(\xi=\eta=0.1\), \(a(t)=t\), \(f(u(t))=\sqrt[3]{u}(t)\) and by simple computation we get \(M<{4.4792}\) and thus by choosing \(M=4.00\), \(b=1\), and \(p=q=2\), we see that (36) satisfy (J1) and (J3). Hence by Theorem 12, the FOBVP with p-Laplacian operator (36) has at least one positive solution.

Example 3

The uniqueness of the solution for FOBVP with the p-Laplacian operator; we have
$$ \begin{aligned} &D^{3.5}\bigl(\phi_{p} \bigl(D^{3.5}z(t)\bigr)\bigr)+t z(t)=0, \\ &z(0)=0=z'''(0),\qquad \frac{1}{3}D^{0.5} z(t)|_{t=1}-z'(0)=0, \qquad \frac {1}{3}z''(1)-z''(0)=0, \\ &\phi_{p}\bigl(D^{3.5}z(t)\bigr)|_{t=0}=0=\bigl( \phi_{p}\bigl(D^{3.5}z(t)\bigr)\bigr)'|_{t=0}, \\ &\bigl(\phi_{p}\bigl(D^{3.5}z(t)\bigr)\bigr)''|_{t=1}= \frac{1}{2} \bigl(\phi_{p}\bigl(D^{3.5}z(t)\bigr) \bigr)'',\qquad \bigl(\phi_{p} \bigl(D^{3.5}z(t)\bigr)\bigr)''|_{t=0}=0. \end{aligned} $$
We apply Theorem 13. In (37), we have \(\theta=\gamma=3.5\), \(\alpha=0.5\), \(\xi=\gamma=\frac{1}{3}\), \(a(t)=t\), \(f(z(t))=z(t)\) it is clear that (37) satisfies conditions (J1), (J4). Also considering \(\beta=\frac {1}{2}\), (J5) is satisfied. Thus by Theorem 13, the fractional differential equation (37) has a unique solution.



We are thankful to the referees and editor for their valuable comments and remarks.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (, 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.

Authors’ Affiliations

Department of Mathematical Sciences, University of South Africa, UNISA, P.O. Box 392, Pretoria, 0003, South Africa
Department of Mathematical Sciences, University of Mazandaran, Babolsar, 47416-95447, Iran
Department of Mathematics Computer Science, Cankaya University, Ankara, 06530, Turkey
Institute of Space Sciences, MG-23, Magurele-Bucharest, 76900, Romania
Department of Mathematics, University of Malakand, Dir Lower, P.O. Box 18000, Chakdara, Khybarpukhtunkhwa, Pakistan
Shaheed Benazir Bhutto University, Dir Upper, P.O. Box 18000, Sheringal, Khybarpakhtunkhwa, Pakistan


  1. Hilfer, R: Applications of Fractional Calculus in Physics. World Scientific, Singapore (2000) MATHView ArticleGoogle Scholar
  2. Kilbas, AA, Srivastava, HM, Trujillo, JJ: Theory and Applications of Fractional Differential Equations. North-Holland Mathematics Studies. Elsevier, Amsterdam (2006) MATHGoogle Scholar
  3. Miller, KS, Ross, B: An Introduction to the Fractional Calculus and Fractional Differential Equations. Wiley, New York (1993) MATHGoogle Scholar
  4. Oldhalm, KB, Spainer, J: The Fractional Calculus. Academic Press, New York (1974) Google Scholar
  5. Podlubny, I: Fractional Differential Equations. Academic Press, New York (1999) MATHGoogle Scholar
  6. Sabatier, J, Agrawal, OP, Machado, JAT: Advances in Fractional Calculus. Springer, Berlin (2007) MATHView ArticleGoogle Scholar
  7. Lv, ZW: Existence results for m-point boundary value problems of nonlinear fractional differential equations with p-Laplacian operator. Adv. Differ. Equ. 2014, 69 (2014) View ArticleGoogle Scholar
  8. Prasad, KR, Krushna, BMB: Existence of multiple positive solutions for p-Laplacian fractional order boundary value problems. Int. J. Anal. Appl. 6(1), 63-81 (2014) Google Scholar
  9. Yuan, Q, Yang, W: Positive solution for q-fractional four-point boundary value problems with p-Laplacian operator. J. Inequal. Appl. 2014, 481 (2014) View ArticleGoogle Scholar
  10. Zhang, JJ, Liu, WB, Ni, JB, Chen, TY: Multiple periodic solutions of p-Laplacian equation with one side Nagumo condition. J. Korean Math. Soc. 45(6), 1549-1559 (2008) MATHMathSciNetView ArticleGoogle Scholar
  11. Xu, X, Xu, B: Sign-changing solutions of p-Laplacian equation with a sub-linear nonlinearity at infinity. Electron. J. Differ. Equ. 2013, 61 (2013) View ArticleGoogle Scholar
  12. Wang, B: Positive solutions for boundary value problems on a half line. Int. J. Math. Anal. 3(5), 221-229 (2009) MATHMathSciNetGoogle Scholar
  13. Herzallah, MAE, Baleanu, D: On fractional order hybrid differential equations. Abstr. Appl. Anal. 2014, Article ID 389386 (2014) MathSciNetView ArticleGoogle Scholar
  14. Guo, D, Lakshmikantham, V: Nonlinear Problems in Abstract Cones. Academic Press, Orland (1988) MATHGoogle Scholar
  15. Han, Z, Lu, H, Sun, S, Yang, D: Positive solution to boundary value problem of p-Laplacian fractional differential equations with a parameter in the boundary. Electron. J. Differ. Equ. 2012, 213 (2012) MathSciNetView ArticleGoogle Scholar