1 Introduction
In this paper, we are investigating the following singular nonlinear higher order fractional boundary value problem (BVP for short):
where \(D_{0+}^{\gamma }\) is the Riemann–Liouville fractional derivative of γ order, \(\lambda >0\) is a parameter, \(n-1<\gamma \leq n\) (\(n\geq 3\)), \(k-1<\nu _{k}\), \(q_{k}\leq k\) (\(k=1, 2,\ldots ,n-2\)), \(\nu _{n-2}-q_{k}\leq n-2-k\) (\(k=1, 2,\ldots ,n-2\)), \(1<\gamma -\nu _{n-2}\leq 2\), \(\nu _{n-2}\leq \gamma _{0}\leq n-1\), \(\gamma -\gamma _{0}\geq 1\), \(a_{i}\geq 0\) (\(i=1,2,\ldots ,p\)), \(\nu _{n-2}\leq \alpha _{i}\leq \gamma _{0}\) (\(i=1, 2, \ldots , p\)), \(b_{j}\geq 0\) (\(j=1,2, \ldots \)), \(\nu _{n-2}\leq \beta _{j}\leq \gamma _{0}\) (\(j=1, 2, \ldots \)), \(0<\xi _{1}<\xi _{2}<\cdots <\xi _{j}<\cdots <1\); \(I_{i}\subseteq [0, 1]\) (\(i=1,2,\ldots ,p\)) is measurable; \(w_{i}: (0, 1)\rightarrow \mathbb{R}_{+}=[0, +\infty )\) is continuous with \(w_{i}\in L^{1}(0, 1)\), and \(\int _{0}^{1}w_{i}(s)z(s)\,dA_{i}(s)\) denotes the Riemann–Stieltjes integral, in which \(A_{i}: I_{i} \rightarrow \mathbb{R}\) (\(i=1,2, \ldots ,p\)) is a function of bounded variation. \(f :(0,1)\times (0, + \infty )^{n-1}\rightarrow \mathbb{R}_{+}\) (\(\mathbb{R}_{+}=[0,+\infty )\)) is continuous. A function \(z\in C[0,1]\) is called a positive solution of BVP (1.1) if it satisfies (1.1) and \(z(t)>0\) for \(t\in (0, 1)\).
$$ \textstyle\begin{cases} D_{0+}^{\gamma }z(t)+\lambda f(t,z(t),D_{0+}^{\nu _{1}}z(t),\ldots ,D _{0+}^{\nu _{n-3}}z(t),D_{0+}^{\nu _{n-2}}z(t))=0,\quad 0< t< 1, \\ z(0)=D _{0+}^{q_{1}}z(0)=\cdots =D_{0+}^{q_{n-2}}z(0)=0,\\ D_{0+}^{\gamma _{0}}z(1)=\sum_{i=1}^{p}a_{i}\int _{I_{i}}w_{i}(s)D_{0+}^{\alpha _{i}}z(s)\,dA _{i}(s)+\sum_{j=1}^{\infty }b_{j}D_{0+}^{\beta _{j}}z(\xi _{j}), \end{cases} $$
(1.1)
In recent years, the fractional differential equations have drawn the attention of many famous researchers, readers can refer to [1‐41] and the references therein. It is caused by the applications of fractional differential equations in a proposed framework for describing significant phenomena, for example, the deflection of an elastic beam, the non-Newtonian fluid theory, the degrading of polymer materials, etc. Some interesting results can be found in [1‐5].
Anzeige
In [7], the authors considered the following nonlinear fractional differential equation:
where \(D_{0+}^{\alpha }\) is the Riemann–Liouville fractional derivative, \(n-1<\alpha \leq n\) (\(n\geq 3\)), \(1\leq \beta \leq n-2\), \(0\leq \gamma \leq \beta \), \(0<\xi _{1}<\xi _{2}<\cdots <\xi _{m}<1\), \(Tz(t)=\int _{0}^{t}K(t, s)z(s)\,ds\), and \(Sz(t)=\int _{0}^{1}H(t, s)z(s)\,ds\). By using the Banach contraction mapping principle and the Krasnosel’skii fixed point theorem, they obtained the existence of nonnegative solutions for this problem.
$$ \textstyle\begin{cases} D_{0^{+}}^{\alpha }z(t)+f(t, z(t), Tz(t), Sz(t))=0, \quad 0< t< 1, \\ z(0)=z'(0)= \cdots = z^{(n-2)}(0)=0, \\ D_{0^{+}}^{\beta }z(1)=\sum_{i=1}^{m}a _{i}D_{0^{+}}^{\gamma }z(\xi _{i}), \end{cases} $$
By using the Banach contraction map principle and the theory of \(u_{0}\)-positive linear operator, Zhang and Zhong in [8] studied the following fractional differential equation:
where \(D_{0^{+}}^{\alpha }\) is the Riemann–Liouville derivative, \(n-1<\alpha \leq n\) (\(n\geq 3\)), \(\beta \geq 1\), \(\alpha -\beta >1\), \(0<\eta \leq 1\), \(\lambda >0\) is a parameter, \(h\in L^{1}[0, 1]\), \(f:[0,1]\times \mathbb{R}\rightarrow \mathbb{R}\) is continuous. They got the existence and uniqueness of solutions for this problem.
$$ \textstyle\begin{cases} D_{0^{+}}^{\alpha }z(t)+f(t, z(t))=0, \quad 0< t< 1, \\ z(0)=z'(0)= \cdots = z^{(n-2)}(0)=0, \\ D_{0^{+}}^{\beta }z(1)=\lambda \int _{0} ^{\eta }h(s)D_{0^{+}}^{\gamma }z(s)\,ds, \end{cases} $$
Based on the reducing method of fractional orders, the Schauder fixed point theorem, and the upper and lower solutions method, Zhang, Liu, and Wu in [9] obtained an eigenvalue interval for the existence of positive solutions of the following fractional differential equation:
where \(D_{0+}^{\alpha }\) is the Riemann–Liouville fractional derivative, \(n-1<\alpha \leq n\) (\(n\geq 3\)), \(n-i-1\leq \alpha -\mu _{i}\leq n-i\) (\(i=1, 2,\ldots , n-2\)), \(\mu -\mu _{n-1}>0\), \(\alpha - \mu _{n-1}\leq 2\), \(\alpha -\mu >1\), \(a_{j}\geq 0\) (\(j=1,2,\ldots , p-2\)), \(0<\xi _{1}<\xi _{2}<\cdots <\xi _{p-2}<1\); \(f:(0, +\infty )^{n}\rightarrow \mathbb{R}_{+}\) is continuous and is nonincreasing in \(x_{i}>0\) for \(i=1,2,\ldots ,n\).
$$ \textstyle\begin{cases} -D_{0+}^{\alpha }u(t)=\lambda f(u(t), D_{0+}^{\mu _{1}}u(t), D_{0+} ^{\mu _{2}}u(t),\ldots , D_{0+}^{\mu _{n-1}}u(t)), \quad 0< t< 1, \\ u(0)=0, \qquad D_{0+}^{\mu _{i}}u(0)=0, \qquad D_{0+}^{\mu }u(1)=\sum_{j=1}^{p-2}a _{j}D_{0^{+}}^{\mu }u(\xi _{j}), \quad 1\leq i\leq n-2, \end{cases} $$
Inspired by the above-mentioned papers, we investigate BVP (1.1). As far as we know, BVP (1.1) has seldom been researched up to now, and the novelty of this paper lies in three aspects. Firstly, the boundary conditions are the combination of sum of Riemann–Stieltjes integral type boundary conditions and nonlocal infinite-point discrete type boundary conditions, which involves fractional derivative of different orders. This fact suggests BVP (1.1) is more general than the above-mentioned literature. For instance, let \(I_{1}=(0, 1)\), \(I_{i}=0\) (\(i= 2, 3,\ldots , p\)), and \(b_{j}= 0\) (\(j=1,2,\ldots \)), then the boundary condition of BVP (1.1) reduces to the boundary condition in [16]; if \(b_{j}=0\) (\(j=1,2,\ldots \)), \(\alpha _{i}=0\)
\(w_{i}=1\) (\(i= 2, 3,\ldots , p\)), then the boundary condition is equal to [37]; and if \(I_{1}=(0, \eta )\) (\(\eta \in (0, 1)\)), \(I_{i}=0\) (\(i= 2, 3,\ldots , p\)), the boundary condition of BVP (1.1) is the same as [18]. Meanwhile the work to check the properties of the corresponding Green’s function is too hard. Secondly, the nonlinearity f contains different orders of fractional derivative of the unknown function. In general, many papers consider these kinds of boundary value problem in the space \(E=\{u\in C[0, 1]: D_{0+} ^{\nu _{i}}u\in C[0, 1], i=1,2,\ldots ,n-2\}\), which makes the study extremely difficult. In this paper, we use the reducing method to transform BVP (1.1) into a relatively low-order equivalent problem, which could be considered in the space \(C[0, 1]\), and is a good way to do this. Some interesting results of the reducing method can be found in [9, 17, 23, 25, 27, 29, 31] and the references therein. Thirdly, there is much to be learned about the theory and applications of mixed monotone operator, recently. Especially, many papers have taken it into the research for fractional boundary value problems. Some interesting results can be found in [10, 11, 15, 21‐23, 25, 27] and the references therein. Thus, in this paper, by using the fixed point theorem of mixed monotone operator, we obtain the uniqueness of positive solution under the assumption that f may be singular with respect to both the time and space variables. It is worth mentioning that some important properties of the unique solution rely on the parameter λ.
Anzeige
The paper is organized as follows. In Sect. 2, we present some preliminary setting, derive the corresponding Green’s function, and transform BVP (1.1) into a relatively low-order equivalent problem, in which the nonlinear term has no fractional derivatives. In Sect. 3, we pay particular attention to establishing the uniqueness of positive solutions and consider some relative properties of the unique positive solution. In Sect. 4, two examples are devoted to our main results.
2 Preliminaries and lemmas
Let E be a Banach space and P be a cone in E. P is said to be normal if there exists a constant \(N>0\) such that, for any \(u, v \in E\), \(\theta \leq u\leq v\) implies \(\|u\|\leq N\|v\|\), the smallest constant, which satisfies this inequality, is called the normality constant of P. Then E is partially ordered by P, i.e., \(u\leq v\) if and only if \(v-u\in P\). For any \(u, v\in E\), the notation \(u\sim v\) means that there exist constants \(\lambda >0\) and \(\mu >0\) such that \(\lambda u\leq v\leq \mu u\). Obviously, ∼ is an equivalence relation. For fixed \(e\in P_{e}\) and \(e>\theta \), we denote \(P_{e}=\{u\in E : u\sim e\}=\{u\in E : \omega e\leq u \leq \frac{1}{\omega }e, 0<\omega <1\}\). It is easy to see that \(P_{e}\subset P\) is a component of P.
Definition 2.1
([11])
Let E be a Banach space and \(D\subset E\). The operator \(A :D\times D \rightarrow E\) is called a mixed monotone operator if \(A(u, v)\) is increasing in \(u\in D\) and decreasing in \(v\in D\), i.e., \(u_{i}\), \(v_{i} \in D\) (\(i=1,2\)), \(u_{1}\leq u_{2}\), \(v_{1}\geq v_{2}\) imply \(A(u_{1}, v_{1})\leq A(u _{2}, v_{2})\). An element \(u\in D\) is called a fixed point of A if \(A(u, u)=u\).
Lemma 2.2
Let
P
be a normal cone in the Banach space
E, and
\(A, B: P_{e}\times P_{e} \rightarrow P_{e}\)
be two mixed monotone operators which satisfy the following conditions:
(i)
For any
\(\mu \in (0, 1)\), there exists
\(\varphi (\mu )\in ( \mu ,1]\)
such that
$$ A\bigl(\mu u,\mu ^{-1}v\bigr)\geq \varphi (\mu )A(u, v), \quad \forall u, v \in P_{e}. $$
(ii)
For any
\(\mu \in (0,1)\), \(u, v\in P_{e}\),
$$ B\bigl(\mu u, \mu ^{-1}v\bigr)\geq \mu B(u, v). $$
(iii)
There exists a constant
\(\kappa >0\)
such that
\(A(u, v)\geq \kappa B(u, v)\), \(\forall u, v\in P_{e}\).
Then there exists a unique fixed point
\(u^{*}\in P_{e}\)
such that
\(A(u^{*}, u^{*})+B(u^{*}, u^{*})=u^{*}\). And for any initial values
\(u_{0}\), \(v_{0}\in P_{e}\), by constructing successively the sequences as follows:
we have
\(u_{n}\rightarrow u^{*} \)
and
\(v_{n}\rightarrow u^{*} \)
in
E, as
\(n\rightarrow \infty \).
$$ u_{n}=A(u_{n-1},v_{n-1})+B(u_{n-1}, v_{n-1}),\qquad v_{n}=A(v_{n-1},u _{n-1})+B(v_{n-1}, u_{n-1}),\quad n=1,2,\ldots , $$
Lemma 2.3
Suppose that operators
A
and
B
satisfy all the conditions of Lemma 2.2. Then the equation
has a unique solution
\(u_{\lambda }\)
in
\(P_{e}\)
for all
\(\lambda >0\), which satisfies:
$$ \lambda A(u, u)+ \lambda B(u, u)= u $$
(i)
If there exists
\(r\in (0, 1)\)
such that
then
\(u_{\lambda }\)
is continuous with respect to
\(\lambda \in (0, + \infty )\). That is, for any
\(\lambda _{0}\in (0,+\infty )\),
$$ \varphi (\mu )\geq \frac{\mu ^{r}-\mu }{\kappa }+\mu ^{r},\quad \forall \mu \in (0, 1), $$
$$ \Vert u_{\lambda }-u_{\lambda _{0}} \Vert \rightarrow 0, \quad \textit{as } \lambda \rightarrow \lambda _{0}. $$
(ii)
If
then
\(0<\lambda _{1}< \lambda _{2}\)
implies
\(u_{\lambda _{1}}< u_{\lambda _{2}}\).
$$ \varphi (\mu )\geq \frac{\mu ^{\frac{1}{2}}-\mu }{\kappa }+ \mu ^{\frac{1}{2}},\quad \forall \mu \in (0, 1), $$
(iii)
If there exists
\(r\in (0, \frac{1}{2})\)
such that
then
$$ \varphi (\mu )\geq \frac{\mu ^{r}-\mu }{\kappa }+\mu ^{r},\quad \forall \mu \in (0, 1), $$
$$ \lim_{\lambda \rightarrow 0^{+}} \Vert u_{\lambda } \Vert =0, \qquad \lim_{\lambda \rightarrow +\infty } \Vert u_{\lambda } \Vert =+\infty . $$
Definition 2.4
([5])
Let \(\alpha >0\). The Riemann–Liouville fractional integral of order α of a function \(z: (0, \infty ) \rightarrow \mathbb{R} \) is given by
provided that the right-hand side is pointwise defined on \((0, \infty )\).
$$ I_{0^{+}}^{\alpha }z(t)=\frac{1}{\varGamma (\alpha )} \int _{0}^{t}(t-s)^{ \alpha -1}z(s)\,ds $$
Definition 2.5
([5])
Let \(\alpha >0\). The Riemann–Liouville fractional derivative of order α of a continuous function \(z: (0, \infty ) \rightarrow \mathbb{R}\) is given by
where \(n = [\alpha ]+1\), \([\alpha ]\) denotes the integer part of α, provided that the right-hand side is pointwise defined on \((0, \infty )\).
$$ D_{0^{+}}^{\alpha }z(t)= \biggl(\frac{d}{dt} \biggr)^{n}I_{0^{+}}^{n- \alpha }z(t)= \frac{1}{\varGamma (n-\alpha )} \biggl( \frac{d}{dt} \biggr) ^{n} \int _{0}^{t}\frac{z(s)}{(t-s)^{\alpha -n+1}}\,ds, $$
Lemma 2.6
([6])
Let
\(z\in C(0, 1) \cap L^{1}(0, 1)\). Then the fractional differential equation
has a unique solution
where
n
is the smallest integer greater than or equal to
α.
$$ D_{0^{+}}^{\alpha }z(t)=0 $$
$$ z(t) = c_{1}t^{\alpha -1}+c_{2}t^{\alpha -2}+ \cdots +c_{n}t^{\alpha -n},\quad c_{i} \in \mathbb{R}, i =1, 2, \ldots , n, $$
Lemma 2.7
([6])
Let
\(z \in C(0, 1) \cap L^{1}(0, 1)\)
and
\(D_{0^{+}}^{\alpha }z\in C(0, 1) \cap L^{1}(0,1)\). Then
where
n
is the smallest integer greater than or equal to
α.
$$ I_{0^{+}}^{\alpha }D_{0^{+}}^{\alpha }z(t)=z(t)+ c_{1}t^{\alpha -1}+c _{2}t^{\alpha -2}+\cdots +c_{n}t^{\alpha -n},\quad c_{i} \in \mathbb{R}, i =1, 2, \ldots , n, $$
Lemma 2.8
([5])
Suppose that
\(z\in C(0, 1) \cap L^{1}(0, 1)\), then
(i)
\(I_{0^{+}}^{\alpha }I_{0^{+}}^{\beta }z(t)=I_{0^{+}} ^{\alpha +\beta }z(t)\)
for
α, \(\beta >0\);
(ii)
\(D_{0^{+}}^{\beta }I_{0^{+}}^{\alpha }z(t)=I_{0^{+}} ^{\alpha -\beta }z(t)\)
for
\(\alpha \geq \beta >0\).
Lemma 2.9
Suppose that BVP (1.1) has a solution
\(z\in C[0, 1]\), then
\(u=D_{0^{+}}^{\nu _{n-2}}z\)
is a solution of the following boundary value problem:
where
\(1<\gamma -\nu _{n-2}\leq 2\). On the other hand, if we assume BVP (2.1) has a solution
\(u\in C[0,1]\), then BVP (1.1) has a solution
\(z=I_{0^{+}}^{\nu _{n-2}}u\).
$$ \textstyle\begin{cases} D_{0+}^{\gamma -\nu _{n-2}} u(t)+\lambda f(t,I_{0+}^{\nu _{n-2}}u(t),I _{0+}^{\nu _{n-2}-\nu _{1}}u(t),\ldots ,I_{0+}^{\nu _{n-2}-\nu _{n-3}}u(t), u(t))=0, \\ D_{0^{+}}^{q_{n-2}-\nu _{n-2}}u(0)=0, \quad 0< t< 1,\\ D_{0+}^{\gamma _{0}-\nu _{n-2}}u(1)=\sum_{i=1}^{p}a_{i} \int _{I_{i}}w_{i}(s)D_{0+}^{\alpha _{i}-\nu _{n-2}}u(s)\,dA_{i}(s)\\ \hphantom{D_{0+}^{\gamma _{0}-\nu _{n-2}}u(1)=}{} +\sum_{j=1}^{\infty }b_{j}D_{0+}^{\beta _{j}-\nu _{n-2}}u(\xi _{j}), \end{cases} $$
(2.1)
Proof
Suppose that \(z\in C[0, 1]\) and satisfies BVP (1.1). Let
It follows from Lemma 2.7 that
where \(c_{i}\in \mathbb{R}\) (\(i=1,2,\ldots , n-2\)). Hence,
Since \(z(0)=0,\ldots \) , \(D_{0^{+}}^{q_{n-3}}z(0)=0\), we immediately obtain that \(c_{1}=\cdots =c_{n-2}=0\). Thus,
By using (ii) in Lemma 2.8, we have
and
Similar to (2.5), we have
It follows from (2.2)–(2.5) that
On the basis of (2.6), (2.8), and (2.9), we have
From (2.7), we have
Combining (2.10)–(2.12), we deduce that \(u =D_{0^{+}}^{\nu _{n-2}}z \) is a solution of BVP (2.1).
$$ u(t)=D_{0^{+}}^{\nu _{n-2}}z(t),\quad t\in [0,1]. $$
(2.2)
$$ I_{0^{+}}^{\nu _{n-2}}u(t)=I_{0^{+}}^{\nu _{n-2}}D_{0^{+}}^{\nu _{n-2}}z(t)=z(t)+c _{1} t^{\nu _{n-2}-1}+\cdots +c_{n-2}t^{\nu _{n-2}-(n-2)}, $$
$$ z(t)=I_{0^{+}}^{\nu _{n-2}}u(t)- c_{1} t^{\nu _{n-2}-1}-\cdots -c_{n-2}t ^{\nu _{n-2}-(n-2)}. $$
$$ z(t)=I_{0^{+}}^{\nu _{n-2}}u(t), \quad t\in [0, 1]. $$
(2.3)
$$ D_{0^{+}}^{\nu _{i}}z(t)=D_{0^{+}}^{\nu _{i}}I_{0^{+}}^{\nu _{n-2}}u(t)=I _{0^{+}}^{\nu _{n-2}-\nu _{i}}u(t),\quad i=1, 2, \ldots , n-3, $$
(2.4)
$$ \begin{aligned}[b] D_{0^{+}}^{\gamma }z(t) &=D_{0^{+}}^{\gamma }I_{0^{+}}^{\nu _{n-2}}u(t)= \frac{d ^{n}}{dt^{n}}I_{0^{+}}^{n-\gamma }I_{0^{+}}^{\nu _{n-2}}u(t) =\frac{d ^{n}}{dt^{n}}I_{0^{+}}^{n-(\gamma -\nu _{n-2})}u(t) \\ &=\frac{d^{2}}{dt ^{2}}\frac{d^{n-2}}{dt^{n-2}}I_{0^{+}}^{n-2}I_{0^{+}}^{2-(\gamma - \nu _{n-2})}u(t) \\ &=D_{0^{+}}^{\gamma -\nu _{n-2}}u(t). \end{aligned} $$
(2.5)
$$\begin{aligned}& D_{0^{+}}^{\gamma _{0}}z(t)=D_{0^{+}}^{\gamma _{0}-\nu _{n-2}}u(t), \end{aligned}$$
(2.6)
$$\begin{aligned}& D_{0^{+}}^{q_{n-2}}z(t)=D_{0^{+}}^{q_{n-2}-\nu _{n-2}}u(t), \end{aligned}$$
(2.7)
$$\begin{aligned}& D_{0^{+}}^{\alpha _{i}}z(t)=D_{0^{+}}^{\alpha _{i}-\nu _{n-2}}u(t), \quad i=1,2, \ldots ,p, \end{aligned}$$
(2.8)
$$\begin{aligned}& D_{0^{+}}^{\beta _{j}}z(t)=D_{0^{+}}^{\beta _{j}-\nu _{n-2}}u(t), \quad j=1,2, \ldots . \end{aligned}$$
(2.9)
$$ \begin{aligned}[b] &D_{0^{+}}^{\gamma -\nu _{n-2}}u(t)+\lambda f\bigl(t, I_{0+}^{\nu _{n-2}}u(t),I _{0+}^{\nu _{n-2}-\nu _{1}}u(t), \ldots ,I_{0+}^{\nu _{n-2}-\nu _{n-3}}u(t), u(t)\bigr) \\ &\quad =D_{0+}^{\gamma }z(t)+\lambda f\bigl(t,z(t),D_{0+}^{\nu _{1}}z(t), \ldots ,D_{0+}^{\nu _{n-3}}z(t),D_{0+}^{\nu _{n-2}}z(t) \bigr) \\ &\quad =0. \end{aligned} $$
(2.10)
$$ \begin{aligned}[b] D_{0+}^{\gamma _{0}-\nu _{n-2}}u(1)&= D_{0^{+}}^{\gamma _{0}}z(1) \\ &= \sum_{i=1}^{p}a_{i} \int _{I_{i}}w_{i}(s)D_{0+}^{\alpha _{i}}z(s) \,dA_{i}(s)+ \sum_{j=1}^{\infty }b_{j}D_{0+}^{\beta _{j}}z( \xi _{j}) \\ &=\sum_{i=1} ^{p}a_{i} \int _{I_{i}}w_{i}(s)D_{0+}^{\alpha _{i}-\nu _{n-2}}u(s) \,dA_{i}(s) +\sum_{j=1}^{\infty }b_{j}D_{0+}^{\beta _{j}-\nu _{n-2}}u( \xi _{j}). \end{aligned} $$
(2.11)
$$ D_{0^{+}}^{q_{n-2}-\nu _{n-2}}u(0)=D_{0^{+}}^{q_{n-2}}z(0)=0. $$
(2.12)
Remark 2.10
Lemma 2.11
Let
\(x\in C(0,1)\cap L^{1}(0,1)\). Then the boundary value problem
is equivalent to
where
in which
Obviously, \(K(t, s)\)
is continuous on
\([0, 1]\times [0, 1]\).
$$ \textstyle\begin{cases} D_{0+}^{\gamma -\nu _{n-2}}u(t)+x(t)=0, \quad 0< t< 1, 1< \gamma - \nu _{n-2}\leq 2, \\ D_{0^{+}}^{q_{n-2}-\nu _{n-2}}u(0)=0, \\ D_{0+} ^{\gamma _{0}-\nu _{n-2}}u(1)=\sum_{i=1}^{p}a_{i}\int _{I_{i}}w_{i}(s)D _{0+}^{\alpha _{i}- \nu _{n-2}}u(s)\,dA_{i}(s)+\sum_{j=1}^{\infty }b_{j}D _{0+}^{\beta _{j}-\nu _{n-2}}u(\xi _{j}), \end{cases} $$
(2.13)
$$ u(t)= \int _{0}^{1}K(t, s)x(s)\,ds, $$
(2.14)
$$ \begin{aligned}[b] K(t, s)&= K_{0}(t, s)+t^{\gamma -\nu _{n-2}-1}\sum_{i=1}^{p} \biggl( \int _{I _{i}}K_{i}(\tau ,s)w_{i}( \tau )\,dA_{i}(\tau ) \biggr) \\ &\quad {}+t^{\gamma - \nu _{n-2}-1}\sum_{j=1}^{\infty }H_{j}( \xi _{j},s), \end{aligned} $$
(2.15)
$$\begin{aligned}& K_{0}(t, s)=\frac{1}{\varGamma (\gamma -\nu _{n-2})} \textstyle\begin{cases} t^{\gamma -\nu _{n-2}-1}(1-s)^{\gamma -\gamma _{0}-1}-(t-s)^{\gamma - \nu _{n-2}-1}, & 0\leq s\leq t\leq 1 , \\ t^{\gamma -\nu _{n-2}-1}(1-s)^{ \gamma -\gamma _{0}-1}, & 0\leq t\leq s\leq 1 , \end{cases}\displaystyle \\& K_{i}(t, s)=\frac{a_{i}}{\sigma \varGamma (\gamma -\nu _{n-2})\varGamma ( \gamma -\alpha _{i})} \textstyle\begin{cases} t^{\gamma -\alpha _{i}-1}(1-s)^{\gamma -\gamma _{0}-1}-(t-s)^{\gamma - \alpha _{i}-1}, & 0\leq s\leq t\leq 1 , \\ t^{\gamma -\alpha _{i}-1}(1-s)^{ \gamma -\gamma _{0}-1}, & 0\leq t\leq s\leq 1 , \end{cases}\displaystyle \\& \quad (i=1, 2, \ldots , p), \\& H_{j}(t, s)=\frac{b_{j}}{\sigma \varGamma (\gamma -\nu _{n-2})\varGamma ( \gamma -\beta _{j})} \textstyle\begin{cases} t^{\gamma -\beta _{j}-1}(1-s)^{\gamma -\gamma _{0}-1}-(t-s)^{\gamma - \beta _{j}-1}, & 0\leq s\leq t\leq 1 , \\ t^{\gamma -\beta _{j}-1}(1-s)^{ \gamma -\gamma _{0}-1}, & 0\leq t\leq s\leq 1 , \end{cases}\displaystyle \\& \quad (j=1,2,\ldots ), \\& \sigma =\frac{1}{\varGamma (\gamma -\gamma _{0})} -\sum_{i=1}^{p} \frac{a _{i}}{\varGamma (\gamma -\alpha _{i})} \int _{I_{i}}s^{\gamma -\alpha _{i}-1}w _{i}(s) \,dA_{i}(s) -\sum_{j=1}^{\infty } \frac{b_{j}}{\varGamma (\gamma - \beta _{j})}\xi _{j}^{\gamma -\beta _{j}-1}\neq 0. \end{aligned}$$
Proof
By using Lemma 2.7, we may express (2.13) as
where \(c_{1}\), \(c_{2}\in \mathbb{R}\). Since \(D_{0^{+}}^{q_{n-2}-\nu _{n-2}}u(0)=0\), we get \(c_{2}=0 \) and rewrite (2.16) as
With the help of (ii) in Lemma 2.8, we have
and
which combined with the boundary condition
yields
where
Applying (2.18) into (2.17), we can obtain
The proof is complete. □
$$ u(t)=- \int _{0}^{t}\frac{(t-s)^{\gamma -\nu _{n-2}-1}}{\varGamma (\gamma -\nu _{n-2})}x(s) \,ds+c_{1}t^{\gamma -\nu _{n-2}-1}+c_{2}t^{\gamma -\nu _{n-2}-2}, $$
(2.16)
$$ u(t)=- \int _{0}^{t}\frac{(t-s)^{\gamma -\nu _{n-2}-1}}{\varGamma (\gamma -\nu _{n-2})}x(s) \,ds+c_{1}t^{\gamma -\nu _{n-2}-1}. $$
(2.17)
$$\begin{aligned}& D_{0+}^{\gamma _{0}-\nu _{n-2}}u(t)=-\frac{1}{(\gamma -\gamma _{0})} \int _{0}^{t}(t-s)^{\gamma -\gamma _{0}-1}x(s)\,ds +c_{1}\frac{\varGamma (\gamma -\nu _{n-2})}{\varGamma (\gamma -\gamma _{0})}t^{\gamma -\gamma _{0}-1}, \\& D_{0+}^{\alpha _{i}-\nu _{n-2}}u(t)=-\frac{1}{(\gamma -\alpha _{i})} \int _{0}^{t}(t-s)^{\gamma -\alpha _{i}-1}x(s)\,ds +c_{1}\frac{\varGamma (\gamma -\nu _{n-2})}{\varGamma (\gamma -\alpha _{i})}t^{\gamma -\alpha _{i}-1},\quad i=1,2, \ldots ,p, \end{aligned}$$
$$ D_{0+}^{\beta _{j}-\nu _{n-2}}u(t)=-\frac{1}{(\gamma -\beta _{j})} \int _{0}^{t}(t-s)^{\gamma -\beta _{j}-1}x(s)\,ds +c_{1}\frac{\varGamma (\gamma -\nu _{n-2})}{\varGamma (\gamma -\beta _{j})}t^{\gamma -\beta _{i}-1},\quad j=1,2, \ldots , $$
$$ D_{0+}^{\gamma _{0}-\nu _{n-2}}u(1) =\sum_{i=1}^{p}a_{i} \int _{I_{i}}w _{i}(s)D_{0+}^{\alpha _{i}-\nu _{n-2}}u(s) \,dA_{i}(s)+\sum_{j=1}^{\infty }b_{j}D_{0+}^{\beta _{j}-\nu _{n-2}}u( \xi _{j}) $$
$$ \begin{aligned}[b] c_{1}&= \frac{1}{\sigma \varGamma (\gamma -\nu _{n-2})} \Biggl\{ \frac{1}{ \varGamma (\gamma -\gamma _{0})} \int _{0}^{1}(1-s)^{\gamma -\gamma _{0}-1}x(s)\,ds \\ &\quad {}-\sum_{i=1}^{p} \frac{a_{i}}{\varGamma (\gamma -\alpha _{i})} \int _{I _{i}}w_{i}(s) \biggl( \int _{0}^{s}(s-\tau )^{\gamma -\alpha _{i}-1}x( \tau )\,d\tau \biggr) \,dA_{i}(s) \\ &\quad {}-\sum_{j=1}^{\infty } \frac{b _{j}}{\varGamma (\gamma -\beta _{j})} \int _{0}^{\xi _{j}}(\xi _{j}-\tau )^{ \gamma -\beta _{j}-1}x(\tau )\,d\tau \Biggr\} , \end{aligned} $$
(2.18)
$$ \begin{aligned}[b] \sigma &=\frac{1}{\varGamma (\gamma -\gamma _{0})}-\sum_{i=1}^{p} \frac{a _{i}}{\varGamma (\gamma -\alpha _{i})} \int _{I_{i}}s^{\gamma -\alpha _{i}-1}w _{i}(s) \,dA_{i}(s) -\sum_{j=1}^{\infty } \frac{b_{j}}{\varGamma (\gamma - \beta _{j})}\xi _{j}^{\gamma -\beta _{j}-1}\\ &\neq 0. \end{aligned} $$
(2.19)
$$\begin{aligned} u(t)&= -\frac{1}{\varGamma (\gamma -\nu _{n-2})} \int _{0}^{t}(t-s)^{\gamma -\nu _{n-2}-1}x(s)\,ds \\ &\quad{} +\frac{t^{\gamma -\nu _{n-2}-1}}{\sigma \varGamma (\gamma -\nu _{n-2})} \Biggl\{ \frac{1}{\varGamma (\gamma -\gamma _{0})} \int _{0}^{1}(1-s)^{\gamma -\gamma _{0}-1}x(s)\,ds \\ &\quad{} -\sum_{i=1}^{p} \frac{a _{i}}{\varGamma (\gamma -\alpha _{i})} \int _{I_{i}} \biggl( \int _{0}^{s}(s- \tau )^{\gamma -\alpha _{i}-1}x(\tau )\,d\tau \biggr)w_{i}(s)\,dA_{i}(s) \\ &\quad{} -\sum_{j=1}^{\infty } \frac{b_{j}}{\varGamma (\gamma -\beta _{j})} \int _{0}^{\xi _{j}}(\xi _{j}-\tau )^{\gamma -\beta _{j}-1}x(\tau )\,d \tau \Biggr\} \\ &= -\frac{1}{\varGamma (\gamma -\nu _{n-2})} \int _{0}^{t}(t-s)^{\gamma - \nu _{n-2}-1}x(s)\,ds \\ &\quad{} + \Biggl\{ \frac{1}{\varGamma (\gamma -\nu _{n-2})} +\frac{1}{ \sigma \varGamma (\gamma -\nu _{n-2})} \Biggl(\sum _{i=1}^{p}\frac{a_{i}}{ \varGamma (\gamma -\alpha _{i})} \int _{I_{i}}w_{i}(s)s^{\gamma -\alpha _{i}-1}\,dA _{i}(s) \\ & \quad{} +\sum_{j=1}^{\infty } \frac{b_{j}}{\varGamma (\gamma -\beta _{j})}\xi _{j}^{\gamma -\beta _{j}-1} \Biggr) \Biggr\} t^{\gamma -\nu _{n-2}-1} \int _{0}^{1}(1-s)^{\gamma -\gamma _{0}-1}x(s)\,ds \\ &\quad{} -\frac{t ^{\gamma -\nu _{n-2}-1}}{\sigma \varGamma (\gamma -\nu _{n-2})} \Biggl\{ \sum_{i=1}^{p} \frac{a_{i}}{\varGamma (\gamma -\alpha _{i})} \int _{I_{i}} \biggl( \int _{0}^{s}(s-\tau )^{\gamma -\alpha _{i}-1}x(\tau )\,d\tau \biggr)w _{i}(s)\,dA_{i}(s) \\ & \quad{} -\sum_{j=1}^{\infty } \frac{b_{j}}{\varGamma (\gamma -\beta _{j})} \int _{0}^{\xi _{j}}(\xi _{j}- \tau )^{\gamma -\beta _{j}-1}x( \tau )\,d\tau \Biggr\} \\ &= -\frac{1}{\varGamma (\gamma -\nu _{n-2})} \int _{0}^{t}(t-s)^{\gamma - \nu _{n-2}-1}x(s)\,ds+ \frac{t^{\gamma -\nu _{n-2}-1}}{\varGamma (\gamma - \nu _{n-2})} \int _{0}^{1}(1-s)^{\gamma -\gamma _{0}-1}x(s)\,ds \\ &\quad{} +\sum_{i=1}^{p} \frac{a_{i}t^{\gamma -\nu _{n-2}-1}}{\sigma \varGamma (\gamma - \nu _{n-2})\varGamma (\gamma -\alpha _{i})} \int _{I_{i}} \biggl( \int _{0} ^{1}s^{\gamma -\alpha _{i}-1}(1-\tau )^{\gamma -\gamma _{0}-1}x(\tau )\,d \tau \biggr)w_{i}(s) \,dA_{i}(s) \\ &\quad{} -\sum_{i=1}^{p} \frac{a_{i}t^{\gamma -\nu _{n-2}-1}}{\sigma \varGamma (\gamma -\nu _{n-2})\varGamma (\gamma -\alpha _{i})} \int _{I_{i}} \biggl( \int _{0}^{s}(s-\tau )^{\gamma -\gamma _{0}-1}x( \tau )\,d\tau \biggr)w_{i}(s)\,dA_{i}(s) \\ &\quad{} +\sum_{j=1}^{\infty } \frac{b _{j}t^{\gamma -\nu _{n-2}-1}}{\sigma \varGamma (\gamma -\nu _{n-2})\varGamma (\gamma -\beta _{j})} \int _{0}^{1}\xi _{j}^{\gamma -\beta _{j}-1}(1-s)^{ \gamma -\gamma _{0}-1}x(s) \,ds \\ &\quad{} -\sum_{j=1}^{\infty } \frac{b_{j}t^{ \gamma -\nu _{n-2}-1}}{\sigma \varGamma (\gamma -\nu _{n-2})\varGamma (\gamma -\beta _{j})} \int _{0}^{\xi _{j}}(\xi _{j}-s)^{\gamma -\beta _{j}-1}x(s) \,ds \\ &= \int _{0}^{1}K_{0}(t,s)x(s)\,ds+ \sum_{i=1}^{p} \int _{I_{i}}t^{\gamma -\nu _{n-2}-1} \biggl( \int _{0}^{1}K_{i}(s,\tau )x( \tau )\,d\tau \biggr)w _{i}(s)\,dA_{i}(s) \\ &\quad{} +\sum_{j=1}^{\infty } \int _{0}^{1}t^{\gamma -\nu _{n-2}-1}H_{j}( \xi _{j},s)x(s)\,ds \\ &= \int _{0}^{1}K_{0}(t,s)x(s)\,ds+ \int _{0}^{1}t^{\gamma -\nu _{n-2}-1}\sum _{i=1}^{p} \biggl( \int _{I_{i}}K_{i}( \tau ,s)w_{i}( \tau )\,dA_{i}(\tau ) \biggr)x(s)\,ds \\ &\quad{} + \int _{0}^{1}t^{ \gamma -\nu _{n-2}-1}\sum _{j=1}^{\infty }H_{j}(\xi _{j},s)x(s)\,ds \\ &= \int _{0}^{1}K(t,s)x(s)\,ds. \end{aligned}$$
Lemma 2.12
Let
\(\sigma >0\) (defined in (2.19) of Lemma 2.11), \(\int _{I_{i}}s^{\gamma -\alpha _{i}-1}w_{i}(s)\,dA_{i}(s)\geq 0\) (\(i=1, 2,\ldots , p\)), and
\(0<\sum_{j=1}^{\infty }\frac{b_{j}}{ \varGamma (\gamma -\beta _{j})}\xi _{j}^{\gamma -\beta _{j}-1}<\infty \). Then the functions
\(K_{0}(t, s)\), \(K_{i}(t, s)\) (\(i=1, 2,\ldots , p\)) and
\(H_{j}(t, s)\) (\(j=1, 2,\ldots \)) given in Lemma 2.11
have the following properties:
(i)
\(t^{\gamma -\nu _{n-2}-1}k_{0}(s)\leq K_{0}(t, s)\leq \frac{1}{ \varGamma (\gamma -\nu _{n-2})}t^{\gamma -\nu _{n-2}-1}\), where
$$ k_{0}(s)=\frac{1}{\varGamma (\gamma -\nu _{n-2})}(1-s)^{\gamma -\gamma _{0}-1} \bigl(1-(1-s)^{\gamma _{0}-\nu _{n-2}}\bigr). $$
(ii)
\(t^{\gamma -\alpha _{i}-1}k_{i}(s)\leq K_{i}(t, s)\leq \frac{a _{i}}{\sigma \varGamma (\gamma -\nu _{n-2})\varGamma (\gamma -\alpha _{i})}t ^{\gamma -\alpha _{i}-1}\) (\(i=1, 2,\ldots , p\)), where
$$ k_{i}(s)=\frac{a_{i}}{\sigma \varGamma (\gamma -\nu _{n-2})\varGamma (\gamma -\alpha _{i})}(1-s)^{\gamma -\gamma _{0}-1} \bigl(1-(1-s)^{\gamma _{0}-\alpha _{i}}\bigr). $$
(iii)
\(t^{\gamma -\beta _{j}-1}h_{j}(s)\leq H_{j}(t, s)\leq \frac{b _{j}}{\sigma \varGamma (\gamma -\nu _{n-2})\varGamma (\gamma -\beta _{j})}t ^{\gamma -\beta _{j}-1}\) (\(j=1, 2,\ldots \)), where
$$ h_{j}(s)=\frac{b_{j}}{\sigma \varGamma (\gamma -\nu _{n-2})\varGamma (\gamma -\beta _{j})}(1-s)^{\gamma -\gamma _{0}-1} \bigl(1-(1-s)^{\gamma _{0}-\beta _{j}}\bigr). $$
Proof
(i) For \(s\leq t\),
For \(t\leq s\),
Using the same argument again, it is straightforward to infer (ii) and (iii). The proof is complete. □
$$\begin{aligned}& \begin{aligned} K_{0}(t, s) &=\frac{1}{\varGamma (\gamma -\nu _{n-2})} \bigl(t^{\gamma - \nu _{n-2}-1}(1-s)^{\gamma -\gamma _{0}-1}-(t-s)^{\gamma -\nu _{n-2}-1} \bigr) \\ &\geq \frac{t^{\gamma -\nu _{n-2}-1}}{\varGamma (\gamma -\nu _{n-2})} \bigl((1-s)^{\gamma -\gamma _{0}-1}-(1-s)^{\gamma -\nu _{n-2}-1} \bigr) \\ &=\frac{t^{\gamma -\nu _{n-2}-1}(1-s)^{\gamma -\gamma _{0}-1}}{ \varGamma (\gamma -\nu _{n-2})}\bigl(1-(1-s)^{\gamma _{0}-\nu _{n-2}}\bigr) \\ &=t^{ \gamma -\nu _{n-2}-1}k_{0}(s), \end{aligned} \\& K_{0}(t, s)\leq \frac{t^{\gamma -\nu _{n-2}-1}}{\varGamma (\gamma -\nu _{n-2})}. \end{aligned}$$
$$\begin{aligned}& \begin{aligned} K_{0}(t, s) &= \frac{t^{\gamma -\nu _{n-2}-1}}{\varGamma (\gamma -\nu _{n-2})}(1-s)^{ \gamma -\gamma _{0}-1} \\ &\geq \frac{t^{\gamma -\nu _{n-2}-1}(1-s)^{ \gamma -\gamma _{0}-1}}{\varGamma (\gamma -\nu _{n-2})}\bigl(1-(1-s)^{\gamma _{0}-\nu _{n-2}}\bigr) \\ &=t^{\gamma -\nu _{n-2}-1}k_{0}(s), \end{aligned} \\& K_{0}(t, s)\leq \frac{t^{\gamma -\nu _{n-2}-1}}{\varGamma (\gamma -\nu _{n-2})}. \end{aligned}$$
Lemma 2.13
Let
\(\sigma >0\) (defined in (2.19) of Lemma 2.11), \(\int _{I_{i}}s^{\gamma -\alpha _{i}-1}w_{i}(s)\,dA_{i}(s)\geq 0\) (\(i=1, 2,\ldots , p\)), and
\(0<\sum_{j=1}^{\infty }\frac{b_{j}}{ \varGamma (\gamma -\beta _{j})}\xi _{j}^{\gamma -\beta _{j}-1}<\infty \). Then the Green’s function
\(K(t, s)\)
defined in Lemma 2.11
satisfies:
(i)
\(K(t, s)\leq Q_{1}e(t)\)
for
\(t, s\in [0, 1]\), where
\(e(t)=t ^{\gamma -\nu _{n-2}-1}\),
$$ \begin{aligned} Q_{1}&= \frac{1}{\varGamma (\gamma -\nu _{n-2})}+\sum _{i=1}^{p}\frac{a_{i}}{ \sigma \varGamma (\gamma -\nu _{n-2})\varGamma (\gamma - \alpha _{i})} \int _{I_{i}}\tau ^{\gamma -\alpha _{i}-1}w_{i}(\tau ) \,dA_{i}(\tau ) \\ &\quad{} + \sum_{j=1}^{\infty } \frac{b_{j}}{\sigma \varGamma (\gamma -\nu _{n-2}) \varGamma (\gamma -\beta _{j})}\xi _{j}^{\gamma -\beta _{j}-1}. \end{aligned} $$
(ii)
\(K(t, s)\geq Q_{2}(s)e(t)\)
for
\(t, s\in [0, 1]\), where
$$ Q_{2}(s)=k_{0}(s)+\sum_{i=1}^{p}k_{i}(s) \int _{I_{i}} \tau ^{\gamma -\alpha _{i}-1}w_{i}(\tau ) \,dA_{i}(\tau )+\sum_{j=1}^{ \infty }H_{j}( \xi _{j},s). $$
(iii)
\(K(t, s)>0\)
for
\(t, s\in (0, 1)\).
Proof
The conclusion can be easily given by Lemma 2.12. So we omit it. □
In this paper, we equip \(E=C[0, 1]\) with the norm \(\|u\|=\sup_{0\leq t\leq 1}|u(t)|\). Then \((E, \|\cdot \|)\) is a Banach space. Let \(P=\{u\in E:u(t)\geq 0, t\in [0, 1]\}\) be a cone in E. Let us define a nonlinear operator \(T_{\lambda }:P\rightarrow P\) by
It is easy to check that BVP (2.1) has a solution if and only if the operator \(T_{\lambda }\) has a fixed point.
$$ \begin{aligned}[b] &(T_{\lambda }u) (t)=\lambda \int _{0}^{1}K(t, s)f\bigl(s,I_{0+}^{\nu _{n-2}}u(s),I _{0+}^{\nu _{n-2}-\nu _{1}}u(s),\ldots ,I_{0+}^{\nu _{n-2}-\nu _{n-3}}u(s),u(s) \bigr)\,ds,\\ & \quad t\in [0, 1]. \end{aligned} $$
(2.20)
3 Main results
Theorem 3.1
Suppose that
\(f\in C((0,1)\times (0,+ \infty )^{n-1}, \mathbb{R}_{+})\)
satisfies:
Then BVP (1.1) has a unique solution
\(z_{\lambda }^{*}\)
in
P, and there exists a constant
\(\eta _{\lambda }\in (0, 1)\)
such that
And at the same time, \(z_{\lambda }^{*}\)
satisfies:
Moreover, for any initial values
\(z_{0}\), \(\tilde{z}_{0}\in P_{e}\), by constructing successively the sequences as follows:
we have
\(z_{n}\rightarrow z_{\lambda }^{*}\)
and
\(\tilde{z}_{n}\rightarrow z_{\lambda }^{*}\)
in
E, as
\(n\rightarrow \infty \).
\((\mathrm{H}_{1})\)
There exist two functions
\(f_{1}\), \(f_{2}\in C((0,1)\times (0,+\infty )^{2(n-1)},\mathbb{R}_{+})\)
such that
$$ \begin{aligned} f(t,x_{1},x_{2},\ldots ,x_{n-1})&= f_{1}(t,x_{1},x_{2}, \ldots ,x_{n-1},x _{1},x_{2},\ldots ,x_{n-1}) \\ &\quad{} +f_{2}(t,x_{1},x_{2},\ldots ,x_{n-1},x _{1},x_{2},\ldots ,x_{n-1}). \end{aligned} $$
\((\mathrm{H}_{2})\)
For all
\(t\in (0,1)\)
and
\((y_{1}, y_{2}, \ldots , y_{n-1})\in (0,+\infty )^{n-1}\), \(f_{1}(t, x _{1}, x_{2}, \ldots , x_{n-1}, y_{1}, y_{2}, [4] \ldots , y_{n-1})\), \(f_{2}(t, x_{1}, x_{2}, \ldots , x_{n-1}, y_{1}, y_{2},\ldots , y_{n-1})\)
are increasing in
\((x_{1}, x_{2}, \ldots , x_{n-1})\in (0,+\infty )^{n-1}\); for all
\(t\in (0,1)\)
and
\((x_{1}, x_{2}, \ldots , x_{n-1})\in (0,+\infty )^{n-1}\), \(f_{1}(t, x_{1}, x_{2}, \ldots , x _{n-1}, y_{1}, y_{2},\ldots , y_{n-1})\), \(f_{2}(t, x_{1}, x_{2}, \ldots , x_{n-1}, y_{1}, y_{2},\ldots , y_{n-1})\)
are decreasing in
\((y_{1}, y_{2}, \ldots , y_{n-1})\in (0,+\infty )^{n-1}\).
\((\mathrm{H}_{3})\)
For all
\(\mu \in (0,1)\), there exists
\(\varphi (\mu )\in (\mu , 1]\)
such that, for all
\(t\in (0,1)\)
and
\((x_{1}, x_{2}, \ldots , x_{n-1})\), \((y_{1}, y_{2}, \ldots , y_{n-1})\in (0,+\infty )^{n-1}\),
$$\begin{aligned}& \begin{aligned} &f_{1}\bigl(t, \mu x_{1}, \mu x_{2}, \ldots , \mu x_{n-1}, \mu ^{-1} y_{1}, \mu ^{-1} y_{2},\ldots , \mu ^{-1} y_{n-1}\bigr) \\ &\quad \geq \varphi (\mu ) f _{1}(t, x_{1}, x_{2}, \ldots , x_{n-1}, y_{1}, y_{2},\ldots , y_{n-1}), \end{aligned} \\& \begin{aligned} &f_{2}\bigl(t, \mu x_{1}, \mu x_{2}, \ldots , \mu x_{n-1}, \mu ^{-1} y_{1}, \mu ^{-1} y_{2},\ldots , \mu ^{-1} y_{n-1}\bigr) \\ &\quad \geq \mu f_{2}(t, x_{1}, x_{2}, \ldots , x_{n-1}, y_{1}, y_{2},\ldots , y_{n-1}). \end{aligned} \end{aligned}$$
\((\mathrm{H}_{4})\)
There exists a constant
\(\kappa >0\)
such that, for all
\(t\in (0,1)\)
and
\((x_{1}, x_{2}, \ldots , x_{n-1})\), \((y_{1}, y_{2}, \ldots , y_{n-1})\in (0,+\infty )^{n-1}\),
$$ f_{1}(t, x_{1}, x_{2}, \ldots , x_{n-1}, y_{1}, y_{2},\ldots , y_{n-1}) \geq \kappa f_{2}(t, x_{1}, x_{2}, \ldots , x_{n-1}, y_{1}, y_{2}, \ldots , y_{n-1}). $$
\((\mathrm{H}_{5})\)
The functions
\(f_{1}\)
and
\(f_{2}\)
satisfy
$$\begin{aligned}& 0< \int _{0}^{1}f_{1}\bigl(s,1,1, \ldots ,1,s^{\gamma -1},s^{\gamma -1}, \ldots ,s^{\gamma -1}\bigr) \,ds< +\infty , \\& 0< \int _{0}^{1}f_{2}\bigl(s,1,1, \ldots ,1,s^{\gamma -1},s^{\gamma -1}, \ldots ,s^{\gamma -1}\bigr) \,ds< +\infty . \end{aligned}$$
$$ \frac{\eta _{\lambda }\varGamma (\gamma -\nu _{n-2})}{\varGamma (\gamma )}t ^{\gamma -1} \leq z_{\lambda }^{*}(t) \leq \frac{\varGamma (\gamma -\nu _{n-2})}{\eta _{\lambda }\varGamma (\gamma )}t^{\gamma -1},\quad t\in [0, 1]. $$
(i)
If there exists
\(r\in (0, 1)\)
such that
then
\(z_{\lambda }^{*}\)
is continuous with respect to
\(\lambda \in (0,+ \infty )\), i.e., for ∀ \(\lambda _{0}\in (0,+\infty )\),
$$ \varphi (\mu )\geq \frac{\mu ^{r}-\mu }{\kappa }+\mu ^{r},\quad \forall \mu \in (0, 1), $$
$$ \bigl\Vert z_{\lambda }^{*}-z_{\lambda _{0}}^{*} \bigr\Vert \rightarrow 0, \quad \textit{as } \lambda \rightarrow \lambda _{0}. $$
(ii)
If
then
\(0<\lambda _{1}< \lambda _{2}\)
implies
\(z_{\lambda _{1}}^{*}< z_{ \lambda _{2}}^{*}\).
$$ \varphi (\mu )\geq \frac{\mu ^{\frac{1}{2}}-\mu }{\kappa }+ \mu ^{\frac{1}{2}},\quad \forall \mu \in (0, 1), $$
(iii)
If there exists
\(r\in (0, \frac{1}{2})\)
such that
then
$$ \varphi (\mu )\geq \frac{\mu ^{r}-\mu }{\kappa }+\mu ^{r},\quad \forall \mu \in (0, 1), $$
$$ \lim_{\lambda \rightarrow 0^{+}} \bigl\Vert z_{\lambda }^{*} \bigr\Vert =0, \qquad \lim_{\lambda \rightarrow +\infty } \bigl\Vert z_{\lambda }^{*} \bigr\Vert =+\infty . $$
$$\begin{aligned}& \begin{aligned} z_{n}(t)&= I_{0^{+}}^{\nu _{n-2}} \biggl\{ \lambda \int _{0}^{1}K(t,s)f _{1} \bigl(s,I_{0^{+}}^{\nu _{n-2}}z_{n-1}(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}}z _{n-1}(s),\ldots , z_{n-1}(s), \\ & \quad I_{0^{+}}^{\nu _{n-2}}\tilde{z}_{n-1}(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}} \tilde{z}_{n-1}(s),\ldots , \tilde{z}_{n-1}(s)\bigr)\,ds \\ &\quad{} +\lambda \int _{0}^{1}K(t,s)f_{2} \bigl(s,I_{0^{+}}^{\nu _{n-2}}z_{n-1}(s),I_{0^{+}} ^{\nu _{n-2}-\nu _{1}}z_{n-1}(s),\ldots , z_{n-1}(s), \\ & \quad I_{0^{+}}^{\nu _{n-2}}\tilde{z}_{n-1}(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}} \tilde{z}_{n-1}(s),\ldots , \tilde{z}_{n-1}(s)\bigr)\,ds \biggr\} , \end{aligned} \\ & \begin{aligned} \tilde{z}_{n}(t)&= I_{0^{+}}^{\nu _{n-2}} \biggl\{ \lambda \int _{0}^{1}K(t,s)f _{1} \bigl(s,I_{0^{+}}^{\nu _{n-2}}\tilde{z}_{n-1}(s),I_{0^{+}}^{\nu _{n-2}- \nu _{1}} \tilde{z}_{n-1}(s),\ldots , \tilde{z}_{n-1}(s), \\ & \quad I_{0^{+}}^{\nu _{n-2}}z_{n-1}(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}}z_{n-1}(s), \ldots , z_{n-1}(s)\bigr)\,ds \\ &\quad{} +\lambda \int _{0}^{1}K(t,s)f_{2}\bigl(s,I _{0^{+}}^{\nu _{n-2}}\tilde{z}_{n-1}(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}} \tilde{z}_{n-1}(s),\ldots , \tilde{z}_{n-1}(s), \\ & \quad I_{0^{+}}^{\nu _{n-2}} z_{n-1}(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}} z_{n-1}(s), \ldots , z_{n-1}(s)\bigr)\,ds \biggr\} , \\ & n=1,2,\ldots , \end{aligned} \end{aligned}$$
Proof
Let \(P_{e}=\{u\in E: u\sim e\}\), where \(e(t)=t^{ \gamma -\nu _{n-2}-1}\). Then \(P_{e}\) is a component of P. Now, we define two operators \(A_{\lambda }, B_{\lambda }: P_{e}\times P_{e} \rightarrow P\) by
Combining the definition of \(T_{\lambda }\) in (2.20) and \((H_{1})\), we have
Then we can conclude that u is the solution of BVP (2.1) if u satisfies \(u=A_{\lambda }(u, u)+B_{\lambda }(u, u)\).
$$\begin{aligned}& \begin{aligned} A_{\lambda }(u, v) (t)&=\lambda \int _{0}^{1}K(t,s) f_{1} \bigl(s,I_{0^{+}} ^{\nu _{n-2}}u(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}}u(s), \ldots , u(s), \\ &\quad I_{0^{+}}^{\nu _{n-2}}v(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}}v(s), \ldots , v(s) \bigr)\,ds, \end{aligned} \\ & \begin{aligned} B_{\lambda }(u, v) (t)&=\lambda \int _{0}^{1}K(t,s) f_{2} \bigl(s,I_{0^{+}} ^{\nu _{n-2}}u(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}}u(s), \ldots , u(s), \\ &\quad I_{0^{+}}^{\nu _{n-2}}v(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}}v(s), \ldots , v(s) \bigr)\,ds. \end{aligned} \end{aligned}$$
$$ \begin{aligned}[b] (T_{\lambda }u) (t)&= \lambda \int _{0}^{1}K(t, s)f\bigl(s,I_{0+}^{\nu _{n-2}}u(s),I _{0+}^{\nu _{n-2}-\nu _{1}}u(s),\ldots ,I_{0+}^{\nu _{n-2}-\nu _{n-3}}u(s),u(s) \bigr)\,ds \\ &=\lambda \int _{0}^{1}K(t,s)f_{1} \bigl(s,I_{0^{+}}^{\nu _{n-2}}u(s),I _{0^{+}}^{\nu _{n-2}-\nu _{1}}u(s), \ldots , u(s), \\ & \quad I_{0^{+}}^{\nu _{n-2}}u(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}}u(s), \ldots , u(s)\bigr)\,ds \\ &\quad{} +\lambda \int _{0}^{1}K(t,s)f_{2} \bigl(s,I_{0^{+}}^{\nu _{n-2}}u(s),I _{0^{+}}^{\nu _{n-2}-\nu _{1}}u(s), \ldots , u(s), \\ & \quad I_{0^{+}}^{\nu _{n-2}}u(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}}u(s), \ldots , u(s)\bigr)\,ds \\ &=A_{\lambda }(u, u) (t)+B_{\lambda }(u, u) (t), \quad t \in [0, 1]. \end{aligned} $$
(3.1)
We prove that \(A_{\lambda }, B_{\lambda }: P_{e}\times P_{e}\rightarrow P \) are well defined at first. For any \(u, v\in P_{e}\), there exists a constant \(\omega \in (0,1)\) such that \(\omega e(t)\leq u(t)\leq \frac{1}{ \omega }e(t)\), \(\omega e(t)\leq v(t)\leq \frac{1}{\omega }e(t)\), \(t\in [0, 1]\). Moreover, by the definition of fractional integral and \(e(t)\leq 1\), for all \(t\in [0,1]\),
and
Thus, by \(\mathrm{(H_{1})}\), \(\mathrm{(H_{2})}\), \(\mathrm{(H_{3})}\), \(\mathrm{(H_{5})}\), (3.2), and (3.3), we know that, for all \(t\in [0,1]\),
where
Similarly, for all \(t\in [0,1]\),
So, \(A_{\lambda }, B_{\lambda }: P_{e}\times P_{e}\rightarrow P\) are well defined.
$$ \begin{aligned}[b] I_{0^{+}}^{\nu _{n-2}}e(t)&= \frac{1}{\varGamma (\nu _{n-2})} \int _{0}^{t}(t-s)^{ \nu _{n-2}-1}s^{\gamma -\nu _{n-2}-1} \,ds \\ &=\frac{\varGamma (\gamma -\nu _{n-2})}{\varGamma (\gamma )}t^{\gamma -1}\leq 1 \end{aligned} $$
(3.2)
$$ \begin{aligned}[b] I_{0^{+}}^{\nu _{n-2}-\nu _{i}}e(t)& = \frac{1}{\varGamma (\nu _{n-2}-\nu _{i})} \int _{0}^{t}(t-s)^{\nu _{n-2}-\nu _{i}-1}s^{\gamma -\nu _{n-2}-1} \,ds \\ &=\frac{\varGamma (\gamma -\nu _{n-2})}{ \varGamma (\gamma -\nu _{i})}t^{\gamma -\nu _{i}-1}\leq 1, \quad i=1,2,\ldots ,n-3. \end{aligned} $$
(3.3)
$$ \begin{aligned}[b] A_{\lambda }(u, v) (t)&= \lambda \int _{0}^{1}K(t,s) f_{1} \bigl(s,I_{0^{+}} ^{\nu _{n-2}}u(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}}u(s), \ldots , u(s), \\ &\quad I_{0^{+}}^{\nu _{n-2}}v(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}}v(s), \ldots , v(s) \bigr)\,ds \\ &\leq \lambda \int _{0}^{1}K(t,s)f _{1} \bigl(s,I_{0^{+}}^{\nu _{n-2}}\omega ^{-1}e(s),I_{0^{+}}^{\nu _{n-2}- \nu _{1}} \omega ^{-1}e(s),\ldots ,\omega ^{-1}e(s), \\ &\quad I_{0^{+}}^{\nu _{n-2}}\omega e(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}} \omega e(s),\ldots ,\omega e(s) \bigr)\,ds \\ &\leq \lambda \int _{0} ^{1}K(t,s) f_{1} \biggl(s, \omega ^{-1} , \omega ^{-1},\ldots , \omega ^{-1}, \\ &\quad \frac{\varGamma (\gamma -\nu _{n-2})}{ \varGamma (\gamma )}\omega s^{\gamma -1}, \frac{\varGamma (\gamma -\nu _{n-2})}{ \varGamma (\gamma -\nu _{1})} \omega s^{\gamma -\nu _{1}-1},\ldots ,\omega s^{\gamma -\nu _{n-2}-1} \biggr)\,ds \\ &\leq \lambda \int _{0}^{1}K(t,s) f _{1} \bigl(s,(\rho \omega )^{-1}, (\rho \omega )^{-1},\ldots , ( \rho \omega )^{-1}, \\ &\quad \rho \omega s^{\gamma -1}, \rho \omega s^{\gamma -\nu _{1}-1},\ldots , \omega s^{\gamma -\nu _{n-3}-1} \bigr)\,ds \\ &\leq \lambda \frac{Q_{1}}{\varphi (\rho \omega )}e(t) \int _{0}^{1}f _{1}\bigl(s,1,1, \ldots ,1,s^{\gamma -1},s^{\gamma -1},\ldots ,s^{\gamma -1}\bigr) \,ds \\ &< +\infty , \end{aligned} $$
(3.4)
$$ \rho =\min \biggl\{ \frac{\varGamma (\gamma -\nu _{n-2})}{\varGamma (\gamma )},\frac{ \varGamma (\gamma -\nu _{n-2})}{\varGamma (\gamma -\nu _{1})}, \ldots , \frac{ \varGamma (\gamma -\nu _{n-2})}{\varGamma (\gamma -\nu _{n-3})}, 1 \biggr\} >0. $$
$$ B_{\lambda }(u, v) (t)\leq \lambda \frac{Q_{1}}{\rho \omega }e(t) \int _{0}^{1}f_{2}\bigl(s,1,1, \ldots ,1,s^{\gamma -1},s^{\gamma -1},\ldots ,s ^{\gamma -1}\bigr) \,ds< +\infty . $$
(3.5)
Now, we prove that \(A_{\lambda }, B_{\lambda }: P_{e}\times P_{e} \rightarrow P_{e}\). Taking a constant \(W>1\) such that
Then from \(\mathrm{(H_{1})}\), \(\mathrm{(H_{2}),}\) and \(\mathrm{(H_{3})}\), for all \(t\in [0,1]\),
and
On the other hand, from (3.4) and (3.5), we know, for all u, \(v\in P_{e}\), \(t\in [0, 1]\),
and
So, \(A_{\lambda }, B_{\lambda }: P_{e}\times P_{e}\rightarrow P_{e}\).
$$ \begin{aligned}[b] W&>\max \biggl\{ \frac{\lambda Q_{1}}{\varphi (\rho \omega )} \int _{0} ^{1}f_{1}\bigl(s,1,1, \ldots ,1,s^{\gamma -1},s^{\gamma -1},\ldots ,s^{ \gamma -1}\bigr) \,ds, \\ &\quad \frac{\lambda Q_{1}}{\rho \omega } \int _{0} ^{1}f_{2}\bigl(s,1,1, \ldots ,1,s^{\gamma -1},s^{\gamma -1},\ldots ,s^{ \gamma -1}\bigr) \,ds, \\ &\quad \biggl(\lambda \varphi (\rho \omega ) \int _{0}^{1}Q _{2}(s)f_{1} \bigl(s,s^{\gamma -1},s^{\gamma -1},\ldots ,s^{\gamma -1},1,1, \ldots ,1\bigr)\,ds \biggr)^{-1}, \\ &\quad \biggl(\lambda \rho \omega \int _{0}^{1}Q_{2}(s)f_{2} \bigl(s,s^{\gamma -1},s^{\gamma -1},\ldots ,s^{\gamma -1},1,1, \ldots ,1\bigr)\,ds \biggr)^{-1} \biggr\} . \end{aligned} $$
(3.6)
$$ \begin{aligned}[b] A_{\lambda }(u, v) (t)&= \lambda \int _{0}^{1}K(t,s) f_{1} \bigl(s,I_{0^{+}} ^{\nu _{n-2}}u(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}}u(s), \ldots , u(s), \\ &\quad I_{0^{+}}^{\nu _{n-2}}v(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}}v(s), \ldots , v(s) \bigr)\,ds \\ &\geq \lambda \int _{0}^{1}K(t,s)f _{1} \bigl(s,I_{0^{+}}^{\nu _{n-2}}\omega e(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}} \omega e(s),\ldots ,\omega e(s), \\ &\quad I_{0^{+}} ^{\nu _{n-2}}\omega ^{-1} e(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}}\omega ^{-1} e(s),\ldots , \omega ^{-1} e(s) \bigr)\,ds \\ &\geq \lambda \int _{0}^{1}K(t,s)f _{1} \bigl(s,\rho \omega s^{\gamma -1}, \rho \omega s^{\gamma -\nu _{1}-1},\ldots , \omega s^{\gamma -\nu -1}, \\ &\quad ( \rho \omega )^{-1}, (\rho \omega )^{-1},\ldots , (\rho \omega )^{-1} \bigr)\,ds \\ &\geq \lambda \varphi (\rho \omega )e(t) \int _{0}^{1}Q_{2}(s)f_{1} \bigl(s,s^{\gamma -1},s^{\gamma -1},\ldots ,s ^{\gamma -1},1,1, \ldots ,1\bigr)\,ds \\ &\geq W^{-1}e(t) \end{aligned} $$
(3.7)
$$ \begin{aligned}[b] B_{\lambda }(u, v) (t) &\geq \lambda \rho \omega e(t) \int _{0}^{1}Q_{2}(s)f _{2}\bigl(s,s^{\gamma -1},s^{\gamma -1},\ldots ,s^{\gamma -1},1,1,\ldots ,1\bigr)\,ds \\ &\geq W^{-1}e(t). \end{aligned} $$
(3.8)
$$ \begin{aligned}[b] A_{\lambda }(u, v) (t) &\leq \frac{\lambda Q_{1}}{\varphi (\rho \omega )}e(t) \int _{0}^{1}f_{1}\bigl(s,1,1, \ldots ,1,s^{\gamma -1},s^{\gamma -1}, \ldots ,s^{\gamma -1}\bigr) \,ds \\ &\leq We(t) \end{aligned} $$
(3.9)
$$ \begin{aligned}[b] B_{\lambda }(u, v) (t)&\leq \frac{\lambda Q_{1}}{\rho \omega }e(t) \int _{0}^{1}f_{2}\bigl(s,1,1, \ldots ,1,s^{\gamma -1},s^{\gamma -1},\ldots ,s ^{\gamma -1}\bigr) \,ds \\ &\leq We(t). \end{aligned} $$
(3.10)
In fact, from \(\mathrm{(H_{2})}\), it is easy to check that \(A_{\lambda }\), \(B_{\lambda }\) are mixed monotone operators. Furthermore, it follows from \(\mathrm{(H_{3})}\) that, for all \(\mu \in (0,1)\), there exists \(\varphi (\mu )\in (\mu , 1]\) such that, for any \(u, v\in P_{e}\), \(t\in [0, 1]\),
and
By \(\mathrm{(H_{4})}\), we infer that there exists \(\kappa >0\) such that, for all u, \(v\in P_{e}\), \(t\in [0, 1]\),
Combining (3.11)–(3.13) and using Lemma 2.2, we infer that there exists a unique fixed point \(u^{*}_{\lambda }\in P_{e}\) such that
That is, BVP (2.1) has a unique solution \(u^{*}_{\lambda }\in P_{e}\). Since \(u^{*}_{\lambda }\in P_{e}\), there exists a constant \(\eta _{\lambda }\in (0, 1)\) such that
Moreover, for any \(\lambda >0\), let \(\tilde{A}= (\lambda ^{-1} A_{ \lambda })\) and \(\tilde{B}= (\lambda ^{-1} B_{\lambda })\). Obviously, Ã and B̃ satisfy all the conditions of Lemma 2.3. With the preceding proof, we can infer that \(u^{*}_{\lambda }\) is the unique positive solution of the following equation:
By means of Lemma 2.3, we know that \(u^{*}_{\lambda }\) satisfies: Furthermore, by Lemma 2.2, we can infer that, for any initial values \(u_{0}, v_{0}\in P_{e}\), by constructing successively the sequences as follows:
we have
$$ \begin{aligned}[b] A_{\lambda }\bigl(\mu u, \mu ^{-1}v\bigr) (t)&= \lambda \int _{0}^{1}K(t,s) f_{1} \bigl(s,I_{0^{+}}^{\nu _{n-2}}\mu u(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}} \mu u(s),\ldots , \mu u(s), \\ &\quad I_{0^{+}}^{\nu _{n-2}}\mu ^{-1}v(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}}\mu ^{-1}v(s),\ldots , \mu ^{-1}v(s) \bigr)\,ds \\ &=\lambda \int _{0}^{1}K(t,s) f_{1} \bigl(s, \mu I_{0^{+}}^{\nu _{n-2}} u(s), \mu I_{0^{+}}^{\nu _{n-2}-\nu _{1}}u(s), \ldots , \mu u(s), \\ &\quad \mu ^{-1}I_{0^{+}}^{\nu _{n-2}}v(s), \mu ^{-1}I_{0^{+}} ^{\nu _{n-2}-\nu _{1}}v(s),\ldots , \mu ^{-1}v(s) \bigr)\,ds \\ &\geq \lambda \varphi (\mu ) \int _{0}^{1}K(t,s) f_{1} \bigl(s,I_{0^{+}}^{ \nu _{n-2}}u(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}}u(s), \ldots , u(s), \\ &\quad I_{0^{+}}^{\nu _{n-2}}v(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}}v(s), \ldots , v(s) \bigr)\,ds \\ &= \varphi (\mu ) A_{\lambda }(u, v) (t) \end{aligned} $$
(3.11)
$$ \begin{aligned}[b] B_{\lambda }\bigl(\mu u, \mu ^{-1}v\bigr) (t)&= \lambda \int _{0}^{1}K(t,s) f_{2} \bigl(s,I_{0^{+}}^{\nu _{n-2}}\mu u(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}} \mu u(s),\ldots , \mu u(s), \\ &\quad I_{0^{+}}^{\nu _{n-2}}\mu ^{-1}v(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}}\mu ^{-1}v(s),\ldots , \mu ^{-1}v(s) \bigr)\,ds \\ &\geq \lambda \mu \int _{0}^{1}K(t,s) f_{2} \bigl(s,I_{0^{+}}^{\nu _{n-2}}u(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}}u(s), \ldots , u(s), \\ &\quad I_{0^{+}}^{\nu _{n-2}}v(s), I _{0^{+}}^{\nu _{n-2}-\nu _{1}}v(s), \ldots , v(s) \bigr)\,ds \\ &=\mu B _{\lambda }(u, v) (t). \end{aligned} $$
(3.12)
$$ \begin{aligned}[b] A_{\lambda }(u, v) (t)&= \lambda \int _{0}^{1}K(t,s) f_{1} \bigl(s,I_{0^{+}} ^{\nu _{n-2}}u(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}}u(s), \ldots , u(s), \\ &\quad I_{0^{+}}^{\nu _{n-2}}v(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}}v(s), \ldots , v(s) \bigr)\,ds \\ &\geq \kappa \lambda \int _{0}^{1}K(t,s)f _{2} \bigl(s,I_{0^{+}}^{\nu _{n-2}}u(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}}z(s), \ldots , u(s), \\ &\quad I_{0^{+}}^{\nu _{n-2}}v(s), I _{0^{+}}^{\nu _{n-2}-\nu _{1}}v(s), \ldots , v(s) \bigr)\,ds \\ &=\kappa B _{\lambda }(u, v) (t). \end{aligned} $$
(3.13)
$$ A_{\lambda }\bigl(u^{*}_{\lambda }, u^{*}_{\lambda }\bigr)+B_{\lambda } \bigl(u^{*} _{\lambda }, u^{*}_{\lambda } \bigr)=u^{*}_{\lambda }. $$
$$ \eta _{\lambda } t^{\gamma -\nu _{n-2}-1} \leq u^{*}_{\lambda }(t) \leq \frac{1}{\eta _{\lambda } }t^{\gamma -\nu _{n-2}-1},\quad t\in [0, 1]. $$
(3.14)
$$ \lambda \tilde{A}(u, u)+\lambda \tilde{B}(u, u)=A(u, u)+ B(u, u)= u. $$
(1)
If there exists \(r\in (0, 1)\) such that
then \(u^{*}_{\lambda }\) is continuous with respect to \(\lambda \in (0, +\infty )\). That is, for ∀ \(\lambda _{0}\in (0,+\infty )\),
$$ \varphi (\mu )\geq \frac{\mu ^{r}-\mu }{\kappa }+\mu ^{r},\quad \forall \mu \in (0, 1), $$
$$ \bigl\Vert u^{*}_{\lambda }-u^{*}_{\lambda _{0}} \bigr\Vert \rightarrow 0, \quad \text{as } \lambda \rightarrow \lambda _{0}. $$
(2)
If
then \(0<\lambda _{1}< \lambda _{2}\) implies \(u^{*}_{\lambda _{1}}< u^{*} _{\lambda _{2}}\).
$$ \varphi (\mu )\geq \frac{\mu ^{\frac{1}{2}}-\mu }{\kappa }+ \mu ^{\frac{1}{2}},\quad \forall \mu \in (0, 1), $$
(3)
If there exists \(r\in (0, \frac{1}{2})\) such that
then
$$ \varphi (\mu )\geq \frac{\mu ^{r}-\mu }{\kappa }+\mu ^{r},\quad \forall \mu \in (0, 1), $$
$$ \lim_{\lambda \rightarrow 0^{+}} \bigl\Vert u^{*}_{\lambda } \bigr\Vert =0, \qquad \lim_{\lambda \rightarrow +\infty } \bigl\Vert u^{*}_{\lambda } \bigr\Vert =+\infty . $$
$$\begin{aligned}& \begin{aligned} u_{n}(t)&= \lambda \int _{0}^{1}K(t,s)f_{1} \bigl(s,I_{0^{+}}^{\nu _{n-2}}u _{n-1}(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}}u_{n-1}(s), \ldots , u_{n-1}(s), \\ & \quad I_{0^{+}}^{\nu _{n-2}}v_{n-1}(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}}v_{n-1}(s), \ldots , v_{n-1}(s)\bigr)\,ds \\ &\quad{} +\lambda \int _{0}^{1}K(t,s)f_{2} \bigl(s,I_{0^{+}} ^{\nu _{n-2}}u_{n-1}(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}}u_{n-1}(s), \ldots , u_{n-1}(s), \\ & \quad I_{0^{+}}^{\nu _{n-2}}v_{n-1}(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}}v_{n-1}(s), \ldots , v_{n-1}(s)\bigr)\,ds, \end{aligned} \\& \begin{aligned} v_{n}(t)&= \lambda \int _{0}^{1}K(t,s)f_{1} \bigl(s,I_{0^{+}}^{\nu _{n-2}}v _{n-1}(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}}v_{n-1}(s), \ldots , v_{n-1}(s), \\ & \quad I_{0^{+}}^{\nu _{n-2}}u_{n-1}(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}}u_{n-1}(s), \ldots , u_{n-1}(s)\bigr)\,ds \\ &\quad{} +\lambda \int _{0}^{1}K(t,s)f_{2} \bigl(s,I_{0^{+}} ^{\nu _{n-2}}v_{n-1}(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}}v_{n-1}(s), \ldots , v_{n-1}(s), \\ & \quad I_{0^{+}}^{\nu _{n-2}}u_{n-1}(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}}u_{n-1}(s), \ldots , u_{n-1}(s)\bigr)\,ds, \\ &\quad t\in [0, 1],\quad n=1,2,\ldots , \end{aligned} \end{aligned}$$
$$ u_{n}\rightarrow u^{*}_{\lambda },\qquad v_{n}\rightarrow u^{*}_{ \lambda }, \quad \text{in } E, \text{ as }n\rightarrow \infty . $$
Finally, by what we have proved in Lemma 2.9, we know \(z^{*}_{\lambda }=I^{\nu _{n-2}}u^{*}_{\lambda }\) is the unique positive solution of BVP (1.1). From (3.14), we know that \(z^{*}\) satisfies
And from the monotonicity and continuity of fractional integral, we get: Furthermore, for any initial values \(z_{0}, \tilde{z}_{0}\in P_{e}\), by constructing successively the sequences as follows:
we have
The proof of Theorem 3.1 is completed. □
$$ \frac{\eta _{\lambda }\varGamma (\gamma -\nu _{n-2})}{\varGamma (\gamma )}t ^{\gamma -1} \leq z^{*}_{\lambda }(t) \leq \frac{\varGamma (\gamma -\nu _{n-2})}{\eta _{\lambda }\varGamma (\gamma )}t^{\gamma -1},\quad t\in [0, 1]. $$
(3.15)
(i)
If there exists \(r\in (0, 1)\) such that
then for ∀ \(\lambda _{0}\in (0,+\infty )\),
$$ \varphi (\mu )\geq \frac{\mu ^{r}-\mu }{\kappa }+\mu ^{r},\quad \forall \mu \in (0, 1), $$
$$ \bigl\Vert z^{*}_{\lambda }-z^{*}_{\lambda _{0}} \bigr\Vert \rightarrow 0, \quad \text{as } \lambda \rightarrow \lambda _{0}. $$
(ii)
If
then \(0<\lambda _{1}< \lambda _{2}\) implies \(z^{*}_{\lambda _{1}}< z^{*} _{\lambda _{2}}\).
$$ \varphi (\mu )\geq \frac{\mu ^{\frac{1}{2}}-\mu }{\kappa }+ \mu ^{\frac{1}{2}},\quad \forall \mu \in (0, 1), $$
(iii)
If there exists \(r\in (0, \frac{1}{2})\) such that
then
$$ \varphi (\mu )\geq \frac{\mu ^{r}-\mu }{\kappa }+\mu ^{r},\quad \forall \mu \in (0, 1), $$
$$ \lim_{\lambda \rightarrow 0^{+}} \bigl\Vert z^{*}_{\lambda } \bigr\Vert =0, \qquad \lim_{\lambda \rightarrow +\infty } \bigl\Vert z^{*}_{\lambda } \bigr\Vert =+\infty . $$
$$\begin{aligned}& \begin{aligned} z_{n}(t)&= I_{0^{+}}^{\nu _{n-2}} \biggl\{ \lambda \int _{0}^{1}K(t,s)f _{1} \bigl(s,I_{0^{+}}^{\nu _{n-2}}z_{n-1}(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}}z _{n-1}(s),\ldots , z_{n-1}(s), \\ & \quad I_{0^{+}}^{\nu _{n-2}}\tilde{z}_{n-1}(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}} \tilde{z}_{n-1}(s),\ldots , \tilde{z}_{n-1}(s)\bigr)\,ds \\ &\quad{} +\lambda \int _{0}^{1}K(t,s)f_{2} \bigl(s,I_{0^{+}}^{\nu _{n-2}}z_{n-1}(s),I_{0^{+}} ^{\nu _{n-2}-\nu _{1}}z_{n-1}(s),\ldots , z_{n-1}(s), \\ & \quad I_{0^{+}}^{\nu _{n-2}}\tilde{z}_{n-1}(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}} \tilde{z}_{n-1}(s),\ldots , \tilde{z}_{n-1}(s)\bigr)\,ds \biggr\} , \end{aligned} \\& \begin{aligned} \tilde{z}_{n}(t)&= I_{0^{+}}^{\nu _{n-2}} \biggl\{ \lambda \int _{0}^{1}K(t,s)f _{1} \bigl(s,I_{0^{+}}^{\nu _{n-2}}\tilde{z}_{n-1}(s),I_{0^{+}}^{\nu _{n-2}- \nu _{1}} \tilde{z}_{n-1}(s),\ldots , \tilde{z}_{n-1}(s), \\ & \quad I_{0^{+}}^{\nu _{n-2}}z_{n-1}(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}}z_{n-1}(s), \ldots , z_{n-1}(s)\bigr)\,ds \\ &\quad{} +\lambda \int _{0}^{1}K(t,s)f_{2}\bigl(s,I _{0^{+}}^{\nu _{n-2}}\tilde{z}_{n-1}(s),I_{0^{+}}^{\nu _{n-2}-\nu _{1}} \tilde{z}_{n-1}(s),\ldots , \tilde{z}_{n-1}(s), \\ & \quad I_{0^{+}}^{\nu _{n-2}} z_{n-1}(s), I_{0^{+}}^{\nu _{n-2}-\nu _{1}} z_{n-1}(s), \ldots , z_{n-1}(s)\bigr)\,ds \biggr\} , \\ & \quad t\in [0, 1],\quad n=1,2,\ldots , \end{aligned} \end{aligned}$$
$$ z_{n}\rightarrow z^{*}_{\lambda },\qquad \tilde{z}_{n}\rightarrow z ^{*}_{\lambda }, \quad \text{in } E, \text{ as } n\rightarrow \infty . $$
4 Examples
Example 4.1
We consider the following problem:
where \(\lambda >0\) is a parameter, and
$$ \textstyle\begin{cases} D_{0+}^{\frac{5}{2}}z(t)+ \lambda (6t^{-\frac{1}{3}}u^{ \frac{1}{8}} (D_{0+}^{\frac{3}{5}}z(t) )^{-\frac{1}{8}}+ 5u ^{\frac{1}{6}} (1-t^{2})^{-\frac{1}{2}} +(1+t^{2})^{-1} \\ \quad {}\times ( (6D_{0+}^{\frac{3}{5}}z(t) )^{ \frac{1}{3}}+ (D_{0+}^{\frac{3}{5}}z(t) )^{\frac{1}{4}}+1 ) + (tu)^{-\frac{1}{5}} (7+ (u+1)^{-\frac{4}{5}} ) )=0, \quad 0< t< 1, \\ z(0)= D_{0+}^{\frac{2}{3}}z(0)=0, \\ D_{0+}^{ \frac{3}{2}}z(1)=2\int _{0}^{1}s^{\frac{3}{4}}(1-s)^{2}D_{0+}^{ \frac{5}{4}}z(s)\,dA_{1}(s) +\frac{1}{2}\int _{0}^{\frac{2}{3}}s^{ \frac{7}{8}}(1+s^{2})^{-1}D_{0+}^{\frac{11}{8}}z(s)\,dA_{2}(s)\\ \hphantom{D_{0+}^{ \frac{3}{2}}z(1)=}{}+\sum_{j=1}^{\infty }(5j-4)^{-1}(5j+1)^{-1}D_{0+}^{\frac{3}{2}- \frac{1}{2^{(7+j)}}}u ((28+2j)^{-1} ), \end{cases} $$
(4.1)
$$ A_{1}(t)= \textstyle\begin{cases} \frac{1}{7}, & t\in [0, \frac{1}{2} ), \\ \frac{8}{7}, & t\in [\frac{1}{2}, 1 ], \end{cases}\displaystyle \qquad A_{2}(t)= \textstyle\begin{cases} \frac{1}{9}, & t\in [0, \frac{1}{2} ), \\ \frac{10}{9}, & t\in [\frac{1}{2}, 1 ]. \end{cases} $$
Let
\(n=3\), \(\gamma =\frac{5}{2}\), \(\gamma _{0}=\frac{3}{2}\), \(\nu _{1}= \frac{3}{5}\), \(q_{1}=\frac{2}{3}\), \(a_{1}=2\), \(a_{2}=\frac{1}{2}\), \(\alpha _{1}=\frac{5}{4}\), \(\alpha _{2}=\frac{11}{8}\), \(I_{1}=[0, 1]\), \(I_{2}=[0, \frac{2}{3}]\), \(b_{j}=(5j-4)^{-1}(5j+1)^{-1}\) (\(j=1,2, \ldots \)), \(\beta _{j}= \frac{3}{2}-\frac{1}{2^{(7+j)}}\) (\(j=1,2, \ldots \)), \(\xi _{j}=(28+2j)^{-1}\) (\(j=1,2,\ldots \)), \(w_{1}(t)=t^{ \frac{3}{4}}(1-t)^{2}\), \(w_{2}(t)=t^{\frac{7}{8}}(1+t^{2})^{-1}\). Then problem (4.1) can be transformed into BVP (1.1) for \(\lambda >0\).
$$ \begin{aligned} f(t,u,v)&= 6t^{-\frac{1}{3}}u^{\frac{1}{8}} v^{-\frac{1}{8}}+ 5u^{ \frac{1}{6}} \bigl(1-t^{2} \bigr)^{-\frac{1}{2 }}+\bigl(6v^{\frac{1}{3}}+v^{ \frac{1}{4}}+1\bigr) \bigl(1+t^{2}\bigr)^{-1} \\ &\quad{} + (tu)^{-\frac{1}{5}} \bigl(7+ (u+1)^{- \frac{4}{5}} \bigr) , \end{aligned} $$
By simple computation, we have a rough estimate
and
which means the properties of Green’s function in Lemma 2.13 are achieved. Let
and
Then
It is easy to check the following conditions: Let \(r=\frac{7}{15}<\frac{1}{2}\). It is easy to check that Therefore, the assumptions of Theorem 3.1 are satisfied. Then BVP (4.1) has a unique solution \(z_{\lambda }^{*}\) in P, and there exists a constant \(\eta _{\lambda }\in (0, 1)\) such that
And at the same time, \(z_{\lambda }^{*}\) satisfies: Moreover, for any initial values \(z_{0}\), \(\tilde{z}_{0}\in P_{e}\), by constructing successively the sequences as follows:
we have \(z_{n}\rightarrow z_{\lambda }^{*}\) and \(\tilde{z}_{n}\rightarrow z_{\lambda }^{*}\) in E, as \(n\rightarrow \infty \).
$$\begin{aligned}& \int _{I_{1}}\tau ^{\gamma -\alpha _{1}-1}w_{1}(\tau ) \,dA_{1}(\tau )= \int _{0}^{1}\tau (1-\tau )^{2} \,dA_{1}(\tau )=0.125>0, \\& \int _{I_{2}}\tau ^{\gamma -\alpha _{2}-1}w_{2}(\tau ) \,dA_{2}(\tau )= \int _{0}^{\frac{2}{3}}\tau \bigl(1+\tau ^{2}\bigr)^{-1}\,dA_{2}(\tau )=0.4>0, \\& \begin{aligned} \sum_{j=1}^{\infty } \frac{b_{j}}{\varGamma (\gamma -\beta _{j})}\xi _{j} ^{\gamma -\beta _{j}-1} &= \sum _{j=1}^{\infty } \frac{1}{\varGamma (1+ 2^{-(7+j)})}(5j-4)^{-1}(5j+1)^{-1}(28+2j)^{- \frac{1}{2^{(7+j)}}} \\ &\leq \sum_{j=1}^{\infty }(5j-4)^{-1}(5j+1)^{-1}=0.2, \end{aligned} \end{aligned}$$
$$ \begin{aligned} \sigma &= \frac{1}{\varGamma (\gamma -\gamma _{0})} -\sum _{i=1}^{p}\frac{a _{i}}{\varGamma (\gamma -\alpha _{i})} \int _{I_{i}}s^{\gamma -\alpha _{i}-1}w _{i}(s) \,dA_{i}(s) -\sum_{j=1}^{\infty } \frac{b_{j}}{\varGamma (\gamma - \beta _{j})}\xi _{j}^{\gamma -\beta _{j}-1} \\ &\approx1-\frac{2}{\varGamma (\frac{5}{4})} \int _{0}^{1}s^{\frac{1}{4}}s^{\frac{3}{4}}(1-s)^{2} \,ds -\frac{1}{2 \varGamma (\frac{9}{8})} \int _{0}^{\frac{2}{3}}s^{\frac{1}{8}}s^{ \frac{7}{8}} \bigl(1+s^{2}\bigr)^{-1}\,ds \\ &\quad{} -\sum_{j=1}^{\infty } \frac{1}{\varGamma (1+ 2^{-(7+j)})} (5j-4)^{-1}(5j+1)^{-1} ( 28+2j )^{- \frac{1}{2^{(7+j)}}} \\ &\geq 1-0.25-0.2-0.2=0.35>0, \end{aligned} $$
$$ f_{1}(t,u,v,w,z)= 5t^{-\frac{1}{3}}u^{\frac{1}{8}} z^{-\frac{1}{8}}+ 4u ^{\frac{1}{6}} \bigl(1-t^{2} \bigr)^{-\frac{1}{2}}+\bigl(6v^{\frac{1}{3}}+1\bigr) \bigl(1+t^{2} \bigr)^{-1} + 7 (tw)^{- \frac{1}{5}} $$
$$ f_{2}(t,u,v,w,z)= t^{-\frac{1}{3}}u^{\frac{1}{8}}z^{-\frac{1}{8}}+u ^{\frac{1}{6}} \bigl(1-t^{2}\bigr)^{-\frac{1}{2}}+ v^{\frac{1}{4}}\bigl(1+t^{2}\bigr)^{-1} + (tw)^{-\frac{1}{5}} (w+1)^{-\frac{4}{5}}. $$
$$ f(t,u,v)=f_{1}(t,u,v, u, v)+f_{2}(t,u,v,u,v). $$
(1)
For all \(t\in (0,1)\) and \((w, z)\in (0,+\infty )^{2}\), \(f_{1}(t, u, v, w, z)\), \(f_{2}(t, u, v, w, z)\) are increasing in \((u, v)\in (0,+\infty )^{2}\); for all \(t\in (0,1)\) and \((u, v)\in (0,+ \infty )^{2}\), \(f_{1}(t, u, v, w, z)\), \(f_{2}(t, u, v, w, z)\) are decreasing in \((w, z)\in (0,+\infty )^{2}\).
(2)
Let \(\varphi (\mu )=\mu ^{\frac{1}{3}}\). Then, for \(\mu \in (0,1)\), \(t\in (0,1)\), and \((u, v, w, z)\in (0,+\infty )^{4}\),
$$\begin{aligned}& \begin{aligned} f_{1}\bigl(t, \mu u, \mu v, \mu ^{-1} w, \mu ^{-1} z\bigr)&= 5\mu ^{\frac{1}{4}} t ^{-\frac{1}{3}}u^{\frac{1}{8}}z^{-\frac{1}{8}}+4\mu ^{\frac{1}{6}}u ^{\frac{1}{6}}\bigl(1-t^{2}\bigr)^{-\frac{1}{2}} \\ &\quad{} +\bigl(6\mu ^{\frac{1}{3}}v^{ \frac{1}{3}}+1\bigr) \bigl(1+t^{2}\bigr)^{-1}+ 7\mu ^{\frac{1}{5}} (tw)^{-\frac{1}{5}} \\ &\geq \mu ^{\frac{1}{3}} f_{1}(t, u, v, w, z), \end{aligned} \\& \begin{aligned} f_{2}\bigl(t, \mu u, \mu v, \mu ^{-1} w, \mu ^{-1} z\bigr)&= \mu ^{\frac{1}{4}}t^{-\frac{1}{3}}u^{\frac{1}{8}}z^{-\frac{1}{8}}+ \mu ^{\frac{1}{6}}u ^{\frac{1}{6}} \bigl(1-t^{2}\bigr)^{-\frac{1}{2}} \\ &\quad{} + \mu ^{\frac{1}{4}}v^{ \frac{1}{4}}\bigl(1+t^{2} \bigr)^{-1}+ \mu ^{\frac{4}{15}} (tw)^{-\frac{1}{5}} (w+ \mu )^{-\frac{4}{5}} \\ &\geq \mu f_{2}(t, u, v, w, z). \end{aligned} \end{aligned}$$
(3)
Let \(\kappa =4\). Then, for all \((u, v, w, z)\in (0,+\infty )^{4}\),
$$ f_{1}(t, u, v, w, z)\geq \kappa f_{2}(t, u, v, w, z). $$
(4)
The functions \(f_{1}\) and \(f_{2}\) satisfy
$$\begin{aligned}& \begin{aligned} 0&< \int _{0}^{1}f_{1}\bigl(s, 1, 1, s^{\gamma -1}, s^{\gamma -1}\bigr)\,ds\\ &= \int _{0}^{1} \bigl(5s^{-\frac{25}{48}}+4 \bigl(1-s^{2}\bigr)^{-\frac{1}{2}}+7\bigl(1+s ^{2} \bigr)^{-1}+7s^{-\frac{1}{2}} \bigr)< +\infty , \end{aligned} \\& 0< \int _{0}^{1}f_{2}\bigl(s, 1, 1, s^{\gamma -1}, s^{\gamma -1}\bigr)\,ds\leq \int _{0}^{1} \bigl(s^{-\frac{25}{48}}+ \bigl(1-s^{2}\bigr)^{-\frac{1}{2}}+\bigl(1+s^{2} \bigr)^{-1}+s ^{-\frac{1}{2}} \bigr)< +\infty . \end{aligned}$$
(i)
$$ \varphi (\mu )=\mu ^{\frac{1}{3}}\geq \frac{\mu ^{r}-\mu }{\kappa }+\mu ^{r}=\frac{5}{4}\mu ^{\frac{7}{15}}- \frac{1}{4}\mu ,\quad \forall \mu \in (0, 1). $$
(ii)
$$ \varphi (\mu )=\mu ^{\frac{1}{3}}\geq \frac{\mu ^{\frac{1}{2}}-\mu }{ \kappa }+\mu ^{\frac{1}{2}}=\frac{5}{4}\mu ^{\frac{1}{2}}- \frac{1}{4} \mu ,\quad \forall \mu \in (0, 1). $$
$$ \frac{6\varGamma (\frac{9}{10})\eta _{\lambda } }{5\varGamma (\frac{1}{2})} t ^{ \frac{3}{2}} \leq z_{\lambda }^{*}(t) \leq \frac{6\varGamma ( \frac{9}{10})}{5\varGamma (\frac{1}{2})\eta _{\lambda }} t^{ \frac{3}{2}} ,\quad t\in [0, 1]. $$
(i)
\(z_{\lambda }^{*}\) is continuous with respect to \(\lambda \in (0,+\infty )\), i.e., for ∀ \(\lambda _{0}\in (0,+\infty )\),
$$ \bigl\Vert z_{\lambda }^{*}-z_{\lambda _{0}}^{*} \bigr\Vert \rightarrow 0, \quad \text{as } \lambda \rightarrow \lambda _{0}. $$
(ii)
\(0<\lambda _{1}< \lambda _{2}\) implies \(z_{\lambda _{1}}^{*}< z _{\lambda _{2}}^{*}\).
(iii)
$$ \lim_{\lambda \rightarrow 0^{+}} \bigl\Vert z_{\lambda }^{*} \bigr\Vert =0, \qquad \lim_{\lambda \rightarrow +\infty } \bigl\Vert z_{\lambda }^{*} \bigr\Vert =+\infty . $$
$$\begin{aligned}& \begin{aligned} z_{n}(t)&= I_{0^{+}}^{\frac{3}{5}} \biggl\{ \lambda \int _{0}^{1}K(t,s)f _{1} \bigl(s,I_{0^{+}}^{\frac{3}{5}}z_{n-1}(s), z_{n-1}(s), I_{0^{+}}^{ \frac{3}{5}}\tilde{z}_{n-1}(s), \tilde{z}_{n-1}(s)\bigr)\,ds \\ &\quad{} + \lambda \int _{0}^{1}K(t,s)f_{2} \bigl(s,I_{0^{+}}^{\frac{3}{5}}z_{n-1}(s), z _{n-1}(s), I_{0^{+}}^{\frac{3}{5}}\tilde{z}_{n-1}(s), \tilde{z}_{n-1}(s)\bigr)\,ds \biggr\} , \end{aligned} \\& \begin{aligned} \tilde{z}_{n}(t)&= I_{0^{+}}^{\frac{3}{5}} \biggl\{ \lambda \int _{0}^{1}K(t,s)f _{1} \bigl(s,I_{0^{+}}^{\frac{3}{5}}\tilde{z}_{n-1}(s), \tilde{z}_{n-1}(s), I_{0^{+}}^{\frac{3}{5}}z_{n-1}(s), z_{n-1}(s)\bigr)\,ds \\ &\quad{} +\lambda \int _{0}^{1}K(t,s)f_{2}( \bigl(s,I_{0^{+}}^{\frac{3}{5}}\tilde{z}_{n-1}(s), \tilde{z}_{n-1}(s), I_{0^{+}}^{\frac{3}{5}}z_{n-1}(s), z_{n-1}(s)\bigr)\,ds \biggr\} , \\ &\quad n=1,2,\ldots , \end{aligned} \end{aligned}$$
Example 4.2
We consider the following problem:
where
$$ \textstyle\begin{cases} D_{0+}^{\frac{8}{3}}z(t)+ z^{\frac{1}{4}}(t) (D_{0+}^{ \frac{2}{3}}z(t) )^{-\frac{1}{4}} (t^{-\frac{1}{3}}+ t ^{-\frac{1}{4}} ) + (2D_{0+}^{\frac{2}{3}}z(t) ) ^{\frac{1}{3}} (1-t^{2})^{-\frac{1}{2}}\\ \quad {}+ (tz(t))^{-\frac{1}{4}} (1+ (z(t)+1)^{-\frac{3}{4}} )=0, \quad 0< t< 1, \\ z(0)=D_{0+}^{\frac{3}{4}}z(0)=0, \\ D_{0+}^{\frac{3}{2}}z(1)=2 \int _{0}^{1}s^{\frac{3}{4}}(1-s)^{2}D_{0+}^{\frac{5}{4}}z(s)\,dA_{1}(s) +\frac{1}{2}\int _{0}^{\frac{2}{3}}s^{\frac{7}{8}}(1+s^{2})^{-1}D_{0+} ^{\frac{11}{8}}z(s)\,dA_{2}(s)\\ \hphantom{D_{0+}^{\frac{3}{2}}z(1)=}{} +\sum_{j=1}^{\infty }(5j-4)^{-1}(5j+1)^{-1}D_{0+}^{\frac{5}{3}-2^{-(7+j)}}u ((28+2j)^{-1} ), \end{cases} $$
(4.2)
$$ A_{1}(t)= \textstyle\begin{cases} \frac{1}{11}, & t\in [0, \frac{1}{2} ), \\ \frac{12}{11}, & t\in [\frac{1}{2}, 1 ], \end{cases}\displaystyle \qquad A_{2}(t)= \textstyle\begin{cases} \frac{1}{13}, & t\in [0, \frac{1}{2} ), \\ \frac{14}{13}, & t\in [\frac{1}{2}, 1 ]. \end{cases} $$
Let
\(\gamma =\frac{8}{3}\) (\(n=3\)), \(\gamma _{0}=\frac{3}{2}\), \(\nu _{1}= \frac{2}{3}\), \(q_{1}=\frac{3}{4}\), \(a_{1}=2\), \(a_{2}=\frac{1}{2}\), \(\alpha _{1}=\frac{5}{4}\), \(\alpha _{2}=\frac{11}{8}\), \(I_{1}=[0, 1]\), \(I_{2}=[0, \frac{2}{3}]\), \(b_{j}= (5j-4)^{-1}(5j+1)^{-1}\) (\(j=1,2, \ldots \)), \(\beta _{j}=\frac{5}{3}- 2^{-(7+j)}\) (\(j=1,2,\ldots \)), \(\xi _{j}=(28+2j)^{-1}\) (\(j=1,2,\ldots \)), \(w_{1}(t)=t^{\frac{3}{4}}(1-t)^{2}\), \(w_{2}(t)=t^{\frac{7}{8}}(1+t ^{2})^{-1}\). Then problem (4.2) can be transformed into BVP (1.1) for \(\lambda =1\). By simple computation, we have a rough estimate:
and
which means the properties of Green’s function in Lemma 2.13 are achieved. Let
and
Then
It is easy to check the following conditions: Thus BVP (4.2) has a unique positive solution \(z_{1}^{*}\). Then BVP (1.1) has a unique solution \(z_{1}^{*}\) in P, and there exists a constant \(\eta _{1}\in (0, 1)\) such that
Moreover, for any initial values \(z_{0}\), \(\tilde{z}_{0}\in P_{e}\), by constructing successively the sequences as follows:
we have \(z_{n}\rightarrow z_{1}^{*}\) and \(\tilde{z}_{n}\rightarrow z _{1}^{*}\) in E, as \(n\rightarrow \infty \).
$$ f(t,u,v)= u^{\frac{1}{4}}v^{-\frac{1}{4}} \bigl( t^{-\frac{1}{3}}+ t ^{-\frac{1}{4}} \bigr)+ 2v^{\frac{1}{3}} \bigl(1-t^{2} \bigr)^{-\frac{1}{2}} + (tu)^{-\frac{1}{4}} \bigl(1+ (u+1)^{-\frac{3}{4}} \bigr) , $$
$$\begin{aligned}& \int _{I_{1}}\tau ^{\gamma -\alpha _{1}-1}w_{1}(\tau ) \,dA_{1}(\tau )= \int _{0}^{1}\tau (1-\tau )^{2} \,dA_{1}(\tau )=0.125>0, \\& \int _{I_{2}}\tau ^{\gamma -\alpha _{2}-1}w_{2}(\tau ) \,dA_{2}(\tau )= \int _{0}^{\frac{2}{3}}\tau \bigl(1+\tau ^{2}\bigr)^{-1}\,dA_{2}(\tau )=0.4>0, \\& \begin{aligned} \sum_{j=1}^{\infty } \frac{b_{j}}{\varGamma (\gamma -\beta _{j})}\xi _{j} ^{\gamma -\beta _{j}-1}&= \sum _{j=1}^{\infty } \frac{1}{\varGamma (1+ 2^{-(7+j)})}(5j-4)^{-1}(5j+1)^{-1} ( 28+2j ) ^{-\frac{1}{2^{(7+j)}}} \\ &\leq \sum_{j=1}^{\infty }(5j-4)^{-1}(5j+1)^{-1}=0.2, \end{aligned} \end{aligned}$$
$$ \begin{aligned} \sigma &= \frac{1}{\varGamma (\gamma -\gamma _{0})} -\sum _{i=1}^{p}\frac{a _{i}}{\varGamma (\gamma -\alpha _{i})} \int _{I_{i}}s^{\gamma -\alpha _{i}-1}w _{i}(s) \,dA_{i}(s) -\sum_{j=1}^{\infty } \frac{b_{j}}{\varGamma (\gamma - \beta _{j})}\xi _{j}^{\gamma -\beta _{j}-1} \\ &\approx1-\frac{2}{\varGamma (\frac{5}{4})} \int _{0}^{1}s^{\frac{1}{4}}s^{\frac{3}{4}}(1-s)^{2} \,ds -\frac{1}{2 \varGamma (\frac{9}{8})} \int _{0}^{\frac{2}{3}}s^{\frac{1}{8}}s^{ \frac{7}{8}} \bigl(1+s^{2}\bigr)^{-1}\,ds \\ &\quad{} -\sum_{j=1}^{\infty } \frac{1}{\varGamma (1+ 2^{-(7+j)})} (5j-4)^{-1}(5j+1)^{-1} ( 28+2j )^{- \frac{1}{2^{(7+j)}}} \\ &\geq 1-0.25-0.2-0.2=0.35>0, \end{aligned} $$
$$ f_{1}(t,u,v,w,z)= t^{-\frac{1}{3}}u^{\frac{1}{4}} z^{-\frac{1}{4} }+ v ^{\frac{1}{3}} \bigl(1-t^{2} \bigr)^{-\frac{1}{2}} + (tw)^{-\frac{1}{4}} $$
$$ f_{2}(t,u,v,w,z)= t^{-\frac{1}{4}}u^{\frac{1}{4}}z^{-\frac{1}{4}}+ v ^{\frac{1}{3}} \bigl(1-t^{2}\bigr)^{-\frac{1}{2 }} + (tw)^{-\frac{1}{4}}(w+1)^{- \frac{3}{4}}. $$
$$ f(t,u,v)=f_{1}(t,u,v, u, v)+f_{2}(t,u,v,u,v). $$
(1)
For all \(t\in (0,1)\) and \((w, z)\in (0,+\infty )^{2}\), \(f_{1}(t, u, v, w, z)\), \(f_{2}(t, u, v, w, z)\) are increasing in \((u, v)\in (0,+\infty )^{2}\); for all \(t\in (0,1)\) and \((u, v)\in (0,+ \infty )^{2}\), \(f_{1}(t, u, v, w, z)\), \(f_{2}(t, u, v, w, z)\) are decreasing in \((w, z)\in (0,+\infty )^{2}\).
(2)
Let \(\varphi (\mu )=\mu ^{\frac{1}{2}}\). Then, for \(\mu \in (0,1)\), \(t\in (0,1)\), and \((u, v, w, z)\in (0,+\infty )^{4}\),
$$\begin{aligned}& \begin{aligned} &f_{1}\bigl(t, \mu u, \mu v, \mu ^{-1} w, \mu ^{-1} z\bigr)\\ &\quad = t^{-\frac{1}{3}}( \mu u)^{\frac{1}{4}}\bigl(\mu ^{-1}z\bigr)^{-\frac{1}{4 }}+ (\mu v)^{\frac{1}{3}} \bigl(1-t^{2}\bigr)^{-\frac{1}{2} } + t^{-\frac{1}{4} }\bigl(\mu ^{-1}w\bigr)^{- \frac{1}{4} } \\ &\quad \geq \mu ^{\frac{1}{2}} f_{1}(t, u, v, w, z), \end{aligned} \\& \begin{aligned}& f_{2}\bigl(t, \mu u, \mu v, \mu ^{-1} w, \mu ^{-1} z\bigr)\\ &\quad = t^{-\frac{1}{4}} ( \mu u)^{\frac{1}{4}} \bigl(\mu ^{-1}z\bigr)^{-\frac{1}{4} }+ (\mu v)^{ \frac{1}{3}} \bigl(1-t^{2}\bigr)^{-\frac{1}{2} } + \mu (tw)^{-\frac{1}{4}}(w+ \mu )^{-\frac{3}{4}} \\ &\quad \geq \mu f_{2}(t, u, v, w, z). \end{aligned} \end{aligned}$$
(3)
Let \(\kappa =1\). Then, for all \((u, v, w, z)\in (0,+\infty )^{4}\),
$$ f_{1}(t, u, v, w, z)\geq f_{2}(t, u, v, w, z). $$
(4)
The functions \(f_{1}\) and \(f_{2}\) satisfy
$$\begin{aligned}& 0< \int _{0}^{1}f_{1}\bigl(s, 1, 1, s^{\gamma -1}, s^{\gamma -1}\bigr)\,ds= \int _{0}^{1} \bigl(s^{-\frac{3}{4}}+ \bigl(1-s^{2}\bigr)^{-\frac{1}{2}}+s^{- \frac{2}{3}} \bigr)< + \infty , \\& 0< \int _{0}^{1}f_{2}\bigl(s, 1, 1, s^{\gamma -1}, s^{\gamma -1}\bigr)\,ds\leq \int _{0}^{1} \bigl(s^{-\frac{2}{3}}+ \bigl(1-s^{2}\bigr)^{-\frac{1}{2}}+s^{- \frac{2}{3}} \bigr)< + \infty . \end{aligned}$$
$$ \frac{9\eta _{1} }{10\varGamma ( \frac{2}{3})}t^{ \frac{5}{3}} \leq z_{1} ^{*}(t)\leq \frac{9}{10\varGamma ( \frac{2}{3})\eta _{1}}t^{ \frac{5}{3}} ,\quad t\in [0, 1]. $$
$$\begin{aligned}& \begin{aligned} z_{n}(t)&= I_{0^{+}}^{\frac{2}{3}} \biggl\{ \int _{0}^{1}K(t,s)f_{1}\bigl(s,I _{0^{+}}^{\frac{2}{3}}z_{n-1}(s), z_{n-1}(s), I_{0^{+}}^{\frac{2}{3}} \tilde{z}_{n-1}(s), \tilde{z}_{n-1}(s)\bigr)\,ds \\ &\quad{} + \int _{0}^{1}K(t,s)f _{2} \bigl(s,I_{0^{+}}^{\frac{2}{3}}z_{n-1}(s), z_{n-1}(s), I_{0^{+}}^{ \frac{2}{3}}\tilde{z}_{n-1}(s), \tilde{z}_{n-1}(s)\bigr)\,ds \biggr\} , \end{aligned} \\& \begin{aligned} \tilde{z}_{n}(t)&= I_{0^{+}}^{\frac{2}{3}} \biggl\{ \int _{0}^{1}K(t,s)f _{1} \bigl(s,I_{0^{+}}^{\frac{2}{3}}\tilde{z}_{n-1}(s), \tilde{z}_{n-1}(s), I_{0^{+}}^{\frac{2}{3}}z_{n-1}(s), z_{n-1}(s)\bigr)\,ds \\ &\quad{} + \int _{0} ^{1}K(t,s)f_{2} \bigl(s,I_{0^{+}}^{\frac{2}{3}}\tilde{z}_{n-1}(s), \tilde{z}_{n-1}(s), I_{0^{+}}^{\frac{2}{3}}z_{n-1}(s), z_{n-1}(s)\bigr)\,ds \biggr\} , \\ &\quad n=1,2,\ldots , \end{aligned} \end{aligned}$$
Acknowledgements
The authors would like to thank the editors for their helpful suggestions.
Availability of data and materials
Data sharing not applicable to this article as no datasets were generated or analysed during the current study.
Ethics approval and consent to participate
Not applicable.
Competing interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Consent for publication
Not applicable.
Open Access 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.
Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.