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Explicit iteration to Hadamard fractional integro-differential equations on infinite domain

Advances in Difference Equations20162016:299

  • Received: 10 September 2016
  • Accepted: 9 November 2016
  • Published:


This paper investigates the existence of the unique solution for a Hadamard fractional integral boundary value problem of a Hadamard fractional integro-differential equation with the monotone iterative technique. Two examples to illustrate our result are given.


  • Hadamard derivative
  • Hadamard integro-differential boundary conditions
  • monotone iterative
  • infinite interval

1 Introduction

Fractional differential equations are becoming more and more popular recently in several journals and books due to their applications in a number of fields such as physics, biophysics, mechanical systems, electrical-analytical, and thermal systems [16]. For some recent development of this topic, see for example [713] and the references therein.

In 1892 [14], Hadamard presented a concept of fractional derivatives, which is different from Caputo and Riemann-Liouville type fractional derivatives and involves a logarithmic function of an arbitrary exponent in the integral kernel. It is significant that the study of Hadamard type fractional differential equations is still in its infancy and deserves further study. A detailed presentation of Hadamard fractional derivative is available in [3] and [1523].

As was pointed out in [22], Hadamard’s construction is more appropriate for problems on half axes. In this situation, we consider the following Hadamard fractional integro-differential equations with Hadamard fractional integral boundary conditions on an infinite interval:
$$ \left \{ \textstyle\begin{array}{ll} {}^{H}D^{\gamma }u(t) +f(t,u(t),{{}^{H}I}^{q}u(t))=0, \quad 2< \gamma < 3, t\in (1,+\infty ), \\ u(1)=u^{\prime}(1)=0, \qquad {{}^{H}D^{\gamma -1}u(\infty )}=\sum_{i=1}^{m}\lambda_{i}{{}^{H}I ^{\beta_{i}}u(\eta ),} \end{array}\displaystyle \right . $$
where \({{}^{H}D}^{\gamma }\) denotes Hadamard fractional derivative of order γ, \(\eta \in (1,\infty )\), and \({{}^{H}I}^{(\cdot )}\) is the Hadamard fractional integral, \(q,\beta_{i}>0\) \((i=1,2,\ldots,m), \lambda _{i}\geq 0\) \((i=1,2,\ldots,m)\) are given constants and γ, η, \(\beta _{i}\), \(\lambda_{i}\) satisfy \(\Gamma (\gamma )>\sum_{i=1}^{m}\frac{ \lambda_{i}\Gamma (\gamma )}{\Gamma (\gamma +\beta_{i})} (\log \eta )^{ \gamma +\beta_{i}-1}\).

We recall that the monotone iterative technique represents a powerful tool for seeking the solution of a nonlinear problem. For more details as regards the application of this method in fractional differential equations, see [2438] and the references therein.

We organize the rest of our manuscript as follows: In Section 2, we show some useful preliminaries and the key lemmas that are used in subsequent part of the manuscript. Then, in Section 3, the main results and proofs are provided. Section 4, exhibits two examples to illustrate our main results.

2 Preliminaries

Below, we will present some useful definitions and related lemmas.

$$ E=\biggl\{ u\in C\bigl([1,\infty ),\mathbb{R}\bigr):\sup_{t\in [1,\infty )} \frac{\vert u(t) \vert }{1+( \log t)^{\gamma -1}}< \infty \biggr\} , $$
then E denotes a Banach space equipped with norm \(\Vert u \Vert _{E}=\sup_{t\in [1,\infty )}\frac{\vert u(t) \vert }{1+(\log t)^{\gamma -1}}\).
$$ \Omega =\Gamma (\gamma )-\sum_{i=1}^{m} \frac{\lambda_{i}\Gamma (\gamma )}{\Gamma (\gamma +\beta_{i})} (\log \eta )^{\gamma +\beta _{i}-1} ; $$
obviously \(\Omega >0\).

Definition 2.1

[3] For a function g, the Hadamard fractional integral of order γ has the following form:
$$ {}^{H}I^{\gamma } g(t)=\frac{1}{\Gamma (\gamma )} \int_{1}^{t}\biggl(\log \frac{t}{s} \biggr)^{\gamma -1}\frac{g(s)}{s}\,ds,\quad \gamma >0, $$
provided the integral exists.

Definition 2.2

[3] The Hadamard fractional derivative of fractional order γ for a function \(g:[1,\infty ) \rightarrow \mathbb{R} \) has the following form:
$$ {}^{H}D^{\gamma } g(t)=\frac{1}{\Gamma (n-\gamma )}\biggl(t\frac{d}{dt} \biggr)^{n} \int_{1}^{t}\biggl(\log \frac{t}{s} \biggr)^{n-\gamma -1}\frac{g(s)}{s}\,ds,\quad n-1< \gamma < n,n=[\gamma ]+1, $$
where \([\gamma ]\) means the integer part of the real number γ and \(\log (\cdot )=\log_{e}(\cdot )\).

Lemma 2.1

[3] If \(a,\gamma,\beta >0\) then
$$ {}^{H}I^{\gamma }_{a} \biggl(\log \biggl( \frac{t}{a}\biggr)^{\beta -1}\biggr) (x)=\frac{\Gamma (\beta )}{\Gamma (\beta +\gamma )}\biggl(\log \frac{x}{a}\biggr)^{\beta +\gamma -1}. $$

Lemma 2.2

[3] If \(a,\gamma,\beta >0\) then
$$ {}^{H}D^{\gamma }_{a} \biggl(\log \biggl( \frac{t}{a}\biggr)^{\beta -1}\biggr) (x)=\frac{\Gamma (\beta )}{\Gamma (\beta -\gamma )}\biggl(\log \frac{x}{a}\biggr)^{\beta -\gamma -1}. $$

Lemma 2.3

[3] Given \(\gamma >0\) and \(x\in C[1,\infty )\cap L^{1}[1,\infty )\), then the solution of the Hadamard fractional differential equation \({{}^{H}D}^{\gamma }x(t)=0\) is
$$ x(t)=\sum_{i=1}^{n}c_{i}(\log t)^{\gamma -i} $$
$$ {{}^{H}I}^{\gamma } {{}^{H}D}^{\gamma }x(t)=x(t)+\sum _{i=1}^{n}c _{i}(\log t)^{\gamma -i} $$
where \(c_{i}\in \mathbb{R}, i=1,2,\ldots,n\), and \(n-1<\gamma <n\).

Lemma 2.4

Let \(h\in C[1,\infty )\) with \(0<\int_{1}^{\infty }h(s)\frac{ds}{s}< \infty \), then the Hadamard fractional integral boundary value problem
$$ \left \{ \textstyle\begin{array}{ll} {{}^{H}D}^{\gamma }u(t)+h(t)=0, \quad 2< \gamma < 3, t\in (1,+\infty ), \\ u(1)=u^{\prime}(1)=0, \qquad {{}^{H}D^{\gamma -1}u(\infty )}=\sum_{i=1}^{m}\lambda_{i}{{}^{H}I ^{\beta_{i}}u(\eta ),} \end{array}\displaystyle \right . $$
has the unique solution
$$ u(t)= \int_{1}^{\infty }G(t,s)h(s)\frac{ds}{s}, $$
$$ G(t,s)=g(t,s)+\sum_{i=1}^{m} \frac{\lambda_{i}(\log t)^{\gamma -1}}{\Omega \Gamma (\gamma +\beta_{i})}g_{i}(\eta,s), $$
$$\begin{aligned}& g(t,s) = \frac{1}{\Gamma (\gamma )} \textstyle\begin{cases} (\log t)^{\gamma -1}-(\log (\frac{t}{s}))^{\gamma -1},&1\leq s\leq t< \infty, \\(\log t)^{\gamma -1},&1\leq t\leq s< \infty, \end{cases}\displaystyle \end{aligned}$$
$$\begin{aligned}& g_{i}(\eta,s) = \textstyle\begin{cases} (\log \eta )^{\gamma +\beta_{i}-1}-(\log (\frac{\eta }{s}))^{\gamma + \beta_{i}-1},&1\leq s\leq \eta < \infty, \\ (\log\eta )^{\gamma +\beta_{i}-1},&1\leq\eta \leq s< \infty. \end{cases}\displaystyle \end{aligned}$$


We apply the Hadamard fractional integral of order γ to
$$\begin{aligned} {{}^{H}D}^{\gamma }u(t)+h(t)=0, \end{aligned}$$
and we conclude that
$$ u(t)=c_{1}(\log t)^{\gamma -1}+c_{2}(\log t)^{\gamma -2}+c_{3}(\log t)^{ \gamma -3}-\frac{1}{\Gamma (\gamma )} \int_{1}^{t}\biggl(\log \frac{t}{s} \biggr)^{ \gamma -1}h(s)\frac{ds}{s}, $$
where \(c_{1},c_{2},c_{3}\in \mathbb{R}\).
Using the fact that \(u(1)=u^{\prime}(1)=0\), we conclude that \(c_{2}=c_{3}=0\). Thus,
$$ u(t)=c_{1}(\log t)^{\gamma -1}-\frac{1}{\Gamma (\gamma )} \int_{1}^{t}\biggl( \log \frac{t}{s} \biggr)^{\gamma -1}h(s)\frac{ds}{s}. $$
Lemma 2.2 implies that
$$ {{}^{H}D}^{\gamma -1}u(t)=c_{1}\Gamma (\gamma )- \int_{1}^{t}h(s) \frac{ds}{s}. $$
Thus, the condition
$$\begin{aligned} {{}^{H}D^{\gamma -1}(\infty )}=\sum_{i=1}^{m} \lambda_{i}{{}^{H}I ^{\beta_{i}}u(\eta )} \end{aligned}$$
leads to
$$ c_{1}=\frac{1}{\Omega } \Biggl( \int_{1}^{\infty} h(s)\frac{ds}{s}- \sum _{i=1}^{m}\frac{\lambda_{i}}{\Gamma (\gamma +\beta_{i})} \int_{1}^{\eta }\biggl(\log \frac{\eta }{s} \biggr)^{\gamma +\beta_{i}-1}h(s) \frac{ds}{s} \Biggr), $$
where Ω is defined by (2.2). Substituting \(c_{2}=c _{3}=0\) and (2.13) into (2.10), we get the unique solution of the Hadamard fractional integral boundary value problem (2.5)
$$ \begin{aligned}[b] u(t)={}& \frac{(\log t)^{\gamma -1}}{\Omega } \Biggl( \int_{1}^{\infty} h(s) \frac{ds}{s} \\ &{}-\sum _{i=1}^{m}\frac{\lambda_{i}}{\Gamma (\gamma + \beta_{i})} \int_{1}^{\eta }\biggl(\log \frac{\eta }{s} \biggr)^{\gamma +\beta_{i}-1}h(s) \frac{ds}{s} \Biggr) \\ &{}-\frac{1}{\Gamma (\gamma )} \int_{1}^{t}\biggl(\log \frac{t}{s} \biggr)^{\gamma -1}h(s) \frac{ds}{s} \\ = {}& \int_{1}^{\infty} g(t,s)h(s)\frac{ds}{s}+\sum _{i=1}^{m}\frac{ \lambda_{i}(\log t)^{\gamma -1}}{\Omega \Gamma (\gamma +\beta_{i})} \int_{1}^{\infty} g_{i}(\eta,s)h(s) \frac{ds}{s} \\ ={} & \int_{1}^{\infty} G(t,s)h(s)\frac{ds}{s}. \end{aligned} $$
The proof is finished. □

Lemma 2.5

The Green’s function \(G(t,s)\) defined by (2.7) has the following properties:

\(G(t,s)\) is continuous and \(G(t,s)\geq 0\) for \((t,s)\in [1,\infty )\times [1,\infty )\).


\(\frac{G(t,s)}{1+(\log t)^{\gamma -1}}\leq \frac{1}{ \Gamma (\gamma )}+\sum_{i=1}^{m} \frac{\lambda_{i} g_{i}( \eta,s)}{\Omega \Gamma (\gamma +\beta_{i})}\) for all \(s,t\in [1, \infty )\).


Since \((A_{1})\) it is easy to prove, we do not present it but only prove the property \((A_{2})\).

For \(\forall s,t\in [1,\infty )\),
$$\begin{aligned} \frac{G(t,s)}{1+(\log t)^{\gamma -1}} &= \frac{1}{1+(\log t)^{\gamma -1}} \Biggl[g(t,s)+\sum_{i=1}^{m} \frac{\lambda_{i} (\log t)^{ \gamma -1} g_{i}(\eta,s)}{\Omega \Gamma (\gamma +\beta_{i})} \Biggr] \\ &\leq \frac{1}{\Gamma (\gamma )}+\sum_{i=1}^{m} \frac{ \lambda_{i} (\log t)^{\gamma -1} g_{i}(\eta,s)}{\Omega \Gamma ( \gamma +\beta_{i})(1+(\log t)^{\gamma -1})} \\ &\leq \frac{1}{\Gamma (\gamma )}+\sum_{i=1}^{m} \frac{ \lambda_{i} g_{i}(\eta,s)}{\Omega \Gamma (\gamma +\beta_{i})}. \end{aligned}$$
We present the following conditions for the sake of convenience:
There exist two positive functions \(p(t)\) and \(q(t)\) such that
$$\begin{aligned}& \Lambda = \int_{1}^{\infty } \bigl[1+(\log t)^{\gamma -1} \bigr] \biggl[p(t)+\frac{q(t)( \log t)^{q}}{\Gamma (q)} \biggr]\frac{dt}{t}< \infty, \\& \bigl\vert f(t,u,v)-f(t,\overline{u},\overline{v}) \bigr\vert \leq p(t) \vert u-\overline{u} \vert +q(t)\vert v-\overline{v} \vert ,\quad t\in [1, \infty ), u,v, \overline{u},\overline{v}\in \mathbb{R}. \end{aligned}$$
$$\begin{aligned} \lambda = \int_{1}^{\infty }\bigl\vert f(t,0,0) \bigr\vert \frac{dt}{t}< \infty. \end{aligned}$$

Lemma 2.6

If \((C_{1})\), \((C_{2})\) hold, then for any \(u\in E \)
$$ \int_{1}^{\infty }\bigl\vert f(t,u(t),{{}^{H}I}^{q}u(t) \bigr\vert \frac{dt}{t}\leq \Lambda \Vert u \Vert _{E}+ \lambda. $$


For any \(u\in E\), taking \(\overline{u}=0\), then \({{}^{H}I}^{q} \overline{u}=0\). Thus, by condition \((C_{1})\) we have
$$\begin{aligned} \bigl\vert f\bigl(t,u(t),{{}^{H}I}^{q}u(t) \bigr) \bigr\vert \leq{} & p(t)\bigl\vert u(t) \bigr\vert +q(t)\bigl\vert {{}^{H}I}^{q}u(t) \bigr\vert +\bigl\vert f(t,0,0) \bigr\vert \\ \leq {}& p(t)\bigl[1+(\log t)^{\gamma -1}\bigr]\frac{\vert u(t) \vert }{1+(\log t)^{\gamma -1}} \\ &{}+q(t)\frac{1}{\Gamma (q)}\int_{1}^{t}\biggl(\log \frac{t}{s}\biggr)^{q-1} \frac{\vert u(s) \vert }{s}\,ds+\bigl\vert f(t,0,0) \bigr\vert \\ \leq {}&p(t)\bigl[1+(\log t)^{\gamma -1}\bigr]\Vert u \Vert _{E} \\ &+q(t)\frac{1+(\log t)^{\gamma -1}}{\Gamma (q)} \\ &{}\times\int_{1}^{t}\frac{( \log \frac{t}{s})^{q-1}}{s}\frac{\vert u(s) \vert }{1+(\log s)^{\gamma -1}} \frac{1+( \log s)^{\gamma -1}}{1+(\log t)^{\gamma -1}}\,ds \\ &{}+\bigl\vert f(t,0,0) \bigr\vert \\ \leq{} & p(t)\bigl[1+(\log t)^{\gamma -1}\bigr]\Vert u \Vert _{E} \\ &{}+q(t)\frac{1+(\log t)^{\gamma -1}}{\Gamma (q)}\Vert u \Vert _{E} \int_{1}^{t}\frac{( \log t)^{q-1}}{s}\,ds+\bigl\vert f(t,0,0) \bigr\vert \\ \leq {}& p(t)\bigl[1+(\log t)^{\gamma -1}\bigr]\Vert u \Vert _{E}+q(t)\frac{1+(\log t)^{ \gamma -1}}{\Gamma (q)}(\log t)^{q}\Vert u \Vert _{E} \\ &{}+\bigl\vert f(t,0,0) \bigr\vert , \end{aligned}$$
from which, combined with \((C_{1})\) and \((C_{2})\), we can obtain
$$\begin{aligned} \int_{1}^{\infty }\bigl\vert f\bigl(t,u(t),{{}^{H}I}^{q}u(t) \bigr) \bigr\vert \frac{dt}{t} \leq{} & \int _{1}^{\infty }p(t)\bigl[1+(\log t)^{\gamma -1} \bigr]\Vert u \Vert _{E}\frac{dt}{t} \\ &{}+ \int_{1}^{\infty }q(t)\frac{1+(\log t)^{\gamma -1}}{\Gamma (q)}( \log t)^{q}\Vert u \Vert _{E}\frac{dt}{t} \\ &{}+ \int_{1}^{\infty }\bigl\vert f(t,0,0) \bigr\vert \frac{dt}{t} \\ = {}&\Lambda \Vert u \Vert _{E}+\lambda. \end{aligned}$$
The proof is done. □

3 Main results

Theorem 3.1

Suppose that the conditions \((C_{1})\) and \((C_{2})\) hold. Let
$$ w=\Lambda \Biggl(\frac{1}{\Gamma (\gamma )}+\sum _{i=1}^{m} \frac{ \lambda_{i} g_{i}(\eta,s)}{\Omega \Gamma (\gamma +\beta_{i})} \Biggr)< 1. $$
Then the Hadamard fractional integral boundary value problem (1.1) admits an unique solution \(\widetilde{u}(t)\) in E. In addition, there exists a monotone iterative sequence \({u_{n}(t)}\) such that \(u_{n}(t)\rightarrow \widetilde{u}(t)\) \((n\rightarrow \infty )\) uniformly on any finite sub-interval of \([1,\infty )\), where
$$ u_{n}(t)= \int_{1}^{\infty }G(t,s)f\bigl(s,u_{n-1}(s),{{}^{H}I}^{q}u_{n-1}(s) \bigr) \frac{ds}{s}. $$
Furthermore, there exists an error estimate for the approximating sequence
$$ \Vert u_{n}-\widetilde{u} \Vert _{E}\leq \frac{w^{n}}{1-w}\Vert u_{1}-u_{0} \Vert _{E} \quad (n=1,2,\ldots). $$


Define the operator T by
$$ (Tu) (t)= \int_{1}^{\infty }G(t,s)f\bigl(s,u(s),{{}^{H}I}^{q}u(s) \bigr)\frac{ds}{s}. $$
By Lemma 2.4, the Hadamard fractional integral boundary value problem (1.1) possesses a solution u iff u is a solution of \(u=Tu\).
First, for any \(t\in [1,\infty )\), by Lemma 2.5 and Lemma 2.6, we have
$$\begin{aligned} \frac{\vert (Tu)(t) \vert }{1+(\log t)^{\gamma -1}}&\leq \int_{1}^{\infty }\frac{G(t,s)}{1+( \log t)^{\gamma -1}} \bigl\vert f \bigl(s,u(s),{{}^{H}I}^{q}u(s)\bigr) \bigr\vert \frac{ds}{s} \\ &\leq \Biggl(\frac{1}{\Gamma (\gamma )}+\sum_{i=1}^{m} \frac{ \lambda_{i} g_{i}(\eta,s)}{\Omega \Gamma (\gamma +\beta_{i})} \Biggr) \bigl[\Lambda \Vert u \Vert _{E}+ \lambda \bigr] \\ &= w\Vert u \Vert _{E}+k. \end{aligned}$$
This means
$$ \Vert Tu \Vert _{E}\leq w\Vert u \Vert _{E}+k,\quad \forall t \in [1,\infty ), $$
where w is defined in (3.1) and
$$ k=\lambda \Biggl(\frac{1}{\Gamma (\gamma )}+\sum_{i=1}^{m} \frac{ \lambda_{i} g_{i}(\eta,s)}{\Omega \Gamma (\gamma +\beta_{i})} \Biggr). $$
In addition, for any \(u, \overline{u}\in E\), we have
$$\begin{aligned} \frac{\vert (Tu)(t)-(T\overline{u})(t) \vert }{1+(\log t)^{\gamma -1}} \leq{} & \int _{1}^{\infty }\frac{G(t,s)}{1+(\log t)^{\gamma -1}} \bigl\vert f(s,u(s),{{}^{H}I}^{q}u(s)-f\bigl(s,\overline{u}(s),{{}^{H}I}^{q}\overline{u}(s)\bigr) \bigr\vert \frac{ds}{s} \\ \leq{} &\int_{1}^{\infty }\frac{G(t,s)}{1+(\log t)^{\gamma -1}} \bigl[p(s)\bigl\vert u(s)-\overline{u}(s)\bigl\vert \mathop{+} q(s) \bigr\vert {{}^{H}I}^{q}u(s)-{{}^{H}I}^{q}\overline{u}(s) \bigr\vert \bigr]\frac{ds}{s} \\ \leq{} & \int_{1}^{\infty }\frac{G(t,s)}{1+(\log t)^{\gamma -1}}p(s)\bigl[1+( \log s)^{\gamma -1}\bigr]\frac{\vert u(s)-\overline{u}(s) \vert }{1+(\log s)^{\gamma -1}}\,ds \\ &{}+ \int_{1}^{\infty }\frac{G(t,s)}{1+(\log t)^{\gamma -1}}q(s)\bigl\vert {{}^{H}I}^{q}u(s)-{{}^{H}I}^{q} \overline{u}(s) \bigr\vert \frac{ds}{s} \\ \leq{} & \int_{1}^{\infty }\frac{G(t,s)}{1+(\log t)^{\gamma -1}}p(s)\bigl[1+( \log s)^{\gamma -1}\bigr]\Vert u-\overline{u} \Vert _{E}\,ds \\ &{}+\int_{1}^{\infty }\frac{G(t,s)}{1+(\log t)^{\gamma -1}}q(s)\frac{[1+(\log s)^{\gamma -1}](\log s)^{\gamma }}{\Gamma (\gamma )}\Vert u-\overline{u} \Vert _{E}\frac{ds}{s} \\ \leq {}& \Biggl(\frac{1}{\Gamma (\gamma )}+\sum_{i=1}^{m} \frac{\lambda_{i}g_{i}(\eta,s)}{\Omega \Gamma (\gamma +\beta_{i})} \Biggr) \\ &{}\times\int_{1}^{\infty }\Vert u-\overline{u} \Vert _{E}\bigl[1+(\log s)^{\gamma -1}\bigr] \biggl[p(s)+ \frac{q(s)( \log s)^{\gamma }}{\Gamma (\gamma )}\biggr]\frac{ds}{s} \\ \leq{} & \Biggl(\frac{1}{\Gamma (\gamma )}+\sum_{i=1}^{m} \frac{ \lambda_{i} g_{i}(\eta,s)}{\Omega \Gamma (\gamma +\beta_{i})} \Biggr) \Lambda \Vert u-\overline{u} \Vert _{E} \\ = {}& w\Vert u-\overline{u} \Vert _{E}. \end{aligned}$$
Then we get
$$ \Vert Tu-T\overline{u} \Vert _{E}\leq w\Vert u- \overline{u} \Vert _{E},\quad\forall u,\overline{u}\in E. $$
Through the Banach fixed point theorem, we can ensure that T has a unique fixed point ũ in E. That is, (1.1) admits a unique solution ũ in E. In addition, for any \(u_{0}\in E\), \(\Vert u_{n}-\widetilde{u} \Vert _{E}\rightarrow 0\) as \(n\rightarrow \infty \), where \(u_{n}=Tu_{n-1}\) \((n=1,2,\ldots)\).
From (3.9), we have
$$ \Vert u_{n}-u_{n-1} \Vert _{E}\leq w^{n-1}\Vert u_{1}-u_{0} \Vert _{E} $$
$$\begin{aligned} \Vert u_{n}-u_{j} \Vert _{E} &\leq \Vert u_{n}-u_{n-1} \Vert _{E}+\Vert u_{n-1}-u_{n-2} \Vert _{E}+\cdots+\Vert u_{j+1}-u_{j} \Vert _{E} \\ &\leq \frac{w^{n}(1-w^{n-j})}{1-w}\Vert u_{1}-u_{0} \Vert _{E}. \end{aligned}$$
Letting \(n\rightarrow \infty \) on both sides of (3.11), we conclude that
$$ \Vert u_{n}-\widetilde{u} \Vert _{E}\leq \frac{w^{n}}{1-w}\Vert u_{1}-u_{0} \Vert _{E}. $$

4 Example

Example 4.1

In the following we discuss the Hadamard fractional integral boundary value problem
$$ \textstyle\begin{cases} {}^{H}D^{\frac{5}{2}} u(t) +\frac{e^{-3t}t}{1+(\log t)^{\frac{3}{2}}}\cos (3t^{2} \\ \quad +u(t))+\frac{3\sqrt{\pi }e^{-3t}t}{8[1+(\log t)^{\frac{3}{2}}](\log t)^{\frac{5}{2}}}\arctan ({}^{H}I^{\frac{5}{2}}u(t))=0,\\ u(1)=u^{\prime}(1)=0, \qquad {}^{H}D^{\frac{3}{2}}u(+\infty )=\lambda_{1}{{}^{H}I^{\beta_{1}}u(\eta ),} \end{cases} $$
where \(\gamma =\frac{5}{2}\), \(m=1\), \(q=\frac{5}{2}\), and \(\lambda_{1}\), \(\beta_{1}\), η satisfy (\(\lambda_{1}\geq 0, \beta_{1}>0, \eta >1\))
$$ 0< \frac{\Gamma (\frac{1}{2}+\beta_{1})}{3\Gamma (\frac{1}{2}+\beta _{1})-4\lambda_{1}(\log \eta )^{\frac{3}{2}+\beta_{1}}}< \frac{e^{3}\sqrt{ \pi }}{2} $$
(see Figure 1).
Figure 1
Figure 1

Set of \(\pmb{\lambda_{1}\geq 0,\beta_{1}>0,\eta >1}\) such that \(\pmb{0< \frac{\Gamma (\frac{1}{2}+\beta_{1})}{3\Gamma (\frac{1}{2}+\beta _{1})-4\lambda_{1}(\log \eta )^{\frac{3}{2}+\beta_{1}}}<\frac{e^{3}\sqrt{ \pi }}{2}}\) .

For example, we can take \(\lambda_{1}=\frac{1}{10},\beta_{1}= \frac{3}{2},\eta =\frac{5}{2}\),
$$\begin{aligned}& \bigl\vert f\bigl(t,u,{{}^{H}I}^{\frac{5}{2}}u(t) \bigr)-f\bigl(t,\overline{u},{{}^{H}I}^{\frac{5}{2}}\overline{u}(t)\bigr) \bigr\vert \\& \quad \leq \frac{e^{-3t}t}{1+(\log t)^{ \frac{3}{2}}}\bigl\vert \cos \bigl(3t^{2}+u(t) \bigr)-\cos \bigl(3t^{2}+\overline{u}(t)\bigr) \bigr\vert \\& \qquad {}+\frac{3\sqrt{\pi }e^{-3t}t}{8[1+(\log t)^{\frac{3}{2}}](\log t)^{ \frac{5}{2}}}\bigl\vert \arctan \bigl({}^{H}I^{\frac{5}{2}}u(t) \bigr)-\arctan \bigl({{}^{H}I}^{\frac{5}{2}}\overline{u}(t)\bigr) \bigr\vert \\& \quad \leq \frac{e^{-3t}t}{1+(\log t)^{\frac{3}{2}}}\bigl\vert u(t)-\overline{u}(t) \bigr\vert \\& \qquad {}+\frac{3\sqrt{\pi }e^{-3t}t}{8[1+(\log t)^{\frac{3}{2}}](\log t)^{ \frac{5}{2}}}\bigl\vert {}^{H}I^{\frac{5}{2}}u(t)-{{}^{H}I}^{\frac{5}{2}} \overline{u}(t) \bigr\vert . \end{aligned}$$
Since \(p(t)=\frac{e^{-3t}t}{1+(\log t)^{\frac{3}{2}}}\) and \(q(t)=\frac{3\sqrt{\pi }e^{-3t}t}{8[1+(\log t)^{\frac{3}{2}}]( \log t)^{\frac{5}{2}}}\), we can show that
$$\begin{aligned}& \Lambda = \int_{1}^{\infty }\bigl[1+(\log t)^{\frac{3}{2}}\bigr] \biggl[\frac{e^{-3t}t}{1+( \log t)^{\frac{3}{2}}}+\frac{3\sqrt{\pi }e^{-3t}t(\log t)^{ \frac{5}{2}}}{8[1+(\log t)^{\frac{3}{2}}](\log t)^{\frac{5}{2}}\Gamma (\frac{5}{2})}\biggr]\frac{dt}{t} \\& \hphantom{\Lambda}= \frac{1}{2e^{3}}< \infty, \\& \lambda = \int_{1}^{\infty }\bigl\vert f(t,0,0) \bigr\vert \frac{dt}{t}= \int_{1}^{\infty }e^{-3t}\,dt=\frac{1}{3e^{3}}< \infty. \end{aligned}$$
Then \((C_{1})\) and \((C_{2})\) hold. At last, by a simple computation, we have
$$\begin{aligned}& \Omega =\Gamma (\gamma )-\frac{\lambda_{1}\Gamma (\gamma )}{\Gamma (\gamma +\beta_{1})}(\log \eta )^{\gamma +\beta_{1}-1}= \frac{3\sqrt{ \pi }}{4}-\frac{\sqrt{\pi }\lambda_{1}(\log \eta )^{\frac{3}{2}+ \beta_{1}}}{\Gamma (\frac{1}{2}+\beta_{1})}>\frac{1}{2e^{3}}>0, \\& w=\Lambda \biggl(\frac{1}{\Gamma (\gamma )}+\frac{\lambda_{1} g_{1}( \eta,s)}{\Omega \Gamma (\gamma +\beta_{i})} \biggr)= \frac{2}{e^{3}\sqrt{ \pi }} \biggl(\frac{\Gamma (\frac{1}{2}+\beta_{1})}{3\Gamma (\frac{1}{2}+ \beta_{1})-4\lambda_{1}(\log \eta )^{\frac{3}{2}+\beta_{1}}} \biggr)< 1. \end{aligned}$$
As a result, the conditions of Theorem 3.1 hold. Thus, the conclusion of Theorem 3.1 implies that (4.1) possesses a unique solution.

Example 4.2

Let us discuss the following Hadamard fractional integral boundary value problem:
$$ \left \{ \textstyle\begin{array}{ll} {{}^{H}D}^{\frac{9}{4}} u(t) +f(t,u(t), {{}^{H}I}^{\frac{9}{2}}u(t))=0, \\ u(1)=u^{\prime} (1)=0, \qquad {{}^{H}D}^{\frac{9}{4}}u(+\infty )=\sum_{i=1}^{3}\lambda_{i} {{{}^{H}I}^{\beta_{i}}u(\eta )}, \end{array}\displaystyle \right . $$
$$\begin{aligned} f\bigl(t,u(t),{}^{H}I^{\frac{9}{2}}u(t)\bigr)=\frac{\sin (t^{2}+u(t))}{16(1+t^{2})[1+( \log t)^{\frac{5}{4}}]}+ \frac{{}^{H}I^{\frac{9}{2}}u(t)-\sin ({}^{H}I^{ \frac{9}{2}}u(t))\cos ({}^{H}I^{\frac{9}{2}}u(t))}{128(1+t^{2})[1+( \log t)^{\frac{5}{4}}](\log t)^{\frac{9}{2}}}. \end{aligned}$$
Take \(\gamma =\frac{9}{4}\), \(m=3\), \(q=\frac{9}{2}\), \(\eta = \frac{9}{4}\), \(\lambda_{1}=\frac{1}{10}\), \(\beta_{1}=\frac{3}{4}\), \(\lambda_{2}=\frac{\sqrt{\pi }}{2}\), \(\beta_{2}=\frac{7}{4}\), \(\lambda_{3}=4 \), and \(\beta_{3}=\frac{11}{4}\).
The function f satisfies the inequality
$$\begin{aligned}& \bigl\vert f\bigl(t,u,{}^{H}I^{\frac{9}{2}}u(t) \bigr)-f\bigl(t,\overline{u},{{}^{H}I}^{\frac{9}{2}}\overline{u}(t)\bigr) \bigr\vert \\& \quad \leq \frac{1}{16(1+t^{2})[1+(\log t)^{\frac{5}{4}}]}\bigl\vert u(t)-\overline{u}(t) \bigr\vert \\& \qquad {}+ \frac{1}{64(1+t^{2})[1+(\log t)^{\frac{5}{4}}]( \log t)^{\frac{9}{2}}}\bigl\vert {}^{H}I^{\frac{9}{2}}u(t)-{{}^{H}I}^{\frac{9}{2}} \overline{u}(t) \bigr\vert . \end{aligned}$$
Since \(p(t)=\frac{1}{16(1+t^{2})[1+(\log t)^{\frac{5}{4}}]}\) and \(q(t)=\frac{1}{64(1+t^{2})[1+(\log t)^{\frac{5}{4}}](\log t)^{ \frac{9}{2}}}\), we can show that
$$\begin{aligned}& \Lambda =\int_{1}^{\infty }\bigl[1+(\log t)^{\frac{5}{4}}\bigr] \\& \qquad {}\times\biggl[\frac{1}{16(1+t^{2})[1+(\log t)^{\frac{5}{4}}]}+\frac{(\log t)^{\frac{9}{2}}}{64 \Gamma (\frac{9}{2})(1+t^{2})[1+(\log t)^{\frac{5}{4}}](\log t)^{ \frac{9}{2}}}\biggr]\frac{dt}{t} \\& \hphantom{\Lambda{}}< \frac{\pi }{32}< \infty, \\& \lambda = \int_{1}^{\infty }\bigl\vert f(t,0,0) \bigr\vert \frac{dt}{t}< \int_{1}^{\infty }\frac{1}{4(1+t^{2})}\,dt= \frac{\pi }{64}< \infty. \end{aligned}$$
Then \((C_{1})\) and \((C_{2})\) hold. At last, by a simple computation, we have
$$\begin{aligned}& \Omega =\Gamma \biggl(\frac{9}{4}\biggr)-\sum _{i=1}^{3}\frac{\lambda_{i} \Gamma (\frac{9}{4})}{\Gamma (\frac{9}{4}+\beta_{i})}(\log \eta )^{ \frac{9}{4}+\beta_{i}-1}\approx 0.8562>0, \\& w=\Lambda \Biggl(\frac{1}{\Gamma (\frac{9}{4})}+\sum_{i=1}^{3} \frac{ \lambda_{i} g_{i}(\eta,s)}{\Omega \Gamma (\frac{9}{4}+\beta_{i})}\Biggr) \approx 0.0452< 1. \end{aligned}$$
Thus, by the application of Theorem 3.1 the Hadamard fractional integral boundary value problem (4.3) admits an unique solution.



Partially supported by National Natural Science Foundation of China (No. 11501342) and the Scientific and Technological Innovation Programs of Higher Education Institutions in Shanxi (Nos. 2014135 and 2014136)

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

School of Mathematics, Shanxi Normal University, Linfen, Shanxi, 041004, People’s Republic of China
Department of Mathematics, Faculty of Art and Sciences, Çankaya University, Balgat, 06530, Turkey
Institute of Space Sciences, Magurele-Bucharest, Romania


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