Positive solutions for nonlinear Caputo-Hadamard fractional differential equations with integral boundary conditions

where \(\mathfrak{D}_{1}^{\alpha }\) is the Caputo-Hadamard fractional derivative of order \(00\), \( f:[1,e]\times \lbrack 0,\infty )\rightarrow \lbrack 0,\infty )\) is a given continuous function. To show the existence and uniqueness of positive solutions, we transform (1) into an integral equation and then by the method of upper and lower solutions and use the Schauder and Banach fixed point theorems.

This paper is organized as follows. In Section 2, we introduce some notations and lemmas, and state some preliminaries results needed in later sections. Also, we present the inversion of (1) and the Banach and Schauder fixed point theorems. For details on the Banach and Schauder theorems we refer the reader to [15]. In Sections 3 and 4, we give and prove our main results on positivity and we provide an example to illustrate our results.

2. Preliminaries

Let \(X=C\left( [1,e]\right) \) be the Banach space of all real-valued continuous functions defined on the compact interval \([1,e]\), endowed with the maximum norm. Define the cone \begin{equation*} \mathcal{E}=\left\{ x\in X:x(t)\geq 0,\ \forall t\in \lbrack 1,e]\right\} . \end{equation*}

We introduce some necessary definitions, lemmas and theorems which will be used in this paper. For more details, see [10, 14].

Definition 1. ([10]) The Hadamard fractional integral of order \(\alpha >0\) for a continuous function \(x:[1,+\infty )\rightarrow \mathbb{R}\) is defined as \begin{equation*} \mathfrak{I}_{1}^{\alpha }x(t)=\frac{1}{\Gamma (\alpha )}\int_{1}^{t}\left( \log \frac{t}{s}\right) ^{\alpha -1}x(s)\frac{ds}{s},\text{ }\alpha >0. \end{equation*}

Definition 2. ([10]) The Caputo-Hadamard fractional derivative of order \(\alpha >0\) for a continuous function \(x:[1,+\infty )\rightarrow \mathbb{R}\) is defined as \begin{equation*} \mathfrak{D}_{1}^{\alpha }x(t)=\frac{1}{\Gamma (n-\alpha )} \int_{1}^{t}\left( \log \frac{t}{s}\right) ^{n-\alpha -1}\delta ^{n}(x)(s) \frac{ds}{s},\text{ }n-1< \alpha < n, \end{equation*} where \(\delta ^{n}=\left( t\frac{d}{dt}\right) ^{n}\), \(n\in \mathbb{N}\).

Lemma 3.([1]) Let \(n-1< \alpha \leq n\), \(n\in \mathbb{N}\) and \(x\in C^{n}\left( [1,T]\right) \). Then \begin{equation*} \left( \mathfrak{I}_{1}^{\alpha }\mathfrak{D}_{1}^{\alpha }x\right) (t)=x(t)-\sum\limits_{k=0}^{n-1}\frac{x^{(k)}(1)}{\Gamma (k+1)}(\log t)^{k}. \end{equation*}

Lemma 4.([1]) For all \(\mu >0\) and \(\nu >-1,\) \begin{equation*} \frac{1}{\Gamma (\mu )}\int_{1}^{t}\left( \log \frac{t}{s}\right) ^{\mu -1}(\log s)^{\nu }\frac{ds}{s}=\frac{\Gamma (\nu +1)}{\Gamma (\mu +\nu +1)} (\log t)^{\mu +\nu }. \end{equation*}

The following lemma is fundamental to our results.

Lemma 5. Let \(x\in C\left( \left[ 1,e\right] \right) \), \(x^{\prime }\) exists,\(\ \)then \(x\) is a solution of (1) if and only if

Proof. Suppose \(x\) satisfies (1), then applying \(\mathfrak{I}_{1}^{\alpha }\) to both sides of (1), we have \begin{equation*} \mathfrak{I}_{1}^{\alpha }\mathfrak{D}_{1}^{\alpha }x\left( t\right) = \mathfrak{I}_{1}^{\alpha }f\left( t,x\left( t\right) \right) . \end{equation*} In view of Lemma 3 and the integral boundary condition, we get \begin{equation*} x\left( t\right) =\frac{1}{\Gamma (\alpha )}\int_{1}^{t}\left( \log \frac{t}{ s}\right) ^{\alpha -1}f\left( s,x\left( s\right) \right) \frac{ds}{s} +\lambda \int_{1}^{e}x\left( s\right) ds+d, 1 \leq t \leq e. \end{equation*} Conversely, suppose \(x\) satisfies (2), then applying \(\mathfrak{D} _{1}^{\alpha }\) to both sides of (2), we obtain \begin{align*} \mathfrak{D}_{1}^{\alpha }x\left( t\right) & =\mathfrak{D}_{1}^{\alpha }\left( \frac{1}{\Gamma (\alpha )}\int_{1}^{t}\left( \log \frac{t}{s}\right) ^{\alpha -1}f\left( s,x\left( s\right) \right) \frac{ds}{s}+\lambda \int_{1}^{e}x\left( s\right) ds+d\right) \\ & =\mathfrak{D}_{1}^{\alpha }\mathfrak{I}_{1}^{\alpha }f\left( t,x\left( t\right) \right) +\mathfrak{D}_{1}^{\alpha }\left( \lambda \int_{1}^{e}x\left( s\right) ds+d\right) \\ & =f\left( t,x\left( t\right) \right) . \end{align*} Moreover, the integral boundary condition \(x\left( 1\right) =\lambda \int_{1}^{e}x\left( s\right) ds+d\) holds.

Lastly in this section, we state the fixed point theorems which enable us to prove the existence and uniqueness of a positive solution of (1).

Definition 6. Let \((X,\left\Vert .\right\Vert )\) be a Banach space and \( \mathcal{A}:X\rightarrow X\). The operator \(\mathcal{A}\) is a contraction operator if there is an \(\lambda \in (0,1)\) such that \(x,y\in X\) imply \begin{equation*} \left\Vert \mathcal{A}x-\mathcal{A}y\right\Vert \leq \lambda \left\Vert x-y\right\Vert . \end{equation*}

Theorem 7.(Banach [15]) Let \(\mathcal{K}\) be a nonempty closed convex subset of a Banach space \(X\) and \(\mathcal{A}:\mathcal{K}\rightarrow \mathcal{K}\) be a contraction operator. Then there is a unique \(x\in \mathcal{K}\) with \( \mathcal{A}x=x\).

Theorem 8.(Schauder [1] Let \(\mathcal{K}\) be a nonempty bounded, closed and convex subset of a Banach space \(X\) and \(\mathcal{A}:\mathcal{K}\rightarrow \mathcal{K}\) be a completely continuous operator. Then \(\mathcal{A}\) has a fixed point in \(\mathcal{K}\).

3. Existence of positive solutions

In this section, we consider the results of existence problems for many cases of (1). We express (2) as where the operators \(\mathcal{A}:\mathcal{E}\rightarrow X\) is defined by \begin{equation*} \left( \mathcal{A}x\right) \left( t\right) =\frac{1}{\Gamma (\alpha )} \int_{1}^{t}\left( \log \frac{t}{s}\right) ^{\alpha -1}f\left( s,x\left( s\right) \right) \frac{ds}{s}+\lambda \int_{1}^{e}x\left( s\right) ds+d. \end{equation*} We need the following lemmas to establish our results.

Lemma 9. Assume that \(f:\left[ 1,e\right] \times \left[ 0,\infty \right) \rightarrow \left[ 0,\infty \right) \) is continuous. Then, the operator \(\mathcal{A}:\mathcal{E}\rightarrow \mathcal{E}\) is completely continuous.

Proof. By taking into account that \(f\) is continuous nonnegative function, we get that \(\mathcal{A}:\mathcal{E}\rightarrow \mathcal{E}\) is continuous. The function \(f:[1,e]\times B_{\eta }\rightarrow \lbrack 0,\infty )\) is bounded, then there exists \(\rho >0\) such that \begin{equation*} 0\leq f\left( t,x\left( t\right) \right) \leq \rho , \end{equation*} where \(B_{\eta }=\left\{ x\in \mathcal{E},\text{ }\left\Vert x\right\Vert \leq \eta \right\} \). We obtain \begin{align*} \left\vert \left( \mathcal{A}x\right) \left( t\right) \right\vert & =\left\vert \frac{1}{\Gamma (\alpha )}\int_{1}^{t}\left( \log \frac{t}{s} \right) ^{\alpha -1}f\left( s,x\left( s\right) \right) \frac{ds}{s}+\lambda \int_{1}^{e}x\left( s\right) ds+d\right\vert \\ & \leq \frac{1}{\Gamma \left( \alpha \right) }\int_{1}^{t}\left( \log \frac{t }{s}\right) ^{\alpha -1}\left\vert f\left( s,x\left( s\right) \right) \right\vert \frac{ds}{s}+\lambda \int_{1}^{e}\left\vert x\left( s\right) \right\vert ds+d \\ & \leq \frac{\rho }{\Gamma \left( \alpha \right) }\int_{1}^{t}\left( \log \frac{t}{s}\right) ^{\alpha -1}\frac{ds}{s}+\lambda \left( e-1\right) \eta +d \\ & \leq \frac{\rho \left( \log t\right) ^{\alpha }}{\Gamma \left( \alpha +1\right) }+\lambda \left( e-1\right) \eta +d. \end{align*} Thus, \begin{equation*} \left\Vert \mathcal{A}x\right\Vert \leq \frac{\rho }{\Gamma \left( \alpha +1\right) }+\lambda \left( e-1\right) \eta +d. \end{equation*} Hence, \(\mathcal{A}(B_{\eta })\) is uniformly bounded. Now, we will prove that \(\mathcal{A}(B_{\eta })\) is equicontinuous. Let \( x\in B_{\eta }\), then for any \(t_{1},t_{2}\in \lbrack 1,e]\), \(t_{2}>t_{1}\), we have \begin{eqnarray*} \left\vert \left( \mathcal{A}x\right) \left( t_{2}\right) -\left( \mathcal{ A}x\right) \left( t_{1}\right) \right\vert & =&\left\vert \frac{1}{\Gamma \left( \alpha \right) }\int_{1}^{t_{2}}\left( \log \frac{t_{2}}{s}\right) ^{\alpha -1}f\left( s,x\left( s\right) \right) \frac{ds}{s}\right. \left. -\frac{1}{\Gamma \left( \alpha \right) }\int_{1}^{t_{1}}\left( \log \frac{t_{1}}{s}\right) ^{\alpha -1}f\left( s,x\left( s\right) \right) \frac{ ds}{s}\right\vert \\ & \leq& \frac{1}{\Gamma \left( \alpha \right) }\int_{1}^{t_{1}}\left\vert \left( \log \frac{t_{2}}{s}\right) ^{\alpha -1}-\left( \log \frac{t_{1}}{s} \right) ^{\alpha -1}\right\vert \left\vert f\left( s,x\left( s\right) \right) \right\vert \frac{ds}{s} \\ && +\frac{1}{\Gamma \left( \alpha \right) }\int_{t_{1}}^{t_{2}}\left( \log \frac{t_{2}}{s}\right) ^{\alpha -1}\left\vert f\left( s,x\left( s\right) \right) \right\vert \frac{ds}{s} \\ & \leq& \frac{\rho }{\Gamma \left( \alpha \right) }\int_{1}^{t_{1}}\left( \log \frac{t_{1}}{s}\right) ^{\alpha -1}-\left( \log \frac{t_{2}}{s}\right) ^{\alpha -1}\frac{ds}{s} +\frac{\rho }{\Gamma \left( \alpha \right) }\int_{t_{1}}^{t_{2}}\left( \log \frac{t_{2}}{s}\right) ^{\alpha -1}\frac{ds}{s} \\ & \leq& \frac{\rho }{\Gamma \left( \alpha +1\right) }\left( \left( \log t_{1}\right) ^{\alpha }+\left( \log \frac{t_{2}}{t_{1}}\right) ^{\alpha }-\left( \log t_{2}\right) ^{\alpha }+\left( \log \frac{t_{2}}{t_{1}}\right) ^{\alpha }\right) \\ & \leq& \frac{2\rho }{\Gamma \left( \alpha +1\right) }\left( \log \frac{t_{2} }{t_{1}}\right) ^{\alpha }. \end{eqnarray*} As \(t_{1}\rightarrow t_{2}\) the right-hand side of the previous inequality is independent of \(x\) and tends to zero. Thus that \(\mathcal{A}(B_{\eta })\) is equicontinuous. So, the compactness of \(\mathcal{A}\) follows by Ascoli Arzela’s theorem.

Now for any \(x\in \lbrack a,b]\subset \mathbb{R}^{+}\), we define respectively the upper and lower control functions as follows \begin{equation*} H\left( t,x\right) =\underset{a\leq y\leq x}{\sup }f\left( t,y\right) ,\text{ }h\left( t,x\right) =\underset{x\leq y\leq b}{\inf }f\left( t,y\right) . \end{equation*} It is clear that these functions are nondecreasing on \([a,b]\).

Definition 10. Let \(\overline{x},\underline{x}\in \mathcal{E},\) \(a\leq \underline{x}\leq \overline{x}\leq b\), satisfying \begin{eqnarray*} \overline{x}\left( t\right) &\geq &\frac{1}{\Gamma \left( \alpha \right) } \int_{1}^{t}\left( \log \frac{t}{s}\right) ^{\alpha -1}H\left( s,\overline{x} \left( s\right) \right) \frac{ds}{s} +\lambda \int_{1}^{e}\overline{x}\left( s\right) ds+d, 1 \leq t \leq e, \end{eqnarray*} and \begin{eqnarray*} \underline{x}\left( t\right) &\leq &\frac{1}{\Gamma \left( \alpha \right) } \int_{1}^{t}\left( \log \frac{t}{s}\right) ^{\alpha -1}h\left( s,\underline{x }\left( s\right) \right) \frac{ds}{s} +\lambda \int_{1}^{e}\underline{x}\left( s\right) ds+d, \ 1 \leq t \leq e. \end{eqnarray*} Then the functions \(\overline{x}\) and \(\underline{x}\) are called a pair of upper and lower solutions for the equation (1).

Theorem 11. Assume that \(f:\left[ 1,e\right] \times \left[ 0,\infty \right) \rightarrow \left[ 0,\infty \right) \) is continuous, and \(\overline{x}\) and \( \underline{x}\) are respectively upper and lower solutions of (1), then the problem (1) has at least one positive solution.

Proof. Let \begin{equation*} \mathcal{K}=\left\{ x\in \mathcal{E},\text{ }\underline{x}\left( t\right) \leq x\left( t\right) \leq \overline{x}\left( t\right) ,\text{ }t\in \left[ 1,e\right] \right\} . \end{equation*} As \(\mathcal{K}\subset \mathcal{E}\) and \(\mathcal{K}\) is a nonempty bounded, closed and convex subset. By Lemma 9, \(\mathcal{A}:\mathcal{K} \rightarrow \mathcal{E}\) is completely continuous. Next, we show that if \( x\in \mathcal{K}\), we have \(\mathcal{A}x\in \mathcal{K}\). For any \(x\in \mathcal{K}\), we have \(\underline{x}\leq x\leq \overline{x}\), then \begin{align} \left( \mathcal{A}x\right) \left( t\right) & =\frac{1}{\Gamma \left( \alpha \right) }\int_{1}^{t}\left( \log \frac{t}{s}\right) ^{\alpha -1}f\left( s,x\left( s\right) \right) \frac{ds}{s}+\lambda \int_{1}^{e}x\left( s\right) ds+d \notag \\ & \leq \frac{1}{\Gamma \left( \alpha \right) }\int_{1}^{t}\left( \log \frac{t }{s}\right) ^{\alpha -1}H\left( s,\overline{x}\left( s\right) \right) \frac{ ds}{s}+\lambda \int_{1}^{e}\overline{x}\left( s\right) ds+d \notag \\ & \leq \overline{x}\left( t\right) , \label{4} \end{align} and \begin{align} \left( \mathcal{A}x\right) \left( t\right) & =\frac{1}{\Gamma \left( \alpha \right) }\int_{1}^{t}\left( \log \frac{t}{s}\right) ^{\alpha -1}f\left( s,x\left( s\right) \right) \frac{ds}{s}+\lambda \int_{1}^{e}x\left( s\right) ds+d \notag \\ & \geq \frac{1}{\Gamma \left( \alpha \right) }\int_{1}^{t}\left( \log \frac{t }{s}\right) ^{\alpha -1}h\left( s,\underline{x}\left( s\right) \right) \frac{ ds}{s}+\lambda \int_{1}^{e}\underline{x}\left( s\right) ds+d \notag \\ & \geq \underline{x}\left( t\right) . \label{5} \end{align} Thus, from (4) and (5), we obtain that \(\mathcal{A}x\in \mathcal{ K}\). We now see that all the conditions of the Schauder fixed point theorem are satisfied. Thus there exists a fixed point \(x\) in \(\mathcal{K}\). Therefore, the problem (1) has at least one positive solution \(x\) in \( \mathcal{K}\).

Corollary 12. Assume that \(\lambda =0\) and \(f:\left[ 1,e\right] \times \left[ 0,\infty \right) \rightarrow \left[ 0,\infty \right) \) is continuous, and there exist \(\lambda _{1},\lambda _{2}>0\) such that

Then the problem (1) has at least one positive solution \(x\in \mathcal{ E}\), moreover and

Proof. From (6) and the definition of control functions, we have

Now, let The above equation (10) has a positive solution \begin{align*} \overline{x}\left( t\right) & =\frac{\lambda _{2}}{\Gamma \left( \alpha \right) }\int_{1}^{t}\left( \log \frac{t}{s}\right) ^{\alpha -1}\frac{ds}{s} +d =\lambda _{2}\frac{\left( \log t\right) ^{\alpha }}{\Gamma \left( \alpha +1\right) }+d. \end{align*} Taking into account (9), we have \begin{align*} \overline{x}\left( t\right) & =\frac{\lambda _{2}}{\Gamma \left( \alpha \right) }\int_{1}^{t}\left( \log \frac{t}{s}\right) ^{\alpha -1}\frac{ds}{s} +d \geq \frac{1}{\Gamma \left( \alpha \right) }\int_{1}^{t}\left( \log \frac{t }{s}\right) ^{\alpha -1}H\left( s,\overline{x}\left( s\right) \right) \frac{ ds}{s}+d. \end{align*} It is clear that \(\overline{x}\) is the upper solution of (1). Now, let \begin{equation*} \mathfrak{D}_{1}^{\alpha }\underline{x}\left( t\right) =\lambda _{1},\text{ } \underline{x}\left( 1\right) =d. \end{equation*} which has also a positive solution \begin{align*} \underline{x}\left( t\right) & =\frac{\lambda _{1}}{\Gamma \left( \alpha \right) }\int_{1}^{t}\left( \log \frac{t}{s}\right) ^{\alpha -1}\frac{ds}{s} +d =\lambda _{1}\frac{\left( \log t\right) ^{\alpha }}{\Gamma \left( \alpha +1\right) }+d. \end{align*} By (9) and the same way that we used to search the upper solution, we conclude also that \(\underline{x}\) is the lower solution of (1). Therefore, from Theorem 11, we conclude that the problem (1) has at least one positive solution \(x\in \mathcal{E}\) which verifies the inequalities (7) and (8).

4. Uniqueness of positive solutions

In this section, we shall prove the uniqueness of positive solutions using the contraction mapping principle.

Theorem 13. Assume that \(f:\left[ 1,e\right] \times \left[ 0,\infty \right) \rightarrow \left[ 0,\infty \right) \) is continuous and there exists \(L>0\) such that

with Then the problem (1) has a unique positive solution \(x\in \mathcal{K}\).

Proof. From Theorem 11, it follows that (1) has at least one positive solution in \(\mathcal{K}\). Hence, we need only to prove that the operator \(\mathcal{A}\) defined in (3) is a contraction on \(\mathcal{K }\). In fact, since for any \(x_{1},x_{2}\in \mathcal{K}\), (11) and (12) are verified, then we have \begin{eqnarray*} \left\vert \left( \mathcal{A}x_{1}\right) \left( t\right) -\left( \mathcal{ A}x_{2}\right) \left( t\right) \right\vert & \leq& \frac{1}{\Gamma \left( \alpha \right) }\int_{1}^{t}\left( \log \frac{t }{s}\right) ^{\alpha -1}\left\vert f\left( s,x_{1}\left( s\right) \right) -f\left( s,x_{2}\left( s\right) \right) \right\vert \frac{ds}{s} +\lambda \int_{1}^{e}\left\vert x_{1}\left( s\right) -x_{2}\left( s\right) \right\vert ds \\ & \leq &\frac{\left( \log t\right) ^{\alpha }}{\Gamma \left( \alpha +1\right) }L\left\Vert x_{1}-x_{2}\right\Vert +\lambda \left( e-1\right) \left\Vert x_{1}-x_{2}\right\Vert \leq \left( \frac{L}{\Gamma \left( \alpha +1\right) }+\lambda \left( e-1\right) \right) \left\Vert x_{1}-x_{2}\right\Vert . \end{eqnarray*} Thus, \begin{equation*} \left\Vert \mathcal{A}x_{1}-\mathcal{A}x_{2}\right\Vert \leq \left( \frac{L}{ \Gamma \left( \alpha +1\right) }+\lambda \left( e-1\right) \right) \left\Vert x_{1}-x_{2}\right\Vert . \end{equation*} Hence, the operator \(\mathcal{A}\) is a contraction mapping by (12). Therefore, by the contraction mapping principle, we conclude that the problem (1) has a unique positive solution \(x\in \mathcal{K}\).

Now, we give an example to illustrate our results.

Example 1. We consider the following nonlinear fractional differential equation

where \(\alpha =2/5\), \(\lambda =0\), \(d=1\) and \(f\left( t,x\right) =\dfrac{1}{ 3+t}\left( \dfrac{t-1}{x+1}+2\right) \). Since \(f\) is continuous and \begin{equation*} \frac{2}{3+e}\leq f\left( t,x\right) \leq \frac{1}{2}, \end{equation*} for \(\left( t,x\right) \in \left[ 1,e\right] \times \left[ 0,\infty \right) \) , then by Corollary 12, (13) has a positive solution which verifies \(\underline{x}\left( t\right) \leq x\left( t\right) \leq \overline{x }\left( t\right) \) where \begin{equation*} \overline{x}\left( t\right) =\frac{1}{2}\frac{\left( \log t\right) ^{2/5}}{ \Gamma \left( 7/5\right) }+1\text{ and }\underline{x}\left( t\right) =\frac{2 }{3+e}\frac{\left( \log t\right) ^{2/5}}{\Gamma \left( 7/5\right) }+1, \end{equation*} are respectively the upper and lower solutions of (13). Also, we have \begin{equation*} \frac{L}{\Gamma \left( \alpha +1\right) }\simeq 0.339< s1, \end{equation*} then by Theorem 13, (13) has a unique positive solution which is bounded by \(\underline{x}\) and \(\overline{x}\).

5. Conclusion

In this paper, the existence and uniqueness of positive solutions for nonlinear Caputo-Hadamard fractional differential equations with integral boundary conditions were obtained using the Schauder fixed point theorem and the Banach contraction mapping principle. The obtained results extend some known results in the literature. An example is introduced to illustrate the main results of this paper.

Acknowledgments

The authors would like to thank the anonymous referee for his/her valuable comments and good advice.

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Competing Interests

The author(s) do not have any competing interests in the manuscript.