Abstract
We deal with the following Riemann–Liouville fractional nonlinear boundary value problem:
Under mild assumptions, we prove the existence of a unique continuous solution v to this problem satisfying
Our results improve those obtained by Zou and He (Appl. Math. Lett. 74:68–73, 2017).
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1 Introduction
Fractional differential equations have attracted great attention due to their ability to model various phenomena in applied sciences. The so-called fractional differential equations are specified by generalizing the standard integer-order derivative to arbitrary order. For more interesting theoretical results and scientific applications of fractional differential equations, we refer to the monographs of Diethelm [2] and Kilbas et al. [3] and references therein.
The existence, uniqueness, and global behavior of solutions for boundary value problems of fractional differential equations have been considered in several recent papers (see, e.g., [1, 4–9] and references therein).
Zou and He [1] investigated the problem
where \(\mathcal{D}^{\alpha }\) denotes the standard Riemann–Liouville fractional derivative, and f satisfies the following conditions:
-
(H1)
\(f\in C((0,1)\times \mathbb{R},\mathbb{R})\) and \(\int _{0}^{1} \vert f(x,0) \vert \,dx<\infty \);
-
(H2)
There exists \(q\in C((0,1),[0,\infty ))\) such that
$$ \bigl\vert f(x,v)-f(x,w) \bigr\vert \leq q(x) \vert v-w \vert ,\quad \forall x\in (0,1), v,w\in \mathbb{R,} $$and
$$ 0< \int _{0}^{1}q(x)\,dx< \infty. $$(1.2)
Let \(L>0\) be the minimum positive constant such that
where \(G_{\alpha }(x,y)\) is the Green’s function (given later in this paper) associated with problem (1.1). By using Banach’s contraction principle on some convenient Banach space they have obtained the following result.
Theorem 1.1
Under assumptions (H1)–(H2) and \(L<1\), problem (1.1) has a unique solution in \(C([0,1])\).
Motivated by this reault, we prove that the conclusion of Theorem 1.1 remains true under the following weaker assumptions:
-
(A1)
\(f\in C((0,1)\times \mathbb{R},\mathbb{R})\) and \(\int _{0}^{1}(1-x)^{\alpha -2} \vert f(x,0) \vert \,dx<\infty \);
-
(A2)
There exists \(q\in C((0,1),[0,\infty ))\) such that
$$ \bigl\vert f(x,v)-f(x,w) \bigr\vert \leq q(x) \vert v-w \vert ,\quad\forall x\in (0,1), v,w\in \mathbb{R,} $$and
$$ 0< M_{q,\alpha }:=\frac{1}{\Gamma (\alpha -1)} \int _{0}^{1}x^{\alpha -1}(1-x)^{ \alpha -1}q(x) \,dx< \infty. $$(1.4)
Remark 1.2
It is clear that conditions (H1)–(H2) imply (A1)–(A2).
Conversely, for \(\beta \in [ 1,\alpha -1)\), the function \(f(x,v):=(1-x)^{-\beta }(1+v)\) satisfies hypotheses (A1)–(A2) but not conditions (H1)–(H2). So assumptions (A1)–(A2) are weaker.
In this paper, for \(\alpha \in [ 2,3)\), we use the following notations:
• \(h(x):=x^{\alpha -1}(1-x)\), \(x\in [ 0,1]\).
• \(G_{\alpha }(x,y)\) denotes the Green’s function of the operator \(v\rightarrow -D^{\alpha }v\) with boundary conditions \(v(0)=v^{\prime }(0)=v(1)\).
• \(E:= \{ a>0: \int _{0}^{1}G_{\alpha }(x,y)h(y)q(y)\,dy \leq ah(x), x\in [ 0,1] \} \) (we will see that \(E \neq \emptyset\)).
We will prove that M is a positive constant satisfying the following range estimation:
• For \(a\in \mathbb{R}\), \(a^{+}:=\max (a,0)\).
• \(C_{h}([0,1]):=\{v\in C([0,1]):\text{ there is }\sigma >0\text{ such that }\vert v(x) \vert \leq \sigma h(x),x\in [ 0,1]\}\).
In the next remark, we list some properties of elements of \(C_{h}([0,1])\).
Remark 1.3
-
(i)
\(C_{h}([0,1])\) is a Banach space equipped with the following h-norm:
$$ \Vert v \Vert _{h}:=\inf \bigl\{ \sigma >0: \bigl\vert v(x) \bigr\vert \leq \sigma h(x), x\in [ 0,1]\bigr\} = \underset{x\in (0,1)}{ \sup }\frac{ \vert v(x) \vert }{h(x)}. $$(1.7) -
(ii)
\(v\in C_{h}([0,1])\) if and only if \(v=h\varphi \), where φ is a bounded continuous function in \((0,1)\).
Our main result is the following:
Theorem 1.4
Assume that (A1) and (A2) hold. If \(M<1\), then problem (1.1) has a unique solution v in \(C_{h}([0,1])\). In addition, for any \(v_{0}\in C_{h}([0,1])\), the iterative sequence \(v_{k}(x):=\int _{0}^{1}G_{\alpha }(x,y)f(y,v_{k-1}(y))\,dy\) converges to v with respect to the h-norm, and we have
Our paper is organized as follows. In Sect. 2, we improve the estimates on Green’s function \(G_{\alpha }\) obtained in [1, Lemma 2.2]. This allows us to obtain the range estimation (1.6). Our main result is proved in Sect. 3. Some examples and approximations are given at the end.
2 Preliminaries
Definition 2.1
([3])
Let \(f: ( 0,\infty ) \rightarrow \mathbb{R} \) be a measurable function.
-
(i)
The Riemann–Liouville fractional integral of order \(\gamma >0 \) for f is defined as
$$ I^{\gamma }f(x):=\frac{1}{\Gamma (\gamma )} \int _{0}^{x}(x-y)^{ \gamma -1}f(y)\,dy, $$where Γ is the Euler gamma function.
-
(ii)
The Riemann–Liouville fractional derivative of order \(\gamma >0\) for f is defined as
$$ \mathcal{D}^{\gamma }f(x):= \frac{1}{\Gamma ( n-\gamma ) } \biggl( \frac{d}{dx} \biggr) ^{n} \int _{0}^{x} ( x-y ) ^{n-\gamma -1}f ( y ) \,dy, $$where \(n=[\gamma ]+1\), and \([\gamma ]\) is the integer part of γ.
By [10, Lemma 2.2] the Green’s function associated with problem (1.1) is given by
Lemma 2.2
The Green’s function \(G_{\alpha } ( x,y ) \) has the following properties:
-
(i)
\(G_{\alpha }(x,y)\) is a nonnegative continuous function on \([ 0,1 ] \times [ 0,1 ] \).
-
(ii)
For all \(x,y\in [0,1]\), we have
$$ H_{\alpha }(x,y)\leq G_{\alpha } ( x,y ) \leq ( \alpha -1 ) H_{\alpha }(x,y), $$(2.2)where \(H_{\alpha }(x,y):=\frac{1}{\Gamma (\alpha )}x^{\alpha -2}(1-y)^{ \alpha -2}\min (x,y)(1-\max (x,y))\).
Proof
It is obvious that (i) holds. Now we prove (ii). From (2.1), for all \(x,y\in (0,1)\), we have
Since for \(\lambda >0\) and \(t\in [0,1]\),
we deduce that
Using this fact and (2.4), we obtain
Hence estimates (2.2) follow from
□
Remark 2.3
In [1, Lemma 2.2], the authors stated that for all \(x,y\in [0,1]\),
-
(i)
\(x^{\alpha -1}(1-x)y(1-y)^{\alpha -1}\leq \Gamma (\alpha )G_{\alpha }(x,y) \leq ( \alpha -1 )y(1-y)^{\alpha -1}\),
-
(ii)
\(x^{\alpha -1}(1-x)y(1-y)^{\alpha -1}\leq \Gamma (\alpha )G_{\alpha }(x,y) \leq ( \alpha -1 )x^{\alpha -1}(1-x)\).
Note that since for all \(x,y\in [0,1]\),
we get
Combining this fact with (2.2), we immediately obtain inequalities (i) and (ii).
Therefore estimates (2.2) improve those stated in [1, Lemma 2.2].
Lemma 2.4
Let \(q\in C((0,1),[0,\infty ))\) and assume that \(0< M_{q,\alpha }<\infty \). Then
where M is the constant defined by (1.5).
Proof
Let
where \(h(x):=x^{\alpha -1}(1-x)\), \(x\in [ 0,1]\).
By (2.2) we obtain
It follows that \(E\neq \emptyset \) and \(M\leq M_{q,\alpha }\), where \(M:=\inf E\).
On the other hand, using again (2.2) and that
\(\min (x,y)(1-\max (x,y))\geq xy(1-x)(1-y)\) for \(x,y\in [ 0,1]\), we deduce that for any \(a\in E\),
Hence for each \(a\in E\),
Therefore \(M\geq M_{q,\alpha +1}\), that is, \(M\in [ M_{q,\alpha +1},M_{q,\alpha }]\). □
Remark 2.5
From Lemma 2.4 it is obvious that if \(M_{q,\alpha }<1\), then
\(M:=\inf E<1\). Note that the inequality \(M_{q,\alpha }<1\) can be verified for a large class of functions q, including the singular cases. For example, let
\(B(a,b):=\int _{0}^{1}t^{a-1}(1-t)^{b -1}\,dt\) for \(a>0\) and \(b>0\).
Then by using MATLAB we obtain
-
(i)
If \(q\in C((0,1))\) with \(q>0\) and \(\Vert q \Vert _{\infty }\leq 1\), then
$$ M_{q,\alpha }\leq \frac{B(\alpha,\alpha )}{\Gamma (\alpha -1)}< 1. $$ -
(ii)
If \(q(x):=(1-x)^{-\frac{\alpha }{2}}\), then
$$ M_{q,\alpha }=\frac{B(\alpha,\frac{\alpha }{2})}{\Gamma (\alpha -1)}< 1. $$ -
(iii)
If \(q(x):=x^{-\frac{\alpha }{3}}(1-x)^{-\frac{\alpha }{2}} \), then
$$ M_{q,\alpha }= \frac{B(\frac{2\alpha }{3},\frac{\alpha }{2})}{\Gamma (\alpha -1)}< 1. $$
3 Existence and uniqueness
We need the following useful lemma.
Lemma 3.1
Let \(2<\alpha <3\), and let φ be a function such that \(x\rightarrow (1-x)^{\alpha -1}\varphi (x)\in C((0,1))\cap L^{1}((0,1))\). Then the unique continuous solution of the problem
is given by
Proof
Let φ be a function such that \(x\rightarrow (1-x)^{\alpha -1}\varphi (x)\in C((0,1))\cap L^{1}((0,1))\). Since by Lemma 2.2, \(G_{\alpha }(x,y)\) belongs to \(C( [ 0,1 ] \times [ 0,1 ] )\) with
we deduce by the dominated convergence theorem that \(V\varphi \in C([0,1])\)
and \(V\varphi (0)=V\varphi (1)=0\). Therefore \(I^{3-\alpha }(V \vert \varphi \vert )\) is bounded on \([0,1]\). By Fubini’s theorem we obtain
where \(K(x,r):=\frac{1}{\Gamma (3-\alpha )}\int _{0}^{x}(x-y)^{2-\alpha }G_{\alpha }(y,r)\,dy\).
Simple calculation gives
Hence, for \(x\in (0,1)\), we have
This implies that
Now, since for each \(y\in (0,1)\),
by the dominated convergence theorem we obtain \((V\varphi )^{\prime }(0)=0\).
To prove the uniqueness, let \(v,w\in C([0,1])\) be two solutions of problem (3.1) and set \(\theta:=v-w\). Then \(\theta \in C([0,1])\), and we have
By [3, Corollary 2.1] there exist \(c_{1},c_{2},c_{3}\in \mathbb{R}\) such that
Applying the boundary conditions, we obtain \(c_{3}=c_{2}=c_{1}=0\), that is, \(v=w\). □
Remark 3.2
The conclusion of Lemma 3.1 remains true for \(\alpha =3\).
Proof of Theorem 1.4
Assume that (A1) and (A2) hold and \(M<1\), where M is given by (1.5). Let us prove that problem (1.1) has a unique solution v in \(C_{h}([0,1])\). In addition, for any \(v_{0}\in C_{h}([0,1])\), the iterative sequence \(v_{k}(x):=\int _{0}^{1}G_{\alpha }(x,y)f(y,v_{k-1}(y))\,dy\) converges to v with respect to the h-norm, and we have
To this end, define the operator T by
We claim that T is a contraction operator from \((C_{h}([0,1]), \Vert \cdot \Vert _{h})\) into itself.
Let \(v\in C_{h}([0,1])\), and let \(\sigma >0\) be such that \(\vert v(x) \vert \leq \sigma h(x)\) for all \(x\in [ 0,1]\).
Since by Lemma 2.2(ii), \(0\leq G_{\alpha }(x,y)\leq \frac{1}{\Gamma (\alpha -1)}(1-y)^{ \alpha -2}\), it follows from \((A2)\) that
Since \(G_{\alpha }(x,y)\) is continuous on \([ 0,1 ] \times [ 0,1 ] \), by (A1)–(A2) and the dominated convergence theorem we deduce that \(Tv\in C([0,1])\).
Furthermore, from Lemma 2.2(ii) we have
Hence by using (3.3) and similar arguments as before we obtain
and thus \(T(C_{h}([0,1]))\subset C_{h}([0,1])\).
Next, for any \(v,w\in C_{h}([0,1])\), by using \((A2)\) we obtain that for \(x\in [0,1]\),
Hence
Since \(M<1\), T becomes a contraction operator in \(C_{h}([0,1])\). So there exists a unique \(v\in C_{h}([0,1])\) satisfying
It remains to prove that v is a solution of problem (1.1). Indeed, it is clear that \(x\rightarrow (1-x)^{\alpha -1}f(x,v(x))\in C((0,1))\). Next, by using \((A2)\) and \(v\in C_{h}([0,1])\) we obtain
Therefore by (A1) and (A2) it follows that \(x\rightarrow (1-x)^{\alpha -1}f(x,v(x))\in L^{1}((0,1))\). Hence from Lemma 3.1 we derive that v is a solution of problem (1.1).
Finally, it is well known that for any \(v_{0}\in \) \(C_{h}([0,1])\), the iterative sequence
\(v_{k}(x):=\int _{0}^{1}G_{\alpha }(x,y)f(y,v_{k-1}(y))\,dy\) converges to v, and we have
□
Example 3.3
Let \(2<\alpha \leq 3\). Consider the problem
where \(q\in C((0,1))\) with \(q>0\) and \(\Vert q \Vert _{\infty }\leq 1\). Let \(f(x,v):=q(x)\cos v\) for \((x,v)\in (0,1)\times \mathbb{R}\). We have \(f\in C((0,1)\times \mathbb{R},\mathbb{R})\) and
So assumption (A1) is verified.
On the other hand, since \(v\rightarrow \cos v\) is a Lipschitz function, we obtain
By Lemma 2.4 and Remark 2.5(i) we have
Hence by Theorem 1.4 problem (3.5) has a unique solution \(v\in C_{h}([0,1])\).
Example 3.4
Let \(2<\alpha \leq 3\) and consider the singular problem
In this case, we have \(f(x,v)=(1-x)^{-\frac{\alpha }{2}}(1+\sin v)\) for \((x,v)\in (0,1)\times \mathbb{R}\).
So \(f\in C((0,1)\times \mathbb{R},\mathbb{R})\) and \(\int _{0}^{1}(1-x)^{\alpha -2} \vert f(x,0) \vert \,dx=\int _{0}^{1}(1-x)^{\frac{\alpha }{2} -2}\,dx< \infty \), that is, assumption (A1) is satisfied.
On the other hand, we clearly have
where \(q(x):=(1-x)^{-\frac{\alpha }{2}}\).
From Lemma 2.4 and Remark 2.5(ii) we deduce that
Hence by Theorem 1.4 this problem has a unique solution \(v\in C_{h}([0,1])\). In particular, for \(\alpha =\frac{5}{2}\), the unique solution is approximated (see Fig. 1) by the iterative sequence \(v_{k}(x):=\int _{0}^{1}G_{\frac{5}{2}}(x,y)(1-y)^{-\frac{5}{4}}(1+ \sin (v_{k-1}(y)))\,dy\) with \(v_{0}(x)=x^{\frac{3}{2}}(1-x)\), \(x\in [ 0,1]\).
Example 3.5
Consider the problem
As in Example 3.4, we verify that assumptions (A1) and (A2) are satisfied. Therefore by Theorem 1.4 problem (3.7) has a unique solution \(v\in C_{h}([0,1])\), and the iterative sequence defined by \(v_{0}(x):=x^{\frac{3}{2}}(1-x)\), \(x\in [ 0,1]\), and
converges to v (see Fig. 2).
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References
Zou, Y., He, G.: On the uniqueness of solutions for a class of fractional differential equations. Appl. Math. Lett. 74, 68–73 (2017)
Diethelm, K.: The Analysis of Fractional Differential Equations. Lecture Notes in Mathematics, vol. 2004. Springer, Berlin (2010)
Kilbas, A.A., Srivastava, H.M., Trujillo, J.J.: Theory and Applications of Fractional Differential Equations. North-Holland Mathematics Studies, vol. 204. Elsevier, Amsterdam (2006)
Atangana, A., Akgül, A.: Analysis of new trends of fractional differential equations. In: Fractional Order Analysis, pp. 91–111. Wiley, USA (2020)
Bai, Z.: On positive solutions of a nonlocal fractional boundary value problem. Nonlinear Anal. 72(2), 916–924 (2010)
Bai, Z., Zhang, S., Sun, S., Yin, C.: Monotone iterative method for fractional differential equations. Electron. J. Differ. Equ. 2016, 6 (2016)
Cui, Y.: Uniqueness of solution for boundary value problems for fractional differential equations. Appl. Math. Lett. 51, 48–54 (2016)
Liang, S., Zhang, J.: Positive solutions for boundary value problems of nonlinear fractional differential equation. Nonlinear Anal. 71(11), 5545–5550 (2009)
Zhang, X., Liu, L., Wu, Y.: The uniqueness of positive solution for a fractional order model of turbulent flow in a porous medium. Appl. Math. Lett. 37, 26–33 (2014)
Zhang, X., Liu, L., Wu, Y.: Multiple positive solutions of a singular fractional differential equation with negatively perturbed term. Math. Comput. Model. 55(3–4), 1263–1274 (2012)
Acknowledgements
The authors express their thanks to the referees for their valuable suggestions and comments, which improved the presentation.
Funding
The authors would like to extend their sincere appreciation to the Deanship of Scientific Research at King Saud University for its funding this Research group NO (RG-1435-043).
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Bachar, I., Mâagli, H. & Eltayeb, H. Existence and uniqueness of solutions for a class of fractional nonlinear boundary value problems under mild assumptions. Adv Differ Equ 2021, 22 (2021). https://doi.org/10.1186/s13662-020-03176-w
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DOI: https://doi.org/10.1186/s13662-020-03176-w