Abstract
In this paper, we consider a class of fractional differential equations with integral boundary conditions which involve two disturbance parameters. By using the Guo-Krasnoselskii fixed point theorem, new results on the existence and nonexistence of positive solutions for the boundary value problem are obtained. And the impact of the disturbance parameters on the existence of positive solutions is also investigated. Finally, we give some examples to illustrate our main results.
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1 Introduction
The theory of boundary value problems for ordinary differential equations and functional differential equations plays an important role in many research fields of science and engineering; for details, see [1–9] and the references therein. Meanwhile, fractional differential equations have also widely appeared in various fields such as physics, mechanics, electricity, biology, control theory, etc. Therefore, the study of fractional differential equations has gained prominence and has been growing rapidly, see[10–18]. Last but not least, as an important part of fractional differential equations, the integral boundary value problems have also been extensively researched, see [19–23].
In [21], Jia and Liu investigated the existence and nonexistence of positive solutions for the following integral boundary value problem of fractional differential equations with a disturbance parameter in the boundary conditions and the impact of the disturbance parameter on the existence of positive solutions
where \(^{\mathrm{C}} D^{\delta}\) is the Caputo fractional derivative, \(1<\delta\leq 2\), \(f\in C([0,1]\times\mathbb{R}^{+},\mathbb{R}^{+})\) and \(g\in C([0,1], \mathbb{R}^{+})\).
In this paper, we are concerned with the Riemann-Liouville fractional differential equation
with the integral boundary conditions
where \(D_{0{+}}^{\alpha}\) is the standard Riemann-Liouville fractional derivative of order α, \(3<\alpha\leq4\), \(f:[0,1]\times[0,+\infty)\rightarrow[0,+\infty)\) is a continuous function, \(g_{1},g_{2}\in L^{1}[0,1]\), and \(a,b\geq0\). The existence and nonexistence of positive solutions for the integral boundary value problem (1.1)-(1.2) and the impact of the disturbance parameters a, b on the existence of positive solutions is also investigated. Finally, we give two examples to illustrate our results.
2 Preliminaries
In this section, we present some useful definitions and related lemmas.
Definition 2.1
(See [12])
Let \(\alpha>0\). The fractional integral operator of a function \(y:(0,+\infty )\rightarrow \mathbb{R}\) is given by
provided the integral exists.
Definition 2.2
(See [12])
Let \(\alpha>0\). The Riemann-Liouville fractional derivative of a function \(y:(0,+\infty)\rightarrow\mathbb{R}\) is given by
where \({z}\in\mathbb{N}\), \({z-1}<\alpha<z\), provided the right-hand side is pointwise defined on \((0,\infty)\).
Lemma 2.1
(See [12])
For \(\alpha>0\), \(z\in\mathbb{N}\) and \(z-1<\alpha<z\), if \(x\in L^{1}[0,1]\) and \(I_{0{+}}^{z-\alpha}x\in AC^{z}[0,1]\), we have the equation
where \(c_{i}\in\mathbb{R}\), \(i=1,2,3,\ldots,z\).
Lemma 2.2
The boundary value problem (1.1)-(1.2) is equivalent to the following integral equation:
where
and
Proof
Assume that \(x=x(t)\) is a solution of (1.1), it follows from Lemma 2.1 that
where \(c_{i}\in\mathbb{R}\), \(i=1,2,3,4\).
From the boundary conditions \(x(0)=x'(0)=0\), we get \(c_{3}=c_{4}=0\).
And from the boundary conditions \(x(1)=\int_{0}^{1}g_{1}(s)x(s)\, \mathrm {d}s+a\) and \(x'(1)=\int_{0}^{1}g_{2}(s)x(s)\,\mathrm{d}s-b\), we can get
Then
Hence, \(x=x(t)\) is a solution of the integral equation (2.1) if it is the solution of the boundary value problem (1.1)-(1.2), and vice versa.
The proof is completed. □
Lemma 2.3
Let G be defined by (2.2), then
-
(1)
\(G(t,s)\in C[0,1]\times[0,1]\), and \(G(t,s)>0\) for any \(t, s\in(0,1)\),
-
(2)
\((\alpha-2)q(t)k(s)\frac{1}{\Gamma(\alpha)}\leq G(t,s)\leq M_{0}k(s)\frac{1}{\Gamma(\alpha)}\leq M_{0}\frac{1}{\Gamma(\alpha)}\) for any \(t, s\in[0,1]\), where
$$M_{0}=\max\bigl\{ \alpha-1,(\alpha-2)^{2}\bigr\} ,\qquad q(t)=t^{\alpha-2}(1-t)^{2},\qquad k(s)=s^{2}(1-s)^{\alpha-2}. $$
Proof
(1) It is easy to show that the result holds.
(2) For \(s\leq t\), we could use the mean value theorem of differential calculus and get
and
For \(s\geq t\), we have that
The proof is completed. □
Now we make the following assumption.
-
(B0)
\(g_{1},g_{2}\in L^{1}[0,1]\) such that \(0\leq\inf_{t,s\in[0,1]}H(t,s)<\sup_{t,s\in[0,1]}H(t,s):=M<1\).
Let the Banach space \(C [0, 1]\) be endowed with the norm \(\|x\|:=\max_{t\in[0,1]}|x(t)|\), and let
then P is a cone in \(C[0,1]\).
We define an operator \(A:P \rightarrow C[0,1] \) by
Lemma 2.4
Assume (B0) holds, then the operator A satisfies the following properties:
-
(1)
A is a bounded linear operator;
-
(2)
\(A(P)\subset P\);
-
(3)
the operator A is reversible;
-
(4)
\(\|(I-A)^{-1}\| \leq\frac{1}{1-M}\).
Proof
By (B0), it is obvious that (1), (2) hold.
(3) Since \(M<1\), we get that \(\|Ax\|\leq M\|x\|<\|x\|\), then \(\|A\|\leq M<1\), so that \(I-A\) is reversible.
(4) Let \(y(t)=x(t)-Ax(t)\), that is, \(x(t)=y(t)+Ax(t)\), \(x(t)=(I-A)^{-1}y(t)\), and \(y\in C[0,1]\) for \(t\in[0,1]\). From the definition of operator A, we have that
Let
We apply the method of iteration to solve the above equation.
According to this method, we can get that
where
Because of \(0\leq H(t,s)< M<1\), we have that
Since \((I-A)^{-1}y(t)=x(t)\), we get
and
So
The proof is completed. □
We define another operator \(T:P\rightarrow C[0,1]\),
It can be easily shown that \(T:P\rightarrow P\). \(x(t)\) is a solution of the boundary value problem (1.1)-(1.2) if and only if it satisfies
Hence,
and
Let
Lemma 2.5
Assume that condition (B0) holds, then the operator \((I-A)^{-1}T:P\rightarrow P_{0}\) is completely continuous.
Proof
From the continuity and the non-negativeness of functions G, R and f, we have that if \(x\in P\), then \((I-A)^{-1}(Tx)(t)\geq0\) and \((I-A)^{-1}(Tx)\in P\).
It follows from (2.4) and Lemma 2.3, for \(x\in P\) and \(t\in[0,1]\),
Hence,
By (2.4), (2.5) and (2.6), we have
It follows from (2.7) that
Hence \((I-A)^{-1}T(P)\subset P_{0}\).
Let \(\{x_{n}\}\subset P\), \(x\in P\), and \(\|x_{n}-x\|\rightarrow0\) as \(n\rightarrow+\infty\), there exists a constant \(r>0\) such that \(\|x_{n}\| \leq r\) and \(\|x\|\leq r\). We have
By the Lebesgue dominated convergence theorem, we get
So
Then the operator \((I-A)^{-1}T\) is continuous.
Let \(\Omega\subset P\) be bounded. Then there exists a positive constant \(l>0\) such that \(\|x\|\leq l\) for all \(x\in\Omega\). Let \(N= \max_{0\leq t\leq1,0\leq x\leq l}|f(t,x)|+1\). By (2.7), for all \(x\in\Omega\), we have
which means \((I-A)^{-1}T(\Omega)\) is bounded in P.
In addition, for any given \(x\in\Omega\), because \(G(t,s)\) is continuous for \((t, s)\in[0,1]\times[0,1]\), then it must be uniformly continuous. So, for any \(\varepsilon>0\), there exists a constant \(\delta_{0}>0\) such that for all \(s\in[0,1]\), as \(|t_{1}-t_{2}|<\delta_{0}\), we have that
For each \(x\in\Omega\),
We have
and \((I-A)^{-1}T(\Omega)\) is equicontinuous in \(P_{0}\).
Now, according to the Arzela-Ascoli theorem, we conclude that \((I-A)^{-1}T(\Omega)\) is relatively compact.
Therefore, \((I-A)^{-1}T:P\rightarrow P_{0}\) is a completely continuous operator.
The proof is completed. □
By Lemma 2.2, we can easily deduce that the following lemma holds.
Lemma 2.6
Assume \(x\in C[0,1]\), \(D_{0^{+}}^{\alpha}x\in L^{1}[0,1]\). Then the boundary value problem (1.1)-(1.2) has a positive solution if and only if the operator \((I-A)^{-1}T\) has a fixed point in P. Furthermore, if x is a positive solution of the fractional boundary value problem (1.1)-(1.2), then \(x\in P_{0}\).
To prove the existence of positive solution for the boundary value problem (1.1)-(1.2), we state the following Guo-Krasnoselskii fixed point theorem, see [24].
Lemma 2.7
Let E be a Banach space and \(P\subset E\) be a cone. Assume that \(\Omega_{1}\), \(\Omega_{2}\) are bounded open subsets of E with \(\theta \in\Omega_{1}\subset\overline{\Omega}_{1}\subset\Omega_{2}\), and let \(T:P\cap (\overline{\Omega}_{2}\setminus\Omega_{1})\rightarrow P\) be a completely continuous operator such that either
-
(1)
\(\|T(x)\|\leq\|x\|\), \(\forall x\in P\cap\partial\Omega_{1}\); and \(\| T(x)\|\geq\|x\|\), \(\forall x\in P\cap\partial\Omega_{2}\), or
-
(2)
\(\|T(x)\|\geq\|x\|\), \(\forall x\in P\cap\partial\Omega_{1}\); and \(\| T(x)\|\leq\|x\|\), \(\forall x\in P\cap\partial\Omega_{2}\).
Then the operator T has at least one fixed point in \(P\cap(\overline {\Omega}_{2}\setminus\Omega_{1})\).
3 Existence and nonexistence of positive solutions
Denote
Theorem 3.1
Suppose (B0) holds, \(f^{0}<\rho_{1}\) and \(f_{\infty}>\rho_{2}\). Then there exist small enough \(a_{0}\) and \(b_{0}\) such that the boundary value problem (1.1)-(1.2) has at least one positive solution for \(0\leq a\leq a_{0}\) and \(0\leq b\leq b_{0}\).
Proof
Since \(f^{0}<\rho_{1}\), there exists a constant \(r_{1}>0\) such that
for all \(t\in[0,1]\) and \(x\in[0,r_{1}]\).
Let \(\Omega_{1}=\{x\in P_{0}: \|x\|< r_{1}\}\), \(0\leq a\leq a_{0}\), \(0\leq b\leq b_{0}\) and \(\max\{a_{0},b_{0}\}\leq\rho_{1}r_{1}\).
By Lemma 2.3, for \(x\in\partial\Omega_{1}\), we have \(\|x\| =r_{1}\) and
So we get \(\|(I-A)^{-1}Tx\|\leq\|x\|\), \(x\in\partial\Omega_{1}\).
Since \(f_{\infty}>\rho_{2}\), there exists a constant \(R>0\) such that
for all \(t\in[\frac{1}{4},\frac{3}{4}]\) and \(x\in[R,+\infty)\).
Let \(r_{2}>\max\{r_{1},\frac{R}{\sigma}\}\) and \(\Omega_{2}=\{x\in P_{0}:\|x\| < r_{2}\}\).
For all \(x\in\partial\Omega_{2}\), we have that \(\|x\|=r_{2}\) and \(x(t)\geq\frac{q(t)(1-M)}{M_{0}}\|x\|\geq\sigma\|x\| =\sigma r_{2}>R\) for \(t\in[\frac{1}{4},\frac{3}{4}]\). Then
So \(\|(I-A)^{-1}Tx\|\geq\|x\|\), \(x\in\partial\Omega_{2}\).
By Lemma 2.7, we conclude that the operator \((I-A)^{-1}T\) has at least one fixed point in \(P_{0}\cap(\overline{\Omega}_{2}\setminus\Omega_{1})\), which implies that the boundary value problem (1.1)-(1.2) has a positive solution.
The proof is completed. □
Theorem 3.2
Assume (B0) holds, \(f^{\infty}<\rho_{1}\) and \(f_{0}>\rho_{2}\). Then there exist small enough \(a_{0}\) and \(b_{0}\) such that the boundary value problem (1.1)-(1.2) has at least one positive solution for \(0\leq a\leq a_{0}\) and \(0\leq b\leq b_{0}\).
Proof
Since \(f^{\infty}<\rho_{1}\), for \(\varepsilon=\frac{\rho_{1}-f^{\infty }}{2}>0\), there exists a constant \(R_{1}>0\) such that \(f(t,x)<(\rho_{1}-\varepsilon)x\) for \(t\in[0,1]\) and \(x\in[R_{1},+\infty)\).
Let \(L=\max_{(t,x)\in[0,1]\times[0,R_{1}]}f(t,x)\), so
Let \(r_{3}>\max\{R_{1},\frac{L}{\varepsilon}\}\), \(\Omega_{3}=\{x\in P_{0}:\| x\|< r_{3}\}\), \(0\leq a\leq a_{0} \), \(0\leq b\leq b_{0}\) and \(\max\{a_{0},b_{0}\} \leq\rho_{1}r_{3}\).
For all \(x\in\partial\Omega_{3}\), we have \(\|x\|=r_{3}\) and
So \(\|(I-A)^{-1}Tx\|\leq\|x\|\), \(x\in\partial\Omega_{3}\).
Since \(f_{0}>\rho_{2}\), there exists a constant \(0< r_{4}< R_{1}\) such that \(f(t,x)>\rho_{2}x\) for \(t\in[\frac{1}{4},\frac{3}{4}]\) and \(x\in[0,r_{4}]\).
Let \(\Omega_{4}=\{x\in P_{0}:\|x\|< r_{4}\}\). Similar to the proof of Theorem 3.1, we show \(\|(I-A)^{-1}Tx\|\geq\|x\|\), \(x\in\partial\Omega_{4}\).
By Lemma 2.7, we conclude that the operator \((I-A)^{-1}T\) has at least one fixed point in \(P_{0}\cap(\overline{\Omega}_{2}\setminus\Omega _{1})\), which implies that the boundary value problem (1.1)-(1.2) has at least one positive solution.
The proof is completed. □
Theorem 3.3
Suppose (B0) holds, \(f_{\infty}>\rho_{2}\). Then there exist large enough positive constants \(a_{1}\) and \(b_{1}\) such that the boundary value problem (1.1)-(1.2) has no positive solution for \(a>a_{1}\) and \(b>b_{1}\).
Proof
Assume that for any large enough \(a>0\) and \(b>0\), the boundary value problem (1.1)-(1.2) has a positive solution \(x(t)\).
Since \(f_{\infty}>\rho_{2}\), there exists a large enough constant \(R_{0}>0\) such that
Let \(\min\{a_{1},b_{1}\}>2^{\alpha-1}R_{0}\), \(a>a_{1}\) and \(b>b_{1}\). So \(\alpha a+b>\alpha a_{1}+b_{1}>(\alpha+1)2^{\alpha-1}R_{0}>2^{\alpha-1}R_{0}\).
By (2.1), Lemma 2.3 and (B0), we have
Hence,
and we get \(\|x\|>R_{0}\).
On the other hand, in view of Lemma 2.6, \(x\in P_{0}\). Then \(x(t)\geq\sigma\|x\|>\sigma R_{0}\) for \(t\in[\frac{1}{4},\frac{3}{4}]\). Therefore,
So \(\|x\|>\|x\|+R_{0}\), which is a contradiction. Thus, there exist large enough positive constants \(a_{1}\) and \(b_{1}\) such that the boundary value problem (1.1)-(1.2) has no positive solution for \(a>a_{1}\) and \(b>b_{1}\). □
4 Examples
To illustrate our main results, we present the following examples.
Example 4.1
We consider the boundary value problem
and we can establish the following results:
-
(1)
The boundary value problem (4.1) has at least one positive solution if parameters \(a\in[0,0.001)\) and \(b\in[0,0.001)\).
-
(2)
The boundary value problem (4.1) has no positive solution if parameters \(a\in(2.15\times10^{10},+\infty)\) and \(b\in(2.15\times 10^{10},+\infty)\).
Proof
The boundary value problem (4.1) can be regarded as the boundary value problem (1.1)-(1.2), where \(\alpha=\frac{10}{3}\), \(g_{1}(s)=\frac{1}{3}\), \(g_{2}(s)=\frac {2}{3}\), \(f(t,x)=x^{\frac{3}{2}}+x^{2}\sin{t}\).
Then \(M=\frac{1}{3}\), \(M_{0}=\frac{7}{3}\), \(\sigma=\min_{t\in [\frac{1}{4},\frac{3}{4}]}\frac{q(t)(1-M)}{M_{0}}=\frac{(1-M)3^{\alpha -2}}{M_{0} 4^{\alpha}}=0.0122\), \(\rho_{1}=\frac{(1-M)\Gamma(\alpha )}{M_{0}+\alpha\Gamma(\alpha)}=0.16\), \(\rho_{2}=\frac{7\text{,}680\Gamma(\alpha )2^{\alpha}}{203\sigma(\alpha-2)}=65\text{,}296.4\) and
(1) Let \(r_{1}=0.0064\), we choose \(\max\{a_{0},b_{0}\}<\rho_{1}r_{1}=0.001\). When \(x\in(0,0.0064]\), \(t\in[0,1]\), we have \(f(t,x)\leq\rho_{1}r_{1}\). Then, by Theorem 3.1, when \(a\in[0,0.001)\) and \(b\in[0,0.001)\), the boundary value problem (4.1) has a positive solution.
(2) Let \(R_{0}=4.26\times10^{9}\), when \(x\in[5.2\times10^{7},+\infty)\) and \(t\in[\frac{1}{4},\frac{3}{4}]\), so we choose \(\min\{a_{1},b_{1}\} >2^{\alpha-1}R_{0}=2.15\times10^{10}\). By Theorem 3.3, for \(a\in (2.15\times10^{10},+\infty)\) and \(b\in(2.15\times10^{10},+\infty)\), the boundary value problem (4.1) has no positive solution. □
Example 4.2
We consider the boundary value problem
The boundary value problem (4.2) has a positive solution for parameters \(a\in[0,8.16)\) and \(b\in[0,8.16)\).
Proof
Where \(\alpha=\frac{13}{4}\), \(g_{1}(s)=\frac{1}{2}\), \(g_{2}(s)=\frac{3}{4}\), \(f(t,x)=\frac{(t\sqrt{x}+1)x^{\frac {1}{3}}}{2+\sqrt{x}}\), we have that \(M_{0}=\frac{9}{4}\), \(M=\frac{1}{2}\), \(\sigma=\min_{t\in[\frac{1}{4},\frac {3}{4}]}\frac{q(t)(1-M)}{M_{0}}=\frac{(1-M)3^{\alpha-2}}{M_{0} 4^{\alpha }}=0.0097\), \(\rho_{1}=\frac{(1-M)\Gamma(\alpha)}{M_{0}+\alpha\Gamma (\alpha)}=0.12\) and \(\rho_{2}=\frac{7\text{,}680\Gamma(\alpha)2^{\alpha}}{203\sigma(\alpha-2)}=75\text{,}722\).
We set \(r_{3}=68.04\), when \(x\in(68.04,+\infty)\), \(t\in[0,1]\), and we choose \(\max\{a_{0},b_{0}\}<\rho_{1}r_{3}=8.16\) for \(a\in[0,8.16)\) and \(b\in [0,8.16)\). So we have
By Theorem 3.2, the boundary value problem (4.2) has a positive solution. □
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Acknowledgements
This work is supported by the National Natural Science Foundation of China (No. 11171220) and the Hujiang Foundation of China (B14005).
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Wang, X., Liu, X. & Deng, X. Existence and nonexistence of positive solutions for fractional integral boundary value problem with two disturbance parameters. Bound Value Probl 2015, 186 (2015). https://doi.org/10.1186/s13661-015-0450-1
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DOI: https://doi.org/10.1186/s13661-015-0450-1