Abstract
In this paper, we consider the linear fractional differential equation
![](http://media.springernature.com/lw300/springer-static/image/art%3A10.1186%2Fs13660-020-02448-z/MediaObjects/13660_2020_2448_Equa_HTML.png)
By obtaining the Green’s function we derive the Lyapunov-type inequality for such a boundary value problem. Furthermore, we use the contraction mapping theorem to study the existence of a unique solution for the corresponding nonlinear problem.
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1 Introduction
In 1907, Lyapunov [1] stated the following outstanding result.
Theorem 1.1
([1])
If the boundary value problem (BVP)
has a nontrivial solution, then we have the following Lyapunov inequality:
Inequality (1) is very useful in various problems related to differential equations. Since the appearance of Lyapunov’s fundamental paper [1], many improvements and generalizations of inequality (1) for integer-order (second- and higher-order) BVPs have appeared in the literature; we refer the reader to the summary reference by Tiryaki [2].
Recently, the studies on Lyapunov’s inequality for fractional boundary value problem (FBVP) have begun, in which fractional derivatives (Riemann–Liouville derivative or Caputo derivative
) are used instead of the classical ordinary derivative. Such a work was initiated by Ferreira [3] in 2013, who obtained a Lyapunov inequality for the following differential equation with Riemann–Liouville fractional derivative:
![](http://media.springernature.com/lw293/springer-static/image/art%3A10.1186%2Fs13660-020-02448-z/MediaObjects/13660_2020_2448_Equ2_HTML.png)
subject to the boundary value condition
Next, in 2014, Ferreira [4] obtained a Lyapunov inequality for the following differential equation with Caputo fractional derivative:
![](http://media.springernature.com/lw294/springer-static/image/art%3A10.1186%2Fs13660-020-02448-z/MediaObjects/13660_2020_2448_Equ4_HTML.png)
subject to boundary value condition (3).
After [3] and [4], many results appeared in the literature; we refer the reader to [5–10], where Lyapunov or Lyapunov-type inequalities are obtained for fractional differential equation subject various boundary value conditions such as
Inspired by the works mentioned, in this paper, we aim to investigate the Lyapunov-type inequality for the following fractional differential equations:
![](http://media.springernature.com/lw300/springer-static/image/art%3A10.1186%2Fs13660-020-02448-z/MediaObjects/13660_2020_2448_Equ5_HTML.png)
where δ and γ are real numbers, and \(q(t)\in L(0,1)\) is not identically zero on any compact subinterval of \((0,1)\). Furthermore, we obtain the existence of a solution for the corresponding nonlinear problem:
![](http://media.springernature.com/lw318/springer-static/image/art%3A10.1186%2Fs13660-020-02448-z/MediaObjects/13660_2020_2448_Equ6_HTML.png)
BVP (6) was recently studied in [11], but we should point out that only the case of \(\delta >1\) and \(0<\gamma <1\) was considered in [11]. In this paper, we give a comprehensive discussion on parameters δ and γ.
2 Preliminaries and lemmas
For convenience, we present some definitions and lemmas from fractional calculus theory in the sense of Riemann–Liouville and Caputo.
Definition 2.1
([12])
Let \(\varGamma (v)=\int _{0}^{\infty }t^{v-1}e^{-t}\,dt\), \(v>0\), be the gamma function. Then the Riemann–Liouville fractional integral of order v for \(y(t)\) is defined as
![](http://media.springernature.com/lw322/springer-static/image/art%3A10.1186%2Fs13660-020-02448-z/MediaObjects/13660_2020_2448_Equd_HTML.png)
Definition 2.2
([12])
Let \(v>0\) and \(n=[v]+1\), where \([v]\) denotes the integer part of a number v. Then the Caputo fractional derivative of order v for \(y(t)\) is defined as
![](http://media.springernature.com/lw447/springer-static/image/art%3A10.1186%2Fs13660-020-02448-z/MediaObjects/13660_2020_2448_Eque_HTML.png)
By Definitions 2.1 and 2.2 we have
![](http://media.springernature.com/lw312/springer-static/image/art%3A10.1186%2Fs13660-020-02448-z/MediaObjects/13660_2020_2448_Equ7_HTML.png)
Lemma 2.1
A function\(u(t)\)is a solution of the boundary value problem (5) if and only if\(u(t)\)satisfies
where
Proof
Let \(u(t)\) be a solution of (5). Then
![](http://media.springernature.com/lw238/springer-static/image/art%3A10.1186%2Fs13660-020-02448-z/MediaObjects/13660_2020_2448_Equg_HTML.png)
By (7) we obtain
Considering \(u(0)=\delta u(1)\), we have
considering \(u'(0)=\gamma u'(1)\), we have
and thus we get
Substituting (10) and (11) into (9), we obtain
where \(G(t,s)\) is the Green’s function:
□
Lemma 2.2
When\(\delta \in (0,1)\)and\(\gamma \in (0,1)\), Green’s function G(t,s) satisfies the following properties:
-
(i)
\(G(t,s)\leq 0\), \((t,s)\in [0,1]\times [0,1]\);
-
(ii)
\(\max_{0\leq t \leq 1} \vert G(t,s) \vert =-G(1,s)= \frac{(1-s)^{\nu -2}}{\varGamma (\nu )(1-\delta )(1-\gamma )}[\gamma ( \nu -1)+(1-\gamma )(1-s)]\)for\(s\in [0,1]\),
-
(iii)
\(\int _{0}^{1} \vert G(t,s) \vert \,ds\leq \frac{\gamma (\nu -1)+1}{\varGamma (\nu +1)(1-\delta )(1-\gamma )}\).
Proof
(i) \(G(t,s)\leq 0\) is obvious since \(\delta \in (0,1)\) and \(\gamma \in (0,1)\).
(ii) For \(s\in [0,1]\) and \(t\in [s,1] \), we have
which means
for \(t\in [0,s]\), we have
which means
Inequalities (12) and (13) show that, for \(s\in [0,1]\),
Therefore, for \(s\in [0,1]\),
(iii) By (ii) we have
□
Lemma 2.3
When\(\delta \in (1,+\infty )\)and\(\gamma \in (0,1)\), Green’s function\(G(t,s)\)satisfies the following properties:
-
(i)
\(G(t,s)\geq 0\), \((t,s)\in [0,1]\times [0,1]\);
-
(ii)
\(\max_{0\leq t \leq 1}|G(t,s)|=G(0,s)= \frac{\delta (1-s)^{\nu -2}}{\varGamma (\nu )(\delta -1)(1-\gamma )}[ \gamma (\nu -1)+(1-\gamma )(1-s)]\)for\(s\in [0,1]\),
-
(iii)
\(\int _{0}^{1}|G(t,s)|\,ds\leq \frac{\delta (\gamma {\nu }+1-\gamma )}{\varGamma (\nu +1)(\delta -1)(1-\gamma )}\).
Proof
(i) When \(0\leq t\leq s\leq 1\),
When \(0\leq s \leq t\leq 1\),
(ii) For \(s\in [0,1]\) and \(t\in [s,1]\), we have
which means
For \(t\in [0,s]\), we have
which means
Inequalities (14) and (15) show us that, for \(s\in [0,1]\),
(iii) From (ii) we have
□
Lemma 2.4
When\(\delta \in (0,1)\)and\(\gamma \in (1,1+\frac{(\nu -1)\delta }{2-\nu }]\), Green’s function\(G(t,s)\)satisfies the following properties:
-
(i)
\(G(t,s)\geq 0\), \((t,s)\in [0,1]\times [0,1]\);
-
(ii)
\(\max_{0\leq t\leq 1}|G(t,s)|=G(1,s)= \frac{(1-s)^{\nu -2}}{\varGamma (\nu )(1-\delta )(\gamma -1)} [ \gamma (\nu -1)-(\gamma -1)(1-s) ]\)for\(s\in [0,1]\),
-
(iii)
\(\int ^{1}_{0}|G(t,s)|\,ds\leq \frac{\gamma (\nu -1)+1}{\varGamma (\nu +1)(1-\delta )(\gamma -1)}\).
Proof
We first prove (i) and (ii). For \(s\in [0,1]\), when \(t\in [0,s]\),
which means that, for \(s\in [0,1]\),
When \(t\in [s,1]\),
Letting \(G_{t}^{\prime }(t,s)=0\), we get \({t^{*}=(\frac{\gamma }{\gamma -1})^{\frac{1}{\nu -2}}(1-s)+s}\in [s,1]\). Combining with (17), for \(s\in [0,1]\), we have
Inequalities (16), (18), and (19) show us that, for \(s\in [0,1]\),
Now we prove \(G(0,s)\), \(G(s,s)\), \(G(t^{*},s)\), and \(G(1,s)\) are all nonnegative.
For \(G(0,s)\), we have
which means that \(G(0, s)\) is increasing for \(s\in [0,1]\). Considering that \(\gamma <\frac{1}{2-\nu }\) in case of \(1<\gamma \leq 1+\frac{(\nu -1)\delta }{2-\nu }\) and \(0<\delta <1\), we have, for \(s\in [0,1]\),
Inequalities(16) and (20) show that \(G(s,s)>0\).
For \(G(t^{*},s)\), we have
where
Obviously, \(g(s)\) is increasing on \([0,1]\), and thus
Let \(k(t)=\frac{\delta (\nu -1)}{1-\delta }t-(2-\nu )t^{\frac{\nu -1}{\nu -2}}-\frac{\delta }{1-\delta }\), \(t\in [1,+\infty )\). Then
which mean that \(k(t)\) is increasing in \([1,+\infty )\). Letting \(k(t_{0})=0\), we get \(t_{0}= \frac{(1-\delta )(2-\nu )t_{0}^{\frac{\nu -1}{\nu -2}}+\delta }{\delta (\nu -1)} \) and
Then
that is, \(t_{0}<\frac{\gamma }{\gamma -1}\). By (23) we obtain \(g(0)=k(\frac{\gamma }{\gamma -1})\geq k(t_{0})=0\), and thus
From (21) and (24) it follows that \(G(t^{*},s)\geq 0\).
Since \(G(t^{*},s)\geq 0\), by (19) it follows that \(G(1,s)\geq 0\)
Above all, we conclude that
and
Since \(\gamma <\frac{1}{2-\nu }\) in the case of \(1<\gamma \leq 1+\frac{(\nu -1)\delta }{2-\nu }\) and \(0<\delta <1\), we get
so
(iii) By (ii) we have
□
Lemma 2.5
When\(\delta \in (1,+\infty )\)and\(\gamma \in (1,\frac{1}{2-\nu }]\), Green’s function\(G(t,s)\)satisfies the following properties:
-
(i)
\(G(t,s)\leq 0\), \((t,s)\in [0,1]\times [0,1]\);
-
(ii)
For\(s\in [0,1]\),
$$\begin{aligned}& \max_{0\leq t\leq 1} \bigl\vert G(t,s) \bigr\vert \\& \quad \leq \frac{(1-s)^{\nu -2}}{\varGamma (\nu )(\delta -1)(\gamma -1)} \\& \qquad {}\times \biggl\{ \delta \gamma (\nu -1)- \biggl[\delta (\gamma -1)-\gamma (2- \nu ) ( \delta -1) \biggl(\frac{\gamma -1}{\gamma }\biggr)^{\frac{1}{2-\nu }} \biggr](1-s) \biggr\} ; \end{aligned}$$ -
(iii)
\(\int ^{1}_{0}|G(t,s)|\,ds\leq \frac{1}{\varGamma (\nu +1)(\delta -1)(\gamma -1)} \{ \delta [1+ \gamma (\nu -1)]+(\delta -1)\gamma (2-\nu )(\frac{\gamma -1}{\gamma })^{ \frac{1}{2-\nu }} \} \).
Proof
We first prove (i) and (ii). For \(s\in [0,1]\) and \(t\in [0,s]\),
which means
When \(t\in [s,1]\),
Letting \(G_{t}^{\prime }(t,s)=0\), we get \(t^{*}={(\frac{\gamma }{\gamma -1})^{\frac{1}{\nu -2}}(1-s)+s}\in [s,1]\). Combining (26), for \(s\in [0,1]\), we have
Inequalities (25), (27), and (28) show that, for \(s\in [0,1]\),
We now prove that \(G(0,s)\), \(G(s,s)\), \(G(t^{*},s)\), \(G(1,s)\) are all nonpositive.
For \(G(s,s)\), we have
where
We have
which means that \(L(s)\leq 0\), \(s\in [0,1]\), and therefore
Inequalities \(G(0,s)\leq 0\) and \(G(t^{*},s)\leq 0\) follow from (25), (27), and \(G(s,s)\leq 0\).
For \(G(1,s)\), we have
and thus, for \(s\in [0,1]\),
Above all, we get that \(G(t,s)\leq 0\) and, for \(s\in [0,1]\),
We can easily compute that
where
and
where
Obviously,
So, if we make a line \(H(s)\) through \((0,h_{1}(0))\) and \((1,h_{2}(1))\), that is,
then we have
Therefore, for \(s\in [0,1]\),
(iii) easily s follows by (ii):
□
3 Main result
Theorem 3.1
Suppose the boundary value problem (5) has a nonzero solution\(u(t)\).
-
(i)
If\(\delta \in (0,1)\)and\(\gamma \in (0,1)\), then
$$ \int _{0}^{1}(1-s)^{\nu -2}\bigl[\gamma (\nu -1)+(1-\gamma ) (1-s)\bigr] \bigl\vert q(s) \bigr\vert \,ds> \varGamma ( \nu ) (1-\delta ) (1-\gamma ); $$ -
(ii)
If\(\delta \in (1,+\infty )\)and\(\gamma \in (0,1)\), then
$$ \int _{0}^{1}(1-s)^{\nu -2}\bigl[\gamma (\nu -1)+(1-\gamma ) (1-s)\bigr] \bigl\vert q(s) \bigr\vert \,ds> \frac{\varGamma (\nu )(\delta -1)(1-\gamma )}{\delta }; $$ -
(iii)
If\(\delta \in (0,1)\)and\(\gamma \in (1,1+\frac{(\nu -1)\delta }{2-\nu }]\), then
$$ \int _{0}^{1}(1-s)^{\nu -2}\bigl[\gamma (\nu -1)-(\gamma -1) (1-s)\bigr] \bigl\vert q(s) \bigr\vert \,ds> \varGamma ( \nu ) (1-\delta ) (\gamma -1); $$ -
(iv)
if\(\delta \in (1,+\infty )\)and\(\gamma \in (1,\frac{1}{2-\nu }]\), then
$$\begin{aligned}& \int _{0}^{1}(1-s)^{\nu -2} \biggl\{ \delta \gamma (\nu -1)- \biggl[ \delta (\gamma -1)-\gamma (2-\nu ) (\delta -1) \biggl( \frac{\gamma -1}{\gamma }\biggr)^{\frac{1}{2-\nu }} \biggr](1-s) \biggr\} \\& \qquad {}\times \bigl\vert q(s) \bigr\vert \,ds \\& \quad > \varGamma (\nu ) (\delta -1) (\gamma -1). \end{aligned}$$
Proof
Let \(u(t)\) be a nonzero solution of the boundary value problem (5). By Lemma 2.1 we have
Let \(m=\max_{t\in [0,1]} \vert u(t) \vert \). Then
Next, since \(|G(t,s)| \vert q(s) \vert \leq \max_{0\leq t\leq 1}|G(t,s)||q(s)|\), but \(|G(t,s)| \vert q(s) \vert \not \equiv \max_{0\leq t\leq 1}|G(t,s)| |q(s)|\), we have
which means
that is,
Substituting Lemma 2.2(ii), Lemma 2.3(ii), Lemma 2.4(ii), and Lemma 2.5(ii) into (29), we easily get statements (i), (ii), (iii), and (iv) of Theorem 3.1. □
By Theorem 3.1 we have the following conclusions.
Theorem 3.2
-
(i)
when\(\delta \in (0,1)\)and\(\gamma \in (0,1)\), if
$$ \int _{0}^{1}(1-s)^{\nu -2}\bigl[\gamma (\nu -1)+(1-\gamma ) (1-s)\bigr] \bigl\vert q(s) \bigr\vert \,ds\leq \varGamma (\nu ) (1-\delta ) (1-\gamma ), $$then the boundary value problem (5) has no nonzero solution.
-
(ii)
when\(\delta \in (1,+\infty) \)and\(\gamma \in (0,1)\), if
$$ \int _{0}^{1}(1-s)^{\nu -2}\bigl[\gamma (\nu -1)+(1-\gamma ) (1-s)\bigr] \bigl\vert q(s) \bigr\vert \,ds\leq \frac{\varGamma (\nu )(\delta -1)(1-\gamma )}{\delta }, $$then the boundary value problem (5) has no nonzero solution.
-
(iii)
when\(\delta \in (0,1)\)and\(\gamma \in (1,1+\frac{(\nu -1)\delta }{2-\nu }]\), if
$$ \int _{0}^{1}(1-s)^{\nu -2}\bigl[\gamma (\nu -1)-(\gamma -1) (1-s)\bigr] \bigl\vert q(s) \bigr\vert \,ds\leq \varGamma (\nu ) (1-\delta ) (\gamma -1), $$then the boundary value problem (5) has no nonzero solution.
-
(iv)
when\(\delta \in (1,+\infty )\)and\(\gamma \in (1,\frac{1}{2-\nu }]\), if
$$\begin{aligned}& \int _{0}^{1}(1-s)^{\nu -2} \biggl\{ \delta \gamma (\nu -1) \\& \qquad {}- \biggl[ \delta (\gamma -1)-\gamma (2-\nu ) (\delta -1) \biggl( \frac{\gamma -1}{\gamma }\biggr)^{\frac{1}{2-\nu }} \biggr](1-s) \biggr\} \bigl\vert q(s) \bigr\vert \,ds \\& \quad \leq \varGamma (\nu ) (\delta -1) (\gamma -1), \end{aligned}$$then the boundary value problem (5) has no nonzero solution.
Now we consider the existence of solutions to the following nonlinear boundary value problem:
![](http://media.springernature.com/lw222/springer-static/image/art%3A10.1186%2Fs13660-020-02448-z/MediaObjects/13660_2020_2448_Equ30_HTML.png)
Theorem 3.3
Let\(f:[0,1]\times R\rightarrow R\)be continuous and satisfy the following Lipschitz condition with Lipschitz constantL:
for all\((t,u_{1}),(t,u_{2})\in [0,1]\times R\). If
then (30) has a unique solution.
Proof
Let E be the Banach space \(C[0,1]\) with norm \(\Vert u\Vert =\max_{t\in [0,1]}|u(t)|\).
By Lemma 2.1, \(u\in E\) is a solution of (30) if and only if it satisfies the integral equation
Define the operator \(T:E\rightarrow E\) by
Then T is completely continuous. We claim that T has a unique fixed point in E. In fact, for any \(u_{1},u_{2}\in E\), we have
Substituting of Lemma 2.2(iii), Lemma 2.3(iii), Lemma 2.4(iii), and Lemma 2.5(iii) into (33), we conclude that T is a contraction mapping and thus obtain the desired result. □
4 Conclusion
In this paper, we study a linear fractional differential equation. Firstly, by obtaining the Green’s function we derive a Lyapunov-type inequality for such a boundary value problem. Furthermore, we use the contraction mapping theorem to study the existence of a unique solution for the corresponding nonlinear problem.
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Ma, D., Yang, Z. Lyapunov-type inequality and solution for a fractional differential equation. J Inequal Appl 2020, 181 (2020). https://doi.org/10.1186/s13660-020-02448-z
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DOI: https://doi.org/10.1186/s13660-020-02448-z