A generalized Lyapunov-type inequality in the frame of conformable derivatives
- Thabet Abdeljawad^{1}Email author,
- Jehad Alzabut^{1} and
- Fahd Jarad^{2}
https://doi.org/10.1186/s13662-017-1383-z
© The Author(s) 2017
Received: 15 June 2017
Accepted: 1 October 2017
Published: 11 October 2017
Abstract
Keywords
Lyapunov inequality conformable derivative Green’s function boundary value problem Sturm-Liouville eigenvalue problemMSC
34A08 26D151 Background
2 Preliminaries on conformable derivatives
This section is devoted to the presentation of some preliminaries about higher order fractional conformable derivatives developed in [14].
Definition 1
In the case of higher order, the following definition becomes true.
Definition 2
([14])
Lemma 1
([13])
In the case of higher order, the following definition is valid.
Definition 3
([14])
Notice that if \(\alpha =n+1\) then \(\gamma =1\) and hence \((I_{\alpha } ^{c} g)(t)=(\textbf{I}_{n+1}^{c} g)(t)=\frac{1}{n!}\int_{c}^{t} (t-x)^{n} g(x)\,dx\), which is the iterative integral of g, \(n+1\) times over \((c,t]\).
Example 1
The following is a generalization of Lemma 1.
Lemma 2
([14])
Theorem 1
([14])
Example 2
The following example shows why it is useful to work in conformable differential systems.
Example 3
3 A Lyapunov-type inequality for a conformable BVP
Lemma 3
Proof
Lemma 4
- (i)
\(H(t,s)\geq 0\) for all \(c\leq t,s \leq d\).
- (ii)
\(\max_{t \in [c,d]} H(t,s)=H(s,s)\) for \(s \in [c,d]\).
- (iii)\(H(s,s)\) has a unique maximum, given by$$ \max_{s \in [c,d]} H(s,s)= H \biggl(\frac{c+(\alpha -1)d}{\alpha }, \frac{c+( \alpha -1)d}{\alpha } \biggr)=\frac{(d-c)^{\alpha -1}(\alpha -1)^{\alpha -1}}{\alpha^{\alpha }}. $$
Proof
- (i)
It is clear that \(h_{1}\geq 0\). To determine the sign of \(h_{2}\), we observe that \((t-s)=\frac{t-c}{d-c} (d- (c+ \frac{(s-c)(d-c)}{(t-c)}) )\) and that \(c+\frac{(s-c)(d-c)}{(t-c)} \geq s\) if and only if \(s\geq c\). Together with \((s-c)^{ \alpha -2} \geq 0 \) we conclude that \(h_{2}\geq 0\) as well. Hence, the proof of the first part is complete.
- (ii)
It is clear that \(h_{1}(t,s)\) is an increasing function in t. Differentiating \(h_{2}\) with respect to t for every fixed s and following similar analysis as in first part we conclude that \(h_{2}\) is a decreasing function.
- (iii)
Let \(g(s)=H(s,s)=\frac{(s-c)^{\alpha -1}(d-s)}{d-c}\). Then it is sufficient to show that \(g^{\prime }(s)=0\) if \(s=\frac{c+( \alpha -1)d}{\alpha }\) and hence the proof is completed.
Theorem 2
Proof
Remark 1
If \(\alpha =2\), then (17) reduces to the classical Lyapunov inequality (3). We also invite the reader to compare the obtained generalized Lyapunov inequality in Theorem 2 and the one obtained recently and independently in [19]. The approach in [19] is different and the authors there proved the existence of solution in the space \(AC^{2}[c,d]=\{u \in C^{1}[c,d]: u^{\prime }\in AC[c,d]\}\). Further, we see that our obtained inequality provides, for example when applied to the Sturm-Liouville eigenvalue problem, sharper lower estimate for the eigenvalues. Indeed, in Section 5 we can see that the lower estimate \(\frac{\alpha^{\alpha }}{(\alpha -1)^{\alpha -1}}\) is bigger than \(4 (\alpha -1)\) for \(1<\alpha \leq 2\). This is due to the observation that \(( \frac{\alpha }{\alpha -1} ) ^{\alpha }\geq 4\), for \(1<\alpha \leq 2\). Further, for convenience, in the next section we prove a sequential type Lyapunov inequality version as well.
4 A Lyapunov-type inequality for a sequential conformable BVP
Lemma 5
Proof
Lemma 6
- (i)
\(G(t,s)\geq 0\) for all \(c\leq t,s \leq d\).
- (ii)
\(\max_{t \in [c,d]} G(t,s)=G(s,s)\) for \(s \in [c,d]\).
- (iii)\(f(s)=G(s,s)\) has a unique maximum, given bywhere$$\begin{aligned} \max_{s \in [c,d]} G(s,s) &= G\bigl(\Lambda (c,d,\alpha ),\Lambda (c,d, \alpha )\bigr) \\ & = \frac{(d-c)^{2\alpha -1}}{3\alpha -1} \biggl( \frac{2\alpha -1}{3 \alpha -1} \biggr) ^{\frac{2\alpha -1}{\alpha }}, \end{aligned}$$$$ \Lambda (c,d,\alpha )=c+ (d-c) \biggl[ \frac{(2\alpha -1)}{(3\alpha -1)} \biggr] ^{\frac{1}{\alpha }}. $$
Proof
- (i)
The proof follows by noting that the function \(g_{1}\geq 0\) since \(g_{1}(t,s)\) is decreasing in t for any s and \(g_{1}(d,s)=0\) for any s. Also, \(g_{2}\geq 0\) since \(g_{2}(t,s)\) is increasing in t for any s and \(g_{2}(c,s)= 0\) for any s.
- (ii)
The proof of this part follows by noting that the function \(g_{1}(t,s)\) is decreasing in t for any s and that \(g_{2}(t,s)\) is increasing in t for any s by realizing that \(( 1-\frac{(s-c)^{ \alpha }}{(d-c)^{\alpha }} ) \geq 0\) for all s.
- (iii)
Let \(f(s)=G(s,s)=(s-c)^{\alpha -1} [ \frac{(s-c)^{ \alpha }}{\alpha }-\frac{(t-c)^{\alpha }(s-c)^{\alpha }}{\alpha (d-c)^{ \alpha }} ] \). Then one can show that \(f^{\prime }(s)=0\) if \(s=\Lambda (c,d,\alpha )\) and hence the proof is concluded.
Theorem 3
Proof
Remark 2
Since \(G(\Lambda (c,d,\alpha ),\Lambda (c,d,\alpha ))\) tends to \(\frac{d-c}{4}\) as \(\alpha \rightarrow 1\) then the classical Lyapunov inequality (3) is obtained again: \(\alpha \rightarrow 1\). In this case, one may also deduce that \(x^{(2\alpha )}(t) \rightarrow x^{\prime \prime }(t)\) as \(\alpha \rightarrow 1\).
5 Application
Declarations
Acknowledgements
The first and the second author would like to thank Prince Sultan University for funding this work through research group Nonlinear Analysis Methods in Applied Mathematics (NAMAM) group number RG-DES-2017-01-17.
Authors’ contributions
All authors read and approved the final manuscript.
Competing interests
The authors declare that they have no competing interests.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), 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
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