1 Introduction
The p-Laplacian operator arises in different mathematical models that describe physical and natural phenomena (see, for example, [1‐6]). In particular, it is used in some models related to turbulent flows (see, for example, [7‐9]).
In this paper, we present some Lyapunov-type inequalities for a fractional-order model for turbulent flow in a porous medium. More precisely, we are interested with the nonlinear fractional boundary value problem where \(2<\alpha\leq3\), \(1<\beta\leq2\), \(D_{a^{+}}^{\alpha}\), \(D_{a^{+}}^{\beta}\) are the Riemann-Liouville fractional derivatives of orders α, β, \(\Phi_{p}(s)=|s|^{p-2}s\), \(p>1\), and \(\chi: [a,b]\to\mathbb{R}\) is a continuous function. Under certain assumptions imposed on the function q, we obtain necessary conditions for the existence of nontrivial solutions to (1.1). Some applications to eigenvalue problems are also presented.
$$ \left \{ \textstyle\begin{array}{l} D_{a^{+}}^{\beta}(\Phi_{p}(D_{a^{+}}^{\alpha}u(t)))+\chi(t)\Phi _{p}(u(t))=0, \quad a< t< b, \\ u(a)=u'(a)=u'(b)=0, \qquad D_{a^{+}}^{\alpha}u(a)=D_{a^{+}}^{\alpha}u(b)=0, \end{array}\displaystyle \right . $$
(1.1)
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For completeness, let us recall the standard Lyapunov inequality [10], which states that if u is a nontrivial solution of the problem where \(a< b\) are two consecutive zeros of u, and \(\chi: [a,b]\to \mathbb{R}\) is a continuous function, then Note that in order to obtain this inequality, it is supposed that a and b are two consecutive zeros of u. In our case, as it will be observed in the proof of our main result, we assume just that u is a nontrivial solution to (1.1).
$$ \left \{ \textstyle\begin{array}{l} u''(t)+\chi(t)u(t)=0,\quad a< t< b, \\ u(a)=u(b)=0, \end{array}\displaystyle \right . $$
$$ \int_{a}^{b} \bigl\vert \chi(t)\bigr\vert \, \mathrm{d}t> \frac{4}{b-a}. $$
(1.2)
Inequality (1.2) is useful in various applications, including oscillation theory, stability criteria for periodic differential equations, and estimates for intervals of disconjugacy.
Several generalizations and extensions of inequality (1.2) to different boundary value problems exist in the literature. As examples, we refer to [11‐16] and the references therein.
Recently, some Lyapunov-type inequalities for fractional boundary value problems have been obtained. Ferreira [17] established a fractional version of inequality (1.2) for a fractional boundary value problem involving the Riemann-Liouville fractional derivative of order \(1<\alpha\leq2\). More precisely, Ferreira [17] studied the fractional boundary value problem where \(D_{a^{+}}^{\alpha}\) is the Riemann-Liouville fractional derivative of order \(1<\alpha\leq2\), and \(\chi: [a,b]\to\mathbb{R}\) is a continuous function. In this case, it was proved that if (1.3) has a nontrivial solution, then where Γ is the Euler gamma function. Observe that if we take \(\alpha=2\) in the last inequality, we obtain the standard Lyapunov inequality (1.2).
$$ \left \{ \textstyle\begin{array}{l} D_{a^{+}}^{\alpha}u(t)+\chi(t)u(t)=0,\quad a< t< b, \\ u(a)=u(b)=0, \end{array}\displaystyle \right . $$
(1.3)
$$\int_{a}^{b} \bigl\vert \chi(t)\bigr\vert \, \mathrm{d}t> \Gamma(\alpha) \biggl(\frac {4}{b-a} \biggr)^{\alpha-1}, $$
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Ferreira [18] established a fractional version of inequality (1.2) for a fractional boundary value problem involving the Caputo fractional derivative of order \(1<\alpha\leq2\). In both papers [17, 18], the author presented nice applications to obtain intervals where certain Mittag-Leffler functions have no real zeros.
Jleli and Samet [19] studied a fractional differential equation involving the Caputo fractional derivative under mixed boundary conditions. More precisely, they considered the fractional differential equation under the mixed boundary conditions or where \({}^{\mathrm{C}}D_{a^{+}}^{\alpha}\) is the Caputo fractional derivative of order \(1<\alpha\leq2\). For the boundary conditions (1.5) and (1.6), the following two Lyapunov-type inequalities were derived respectively: and
$$ {}^{\mathrm{C}}D_{a^{+}}^{\alpha}u(t)+ \chi(t)u(t)=0,\quad a< t< b, $$
(1.4)
$$ u(a)=u'(b)=0 $$
(1.5)
$$ u'(a)=u(b)=0, $$
(1.6)
$$\int_{a}^{b} (b-s)^{\alpha-2}\bigl\vert q(s) \bigr\vert \,\mathrm{d}s\geq\frac{\Gamma(\alpha )}{\max\{\alpha-1,2-\alpha\}(b-a)} $$
$$\int_{a}^{b} (b-s)^{\alpha-1}\bigl\vert q(s) \bigr\vert \,\mathrm{d}s\geq\Gamma(\alpha). $$
The same equation (1.4) was considered by Rong and Bai [20] with the fractional boundary condition where \(0<\beta\leq1\).
$$u(a)= {}^{\mathrm{C}}D_{a^{+}}^{\beta}u(b)=0, $$
The paper is organized as follows. In Section 2, we recall some basic concepts on fractional calculus and establish some preliminary results that will be used in Section 3, where we state and prove our main result. In Section 4, we present some applications of the obtained Lyapunov-type inequalities to eigenvalue problems.
2 Preliminaries
For the convenience of the reader, we recall some basic concepts on fractional calculus to make easy the analysis of (1.1). For more details, we refer to [24].
Let \(C[a,b]\) be the set of real-valued and continuous functions in \([a,b]\). Let \(f\in C[a,b]\). Let \(\alpha\geq0\). The Riemann-Liouville fractional integral of order α of f is defined by \(I_{a}^{0} f\equiv f\) and where Γ is the gamma function.
$$\bigl(I_{a^{+}}^{\alpha}f\bigr) (t)=\frac{1}{\Gamma(\alpha)} \int_{a}^{t} (t-s)^{\alpha-1}f(s) \,\mathrm{d}s, \quad \alpha>0, t\in[a,b], $$
The Riemann-Liouville fractional derivative of order \(\alpha>0\) of f is defined by where \(n=[\alpha]+1\).
$$\bigl(D_{a^{+}}^{\alpha}f\bigr) (t)=\frac{1}{\Gamma(n-\alpha)} \biggl( \frac {\mathrm{d}}{\mathrm{d}t} \biggr)^{n} \int_{a}^{t} \frac{f(s)}{(t-s)^{\alpha-n+1}} \,\mathrm{d}s,\quad t \in[a,b], $$
Lemma 2.1
(see [24])
Let
\(\alpha>0\). If
\(D_{a^{+}}^{\alpha}u\in C[a,b]\), then
where
\(n=[\alpha]+1\).
$$I_{a^{+}}^{\alpha}D_{a^{+}}^{\alpha}u(t)=u(t)+ \sum _{k=1}^{n} c_{k}(t-a)^{\alpha-k}, $$
Now, in order to obtain an integral formulation of (1.1), we need the following results.
Lemma 2.2
Let
\(2<\alpha\leq3\)
and
\(y\in C[a,b]\). Then the problem
has a unique solution
where
$$ \left \{ \textstyle\begin{array}{l} D_{a^{+}}^{\alpha}u(t)+y(t)=0, \quad a< t< b, \\ u(a)=u'(a)=u'(b)=0 \end{array}\displaystyle \right . $$
$$u(t)= \int_{a}^{b} G(t,s)y(s) \,\mathrm{d}s, $$
$$ G(t,s)=\frac{1}{\Gamma(\alpha)} \left \{ \textstyle\begin{array}{l@{\quad}l} (\frac{b-s}{b-a} )^{\alpha-2}(t-a)^{\alpha -1}-(t-s)^{\alpha-1}, & a\leq s\leq t\leq b, \\ (\frac{b-s}{b-a} )^{\alpha-2}(t-a)^{\alpha-1},& a\leq t\leq s\leq b. \end{array}\displaystyle \right . $$
Proof
From Lemma 2.1 we have for some real constants \(c_{i}\), \(i=1,2,3\), that is, The condition \(u(a)=0\) yields \(c_{3}=0\). Therefore, The condition \(u'(a)=0\) implies that \(c_{2}=0\). Then Since \(u'(b)=0\), we get Thus, For the uniqueness, suppose that \(u_{1}\) and \(u_{2}\) are two solutions of the considered problem. Define \(u=u_{1}-u_{2}\). By linearity, u solves the boundary value problem which has as a unique solution \(u=0\). Therefore, \(u_{1}=u_{2}\), and the uniqueness follows. □
$$u(t)=-\bigl(I_{a^{+}}^{\alpha}y\bigr) (t)+c_{1}(t-a)^{\alpha-1}+c_{2}(t-a)^{\alpha -2}+c_{3}(t-a)^{\alpha-3} $$
$$u(t)=-\frac{1}{\Gamma(\alpha)} \int_{a}^{t}(t-s)^{\alpha-1}y(s) \, \mathrm{d}s+c_{1}(t-a)^{\alpha-1}+c_{2}(t-a)^{\alpha-2} +c_{3}(t-a)^{\alpha-3}. $$
$$u'(t)=-\frac{(\alpha-1)}{\Gamma(\alpha)} \int_{a}^{t} (t-s)^{\alpha -2}y(s) \, \mathrm{d}s+c_{1}(\alpha-1) (t-a)^{\alpha-2}+c_{2}(\alpha -2) (t-a)^{\alpha-3}. $$
$$u'(b)=-\frac{(\alpha-1)}{\Gamma(\alpha)} \int_{a}^{b} (b-s)^{\alpha -2}y(s) \, \mathrm{d}s+c_{1}(\alpha-1) (b-a)^{\alpha-2}. $$
$$c_{1}=\frac{1}{(b-a)^{\alpha-2}\Gamma(\alpha)} \int_{a}^{b} (b-s)^{\alpha -2}y(s) \,\mathrm{d}s. $$
$$u(t)=- \int_{a}^{t}\frac{(t-s)^{\alpha-1}y(s)}{\Gamma(\alpha)} \,\mathrm{d}s+ \int_{a}^{b} \frac{1}{\Gamma(\alpha)} \biggl( \frac {b-s}{b-a} \biggr)^{\alpha-2}(t-a)^{\alpha-1}y(s) \,\mathrm{d}s. $$
$$ \left \{ \textstyle\begin{array}{l} D_{a^{+}}^{\alpha}u(t)=0, \quad a< t< b, \\ u(a)=u'(a)=u'(b)=0, \end{array}\displaystyle \right . $$
Lemma 2.3
Let
\(y\in C[a,b]\), \(2<\alpha\leq3\), \(1<\beta\leq2\), \(p>1\), and
\(\frac{1}{p}+ \frac{1}{q}=1\). Then the problem
has a unique solution
where
$$ \left \{ \textstyle\begin{array}{l} D_{a^{+}}^{\beta}(\Phi_{p}(D_{a^{+}}^{\alpha}u(t)))+y(t)=0,\quad a< t< b, \\ u(a)=u'(a)=u'(b)=0,\qquad D_{a^{+}}^{\alpha}u(a)=D_{a^{+}}^{\alpha}u(b)=0 \end{array}\displaystyle \right . $$
$$u(t)=- \int_{a}^{b} G(t,s)\Phi_{q} \biggl( \int_{a}^{b} H(s,\tau)y(\tau) \, \mathrm{d}\tau \biggr) \,\mathrm{d}s, $$
$$ H(t,s)=\frac{1}{\Gamma(\beta)} \left \{ \textstyle\begin{array}{l@{\quad}l} (\frac{b-s}{b-a} )^{\beta-1}(t-a)^{\beta-1}-(t-s)^{\beta -1}, & a\leq s\leq t\leq b, \\ (\frac{b-s}{b-a} )^{\beta-1}(t-a)^{\beta-1}, & a\leq t\leq s\leq b. \end{array}\displaystyle \right . $$
Proof
From Lemma 2.1 we have where \(c_{i}\), \(i=1,2\), are real constants. The condition \(D_{a^{+}}^{\alpha}u(a)=0\) implies that \(\Phi_{p}(D_{a^{+}}^{\alpha}u)(a)=0\), which yields \(c_{2}=0\). The condition \(D_{a^{+}}^{\alpha}u(b)=0\) implies that \(\Phi _{p}(D_{a^{+}}^{\alpha}u)(b)=0\), which yields Therefore, that is, Then we have Setting we obtain Finally, applying Lemma 2.2, we obtain the desired result. □
$$\Phi_{p}\bigl(D_{a^{+}}^{\alpha}u\bigr) (t)=- \int_{a}^{t} \frac{(t-s)^{\beta-1}}{\Gamma (\beta)}y(s) \, \mathrm{d}s+c_{1}(t-a)^{\beta-1} +c_{2}(t-a)^{\beta-2}, $$
$$c_{1}=\frac{1}{(b-a)^{\beta-1}} \int_{a}^{b} \frac{(b-s)^{\beta -1}}{\Gamma(\beta)}y(s) \,\mathrm{d}s. $$
$$\Phi_{p}\bigl(D_{a^{+}}^{\alpha}u\bigr) (t)=- \int_{a}^{t} \frac{(t-s)^{\beta-1}}{\Gamma (\beta)}y(s) \,\mathrm{d}s+ \frac{1}{(b-a)^{\beta-1}} \int_{a}^{b} \frac {(b-s)^{\beta-1}(t-a)^{\beta-1}}{\Gamma(\beta)}y(s) \,\mathrm{d}s, $$
$$\Phi_{p}\bigl(D_{a^{+}}^{\alpha}u\bigr) (t)= \int_{a}^{b} H(t,\tau)y(\tau) \,\mathrm{d}\tau. $$
$$D_{a^{+}}^{\alpha}u(t)-\Phi_{q} \biggl( \int_{a}^{b} H(t,\tau)y(\tau) \,\mathrm{d}s \biggr)=0. $$
$$\tilde{y}(t)=-\Phi_{q} \biggl( \int_{a}^{b} H(t,\tau)y(\tau) \,\mathrm{d}\tau \biggr), $$
$$ \left \{ \textstyle\begin{array}{l} D_{a^{+}}^{\alpha}u(t)+\tilde{y}(t)=0,\quad a< t< b, \\ u(a)=u'(a)=u'(b)=0. \end{array}\displaystyle \right . $$
The following estimates will be useful later.
Lemma 2.4
We have
$$0\leq G(t,s)\leq G(b,s),\quad (t,s)\in[a,b]\times[a,b]. $$
Proof
Differentiating with respect to t, we obtain Set and Clearly, On the other hand, using the inequality and the fact that \(\alpha>2\), we obtain which yields As consequence, we have Then \(G(\cdot,s)\) is a nondecreasing function for all \(s\in[a,b]\), which yields The proof is complete. □
$$ G_{t}(t,s)=\frac{(\alpha-1)}{\Gamma(\alpha)} \left \{ \textstyle\begin{array}{l@{\quad}l} (\frac{b-s}{b-a} )^{\alpha-2}(t-a)^{\alpha -2}-(t-s)^{\alpha-2}, & a\leq s\leq t\leq b, \\ (\frac{b-s}{b-a} )^{\alpha-2}(t-a)^{\alpha-2},& a\leq t\leq s\leq b. \end{array}\displaystyle \right . $$
$$g_{1}(t,s)= \biggl(\frac{b-s}{b-a} \biggr)^{\alpha-2}(t-a)^{\alpha -2}-(t-s)^{\alpha-2}, \quad a\leq s\leq t\leq b $$
$$g_{2}(t,s)= \biggl(\frac{b-s}{b-a} \biggr)^{\alpha-2}(t-a)^{\alpha -2}, \quad a\leq t\leq s\leq b. $$
$$g_{2}(t,s)\geq0,\quad a\leq t\leq s\leq b. $$
$$(b-s) (t-a)\geq(t-s) (b-a), \quad a\leq s\leq t\leq b, $$
$$(b-s)^{\alpha-2}(t-a)^{\alpha-2}\geq(t-s)^{\alpha-2}(b-a)^{\alpha -2}, \quad a\leq s\leq t\leq b, $$
$$g_{1}(t,s)\geq0,\quad a\leq s\leq t\leq b. $$
$$G_{t}(t,s)\geq0,\quad (t,s)\in[a,b]\times[a,b]. $$
$$0=G(a,s)\leq G(t,s)\leq G(b,s),\quad (t,s)\in[a,b]\times[a,b]. $$
Lemma 2.5
We have
$$0\leq H(t,s)\leq H(s,s),\quad (t,s)\in[a,b]\times[a,b]. $$
Proof
Observe that \(H(t,s)=G_{t}(t,s)\) for \(\alpha =\beta+1\). Then, from the proof of Lemma 2.4 we have On the other hand, for all \(s\in[a,b]\), we have For \(a\leq t\leq s\leq b\), we have For \(a\leq s< t\leq b\), we have Let \(s\in[a,b)\) be fixed. Define the function \(\psi:(s,b]\to\mathbb {R}\) by We have Using the inequalities we get Thus, for all \(t\in(s,b]\), we have that is, The proof is complete. □
$$H(t,s)\geq0,\quad (t,s)\in[a,b]\times[a,b]. $$
$$\Gamma(\beta)H(s,s)= \biggl(\frac{b-s}{b-a} \biggr)^{\beta -1}(s-a)^{\beta-1}. $$
$$\begin{aligned} \Gamma(\beta)H(t,s) =& \biggl(\frac{b-s}{b-a} \biggr)^{\beta -1}(t-a)^{\beta-1} \\ \leq& \biggl(\frac{b-s}{b-a} \biggr)^{\beta-1}(s-a)^{\beta-1} \\ =&\Gamma(\beta)H(s,s). \end{aligned}$$
$$\Gamma(\beta)H(t,s)= \biggl(\frac{b-s}{b-a} \biggr)^{\beta -1}(t-a)^{\beta-1}-(t-s)^{\beta-1}. $$
$$\psi(t)=\Gamma(\beta)H(t,s), \quad t\in(s,b]. $$
$$\psi'(t)=(\beta-1) \biggl[ \biggl(\frac{b-s}{b-a} \biggr)^{\beta -1}(t-a)^{\beta-2}-(t-s)^{\beta-2} \biggr], \quad t \in(s,b]. $$
$$\biggl(\frac{b-s}{b-a} \biggr)^{\beta-1}\leq1,\qquad \beta-2\leq0,\qquad (t-a)^{\beta-2}\leq(t-s)^{\beta-2}, $$
$$\psi'(t)\leq0, \quad t\in(s,b]. $$
$$\psi(t)\leq\psi(s), $$
$$\Gamma(\beta)H(t,s)\leq\Gamma(\beta)H(s,s),\quad t\in(s,b]. $$
Now, we are ready to state and prove our main result.
3 Main result
Our main result is the following Lyapunov-type inequality.
Theorem 3.1
Suppose that
\(2<\alpha\leq3\), \(1<\beta\leq2\), \(p>1\), and
\(\chi: [a,b]\to\mathbb{R}\)
is a continuous function. If (1.1) has a nontrivial continuous solution, then
$$\begin{aligned}& \int_{a}^{b} (b-s)^{\beta-1}(s-a)^{\beta-1} \bigl\vert \chi(s)\bigr\vert \,\mathrm{d}s \\& \quad \geq\bigl[\Gamma(\alpha) \bigr]^{p-1}\Gamma(\beta) (b-a)^{\beta-1} \biggl( \int_{a}^{b}(b-s)^{\alpha-2}(s-a) \,\mathrm{d}s \biggr)^{1-p}. \end{aligned}$$
(3.1)
Proof
We endow the set \(C[a,b]\) with the Chebyshev norm \(\|\cdot\|_{\infty}\) given by Suppose that \(u\in C[a,b]\) is a nontrivial solution of (1.1). From Lemma 2.3 we have Let \(t\in[a,b]\) be fixed. We have where Using Lemma 2.4 and Lemma 2.5, we obtain Since the last inequality holds for every \(t\in[a,b]\), we obtain which yields the desired result. □
$$\|u\|_{\infty}=\max\bigl\{ \bigl\vert u(t)\bigr\vert : a\leq t\leq b \bigr\} , \quad u\in C[a,b]. $$
$$u(t)=- \int_{a}^{b} G(t,s)\Phi_{q} \biggl( \int_{a}^{b} H(s,\tau)\chi(\tau )\Phi_{p} \bigl(u(\tau)\bigr) \,\mathrm{d}\tau \biggr) \,\mathrm{d}s, \quad t\in[a,b]. $$
$$\begin{aligned} \bigl|u(t)\bigr| \leq& \int_{a}^{b} \bigl\vert G(t,s)\bigr\vert \biggl\vert \Phi_{q} \biggl( \int_{a}^{b} H(s,\tau)\chi(\tau)\phi_{p} \bigl(u(\tau)\bigr) \,\mathrm{d}\tau \biggr) \biggr\vert \,\mathrm{d}s \\ =& \int_{a}^{b} \bigl\vert G(t,s)\bigr\vert \biggl\vert \int_{a}^{b} H(s,\tau)\chi(\tau)\Phi _{p} \bigl(u(\tau)\bigr) \,\mathrm{d}\tau\biggr\vert ^{q-1} \,\mathrm{d}s \\ \leq& \int_{a}^{b} \bigl\vert G(t,s)\bigr\vert \theta(s) \,\mathrm{d}s, \end{aligned}$$
$$\theta(s)= \biggl( \int_{a}^{b} \bigl\vert H(s,\tau)\bigr\vert \bigl\vert \chi(\tau)\bigr\vert \bigl\vert u(\tau)\bigr\vert ^{p-1} \,\mathrm{d}\tau \biggr)^{q-1}, \quad s\in[a,b]. $$
$$\bigl\vert u(t)\bigr\vert \leq\|u\|_{\infty}^{(p-1)(q-1)} \biggl( \int_{a}^{b}G(b,s) \,\mathrm{d}s \biggr) \biggl( \int_{a}^{b} H(s,s)\bigl\vert \chi(s)\bigr\vert \,\mathrm{d}s \biggr)^{q-1}. $$
$$1\leq \biggl( \int_{a}^{b}G(b,s) \,\mathrm{d}s \biggr) \biggl( \int_{a}^{b} H(s,s)\bigl\vert \chi(s)\bigr\vert \,\mathrm{d}s \biggr)^{q-1}, $$
Corollary 3.2
Suppose that
\(2<\alpha\leq3\), \(1<\beta\leq2\), \(p>1\), and
\(\chi: [a,b]\to\mathbb{R}\)
is a continuous function. If (1.1) has a nontrivial continuous solution, then
$$ \int_{a}^{b} \bigl\vert \chi(s)\bigr\vert \, \mathrm{d}s\geq\frac{4^{\beta-1}[\Gamma (\alpha)]^{p-1}\Gamma(\beta)}{(b-a)^{\beta-1}} \biggl( \int _{a}^{b}(b-s)^{\alpha-2}(s-a) \,\mathrm{d}s \biggr)^{1-p}. $$
(3.2)
Proof
Let Observe that the function ψ has a maximum at the point \(s^{*}=\frac {a+b}{2}\), that is, The desired result follows immediately from the last equality and inequality (3.1). □
$$\psi(s)=(b-s) (s-a), \quad s\in[a,b]. $$
$$\|\psi\|_{\infty}=\psi\bigl(s^{*}\bigr)=\frac{(b-a)^{2}}{4}. $$
For \(p=2\), problem (1.1) becomes where \(2<\alpha\leq3\), \(1<\beta\leq2\), and \(\chi: [a,b]\to\mathbb {R}\) is a continuous function. In this case, taking \(p=2\) in Theorem 3.1, we obtain the following result.
$$ \left \{ \textstyle\begin{array}{l} D_{a^{+}}^{\beta}(D_{a^{+}}^{\alpha}u(t))+\chi(t)u(t)=0,\quad a< t< b, \\ u(a)=u'(a)=u'(b)=0, \qquad D_{a^{+}}^{\alpha}u(a)=D_{a^{+}}^{\alpha}u(b)=0, \end{array}\displaystyle \right . $$
(3.3)
Corollary 3.3
Suppose that
\(2<\alpha\leq3\), \(1<\beta\leq2\), and
\(\chi: [a,b]\to \mathbb{R}\)
is a continuous function. If (3.3) has a nontrivial continuous solution, then
$$\int_{a}^{b} (b-s)^{\beta-1}(s-a)^{\beta-1} \bigl\vert \chi(s)\bigr\vert \,\mathrm{d}s\geq\Gamma(\alpha)\Gamma(\beta) (b-a)^{\beta-1} \biggl( \int_{a}^{b}(b-s)^{\alpha-2}(s-a) \,\mathrm{d}s \biggr)^{-1}. $$
Taking \(p=2\) in Corollary 3.2, we obtain the following result.
Corollary 3.4
Suppose that
\(2<\alpha\leq3\), \(1<\beta\leq2\), and
\(\chi: [a,b]\to \mathbb{R}\)
is a continuous function. If (3.3) has a nontrivial continuous solution, then
$$\int_{a}^{b} \bigl\vert \chi(s)\bigr\vert \, \mathrm{d}s\geq\frac{4^{\beta-1}\Gamma (\alpha)\Gamma(\beta)}{(b-a)^{\beta-1}} \biggl( \int _{a}^{b}(b-s)^{\alpha-2}(s-a) \,\mathrm{d}s \biggr)^{-1}. $$
4 Applications to eigenvalue problems
In this section, we present some applications of the obtained results to eigenvalue problems.
Corollary 4.1
Let
λ
be an eigenvalue of the problem
where
\(2<\alpha\leq3\), \(1<\beta\leq2\), and
\(p>1\). Then
$$ \left \{ \textstyle\begin{array}{l} D_{0^{+}}^{\beta}(\Phi_{p}(D_{0^{+}}^{\alpha}u(t)))+\lambda\Phi _{p}(u(t))=0, \quad 0< t< 1, \\ u(0)=u'(0)=u'(1)=0, \qquad D_{0^{+}}^{\alpha}u(0)=D_{0^{+}}^{\alpha}u(1)=0, \end{array}\displaystyle \right . $$
(4.1)
$$ |\lambda|\geq\frac{\Gamma(2\beta)}{\Gamma(\beta)} \biggl(\frac {\Gamma(\alpha)\Gamma(\alpha+1)}{\Gamma(\alpha-1)} \biggr)^{p-1}. $$
(4.2)
Proof
Let λ be an eigenvalue of (4.1). Then there exists a nontrivial solution \(u=u_{\lambda}\) to (4.1). Using Theorem 3.1 with \((a,b)=(0,1)\) and \(\chi(s)=\lambda\), we obtain Observe that and where B is the beta function defined by Using the identity we get the desired result. □
$$|\lambda| \int_{0}^{1} (1-s)^{\beta-1}s^{\beta-1} \, \mathrm{d}s\geq \bigl[\Gamma(\alpha)\bigr]^{p-1}\Gamma(\beta) \biggl( \int_{0}^{1}(1-s)^{\alpha -2}s \,\mathrm{d}s \biggr)^{1-p}. $$
$$\int_{0}^{1} (1-s)^{\beta-1}s^{\beta-1} \, \mathrm{d}s=B(\beta,\beta) $$
$$\int_{0}^{1}(1-s)^{\alpha-2}s \,\mathrm{d}s= \int_{0}^{1}s^{2-1}(1-s)^{(\alpha -1)-1} \, \mathrm{d}s=B(2,\alpha-1), $$
$$B(x,y)= \int_{0}^{1} s^{x-1}(1-s)^{y-1} \, \mathrm{d}s, \quad x,y>0. $$
$$B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}, $$
Corollary 4.2
Let
λ
be an eigenvalue of the problem
where
\(2<\alpha\leq3\)
and
\(1<\beta\leq2\). Then
$$ \left \{ \textstyle\begin{array}{l} D_{0^{+}}^{\beta}(D_{0^{+}}^{\alpha}u(t))+\lambda u(t)=0,\quad 0< t< 1, \\ u(0)=u'(0)=u'(1)=0, \qquad D_{0^{+}}^{\alpha}u(0)=D_{0^{+}}^{\alpha}u(1)=0, \end{array}\displaystyle \right . $$
$$ |\lambda|\geq\frac{\Gamma(\alpha)\Gamma(\alpha+1)\Gamma(2\beta )}{\Gamma(\alpha-1)\Gamma(\beta)}. $$
(4.3)
Proof
It follows from inequality (4.2) by taking \(p=2\). □
Acknowledgements
The authors extend their appreciation to Distinguished Scientist Fellowship Program (DSFP) at King Saud University (Saudi Arabia).
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.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors carried out the proof. All authors conceived of the study and participated in its design and coordination. All authors read and approved the final manuscript.