We begin to write problem (1.12) in its equivalent integral form.
Lemma 3.1
If
\(u \in C[a, b]\), \(1 < \alpha, \beta\leq2\), \(p > 1\), and
\(\frac{1}{p} + \frac{1}{q} = 1\). Then BVP (1.12) has a unique solution
$$ u(t) = \int_{a}^{b} G(t,s) \Phi_{q} \biggl( \int_{a}^{b} H(s, \tau) k(\tau)f \bigl(u(\tau) \bigr) \,\mathrm{d}\tau \biggr) \,\mathrm{d}s, $$
(3.1)
where
$$ G(t,s) = \frac{1}{\Gamma(\alpha)} \textstyle\begin{cases} (b-s)^{\alpha-1} - (t-s)^{\alpha-1}, & a\leq s\leq t\leq b, \\ (b-s)^{\alpha-1}, & a \leq t \le s \leq b, \end{cases} $$
(3.2)
and
$$ H(s,\tau) = \frac{1}{\Gamma(\beta)} \textstyle\begin{cases} (\frac{s-a}{b-a} )^{\beta-1}(b-\tau)^{\beta-1} - (s-\tau)^{\beta-1}, & a \leq\tau\leq s \le b, \\ (\frac{s-a}{b-a} )^{\beta-1}(b-\tau)^{\beta-1}, & a \leq s \le\tau \le b. \end{cases} $$
(3.3)
Proof
Set \(\Phi_{p}(^{c} D_{a^{+}}^{\alpha} u(t))=v(t) \). Then BVP (1.12) can be turned into the following coupled boundary value problems:
$$ \textstyle\begin{cases} D_{a^{+}}^{\beta} v(t) = k(t) f(u(t)),\quad a < t < b, \\ v(a) = v(b) = 0, \end{cases} $$
(3.4)
and
$$ \textstyle\begin{cases} {}^{c} D_{a^{+}}^{\alpha} u(t) = \Phi_{q}(v(t)),\quad a < t < b, \\ u^{\prime}(a) = u(b) = 0. \end{cases} $$
(3.5)
From Lemma 2 of [7], we see that BVP (3.4) has a unique solution, which is given by
$$ v(t) = - \int_{a}^{b} H(t, s) k(s) f \bigl(u(s) \bigr) \, \mathrm{d}s, $$
(3.6)
where \(H(t, s)\) is as in (3.3). Moreover, by Lemma 5 of [8], we see that BVP (3.5) has a unique solution, which is given by
$$ u(t) = - \int_{a}^{b} G(t, s) \Phi_{q} \bigl(v(s) \bigr) \,\mathrm{d}s, $$
(3.7)
where \(G(t, s)\) is as in (3.2). Substitute (3.6) into (3.7), we see that BVP (1.12) has a unique solution which is given by (3.1). □
Lemma 3.2
The Green’s function
H
defined by (3.3) satisfies the following properties:
-
(1)
\(H(t, s) \ge0\)
for all
\(a \le t, s \le b\);
-
(2)
\(\max_{t \in[a, b]} H(t, s) = H(s, s), s \in[a, b]\);
-
(3)
\(H(s, s)\)
has a unique maximum given by
$$\max_{s \in[a, b]} H(s, s) = \frac{(b-a)^{\beta-1}}{4^{\beta-1} \Gamma (\beta)}; $$
-
(4)
\({ \min_{t \in [\frac{3a+b}{4}, \frac{a+3b}{4} ]} H(t, s) \ge \sigma(s) H(s, s), a < s < b,}\)
where
$$\begin{aligned} \begin{aligned} &\sigma(s) = \textstyle\begin{cases} \frac{ (\frac{3(b-a)(b-s)}{4} )^{\beta-1} - (b-a)^{\beta-1} (\frac {a+3b}{4}-s )^{\beta-1}}{(s-a)^{\beta-1} (b-s)^{\beta-1}} & \textit{if } s \in (a, c_{\beta} ], \\ (\frac{b-a}{4(s-a)} )^{\beta-1} & \textit{if } s \in [c_{\beta}, b ), \end{cases}\displaystyle \\ &c_{\beta}:= \frac{\frac{a+3b}{4} - b A_{\beta}}{1-A_{\beta}},\qquad A_{\beta } = \biggl( \biggl( \frac{3}{4} \biggr)^{\beta-1} - \biggl(\frac{1}{4} \biggr)^{\beta-1} \biggr)^{\frac{1}{\beta-1}}. \end{aligned} \end{aligned}$$
(3.8)
Proof
The first three properties are proved in [7]. For convenience, we set
$$h_{1}(t, s) = \frac{1}{\Gamma(\beta)} \biggl( \biggl(\frac{t-a}{b-a} \biggr)^{\beta-1}(b-s)^{\beta-1} - (t-s)^{\beta-1} \biggr),\quad s \le t $$
and
$$h_{2}(t, s) = \frac{1}{\Gamma(\beta)} \biggl(\frac{t-a}{b-a} \biggr)^{\beta-1}(b-s)^{\beta-1},\quad t \le s. $$
From [7], we know that \(h_{1}(t, s)\) is decreasing with respect to t for \(s \le t\), and \(h_{2}(t, s)\) is increasing with respect to t for \(t \le s\). Thus
$$\min_{t \in [\frac{3a+b}{4}, \frac{a+3b}{4} ]} H(t, s) = \textstyle\begin{cases} h_{1} (\frac{a+3b}{4}, s ) & \text{if } s \in (a, \frac{3a+b}{4} ], \\ \min \{h_{1} (\frac{a+3b}{4}, s ), h_{2} (\frac{3a+b}{4}, s ) \} & \text{if } s \in [\frac{3a+b}{4}, \frac{a+3b}{4} ], \\ h_{2} (\frac{3a+b}{4}, s ) & \text{if } s \in [\frac{a+3b}{4}, b ). \end{cases} $$
From
$$h_{1} \biggl(\frac{a+3b}{4}, s \biggr) = h_{2} \biggl( \frac{3a+b}{4}, s \biggr) $$
we have
$$\biggl(\frac{\frac{a+3b}{4}-s}{b-s} \biggr)^{\beta-1} = \biggl(\frac {3}{4} \biggr)^{\beta-1} - \biggl(\frac{1}{4} \biggr)^{\beta-1}, $$
which implies that
$$s = \frac{\frac{a+3b}{4} - b A_{\beta}}{1-A_{\beta}} = c_{\beta}, $$
where \(c_{\beta}\) and \(A_{\beta}\) are as in (3.8). It is easy to check that \(A_{\beta} < \frac{3}{4}\) and \(c_{\beta} < \frac{a+3b}{4}\). On the other hand, since
$$3^{\beta-1} + 8^{\beta-1} \ge2 \sqrt{3^{\beta-1} 8^{\beta-1}} \ge4^{\frac{\beta-1}{2}} 3^{\frac{\beta-1}{2}} 8^{\frac {\beta-1}{2}} = 96^{\frac{\beta-1}{2}} > 9^{\beta-1}, $$
we have
$$\biggl(\frac{3}{4} \biggr)^{\beta-1} < \biggl(\frac{2}{3} \biggr)^{\beta-1} + \biggl(\frac{1}{4} \biggr)^{\beta-1}, $$
from which we deduce that \(A_{\beta} < \frac{2}{3}\) and \(c_{\beta} > \frac{3a+b}{4}\). So \(c_{\beta} \in (\frac{3a+b}{4}, \frac{a+3b}{4} )\) is the unique solution of the equation \(h_{1} (\frac {a+3b}{4}, s ) = h_{2} (\frac{3a+b}{4}, s )\). Hence
$$\begin{aligned} \min_{t \in [\frac{3a+b}{4}, \frac{a+3b}{4} ]} H(t, s) = {}& \textstyle\begin{cases} h_{1} (\frac{a+3b}{4}, s ) & \text{if } s \in (a, c_{\beta} ], \\ h_{2} (\frac{3a+b}{4}, s ) & \text{if } s \in [c_{\beta}, b ) \end{cases}\displaystyle \\={}& \frac{1}{\Gamma(\beta)} \textstyle\begin{cases} (\frac{3(b-s)}{4} )^{\beta-1} - (\frac{a+3b}{4} - s )^{\beta-1} & \text{if } s \in (a, c_{\beta} ], \\ (\frac{b-s}{4} )^{\beta-1} & \text{if } s \in [c_{\beta}, b ) \end{cases}\displaystyle \\ \ge{}&\sigma(s) H(s, s). \end{aligned}$$
□
Remark 3.1
Since \(\frac{2a+b}{3} < \frac{2b-a}{3}\) implies \(3a < b\), we see that the conclusion of Lemma 7(4) in [13] only holds for \(a < \frac{b}{3}\).
Lemma 3.3
[8]
The Green’s function
G
defined by (3.2) satisfies the following properties:
-
(i)
\(0 \le G(t, s) \le G(s, s) = \frac{1}{\Gamma(\alpha)}(b-s)^{\alpha -1} \)
for all
\(a \le t, s \le b\);
-
(ii)
\(G(s, s)\)
has a unique maximum given by
$$\max_{s \in[a, b]} G(s, s) = \frac{1}{\Gamma(\alpha)}(b-a)^{\alpha-1}; $$
-
(iii)
\({ \min_{t \in [\frac{3a+b}{4}, \frac{a+3b}{4} ]} G(t, s) \ge \mu(s) G(s, s), a < s < b,}\)
where
$$\mu(s) = \textstyle\begin{cases} 1- (\frac{\frac{a+3b}{4}-s}{b-s} )^{\alpha-1} & \textit{if } s \in (a, \frac{a+3b}{4} ], \\ 1 & \textit{if } s \in [\frac{a+3b}{4}, b ). \end{cases} $$
Let
\(E = C[a, b]\)
be endowed with the norm
\(\Vert x \Vert = \max_{t \in[a, b]} \vert x(t) \vert \). Define the cone
\(P \subset E\)
by
$$P = \bigl\{ x \in E| x(t) \ge0 \ \forall t \in[a, b] \textit{ and } \Vert x \Vert \neq0 \bigr\} . $$
Theorem 3.4
Let
\(k: [a, b] \to\mathbb{R}_{+} = [0, + \infty)\)
be a nontrivial Lebesgue integrable function. Suppose that there exist two positive constants
\(r_{2} > r_{1} > 0\)
such that the following assumptions:
-
(H1)
\(f(x) \ge\rho\Phi_{p}(r_{1})\)
for
\(x \in[0, r_{1}]\),
-
(H2)
\(f(x) \le\omega\Phi_{p}(r_{2})\)
for
\(x \in[0, r_{2}]\),
are satisfied, where
$$\rho= \biggl[ \int_{a}^{b} \sigma(\tau) H(\tau, \tau) k(\tau) \, \mathrm{d}\tau\times\Phi_{p} \biggl( \int_{\frac{3a+b}{4}}^{\frac{a+3b}{4}} \mu(s) G(s, s) \,\mathrm{d}s \biggr) \biggr]^{-1} $$
and
$$\omega= \biggl[ \int_{a}^{b} H(\tau, \tau) k(\tau) \,\mathrm{d}\tau \times\Phi_{p} \biggl( \int_{a}^{b} G(s,s) \,\mathrm{d}s \biggr) \biggr]^{-1}. $$
Then FBVP (1.12) has at least one nontrivial positive solution
u
belonging to
E
such that
\(r_{1} \le \Vert u \Vert \le r_{2}\).
Proof
Let \(T: P \to E\) be the operator defined by
$$Tu(t) = \int_{a}^{b} G(t,s) \Phi_{q} \biggl( \int_{a}^{b} H(s, \tau) k(\tau)f\bigl(u(\tau)\bigr) \,\mathrm{d}\tau \biggr) \,\mathrm{d}s. $$
By using the Arzela-Ascoli theorem, we can prove that \(T: P \to P\) is completely continuous. Let \(\Omega_{i} = \{u \in P: \Vert u \Vert \le r_{i}\}\), \(i = 1, 2\). From (H1), and Lemmas 3.2 and 3.3, we obtain for \(t \in [\frac{3a+b}{4}, \frac{a+3b}{4} ]\) and \(u \in P \cap \partial\Omega_{1}\)
$$\begin{aligned} (Tu) (t) &\ge \int_{a}^{b} \min_{t \in [\frac{3a+b}{4}, \frac{a+3b}{4} ]} G(t,s) \Phi_{q} \biggl( \int_{a}^{b} H(s, \tau) k(\tau)f \bigl(u(\tau) \bigr) \,\mathrm{d} \tau \biggr) \,\mathrm{d}s \\ &\ge \int_{a}^{b} \mu(s) G(s, s) \Phi_{q} \biggl( \int_{a}^{b} H(s, \tau) k(\tau)f \bigl(u(\tau) \bigr) \,\mathrm{d} \tau \biggr) \,\mathrm{d}s \\ &\ge \int_{\frac{3a+b}{4}}^{\frac{a+3b}{4}} \mu(s) G(s, s) \,\mathrm{d}s \cdot \Phi_{q} \biggl( \int_{a}^{b} \min_{s \in [\frac{3a+b}{4}, \frac{a+3b}{4} ]} H(s, \tau) k(\tau)f \bigl(u(\tau) \bigr) \,\mathrm{d}\tau \biggr) \\ &\ge \int_{\frac{3a+b}{4}}^{\frac{a+3b}{4}} \mu(s) G(s, s) \,\mathrm{d}s \cdot \Phi_{q} \biggl( \int_{a}^{b} \sigma(\tau) H(\tau, \tau) k(\tau)f \bigl(u(\tau) \bigr) \,\mathrm{d}\tau \biggr) \\ &\ge \int_{\frac{3a+b}{4}}^{\frac{a+3b}{4}} \mu(s) G(s, s) \,\mathrm{d}s \cdot \Phi_{q} \biggl( \int_{a}^{b} \sigma(\tau) H(\tau, \tau) k(\tau) \, \mathrm{d}\tau \biggr) \Phi_{q}(\rho) \cdot r_{1} \\ &= \Vert u \Vert . \end{aligned}$$
Hence, \(\Vert Tu \Vert \ge \Vert u \Vert \) for \(u \in P \cap\partial\Omega_{1}\). On the other hand, from (H2), Lemmas 3.2 and 3.3, we have
$$\begin{aligned} \Vert Tu \Vert &= \max_{t \in[a, b]} \int_{a}^{b} G(t,s) \Phi_{q} \biggl( \int_{a}^{b} H(s, \tau) k(\tau)f \bigl(u(\tau) \bigr) \,\mathrm{d} \tau \biggr) \,\mathrm{d}s \\ &\le \int_{a}^{b} G(s,s) \,\mathrm{d}s \cdot \Phi_{q} \biggl( \int_{a}^{b} H(\tau, \tau) k(\tau)f \bigl(u(\tau) \bigr) \,\mathrm{d}\tau \biggr) \\ &\le \int_{a}^{b} G(s,s) \,\mathrm{d}s \cdot \Phi_{q} \biggl( \int_{a}^{b} H(\tau, \tau) k(\tau) \,\mathrm{d}\tau \biggr) \Phi_{q}(\omega) r_{2} = \Vert u \Vert \end{aligned}$$
for \(u \in P \cap\partial\Omega_{2}\). Thus, by Lemma 2.1, we see that the operator T has a fixed point in \(u \in P \cap(\bar{\Omega}_{2} \setminus\Omega_{1})\) with \(r_{1} \le \Vert u \Vert \le r_{2}\), and clearly u is a positive solution for FBVP (1.12).
Next, we will give two Lyapunov inequalities for FBVP (1.12). □
Theorem 3.5
Let
\(k: [a, b] \to\mathbb{R}_{+}\)
be a real nontrivial Lebesgue function. Suppose that there exists a positive constant
M
satisfying
\(0 \le f(x) \le M \Phi_{p}(x)\)
for any
\(x \in\mathbb{R}_{+}\). If (1.12) has a nontrivial solution in
P, then the following Lyapunov inequality holds:
$$\int_{a}^{b} k(\tau) \,\mathrm{d}\tau> \frac{4^{\beta-1} \Gamma(\beta)}{M (b-a)^{\beta-1}} \Phi_{p} \biggl(\frac {\Gamma(\alpha+1)}{(b-a)^{\alpha}} \biggr). $$
Proof
Assume \(u \in P\) is a nontrivial solution for (1.12), then \(\Vert u \Vert \neq0\). From (3.1), and Lemmas 3.2 and 3.3, \(\forall t \in[a, b]\), we have
$$\begin{aligned} 0 &\le u(t) \le \int_{a}^{b} G(s,s) \Phi_{q} \biggl( \int_{a}^{b} H(\tau, \tau) k(\tau)f \bigl(u(\tau) \bigr) \,\mathrm{d}\tau \biggr) \,\mathrm{d}s \\ &< \int_{a}^{b} G(s,s) \,\mathrm{d}s \cdot \Phi_{q} \biggl( \int_{a}^{b} H(\tau, \tau) k(\tau) \,\mathrm{d}\tau \biggr) \Phi_{q}(M) \Vert u \Vert \\ &\le\frac{1}{\Gamma(\alpha)} \int_{a}^{b} (b-s)^{\alpha-1} \,\mathrm{d}s \cdot \Phi_{q} \biggl( \int_{a}^{b} \frac{(b-a)^{\beta-1}}{4^{\beta-1} \Gamma(\beta)} k(\tau) \,\mathrm{d} \tau \biggr) \Phi_{q}(M) \Vert u \Vert \\ &= \frac{1}{\Gamma(\alpha+1)} (b-a)^{\alpha} \cdot\Phi_{q} \biggl( \frac{(b-a)^{\beta-1}}{4^{\beta-1} \Gamma(\beta)} \biggr) \Phi_{q} \biggl( \int_{a}^{b} k(\tau) \,\mathrm{d}\tau \biggr) \Phi_{q}(M) \Vert u \Vert , \end{aligned}$$
which implies that
$$\int_{a}^{b} k(\tau) \,\mathrm{d}\tau> \frac{4^{\beta-1} \Gamma(\beta)}{M (b-a)^{\beta-1}} \Phi_{p} \biggl(\frac {\Gamma(\alpha+1)}{(b-a)^{\alpha}} \biggr). $$
□
Theorem 3.6
Let
\(k: [a, b] \to\mathbb{R}_{+}\)
be a real nontrivial Lebesgue function. Assume that
\(f \in C(\mathbb{R}_{+}, \mathbb{R}_{+})\)
is a concave and nondecreasing function. If (1.12) has a nontrivial solution
\(u \in P\), then
$$\int_{a}^{b} k(\tau) \,\mathrm{d}\tau> \frac{4^{\beta-1} \Gamma(\beta) \Phi_{p}(\Gamma(\alpha+1)) \Phi_{p}(\eta)}{(b-a)^{\alpha p + \beta-\alpha -1} f(\eta)}, $$
where
\(\eta= \max_{t \in[a, b]} u(t)\).
Proof
By (3.1), Lemmas 3.2 and 3.3, we get
$$\begin{aligned} &u(t) \le \int_{a}^{b} G(s,s) \Phi_{q} \biggl( \int_{a}^{b} H(\tau, \tau) k(\tau)f \bigl(u(\tau) \bigr) \,\mathrm{d} \tau \biggr) \,\mathrm{d}s, \\ &\Vert u \Vert < \frac{1}{\Gamma(\alpha)} \int_{a}^{b} (b-s)^{\alpha-1} \,\mathrm{d}s \cdot \Phi_{q} \biggl( \frac{(b-a)^{\beta-1}}{4^{\beta-1} \Gamma(\beta)} \biggr) \Phi_{q} \biggl( \int_{a}^{b} k(\tau)f \bigl(u(\tau) \bigr) \, \mathrm{d} \tau \biggr) \\ &\phantom{\Vert u \Vert }= \frac{(b-a)^{\alpha}}{\Gamma(\alpha+1)} \cdot\Phi_{q} \biggl( \frac {(b-a)^{\beta-1}}{4^{\beta-1} \Gamma(\beta)} \biggr) \Phi_{q} \biggl( \int_{a}^{b} k(\tau)f \bigl(u(\tau) \bigr) \, \mathrm{d} \tau \biggr). \end{aligned}$$
Using Lemma 2.2, and taking into account that f is concave and nondecreasing, we see that
$$\begin{aligned} \Vert u \Vert &< \frac{(b-a)^{\alpha}}{\Gamma(\alpha+1)} \cdot\Phi _{q} \biggl( \frac{(b-a)^{\beta-1}}{4^{\beta-1} \Gamma(\beta)} \biggr) \Phi_{q} \biggl( \int_{a}^{b} k(s) \,\mathrm{d}s \biggr) \Phi_{q} \biggl( \int_{a}^{b} \frac{k(\tau)f(u(\tau)) \,\mathrm{d} \tau}{\int_{a}^{b} k(s) \,\mathrm {d}s} \biggr) \\ &< \frac{(b-a)^{\alpha}}{\Gamma(\alpha+1)} \cdot\Phi_{q} \biggl( \frac {(b-a)^{\beta-1}}{4^{\beta-1} \Gamma(\beta)} \biggr) \Phi_{q} \biggl( \int_{a}^{b} k(s) \,\mathrm{d}s \biggr) \Phi_{q} \bigl(f(\eta) \bigr), \end{aligned}$$
where \(\eta= \max_{t \in[a, b]} u(t)\). Hence,
$$\int_{a}^{b} k(s) \,\mathrm{d}s > \frac{4^{\beta-1} \Gamma(\beta) \Phi_{p}(\Gamma(\alpha+1)) \Phi_{p}(\eta)}{(b-a)^{\alpha p + \beta-\alpha-1} f(\eta)}. $$
The proof is completed. □