Theorem 3.1
Assume that the following conditions
\((\mathrm{H}_{1})\)–\((\mathrm{H}_{3})\)
are satisfied:
-
\((\mathrm{H}_{1})\)
:
-
\(f(t,u)\in C((0,1)\times(0,+\infty),[0,+\infty))\)
and
\(f(t,u)\)
is nonincreasing relative to
u.
-
\((\mathrm{H}_{2})\)
:
-
For any constant
\(\lambda>0\),
$$0< \int^{1}_{0} (1-s)^{\alpha-1}f\bigl(s,\lambda s^{\beta-2}(1-s)^{2}\bigr)\,ds< +\infty. $$
-
\((\mathrm{H}_{3})\)
:
-
There exist a function
\(a\in C[0,1]\)
and a constant
\(k>0\)
such that
\(a(t)\geq kt^{\beta-2}(1-t)^{2}\), \(t\in[0,1]\), and
$$\begin{aligned} & \int^{1}_{0}G(t,r)\varphi_{q} \biggl( \int^{1}_{0}H(r,s)f\bigl(s,a(s)\bigr)\,ds \biggr)\,dr=b(t)\geq a(t), \\ & \int^{1}_{0}G(t,r)\varphi_{q} \biggl( \int^{1}_{0}H(r,s)f\bigl(s,b(s)\bigr)\,ds \biggr)\,dr \geq a(t). \end{aligned}$$
Then BVP (1.1) has at least one positive solution
w
which satisfies
\(w(t)\geq mt^{\beta-2}\times(1-t)^{2}\)
for some
\(m>0\).
Proof
Let
$$P= \bigl\{ u\in C[0,1]: \mbox{there exists }k_{u}>0\mbox{ such that }u(t) \geq k_{u}t^{\beta-2}(1-t)^{2}, t\in[0,1] \bigr\} . $$
Define an operator T by
$$Tu(t)= \int^{1}_{0}G(t,r)\varphi_{q} \biggl( \int^{1}_{0}H(r,s)f\bigl(s,u(s)\bigr)\,ds \biggr)\,dr, \quad u\in P. $$
For \(u\in P\), there exists \(k_{u}>0\) such that \(u(t)\geq k_{u}t^{\beta-2}(1-t)^{2}\), \(t\in[0,1]\). By \((H_{1})\) and \((H_{2})\), we have
$$\int^{1}_{0}H(t,s)f\bigl(s,u(s)\bigr)\,ds\leq d_{1} \int^{1}_{0}(1-s)^{\alpha-1}f \bigl(s,k_{u}s^{\beta-2}(1-s)^{2}\bigr)\,ds< +\infty. $$
Hence
$$\begin{aligned} Tu(t) =& \int^{1}_{0}G(t,r)\varphi_{q} \biggl( \int^{1}_{0}H(r,s)f\bigl(s,u(s)\bigr)\,ds \biggr)\,dr \\ \leq& \int^{1}_{0}\frac{M_{0}}{\Gamma(\beta)}q(r) \varphi_{q} \biggl( \int^{1}_{0}d_{1}(1-s)^{\alpha-1}f \bigl(s,k_{u}s^{\beta-2}(1-s)^{2}\bigr)\,ds \biggr)\,dr \\ =&\frac{M_{0}}{\Gamma(\beta)}d_{1}^{q-1} \int^{1}_{0}q(r)\,dr\varphi_{q} \biggl( \int^{1}_{0}(1-s)^{\alpha-1}f \bigl(s,k_{u}s^{\beta-2}(1-s)^{2}\bigr)\,ds \biggr)< + \infty. \end{aligned}$$
(3.1)
On the other hand, choose \(t_{0}\in(0,1)\) such that \(Tu(t_{0})=k_{T_{u}}>0\). It follows from Lemma 2.3 that
$$\begin{aligned} Tu(t) =& \int^{1}_{0}G(t,r)\varphi_{q} \biggl( \int^{1}_{0}H(r,s)f\bigl(s,u(s)\bigr)\,ds \biggr)\,dr \\ \geq&\frac{\beta-2}{\Gamma(\beta)}k(t) \int^{1}_{0}q(r)\varphi_{q} \biggl( \int^{1}_{0}H(r,s)f\bigl(s,u(s)\bigr)\,ds \biggr)\,dr \\ \geq&\frac{\beta-2}{M_{0}}k(t)Tu(t_{0}) =\frac{\beta-2}{M_{0}}k_{T_{u}}t^{\beta-2}(1-t)^{2}, \quad t\in[0,1]. \end{aligned}$$
(3.2)
It follows from (3.1) and (3.2) that T is well defined and \(T(P)\subset P\).
Next, we determine upper and lower solutions of BVP (1.1). In fact, by simple computations, we have
$$\begin{aligned} &{-}D^{\alpha}_{0^{+}}\bigl(\varphi_{p} \bigl(D^{\beta}_{0^{+}}\bigl(Tu(t)\bigr)\bigr)\bigr)=f\bigl(t,u(t) \bigr),\quad t\in(0,1), \end{aligned}$$
(3.3)
$$\begin{aligned} & \textstyle\begin{cases} (Tu)(0)=(Tu)(1)=(Tu)'(0)=(Tu)'(1)=0,\\ D^{\beta}_{0^{+}}(Tu)(0)=0,\qquad D^{\beta}_{0^{+}}(Tu)(1)=bD^{\beta}_{0^{+}}(Tu)(\eta). \end{cases}\displaystyle \end{aligned}$$
(3.4)
Let \(b(t)=Ta(t)\), then by \((H_{1})\) and \((H_{3})\), we have
$$ a(t)\leq Ta(t)=b(t),\qquad b(t)=Ta(t)\geq Tb(t),\quad t\in[0,1]. $$
(3.5)
Since \(a(t)\in P\), from (3.2), we obtain \(Ta(t),Tb(t)\in P\). Thus, by (3.3) and (3.5),
$$\begin{aligned} &{-}D^{\alpha}_{0^{+}}\bigl(\varphi_{p} \bigl(D^{\beta}_{0^{+}}(Tb) (t)\bigr)\bigr)-f\bigl(t,(Tb) (t)\bigr) \leq-D^{\alpha}_{0^{+}}\bigl(\varphi_{p} \bigl(D^{\beta}_{0^{+}}(Tb) (t)\bigr)\bigr)-f\bigl(t,b(t)\bigr)=0, \end{aligned}$$
(3.6)
$$\begin{aligned} &{-}D^{\alpha}_{0^{+}}\bigl(\varphi_{p} \bigl(D^{\beta}_{0^{+}}(Ta) (t)\bigr)\bigr)-f\bigl(t,(Ta) (t)\bigr) \geq-D^{\alpha}_{0^{+}}\bigl(\varphi_{p} \bigl(D^{\beta}_{0^{+}}(Ta) (t)\bigr)\bigr)-f\bigl(t,a(t)\bigr)=0. \end{aligned}$$
(3.7)
Meanwhile, (3.4) implies that \(Ta(t), Tb(t)\) satisfy the boundary conditions of BVP (1.1). Then, from (3.5)–(3.7), \(\varphi(t)=Tb(t)\) and \(\psi(t)=Ta(t)\) are lower and upper solutions of BVP (1.1), respectively.
Next, we shall show that the BVP
$$ \textstyle\begin{cases} -D^{\alpha}_{0^{+}}(\varphi_{p}(D^{\beta}_{0^{+}}u(t)))=g(t,u(t)),\quad 0< t< 1, \\ u(0)=u(1)=u'(0)=u'(1)=0,\qquad D^{\beta}_{0^{+}}u(0)=0,\qquad D^{\beta}_{0^{+}}u(1)=bD^{\beta}_{0^{+}}u(\eta) \end{cases} $$
(3.8)
has a positive solution, where
$$ g\bigl(t,u(t)\bigr)= \textstyle\begin{cases} f(t,\varphi(t)),& u(t)< \varphi(t), \\ f(t,u(t)),& \varphi(t)\leq u(t)\leq \psi(t),\\ f(t,\psi(t)),& u(t)>\psi(t). \end{cases} $$
(3.9)
To see this, we consider the operator \(A:C[0,1]\rightarrow C[0,1]\) defined as follows:
$$Au(t)= \int^{1}_{0}G(t,r)\varphi_{q} \biggl( \int^{1}_{0}H(r,s)g\bigl(s,u(s)\bigr)\,ds \biggr)\,dr. $$
It is well known that a fixed point of the operator A is a solution of BVP (3.8).
It is clear that A is continuous. Since \(\varphi(t)\in P\), there exists \(k_{\varphi}>0\) such that \(\varphi(t)\geq k_{\varphi}t^{\beta-2}(1-t)^{2}\), \(t\in[0,1]\). It follows from \((H_{2})\) that
$$\begin{aligned} \int^{1}_{0}H(t,s)g\bigl(s,u(s)\bigr)\,ds&\leq d_{1} \int^{1}_{0}(1-s)^{\alpha-1}f\bigl(s,\varphi(s) \bigr)\,ds \\ &\leq d_{1} \int^{1}_{0}(1-s)^{\alpha-1}f \bigl(s,k_{\varphi}s^{\beta-2}(1-s)^{2}\bigr)\,ds< +\infty. \end{aligned}$$
(3.10)
Consequently, for \(u\in C[0,1]\) and \(t\in [0,1]\), by (3.9) and (3.10), we have
$$\begin{aligned} Au(t) =& \int^{1}_{0}G(t,r)\varphi_{q} \biggl( \int^{1}_{0}H(r,s)g\bigl(s,u(s)\bigr)\,ds \biggr)\,dr \\ \leq& M_{0} \int^{1}_{0}q(r)\varphi_{q} \biggl( \int^{1}_{0}d_{1}(1-s)^{\alpha-1}g \bigl(s,u(s)\bigr)\,ds \biggr)\,dr \\ \leq& M_{0}d_{1}^{q-1} \int^{1}_{0} q(r)\,dr\varphi_{q} \biggl( \int^{1}_{0}(1-s)^{\alpha-1}f \bigl(s,k_{\varphi}s^{\beta-2}(1-s)^{2}\bigr)\,ds \biggr)< + \infty, \end{aligned}$$
which implies that A is uniformly bounded.
On the other hand, since \(G(t,s)\) is continuous on \([0,1]\times[0,1]\), it is uniformly continuous on \([0,1]\times[0,1]\). Thus, for \(s\in[0,1]\) and for each \(\varepsilon>0\), there exists \(\delta>0\) such that \(\vert t_{1}-t_{2} \vert <\delta\) implies
$$\bigl\vert G(t_{1},s)-G(t_{2},s) \bigr\vert < \frac{\varepsilon}{\varphi_{q} (d_{1}\int^{1}_{0}(1-s)^{\alpha-1}f(s,k_{\varphi}s^{\beta-2}(1-s)^{2})\,ds )}. $$
Furthermore, for \(u\in C[0,1]\),
$$\begin{aligned} & \bigl\vert Au(t_{1})-Au(t_{2}) \bigr\vert \\ &\quad \leq \int^{1}_{0} \bigl\vert G(t_{1},r)-G(t_{2},r)) \bigr\vert \varphi_{q} \biggl( \int^{1}_{0}H(r,s)g\bigl(s,u(s)\bigr)\,ds \biggr)\,dr \\ &\quad \leq \int^{1}_{0} \bigl\vert G(t_{1},r)-G(t_{2},r) \bigr\vert \varphi_{q} \biggl( \int^{1}_{0}d_{1}(1-s)^{\alpha-1}f \bigl(s,\varphi(s)\bigr)\,ds \biggr)\,dr \\ &\quad \leq \int^{1}_{0} \bigl\vert G(t_{1},r)-G(t_{2},r) \bigr\vert \,dr\varphi_{q} \biggl(d_{1} \int^{1}_{0}(1-s)^{\alpha-1}f \bigl(s,k_{\varphi}s^{\beta-2}(1-s)^{2}\bigr)\,ds \biggr)< \varepsilon, \end{aligned}$$
which implies that A is equicontinuous. Thus, the Ascoli–Arzela theorem guarantees A is a compact operator. It follows from Schauder’s fixed point theorem that A has a fixed point w, i.e., \(w=Aw\). Consequently, (3.8) has a solution.
Finally, we will show that BVP (1.1) has at least one positive solution. In fact, we only need to prove that \(\varphi(t)\leq w(t)\leq\psi(t)\), \(t\in[0,1]\). By \((H_{1})\), we have
$$ f\bigl(t,\psi(t)\bigr)\leq g\bigl(t,w(t)\bigr)\leq f\bigl(t, \varphi(t)\bigr),\quad t\in[0,1]. $$
(3.11)
It follows from (3.5) and \((H_{3})\) that
$$ f\bigl(t,b(t)\bigr)\leq g\bigl(t,w(t)\bigr)\leq f\bigl(t,a(t) \bigr),\quad t\in[0,1]. $$
(3.12)
Since \(a(t)\in P\), by (3.3), we have
$$-D^{\alpha}_{0^{+}}\bigl(\varphi_{p} \bigl(D^{\beta}_{0^{+}}\psi(t)\bigr)\bigr)=-D^{\alpha}_{0^{+}} \bigl(\varphi_{p}\bigl(D^{\beta}_{0^{+}}(Ta) (t)\bigr) \bigr)=f\bigl(t,a(t)\bigr),\quad t\in(0,1). $$
By (3.4), (3.5), (3.11), and (3.12), we have
$$\begin{aligned} &{-}D^{\alpha}_{0^{+}}\bigl(\varphi_{p} \bigl(D^{\beta}_{0^{+}}\psi(t)\bigr)\bigr)-\bigl[-D^{\alpha}_{0^{+}} \bigl(\varphi_{p}\bigl(D^{\beta}_{0^{+}}w(t)\bigr)\bigr) \bigr]=f\bigl(t,a(t)\bigr)-g\bigl(t,w(t)\bigr)\geq0,\quad t\in[0,1], \\ &(\psi-w) (0)=(\psi-w) (1)=(\psi-w)'(0)=(\psi-w)'(1)=0, \\ &D^{\beta}_{0^{+}}(\psi-w) (0)=0,\qquad D^{\beta}_{0^{+}}( \psi-w) (1)=bD^{\beta}_{0^{+}}(\psi-w) (\eta). \end{aligned}$$
Setting \(z=\varphi_{p}(D^{\beta}_{0^{+}}\psi(t))-\varphi_{p}(D^{\beta}_{0^{+}}w(t))\), then
$$\begin{aligned} &{-}D^{\alpha}_{0^{+}}z(t)=-D^{\alpha}_{0^{+}}\bigl( \varphi_{p}\bigl(D^{\beta}_{0^{+}}\psi(t)\bigr)\bigr)- \bigl[-D^{\alpha}_{0^{+}}\bigl(\varphi_{p} \bigl(D^{\beta}_{0^{+}}w(t)\bigr)\bigr)\bigr]\geq0, \\ &z(0)=0,\quad z(1)=\varphi_{p}(b)z(\eta). \end{aligned}$$
Hence, by Lemma 2.1, we get \(z(t)\geq0\), \(t\in[0,1]\). Since \(\varphi_{p}\) is monotone increasing, we have \(D^{\beta}_{0^{+}}\psi(t)\geq D^{\beta}_{0^{+}}w(t)\), that is, \(D^{\beta}_{0^{+}}(\psi(t)-w(t))\geq0\), \(t\in[0,1]\). By Remark 2.1, we have \(w(t)\leq\psi(t)\) for \(t\in[0,1]\). Similarly, \(w(t)\geq\varphi(t)\) on \([0,1]\). Therefore, \(w(t)\) is a positive solution of BVP (1.1). And \(\varphi(t)\in P\) implies that there exists \(m>0\) such that \(w(t)\geq\varphi(t)\geq mt^{\beta-2}(1-t)^{2}\), \(t\in[0,1]\). This completes the proof. □