Let \(E=C^{n-2}[0,1]\) and define the cone K in E as
$$K= \bigl\{ u\in E, u^{(i)}(0)=0, u^{(i)}(t)\geq0, t\in[0,1], i=0,1,2,\ldots,n-2 \bigr\} . $$
By Lemma 2.4 and (2.1) we have
$$ D^{\mu_{i}}_{0^{+}}u\in C[0,1], \qquad D^{\mu_{i}}_{0^{+}}u(t) \geq0, \quad u\in K, t\in[0,1], i=0,1,2,\ldots,n-2 . $$
(3.1)
For any \(m\in\mathbb{N}\), define the operator \(Q_{m}:K\rightarrow E\) as follows:
$$ (Q_{m}u) (t)= \int_{0}^{1}G(t,s)f_{m} \bigl(s,u(s),D^{\mu _{1}}_{0^{+}}u(s),D^{\mu_{2}}_{0^{+}}u(s), \ldots,D^{\mu _{n-2}}_{0^{+}}u(s) \bigr)\,ds. $$
(3.2)
Lemma 3.1
Let (H0)-(H3) hold. Then, for any
\(m\in\mathbb{N}\), \(Q_{m}: K\rightarrow K\)
is a completely continuous operator.
Proof
First, we show that \(Q_{m}:K\rightarrow K\). Given \(u\in K\), by Lemma 2.6 we get that
$$\frac{\partial^{j}}{\partial t^{j}}G(t,s), \quad j=0,1,2,\ldots,n-2, $$
are nonnegative and continuous on \([0,1]\times[0,1]\) and \(G(0,s)=0\) for \(s\in[0,1]\). So we have
$$\begin{aligned}& Q_{m}u\in C^{n-2}[0,1],\qquad (Q_{m}u)^{(j)}(0)=0,\quad j=0,1,2,\ldots,n-2, \\& (Q_{m}u)^{(j)}(t)\geq0,\quad t\in[0,1], j=0,1,2,\ldots,n-2. \end{aligned}$$
As a result, \(Q_{m}:K\rightarrow K\).
In order to prove that \(Q_{m}\) is a continuous operator, let \(\{u_{k}\} \subset K\) be a convergent sequence. Suppose that \(\lim_{k\rightarrow \infty}u_{k}=u\in K\). Then
$$\lim_{k\rightarrow\infty}u_{k}^{(j)}(t)=u^{(j)}(t), \quad j=0,1,2,\ldots,n-2, $$
uniformly for \(t\in[0,1]\). For \(i-1<\mu_{i}\leq i\) (\(i=1,2,\ldots,n-2\)) and \(t\in[0,1]\), we get
$$\bigl\vert D_{0^{+}}^{\mu_{i}}u_{k}(t)-D_{0^{+}}^{\mu_{i}}u(t) \bigr\vert \leq\frac{\| u^{(n-2)}_{k}-u^{(n-2)}\|}{\Gamma(n-2-\mu_{i})} \int ^{t}_{0}(t-s)^{n-\mu _{i}-3}\,ds\leq \frac{\|u^{(n-2)}_{k}-u^{(n-2)}\|}{\Gamma(n-1-\mu_{i})}, $$
so we get
$$\lim_{k\rightarrow\infty}D_{0^{+}}^{\mu_{i}}u_{k}(t)=D_{0^{+}}^{\mu _{i}}u(t) ,\quad i=1,2,\ldots,n-2 $$
uniformly for \(t\in[0,1]\). Moreover, \(\{u_{k}\}\subset K\) is a convergent sequence. There exists \(r>0\) such that \(\|u_{k}\|_{2}\leq r\) (\(k\in\mathbb{N}\)). Then \(\| u^{(j)}_{k}\|\leq r\) (\(j=0,1,2,\ldots,n-2\); \(k\in\mathbb{N}\)). For \(i-1<\mu_{i}\leq i\) (\(i=1,2,\ldots,n-2\)), by (2.1), for any \(t\in [0,1]\), we have
$$\begin{aligned} 0&\leq D_{0^{+}}^{\mu_{i}}u_{k}(t)\leq\frac{\|u^{(n-2)}_{k}\| }{\Gamma (n-2-\mu_{i})} \int^{t}_{0}(t-s)^{n-\mu_{i}-3}\,ds \\ &\leq\frac{\|u^{(n-2)}_{k}\|}{\Gamma(n-1-\mu_{i})}\leq\frac {r}{\Gamma (n-1-\mu_{i})},\quad i=0,1,2,\ldots,n-2,k\in \mathbb{N}. \end{aligned}$$
(3.3)
Let
$$ \begin{aligned} &\rho_{k}(t)=f_{m} \bigl(t,u_{k}(t),D^{\mu _{1}}_{0^{+}}u_{k}(t),D^{\mu_{2}}_{0^{+}}u_{k}(t), \ldots,D^{\mu _{n-2}}_{0^{+}}u_{k}(t) \bigr), \\ &\rho(t)=f_{m} \bigl(t,u(t),D^{\mu_{1}}_{0^{+}}u(t),D^{\mu _{2}}_{0^{+}}u(t), \ldots,D^{\mu_{n-2}}_{0^{+}}u(t) \bigr),\quad t\in[0,1], k\in\mathbb{N}. \end{aligned} $$
(3.4)
For \(s\in[0,1]\setminus\Gamma\), where \(\operatorname{mes}(\Gamma)=0\), \(f_{m}(s,x_{1},x_{2},\ldots,x_{n-1})\) is continuous on \(\mathbb {R}^{n-1}_{+} \) with respect to \(x_{i}\), so \(f_{m}(s,x_{1},x_{2},\ldots ,x_{n-1})\) is uniformly continuous with respect to \(x_{i}\) on
$$[0,r]\times \biggl[0,\frac{r}{\Gamma(n-1-\mu_{1})} \biggr]\times \cdots \times \biggl[0, \frac{r}{\Gamma(n-1-\mu_{n-2})} \biggr]. $$
Hence, for any \(\varepsilon>0\), there exists \(\delta>0\) such that, for any \(x_{1}^{1}, x_{1}^{2}\in[0,r], x_{2}^{1}, x_{2}^{2}\in [0,\frac{r}{\Gamma(n-1-\mu_{1})}], \ldots, x_{n-1}^{1}, x_{n-1}^{2}\in[0,\frac{r}{\Gamma(n-1-\mu_{n-2})}]\), \(|x_{1}^{1}-x_{1}^{2}|<\delta, |x_{2}^{1}-x_{2}^{2}|<\delta, \ldots , |x_{n-1}^{1}-x_{n-1}^{2}|<\delta\), we have
$$ \bigl\vert f_{m} \bigl(s,x_{1}^{1},x_{2}^{1}, \ldots ,x_{n-1}^{1} \bigr)-f_{m} \bigl(s,x_{1}^{2},x_{2}^{2},\ldots ,x_{n-1}^{2} \bigr) \bigr\vert < \varepsilon. $$
(3.5)
Since \(\|u_{k}-u\|_{2}\rightarrow0\), for the above \(\delta>0\), there exists \(N\in\mathbb{N}\) such that, for \(k>N\),
$$\begin{aligned}& \bigl\vert u_{k}(t)-u(t) \bigr\vert , \bigl\vert D_{0^{+}}^{\mu_{1}}u_{k}(t)-D_{0^{+}}^{\mu _{1}}u(t) \bigr\vert ,\ldots, \bigl\vert D_{0^{+}}^{\mu_{n-2}}u_{k}(t)-D_{0^{+}}^{\mu _{n-2}}u(t) \bigr\vert \\& \quad \leq\|u_{k}-u\|_{2}< \delta, \quad \forall t \in[0,1]. \end{aligned}$$
Therefore, for \(k>N\), by (3.5) we have
$$\begin{aligned} \bigl\vert \rho_{k}(s)-\rho(s) \bigr\vert \leq& \bigl\vert f_{m} \bigl(s,u_{k}(s),D^{\mu _{1}}_{0^{+}}u_{k}(s),D^{\mu_{2}}_{0^{+}}u_{k}(s), \ldots,D^{\mu_{n-2}}_{0^{+}}u_{k}(s) \bigr) \\ &{}-f_{m} \bigl(s,u(s),D^{\mu_{1}}_{0^{+}}u(s),D^{\mu _{2}}_{0^{+}}u(s), \ldots,D^{\mu_{n-2}}_{0^{+}}u(s) \bigr) \bigr\vert < \varepsilon. \end{aligned}$$
(3.6)
It follows from (3.6) that
$$ \rho_{k}(s)\rightarrow\rho(s)\quad \mbox{for a.e. } s \in[0,1]. $$
(3.7)
By (1.4) we have
$$\begin{aligned} 0 \leq&\rho_{k}(t)=f_{m} \bigl(t,u_{k}(t),D^{\mu _{1}}_{0^{+}}u_{k}(t),D^{\mu_{2}}_{0^{+}}u_{k}(t), \ldots,D^{\mu _{n-2}}_{0^{+}}u_{k}(t) \bigr) \\ \leq&\beta(t) p \biggl(\max \biggl\{ u_{k}(t),\frac{1}{m} \biggr\} ,\max \biggl\{ D^{\mu_{1}}_{0^{+}}u_{k}(t), \frac{1}{m} \biggr\} , \ldots,\max \biggl\{ D^{\mu_{n-2}}_{0^{+}}u_{k}(t), \frac{1}{m} \biggr\} \biggr) \\ &{}+\gamma(t)h \biggl(u_{k}(t)+\frac{1}{m},D^{\mu _{1}}_{0^{+}}u_{k}(t)+ \frac {1}{m},\ldots, D^{\mu_{n-2}}_{0^{+}}u_{k}(t)+ \frac{1}{m} \biggr) \\ \leq&\beta(t)p \biggl(\frac{1}{m},\frac{1}{m},\ldots, \frac {1}{m} \biggr) \\ &{}+\gamma(t)h \biggl(r+1,\frac{r}{\Gamma(n-1-\mu_{1})}+1,\ldots, \frac{r}{\Gamma(n-1-\mu_{n-2})}+1 \biggr) \\ =&\varphi_{r}(t),\quad k=1,2,\ldots. \end{aligned}$$
(3.8)
By the integrability of \(\beta(t)\), \(\gamma(t)\) on \([0,1]\) we get that \(\varphi_{r}\in L^{1}(0,1)\), and by (3.7) and (3.8) we have
$$\bigl\vert \rho_{k}(t)-\rho(t) \bigr\vert \leq2 \varphi_{r}(t)\quad \mbox{for a.e. } t\in[0,1], k=1,2,3,\ldots. $$
It follows from the relations in Lemma 2.6 and the Lebesgue dominated convergence theorem that, for any \(m\in\mathbb{N}\),
$$\begin{aligned}& \bigl\vert (Q_{m}u_{k})^{(j)}(t)-(Q_{m}u)^{(j)}(t) \bigr\vert \\& \quad = \biggl\vert \int_{0}^{1}\frac{\partial^{j}}{\partial t^{j}}G(t,s) \bigl[f_{m} \bigl(s,u_{k}(s),D^{\mu_{1}}_{0^{+}}u_{k}(s),D^{\mu _{2}}_{0^{+}}u_{k}(s), \ldots,D^{\mu_{n-2}}_{0^{+}}u_{k}(s) \bigr) \\& \qquad {} -f_{m} \bigl(s,u(s),D^{\mu_{1}}_{0^{+}}u(s),D^{\mu _{2}}_{0^{+}}u(s), \ldots,D^{\mu_{n-2}}_{0^{+}}u(s) \bigr) \bigr]\,ds \biggr\vert \\& \quad \leq \frac{\Theta}{P(0)\Gamma(\alpha-j)} \int_{0}^{1} \bigl\vert \rho _{k}(s)- \rho(s) \bigr\vert \,ds\rightarrow0, \\& \qquad k\rightarrow\infty, t\in[0,1], \quad j=0,1,2,\ldots,n-2. \end{aligned}$$
Hence, for any \(m\in\mathbb{N}\), we have
$$\lim_{k\rightarrow\infty}(Q_{m}u_{k})^{(j)}(t)=(Q_{m}u)^{(j)}(t), \quad j=0,1,2,\ldots,n-2 $$
uniformly for \(t\in[0,1]\). Therefore, for any \(m\in\mathbb{N}\), \(Q_{m}\) is a continuous operator.
Now, for any bounded set \(D\subset K\), we need to prove that \(\{ Q_{m}(D)\}\) is relatively compact in E. In order to apply the Arzelà-Ascoli theorem, we have to prove that \(\{Q_{m}(D)\}\) is bounded in E and that,for any \(m\in\mathbb{N}\), \(\{ (Q_{m}(D))^{(n-2)}(t) \}\) is equicontinuous on \([0,1]\). By the boundedness of \(D\subset K\) there exists a positive number \(R>0\) such that
$$\bigl\Vert u^{(j)} \bigr\Vert \leq R,\quad \forall u\in D, j=0,1,2, \ldots,n-2. $$
Then (3.3) means that
$$\bigl\Vert D_{0^{+}}^{\mu_{i}}u \bigr\Vert \leq \frac{R}{\Gamma(n-1-\mu_{i})},\quad \forall u\in D, i=1,2,\ldots,n-2. $$
Put ρ as in (3.4) and \(0\leq\rho(t)\leq\varphi_{R}(t)\). Then, for \(u\in D\), we have
$$\begin{aligned} 0&\leq(Q_{m}u)^{(j)}(t)= \int_{0}^{1}\frac{\partial ^{j}G(t,s)}{\partial t^{j}}\rho(s)\,ds \\ &\leq\frac{\Theta}{P(0)\Gamma(\alpha-j)} \int_{0}^{1}\varphi _{R}(s)\,ds \\ &=\frac{\Theta\|\varphi_{R}\|_{1}}{P(0)\Gamma(\alpha-j)},\quad j=0,1,\ldots,n-2, \end{aligned}$$
which shows that, for any \(m\in\mathbb{N}\), \(\{Q_{m}(D)\}\) is bounded in E. Moreover, for \(0\leq t_{1}\leq t_{2}\leq1\) and \(u\in D\), we have
$$\begin{aligned}& \bigl\vert (Q_{m}u)^{(n-2)}(t_{2})-(Q_{m}u)^{(n-2)}(t_{1}) \bigr\vert \\& \quad = \biggl\vert \int_{0}^{1} \biggl(\frac{\partial^{n-2}G(t_{2},s)}{\partial t^{n-2}}- \frac{\partial^{n-2}G(t_{1},s)}{\partial t^{n-2}} \biggr)f_{m} \bigl(s,u(s),D^{\mu_{1}}_{0^{+}}u(s),D^{\mu _{2}}_{0^{+}}u(s), \ldots,D^{\mu_{n-2}}_{0^{+}}u(s) \bigr) \,ds \biggr\vert \\& \quad = \biggl\vert \int_{0}^{1} \biggl(\frac{\partial^{n-2}G(t_{2},s)}{\partial t^{n-2}}- \frac{\partial^{n-2}G(t_{1},s)}{\partial t^{n-2}} \biggr)\rho (s)\,ds \biggr\vert \\& \quad \leq e \bigl(t_{2}^{\alpha-n+1}-t_{1}^{\alpha-n+1} \bigr) \int _{0}^{1}(1-s)^{\alpha-n+1}\rho(s)\,ds \\& \qquad {}+\frac{1}{\Gamma(\alpha-n+2)} \biggl\vert \int_{0}^{t_{2}}(t_{2}-s)^{\alpha-n+1} \rho(s)\,ds- \int _{0}^{t_{1}}(t_{1}-s)^{\alpha-n+1} \rho(s)\,ds \biggr\vert \\& \quad \leq \|\rho\|_{1}e \bigl(t_{2}^{\alpha-n+1}-t_{1}^{\alpha-n+1} \bigr)+\frac {1}{\Gamma(\alpha-n+2)} \biggl\vert \int_{t_{1}}^{t_{2}}(t_{2}-s)^{\alpha-n+1} \rho(s)\,ds \\& \qquad {} + \int_{0}^{t_{1}} \bigl((t_{2}-s)^{\alpha -n+1}-(t_{1}-s)^{\alpha -n+1} \bigr)\rho(s)\,ds \biggr\vert \\& \quad \leq e\|\varphi_{R}\|_{1} \bigl(t_{2}^{\alpha-n+1}-t_{1}^{\alpha -n+1} \bigr)+\frac {1}{\Gamma(\alpha-n+2)} \biggl[(t_{2}-t_{1})^{\alpha-n+1} \|\varphi_{R}\|_{1} \\& \qquad {}+ \int _{0}^{t_{1}} \bigl((t_{2}-s)^{\alpha-n+1}-(t_{1}-s)^{\alpha -n+1} \bigr) \varphi_{R}(s)\,ds \biggr] \\& \quad \leq e\|\varphi_{R}\|_{1} \bigl(t_{2}^{\alpha-n+1}-t_{1}^{\alpha-n+1} \bigr)+e \biggl[(t_{2}-t_{1})^{\alpha-n+1}\| \varphi_{R}\|_{1} \\& \qquad {} + \int_{0}^{t_{1}} \bigl((t_{2}-s)^{\alpha-n+1}-(t_{1}-s)^{\alpha -n+1} \bigr) \varphi_{R}(s)\,ds \biggr], \end{aligned}$$
where e is from (H3). Since \((t-s)^{\alpha-n+1}\) is uniformly continuous on \([0,1]\times[0,1]\) and \(t^{\alpha-n+1}\) is uniformly continuous on \([0,1]\), for any \(\varepsilon>0\), there exists \(\delta>0\) such that, for \(0\leq t_{1}\leq t_{2}\leq1\), \(t_{2}-t_{1}<\delta\), \(0< s\leq t_{1}\),
$$\begin{aligned}& t_{2}^{\alpha-n+1}-t_{1}^{\alpha-n+1}< \varepsilon, \\& (t_{2}-s)^{\alpha-n+1}-(t_{1}-s)^{\alpha-n+1}< \varepsilon. \end{aligned}$$
Consequently, for all \(u\in D\), \(0\leq t_{1}\leq t_{2}\leq1\), and \(t_{2}-t_{1}<\min \{\delta,\sqrt[\alpha-n+1]{\varepsilon } \}\), we have the inequality
$$\bigl\vert (Q_{m}u)^{(n-2)}(t_{2})-(Q_{m}u)^{(n-2)}(t_{1}) \bigr\vert \leq3e\| \varphi_{R}\|_{1}\varepsilon. $$
Hence, for any \(m\in\mathbb{N}\), \(\{(Q_{m}(D))^{(n-2)}(t) \}\) is equicontinuous on \([0,1]\). Therefore, for any \(m\in\mathbb{N}\), \(Q_{m}: K\rightarrow K\) is a completely continuous operator. □
To prove the main results, we need the following well-known fixed point theorem.
Lemma 3.2
[20]
Let
K
be a positive cone in a Banach space
E, \(\Omega_{1}\)
and
\(\Omega_{2}\)
be two bounded open sets in
E
such that
\(\theta\in\Omega_{1}\)
and
\(\overline{\Omega}_{1}\subset \Omega _{2}\), and
\(A:K\cap(\overline{\Omega}_{2}\setminus\Omega _{1})\rightarrow K\)
be a completely continuous operator, where
θ
denotes the zero element of
E. Suppose that one of the following two conditions holds:
-
(i)
\(\|Au\|\leq\|u\|\), \(\forall u\in K\cap\partial\Omega_{1}\); \(\|Au\| \geq\| u\|\), \(\forall u\in K\cap\partial\Omega_{2}\);
-
(ii)
\(\|Au\|\geq\|u\|\), \(\forall u\in K\cap\partial\Omega_{1}\); \(\|Au\| \leq \|u\|\), \(\forall u\in K\cap\partial\Omega_{2} \).
Then
A
has a fixed point in
\(P\cap(\overline{\Omega}_{2}\setminus \Omega_{1})\).
Theorem 3.1
Let (H0)-(H3) hold. Then problem (1.5) has a solution
\(x_{m}\in K\), and
$$ \begin{aligned} &x_{m}(t)\geq Mt^{\alpha-1}, \\ &D_{0^{+}}^{\mu_{i}}x_{m}(t) \geq\frac {(n-2-\mu_{i})M}{(n-i+1)!}t^{n-1-\mu_{i}}, \quad i=1,2,\ldots,n-2, t\in [0,1], m\in \mathbb{N}, \end{aligned} $$
(3.9)
where
M
is defined by (H2).
Proof
By Lemma 3.1, \(Q_{m}:K\rightarrow K\) is a completely continuous operator. Then by (1.2) and Lemma 2.6 we have
$$ (Q_{m}u) (t)\geq C \int_{0}^{1}G(t,s)\,ds\geq Mt^{\alpha-1}, $$
(3.10)
and hence, \(\|Q_{m}u\|\geq M\) and \(\|Q_{m}u\|_{2}\geq M\) for \(u\in K\). Let \(\Omega_{1}=\{u\in E:\|u\|_{2}< M\}\). Then
$$\bigl\Vert (Q_{m}u) \bigr\Vert _{2}\geq \Vert u \Vert _{2} \quad \mbox{for } u\in K\cap\partial\Omega_{1}. $$
Let \(W_{m}=p (\frac{1}{m},\frac{1}{m},\ldots,\frac{1}{m} )\). For any \(u\in K\) and \(t\in[0,1]\), by Lemma 2.6 and (1.4) we have
$$\begin{aligned} 0 \leq&(Q_{m}u)^{(i)}(t) \\ \leq& e \int_{0}^{1} \biggl(\beta(t)W_{m}+ \gamma(s)h \biggl(u(s)+\frac {1}{m},D^{\mu_{1}}_{0^{+}}u(s)+ \frac{1}{m},D^{\mu_{2}}_{0^{+}}u(s)+\frac{1}{m}, \ldots, \\ &D^{\mu_{n-2}}_{0^{+}}u(s)+\frac {1}{m} \biggr) \biggr)\,ds \\ \leq& e \biggl(\Vert \beta \Vert _{1}W_{m}+h \biggl( \Vert u\Vert +\frac{1}{m}, \bigl\Vert D^{\mu _{1}}_{0^{+}}u \bigr\Vert +\frac{1}{m}, \bigl\Vert D^{\mu_{2}}_{0^{+}} u \bigr\Vert +\frac{1}{m},\ldots, \\ & \bigl\Vert D^{\mu_{n-2}}_{0^{+}}u \bigr\Vert +\frac{1}{m} \biggr)\Vert \gamma \Vert _{1} \biggr), \quad i=0,1,2,\ldots,n-3,n-2, \end{aligned}$$
(3.11)
where e is from (H3).
For any \(u\in K\) such that \(\|u\|_{2}\leq S\) (\(S>M\)), by (3.3) and (3.11) we have
$$\begin{aligned} \|Q_{m}u\|_{2} \leq& e \biggl(\|\beta \|_{1}W_{m}+h \biggl(\|u\| _{2}+1, \frac {\|u\|_{2}}{\Gamma(n-1-\mu_{1})}+1,\ldots, \\ &\frac{\|u\|_{2}}{\Gamma(n-1-\mu_{n-2})}+1 \biggr)\|\gamma\| _{1} \biggr) \\ \leq& e \biggl(\|\beta\|_{1}W_{m} +h \biggl(\frac{S}{ \Gamma(n-1-\mu_{n-2})}+1,\frac{S}{ \Gamma(n-1-\mu_{n-2})}+1,\ldots, \\ & \frac{S}{ \Gamma(n-1-\mu_{n-2})}+1 \biggr)\|\gamma\|_{1} \biggr). \end{aligned}$$
(3.12)
By (H3), taking \(\overline{\lambda}>0\) such that
$$\limsup_{x\rightarrow\infty}\frac{h(x,x,\ldots,x)}{x}= \lambda< \overline{\lambda}< \frac{\Gamma(n-1-\mu_{n-2})}{e\|\gamma\|_{1}}, $$
we have that there exists \(G>M+1\) such that, for any \(x>G\),
$$ h(x,x,\ldots,x)< \overline{\lambda}x. $$
(3.13)
Taking
$$S>\max \biggl\{ M+1,(G-1)\Gamma(n-1-\mu_{n-2}), \frac{e\|\beta\| _{1}W_{m}+e\overline{\lambda}\|\gamma\|_{1}}{1-e\overline{\lambda }\| \gamma\|_{1}\Gamma^{-1}(n-1-\mu_{n-2})} \biggr\} , $$
we have \(\frac{S}{\Gamma(n-1-\mu_{n-2})}+1>G\) and \(S>M\). Let \(\Omega _{2}= \{u\in E:\|u\|_{2}< S \}\). Then, for any \(u\in K\cap \partial\Omega_{2}\), by (3.12) and (3.13) we get
$$\begin{aligned} \begin{aligned} \|Q_{m}u\|_{2}&\leq e \biggl(\|\beta\|_{1}W_{m}+ \frac{\overline {\lambda }S\|\gamma\|_{1}}{ \Gamma(n-1-\mu_{n-2})}+\overline{\lambda}\|\gamma\|_{1} \biggr) \\ &=e\|\beta\|_{1}W_{m}+e\overline{\lambda}\|\gamma \|_{1}+\frac {e\overline{\lambda}\|\gamma\|_{1}}{\Gamma(n-1-\mu_{n-2})}S \leq S. \end{aligned} \end{aligned}$$
Hence,
$$ \|Q_{m}u\|_{2}\leq\|u\|_{2} \quad \mbox{for } u\in K\cap\partial\Omega _{2}. $$
(3.14)
By Lemma 3.2 we get that the operator \(Q_{m}\) has a fixed point in \(K\cap(\overline{\Omega}_{2}\setminus\Omega_{1})\), and, as a result, \(x_{m}\) is a solution of problem (1.5). Since \(x_{m}\) is a solution of problem (1.5),
$$\begin{aligned}& x_{m}(t)= \int_{0}^{1}G(t,s)f_{m} \bigl(s,x_{m}(s),D^{\mu _{1}}_{0^{+}}x_{m}(s),D^{\mu_{2}}_{0^{+}}x_{m}(s), \ldots, D^{\mu_{n-2}}_{0^{+}}x_{m}(s) \bigr)\,ds, \\& \quad t \in[0,1], m\in\mathbb {N}, \end{aligned}$$
(3.15)
and \(x_{m}\) satisfies \(x_{m}(t)\geq Mt^{\alpha-1}\). In addition, Lemma 2.6 and (1.2) imply
$$\begin{aligned}& x_{m}^{(j)}(t)\geq C \int_{0}^{1}\frac{\partial^{j}}{\partial t^{j}}G(t,s)\,ds \\& \hphantom{x_{m}^{(j)}(t)}\geq \frac{C t^{\alpha-j-1}}{(\alpha-n+2)\Gamma (\alpha -j+1)},\quad t\in[0,1], m\in\mathbb{N}, j=0,1,2,\ldots,n-3, \\& x_{m}^{(n-2)}(t)\geq C \int_{0}^{1}\frac{\partial^{n-2}}{\partial t^{n-2}}G(t,s)\,ds\geq \frac{C t(1-t)}{\Gamma(\alpha-n+3)},\quad t\in[0,1], m\in\mathbb{N}. \end{aligned}$$
By (2.1) we get
$$\begin{aligned} D^{\mu_{i}}_{0^{+}}x_{m}(t) =&\frac{1}{\Gamma(n-2-\mu_{i})} \int _{0}^{t}(t-s)^{n-3-\mu_{i}}x_{m}^{(n-2)}(s) \,ds \\ \geq&\frac{ C}{\Gamma(n-2-\mu_{i})\Gamma(\alpha-n+3)} \int _{0}^{t}(t-s)^{n-\mu_{i}-3}s(1-s)\,ds,\quad i=1,2,\ldots,n-2. \end{aligned}$$
Further, since
$$\begin{aligned} \int_{0}^{t}(t-s)^{n-\mu_{i}-3}s(1-s)\,ds =& \frac{1}{(n-2-\mu _{i})} \int _{0}^{t}(t-s)^{n-2-\mu_{i}}(1-2s)\,ds \\ =&\frac{1}{(n-2-\mu_{i})} \biggl(\frac{t^{n-1-\mu_{i}}}{n-1-\mu _{i}}-\frac {2t^{n-\mu_{i}}}{(n-1-\mu_{i})(n-\mu_{i})} \biggr) \\ =&\frac{t^{n-1-\mu_{i}}}{(n-2-\mu_{i})}\frac{n-\mu _{i}-2t}{(n-1-\mu _{i})(n-\mu_{i})} \\ \geq&\frac{t^{n-1-\mu_{i}}}{(n-2-\mu_{i})}\frac{n-2-\mu _{i}}{(n-1-\mu _{i})(n-\mu_{i})} \\ =&\frac{t^{n-1-\mu_{i}}}{(n-1-\mu_{i})(n-\mu_{i})}, \quad i=1,2,\ldots,n-2, \end{aligned}$$
we get
$$\begin{aligned} D^{\mu_{i}}_{0^{+}}x_{m}(t) \geq& \frac{ C}{\Gamma(n-2-\mu _{i})\Gamma (\alpha-n+3)} \frac{t^{n-1-\mu_{i}}}{(n-1-\mu_{i})(n-\mu_{i})} \\ \geq&\frac{ C(n-2-\mu_{i})}{\Gamma(n+1-\mu_{i})\Gamma(\alpha -n+3)}t^{n-1-\mu_{i}}, \quad i=1,2,\ldots,n-2. \end{aligned}$$
(3.16)
Furthermore,
$$ \Gamma(n+1-\mu_{i})< \Gamma \bigl(n-(i-2) \bigr)=(n-i+1)!,\quad i=1,2, \ldots ,n-2, $$
(3.17)
and then it follows from \(x_{m}(t)\geq Mt^{\alpha-1}\), (3.16) and (3.17) that, for \(t\in[0,1]\) and \(m\in\mathbb{N}\), we have
$$ x_{m}(t)\geq Mt^{\alpha-1},\qquad D_{0^{+}}^{\mu_{i}}x_{m}(t) \geq\frac {(n-2-\mu_{i})M}{(n-i+1)!}t^{n-1-\mu_{i}},\quad i=1,2,\ldots,n-2, $$
(3.18)
where M is defined by (H2). □
In order to finish the main result, we also need the following lemma.
Lemma 3.3
Let
\(x_{m}\)
be a solution of problem (1.5). If (H0)-(H3) hold, then the sequence
\(\{x_{m}\}\)
is relatively compact in
E.
Proof
For \(t\in[0,1]\) and \(m\in\mathbb{N}\), since p is nondecreasing, by (3.18) we have
$$\begin{aligned}& p \bigl(x_{m}(t),D_{0^{+}}^{\mu_{1}}x_{m}(t), \ldots,D_{0^{+}}^{\mu _{n-2}}x_{m}(t) \bigr) \\& \quad \leq p \biggl(Mt^{\alpha-1},\frac{(n-2-\mu_{1})M}{n!}t^{n-1-\mu _{1}},\ldots, \frac{(n-2-\mu_{n-2})M}{3!}t^{n-1-\mu_{n-2}} \biggr), \end{aligned}$$
(3.19)
and by Lemma 2.6, (1.4), (3.3), (3.15), and (3.19), for \(t\in [0,1]\) and \(m\in\mathbb{N}\), we get
$$\begin{aligned} 0 \leq& x_{m}^{(i)}(t) \\ \leq& \int_{0}^{1}\frac{\partial^{n-2}}{\partial t^{n-2}}G(t,s)f_{m} \bigl(s,x_{m}(s),D^{\mu_{1}}_{0^{+}}x_{m}(s), \ldots ,D^{\mu _{n-2}}_{0^{+}}x_{m}(s) \bigr)\,ds \\ \leq& e \int_{0}^{1}\beta(t)p \biggl(Mt^{\alpha-1}, \frac{(n-2-\mu _{1})M}{n!}t^{n-1-\mu_{1}},\ldots,\frac{(n-2-\mu _{n-2})M}{3!}t^{n-1-\mu _{n-2}} \biggr)\,ds \\ &{}+ eh \biggl(\|x_{m}\|_{2}+\frac{1}{m}, \frac{\| x_{m}\| _{2}}{\Gamma(n-1-\mu_{1})}+\frac{1}{m},\ldots, \frac{\|x_{m}\|_{2}}{\Gamma(n-1-\mu_{n-2})}+\frac{1}{m} \biggr) \int _{0}^{1}\gamma(s)\,ds \\ =& e \biggl(\Upsilon+h \biggl(\|x_{m}\|_{2}+1, \frac{\|x_{m}\| _{2}}{\Gamma (n-1-\mu_{1})}+1,\ldots, \frac{\|x_{m}\|_{2}}{\Gamma(n-1-\mu_{n-2})}+1 \biggr)\|\gamma\| _{1} \biggr), \\ &i= 0,1,2,\ldots,n-3,n-2, \end{aligned}$$
(3.20)
where
$$\Upsilon= \int_{0}^{1}\beta(t)p \biggl(Mt^{\alpha-1}, \frac{(n-2-\mu _{1})M}{n!}t^{n-1-\mu_{1}} ,\ldots,\frac{(n-2-\mu_{n-2})M}{3!}t^{n-1-\mu_{n-2}} \biggr)\,ds. $$
By (H2) and (3.20) we have
$$\begin{aligned}& \|x_{m}\|_{2}\leq e \biggl(\Upsilon+h \biggl( \|x_{m}\| _{2}+1,\frac {\|x_{m}\|_{2}}{\Gamma(n-1-\mu_{1})}+1 ,\ldots, \frac{\|x_{m}\|_{2}}{\Gamma(n-1-\mu_{n-2})}+1 \biggr)\| \gamma\| _{1}\,ds \biggr), \\& \quad m\in \mathbb{N}, \end{aligned}$$
(3.21)
and \(\Upsilon<\infty\).
By (H3), taking \(\lambda_{1}>0\) such that
$$\limsup_{x\rightarrow\infty}\frac{h(x,x,\ldots,x)}{x}= \lambda< \lambda _{1}< \frac{\Gamma(n-1-\mu_{n-2})}{e\|\gamma\|_{1}}, $$
we have that there exists \(A>M+1\) such that, for any \(x>A\),
$$ h(x,x,\ldots,x)< \lambda_{1}x. $$
(3.22)
In order to prove that \(\{x_{m}\}\subset K\) is relatively compact in \(E=C^{(n-2)}[0,1]\), we need to prove that \(\{x_{m}\}\) is bounded in E and \(\{x_{m}\}\) is equicontinuous on \([0,1]\). First, we prove that \(\{ x_{m}\}\) is bounded in E. If \(\{x_{m}\}\) is unbounded, then there exists a subsequence \(\{x_{m_{j}}\}\subset\{x_{m}\}\) such that \(\| x_{m_{j}}\|_{2}\rightarrow+\infty\), and then there exists \(j_{0}\) such that
$$\|x_{m_{j_{0}}}\|_{2}>\max \biggl\{ M+1,(A-1)\Gamma(n-1- \mu_{n-2}), \frac {e\Upsilon+e\lambda_{1}\|\gamma\|_{1}}{1-e\lambda_{1}\|\gamma\| _{1}\Gamma^{-1}(n-1-\mu_{n-2})} \biggr\} , $$
and then \(\frac{\|x_{m_{j_{0}}}\|_{2}}{\Gamma(n-1-\mu_{n-2})}+1>A\), and by (3.21) and (3.22) we get
$$\begin{aligned} \|x_{m_{j_{0}}}\|_{2}&\leq e \biggl(\Upsilon+\frac{\lambda_{1}\| \gamma\| _{1}\|x_{m_{j_{0}}}\|_{2}}{ \Gamma(n-1-\mu_{n-2})}+ \lambda_{1}\|\gamma\|_{1} \biggr) \\ &=e\Upsilon+e\lambda_{1}\|\gamma\|_{1}+\frac{b\lambda_{1}\|\gamma \| _{1}}{\Gamma(n-1-\mu_{n-2})} \|x_{m_{j_{0}}}\|_{2} \\ & < \|x_{m_{j_{0}}}\|_{2}. \end{aligned}$$
This is a contradiction, which means that \(\{x_{m}\}\) is bounded in E. Next, we will prove that \(\{x_{m}^{(n-2)}(t)\}\) is equicontinuous on \([0,1]\). Since \(\{x_{m}\}\) is bounded in E, there exists \(\Lambda >0\) such that \(\|x_{m}\|\leq\Lambda\). Let
$$\begin{aligned}& V_{1}=h \biggl(\Lambda+1,\frac{\Lambda}{\Gamma(n-1-\mu _{1})}+1,\ldots , \frac{\Lambda}{\Gamma(n-1-\mu_{n-2})}+1 \biggr), \\& \phi(t)=\beta(t)p \biggl(Mt^{\alpha-1},\frac{(n-2-\mu _{1})M}{n!}t^{n-1-\mu _{1}}, \ldots, \frac{(n-2-\mu_{n-2})M}{3!}t^{n-1-\mu_{n-2}} \biggr), \\& \quad t\in (0,1]. \end{aligned}$$
(3.23)
Then \(\Upsilon=\int_{0}^{1}\phi(t)\, dt\), and for any \(m\in\mathbb{N}\) and a.e. \(t\in[0,1]\),
$$ f_{m} \bigl(t,x_{m}(t),D^{\mu_{1}}_{0^{+}}x_{m}(t),D^{\mu _{2}}_{0^{+}}x_{m}(t), \ldots,D^{\mu_{n-2}}_{0^{+}}x_{m}(t) \bigr)\leq \phi(t)+V_{1}\gamma (t). $$
(3.24)
Assume that \(0\leq t_{1}< t_{2}\leq1\). Then by (3.24), for any \(m\in \mathbb{N}\), we have
$$\begin{aligned}& \bigl\vert x_{m}^{(n-2)}(t_{2})-x_{m}^{(n-2)}(t_{1}) \bigr\vert \\& \quad = \biggl\vert \int_{0}^{1} \biggl(\frac{\partial^{n-2}}{\partial t^{n-2}}G(t_{2},s)- \frac{\partial^{n-2}}{\partial t^{n-2}}G(t_{1},s) \biggr) \\& \qquad {}\times f_{m} \bigl(s,x_{m}(s),D^{\mu _{1}}_{0^{+}}x_{m}(s),D^{\mu_{2}}_{0^{+}}x_{m}(t), \ldots,D^{\mu_{n-2}}_{0^{+}}x_{m}(t) \bigr)\,ds \biggr\vert \\& \quad \leq e \bigl(t_{2}^{\alpha-n+1}-t_{1}^{\alpha-n+1} \bigr) \int _{0}^{1} \bigl(\phi(s)+V_{1}\gamma(s) \bigr)\,ds \\& \qquad {}+e \biggl\vert \int _{0}^{t_{2}}(t_{2}-s)^{\alpha-n+1} \bigl(\phi(s)+V_{1}\gamma (s) \bigr)\,ds- \int_{0}^{t_{1}}(t_{1}-s)^{\alpha-n+1} \bigl(\phi (s)+V_{1}\gamma(s) \bigr)\,ds \biggr\vert \\& \quad \leq e \bigl(\Upsilon+V_{1}\Vert \gamma \Vert _{1} \bigr) \bigl(t_{2}^{\alpha -n+1}-t_{1}^{\alpha-n+1} \bigr)+e \biggl[ \int_{t_{1}}^{t_{2}}(t_{2}-s)^{\alpha-n+1} \bigl(\Upsilon +V_{1}\Vert \gamma \Vert _{1} \bigr)\,ds \\& \qquad {}+ \int_{0}^{t_{1}} \bigl((t_{2}-s)^{\alpha -n+1}-(t_{1}-s)^{\alpha-n+1} \bigr) \bigl(\Upsilon+V_{1}\Vert \gamma \Vert _{1} \bigr) \,ds \biggr] \\& \quad \leq e \bigl(\Upsilon+V_{1}\Vert \gamma \Vert _{1} \bigr) \bigl(t_{2}^{\alpha -n+1}-t_{1}^{\alpha-n+1} \bigr)+e \biggl[(t_{2}-t_{1})^{\alpha-n+1} \bigl( \Upsilon+V_{1}\Vert \gamma \Vert _{1} \bigr) \\& \qquad {}+ \int_{0}^{t_{1}} \bigl((t_{2}-s)^{\alpha-n+1}-(t_{1}-s)^{\alpha -n+1} \bigr) \bigl(\Upsilon+V_{1}\Vert \gamma \Vert _{1} \bigr) \,ds \biggr]. \end{aligned}$$
Hence, we can prove that \(\{x_{m}^{(n-2)}(t)|m=1,2,\ldots \}\) is equicontinuous on \([0,1]\). □