In this section, we always assume that (D_{1})(D_{3}) are satisfied and \(\Gamma_{*}=0\).
Theorem 3.1
Given a vector
\(v\in {\mathbb {R}}_{+}^{N}\), suppose that there exists a constant
\(r>0\)
such that
 (H_{1}):

there exists a continuous nonnegative function
\(\phi_{r+\Lambda^{*}}(t)\)
on
\([0,1]\)
such that
$$\bigl\langle v,f(t,u) \bigr\rangle \geq\phi_{r+\Lambda^{*}}(t) $$
for all
\(t\in(0,1)\)
and
\(u\in {\mathbb {R}}_{+}^{N}\)
with
\(0<u_{v}\leq r+\Lambda^{*}\);
 (H_{2}):

there exist two continuous nonnegative functions
\(g(\cdot)\)
and
\(h(\cdot)\)
on
\((0,\infty)\)
such that
$$0 \leq \bigl\langle v,f(t,u) \bigr\rangle \leq g\bigl(u_{v}\bigr)+h\bigl(u_{v}\bigr) $$
for all
\(t\in(0,1)\)
and
\(u\in {\mathbb {R}}_{+}^{N}\)
with
\(0<u_{v}\leq r+\Lambda^{*}\), where
\(g(\cdot)>0\)
is nonincreasing and
\(h(\cdot)/g(\cdot)\)
is nondecreasing;
 (H_{3}):

the following inequality is satisfied:
$$\biggl\{ 1+\frac{h(r+\Lambda^{*})}{g(r+\Lambda^{*})} \biggr\} b< \int_{0}^{r}\frac{1}{g(x)}\,\mathrm{d}x, $$
where
$$b=\max \biggl\{ 2 \int_{0}^{1/2}t(1t)q(t)\,\mathrm{d}t, 2 \int^{1}_{1/2}t(1t)q(t)\,\mathrm{d}t \biggr\} . $$
Then (1.1) has at least one nontrivial solution
u
with
\(0<u\gamma_{v}<r\).
Proof
First, we show that the system
$$ \left \{ \textstyle\begin{array}{l}\ddot{u}+q(t)f(t,u(t)+\gamma(t))=0, \quad0< t< 1,\\ u(0)=0,\qquad u(1)=0, \end{array}\displaystyle \right . $$
(3.1)
has a nontrivial solution u satisfying \(u(t)+\gamma(t)_{v}>0\) for \(t\in(0,1)\) and \(0<u_{v}<r\). If this is true, by calculating we get
$$\ddot{u}+\ddot{\gamma}+q(t)f \bigl(t,u(t)+\gamma(t) \bigr)+e(t)=0, $$
that is, \(y(t)=u(t)+\gamma(t)\) is a nontrivial solution of (1.1) with \(0<y\gamma_{v}<r\).
Since (H_{3}) holds, we can choose a positive constant ϵ with \(\epsilon< r\) such that
$$ \biggl\{ 1+\frac{h(r+\Lambda^{*})}{g(r+\Lambda^{*})} \biggr\} b< \int_{\epsilon}^{r}\frac{1}{g(x)}\,\mathrm{d}x. $$
(3.2)
Choose a positive integer \(n_{0}\in\{1,2,\ldots\}\) such that \(\frac{1}{n_{0}}<\frac{\epsilon}{2}\). Next, we set \(N_{0}= \{n_{0},n_{0}+1,\ldots\}\) and fix \(n\in N_{0}\). To this end, we consider the family of systems
$$ \left \{ \textstyle\begin{array}{l}\ddot{u}+\lambda q(t)f^{n}(t,u(t)+\gamma(t))=0, \quad0< t< 1,\\ u(0)=\mathbf{\frac{1}{n}},\qquad u(1)=\mathbf{\frac{1}{n}}, \end{array}\displaystyle \right . $$
(3.3)
where \(\lambda\in[0,1]\), \(\mathbf{\frac{1}{n}}=(\frac {1}{n},\ldots,\frac{1}{n}) \in {\mathbb {R}}_{+}^{N}\), and
$$f^{n}(t,u)=\left \{ \textstyle\begin{array}{l@{\quad}l}f(t,u) & \mbox{if } u_{v}\geq\frac{1}{n},\\ f(t,u_{1},\ldots,u_{i1},\frac{1}{n},u_{i+1},\ldots,u_{N}) &\mbox{if } u_{v}< \frac{1}{n}. \end{array}\displaystyle \right . $$
It is immediate that a nontrivial solution of (3.3) is exactly a fixed point of the operator equation
$$\begin{aligned} u=\lambda T u+(1\lambda)p, \end{aligned}$$
(3.4)
where \(p=\mathbf{\frac{1}{n}}\), and T stands for the operator
$$(T u) (t)= \int_{0}^{1}G(t,s)q(s)f^{n} \bigl(s,u(s)+ \gamma(s) \bigr)\,\mathrm{d}s+p. $$
Next, we show that any fixed point u of (3.4) for all \(\lambda \in [0,1]\) must satisfy
$$ u_{v}\ne r. $$
(3.5)
Assume on the contrary that there exists \(\lambda\in[0,1]\) such that u, a fixed point of (3.4), satisfies \(u_{v}= r\). We conclude from (3.3) that, for all \(t\in[0,1]\),
$$\bigl\langle v,\ddot{u}(t) \bigr\rangle = \bigl\langle v,\lambda q(t)f^{n} \bigl(t,u(t)+\gamma(t) \bigr) \bigr\rangle \leq0. $$
Then we have \(\langle v,u(t) \rangle\geq\frac{1}{n}\) for \(0\leq t\leq 1\). Furthermore, from Lemma 2.1 we have
$$ \bigl\langle v,u(t) \bigr\rangle \geq t(1t)u_{v}, \quad0\leq t \leq1. $$
It is obvious that there exists \(t_{n}\in(0,1)\) such that \(\langle v,\dot{u}(t) \rangle\geq0\) on \((0,t_{n}), \langle v,\dot{u}(t) \rangle\leq0\) on \((t_{n},1)\), and \(\langle v,u(t_{n}) \rangle=u_{v}=r\). Hence, for all \(z\in (0,1)\), we have
$$ \begin{aligned}[b]\bigl\langle v,\ddot{u}(z) \bigr\rangle & = \bigl\langle v,\lambda q(z)f^{n}\bigl( z,u(z)+\gamma(z)\bigr) \bigr\rangle \\ & = \lambda q(z)\bigl\langle v,f\bigl(z,u(z)+\gamma(z)\bigr)\bigr\rangle \\ & \le q(z)\bigl\langle v,f\bigl(z,u(z)+\gamma(z)\bigr) \bigr\rangle \\ & \le q(z)g\bigl(\bigu(z)+\gamma(z)\big_{v}\bigr) \biggl\{ 1+ \frac {h(u(z)+\gamma(z)_{v})}{g(u(z)+\gamma(z)_{v})} \biggr\} .\end{aligned} $$
(3.6)
Since \(\Gamma_{*}=0\), we have
$$\bigl\langle v,u(t) \bigr\rangle \leq \bigl\langle v,u(t)+\gamma(t) \bigr\rangle \leq\bigu(t)+\gamma(t)\big_{v} \leq\bigu(t)\big_{v}+\big \gamma(t)\big_{v}\leq r+\Lambda^{*}. $$
Calculating the integral for (3.6) from t (\(t\leq t_{n}\)) to \(t_{n}\), we have
$$ \begin{aligned} \bigl\langle v,\dot{u}(t) \bigr\rangle &\leq g \bigl(\bigu(z)+\gamma(z)\big_{v} \bigr) \biggl\{ 1+\frac {h (u(z)+ \gamma(z)_{v} )}{g (u(z)+\gamma(z)_{v} )} \biggr\} \int_{t}^{t_{n}}q(z)\,\mathrm{d}z \\ &\leq g \bigl( \bigl\langle v,u(t) \bigr\rangle \bigr) \biggl\{ 1+ \frac {h (r+ \Lambda^{*} )}{g (r+\Lambda^{*} )} \biggr\} \int_{t}^{t_{n}}q(z)\,\mathrm{d}z. \end{aligned} $$
Thus, for \(t\leq t_{n}\), we have
$$ \frac {\langle v,\dot{u}(t) \rangle}{g(\langle v,u(t) \rangle)}\leq \biggl\{ 1+\frac {h(r+\Lambda^{*})}{g(r+\Lambda^{*})} \biggr\} \int_{t}^{t_{n}}q(z)\,\mathrm{d}z. $$
(3.7)
Integrating (3.7) from 0 to \(t_{n}\), we have
$$ \int_{\frac{1}{n}}^{r}\frac {\mathrm{d}x}{g(x)}\leq \biggl\{ 1+ \frac {h (r+ \Lambda^{*} )}{g (r+\Lambda^{*} )} \biggr\} \int_{0}^{t_{n}} t q(t)\,\mathrm{d}t. $$
Accordingly,
$$ \int_{\epsilon}^{r}\frac {\mathrm{d}x}{g(x)}\leq \biggl\{ 1+ \frac {h(r+\Lambda^{*})}{g(r+\Lambda^{*})} \biggr\} \frac{1}{1t_{n}} \int _{0}^{t_{n}}t(1t)q(t)\,\mathrm{d}t. $$
(3.8)
Applying this calculation method again and integrating (3.6) from \(t_{n}\) to t (\(t\geq t_{n}\)) and then from \(t_{n}\) to 1, we get
$$ \int_{\epsilon}^{r}\frac {\mathrm{d}x}{g(x)}\leq \biggl\{ 1+ \frac {h(r+\Lambda^{*})}{g(r+\Lambda^{*})} \biggr\} \frac{1}{t_{n}} \int ^{1}_{t_{n}}t(1t)q(t)\,\mathrm{d}t. $$
(3.9)
According to (3.8) and (3.9), we have
$$ \int_{\epsilon}^{r}\frac {\mathrm{d}x}{g(x)}\leq b \biggl\{ 1+ \frac {h (r+\Lambda^{*} )}{g (r+\Lambda^{*} )} \biggr\} , $$
which is a contradiction to (3.2), and so the assertion is proved.
Under the assertion above, using Lemma 2.2, we get that
$$\begin{aligned} u=T u \end{aligned}$$
(3.10)
has a fixed point denoted by \(u_{n}\). In other words, the system
$$ \left \{ \textstyle\begin{array}{l}\ddot{u}+q(t)f^{n}(t,u(t)+\gamma(t))=0, \quad0< t< 1,\\ u(0)=\mathbf{\frac{1}{n}},\qquad u(1)=\mathbf{\frac{1}{n}}, \end{array}\displaystyle \right . $$
(3.11)
has a solution \(u_{n}\) satisfying \(u_{n}_{v}< r\). For all \(t\in[0,1]\), since \(\langle v,u_{n}(t) \rangle\geq \frac{1}{n}>0\), \(u_{n}\) is certainly a nontrivial solution of (3.11).
Next, we claim that \(\langle v,u_{n}(t)+\gamma(t) \rangle\) has a uniform positive lower bound. To get the claim above, we need to prove that there exists a constant \(\delta>0\), independent of \(n\in N_{0}\), such that, for any \(t\in[0,1]\),
$$ \bigl\langle v,u_{n}(t)+\gamma(t) \bigr\rangle \geq\delta t(1t). $$
Since \(\Gamma_{*}=0\), we only need to show that
$$ \bigl\langle v,u_{n}(t) \bigr\rangle \geq\delta t(1t) $$
(3.12)
for all \(n\in N_{0}\) and \(t\in[0,1]\). Since (H_{1}) holds, there exists a continuous nonnegative function \(\phi_{r+\Lambda^{*}}\) such that
$$\bigl\langle v,f(t,u) \bigr\rangle \geq\phi_{r+\Lambda^{*}}(t) $$
for all \(t\in(0,1)\) and u with \(0<u_{v}\leq r+\Lambda^{*}\). Let \(u^{r+\Lambda^{*}}\) be the unique solution of the problem
$$\left \{ \textstyle\begin{array}{l}\ddot{u}+q(t)\Phi(t)=0,\quad0< t< 1,\\ u(0)=0,\qquad u(1)=0, \end{array}\displaystyle \right . $$
with \(\Phi=(\phi_{r+\Lambda^{*}},\ldots,\phi_{r+\Lambda^{*}})^{T}\). Then we have
$$\bigl\langle v,u^{r+\Lambda^{*}}(t) \bigr\rangle = \int_{0}^{1}G(t,s)q(s) \bigl\langle v, \phi_{r+\Lambda^{*}}(s) \bigr\rangle \,\mathrm{d}s. $$
Moreover, for \(t\in[0,1]\),
$$ \begin{aligned} \bigl\langle v,\dot{u}^{r+\Lambda^{*}}(t) \bigr\rangle ={}& \int_{t}^{1}(1s)q(s) \bigl\langle v, \phi_{r+\Lambda^{*}}(s) \bigr\rangle \,\mathrm{d}s \\ & \int_{0}^{t} sq(s) \bigl\langle v, \phi_{r+\Lambda^{*}}(s) \bigr\rangle \,\mathrm{d}s \end{aligned} $$
and
$$\bigl\langle v,u^{r+\Lambda^{*}}(0) \bigr\rangle =0,\qquad \bigl\langle v,u^{r+\Lambda ^{*}}(1) \bigr\rangle =0. $$
Assume that there exists a constant \(k_{0}=\langle v,\dot{u}^{r+\Lambda ^{*}}(0) \rangle=\int_{0}^{1}(1s)q(s)\langle v,\phi_{r+\Lambda^{*}}(s) \rangle \,\mathrm{d}s\), and if not, then \(\langle v,\dot{u}^{r+\Lambda^{*}}(0) \rangle=\infty \). Regardless of the two cases above, there exists a positive constant \(\delta_{1}\) independent of n such that \(\langle v,\dot{u}^{r+\Lambda^{*}}(0) \rangle\geq2\delta_{1}\). Hence, there exists a positive constant \(\epsilon_{1}\) such that \(\langle v,u^{r+\Lambda^{*}}(t) \rangle\geq\delta_{1}t(1t)\) for all \(t\in [0,\epsilon_{1}]\). Analogously, there exists a positive constant \(\delta_{2}\), independent of n, and \(\epsilon_{2}>0\) such that \(\langle v,u^{r+\Lambda^{*}}(t) \rangle\geq\delta_{2}t(1t)\) for all \(t\in [1\epsilon_{2},1]\).
Besides, for \(t\in[\epsilon_{1},1\epsilon_{2}]\), it is easily seen that
$$\frac{\langle v,u^{r+\Lambda^{*}}(t) \rangle}{t(1t)} \mbox{ is continuous}. $$
Then there exists a positive constant \(\delta_{3}\), independent of n, such that
$$\bigl\langle v,u^{r+\Lambda^{*}}(t) \bigr\rangle \geq\delta_{3}t(1t). $$
So, if we choose a positive constant \(\delta=\min\{\delta_{1},\ldots ,\delta_{N}\}\), then (3.12) is true.
To pass from the solution \(u_{n}\) of (3.11) to that of (3.1), it is necessary to prove that
$$ \{u_{n}\}_{n\in N_{0}}\mbox{ is bounded and equicontinuous on }[0,1]. $$
(3.13)
Recalling the argument to establish (3.7) and applying it again with u replaced by \(u_{n}\), we obtain the inequalities
$$ \frac {\langle v,\dot{u}_{n}(t) \rangle}{g(\langle v,u_{n} (t) \rangle )}\leq \biggl\{ 1+\frac {h(r+\Lambda^{*})}{g(r+\Lambda^{*})} \biggr\} \int^{t_{n}}_{t} q(z)\,\mathrm{d}z $$
(3.14)
and
$$ \frac { \langle v,\dot{u}_{n}(t) \rangle}{g ( \langle v,u_{n} (t) \rangle )}\geq \biggl\{ 1+\frac {h (r+\Lambda^{*} )}{g (r+\Lambda^{*} )} \biggr\} \int_{t_{n}}^{t}q(z)\,\mathrm{d}z. $$
Accordingly,
$$ \frac {\langle v,\dot{u}_{n}(t) \rangle}{g(u_{n} (t))}\leq \biggl\{ 1+\frac {h(r+\Lambda^{*})}{g(r+\Lambda^{*})} \biggr\} \bigg \int^{t_{n}}_{t} q(z)\,\mathrm{d}z\bigg. $$
(3.15)
Under this claim, we have to show that there exist two constants a, b satisfying \(0< a< b<1\) such that
$$ a< \inf\{t_{n},n\in N_{0}\}\leq\sup \{t_{n},n\in N_{0}\}< b. $$
(3.16)
Hence, we just need to prove the following two inequalities: \(\inf\{t_{n},n\in N_{0}\}>0\) and \(\sup\{t_{n},n\in N_{0}\}<1\). First, assume that the inequality \(\inf\{t_{n},n\in N_{0}\}>0\) is incorrect. Let A be a subsequence of \(N_{0}\) with \(t_{n}\rightarrow0\) as \(n\rightarrow \infty \) in A. Integrating (3.14) from 0 to \(t_{n}\), we have
$$\int_{0}^{\langle v,u_{n}(t_{n}) \rangle}\frac {\mathrm{d}x}{g(x)}\leq \biggl\{ 1+ \frac {h (r+\Lambda^{*} )}{g (r+\Lambda^{*} )} \biggr\} \int_{0} ^{t_{n}}tq(t)\,\mathrm{d}t+ \int_{0}^{\frac{1}{n}}\frac {\mathrm{d}x}{g(x)} $$
for \(n\in A\). Since \(t_{n}\rightarrow0\) as \(n\rightarrow \infty \) in A, from this inequality we get that \(u_{n}(t_{n})\rightarrow0\) as \(n\rightarrow \infty \) in A. Furthermore, \(\langle v,\dot{u}(t_{n}) \rangle =0\), and \(u_{n}\) has a local maximum at \(t_{n}\). Then we obtain that \(u_{n}\rightarrow0\) in \({\mathbb {C}}[0,1]\) as \(n\rightarrow \infty \) in A, which contradicts our claim. So, \(\inf\{t_{n},n\in N_{0}\}>0\). Analogously, we can also prove that \(\sup \{t_{n},n\in N_{0}\}<1\).
According to (3.15) and (3.16), we obtain that
$$ \frac { \langle v,\dot{u}_{n}(t) \rangle}{g ( \langle v,u_{n} (t) \rangle )}\leq \biggl\{ 1+\frac {h (r+\Lambda^{*} )}{g (r+\Lambda^{*} )} \biggr\} V(t), $$
where
$$V(t)= \int_{\min\{t, a\}}^{\max\{t,b\}}q(z)\,\mathrm{d}z. $$
Obviously, \(V\in L^{1}[0,1]\). Let us define \(I:[0,\infty)\rightarrow[0,\infty)\) by
$$I(z)= \int_{0}^{z}\frac{1}{g(x)}\,\mathrm{d}x. $$
Note that \(g(x)>0\) is nonincreasing on \((0,\infty)\). Then the map \(I: [0,\infty)\rightarrow[0,\infty)\) is increasing, and \(I(\infty)=\infty\). Analogously, for any \(D>0\), the map I is continuous. Furthermore, we have
$$\begin{aligned} \bigI\bigl(u_{n}(t)\bigr)I\bigl(u_{n}(s)\bigr)\big& = \bigg \int_{s}^{t}\frac {\langle v,\dot{u}_{n}(z) \rangle }{g(\langle v,u_{n}(z) \rangle)} \,\mathrm{d}z\bigg \\ & \le \biggl\{ 1+\frac {h(r+\Lambda^{*})}{g(r+\Lambda^{*})} \biggr\} \bigg \int_{s}^{t}V(z)\,\mathrm{d}z\bigg,\end{aligned} $$
which implies that
$$\bigl\{ I(u_{n}) \bigr\} _{n\in N_{0}}\mbox{ is bounded and equicontinuous on }[0,1]. $$
Due to the uniform continuity of the inverse map \(I^{1}\) on \([0,I(r+\Lambda^{*})]\) and the equality
$$\bigu_{n}(t)u_{n}(s)\big=\bigI^{1} \bigl(I \bigl(u_{n}(t) \bigr) \bigr)I^{1} \bigl(I \bigl(u_{n}(s) \bigr) \bigr)\big, $$
we have that (3.13) is certainly true.
Now the ArzelàAscoli theorem guarantees that \(\{u_{n} \}_{n\in N_{0}}\) has a subsequence that converges uniformly on \([0,1]\) to a function \(u\in {\mathbb {C}}[0,1]\). It is easy to verify that
$$\ddot{u}(t)+q(t)f \bigl(t,u(t)+\gamma(t) \bigr)=0. $$
Moreover, we have \(\langle v,u(0) \rangle=\langle v,u(1) \rangle=0\), \(0<u_{v}\leq r\), and \(\langle v,u(t) \rangle\geq\delta t(1t)\) for all \(0\leq t\leq1\). Then u is a nontrivial solution of (3.1) satisfying \(0<u_{v}<r\). □
Theorem 3.2
Suppose that (H_{2})(H_{3}) hold. Assume further that
 (H_{4}):

there exist two continuous nonnegative functions
\(g_{1}(\cdot)\), \(h_{1}(\cdot )\)
on
\((0,\infty)\)
such that
$$\bigl\langle v,f(t,u) \bigr\rangle \geq g_{1}\bigl(u_{v}\bigr)+h_{1}\bigl(u_{v}\bigr) $$
for all
\(t\in(0,1)\)
and
\(u\in {\mathbb {R}}_{+}^{N}\), where
\(g_{1}(\cdot)>0\)
is nonincreasing, and
\(h_{1}(\cdot)/g_{1}(\cdot)\)
is nondecreasing;
 (H_{5}):

there exists a positive constant
\(R>r\)
such that
$$ \frac{R}{g_{1}(R+\Lambda^{*})(1+\frac{h_{1}(\sigma R)}{g_{1}(\sigma R)})} \leq \int_{a}^{1a} G(\xi,s)q(s)\,\mathrm{d}s, $$
where
\(a\in(0,\frac{1}{2})\)
is fixed, \(\sigma=a(1a)\), and
\(0\leq\xi\leq1\)
is such that
$$ \int_{a}^{1a}G(\xi,s)q(s)\,\mathrm{d}s=\sup _{0\leq t\leq1} \int_{a}^{1a}G(t,s)q(s)\,\mathrm{d}s. $$
Then (1.1) has a nontrivial solution
u
with
\(r<u\gamma _{v}\leq R\).
Proof
First, we return to the beginning of the proof of Theorem 3.1. Similarly, we only need to prove that (3.1) has a nontrivial solution u, which satisfies \(r<u_{v}\leq R\) and \(\langle v,u(t)+\gamma(t) \rangle>0\) for all \(t\in(0,1)\).
Since (H_{3}) holds, we can choose a positive constant ϵ with \(\epsilon< r\) such that inequality (3.2) holds. Obviously, there exists a positive integer \(n_{1}\in\{1,2,\ldots\}\) such that
$$\frac{1}{n_{1}}< \min \biggl\{ \frac{\epsilon}{2},\sigma R \biggr\} . $$
Let \(N_{1}= \{n_{1},n_{1}+1,\ldots\}\). Fix \(n\in N_{1}\). Let us reconsider system (3.11) and define the set
$$ K= \bigl\{ u \in X: \bigl\langle v,u(t) \bigr\rangle \geq t(1t)u_{v} \mbox{ for } t\in[0,1] \bigr\} . $$
We can easily see that K is a cone in X. Set
$$\Omega^{1}=\bigl\{ u\in X: u_{v}< r\bigr\} , \qquad \Omega^{2}=\bigl\{ u\in X: u_{v}< R\bigr\} . $$
Define the operator \(S: \overline{\Omega}_{K}^{2} \setminus \Omega_{K}^{1}\rightarrow K\) as
$$ (Su) (t)= \int_{0}^{1}G(t,s)q(s)f^{n} \bigl(s,u(s)+ \gamma(s) \bigr)\,\mathrm{d}s+p. $$
A standard argument shows that the operator \(S:\overline{\Omega}_{K}^{2} \setminus\Omega_{K}^{1}\rightarrow X\) is continuous and completely continuous. It is easily seen that the operator \(S:\overline{\Omega}_{K}^{2} \setminus \Omega_{K}^{1}\rightarrow K\) is well defined by Lemma 2.1. To get the desired result, we need to make the following two assertions:

(i)
\(u\neq\lambda Su\) for \(\lambda\in[0,1]\) and \(u \in\partial_{K}\Omega^{1}\), and

(ii)
there exists a vector \(w\in K\setminus\{0\}\) such that \(u \neq Su+\lambda w\) for all \(\lambda> 0\) and all \(u \in\partial_{K}\Omega^{2}\).
We start with (i). Assume that there exiss \(\lambda\in[0,1]\) and \(u\in \partial_{K}\Omega^{1}\) such that \(u=\lambda Su\). Suppose that \(\lambda \neq0\). Now \(u=\lambda Su\) can lead to a contradiction following the same ideas in proving (3.5), and so (i) holds. We omit the details.
Next, we consider assertion (ii). Let \(w(t)=(1,1,\ldots,1)^{\mathrm{T}}\). Then \(w\in K\setminus\{0\}\). Let us prove that \(u \neq Su+\lambda w\) for all \(u \in \partial_{K}\Omega^{2}\) and \(\lambda> 0\). If not, there would exist \(u\in \partial_{K}\Omega^{2}\) and \(\lambda>0\) such that \(u=S u+\lambda w\). Now since \(u\in\partial_{K}\Omega^{2}\), we have that \(u_{v}=R\). It is obvious that \(\langle v,u(t)\rangle\) is concave on \([0,1]\). By Lemma 2.1, for all \(t\in[0,1]\), we have
$$\bigl\langle v,u(t) \bigr\rangle \geq t(1t)R. $$
We suppose that there exists \(t\in[a,1a]\) such that
$$\sigma R=a(1a)R\leq \bigl\langle v,u(t) \bigr\rangle \leq R. $$
Hence, for \(t\in[a,1a]\), we have
$$\sigma R\leq \bigl\langle v,u(t)+\gamma(t) \bigr\rangle \leq R+\Lambda^{*}. $$
Therefore, for \(t\in[a,1a]\), we obtain \(f^{n}(u(s)+\gamma(s))=f(u(s)+\gamma(s))\). Consequently, from (H_{4}) we have
$$ \begin{aligned}[b] R&\geq\bigl\langle v,u(\xi) \bigr\rangle = \bigl\langle v,(Su) (\xi) \bigr\rangle +\langle v,\lambda w \rangle \\ & = \int_{0}^{1}G(\xi,s)q(s)\bigl\langle v,f^{n}\bigl(s,u(s)+\gamma(s)\bigr) \bigr\rangle \,\mathrm{d}s+\langle v,p \rangle+\langle v,\lambda w \rangle \\ & \ge \int_{0}^{1}G(\xi,s)q(s)\bigl\langle v,f \bigl(s,u(s)+\gamma(s)\bigr) \bigr\rangle \,\mathrm{d}s \\ & \ge \int_{0}^{1}G(\xi,s)q(s)\bigl[g_{1} \bigl(\bigu(s)+\gamma (s)\big_{v}\bigr)+h_{1}\bigl(\bigu(s)+ \gamma(s)\big_{v}\bigr)\bigr]\,\mathrm{d}s \\ & \ge \int_{0}^{1}G(\xi,s)q(s)g_{1}\bigl(\bigu(s)+ \gamma(s)\big_{v}\bigr) \biggl\{ 1+\frac {h_{1}(u(s) +\gamma(s)_{v})}{ g_{1}(u(s)+\gamma(s)_{v})} \biggr\} \,\mathrm{d}s \\ & \ge g_{1}\bigl(R+\Lambda^{*}\bigr) \biggl\{ 1+\frac{h_{1}(\sigma R)}{ g_{1}(\sigma R)} \biggr\} \int_{a}^{1a}G(\xi,s)q(s)\,\mathrm{d}s,\end{aligned} $$
which is a contradiction to (H_{5}). So assertion (ii) is proved.
Now it follows from Lemma 2.3 that S has at least one fixed point \(u_{n}\in\overline{\Omega}^{2}_{K}\setminus\Omega^{1}_{K}\) with \(r\lequ_{n}_{v}\leq R\). By assertion (i) we can further get that \(u_{n}_{v}>r\). Therefore, system (3.11) has a solution \(u_{n}\) with \(\langle v,u_{n}(t) \rangle\geq\frac{1}{n}\) for all \(t\in[0,1]\), which implies that system (3.1) has a nontrivial solution \(u_{n}\) with
$$ \bigl\langle v,u_{n}(t) \bigr\rangle \geq\frac{1}{n},\quad 0\leq t\leq 1, \qquad r< u_{n}_{v}\leq R, $$
and
$$ \bigl\langle v,u_{n}(t) \bigr\rangle \geq t(1t)r,\quad 0\leq t \leq1. $$
Now, using a similar argument as in the proof of Theorem 3.1, we can show that
$$\{u_{n}\}_{n\in N_{0}}\mbox{ is bounded and equicontinuous on }[0,1], $$
and the ArzelàAscoli theorem guarantees that \(\{u_{n} \} _{n\in N_{0}}\) has a subsequence that converges uniformly on \([0,1]\) to a function \(u\in {\mathbb {C}}[0,1]\), which is a nontrivial solution of
$$\ddot{u}(t)+q(t)f \bigl(t,u(t)+\gamma(t) \bigr)=0 $$
and satisfies \(r<u_{v}\leq R\). □
The following multiplicity result is a direct consequence of Theorems 3.1 and 3.2.
Theorem 3.3
Assume that (H_{2})(H_{5}) are satisfied. Then (1.1) has at least two nontrivial solutions
u, ũ
with
\(\langle v,u(t) \rangle>0\), \(\langle v,\tilde{u}(t) \rangle>0\)
for
\(t\in(0,1)\)
and
\(u\gamma_{v}< r<\tilde{u}\gamma_{v}\leq R\).
Corollary 3.4
Suppose
\(\alpha>0\), \(\beta\geq0\), \(\Gamma _{*}=0\), and
\(e_{1},e_{2}\in {\mathbb {C}}([0,1],{\mathbb {R}})\).

(i)
For each
\(\mu>0\), system (1.2) has at least one nontrivial solution if
\(\beta<1\).

(ii)
For each
\(0< \mu< \mu_{1}\), system (1.2) has at least one nontrivial solution if
\(\beta\ge1\), where
\(\mu_{1}\)
is a positive constant.

(iii)
For each
\(0< \mu< \mu_{1}\), system (1.2) has at least two nontrivial solutions if
\(\beta>1\).
Proof
We will apply Theorem 3.3. Let \(v=(1,1)^{\mathrm{T}}\). Let
$$g(y)=2^{1+\frac{\alpha}{2}} y^{\alpha},\qquad h(y)=2\mu y^{\beta},\qquad g_{1} (y)=2y^{\alpha}, \qquad h_{1} (y)=2^{1\frac{\beta}{2}} \mu y^{\beta}. $$
Then it is easily seen that (H_{2}) and (H_{4}) are satisfied by using the inequalities
$$\frac{(u+w)^{2}}{2}\leq u^{2}+w^{2}\leq\bigl(u+w\bigr)^{2}\quad \mbox{for all } u, w\in {\mathbb {R}}. $$
Note that
$$b=\max \biggl\{ 2 \int_{0}^{1/2}t(1t)\,\mathrm{d}t, 2 \int^{1}_{1/2}t(1t)\,\mathrm{d}t \biggr\} = \frac{1}{6}. $$
Now condition (H_{3}) holds if there exists a positive constant r such that
$$\mu< \frac{3r^{\alpha+1}(\alpha+1)\cdot2^{\frac{\alpha }{2}}}{(\alpha+1) (r+\Lambda^{*})^{\alpha+\beta}}, $$
which can be deduced to
$$0< \mu< \mu_{1}:=\sup_{r>0}\frac{3r^{\alpha+1}(\alpha+1)\cdot 2^{\frac {\alpha}{2}}}{(\alpha+1) (r+\Lambda^{*})^{\alpha+\beta}}. $$
Notice that \(\mu_{1}=\infty\) since \(\beta<1\) and \(\mu_{1} < \infty\) since \(\beta\ge1\). We have (i) and (ii). The other existence condition (H_{5}) becomes
$$ \mu\geq\frac{R(R+\Lambda^{*})^{\alpha}2L}{L\cdot 2^{1\frac {\beta}{2}} (\sigma R)^{\alpha+\beta}}, $$
(3.17)
where
$$L=\max_{0\leq t\leq1} \int_{\frac{1}{5}}^{\frac{4}{5}}G(t,s)\,\mathrm{d}s. $$
Since \(\beta>1\), we obtain that the righthand side of (3.17) tends to zero as \(R\to+\infty\). Therefore, for any \(0< \mu< \mu_{1}\), we can find R large enough such that inequality (3.17) is satisfied. Therefore, system (1.2) has another nontrivial solution. □
Similarly, we can prove the following result for system (1.3).
Corollary 3.5
Suppose that
\(\alpha>0\), \(\beta>1\), \(\Gamma _{*}=0\), and
\(e_{1},e_{2}\in {\mathbb {C}}([0,1],{\mathbb {R}})\). Then there exists a positive constant
\(\mu_{2}\)
such that system (1.3) has at least two nontrivial solutions for each
\(0< \mu< \mu_{2}\).