We make the following assumptions.

(H1) *f* : ℝ → ℝ is continuous and *sf*(*s*) > 0 for *s* ≠ 0.

(H2) {f}_{0}={\text{lim}}_{\left|s\right|\to 0}\frac{f\left(s\right)}{s}\in \left(0,\infty \right),\phantom{\rule{1em}{0ex}}{f}_{\infty}={\text{lim}}_{\left|s\right|\to +\infty}\frac{f\left(s\right)}{s}\in \left(0,\infty \right).

**Theorem 4.1**. *Suppose that (H1) and (H2) hold Assume that*

r\in \left(\frac{{\lambda}_{m,+}}{{f}_{\infty}},\frac{{\lambda}_{m,+}}{{f}_{0}}\right)\cup \left(\frac{{\lambda}_{m,-}}{{f}_{0}},\frac{{\lambda}_{m,-}}{{f}_{\infty}}\right),

(4.1)

*or*

r\in \left(\frac{{\lambda}_{m,+}}{{f}_{0}},\frac{{\lambda}_{m,+}}{{f}_{\infty}}\right)\cup \left(\frac{{\lambda}_{m,-}}{{f}_{\infty}},\frac{{\lambda}_{m,-}}{{f}_{0}}\right).

(4.2)

*Then* (1.1) *and* (1.2) *has two solutions u*^{+} *and u*^{-} *such that u*^{+} *is positive on* T *and u*^{-} *is negative on* T.

Obviously, we can get the following lemma with ease.

**Lemma 4.1**. *Suppose that u* ∈ *X and* u\not\equiv 0 *on* T *satisfies* (1.1) *(or* (1.3)*) and there exists* {t}_{0}\in T *such that u*(*t*_{0}) = 0, *then u*(*t*_{0} - 1)*u*(*t*_{0} + 1) < 0.

**Proof of Theorem 4.1**. First, we deal with the case *r* > 0.

Let *ζ, ξ* ∈ *C*(ℝ, ℝ) such that

f\left(u\right)={f}_{0}u+\zeta \left(u\right),\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}f\left(u\right)={f}_{\infty}u+\xi \left(u\right).

Clearly

\underset{\left|u\right|\to 0}{\text{lim}}\frac{\zeta \left(u\right)}{u}=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\underset{\left|u\right|\to \infty}{\text{lim}}\frac{\xi \left(u\right)}{u}=0.

(4.3)

Let

\stackrel{\u0303}{\xi}\left(u\right)=\underset{0\le \left|s\right|\le u}{\text{max}}\left|\xi \left(s\right)\right|.

Then \stackrel{\u0303}{\xi} is nondecreasing and

\underset{\left|u\right|\to \infty}{\text{lim}}\frac{\stackrel{\u0303}{\xi}\left(u\right)}{u}=0.

(4.4)

Let us consider

Lu-\lambda m\left(t\right)r{f}_{0}u-\lambda m\left(t\right)r\zeta \left(u\right)=0,

(4.5)

as a bifurcation problem from the trivial solution *u* ≡ 0.

Equation (4.5) can be converted to the equivalent equation

\begin{array}{ll}\hfill u\left(t\right)& =\lambda {L}^{-1}\left[m\left(\cdot \right)r{f}_{0}u\left(\cdot \right)+m\left(\cdot \right)r\zeta \left(u\left(\cdot \right)\right)\right]\left(t\right)\phantom{\rule{2em}{0ex}}\\ :=\lambda Tu+H\left(\lambda ,u\right).\phantom{\rule{2em}{0ex}}\end{array}

(4.6)

It is easy to see that *T* : *X* → *X* is compact. Further we note that *H*(λ, *u*) = λ*L*^{-1}[*m*(·)*ζ*( *u* (·))] = *o*⊠*u*⊠near λ = 0 uniformly on bounded λ intervals, since

\begin{array}{ll}\hfill \left|\right|{L}^{-1}\left[m\left(\cdot \right)\zeta \left(u\left(\cdot \right)\right)\right]\left|\right|& =\underset{t\in \mathsf{\text{T}}}{\text{max}}\left|\sum _{s=1}^{T}G\left(t,s\right)m\left(s\right)\zeta \left(u\left(s\right)\right)\right|\phantom{\rule{2em}{0ex}}\\ \le C\cdot \underset{s\in \mathsf{\text{T}}}{\text{max}}\left|m\left(s\right)\right|\left|\right|\zeta \left(u\left(\cdot \right)\right)\left|\right|,\phantom{\rule{2em}{0ex}}\end{array}

where C=\underset{t\in \widehat{T}}{\text{max}}{\sum}_{s=1}^{T}G\left(t,s\right) and

G\left(t,s\right)=\frac{1}{T+1}\left\{\begin{array}{cc}\hfill \left(T+1-t\right)s,\hfill & \hfill 0\le s\le t\le T+1,\hfill \\ \hfill \left(T+1-t\right)t,\hfill & \hfill 0\le t\le s\le T+1.\hfill \end{array}\right.

Let E=\mathbb{R}\times X under the product topology. Let {S}^{+}:=\left\{u\in X|u\left(t\right)>0\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}t\in T\right\}. Set *S*^{-} = -*S*^{+}, *S* = *S*^{+} ∪ *S*^{-}. Then *S*^{+} and *S*^{-} are disjoint in *X*. Finally let Ψ^{±} = ℝ × *S*^{±} and Ψ = ℝ × *S*. Let Σ be the closure of the set of nontrivial solutions of (1.1) and (1.2).

It is easy to see that \frac{{\lambda}_{m,+}}{r{f}_{0}}\in r\left(T\right) is simple. Now applying Theorems 2.1 and 2.2, we get the result as follows: Σ contains a maximum subcontinuum {\mathcal{C}}_{+} which is composed of two distinct connected set {\mathcal{C}}_{+}^{+} and {\mathcal{C}}_{+}^{-} such that {\mathcal{C}}_{+}={\mathcal{C}}_{+}^{+}\cup {\mathcal{C}}_{+}^{-} and {\mathcal{C}}_{+}^{+}\cap {\mathcal{C}}_{+}^{-}=\left\{\left(\frac{{\lambda}_{m,+}}{r{f}_{0}},0\right)\right\}. Moreover, Lemma 4.1 guarantees the second case in Theorems 2.1 and 2.2 cannot happen. Otherwise, there will exist \left(\eta ,y\right)\in {C}_{+}^{v}, such that *y* has a multiple zero point *t*_{0}, (i.e., *t*_{0} satisfies *y*(*t*_{0}) = 0 and *y*(*t*_{0} - 1)*y*(*t*_{0} + 1) > 0). However, this contradicts Lemma 4.1. Thus, for each \nu \in \left\{+,-\right\},{\mathcal{C}}_{+}^{\nu} joins \left(\frac{{\lambda}_{m,+}}{r{f}_{0}},0\right) to infinity in Ψ^{v}and {\mathcal{C}}_{+}^{\nu}\backslash \left\{\left(\frac{{\lambda}_{m,+}}{r{f}_{0}},0\right)\right\}\subset {\Psi}^{\nu}.

It is obvious that any solution to (4.5) of the form (1, *u*) yields a solution *u* to (1.1) and (1.2). We will show that {\mathcal{C}}_{+}^{\nu} crosses the hyperplane {1} × *X* in ℝ × *X*. To achieve this goal, it will be enough to show that

\left(\frac{{\lambda}_{m,+}}{r{f}_{\infty}},\frac{{\lambda}_{m,+}}{r{f}_{0}}\right)\subseteq {\text{Proj}}_{\mathbb{R}}{\mathcal{C}}_{+}^{\nu},

(4.7)

or

\left(\frac{{\lambda}_{m,+}}{r{f}_{0}},\frac{{\lambda}_{m,+}}{r{f}_{\infty}}\right)\subseteq {\text{Proj}}_{\mathbb{R}}{\mathcal{C}}_{+}^{\nu},

(4.8)

where {\mu}_{n}+\left|\right|{y}_{n}|{|}_{X}\to \infty . denotes the projection of {\mathcal{C}}_{+}^{\nu} on ℝ.

Let \left({\mu}_{n},{y}_{n}\right)\in {\mathcal{C}}_{+}^{\nu} satisfy

{\mu}_{n}+\left|\right|{y}_{n}|{|}_{X}\to \infty .

We note that *μ*_{
n
}> 0 for all *n* ∈ **ℕ** since (0,0) is the only solution of (4.5) for λ = 0 and {\mathcal{C}}_{+}^{\nu}\cap \left(\left\{0\right\}\times X\right)=\varnothing.

*Case 1*. \frac{{\lambda}_{m,+}}{r{f}_{\infty}}<1<\frac{{\lambda}_{m,+}}{r{f}_{0}}.

We divide the proof into two steps.

*Step 1*. We show that if there exists a constant number *M* > 0 such that

{\mu}_{n}\in \left(0,M\right],

(4.9)

then (4.7) holds.

In this case it follows that

\left|\right|{y}_{n}|{|}_{X}\to \infty .

(4.10)

We divide the equation

L{y}_{n}-{\mu}_{n}m\left(t\right)r{f}_{\infty}{y}_{n}-{\mu}_{n}m\left(t\right)r\xi \left({y}_{n}\right)=0

(4.11)

by ⊠*y*_{
n
}⊠_{
x
}and set {\u0233}_{n}=\frac{{y}_{n}}{\left|\right|{y}_{n}|{|}_{X}}. Since {\u0233}_{n} is bounded in *X* and *μ*_{
n
}is bounded in ℝ, after taking the subsequence if necessary, we have that {\u0233}_{n}\to \u0233 for some \u0233\in X with \left|\right|\u0233|{|}_{X}=1 and {\mu}_{n}\to \stackrel{\u0304}{\mu} for some *μ* ∈ ℝ. Moreover, from (4.4) and the fact that \stackrel{\u0303}{\xi} is nondecreasing, we have that

\underset{n\to \infty}{\text{lim}}\frac{\left|\xi \left({y}_{n}\left(t\right)\right)\right|}{\parallel {y}_{n}{\parallel}_{X}}=0,

(4.12)

since \frac{\left|\xi \left({y}_{n}\left(t\right)\right)\right|}{\parallel {y}_{n}{\parallel}_{X}}\le \frac{\stackrel{\u0303}{\xi}\left(\left|{y}_{n}\left(t\right)\right|\right)}{\parallel {y}_{n}{\parallel}_{X}}\le \frac{\xi \left(\parallel {y}_{n}{\parallel}_{X}\right)}{\parallel {y}_{n}{\parallel}_{X}}. Thus,

\u0233\left(t\right)=\sum _{s=1}^{T}G\left(t,s\right)\stackrel{\u0304}{\mu}m\left(s\right)r{f}_{\infty}\u0233\left(s\right),

which implies that

L\u0233-\stackrel{\u0304}{\mu}m\left(t\right)r{f}_{\infty}\u0233=0.

(4.13)

We claim that

\left(\stackrel{\u0304}{\mu},\u0233\right)\in {\mathcal{C}}_{+}^{\nu}.

We only prove that if {y}_{n}\in {\mathcal{C}}_{+}^{+}, then {\u0233}_{n}\in {\mathcal{C}}_{+}^{+}. The other case that if {y}_{n}\in {\mathcal{C}}_{+}^{-}, then {\u0233}_{n}\in {\mathcal{C}}_{+}^{-} can be treated similarly.

Obviously when {y}_{n}\in {\mathcal{C}}_{+}^{+}, then \u0233\left(t\right)\ge 0 on \widehat{T}. Furthermore, \u0233\left(t\right)>0 on T. In fact, if there exists a {t}_{0}\in T such that \u0233\left({t}_{0}\right)=0, then, by Lemma 4.1, we obtain \u0233\left({t}_{0}-1\right)\u0233\left({t}_{0}+1\right)<0 which contradicts the fact that \u0233\left(t\right)\ge 0 on \widehat{T}. Thus, \u0233\left(t\right)>0 on T. This together with the fact {\mathcal{C}}_{+} is a closed set in E implies that \u0233\in {\mathcal{C}}_{+}^{+}. Moreover, \stackrel{\u0304}{\mu}r{f}_{\infty}={\lambda}_{m,+}, so that

\stackrel{\u0304}{\mu}=\frac{{\lambda}_{m,+}}{r{f}_{\infty}}.

Thus, (4.7) holds.

*Step 2*. We show that there exists a constant *M* > 0 such that *μ*_{
n
}∈ (0, *M*] for all *n*.

Since {(*μ*_{
n
}, *y*_{
n
})} are the solutions to (4.5), they follow that

L{y}_{n}={\mu}_{n}rm\left(t\right){\Gamma}_{n}\left(t\right){y}_{n},

(4.14)

where {\Gamma}_{n}\left(t\right):=\frac{f\left({y}_{n}\left(t\right)\right)}{{y}_{n}\left(t\right)}. From (H1) and (H2), there exist two positive constants *ρ*_{1} and *ρ*_{2}, such that

{\rho}_{1}<\frac{f\left({y}_{n}\right)}{{y}_{n}}<{\rho}_{2}.

(4.15)

Let *η*_{*} > 0 be the positive principal eigenvalue of the following linear eigenvalue problem

Lv=\eta {\chi}_{1}\left(t\right)m\left(t\right)v

(4.16)

and *η** > 0 the positive principal eigenvalue of the following linear eigenvalue problem

Lv=\eta {\chi}_{2}\left(t\right)m\left(t\right)v,

(4.17)

where

\begin{array}{cc}\hfill {\chi}_{1}\left(t\right)=\left\{\begin{array}{cc}\hfill {\rho}_{1}\hfill & \hfill \mathsf{\text{if}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}m\left(t\right)>0,\phantom{\rule{0.3em}{0ex}}t\in T,\hfill \\ \hfill {\rho}_{2}\hfill & \hfill \mathsf{\text{if}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}m\left(t\right)<0,\phantom{\rule{0.3em}{0ex}}t\in T,\hfill \end{array}\right.\hfill & \hfill {\chi}_{2}\left(t\right)=\left\{\begin{array}{cc}\hfill {\rho}_{2}\hfill & \hfill \mathsf{\text{if}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}m\left(t\right)>0,\phantom{\rule{0.3em}{0ex}}t\in T,\hfill \\ \hfill {\rho}_{1}\hfill & \hfill \mathsf{\text{if}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}m\left(t\right)<0,\phantom{\rule{0.3em}{0ex}}t\in T.\hfill \end{array}\right.\hfill \end{array}

By Theorem 3.2, (4.14), (4.15), (4.16), and (4.17), we get

\frac{{\eta}^{*}}{r}<{\mu}_{n}<\frac{{\eta}_{*}}{r}.

*Case 2*. \frac{{\lambda}_{m,+}}{r{f}_{0}}<1<\frac{{\lambda}_{m,+}}{r{f}_{\infty}}.

From Step 2 of Case 1, there exists *M* > 0 such that for all *n* ∈ **ℕ**,

{\mu}_{n}\in \left(0,M\right].

Applying a similar argument to that used in Step 1 of Case 1 (after taking a subsequence and relabeling, if necessary), we get

{\mu}_{n}\to \frac{{\lambda}_{m,+}}{r{f}_{\infty}},\phantom{\rule{1em}{0ex}}{y}_{n}\to \infty \phantom{\rule{1em}{0ex}}\mathsf{\text{as}}\phantom{\rule{2.77695pt}{0ex}}n\to \infty ,

which implies that (4.8) holds.

At last, we deal with the case *r* < 0.

Let us consider

Lu-\lambda rm\left(t\right){f}_{0}u-\lambda rm\left(t\right)\zeta \left(u\right)=0,

(4.18)

as a bifurcation problem from the trivial solution *u* ≡ 0. Now, applying Theorems 2.1 and 2.2, we get the following results: Σ contains a maximum subcontinuum {\mathcal{C}}_{-} which is composed of two distinct connected set {\mathcal{C}}_{-}^{+} and {\mathcal{C}}_{-}^{-} such that {\mathcal{C}}_{-}={\mathcal{C}}_{-}^{+}\cup {\mathcal{C}}_{-}^{-} and {\mathcal{C}}_{+}^{-}\cap {\mathcal{C}}_{-}^{-}=\left\{\left({\scriptscriptstyle \frac{{\lambda}_{m,-}}{-r{f}_{0}}},0\right)\right\}. Moreover, by Lemma 4.1, for each \nu \in \left\{+,-\right\},{\mathcal{C}}_{-}^{\nu} joins \left(\frac{{\lambda}_{m,-}}{-r{f}_{0}},0\right) to infinity in Ψ^{v}and {\mathcal{C}}_{-}^{\nu}\backslash \left\{\left(\frac{{\lambda}_{m,-}}{-r{f}_{0}},0\right)\right\}\subset {\Psi}^{\nu}, where Σ and Ψ^{v}are defined as in the case *r* > 0.

It is clear that any solution to (4.18) of the form (-1, *u*) yields a solutions *u* of (1.1) and (1.2). We will show {\mathcal{C}}_{-}^{\nu} crosses the hyperplane {-1} × *X* in ℝ × *X*. To achieve this goal, it will be enough to show that

\left(\frac{{\lambda}_{m,-}}{-r{f}_{\infty}},\frac{{\lambda}_{m,-}}{-r{f}_{0}}\right)\subseteq {\text{Proj}}_{\mathbb{R}}{\mathcal{C}}_{-}^{\nu},

(4.19)

or

\left(\frac{{\lambda}_{m,-}}{-r{f}_{0}},\frac{{\lambda}_{m,-}}{-r{f}_{\infty}}\right)\subseteq {\text{Proj}}_{\mathbb{R}}{\mathcal{C}}_{-}^{v}.

(4.20)

Let \left({\mu}_{n},{y}_{n}\right)\in {\mathcal{C}}_{-}^{\nu} satisfy

\left|{\mu}_{n}\right|+\parallel {y}_{n}{\parallel}_{X}\to \infty .

We note that *μ*_{
n
}< 0 for all *n* ∈ **ℕ** since (0, 0) is the only solution to (4.18) for λ = 0 and {\mathcal{C}}_{-}^{\nu}\cap \left(\left\{0\right\}\times X\right)=\varnothing.

The rest of the proof is similar to the proof of the case *r* > 0, so we omit it.