For any

*u* ∈

*X*, we define

*e*(

*u*)(·): [0, 1] → ℝ by

$e\left(u\right)\left(x\right)=p\left(x,u\left(x\right),{u}^{\prime}\left(x\right)\right),\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}x\in \left[0,1\right].$

It follows from (1.5) that

$\left|e\left(u\right)\left(x\right)\right|\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\le C+\beta \left|u\left(x\right)\right|,\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}x\in \left[0,1\right].$

(3.1)

For any

*s* ∈ ℝ, let

$G\left(s\right)=\underset{0}{\overset{s}{\int}}g\left(\tau \right)\mathsf{\text{d}}\tau \ge 0,$

and for any

*s* ≥ 0, let

$\gamma \left(s\right)=max\left\{\left|g\left(r\right)\right|:\left|r\right|\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\le s\right\},\phantom{\rule{1em}{0ex}}\Gamma \left(s\right)=max\left\{G\left(r\right):\left|r\right|\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\le s\right\}.$

We now consider the boundary value problem

${u}^{\u2033}+\lambda u+\alpha \left(g\left(u\right)+e\left(u\right)\right)=0,\phantom{\rule{1em}{0ex}}u\in X,$

(3.2)

where *α* ∈ [0, 1] is an arbitrary fixed number and *λ* ∈ ℝ. In the following lemma (*λ*, *u*) ∈ ℝ × *X* will be an arbitrary solution of (3.2).

By (H2), we can choose

*b*_{1} ≥ 1 such that

$\left|s\right|\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\ge {b}_{1}\phantom{\rule{2.77695pt}{0ex}}\Rightarrow \phantom{\rule{2.77695pt}{0ex}}\left|g\left(s\right)\right|\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\ge C+\beta \left|s\right|.$

(3.3)

By (1.2), we have the following

**Lemma 3.1**. Let (H1) hold and let

*u* ∈

*X*. Then

$|u{|}_{0}\le \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}|{u}^{\prime}{|}_{0}\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}|{u}^{\u2033}{|}_{0}.$

(3.4)

**Lemma 3.2**. Let

*u* be a solution of (3.2). Then for any

*x*_{0},

*x*_{1} ∈ [0, 1],

$\begin{array}{c}{u}^{\prime}{\left({x}_{1}\right)}^{2}+\lambda u{\left({x}_{1}\right)}^{2}+2\alpha G\left(u\left({x}_{1}\right)\right)={u}^{\prime}{\left({x}_{0}\right)}^{2}+\lambda u{\left({x}_{0}\right)}^{2}+2\alpha G\left(u\left({x}_{0}\right)\right)\\ \phantom{\rule{1em}{0ex}}-2\alpha \underset{{x}_{0}}{\overset{{x}_{1}}{\int}}e\left(u\right)\left(s\right){u}^{\prime}\left(s\right)\mathsf{\text{d}}s.\end{array}$

**Proof**. Multiply (3.2) by *u*' and integrate from *x*_{0} to *x*_{1}, then we get the desired result. ■

In the following, let us fix

*R* ∈ (0, ∞) so large that

*R* ≥

*b*_{1} and

$\begin{array}{c}g\left(r\right)+p\left(t,r,v\right)>0,\phantom{\rule{1em}{0ex}}t\in \left[0,1\right],v\in \mathbb{R},\phantom{\rule{2.77695pt}{0ex}}r>R,\\ g\left(r\right)+p\left(t,r,v\right)<0,\phantom{\rule{1em}{0ex}}t\in \left[0,1\right],v\in \mathbb{R},\phantom{\rule{2.77695pt}{0ex}}r<-R.\end{array}$

(3.5)

**Lemma 3.3**. There exists an increasing function

*ζ*_{1} : [0, ∞) → [0, ∞), such that for any solution

*u* of (3.2) with 0 ≤

*λ* ≤

*R* and |

*u*(

*x* 0)| + |

*u*'(

*x*_{0})| ≤

*R* for some

*x*_{0} ∈ [0, 1], we have

$|{u}^{\prime}{|}_{0}\le {\zeta}_{1}\left(R\right).$

**Proof**. Choose

*x*_{1} ∈ [0, 1] such that |

*u*'|

_{0} = |

*u*'(

*x*_{1})|. We obtain from Lemma 3.2 that

$\begin{array}{lll}\hfill |{u}^{\prime}{|}_{0}^{2}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}& ={u}^{\prime}{\left({x}_{1}\right)}^{2}\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ \le {u}^{\prime}{\left({x}_{1}\right)}^{2}+\lambda u{\left({x}_{1}\right)}^{2}+2\alpha G\left(u\left({x}_{1}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ ={u}^{\prime}{\left({x}_{0}\right)}^{2}+\lambda u{\left({x}_{0}\right)}^{2}+2\alpha G\left(u\left({x}_{0}\right)\right)-2\alpha \underset{{x}_{0}}{\overset{{x}_{1}}{\int}}e\left(u\right)\left(\xi \right){u}^{\prime}\left(\xi \right)\mathsf{\text{d}}\xi .\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ \hfill \text{(4)}\end{array}$

Combining this with (3.1), (3.4), it concludes that

$|{u}^{\prime}{|}_{0}^{2}\phantom{\rule{0.3em}{0ex}}\le {R}^{2}+{R}^{3}+2\Gamma \left(R\right)+2\left(C+\beta |u{|}_{0}\right)|{u}^{\prime}{|}_{0}\phantom{\rule{0.3em}{0ex}}\le K\left(R\right)+2C|{u}^{\prime}{|}_{0}+2\beta |{u}^{\prime}{|}_{0}^{2},$

with

$K\left(R\right)={R}^{2}+{R}^{3}+2\Gamma \left(R\right).$

This implies

$|{u}^{\prime}{|}_{0}\phantom{\rule{0.3em}{0ex}}\le {\zeta}_{1}\left(R\right):=\frac{2C+\sqrt{4{C}^{2}+4\left(1-2\beta \right)K\left(R\right)}}{2\left(1-2\beta \right)}.$

■

Define

${\zeta}_{2}\left(s\right)={\zeta}_{1}\left(s+{s}^{2}\right)+1,\phantom{\rule{1em}{0ex}}s>0.$

(3.6)

Clearly, the function is nondecreasing.

**Lemma 3.4** Let *u* be a solution of (3.2) with 0 ≤ *λ* ≤ *R* and |*u*'|_{0} ≥ *ζ*_{2}(*R*) for some *R >* 0. Then, for any *x* ∈ [0, 1] with |*u*(*x*)| ≤ *R*, we have |*u*'(*x*)| ≥ *R*^{2}.

**Proof**. Suppose, on the contrary that there exists

*x*_{0} ∈ (0, 1) such that |

*u*(

*x*_{0})| ≤

*R* and |

*u*'(

*x*_{0})|

*< R*^{2}. Then

$\left|u\left({x}_{0}\right)\right|+\left|{u}^{\prime}\left({x}_{0}\right)\right|\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}<R+{R}^{2}.$

Combining this with

*λ* ≤

*R < R* +

*R*^{2} and using Lemma 3.3, it concludes that

$|{u}^{\prime}{|}_{0}\phantom{\rule{0.3em}{0ex}}\le {\zeta}_{1}\left(R+{R}^{2}\right).$

However, this is impossible if |*u*'|_{0} ≥ *ζ*_{2}(*R*). ■

For fixed

*R > b*_{1}, let us define

$\chi \left(s\right):={\zeta}_{2}\left(R\right)+{\zeta}_{2}\left(s\right),\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}s\ge 0.$

(3.7)

Let us now consider the problem

${u}^{\u2033}+\lambda u+\theta \left(|{u}^{\prime}{|}_{0}\u2215\chi \left(\lambda \right)\right)\left(g\left(u\right)+e\left(u\right)\right)=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}u\in X,$

(3.8)

where *θ* : ℝ → ℝ is a strictly increasing, *C*^{∞}-function with *θ*(*s*) = 0, *s* ≤ 1 and *θ*(*s*) = 1, *s* ≥ 2. The nonlinear term in (3.8) is a continuous function of (*λ*, *u*) ∈ ℝ × *X* and is zero for *λ* ∈ ℝ, |*u*'|_{0} ≤ *χ*(*λ*), so (3.8) becomes a linear eigenvalue problem in this region, and overall the problem can be regarded as a bifurcation (from *u* = 0) problem.

The next lemma now follows immediately.

**Lemma 3.5** The set of solutions (

*λ*,

*u*) of (3.8) with |

*u*'|

_{0} ≤

*χ*(

*λ*) is

$\left\{\left(\lambda ,0\right):\lambda \in \mathbb{R}\right\}\cup \left\{\left({\lambda}_{k},t{\varphi}_{k}\right):\phantom{\rule{2.77695pt}{0ex}}k\ge 1,\left|t\right|\phantom{\rule{0.3em}{0ex}}\le \chi \left({\lambda}_{k}\right)\u2215|{\varphi}^{\prime}k{|}_{0}\right\}.$

We also have the following global bifurcation result for (3.8).

**Lemma 3.6** For each

*k* ≥ 1 and

*ν* ∈ {+, -}, there exists a connected set

${\mathcal{C}}_{k}^{\nu}\subset \mathbb{R}\times E$ of nontrivial solutions of (3.8) such that

${\mathcal{C}}_{k}^{\nu}\cup \left({\lambda}_{k},0\right)$ is closed and connected and:

- (i)
there exists a neighborhood *N*_{
k
} of (*λ*_{
k
} , 0) in ℝ × *E* such that ${N}_{k}\cap {\mathcal{C}}_{k}^{\nu}\subset \mathbb{R}\times {\Gamma}_{k}^{\nu}$,

- (ii)
${\mathcal{C}}_{k}^{\nu}$ meets infinity in ℝ × *E* (that is, there exists a sequence $\left({\lambda}_{n},{u}_{n}\right)\in {\mathcal{C}}_{k}^{\nu},n=1,2,\dots $, such that |*λ*_{
n
} | + |*u*_{
n
} | _{
E
} → ∞).

**Proof**. Since

*L*^{-1} :

*Y* →

*X* exists and is bounded, (3.8) can be rewritten in the form

$u=\lambda {L}^{-1}u+\theta \left(|{u}^{\prime}{|}_{0}\u2215\chi \left(\lambda \right)\right){L}^{-1}\left(g\left(u\right)+e\left(u\right)\right),$

(3.9)

and since *L*^{-1} can be regarded as a compact operator from *Y* to *E*, it is clear that finding a solution (*λ*, *u*) of (3.8) in ℝ × *E* is equivalent to finding a solution of (3.9) in ℝ × *E*. Now, by the similar method used in the proof of [1, Theorem 4.2]), we may deduce the desired result.

■

Since *e*(*u*)(*t*) σ 0 in (3.8), nodal properties need not be preserved. However, we will rely on preservation of nodal properties for "large" solutions, encapsulated in the following result.

**Lemma 3.7** If (*λ*, *u*) is a solution of (3.8) with *λ* ≥ 0 and |*u*'|_{0}*> χ*(*λ*), then $u\in {\Gamma}_{k}^{\nu}$, for some *k* ≥ 1 and *ν* ∈ {+, -}.

**Proof**. If $u\notin {\Gamma}_{k}^{\nu}$ for any *k* ≥ 1 and *ν*, then one of the following cases must occur:

*Case 1. u*'(0) = 0;

*Case 2. u*' (*τ*) = *u*″(*τ*) = 0 for some *τ* ∈ (0, 1].

In the Case 1, *u*(*t*) ≡ 0 on [0, 1]. This contradicts the assumption |*u*'|_{0}*> χ*(*λ*) ≥ *ζ*_{2}(*λ*). So this case cannot occur.

In the Case 2, we have from (3.8) that

$\lambda u\left(\tau \right)+\theta \left(|{u}^{\prime}{|}_{0}\u2215\chi \left(\lambda \right)\right)\left(g\left(u\left(\tau \right)\right)+e\left(u\right)\left(\tau \right)\right)=0.$

(3.10)

Since |

*u*'|

_{0}*> χ*(

*λ*), we have from the definition of

*θ* that

$\theta \left(|{u}^{\prime}{|}_{0}\u2215\chi \left(\lambda \right)\right)>0.$

(3.11)

It follows from Lemma 3.4 that |

*u*(

*τ*)|

*> R* ≥

*b*_{1}. Combining this with (3.11) and (3.3), it concludes that

$\lambda u\left(\tau \right)+\theta \left(|{u}^{\prime}{|}_{0}\u2215\chi \left(\lambda \right)\right)\left(g\left(u\left(\tau \right)\right)+e\left(u\right)\left(\tau \right)\right)\ne 0,$

(3.12)

which contradicts (3.10). So, Case 2 cannot occur.

Therefore, $u\in {\Gamma}_{k}^{\nu}$ for any *k* ≥ 1 and *ν* ∈ {+, -}. ■

In view of Lemmas 3.5 and 3.7, in the following lemma, we suppose that (*λ*, *u*) is an arbitrary nontrivial solution of (3.8) with *λ* ≥ 0 and $u\in {\Gamma}_{k}^{\nu}$, for some *k* ≥ 1 and *ν*.

**Lemma 3.8**. There exists an integer

*k*_{0} ≥ 1 (depending only on

*χ*(0)) such that for any nontrivial solution

*u* of (3.8) with

*λ* = 0 and

*χ*(0) ≤ |

*u*'|

_{0} ≤ 2

*χ*(0), we have

$k\mathsf{\text{}}{k}_{\mathsf{\text{0}}}\mathsf{\text{.}}$

(3.13)

**Proof**. Let

*x*_{1},

*x*_{2} be consecutive zeros of

*u*. Then there exists

*x*_{3} ∈ (

*x*_{1},

*x*_{2}) such that

*u*'(

*x*_{3}) = 0, and hence, Lemma 3.4, (3.3), and (3.7) yield that |

*u*(

*x*_{3})|

*>* 1. Since

$\begin{array}{lll}\hfill 2\phantom{\rule{1em}{0ex}}& <\phantom{\rule{1em}{0ex}}|u\left({x}_{2}\right)-u\left({x}_{3}\right)|+|u\left({x}_{3}\right)-u\left({x}_{1}\right)|\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =\phantom{\rule{1em}{0ex}}\left|\left({x}_{2}-{x}_{3}\right){u}^{\prime}\left({\tau}_{1}\right)\right|+\left|\left({x}_{3}-{x}_{1}\right){u}^{\prime}\left({\tau}_{2}\right)\right|\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \le \phantom{\rule{1em}{0ex}}\left({x}_{2}-{x}_{3}\right)|{u}^{\prime}{|}_{0}+\left({x}_{3}-{x}_{1}\right)|{u}^{\prime}{|}_{0}\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ =\phantom{\rule{1em}{0ex}}\left({x}_{2}-{x}_{1}\right)|{u}^{\prime}{|}_{0}\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}$

for some

*τ*_{1} ∈ (

*x*_{3},

*x*_{2}),

*τ*_{2} ∈ (

*x*_{1},

*x*_{3}), it follows that

$|{x}_{2}-{x}_{1}|\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}>2\u2215|{u}^{\prime}{|}_{0}.$

(3.14)

Notice that |

*u*'|

_{0}*> χ*(0) ≥

*ζ*_{2}(

*R*) implies that

$u\in {\Gamma}_{k}^{\nu}$ for some

*k* ∈ ∞ and

*ν* ∈ {+, -}, and subsequently, there exist 0

*< r*_{1}*< r*_{2}*<* · · ·

*< r*_{k-1}, such that

$u\left({r}_{j}\right)=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}j=1,\dots ,k-1.$

This together with (3.14) imply that

$1>\left(k-1\right)\cdot 2\u2215|{u}^{\prime}{|}_{0},$

and accordingly, *k <* |*u*'|_{0}/2 + 1 ≤ *χ*(0) + 1. ■

Now let

${V}_{R}\left(u\right)=\left\{x\in \left[0,1\right]:\phantom{\rule{2.77695pt}{0ex}}\left|u\left(x\right)\right|\phantom{\rule{2.77695pt}{0ex}}\ge R\right\},\phantom{\rule{1em}{0ex}}{W}_{R}\left(u\right)=\left\{x\in \left[0,1\right]:\phantom{\rule{2.77695pt}{0ex}}\left|u\left(x\right)\right|\phantom{\rule{2.77695pt}{0ex}}<R\right\}.$

**Lemma 3.9**. Suppose that 0 ≤ *λ* ≤ *R* and |*u*'|_{0} ≥ *χ*(*R*). Then *W*_{
R
} (*u*) consists of at least *k* intervals and at most *k* + 1 intervals, each of length less than 2*/R*, and *V*_{
R
} (*u*) consists of at least *k* intervals and at most *k* + 1 intervals.

**Proof**. Lemma 3.4 implies that |

*u*'(

*x*)| ≥

*R*^{2} for all

*x* ∈

*W*_{
R
} (

*u*). For any interval

*I* ⊂

*W*_{
R
} (

*u*),

*u*' does not change sign on

*I*, say,

${u}^{\prime}\left(x\right)\ge {R}^{2},\phantom{\rule{1em}{0ex}}x\in I.$

We claim that the length of *I* is less than 2/*R*.

In fact, for

*x*,

*y* ∈

*I* with

*x > y*, say,

$u\left(x\right)-u\left(y\right)=\underset{y}{\overset{x}{\int}}{u}^{\prime}\left(s\right)\mathsf{\text{d}}s\ge {R}^{2}\left(x-y\right).$

Thus,

$x-y\le \frac{R-\left(-R\right)}{{R}^{2}}=\frac{2}{R},$

which implies

$\left|I\right|\phantom{\rule{2.77695pt}{0ex}}\le \frac{2}{R}.$

The case

${u}^{\prime}\left(x\right)\le -{R}^{2},\phantom{\rule{1em}{0ex}}x\in I$

can be treated by the similar method. Since *u* is monotonic in any subinterval containing in *W*_{
R
} (*u*), the desired result is followed. ■

**Lemma 3.10**. There exists

*ζ*_{3} with lim

_{R→∞}*ζ*_{3}(

*R*) = 0, and

*η*_{1} ≥ 0 such that, for any

*R* ≥

*η*_{1}, if either

- (a)
0 ≤ *λ* ≤ *R* and |*u*'|_{0} = 2*χ*(*R*), or

- (b)
*λ* = *R* and *χ*(*R*) ≤ |*u*'|_{0} ≤ 2*χ*(*R*),

then the length of each interval of *V*_{
R
} (*u*) is less than *ζ*_{3}(*R*).

**Proof**. Define

*H* =

*H*(

*R*) by

$H{(R)}^{2}:=\mathrm{min}\{R,\mathrm{min}\{g(\xi )/\xi :\left|\xi \right|\ge R\}-(C/R+\beta )\},$

and let *ζ*_{3}(*R*) := 2*π*/*H*(*R*). By (1.4), lim_{R→∞}*H*(*R*) = ∞, so lim_{R→∞}*ζ*_{3}(*R*) = 0, and we may choose *η*_{1} ≥ *b*_{1} sufficiently large that *H*(*R*) *>* 0 for all *R* ≥ *η*_{1}.

We firstly show that

$\left|u\left(\tau \right)\right|\phantom{\rule{2.77695pt}{0ex}}>R,\phantom{\rule{1em}{0ex}}\mathsf{\text{forsome}}\phantom{\rule{2.77695pt}{0ex}}\tau \in \left(0,1\right).$

(3.15)

In fact, if |

*u*(

*x*)| ≤

*R* on [0, 1], then Lemma 3.4 yields that either

${u}^{\prime}\left(x\right)\ge {R}^{2},\phantom{\rule{1em}{0ex}}x\in \left[0,\phantom{\rule{2.77695pt}{0ex}}1\right],$

or

${u}^{\prime}\left(x\right)\le -{R}^{2},\phantom{\rule{1em}{0ex}}x\in \left[0,\phantom{\rule{2.77695pt}{0ex}}1\right].$

However, these contradict the boundary conditions (1.2), since (H1) implies *u*'(*s*_{0}) = 0 for some *s*_{0} ∈ (0, 1). Therefore, (3.15) is valid.

Now, Let us choose

*x*_{0},

*x*_{2} such that either

- (1)
*u*(*x* _{0}) = *u*(*x* _{2}) = *R* and *u > R* on (*x* _{0}, *x* _{2}) or

- (2)
*u*(*x* _{0}) = *R*, *x* _{2} = 1 and *u > R* on (*x* _{0}, 1].

(the case of intervals on which

*u <* 0 is similar). Let

$I=\mathsf{\text{[}}{x}_{\mathsf{\text{0}}},\phantom{\rule{2.77695pt}{0ex}}{x}_{\mathsf{\text{2}}}\mathsf{\text{]}}\mathsf{\text{.}}$

By (3.8) and the construction of

*H*(

*R*), if either (a) or (b) holds then

$-{u}^{\u2033}\left(x\right)\ge H{\left(R\right)}^{2}u\left(x\right)>0,\phantom{\rule{1em}{0ex}}x\in I,$

and by Lemma 3.4, *u*'(*x*_{0}) *>* 0, and *u*'(*x*_{2}) *<* 0, if *x*_{2}*<* 1.

Suppose, now on the contrary that

*x*_{2} -

*x*_{0}*> ζ*_{3}(

*R*), that is,

*l* := 2

*π*/(

*x*_{2} -

*x*_{0})

*< H*(

*R*). Defining

*x*_{1} = (

*x*_{0} +

*x*_{2})/2 and

$v\left(x\right)=1+cosl\left(x-{x}_{1}\right),\phantom{\rule{1em}{0ex}}x\in I,$

we have

$\begin{array}{lll}\hfill v\left({x}_{0}\right)& =v\left({x}_{2}\right)=0,\phantom{\rule{1em}{0ex}}{v}^{\prime}\left({x}_{0}\right)={v}^{\prime}\left({x}_{2}\right)=0,\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ \hfill {v}^{\u2033}\left(x\right)& =-{l}^{2}\left(v\left(x\right)-1\right),\phantom{\rule{1em}{0ex}}x\in I,\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$

and hence

$\begin{array}{lll}\hfill 0\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}\underset{{x}_{0}}{\overset{{x}_{2}}{\int}}\frac{d}{dx}\left({u}^{\prime}v-u{v}^{\prime}\right)\mathsf{\text{d}}x\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =\phantom{\rule{1em}{0ex}}\underset{{x}_{0}}{\overset{{x}_{2}}{\int}}\left({u}^{\u2033}v-u{v}^{\u2033}\right)\mathsf{\text{d}}x\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \le \phantom{\rule{1em}{0ex}}\underset{{x}_{0}}{\overset{{x}_{2}}{\int}}\left(-{H}^{2}uv+{l}^{2}\left(v-1\right)u\right)\mathsf{\text{d}}x\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ =\phantom{\rule{1em}{0ex}}-{l}^{2}\underset{{x}_{0}}{\overset{{x}_{2}}{\int}}u\mathsf{\text{d}}x\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\\ \le \phantom{\rule{1em}{0ex}}-{l}^{2}R<0,\phantom{\rule{2em}{0ex}}& \hfill \text{(5)}\\ \hfill \text{(6)}\end{array}$

and this contradiction shows that *x*_{2} - *x*_{0} ≤ *ζ*_{3}(*R*), which proves the lemma.

■

Now, we are in the position to prove Theorem 1.1.

**Proof of Theorem 1.1** Now, choose an arbitrary integer

*k* ≥

*k*_{0} and

*ν* ∈ {+, -}, and choose Λ

*>* max{

*η*_{1},

*μ*_{
k
} } (Here, we assume Λ

*> η*_{1}, so that Lemma 3.10 could be applied!) such that

$\left(k+1\right)\phantom{\rule{2.77695pt}{0ex}}\frac{2}{\Lambda}+\left(k+1\right)\phantom{\rule{2.77695pt}{0ex}}{\zeta}_{3}\left(\Lambda \right)<1.$

(3.16)

Notice that Lemma 3.9 implies that if |

*u*'|

_{0} ≥

*χ*(Λ), then the length of each interval of

*W*_{Λ}(

*u*) is less than

$\frac{2}{\Lambda}$ for 0 ≤

*λ* ≤ Λ. This together with (3.16) and Lemma 3.10 imply that there exists no solution (

*λ*,

*u*) of (3.8), which satisfies either

- (a)
0 ≤ *λ* ≤ Λ and |*u*'|_{0} = 2*χ*(Λ) or

- (b)
*λ* = Λ and *χ*(Λ) ≤ |*u*'|_{0} ≤ 2*χ*(Λ).

Now, let us denote

$\begin{array}{lll}\hfill B\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}\left\{\left(\lambda ,u\right):0\le \lambda \le \Lambda ,\chi \left(\lambda \right)\le \phantom{\rule{2.77695pt}{0ex}}|{u}^{\prime}{|}_{0}\phantom{\rule{2.77695pt}{0ex}}\le 2\chi \left(\Lambda \right)\right\},\phantom{\rule{2em}{0ex}}& \hfill \\ \hfill {D}_{1}\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}\left\{\left(\lambda ,u\right):0\le \lambda \le \Lambda ,\phantom{\rule{2.77695pt}{0ex}}|{u}^{\prime}{|}_{0}\phantom{\rule{2.77695pt}{0ex}}=\chi \left(\lambda \right)\right\},\phantom{\rule{2em}{0ex}}& \hfill \\ \hfill {D}_{2}\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}\left\{\left(0,u\right):2\chi \left(0\right)\le \phantom{\rule{2.77695pt}{0ex}}|{u}^{\prime}{|}_{0}\phantom{\rule{2.77695pt}{0ex}}\le 2\chi \left(\Lambda \right)\right\}.\phantom{\rule{2em}{0ex}}& \hfill \\ \hfill \text{(4)}\end{array}$

It follows from Lemma 3.5 that

${\mathcal{C}}_{k}^{\nu}$ "enters"

*B* through the set

*D*_{1}, while from Lemma 3.7,

${\mathcal{C}}_{k}^{\nu}\cap B\subset \mathbb{R}\times {\Gamma}_{k}^{\nu}$. Thus, by Lemma 3.6 and the fact

$|u{|}_{0}\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}|{u}^{\prime}{|}_{0},$

${\mathcal{C}}_{k}^{\nu}$ must "leave"

*B*. (Suppose, on the contrary that

${\mathcal{C}}_{k}^{\nu}$ does not "leave"

*B*, then

$|u{|}_{0}\phantom{\rule{2.77695pt}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}|{u}^{\prime}{|}_{0}\phantom{\rule{2.77695pt}{0ex}}\le 2\chi \left(\Lambda \right),$

which contradicts the fact that ${\mathcal{C}}_{k}^{\nu}$ joins (*μ*_{
k
} , 0) to infinity in ℝ × *E*.) Since ${\mathcal{C}}_{k}^{\nu}$ is connected, it must intersect *∂B*. However, Lemmas 3.8-3.10 (together with (3.16)) show that the only portion of *∂B* (other than *D*_{1}), which ${\mathcal{C}}_{k}^{\nu}$ can intersect is *D*_{2}. Thus, there exists a point $\left(0,\phantom{\rule{2.77695pt}{0ex}}{u}_{k}^{\nu}\right)\in {\mathcal{C}}_{k}^{\nu}\cap {D}_{2}$, and clearly ${u}_{k}^{\nu}$ provides the desired solution of (1.1)-(1.2).