Let us start with the problem (3) with *λ* = *λ*_{1} and *h* = 0, i.e.

$$\begin{array}{}\left\{\begin{array}{rl}& {u}^{(4)}(x)-{\lambda}_{1}u(x)=f(x,u(x)),x\in (0,1),\\ & u(0)=u(1)={u}^{\prime}(0)={u}^{\prime}(1)=0.\end{array}\right.\end{array}$$(6)

#### Definition 3.1.

*A function α* ∈ *C*^{4}[0,1] *is said to be a lower solution of problem (6) if*

$$\begin{array}{}{\alpha}^{(4)}(x)-{\lambda}_{1}\alpha (x)\le f(x,\alpha (x))\phantom{\rule{thinmathspace}{0ex}}\text{for}\phantom{\rule{thinmathspace}{0ex}}x\in (0,1),\end{array}$$(7)

and

$$\begin{array}{}\alpha (0)=\alpha (1)=0,{\alpha}^{\prime}(0)={\alpha}^{\prime}(1)=0.\end{array}$$

*Similarly, an upper solution β* ∈ *C*^{4}[0,1] *is defined by reversing the inequality in (7). Such a lower or upper solution is called \itstrict if the inequality is strict for x* ∈ (0,1).

#### Lemma 3.2.

*Assume that f* : [0,1] × ℝ → ℝ *is a continuous function and*

$$\begin{array}{r}\underset{\mathrm{s}\to -\mathrm{\infty}}{lim\u2006inf}\mathrm{f}(\mathrm{x},\mathrm{s})\ge \underset{\_}{\mathrm{c}}(\mathrm{x})\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}uniformly\phantom{\rule{thickmathspace}{0ex}}for\phantom{\rule{thickmathspace}{0ex}}\end{array}x\in (0,1),$$(8)

*where* *c* − ∈ *L*^{1} (0,1) *and satisfies* $\begin{array}{}{\int}_{0}^{1}\underset{\_}{c}(x){\phi}_{1}(x)dx>0\end{array}$. *Then there exist R* > 0 *and d* ∈ *C*[0,1] *with* $\begin{array}{}{\int}_{0}^{1}d(x){\phi}_{1}(x)dx>0\end{array}$ such that

$$\begin{array}{}f(x,u)\ge d\end{array}$$

*for all x* ∈ (0,1)*, whenever u* ≤ − *Rφ*_{1} in (0,1).

#### Proof

Let *c*_{ε} = *c* − *ε*, where *ε* > 0 is small enough to keep $\begin{array}{}{\int}_{0}^{1}{c}_{\epsilon}(x){\phi}_{1}(x)dx>0\end{array}$. Then we can assume that the strict inequality holds in (8) with *c* (*x*) = *c*_{ε}(*x*). And subsequently, there exists *R*_{0} > 0 such that *s* ≤ − *R*_{0} implies

$$\begin{array}{}f(x,s)>\underset{\_}{c}(x)\end{array}$$(9)

for *x* ∈ (0,1).

Since *f* is bounded on (0,1) × *I*, where *I* ⊂ R is a bounded interval. Combining this fact with above (9) show that there exists *m* ∈ *L*_{1}(0,1) such that

$$\begin{array}{}f(x,s)\ge m(x)\end{array}$$

for all *s* ≤ 0 and *x* ∈ (0,1).

Let *a*, *b* ∈ ℝ with 0 < *a* < *b* < 1. We can choose *a*, *b* so large such that

$$\begin{array}{}|{\int}_{0}^{a}(m(x)-\underset{\_}{c}(x)){\phi}_{1}(x)dx+{\int}_{b}^{1}(m(x)-\underset{\_}{c}(x)){\phi}_{1}(x)dx|<{\int}_{0}^{1}\underset{\_}{c}(x){\phi}_{1}(x)dx.\end{array}$$

Therefore, it follows from the fact

$$\begin{array}{}{\int}_{a}^{b}\underset{\_}{c}(x){\phi}_{1}(x)dx+{\int}_{0}^{a}m(x){\phi}_{1}(x)dx& +{\int}_{b}^{1}m(x){\phi}_{1}(x)dx={\int}_{0}^{1}\underset{\_}{c}(x){\phi}_{1}(x)dx\\ & +{\int}_{0}^{a}(m(x)-\underset{\_}{c}(x)){\phi}_{1}(x)dx+{\int}_{b}^{1}(m(x)-\underset{\_}{c}(x)){\phi}_{1}(x)dx\end{array}$$

that

$$\begin{array}{}{\int}_{a}^{b}\underset{\_}{c}(x){\phi}_{1}(x)dx+{\int}_{0}^{a}m(x){\phi}_{1}(x)dx+{\int}_{b}^{1}m(x){\phi}_{1}(x)dx>0.\end{array}$$

Let us define *d*_{0} :(0,1) → ℝ by setting

$$\begin{array}{}{d}_{0}(x)=\left\{\begin{array}{rl}\underset{\_}{c}(x),& x\in (a,b),\\ m(x),& x\in (0,1)\mathrm{\setminus}(a,b).\end{array}\right.\end{array}$$

Observe that *d*_{0} ∈ *L*_{1}(0,1) and $\begin{array}{}{\int}_{0}^{1}{d}_{0}(x){\phi}_{1}(x)dx>0.\end{array}$

Let *x* ∈ (0,1) and *s* ∈ R such that *s* ≤ − *Rφ*_{1}(*x*). We claim that *f*(*x*, *s*) ≥ *d*(*x*).

Indeed, for *x* ∈ (*a*, *b*), since $\begin{array}{}\underset{x\in [a,b]}{min}{\phi}_{1}(x)>0\end{array}$, then there exists *R* > 0 such that *Rφ*_{1} ≥ *R*_{0} for *x* ∈ (*a*, *b*).

Therefore, we have *s* ≤ − *R*_{0}, and by (9), *f*(*x*, *s*) > *c* − (*x*). For *x* ∈ (0,1)∖(*a*, *b*), since *s* ≤ 0, we conclude that *f*(*x*, *s*) ≥ *m*(*x*). The conclusion now follows by taking a *d* ∈ *C*[0,1] with *d*_{0}(*x*) ≥ *d*(*x*), *x* ∈ (0,1) but still satisfies $\begin{array}{}{\int}_{0}^{1}d(x){\phi}_{1}(x)dx>0\end{array}$.

#### Lemma 3.3.

*Let the assumptions of Lemma 3.2 be satisfied. Then (6) has a strict lower solution α with α* < 0 *for all x* ∈ (0,1) *and such that u* ≥ *α for all possible solutions u of (6)*.

#### Proof

We divide the proof into two steps.

(i) Let *R* > 0 and *d* = *d*(*x*) be as in Lemma 3.2, such that $\begin{array}{}{\int}_{0}^{1}d(x){\phi}_{1}(x)dx>0\end{array}$ and *f*(*x*, *u*) ≥ *d* whenever *u* ≤ − *Rφ*_{1}.

Consider the linear problem

$$\begin{array}{}\left\{\begin{array}{rl}& {u}^{(4)}(x)-{\lambda}_{1}u(x)=d(x)-({\int}_{0}^{1}d(x){\phi}_{1}(x)dx){\phi}_{1}(x),x\in (0,1),\\ & u(0)=u(1)={u}^{\prime}(0)={u}^{\prime}(1)=0.\end{array}\right.\end{array}$$(10)

It’s worth pointing out that the right-hand member $\begin{array}{}v:=d(x)-({\int}_{0}^{1}d(x){\phi}_{1}(x)dx){\phi}_{1}(x)\end{array}$ satisfies the orthogonality condition $\begin{array}{}{\int}_{0}^{1}v(x){\phi}_{1}(x)dx=0\end{array}$. Then any *α* = *sφ*_{1}(*x*)+*α*_{0}, *s* ∈ ℝ is a solution of (10), where *α*_{0} is the unique solution of the problem

$$\begin{array}{}\left\{\begin{array}{rl}& {\alpha}_{0}^{(4)}(x)-{\lambda}_{1}{\alpha}_{0}(x)=v(x),x\in (0,1),\\ & {\alpha}_{0}(0)={\alpha}_{0}(1)={\alpha}_{0}^{\prime}(0)={\alpha}_{0}^{\prime}(1)=0,\end{array}\right.\end{array}$$

and satisfies $\begin{array}{}{\int}_{0}^{1}({\alpha}_{0}^{(4)}(x)-{\lambda}_{1}{\alpha}_{0}(x)){\phi}_{1}(x)dx=0.\end{array}$

Therefore, there exist constants *a*, *A* such that *aφ*_{1} ≤ *α*_{0} ≤ *Aφ*_{1} for *x* ∈ (0,1), and accordingly, taking *s* negative sufficiently large (precisely, *s* < − (*R* + *A*)), we can arrange such that *α* = *sφ*_{1}(*x*) + *α*_{0} < −*Rφ*_{1} for *x* ∈ (0,1). But then *f*(*x*, *α*) ≥ *d* for all *x* ∈ (0,1), and since $\begin{array}{}{\int}_{0}^{1}d(x){\phi}_{1}(x)dx>0\end{array}$, we conclude that

$$\begin{array}{}\left\{\begin{array}{rl}& {\alpha}^{(4)}(x)-{\lambda}_{1}\alpha (x)<d(x)\le f(x,\alpha (x)),x\in (0,1),\\ & \alpha (0)=\alpha (1)={\alpha}^{\prime}(0)={\alpha}^{\prime}(1)=0.\end{array}\right.\end{array}$$(11)

This implies *α* is a strict lower solution of (6).

(ii) Let *u* be a solution of (6). To prove that *u* ≥ *α* we set *w* = *u* − *α* and by (11), we observe that *w* satisfies

$$\begin{array}{}\left\{\begin{array}{rl}& {w}^{(4)}(x)-{\lambda}_{1}w(x)>f(x,u)-d(x),x\in (0,1),\\ & w(0)=w(1)={w}^{\prime}(0)={w}^{\prime}(1)=0.\end{array}\right.\end{array}$$(12)

Multiplying both sides of the equation in (12) by *v* ∈ {*v* ∈ *C*_{4}[0,1] : *v* ≥ 0, *v*(0) = *v*(1) = *v*′(0) = *v*′(1) = 0} and integrating from 0 to 1, we get that for all *x* ∈ (0,1),

$$\begin{array}{}{\int}_{0}^{1}{w}^{\u2033}(x){v}^{\u2033}(x)dx-{\lambda}_{1}{\int}_{0}^{1}w(x)v(x)dx>{\int}_{0}^{1}[f(x,u)-d(x)]v(x)dx.\end{array}$$(13)

Let *w*_{+} = max{*w*,0} and *w*_{−} = max{−*w*, 0} denote the positive and negative parts of *w*.

We claim *u* ≥ *α*, i.e. *w*_{−} = 0. Assume on the contrary that w_{−} ≠ 0, then choosing *v* = *w*_{−} in (13) we obtain

$$\begin{array}{}\begin{array}{rl}-{\int}_{0}^{1}({w}^{-}(x){)}^{\u20332}dx+{\lambda}_{1}{\int}_{0}^{1}({w}^{-}(x){)}^{2}dx& >{\int}_{0}^{1}[f(x,u)-d(x)]{w}^{-}(x)dx.\end{array}\end{array}$$

But *w*_{−}(*x*) > 0 means *u*(*x*) < *α*(*x*), which in turn implies *u*(*x*) < − *Rφ*_{1} and thus *f*(*x*, *u*) ≥ *d*(*x*). Therefore, the last integral is nonnegative and

$$\begin{array}{}{\int}_{0}^{1}({w}^{-}(x){)}^{\u20332}dx<{\lambda}_{1}{\int}_{0}^{1}({w}^{-}(x){)}^{2}dx.\end{array}$$

However, this contradicts the one-dimensional Poincaré inequality

$$\begin{array}{}{\int}_{0}^{1}({w}^{-}(x){)}^{\u20332}dx\ge {\lambda}_{1}{\int}_{0}^{1}({w}^{-}(x){)}^{2}dx.\end{array}$$ □

Now, we extend the above result to the problem

$$\begin{array}{}\left\{\begin{array}{rl}& {u}^{(4)}(x)-\lambda u(x)=f(x,u(x)),x\in (0,1),\\ & u(0)=u(1)={u}^{\prime}(0)={u}^{\prime}(1)=0.\end{array}\right.\end{array}$$(14)

#### Lemma 3.4.

*Under the same assumptions as in Lemma 3.2, there exists a strict lower solution α* < 0 *of (14) such that u* ≥ *α for all possible solutions u of (14) for λ* ≤ *λ*_{1}.

#### Proof

Let *α* be the lower solution for the (14) with *λ* = *λ*_{1} determined in Lemma 3.3. Since *α* < − *Rφ*_{1} < 0 for all *x* ∈ (0,1), we have from (11) if *λ* ≤ *λ*_{1},

$$\begin{array}{}{\alpha}^{(4)}(x)-{\lambda}_{1}\alpha (x)<d(x)+(\lambda -{\lambda}_{1})\alpha (x)\le f(x,\alpha )+(\lambda -{\lambda}_{1})\alpha (x)\end{array}$$

for all *x* ∈ (0,1), and

$$\begin{array}{}\alpha (0)=\alpha (1)={\alpha}^{\prime}(0)={\alpha}^{\prime}(1)=0,\end{array}$$

therefore, *α* is a strict lower solution of (14) for all *λ* ≤ *λ*_{1}. Moreover, for any solution *u* of (14), setting *w* = *u* − *α* we have

$$\begin{array}{}{w}^{(4)}(x)-\lambda w(x)>f(x,u)-d(x)\end{array}$$

for all *x* ∈ (0,1), and

$$\begin{array}{}w(0)=w(1)={w}^{\prime}(0)={w}^{\prime}(1)=0.\end{array}$$

Now to prove that *w* ≥ 0, one has only to remark that the argument in the proof of Lemma 3.3, part (ii), works equally well for any *λ* ≤ *λ*_{1}.

A result similar to the Lemma 3.4 holds for positive strict upper solution of (14) if we impose a symmetric condition on *f* = *f*(*x*, *s*) as *s* → + ∞.

#### Lemma 3.5.

*Assume that f* : [0,1] × ℝ → ℝ is a continuous function and

$$\begin{array}{}\underset{s\to +\mathrm{\infty}}{lim\u2006sup}f(x,s)\le \overline{c}(x)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{uniformly for}\phantom{\rule{thinmathspace}{0ex}}x\in (0,1),\end{array}$$(15)

*where c*¯ ∈ *L*_{1}(0,1) *and satisfies* $\begin{array}{}{\int}_{0}^{1}\overline{c}(x){\phi}_{1}(x)dx<0.\end{array}$ *Then there exists a strict upper solution β* > 0 *of (14) such that u* ≤ *β for all possible solutions u of (14) for λ* ≤ *λ*_{1}.

At this point, although existence of a strict lower solution *α* < 0 and an strict upper solution *β* > 0 of (14) have been obtained for all *λ* ≤ *λ*_{1}, this is no longer true for existence of a solution of (14) in the sector enclosed by [*α*(*x*), *β*(*x*)]. Hence, we also assume that there exists $\begin{array}{}\mu \in (0,\frac{3{\lambda}_{1}}{4}]\end{array}$, such that

$$\begin{array}{}f(x,{u}_{2})+\mu {u}_{2}-(f(x,{u}_{1})+\mu {u}_{1})\ge 0\phantom{\rule{thinmathspace}{0ex}}\text{for}\phantom{\rule{thinmathspace}{0ex}}\alpha (x)\le {u}_{1}\le {u}_{2}\le \beta (x),\text{and}\phantom{\rule{thinmathspace}{0ex}}x\in [0,1].\end{array}$$(16)

Based on Corollary 3.3 of [27], it allows us to present a maximum principle for the operator *L*_{λ}:D → *C*[0,1] defined by

$$\begin{array}{}({L}_{\lambda}u)(x)={u}^{(4)}(x)-\lambda u(x),x\in [0,1],\end{array}$$(17)

where *u* ∈ D and D = {*u* ∈ *C*_{4}[0,1] : *u*(0) = *u*(1) = *u*′(0) = *u*′(1) = 0}.

#### Lemma 3.6.

Let *λ* ∈ [0, *λ*_{1}). If *u* ∈ *D* satisfies *L*_{λ} *u* ≥ 0, then *u* ≥ 0 in [0,1].

#### Theorem 3.7.

*Let (8) and (15) hold. Then there exist **α* and *β*, strict lower and upper solutions, respectively, for the problem (14) which satisfy

$$\begin{array}{}\alpha (x)<0<\beta (x)\end{array}$$

*for all x* ∈ [0,1] *and λ* ≤ *λ*_{1}, *and if f satisfies (16), then (14) has a solution u such that*

$$\begin{array}{}\alpha (x)\le u(x)\le \beta (x)\end{array}$$

*for all x* ∈ [0,1] *and λ* ∈ [3*λ*_{1}/4, *λ*_{1}].

#### Proof

Let *α* < 0 and *β* > 0 be as in Lemma 3.4 and Lemma 3.5. Let *λ* = *λ*_{1} + *ε* and *ε* ∈ [−*λ*_{1}/4,0].

Let us consider the auxiliary problem

$$\begin{array}{}\left\{\begin{array}{rl}& {u}^{(4)}(x)-[({\lambda}_{1}+\epsilon )-\mu ]u(x)=f(x,\eta (x))+\mu \eta (x),x\in (0,1),\\ & u(0)=u(1)={u}^{\prime}(0)={u}^{\prime}(1)=0,\end{array}\right.\end{array}$$(18)

with *η* ∈ *C*[0,1] and *μ* ∈ (0,3*λ*_{1}/4] is a fixed constant. For *ε* = 0, (18) reduces to (14).

Define *T*_{λ} : *C*[0,1] → *C*[0,1] by

$$\begin{array}{}{T}_{\lambda}\eta =u,\end{array}$$

where *u* is the unique solution of (18). Clearly the operator *T*_{λ} is compact.

*Step 1*. We show *T*_{λ}*C* ⊆ *C*.

Here *C* = {*η* ∈ *C*[0,1] : *α* ≤ *η* ≤ *β*} is a nonempty bounded closed subset in *C*[0,1].

In fact, for *ξ* ∈ *C*, set *y* = *T*_{λ}*ξ*. From the definition of *α* and *C*, and using (16), we have that

$$\begin{array}{}\begin{array}{rl}& (y-\alpha {)}^{(4)}(x)-[({\lambda}_{1}+\epsilon )-\mu ](y-\alpha )(x)\\ \ge & (f(x,\xi (x))+\mu \xi (x))-(f(x,\alpha (x))+\mu \alpha (x))\\ \ge & 0,\end{array}\end{array}$$

and

$$\begin{array}{}(y-\alpha )(0)=(y-\alpha )(1)=(y-\alpha {)}^{\prime}(0)=(y-\alpha {)}^{\prime}(1)=0.\end{array}$$

Since (*λ*_{1} + *ε*) − *μ* ∈ [0, *λ*_{1}), therefore, by Lemma 3.6, we have *y* ≥ *α*. Analogously, we can show that *y* ≤ *β*.

*Step 2. T*_{λ} : *C*[0,1] → *C*[0,1] is nondecreasing.

Let *η*_{1}, *η*_{2} ∈ *C*[0,1] with *η*_{1} ≤ *η*_{2} and put *u*_{i} = *T*_{λ}*η*_{i}, *i* = 1,2. Then from (16), *w* = *u*_{2} − *u*_{1}satisfies

$$\begin{array}{}\left\{\begin{array}{rl}& {w}^{(4)}(x)-[({\lambda}_{1}+\epsilon )-\mu ]w(x)=(f(x,{\eta}_{2}(x))+\mu {\eta}_{2}(x))-(f(x,{\eta}_{1}(x))+\mu {\eta}_{1}(x))\ge 0,x\in (0,1),\\ & w(0)=w(1)={w}^{\prime}(0)={w}^{\prime}(1)=0.\end{array}\right.\end{array}$$(19)

From Lemma 3.6, it follows that *w* ≥ 0 and hence *u*_{1} ≤ *u*_{2}.

*Step 3*. *α* ≤ *T*_{λ}*α* and *T*_{λ}*β* ≤ *β*.

Since *α* is a lower solution we have that

$$\begin{array}{}\begin{array}{rl}({T}_{\lambda}\alpha {)}^{(4)}(x)-[({\lambda}_{1}+\epsilon )-\mu ]({T}_{\lambda}\alpha )(x)& =f(x,\alpha (x))+\mu \alpha (x)\\ & \ge {\alpha}^{(4)}(x)-({\lambda}_{1}+\epsilon )\alpha (x)+\mu \alpha (x)\\ & ={\alpha}^{(4)}(x)-[({\lambda}_{1}+\epsilon )-\mu ]\alpha (x).\end{array}\end{array}$$(20)

Thus *w* = *T*_{λ}*α* − *α* satisfies that

$$\begin{array}{}\left\{\begin{array}{rl}& {w}^{(4)}(x)-[({\lambda}_{1}+\epsilon )-\mu ]w(x)\ge 0,x\in (0,1),\\ & w(0)=w(1)={w}^{\prime}(0)={w}^{\prime}(1)=0,\end{array}\right.\end{array}$$(21)

and then by Lemma 3.6 we deduce that *w* = *T*_{λ}*α* − *α* ≥ 0. Analogously, we can prove that *T*_{λ}*β* ≤ *β*.

The interval [*α*, *β*] is a closed, convex, bounded and nonempty subset of the Banach space *C*[0,1]. Then by *Step 1* we can apply Schauder’s fixed point theorem to obtain the existence of a fixed point of *T*_{λ}, which obviously is a solution of problem (14) in [*α*, *β*].

#### Lemma 3.8.

*Assume that the assumptions of Theorem 3.7 are satisfied. Let u be a solution of (14). Then for any λ*_{0} ∈ (3*λ*_{1}/4, *λ*_{1})*, there exists ρ* > 0 *such that* ∥*u*∥_{C3} < *ρ for λ* ∈ [*λ*_{0}, *λ*_{1}].

#### Proof

Let

$$\begin{array}{}{M}_{1}:=sup\{|\lambda u+f(x,u)|:{\lambda}_{0}\le \lambda \le {\lambda}_{1},\alpha (x)\le u(x)\le \beta (x),x\in [0,1]\}.\end{array}$$(22)

By Theorem 3.7, we conclude *M*_{1} is finite and then by (14) we know |*u*^{(4)}| ≤ *M*_{1} for all *x* ∈ (0,1).

Combine the boundary conditions

$$\begin{array}{}u(0)=u(1)={u}^{\prime}(0)={u}^{\prime}(1)=0,\end{array}$$

we know that there exists *t*_{0} ∈ (0,1) such that *u*′′′(*t*_{0}) = 0, see [23, P. 1212], and subsequently,

$$\begin{array}{}|{u}^{\u2034}(t)|\le {\int}_{{t}_{0}}^{t}|{u}^{(4)}(s)|ds\le {M}_{1}.\end{array}$$

Using the similar argument of above, we can prove that there exist constants *M*_{2}, *M*_{3} and *M*_{4} such that for all possible solutions *u* of (14) for *λ*_{0} ≤ *λ* ≤ *λ*_{1},

$$\begin{array}{}|{u}^{\u2033}(t)|\le {M}_{2},|{u}^{\prime}(t)|\le {M}_{3},|u(t)|\le {M}_{4}\end{array}$$

for *t* ∈ [0,1]. Clearly, ∥*u*∥_{C3} < *ρ* for *λ*_{0} ≤ *λ* ≤ *λ*_{1} as long as *ρ*:= max{*M*_{1}, *M*_{2}, *M*_{3}, *M*_{4}} + 1. □

#### Lemma 3.9.

*Assume that the assumptions of Theorem 3.7 are satisfied. Let*

$$\begin{array}{}{\mathrm{\Omega}}_{\alpha ,\beta}=\{u\in {C}^{4}[0,1]:\alpha (x)<u(x)<\beta (x),x\in [0,1]\},\end{array}$$

*and*

$$\begin{array}{}\mathrm{\Omega}={B}_{\rho}\cap {\mathrm{\Omega}}_{\alpha ,\beta},\end{array}$$

*where B*_{ρ} = {*u* ∈ *C*^{3}[0,1]:∥*u*∥_{C3} < *ρ*} *and ρ is given in Lemma 3.8. Then there exists λ*_{0} ∈ (3*λ*_{1}/4, *λ*_{1})*, such that*

$$\begin{array}{}\mathrm{deg}[I-{T}_{\lambda},\mathrm{\Omega},0]=1\end{array}$$

*for λ* ∈ [*λ*_{0}, *λ*_{1}].

#### Proof

For any *μ* ∈ [0,1], consider now the homotopy

$$\begin{array}{}\left\{\begin{array}{rl}& {u}^{(4)}(x)=\mu (\lambda u(x)+f(x,u(x))),x\in (0,1),\\ & u(0)=u(1)={u}^{\prime}(0)={u}^{\prime}(1)=0.\end{array}\right.\end{array}$$(23)

Reasoning as in Lemma 3.8, we observe that all possible solutions of the problems (23) satisfy

$$\begin{array}{}\parallel u{\parallel}_{{C}^{3}}<\rho \end{array}$$

for any *λ* ∈ [*λ*_{0}, *λ*_{1}]. Therefore, if we write (23) as $\begin{array}{}u=\mu \hat{T}(u)\end{array}$, and set

$$\begin{array}{}H(\mu ,u)=u-\mu \hat{T}(u).\end{array}$$

Then H(*μ*, u) ≠ 0 for all *μ* ∈ [0,1] and ∥*u*∥_{C3} = *ρ*, so that by the homotopy invariance of the Leray-Schauder degree

$$\begin{array}{}\mathrm{deg}[I-{T}_{\lambda},{B}_{\rho},0]=\mathrm{deg}[I-\hat{T},{B}_{\rho},0]=\mathrm{deg}[I,{B}_{\rho},0]=1.\end{array}$$(24)

On the other hand, by Theorem 3.7, all zeros of *I* − *T*_{λ} belong to Ω_{α, β}. Therefore, if *ρ* is large enough, then by the excision property of the Leray-Schauder degree, we have

$$\begin{array}{}\\ deg[I-{T}_{\lambda},\mathrm{\Omega},0]=\mathrm{deg}[I-{T}_{\lambda},{\mathrm{\Omega}}_{\alpha ,\beta},0]=\mathrm{deg}[I-{T}_{\lambda},{B}_{\rho},0]=1.\end{array}$$ □

#### Lemma 3.10.

*Let* Ω = Ω_{α, β} ∩ *B*_{ρ}. Then there exists *δ* > 0 such that

$$\begin{array}{}\mathrm{deg}[I-{T}_{\lambda},\mathrm{\Omega},0]=1\end{array}$$

*for all λ* ∈ [*λ*_{1}, *λ*_{1} + *δ*].

#### Proof

The proof is trivial, so we omit it. □

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