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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 16, Issue 1

# Nonlinear elastic beam problems with the parameter near resonance

Man Xu
/ Ruyun Ma
Published Online: 2018-10-29 | DOI: https://doi.org/10.1515/math-2018-0097

## Abstract

In this paper, we consider the nonlinear fourth order boundary value problem of the form

$u(4)(x)−λu(x)=f(x,u(x))−h(x),x∈(0,1),u(0)=u(1)=u′(0)=u′(1)=0,$

which models a statically elastic beam with both end-points cantilevered or fixed. We show the existence of at least one or two solutions depending on the sign of λλ1, where λ1 is the first eigenvalue of the corresponding linear eigenvalue problem and λ is a parameter. The proof of the main result is based upon the method of lower and upper solutions and global bifurcation techniques.

MSC 2010: 34B27; 34B15; 34B05; 34A40

## 1 Introduction

The existence and multiplicity of solutions to the nonlinear second order ordinary differential equation boundary value problem with the parameter near resonance of the form

$−u″(x)−λu(x)=f(x,u(x))−h(x),x∈(0,1),u(0)=u(1)=0$(1)

have been extensively studied by many authors, see Mawhin and Schmitt et al. [1-3], Iannacci and Nkashama [4], Costa and Goncalves [5], Ambrosetti and Mancini [6], Fonda and Habets [7], Các [8], Ahmad [9] and Ma [10], and the references therein. In particular, Chiappinelli, Mawhin and Nugari [1] proved that there exists ν>0 such that problem (1), with λ near λ1, had at least one solution for λλ1 and two solutions for λ1 < λ < λ1 + ν under the assumption $\begin{array}{}\underset{s\to +\mathrm{\infty }}{lim}\frac{f\left(x,s\right)}{s}=0\end{array}$ and a Landesman-Lazer type condition.

However, relatively little is known about the related work on the existence of solutions of the fourth order boundary value problems. The likely reason is that fewer techniques are available for the fourth order operators and the results known for the second order case do not necessarily hold for the corresponding fourth order problem. A natural motivation for studying higher order boundary value problems exists in their applications. For example, it is well-known that the deformation of an elastic beam in equilibrium state, whose both end-points are cantilevered or fixed, can be described by the fourth order boundary value problem

$u(4)(x)=f(x,u(x),u″(x)),x∈(0,1),u(0)=u(1)=u′(0)=u′(1)=0,$(2)

where f : [0,1]× ℝ × ℝ → ℝ is a continuous function, see [11]. There are some papers discussing the existence of solutions of the problem by using various methods, such as the lower and upper solution method, the Leray-Schauder continuation method, fixed-point theory, and the monotone iterative method, see Rynne [12], Korman [13], Gupta and Kwong [14], Jurkiewicz [15], Vrabel [16], Cabada et al. [17], Bai and Wang [18] and Ma et al. [19], and the references therein.

But to the best of our knowledge, the analogue of (1) has not been established for fourth order boundary value problems.

The purpose of this paper is to establish the similar existence result for the corresponding fourth order analogue of (1) of the form

$u(4)(x)−λu(x)=f(x,u(x))−h(x),x∈(0,1),u(0)=u(1)=u′(0)=u′(1)=0,$(3)

where λ is a parameter, hC[0,1] and fC([0,1] × ℝ, ℝ). The proof of our main result is based upon the method of lower and upper solutions and global bifurcation techniques.

Particular significance in these points lie in the fact that for a second-order differential equation, with Neumann or Dirichlet boundary conditions, the existence of a lower solution α and an upper solution β with α(x) ≤ β(x) in [0,1] can ensure the existence of solutions in the order interval [α(x), β(x)], see Coster and Habets [20]. However, this result is not true for fourth-order boundary value problems, see the counterexample in Cabada, Cid and Sanchez [17, P. 1607]. Thus, new challenges are faced and innovation is required.

To apply the bifurcation techniques to study the existence of solutions of (3), we state and prove a spectrum result for fourth order linear eigenvalue problem

$u(4)(x)=λu(x),x ∈ (0,1),u(0)=u(1)=u′(0)=u′(1)=0.$(4)

More precisely, we can show that the eigenvalues of (4) form a sequence

$0<λ1<λ2<λ3<⋯→+∞.$(5)

Moreover, for each j ∈ ℕ, λj (λj = m4j, mjis the simple root of the equation cos m cosh m − 1 = 0) is simple. In particular, λ1 ≈ (4.73004)4 ≈ 500.564 is simple and the corresponding eigenspace is spanned by $\begin{array}{}{\phi }_{1}\left(x\right)=\mathrm{sin}{m}_{1}x-\mathrm{sinh}{m}_{1}x+\frac{\mathrm{sin}{m}_{1}-\mathrm{sinh}{m}_{1}}{\mathrm{cos}{m}_{1}-\mathrm{cosh}{m}_{1}}\left(\mathrm{cosh}{m}_{1}x-\mathrm{cos}{m}_{1}x\right)\end{array}$ with φ1(x) > 0 in (0, 1), and we normalize by $\begin{array}{}{\int }_{0}^{1}\left({\phi }_{1}\left(x\right){\right)}^{2}dx=1\end{array}$.

We shall make the following assumptions:

(H1) f : [0,1] × ℝ → ℝ is a continuous function and satisfies

$f(x,u2)−f(x,u1)+34λ1(u2−u1)≥0for−∞

(H2) hC[0,1] and satisfies

$∫01C(x)φ1(x)dx<∫01h(x)φ1(x)dx<∫01c(x)φ1(x)dx,$

where c, CL1(0,1) with $\begin{array}{}\underset{s\to -\mathrm{\infty }}{lim inf}f\left(x,s\right)\ge c\left(x\right),\underset{s\to +\mathrm{\infty }}{lim sup}f\left(x,s\right)\le C\left(x\right)\end{array}$ uniformly for x ∈ (0,1);

(H3)

$lims→+∞f(x,s)s=0uniformly forx ∈ [0,1].$

The main result of this paper is the following

#### Theorem 1.1.

Assume (Hl)-(H3) hold. Then there exist δ1 > 0 and λ0 ∈ (3λ1/4, λ1) such that (3) has at least one or at least two solutions according to λ0λλ1 or λ1 < λλ1 + δ1. Moreover, one of these two solutions is a positive solution.

We first prove the existence of a lower solution α and an upper solution β of (3) for λλ1, which are well ordered, that is αβ (in fact α < 0 and β > 0) under condition (H2). But this is not enough to ensure the existence of a solution in the order interval [α, β], so we also make the assumption (H1). It is precisely this circumstance which gives a priori bound for λλ1. Once this is done, since λ1 is simple and the assumption (H3) hold, the Rabinowitz global bifurcation techniques [21] can be used to obtain the second solution following very much the same lines as in [10]. More precisely, there exists an unbounded connected component Σ that is bifurcating from infinity. Since we have established a prior bound for solutions of (3) when λλ1, the connected component Σ, must do so for λ > λ1.

For other results concerning the existence of solutions of the nonlinear fourth order differential or difference equations via the bifurcation techniques, we refer the reader to [22, 23].

The rest of the paper is arranged as follows. In Section 2, we investigate the spectrum structure of the linear eigenvalue problem (4). In Section 3, we give some preliminary results and develop the method of lower and upper solutions for (3). Finally Section 4 is devoted to proving our main result by the well-known Rabinowitz bifurcation techniques and the lower and upper solutions arguments. We also give some examples to illustrate our main result.

## 2 Spectrum of the linear eigenvalue problem

In this section we state a spectrum result of the linear eigenvalue problem (4).

#### Lemma 2.1. ([24, Lemma 1]).

The equation

$cos⁡mcosh⁡m−1=0,m ∈ R+$

has infinitely many simple roots

$0

Moreover,

$m2k−1 ∈ ((2k−12)π,2kπ),m2k ∈ (2kπ,(2k+12)π)$

for k ∈ ℕ.

It is well known that linear eigenvalue problem (4) is completely regular Sturmian system and therefore, has infinitely many simple and positive eigenvalues 0 < λ1 < λ2 < ⋯ → + ∞. The eigenfunction φj, corresponding to λj, has exactly j − 1 simple zeros in (0,1). The eigenvalues λk, k ∈ N are the roots of the transcendental equation cos m cosh m − 1 = 0. See Rynne [12, P. 308], Janczewsky [25] and Courant and Hilbert [26].

Moreover, we have the following

#### Lemma 2.2 ([24, Lemma 2]).

The linear eigenvalue problem (4) has infinitely many eigenvalues

$λj=mj4,j ∈ N,$

and the eigenfunction corresponding to λj is given by

$φj(x)=sin⁡mjx−sinh⁡mjx+sin⁡mj−sinh⁡mjcos⁡mj−cosh⁡mj(cosh⁡mjx−cos⁡mjx).$

Moreover, φjSj,+, where Sj,+ denote the set of uC3[0,1] such that:

1. u has only simple zeros in (0,1) and has exactly j − 1 such zeros;

2. u″″(0) > 0 and u″(1) ≠ 0.

## 3 The existence of lower and upper solutions

Let us start with the problem (3) with λ = λ1 and h = 0, i.e.

$u(4)(x)−λ1u(x)=f(x,u(x)),x ∈ (0,1),u(0)=u(1)=u′(0)=u′(1)=0.$(6)

#### Definition 3.1.

A function αC4[0,1] is said to be a lower solution of problem (6) if

$α(4)(x)−λ1α(x)≤f(x,α(x))forx∈(0,1),$(7)

and

$α(0)=α(1)=0,α′(0)=α′(1)=0.$

Similarly, an upper solution βC4[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

$lim infs→−∞f(x,s)≥c_(x)uniformlyforx∈(0,1),$(8)

where c − ∈ L1 (0,1) and satisfies $\begin{array}{}{\int }_{0}^{1}\underset{_}{c}\left(x\right){\phi }_{1}\left(x\right)dx>0\end{array}$. Then there exist R > 0 and dC[0,1] with $\begin{array}{}{\int }_{0}^{1}d\left(x\right){\phi }_{1}\left(x\right)dx>0\end{array}$ such that

$f(x,u)≥d$

for all x ∈ (0,1), whenever u ≤ − 1 in (0,1).

#### Proof

Let cε = cε, where ε > 0 is small enough to keep $\begin{array}{}{\int }_{0}^{1}{c}_{\epsilon }\left(x\right){\phi }_{1}\left(x\right)dx>0\end{array}$. Then we can assume that the strict inequality holds in (8) with c (x) = cε(x). And subsequently, there exists R0 > 0 such that s ≤ − R0 implies

$f(x,s)>c_(x)$(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 mL1(0,1) such that

$f(x,s)≥m(x)$

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

$|∫0a(m(x)−c_(x))φ1(x)dx+∫b1(m(x)−c_(x))φ1(x)dx|<∫01c_(x)φ1(x)dx.$

Therefore, it follows from the fact

$∫abc_(x)φ1(x)dx+∫0am(x)φ1(x)dx+∫b1m(x)φ1(x)dx=∫01c_(x)φ1(x)dx+∫0a(m(x)−c_(x))φ1(x)dx+∫b1(m(x)−c_(x))φ1(x)dx$

that

$∫abc_(x)φ1(x)dx+∫0am(x)φ1(x)dx+∫b1m(x)φ1(x)dx>0.$

Let us define d0 :(0,1) → ℝ by setting

$d0(x)=c_(x),x ∈ (a,b),m(x),x ∈ (0,1)∖(a,b).$

Observe that d0L1(0,1) and $\begin{array}{}{\int }_{0}^{1}{d}_{0}\left(x\right){\phi }_{1}\left(x\right)dx>0.\end{array}$

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

Indeed, for x ∈ (a, b), since $\begin{array}{}\underset{x\in \left[a,b\right]}{min}{\phi }_{1}\left(x\right)>0\end{array}$, then there exists R > 0 such that 1R0 for x ∈ (a, b).

Therefore, we have s ≤ − R0, 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 dC[0,1] with d0(x) ≥ d(x), x ∈ (0,1) but still satisfies $\begin{array}{}{\int }_{0}^{1}d\left(x\right){\phi }_{1}\left(x\right)dx>0\end{array}$.

#### 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 uα 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 $\begin{array}{}{\int }_{0}^{1}d\left(x\right){\phi }_{1}\left(x\right)dx>0\end{array}$ and f(x, u) ≥ d whenever u ≤ − 1.

Consider the linear problem

$u(4)(x)−λ1u(x)=d(x)−(∫01d(x)φ1(x)dx)φ1(x),x ∈ (0,1),u(0)=u(1)=u′(0)=u′(1)=0.$(10)

It’s worth pointing out that the right-hand member $\begin{array}{}v:=d\left(x\right)-\left({\int }_{0}^{1}d\left(x\right){\phi }_{1}\left(x\right)dx\right){\phi }_{1}\left(x\right)\end{array}$ satisfies the orthogonality condition $\begin{array}{}{\int }_{0}^{1}v\left(x\right){\phi }_{1}\left(x\right)dx=0\end{array}$. Then any α = 1(x)+α0, s ∈ ℝ is a solution of (10), where α0 is the unique solution of the problem

$α0(4)(x)−λ1α0(x)=v(x),x ∈ (0,1),α0(0)=α0(1)=α0′(0)=α0′(1)=0,$

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

Therefore, there exist constants a, A such that 1α01 for x ∈ (0,1), and accordingly, taking s negative sufficiently large (precisely, s < − (R + A)), we can arrange such that α = 1(x) + α0 < −1 for x ∈ (0,1). But then f(x, α) ≥ d for all x ∈ (0,1), and since $\begin{array}{}{\int }_{0}^{1}d\left(x\right){\phi }_{1}\left(x\right)dx>0\end{array}$, we conclude that

$α(4)(x)−λ1α(x)(11)

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

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

$w(4)(x)−λ1w(x)>f(x,u)−d(x),x ∈ (0,1),w(0)=w(1)=w′(0)=w′(1)=0.$(12)

Multiplying both sides of the equation in (12) by v ∈ {vC4[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),

$∫01w″(x)v″(x)dx−λ1∫01w(x)v(x)dx>∫01[f(x,u)−d(x)]v(x)dx.$(13)

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

We claim uα, i.e. w = 0. Assume on the contrary that w ≠ 0, then choosing v = w in (13) we obtain

$−∫01(w−(x))″2dx+λ1∫01(w−(x))2dx>∫01[f(x,u)−d(x)]w−(x)dx.$

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

$∫01(w−(x))″2dx<λ1∫01(w−(x))2dx.$

However, this contradicts the one-dimensional Poincaré inequality

$∫01(w−(x))″2dx≥λ1∫01(w−(x))2dx.$

Now, we extend the above result to the problem

$u(4)(x)−λu(x)=f(x,u(x)),x ∈ (0,1),u(0)=u(1)=u′(0)=u′(1)=0.$(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 uα for all possible solutions u of (14) for λλ1.

#### Proof

Let α be the lower solution for the (14) with λ = λ1 determined in Lemma 3.3. Since α < − 1 < 0 for all x ∈ (0,1), we have from (11) if λλ1,

$α(4)(x)−λ1α(x)

for all x ∈ (0,1), and

$α(0)=α(1)=α′(0)=α′(1)=0,$

therefore, α is a strict lower solution of (14) for all λλ1. Moreover, for any solution u of (14), setting w = uα we have

$w(4)(x)−λw(x)>f(x,u)−d(x)$

for all x ∈ (0,1), and

$w(0)=w(1)=w′(0)=w′(1)=0.$

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 λλ1.

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

$lim sups→+∞f(x,s)≤c¯(x)uniformly forx ∈ (0,1),$(15)

where c¯ ∈ L1(0,1) and satisfies $\begin{array}{}{\int }_{0}^{1}\overline{c}\left(x\right){\phi }_{1}\left(x\right)dx<0.\end{array}$ Then there exists a strict upper solution β > 0 of (14) such that uβ for all possible solutions u of (14) for λλ1.

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 λλ1, 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 $\begin{array}{}\mu \in \left(0,\frac{3{\lambda }_{1}}{4}\right]\end{array}$, such that

$f(x,u2)+μu2−(f(x,u1)+μu1)≥0forα(x)≤u1≤u2≤β(x),andx ∈ [0,1].$(16)

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

$(Lλu)(x)=u(4)(x)−λu(x),x ∈ [0,1],$(17)

where u ∈ D and D = {uC4[0,1] : u(0) = u(1) = u′(0) = u′(1) = 0}.

#### Lemma 3.6.

Let λ ∈ [0, λ1). If uD satisfies Lλ 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

$α(x)<0<β(x)$

for all x ∈ [0,1] and λλ1, and if f satisfies (16), then (14) has a solution u such that

$α(x)≤u(x)≤β(x)$

for all x ∈ [0,1] and λ ∈ [3λ1/4, λ1].

#### Proof

Let α < 0 and β > 0 be as in Lemma 3.4 and Lemma 3.5. Let λ = λ1 + ε and ε ∈ [−λ1/4,0].

Let us consider the auxiliary problem

$u(4)(x)−[(λ1+ε)−μ]u(x)=f(x,η(x))+μη(x),x ∈ (0,1),u(0)=u(1)=u′(0)=u′(1)=0,$(18)

with ηC[0,1] and μ ∈ (0,3λ1/4] is a fixed constant. For ε = 0, (18) reduces to (14).

Define Tλ : C[0,1] → C[0,1] by

$Tλη=u,$

where u is the unique solution of (18). Clearly the operator Tλ is compact.

Step 1. We show TλCC.

Here C = {ηC[0,1] : αηβ} is a nonempty bounded closed subset in C[0,1].

In fact, for ξC, set y = Tλξ. From the definition of α and C, and using (16), we have that

$(y−α)(4)(x)−[(λ1+ε)−μ](y−α)(x)≥(f(x,ξ(x))+μξ(x))−(f(x,α(x))+μα(x))≥0,$

and

$(y−α)(0)=(y−α)(1)=(y−α)′(0)=(y−α)′(1)=0.$

Since (λ1 + ε) − μ ∈ [0, λ1), therefore, by Lemma 3.6, we have yα. Analogously, we can show that yβ.

Step 2. Tλ : C[0,1] → C[0,1] is nondecreasing.

Let η1, η2C[0,1] with η1η2 and put ui = Tληi, i = 1,2. Then from (16), w = u2u1satisfies

$w(4)(x)−[(λ1+ε)−μ]w(x)=(f(x,η2(x))+μη2(x))−(f(x,η1(x))+μη1(x))≥0,x ∈ (0,1),w(0)=w(1)=w′(0)=w′(1)=0.$(19)

From Lemma 3.6, it follows that w ≥ 0 and hence u1u2.

Step 3. αTλα and Tλββ.

Since α is a lower solution we have that

$(Tλα)(4)(x)−[(λ1+ε)−μ](Tλα)(x)=f(x,α(x))+μα(x)≥α(4)(x)−(λ1+ε)α(x)+μα(x)=α(4)(x)−[(λ1+ε)−μ]α(x).$(20)

Thus w = Tλαα satisfies that

$w(4)(x)−[(λ1+ε)−μ]w(x)≥0,x ∈ (0,1),w(0)=w(1)=w′(0)=w′(1)=0,$(21)

and then by Lemma 3.6 we deduce that w = Tλαα ≥ 0. Analogously, we can prove that Tλββ.

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 Tλ, 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 λ0 ∈ (3λ1/4, λ1), there exists ρ > 0 such thatuC3 < ρ for λ ∈ [λ0, λ1].

#### Proof

Let

$M1:=sup{|λu+f(x,u)|:λ0≤λ≤λ1,α(x)≤u(x)≤β(x),x ∈ [0,1]}.$(22)

By Theorem 3.7, we conclude M1 is finite and then by (14) we know |u(4)| ≤ M1 for all x ∈ (0,1).

Combine the boundary conditions

$u(0)=u(1)=u′(0)=u′(1)=0,$

we know that there exists t0 ∈ (0,1) such that u′′′(t0) = 0, see [23, P. 1212], and subsequently,

$|u‴(t)|≤∫t0t|u(4)(s)|ds≤M1.$

Using the similar argument of above, we can prove that there exist constants M2, M3 and M4 such that for all possible solutions u of (14) for λ0λλ1,

$|u″(t)|≤M2,|u′(t)|≤M3,|u(t)|≤M4$

for t ∈ [0,1]. Clearly, ∥uC3 < ρ for λ0λλ1 as long as ρ:= max{M1, M2, M3, M4} + 1. □

#### Lemma 3.9.

Assume that the assumptions of Theorem 3.7 are satisfied. Let

$Ωα,β={u ∈ C4[0,1]:α(x)

and

$Ω=Bρ∩Ωα,β,$

where Bρ = {uC3[0,1]:∥uC3 < ρ} and ρ is given in Lemma 3.8. Then there exists λ0 ∈ (3λ1/4, λ1), such that

$deg⁡[I−Tλ,Ω,0]=1$

for λ ∈ [λ0, λ1].

#### Proof

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

$u(4)(x)=μ(λu(x)+f(x,u(x))),x ∈ (0,1),u(0)=u(1)=u′(0)=u′(1)=0.$(23)

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

$∥u∥C3<ρ$

for any λ ∈ [λ0, λ1]. Therefore, if we write (23) as $\begin{array}{}u=\mu \stackrel{^}{T}\left(u\right)\end{array}$, and set

$H(μ,u)=u−μT^(u).$

Then H(μ, u) ≠ 0 for all μ ∈ [0,1] and ∥uC3 = ρ, so that by the homotopy invariance of the Leray-Schauder degree

$deg⁡[I−Tλ,Bρ,0]=deg⁡[I−T^,Bρ,0]=deg⁡[I,Bρ,0]=1.$(24)

On the other hand, by Theorem 3.7, all zeros of ITλ belong to Ωα, β. Therefore, if ρ is large enough, then by the excision property of the Leray-Schauder degree, we have

$deg[I−Tλ,Ω,0]=deg⁡[I−Tλ,Ωα,β,0]=deg⁡[I−Tλ,Bρ,0]=1.$

#### Lemma 3.10.

Let Ω = Ωα, βBρ. Then there exists δ > 0 such that

$deg⁡[I−Tλ,Ω,0]=1$

for all λ ∈ [λ1, λ1 + δ].

#### Proof

The proof is trivial, so we omit it. □

## 4 Bifurcation from infinity and the multiplicity of solutions

It follows from Lemmas 3.9 and 3.10, and using the similar arguments of [28], we have the following

#### Lemma 4.1.

Let λ0 ∈ (3λ1/4, λ1), ρ > 0 and δ > 0 are sufficiently small constants. Then the set

$Γ:={(λ,u):λ ∈ [λ0,λ1+δ],∥u∥C3<ρ,(λ,u)satisfies(14)}$

contains a connected component C = {(λ, uλ)}.

To apply the argument of [21], let us extend f(x, ⋅) to all of ℝ by setting

$f~(x,u)=f(x,u),u>0,f(x,0),u≤0,$(25)

and deduce that

#### Lemma 4.2.

Since f : [0,1] × ℝ → ℝ is continuous and satisfies (H3), and λ1 is a simple eigenvalue, then there exists an unbounded connected component Σ ⊂ ℝ × C3 [0,1] of solutions of (14) such that for all sufficiently small r > 0,

$Σ∞∩Ur≠∅,$(26)

where $\begin{array}{}{U}_{r}:=\left\{\left(\lambda ,u\right)\in \mathbb{R}×{C}^{3}\left[0,1\right]:|\lambda -{\lambda }_{1}|\frac{1}{r}\right\}.\end{array}$

#### Proof of Theorem

For any given hC[0,1], let

$fh(x,s)=f(x,s)−h(x).$

Notice that fh : [0,1] × ℝ → ℝ is a continuous function and satisfies (H3), and by (H2) and (H1), we know fh satisfies (8) and (15) with c − = ch, c = Ch, and (16). Therefore, we may apply Theorem 3.7 and Lemma 4.1, to deduce that the problem (3) has at least one solution u1 in

$Bρ={u ∈ C3[0,1]:∥u∥C3<ρ}$

for λ ∈ [λ0, λ1 + δ].

On the other hand, since λ1 is a simple eigenvalue, by Lemma 4.2, we conclude that there exists an unbounded connected component Σ ⊂ ℝ × C3[0,1] of solutions of (3) bifurcating from infinity at λ = λ1.

In Lemma 3.8, we have established a priori bound for solutions of (3) when λ ∈ [λ0, λ1], hence the connected component Σ must bifurcate to right. More precisely, we infer Σ must satisfy

$Σ∞⊂{(λ,u) ∈ R×C3[0,1]:λ1<λ<λ1+r,∥u∥C3>1r},$

and hence, if $\begin{array}{}\frac{1}{r}>\mathrm{\rho ,}\end{array}$ i.e. $\begin{array}{}r<\frac{1}{\rho },\end{array}$ we obtain the second solution u2 with

$∥u2∥C3>1r>ρ$

for λ ∈ [λ1, λ1 + δ1], where δ1 = min{r, δ}.

Next, we will show that u2 > 0. It suffices to show that if (μn, un) ∈ Σ with μnλ1 and ||un||C3 → ∞ then un > 0 in (0,1) for n large. In fact, let $\begin{array}{}{w}_{n}=\frac{{u}_{n}}{||{u}_{n}|{|}_{{C}^{3}}}.\end{array}$ Then

$wn(4)(x)−μnwn(x)=fh(x,un(x))un(x)wn(x),x ∈ (0,1),wn(0)=wn(1)=wn′(0)=wn′(1)=0.$(27)

Assumption (H3) yields that, up to a subsequence, wnw in C1[0,1], where w is such that ||w||C1 = 1 and satisfies

$w(4)(x)−λ1w(x)=0,x ∈ (0,1),w(0)=w(1)=w′(0)=w′(1)=0,$

it follows that w ≥ 0, and hence there exists β > 0 such that w = βφ1. Then, it follows that un > 0 in (0,1), for n large.

#### Example 4.3.

Let us consider the following nonlinear fourth order boundary value problem

$u(4)(x)−λu(x)=f(x,u(x))−h(x),x ∈ (0,1),u(0)=u(1)=u′(0)=u′(1)=0$(28)

with

$f(x,u)=−arctan⁡u,|u|≥1,−π4u,0≤|u|<1,$

and

$h(x)≡12.$

Obviously, the function $\begin{array}{}f\left(x,u\right)+\frac{3{\lambda }_{1}}{4}u\end{array}$ is increasing in u.

Let c(x)≡1 and c(x)≡ −1. Then $\begin{array}{}c\left(x\right)\equiv \frac{3}{2}\end{array}$ and $\begin{array}{}C\left(x\right)\equiv -\frac{1}{2}\end{array}$. And it is easy to check that all assumptions of Theorem 1.1 are valid. Therefore, from Theorem 1.1, we know there exist δ1 > 0 and λ0 ∈ (3λ1/4, λ1) such that (28) has at least one solution for λ ∈ [λ0, λ1], and two solutions for λ ∈ (λ1, λ1 + δ1]. Moreover, one of these two solutions is a positive solution.

#### Example 4.4.

The functions f and h can be given respectively by

$f(x,u)=−u,u≥100,−110u,|u|<100,−u,u≤−100,$

and

$h(x)=50φ1(x),x ∈ [0,1],$

for which if we take c(x) = 100φ1(x) and C(x) = − 200φ1(x) for x ∈ [0,1], then c(x) = 50φ1(x) and c¯(x) = − 250φ1(x).

## Acknowledgement

The authors are very grateful to the anonymous referees for their valuable suggestions. This work was supported by the NSFC (No.11671322).

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Accepted: 2018-08-20

Published Online: 2018-10-29

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 1176–1186, ISSN (Online) 2391-5455,

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