Nonlinear elastic beam problems with the parameter near resonance

Man Xu; Ruyun Ma

doi:10.1515/math-2018-0097

Article Open Access

Nonlinear elastic beam problems with the parameter near resonance

Man Xu and Ruyun Ma

Published/Copyright: October 29, 2018

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$Open Mathematics$

From the journal Open Mathematics Volume 16 Issue 1

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 λ−λ₁, 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.

Keywords: Elastic beam; At resonance; Multiplicity; Eigenvalue; Lower and upper solutions; Bifurcation

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 λ ≤ λ₁ and two solutions for λ₁ < λ < λ₁ + ν under the assumption lims→+∞f(x,s)s=0 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, h ∈ C[0,1] and f ∈ C([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, λ_j (λ_j = m_4j, m_jis 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 φ1(x)=sin⁡m1x−sinh⁡m1x+sin⁡m1−sinh⁡m1cos⁡m1−cosh⁡m1(cosh⁡m1x−cos⁡m1x) with φ₁(x) > 0 in (0, 1), and we normalize by ∫01(φ1(x))2dx=1.

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−∞<u1≤u2<+∞andx ∈ [0,1];

(H2) h ∈ C[0,1] and satisfies

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

where c, C ∈ L¹(0,1) with lim infs→−∞f(x,s)≥c(x),lim sups→+∞f(x,s)≤C(x) 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 λ₀ ≤ λ ≤ λ₁or λ₁ < λ ≤ λ₁ + 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 λ ≤ λ₁, which are well ordered, that is α ≤ β (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 λ ≤ λ₁. 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 λ ≤ λ₁, 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<m1<m2<m3<⋯→+∞.

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 φ_j, corresponding to λ_j, has exactly j − 1 simple zeros in (0,1). The eigenvalues λ_k, 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 λ_j is given by

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

Moreover, φ_j ∈ S_j,+, where S_j,+ denote the set of u ∈ C³[0,1] such that:

u has only simple zeros in (0,1) and has exactly j − 1 such zeros;
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 α ∈ C⁴[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 β ∈ C⁴[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)

wherec − ∈ L¹ (0,1) and satisfies∫01c_(x)φ1(x)dx>0. Then there exist R > 0 and d ∈ C[0,1] with∫01d(x)φ1(x)dx>0 such that

f(x,u)≥d

for all x ∈ (0,1), whenever u ≤ − Rφ₁ in (0,1).

Proof

Let c_ε = c − ε, where ε > 0 is small enough to keep ∫01cε(x)φ1(x)dx>0. Then we can assume that the strict inequality holds in (8) with c (x) = c_ε(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 m ∈ L₁(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 d₀ ∈ L₁(0,1) and ∫01d0(x)φ1(x)dx>0.

Let x ∈ (0,1) and s ∈ R such that s ≤ − Rφ₁(x). We claim that f(x, s) ≥ d(x).

Indeed, for x ∈ (a, b), since minx ∈ [a,b]φ1(x)>0, then there exists R > 0 such that Rφ₁ ≥ R₀ 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 d ∈ C[0,1] with d₀(x) ≥ d(x), x ∈ (0,1) but still satisfies ∫01d(x)φ1(x)dx>0.

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 ∫01d(x)φ1(x)dx>0 and f(x, u) ≥ d whenever u ≤ − Rφ₁.

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 v:=d(x)−(∫01d(x)φ1(x)dx)φ1(x) satisfies the orthogonality condition ∫01v(x)φ1(x)dx=0. Then any α = sφ₁(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 ∫01(α0(4)(x)−λ1α0(x))φ1(x)dx=0.

Therefore, there exist constants a, A such that aφ₁ ≤ α₀ ≤ Aφ₁ for x ∈ (0,1), and accordingly, taking s negative sufficiently large (precisely, s < − (R + A)), we can arrange such that α = sφ₁(x) + α₀ < −Rφ₁ for x ∈ (0,1). But then f(x, α) ≥ d for all x ∈ (0,1), and since ∫01d(x)φ1(x)dx>0, we conclude that

α(4)(x)−λ1α(x)<d(x)≤f(x,α(x)),x ∈ (0,1),α(0)=α(1)=α′(0)=α′(1)=0.(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

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 ∈ {v ∈ C₄[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 u ≥ α, 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) < − Rφ₁ 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 u ≥ α for all possible solutions u of (14) for λ ≤ λ₁.

Proof

Let α be the lower solution for the (14) with λ = λ₁ determined in Lemma 3.3. Since α < − Rφ₁ < 0 for all x ∈ (0,1), we have from (11) if λ ≤ λ₁,

α(4)(x)−λ1α(x)<d(x)+(λ−λ1)α(x)≤f(x,α)+(λ−λ1)α(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 λ ≤ λ₁. Moreover, for any solution u of (14), setting w = u − α 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 λ ≤ λ₁.

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∫01c¯(x)φ1(x)dx<0.Then there exists a strict upper solution β > 0 of (14) such that u ≤ β for all possible solutions u of (14) for λ ≤ λ₁.

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 ≤ λ₁, 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 μ ∈ (0,3λ14], 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 L_λ:D → C[0,1] defined by

(Lλu)(x)=u(4)(x)−λu(x),x ∈ [0,1],(17)

where u ∈ D and D = {u ∈ C₄[0,1] : u(0) = u(1) = u′(0) = u′(1) = 0}.

Lemma 3.6.

Let λ ∈ [0, λ₁). If u ∈ D satisfies L_λu ≥ 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 λ ≤ λ₁, 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 η ∈ C[0,1] and μ ∈ (0,3λ₁/4] is a fixed constant. For ε = 0, (18) reduces to (14).

Define T_λ : C[0,1] → C[0,1] by

Tλη=u,

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

Step 1. We show T_λC ⊆ 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_λξ. 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 y ≥ α. Analogously, we can show that y ≤ β.

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

Let η₁, η₂ ∈ C[0,1] with η₁ ≤ η₂ and put u_i = T_λη_i, i = 1,2. Then from (16), w = u₂ − u₁satisfies

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 u₁ ≤ u₂.

Step 3. α ≤ T_λα and T_λβ ≤ β.

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 = T_λα − α 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 = T_λα − α ≥ 0. Analogously, we can prove that T_λβ ≤ 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 λ₀ ∈ (3λ₁/4, λ₁), there exists ρ > 0 such that ∥u∥_C³ < 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 λ₀ ≤ λ ≤ λ₁,

|u″(t)|≤M2,|u′(t)|≤M3,|u(t)|≤M4

for t ∈ [0,1]. Clearly, ∥u∥_C³ < ρ for λ₀ ≤ λ ≤ λ₁ 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)<u(x)<β(x),x ∈ [0,1]},

and

Ω=Bρ∩Ωα,β,

where B_ρ = {u ∈ C³[0,1]:∥u∥_C³ < 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 u=μT^(u), and set

H(μ,u)=u−μT^(u).

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

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 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

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

□

Lemma 3.10.

Let Ω = Ω_{α, β} ∩ B_ρ. 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, u_λ)}.

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 λ₁is 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)

whereUr:={(λ,u) ∈ R×C3[0,1]:|λ−λ1|<r,∥u∥C3>1r}.

Proof of Theorem

For any given h ∈ C[0,1], let

fh(x,s)=f(x,s)−h(x).

Notice that f_h : [0,1] × ℝ → ℝ is a continuous function and satisfies (H3), and by (H2) and (H1), we know f_h satisfies (8) and (15) with c − = c − h, c = C − h, 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 1r>ρ, i.e. r<1ρ, 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 (μ_n, u_n) ∈ Σ_∞ with μ_n → λ₁ and ||u_n||_C³ → ∞ then u_n > 0 in (0,1) for n large. In fact, let wn=un||un||C3. 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, w_n → w 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 u_n > 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 f(x,u)+3λ14u is increasing in u.

Letc(x)≡1 andc(x)≡ −1. Then c(x)≡32 and C(x)≡−12. 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], thenc(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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Received: 2018-06-07

Accepted: 2018-08-20

Published Online: 2018-10-29

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License.

Articles in the same Issue

https://doi.org/10.1515/math-2018-0097

Keywords for this article

Elastic beam; At resonance; Multiplicity; Eigenvalue; Lower and upper solutions; Bifurcation

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