Existence Results for a Fully Fourth-Order Boundary Value Problem

Abstract

We discuss the existence of solution for the fully fourth-order boundary value problem 𝑢⁽⁴⁾=𝑓(𝑡,𝑢,𝑢^′,𝑢^′′,𝑢^′′′), 0≤𝑡≤1, 𝑢(0)=𝑢(1)=𝑢^′′(0)=𝑢^′′(1)=0. A growth condition on 𝑓 guaranteeing the existence of solution is presented. The discussion is based on the Fourier analysis method and Leray-Schauder fixed point theorem.

1. Introduction and Main Results

In this paper we deal with the existence of solution for the fully fourth-order ordinary differential equation boundary value problem (BVP) 𝑢⁽⁴⁾(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡),𝑢^′′(𝑡),𝑢^′′′(𝑡)𝑢(0)(1),0≤𝑡≤1,=𝑢=𝑢^′′(0)=𝑢^′′(1)=0, (1) where 𝑓∶[0,1]×ℝ⁴→ℝ is continuous. This problem models deformations of an elastic beam whose two ends are simply supported in equilibrium state, and its research has important significance in mechanics.

For the special case of BVP(1) that 𝑓 does not contain derivative terms 𝑢^′ and 𝑢^′′′, namely, simply fourth-order boundary value problem 𝑢⁽⁴⁾(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢^′′(𝑡)𝑢(0)(1),0≤𝑡≤1,=𝑢=𝑢^′′(0)=𝑢^′′(1)=0, (2)

the existence of solution has been studied by many authors; see [1–8]. In [1], Aftabizadeh showed the existence of a solution to PBV(2) under the restriction that 𝑓 is a bounded function. In [2, Theorem 1], Yang extended Aftabizadeh’s result and showed the existence for BVP(2) under the growth condition of the form 𝑓(𝑡,𝑢,𝑣)≤𝑎|𝑢|+𝑏|𝑣|+𝑐, (3) where 𝑎, 𝑏, and 𝑐 are positive constants such that 𝑎𝜋⁴+𝑏𝜋²<1. (4)

In [3], under a more general linear growth condition of two-parameter nonresonance, del Pino and Manásevich also discussed the existence of BVP(2) and the result of Yang was further extended. For more results involving two-parameter nonresonance condition see [4, 7]. All these works are based on Leray-Schauder degree theory. In [5, 6], the upper and lower solutions method is applied to discuss the existence of BVP(2). Recently, in [8] the fixed point index theory in cones is employed to BVP(2) and some existence results of positive are obtained, where 𝑓 may be super-linear growth.

For the more simple case of BVP(1) that 𝑓 does not contain any derivative terms, the following fourth-order boundary value problem 𝑢⁽⁴⁾(𝑡)((𝑡𝑢(0)(1)=𝑓𝑡,𝑢)),0≤𝑡≤1,=𝑢=𝑢^′′(0)=𝑢^′′(1)=0, (5)

has been studied by more researchers, and various theorems and methods of nonlinear analysis have been applied; see [9–13] and reference therein.

However, few researchers consider the fully fourth-order boundary value problem BVP(1). The purpose of this paper is to discuss the existence of solution of BVP(1). We will extend the Yang’s result previously mentioned from BVP(2) to the general BVP(1). Our results are as follows.

Theorem 1. Assume that 𝑓∈𝐶([0,1]×ℝ⁴,ℝ) and it satisfies the growth condition 𝑓𝑡,𝑥₀,𝑥₁,𝑥₂,𝑥₃≤𝑐₀𝑥₀+𝑐₁𝑥₁+𝑐₂𝑥₂+𝑐₃𝑥₃+𝑀, (6)
for all 𝑡∈[0,1] and (𝑥₀,𝑥₁,𝑥₂,𝑥₃)∈ℝ⁴, where 𝑐₀,𝑐₁,𝑐₂,𝑐₃≥0 and 𝑀>0 are constants and 𝑐₀,𝑐₁,𝑐₂,𝑐₃ satisfy the restriction 𝑐₀𝜋⁴+𝑐₁𝜋³+𝑐₂𝜋²+𝑐₃𝜋<1. (7)
Then the BVP(1) possesses at least one solution.

Theorem 1 is a directly extension of Yang’s result previously mentioned. In Theorem 1, the condition (7) is optimal. If the condition (7) does not hold, the existence of solution of BVP(1) cannot be guaranteed. Strengthening the condition (6) of Theorem 1, we can obtain the following uniqueness result.

Theorem 2. Assume that 𝑓∈𝐶([0,1]×ℝ⁴,ℝ) and it satisfies the Lipschitz-type condition 𝑓𝑡,𝑥₀,𝑥₁,𝑥₂,𝑥₃−𝑓𝑡,𝑦₀,𝑦₁,𝑦₂,𝑦₃≤³_𝑖=0𝑐_𝑖𝑥_𝑖−𝑦_𝑖, (8)
for any (𝑡,𝑥₀,𝑥₁,𝑥₂,𝑥₃) and (𝑡,𝑦₀,𝑦₁,𝑦₂,𝑦₃)∈[0,1]×ℝ⁴, where 𝑐₀, 𝑐₁, 𝑐₂, 𝑐₃≥0 are constants and satisfy (7). Then BVP(1) has a unique solution.

If the partial derivatives 𝑓_𝑥0, 𝑓_𝑥1, 𝑓_𝑥2, and 𝑓_𝑥3 exist, then from Theorem 2 and the theorem of differential mean value, we have the following.

Corollary 3. Let 𝑓∈𝐶([0,1]×ℝ⁴,ℝ) and the partial derivatives 𝑓_𝑥0, 𝑓_𝑥1, 𝑓_𝑥2, and 𝑓_𝑥3 exist. If there exist positive constants 𝑐₀,𝑐₁,𝑐₂,𝑐₃ such that 𝑓_𝑥𝑖𝑡,𝑥₀,𝑥₁,𝑥₂,𝑥₃≤𝑐_𝑖,𝑖=0,1,2,3, (9)
and the constants 𝑐₀,𝑐₁,𝑐₂,𝑐₃ satisfy (7), then BVP(1) has one unique solution.

The proofs of Theorems 1 and 2 are based on the Fourier analysis method and Leray-Schauder fixed point theorem, which will be given in Section 2.

2. Proof of the Main Results

Let 𝐼=[0,1] and 𝐻=𝐿²(𝐼) be the usual Hilbert space with the interior product (𝑢,𝑣)=∫¹₀𝑢(𝑡)𝑣(𝑡)𝑑𝑡 and the norm ‖𝑢‖₂=(∫¹₀|𝑢(𝑡)|²𝑑𝑡)^1/2. For 𝑚∈ℕ, let 𝑊^𝑚,2(𝐼) be the usual Sobolev space with the norm ‖𝑢‖_𝑚,2=∑^𝑚_𝑖=0‖𝑢^(𝑖)‖²₂. 𝑢∈𝑊^𝑚,2(𝐼) means that 𝑢∈𝐶^𝑚−1(𝐼), 𝑢^(𝑚−1)(𝑡) is absolutely continuous on 𝐼 and 𝑢^(𝑚)∈𝐿²(𝐼).

Given ℎ∈𝐿²(𝐼), we consider the linear fourth-order boundary value problem (LBVP) 𝑢⁽⁴⁾(𝑡)(𝑡)𝑢(0)(1)=ℎ,𝑡∈𝐼,=𝑢=𝑢^′′(0)=𝑢^′′(1)=0. (10)

Let 𝐺(𝑡,𝑠) be the Green’s function to the second-order linear boundary value problem −𝑢^′′=0,𝑢(0)=𝑢(1)=0, (11)

which is explicitly expressed by 𝐺()=𝑡()𝑠()𝑡,𝑠1−𝑠,0≤𝑡≤𝑠≤1,1−𝑡,0≤𝑠≤𝑡≤1. (12)

For every given ℎ∈𝐿²(𝐼), it is easy to verify that the LBVP(10) has a unique solution 𝑢∈𝑊^4,2(𝐼) in Carathéodory sense, which is given by 𝑢(𝑡)=¹₀𝐺(𝑡,𝜏)𝐺(𝜏,𝑠)ℎ(𝑠)𝑑𝑠𝑑𝜏∶=𝑆ℎ(𝑡). (13)

If ℎ∈𝐶(𝐼), the solution is in 𝐶⁴(𝐼) and is a classical solution. Moreover, the solution operator of LBVP(10), 𝑆∶𝐿²(𝐼)→𝑊^4,2(𝐼) is a linearly bounded operator. By the compactness of the Sobolev embedding 𝑊^4,2(𝐼)↪𝐶³(𝐼) and the continuity of embedding 𝐶³(𝐼)↪𝑊^3,2(𝐼), we see that 𝑆 maps 𝐿²(𝐼) into 𝑊^3,2(𝐼) and 𝑆∶𝐿²(𝐼)→𝑊^3,2(𝐼) is a completely continuous operator.

Choose a subspace of 𝑊^3,2(𝐼) by 𝐷=𝑢∈𝑊^3,2(𝐼)∣𝑢(0)=𝑢(1)=0,𝑢^′′(0)=𝑢^′′(1)=0. (14)

Clearly, 𝐷 is a closed subspace, and hence 𝐷 is a Banach space by the norm ‖𝑢‖_3,2 of 𝑊^3,2(𝐼). Define another norm on 𝐷 by ‖𝑢‖_𝑋=𝑢^′′′₂,𝑢∈𝐷. (15)

One easily verifies that ‖𝑢‖_𝑋 is equivalent to ‖𝑢‖_3,2. Hereafter, we use 𝑋 to denote the Banach space 𝐷 endowed the norm ‖𝑢‖_𝑋, namely, 𝑋=𝐷,‖⋅‖_𝑋. (16)

By the boundary condition of LBVP(10), the solution operator 𝑆 maps 𝐻 into 𝐷. Hence 𝑆∶𝐻→𝑋 is completely continuous.

Lemma 4. For LBVP(10), the following two conclusions hold.(a)The norm of the solution operator of LBVP(10) 𝑆∶𝐻→𝑋 satisfies ‖𝑆‖_{ℒ(𝐻,𝑋)}≤1/𝜋.(b)For every ℎ∈𝐻, the unique solution of LBVP(10) 𝑢∈𝑊^4,2(𝐼) satisfies the inequalities ‖𝑢‖₂≤1𝜋³𝑢^′′′₂,𝑢^′₂≤1𝜋²𝑢^′′′₂,𝑢^′′₂≤1𝜋𝑢^′′′₂. (17)

Proof. Since sine system {sin𝑘𝜋𝑡∣𝑘∈ℕ} is a complete orthogonal system of 𝐿²(𝐼), every ℎ∈𝐿²(𝐼) can be expressed by the Fourier series expansion ℎ(𝑡)=^∞_𝑘=1ℎ_𝑘sin𝑘𝜋𝑡, (18) where ℎ_𝑘=2∫¹₀ℎ(𝑠)sin𝑘𝜋𝑠𝑑𝑠, 𝑘=1,2,…, and the Parseval equality ‖ℎ‖²₂=12^∞_𝑘=1ℎ_𝑘², (19)
holds. Let 𝑢=𝑆ℎ; then 𝑢∈𝑊^4,2(𝐼) is the unique solution of LBVP(10), and 𝑢, 𝑢^′′, and 𝑢⁽⁴⁾ can be expressed by the Fourier series expansion of the sine system. Since 𝑢⁽⁴⁾=ℎ, by the integral formula of Fourier coefficient, we obtain that 𝑢(𝑡)=^∞_𝑘=1ℎ_𝑘𝑘⁴𝜋⁴𝑢sin𝑘𝜋𝑡,_′′(𝑡)=−^∞_𝑘=1ℎ_𝑘𝑘²𝜋²sin𝑘𝜋𝑡. (20)
On the other hand, since cosine system {cos𝑘𝜋𝑡∣𝑘=0,1,2,…} is another complete orthogonal system of 𝐿²(𝐼), every 𝑣∈𝐿²(𝐼) can be expressed by the cosine series expansion 𝑣(𝑡)=𝑎₀2+^∞_𝑘=1𝑎_𝑘cos𝑘𝜋𝑡, (21) where 𝑎_𝑘=2∫¹₀ℎ(𝑠)cos𝑘𝜋𝑠𝑑𝑠, 𝑘=0,1,2,…. For the above 𝑢=𝑆ℎ, by the integral formula of the coefficient of cosine series, we obtain the cosine series expansions of 𝑢^′ and 𝑢^′′′: 𝑢^′(𝑡)=^∞_𝑘=1ℎ_𝑘𝑘³𝜋³𝑢cos𝑘𝜋𝑡,_′′′(𝑡)=−^∞_𝑘=1ℎ_𝑘𝑘𝜋cos𝑘𝜋𝑡. (22)(23)
Now from (23), (19), and Parseval equality, it follows that ‖𝑆ℎ‖²_𝑋=𝑢^′′′2₂=12^∞_𝑘=1ℎ_𝑘𝑘𝜋²≤12𝜋^2∞_𝑘=−∞ℎ_𝑘²=1𝜋²‖ℎ‖²₂. (24)
This means that ‖𝑆‖_{ℒ(𝐻,𝑋)}≤1/𝜋, namely, (a) holds.
By (20)–(22) and Paserval equality, we have that ‖𝑢‖²₂=12^∞_𝑘=1ℎ_𝑘𝑘⁴𝜋⁴²≤12𝜋^6∞_𝑘=1ℎ_𝑘𝑘𝜋²=1𝜋⁶𝑢^′′′2₂,𝑢^′2₂=12^∞_𝑘=1ℎ_𝑘𝑘³𝜋³²≤12𝜋^4∞_𝑘=1ℎ_𝑘𝑘𝜋²=1𝜋⁴𝑢^′′′2₂,𝑢^′′2₂=12^∞_𝑘=1ℎ_𝑘𝑘²𝜋²²≤12𝜋^2∞_𝑘=1ℎ_𝑘𝑘𝜋²=1𝜋²𝑢^′′′2₂. (25)(26)(27)

This shows that the conclusion (b) holds.

Proof of Theorem 1. We define a mapping 𝐹∶𝑋→𝐻 by 𝐹(𝑢)(𝑡)∶=𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡),𝑢^′′(𝑡),𝑢^′′′(𝑡),𝑢∈𝑋. (28)
From the assumption (6) and the property of Carathéodory mapping it follows that 𝐹∶𝑋→𝐻 is continuous and it maps every bounded set of 𝑋 into a bounded set of 𝐻. Hence, the composite mapping 𝑆∘𝐹∶𝑋→𝑋 is completely continuous. We use the Leray-Schauder fixed-point theorem to show that 𝑆∘𝐹 has at least one fixed-point. For this, we consider the homotopic family of the operator equations: 𝑢=𝜆(𝑆∘𝐹)(𝑢),0<𝜆<1. (29)
We need to prove that the set of the solutions of (29) is bounded in 𝑋. See [14].
Let 𝑢∈𝑋 be a solution of an equation of (29) for 𝜆∈(0,1). Set ℎ=𝜆𝐹(𝑢); then by the definition of 𝑆, 𝑢=𝑆ℎ∈𝑊^4,2(𝐼) is the unique solution of LBVP(10). By (a) of Lemma 4, we have ‖𝑢‖_𝑋=‖𝑆ℎ‖_𝑋≤‖𝑆‖_{ℒ(𝑋,𝐻)}‖ℎ‖₂≤1𝜋‖ℎ‖₂≤1𝜋‖𝐹(𝑢)‖₂. (30)
From (28), (6), and (b) of Lemma 4, it follows that ‖𝐹(𝑢)‖₂≤𝑐₀‖𝑢‖₂+𝑐₁𝑢^′₂+𝑐₂𝑢^′′₂+𝑐₃𝑢^′′′₂≤𝑐+𝑀₀𝜋³+𝑐₁𝜋²+𝑐₂𝜋+𝑐₃𝑢^′′′₂𝑐+𝑀=𝜋⁰𝜋⁴+𝑐₁𝜋³+𝑐₂𝜋²+𝑐₃𝜋‖𝑢‖_𝑋+𝑀. (31)
Combining this inequality with (30), we obtain that ‖𝑢‖_𝑋≤𝑀1−𝑐₀/𝜋⁴+𝑐₁/𝜋³+𝑐₂/𝜋²+𝑐₃/𝜋∶=𝐶₀. (32)
This means that the set of the solutions for (29) is bounded in 𝑋. Therefore, by the Leray-Schauder fixed-point theorem [14], 𝑆∘𝐹 has a fixed-point 𝑢₀∈𝑋. Let ℎ₀=𝐹(𝑢₀). By the definition of 𝑆, 𝑢₀=𝑆ℎ₀∈𝑊^4,2(𝐼) is a solution of LBVP(10) for ℎ=ℎ₀. Since 𝑊^4,2(𝐼)↪𝐶³(𝐼), from (28) it follows that ℎ₀∈𝐶(𝐼). Hence 𝑢₀∈𝐶⁴(𝐼) is a classical solution of LBVP(10), and by (28) 𝑢₀ is also a solution of BVP(1).
The proof of Theorem 1 is completed.

Proof of Theorem 2. Let 𝑀=max{|𝑓(𝑡,0,0,0,0)|∶𝑡∈𝐼}+1. From condition (8) of Theorem 2 we easily see that Condition (6) of Theorem 1 holds. By Theorem 1, the BVP(1) has at least one solution.
Now, let 𝑢₁,𝑢₂∈𝐶⁴(𝐼) be two solutions of BVP(1); then 𝑢_𝑖=𝑆(𝐹(𝑢_𝑖)), 𝑖=1,2. From (8) and (28), we obtain that 𝐹𝑢₂(𝑡)−𝐹𝑢₁(𝑡)≤+³_𝑖=0𝑐_𝑖𝑢₂^(𝑖)(𝑡)−𝑢₁^(𝑖)(𝑡), (33)
for 𝑡∈𝐼. Since 𝑢₂−𝑢₁ is the solution of LBVP(10) for ℎ=𝐹(𝑢₂)−𝐹(𝑢₁), by (33) and (b) of Lemma 4, we have 𝐹𝑢₂−𝐹𝑢₁₂≤³_𝑖=0𝑐_𝑖𝑢₂^(𝑖)−𝑢₁^(𝑖)₂≤𝑐₀𝜋³+𝑐₁𝜋²+𝑐₂𝜋+𝑐₃𝑢₂^′′′−𝑢₁^′′′₂𝑐=𝜋⁰𝜋⁴+𝑐₁𝜋³+𝑐₂𝜋²+𝑐₃𝜋𝑢₂−𝑢_1𝑋. (34)
From this and (a) of Lemma 4, it follows that 𝑢₂−𝑢_1𝑋=𝑆𝐹𝑢₂−𝐹𝑢_1𝑋≤‖𝑆‖_{ℒ(𝑋,𝐻)}𝐹𝑢₂−𝐹𝑢₁₂≤𝑐₀𝜋⁴+𝑐₁𝜋³+𝑐₂𝜋²+𝑐₃𝜋𝑢₂−𝑢_1𝑋. (35)
Since 𝑐₀/𝜋⁴+𝑐₁/𝜋³+𝑐₂/𝜋²+𝑐₃/𝜋<1, from (35) we see that ‖𝑢₂−𝑢₁‖_𝑋=0, that is 𝑢₂=𝑢₁. Therefore, BVP(1) has only one solution.
The proof of Theorem 2 is completed.

Example 5. Consider the following fully linear fourth-order boundary value problem 𝑢⁽⁴⁾(𝑡)=𝑎₀(𝑡)𝑢(𝑡)+𝑎₁(𝑡)𝑢^′(𝑡)+𝑎₂(𝑡)𝑢^′′(𝑡)+𝑎₃(𝑡)𝑢^′′′(𝑡)(𝑡)𝑢(0)(1)+ℎ,𝑡∈𝐼,=𝑢=𝑢^′′(0)=𝑢^′′(1)=0, (36) where the coefficient functions 𝑎₀,𝑎₁,𝑎₂,𝑎₃∈𝐶(𝐼) and the inhomogeneous term ℎ∈𝐶(𝐼). All the known results of [1–13] are not applicable to this equation. Let 𝑓𝑡,𝑥₀,𝑥₁,𝑥₂,𝑥₃=𝑎₀(𝑡)𝑥₀+𝑎₁(𝑡)𝑥₁+𝑎₂(𝑡)𝑥₂+𝑎₃(𝑡)𝑥₃(𝑡),𝑐+ℎ_𝑖=max_𝑡∈𝐼𝑎_𝑖(𝑡),𝑖=0,1,2,3. (37)
It is easy to see that the partial derivatives 𝑓_𝑥0, 𝑓_𝑥1, 𝑓_𝑥2, and 𝑓_𝑥3 exist and 𝑓_𝑥𝑖𝑡,𝑥₀,𝑥₁,𝑥₂,𝑥₃=𝑎_𝑖(𝑡)≤𝑐_𝑖,𝑖=0,1,2,3. (38)
Assume that the constants 𝑐₀,𝑐₁,𝑐₂,𝑐₃ satisfy (7). Then by Corollary 3, (36) has a unique solution.

Example 6. Consider the following nonlinear fourth-order boundary value problem 𝑢⁽⁴⁾(𝑡)=³_𝑖=0𝑏_𝑖(𝑡)𝑢^(𝑖)(𝑡)^𝛼𝑖𝑢(0)(1)+sin𝜋𝑡,𝑡∈𝐼,=𝑢=𝑢^′′(0)=𝑢^′′(1)=0, (39) where 𝑏_𝑖∈𝐶(𝐼), 𝛼_𝑖∈(0,1),𝑖=0,1,2,3. Let 𝑓𝑡,𝑥₀,𝑥₁,𝑥₂,𝑥₃=³_𝑖=0𝑏_𝑖(𝑡)𝑥_𝑖^𝛼𝑖+sin𝜋𝑡. (40)
Then 𝑓∈𝐶([0,1]×ℝ⁴,ℝ) and it satisfies that lim_{|𝑥0|+|𝑥1|+|𝑥2|+|𝑥3|→∞}max_𝑡∈𝐼𝑓𝑡,𝑥₀,𝑥₁,𝑥₂,𝑥₃𝑥₀+𝑥₁+𝑥₂+𝑥₃=0. (41)
From this one easily proves that there exists a positive constant 𝑀>0 such that 𝑓𝑡,𝑥₀,𝑥₁,𝑥₂,𝑥₃≤𝑥₀+𝑥₁+𝑥₂+𝑥₃+𝑀. (42)
Since (7) holds for the constants 𝑐₀=𝑐₁=𝑐₂=𝑐₃=1, by (42) 𝑓 satisfies the conditions of Theorem 1. Hence by Theorem 1, (39) has at least one solution. This conclusion cannot be obtained from the results in [1–13].

Acknowledgments

This research is supported by NNSFs of China (11261053, 11061031) and the NFS of Gansu province (1208RJZA129).