Unified approach to some boundary value problems ...

Introduction

Existence results

Iterative method

Some particular cases

Concluding remarks

Unified approach to some boundary value problems for fully fourth order nonlinear differential equations Dang Quang A Center for Informatics and Computing, VAST 18 Hoang Quoc Viet, Cau giay, Hanoi, Vietnam E-mail: [email protected] Presented at Seminar in National Sun Yat-sen University, Kaohsiung, Taiwan , July 2018

Introduction

Existence results

Iterative method

Contents

1

Introduction

2 Existence results 3 Iterative method 4 Some particular cases 5 Concluding remarks


Concluding remarks

Introduction

Existence results

Iterative method


Concluding remarks

Consider the BVP u (4) (x) = f (x, u(x), u 0 (x), u 00 (x), u 000 (x)),

0 < x < 1, (1.1)

B1 (u) ¯ = 0, B2 (u) ¯ = 0, B3 (u¯) = 0, B4 (u¯) = 0,

(1.2)

where u¯ = u(0), u(1), u 0 (0), u 0 (1) , u¯ = u 00 (0), u 00 (1), u 000 (0), u 000 (1) , B1 , B2 , B3 , B4 are linear combinations of the components of arguments. The BVP (1.1), (1.2) is a generalization of numerous BVPs studied before by many authors. The main interest was the existence of positive solutions.

Introduction

Existence results

Iterative method


Concluding remarks

For functions ϕ(x) ∈ C [0, 1] define the nonlinear operator A by (Aϕ)(x) = f (x, u(x), u 0 (x), u 00 (x), u 000 (x)),

(2.1)

where u(x) is a solution of the BVP u (4) (x) = ϕ(x),

0 < x < 1,

B1 (u) ¯ = 0, B2 (u) ¯ = 0, B3 (u¯) = 0, B4 (u¯) = 0,

(2.2)

Proposition 2.1 If the function ϕ(x) is a fixed point of the operator A, i.e., ϕ(x) is a solution of the operator equation ϕ = Aϕ then the function u(x) determined from the boundary value problem (2.2) solves the problem (1.1)-(1.2). Conversely, if u(x) is a solution of the boundary value problem (1.1)-(1.2) then the function ϕ(x) = f (x, u(x), u 0 (x), u 00 (x), u 000 (x)) is a fixed point of the operator A

Introduction

Existence results

Iterative method


Concluding remarks

Remark 2.2 By the above proposition, the solution of the original problem (1.1), (1.2) is reduced to the solution of the operator equation. This is the technique used in our previous works for some particular cases of the problem (1.1)-(1.2) but not formulated as a separate proposition. Now in the problem (1.1)-(1.2) we set v (x) = u 00 (x). Then it is decomposed into two second order problems 00 v (x) = ϕ(x), 0 < x < 1, (2.3) B3 (¯ v ) = 0, B4 (¯ v ) = 0, u 00 (x) = v (x), 0 < x < 1, B1 (u) ¯ = 0, B2 (u) ¯ = 0, where v¯ = v (0), v (1), v 0 (0), v 0 (1) .

Introduction

Existence results

Iterative method


Concluding remarks

Suppose the Green functions of the problem (1.1), (1.2) and the problem (2.3) exist and are denoted by G (x, t) and Gb(x, t), respectively. Then their solutions are represented in the forms Z 1 Z 1 u(x) = G (x, t)ϕ(t)dt, v (x) = Gb(x, t)ϕ(t)dt. (2.4) 0

0

From the differentiability property of Green functions we have Z 1 Z 1 u 0 (x) = G1 (x, t)ϕ(t)dt, v 0 (x) = Gb1 (x, t)ϕ(t)dt, (2.5) 0

0

where G1 (x, t) = Gx0 (x, t) is a function continuous in the square [0, 1]2 and Gb1 (x, t) = Gbx0 (x, t) is continuous in the square [0, 1]2 except for the line t = x.

Introduction

Existence results

Iterative method


Concluding remarks

Suppose that for any 0 ≤ x ≤ 1 there hold the estimates Z 1 Z 1 |G (x, t)|dt ≤ M0 , |G1 (x, t)|dt ≤ M1 , 0 0 Z 1 Z 1 b |G (x, t)|dt ≤ M2 , |Gb1 (x, t)|dt ≤ M3 , 0

0

where M0 , M1 , M2 and M3 are some constants. Also, denote y (x) = u 0 (x), z(x) = v 0 (x). Then from (2.5) we have Z y (x) =

1

Z G1 (x, t)ϕ(t)dt,

0

z(x) =

1

Gb1 (x, t)ϕ(t)dt. 0

(2.6)

Introduction

Existence results

Iterative method


Concluding remarks

Next, for each real number M > 0 introduce the domain DM = {(x, u, y , v , z)| 0 ≤ x ≤ 1, |u| ≤ M0 M, |y | ≤ M1 M, |v | ≤ M2 M, |z| ≤ M3 M}, and as usual, by B[O, M] we denote B[O, M] = {ϕ ∈ C [0, 1]| kϕk ≤ M}, where kϕk = max |ϕ(x)|. 0≤x≤1

Introduction

Existence results

Iterative method


Concluding remarks

Theorem 2.3 (Existence of solution) Suppose that there exists a number M > 0 such that the function f (x, u, y , v , z) is continuous and bounded by M in the domain DM , i.e., |f (x, u, y , v , z)| ≤ M, (2.7) for any (x, u, y , v , z) ∈ DM . Then, the problem (1.1)-(1.2) has a solution u(x) satisfying |u(x)| ≤ M0 M, |u 0 (x)| ≤ M1 M, |u 00 (x)| ≤ M2 M, |u 000 (x)| ≤ M3 M for any 0 ≤ x ≤ 1.

Introduction

Existence results

Iterative method


Concluding remarks

Proof. Sketch of proof: - Operator A maps any closed ball B[O, M] into itself. - Operator A is compact one in C [0, 1] due to the continuity of the function f (x, u, y , v , z) and the fact that integral operators (2.4),(2.5) which put each function ϕ ∈ C [0, 1] in correspondence to the functions u, u 0 , u 00 , u 000 , respectively, are compact operators. By the Schauder Fixed Point Theorem the operator equation Aϕ = ϕ has a solution. This solution generates a solution of the problem (1.1)-(1.2).

Introduction

Existence results

Iterative method


Concluding remarks

Now suppose that the Green functions G (x, t), Gb(x, t) and their derivatives G1 (x, t), Gb1 (x, t) are of constant signs in the square Q = [0, 1]2 . Let’s make the following convention: For a function H(x, t) defined and having a constant sign in Q 1, if H(x, t) ≥ 0, σ(H) = sign(H(x, t)) = −1, if H(x, t) ≤ 0. In order to investigate the existence of positive solutions of the problem (1.1)-(1.2) we introduce the notations + DM ={(x, u, y , v , z)| 0 ≤ x ≤ 1, 0 ≤ σ(G )u ≤ M0 M, c1 )z ≤ M3 M} 0 ≤ σ(G1 )y ≤ M1 M, 0 ≤ σ(Gb)v ≤ M2 M, 0 ≤ σ(G (2.8)

and SM = {ϕ ∈ C [0, 1]| 0 ≤ ϕ ≤ M}.

Introduction

Existence results

Iterative method


Concluding remarks

Theorem 2.4 (Existence of constant sign solution) Suppose that there exists a number M > 0 such that the function f (x, u, y , v , z) is continuous and 0 ≤ f (x, u, y , v , z) ≤ M

+ ∀(x, u, y , v , z) ∈ DM .

(2.9)

Then, the problem (1.1)-(1.2) has a solution u(x) satisfying 0 ≤ σ(G )u(x) ≤ M0 M, 0 ≤ σ(G1 )u 0 (x) ≤ M1 M, c1 )u 000 (x) ≤ M3 M 0 ≤ σ(Gb)u 00 (x) ≤ M2 M, 0 ≤ σ(G

(2.10)

In the case if the Green function G (x, t) is positive then the problem has a positive solution. Proof. The proof is similar to that of the previous theorem, where instead of DM + and B[O, M] there stand DM and SM .

Introduction

Existence results

Iterative method


Concluding remarks

Theorem 2.5 (Uniqueness of solution) Suppose that the function f (x, u, y , v , z) satisfies the Lipschitz condition in variables u, y , v , z, namely, there exist numbers c0 , c1 , c2 , c3 ≥ 0 such that |f (x, u2 , y2 , v2 , z2 ) − f (x, u1 , y1 , v1 , z1 )| ≤ c0 |u2 − u1 | + c1 |y2 − y1 | + c2 |v2 − v1 | + c3 |z2 − z1 | (2.11) for any (x, ui , yi , vi , zi ) ∈ [0, 1] × R4 (i = 1, 2) and q := c0 M0 + c1 M1 + c2 M2 + c3 M3 < 1.

(2.12)

Then the solution of the problem (1.1)-(1.2) is unique if it exists.

Introduction

Existence results

Iterative method


Concluding remarks

Theorem 2.6 (Existence and uniqueness of solution) Assume that there exist numbers M, c0 , c1 , c2 , c3 ≥ 0 such that |f (x, u, y , v , z)| ≤ M,

(2.13)

|f (x, u2 , y2 , v2 , z2 , ) − f (x, u1 , y1 , v1 , z1 )| ≤ c0 |u2 − u1 | + c1 |y2 − y1 | + c2 |v2 − v1 | + c3 |z2 − z1 | (2.14) for any (x, u, y , v , z), (x, ui , yi , vi , zi ) ∈ DM (i = 1, 2) and q := c0 M0 + c1 M1 + c2 M2 + c3 M3 < 1. Then, the problem (1.1)-(1.2) has a unique solution u(x) such that |u(x)| ≤ M0 M, |u 0 (x)| ≤ M1 M, |u 00 (x)| ≤ M2 M, |u 000 (x)| ≤ M3 M for any 0 ≤ x ≤ 1. Proof.

Introduction

Existence results

Iterative method


Concluding remarks

Theorem 2.7 (Existence and uniqueness of constant sign solution) Assume that all the conditions of Theorem 2.6 are satisfied in the + domain DM . Then, the problem (1.1)-(1.2) has a unique constant sign solution u(x) satisfying (2.10). Remark 2.8 Remark that in Theorem 2.5 the Lipschitz condition is required to be satisfied in [0, 1] × R4 , while in Theorem 2.6 or Theorem 2.7 + under the condition (2.13) it is required only in DM or DM , respectively.

Introduction

Existence results

Iterative method


Concluding remarks

Iterative method: Given ϕ0 = f (x, 0, 0, 0, 0) Knowing ϕk , k = 0, 1, ... consecutively solve vk ” = ϕk (x), 0 < x < 1, uk ” = vk (x) 0 < x < 1, B3 (¯ vk ) = 0, B4 (¯ vk ) = 0, B1 (u¯k ) = 0, B2 (u¯k ) = 0, Update ϕk+1 = f (x, uk , uk0 , vk , vk0 )

Introduction

Existence results

Iterative method


Concluding remarks

Theorem 3.1 Under the assumptions of Theorem 2.6 the above iterative method converges with the rate of geometric progression and there hold the estimates kuk − uk ≤ M0 pk , kuk00 − u 00 k ≤ M2 pk ,

kuk0 − u 0 k ≤ M1 pk , kuk000 − u 000 k ≤ M3 pk ,

where u is the exact solution of the problem (1.1)-(1.2), qk kϕ1 − ϕ0 k. pk = 1−q

(3.1)

Introduction

Existence results

Iterative method


Concluding remarks

Case 1. u (4) (x) = f (x, u(x), u 0 (x), u 00 (x), u 000 (x)), 00

0 < x < 1,

00

u(0) = u(1) = u (0) = u (1) = 0.

(4.1)

This problem models the bending equilibrium of a beam on an elastic foundation, whose two ends are simply supported. • The case when f = f (x, u, u 00 ) was studied by Aftabizadeh (1986), Ma (1997), Bai (2004), Li (2010) by the lower and upper solution method. In 2017 in Numer Algor we established the existence and uniqueness of solution and investigated iterative method.

Introduction

Existence results

Iterative method


Concluding remarks

• The case when f (x, u(x), u 0 (x), u 00 (x), u 000 (x)) Li and Liang (2013) by the Fourier analysis method and Leray-Schauder fixed point theorem under growth condition and Lipschitz condition of the function f (x, u, y , v , z) established existence and uniqueness of solution. In 2018 in Bol. Soc. Paran. Mat. we by reduction of BVP to operator equation for nonlinear term established the existence and uniqueness of solution and studied iterative method.

Introduction

Existence results

Iterative method


Concluding remarks

For this problem 1 G (x, t) = 6

x(t − 1)(t 2 + x 2 − 2t), t(x − 1)(t 2 + x 2 − 2x),

Gb(x, t) =

0 ≤ x ≤ t ≤ 1, 0 ≤ t ≤ x ≤ 1,

x(t − 1), 0 ≤ x ≤ t ≤ 1, t(x − 1), 0 ≤ t ≤ x ≤ 1,

G1 (x, t) = Gx0 (x, t), Gb1 (x, t) = Gbx0 (x, t), 1 1 1 5 , M1 = , M2 = , M3 = . 384 24 8 2 Therefore, with a suitable selection of M > 0 the above Theorem 2.3 guarantees the existence of a solution of the problem (1.1), (2.1) without the Lipschitz condition and any growth conditions of f at infinity. M0 =

Introduction

Existence results

Iterative method


Concluding remarks

Remarks: • The iterative method reduces the solution of BVP for fully fourth order equation to a sequence of second order BVPs. • The Green function is absent in the iterative method. It is used only for studying the existence, uniqueness and the convergence of the iterative method.

Introduction

Existence results

Iterative method


Concluding remarks

Example 1 Consider the boundary value problem  u 000 (x) sin πx  (4) 00  u (x) = − + cos − 2 − u (x) − u 0 (x) − u 2 (x)   3 π  cos πx sin2 πx cos πx  + sin πx + + − − 1, 0 < x < 1,   3 8  π π 3π  00 00 u(0) = u(1) = u (0) = u (1) = 0. The exact solution of the problem is sin(πx) u(x) = π4 In this example z sin πx f (x, u, y , v , z) = − + cos − 2 − v − y − u 2 + sin πx 3 π 2 cos πx sin πx cos πx + + − − 1. 3 8

Introduction

Existence results

Iterative method


Concluding remarks

In D5 all the conditions of Theorem 2.6 are satisfied with 1 5 c0 = , c1 = c2 = 1, c3 = and q = 0.3354. Therefore, the 32 3 problem has a unique solution and the iterative method converges. The convergence of the iterative method for the example is given in Table 1. N 30 50 100 1000

k 11 11 11 11

error 1.0339e-8 1.3400e-9 8.3781e-11 3.0210e-15

error 1 7.3102e-7 9.5834e-8 6.0186e-9 6.1669e-13

error 2 5.2161e-8 6.7722e-9 4.2351e-10 2.3273e-14

error 3 7.3898e-6 9.7008e-7 6.0994e-8 6.1727e-12

Table 1: The convergence in Example 1.

error = kuk − uk, error 1 = kuk0 − u 0 k, error 2 = kuk00 − u 00 k, error 3 = kuk000 − u 000 k, where u is the exact solution.

Introduction

Existence results

Iterative method


Concluding remarks

Other examples when exact solutions are not known are available in the paper Dang Quang A and Ngo Thi Kim Quy, New Fixed Point Approach For a Fully Nonlinear Fourth Order Boundary Value Problem, Bol. Soc. Paran. Mat., v. 36 4 (2018): 209-223

Introduction

Existence results

Iterative method


Concluding remarks


00

000

u(0) = u (0) = u (1) = u (1) = 0.

0 < x < 1,

(4.2)

• Li (NONRWA 2016) established the existence of a positive solution under many conditions posed on the positive function f (x, u, y , v , z), including a growth condition on u, y , v , z at infinity and a Nagumo-type condition on v and z. • Dang and Ngo (NONRWA 2017) freeing all these conditions but requiring the Lipschitz condition in a bounded domain established the existence and uniqueness of positive solution and proposed iterative method for finding solution.

Introduction

Existence results

Iterative method


For this example   t 3 t 2x   − + , 0≤t≤x ≤1 6 2 G (x, t) =  x 3 x 2t   − + , 0 ≤ x ≤ t ≤ 1. 6 2 ( 0, 0 ≤ t ≤ x ≤ 1, Gb(x, t) = t − x, 0 ≤ x ≤ t ≤ 1. G1 (x, t) = Gx0 (x, t), Gb1 (x, t) = Gbx0 (x, t), 1 1 1 M0 = , M1 = , M2 = , M3 = 1. 8 6 2 c1 (x, t) ≤ 0. G (x, t), Gb(x, t), G1 (x, t) ≥ 0, G

Concluding remarks

Introduction

Existence results

Iterative method


Concluding remarks

Example 2 Consider the problem   (4) 3u 0 u 000 (u 00 )2 x 95 u (x) = − + + + , 576 4 4  u(0) = u 0 (0) 1152 = u 00 (1) = u 000 (1) = 0.

0 < x < 1,

The exact solution of the problem is the positive function u(x) = x 4 − 4x 3 + 6x 2 . The function f (x, u, y , v , z) = −

3yz v2 x 95 + + + , 1152 576 4 4

satisfies neither the condition F 1 nor the condition F 4 of Li (2016), so Li’s theorems cannot guarantee the existence of a solution.

Introduction

Existence results

Iterative method


Concluding remarks

Theorem 2.7 for M = 25 with Lipschitz coefficients c0 = 0, c1 =

25 25 25 , c2 = , c3 = , q ≈ 0.043 384 576 2304

confirms the existence and uniqueness of positive solution. Table 2: The convergence in Example 2.

N 30 50 100 900

k 7 7 7 7

error 4.4409e − 16 1.7764e − 15 3.1086e − 15 3.1086e − 15

error 1 1.3323e − 15 3.9968e − 15 4.8850e − 15 6.2172e − 15

error 2 1.8874e − 15 5.3291e − 15 8.8818e − 15 1.1546e − 14

error 3 3.5527e − 15 1.4211e − 14 1.7764e − 14 1.2434e − 13

where error = kuk − uk, error 1 = kuk0 − u 0 k, error 2 = kuk00 − u 00 k, error 3 = kuk000 − u 000 k, u is the exact solution. The iterations are performed until ek = kuk − uk−1 k ≤ 10−16 .

Introduction

Existence results

Iterative method


Concluding remarks

Other examples when exact solutions are not known are available in the paper 1. Dang Quang A, Ngo Thi Kim Quy, Existence results and iterative method for solving the cantilever beam equation with fully nonlinear term, Nonlinear Analysis: Real World Applications 36 (2017) 56-68.

Introduction

Existence results

Iterative method


Concluding remarks


00

0 < x < 1,

000

u(0) = 0, u (1) = 0, au (0) − bu (0) = 0, cu 00 (1) + du 000 (1) = 0, (4.3) where a, b, c, d ≥ 0, ρ := ad + bc + ac > 0. Feng, Ji and Ge (Nonlin Anal 2009) by using the upper and lower solutions method, established the existence and uniqueness of solution of this problem under conditions that the function f (x, u, y , v , z) satisfies a Nagumo-type condition with respect to α00 , β 00 (where α, β are the upper and lower solutions of the problem, α00 ≤ β 00 ) and is decreasing in x, y and strictly increasing in z.

Introduction

Existence results

Iterative method


Concluding remarks

For this case  2 2 2   x(ac(1 − t)(3t − x ) + ad(6t − x − 3t ) 1  +bc(1 − t)(3 + 3t − 3x) + bd(6 − 3x)), 0 ≤ x ≤ t ≤ 1, G (x, t) =  ac(3xt − 3x 2 t + x 3 t − t 3 ) + ad(6xt − 3x 2 t − t 3 ) 6ρ   +bc(3x − 3x 2 + x 3 − t 3 ) + bd(6x − 3x 2 ), 0 ≤ t ≤ x ≤ 1,

(ct − c − d)(b + ax), 0 ≤ x ≤ t ≤ 1, (b + at)(cx − c − d), 0 ≤ t ≤ x ≤ 1,  ac(1 − t)(t − x 2 ) + ad(2t − x 2 − t 2 )     +bc(1 − t)(1 + t − 2x) + bd(2 − 2x), 1  G1 (x, t) = Gx0 (x, t) = 0 ≤ x ≤ t ≤ 1, 2ρ    act(1 − x)2 + ad(2t − 2tx) + bc(1 − x)2   +bd(2 − 2x), 0 ≤ t ≤ x ≤ 1, 1 Gb(x, t) = ρ

Introduction

Existence results

Iterative method

1 Gb1 (x, t) = Gbx0 (x, t) = ρ


Concluding remarks

a(ct − c − d), 0 ≤ x ≤ t ≤ 1, c(b + at), 0 ≤ t ≤ x ≤ 1,

1 2ad + bc + 6bd 1 ad + bc + 4bd + , M1 = + , 24 12ρ 12 4ρ 1 ac 1 a(d + c/2) 2 b(d + c/2) + , M3 = +max(ad, bc) . M2 = 2 ρ ρ ρ 2 M0 =

Therefore, with a suitable selection of M > 0 the above Theorem 2.3 guarantees the existence of a solution of the problem (4.3) without the Lipschitz condition and any growth conditions of f at infinity.

Introduction

Existence results

Iterative method


Concluding remarks

Example 3 Consider the problem   (4) u 2 (x) u 0 (x) sin u 000 (x) x u (x) = + − u 00 (x) + + + 0.1, 0 < x < 10 10 10  u(0) = 0, u40 (1) = 0, 00 000 00 2u (0) − u (0) = 0, u (1) + u 000 (1) = 0. In this example a = 2, b = 1, c = 1, d = 1 and f (x, u, y , v , z) = −

u2 y sin z x + −v + + + 0.1. 4 10 10 10

Introduction

Existence results

Iterative method


Concluding remarks

In the domain n 9M 13M DM = (x, u, y , v , z) | 0 ≤ x ≤ 1, |u| ≤ , |y | ≤ , 40 30 3M o 12M , |z| ≤ |v | ≤ 25 5 with M = 2 all the conditions of Theorem 2.6 are satisfied. Therefore, the problem has a unique solution. Meanwhile, this result is not guaranteed by Feng, Ji and Ge because the function f (x, u, y , v , z) is not decreasing in u, y and strictly increasing in z.

Introduction

Existence results

Iterative method


Concluding remarks

Remark 4.1 The boundary conditions u(0) = u 0 (1) = u 00 (0) = u 000 (1) = 0, u(0) = u 0 (1) = u 00 (1) = u 000 (0) = 0, considered by Minhos, Gyulov and Santos (Nonlin Anal 2009) with the use of the lower and upper solutions method are particular cases of the boundary conditions in Case 3.

Introduction

Existence results

Iterative method


Concluding remarks

Case 4. (4) u (x) = f (x, u(x), u 0 (x), u 00 (x), u 000 (x)), 0 < x < 1, u(1) = 0, u 0 (0) = 0, au 00 (0) − bu 000 (0) = 0, cu 00 (1) + du 000 (1) = 0,

(4.4)

where a, b, c, d ≥ 0, ρ := ad + bc + ac > 0. In this case  ac(1 − t)(2t − t 2 − x 3 ) + ad(t 3 − x 3 + 3t − 3t 2 )    2 2 2  +bc(1 −  t)(2 + 2t − t − 3x ) + bd(3 − 3x ), 0 ≤ x ≤ t ≤ 1, 1 2 2 G (x, t) = + 3xt − 3t 2 )  (1 − x) ac(2t + 2xt − 3t − x t) + ad(3t 6ρ     +bc(2 + 2x − 3t 2 − x 2 ) + bd(3 + 3x) , 0 ≤ t ≤ x ≤ 1, 1 Gb(x, t) = ρ

(ct − c − d)(b + ax), 0 ≤ x ≤ t ≤ 1, (b + at)(cx − c − d), 0 ≤ t ≤ x ≤ 1,

Introduction

Existence results

Iterative method


Remark 4.2 The boundary conditions u(1) = u 0 (0) = u 00 (0) = u 000 (1) = 0, mentioned in (Nonlin Anal 2009) is a particular case of the boundary conditions in Case 4.

Concluding remarks

Introduction

Existence results

Iterative method


Concluding remarks

Other cases: (4) u (x) = f (x, u(x), u 0 (x), u 00 (x), u 000 (x)), 0 < x < 1, u(0) = 0, u 0 (0) = 0, au 00 (0) − bu 000 (0) = 0, cu 00 (1) + du 000 (1) = 0,   x 2 acx(t − 1) − adx + 3bc(t − 1) − 3bd , 0 ≤ x ≤ t ≤ 1,  1 2 3 3 2 2 2 3 G (x, t) =  ac(3xt + x t − t − 3x t) + ad(3xt − 3x t − t ) 6ρ  2 3 2 3 2 +bc(3xt − t − 3x + x ) − 3bdx , 0 ≤ t ≤ x ≤ 1,

Introduction

Existence results

Iterative method


Concluding remarks

u (4) (x) = f (x, u(x), u 0 (x), u 00 (x), u 000 (x)), 0 < x < 1, u(1) = 0, u 0 (1) = 0, au 00 (0) − bu 000 (0) = 0, cu 00 (1) + du 000 (1) = 0,

 ac(t − 1)(t 2 + t + x 3 − 3xt) + ad(t 3 − 3xt 2 − 3t + 6xt − x 3 )     +bc(t − 1)(t 2 + t + 1 + 3x 2 − 3x − 3xt) − 3bd(x − 1)2 ,  1  0 ≤ x ≤ t ≤ 1, G (x, t) = 6ρ  2   (x − 1) act(x − 1) − 3adt + bc(x − 1) − 3bd ,    0 ≤ t ≤ x ≤ 1,

Introduction

Existence results

Iterative method


Concluding remarks

The BVP u (4) (x) = f (x, u(x), u 0 (x), u 00 (x), u 000 (x)),

0 < x < 1,

B1 (u) ¯ = 0, B2 (u) ¯ = 0, ¯ ¯ B3 (u ) = 0, B4 (u ) = 0, contains almost cases considered by many authors by different methods but they obtained results on existence of solution under strong assumptions on the nonlinear term. Our approach is based on the reduction of the BVP to operator equation for the right-hand side but not for the function to be sought. The conditions required for the nonlinear term is weaker since it is needed to consider it in bounded domains and does not require the monotony of it. There are examples when the methods of other authors do not work while our method works well.

Introduction

Existence results

Iterative method


Concluding remarks

When the BVP has a unique solution we propose an iterative method for solution. This method reduces the solution of fully fourth order nonlinear BVPs to a sequence of linear second order BVPs. The efficiency of the method is shown on examples. The proposed approach can be applicable to other nonlinear BVPs when equation and boundary conditions may be nonlocal, the order of equations may be 3, 4 or higher, ODE and PDE.

Introduction

Existence results

Iterative method


Concluding remarks

LIST OF OUR WORKS RELATED TO THE APPROACH Dang Q.A, Ngo T.K.Q, Existence results and iterative method for solving the cantilever beam equation with fully nonlinear term, Nonlinear Analysis: Real World Applications 36 (2017) 56-68. Dang Q.A, Dang Q.L. & Ngo TT.K.Q., A novel efficient method for nonlinear boundary value problems, Numerical Algorithms 76 (2017) 427-439. Dang Quang A and Ngo T.K.Q., New Fixed Point Approach For a Fully Nonlinear Fourth Order Boundary Value Problem, Bol. Soc. Paran. Mat., v. 36 4 (2018): 209-223, doi:10.5269/bspm.v36i4.33584. Dang Q. A and Nguyen T.H., The Unique Solvability and Approximation of BVP for a Nonlinear Fourth Order Kirchhoff Type Equation, East Asian Journal on Applied Mathematics, Vol. 8 (2018), No. 2, pp. 323-335.

Introduction

Existence results

Iterative method


Concluding remarks

Dang Q. A, Dang Q.L., A simple efficient method for solving sixth-order nonlinear boundary value problems, Comp. Appl. Math. DOI 10.1007/s40314-018-0643-1, Published online 16 May 2018. Dang Q. A and Nguyen T.H., Existence results and iterative method for solving a nonlinear biharmonic equation of Kirchhoff type, Computers and Mathematics with Applications 76 (2018) 11-22. Dang Q.A, et al., Solving a Nonlinear Biharmonic Boundary Value Problem, Journal of Computer Science and Cybernetics, V.33, N.4 (2017), 308-324. Dang Q.A, Nguyen T.H., Existence results and iterative methods for solving a nonlocal fourth order boundary value problem, Journal of Mathematical Applications, Vol. XIV, No 2, 2016,63-78.

Introduction

Existence results

Iterative method


Concluding remarks

Thanks for your attention!