Solving some nonlinear boundary value problems for fourth order differential equations

In this chapter, we investigate the unique solvability and iterative method

for five boundary value problems for local or nonlocal nonlinear fourth order

differential equations with different boundary conditions: The case of boundary

conditions of simply supported type, combined boundary conditions, Dirichlet

boundary condition, nonlinear boundary conditions. By using the reduction of

these problems to the operator equations for the function to be sought or for an

intermediate function, we prove that under some assumptions, which are easy to

verify, the operator is contractive. Then, the uniqueness of a solution is established, and the iterative method for solving the problem converges. We also give

some examples for illustrating the applicability of the obtained theoretical results,

including examples showing the advantages of the method in the thesis compared

with the methods of other authors.

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of iterative methods. The thesis is written on the basis of articles [A1]-[A8] in the list of works of the author related to the thesis. Besides the introduction, conclusion and references, the contents of the thesis are presented in three chapters. The results in the thesis were reported and discussed at: 1. 11th Workshop on Optimization and Scientific Computing, Ba Vi, 24-27/4/2013. 4 2. 4th National Conference on Applied Mathematics, Hanoi, 23-25/12/2015. 3. 14th Workshop on Optimization and Scientific Computing, Ba Vi, 21-23/4/2016. 4. Conference of Applied Mathematics and Informatics, Hanoi University of Science and Technology, 12-13/11/2016. 5. 10th National Conference on Fundamental and Applied Information Tech- nology Research (FAIR’10), Da Nang, 17-18/8/2017. 6. The second Vietnam International Applied Mathematics Conference (VI- AMC 2017), Ho Chi Minh, December 15 to 18, 2017. 7. Scientific Seminar of the Department of Mathematical methods in Informa- tion Technology, Institute of Information Technology, Vietnam Academy of Science and Technology. 5 Chapter 1 Preliminary knowledge This chapter presents some preparation knowledge needed for subsequent chapters referenced from the literatures of A.N. Kolmogorov and S.V. Fomin (1957), E. Zeidler (1986), A.A. Sammarskii (1989, 2001), A. Granas and J. Dugundji (2003), J. Li (2005), Dang Quang A (2009), R.L. Burden (2011). • Section 1.1 recalls three fixed point theorems: Brouwer fixed point theorem, Schauder fixed point theorem, Banach fixed point theorem. • Section 1.2 presents the definition of the Green function for the boundary value problem for linear differential equations of order n and some specific examples of how to define the Green function of boundary problems for second order and fourth order differential equations with different boundary conditions. • Section 1.3 gives some formulas for approximation derivatives and integrals with second order and fourth order accuracy. • Section 1.4 presents the formula for approximation of Poisson equation with fourth order accuracy. • Section 1.5 mentions the elimination method for three-point equations and the cyclic reduction method for three-point vector equations. 6 Chapter 2 The existence and uniqueness of a solution and the iterative method for solving boundary value problems for nonlinear fourth order ordinary equations Chapter 2 investigates the unique solvability and an iterative method for solv- ing five boundary value problems for nonlinear fourth order ordinary differential equations with different types of boundary conditions: simply supported type, Dirichlet boundary condition, combined boundary conditions, nonlinear bound- ary conditions. By using the reduction of these problems to the operator equations for the function to be sought or for an intermediate function, we prove that under some assumptions, which are easy to verify, the operator is contractive. Then, the uniqueness of a solution is established, and the iterative method for solving the problem converges. This chapter is written on the basis of articles [A2]-[A4], [A6]-[A8] in the list of works of the author related to the thesis. 2.1. The boundary value problem for the local nonlinear fourth order differential equation 2.1.1. The case of combined boundary conditions The thesis presents in detail the results of the work [A4] for the problem u(4)(x) = f(x, u(x), u′(x), u′′(x), u′′′(x)), 0 < x < 1, u(0) = 0, u′(1) = 0, au′′(0)− bu′′′(0) = 0, cu′′(1) + du′′′(1) = 0, (2.1.1) where a, b, c, d ≥ 0, ρ := ad + bc + ac > 0 and f : [0, 1]× R4 → R is a continuous function. 2.1.1.1. The existence and uniqueness of a solution For function ϕ(x) ∈ C[0, 1], consider the nonlinear operator A : C[0, 1] → C[0, 1] defined by (Aϕ)(x) = f(x, u(x), u′(x), u′′(x), u′′′(x)), (2.1.2) 7 where u(x) is a solution of the problem u(4)(x) = ϕ(x), 0 < x < 1, u(0) = 0, u′(1) = 0, au′′(0)− bu′′′(0) = 0, cu′′(1) + du′′′(1) = 0. (2.1.3) Proposition 2.1. A function ϕ(x) is a fixed point of the operator A, i.e., ϕ(x) is a solution of the operator equation ϕ = Aϕ if and only if the function u(x) determined from the boundary value problem (2.1.3) satisfiesthe problem (2.1.1). Set v(x) = u′′(x), the problem (2.1.3) can be decomposed into two second problems{ v′′(x) = ϕ(x), 0 < x < 1, av(0)− bv′(0) = 0, cv(1) + dv′(1) = 0, { u′′(x) = v(x), 0 < x < 1, u(0) = 0, u′(1) = 0. Then (Aϕ)(x) = f(x, u(x), y(x), v(x), z(x)), y(x) = u′(x), z(x) = v′(x). For any number M > 0, we define the set DM = { (x, u, y, v, z) | 0 ≤ x ≤ 1, |u| ≤ ρ1M, |y| ≤ ρ2M, |v| ≤ ρ3M, |z| ≤ ρ4M } , where ρ1 = 1 24 + 2ad+ bc+ 6bd 12ρ , ρ2 = 1 12 + ad+ bc+ 4bd 4ρ , ρ3 = 1 2 (a(d+ c/2) ρ )2 + b(d+ c/2) ρ , ρ4 = 1 ρ (ac 2 + max(ad, bc) ) . Denote the closed ball in the space C[0, 1] by B[O,M ]. Lemma 2.1. Assume that there exist constants M > 0, K1, K2, K3, K4 ≥ 0 such that |f(x, u, y, v, z)| ≤ M for all (x, u, y, v, z) ∈ DM . Then, the operator A maps B[O,M ] into itself. Furthermore, if |f(x, u2,y2, v2, z2)− f(x, u1, y1, v1, z1)| ≤ K1|u2 − u1|+K2|y2 − y1|+K3|v2 − v1|+K4|z2 − z1| (2.1.4) for all (t, ui, yi, vi, zi) ∈ DM (i = 1, 2) and q = K1ρ1 +K2ρ2 +K3ρ3 +K4ρ4 < 1 (2.1.5) then A is a contraction operator in B[O,M ]. Theorem 2.1. Assume that all the conditions of Lemma 2.1 are satisfied. Then the problem (2.1.1) has a unique solution u and ‖u‖ ≤ ρ1M, ‖u′‖ ≤ ρ2M, ‖u′′‖ ≤ ρ3M, ‖u′′′‖ ≤ ρ4M. 8 Denote D+M = { (x, u, y, v, z) | 0 ≤ x ≤ 1, 0 ≤ u ≤ ρ1M, 0 ≤ y ≤ ρ2M, −ρ3M ≤ v ≤ 0, −ρ4M ≤ z ≤ ρ4M } . Theorem 2.2. (Positivity of solution) Suppose that in D+M the function f is such that 0 ≤ f(x, u, y, v, z) ≤ M and the conditions (2.1.4), (2.1.5) of Lemma 2.1 are satisfied. Then the problem (2.1.1) has a unique nonnegative solution. 2.1.1.2. Solution method The iterative method for solving the problem (2.1.1) is proposed as follows: Iterative method 2.1.1a i) Given an initial approximation ϕ0(x), for example, ϕ0(x) = f(x, 0, 0, 0, 0). ii) Knowing ϕk(x) (k = 0, 1, 2, ...) solve consecutively two problems{ (vk)′′(x) = ϕk(x), 0 < x < 1, avk(0)− b(vk)′(0) = 0, cvk(1) + d(vk)′(1) = 0, { (uk)′′(x) = vk(x), 0 < x < 1, uk(0) = (uk)′(1) = 0. iii) Update ϕk+1(x) = f(x, uk(x), (uk)′(x), vk(x), (vk)′(x)). Set pk = qk 1− q‖ϕ 1 − ϕ0‖. We have the following result: Theorem 2.3. Under the assumptions of Lemma 2.1, Iterative method 2.1.1a converges and there hold the estimates ‖uk − u‖ ≤ ρ1pk, ‖(uk)′ − u′‖ ≤ ρ2pk, ‖(uk)′′ − u′′‖ ≤ ρ3pk, ‖(uk)′′′ − u′′′‖ ≤ ρ4pk, where u is the exact solution of the problem (2.1.1). Consider the second order boundary value problem{ v′′(x) = g(x), x ∈ (0, 1), c0v(0)− c1v′(0) = C, d0v(1) + d1v′(1) = D, where c0, c1, d0, d1 ≥ 0, c20 + c21 > 0, d20 + d21 > 0, C,D ∈ R. Based on the results in the work [A8], we construct a difference scheme of fourth order accuracy for solving this problem as follows c0v0 − c1 12h (−25v0 + 48v1 − 36v2 + 16v3 − 3v4) = F0, vi−1 − 2vi + vi+1 = Fi, i = 1, 2, ..., N − 1, d0vN + d1 12h (25vN − 48vN−1 + 36vN−2 − 16vN−3 + 3vN−4) = FN , where F0 = C, FN = D, Fi = h 2 ( gi + h2 12 Λgi + h4 360 Λ2gi ) , i = 1, 2, ..., N − 1. 9 We introduce the uniform grid ωh = {xi = ih, i = 0, 1, ..., N ; h = 1/N} in the interval [0, 1]. Denote by V k, Uk,Φk the grid functions. For the general grid function V on ωh we denote Vi = V (xi) and denote by V ′ i the first difference derivative with fourth order accuracy. Consider the following iterative method at discrete level for solving the problem (2.1.1): Iterative method 2.1.1b i) Given Φ0i = f(xi, 0, 0, 0, 0), i = 0, 1, 2, ..., N. ii) Knowing Φk (k = 0, 1, 2, ...) solve consecutively two problems aV k0 − b 12h (−25V k0 + 48V k1 − 36V k2 + 16V k3 − 3Uk4 ) = 0, ΛV ki = Φ k i + h2 12 ΛΦki + h4 360 Λ2Φki , i = 1, 2, ..., N − 1, cV kN + d 12h (25V kN − 48V kN−1 + 36V kN−2 − 16V kN−3 + 3V kN−4) = 0, Uk0 = 0, ΛUki = V k i + h2 12 ΛV ki + h4 360 Λ2V ki , i = 1, 2, ..., N − 1, 25UkN − 48UkN−1 + 36UkN−2 − 16UkN−3 + 3UkN−4 12h = 0. iii) Update Φk+1i = f(xi, U k i , (U k)′i, V k i , (V k)′i), i = 0, 1, 2, ..., N. We give some examples for illustrating the applicability of the obtained the- oretical results, including examples of advantages of the method in the thesis compared to the methods of H. Feng, D. Ji, W. Ge (2009): According to the proposed method, the problem has a unique solution meanwhile Feng’s method cannot ensure the existence of a solution. 2.1.2. The case of Dirichlet boundary condition The thesis presents in detail the results of the work [A3] for the problem u(4)(x) = f(x, u(x), u′(x), u′′(x), u′′′(x)), a < x < b, u(a) = u(b) = 0, u′(a) = u′(b) = 0, (2.1.6) 2.1.2.1. The existence and uniqueness of a solution For function ϕ(x) ∈ C[0, 1], consider the nonlinear problem A : C[a, b] → C[a, b] defined by (Aϕ)(x) = f(x, u(x), u′(x), u′′(x), u′′′(x)), (2.1.7) 10 where u(x) is the solution of the problem u(4)(x) = ϕ(x), a < x < b, u(a) = u(b) = 0, u′(a) = u′(b) = 0. (2.1.8) Proposition 2.2. 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 (2.1.9) then the function u(x) determined from the boundary value problem (2.1.8) solves the problem (2.1.6). Conversely, ifu(x) is a solution of the boundary value problem (2.1.6) then the function ϕ(x) = f(x, u(x), u′(x), u′′(x), u′′′(x)) is a fixed point of the operator A defined above by (2.1.7) and (2.1.8). Thus, the solution of the problem (2.1.6) is reduced to the solution of the operator equation (2.1.9). For any number M > 0, we define the set DM = { (x, u, y, v, z) | a ≤ x ≤ b, |u| ≤ C4,0(b− a)4M, |y| ≤ C4,1(b− a)3M, |v| ≤ C4,2(b− a)2M, |z| ≤ C4,3(b− a)M } , where C4,0 = 1/384, C4,1 = 1/72 √ 3, C4,2 = 1/12, C4,3 = 1/2. By using Schauder fixed point theorem and Bannach fixed point theorem for the operator A, we establish the existence and uniqueness theorems of the problem (2.1.6). Theorem 2.4. Suppose that the function f is continuous and there exists constant M > 0 such that |f(x, u, y, v, z)| ≤ M for all (x, u, y, v, z) ∈ DM . Then, the problem (2.1.6) has at least a solution. Theorem 2.5. Suppose that the assumptions of Theorem 2.4 hold. Additionally, assume that there exist constants K0, K1, K2, K3 ≥ 0 such that |f(x, u2, y2, v2, z2)− f(x, u1, y1, v1, z1)| ≤ K0|u2 − u1|+K1|y2 − y1| +K2|v2 − v1|+K3|z2 − z1|, (2.1.10) for all (x, ui, yi, vi, zi) ∈ DM (i = 1, 2) and q = 3∑ k=0 KiC4,k(b− a)4−k < 1. (2.1.11) Then the problem (2.1.6) has a unique solution u and ‖u‖ ≤ C4,0(b− a)4M, ‖u′‖ ≤ C4,1(b− a)3M, ‖u′′‖ ≤ C4,2(b− a)2M, ‖u′′′‖ ≤ C4,3(b− a)M. 11 Denote D+M = { (x, u, y, v, z) | a ≤ x ≤ b, 0 ≤ u ≤ C4,0(b− a)4M, |y| ≤ C4,1(b− a)3M, |v| ≤ C4,2(b− a)2M, |z| ≤ C4,3(b− a)M } . Theorem 2.6. (Positivity of solution) Suppose that in D+M the function f is such that 0 ≤ f(t, x, y, u, z) ≤ M and the conditions (2.1.10), (2.1.11) of Theorem 2.5 are satisfied. The the problem (2.1.6) has a unique nonnegative solution. 2.1.2.2. Solution method and numerical examples The iterative method for solving the problem (2.1.6) is proposed as follows: Iterative method 2.1.2 i) Given ϕ0(x), for example, ϕ0(x) = f(x, 0, 0, 0, 0). ii) Knowing ϕk(x), (k = 0, 1, 2, ...) calculate uk(x) = ∫ b a G(x, t)ϕk(t)dt and the derivatives u (m) k (x) of uk(x) u (m) k (x) = ∫ b a ∂mG(x, t) ∂xm ϕk(t)dt (m = 1, 2, 3). iii) Update ϕk+1(x) = f(x, uk(x), u ′ k(x), u ′′ k(x), u ′′′ k (x)). Set pk = qk 1− q‖ϕ1 − ϕ0‖. We have the following result: Theorem 2.7. Under the assumptions of Theorem 2.5, Iterative method 2.1.2 converges with the rate of geometric progression and there hold the estimates ‖uk − u‖ ≤ C4,0(b− a)4pk, ‖u′k − u′‖ ≤ C4,1(b− a)3pk, ‖u′′k − u′′‖ ≤ C4,2(b− a)2pk, ‖u′′′k − u′′′‖ ≤ C4,3(b− a)pk, where u is the exact solution of the problem (2.1.6). Ch 2.1. Consider the problem u(4)(x) = f(x, u(x), u′(x), u′′(x), u′′′(x)), a < x < b, u(a) = A1, u(b) = B1, u ′(a) = A2, u′(b) = B2. (2.1.12) Set v(x) = u(x) − P (x), where P (x) is the third degree polynomial satisfying the boundary conditions in this problem and denote F (x, v(x), v′(x), v′′(x), v′′′(x)) = f(x, v(x) + P (x), (v(x) + P (x))′, (v(x) + P (x))′′, (v(x) + P (x))′′′). Then, the problem (2.1.12) becomes{ v(4)(x) = F (x, v(x), v′(x), v′′(x), v′′′(x)), a < x < b, v(a) = v(b) = 0, v′(a) = v′(b) = 0. Therefore, we can apply the results derived above to this problem. 12 Theorem 2.8. Suppose that the function f is continuous and there exists constan M > 0 such that |f(x, v0, v1, v2, v3)| ≤M for all (x, v0, v1, v2, v3) ∈ DM , where DM = { (x, v0, v1, v2, v3) | a ≤ x ≤ b, |vi| ≤ max x∈[a,b] |P (i)(x)| + C4,i(b− a)4−iM, i = 0, 1, 2, 3 } . Then, the problem (2.1.12) has at least a solution. We give some examples for illustrating the applicability of the obtained the- oretical results, including examples of advantages of the method in the thesis compared to the methods of R.P. Agarwal (1984): Agarwal can only establish the existence of a solution of the problem or does not guarantee the existence of a solution of the problem meanwhile according to the proposed method, the problem has a unique solution or a unique positive solution. 2.1.3. The case of nonlinear boundary conditions The thesis presents in detail the results of the work [A7] for the problem{ u(4)(x) = f(x, u, u′), 0 < x < L, u(0) = 0, u(L) = 0, u′′(0) = g(u′(0)), u′′(L) = h(u′(L)). (2.1.13) Set u′ = v, u′′ = w. Then, the problem (2.1.13) is decomposed to the problems for w v u w ′′(x) = f ( x, x∫ 0 v(t)dt, v(x) ) , 0 < x < L, w(0) = g(v(0)), w(L) = h(v(L)), { u′′(x) = w(x), 0 < x < L, u(0) = 0, u(L) = 0. The solution u(x) from these problems depends on the function v. Conse- quently, its derivative u′ also depends on v. Therefore, we can represent this dependence by an operator T : C[0, L]→ C[0, L] defined by Tv = u′. Combining with u′ = v we get the operator equation v = Tv, i.e., v is a fixed point of T . To consider properties of the operator T, we introduce the space S = { v ∈ C[0, L], L∫ 0 v(t)dt = 0 } . We make the following assumptions on the given functions in the problem (2.1.13): there exist constants λf , λg, λh ≥ 0 such that |f(x, u, v)− f(x, u, v)| ≤ λf max |u− u|, |v − v|, |g(u)− g(u)| ≤ λg|u− u|, |h(u)− h(u)| ≤ λh|u− u|, (2.1.14) for any u, u, v, v. Applying Banach fixed point theorem for T, we establish the existence and uniqueness of a solution of the problem. 13 Proposition 2.3. With assumption (2.1.14), the problem (2.1.13) has a unique solution if q = L3 16 λf max (L 2 , 1 ) + L 2 (λg + λh) < 1. (2.1.15) The iterative method for solving the problem (2.1.13) is proposed as follows: Iterative method 2.1.3 (i) Given an initial approximation v0(x), for example, v0(x) = 0. (ii) Knowing vk(x) (k = 0, 1, 2, ...) solve consecutively two problems w′′k(x) = f ( x, ∫ x 0 vk(t)dt, vk(x) ) , 0 < x < L, wk(0) = g(vk(0)), wk(L) = h(vk(L)), { u′′k(x) = wk(x), 0 < x < L, uk(0) = uk(L) = 0. (iii) Update vk+1(x) = u ′ k(x). Theorem 2.9. Under the assumptions (2.1.14), (2.1.15), Iterative method 2.1.3 converges with rate of geometric progression with the quotient q, and there hold the estimates ‖u′k − u′‖ ≤ qk 1− q‖v1 − v0‖, ‖uk − u‖ ≤ L 2 ‖u′k − u′‖, where u is the exact solution of the original problem (2.1.13). For testing the convergence of the method, we perform some experiments for the case of the known exact solutions and also for the case of the unknown exact solutions. 2.2. The boundary value problem for the nonlocal nonlinear fourth order differential equation 2.2.1. The case of boundary conditions of simply supported type The thesis presents in detail the results of the work [A2] for the problem u(4)(x)− M̂ (∫ L 0 |u′(s)|2ds ) u′′(x) = f(x, u(x), u′(x), u′′(x), u′′′(x)), 0 < x < L, u(0) = u(L) = 0, u′′(0) = u′′(L) = 0. (2.2.1) 2.2.1.1. The existence and uniqueness of a solution For function ϕ(x) ∈ C[0, L], consider the nonlinear operator A : C[0, L] → C[0, L] defined by (Aϕ)(x) = M̂(‖u′‖22)u′′(x) + f(x, u(x), u′(x), u′′(x), u′′′(x)), (2.2.2) 14 where ‖.‖2 is the norm in L2[0, L], u(x) is a solution of the problem u(4)(x) = ϕ(x), 0 < x < L, u(0) = u(L) = 0, u′′(0) = u′′(L) = 0. (2.2.3) Proposition 2.4. A function ϕ(x) is a fixed point of the operator A, i.e., ϕ(x) is a solution of the operator equation ϕ = Aϕ if and only if the function u(x) determined from the boundary value problem (2.2.3) satisfies the problem (2.2.1). By setting v(x) = u′′(x), the problem (2.2.3) is decomposed to the problems{ v′′(x) = ϕ(x), 0 < x < L, v(0) = v(L) = 0, { u′′(x) = v(x), 0 < x < L, u(0) = u(L) = 0. Then the operator A is represented in the form (Aϕ)(x) := M̂(‖y‖22)v(x)+f(x, u(x), y(x), v(x), z(x)), y(x) = u′(x), z(x) = v′(x). For any number R > 0, we define the set DR := { (x, u, y, v, z) | 0 ≤ x ≤ L, |u| ≤ 5L 4R 384 , |y| ≤ L 3R 24 , |v| ≤ L 2R 8 , |z| ≤ LR 2 } . Let B[O,R] denote the closed ball in the space C[0, L]. Lemma 2.2. If there are constants R > 0, 0 ≤ m ≤ 8 L2 , λ M̂ , K1, K2, K3, K4 ≥ 0 such that |M̂(s)| ≤ m, |f(x, u, y, v, z)| ≤ R ( 1− mL 2 8 ) , for all (x, u, y, v, z) ∈ DR and 0 ≤ s ≤ R 2L7 576 , then, the operator A maps B[O,R] into itself. If, in addition, |M̂(s2)− M̂(s1)| ≤ λM̂ |s2 − s1|, |f(x, u2, y2, v2, z2)− f(x, u1, y1, v1, z1)| ≤ K1|u2 − u1|+K2|y2 − y1|+K3|v2 − v1|+K4|z2 − z1|, for all (x, ui, yi, vi, zi) ∈ DR, 0 ≤ si ≤ R 2L7 576 (i = 1, 2) and q = K1 5L4 384 +K2 L3 24 +K3 L2 8 +K4 L 2 + mL2 8 + λ M̂ R2L9 2304 < 1 then A is a contraction operator in B[O,R]. Theorem 2.10. In conditions of Lemma 2.2, the problem (2.2.1) has a unique solution u such that ‖u‖ ≤ 5L 4 384 R, ‖u′‖ ≤ L 3 24 R, ‖u′′‖ ≤ L 2 8 R, ‖u′′′‖ ≤ L 2 R. 15 2.2.1.2. Iterative method and numerical examples The iterative method for solving the problem (2.2.1) is proposed as follows: Iterative method 2.2.1 i) Given an initial approximation ϕ0(x), for example, ϕ0(x) = f(x, 0, 0, 0, 0). ii) Knowing ϕk(x) (k = 0, 1, 2, ...) solve successively the problems{ v′′k(x) = ϕk(x), 0 < x < L, vk(0) = vk(L) = 0, { u′′k(x) = vk(x), 0 < x < L, uk(0) = uk(L) = 0. iii) Update ϕk+1(x) = M̂(‖u′k‖22)u′′k(x) + f(x, uk(x), u′k(x), u′′k(x), u′′′k (x)). Set pk = qk 1− q‖ϕ1 − ϕ0‖. We have the following theorem: Theorem 2.11. In conditions of Lemma 2.2, Iterative method 2.2.1 converges to the exact solution u of the problem (2.2.1) and ‖uk − u‖ ≤ 5L 4 384 pk, ‖u′k − u′‖ ≤ L3 24 pk, ‖u′′k − u′′‖ ≤ L2 8 pk, ‖u′′′k − u′′′‖ ≤ L 2 pk. We give some examples for illustrating the applicability of the obtained the- oretical results, including examples of advantages of the method in the thesis compared to the methods of P. Amster, P.P. Ca´rdenas Alzate (2008): Accord- ing to the method proposed, the problem has a unique solution meanwhile these authors’s method cannot ensure the existence of a solution. 2.2.2. The case of nonlinear boundary conditions The thesis presents in detail the results of the work [A6] for the problem u(4)(x)− M̂ (∫ L 0 |u′(s)|2ds ) u′′(x) = f(x, u(x)), 0 < x < L, u(0) = u′(0) = u′′(L) = 0, u′′′(L)− M̂ (∫ L 0 |u′(s)|2ds ) u′(L) = g(u(L)). (2.2.4) 2.2.2.1. The existence and uniqueness of a solution By setting v(x) = u′′(x) − M̂(||u′||22)u(x), where ‖.‖2 denotes the norm of L2[0, L], the problem (2.2.4) is reduced to the problems v′′(x) = f(x, u(x)), 0 < x < L, v(L) = −M ( ‖u′‖22 ) u(L), v′(L) = g(u(L)) u′′(x) = M̂ ( ‖u′‖22 ) u(x) + v(x), 0 < x < L, u(0) = u′(0) = 0. 16 We can see that u is a solution of the problem (2.2.4) if and only if it is a solution of the integral equation u(x) = (Tu)(x), where (Tu)(x) = ∫ L 0 G(x, t) [ M̂ ( ‖u′‖22 ) u(t) + ∫ L 0 G(t, s)f(s, u(s))ds+ g(u(L))(t− L)− M̂ ( ‖u′‖22 ) u(L) ] dt. Applying Schauder fixed point theorem and Banach fixed point theorem for the operator T , we establish the existence and uniqueness theorams of the problem (2.2.4). Theorem 2.12. Suppose that f, g, M̂ are continuous functions and there exist constants R,A,B,m > 0 such that |f(t, u)| ≤ A, ∀(t, u) ∈ [0, L]× [−L2R,L2R], |g(u)| ≤ B, ∀u ∈ [−L2R,L2R], |M̂(s)| ≤ m, ∀s ∈ [0, L3R2]. Then, if L 2 2 A+ LB ≤ R(1−mL2), the problem (2.2.4) has at least a solution. Theorem 2.13. Suppose that the assumptions of Theorem 2.12 hold. Further assume that there exist constants λf , λg, λM̂ > 0 such that |f(x, u)− f(x, v)| ≤ λf |u− v|, ∀(x, u), (x, v) ∈ [0, L]× [−L2R,L2R], |g(u)− g(v)| ≤ λg|u− v|, ∀u, v ∈ [−L2R,L2R], |M̂(u)− M̂(v)| ≤ λ M̂ |u− v|, ∀u, v ∈ [0, L3R2]. Then, if q = 4L5R 2 λ M̂ + L 4 2 λf + L 3λg + 2mL 2 < 1, the problem (2.2.4) has a unique solution. 2.2.2.2. Iterative method and numerical examples The iterative method for solving the problem (2.2.4) is proposed as follows: Iterative method 2.2.2 i) Given an initial approximation u0(x), example, u0(x) = 0, in [0, L]. ii) Knowing uk(x) (k = 0, 1, 2, ...) solve consecutively the final value problem v′′k(x) = f(x, uk(x)), 0 < x < L, vk(L) = −M̂ ( ‖u′k‖22 ) uk(L), v ′ k(L) = g(uk(L)), u′′k+1(x) = vk(x) + M̂ ( ‖u′k‖22 ) uk(x), 0 < x < L, uk+1(0) = u ′ k+1(0) = 0. 17 Theorem 2.14. Under the assumptions of Theorem 2.13, Iterative method 2.2.2 converges with rate of geometric progression with the quotient q and there hold the estimates ‖uk − u‖∞ ≤ L‖u′k − u′‖∞ ≤ L2‖u′′k − u′′‖∞ ≤ L2 qk 1− q‖u ′′ 1 − u′′0‖∞, where u is the exact solution of the original problem (2.2.4). We give some examples for illustrating the applicability of the obtained the- oretical results, including examples of advantages of the method in the thesis compared to the methods of T.F. Ma (2003): According to the proposed method, the problem has a unique solution meanwhile Ma’s method can only establish the existence of a solution or cannot ensure the existence of a solution. CONCLUSION OF CHAPTER 2 In this chapter, we investigate the unique solvability and iterative method for five boundary value problems for local or nonlocal nonlinear fourth order differential equations with different boundary conditions: The case of boundary conditions of simply supported type, combined boundary conditions, Dirichlet boundary condition, nonlinear boundary conditions. By using the reduction of these problems to the operator equations for the function to be sought or for an intermediate function, we prove that under some assumptions, which are easy to verify, the operator is contractive. Then, the uniqueness of a solution is estab- lished, and the iterative method for solving the problem converges. We also give some examples for illustrating the applicability of the obtained theoretical results, including examples showing the advantages of the method in the thesis compared with the methods of other authors. 18 Chapter 3 The existence and uniqueness of a solution and the iterative method for solving boundary value problems for nonlinear fourth order partial equations Continuing the development of the techniques in Chapter 2, in Chapter 3, we also obtain the results of the existence and uniqueness of a solution, and the convergence of iterative methods for solving two boundary value problems for a nonlinear biharmonic equation and a nonlinear biharmonic equation of Kirchhoff type. The results of this chapter are presented in articles [A1], [A4] in the list of works of the author related to the thesis. 3.1. The nonlinear boundary value problem for the bihar- monic equation The thesis presents in detail the results of the work [A5] for the problem ∆2u = f(x, u,∆u), x ∈ Ω, u = 0, ∆u = 0, x ∈ Γ, (3.1.1) where Ω is a connected bounded domain in R2 with a smooth (or piecewise smooth) boundary Γ. 3.1.1. The existence and uniqueness of a solution For function ϕ(x) ∈ C(Ω), consider the nonlinear operation A : C(Ω)→ C(Ω) defined by (Aϕ)(x) = f(x, u(x),∆u(x)), (3.1.2) where u(x) is a solution of the problem ∆2u = ϕ(x), x ∈ Ω, u = ∆u = 0, x ∈ Γ. (3.1.3) Proposition 3.1. A function ϕ(x) is a solution of the operator equation Aϕ = ϕ, i.e., ϕ(x) is a fixed point of the operator A defined by (3.1.2)-(3.1.3) if and only if the function u(x) being the solution of the boundary value problem (3.1.3) solves the problem (3.1.1). 19 Lemma 3.1. Suppose that Ω is a connected bounded domain in RK (K ≥ 2) with a smooth boundary (or smooth of each piece) Γ. Then, fo

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