# The Solution of Nonlinear Fourth-Order Differential Equation with Integral Boundary Conditions

Authors: Yanli Fu; Huanmin Yao
Journal: Journal of Function Spaces
 (2014)

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## Abstract

An iterative algorithm is proposed for solving the solution of a nonlinear fourth-order differential equation with integral boundary conditions. Its approximate solutionun(x) is represented in the reproducing kernel space. It is proved that un(x) converges uniformly to the exact solution u(x). Moreover, the derivatives of un(x) are also convergent to the derivatives of u(x). Numerical results show that the method employed in the paper is valid.


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## Body

## 1. Introduction

Integral boundary conditions for evolution problems have various applications in chemical engineering, thermoelasticity, underground water flow, and population dynamics. In fact, boundary value problems (BVPs) involving integral boundary conditions have received considerable attention. For BVPs with integral boundary conditions and comments on their importance, we refer the reader to the papers by Gallardo [1], Karakostas and Tsamatos [2], Lomtatidze and Malaguti [3], and the references therein.In this paper, we study the nonlinear fourth-order differential equation with integral boundary conditions in the reproducing kernel space:(1)u(4)(x)-λf(x,u(x))=0,0<x<1,u(0)=u(1)=∫01‍h1(s)u(s)ds,u′′(0)=u′′(1)=∫01‍h2(s)u′′(s)ds,
where λ is a nonnegative constant. The existence of (1) was obtained in [4, 5]. However, the literature of numerical analysis contains little on the solution of (1).In the reproducing kernel spaces, many classical problems such as population models and complex dynamics have been solved [6, 7]. For more details of the reproducing kernel spaces, we refer the readers to [8–11]. In [8–11], two-point BVPs were solved in the reproducing kernel space which satisfied two-point boundary conditions. In this paper, however, we can solve integral boundary problems in the reproducing kernel space which satisfies integral boundary conditions. Reference [12] investigated the existence and multiplicity of symmetric positive solutions for a class of Laplacian fourth-order differential equations with integral boundary conditions. The arguments were based upon a specially constructed cone and the fixed point theory for cones. In [13], the authors are concerned with a new algorithm for giving the analytical and approximate solutions of a class of fourth-order in the new reproducing kernel space. Theorem on the completeness of the system of eigenvector and associated vectors of the operator were proved. Based upon a specially constructed cone and the fixed point theory in a cone, [14] established various results on the existence and nonexistence of symmetric positive solutions to fourth-order boundary value problems with integral boundary conditions.In the paper, the representation of the exact and approximate solutions of (1) in the reproducing kernel space is given. The advantages of this method are as follows. First, the conditions for determining solution in (1) can be imposed on the reproducing kernel space and therefore the reproducing kernel satisfying the conditions for determining solution can be calculated. We will use the kernel to solve problems. Second, the iterative sequence un(x) of approximate solutions converges in C4 to the solution u(x).This paper is organized as follows. Several reproducing spaces and a linear operator are introduced in Section2. Section 3 provides the main results; the exact and approximate solutions of (1) and an iterative method are developed for the kind of problems in the reproducing kernel space. We verify that the approximate solution converges to the exact solution uniformly. Some numerical experiments are illustrated in Section 4. Finally, Section 5 is the conclusions.


## 2. Preliminaries

### 2.1. The Reproducing Kernel SpaceW5[0,1]

The inner product spaceW5[0,1] is defined as follows: W5[0,1]={u(x)∣u(4)(x) is a absolutely continuous real-valued function, u(5)(x)∈L2[0,1], u(0)=u(1)=∫01‍h1(x)u(x)dx, u′′(0)=u′′(1)=∫01‍h2(x)u′′(x)dx}.The inner product and norm inW5[0,1] are defined, respectively, by
(2)〈u(x),v(x)〉W5=∑i=04‍u(i)(0)v(i)(0)+∫01‍u(5)(x)v(5)(x)dx,(3)∥u∥W5=〈u,u〉W5,
where u,v∈W5[0,1].Theorem 1.
The spaceW5[0,1] is a complete reproducing kernel space. That is, for each fixed x∈[0,1], there exists K5(y,x)∈W5[0,1], such that 〈u(y),K5(y,x)〉W5=u(x) for any u(y)∈W5[0,1] and y∈[0,1]. The reproducing kernel K5(y,x) can be written as
(4)K5(y,x)={∑i=110‍ai(x)yi-1-∑i=12‍ci(x)Hi(y),y≤x,∑i=110‍bi(x)yi-1-∑i=12‍ci(x)Hi(y),y>x,
where(5)H1(y)=∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍h1(y)dydydydydydydydydydy,H2(y)=∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍h2′′(y)dydydydydydydydydydy.ai(x), bi(x)(i=1,2,…10) and c1(x), c2(x), which are related to h1(x) and h2(x), will be given in the following proof.Proof.
(i) The proof of the completeness and reproducing property ofW5[0,1] is similar to the proof of Theorem  1.3.1 in [15].
(ii) Now, let us find out the expression form of the reproducing kernel functionK5(y,x) in W5[0,1].
Through several integration by parts for (2), we have
(6)〈u(y),K5(y,x)〉W5=∑i=04‍u(i)(0)[∂yiK5(0,x)-(-1)4-i∂y9-iK5(0,x)]+∑i=04‍(-1)4-iu(i)(1)∂y9-iK5(1,x)-∫01‍u(y)∂y10K5(y,x)dy,
and since u(x)∈W5[0,1], it follows that
(7)u(0)=u(1)=∫01‍h1(x)u(x)dx,u′′(0)=u′′(1)=∫01‍h2(x)u′′(x)dx.
Then, we have
(8)〈u(y),K5(y,x)〉W5=∑i=04‍u(i)(0)[∂yiK5(0,x)-(-1)4-i∂y9-iK5(0,x)]+∑i=04‍(-1)4-iu(i)(1)∂y9-iK5(1,x)-∫01‍u(y)∂y10K5(y,x)dy+c1[u(0)-∫01‍h1(y)u(y)dy]+c2[u′′(0)-∫01‍h2(y)u′′(y)dy]=∑i=04‍u(i)(0)[∂yiK5(0,x)-(-1)4-i∂y9-iK5(0,x)]+∑i=04‍(-1)4-iu(i)(1)∂y9-iK5(1,x)-∫01‍u(y)∂y10K5(y,x)dy+c1[u(0)-∫01‍h1(y)u(y)dy]+c2[∫01‍h2′′(y)u(y)dyu′′(0)-u′(1)h2(1)+u′(0)h2(0)iiiiiiiiiiiiiiiiiiiiii+h2′(1)u(1)-h2′(0)u(0)iiiiiiiiiiiiiiiiiii-∫01‍h2′′(y)u(y)dy].
Note that the property of the reproducing kernel〈u(y),K5(y,x)〉W5=u(x) and u(0)=u(1), u′′(0)=u′′(1), and K5(y,x) is the solution of the following generalized differential equation:
(9)-(∂y10K5(y,x)+c1h1(y)+c2h2′′(y))=δ(y-x),K5(0,x)-∂y9K5(0,x)+c1+∂y9K5(1,x)-c2h2′(0)+c2h2′(1)=0,∂y1K5(0,x)+∂y8K5(0,x)+c2h2(0)=0,∂y2K5(0,x)-∂y7K5(0,x)+c2+∂y7K5(1,x)=0,∂y3K5(0,x)+∂y6K5(0,x)=0,∂y4K5(0,x)-∂y5K5(0,x)=0,-∂y8K5(1,x)-c2h2(1)=0,-∂y6K5(1,x)=0,∂y5K5(1,x)=0.
Whiley≠x, it is easy to know that K5(y,x) is the solution of the following constant linear homogeneous differential equation with 10 orders; that is,
(10)-(∂y10K5(y,x)+c1h1(y)+c2h2′′(y))=0,
with the following boundary conditions:
(11)K5(0,x)-∂y9K5(0,x)+c1+∂y9K5(1,x)-c2h2′(0)+c2h2′(1)=0,∂y1K5(0,x)+∂y8K5(0,x)+c2h2(0)=0,∂y2K5(0,x)-∂y7K5(0,x)+c2+∂y7K5(1,x)=0,∂y3K5(0,x)+∂y6K5(0,x)=0,∂y4K5(0,x)-∂y5K5(0,x)=0,-∂y8K5(1,x)-c2h2(1)=0,-∂y6K5(1,x)=0,∂y5K5(1,x)=0.
We know that (9) has the characteristic equation λ10=0, and the eigenvalue λ=0 is a root whose multiplicity is 10. Therefore, the general solution of (9) is
(12)K5(y,x)={∑i=110‍ai(x)yi-1-∑i=12‍ci(x)Hi(y),y≤x,∑i=110‍bi(x)yi-1-∑i=12‍ci(x)Hi(y),y>x,
where(13)H1(y)=∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍h1(y)dydydydydydydydydydy,H2(y)=∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍h2′′(y)dydydydydydydydydydy. Now, we are ready to calculate the coefficient ai(x), bi(x)(i=1,2,…,10) and c1(x), c2(x).
Since(14)-(∂y10K5(y,x)+c1h1(y)+c2h2′′(y))=δ(y-x),
we have
(15)∂ykK5(x+0,x)=∂ykK5(x-0,x),k=0,1,2,…,8,(16)∂y9K5(x-0,x)-∂y9K5(x+0,x)=1.
SinceK5(y,x)∈W5[0,1], it follows that
(17)K5(0,x)=K5(1,x),K5(0,x)=∫01‍h1(y)K5(y,x)dy,∂y2K5(0,x)=∂y2K5(1,x),∂y2K5(0,x)=∫01‍h2(y)∂y2K5(y,x)dy.
The above 10 equations in (12) and (15) provided 10 conditions for solving the coefficients ai(x), bi(x)(i=1,2,…10) in (11). Noting that (10) and (16) provided 12 boundary conditions, we have 22 equations, that is, (12), (15), (10), and (16). It is easy to know that these 22 equations are linear equations with the variables ai(x), bi(x)(i=1,2,…,10) and c1(x), c2(x). Therefore, ai(x), bi(x)(i=1,2,…,10) and c1(x), c2(x) could be calculated by many methods. As long as the coefficients ai(x), bi(x), i=1,2,…,10, and c1(x), c2(x) are known, the exact expression of the reproducing kernel function K5(y,x) could be calculated from (11).


### 2.2. The Reproducing Kernel SpaceW1[0,1]

The inner product space is defined byW1[0,1]={u(x)∣u is a absolutely continuous real-value function, u′∈L2[0,1]}.The inner product and norm inW1[0,1] are given, respectively, by
(18)〈u(x),v(x)〉W1=u(0)v(0)+∫01‍u′(x)v′(x)dx,∥u∥W1=〈u,u〉W1,
where u(x),v(x)∈W1[0,1]. In [15], it has been proved that W1[0,1] is also a complete reproducing kernel space and its reproducing kernel is
(19)K1(y,x)={1+y,y≤x,1+x,y>x.


### 2.3. Introduction into a Linear Operator

LetLu=u(4) and L:W5[0,1]→W1[0,1]; then (1) can be converted into the form as follows:
(20)Lu=F(x,u(x)),0<x<1,
where u(x)∈W5[0,1] and F(x,u(x))=λf(x,u(x))∈W1[0,1] as u(x)∈W1[0,1]. Therefore, we have
(21)u(0)=u(1)=∫01‍h1(x)u(x)dx,u′′(0)=u′′(1)=∫01‍h2(x)u′′(x)dx.
It is easy to prove that L is a bounded linear operator.Now, we construct an orthogonal function system. Letφi(x)=K1(x,xi) and ψi(x)=L*φi(x), where L* is the conjugate operator of L. In terms of the properties of reproducing kernel K1(y,x), one obtains
(22)〈u(x),ψi(x)〉W5=〈Lu(x),φi(x)〉W1=Lu(xi),oooooooooooooooooi=1,2,….The normal orthogonal system of functions{ψ¯i(x)}i=1∞ in W5[0,1] can be derived from Gram-Schmidt orthogonalization process of {ψi(x)}i=1∞:
(23)ψ¯i(x)=∑k=1i‍βikψk(x),
where βik are orthogonalization coefficients, βii>0, i=1,2,….We collect two lemmas in [9] for future use.Lemma 2.
If{xi}i=1∞ is dense on [0,1], then {ψi(x)}i=1∞ is a complete system of W5[0,1] and ψi(x)=LyK5(y,x)|y=xi. The subscript y by the operator L indicates that the operator L applies to the function of y.Lemma 3.
Ifu(x)∈W5[0,1], then there exists M1>0, such that
(24)∥u∥C4[0,1]≤M1∥u∥W5,
where ∥u∥C4[0,1]=max⁡x∈[0,1](|u(x)|+|u′(x)|+|u′′(x)|+|u′′′(x)|+|u(4)(x)|).Lemma 4.
If∥un-u∥W5→0, xn→x, (n→∞), and f(x,y) is continuous with respect to x,y for x∈[0,1], y∈(-∞,+∞), then
(25)f(xn,un-1(xn))⟶f(x,u(x))asn⟶∞.Proof.
Since∥un-u∥W5→0, (n→∞), by Lemma 3, we know that un(x) is convergent uniformly to u(x); therefore, the proof is complete.


## 2.1. The Reproducing Kernel SpaceW5[0,1]

The inner product spaceW5[0,1] is defined as follows: W5[0,1]={u(x)∣u(4)(x) is a absolutely continuous real-valued function, u(5)(x)∈L2[0,1], u(0)=u(1)=∫01‍h1(x)u(x)dx, u′′(0)=u′′(1)=∫01‍h2(x)u′′(x)dx}.The inner product and norm inW5[0,1] are defined, respectively, by
(2)〈u(x),v(x)〉W5=∑i=04‍u(i)(0)v(i)(0)+∫01‍u(5)(x)v(5)(x)dx,(3)∥u∥W5=〈u,u〉W5,
where u,v∈W5[0,1].Theorem 1.
The spaceW5[0,1] is a complete reproducing kernel space. That is, for each fixed x∈[0,1], there exists K5(y,x)∈W5[0,1], such that 〈u(y),K5(y,x)〉W5=u(x) for any u(y)∈W5[0,1] and y∈[0,1]. The reproducing kernel K5(y,x) can be written as
(4)K5(y,x)={∑i=110‍ai(x)yi-1-∑i=12‍ci(x)Hi(y),y≤x,∑i=110‍bi(x)yi-1-∑i=12‍ci(x)Hi(y),y>x,
where(5)H1(y)=∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍h1(y)dydydydydydydydydydy,H2(y)=∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍h2′′(y)dydydydydydydydydydy.ai(x), bi(x)(i=1,2,…10) and c1(x), c2(x), which are related to h1(x) and h2(x), will be given in the following proof.Proof.
(i) The proof of the completeness and reproducing property ofW5[0,1] is similar to the proof of Theorem  1.3.1 in [15].
(ii) Now, let us find out the expression form of the reproducing kernel functionK5(y,x) in W5[0,1].
Through several integration by parts for (2), we have
(6)〈u(y),K5(y,x)〉W5=∑i=04‍u(i)(0)[∂yiK5(0,x)-(-1)4-i∂y9-iK5(0,x)]+∑i=04‍(-1)4-iu(i)(1)∂y9-iK5(1,x)-∫01‍u(y)∂y10K5(y,x)dy,
and since u(x)∈W5[0,1], it follows that
(7)u(0)=u(1)=∫01‍h1(x)u(x)dx,u′′(0)=u′′(1)=∫01‍h2(x)u′′(x)dx.
Then, we have
(8)〈u(y),K5(y,x)〉W5=∑i=04‍u(i)(0)[∂yiK5(0,x)-(-1)4-i∂y9-iK5(0,x)]+∑i=04‍(-1)4-iu(i)(1)∂y9-iK5(1,x)-∫01‍u(y)∂y10K5(y,x)dy+c1[u(0)-∫01‍h1(y)u(y)dy]+c2[u′′(0)-∫01‍h2(y)u′′(y)dy]=∑i=04‍u(i)(0)[∂yiK5(0,x)-(-1)4-i∂y9-iK5(0,x)]+∑i=04‍(-1)4-iu(i)(1)∂y9-iK5(1,x)-∫01‍u(y)∂y10K5(y,x)dy+c1[u(0)-∫01‍h1(y)u(y)dy]+c2[∫01‍h2′′(y)u(y)dyu′′(0)-u′(1)h2(1)+u′(0)h2(0)iiiiiiiiiiiiiiiiiiiiii+h2′(1)u(1)-h2′(0)u(0)iiiiiiiiiiiiiiiiiii-∫01‍h2′′(y)u(y)dy].
Note that the property of the reproducing kernel〈u(y),K5(y,x)〉W5=u(x) and u(0)=u(1), u′′(0)=u′′(1), and K5(y,x) is the solution of the following generalized differential equation:
(9)-(∂y10K5(y,x)+c1h1(y)+c2h2′′(y))=δ(y-x),K5(0,x)-∂y9K5(0,x)+c1+∂y9K5(1,x)-c2h2′(0)+c2h2′(1)=0,∂y1K5(0,x)+∂y8K5(0,x)+c2h2(0)=0,∂y2K5(0,x)-∂y7K5(0,x)+c2+∂y7K5(1,x)=0,∂y3K5(0,x)+∂y6K5(0,x)=0,∂y4K5(0,x)-∂y5K5(0,x)=0,-∂y8K5(1,x)-c2h2(1)=0,-∂y6K5(1,x)=0,∂y5K5(1,x)=0.
Whiley≠x, it is easy to know that K5(y,x) is the solution of the following constant linear homogeneous differential equation with 10 orders; that is,
(10)-(∂y10K5(y,x)+c1h1(y)+c2h2′′(y))=0,
with the following boundary conditions:
(11)K5(0,x)-∂y9K5(0,x)+c1+∂y9K5(1,x)-c2h2′(0)+c2h2′(1)=0,∂y1K5(0,x)+∂y8K5(0,x)+c2h2(0)=0,∂y2K5(0,x)-∂y7K5(0,x)+c2+∂y7K5(1,x)=0,∂y3K5(0,x)+∂y6K5(0,x)=0,∂y4K5(0,x)-∂y5K5(0,x)=0,-∂y8K5(1,x)-c2h2(1)=0,-∂y6K5(1,x)=0,∂y5K5(1,x)=0.
We know that (9) has the characteristic equation λ10=0, and the eigenvalue λ=0 is a root whose multiplicity is 10. Therefore, the general solution of (9) is
(12)K5(y,x)={∑i=110‍ai(x)yi-1-∑i=12‍ci(x)Hi(y),y≤x,∑i=110‍bi(x)yi-1-∑i=12‍ci(x)Hi(y),y>x,
where(13)H1(y)=∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍h1(y)dydydydydydydydydydy,H2(y)=∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍∫0y‍h2′′(y)dydydydydydydydydydy. Now, we are ready to calculate the coefficient ai(x), bi(x)(i=1,2,…,10) and c1(x), c2(x).
Since(14)-(∂y10K5(y,x)+c1h1(y)+c2h2′′(y))=δ(y-x),
we have
(15)∂ykK5(x+0,x)=∂ykK5(x-0,x),k=0,1,2,…,8,(16)∂y9K5(x-0,x)-∂y9K5(x+0,x)=1.
SinceK5(y,x)∈W5[0,1], it follows that
(17)K5(0,x)=K5(1,x),K5(0,x)=∫01‍h1(y)K5(y,x)dy,∂y2K5(0,x)=∂y2K5(1,x),∂y2K5(0,x)=∫01‍h2(y)∂y2K5(y,x)dy.
The above 10 equations in (12) and (15) provided 10 conditions for solving the coefficients ai(x), bi(x)(i=1,2,…10) in (11). Noting that (10) and (16) provided 12 boundary conditions, we have 22 equations, that is, (12), (15), (10), and (16). It is easy to know that these 22 equations are linear equations with the variables ai(x), bi(x)(i=1,2,…,10) and c1(x), c2(x). Therefore, ai(x), bi(x)(i=1,2,…,10) and c1(x), c2(x) could be calculated by many methods. As long as the coefficients ai(x), bi(x), i=1,2,…,10, and c1(x), c2(x) are known, the exact expression of the reproducing kernel function K5(y,x) could be calculated from (11).


## 2.2. The Reproducing Kernel SpaceW1[0,1]

The inner product space is defined byW1[0,1]={u(x)∣u is a absolutely continuous real-value function, u′∈L2[0,1]}.The inner product and norm inW1[0,1] are given, respectively, by
(18)〈u(x),v(x)〉W1=u(0)v(0)+∫01‍u′(x)v′(x)dx,∥u∥W1=〈u,u〉W1,
where u(x),v(x)∈W1[0,1]. In [15], it has been proved that W1[0,1] is also a complete reproducing kernel space and its reproducing kernel is
(19)K1(y,x)={1+y,y≤x,1+x,y>x.


## 2.3. Introduction into a Linear Operator

LetLu=u(4) and L:W5[0,1]→W1[0,1]; then (1) can be converted into the form as follows:
(20)Lu=F(x,u(x)),0<x<1,
where u(x)∈W5[0,1] and F(x,u(x))=λf(x,u(x))∈W1[0,1] as u(x)∈W1[0,1]. Therefore, we have
(21)u(0)=u(1)=∫01‍h1(x)u(x)dx,u′′(0)=u′′(1)=∫01‍h2(x)u′′(x)dx.
It is easy to prove that L is a bounded linear operator.Now, we construct an orthogonal function system. Letφi(x)=K1(x,xi) and ψi(x)=L*φi(x), where L* is the conjugate operator of L. In terms of the properties of reproducing kernel K1(y,x), one obtains
(22)〈u(x),ψi(x)〉W5=〈Lu(x),φi(x)〉W1=Lu(xi),oooooooooooooooooi=1,2,….The normal orthogonal system of functions{ψ¯i(x)}i=1∞ in W5[0,1] can be derived from Gram-Schmidt orthogonalization process of {ψi(x)}i=1∞:
(23)ψ¯i(x)=∑k=1i‍βikψk(x),
where βik are orthogonalization coefficients, βii>0, i=1,2,….We collect two lemmas in [9] for future use.Lemma 2.
If{xi}i=1∞ is dense on [0,1], then {ψi(x)}i=1∞ is a complete system of W5[0,1] and ψi(x)=LyK5(y,x)|y=xi. The subscript y by the operator L indicates that the operator L applies to the function of y.Lemma 3.
Ifu(x)∈W5[0,1], then there exists M1>0, such that
(24)∥u∥C4[0,1]≤M1∥u∥W5,
where ∥u∥C4[0,1]=max⁡x∈[0,1](|u(x)|+|u′(x)|+|u′′(x)|+|u′′′(x)|+|u(4)(x)|).Lemma 4.
If∥un-u∥W5→0, xn→x, (n→∞), and f(x,y) is continuous with respect to x,y for x∈[0,1], y∈(-∞,+∞), then
(25)f(xn,un-1(xn))⟶f(x,u(x))asn⟶∞.Proof.
Since∥un-u∥W5→0, (n→∞), by Lemma 3, we know that un(x) is convergent uniformly to u(x); therefore, the proof is complete.


## 3. An Iterative Algorithm and Its Convergence

In this section, the exact solution of (1) is given in the reproducing kernel space W5[0,1].Theorem 5.
If{xi}i=1∞ is dense on [0,1] and u(x)∈W5[0,1] is the solution of (20), then u(x) satisfies the form
(26)u(x)=∑i=1∞∑k=1i‍βikF(xk,u(xk))ψi¯(x).Proof.
u
(
x
) can be expanded to the Fourier series in terms of normal orthogonal basis ψ¯i(x) in W25[0,1]:
(27)u(x)=∑i=1∞‍〈u(x),ψ¯i(x)〉W5ψ¯i(x)=∑i=1∞‍∑k=1i‍βik〈u(x),ψk(x)〉W5ψ¯i(x)=∑i=1∞‍∑k=1i‍βik〈u(x),L*φk(x)〉W5ψ¯i(x)=∑i=1∞‍∑k=1i‍βik〈Lu(x),φk(x)〉W1ψ¯i(x)=∑i=1∞‍∑k=1i‍βik〈F(x,u(x)),φk(x)〉W1ψ¯i(x)=∑i=1∞‍∑k=1i‍βikf(xk,u(xk))ψ¯i(x).Remark 6.
(i) If (20) is linear, that is, F(x,u(x))=F(x), then the analytical solution of (20) can be obtained directly by (26).
(ii) If (20) is nonlinear, that is, F depends on u, then the solution of (20) can be obtained by the following iterative method.
We construct an iterative sequenceun(x), putting
(28)any fixedu0(x)∈W5[0,1],un(x)=∑i=1n‍Aiψ¯i(x),
where
(29)A1=β11F(x1,u0(x1)),A2=∑k=12‍β2kF(xk,uk-1(xk)),⋮An=∑k=1n‍βnkF(xk,uk-1(xk)).
Next, we prove thatun(x) in iterative formula (28) is convergent to the exact solution of (20).Theorem 7.
Suppose the following conditions are satisfied:(i)
∥un∥W5 is bounded;(ii)
{xi}i=1∞ is dense in [0,1];(iii)
f(x,y(x))∈W1[0,1] for any y(x)∈W5[0,1].
Then,un(x) in iterative formula (28) converges to the exact solution u(x) of (20) in W5[0,1] and
(30)u(x)=∑i=1∞‍Aiψ¯i,
where Ai are given by (29).Proof.
(i) First, we will prove the convergence ofun(x).
By (28), we have
(31)un+1(x)=un(x)+An+1ψ¯n+1(x).
From the orthogonality of {ψ¯(x)}i=1∞, it follows that
(32)∥un+1∥W52=∥un∥W52+(An+1)2=∥un-1∥W52+(An)2+(An+1)2=⋯=∑i=1n+1‍(Ai)2.
From boundedness of ∥un∥W5, we have ∑i=1∞(Ai)2<∞; thatis, {Ai}∈l2(i=1,2,…).
Letm>n; in view of (um-um-1)⊥(um-1-um-2)⊥⋯⊥(un+1-un), it follows that
(33)∥um-un∥W52=∥um-um-1+um-1-um-2+⋯+un+1-un∥W52≤∥um-um-1∥W52+⋯+∥un+1-un∥W52=∑i=n+1m‍(Ai)2⟶0,(m,n⟶∞).
Considering the completeness of W5[0,1], there exists u(x)∈W5[0,1], such that
(34)un(x)⟶∥·∥W5u(x),asn⟶∞.
(ii) Second, we will prove thatu(x) is the solution of (20).
By Lemma3 and (i) of Theorem 7, we know that un(x) converge uniformly to u(x). It follows that, on taking limits in (28), we have
(35)u(x)=∑i=1∞‍Aiψ¯i.
Since(36)(Lu)(xj)=∑i=1∞‍Ai〈Lψ¯i(x),φj(x)〉W1=∑i=1∞‍Ai〈ψ¯i(x),L*φj(x)〉W5=∑i=1∞‍Ai〈ψ¯i(x),ψj(x)〉W5,
it follows that
(37)∑j=1n‍βnj(Lu)(xj)=∑i=1∞‍Ai〈ψ¯i(x),∑j=1n‍βnjψj(x)〉W5=∑i=1∞‍Ai〈ψ¯i(x),ψ¯n(x)〉W5=An.
Ifn=1, then
(38)(Lu)(x1)=F(x1,u0(x1)).
Ifn=2, then
(39)β21(Lu)(x1)+β22(Lu)(x2)=β21F(x1,u0(x1))+β22F(x2,u1(x2)).
From (38) and (39), it is clear that (Lu)(x2)=F(x2,u1(x2)); furthermore, it is easy to see by induction that
(40)(Lu)(xj)=F(xj,uj-1(xj)).
Since{xi}i=1∞ is dense on interval [0,1], for any y∈[0,1], there exists subsequence {xnj}, such that xnj→y, as j→∞. Hence, let j→∞ in (40); by the convergence of un(x) and Lemma 4, we have
(41)(Lu)(y)=F(y,u(y));
that is, u(x) is the solution of (20) and
(42)u(x)=∑i=1∞‍Aiψ¯i,
where Ai are given by (29).From Lemma3, we have Corollary 8.Corollary 8.
Assume that the conditions of Theorem7 hold; then un(x) in (28) satisfies ∥un-u∥C4[0,1]→0, n→∞, where u(x) is the solution of (20).


## 4. Numerical Experiment

In this section, the method in the paper will be applied to some numerical examples. All computations are performed in Mathematica 5.0.Example 1.
Consider(43)u(4)(x)=u2(x)+f(x),0<x<1,u(0)=u(1)=12∫01‍xu(x)dx,u′′(0)=u′′(1)=34∫01‍u′′(x)dx.The exact solution isu(x)=1+(120x/7)-(90x2/7)-(60x3/7)+(30x4/7), f(x)=(720/7)-(1+(120x/7)-(90x2/7)-(60x3/7)+(30x4/7))2. The numerical results and the root-mean-square errors for the partial derivatives are displayed in Tables 1 and 2, respectively. The superimposed images of u(x) with u100(x) and |u(x)-u100(x)| are displayed in Figure 1.Table 1
The numerical results for Example1.
Node
True solutionu(x)
Approximate solutionu100(x)
Absolute error
Relative error
0.1
2.57757
2.57756
1.39670E −5
5.41865E −6
0.2
3.85257
3.85255
2.08943E −5
5.42347E −6
0.3
4.78900
4.78897
2.59723E −5
5.42333E −6
0.4
5.36114
5.36111
2.90522E −5
5.41903E −6
0.5
5.55357
5.55354
3.00501E −5
5.41096E −6
0.6
5.36114
5.36111
2.89504E −5
5.40005E −6
0.7
4.78900
4.78897
2.58036E −5
5.38810E −6
0.8
3.85257
3.85255
2.07194E −5
5.37806E −6
0.9
2.57757
2.57756
1.38544E −5
5.37499E −6
1.0
1
0.999995
5.39775E −6
5.39775E −6Table 2
The root-mean-square errors for the partial derivatives for Example1.
1-order derivative
2-order derivative
3-order derivative
4-order derivative
5.65090E −5
1.85920E −4
1.56359E −4
5.80200E −4Figure 1
The congruent figure ofu(x) with u100(x) and |u(x)-u100(x)| for Example 1.Example 2.
Consider(44)u(4)(x)=sin⁡[u(x)]+f(x),0<x<1,u(0)=u(1)=23∫01‍xu(x)dx,u′′(0)=u′′(1)=56∫01‍(x+3)u′′(x)dx.The exact solution isu(x)=(1/(1769+2401e))(36(1769-8280ex-60x(74+x(41+23x))+e(2401+60x(80+x(35+23x))))), f(x)=-(298080ex/(1769+2401e))-sin[(1/(1769+2401e))(36(1769-8280ex-60x(74+x(41+23x))+e(2401+60x(80+x(35+23x)))))]. Similar to Example 1, the numerical results are displayed in Tables 3 and 4 and Figure 2.Table 3
The numerical results for Example2.
Node
True solutionu(x)
Approximate solutionu100(x)
Absolute error
Relative error
0.1
0.175365
0.175366
1.19292E −6
6.80252E −6
0.2
0.229289
0.229291
1.57610E −6
6.87384E −6
0.3
0.249391
0.249393
1.71129E −6
6.86188E −6
0.4
0.251217
0.251218
1.70213E −6
6.77554E −6
0.5
0.245453
0.245455
1.62740E −6
6.63021E −6
0.6
0.237419
0.237420
1.53473E −6
6.46421E −6
0.7
0.226498
0.226500
1.43357E −6
6.32929E −6
0.8
0.205517
0.205519
1.28776E −6
6.26595E −6
0.9
0.160054
0.160055
1.00724E −6
6.29316E −6
1
0.0676755
0.0676759
4.39185E −7
6.48958E −6Table 4
The root-mean-square errors for the partial derivatives for Example2.
1-order derivative
2-order derivative
3-order derivative
4-order derivative
3.45285E −6
2.39862E −5
1.28849E −4
4.55038E −4Figure 2
The congruent figure ofu(x) with u100(x) and |u(x)-u100(x)| for Example 2.


## 5. Conclusions

In summary, we use an iterative method to find the approximate solution of the nonlinear fourth-order equation with integral boundary conditions in the reproducing kernel space. Using this method, we obtain the sequence which is proved to converge to the exact solution uniformly. Numerical results show that the method employed in the paper is valid. It is worthy to note that the new method can be used as a very accurate algorithm for solving linear and nonlinear integral boundary problems.



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