/ / Article

Research Article | Open Access

Volume 2012 |Article ID 891519 | https://doi.org/10.1155/2012/891519

Dang Quang A., Nguyen Van Thien, "Iterative Method for Solving the Second Boundary Value Problem for Biharmonic-Type Equation", Journal of Applied Mathematics, vol. 2012, Article ID 891519, 18 pages, 2012. https://doi.org/10.1155/2012/891519

# Iterative Method for Solving the Second Boundary Value Problem for Biharmonic-Type Equation

Revised11 Jun 2012
Accepted11 Jun 2012
Published04 Sep 2012

#### Abstract

Solving boundary value problems (BVPs) for the fourth-order differential equations by the reduction of them to BVPs for the second-order equations with the aim to use the achievements for the latter ones attracts attention from many researchers. In this paper, using the technique developed by ourselves in recent works, we construct iterative method for the second BVP for biharmonic-type equation, which describes the deflection of a plate resting on a biparametric elastic foundation. The convergence rate of the method is established. The optimal value of the iterative parameter is found. Several numerical examples confirm the efficiency of the proposed method.

#### 1. Introduction

Solving BVPs for the fourth-order differential equations by the reduction of them to BVPs for the second-order equations with the aim to use a lot of efficient algorithms for the latter ones attracts attention from many researchers. Namely, for the biharmonic equation with the Dirichlet boundary condition, there is intensively developed the iterative method, which leads the problem to two problems for the Poisson equation at each iteration (see e.g., ). Recently, Abramov and Ul’yanova  proposed an iterative method for the Dirichlet problem for the biharmonic-type equation, but the convergence of the method is not proved. In our previous works [5, 6] with the help of boundary or mixed boundary-domain operators appropriately introduced, we constructed iterative methods for biharmonic and biharmonic-type equations associated with the Dirichlet boundary condition. For the biharmonic-type equation with Neumann boundary conditions in  an iterative method also was proposed. It is proved that the iterative methods are convergent with the rate of geometric progression. In this paper we develop our technique in the above-mentioned papers for the second BVP for the biharmonic-type equation. Namely, we consider the following problem

where is the Laplace operator, is a bounded domain in is the sufficiently smooth boundary of , and are nonnegative constants. This problem has not yet considered in .

It should be noticed that when (1.1) is the equation for a thin plate, and problem (1.1)–(1.3) is decomposed into two consecutive problems for the Poisson equations.

In this paper we suppose that . Then (1.1) describes the deflection of a plate resting on biparametric elastic foundation. For solving this equation several methods such as the boundary element, the finite element methods , domain/boundary element technique , and boundary differential integral equation (BDIE) method  were used. It should be noticed that at present the boundary element method is intensively developed and is used for solving more complex problems of plates and shells (see e.g., ).

In this paper we use completely different approach to (1.1). Two cases will be treated in dependence on the sign of . In the case if we can immediately decompose the problem into two problems for second-order equations. In the opposite case we propose an iterative method for reducing problem (1.1)–(1.3) to a sequence of second-order problems. The convergence of the method is established and verified on examples.

#### 2. Case 𝑎2−4𝑏≥0

In this case we always can lead the original problem (1.1)–(1.3) to two problems for second-order equations. To do this, we put Then problem (1.1)–(1.3) is reduced to the following problems: These Dirichlet problems can be solved by known methods such as finite element method, boundary element method, or finite difference method. Some fast Poisson solvers in [16, 17] can be applied for the above problems.

#### 3. Case 𝑎2−4𝑏<0

This case is very important in mechanics because (1.1) describes the bending plate on elastic foundation (see ).

##### 3.1. Construction of Iterative Method on Continuous Level

As in , we set

Then problem (1.1)–(1.3) leads to the following second-order problems where all the functions , and are unknown but they are related with each other by (3.2).

Now consider the following iterative process for finding and simultaneously for finding . (1)Given , for example, (2)Knowing on solve consecutively two problems (3)Compute the new approximation where is an iterative parameter to be chosen later.

##### 3.2. Investigation of Convergence

In order to investigate the convergence of the iterative process (3.4)–(3.7), firstly we rewrite (3.7) in the canonical form of two-layer iterative scheme 

Now, we introduce the operator defined by the formula where is the function determined from the problems

The properties of the operator will be investigated in the sequel.

Now, let us return to the problem (3.3). We represent their solutions in the form where are the solutions of the problems According to the definition of we have Since should satisfy the relation (3.2), taking into account the representation (3.12) we obtain the equation where is the identity operator, and

Thus, we have reduced the original problem (1.1)–(1.3) to the operator equation (3.16), whose right-hand side is completely defined by the data functions , and , and coefficients .

Proposition 3.1 3.1. The iterative process (3.4)–(3.7) is a realization of the two-layer iterative scheme for the operator equation (3.16).

Proof. Indeed, if in (3.5), (3.6) we put where are the solutions of problem (3.14), then we get From here it is easy to see that Therefore, taking into account the first relation of (3.19), the above equality, and (3.17), from (3.8) we obtain (3.18). Thus, the proposition is proved.

Proposition 3.1 enables us to lead the investigation of convergence of processs (3.4)–(3.7) to the study of scheme (3.18). For this reason we need some properties of the operator .

Proposition 3.2. The operator defined by (3.9)–(3.11) is linear, symmetric, positive, and compact operator in the space .

Proof. The linearity of is obvious. To establish the other properties of let us consider the inner product for two arbitrary functions and in . Recall that the operator is defined by (3.9)–(3.11). We denote by and the solutions of (3.10) and (3.11), where instead of there stands . We have Applying the Green formula for the functions and , vanishing on the boundary , we obtain Hence, Clearly, are due to . Moreover, it is easy seen that if and only if . So, we have shown that the operator is symmetric and positive in .
It remains to show the compactness of . As is well known that if then problem (3.10) has a unique solution , therefore, problem (3.11) has a unique solution . Consequently, the operator maps into . In view of the compact imbedding into it follows that is compact operator in .
Thus, the proof of Proposition 3.2 is complete.

Due to the above proposition the operator is linear, symmetric, positive definite, and bounded operator in the space . More precisely, we have where Notice that since the operator is defined by (3.9)–(3.11) its norm depends on the value of but not on in (1.1).

From the theory of elliptic problems  we have the following estimates for the functions given by (3.10), (3.11): where are constants. Therefore, Before stating the result of convergence of the iterative process (3.5)–(3.7) we assume the following regularity of the data of the original problem (1.1)–(1.3): Then the problem (1.1)–(1.3) has a unique solution . For the function determined by (3.14) we have also .

Theorem 3.3. Let be the solution of problem (1.1)–(1.3) and be the solution of (3.16). Then, if is the Chebyshev collection of parameters, constructed by the bounds and of the operator , namely we have where with being the constant in (3.30), In the case of the stationary iterative process, we have

Proof. According to the general theory of the two-layer iterative schemes (see ) for the operator equation (3.16) we have if the parameter is chosen by the formulae (3.32) and if . In view of these estimates the corresponding estimates (3.33) and (3.36) follow from (3.30) applied to the problems which are obtained in the result of the subtraction of (3.3) from (3.5) and (3.6), respectively. The theorem is proved.

Remark 3.4. From the above theorem it is easy to see that for each fixed value of the numbers and characterizing the rate of convergence of the iterative method decrease with the decrease of . So, the smaller is, the faster the iterative process converges. In the special case when the mentioned above numbers also are zero, hence the iterative process converges at once and the original problem (1.1)–(1.3) is decomposed into two second-order problems.

##### 3.3. Computation of the Norm ‖𝐴‖

As we see in Theorem 3.3 for determining the iterative parameter we need the bounds and of the operator , and in its turn the latter bound requires to compute . Therefore, below we consider the problem of computation .

Suppose the domain in the plane . In this case by Fourier method we found that the system of functions is the eigenfunctions of the spectral problem corresponding to the eigenvalues Moreover, this system is orthogonal and complete in .

Now let a function have the expansion Then we have Representing the solution of (3.10) in the form of the double series we found Next, we seek the solution of (3.11) in the form Then from (3.45) we find From the definition of the operator by (3.9)–(3.11) we have due to the orthogonality of the system and (3.46), (3.48), and (3.42). Thus, there holds the estimate The sign of equality occurs for . Since is shown to be symmetric and positive in it follows:

#### 4. Numerical Realization of the Iterative Method

In the previous section we proposed and investigated an iterative method for problem (1.1)–(1.3) in the case if . Now we study numerical realization of the method.

For simplicity we consider the case, where the domain is a rectangle, in the plane . In this domain we construct an uniform grid where .

Denote by the set of inner nodes, by the set of boundary nodes of the grid, and by the grid functions defined on .

Now consider a discrete version of the iterative method (3.4)–(3.7) when . (1)Given a starting , for example, (2)Knowing on solve consecutively two difference problems where is the discrete Laplace operator, (3)Compute the new approximation It is easy to see that the convergence of the above iterative method is related to the discrete version of the operator , defined by the formula where is determined from the difference problems Using the results of the spectral problem for the discrete Laplace operator (see ) we find the bounds of : where Therefore, for the operator , the discrete version of , we obtain the estimate where Hence, we choose which is the optimal parameter of the iterative process (4.2)–(4.6).

Now we study the deviation of from obtained by the iterative process (3.4)–(3.7). In the future for short we will write instead of .

Proposition 4.1. For any there holds the estimate where ,   are computed by the process (3.4)–(3.7) and are computed by (4.2)–(4.6).

Proof. We shall prove this proposition by induction in .
For we have and the second estimate in (4.14) is valid due to the the second-order accuracy of the iterative schemes (4.3) and (4.4) for the problems (3.5) and (3.6) (see ).
Now suppose (4.14) is valid for . We shall show that it also is valid for . For this purpose we recall that is computed by the formula where and is computed by the formula being given by (4.13) and (4.12).
From (4.10)–(4.13), (4.16), and (3.51) it is possible to obtain the estimate Next, subtracting (4.15) from (4.17) and taking into account the above formula we get By the assumptions of the induction from (4.19) it follows . Now, having in mind this estimate due to the second-order approximation of the difference operators in (4.3) and (4.4) we get the second estimate in (4.14). Thus, the proof of the proposition is complete.

In realization of the discrete iterative process (4.2)–(4.6) we shall stop iterations when , where is a some given accuracy. Then for the total error of the discrete solution there there holds the following theorem.

Theorem 4.2. For the total error of the discrete solution from the exact solution of the original problem (1.1)–(1.3) there holds the estimate where are some constants and is the number in (3.36).

Proof. We have the following estimate Since the space is continuously embedded to the space (see ) from (3.36) we have for some constant . Now using Proposition 4.1 and the above estimate, from (4.22) we obtain (4.21). Thus, the theorem is proved.

Below we report the results of some numerical examples for testing the convergence of the iterative method. In all examples we test the iterative method for some values of and with . The results of convergence of the method are given in tables, where is the number of iterations, is the error of approximate solution , .

Example 4.3. We take an exact solution in the rectangle and corresponding boundary conditions. The right-hand side of (1.1) in this case is .
The results of convergence in the case of the uniform grids including , and nodes for and are presented in Tables 1, 2, and 3, respectively.

 𝑏 Grid 6 5 × 6 5 Grid 1 2 9 × 1 2 9 Grid 2 5 7 × 2 5 7 𝐸 1 / 𝐸 2 𝐸 2 / 𝐸 3 𝑘 𝐸 1 𝑘 𝐸 2 𝑘 𝐸 3 0 . 3 5 3 . 7 3 7 8 𝑒 − 4 5 9 . 3 5 2 7 𝑒 − 5 5 2 . 3 4 7 7 𝑒 − 5 3 . 9 9 6 5 3.9838 0.7 6 3 . 4 1 2 5 𝑒 − 4 6 8 . 4 8 5 9 𝑒 − 5 6 2 . 0 7 7 1 𝑒 − 5 4 . 0 2 1 4 4 . 0 8 5 5 1.0 7 3 . 2 1 7 9 𝑒 − 4 7 8 . 0 7 9 3 𝑒 − 5 7 2 . 0 5 5 0 𝑒 − 5 3 . 9 8 2 9 3.9315 1.5 8 2 . 9 1 1 8 𝑒 − 4 8 7 . 2 1 0 1 𝑒 − 5 8 1 . 7 3 3 7 𝑒 − 5 4 . 0 3 8 5 4.1588 2.0 9 2 . 6 9 0 4 𝑒 − 4 9 6 . 8 2 1 5 𝑒 − 5 9 1 . 8 0 1 3 𝑒 − 5 3 . 9 4 4 0 3.7870 2.5 10 2 . 4 5 6 1 𝑒 − 4 1 0 6 . 0 2 4 7 𝑒 − 5 1 0 1 . 3 9 0 9 𝑒 − 5 4 . 0 7 6 7 4.3315
 𝑏 Grid 6 5 × 6 5 Grid 1 2 9 × 1 2 9 Grid 2 5 7 × 2 5 7 𝐸 1 / 𝐸 2 𝐸 2 / 𝐸 3 𝑘 𝐸 1 𝑘 𝐸 2 𝑘 𝐸 3 0 . 3 5 3 . 7 3 7 8 𝑒 − 4 5 8 . 5 2 9 1 𝑒 − 5 5 2 . 1 3 5 4 𝑒 − 5 3 . 9 9 9 0 3.9941 0.7 6 3 . 4 1 2 5 𝑒 − 4 6 7 . 9 0 9 7 𝑒 − 5 6 1 . 9 6 4 8 𝑒 − 5 4 . 0 0 6 9 4 . 0 2 5 7 1.0 7 3 . 2 1 7 9 𝑒 − 4 7 7 . 5 4 1 5 𝑒 − 5 7 1 . 8 9 3 8 𝑒 − 5 3 . 9 9 5 9 3.9822 1.5 8 2 . 9 1 1 8 𝑒 − 4 8 6 . 9 3 1 6 𝑒 − 5 8 1 . 7 1 8 4 𝑒 − 5 4 . 0 0 8 6 4.0338 2.0 9 2 . 6 9 0 4 𝑒 − 4 9 6 . 4 7 8 2 𝑒 − 5 9 1 . 6 3 7 4 𝑒 − 5 3 . 9 8 9 2 3.9564 2.5 9 2 . 4 5 6 1 𝑒 − 4 9 6 . 1 5 2 1 𝑒 − 5 9 1 . 6 3 4 0 𝑒 − 5 3 . 9 3 7 8 3.7651
 𝑏 Grid 6 5 × 6 5 Grid 1 2 9 × 1 2 9 Grid 2 5 7 × 2 5 7 𝐸 1 / 𝐸 2 𝐸 2 / 𝐸 3 𝑘 𝐸 1 𝑘 𝐸 2 𝑘 𝐸 3 0.3 5 3 . 1 8 7 9 𝑒 − 4 5 7 . 9 7 0 2 𝑒 − 5 5 1 . 9 9 3 8 𝑒 − 5 3 . 9 9 9 8 3.9975 0.7 6 2 . 9 9 6 8 𝑒 − 4 6 7 . 4 8 6 8 𝑒 − 5 6 1 . 8 6 7 2 𝑒 − 5 4 . 0 0 2 8 4 . 0 0 9 6 1.0 6 2 . 8 6 4 4 𝑒 − 4 6 7 . 1 2 6 7 𝑒 − 5 6 1 . 7 4 8 0 𝑒 − 5 4 . 0 1 9 3 4.0771 1.5 7 2 . 6 8 2 3 𝑒 − 4 7 6 . 7 4 0 5 𝑒 − 5 7 1 . 7 2 0 4 𝑒 − 5 3 . 9 7 9 4 3.9180 2.0 8 2 . 5 0 6 1 𝑒 − 4 8 6 . 2 3 4 6 𝑒 − 5 8 1 . 5 2 8 3 𝑒 − 5 4 . 0 1 9 7 4.0794 2.5 9 2 . 3 6 5 8 𝑒 − 4 9 5 . 9 3 8 5 𝑒 − 5 9 1 . 5 0 9 0 𝑒 − 5 3 . 9 8 3 8 3.9354

Example 4.4. We take an exact solution in the rectangle with corresponding boundary conditions. The the right-hand side of (1.1) is .
The results of convergence in the case of the grids including , and nodes for , and are presented in Tables 4, 5, and 6, respectively.
In the two above examples the grid step sizes are . Therefore, ,  . According to the estimate (4.21) the total error of the discrete approximate solution depends on . The columns in Tables 16 show this fact. It is interesting to notice that in these tables and . It means that if the number of grid nodes increases in 4 times then it is expected that the accuracy of the approximate solution increases in the same times. From the tables we also observe that the number of iterations increases with the increase of the parameter for fixed values of parameter . This confirms Remark 3.4. We also remark that for each pair of the parameters the number of iterations for achieving an accuracy corresponding to the grid step size (or density of grid) does not depend on the grid step size.
In general the error of discrete approximate solution strongly depends on the step size of grid. So, it is not expected to get an approximate solution of higher accuracy on a grid of low density. However, in some exceptional cases we can obtain very accurate approximate solution on sparse grid. Below is an example showing this fact.

 𝑏 Grid 6 5 × 6 5 Grid 1 2 9 × 1 2 9 Grid 2 5 7 × 2 5 7 𝐸 1 / 𝐸 2 𝐸 2 / 𝐸 3 𝑘 𝐸 1 𝑘 𝐸 2 𝑘 𝐸 3 0.3 6 5 . 4 8 1 5 𝑒 − 4 6 1 . 3 7 0 3 𝑒 − 4 6 3 . 4 2 4 7 𝑒 − 5 4 . 0 0 0 2 4.0012 0.7 7 5 . 0 1 3 6 𝑒 − 4 7 1 . 2 5 5 5 𝑒 − 4 7 3 . 1 5 2 5 𝑒 − 5 3 . 9 9 3 3 3 . 9 8 2 6 1.0 8 4 . 7 1 0 1 𝑒 − 4 8 1 . 1 7 6 3 𝑒 − 4 8 2 . 9 2 5 7 𝑒 − 5 4 . 0 0 4 2 4.0206 1.5 9 4 . 2 9 0 9 𝑒 − 4 9 1 . 0 7 7 0 𝑒 − 4 9 2 . 7 3 4 1 𝑒 − 5 3 . 9 8 4 1 3.9391 2.0 1 0 3 . 9 1 9 1 𝑒 − 4 1 0 9 . 7 2 6 4 𝑒 − 5 1 0 2 . 3 5 8 2 𝑒 − 5 4 . 0 2 9 3 4.1245 2.5 1 1 3 . 6 4 2 0 𝑒 − 4 1 1 9 . 2 1 4 1 𝑒 − 5 1 1 2 . 4 0 7 8 𝑒 − 5 3 . 9 5 2 6 3.8268
 𝑏 Grid 6 5 × 6 5 Grid 1 2 9 × 1 2 9 Grid 2 5 7 × 2 5 7 𝐸 1 / 𝐸 2 𝐸 2 / 𝐸 3 𝑘 𝐸 1 𝑘 𝐸 2 𝑘 𝐸 3 0.3 5 5 . 1 8 7 5 𝑒 − 4 5 1 . 2 9 8 5 𝑒 − 4 5 3 . 2 5 8 9 𝑒 − 5 3 . 9 9 5 0 3.9845 0.7 7 4 . 8 2 0 9 𝑒 − 4 7 1 . 2 0 6 1 𝑒 − 4 7 3 . 0 1 8 4 𝑒 − 5 3 . 9 9 7 1 3 . 9 9 5 8 1.0 7 4 . 5 8 4 6 𝑒 − 4 7 1 . 1 4 9 5 𝑒 − 4 7 2 . 9 0 6 7 𝑒 − 5 3 . 9 8 8 3 3.9547 1.5 8 4 . 2 2 1 3 𝑒 − 4 8 1 . 0 4 9 8 𝑒 − 4 8 2 . 5 6 8 7 𝑒 − 5 4 . 0 2 1 1 4.0869 2.0 9 3 . 9 3 6 2 𝑒 − 4 9 9 . 9 0 9 9 𝑒 − 5 9 2 . 5 4 5 8 𝑒 − 5 3 . 9 7 2 0 3.8926 2.5 10 3 . 6 5 5 5 𝑒 − 4 1 0 9 . 0 7 0 9 𝑒 − 5 1 0 2 . 1 9 4 5 𝑒 − 5 4 . 0 2 9 9 4.1335
 𝑏 Grid 6 5 × 6 5 Grid 1 2 9 × 1 2 9 Grid 2 5 7 × 2 5 7 𝐸 1 / 𝐸 2 𝐸 2 / 𝐸 3 𝑘 𝐸 1 𝑘 𝐸 2 𝑘 𝐸 3 0 . 3 5 4 . 9 8 4 3 𝑒 − 4 5 1 . 2 4 7 2 𝑒 − 4 5 3 . 1 2 3 2 𝑒 − 5 3 . 9 9 6 4 3.9933 0.7 6 4 . 6 8 3 9 𝑒 − 4 6 1 . 1 6 9 6 𝑒 − 4 6 2 . 9 0 6 7 𝑒 − 5 4 . 0 0 4 7 4 . 0 2 3 8 1.0 7 4 . 4 8 7 1 𝑒 − 4 7 1 . 1 2 2 9 𝑒 − 4 7 2 . 8 1 7 4 𝑒 − 5 3 . 9 9 6 0 3.9856 1.5 8 4 . 1 8 5 2 𝑒 − 4 8 1 . 0 4 4 9 𝑒 − 4 8 2 . 5 9 7 0 𝑒 − 5 4 . 0 0 5 4 4.0235 2.0 9 3 . 9 2 7 8 𝑒 − 4 9 9 . 8 3 6 9 𝑒 − 5 9 2 . 4 7 6 0 𝑒 − 5 3 . 9 9 2 9 3.9729 2.5 9 3 . 7 0 6 8 𝑒 − 4 9 9 . 3 6 3 6 𝑒 − 5 9 2 . 4 3 3 8 𝑒 − 5 3 . 9 5 8 7 3.8473

Example 4.5. We take an exact solution in the rectangle . The the right-hand side of (1.1) is .
The results of convergence in the case of the grids including and nodes are presented in Tables 7 and 8, respectively.
From Tables 7 and 8 we see a high accuracy even on the grid . The reason of this fact is that for quadratic function the approximation error of the central difference scheme is zero for any grid step size.
The above three numerical examples demonstrate the fast convergence of the iterative method (3.4)–(3.7) for problem (1.1)–(1.3).
Below, we consider an example for examining the variation of the solution of Prob. (1.1)–(1.3) in dependence of the parameters and .

 𝑏 𝑎 = 0 𝑎 = 0 . 5 𝑎 = 1 𝑘 error 𝑘 error 𝑘 error 0.3 5 7 . 8 0 5 2 𝑒 − 9 5 3 . 7 1 4 3 𝑒 − 9 5 1 . 9 4 6 0 𝑒 − 9 0.7 6 1 . 7 7 1 0 𝑒 − 8 5 2 . 3 9 3 3 𝑒 − 7 5 1 . 2 6 4 6 𝑒 − 7 1.0 6 1 . 3 9 5 9 𝑒 − 7 6 5 . 8 6 7 5 𝑒 − 8 6 2 . 7 5 1 9 𝑒 − 8 1.5 7 1 . 0 2 5 1 𝑒 − 7 7 3 . 8 0 7 9 𝑒 − 8 6 2 . 8 5 1 7 𝑒 − 7 2.0 7 6 . 6 4 6 6 𝑒 − 7 7 2 . 5 1 5 5 𝑒 − 7 7 1 . 0 7 2 0 𝑒 − 7 2.5 8 3 . 2 0 3 8 𝑒 − 7 8 1 . 0 7 9 5 𝑒 − 7 7 4 . 5 8 2 2 𝑒 − 7
 𝑏 𝑎 = 0 𝑎 = 0 . 5 𝑎 = 1 𝑘 error 𝑘 error 𝑘 error 0.3 5 7 . 7 9 3 9 𝑒 − 9 5 3 . 7 0 9 4 𝑒 − 9 5 1 . 9 4 3 7 𝑒 − 9 0.7 6 1 . 7 6 7 9 𝑒 − 8 5 2 . 3 9 0 1 𝑒 − 7 5 1 . 2 6 3 0 𝑒 − 7 1.0 6 1 . 3 9 3 5 𝑒 − 7 6 5 . 8 5 7 9 𝑒 − 8 6 2 . 7 4 7 6 𝑒 − 8 1.5 7 1 . 0 2 3 1 𝑒 − 7 7 3 . 8 0 0 9 𝑒 − 8 6 2 . 8 4 7 4 𝑒 − 7 2.0 7 6 . 6 3 3 8 𝑒 − 7 7 2 . 5 1 0 9 𝑒 − 7 7 1 . 0 7 0 1 𝑒 − 7 2.5 8 3 . 1 9 6 8 𝑒 − 7 8 1 . 0 7 7 3 𝑒 − 7 7 4 . 5 7 4 4 𝑒 − 7

Example 4.6. We take the the right-hand side of (1.1) and the boundary conditions and use the proposed method for finding approximate solution for different values of and . In all experiments we use the grid of nodes in the domain and . It turns out that the number of iterations in all cases of and does not exceed and the value of the solution at any fixed point is decreasing with the growth of and . This fact is obvious from Figure 1. The graph of the solution in the case of and is given in Figure 2.

#### 5. Concluding Remark

In the paper an iterative method was proposed for reducing the second problem for biharmonic-type equation to a sequence of Dirichlet problems for second-order equations. The convergence of the method was proved. In the case when the computational domain is a rectangle the optimal iterative parameter was given. Several numerical examples in this case show fast convergence of the method. When the computational domain consists of rectangles the proposed iterative method can be applied successfully if combining with the domain decomposition method.

#### Acknowledgments

This work is supported by Vietnam National Foundation for Science and Technology Development (NAFOSTED) under the Grant 102.99–2011.24. The authors would like to thank the anonymous reviewers sincerely for their helpful comments and remarks to improve the original manuscript.

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