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Linear barycentric rational collocation method for solving biharmonic equation

  • Jin Li EMAIL logo
Published/Copyright: September 23, 2022
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Abstract

Two-dimensional biharmonic boundary-value problems are considered by the linear barycentric rational collocation method, and the unknown function is approximated by the barycentric rational polynomial. With the help of matrix form, the linear equations of the discrete biharmonic equation are changed into a matrix equation. From the convergence rate of barycentric rational polynomial, we present the convergence rate of linear barycentric rational collocation method for biharmonic equation. Finally, several numerical examples are provided to validate the theoretical analysis.

MSC 2010: 65D05; 65L60; 31A30

1 Introduction

In this article, we pay our attention to the numerical solution of biharmonic equation:

(1) 4 u = 2 2 u = 4 u ( x , y ) x 4 + 2 4 u ( x , y ) x 2 y 2 + 4 u ( x , y ) y 4 = f ( x , y ) , ( x , y ) Ω ,

with boundary condition

(2) u ( x , y ) = g ( x , y ) , u ( x , y ) n = h 1 ( x , y ) , ( x , y ) Ω

or

(3) u ( x , y ) = g ( x , y ) , 2 u ( x , y ) x 2 + 2 u ( x , y ) y 2 = h 2 ( x , y ) , ( x , y ) Ω ,

where Ω = [ a , b ] × [ c , d ] and f ( x , y ) C ( Ω ) , g ( x , y ) , h 1 ( x , y ) , h 2 ( x , y ) C ( Ω ) . Biharmonic equation [1] is widely used in electrostatics, mechanical engineering, theoretical physics, and so on. There are a lot of numerical methods to solve biharmonic equation such as finite difference method, finite element method, boundary element method, spectral method, and so on.

There are some advantages of the collocation method [2] to solve the partial differential equation, such as meshless, no integrals, and easy to program. Barycentric formula can be obtained by the Lagrange interpolation formulae [3,4,5], which have been used to solve certain problems such as delay Volterra integro-differential equations [6,7], Volterra integral equations [8], boundary value problems [9], convection-diffusion equations [10,11], and so on. Generally, the interpolation nodes of barycentric Lagrange interpolation such as second kind of Chebyshev point is not the equidistant node. In order to obtain the equidistant node of the barycentric formulae, Floater and Hormann [12], Floater et al. [13], Klein and Berrut [14,15] have proposed a rational interpolation scheme which has high numerical stability and interpolation accuracy on both equidistant nodes and non-equidistant nodes. In recent articles, Wang et al. [16,17, 18,19] successfully applied the barycentric interpolation collocation method to solve initial value problems, plane elasticity problems [20], incompressible plane problems, telegraph equation [21], beam force vibration equation [22], and non-linear problems, which have expanded the application fields of the collocation method. For the two-dimensional biharmonic boundary problems, a new spectral collocation method [23] and depression of order [24] are reported to numerically solve it.

Based on the one-dimensional linear barycentric rational interpolation, two-dimensional barycentric rational interpolation polynomial is constructed, then barycentric rational interpolation collocation method is obtained to solve biharmonic equation. With the help of vector form, the discrete linear equation of two-dimensional biharmonic equation is changed into matrix equation which can be coded easily. Moreover, the error estimate of linear barycentric rational interpolation for biharmonic equation is obtained and the convergence rate is also presented.

This article is organized as follows: In Section 2, the differentiation matrix and barycentric rational interpolation collocation scheme for biharmonic equation are presented, then the matrix form of collocation scheme is obtained. In Section 3, the convergence rate is proved. Finally, some numerical examples are listed to illustrate our theorem.

2 Differentiation matrix of biharmonic equation

Let a = x 1 < x 2 < < x m = b , h = b a m 1 , and c = y 1 < y 2 < < y n = d , τ = d c n 1 with mesh point ( x j , y k ) , j = 1 , 2 , , m ; k = 1 , 2 , , n , Ω i j = [ x i , x i + 1 ] × [ y j , y j + 1 ] , then we have

(4) u ( x j , y ) = u j ( y ) , j = 1 , 2 , , m

on the interval [ a , b ] , and

(5) u ( x , y ) = j = 1 m L j ( x ) u j ( y ) ,

where L j ( x ) = w j x x j k = 1 n w k x x k , w k = i J k ( 1 ) i j = i , j k i + d 1 x k x j is the basis function, and J k = { i I ; k d i k } , I = { 0 , 1 , n d } . Similarly on the interval [ c , d ] , we have

(6) u j ( y k ) = u ( x j , y k ) = u j k , j = 1 , 2 , , m ; k = 1 , 2 , , n

and

(7) u j ( y ) = k = 1 n L k ( y ) u j k ,

with L k ( y ) = w k y y k i = 1 n w i y y i , w i = k J i ( 1 ) k j = k , j i k + d 1 y i y j , then we have

(8) u ( x , y ) = j = 1 m L j ( x ) k = 1 n L k ( y ) u j k = j = 1 m k = 1 n L j ( x ) L k ( y ) u j k .

Taking (8) into equation (1), we obtain

(9) j = 1 m k = 1 n L j ( 4 ) ( x ) L k ( y ) u j k + j = 1 m k = 1 n L j ( x ) L k ( y ) u j k + j = 1 m k = 1 n L j ( x ) L k ( 4 ) ( y ) u j k = f ( x , y ) .

By taking x = x j in (9), then we change equation (9) into the matrix equation as

(10) k = 1 n L k ( 4 ) ( y ) u 1 k k = 1 n L k ( 4 ) ( y ) u m k + 2 C 11 ( 2 ) C 1 m ( 2 ) C m 1 ( 2 ) C m m ( 2 ) k = 1 n L k ( y ) u 1 k k = 1 n L k ( y ) u m k + C 11 ( 4 ) C 1 m ( 4 ) C m 1 ( 4 ) C m m ( 4 ) k = 1 n L k ( y ) u 1 k k = 1 n L k ( y ) u m k = f 1 ( y ) f m ( y ) ,

and equation (10) is found on the point y k , k = 1 , 2 , , n and we obtain

(11) k = 1 n L k ( 4 ) ( y j ) u 1 k k = 1 n L k ( 4 ) ( y j ) u m k + 2 C 11 ( 2 ) C 1 m ( 2 ) C m 1 ( 2 ) C m m ( 2 ) k = 1 n L k ( y j ) u 1 k k = 1 n L k ( y j ) u m k + C 11 ( 4 ) C 1 m ( 4 ) C m 1 ( 4 ) C m m ( 4 ) k = 1 n L k ( y j ) u 1 k k = 1 n L k ( y j ) u m k = f 1 ( y j ) f m ( y j ) ,

where f ( x j , y k ) = f j k , f j = [ f j 1 , f j 2 , , f j n ] T = [ f j ( y 1 ) , f j ( y 2 ) , , f j ( y n ) ] T and C j k ( 2 ) = L k ( x j ) , D j k ( 2 ) = L k ( y j ) , C j k ( 4 ) = L k ( 4 ) ( x j ) , D j k ( 4 ) = L k ( 4 ) ( y j ) , u j = [ u j 1 , u j 2 , u j n ] T , j = 1 , 2 , , m , k = 1 , 2 , , n .

According to mathematical induction, we obtain the recurrence formula of k -order ( k 2 ) differential matrix as

(12) C i j ( k ) = k ( C i i ( k 1 ) C i j ( 1 ) C i j ( k 1 ) x i x j ) , i j C i i ( k ) = j = 1 , j i n C i j ( k )

and

(13) D i j ( k ) = k ( D i i ( k 1 ) D i j ( 1 ) D i j ( k 1 ) y i y j ) , i j D i i ( k ) = j = 1 , j i n D i j ( k ) .

First, the Kronecker product of matrix A = ( a i j ) m × n and B = ( b i j ) k × l is defined as

A B = ( a i j B ) m k × n l ,

where the matrix A B is order of m k × n l and

a i j B = a i j b 11 a i j b 12 a i j b 1 l a i j b 21 a i j b 22 a i j b 2 l a i j b k 1 a i j b k 2 a i j b k l .

Then matrix A and B can be changed to ( m × n ) column vector as

x = [ x 1 , , x 1 , x 2 , , x 2 , , x m , , x m ] y = [ y 1 , y 2 , , y n , y 1 , y 2 , , y n , , y 1 , y 2 , , y n ] .

With the help of the matrix equation, the linear equation systems (11) can be written as

(14) ( C ( 4 ) I n ) u 1 u m + 2 ( C ( 2 ) D ( 2 ) ) u 1 u m + ( I m D ( 4 ) ) u 1 u m = f 1 f m ,

where u j = u j ( y k ) , f j = f j ( y k ) , j = 1 , 2 , , m ; k = 1 , 2 , , n and I m , I n and identity matrix of m , n , then we have

(15) ( C ( 4 ) I n + 2 C ( 2 ) D ( 2 ) + I m D ( 4 ) ) U = F

and

(16) L U = F ,

where

(17) L = C ( 4 ) I n + 2 C ( 2 ) D ( 2 ) + I m D ( 4 )

and U = [ u 1 T , u 2 T , , u n T ] T = [ u 11 , , u 1 n , u 21 , , u 2 n , , u m 1 , , u m n ] T , F = [ f 1 T , f 2 T , , f m T ] T = [ f 11 , , f 1 n , f 21 , , f 2 n , , f m 1 , , f m n ] T , j = 1 , 2 , , m , k = 1 , 2 , , n , and is the Kronecker product of matrix.

For the boundary condition of (2), u ( x , y ) = g ( x , y ) , u ( x , y ) x + u ( x , y ) y = h 1 ( x , y ) , we obtain the discrete formulae as u 1 k = g 1 ( y k ) , j = 1 m k = 1 n C m i ( 1 ) δ k j u i j = h 1 ( y k ) , k = 1 , 2 , , n , and u ( k 1 ) n + 1 , 1 = g 1 ( x k ) , j = 1 m k = 1 n δ k i D n j ( 1 ) u i j = h 1 ( x k ) , k = 1 , 2 , , m .

Now we have finished the barycentric rational discrete form of the biharmonic equation, and the matrix equation of the biharmonic equation is also obtained. Some remarks of the Kronecker product and how to choose collocation points are given in the following.

Second, the equidistant node and the second kind of Chebyshev point are chosen as the collocation point. The equidistant node is

x j = a + b a n 1 ( j 1 ) , j = 1 , , n ,

and its weight function is

w j = ( 1 ) n j 1 C n 1 j 1 ,

where C n 1 j 1 = ( n 1 ) ! ( n j ) ! ( j 1 ) ! .

And the second kind of Chebyshev point is

(18) x j = cos ( j 1 ) π n 1 , j = 1 , , n

and its weight function is

(19) w j = ( 1 ) j δ j , δ j = 1 2 , j = 1 , n 1 , ortherwise .

3 Convergence and error analysis

For one-dimensional function u ( x ) is approximated by rational function r ( x ) . Its error functional is defined as

(20) e ( x ) u ( x ) r ( x ) = ( x x i ) ( x x i + d 0 ) u [ x i , x i + 1 , , x i + d 0 , x ] ,

where d 0 [ 0 , n ] , d 0 N , and

(21) e ( x ) = i = 1 n 1 d 0 λ i ( x ) [ u ( x ) r ( x ) ] i = 1 n 1 d 0 λ i ( x ) = A ( x ) B ( x ) = O ( h d 0 + 1 ) ,

where

(22) A ( x ) i = 1 n 1 d 0 ( 1 ) i u [ x i , , x i + d 0 , x ]

and

(23) B ( x ) i = 1 n 1 d 0 λ i ( x ) ,

where

(24) λ i ( x ) = ( 1 ) i ( x x i ) ( x x i + d 0 ) .

The following theorem has been proved by Jean-Paul Berrut, see [13].

Theorem 3.1

If n , d 0 , d 0 n 1 , and k d 0 are positive integers and u ( x ) C d 0 + 1 + k [ a , b ] . If the nodes x i , i = 1 , n , are equidistant or quasi-equidistant, then we have

e ( k ) ( x i ) C h d 0 + 1 k .

For the barycentric rational interpolation of function u ( x , y ) by u n , m ( x , y ) , we can obtain the barycentric rational interpolation

(25) u n , m ( x , y ) = j = 1 m k = 1 n w j , k ( x x j ) ( y y k ) u j k j = 1 m k = 1 n w j , k ( x x j ) ( y y k ) ,

where u j k = u ( x j , y k ) , d 1 [ 0 , m 1 ] , d 2 [ 0 , n 1 ] , and

(26) w j , k = ( 1 ) j d 1 + k d 2 k 1 J j k 2 J k h 1 = k 1 , h 1 k k 1 + d 1 1 x j x h 1 h 2 = k 2 , h 2 k k 2 + d 2 1 y k y h 2 ,

and J j = { k 1 I m : j d 1 k 1 j } , I m = { 0 , , m 1 d 1 } , J k = { k 2 I n : k d 2 k 2 k } , I n = { 0 , , n 1 d 2 } .

By the error term of barycentric rational interpolation for two-dimensional function, we have

(27) e ( x , y ) u ( x , y ) u n , m ( x , y ) = ( x x j ) ( x x j + d 1 ) u [ x j , x j + 1 , , x j + d 1 , x ] + ( y y k ) ( y y k + d 2 ) u [ y k , y k + 1 , , y k + d 2 , y ] .

The following theorem can be proved similarly as Li and Cheng [11].

Theorem 3.2

For the e ( x , y ) defined in (27) and u ( x , y ) C d 1 + 2 [ a , b ] × C d 2 + 2 [ c , d ] , we have

(28) e ( x , y ) C ( h d 1 + 1 + τ d 2 + 1 ) .

Proof

For the function u ( x , y ) with w j , k ( x , y ) defined as (26), the error functional can be expressed as (see Theorem 1 [24])

(29) u ( x , y ) u n , m ( x , y ) = j = 1 m 1 d 1 k = 1 n 1 d 2 λ j ( x ) λ k ( y ) [ u ( x , y ) u n , m ( x , y ) ] j = 1 m 1 d 1 k = 1 n 1 d 2 λ j ( x ) λ k ( y ) .

By the error formulae of barycentric rational interpolation

(30) u ( x , y ) u n , m ( x , y ) = u ( x , y ) u n , m ( x i , y ) + u n , m ( x i , y ) u n , m ( x , y ) = ( x x j ) ( x x j + d 1 ) u [ x j , x j + 1 , , x j + d 1 , x , y ] + ( y y k ) ( y y k + d 2 ) u [ y k , y k + 1 , , y k + d 2 , x i , y ] .

We have

(31) u ( x , y ) u n , m ( x , y ) = j = 1 m 1 d 1 ( 1 ) j u [ x j , x j + 1 , , x j + d 1 , x , y ] j = 1 m 1 d 1 λ j ( x ) + k = 1 n 1 d 2 ( 1 ) k u [ y k , y k + 1 , , y k + d 2 , x i , y ] k = 1 n 1 d 2 λ k ( y ) .

By the similar analysis in Floater and Hormann [12], we have

(32) j = 1 m 1 d 1 λ j ( x ) 1 d 1 ! h d 1 + 1

and

(33) k = 1 n 1 d 2 λ k ( y ) 1 d 2 ! τ d 2 + 1 .

Combining (31), (32), and (33) together, the proof of Theorem 3.2 is completed.□

Corollary 3.3

For the e ( x , y ) defined in (27), we have

(34) e x ( x , y ) C ( h d 1 + τ d 2 + 1 ) , u ( x , y ) C d 1 + 3 [ a , b ] × C d 2 + 2 [ c , d ] , e y ( x , y ) C ( h d 1 + 1 + τ d 2 ) , u ( x , y ) C d 1 + 2 [ a , b ] × C d 2 + 3 [ c , d ] , e x x ( x , y ) C ( h d 1 1 + τ d 2 + 1 ) , u ( x , y ) C d 1 + 4 [ a , b ] × C d 2 + 2 [ c , d ] , d 1 2 , e y y ( x , y ) C ( h d 1 + 1 + τ d 2 1 ) , u ( x , y ) C d 1 + 2 [ a , b ] × C d 2 + 4 [ c , d ] , d 2 2 , e x x x x ( x , y ) C ( h d 1 3 + τ d 2 + 1 ) , u ( x , y ) C d 1 + 6 [ a , b ] × C d 2 + 2 [ c , d ] , d 1 4 , e y y y y ( x , y ) C ( h d 1 + 1 + τ d 2 3 ) , u ( x , y ) C d 1 + 2 [ a , b ] × C d 2 + 6 [ c , d ] , d 2 4 .

This corollary can be obtained similarly as Theorem 3.2, here we omit it.

Combining (8) and (1), we have

(35) T e ( x , y ) e x x x x ( x , y ) + 2 e x x y y ( x , y ) + e y y y y ( x , y ) .

In the following theorem, the main result is presented.

Theorem 3.4

Let

(36) 4 u ( x , y ) = 2 2 u ( x , y ) = 4 u ( x , y ) x 4 + 2 4 u ( x , y ) x 2 y 2 + 4 u ( x , y ) y 4 = f ( x , y ) , ( x , y ) Ω

and

(37) u ( x , y ) = g ( x , y ) , u ( x , y ) x + u ( x , y ) y = h 1 ( x , y ) , ( x , y ) ( Ω ) ,

where Ω = [ a , b ] × [ c , d ] and f ( x , y ) , g ( x , y ) , h ( x , y ) are consistent on Ω . Then we have

(38) max Ω i j u ( x , y ) u n , m ( x i , y j ) C ( h d 1 3 + τ d 2 3 ) ,

where u n , m ( x i , y j ) is defined as (25) and u ( x , y ) C d 1 + 6 [ a , b ] × C d 2 + 6 [ c , d ] , d 1 3 , d 2 3 .

Proof

As L C ( 4 ) I n + 2 C ( 2 ) D ( 2 ) + I m D ( 4 ) and u ( x , y ) = j = 1 m L j ( x ) k = 1 n L k ( y ) u i k = j = 1 m k = 1 n L j ( x ) L k ( y ) u i k .

By

U U n , m = U L 1 F = L 1 ( L U F ) = L 1 T e ( x , y ) ,

where we have used L U n , m = F , we assume that matrix L is the inverse matrix which means

(39) u ( x , y ) u n , m ( x , y ) = j = 1 m n M j k ( x , y ) T e ( x , y ) ,

where matrix L is the inverse matrix and M j k ( x , y ) is the element of matrix L 1 .

By the definition of T e ( x , y ) and

(40) 4 u ( x , y ) 4 u m , n ( x i , y j ) = 4 u ( x , y ) x 4 + 2 4 u ( x , y ) x 2 y 2 + 4 u ( x , y ) y 4 4 u ( x m , y n ) x 4 + 2 4 u ( x m , y n ) x 2 y 2 + 4 u ( x m , y n ) y 4 = 4 u ( x , y ) x 4 4 u ( x m , y n ) x 4 + 2 4 u ( x , y ) x 2 y 2 4 u ( x m , y n ) x 2 y 2 + 4 u ( x , y ) y 4 4 u ( x m , y n ) y 4 R 1 ( x , y ) + R 2 ( x , y ) + R 3 ( x , y ) ,

we have

R 1 ( x , y ) = e x x x x ( x , y ) = u x x x x ( x , y ) u x x x x ( x m , y n ) , R 2 ( x , y ) = e y y y y ( x , y ) = u y y y y ( x , y ) u y y y y ( x m , y n ) , R 3 ( x , y ) = 2 e x x y y ( x , y ) = 2 u x x y y ( x , y ) 2 u x x y y ( x m , y n ) .

Then, for R 1 ( x , y ) we have

R 1 ( x , y ) = u x x x x ( x , y ) u x x x x ( x m , y n ) = u x x x x ( x , y ) u x x x x ( x m , y ) + u x x x x ( x m , y ) u x x x x ( x m , y n ) = j = 1 m 1 d 1 ( 1 ) j u x x x x [ x j , x j + 1 , , x j + d 1 , x , y ] j = 1 m 1 d 1 λ j ( x ) + k = 1 n 1 d 2 ( 1 ) k u x x x x [ y k , y k + 1 , , y k + d 2 , x m , y ] k = 1 n 1 d 2 λ k ( y ) = e x x x x ( x , y n ) + e x x x x ( x m , y n ) ,

where

(41) R 1 ( x , y ) e x x x x ( x , y n ) + e x x x x ( x m , y n ) C ( h d 1 3 + τ d 2 + 1 ) .

Similarly, for R 2 ( x , y ) we have

(42) R 2 ( x , y ) = u y y y y ( x , y ) u y y y y ( x m , y n ) = e y y y y ( x , y n ) + e y y y y ( x m , y n )

and

(43) R 2 ( x , y ) e y y y y ( x , y n ) + e y y y y ( x m , y n ) C ( h d 1 + 1 + τ d 2 3 ) .

Similarly, for R 3 ( x , y ) we have

(44) R 3 ( x , y ) e x x y y ( x , y n ) + e x x y y ( x m , y n ) C ( h d 1 3 + τ d 2 3 ) .

Combining (41), (43), and (44), the proof of Theorem 3.4 is completed.□

4 Numerical examples

In the following section, we present some examples to illustrate our theorem analysis. The boundary condition has been considered by replacing the first, last row and column of discrete equation by the replacement method.

Example 4.1

Consider the biharmonic equation with Ω = [ 0 , 1 ] × [ 0 , 1 ] and f ( x , y ) = ( x 4 + y 4 + 2 x 2 y 2 + 8 x y + 4 ) e x y , with the boundary condition

u ( 0 , y ) = u ( 1 , y ) = u ( x , 0 ) = u ( x , 1 ) = 0

and

u x ( 0 , y ) = y , u x ( 1 , y ) = y e y , u y ( x , 0 ) = x , u y ( x , 1 ) = x e x .

Its analytical solution is

u ( x , y ) = e x y .

In this example, we test the linear barycentric rational collocation method with the equidistant nodes, and Table 1 shows that the convergence rate is O ( h d 1 1 ) with d 1 = d 2 = 3 , 4 , 5 . In Table 2, for the Chebyshev nodes, the convergence rate of times is O ( τ d 2 ) with d 1 = d 2 = 3 , 4 , 5 , which agree with our theorem analysis. For d 1 = d 2 = 2 , there are convergence rates O ( h ) and O ( h 1.5 ) in Tables 1 and 2, which is out of our goal of this article.

Table 1

Convergence rate of equidistant nodes with different d 1 and d 2 by barycentric rational interpolation

m × n d 1 = d 2 = 2 d 1 = d 2 = 3 d 1 = d 2 = 4 d 1 = d 2 = 5
8 × 8 1.5128 × 1 0 4 4.7103 × 1 0 5 7.2316 × 1 0 7 3.7284 × 1 0 7
16 × 16 5.8896 × 1 0 5 1.3610 9.4358 × 1 0 6 2.3196 1.1067 × 1 0 7 2.7081 2.2959 × 1 0 8 4.0215
32 × 32 2.4460 × 1 0 5 1.2677 1.3426 × 1 0 6 2.8131 1.3944 × 1 0 7 1.4910 × 1 0 7
64 × 64 9.5489 × 1 0 6 1.3570 8.2261 × 1 0 6 6.7989 × 1 0 6 6.1683 × 1 0 6
Table 2

Convergence rate of Chebyshev nodes with different d 1 and d 2 by barycentric rational interpolation

m × n d 1 = d 2 = 2 d 1 = d 2 = 3 d 1 = d 2 = 4 d 1 = d 2 = 5
8 × 8 1.8264 × 1 0 4 6.7627 × 1 0 5 3.5778 × 1 0 7 1.0981 × 1 0 7
16 × 16 9.0841 × 1 0 5 1.0076 3.4022 × 1 0 6 4.3131 3.7635 × 1 0 8 3.2489 1.1783 × 1 0 8 3.2202
32 × 32 3.1179 × 1 0 5 1.5428 6.4577 × 1 0 6 1.1016 × 1 0 5 8.8547 × 1 0 5
64 × 64 2.6024 × 1 0 3 3.4060 × 1 0 3 9.3608 × 1 0 2 6.9634 × 1 0 1

We choose m = 20 ; n = 20 ; d 1 = d 2 = 10 ; to test our algorithm. Figure 1 shows the error estimate of equidistant nodes with barycentric Lagrange interpolation collocation method, and Figure 2 shows the error estimate of Chebyshev nodes with barycentric Lagrange interpolation collocation method.

Figure 1 
               Error estimate of equidistant nodes with Lagrange interpolation 
                     
                        
                        
                           m
                           =
                           20
                           ;
                           
                           n
                           =
                           20
                           ;
                           
                           
                              
                                 d
                              
                              
                                 1
                              
                           
                           =
                           
                              
                                 d
                              
                              
                                 2
                              
                           
                           =
                           10
                           .
                        
                        m=20;\hspace{0.33em}n=20;\hspace{0.33em}{d}_{1}={d}_{2}=10.
Figure 1

Error estimate of equidistant nodes with Lagrange interpolation m = 20 ; n = 20 ; d 1 = d 2 = 10 .

Figure 2 
               Error estimate of Chebyshev nodes with Lagrange interpolation 
                     
                        
                        
                           m
                           =
                           20
                           ;
                           
                           n
                           =
                           20
                           ;
                           
                           
                              
                                 d
                              
                              
                                 1
                              
                           
                           =
                           
                              
                                 d
                              
                              
                                 2
                              
                           
                           =
                           10
                           .
                        
                        m=20;\hspace{0.33em}n=20;\hspace{0.33em}{d}_{1}={d}_{2}=10.
Figure 2

Error estimate of Chebyshev nodes with Lagrange interpolation m = 20 ; n = 20 ; d 1 = d 2 = 10 .

Figure 3 shows the error estimate of barycentric rational Lagrange interpolation collocation method with equidistant nodes, and Figure 4 shows the barycentric rational interpolation collocation method of the error estimate of Chebyshev nodes. From Figures 3 and 4, we know that the barycentric rational interpolation collocation method has higher accuracy under the condition of Chebyshev nodes.

Figure 3 
               Error estimate of equidistant nodes with barycentric rational interpolation 
                     
                        
                        
                           m
                           =
                           20
                           ;
                           
                           n
                           =
                           20
                           ;
                           
                           
                              
                                 d
                              
                              
                                 1
                              
                           
                           =
                           
                              
                                 d
                              
                              
                                 2
                              
                           
                           =
                           10
                        
                        m=20;\hspace{0.33em}n=20;\hspace{0.33em}{d}_{1}={d}_{2}=10
                     
                  .
Figure 3

Error estimate of equidistant nodes with barycentric rational interpolation m = 20 ; n = 20 ; d 1 = d 2 = 10 .

Figure 4 
               Error estimate of Chebyshev nodes with barycentric rational interpolation 
                     
                        
                        
                           m
                           =
                           20
                           ;
                           
                           n
                           =
                           20
                           ;
                           
                           
                              
                                 d
                              
                              
                                 1
                              
                           
                           =
                           
                              
                                 d
                              
                              
                                 2
                              
                           
                           =
                           10
                        
                        m=20;\hspace{0.33em}n=20;\hspace{0.33em}{d}_{1}={d}_{2}=10
                     
                  .
Figure 4

Error estimate of Chebyshev nodes with barycentric rational interpolation m = 20 ; n = 20 ; d 1 = d 2 = 10 .

Table 3 shows condition number of equidistant nodes with different d 1 and d 2 . In Table 4, condition number of Chebyshev nodes with different d 1 and d 2 is shown.

Table 3

Condition number of equidistant nodes with different d 1 and d 2 by barycentric rational interpolation

m × n d 1 = d 2 = 2 d 1 = d 2 = 3 d 1 = d 2 = 4 d 1 = d 2 = 5
8 × 8 1.5128 × 1 0 4 1.0458 × 1 0 6 4.7103 × 1 0 5 8.8260 × 1 0 5 7.2316 × 1 0 7 7.7385 × 1 0 5 3.7284 × 1 0 7 1.0259 × 1 0 6
16 × 16 5.8896 × 1 0 5 4.2634 × 1 0 7 9.4358 × 1 0 6 4.3390 × 1 0 7 1.1067 × 1 0 7 4.8265 × 1 0 7 2.2959 × 1 0 8 8.4677 × 1 0 7
32 × 32 2.4460 × 1 0 5 1.2187 × 1 0 9 1.3426 × 1 0 6 1.2153 × 1 0 9 1.3944 × 1 0 7 1.2483 × 1 0 9 1.4910 × 1 0 7 2.1078 × 1 0 9
64 × 64 9.5489 × 1 0 6 3.0767 × 1 0 10 8.2261 × 1 0 6 3.0609 × 1 0 10 6.7989 × 1 0 6 3.0648 × 1 0 10 6.1683 × 1 0 6 3.7388 × 1 0 10
Table 4

Condition number of Chebyshev nodes with different d 1 and d 2 by barycentric rational interpolation

m × n d 1 = d 2 = 2 d 1 = d 2 = 3 d 1 = d 2 = 4 d 1 = d 2 = 5
8 × 8 1.8264 × 1 0 4 5.2010 × 1 0 5 6.7627 × 1 0 5 4.9504 × 1 0 5 3.5778 × 1 0 7 3.9983 × 1 0 5 1.0981 × 1 0 7 3.5166 × 1 0 5
16 × 16 9.0841 × 1 0 5 6.8575 × 1 0 7 3.4022 × 1 0 6 6.0817 × 1 0 7 3.7635 × 1 0 8 7.3456 × 1 0 7 1.1783 × 1 0 8 9.0076 × 1 0 7
32 × 32 3.1179 × 1 0 5 2.2982 × 1 0 10 6.4577 × 1 0 6 2.6232 × 1 0 10 1.1016 × 1 0 5 3.6369 × 1 0 11 8.8547 × 1 0 5 6.0085 × 1 0 12
64 × 64 2.6024 × 1 0 3 1.2637 × 1 0 13 3.4060 × 1 0 3 3.5042 × 1 0 13 9.3608 × 1 0 2 6.8143 × 1 0 15 6.9634 × 1 0 1 1.6731 × 1 0 18

Example 4.2

Consider the biharmonic equation with Ω = [ 1 , 1 ] × [ 1 , 1 ] and f ( x , y ) = 0 , with the boundary condition

u ( 1 , y ) = u ( 1 , y ) = u ( x , 1 ) = u ( x , 1 ) = 0

and

u x ( 1 , y ) = 1 2 ( sin ( 1 ) cosh ( y ) cos ( 1 ) sinh ( y ) ) + 1 2 ( cos ( 1 ) cosh ( y ) + sin ( 1 ) sinh ( y ) ) , u x ( 1 , y ) = 1 2 ( sin ( 1 ) cosh ( y ) cos ( 1 ) sinh ( y ) ) + 1 2 ( cos ( 1 ) cosh ( y ) + sin ( 1 ) sinh ( y ) ) , u y ( x , 1 ) = x 2 ( sin ( x ) sinh ( 1 ) cos ( x ) cosh ( 1 ) ) u y ( x , 1 ) = x 2 ( sin ( x ) sinh ( 1 ) cos ( x ) cosh ( 1 ) ) .

Its analytical solution is

u ( x , y ) = x 2 ( sin ( x ) cosh ( y ) cos ( x ) sinh ( y ) ) .

In this example, we test the linear barycentric rational collocation method with the equidistant nodes, Table 5 shows that the convergence rate is O ( h d 1 ) with d 1 = d 2 = 3 , 4 , 5 . In Table 6, for the Chebyshev nodes, the convergence rate of times is O ( τ d 2 ) with d 1 = d 2 = 3 , 4 , 5 , which agrees with our theorem analysis. For d 1 = d 2 = 2 , there are the convergence rates O ( h 1.5 ) and O ( h 2 ) in Tables 5 and 6, which is out of our goal of this article.

Table 5

Convergence rate of equidistant nodes with different d 1 and d 2 by barycentric rational interpolation

m × n d 1 = d 2 = 2 d 1 = d 2 = 3 d 1 = d 2 = 4 d 1 = d 2 = 5
8 × 8 1.1279 × 1 0 2 4.3323 × 1 0 3 1.0184 × 1 0 3 5.8208 × 1 0 4
16 × 16 4.8796 × 1 0 3 1.2088 4.5554 × 1 0 4 3.2495 8.6515 × 1 0 5 3.5572 1.4807 × 1 0 5 5.2969
32 × 32 1.8411 × 1 0 3 1.4062 4.9607 × 1 0 5 3.1989 5.7439 × 1 0 6 3.9129 3.8031 × 1 0 7 5.2830
64 × 64 6.2066 × 1 0 4 1.5687 5.6875 × 1 0 6 3.1247 3.6380 × 1 0 7 3.9808 1.0567 × 1 0 7 1.8476
Table 6

Convergence rate of Chebyshev nodes with different d 1 and d 2 by barycentric rational interpolation

m × n d 1 = d 2 = 2 d 1 = d 2 = 3 d 1 = d 2 = 4 d 1 = d 2 = 5
8 × 8 1.0182 × 1 0 2 6.0338 × 1 0 3 5.2497 × 1 0 4 9.9132 × 1 0 5
16 × 16 7.7462 × 1 0 3 0.3945 1.6321 × 1 0 4 5.2083 1.4431 × 1 0 5 5.1850 3.3800 × 1 0 6 4.8743
32 × 32 2.3755 × 1 0 3 1.7053 2.0014 × 1 0 5 3.0277 6.8929 × 1 0 7 4.3879 2.4726 × 1 0 7 3.7729
64 × 64 6.4680 × 1 0 4 1.8768 8.8375 × 1 0 5 2.5854 × 1 0 3 2.3854 × 1 0 2

We choose m = 10 ; n = 10 ; d 1 = d 2 = 8 ; to test our algorithm. Figure 5 shows the error estimate of equidistant nodes with barycentric Lagrange interpolation collocation method, and Figure 6 shows the error estimate of Chebyshev nodes with barycentric Lagrange interpolation collocation method.

Figure 5 
               Error estimate of equidistant nodes with Lagrange interpolation 
                     
                        
                        
                           m
                           =
                           10
                           ;
                           
                           n
                           =
                           10
                           ;
                           
                           
                              
                                 d
                              
                              
                                 1
                              
                           
                           =
                           
                              
                                 d
                              
                              
                                 2
                              
                           
                           =
                           8
                        
                        m=10;\hspace{0.33em}n=10;\hspace{0.33em}{d}_{1}={d}_{2}=8
                     
                  .
Figure 5

Error estimate of equidistant nodes with Lagrange interpolation m = 10 ; n = 10 ; d 1 = d 2 = 8 .

Figure 6 
               Error estimate of Chebyshev nodes with Lagrange interpolation 
                     
                        
                        
                           m
                           =
                           10
                           ;
                           
                           n
                           =
                           10
                           ;
                           
                           
                              
                                 d
                              
                              
                                 1
                              
                           
                           =
                           
                              
                                 d
                              
                              
                                 2
                              
                           
                           =
                           8
                        
                        m=10;\hspace{0.33em}n=10;\hspace{0.33em}{d}_{1}={d}_{2}=8
                     
                  .
Figure 6

Error estimate of Chebyshev nodes with Lagrange interpolation m = 10 ; n = 10 ; d 1 = d 2 = 8 .

Figure 7 shows the error estimate of equidistant nodes with rational barycentric rational interpolation collocation method, and Figure 8 shows the error estimate of Chebyshev nodes. From Figures 7 and 8, we know that the barycentric rational interpolation collocation method has higher accuracy under the condition of Chebyshev nodes.

Figure 7 
               Error estimate of equidistant nodes with barycentric rational interpolation 
                     
                        
                        
                           m
                           =
                           10
                           ;
                           
                           n
                           =
                           10
                           ;
                           
                           
                              
                                 d
                              
                              
                                 1
                              
                           
                           =
                           
                              
                                 d
                              
                              
                                 2
                              
                           
                           =
                           8
                        
                        m=10;\hspace{0.33em}n=10;\hspace{0.33em}{d}_{1}={d}_{2}=8
                     
                  .
Figure 7

Error estimate of equidistant nodes with barycentric rational interpolation m = 10 ; n = 10 ; d 1 = d 2 = 8 .

Figure 8 
               Error estimate of Chebyshev nodes with barycentric rational interpolation 
                     
                        
                        
                           m
                           =
                           10
                           ;
                           
                           n
                           =
                           10
                           ;
                           
                           
                              
                                 d
                              
                              
                                 1
                              
                           
                           =
                           
                              
                                 d
                              
                              
                                 2
                              
                           
                           =
                           8
                        
                        m=10;\hspace{0.33em}n=10;\hspace{0.33em}{d}_{1}={d}_{2}=8
                     
                  .
Figure 8

Error estimate of Chebyshev nodes with barycentric rational interpolation m = 10 ; n = 10 ; d 1 = d 2 = 8 .

Example 4.3

Consider the biharmonic equation with the Ω = [ 0 , 1 ] × [ 0 , 1 ] and f ( x , y ) = 384 π 4 640 π 4 cos ( 2 π y ) 2 640 π 4 cos ( 2 π x ) 2 + 1024 cos ( 2 π x ) 2 π 4 cos ( 2 π y ) 2 , with the boundary condition

u ( 0 , y ) = u ( 1 , y ) = u ( x , 0 ) = u ( x , 1 ) = 0

and

u x ( 0 , y ) = u x ( 1 , y ) = u y ( x , 0 ) = u y ( x , 1 ) = 0 .

Its analytical solution is

u ( x , y ) = ( sin ( 2 π x ) ) 2 ( sin ( 2 π y ) ) 2 .

In this example, we test the linear barycentric rational with the equidistant nodes, and Table 7 shows the convergence rate is O ( h d 1 ) with d 1 = d 2 = 2 , 3 , 4 , 5 . In Table 8, for the Chebyshev nodes, the convergence rate of times is O ( τ d 2 ) with d 1 = d 2 = 2 , 3 , 4 , 5 , which agree with our theorem analysis.

Table 7

Convergence rate of equidistant nodes with different d 1 and d 2 by barycentric rational interpolation

m × n d 1 = d 2 = 2 d 1 = d 2 = 3 d 1 = d 2 = 4 d 1 = d 2 = 5
8 × 8 4.4289 × 1 0 1 4.1219 × 1 0 1 6.9430 × 1 0 1 3.6467 × 1 0 0
16 × 16 1.9631 × 1 0 1 1.1738 9.3822 × 1 0 2 2.1353 1.2504 × 1 0 1 2.4732 3.6911 × 1 0 2 6.6264
32 × 32 6.7432 × 1 0 2 1.5416 1.6326 × 1 0 2 2.5227 5.6904 × 1 0 3 4.4577 1.6090 × 1 0 3 4.5198
64 × 64 1.9433 × 1 0 2 1.7949 2.1067 × 1 0 3 2.9541 2.5213 × 1 0 4 4.4963 7.9457 × 1 0 5 4.3399
Table 8

Convergence rate of Chebyshev nodes with different d 1 and d 2 by barycentric rational interpolation

m × n d 1 = d 2 = 2 d 1 = d 2 = 3 d 1 = d 2 = 4 d 1 = d 2 = 5
8 × 8 1.6656 × 1 0 0 1.5281 × 1 0 0 2.0820 × 1 0 1 4.2919 × 1 0 1
16 × 16 2.6816 × 1 0 1 2.6349 3.9003 × 1 0 2 5.2920 1.6147 × 1 0 2 3.6886 7.9025 × 1 0 3 5.7632
32 × 32 8.6146 × 1 0 2 1.6382 6.7989 × 1 0 3 2.5202 4.2653 × 1 0 4 5.2425 2.9583 × 1 0 5 8.0614
64 × 64 2.2906 × 1 0 2 1.9111 4.6030 × 1 0 4 3.8847 4.0683 × 1 0 5 3.3901 3.1813 × 1 0 4

We choose m = 20 ; n = 20 ; d 1 = d 2 = 8 ; to test our algorithm. Figure 9 shows the error estimate of equidistant nodes with barycentric Lagrange interpolation collocation method, and Figure 10 shows the error estimate of Chebyshev nodes with barycentric Lagrange interpolation collocation method. Figure 11 shows the error estimate of equidistant nodes with rational barycentric rational interpolation collocation method, and Figure 12 shows the error estimate of Chebyshev nodes. From Figures 11 and 12, we know that the barycentric rational interpolation collocation method has higher accuracy under the condition of Chebyshev nodes.

Figure 9 
               Error estimate of equidistant nodes with Lagrange interpolation 
                     
                        
                        
                           m
                           =
                           10
                           ;
                           
                           n
                           =
                           10
                           ;
                           
                           
                              
                                 d
                              
                              
                                 1
                              
                           
                           =
                           
                              
                                 d
                              
                              
                                 2
                              
                           
                           =
                           8
                        
                        m=10;\hspace{0.33em}n=10;\hspace{0.33em}{d}_{1}={d}_{2}=8
                     
                  .
Figure 9

Error estimate of equidistant nodes with Lagrange interpolation m = 10 ; n = 10 ; d 1 = d 2 = 8 .

Figure 10 
               Error estimate of Chebyshev nodes with Lagrange interpolation 
                     
                        
                        
                           m
                           =
                           20
                           ;
                           
                           n
                           =
                           20
                           ;
                           
                           
                              
                                 d
                              
                              
                                 1
                              
                           
                           =
                           
                              
                                 d
                              
                              
                                 2
                              
                           
                           =
                           8
                        
                        m=20;\hspace{0.33em}n=20;\hspace{0.33em}{d}_{1}={d}_{2}=8
                     
                  .
Figure 10

Error estimate of Chebyshev nodes with Lagrange interpolation m = 20 ; n = 20 ; d 1 = d 2 = 8 .

Figure 11 
               Error estimate of equidistant nodes with barycentric rational interpolation 
                     
                        
                        
                           m
                           =
                           20
                           ;
                           
                           n
                           =
                           20
                           ;
                           
                           
                              
                                 d
                              
                              
                                 1
                              
                           
                           =
                           
                              
                                 d
                              
                              
                                 2
                              
                           
                           =
                           8
                        
                        m=20;\hspace{0.33em}n=20;\hspace{0.33em}{d}_{1}={d}_{2}=8
                     
                  .
Figure 11

Error estimate of equidistant nodes with barycentric rational interpolation m = 20 ; n = 20 ; d 1 = d 2 = 8 .

Figure 12 
               Error estimate of Chebyshev nodes with barycentric rational interpolation 
                     
                        
                        
                           m
                           =
                           20
                           ;
                           
                           n
                           =
                           20
                           ;
                           
                           
                              
                                 d
                              
                              
                                 1
                              
                           
                           =
                           
                              
                                 d
                              
                              
                                 2
                              
                           
                           =
                           8
                        
                        m=20;\hspace{0.33em}n=20;\hspace{0.33em}{d}_{1}={d}_{2}=8
                     
                  .
Figure 12

Error estimate of Chebyshev nodes with barycentric rational interpolation m = 20 ; n = 20 ; d 1 = d 2 = 8 .

5 Conclusion

In this article, we have presented the linear barycentric collocation methods to solve the two-dimensional elliptic boundary value problems. With the help of error estimation of error functional, the convergence rate of the biharmonic equation is proved with the constant coefficient, we have presented the convergence rate O ( h d 1 3 + τ d 2 3 ) , while the numerical examples show that the convergence rate is O ( h d 1 1 + τ d 2 1 ) for the equidistant nodes and O ( h d 1 + τ d 2 ) for the Chebyshev nodes with barycentric collocation methods. Particularly for the case d 1 = d 2 = 2 , there are still the convergence rate which can reach O ( h ) or even O ( h 2 ) , this is an interesting phenomenon which will be investigated in future works.

Acknowledgments

The author gratefully acknowledges the helpful comments and suggestions of the reviewers, which have improved the presentation.

  1. Funding information: The work of Jin Li was supported by Natural Science Foundation of Shandong Province (Grant No. ZR2016JL006), Natural Science Foundation of Hebei Province (Grant No. A2019209533), National Natural Science Foundation of China (Grant Nos 11471195 and 11771398), and China Postdoctoral Science Foundation (Grant No. 2015T80703).

  2. Conflict of interest: The author declares that he has no conflicts of interest.

  3. Data availability statement: The data that support the findings of this study are available from the corresponding author upon reasonable request.

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Received: 2022-01-06
Revised: 2022-06-18
Accepted: 2022-08-01
Published Online: 2022-09-23

© 2022 Jin Li, published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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