Measuring consumer' willingness to pay for food safety certification labels of packaged rice

Chung-Te Ting; Yu-Sheng Huang; Cheng-Te Lin; Yun Hsieh; Chung-Te Ting; Yu-Sheng Huang; Cheng-Te Lin; Yun Hsieh

doi:10.3934/agrfood.2021060

AIMS Agriculture and Food

2021, Volume 6, Issue 4: 1000-1010. doi: 10.3934/agrfood.2021060

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Research article Topical Sections

Measuring consumer' willingness to pay for food safety certification labels of packaged rice

1.
Department of Tourism, Food & Beverage Management, Chang Jung Christian University, Tainan City 71101, Taiwan, R.O.C.
2.
Business School of Yulin Normal University, Yulin 537000, Guangxi, China
3.
Department of Business Administration, Chang Jung Christian University, Tainan City 71101, Taiwan, R.O.C

Received: 13 August 2021 Accepted: 22 November 2021 Published: 01 December 2021

Food safety is vital for public health, influencing a country's economic development and international reputation. In recent years, Taiwan has encountered several food quality problems, with consequent public questioning and mistrust of food safety. If consumers are unaware of the quality of a product, their perception and demands can be influenced by the credibility of the food safety certification labels. This study uses the contingent valuation method (CVM) to analyze the factors influencing consumer willingness to pay (WTP) for food safety certification labels of packaged rice. Questionnaires are randomly adopted and assigned to 434 participants. The results indicate that product factors should be presented clearly on their packaging and advertisements. Consumers focus on certification labels for food safety perception. Gender, education level, place of purchase, certification cognition, and health cognition are all found to affect willingness to pay. These significant results indicate that women were more willing to pay than men; people were willing to pay a higher amount for purchases of packaged rice at supermarkets or hypermarkets; health cognition was estimated to be negative and significant, showing that health cognition negatively affected WTP, perhaps because participants did not sufficiently trust the foundation's assurance that the rice was safe. In the valuation of WTP for food safety certification, each person was willing to pay on average NT31.07 annually to reduce the risk of food safety problems.

Keywords:

food safety,
health cognition,
willingness to pay,
contingent valuation method, certification label

Citation: Chung-Te Ting, Yu-Sheng Huang, Cheng-Te Lin, Yun Hsieh. Measuring consumer' willingness to pay for food safety certification labels of packaged rice[J]. AIMS Agriculture and Food, 2021, 6(4): 1000-1010. doi: 10.3934/agrfood.2021060

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Abstract

1. Introduction

The differential equations of some problems are defined on infinite intervals of some engineering problems. Because of the infinite of calculation interval, how to calculate the upper boundary value problem of infinite interval becomes an important research subject for many numerical analysts.

Semi-infinite domain problems defined on $(0, \infty)$ as

$\begin{eqnarray} F(x, u(x), u'(x), u''(x)) = 0, 0 < x < \infty, \end{eqnarray}$

(1.1)

with the boundary condition

$\begin{eqnarray} u(0) = c_{0}, u(\infty) = u_{\infty} \end{eqnarray}$

(1.2)

is considered, where $F$ if continues function $c_{0}, c_{\infty}$ are constants and $u(\infty) = \lim_{x\rightarrow \infty}u(x)$ .

It is easier to get the solution of differential equation under mathematical theory in infinite interval than in finite interval by Fourier transform and Laplace transform. Numerical method cannot directly solve the differential equation problem in infinite interval. In order to solve the differential equations in infinite interval, we need to develop new methods such as truncation method and transformation method. In the paper ^[1], strictly monotonic transformation is transform the $[0, \infty)$ into $[-1, 1)$ , two-point boundary value problem is solved by Chebyshev-Gauss collocation. In the paper ^[2], the method of weighted residuals is used to solve some problems involving boundary condition at infinity. In the paper ^[3], an original Petrov-Galerkin formulation of the Falkner-Skan equation is presented which is based on a judiciously chosen special basis function to capture the asymptotic behavior of the unknown. In the paper ^[4], based on the combination of Laplace transformation method and weighted residual method, an numerical method for the approximate solution of problems involving boundary condition at infinity is presented. Schrdinger-boussinesq system ^[5], nonlinear fractional $K(m, n)$ type equation ^[6], nanotechnology and fractional ^[7,8], hirota-maccari system ^[9] and eneralized calogero-bogoyavlenskii-schiff equation ^[10] are studied. Generalized $\phi$ -convex functions ^[11], fractional integral operator ^[12], fractional-calculus theory ^[13], fractional inequalities ^[14], non-singular fractional integral operator ^[15] and differentiability in fractional calculus ^[16] are studied by Rashid and so on. In references ^[17,18], infinite cell method, the method of reconstructed kernels, Howarth's numerical solution and Runge-Kutta Fehlberg method have been used to numerically solve semi-infinite problems, respectively.

Barycentrix interpolation collocation ^{[19,20,21,22,23]} have been developed to avoid the Runge phenomenon which is a meshfree method ^[24,25,26] to find approximate solutions to partial differential equations without integration. Meshless approach for the numerical solution of the nonlinear equal width equation, sine-Gordon system and sinh-Gordon equation were presented in ^[27,28,29], soliton wave solutions of nonlinear mathematical models, nonlinear sine-Gordon model and generalized Rosenau-KdV-RLW Equation were presented in ^[30,31,32]. In the recent paper, heat conduction equation ^[33], integral-differential equation ^[34], differential equation ^[35] and biharmonic equation ^[36] have been solved by linear barycentrix rational collocation methods. In the paper ^[37,38,39], barycentric interpolation collocation method for nonlinear problems, incompressible plane elastic problems and plane elastic problems and so on are presented.

In this paper, we first consider the boundary value problem of linear differential equation on infinite interval for certain strictly monotone differentiable functions. By transformation of algebraic and Logarithmic, the infinite interval $[0, \infty)$ is transformed to finite interval $[-1, 1)$ , then the linear barycentric rational collocation methods (LBRCM) of finite interval problem is illustrated. We also give the truncation method, which is to cut the infinite interval into a finite interval solution. Thirdly, the LBRCM is extended to nonlinear problem by the linearized iterative collocation method for solving nonlinear boundary value problems on infinite interval.

This paper is organized as following: In Section 2, linear boundary problems transform from semi-infinite domain into $[-1, 1)$ . In Section 3, the convergence and error analysis of LBRCM is proved. At last, three numerical examples are listed to illustrated our theorem.

2. Linear boundary problems

In order to compute the semi-infinite domain problems easily, we give the transform as

$\begin{eqnarray} x = \phi(t), t\in[-1, 1), x\in(0, \infty) \end{eqnarray}$

(2.1)

where $\phi(t)$ is the strictly monotone differentiable functions, then we have $u(x) = u(\phi(t)) = v(t)$ . As we have

$\begin{eqnarray*} \frac{dx}{dt} = \phi'(t), \frac{dt}{dx} = \frac{1}{\phi'(t)}, \end{eqnarray*}$

by the rule of derivation, we have

$\begin{eqnarray} \frac{du}{dx} = \frac{1}{\phi'(t)}\frac{dv}{dt}, \end{eqnarray}$

(2.2)

$\begin{eqnarray} \frac{d^2u}{dx^2} = \frac{1}{\phi'^2(t)}\frac{d^2v}{dt^2}-\frac{\phi''(t)}{\phi'^3(t)}\frac{dv}{dt}, \end{eqnarray}$

(2.3)

$\begin{eqnarray} \frac{d^3u}{dx^3} = \frac{1}{\phi'^3(t)}\frac{d^3v}{dt^3}-\frac{3\phi''^2(t)}{\phi'^4(t)} \frac{d^2v}{dt^2}-\frac{\phi''(t)\phi'(t)-3\phi''^2(t)}{\phi'^5(t)}\frac{dv}{dt}. \end{eqnarray}$

(2.4)

Then we transform the $(0, \infty)$ into $(-1, 1)$ as the boundary value problems

$\begin{eqnarray} F(\phi(t), v(t), \frac{1}{\phi'(t)}\frac{dv}{dt}, \frac{1}{\phi'^2(t)}\frac{d^2v}{dt^2}-\frac{\phi''(t)}{\phi'^3(t)}\frac{dv}{dt} ) = 0, -1\le t\le 1 \end{eqnarray}$

(2.5)

and

$v(-1) = c_{0}, v(1) = c_{\infty}.$

In the following, the interval $[-1, 1]$ can be partitioned $t_{0} = -1, t_{1}, \ldots, t_{n} = 1$ and its semi-infinite domain $c_{0}, c_{\infty}$ as $x_{0} = -1, x_{1}, \ldots, x_{n} = \infty$ with

$\begin{eqnarray*} x_{k} = \phi(t_{k}), k = 0, 1, \ldots, n \end{eqnarray*}$

and

$\begin{eqnarray*} u(x_{k}) = v(t_{k}), -1\le t\le 1, \end{eqnarray*}$

then we get the value at the mesh-point

$\begin{eqnarray} u'(x_{k}) = \frac{1}{\phi'(t_{k})}v'(t_{k}), \end{eqnarray}$

(2.6)

$\begin{eqnarray} u''(x_{k}) = \frac{1}{\phi'^2(t_{k})}v''(t_{k})-\frac{\phi''(t)}{\phi'^3(t)}v'(t_{k}), \end{eqnarray}$

(2.7)

and

$\begin{eqnarray} u'''(x_{k}) = \frac{1}{\phi'^3(t)}v'''(t_{k})-\frac{3\phi''^2(t)}{\phi'^4(t)}v''(t_{k})-\frac{\phi''(t)\phi'(t)-3\phi''^2(t)}{\phi'^5(t)}v'(t_{k}) \end{eqnarray}$

(2.8)

with the help of vector form

$\begin{eqnarray} \mathbf{u}^{(1)} = \mathrm{diag}\left(\frac{1}{\phi'(t_{k})}\right)\mathbf{v}^{(1)}, \end{eqnarray}$

(2.9)

$\begin{eqnarray} \mathbf{u}^{(2)} = \mathrm{diag}\left(\frac{1}{\phi'^2(t_{k})}\right)\mathbf{v}^{(2)}-\mathrm{diag}\left(\frac{\phi''(t)}{\phi'^3(t)}\right)\mathbf{v}^{(1)}, \end{eqnarray}$

(2.10)

$\begin{eqnarray} \mathbf{u}^{(3)}& = &\mathrm{diag}\left(\frac{1}{\phi'^3(t)}\right)\mathbf{v}^{(3)}-\mathrm{diag}\left(\frac{3\phi''^2(t)}{\phi'^4(t)}\right)\mathbf{v}^{(2)} -\mathrm{diag}\left(\frac{\phi''(t)\phi'(t)-3\phi''^2(t)}{\phi'^5(t)}\right)\mathbf{v}^{(1)}, \end{eqnarray}$

(2.11)

where

$\mathbf{u} = [u(x_{0}), u(x_{1}), \ldots, u(x_{n})]^{\mathrm{T}}, \mathbf{v} = [v(t_{0}), v(t_{1}), \ldots, v(t_{n})]^{\mathrm{T}},$

$\mathbf{u}^{(1)} = [u'(x_{0}), u'(x_{1}), \ldots, u'(x_{n})]^{\mathrm{T}}, \mathbf{v}^{(1)} = [v'(t_{0}), v'(t_{1}), \ldots, v'(t_{n})]^{\mathrm{T}},$

$\mathbf{u}^{(2)} = [u''(x_{0}), u''(x_{1}), \ldots, u''(x_{n})]^{\mathrm{T}}, \mathbf{v}^{(2)} = [v''(t_{0}), v''(t_{1}), \ldots, v''(t_{n})]^{\mathrm{T}},$

$\mathbf{u}^{(3)} = [u'''(x_{0}), u'''(x_{1}), \ldots, u'''(x_{n})]^{\mathrm{T}}, \mathbf{t} = [t_{0}, t_{1}, \ldots, t_{n}]^{\mathrm{T}}.$

By the relationship of barycentric matrix at the meshpoint $t_{0}, t_{1}, \ldots, t_{n}$ , we have

$\begin{eqnarray} \mathbf{v}^{(1)} = \mathbf{D}^{(1)}\mathbf{v}, \mathbf{v}^{(2)} = \mathbf{D}^{(2)}\mathbf{v}, \mathbf{v}^{(3)} = \mathbf{D}^{(3)}\mathbf{v}, \end{eqnarray}$

(2.12)

where $\mathbf{D}^{(m)} = D_{i j}^{(m)} = R_{j}^{(m)}\left(x_{i}\right)$ is the element of the differentiation matrices and $R_{j}(x) = \frac{ \frac{w_{j}}{x-x_{j}} }{ \sum\limits_{k = 0}^{n} \frac{w_{k}}{x-x_{k}}}$ is basis function, see reference^[37]. For $m = 2$ , we have

$\begin{eqnarray} R_{j}^{\prime \prime}\left(x_{i}\right) = -2 \frac{w_{j} / w_{i}}{x_{i}-x_{j}}\left(\sum\limits_{k \neq i} \frac{w_{k} / w_{i}}{x_{i}-x_{k}}+\frac{1}{x_{i}-x_{j}}\right), \quad j \neq i, \end{eqnarray}$

(2.13)

where

$\begin{eqnarray*} w_{k} = \sum\limits_{i \in J_{k}}(-1)^{i} \prod\limits_{j = i, j \neq k}^{i+d} \frac{1}{x_{k}-x_{j}} \end{eqnarray*}$

is the weight function with $J_{k} = \left\{i \in I : k-d \le i \le k\right\}, 0\le d\le n$ and

$\begin{eqnarray} R_{i}^{\prime \prime}\left(x_{i}\right) = -\sum\limits_{j \neq i} R_{j}^{\prime \prime}\left(x_{i}\right). \end{eqnarray}$

(2.14)

Then we get the differentiable matrices as

$\begin{eqnarray} D_{i j}^{(1)} = R_{j}^{\prime}\left(x_{i}\right), \quad D_{i j}^{(2)} = R_{j}^{\prime \prime}\left(x_{i}\right), \quad D_{i j}^{(3)} = R_{j}^{(3)}\left(x_{i}\right). \end{eqnarray}$

(2.15)

Combining the Eqs (2.10) and (2.12), we get

$\begin{eqnarray} \mathbf{u}^{(1)} = \mathrm{diag}\left(\frac{1}{\phi'(t)}\right)\mathbf{D}^{(1)}\mathbf{v}, \end{eqnarray}$

(2.16)

$\begin{eqnarray} \mathbf{u}^{(2)} = \left[\mathrm{diag}\left(\frac{1}{\phi'^2(t)}\right)\mathbf{D}^{(2)}- \mathrm{diag}\left(\frac{\phi''(t)}{\phi'^3(t)}\right)\mathbf{D}^{(1)}\right]\mathbf{v}, \end{eqnarray}$

(2.17)

$\begin{eqnarray} \mathbf{u}^{(3)}& = &\left[\mathrm{diag}\left(\frac{1}{\phi'^3(t)}\right)\mathbf{D}^{(3)}-\mathrm{diag}\left(\frac{3\phi''^2(t)}{\phi'^4(t_{k})}\right)\mathbf{D}^{(2)} \right]\mathbf{v} \\&-&\mathrm{diag}\left(\frac{\phi''(t)\phi'(t)-3\phi''^2(t)}{\phi'^5(t)}\right)\mathbf{D}^{(1)}\mathbf{v}. \end{eqnarray}$

(2.18)

Taking following linear boundary value differential problems as example,

$\begin{eqnarray} u''(x)+p(x)u'(x)+q(x)u(x) = f(x), 0 < x < \infty, \end{eqnarray}$

(2.19)

by the transformation of $(2.1)$ and (2.2), we have

$\begin{eqnarray} \frac{1}{\phi'^{(2)}}v''(t)-\frac{\phi''(t)}{\phi'^{3}(t)}v'(t)+\frac{p(\phi(t))}{\phi'(t)}v'(t)+q(\phi(t))u(\phi(t)) = f(\phi(t)), -1\le t\le 1. \end{eqnarray}$

(2.20)

Taking the meshpoint $x_{0}, x_{1}, \ldots, x_{n}$ in the Eq (2.12), we have

$\begin{eqnarray} u''(x_{k})+p(x_{k})u'(x_{k})+q(x_{k})u(x_{k}) = f(x_{k}), k = 0, 1, 2, \ldots, N, \end{eqnarray}$

(2.21)

and its matrix form can be written as

$\begin{eqnarray} \mathbf{u}^{(2)}+\mathrm{diag}(p(x_{k}))\mathbf{u}^{(1)}+\mathrm{diag}(q(x_{k}))\mathbf{u} = f(\mathbf{x}), \end{eqnarray}$

(2.22)

Combining Eqs (2.16)–(2.18) and (2.22), we have

$\begin{eqnarray} &&\mathrm{diag}\left(\frac{1}{\phi'^{(2)}}\right)\mathbf{D}^{(2)}\mathbf{v}-\mathrm{diag}\left(\frac{\phi''(t)}{\phi'^{3}(t)}\right)\mathbf{D}^{(1)}\mathbf{v} +\mathrm{diag}\left(\frac{p(\phi(t))}{\phi'(t)}\right)\mathbf{D}^{(1)}\mathbf{v} \\&+&\mathrm{diag}\left(q(\phi(t))\right)\mathbf{v} = \mathbf{f}(\phi(t)), \end{eqnarray}$

(2.23)

where $\mathbf{D}^{(2)}$ and $\mathbf{D}^{(1)}$ are the barycentrix matrix, so we can not need to get the differential equation.

In the actual calculation, we take the algebraic transformation as

$\begin{eqnarray} x = L\frac{1+t}{1-t} \end{eqnarray}$

(2.24)

and Logarithmic transformation

$\begin{eqnarray} x = -L\ln\frac{1-t}{2} = L\ln\frac{2}{1-t}, \end{eqnarray}$

(2.25)

where $L$ is the positive constant called as amplification factor which determine the meshpoint of the semi-infinite domain.

3. Convergence and error analysis

In order to complete the proof of convergence rate, some lemmas are given as below. Firstly, we define the error function

$\begin{eqnarray} e(x) = u(x)-r_{n}(x) = \left(x-x_{i}\right) \ldots\left(x-x_{i+d}\right)u[x_{i}, x_{i+1}, \ldots, x_{i+d}; x] \end{eqnarray}$

(3.1)

and

$\begin{eqnarray} e(x) = \frac{ \sum\limits_{i = 0}^{n-d} \lambda_{i}(x)\left[u(x)-r_{n}(x)\right]}{ \sum\limits_{i = 0}^{n-d} \lambda_{i}(x)} = \frac{A(x)}{B(x)} = O(h^{d+1}), \end{eqnarray}$

(3.2)

where $A(x) : = \sum_{i = 0}^{n-d}(-1)^{i}u\left[x_{i}, \ldots, x_{i+d}; x\right], B(x) : = \sum_{i = 0}^{n-d} \lambda_{i}(x),$ and

$\begin{eqnarray} \lambda_{i}(x) = \frac{ (-1)^{i}}{ (x-x_{i})\cdots(x-x_{i+d})}. \end{eqnarray}$

(3.3)

Taking the numerical scheme

$\begin{eqnarray} \sum\limits_{j = 0}^{n} u_{j} R_{j}^{\prime \prime}(x)+p\sum\limits_{j = 0}^{n} u_{j} R_{j}^{\prime }(x)+q \sum\limits_{j = 0}^{n} u_{j} R_{j}(x) = f(x). \end{eqnarray}$

(3.4)

Combining (3.4) and (2.19), we have

$\begin{eqnarray} \mathcal{L} e(x): = e^{\prime\prime}\left(x\right)+p e^{\prime}\left(x\right)+q e\left(x\right)-R_{f}(x), \end{eqnarray}$

(3.5)

where $R_{f}(x) = f(x)-f(x_{k}), k = 0, 1, 2, \ldots, n$ .

Lemma have been proved by Jean-Paul Berrut.

Lemma 1. (see reference ^[19]) For $e(x)$ defined as (3.1), there holds

$\begin{equation} \left|e^{(k)}(x)\right| \leq C h^{d+1-k}, u\in C^{d+k+2}[a, b], k = 0, 1, \ldots. \end{equation}$

(3.6)

Let $u(x)$ to be the solution of (2.19) and $u_{n}(x)$ is the numerical solution, then we have

$\mathcal{L} u_{n}(x_{k}) = f(x_{k}), k = 0, 1, 2, \ldots, n,$

and

$\lim\limits_{n\rightarrow \infty}u_{n}(x) = u(x).$

Based on the above lemma, we get the following theorem.

Theorem 1. Let $u_{n}(x):\mathcal{L} u_{n}(x_{k}) = f(x_{k}), f(x)\in C[a, b]$ and suppose $\mathcal{L}$ be the invertible operator, we have

$|u_{n}(x)-u(x)|\le Ch^{d-1}.$

Proof. As

${u}_{n}(x) = \sum\limits_{j = 0}^{n} R_{j}(x) u_{j}.$

Combining the Lemma 1 and Eq (3.5), we have

$\begin{eqnarray} |\mathcal{L} e(x)|& = &|e^{\prime\prime}(x)+q e^{\prime}(x)+q e(x) -R_{f}(x)| \\&\le &|e^{\prime\prime}(x)|+|q e^{\prime}(x)|+|q e(x)| +|R_{f}(x)| \\&\le & Ch^{d-1}+Ch^{d}+Ch^{d+1} \\&\le & Ch^{d-1}. \end{eqnarray}$

(3.7)

As $\mathcal{L}$ is invertible operator. Then we have

$|{u}_{n}(x)-{u}(x)| \le Ch^{d-1}.$

The proof is completed.

4. Numerical examples

Three examples are presented to valid our theorem. All the examples were performed on personal computer by Matlab r2013a with a (Confguration: Intel(R) Core(TM) i5-8265U CPU @ 1.60GHz 1.80 GHz).

Example 1. Consider the boundary value problems

$\begin{eqnarray} u''+2u'-2u = -e^{2x}, 0 < x < \infty, \end{eqnarray}$

(4.1)

$\begin{eqnarray} u(0) = 1, u(\infty) = 0 \end{eqnarray}$

(4.2)

with analysis solution

$u(x) = \frac{1}2 (e^{-(1+\sqrt{3})x}+e^{-2x}).$

In , CPU running times of algebraic transformation with equidistant nodes $S = 5$ of linear barycentric rational collocation methods are presented. From Table 1, we know that the running times is less than 3 second.

Table 1. CPU running times of algebraic transformation with equidistant nodes

$S = 5$ .

$n$	$d=2$	$d=3$	$d=4$	$d=5$
$10$	8.1940e-03	4.7143e-03	2.0530e-03	6.5745e-04
$20$	2.9926e-03	8.2744e-04	1.5364e-04	3.4327e-06
$40$	5.8635e-04	8.0910e-05	6.7069e-06	1.5838e-07
$80$	9.1053e-05	6.3050e-06	2.4638e-07	4.8568e-09
$160$	1.2704e-05	4.4188e-07	8.4039e-09	9.7578e-11
$320$	1.6815e-06	2.9356e-08	2.7601e-10	1.6812e-12
time(second)	2.277	3.116	2.260	2.283

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In and , the convergence of algebraic transformation with equidistant nodes and quasi-equidistant nodes $S = 5$ of linear barycentric rational collocation methods are presented. In , with $S = 5, d = 2, 3, 4, 5$ , errors of equidistant nodes are $O(h^{d+1})$ . In , with $S = 5, d = 2, 3, 4, 5$ , errors of quasi-equidistant nodes are $O(h^{d+2})$ .

Table 2. Errors of algebraic transformation with equidistant nodes

$S = 5$ .

$n$	$d=2$		$d=3$		$d=4$		$d=5$
$10$	8.1940e-03		4.7143e-03		2.0530e-03		6.5745e-04
$20$	2.9926e-03	1.4532	8.2744e-04	2.5103	1.5364e-04	3.7401	3.4327e-06	7.5814
$40$	5.8635e-04	2.3516	8.0910e-05	3.3543	6.7069e-06	4.5178	1.5838e-07	4.4379
$80$	9.1053e-05	2.6870	6.3050e-06	3.6817	2.4638e-07	4.7667	4.8568e-09	5.0272
$160$	1.2704e-05	2.8414	4.4188e-07	3.8348	8.4039e-09	4.8737	9.7578e-11	5.6373
$320$	1.6815e-06	2.9175	2.9356e-08	3.9119	2.7601e-10	4.9283	1.6812e-12	5.8590

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Table 3. Errors of algebraic transformation with quasi-equidistant nodes

$S = 5$ .

$n$	$d=2$		$d=3$		$d=4$		$d=5$
$10$	2.6753e-02		1.2130e-02		3.6934e-03		7.1962e-04
$20$	2.2977e-03	3.5414	5.7573e-04	4.3971	9.2680e-05	5.3166	2.9350e-06	7.9377
$40$	1.7519e-04	3.7132	2.1363e-05	4.7522	1.6545e-06	5.8078	1.3132e-08	7.8042
$80$	1.2976e-05	3.7551	7.5147e-07	4.8293	2.8327e-08	5.8681	1.0991e-10	6.9005
$160$	9.3708e-07	3.7915	2.5684e-08	4.8708	4.7511e-10	5.8978	2.1025e-11	2.3862
$320$	6.6037e-08	3.8268	8.6168e-10	4.8976	5.7275e-11	3.0523	8.8952e-10	-

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In and , errors of equidistant nodes and quasi-equidistant nodes log transformation $d = 3$ of linear barycentric rational collocation methods are presented. In , errors of equidistant nodes with $d = 3, S = 5, 15, 25, 40, 50, 60$ are $O(h^{d+1})$ . In , errors of quasi-equidistant nodes with $d = 3, S = 5, 15, 25, 40, 50, 60$ are $O(h^{d+1})$ .

Table 4. Errors of equidistant nodes with log transformation

$d = 3$ .

$n$	$S=5$	$S=15$	$S=25$	$S=40$	$S=50$	$S=60$
$10$	4.3819e-05	5.0803e-04	2.1984e-03	4.9277e-03	7.8668e-03	1.0289e-02
$20$	5.4907e-06	3.9227e-05	2.3810e-04	7.2881e-04	1.5591e-03	2.7012e-03
$40$	7.5950e-07	2.7109e-06	1.9208e-05	6.7990e-05	1.6734e-04	3.3250e-04
$80$	1.0979e-07	1.7834e-07	1.3619e-06	5.1736e-06	1.3642e-05	2.9016e-05
$160$	1.6204e-08	1.1450e-08	9.0754e-08	3.5710e-07	9.7447e-07	2.1444e-06
$320$	2.4156e-09	7.2589e-10	5.8623e-09	2.3479e-08	6.5185e-08	1.4592e-07

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Table 5. Errors of quasi-equidistant nodes with log transformation

$d = 3$ .

$n$	$S=5$	$S=15$	$S=25$	$S=40$	$S=50$	$S=60$
$10$	1.9110e-05	3.0625e-04	1.9235e-03	6.5161e-03	1.5280e-02	2.8619e-02
$20$	1.2327e-06	7.3967e-06	5.3852e-05	2.1490e-04	5.9851e-04	1.3533e-03
$40$	5.5047e-08	1.7798e-07	1.4130e-06	5.9639e-06	1.7865e-05	4.2421e-05
$80$	2.5152e-09	4.6691e-09	3.9511e-08	1.7260e-07	5.2290e-07	1.2650e-06
$160$	1.1602e-10	1.3339e-10	1.1640e-09	5.1541e-09	1.5723e-08	3.8136e-08
$320$	5.7945e-12	6.0677e-12	3.5670e-11	1.5699e-10	4.8081e-10	1.1682e-09

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In and , errors of truncation method with equidistant nodes and quasi-equidistant nodes $d = 3$ of linear barycentric rational collocation methods are presented. In , errors of equidistant nodes with $d = 3, S = 5, 15, 25, 40, 50, 60$ are $O(h^{d})$ . In , errors of quasi-equidistant nodes with $d = 3, S = 5, 15, 25, 40, 50, 60$ are $O(h^{d})$ .

Table 6. Errors of truncation method with equidistant nodes

$d = 3$ .

$n$	$S=5$	$S=15$	$S=25$	$S=40$	$S=50$	$S=60$
$10$	2.1223e-02	5.6913e-02	3.9768e-02	2.1094e-02	2.0192e-02	1.9600e-02
$20$	3.5188e-03	3.4810e-02	4.9520e-02	4.1346e-02	3.2513e-02	2.5230e-02
$40$	3.8273e-04	8.9325e-03	2.4157e-02	4.0168e-02	4.3242e-02	4.2114e-02
$80$	3.5450e-05	1.3067e-03	5.2897e-03	1.5028e-02	2.2010e-02	2.8100e-02
$160$	2.3284e-05	1.4200e-04	7.0907e-04	2.7523e-03	4.9296e-03	7.6340e-03
$320$	2.3284e-05	1.3380e-05	7.3950e-05	3.3526e-04	6.6710e-04	1.1485e-03

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Table 7. Errors of truncation method with quasi-equidistant nodes

$d = 3$ .

$n$	$S=5$	$S=15$	$S=25$	$S=40$	$S=50$	$S=60$
$10$	2.7030e-02	3.0280e-01	4.2518e-01	3.6833e-01	3.0452e-01	2.5081e-01
$20$	1.8985e-03	8.9395e-02	3.5376e-01	8.9282e-01	1.1958e+00	1.4081e+00
$40$	1.0692e-04	7.2929e-03	4.3260e-02	1.8903e-01	3.5525e-01	5.7148e-01
$80$	2.3284e-05	4.4342e-04	2.9492e-03	1.5790e-02	3.3935e-02	6.2143e-02
$160$	2.3284e-05	2.6249e-05	1.7739e-04	9.9627e-04	2.2364e-03	4.3010e-03
$320$	2.3284e-05	1.5817e-06	1.0630e-05	5.9696e-05	1.3464e-04	2.6112e-04

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Example 2. Consider the boundary value problems

$\begin{eqnarray} 8u''+2u'-u = 8e^{-\frac{1}{4}x}, 0 < x < \infty, \end{eqnarray}$

(4.3)

$\begin{eqnarray} u(0) = 1, u(\infty) = 0 \end{eqnarray}$

(4.4)

and its analysis solution is

$u(x) = -3e^{-\frac{1}{2}x}+4e^{-\frac{3}{4}x}.$

In and , errors of log transformation with equidistant nodes and quasi-equidistant nodes $S = 8$ of linear barycentric rational collocation methods are presented. In , errors of log transformation with equidistant nodes $S = 8, d = 2, 3, 4, 5$ are $O(h^{d+1})$ . In , errors of log transformation with quasi-equidistant nodes $S = 8, d = 2, 3, 4, 5$ are $O(h^{d+2})$ .

Table 8. Errors of log transformation with equidistant nodes

$S = 8$ .

$n$	$d=2$		$d=3$		$d=4$		$d=5$
$10$	1.7920e-02		6.1866e-03		8.0099e-04		2.9488e-04
$20$	2.6669e-03	2.7483	4.3639e-04	3.8255	3.2205e-05	4.6364	4.4835e-06	6.0394
$40$	3.5692e-04	2.9015	2.8675e-05	3.9277	1.1054e-06	4.8647	6.8605e-08	6.0302
$80$	4.5963e-05	2.9570	1.8342e-06	3.9666	3.6022e-08	4.9395	1.0604e-09	6.0156
$160$	5.8252e-06	2.9801	1.1594e-07	3.9837	1.1488e-09	4.9707	1.6497e-11	6.0062
$320$	7.3298e-07	2.9905	7.2873e-09	3.9919	3.6425e-11	4.9790	1.6509e-13	6.6428

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Table 9. Errors of log transformation with quasi-equidistant nodes

$S = 8$ .

$n$	$d=2$		$d=3$		$d=4$		$d=5$
$10$	1.2884e-02		3.1184e-03		3.7630e-04		5.0897e-05
$20$	8.9366e-04	3.8497	6.6520e-05	5.5509	4.2951e-06	6.4530	3.8046e-07	7.0637
$40$	5.1689e-05	4.1118	1.4461e-06	5.5236	4.7299e-08	6.5048	2.1039e-09	7.4986
$80$	3.0710e-06	4.0731	3.4891e-08	5.3731	6.1916e-10	6.2553	1.9563e-11	6.7488
$160$	1.8483e-07	4.0545	9.4700e-10	5.2033	1.3292e-11	5.5417	3.9667e-11	-
$320$	1.1294e-08	4.0326	2.7845e-11	5.0879	9.5279e-11	-	1.9074e-09	-

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In and , errors of equidistant nodes and quasi-equidistant nodes log transformation $d = 3$ of linear barycentric rational collocation methods are presented. In , errors of equidistant nodes log transformation with $d = 3, S = 5, 15, 25, 40, 50, 60$ are $O(h^{d+1})$ . In , errors of quasi-equidistant nodes log transformation with $d = 3, S = 5, 15, 25, 40, 50, 60$ are $O(h^{d+1})$ .

Table 10. Errors of equidistant nodes log transformation with

$d = 3$ .

$n$	$S=5$	$S=15$	$S=25$	$S=40$	$S=50$	$S=60$
$10$	1.2754e-01	4.7239e-03	3.3028e-03	1.1102e-15	9.6949e-04	2.0177e-03
$20$	1.0096e-01	1.5512e-03	1.0035e-03	1.5543e-15	6.7291e-05	1.2162e-04
$40$	7.7020e-02	5.2958e-04	3.2856e-04	9.7700e-15	7.1866e-06	7.4030e-06
$80$	5.7531e-02	1.8400e-04	1.1148e-04	1.4821e-14	1.1037e-06	4.5563e-07
$160$	4.2389e-02	6.4485e-05	3.8532e-05	1.2784e-13	1.8340e-07	2.8240e-08
$320$	3.0936e-02	2.2697e-05	1.3453e-05	5.0154e-13	3.1563e-08	1.7571e-09

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Table 11. Errors of quasi-equidistant nodes log transformation with

$d = 3$ .

$n$	$S=5$	$S=15$	$S=25$	$S=40$	$S=50$	$S=60$
$10$	1.1024e-01	4.3445e-03	2.6631e-03	3.3307e-16	5.8846e-04	9.6881e-04
$20$	1.6316e-01	1.1904e-03	6.5698e-04	2.9421e-15	2.3458e-05	1.9189e-05
$40$	1.8549e-01	3.1977e-04	1.5566e-04	8.4377e-15	1.0472e-06	3.9498e-07
$80$	1.4828e-01	6.4136e-05	3.0793e-05	1.6739e-13	5.4456e-08	9.3150e-09
$160$	1.1057e-01	1.2704e-05	6.3129e-06	1.0436e-12	3.1336e-09	2.4767e-10
$320$	8.4925e-02	2.6665e-06	1.3904e-06	4.1613e-12	1.8558e-10	7.7586e-12

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We consider the nonlinear boundary problems on semi-infinite domain with the Logarithmic transformation

$\begin{eqnarray} x = -L\ln\frac{1-t}{2} = L\ln\frac{2}{1-t}, \end{eqnarray}$

(4.5)

where $L$ is the positive constant called as amplification factor

$\begin{eqnarray} \mathbf{u}^{(1)} = \mathrm{diag}\left(\frac{1-t}{L}\right)\mathbf{D}^{(1)}\mathbf{v}, \end{eqnarray}$

(4.6)

$\begin{eqnarray} \mathbf{u}^{(2)} = \left[\mathrm{diag}\left(\frac{(1-t)^2}{L^2}\right)\mathbf{D}^{(2)}-\mathrm{diag}\left(\frac{1-t}{L^3}\right)\mathbf{D}^{(1)}\right]\mathbf{v}, \end{eqnarray}$

(4.7)

$\begin{eqnarray} \mathbf{u}^{(3)} = \left[\mathrm{diag}\left(\frac{(1-t)^3}{L^3}\right)\mathbf{D}^{(3)}-\mathrm{diag}\left(\frac{3(1-t)^2}{L^3}\right)\mathbf{D}^{(2)} +\mathrm{diag}\left(\frac{1-t}{L^3}\right)\mathbf{D}^{(1)}\right]\mathbf{v}. \end{eqnarray}$

(4.8)

Taking the notation as below,

$\begin{eqnarray} \mathbf{C}^{(1)} = \mathrm{diag}\left(\frac{1-t}{L}\right)\mathbf{D}^{(1)}, \\ \mathbf{C}^{(2)} = \mathrm{diag}\left(\frac{(1-t)^2}{L^2}\right)\mathbf{D}^{(2)}-\mathrm{diag}\left(\frac{1-t}{L^3}\right)\mathbf{D}^{(1)}, \\ \mathbf{C}^{(3)} = \mathrm{diag}\left(\frac{(1-t)^3}{L^3}\right)\mathbf{D}^{(3)}-\mathrm{diag}\left(\frac{3(1-t)^2}{L^3}\right)\mathbf{D}^{(2)} +\mathrm{diag}\left(\frac{1-t}{L^3}\right)\mathbf{D}^{(1)}, \end{eqnarray}$

(4.9)

we have

$\begin{eqnarray} \mathbf{u}^{(1)} = \mathbf{C}^{(1)}\mathbf{v}, \quad \mathbf{u}^{(2)} = \mathbf{C}^{(2)}\mathbf{v}, \quad \mathbf{u}^{(3)} = \mathbf{C}^{(3)}\mathbf{v}. \end{eqnarray}$

(4.10)

Example 3. Consider the nonlinear boundary value problems

$\begin{eqnarray} f^{(3)}+ff^{(2)}-\beta f^{(1)}-(1+\alpha)(f')^{2} = 0, 0 < x < \infty, \end{eqnarray}$

(4.11)

$\begin{eqnarray} f(0) = 0, f^{(1)}(0) = 0, f(\infty) = 0 \end{eqnarray}$

(4.12)

and its analysis solution is

$f(\eta) = \frac{1}{\sqrt{1+\beta}}(1-e^{-\eta\sqrt{1+\beta}}).$

For the know function $f_{0}(\eta)$ , (4.11) can be linearized as

$\begin{eqnarray} f'''+f_{0}f''-\beta f'-(1+\alpha)f'_{0}f' = 0, 0 < x < \infty, \end{eqnarray}$

(4.13)

then we get the linearized scheme as

$\begin{eqnarray} f^{(3)}_{n}+f_{n-1}f^{(2)}_{n}-\beta f^{(1)}_{n}-(1+\alpha)(f'_{n-1})f'_{n} = 0, 0\le \eta\le \infty, \end{eqnarray}$

(4.14)

$\begin{eqnarray} f_{n}(0) = 0, f_{n}^{(1)}(0) = 0, f_{n}(\infty) = 0. \end{eqnarray}$

(4.15)

We take the transformation as

$\eta = -L\ln \frac{1-t}{2}.$

Then we get the calculation of barycentrix rational interpolation formulae as

$\begin{eqnarray} \left[\mathbf{C}^{(3)}+\mathrm{diag}(\mathbf{v}_{n-1})\mathbf{C}^{(2)}-\beta \mathbf{C}^{(1)}-(1+\alpha)\mathrm{diag}(\mathbf{C}^{(1)}\mathbf{v}_{n-1})\mathbf{C}^{(1)}\right]\mathbf{v}_{n} = 0, \end{eqnarray}$

(4.16)

$\begin{eqnarray} e_{1}^{\mathbf{T}}\mathbf{v}_{n} = 0, (d_{1}^{(1)})^{\mathbf{T}}\mathbf{v}_{n} = \frac{L}{2}, (d_{N}^{(1)})^{\mathbf{T}}\mathbf{v}_{n} = 0. \end{eqnarray}$

(4.17)

In and , errors of log transformation with equidistant nodes and quasi-equidistant nodes $d = 4$ of linear barycentric rational collocation methods are presented. In , errors of log transformation with equidistant nodes $d = 4, S = 5, 15, 25, 40, 50, 60$ are $O(h^{d+1})$ . In , errors of log transformation with quasi-equidistant nodes $d = 4, S = 5, 15, 25, 40, 50, 60$ are $O(h^{d+1})$ .

Table 12. Errors of log transformation with equidistant nodes

$d = 4$ .

$n$	$S=5$	$S=15$	$S=25$	$S=40$	$S=50$	$S=60$
$10$	1.1417e-01	1.1643e-03	3.8685e-04	2.6218e-03	1.4407e-02	3.3166e-02
$20$	8.6017e-02	3.4701e-04	5.7745e-05	2.8848e-04	1.5798e-03	3.9883e-03
$40$	6.4617e-02	9.8838e-05	7.5309e-06	2.7478e-05	1.5133e-04	4.0728e-04
$80$	4.8719e-02	2.7898e-05	9.3159e-07	2.5008e-06	1.3783e-05	3.8393e-05
$160$	3.6883e-02	7.8585e-06	1.1180e-07	2.2369e-07	1.2320e-06	3.4918e-06
$320$	2.7996e-02	2.2227e-06	1.0890e-08	2.2044e-08	1.0968e-07	3.1330e-07

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Table 13. Errors of log transformation with quasi-equidistant nodes

$d = 4$ .

$n$	$S=5$	$S=15$	$S=25$	$S=40$	$S=50$	$S=60$
$10$	1.5523e-01	5.4277e-04	9.8075e-05	5.4433e-04	3.0401e-03	7.5620e-03
$20$	1.6951e-01	6.0893e-05	1.8383e-06	4.4380e-06	2.4863e-05	7.1834e-05
$40$	3.6280e-01	2.0078e-05	9.6699e-08	1.1358e-07	6.0548e-07	1.7855e-06
$80$	3.3942e-01	1.4406e-05	1.8173e-07	4.4643e-08	1.2382e-08	4.1055e-08
$160$	7.0711e-01	7.0711e-01	7.0840e-01	7.1450e-01	7.2004e-01	7.2419e-01
$320$	7.5314e-01	7.5368e-01	7.0729e-01	9.7077e-01	8.2649e-01	7.4239e-01

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In and , errors of truncation method with equidistant nodes and quasi-equidistant nodes truncation method $d = 4$ of linear barycentric rational collocation methods are presented. In , errors of truncation method with equidistant nodes $d = 4, S = 5, 15, 25, 40, 50, 60$ are $O(h^{d})$ . In , errors of truncation method with quasi-equidistant nodes $d = 4, S = 5, 15, 25, 40, 50, 60$ are $O(h^{d})$ .

Table 14. Errors of truncation method with equidistant nodes

$d = 4$ .

$n$	$S=5$	$S=15$	$S=25$	$S=40$	$S=50$	$S=60$
$10$	4.0047e-02	4.2878e-01	9.6623e-01	1.8391e+00	2.4362e+00	3.0389e+00
$20$	5.6161e-03	1.0508e-01	2.9263e-01	6.4704e-01	9.0921e-01	1.1798e+00
$40$	5.8518e-04	1.8933e-02	6.8111e-02	1.8598e-01	2.8381e-01	3.9150e-01
$80$	1.0416e-03	2.5582e-03	1.1535e-02	3.9660e-02	6.7434e-02	1.0125e-01
$160$	1.1000e-03	2.8311e-04	1.4849e-03	6.1589e-03	1.1596e-02	1.9000e-02
$320$	1.1055e-03	2.8011e-05	1.5973e-04	7.4542e-04	1.5086e-03	2.6428e-03

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Table 15. Errors of truncation method with quasi-equidistant nodes

$d = 4$ .

$n$	$S=5$	$S=15$	$S=25$	$S=40$	$S=50$	$S=60$
$10$	8.7532e-03	1.5797e-01	3.9943e-01	8.4003e-01	1.1639e+00	1.4953e+00
$20$	1.0564e-03	5.2940e-03	2.1988e-02	9.7691e-02	2.0783e-01	3.7063e-01
$40$	1.1047e-03	1.7985e-04	1.1002e-03	5.7232e-03	1.4762e-02	3.3025e-02
$80$	1.1060e-03	4.8801e-06	3.3783e-05	1.8744e-04	4.1176e-04	8.0327e-04
$160$	1.1050e-03	1.4693e-07	9.3971e-07	5.4925e-06	1.2524e-05	2.4385e-05
$320$	9.5026e-04	3.3705e-06	2.3864e-06	4.9717e-07	5.3707e-07	7.2421e-07

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5. Conclusions

Semi-infinite domain problems have been considered by linear barycentric interpolation method with the truncation method and transformation method. By transformation method, the semi-infinite domain $[0, \infty)$ was transformed into $[-1, 1]$ with the function become complex. The proof of the convergence rate have been presented and numerical examples conforms the theorem analysis for both the linear and nonlinear differential equation. In the future works, infinite domain problems will be considered by the linear barycentric interpolation method.

Acknowledgements

The work of Jin Li was supported by Natural Science Foundation of Shandong Province (Grant No. ZR2022MA003). The authors would also like to thank the referees for their constructive comments and remarks which helped improve the quality of the paper.

Conflict of interest

The authors declare that they have no conflicts of interest.

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AIMS Agriculture and Food

Measuring consumer' willingness to pay for food safety certification labels of packaged rice

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1. Introduction

2. Linear boundary problems

3. Convergence and error analysis

4. Numerical examples

5. Conclusions

Acknowledgements

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AIMS Agriculture and Food

Measuring consumer' willingness to pay for food safety certification labels of packaged rice

Related Papers:

Abstract

1. Introduction

2. Linear boundary problems

3. Convergence and error analysis

4. Numerical examples

5. Conclusions

Acknowledgements

Conflict of interest

References

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