Numerical Solution of Fractional Order Burgers’ Equation with Dirichlet and Neumann Boundary Conditions by Reproducing Kernel Method

Abstract

In this research, obtaining of approximate solution for fractional-order Burgers’ equation will be presented in reproducing kernel Hilbert space (RKHS). Some special reproducing kernel spaces are identified according to inner products and norms. Then an iterative approach is constructed by using kernel functions. The convergence of this approach and its error estimates are given. The numerical algorithm of the method is presented. Furthermore, numerical outcomes are shown with tables and graphics for some examples. These outcomes demonstrate that the proposed method is convenient and effective.

1. Introduction

In this article, produced from a part of PhD thesis number 519846 from the Council of Higher Education, an iterative approach of reproducing kernel method (RKM) is considered for obtaining an approximate solution of the Burgers’ equation with fractional order as follows:

$$\begin{array}{c}{}^c{D}_{\xi }^{\mathsf{\alpha }}u(z,\xi )+c_1(z,\xi )u_{zz}(z,\xi )+c_2(z,\xi )u(z,\xi )+c_3(z,\xi )u_z(z,\xi )+c_4(z,\xi )u(z,\xi )u_z(z,\xi )=f(z,\xi )\\ 0\le z\le 1,0\le \xi \le 1,0<\mathsf{\alpha }\le 1,\end{array}$$

(1)

Here, ${}^c{D}_{\xi }^{\mathsf{\alpha }}$ is fractional differential operator in Caputo sense with respect to time variable $\xi$ and also $f(z,\xi )$, $c_1(z,\xi )$, $c_2(z,\xi )$, $c_3(z,\xi )$, $c_4(z,\xi )$ are continuous functions. For this model problem, initial-Neumann boundary conditions:

$$\begin{array}{c}\hfill \left\{\begin{array}{l}u(z,0)=0\hfill \\ u_z(0,\xi )=u_z(1,\xi )=0\hfill \end{array}\right.\end{array}$$

(2a)

and initial-Dirichlet boundary conditions:

$$\begin{array}{l}\hfill \left\{\begin{array}{c}u(z,0)=0\hfill \\ u(0,\xi )=u(1,\xi )=0\hfill \end{array}\right.\end{array}$$

(2b)

will be taken as above.

The Burgers’ equation is a simplified version of the Navier–Stokes equation. It was obtained by use of removing the pressure term from the Navier–Stokes equation by Burgers [1] in 1939. In other words, the Burgers’ equation can be expressed as a result of combining nonlinear wave motion with linear diffusion. Lately, many scientists have focused on Burgers’ equation by using several methods and different approaches. For instance, existence and uniqueness of local and global solution for Burgers’ equation was presented in [2] by Guesmia and Daili. Lombard and Matignon used a diffusive approximation for fractional-order Burgers’ equation in [3]. The averaging principle was proposed by Dong et al. for stochastic Burgers’ equation in [4]. Nojavan et al. obtained a numerical solution of Burgers’ equation by using discretization in reproducing kernel Hilbert space [5]. The Chebyshev wavelet method was developed by Oruc et al. for the numerical solution of time-fractional Burgers’ equation [6]. Pei et al. presented the local discontinuous Galerkin method for modified Burgers’ equation in [7]. The Petrov–Galerkin method was used by Roshan and Bhamra for modified Burgers’ equation in [8]. The collocation method was presented by Ramadan and Danaf for modified Burgers’ equation in [9]. Bahadir and Saglam constructed a mixed method for one dimensional Burgers’ equation [10]. Dag et al. used the cubic B-splines method [11]. Caldwell et al. proposed a finite element approximation for Burgers’ equation [12]. A finite difference method was used by Kutluay et al. for one-dimensional Burgers’ equation [13]. An approximate solution obtained by using the reproducing kernel method for Burgers’ equation [14]. A hybrid technique for the unsteady flow of a Burgers’ fluid is given by Raza et al. [15]. Laplace and finite Hankel transformations were proposed by Safdar et al. for generalized Burgers’ fluid with fractional derivative [16]. Time-fractional coupled Burgers’ equations were solved with generalized differential transform method by Liu and Hou [17]. Zhang et al proposed an analytical and numerical approach for multi-term time-fractional Burgers’ fluid model [18]. The Adomian decomposition method was applied to space-and time-fractional Burgers’ equation by Momani [19]. A generalized Taylor series technique was proposed by Ajou et al. for fractional nonlinear KdV-Burgers’ equation [20]. Mittal and Arora presented a numerical approach by using cubic B-spline functions for coupled viscous Burgers’ equation [21]. Jiwari used a hybrid numerical scheme for Burgers’ equation [22]. Kutluay et al. proposed a B-spline finite element method for Burgers’ equation [23].

Reproducing kernel concept is introduced by Zaremba [24]. In his study, Zaremba focused on the boundary value problem, which includes the Dirichlet boundary condition. Furthermore, the theoretical concept of reproducing kernel is developed in [25,26]. Reproducing kernel spaces of polynomial and trigonometric functions are constructed in [27]. Many studies have been conducted by using reproducing kernel method. For instance, eighth order boundary value problems [28], fractional advection-dispersion equation [29], fractional order systems of Dirichlet function types [30], fractional order Bagley–Torvik equation [31], time fractional telegraph equation [32], a local reproducing kernel method for Burgers’ equation [33], time-fractional partial integro differential equations [34], Riccati differential equations [35], nonlinear hyperbolic telegraph equation [36], time-fractional Tricomi and Keldysh equations [37], one-dimensional sine–Gordon equation [38], reaction-diffusion equations [39], integro differential equations of Fredholm operator type [40], fredholm integro-differential equations [41], nonlinear system of PDEs [42], class of fractional partial differential equation [43], Bagley–Torvik and Painlevé equations [44], nonlinear coupled Burgers equations [45] and so on [46,47,48,49,50,51,52,53,54,55,56,57,58,59].

This research is organized as: Specific definitions and Hilbert spaces are demonstrated in Section 2. Reproducing kernel solution is identified by RKM in Section 3. Convergence analysis of the approximate solution is proved in Section 4. Error estimation of the method is presented in Section 5. Two examples of fractional order Burgers’ equation are examined by the RKM and the algorithm of the process is given in Section 6. Finally, a short conclusion is given in Section 7.

The notation table

Table 1. Notation table.

Symbol	Explanation
${}^c{D}_{\xi }^{\mathsf{\alpha }}$	Caputo derivative operator with arbitrary real order
$\mathsf{\Theta }$	Region of $[0,1]\times [0,1]$
$W_2^3[0,1]$	Hilbert space with one variable function
$W_2^{(3,2)}\left(\mathsf{\Theta }\right)$	Hilbert Space with two variable function
$K_{(\tau ,\beta )}(z,\xi )$	Reproducing kernel function of $W_2^{(3,2)}\left(\mathsf{\Theta }\right)$
$u(z,\xi )$	Exact solution
$u_n(z,\xi )$	Reproducing kernel solution
L	Linear operator
$\langle .,.\rangle$	Inner (scalar) product

is given as follow:

2. Some Specific Definitions and Hilbert Spaces

In this section, some basic definitions and significant reproducing kernel spaces will be given.

Definition 1.

([58,59]) Fractional α order Caputo derivative is defined as:

$$\begin{array}{c}\hfill {}^c{D}_{\xi }^{\mathsf{\alpha }}u(z,\xi )=\frac{1}{\mathsf{\Gamma }(n-\mathsf{\alpha })}{\int }_0^{\xi }\frac{{\partial }_{r}u(z,r)}{{(\xi -r)}^{1+\mathsf{\alpha }-n}}dr,n-1<\mathsf{\alpha }\le n,\xi >0.\end{array}$$

Definition 2.

Let H be Hilbert space and $T\ne \varnothing$ an abstract set. If following conditions are provide, then $S:T\times T\to \mathbb{C}$ is is called as reproducing kernel function:

$$\begin{array}{c}i.S(.,\tau )\in H,\forall \tau \in T,\hfill \\ ii.\langle \mu (.),S(.,\tau )\rangle =\mu \left(\tau \right),\forall \tau \in T,\forall \mu \in H.\hfill \end{array}$$

2.1. Reproducing Kernel Spaces with One Variable

In this subsection, reproducing kernel functions will be presented for some special Hilbert spaces. Definitions and kernel functions of these spaces will be given for z and $\xi$ variables. $W_2^n[a,b]$ shows the general reproducing kernel space for one variable. Equations (1) and (2a,b) has second order derivative for z and first order derivative for $\xi$. Therefore, the kernel function of $W_2^3[0,1]$ will be given for $u_{zz}$ and the kernel function of $W_2^2[0,1]$ will be given for $u_{\xi }$. Furthermore, $W_2^1[0,1]$ space will be given for general function (without derivative). For the obtaining procedure of reproducing kernel functions, please see [47].

${\mathbf{W}}_{\mathbf{2}}^{\mathbf{1}}[\mathbf{0},\mathbf{1}]$ Hilbert space

$$\begin{array}{c}\hfill W_2^1[0,1]=\{g\left(z\right)|g\mathrm{is}\mathrm{absolutely}\mathrm{continuous}\mathrm{function},g^{\prime }\in L^2[0,1]\}.\end{array}$$

• The inner product of $W_2^1\left[0,1\right]$ can be taken as follows:

  $${\langle g\left(z\right),f\left(z\right)\rangle }_{W_2^1}=g\left(0\right)f\left(0\right)+\underset{0}{\overset{1}{\int }}{g}^{\prime }\left(z\right)f^{\prime }\left(z\right)dz.$$

  (3)
• The norm of $W_2^1\left[0,1\right]$ can be taken as follows:

  $${∥g∥}_{W_2^1}^2={\langle g,g\rangle }_{W_2^1},g,f\in W_2^1\left[0,1\right].$$
• The kernel function of $W_2^1\left[0,1\right]$ is as follows:

  $$R_{\tau }^{\left\{1\right\}}\left(z\right)=\left\{\begin{array}{c}1+z,z\le \tau ,\hfill \\ 1+\tau ,\tau >z.\hfill \end{array}\right.$$

  (4)

${\mathbf{W}}_{\mathbf{2}}^{\mathbf{2}}[\mathbf{0},\mathbf{1}]$ Hilbert Space

$$\begin{array}{c}\hfill W_2^2[0,1]=\{g\left(\xi \right)|g,g^{\prime }\mathrm{are}\mathrm{absolutely}\mathrm{continuous}\mathrm{functions},g^{\prime \prime }\in L^2[0,1],g\left(0\right)=0\}.\end{array}$$

Here, $L^2\left[0,1\right]=\left\{\left.g\right|\underset{0}{\overset{1}{\int }}{g}^2\left(\xi \right)d\xi <\infty \right\}$.

• The inner product of $W_2^2\left[0,1\right]$ can be taken as follows:

  $${\langle g\left(\xi \right),f\left(\xi \right)\rangle }_{W_2^2}=g\left(0\right)f\left(0\right)+g^{\prime }\left(0\right)f^{\prime }\left(0\right)+\underset{0}{\overset{1}{\int }}{g}^{\prime \prime }\left(\xi \right)f^{\prime \prime }\left(\xi \right)d\xi .$$

  (5)
• The norm of $W_2^2[0,1]$ can be taken as follows:

  $$\begin{array}{c}\hfill {∥g∥}_{W_2^2}^2={\langle g,g\rangle }_{W_2^2},f,g\in W_2^2\left[0,1\right].\end{array}$$
• The kernel function of $W_2^2[0,1]$ is as follows:

  $$\begin{array}{c}\hfill R_{\beta }^{\left\{2\right\}}\left(\xi \right)=\left\{\begin{array}{c}\xi \beta +\frac{1}{2}\beta {\xi }^2-\frac{1}{6}{\xi }^3,\xi \le \beta ,\hfill \\ -\frac{1}{6}{\beta }^3+\frac{1}{2}\xi {\beta }^2+\beta \xi ,\xi >\beta .\hfill \end{array}\right.\end{array}$$

  (6)

${\mathbf{W}}_{\mathbf{2}}^{\mathbf{3}}[\mathbf{0},\mathbf{1}]$ Hilbert Space

$W_2^3[0,1]$ space with Dirichlet boundary condition:

$$\begin{array}{c}\hfill W_2^3[0,1]=\{g\left(z\right)|g,g^{\prime },g^{\prime \prime }\mathrm{are}\mathrm{absolutely}\mathrm{continuous}\mathrm{functions},g^{\left(3\right)}\in L^2[0,1],g\left(0\right)=g\left(1\right)=0\}.\end{array}$$

• The inner product of $W_2^3[0,1]$ can be taken as follows:

  $${\langle g\left(z\right),f\left(z\right)\rangle }_{W_2^3}=g\left(0\right)f\left(0\right)+g^{\prime }\left(0\right)f^{\prime }\left(0\right)+g\left(1\right)f\left(1\right)+\underset{0}{\overset{1}{\int }}{g}^{\left(3\right)}\left(z\right)f^{\left(3\right)}\left(z\right)dz.$$

  (7)
• The norm of $W_2^3[0,1]$ can be taken as follows:

  $${∥g∥}_{W_2^3}^2={\langle g,g\rangle }_{W_2^3},f,g\in W_2^3\left[0,1\right].$$
• The kernel function of $W_2^3[0,1]$ is as follows:

  $$R_{\tau }^{\left\{3\right\}}\left(z\right)=\left\{\begin{array}{c}\frac{-1}{120}(\tau -1)z(z{\tau }^4-4z{\tau }^3+6z{\tau }^2+\tau z^4-5\tau z^3-120\tau z+120\tau +z^4),z\le \tau ,\hfill \\ \frac{-1}{120}(z-1)\tau (\tau z^4-4\tau z^3+6\tau z^2+z{\tau }^4-5z{\tau }^3-120z\tau +120z+{\tau }^4),z>\tau .\hfill \end{array}\right.$$

  (8)

$W_2^3[0,1]$ space with Neumann boundary condition:

• The inner product of $W_2^3[0,1]$ can be taken as follows:

  $${\langle g\left(z\right),f\left(z\right)\rangle }_{W_2^3}=g\left(0\right)f\left(0\right)+g^{\prime }\left(0\right)f^{\prime }\left(0\right)+g^{\prime \prime }\left(0\right)f^{\prime \prime }\left(0\right)+\underset{0}{\overset{1}{\int }}{g}^{\left(3\right)}\left(z\right)f^{\left(3\right)}\left(z\right)dz.$$

  (9)
• The norm of $W_2^3[0,1]$ can be taken as follows:

  $${∥g∥}_{W_2^3}^2={\langle g,g\rangle }_{W_2^3},f,g\in W_2^3\left[0,1\right].$$
• The kernel function of $W_2^3[0,1]$ is as follows:
  For $z\le \tau$, the kernel function:

  $$\begin{array}{ccc}{R}_{\tau }^{\left\{3\right\}}\left(z\right)\hfill & =\hfill & z(\frac{1}{56}{\tau }^4-\frac{1}{14}{\tau }^3-\frac{3}{14}{\tau }^2+\frac{4}{7}\tau )+z^2(\frac{1}{7}{\tau }^2-\frac{1}{112}{\tau }^4-\frac{1}{28}{\tau }^3+\frac{3}{14}\tau )\hfill \\ \hfill & +\hfill & z^3(\frac{1}{21}{\tau }^2+\frac{1}{336}{\tau }^4-\frac{1}{84}{\tau }^3-\frac{1}{14}\tau )+z^4(\frac{-1}{1344}{\tau }^4+\frac{1}{36}{\tau }^3+\frac{1}{112}{\tau }^2-\frac{1}{42}\tau )+\frac{1}{120}z,\hfill \end{array}$$

  (10)

  and for $z>\tau$ the kernel function:

  $$\begin{array}{ccc}{R}_{\tau }^{\left\{3\right\}}\left(z\right)\hfill & =\hfill & \tau (\frac{1}{56}{z}^4-\frac{1}{14}{z}^3-\frac{3}{14}{z}^2+\frac{4}{7}z)+{\tau }^2(\frac{1}{7}{z}^2-\frac{1}{112}{z}^4-\frac{1}{28}{z}^3+\frac{3}{14}z)\hfill \\ \hfill & +\hfill & {\tau }^3(\frac{1}{21}{z}^2+\frac{1}{336}{z}^4-\frac{1}{84}{z}^3-\frac{1}{14}z)+{\tau }^4(\frac{-1}{1344}{z}^4+\frac{1}{36}{z}^3+\frac{1}{112}{z}^2-\frac{1}{42}z)+\frac{1}{120}\tau \hfill \end{array}$$

  (11)

2.2. Reproducing Kernel Spaces for Two Variable

The problem (1) and (2a,b) has two variables z and $\xi$. For this reason, we should give the spaces, inner products, and kernel functions according to these variables. Because the highest order derivatives z and $\xi$ to be considered, reproducing kernel spaces will be given for both z and $\xi$ variables. The region which we consider is $\mathsf{\Theta }=[0,1]\times [0,1]$. In this part, $W_2^{(3,2)}\left(\mathsf{\Theta }\right)$ space is given for Dirichlet boundary conditions. These reproducing kernel spaces are also determined in the same way for Neumann boundary conditions.

${\mathbf{W}}_{\mathbf{2}}^{(\mathbf{3},\mathbf{2})}\left(\mathsf{\Theta }\right)$ Hilbert Space

$$\begin{array}{c}{W}_2^{(3,2)}\left(\mathsf{\Theta }\right)=\{u(z,\xi )|\frac{{\partial }^{3}u}{\partial z^2\partial \xi }\mathrm{is}\mathrm{completely}\mathrm{continuous}\mathrm{in}\mathsf{\Theta },\\ \frac{{\partial }^{5}u}{\partial z^3\partial {\xi }^2}\in L^2\left(\mathsf{\Theta }\right),u(z,0)=u(0,\xi )=u(1,\xi )=0\}.\end{array}$$

(12)

• The inner product of $W_2^{(3,2)}\left(\mathsf{\Theta }\right)$ can be taken as follows:

  $$\begin{array}{ccc}{\langle u(z,\xi ),v(z,\xi )\rangle }_{W_2^{(3,2)}}\hfill & =\hfill & \sum _{i=0}^1\underset{0}{\overset{1}{\int }}\left[\frac{{\partial }^2}{\partial {\xi }^2}\frac{{\partial }^i}{\partial z^i}u(0,\xi )\frac{{\partial }^2}{\partial {\xi }^2}\frac{{\partial }^i}{\partial z^i}v(0,\xi )\right]d\xi +{\int }_0^1\frac{{\partial }^2}{\partial {\xi }^2}u(1,\xi )\frac{{\partial }^2}{\partial {\xi }^2}v(1,\xi )d\xi \hfill \\ \hfill & +\hfill & \sum _{j=0}^1{\langle \frac{{\partial }^j}{\partial {\xi }^j}u(z,0),\frac{{\partial }^j}{\partial {\xi }^j}v(z,0)\rangle }_{W_2^3}\hfill \\ \hfill & +\hfill & \underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\left[\frac{{\partial }^3}{\partial z^3}\frac{{\partial }^2}{\partial {\xi }^2}u(z,\xi )\frac{{\partial }^3}{\partial z^3}\frac{{\partial }^2}{\partial {\xi }^2}v(z,\xi )\right]dzd\xi .\hfill \end{array}$$

  (13)
• The norm of $W_2^{(3,2)}\left(\mathsf{\Theta }\right)$ can be taken as follows:

  $$\begin{array}{c}\hfill {∥u∥}_{W_2^{(3,2)}}^2={\langle u,u\rangle }_{W_2^{(3,2)}},u,v\in W_2^{(3,2)}\left(\mathsf{\Theta }\right).\end{array}$$

Theorem 1.

$K_{(\tau ,\beta )}(z,\xi )$ is the kernel function of $W_2^{(3,2)}\left(\mathsf{\Theta }\right)$ and also it is obtained by multiplying kernel functions of $W_2^3[0,1]$ and $W_2^2[0,1]$, respectively. So, it can be written that

$$\begin{array}{c}\hfill K_{(\tau ,\beta )}(z,\xi )=R_{\tau }^{\left\{3\right\}}\left(z\right)R_{\beta }^{\left\{2\right\}}\left(\xi \right).\end{array}$$

For any $u(z,\xi )\in W_2^{(3,2)}\left(\mathsf{\Theta }\right)$

$$\begin{array}{c}\hfill u(\tau ,\beta )={\langle u(z,\xi ),K_{(\tau ,\beta )}(z,\xi )\rangle }_{W_2^{(3,2)}}\end{array}$$

and

$$\begin{array}{c}\hfill K_{(z,\xi )}(\tau ,\beta )=K_{(\tau ,\beta )}(z,\xi ).\end{array}$$

(14)

Proof. 

Inner product of $W_2^{(3,2)}$ space (Equation (13)) will be used to prove the theorem.

$$\begin{array}{cc}& \langle u(z,\xi ),R_{\tau }^{\left\{3\right\}}\left(z\right)R_{\beta }^{\left\{2\right\}}\left(\xi \right)){\rangle }_{W_2^{(3,2)}}\hfill \\ =& \sum _{j=0}^1{\langle \frac{{\partial }^j}{\partial {\xi }^j}u(z,0),R_{\tau }^{\left\{3\right\}}\left(z\right)\frac{{\partial }^j}{\partial {\xi }^j}{R}_{\beta }^{\left\{2\right\}}\left(0\right)\rangle }_{W_2^3}+\sum _{i=0}^1\underset{0}{\overset{1}{\int }}\left[\frac{{\partial }^2}{\partial {\xi }^2}\frac{{\partial }^i}{\partial z^i}u(0,\xi )\frac{{\partial }^2}{\partial {\xi }^2}{R}_{\beta }^{\left\{2\right\}}\left(xi\right)\frac{{\partial }^i}{\partial z^i}{R}_{\tau }^{\left\{3\right\}}\left(0\right)\right]d\xi \hfill \\ +& {\int }_0^1\frac{{\partial }^2}{\partial {\xi }^2}u(1,\xi )R_{\tau }^{\left\{3\right\}}\left(1\right)\frac{{\partial }^2}{\partial {\xi }^2}{R}_{\beta }^{\left\{2\right\}}\left(\xi \right)d\xi +\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\left[\frac{{\partial }^3}{\partial z^3}\frac{{\partial }^2}{\partial {\xi }^2}u(z,\xi )\frac{{\partial }^3}{\partial z^3}{R}_{\tau }^{\left\{3\right\}}\left(z\right)\frac{{\partial }^2}{\partial {\xi }^2}{R}_{\beta }^{\left\{2\right\}}\left(\xi \right)\right]dzd\xi \hfill \\ =& \underset{0}{\overset{1}{\int }}\frac{{\partial }^2}{\partial {\xi }^2}{R}_{\beta }^{\left\{2\right\}}\left(\xi \right)\frac{{\partial }^2}{\partial {\xi }^2}\left[\underset{0}{\overset{1}{\int }}\frac{{\partial }^3}{\partial z^3}u(z,\xi )\frac{{\partial }^3}{\partial z^3}{R}_{\tau }^{\left\{3\right\}}\left(z\right)dz+\sum _{i=0}^1\frac{{\partial }^i}{\partial z^i}u(0,\xi )\frac{{\partial }^i}{\partial z^i}{R}_{\tau }^{\left\{3\right\}}\left(0\right)+u(1,\xi )R_{\tau }^{\left\{3\right\}}\left(1\right)\right]d\xi \hfill \\ +& \sum _{j=0}^1\frac{{\partial }^j}{\partial {\xi }^j}u(\tau ,0)\frac{{\partial }^j}{\partial {\xi }^j}{R}_{\beta }^{\left\{2\right\}}\left(0\right)\hfill \\ =& \sum _{j=0}^1\frac{{\partial }^j}{\partial {\xi }^j}u(\tau ,0)\frac{{\partial }^j}{\partial {\xi }^j}{R}_{\beta }^{\left\{2\right\}}\left(0\right)+\underset{0}{\overset{1}{\int }}\frac{{\partial }^2}{\partial {\xi }^2}{R}_{\beta }^{\left\{2\right\}}\left(\xi \right)\frac{{\partial }^2}{\partial {\xi }^2}{\langle u(z,\xi ),R_{\tau }^{\left\{3\right\}}\left(z\right)\rangle }_{W_2^3}d\xi \hfill \\ =& \sum _{j=0}^1\frac{{\partial }^j}{\partial {\xi }^j}u(\tau ,0)\frac{{\partial }^j}{\partial {\xi }^j}{R}_{\beta }^{\left\{2\right\}}\left(0\right)+\underset{0}{\overset{1}{\int }}\frac{{\partial }^2}{\partial {\xi }^2}{R}_{\beta }^{\left\{2\right\}}\left(\xi \right)\frac{{\partial }^2}{\partial {\xi }^2}u(\tau ,\xi )d\xi \hfill \\ =& {\langle u(\tau ,\xi ),R_{\beta }^{\left\{2\right\}}\left(\xi \right)\rangle }_{W_2^2}=u(\tau ,\beta ).\hfill \end{array}$$

So, $\langle u(\tau ,\xi ),R_{\beta }^{\left\{2\right\}}\left(\xi \right)\rangle =u(\tau ,\beta )$, and

$$\begin{array}{c}\hfill K_{(\tau ,\beta )}(z,\xi )={\langle K_{(\tau ,\beta )}(x,y),K_{(z,\xi )}(x,y)\rangle }_{W_2^{(3,2)}}={\langle K_{(z,\xi )}(x,y),K_{(\tau ,\beta )}(x,y)\rangle }_{W_2^{(3,2)}}=K_{(z,\xi )}(\tau ,\beta ).\end{array}$$

Therefore, the proof is completed. □

${\mathbf{W}}_{\mathbf{2}}^{(\mathbf{1},\mathbf{1})}\left(\mathsf{\Theta }\right)$ Hilbert Space

$$\begin{array}{c}\hfill W_2^{(1,1)}\left(\mathsf{\Theta }\right)=\{u(z,\xi )|u\mathrm{is}\mathrm{completely}\mathrm{continuous}\mathrm{in}\mathsf{\Theta },\frac{{\partial }^{2}u}{\partial z\partial \xi }\in L^2\left(\mathsf{\Theta }\right)\}.\end{array}$$

• The inner product of $W_2^{(1,1)}\left(\mathsf{\Theta }\right)$ can be taken as follows:

  $$\begin{array}{ccc}\hfill {\langle u(z,\xi ),v(z,\xi )\rangle }_{W_2^{(1,1)}}& =& \underset{0}{\overset{1}{\int }}\left[\frac{\partial }{\partial \xi }u(0,\xi )\frac{\partial }{\partial \xi }v(0,\xi )\right]d\xi +{\langle u(z,0),v(z,0)\rangle }_{W_2^1}\hfill \\ & +& \underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\left[\frac{\partial }{\partial z}\frac{\partial }{\partial \xi }u(z,\xi )\frac{\partial }{\partial z}\frac{\partial }{\partial \xi }v(z,\xi )\right]dzd\xi \hfill \end{array}$$
• The norm of $W_2^{(1,1)}\left(\mathsf{\Theta }\right)$ can be taken as follows:

  $$\begin{array}{c}\hfill {∥u∥}_{W_2^{(1,1)}}^2={\langle u,u\rangle }_{W_2^{(1,1)}},u,v\in W_2^{(1,1)}\left(\mathsf{\Theta }\right).\end{array}$$
• The kernel function of $W_2^{(1,1)}\left(\mathsf{\Theta }\right)$ is as follows:

  $$\begin{array}{c}\hfill {\tilde{K}}_{(\tau ,\beta )}(z,\xi )=R_{\tau }^{\left\{1\right\}}\left(z\right)R_{\beta }^{\left\{1\right\}}\left(\xi \right).\end{array}$$

Remark 1.

In the next sections all analysis will be given for the Dirichlet boundary conditions. A similar analysis can be made for Neumann boundary conditions.

3. Obtaining of Reproducing Kernel Solution for Equations (1) and (2a,b) in ${\mathit{W}}_{\mathbf{2}}^{(\mathbf{3},\mathbf{2})}\left(\mathsf{\Theta }\right)$

In the reproducing kernel method, an approximate solution will be obtained with the help of kernel function and linear operator L. The choosing of L is arbitrary. One can choose the whole linear part of the model problem or any linear part of it. Here, the whole linear part of the model problem is chosen as follow:

$$\begin{array}{c}\hfill L:W_2^{(3,2)}\left(\mathsf{\Theta }\right)\to W_2^{(1,1)}\left(\mathsf{\Theta }\right)\end{array}$$

$$\begin{array}{c}\hfill Lu(z,\xi )={}^c{D}_{\xi }^{\mathsf{\alpha }}u+c_1(z,\xi )u_{zz}+c_2(z,\xi )u+c_3(z,\xi )u_z.\end{array}$$

(15)

The new statement of Equations (1)-(2a-2b) can be expressed as:

$$\begin{array}{c}\hfill Lu(z,\xi )=F(z,\xi ,u(z,\xi ),u_z(z,\xi )),\xi ,z\in [0,1]\end{array}$$

(16)

and $F(z,\xi ,u(z,\xi ),u_z(z,\xi ))=f(z,\xi )-c_4(z,\xi )u(z,\xi )u_z(z,\xi )$.

Let ${\left\{(z_i,{\xi }_i)\right\}}_{i=1}^{\infty }$ be a countable dense subset in $\mathsf{\Theta }$. Now, ${\psi }_i(z,\xi )$ basis function will be defined by applying the kernel function to the operator L.

$$\begin{array}{ccc}{\psi }_i(z,\xi )\hfill & =\hfill & L_{(\tau ,\beta )}{K}_{(\tau ,\beta )}{(z,\xi )|}_{(\tau ,\beta )=(z_i,{\xi }_i)}\hfill \\ \hfill & =\hfill & \{{}^c{D}_t^{\mathsf{\alpha }}{K}_{(\tau ,\beta )}(z,\xi )+c_1(\tau ,\beta )\frac{{\partial }^2}{\partial x^2}{K}_{(\tau ,\beta )}(z,\xi )+c_2(\tau ,\beta )K_{(\tau ,\beta )}(z,\xi )\hfill \\ \hfill & +\hfill & c_3(\tau ,\beta )\frac{\partial }{\partial x}{K}_{(\tau ,\beta )}(z,\xi ){\}|}_{(\tau ,\beta )=(z_i,{\xi }_i)}\hfill \\ \hfill & =\hfill & \frac{1}{\Gamma (1-\mathsf{\alpha })}{\int }_0^{{\xi }_i}\frac{{\partial }_r{K}_{(z,r)}(z,\xi )}{{({\xi }_i-r)}^{\mathsf{\alpha }}}dr+c_1(z_i,{\xi }_i)\frac{{\partial }^2}{\partial x^2}{K}_{(z_i,{\xi }_i)}(z,\xi )+c_2(z_i,{\xi }_i)K_{(z_i,{\xi }_i)}(z,\xi )\hfill \\ \hfill & +\hfill & c_3(z_i,{\xi }_i)\frac{\partial }{\partial x}{K}_{(z_i,{\xi }_i)}(z,\xi ),i=1,2,\dots \hfill \end{array}$$

(17)

Now, it will be shown that ${\psi }_i(z,\xi )$ basis function belong to $W_2^{(3,2)}\left(\mathsf{\Theta }\right)$ space and ${\psi }_i(z,\xi )$ satisfies the initial-boundary condition of $W_2^{(3,2)}\left(\mathsf{\Theta }\right)$ space. For this purpose, the following theorem will be given.

Theorem 2.

The basis function ${\psi }_i(z,\xi )$ is belong to $W_2^{(3,2)}\left(\mathsf{\Theta }\right)$ reproducing space for $i=1,2,...$.

Proof. 

To prove the theorem, we must show that the following conditions are provided.

• It should be shown that $\frac{{\partial }^5{\psi }_i(z,\xi )}{\partial z^3\partial {\xi }^2}\in L^2\left(\mathsf{\Theta }\right)$.
• $\frac{{\partial }^3{\psi }_i(z,\xi )}{\partial z^2\partial \xi }$ is completely continuous function.
• ${\psi }_i(z,\xi )$ basis function satisfies the initial and boundary conditions.

One can see that any elements of $W_2^{(3,2)}\left(\mathsf{\Theta }\right)$ satisfy the above conditions. Now, the following equation can be written using the property of the kernel function $K_{(\tau ,\beta )}(z,\xi )$

$$\begin{array}{c}\hfill {\partial }_{{\tau }^2{z}^3{\xi }^2}^7{K}_{(\tau ,\beta )}(z,\xi )={\partial }_{{\tau }^2{z}^3}^5{R}_{\tau }^{\left\{3\right\}}\left(z\right){\partial }_{{\xi }^2}^2{R}_{\beta }^{\left\{2\right\}}\left(\xi \right).\end{array}$$

Here, both ${\partial }_{{\tau }^2{z}^3}^5{R}_{\tau }^{\left\{3\right\}}\left(z\right)$ and ${\partial }_{{\xi }^2}^2{R}_{\beta }^{\left\{2\right\}}\left(\xi \right)$ are continuous in $[0,1]$. These functions are bounded because they are continuous in $[0,1]$. So, it can be written

$$\begin{array}{c}\hfill |{\partial }_{{\tau }^2{z}^3{\xi }^2}^7{K}_{(\tau ,\beta )}(z,\xi )|\le M_1.\end{array}$$

In the same way, one can write that

$$\begin{array}{c}\hfill |{\partial }_{\beta z^3{\xi }^2}^6{K}_{(\tau ,\beta )}(z,\xi )|\le M_2\\ \hfill |{\partial }_{z^3{\xi }^2}^5{K}_{(\tau ,\beta )}(z,\xi )|\le M_3\\ \hfill |{\partial }_{\tau z^3{\xi }^2}^6{K}_{(\tau ,\beta )}(z,\xi )|\le M_4.\end{array}$$

Here, $M_1,M_2,M_3$ and $M_4$ are positive constants. From (17),

$$\begin{array}{ccc}\hfill |\frac{{\partial }^5{\psi }_i(z,\xi )}{\partial z^3\partial {\xi }^2}|& \le & |\frac{1}{\Gamma (1-\mathsf{\alpha })}{\int }_0^{{\xi }_i}\frac{M_2}{{({\xi }_i-r)}^{\mathsf{\alpha }}}dr+c_1(z_i,{\xi }_i)M_1\hfill \\ & +& c_2(z_i,{\xi }_i)M_3+c_3(z_i,{\xi }_i)M_4|\hfill \\ & \le & \frac{M_2}{\Gamma (2-\mathsf{\alpha })}{\xi }_i^{1-\mathsf{\alpha }}+|c_1(z_i,{\xi }_i)|M_1+|c_2(z_i,{\xi }_i)|M_3+|c_3(z_i,{\xi }_i)|M_4.\hfill \end{array}$$

Therefore, $\frac{{\partial }^5{\psi }_i(z,\xi )}{\partial z^3\partial {\xi }^2}\in L^2\left(\mathsf{\Theta }\right)$. Furthermore, $\frac{{\partial }^3{\psi }_i(z,\xi )}{\partial z^2\partial \xi }$ is completely continuous in $\mathsf{\Theta }$ since $\mathsf{\Theta }$ is closed region. Finally, basis function ${\psi }_i(z,\xi )$ satisfies initial-boundary conditions such that $K_{(\tau ,\beta )}(z,0)=0$ and $K_{(\tau ,\beta )}(0,\xi )=K_{(\tau ,\beta )}(1,\xi )=0$. Therefore, ${\psi }_i(z,\xi )\in W_2^{(3,2)}\left(\mathsf{\Theta }\right)$. □

Theorem 3.

${\left\{{\psi }_i(z,\xi )\right\}}_{i=1}^{\infty }$ is a complete system in $W_2^{(3,2)}\left(\mathsf{\Theta }\right)$.

Proof. 

It is known that

$$\begin{array}{ccc}{\psi }_i(z,\xi )\hfill & =\hfill & \left(L^{*}{\phi }_i\right)(z,\xi )={\langle \left(L^{*}{\phi }_i\right)(\tau ,\beta ),K_{(z,\xi )}(\tau ,\beta )\rangle }_{W_2^{(3,2)}}\hfill \\ \hfill & =\hfill & {\langle {\phi }_i(\tau ,\beta ),L_{(\tau ,\beta )}{K}_{(z,\xi )}(\tau ,\beta )\rangle }_{W_2^{(1,1)}}=L_{(\tau ,\beta )}{K}_{(z,\xi )}{(\tau ,\beta )|}_{(\tau ,\beta )=(z_i,{\xi }_i)}\hfill \\ \hfill & =\hfill & L_{(\tau ,\beta )}{K}_{(\tau ,\beta )}{(z,\xi )|}_{(\tau ,\beta )=(z_i,{\xi }_i)}.\hfill \end{array}$$

(18)

Clearly, for each fixed $u(z,\xi )\in W_2^{(3,2)}\left(\mathsf{\Theta }\right)$, if ${\langle u(z,\xi ),{\psi }_i(z,\xi )\rangle }_{W_2^{(3,2)}}=0$ then ${\psi }_i(z,\xi )\in W_2^{(3,2)}\left(\mathsf{\Theta }\right)$, $i=1,2,\dots$. Therefore,

$$\begin{array}{c}\hfill {\langle u(z,\xi ),\left(L^{*}{\phi }_i\right)(z,\xi )\rangle }_{W_2^{(3,2)}}={\langle Lu(z,\xi ),{\phi }_i(z,\xi )\rangle }_{W_2^{(1,1)}}=\left(Lu\right)(z_i,{\xi }_i)=0,i=1,2,\dots \end{array}$$

(19)

${\left\{(z_i,{\xi }_i)\right\}}_{i=1}^{\infty }$ is dense in $\mathsf{\Theta }$. Hence, $\left(Lu\right)(z,\xi )=0$. By using of inverse operator $L^{-1}$, it can be seen that $u=0$. So, theorem is proven. □

The orthonormal basis system ${\left\{\overline{{\psi }_i}(z,\xi )\right\}}_{i=1}^{\infty }$ of $W_2^{(3,2)}\left(\mathsf{\Theta }\right)$ can be obtained by the way of Gram–Schmidt orthogonalization process of ${\left\{{\psi }_i(z,\xi )\right\}}_{i=1}^{\infty }$ as follow:

$$\begin{array}{c}\hfill \overline{{\psi }_i}(z,\xi )=\sum _{k=1}^i{\eta }_{ik}{\psi }_k(z,\xi ),i=1,2,\dots \end{array}$$

(20)

In Equation (20), ${\eta }_{ii}>0$ and ${\eta }_{ik}$ are orthogonalization coefficients.

Theorem 4.

If ${\left\{(z_i,{\xi }_i)\right\}}_{i=1}^{\infty }$ is dense in Θ, then the solution (16) is

$$\begin{array}{c}\hfill u(z,\xi )=\sum _{i=1}^{\infty }\sum _{k=1}^i{\eta }_{ik}F(z_k,{\xi }_k,u(z_k,{\xi }_k),{\partial }_{z}u(z_k,{\xi }_k)){\overline{\psi }}_i(z,\xi ).\end{array}$$

(21)

Proof. 

It is known that ${\left\{{\psi }_i(z,\xi )\right\}}_{i=1}^{\infty }$ system is complete in $W_2^{(3,2)}\left(\mathsf{\Theta }\right)$ from the previous theorem. So, it can be written

$$\begin{array}{ccc}u(z,\xi )\hfill & =\hfill & {\displaystyle \sum _{i=1}^{\infty }\langle u(z,\xi ),{\overline{\psi }}_i{(z,\xi \rangle }_{W_2^{(3,2)}}{\overline{\psi }}_i(z,\xi )=\sum _{i=1}^{\infty }\sum _{k=1}^i{\eta }_{ik}{\langle u(z,\xi ),{\psi }_k(z,\xi )\rangle }_{W_2^{(3,2)}}{\overline{\psi }}_i(z,\xi )}\hfill \\ \hfill & =\hfill & {\displaystyle \sum _{i=1}^{\infty }\sum _{k=1}^i{\eta }_{ik}{\langle u(z,\xi ),L^{*}{\phi }_k(z,\xi )\rangle }_{W_2^{(3,2)}}{\overline{\psi }}_i(z,\xi )=\sum _{i=1}^{\infty }\sum _{k=1}^i{\eta }_{ik}{\langle Lu(z,\xi ),{\phi }_k(z,\xi )\rangle }_{W_2^{(1,1)}}{\overline{\psi }}_i(z,\xi )}\hfill \\ \hfill & =\hfill & {\displaystyle \sum _{i=1}^{\infty }\sum _{k=1}^i{\eta }_{ik}{\langle Lu(z,\xi ),{\tilde{K}}_{(z_k,{\xi }_k)}(z,\xi )\rangle }_{W_2^{(1,1)}}{\overline{\psi }}_i(z,\xi )=\sum _{i=1}^{\infty }\sum _{k=1}^i{\eta }_{ik}Lu(z_k,{\xi }_k){\overline{\psi }}_i(z,\xi )}\hfill \\ \hfill & =\hfill & {\displaystyle \sum _{i=1}^{\infty }\sum _{k=1}^i{\eta }_{ik}F(z_k,{\xi }_k,u(z_k,{\xi }_k),{\partial }_{z}u(z_k,{\xi }_k)){\overline{\psi }}_i(z,\xi ).}\hfill \end{array}$$

(22)

So, theorem is proven. □

In Equation (21), $u(z,\xi )$ is described as infinite term sum. In the next equation, finitely n-terms solution will be given as $u_n(z,\xi )$:

$$\begin{array}{c}\hfill u_n(z,\xi )=\sum _{i=1}^n\sum _{k=1}^i{\eta }_{ik}F(z_k,{\xi }_k,u(z_k,{\xi }_k),{\partial }_{z}u(z_k,{\xi }_k)){\overline{\psi }}_i(z,\xi ).\end{array}$$

(23)

4. Convergence of Reproducing Kernel Solution

In this section, it will be shown that

$$\begin{array}{c}\hfill \parallel u(z,\xi )-u_n(z,\xi )\parallel \to 0\mathrm{as}n\to \infty .\end{array}$$

(24)

If we take

$$\begin{array}{c}\hfill A_i=\sum _{k=1}^i{\eta }_{ik}F(z_k,{\xi }_k,u(z_k,{\xi }_k),{\partial }_{z}u(z_k,{\xi }_k)),\end{array}$$

(25)

then (21) can be described as

$$\begin{array}{c}\hfill u(z,\xi )=\sum _{i=1}^{\infty }{A}_i{\overline{\psi }}_i(z,\xi ).\end{array}$$

(26)

Now, $u(z_1,{\xi }_1)$ is found by taking $(z_1,{\xi }_1)=0$ from the initial conditions of problem. Furthermore, by choosing $u_0(z_1,{\xi }_1)=u(z_1,{\xi }_1)$, the n-term approximation to $u(z,\xi )$ is expressed as follows:

$$\begin{array}{c}\hfill u_n(z,\xi )=\sum _{i=1}^n{B}_i{\overline{\psi }}_i(z,\xi ),\end{array}$$

(27)

here

$$\begin{array}{c}\hfill B_i=\sum _{k=1}^i{\eta }_{ik}F(z_k,{\xi }_k,u_{k-1}(z_k,{\xi }_k),{\partial }_z{u}_{k-1}(z_k,{\xi }_k)).\end{array}$$

(28)

Now, some theoretical results will be given for convergence of $F(z_n,{\xi }_n,u_{n-1}(z_n,{\xi }_n),{\partial }_z{u}_{n-1}(z_n,{\xi }_n))$ and $u_n(z,\xi )$, respectively.

Lemma 1.

If $F(z,\xi ,u(z,\xi ),u_z(z,\xi ))$ is continuous and $u_n\to \widehat{u}$ for $(z_n,{\xi }_n)\to (\tau ,\beta )$, then

$$\begin{array}{c}\hfill F(z_n,{\xi }_n,u_{n-1}(z_n,{\xi }_n),{\partial }_z{u}_{n-1}(z_n,{\xi }_n))\to F(\tau ,\beta ,\widehat{u}(\tau ,\beta ),{\partial }_z\widehat{u}(\tau ,\beta )).\end{array}$$

(29)

Proof. 

Since

$$\begin{array}{cc}\hfill |u_{n-1}(z_n,{\xi }_n)-\widehat{u}(\tau ,\beta )|& =|u_{n-1}(z_n,{\xi }_n)-u_{n-1}(\tau ,\beta )+u_{n-1}(\tau ,\beta )-\widehat{u}(\tau ,\beta )|\\ & \le |u_{n-1}(z_n,{\xi }_n)-u_{n-1}(\tau ,\beta )|+|u_{n-1}(\tau ,\beta )-\widehat{u}(\tau ,\beta )|.\end{array}$$

(30)

Using reproducing kernel feature, it can be written that

$$\begin{array}{c}\hfill u_{n-1}(z_n,{\xi }_n)={\langle u_{n-1}(z,\xi ),K_{(z_n,{\xi }_n)}(z,\xi )\rangle }_{W_2^{(3,2)}},u_{n-1}(\tau ,\beta )={\langle u_{n-1}(z,\xi ),K_{(\tau ,\beta )}(z,\xi )\rangle }_{W_2^{(3,2)}}.\end{array}$$

(31)

It follows that

$$\begin{array}{c}\hfill |u_{n-1}(z_n,{\xi }_n)-u_{n-1}(\tau ,\beta )|=|\langle u_{n-1}(z,\xi ),K_{(z_n,{\xi }_n)}(z,\xi )-K_{(\tau ,\beta )}(z,\xi )\rangle |.\end{array}$$

(32)

It can be said that there exists $M>0$ from the convergence of $u_{n-1}(z,\xi )$ such that

$$\begin{array}{c}\hfill \parallel u_{n-1}{(z,\xi )\parallel }_{W_2^{(3,2)}}\le M{\parallel \widehat{u}(\tau ,\beta )\parallel }_{W_2^{(3,2)}},\mathrm{as}n\ge M.\end{array}$$

(33)

In a similar way, it can be proven

$$\begin{array}{c}\hfill \parallel K_{(z_n,{\xi }_n)}(z,\xi )-K_{(\tau ,\beta )}{(z,\xi )\parallel }_{W_2^{(3,2)}}\to 0,\mathrm{for}n\to \infty \end{array}$$

(34)

by using Equation (14). So,

$$\begin{array}{c}\hfill u_{n-1}(z_n,{\xi }_n)\to \widehat{u}(\tau ,\beta ),\mathrm{as}(z_n,{\xi }_n)\to (\tau ,\beta ).\end{array}$$

(35)

In a similar way it can be shown that

$$\begin{array}{c}\hfill {\partial }_z{u}_{n-1}(z_n,{\xi }_n)\to {\partial }_z\widehat{u}(\tau ,\beta ),\mathrm{as}(z_n,{\xi }_n)\to (\tau ,\beta ).\end{array}$$

(36)

Therefore,

$$\begin{array}{c}\hfill F(z_n,{\xi }_n,u_{n-1}(z_n,{\xi }_n),{\partial }_z{u}_{n-1}(z_n,{\xi }_n))\to F(\tau ,\beta ,\widehat{u}(\tau ,\beta ),{\partial }_z\widehat{u}(\tau ,\beta )).\end{array}$$

(37)

So, lemma is proven. □

Theorem 5.

Assume that (16) has a unique solution, $\parallel u_n\parallel$ is a bounded and ${\left\{(z_i,{\xi }_i)\right\}}_{i=1}^{\infty }$ is dense in Θ. Then, $u_n(z,\xi )$ converges to $u(z,\xi )$ and

$$\begin{array}{c}\hfill u(z,\xi )=\sum _{i=1}^{\infty }{B}_i{\overline{\psi }}_i(z,\xi ).\end{array}$$

(38)

Proof. 

Firstly, we aim to show that $u_n(z,\xi )$ is convergence. Following equality can be written

$$\begin{array}{c}\hfill u_{n+1}(z,\xi )=u_n(z,\xi )+B_{n+1}{\overline{\psi }}_{n+1}(z,\xi ).\end{array}$$

(39)

from the Equation (27). Using the orthonormality of ${\left\{{\overline{\psi }}_i\right\}}_{i=1}^{\infty }$, we have

$$\begin{array}{c}\hfill \parallel u_{n+1}{\parallel }^2={\parallel u_n\parallel }^2+B_{n+1}^2=\sum _{i=1}^{n+1}{B}_i^2.\end{array}$$

(40)

Therefore, $\parallel u_{n+1}\parallel >\parallel u_n\parallel$ satisfies from (40). Here, it seems that $\parallel u_n\parallel$ is bounded. So, one can know that $\parallel u_n\parallel$ is convergent. Therefore, there exists a constant b so that

$$\begin{array}{c}\hfill \sum _{i=1}^{\infty }{B}_i^2=b.\end{array}$$

(41)

So, above equation shows that ${\left\{B_i\right\}}_{i=1}^{\infty }\in l^2$. If $m>n$, then

$$\begin{array}{ccc}\parallel u_m-u_n{\parallel }^2\hfill & =\hfill & \parallel u_m-u_{m-1}+u_{m-1}-u_{m-2}+\cdots +u_{n+1}-u_n{\parallel }^2\hfill \\ \hfill & =\hfill & \parallel u_m-u_{m-1}{\parallel }^2+\parallel u_{m-1}-u_{m-2}{\parallel }^2+\cdots +{\parallel u_{n+1}-u_n\parallel }^2.\hfill \end{array}$$

(42)

The following equation is obtained

$$\begin{array}{c}\hfill \parallel u_m-u_{m-1}{\parallel }^2=B_m^2,\end{array}$$

(43)

and consequently

$$\begin{array}{c}\hfill \parallel u_m-u_n{\parallel }^2=\sum _{l=n+1}^m{B}_l^2\to 0,\mathrm{as}n\to \infty .\end{array}$$

(44)

The completeness of $W_2^{(3,2)}\left(\mathsf{\Theta }\right)$ shows that $u_n\to \widehat{u}$ for $n\to \infty$. Next, it will be shown that $\widehat{u}$ is the representation solution of (16). If the limit is taken both sides of Equation (27), the following equation can be written:

$$\begin{array}{c}\hfill \widehat{u}(z,\xi )=\sum _{i=1}^{\infty }{B}_i{\overline{\psi }}_i(z,\xi ).\end{array}$$

(45)

Note that

$$\begin{array}{c}\hfill \left(L\widehat{u}\right)(z,\xi )=\sum _{i=1}^{\infty }{B}_{i}L{\overline{\psi }}_i(z,\xi ),\end{array}$$

(46)

$$\begin{array}{ccc}\left(L\widehat{u}\right)(z_l,{\xi }_l)\hfill & =\hfill & \sum _{i=1}^{\infty }{B}_{i}L{\overline{\psi }}_i(z_l,{\xi }_l)=\sum _{i=1}^{\infty }{B}_i{\langle L{\overline{\psi }}_i(z,\xi ),{\phi }_l(z,\xi )\rangle }_{W_2^{(1,1)}}\hfill \\ \hfill & =\hfill & \sum _{i=1}^{\infty }{B}_i{\langle {\overline{\psi }}_i(z,\xi ),L^{*}{\phi }_l(z,\xi )\rangle }_{W_2^{(3,2)}}=\sum _{i=1}^{\infty }{B}_i{\langle {\overline{\psi }}_i(z,\xi ),{\psi }_l(z,\xi )\rangle }_{W_2^{(3,2)}}.\hfill \end{array}$$

(47)

Therefore,

$$\begin{array}{ccc}\hfill \sum _{l=1}^i{\eta }_{il}\left(L\widehat{u}\right)(z_l,{\xi }_l)& =& \sum _{i=1}^{\infty }{B}_i{\langle {\overline{\psi }}_i(z,\xi ),\sum _{l=1}^i{\eta }_{il}{\psi }_l(z,\xi )\rangle }_{W_2^{(3,2)}}\hfill \end{array}$$

(48)

$$\begin{array}{cc}=& \sum _{i=1}^{\infty }{B}_i{\langle {\overline{\psi }}_i(z,\xi ),{\overline{\psi }}_l(z,\xi )\rangle }_{W_2^{(3,2)}}=B_l.\hfill \end{array}$$

(49)

From (28), the following equation can be expressed

$$\begin{array}{c}\hfill L\widehat{u}(z_l,{\xi }_l)=F(z_l,{\xi }_l,u_{l-1}(z_l,{\xi }_l),{\partial }_z{u}_{l-1}(z_l,{\xi }_l)).\end{array}$$

(50)

For each $(\tau ,\beta )\in \mathsf{\Theta }$, ${\left\{(z_i,{\xi }_i)\right\}}_{i=1}^{\infty }$ is dense in $\mathsf{\Theta }$. Therefore, there exists a subsequence ${\left\{(z_{n_j},{\xi }_{n_j})\right\}}_{j=1}^{\infty }$ such that $(z_{n_j},{\xi }_{n_j})\to (\tau ,\beta )$, $j\to \infty$. It is known that

$$\begin{array}{c}\hfill L\widehat{u}(z_{n_j},{\xi }_{n_j})=F(z_{n_j},{\xi }_{n_j},u_{n_{j-1}}(z_{n_j},{\xi }_{n_j}),{\partial }_z{u}_{n_{j-1}}(z_{n_j},{\xi }_{n_j})).\end{array}$$

(51)

By using Lemma 1 and continuity of F, it can be seen that

$$\begin{array}{c}\hfill \left(L\widehat{u}\right)(\tau ,\beta )=F(\tau ,\beta ,\widehat{u}(\tau ,\beta ),{\partial }_z\widehat{u}(\tau ,\beta )),\mathrm{for}j\to \infty .\end{array}$$

(52)

Equation (52) implies that $\widehat{u}(z,\xi )$ satisfies Equation (16). So, proof is completed. □

Theorem 6.

${\partial }_{z^i}^i{\partial }_{{\xi }^j}^j{u}_n(z,\xi )$ uniformly converges to ${\partial }_{z^i}^i{\partial }_{{\xi }^j}^{j}u(z,\xi )$ for $j=0,1$ and $i=0,1,2$.

Proof. 

The convergence of $u_n$ is given in the previous theorem. Now,

$$\begin{array}{ccc}\hfill |{\partial }_{z^i}^i{\partial }_{{\xi }^j}^{j}u(z,\xi )-{\partial }_{z^i}^i{\partial }_{{\xi }^j}^j{u}_n(z,\xi )|& =& |\langle u(y,s)-u_n(y,s),{\partial }_{z^i}^i{\partial }_{{\xi }^j}^{j}L{K}_{(z,\xi )}(y,s)\rangle |\hfill \\ & \le & \parallel u-u_n\parallel \parallel {\partial }_{z^i}^i{\partial }_{{\xi }^j}^{j}L{K}_{(z,\xi )}(y,s)\parallel \hfill \\ & \le & C_{i,j}\parallel u-u_n\parallel .\hfill \end{array}$$

So,

$$\begin{array}{c}\hfill |{\partial }_{z^i}^i{\partial }_{{\xi }^j}^j{u}_n(z,\xi )-{\partial }_{z^i}^i{\partial }_{{\xi }^j}^{j}u(z,\xi )|\to 0\mathrm{as}n\to \infty .\end{array}$$

 □

5. Error Estimation of Method

In this section, error analysis for the presented method will be given. In this analysis, one can understand that the error estimation varies depending on the selected step size. Now, the step size, chosen of points, and norm will be taken as follow:

$$\begin{array}{c}\hfill z_i=i{h}_z,h_z=1/n,{\xi }_j=j{h}_{\xi },h_{\xi }=1/n,i,j=1,\dots ,n.\end{array}$$

$$\begin{array}{c}\hfill {\parallel u(z,\xi )\parallel }_{\infty }=max|u(z,\xi )|\mathrm{for}(z,\xi )\in \mathsf{\Theta }\end{array}$$

Furthermore, $u(z,\xi )-u_n(z,\xi )$ can be written in two ways for each variable as follow:

$$\begin{array}{c}\hfill u(z,\xi )-u_n(z,\xi )=u(z_i,\xi )-u_n(z_i,\xi )+{\int }_{z_i}^z({\partial }_{\tau }u(\tau ,\xi )-{\partial }_{\tau }{u}_n(\tau ,\xi ))d\tau \end{array}$$

and

$$\begin{array}{c}\hfill u(z,\xi )-u_n(z,\xi )=u(z,{\xi }_i)-u_n(z,{\xi }_i)+{\int }_{{\xi }_i}^{\xi }({\partial }_{\beta }u(z,\beta )-{\partial }_{\beta }{u}_n(z,\beta ))d\beta .\end{array}$$

The following two theorems will be given for error estimation considering the two equations above.

Theorem 7.

Let $u(z,\xi )-u_n(z,\xi )$ be error in $W_2^{(6,2)}\left(\mathsf{\Theta }\right)$. Therefore, there exist a $C>0$ so that

$$\begin{array}{c}\hfill \parallel u(z,\xi )-u_n{(z,\xi )\parallel }_{\infty }\le C(h_z^2+h_z{h}_{\xi }).\end{array}$$

Proof. 

For $[z_i,z_{i+1}]\times [{\xi }_j,{\xi }_{j+1}]\subset \mathsf{\Theta }$, the following equality can be written:

$$\begin{array}{c}\hfill {\partial }_{z}u(z,\xi )-{\partial }_z{u}_n(z,\xi )={\partial }_{z}u(z,\xi )-{\partial }_{z}u(z_i,{\xi }_j)+{\partial }_z{u}_n(z_i,{\xi }_j)-{\partial }_z{u}_n(z,\xi )+{\partial }_{z}u(z_i,{\xi }_j)-{\partial }_z{u}_n(z_i,{\xi }_j)\end{array}$$

Here, Taylor expansion will be used to show error estimate for the function ${\partial }_{z}u(z,\xi )$ around the point $(z_i,{\xi }_j)$. This equality will be analyzed in three cases. It is known that

$$\begin{array}{c}\hfill {\partial }_{z}u(z,\xi )={\partial }_{z}u(z_i,{\xi }_j)+(h_z{\partial }_z+h_{\xi }{\partial }_{\xi }){\partial }_{z}u(z_i+ϵh_z,{\xi }_j+ϵh_{\xi }),ϵ\in [0,1].\end{array}$$

Firstly, the continuity of ${\partial }_{z^2}^{2}u(z,\xi )$ and ${\partial }_{z\xi }^{2}u(z,\xi )$ on $\mathsf{\Theta }$ is considered. So, one can write that

$$\begin{array}{c}\hfill \parallel {\partial }_{z}u(z,\xi )-{\partial }_{z}u(z_i,{\xi }_j){\parallel }_{\infty }=O(h_z+h_{\xi }).\end{array}$$

Secondly, one know that

$$\begin{array}{c}\hfill |{\partial }_z{u}_n(z_i,{\xi }_j)-{\partial }_z{u}_n(z,\xi )|\le {\int }_{z_i}^z|{\partial }_{y^2}^2{u}_n(y,{\xi }_j)|dy+{\int }_{{\xi }_j}^{\xi }|{\partial }_{zs}^2{u}_n(z,s)|ds.\end{array}$$

(53)

The next equation can be stated by using of maximum norm

$$\begin{array}{c}\hfill \parallel {\partial }_z{u}_n(z_i,{\xi }_j)-{\partial }_z{u}_n(z,\xi ){\parallel }_{\infty }=O(h_z+h_{\xi }).\end{array}$$

Finally, for sufficiently large n, any $\epsilon >0$ and using Theorem 6 such that

$$\begin{array}{c}\hfill \parallel {\partial }_{z}u(z_i,{\xi }_j)-{\partial }_z{u}_n(z_i,{\xi }_j){\parallel }_{\infty }<\epsilon .\end{array}$$

(54)

From Equation (54), it is known that $\epsilon$ is arbitrary constant and the chosen of n, the following equality can be expressed

$$\begin{array}{c}\hfill \parallel {\partial }_{z}u(z_i,{\xi }_j)-{\partial }_z{u}_n(z_i,{\xi }_j){\parallel }_{\infty }=O(h_z+h_{\xi }).\end{array}$$

In light of the information given above, error estimation will be given as follows:

$$\begin{array}{c}\hfill u(z,\xi )-u_n(z,\xi )=u(z_i,\xi )-u_n(z_i,\xi )+{\int }_{z_i}^z({\partial }_{y}u(y,\xi )-{\partial }_y{u}_n(y,\xi ))dy\end{array}$$

$$\begin{array}{ccc}\hfill \parallel u(z,\xi )-u_n{(z,\xi )\parallel }_{\infty }& \le & \parallel u(z_i,\xi )-u_n(z_i,\xi ){\parallel }_{\infty }+{\int }_{z_i}^z{\parallel {\partial }_{y}u(y,\xi )-{\partial }_y{u}_n(y,\xi )\parallel }_{\infty }dy\hfill \\ & \le & C(h_z^2+h_{\xi }{h}_z).\hfill \end{array}$$

 □

So far, error estimation analysis is done for z variable. The error estimation analysis for variable $\xi$ will be given by the next theorem.

Theorem 8.

Assume that ${\partial }_{z^3}^3{\partial }_{\xi }u(z,\xi )$ and ${\partial }_{z^2}^2{\partial }_{{\xi }^2}^{2}u(z,\xi )$ are continuous and also, $\parallel {\partial }_{z^3}^3{\partial }_{\xi }{u(z,\xi )\parallel }_{\infty }$ and $\parallel {\partial }_{z^2}^2{\partial }_{{\xi }^2}^2{u(z,\xi )\parallel }_{\infty }$ are bounded. So, there exist a $C>0$ such that the error estimate can be expressed as:

$$\begin{array}{c}\hfill \parallel u_n{(z,\xi )-u(z,\xi )\parallel }_{\infty }\le C(h_{\xi }{h}_z^3+h_{\xi }^2{h}_z^2).\end{array}$$

Proof. 

For $[z_i,z_{i+1}]\times [{\xi }_i,{\xi }_{i+1}]\in \mathsf{\Theta }$, it can be written

$$\begin{array}{ccc}\hfill {\partial }_{z^2}^2{\partial }_{\xi }u(z,\xi )-{\partial }_{z^2}^2{\partial }_{\xi }{u}_n(z,\xi )& =& {\partial }_{z^2}^2{\partial }_{\xi }u(z,\xi )-{\partial }_{z^2}^2{\partial }_{\xi }u(z_i,{\xi }_j)+{\partial }_{z^2}^2{\partial }_{\xi }{u}_n(z_i,{\xi }_j)\hfill \\ & -& {\partial }_{z^2}^2{\partial }_{\xi }{u}_n(z,\xi )+{\partial }_{z^2}^2{\partial }_{\xi }u(z_i,{\xi }_j)-{\partial }_{z^2}^2{\partial }_{\xi }{u}_n(z_i,{\xi }_j).\hfill \end{array}$$

Taylor series expansion for ${\partial }_{z^2}^2{\partial }_{\xi }u(z,\xi )$ function around the point $(z_i,{\xi }_j)$ as follow:

$$\begin{array}{c}\hfill {\partial }_{z^2}^2{\partial }_{\xi }u(z,\xi )={\partial }_{z^2}^2{\partial }_{\xi }u(z_i,{\xi }_j)+(h_z{\partial }_z+h_{\xi }{\partial }_{\xi }){\partial }_{z^2}^2{\partial }_{\xi }u(z_i+h_zϵ,{\xi }_j+h_{\xi }ϵ)+...,ϵ\in [0,1].\end{array}$$

Firstly, the following expression can be written by considering the continuation of ${\partial }_{z^3}^3{\partial }_{\xi }u(z,\xi )$ and ${\partial }_{z^2}^2{\partial }_{{\xi }^2}^{2}u(z,\xi )$ on $\mathsf{\Theta }$,

$$\begin{array}{c}\hfill \parallel {\partial }_{z^2}^2{\partial }_{\xi }u(z,\xi )-{\partial }_{z^2}^2{\partial }_{\xi }u(z_i,{\xi }_j){\parallel }_{\infty }=O(h_z+h_{\xi }).\end{array}$$

Secondly, one knows that

$$\begin{array}{c}\hfill |{\partial }_{z^2}^2{\partial }_{\xi }{u}_n(z_i,{\xi }_j)-{\partial }_{z^2}^2{\partial }_{\xi }{u}_n(z,\xi )|\le {\int }_{z_i}^z|{\partial }_{y^3}^3{\partial }_{\xi }{u}_n(y,{\xi }_j)|dy+{\int }_{{\xi }_i}^{\xi }|{\partial }_{z^2}^2{\partial }_{s^2}^2{u}_n(z,s)|ds.\end{array}$$

The follow equality can be written by using of maximum norm:

$$\begin{array}{c}\hfill \parallel {\partial }_{z^2}^2{\partial }_{\xi }{u}_n(z_i,{\xi }_j)-{\partial }_{z^2}^2{\partial }_{\xi }{u}_n(z,\xi ){\parallel }_{\infty }=O(h_z+h_{\xi }).\end{array}$$

The following statement can be written using Theorem 6 and for any arbitrary $\epsilon >0$, and sufficiently large n:

$$\begin{array}{c}\hfill \parallel {\partial }_{z^2}^2{\partial }_{\xi }u(z_i,{\xi }_j)-{\partial }_{z^2}^2{\partial }_{\xi }{u}_n(z_i,{\xi }_j){\parallel }_{\infty }<\epsilon .\end{array}$$

(55)

The following equation can be written from Equation (55) by choosing of n and using arbitrary constant $\epsilon$. So, we have

$$\begin{array}{c}\hfill \parallel {\partial }_{z^2}^2{\partial }_{\xi }u(z,\xi )-{\partial }_{z^2}^2{\partial }_{\xi }{u}_n{(z,\xi )\parallel }_{\infty }=O(h_z+h_{\xi }).\end{array}$$

(56)

One can know that the following equations can be written from the integral property for differentiable functions:

$$\begin{array}{c}\hfill {\partial }_z{\partial }_{\xi }u(z,\xi )-{\partial }_z{\partial }_{\xi }{u}_n(z,\xi )={\partial }_z{\partial }_{\xi }u(z_i,\xi )-{\partial }_z{\partial }_{\xi }{u}_n(z,\xi )+{\int }_{z_i}^z({\partial }_{y^2}^2{\partial }_{\xi }u(y,\xi )-{\partial }_{y^2}^2{\partial }_{\xi }{u}_n(y,\xi ))dy\end{array}$$

(57)

$$\begin{array}{c}\hfill {\partial }_{\xi }u(z,\xi )-{\partial }_{\xi }{u}_n(z,\xi )={\partial }_{\xi }u(z_i,\xi )-{\partial }_{\xi }{u}_n(z_i,\xi )+{\int }_{z_i}^z({\partial }_y{\partial }_{\xi }u(y,\xi )-{\partial }_y{\partial }_{\xi }{u}_n(y,\xi ))dy\end{array}$$

(58)

$$\begin{array}{c}\hfill u(z,\xi )-u_n(z,\xi )=u(z,{\xi }_i)-u_n(z,{\xi }_i)+{\int }_{{\xi }_i}^{\xi }({\partial }_{s}u(z,s)-{\partial }_s{u}_n(z,s))ds\end{array}$$

(59)

The following inequality can be written from Equations (56)–(59) and Theorem 6:

$$\begin{array}{c}\hfill \parallel u(z,\xi )-u_n{(z,\xi )\parallel }_{\infty }\le C(h_{\xi }{h}_z^3+h_{\xi }^2{h}_z^2).\end{array}$$

 □

6. Numerical Applications and Algorithm of Method

In this section, two fractional Burgers’ problems with variable and constant coefficient are considered. Exact solutions of problems include the fractional parameter $\mathsf{\alpha }$. Reproducing kernel method will be applied for these problems and outcomes will be presented with tables and graphics.

6.1. Algorithm Process of RKM

The algorithm process of RKM is given as follow:

Case 1. Choosing of iteration number as $n=a\times b$.

Case 2. Start ${\psi }_i(z,\xi )=L_{(\tau ,\beta )}{K}_{(\tau ,\beta )}{(z,\xi )|}_{(\tau ,\beta )=(z_i,{\xi }_i)}$.

Case 3. Obtaining of ${\eta }_{ik}$ coefficients.

Case 4. Set ${\overline{\psi }}_i(z,\xi )={\displaystyle \sum _{k=1}^i}{\eta }_{ik}{\psi }_k(z,\xi )$ for $i=1,2,...,n$.

Case 5. Start initial approximation $u_0(z_i,{\xi }_i)$.

Case 6. Calculate $B_i={\displaystyle \sum _{k=1}^i}{\eta }_{ik}F(z_k,{\xi }_k,u_{k-1}(z_k,{\xi }_k),{\partial }_z{u}_{k-1}(z_k,{\xi }_k))$ for $i=1,2,...,n$.

Case 7. Calculate $u_i(z,\xi )={\displaystyle \sum _{k=1}^i}{B}_k{\overline{\psi }}_k(z_k,{\xi }_k)$ for $i=1,2,...,n$.

6.2. Numerical Applications

Example 1.

It will be examined that the following fractional-order Burgers’ problem with Dirichlet boundary condition:

$${}^c{D}_{\xi }^{\mathsf{\alpha }}u+(1+z\xi )u_{zz}+z^{2}u+(z+1)u_z-\xi sin\left(z\right)u{u}_z=f(z,\xi )$$

(60)

$$\begin{array}{c}\hfill 0\le \xi \le 1,0\le z\le 1,0<\mathsf{\alpha }\le 1,\end{array}$$

$$\begin{array}{c}\hfill u(z,0)=0=u(0,\xi )=u(1,\xi )=0.\end{array}$$

(61)

The exact solution of problem:

$$u(z,\xi )=(z^2-z){\xi }^{1+\mathsf{\alpha }},$$

(62)

and $f(z,\xi )$ is the function that provides the Equation (62). Taking $z_i=\frac{i}{a},i=1,2,...,a$, ${\xi }_i=\frac{i}{b},i=1,2,...,b$ and n-th term of approximate solution is selected as $n=a\times b$. Absolute error values for Example 1 is computed for $\mathsf{\alpha }=0.9$, $\mathsf{\alpha }=0.8$, $\mathsf{\alpha }=0.7$ and $n=25(a=b=5)$. Error values are given in

Table 2. Absolute error values of Example 1 for Burgers’ equation with $\mathsf{\alpha }=0.9$.

$\mathit{z}/\mathit{\xi }$	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$	$0.7$	$0.8$	$0.9$
0.1	$1.31\times {10}^{-4}$	$3.13\times {10}^{-5}$	$3.36\times {10}^{-5}$	$3.90\times {10}^{-5}$	$5.67\times {10}^{-5}$	$5.06\times {10}^{-5}$	$6.67\times {10}^{-5}$	$2.92\times {10}^{-5}$	$1.31\times {10}^{-4}$
0.2	$2.34\times {10}^{-4}$	$5.74\times {10}^{-5}$	$6.39\times {10}^{-5}$	$7.70\times {10}^{-5}$	$1.13\times {10}^{-4}$	$1.08\times {10}^{-4}$	$1.44\times {10}^{-4}$	$1.55\times {10}^{-5}$	$1.83\times {10}^{-4}$
0.3	$3.07\times {10}^{-4}$	$7.60\times {10}^{-5}$	$8.59\times {10}^{-5}$	$1.05\times {10}^{-4}$	$1.55\times {10}^{-4}$	$1.53\times {10}^{-4}$	$2.07\times {10}^{-4}$	$7.90\times {10}^{-6}$	$1.96\times {10}^{-4}$
0.4	$3.49\times {10}^{-4}$	$8.65\times {10}^{-5}$	$9.88\times {10}^{-5}$	$1.22\times {10}^{-4}$	$1.82\times {10}^{-4}$	$1.81\times {10}^{-4}$	$2.51\times {10}^{-4}$	$3.54\times {10}^{-5}$	$1.75\times {10}^{-4}$
0.5	$3.62\times {10}^{-4}$	$8.76\times {10}^{-5}$	$1.00\times {10}^{-4}$	$1.24\times {10}^{-4}$	$1.87\times {10}^{-4}$	$1.88\times {10}^{-4}$	$2.66\times {10}^{-4}$	$5.70\times {10}^{-5}$	$1.33\times {10}^{-4}$
0.6	$3.45\times {10}^{-4}$	$8.13\times {10}^{-5}$	$9.31\times {10}^{-5}$	$1.16\times {10}^{-4}$	$1.75\times {10}^{-4}$	$1.76\times {10}^{-4}$	$2.57\times {10}^{-4}$	$7.18\times {10}^{-5}$	$7.86\times {10}^{-5}$
0.7	$2.99\times {10}^{-4}$	$6.69\times {10}^{-5}$	$7.60\times {10}^{-5}$	$9.49\times {10}^{-5}$	$1.45\times {10}^{-4}$	$1.44\times {10}^{-4}$	$2.19\times {10}^{-4}$	$7.05\times {10}^{-5}$	$3.03\times {10}^{-5}$
0.8	$2.25\times {10}^{-4}$	$4.73\times {10}^{-5}$	$5.30\times {10}^{-5}$	$6.56\times {10}^{-5}$	$1.01\times {10}^{-4}$	$9.98\times {10}^{-5}$	$1.58\times {10}^{-4}$	$5.56\times {10}^{-5}$	$3.62\times {10}^{-6}$
0.9	$1.25\times {10}^{-4}$	$2.41\times {10}^{-5}$	$2.63\times {10}^{-5}$	$3.22\times {10}^{-5}$	$5.09\times {10}^{-5}$	$4.82\times {10}^{-5}$	$8.20\times {10}^{-5}$	$2.96\times {10}^{-5}$	$1.49\times {10}^{-5}$

,

Table 3. Absolute error values of Example 1 for Burgers’ equation with $\mathsf{\alpha }=0.8$.

$\mathit{z}/\mathit{\xi }$	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$	$0.7$	$0.8$	$0.9$
0.1	$1.26\times {10}^{-4}$	$1.29\times {10}^{-5}$	$2.37\times {10}^{-5}$	$1.99\times {10}^{-5}$	$4.07\times {10}^{-5}$	$3.53\times {10}^{-5}$	$6.82\times {10}^{-5}$	$1.27\times {10}^{-5}$	$8.96\times {10}^{-6}$
0.2	$2.25\times {10}^{-4}$	$2.58\times {10}^{-5}$	$4.78\times {10}^{-5}$	$4.48\times {10}^{-5}$	$8.67\times {10}^{-5}$	$8.28\times {10}^{-5}$	$1.48\times {10}^{-4}$	$5.90\times {10}^{-5}$	$3.22\times {10}^{-5}$
0.3	$2.96\times {10}^{-4}$	$3.61\times {10}^{-5}$	$6.66\times {10}^{-5}$	$6.49\times {10}^{-5}$	$1.23\times {10}^{-4}$	$1.21\times {10}^{-4}$	$2.13\times {10}^{-4}$	$1.03\times {10}^{-4}$	$8.09\times {10}^{-5}$
0.4	$3.39\times {10}^{-4}$	$4.36\times {10}^{-5}$	$7.94\times {10}^{-5}$	$7.88\times {10}^{-5}$	$1.47\times {10}^{-4}$	$1.48\times {10}^{-4}$	$2.57\times {10}^{-4}$	$1.40\times {10}^{-4}$	$1.30\times {10}^{-4}$
0.5	$3.53\times {10}^{-4}$	$4.62\times {10}^{-5}$	$8.30\times {10}^{-5}$	$8.22\times {10}^{-5}$	$1.54\times {10}^{-4}$	$1.55\times {10}^{-4}$	$2.72\times {10}^{-4}$	$1.58\times {10}^{-4}$	$1.67\times {10}^{-4}$
0.6	$3.39\times {10}^{-4}$	$4.54\times {10}^{-5}$	$7.99\times {10}^{-5}$	$7.82\times {10}^{-5}$	$1.46\times {10}^{-4}$	$1.47\times {10}^{-4}$	$2.61\times {10}^{-4}$	$1.60\times {10}^{-4}$	$1.89\times {10}^{-4}$
0.7	$2.96\times {10}^{-4}$	$3.94\times {10}^{-5}$	$6.77\times {10}^{-5}$	$6.43\times {10}^{-5}$	$1.22\times {10}^{-4}$	$1.20\times {10}^{-4}$	$2.21\times {10}^{-4}$	$1.39\times {10}^{-4}$	$1.85\times {10}^{-4}$
0.8	$2.25\times {10}^{-4}$	$2.95\times {10}^{-5}$	$4.93\times {10}^{-5}$	$4.46\times {10}^{-5}$	$8.69\times {10}^{-5}$	$8.34\times {10}^{-5}$	$1.59\times {10}^{-4}$	$1.01\times {10}^{-4}$	$1.52\times {10}^{-4}$
0.9	$1.26\times {10}^{-4}$	$1.60\times {10}^{-5}$	$2.59\times {10}^{-5}$	$2.17\times {10}^{-5}$	$4.40\times {10}^{-5}$	$4.03\times {10}^{-5}$	$8.22\times {10}^{-5}$	$5.19\times {10}^{-5}$	$8.97\times {10}^{-5}$

and

Table 4. Absolute error values of Example 1 for Burgers’ equation with $\mathsf{\alpha }=0.7$.

$\mathit{z}/\mathit{\xi }$	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$	$0.7$	$0.8$	$0.9$
0.1	$1.29\times {10}^{-4}$	$7.61\times {10}^{-6}$	$2.09\times {10}^{-5}$	$9.72\times {10}^{-6}$	$3.07\times {10}^{-5}$	$2.01\times {10}^{-5}$	$4.75\times {10}^{-5}$	$1.57\times {10}^{-5}$	$4.67\times {10}^{-5}$
0.2	$2.30\times {10}^{-4}$	$9.60\times {10}^{-6}$	$4.42\times {10}^{-5}$	$2.82\times {10}^{-5}$	$7.08\times {10}^{-5}$	$5.80\times {10}^{-5}$	$1.13\times {10}^{-4}$	$6.51\times {10}^{-5}$	$1.30\times {10}^{-4}$
0.3	$3.03\times {10}^{-4}$	$8.97\times {10}^{-6}$	$6.33\times {10}^{-5}$	$4.46\times {10}^{-5}$	$1.04\times {10}^{-4}$	$9.12\times {10}^{-5}$	$1.69\times {10}^{-4}$	$1.11\times {10}^{-4}$	$2.07\times {10}^{-4}$
0.4	$3.48\times {10}^{-4}$	$5.73\times {10}^{-6}$	$7.73\times {10}^{-5}$	$5.69\times {10}^{-5}$	$1.27\times {10}^{-4}$	$1.15\times {10}^{-4}$	$2.08\times {10}^{-4}$	$1.48\times {10}^{-4}$	$2.68\times {10}^{-4}$
0.5	$3.64\times {10}^{-4}$	$2.22\times {10}^{-6}$	$8.27\times {10}^{-5}$	$6.07\times {10}^{-5}$	$1.35\times {10}^{-4}$	$1.24\times {10}^{-4}$	$2.23\times {10}^{-4}$	$1.64\times {10}^{-4}$	$3.03\times {10}^{-4}$
0.6	$3.53\times {10}^{-4}$	$2.36\times {10}^{-6}$	$8.15\times {10}^{-5}$	$5.87\times {10}^{-5}$	$1.30\times {10}^{-4}$	$1.20\times {10}^{-4}$	$2.16\times {10}^{-4}$	$1.64\times {10}^{-4}$	$3.09\times {10}^{-4}$
0.7	$3.11\times {10}^{-4}$	$5.08\times {10}^{-6}$	$7.10\times {10}^{-5}$	$4.82\times {10}^{-5}$	$1.10\times {10}^{-4}$	$1.00\times {10}^{-4}$	$1.84\times {10}^{-4}$	$1.40\times {10}^{-4}$	$2.80\times {10}^{-4}$
0.8	$2.39\times {10}^{-4}$	$6.20\times {10}^{-6}$	$5.32\times {10}^{-5}$	$3.31\times {10}^{-5}$	$7.91\times {10}^{-5}$	$7.04\times {10}^{-5}$	$1.33\times {10}^{-4}$	$1.01\times {10}^{-4}$	$2.17\times {10}^{-4}$
0.9	$1.35\times {10}^{-4}$	$4.68\times {10}^{-6}$	$2.90\times {10}^{-5}$	$1.57\times {10}^{-5}$	$4.05\times {10}^{-5}$	$3.47\times {10}^{-5}$	$6.90\times {10}^{-5}$	$5.18\times {10}^{-5}$	$1.22\times {10}^{-4}$

in order to observe of applicability and influence of method. The graphics of absolute errors are given for $\mathsf{\alpha }=0.7$, $\mathsf{\alpha }=0.8$, and $\mathsf{\alpha }=0.9$ in Figure 1.

Example 2.

It will be examined that the fractional-order Burgers’ equation with Neumann boundary condition as follow:

$${}^c{D}_{\xi }^{\mathsf{\alpha }}u-u_{zz}-u{u}_z=f(z,\xi )$$

(63)

$$\begin{array}{c}\hfill 0\le z\le 1,0\le \xi \le 1,\frac{1}{2}<\mathsf{\alpha }\le 1,\end{array}$$

$$\begin{array}{c}\hfill u(z,0)=0,u_z(0,\xi )=u_z(1,\xi )=0.\end{array}$$

(64)

The exact solution of problem is:

$$u(z,\xi )=(\frac{z^3}{3}-\frac{z^2}{2}){\xi }^{2\mathsf{\alpha }},$$

(65)

and $f(z,\xi )$ is the function that provides the Equations (65). Taking $z_i=\frac{i}{a},i=1,2,...,a$, ${\xi }_i=\frac{i}{b},$$i=1,2,...,b$ and $n=a\times b$. Absolute error of Example 2 is computed for $\mathsf{\alpha }=0.9$, $\mathsf{\alpha }=0.8$, $\mathsf{\alpha }=0.7$ and $n=64(a=b=8)$. Error values are given in

Table 5. Absolute error values of Example 2 for Burgers’ equation with $\mathsf{\alpha }=0.9$.

$\mathit{z}/\mathit{\xi }$	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$	$0.7$	$0.8$	$0.9$
0.1	$4.76\times {10}^{-4}$	$3.40\times {10}^{-4}$	$2.88\times {10}^{-4}$	$2.92\times {10}^{-4}$	$3.24\times {10}^{-4}$	$3.84\times {10}^{-4}$	$4.64\times {10}^{-4}$	$5.77\times {10}^{-4}$	$6.89\times {10}^{-4}$
0.2	$9.30\times {10}^{-4}$	$6.16\times {10}^{-4}$	$4.62\times {10}^{-4}$	$4.00\times {10}^{-4}$	$3.81\times {10}^{-4}$	$4.03\times {10}^{-4}$	$4.50\times {10}^{-4}$	$5.49\times {10}^{-4}$	$6.38\times {10}^{-4}$
0.3	$1.36\times {10}^{-3}$	$8.71\times {10}^{-4}$	$6.16\times {10}^{-4}$	$4.88\times {10}^{-4}$	$4.16\times {10}^{-4}$	$3.97\times {10}^{-4}$	$4.09\times {10}^{-4}$	$4.88\times {10}^{-4}$	$5.54\times {10}^{-4}$
0.4	$1.77\times {10}^{-3}$	$1.10\times {10}^{-3}$	$7.50\times {10}^{-4}$	$5.58\times {10}^{-4}$	$4.33\times {10}^{-4}$	$3.72\times {10}^{-4}$	$3.48\times {10}^{-4}$	$4.06\times {10}^{-4}$	$4.49\times {10}^{-4}$
0.5	$2.15\times {10}^{-3}$	$1.30\times {10}^{-3}$	$8.62\times {10}^{-4}$	$6.09\times {10}^{-4}$	$4.34\times {10}^{-4}$	$3.33\times {10}^{-3}$	$2.74\times {10}^{-4}$	$3.09\times {10}^{-4}$	$3.30\times {10}^{-4}$
0.6	$2.47\times {10}^{-3}$	$1.46\times {10}^{-3}$	$9.49\times {10}^{-4}$	$6.42\times {10}^{-4}$	$4.20\times {10}^{-4}$	$2.82\times {10}^{-4}$	$1.91\times {10}^{-4}$	$2.02\times {10}^{-4}$	$2.01\times {10}^{-4}$
0.7	$2.74\times {10}^{-3}$	$1.59\times {10}^{-3}$	$1.01\times {10}^{-3}$	$6.56\times {10}^{-4}$	$3.94\times {10}^{-4}$	$2.23\times {10}^{-4}$	$1.01\times {10}^{-4}$	$9.05\times {10}^{-5}$	$6.96\times {10}^{-4}$
0.8	$2.93\times {10}^{-3}$	$1.68\times {10}^{-3}$	$1.04\times {10}^{-3}$	$6.56\times {10}^{-4}$	$3.62\times {10}^{-4}$	$1.63\times {10}^{-4}$	$1.72\times {10}^{-5}$	$1.43\times {10}^{-5}$	$5.21\times {10}^{-5}$
0.9	$3.05\times {10}^{-3}$	$1.73\times {10}^{-3}$	$1.06\times {10}^{-3}$	$6.49\times {10}^{-4}$	$3.32\times {10}^{-4}$	$1.15\times {10}^{-4}$	$4.78\times {10}^{-5}$	$9.40\times {10}^{-5}$	$1.43\times {10}^{-4}$

,

Table 6. Absolute error values of Example 2 for Burgers’ equation with $\mathsf{\alpha }=0.8$.

$\mathit{z}/\mathit{\xi }$	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$	$0.7$	$0.8$	$0.9$
0.1	$3.88\times {10}^{-4}$	$2.19\times {10}^{-4}$	$2.12\times {10}^{-4}$	$2.49\times {10}^{-4}$	$3.02\times {10}^{-4}$	$3.77\times {10}^{-4}$	$4.63\times {10}^{-4}$	$5.77\times {10}^{-4}$	$6.84\times {10}^{-4}$
0.2	$7.52\times {10}^{-4}$	$3.60\times {10}^{-4}$	$2.88\times {10}^{-4}$	$2.86\times {10}^{-4}$	$3.07\times {10}^{-4}$	$3.59\times {10}^{-4}$	$4.23\times {10}^{-4}$	$5.31\times {10}^{-4}$	$6.22\times {10}^{-4}$
0.3	$1.10\times {10}^{-3}$	$4.88\times {10}^{-4}$	$3.52\times {10}^{-4}$	$3.10\times {10}^{-4}$	$2.96\times {10}^{-4}$	$3.21\times {10}^{-4}$	$3.61\times {10}^{-4}$	$4.60\times {10}^{-4}$	$5.34\times {10}^{-4}$
0.4	$1.44\times {10}^{-3}$	$6.01\times {10}^{-4}$	$4.05\times {10}^{-4}$	$3.23\times {10}^{-4}$	$2.74\times {10}^{-4}$	$2.73\times {10}^{-4}$	$2.88\times {10}^{-4}$	$3.74\times {10}^{-4}$	$4.32\times {10}^{-4}$
0.5	$1.75\times {10}^{-3}$	$6.97\times {10}^{-4}$	$4.45\times {10}^{-4}$	$3.26\times {10}^{-4}$	$2.43\times {10}^{-4}$	$2.16\times {10}^{-4}$	$2.07\times {10}^{-4}$	$2.79\times {10}^{-4}$	$3.21\times {10}^{-4}$
0.6	$2.01\times {10}^{-3}$	$7.75\times {10}^{-4}$	$4.74\times {10}^{-4}$	$3.19\times {10}^{-4}$	$2.05\times {10}^{-4}$	$1.53\times {10}^{-4}$	$1.22\times {10}^{-4}$	$1.79\times {10}^{-4}$	$2.06\times {10}^{-4}$
0.7	$2.23\times {10}^{-3}$	$8.31\times {10}^{-4}$	$4.90\times {10}^{-4}$	$3.04\times {10}^{-4}$	$1.62\times {10}^{-4}$	$8.89\times {10}^{-5}$	$3.63\times {10}^{-5}$	$7.85\times {10}^{-5}$	$9.18\times {10}^{-5}$
0.8	$2.40\times {10}^{-3}$	$8.68\times {10}^{-4}$	$4.95\times {10}^{-4}$	$2.85\times {10}^{-4}$	$1.20\times {10}^{-4}$	$2.81\times {10}^{-5}$	$4.19\times {10}^{-5}$	$1.33\times {10}^{-5}$	$1.18\times {10}^{-5}$
0.9	$2.50\times {10}^{-3}$	$8.87\times {10}^{-4}$	$4.94\times {10}^{-4}$	$2.67\times {10}^{-4}$	$8.76\times {10}^{-5}$	$1.80\times {10}^{-5}$	$1.00\times {10}^{-4}$	$8.19\times {10}^{-5}$	$8.88\times {10}^{-4}$

and

Table 7. Absolute error values of Example 2 for Burgers’ equation with $\mathsf{\alpha }=0.7$.

$\mathit{z}/\mathit{\xi }$	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$	$0.7$	$0.8$	$0.9$
0.1	$2.86\times {10}^{-4}$	$1.42\times {10}^{-4}$	$1.89\times {10}^{-4}$	$2.48\times {10}^{-4}$	$3.17\times {10}^{-4}$	$4.00\times {10}^{-4}$	$4.88\times {10}^{-4}$	$5.96\times {10}^{-4}$	$6.96\times {10}^{-4}$
0.2	$5.41\times {10}^{-4}$	$1.88\times {10}^{-4}$	$2.15\times {10}^{-4}$	$2.53\times {10}^{-4}$	$3.01\times {10}^{-4}$	$3.73\times {10}^{-4}$	$4.48\times {10}^{-4}$	$5.53\times {10}^{-4}$	$6.41\times {10}^{-4}$
0.3	$7.93\times {10}^{-4}$	$2.26\times {10}^{-4}$	$2.34\times {10}^{-4}$	$2.50\times {10}^{-4}$	$2.77\times {10}^{-4}$	$3.35\times {10}^{-4}$	$3.95\times {10}^{-4}$	$4.95\times {10}^{-4}$	$5.71\times {10}^{-4}$
0.4	$1.03\times {10}^{-3}$	$2.57\times {10}^{-4}$	$2.49\times {10}^{-4}$	$2.42\times {10}^{-4}$	$2.48\times {10}^{-4}$	$2.91\times {10}^{-4}$	$3.36\times {10}^{-4}$	$4.30\times {10}^{-4}$	$4.95\times {10}^{-4}$
0.5	$1.26\times {10}^{-3}$	$2.81\times {10}^{-4}$	$2.58\times {10}^{-4}$	$2.30\times {10}^{-4}$	$2.16\times {10}^{-4}$	$2.45\times {10}^{-4}$	$2.75\times {10}^{-4}$	$3.61\times {10}^{-4}$	$4.16\times {10}^{-4}$
0.6	$1.45\times {10}^{-3}$	$2.98\times {10}^{-4}$	$2.62\times {10}^{-4}$	$2.14\times {10}^{-4}$	$1.81\times {10}^{-4}$	$1.96\times {10}^{-4}$	$2.12\times {10}^{-4}$	$2.91\times {10}^{-4}$	$3.36\times {10}^{-4}$
0.7	$1.62\times {10}^{-3}$	$3.06\times {10}^{-4}$	$2.60\times {10}^{-4}$	$1.95\times {10}^{-4}$	$1.45\times {10}^{-4}$	$1.48\times {10}^{-4}$	$1.52\times {10}^{-4}$	$2.22\times {10}^{-4}$	$2.58\times {10}^{-4}$
0.8	$1.74\times {10}^{-3}$	$3.09\times {10}^{-4}$	$2.55\times {10}^{-4}$	$1.76\times {10}^{-4}$	$1.12\times {10}^{-4}$	$1.04\times {10}^{-4}$	$9.79\times {10}^{-5}$	$1.59\times {10}^{-4}$	$1.89\times {10}^{-4}$
0.9	$1.82\times {10}^{-3}$	$3.08\times {10}^{-4}$	$2.50\times {10}^{-4}$	$1.60\times {10}^{-4}$	$8.69\times {10}^{-5}$	$7.20\times {10}^{-5}$	$5.79\times {10}^{-5}$	$1.14\times {10}^{-4}$	$1.38\times {10}^{-4}$

in order to observe of applicability and influence of method. The graphics of absolute errors are given for $\mathsf{\alpha }=0.7$, $\mathsf{\alpha }=0.8$, and $\mathsf{\alpha }=0.9$ in Figure 2.

7. Conclusions

In this research, some special Hilbert spaces with inner products and the kernel function of these spaces are introduced. Then the iterative solution is obtained by reproducing kernel theory. Error estimation of the approximate solution and convergence analysis are verified with lemma and theorems. Numerical outcomes demonstrate that the iterative approximation is applicable, convenient, and powerful for fractional-order Burgers’ equation with Dirichlet and Neumann conditions. Therefore, iterative RKM is successfully implemented for fractional-order Burgers’ equation and so this study will contribute to the science.