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

: 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.


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: c D α ξ u(z, ξ) + c 1 (z, ξ)u zz (z, ξ) + c 2 (z, ξ)u(z, ξ) + c 3 (z, ξ)u z (z, ξ) + c 4 (z, ξ)u(z, ξ)u z (z, ξ) = f (z, ξ) Here, c D α ξ is fractional differential operator in Caputo sense with respect to time variable ξ and also f (z, ξ), c 1 (z, ξ), c 2 (z, ξ), c 3 (z, ξ), c 4 (z, ξ) are continuous functions. For this model problem, initial-Neumann boundary conditions: and initial-Dirichlet boundary conditions: u(z, 0) = 0 u(0, ξ) = u(1, ξ) = 0 (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 Exact solution u n (z, ξ) Reproducing kernel solution L Linear operator ., .

Some Specific Definitions and Hilbert Spaces
In this section, some basic definitions and significant reproducing kernel spaces will be given.

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 ξ variables. W n 2 [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 ξ. Therefore, the kernel function of W 3 2 [0, 1] will be given for u zz and the kernel function of W 2 2 [0, 1] will be given for u ξ . Furthermore, W 1 2 [0, 1] space will be given for general function (without derivative). For the obtaining procedure of reproducing kernel functions, please see [47].

Reproducing Kernel Spaces for Two Variable
The problem (1) and (2a,b) has two variables z and ξ. For this reason, we should give the spaces, inner products, and kernel functions according to these variables. Because the highest order derivatives z and ξ to be considered, reproducing kernel spaces will be given for both z and ξ variables.
The region which we consider is Θ = [0, 1] × [0, 1]. In this part, W (3,2) 2 (Θ) space is given for Dirichlet boundary conditions. These reproducing kernel spaces are also determined in the same way for Neumann boundary conditions.
Therefore, the proof is completed.

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.

(Θ)
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: The new statement of Equations (1)-(2a-2b) can be expressed as: and be a countable dense subset in Θ. Now, ψ i (z, ξ) basis function will be defined by applying the kernel function to the operator L.
Now, it will be shown that ψ i (z, ξ) basis function belong to W For this purpose, the following theorem will be given.

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

It should be shown that
∂z 2 ∂ξ is completely continuous function. 3. ψ i (z, ξ) basis function satisfies the initial and boundary conditions.
One can see that any elements of W (3,2) 2 (Θ) satisfy the above conditions. Now, the following equation can be written using the property of the kernel function K (τ,β) (z, ξ) . These functions are bounded because they are continuous in [0, 1]. So, it can be written In the same way, one can write that Here, M 1 , M 2 , M 3 and M 4 are positive constants. From (17), Therefore, Proof. It is known that Clearly, for each fixed u(z, ξ) ∈ W Hence, (Lu)(z, ξ) = 0. By using of inverse operator L −1 , it can be seen that u = 0. So, theorem is proven.
So, theorem is proven.
In Equation (21), u(z, ξ) is described as infinite term sum. In the next equation, finitely n-terms solution will be given as u n (z, ξ):
Proof. Since Using reproducing kernel feature, it can be written that It follows that It can be said that there exists M > 0 from the convergence of u n−1 (z, ξ) such that In a similar way, it can be proven by using Equation (14). So, u n−1 (z n , ξ n ) →û(τ, β), as (z n , ξ n ) → (τ, β).
In a similar way it can be shown that Therefore, So, lemma is proven.
Theorem 5. Assume that (16) has a unique solution, u n is a bounded and Then, u n (z, ξ) converges to u(z, ξ) and Proof. Firstly, we aim to show that u n (z, ξ) is convergence. Following equality can be written from the Equation (27). Using the orthonormality of {ψ i } ∞ i=1 , we have Therefore, u n+1 > u n satisfies from (40). Here, it seems that u n is bounded. So, one can know that u n is convergent. Therefore, there exists a constant b so that So, above equation shows that The following equation is obtained and consequently The completeness of W (3,2) 2 (Θ) shows that u n →û for n → ∞. Next, it will be shown thatû is the representation solution of (16). If the limit is taken both sides of Equation (27), the following equation can be written:û Note that Therefore, From (28), the following equation can be expressed Lû(z l , ξ l ) = F(z l , ξ l , u l−1 (z l , ξ l ), ∂ z u l−1 (z l , ξ l )).
Proof. The convergence of u n is given in the previous theorem. Now,

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: z i = ih z , h z = 1/n, ξ j = jh ξ , h ξ = 1/n, i, j = 1, ..., n. u(z, ξ) ∞ = max |u(z, ξ)| for (z, ξ) ∈ Θ Furthermore, u(z, ξ) − u n (z, ξ) can be written in two ways for each variable as follow: The following two theorems will be given for error estimation considering the two equations above. Theorem 7. Let u(z, ξ) − u n (z, ξ) be error in W (6,2) 2 (Θ). Therefore, there exist a C > 0 so that ] ⊂ Θ, the following equality can be written: Here, Taylor expansion will be used to show error estimate for the function ∂ z u(z, ξ) around the point (z i , ξ j ). This equality will be analyzed in three cases. It is known that Firstly, the continuity of ∂ 2 z 2 u(z, ξ) and ∂ 2 zξ u(z, ξ) on Θ is considered. So, one can write that Secondly, one know that The next equation can be stated by using of maximum norm Finally, for sufficiently large n, any ε > 0 and using Theorem 6 such that From Equation (54), it is known that ε is arbitrary constant and the chosen of n, the following equality can be expressed In light of the information given above, error estimation will be given as follows: So far, error estimation analysis is done for z variable. The error estimation analysis for variable ξ will be given by the next theorem.

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 α. Reproducing kernel method will be applied for these problems and outcomes will be presented with tables and graphics.

Algorithm Process of RKM
The algorithm process of RKM is given as follow: Case 1. Choosing of iteration number as n = a × b.

Numerical Applications
Example 1. It will be examined that the following fractional-order Burgers' problem with Dirichlet boundary condition: The exact solution of problem: and f (z, ξ) is the function that provides the Equation (62). Taking z i = i a , i = 1, 2, ..., a, ξ i = i b , i = 1, 2, ..., b and n-th term of approximate solution is selected as n = a × b. Absolute error values for Example 1 is computed for α = 0.9, α = 0.8, α = 0.7 and n = 25 (a = b = 5). Error values are given in Tables 2-4 in order to observe of applicability and influence of method. The graphics of absolute errors are given for α = 0.7, α = 0.8, and α = 0.9 in Figure 1.     Example 2. It will be examined that the fractional-order Burgers' equation with Neumann boundary condition as follow: The exact solution of problem is : and f (z, ξ) is the function that provides the Equations (65). Taking z i = i a , i = 1, 2, ..., a, ξ i = i b , i = 1, 2, ..., b and n = a × b. Absolute error of Example 2 is computed for α = 0.9, α = 0.8, α = 0.7 and n = 64 (a = b = 8). Error values are given in Tables 5-7 in order to observe of applicability and influence of method. The graphics of absolute errors are given for α = 0.7, α = 0.8, and α = 0.9 in Figure 2.

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.