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# Edelweiss Applied Science and Technology (ISSN: 2576-8484)

## Metonou Richard and Weidong Zhao

#### DOI Number: https://doi.org/10.33805/2576-8484.193 PDF

### Introduction

The [L,M]-order Padé approximation of the function f denote by Px[L,M] [f], is the quotient of two polynomials RL(x) and QM(x) of degrees L and M, respectively: Remark 1: The [L,M]-order Padé approximation of the function f(x) is in the form: If according to the signe of .

Definition 1: Let f be function of two variables x and t. We defined two dimensional Padé approximation P[L/M][f](x,t) of the function f as

P[L/M] [f] (x,t) = Px[L,M] [Pt[L,M] [f]] (x,t),                         (2)

where Pt[L,M] [f] (x, t) denote the [L,M]-order Padé approximation of f(x, t) with respect to the variable t, and Px[L,M] [f] (x, t) denote the [L,M]-order Padé approximation of f(x, t) with respect to the variable x.

If M=L, we will denote the diagonal Padé approximation of order M by P[M/M] [f] (x, t), and called [M,M]-order Padé approximation or M Padé approximation of f(x, t).

By replacing the Sumudu transform by Laplce transform and using the same procedure we will obtain the Laplace Adomian Decomposition Methods instead of Sumudu Adomian decomposition method.

We consider the PDE in the form as following:

Ltu(x,t)+Lxu(x,t)+R(u(x,t))+G(u(x,t))=F(x,t)             (3)

with u(x,0)=h(x) the initial condition, Lx the highest order differential respect to x, Lt the first order differential respect to t, G(u(x,y)) is the nonlinear term, F(x,t) is the inhomogeneous term, and R the remaining linear terms of lower order derivative.

The procedure of PSADM for solving (3) can be write as follows.

Step 1: Take the Sumudu transform to the equation (3) and apply the differentiation property of Sumudu transform to obtain

St[u(x,t)](v)=h(x)+v.St[-Lxu(x,t)-R(u(x,t))-G(u(x,t))+F(x,t)](v).  (4)

Step 2: Apply the inverse of the Sumudu transform to the above equation to obtain

u(x,t)=Sv-1[h(x)](t)+Sv-1[v.St[-Lxu(x,t)-R(u(x,t))-G(u(x,t))+F(x,t)](v)](t)..     (5)

Step 3: Use Adomian decomposition method to decomposite the nonlinear function G(u) and the u, respectively, as Step 4: Write the equation in the form  u(x,t,j)=u0+u1+…+ uj.

if L=M, we denote M-PSADM solution by

Remark 2: In step 5 we obtain the Sumudu Adomian Decomposition Method (SADM), instead of Sumudu transform we can use other integral transform like Laplace transform to obtain in step 5 the Laplace Adomian Decomposition Method (LADM). The Sumudu Adomian Decomposition Method and Laplace Adomian Decomposition Methods give similar result. The Sumudu transform due to the unit preserving properties (S(1)=1, provided some advantages in calculation.

For different type of Padé approximation and different order of the Padé approximation we will analyse the behaviour of the PADM solutions.

### Example 1

In the first case of the following example, we will show that for different type of Padé approximation:

u(x, t, j, [L,M]) , for L > M,

u(x, t, j, [L,M]), for L = M, and

u(x, t, j, [L,M]), for L < M, we have different solutions and one of them is more accurate.

In the second case of the following example, we will show that for diagonal Padé approximation u(x,t,j, [L,M]), for L = M, we can increase the accuracy of the method by increasing or reducing de value of M accordingly to the topologie of the solution.

Case 1: Consider the equation: We can easily deduice the SADM solution : The algorithm is coded by the symbolic computation software Mathematica. We know u(x,t) = ei(x+t) is the exact solution for the Problem.

Figure (1a) and Figure (1b) show the real part and imaginary part of SADM solution uSADM = u(x,t,15) in Domain D = [0,2] × [0,2].

For different values of L and M we plot different orders of the PSADM solutions to see the behaviour of the methods.

Figure (1c) and Figure (1d) show respectively, the real part and imaginary part of PSADM solution uPSADM = u(x,t,15,[7,0]) in Domain D = [0,2] × [0,2].

Figure (1e) and Figure (1f) show respectively, the real part and imaginary part of PSADM solution uPSADM = u(x,t,15,[1,7]) in Domain D = [0,2] × [0,2].

Figure (2a) and Figure (2b) show respectively, the real part and imaginary part of PSADM solution uPSADM = u(x,t,15,[7,1]) in Domain D = [0,2] × [0,2].

Figure (2c) and Figure (2d) show respectively, the real part and imaginary part of PSADM solution uPSADM = u(x,t,15,[7,7]) or in short u(x,t,15,7) in Domain D = [0,2] × [0,2].

Figure (2e) and Figure (2f) show the real part and imaginary part of the exact solution u(x,t) in Domain D = [0,2] × [0,2].  In domain D = [0, 2] × [0, 2] we can see that the u(x, t, 15, [1, 7]) and u(x, t, 15, [7, 1]) are not smooth compare to the other solutions. Next we plots the absolute errors.

Figure (3a) and Figure (3b) show respectively the absolute error for real part and imaginary part of the Sumudu Adomian Decomposition solution u(x, t, 15).

Figure (3c) and Figure (3d) show respectively the absolute error for real part and imaginary part of the Padé Sumudu Adomian Decomposition solution u(x, t, 15, [7, 0]).

Figure (3e) and Figure (3f) show respectively the absolute error for real part and imaginary part of the Sumudu Adomian Decomposition solution u(x, t, 15).

Now let see the behaviour of the SADM, PSADM solutions in domain D = [0, 10] × [0, 10].

Figure (4a) and Figure (4b) show the real part and imaginary part of SADM solution uSADM = u(x, t, 15) in Domain D = [0, 10] × [0, 10].

Figure (4c) and Figure (4d) show respectively, the real part and imaginary part of PSADM solution uPSADM = u(x, t, 15, [7, 0]) in Domain D = [0, 10] × [0, 10].

Figure (4e) and Figure (4f) show respectively, the real part and imaginary part of PSADM solution uPSADM = u(x, t, 15, [1, 7]) in Domain D = [0, 10] × [0, 10].

Figure (5a) and Figure (5b) show respectively, the real part and imaginary part of PSADM solution uPSADM = u(x, t, 15, [7, 1]) in Domain D = [0, 10] × [0, 10].

Figure (5c) and Figure (5d) show respectively, the real part and imaginary part of PSADM solution uPSADM = u(x, t, 15, [7, 7]) or in short u(x, t, 15, 7) in Domain D = [0, 10] × [0, 10].

Figure (5e) and Figure (5f) show the real part and imaginary part of the exact solution u(x, t) in Domain D = [0, 10] × [0, 10]. Figure 3: Absolute errors.  We can see in this example, in domain D = [0, 2] × [0, 2] the SADM behave well, but in domain D = [0, 10] × [0, 10] the SADM give bad result. Only the u(x, t, 15, 7) approaching well the exact solution in the domain D = [0, 2] × [0, 2] and D = [0, 10] × [0, 10]. Even in domain

D = [0, 2] × [0, 2] the graph of absolute error show the [7, 7]-order PSADM (in short 7-PASDM) solution is better than SADM solution and other type of PSADM solutions. The solution can be perfom by choosing different value of M.

Remark 3

The real part and imaginary part of the exact solution are bounded,

and the real part and imaginary part of the SADMs are not bounded, in this case in better to use diagonal Padé approximatin to make bounded the Approximate solution.

Case 2: Consider the equation: We can easily deduce the SADM solution : And the [L,M] order PSADM solution:

uPSADM = u(x, t, j, [L,M]) = P[L/M] [u(x, t, j)] (x, t),             (14)

The algorithm is coded by the symbolic computation software Mathematica.

We know u(x, t) = e3i(x-t) is the exact solution of the Problem.

Figure (6c) and Figure (6d) show respectively the absolute error for real part and imaginary part of the PSADM solution u(x, t, 20, 7).

Figure (6e) and Figure (6f) show respectively the absolute error for real part and imaginary part of the PSADM solution u(x, t, 20, 8).

Figure (6a) and Figure (6b) show respectively the absolute error for real part and imaginary part of the SADM solution u(x, t, 20). The graph of the absolute errors show that the SADM solution give better approximation than 7-order PSADM solution, and the 8-order PSADM solution give better approximation than SADM solution. The solution can be perform by choosing different value of M.

### Example 2

Case 1: Consider the equation: Using the SADMs, we can deduice: The solution is given by: The Sumudu Adomian Decomposition solution is given by  By using the symbolic computation software Mathematica.

The figure (7a), Figure (7b), Figure (7c) and Figure (7d) show respectively the curve of SADM solution u(x, t, 3), PSADM solution uPSADM = u(x, t, 3, [2,2]), uPSADM = u(x, t, 3, [1, 2]) and the exact solution u = u(x, t), in domain D = [-0.5, 0.5] × [0, 0.1]. The exact solution is given by u(x, t) = -2sech2(x-4t) Figure (8a), Figure (8b), Figure (8c) and Figure (8d) show respectively the curve of SADM solution u(x, t, 3), PSADM solution uPSADM = u(x, t, 3, [1, 2]) and the exact solution u = u(x, t), in domain D = [-1, 1] × [0, 1] We can see the SADM, [2, 2]- order PSADM and [1, 2]-order PSADM solutions behave well in domain D = [-0.5, 0.5] × [0, 0.1], but only [1, 2]-order PSADM solution give better result in domain D = [-1, 1] × [0, 1]. The [2, 2]-order PSADM solution in this case in not better than [1, 2]-order PSADM solution. Then the diagonal Padé approximation are not accurate in this case. It is recommended in case to use [M,N]-order Padé approximation with M ≠ N.

Case 2: Consider the equation: We know for q = 6, β = -1, and subject to the initial condition the exact solution is given by: For k = 0,5, the figure (9a), Figure (9b), Figure (9c), Figure (9d), and Figure (9e) show respectively the curve of the SADM solution uSADM = u(x, t, 4), PSADM solutions u(x, t, 4, 2), u(x, t, 4, [0, 2]), u(x, t, 4, [2, 0]) and the exact solution uexact in domain D = [0,1] × [0, 2] For k = 0.5, the figures (10a), (10b), (10c), and (10d) show respectively the absolute error curve for the SADM solution uSADM = u(x,t,4), PSADM solutions u(x,t,4,2), u(x,t,4,[0,2]), and u(x,t,4,[2,0]) in domain D = [0,1] × [0,2]

The SADM and [2, 2]-order PSADM (or in short 2-PSADM) provide same results. The [0, 2]-order PSADM solution in this case providing better error than [2, 2]-order PSADM and [2, 0]-order PSADM solutions. The diagonal Padé approximations are not recommended in this case.

Remark 4

The following conditions can help to choose the best PSADM solution.

Condition (*)

If  Figure 10: Absolute errors.

We comput the new solution by

u(x, t, j, [L,M]) = P[L/M] [uSADM] (x, t),with L < M

Then Condition (**)

If We comput the new solution by

u(x, t, j, [L,M]) = P[L/M] [uSADM] (x, t),with L >M

then Condition (***)

If we are not in the case mentioning in conditions (*) and (**), we comput the new solution by

u(x, t, j, [M,M]) = u(x, t, j, M) = P[M/M] [uSADM] (x, t),with L = M

### Conclusion

In this work, we show the behaviour et of the function P[L/M][.](x, t) using to obtain the Padé Sumudu Adomian Decomposition Methods solution for nonlinear partial differential equations such as the Schrödinger equations, and the KdV-Burger's equations. The proposed function provide us a suitable way for controlling the convergences of series solutions with high accuracy by using different order of Padé approximation and different type of the Padé approximation according to the topology of the exact solution u(x, t) and the topology of the SADM solution u(x, t, j). When the exact solutions are unknown, we have some mathematical approach to obtain more information about the topology of the exact solution. This approach can be generalized to investigate more complicated nonlinear partial differential equations that can only be solved by numerically.

### Acknowledgement

This research is partially supported by the Foundations of China (Nos. 12071261, 12001539, 11831010, 11871068), the Science Challenge Project (No. TZ2018001), and the national key basic research program (No. 2018YFA0703903). The first author also acknowledges the financial support of the Chinese Scholarship Council (CSC) in Shandong University with grand (CSC No: 2020DFJ001523).

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### Corresponding author

Metonou Richard, School of Mathematics, Shandong University, China, E-mail: metonourichard@yahoo.fr

Weidong Zhao, School of Mathematics, Shandong University, China, E-mail: wdzhao@sdu.edu.cn

### Citation

Richard M and Zhao W. Behaviour analysis of the Padé sumudu adomian decomposition method solution (2021) Edelweiss Appli Sci Tech 5: 39-45.

#### Keywords 