Approximate Numerical Procedures for the Navier-Stokes System Through the Generalized Method of Lines

1. Introduction

In this article, we develop approximate solutions for the time independent incompressible Navier-Stokes system, through the generalized method of lines. We recall again, for such a method, the domain of the partial differential equation in question is discretized in lines and the concerning solution is written on these lines as functions of the boundary conditions and boundary shape. We emphasize the first article part concerns the application and extension of an approximate proximal approach published in [3]. We develop an analogous algorithm as those presented in [3] but now for a Navier-Stokes system, which is more complex than the systems previously addressed. In this first step we present an algorithm and respective software in MAT-LAB.

Furthermore, we have developed and presented related softwares in MATHEMATICA for a simpler type of domain but also concerning the mentioned proximal approach. Finally, in the last section, we present a software and related line expressions through the original conception of the generalized method of lines, so that in such related numerical examples, the main results are established through applications of the Banach fixed point theorem.

Remark 1. 

We also highlight the next two paragraphs in this article ( a relatively small part) overlaps with the Chapter 28, starting page 526, in the book by F.S. Botelho, [2], published in 2020, by CRC Taylor and Francis. However, we emphasize the present article includes substantial new parts, including a concerning software not included in the previous version of 2020. Another novelty in the present version is the establishment of appropriate boundary conditions for an elliptic system equivalent to original Navier-Stokes one. Such new boundary conditions and concerning results are indicated in Section 2.

At this point we describe the system in question.

Consider $\Omega \subset {\mathbb{R}}^2$ an open, bounded and connected set with a regular (Lipschitzian) internal boundary denoted ${\Gamma }_0$, and a regular external one denoted by ${\Gamma }_1$. For a two-dimensional motion of a fluid on $\Omega$, we denote by $u:\Omega \to \mathbb{R}$ the velocity field in the direction x of the Cartesian system $(x,y)$, by $v:\Omega \to \mathbb{R}$, the velocity field in the direction y and by $p:\Omega \to \mathbb{R}$, the pressure one. Moreover, $\rho$ denotes the fluid density, $\mu$ is the viscosity coefficient and g denotes the gravity field. Under such notation and statements, the time-independent incompressible Navier-Stokes system of partial differential equations stands for,

$$\begin{array}{c}\left\{\begin{array}{cc}\mu {\nabla }^{2}u-\rho u{u}_x-\rho v{u}_y-p_x+\rho g_x=0,\hfill & \mathrm{in}\Omega ,\hfill \\ \mu {\nabla }^{2}v-\rho u{v}_x-\rho v{v}_y-p_y+\rho g_y=0,\hfill & \mathrm{in}\Omega ,\hfill \\ u_x+v_y=0,\hfill & \mathrm{in}\Omega ,\hfill \end{array}\right.\end{array}$$

(1)

$$\begin{array}{c}\left\{\begin{array}{cc}u=v=0,\hfill & \mathrm{on}{\Gamma }_0,\hfill \\ u=u_{\infty },v=0,p=p_{\infty },\hfill & \mathrm{on}{\Gamma }_1\hfill \end{array}\right.\end{array}$$

(2)

At first we look for solutions $(u,v,P)\in W^{2,2}(\Omega )\times W^{2,2}(\Omega )\times W^{1,2}(\Omega )$. We emphasize details about such Sobolev spaces may be found in [1].

About the references, we emphasize that related existence, numerical and theoretical results for similar systems may be found in [6,7,8,9] and [11], respectively. In particular [11] addresses extensively both theoretical and numerical methods and an interesting interplay between them. Moreover, related finite difference schemes are addressed in [10].

2. Details about an equivalent elliptic system

Defining now $P=p/\rho$ and $\nu =\mu /\rho ,$ consider again the Navier-Stokes system in the following format

$$\begin{array}{c}\left\{\begin{array}{cc}\nu {\nabla }^{2}u-u{\partial }_{x}u-v{\partial }_{y}u-{\partial }_{x}P+g_x=0,\hfill & \mathrm{in}\Omega ,\hfill \\ \nu {\nabla }^{2}v-u{\partial }_{x}v-v{\partial }_{y}v-{\partial }_{y}P+g_y=0,\hfill & \mathrm{in}\Omega ,\hfill \\ {\partial }_{x}u+{\partial }_{y}v=0,\hfill & \mathrm{in}\Omega ,\hfill \end{array}\right.\end{array}$$

(3)

$$\begin{array}{c}\left\{\begin{array}{cc}u=v=0,\hfill & \mathrm{on}{\Gamma }_0,\hfill \\ u=u_{\infty },v=0,P=P_{\infty },\hfill & \mathrm{on}{\Gamma }_1\hfill \end{array}\right.\end{array}$$

(4)

As previously mentioned, at first we look for solutions $(u,v,P)\in W^{2,2}(\Omega )\times W^{2,2}(\Omega )\times W^{1,2}(\Omega )$.

We are going to obtain an equivalent Elliptic system with appropriate boundary conditions.

Our main result is summarized by the following theorem.

Theorem 1. 

Let $\Omega \subset {\mathbb{R}}^2$ be an open, bounded, connected set with a regular (Lipschitzian) boundary.

Assume $u,v,P\in W^{2,2}(\Omega )$ are such that

$$\begin{array}{c}\left\{\begin{array}{cc}\nu {\nabla }^{2}u-u{u}_x-v{u}_y-P_x+g_x=0,\hfill & \mathrm{in}\Omega ,\hfill \\ \nu {\nabla }^{2}v-u{v}_x-v{v}_y-P_y+g_y=0,\hfill & \mathrm{in}\Omega ,\hfill \\ {\nabla }^{2}P+u_x^2+v_y^2+2u_y{v}_x-\mathrm{div}g=0,\hfill & \mathrm{in}\Omega ,\hfill \end{array}\right.\end{array}$$

(5)

$$\begin{array}{c}\left\{\begin{array}{cc}u=u_0,v=v_0,\hfill & \mathrm{on}\partial \Omega ,\hfill \\ u_x+v_y=0,\hfill & \mathrm{on}\partial \Omega .\hfill \end{array}\right.\end{array}$$

(6)

Suppose also the unique solution of equation in w

$$\nu {\nabla }^{2}w-u{w}_x-v{w}_y=0,\mathrm{in}\Omega$$

with the boundary conditions

$$w=0\mathrm{on}\partial \Omega ,$$

is

$$w=0,\mathrm{in}\Omega .$$

Under such hypotheses, $u,v,P$ solve the following Navier-Stokes system

$$\begin{array}{c}\left\{\begin{array}{cc}\nu {\nabla }^{2}u-u{u}_x-v{u}_y-P_x+g_x=0,\hfill & \mathrm{in}\Omega ,\hfill \\ \nu {\nabla }^{2}v-u{v}_x-v{v}_y-P_y+g_y=0,\hfill & \mathrm{in}\Omega ,\hfill \\ u_x+v_y=0,\hfill & \mathrm{in}\Omega ,\hfill \end{array}\right.\end{array}$$

(7)

$$\begin{array}{c}\left\{\begin{array}{cc}u=u_0,v=v_0,\hfill & \mathrm{on}\partial \Omega ,\hfill \\ u_x+v_y=0,\hfill & \mathrm{on}\partial \Omega .\hfill \end{array}\right.\end{array}$$

(8)

Proof. 

In (5), taking the derivative in x of the first equation and adding with the derivative in y of the second equation, we obtain

$$\begin{array}{ccc}\hfill & & \nu {\nabla }^2(u_x+v_y)-u{(u_x+v_y)}_x-v{(u_x+v_y)}_y\hfill \\ & & -{\nabla }^{2}P-u_x^2-v_y^2-2u_y{v}_x+\mathrm{div}g=0,\mathrm{in}\Omega \hfill \end{array}$$

(9)

From the hypotheses, $u,v,P$ are such that

$${\nabla }^{2}P+u_x^2+v_y^2+2u_y{v}_x-\mathrm{div}g=0,\mathrm{in}\Omega ,$$

From this and (9), we get

$$\nu {\nabla }^2(u_x+v_y)-u{(u_x+v_y)}_x-v{(u_x+v_y)}_y=0,\mathrm{in}\Omega .$$

(10)

Denoting $w=u_x+v_y$, from this last equation we obtain

$$\nu {\nabla }^{2}w-u{w}_x-v{w}_y=0,\mathrm{in}\Omega .$$

From the hypothesis, the unique solution of this last equation with the boundary conditions $w=0,\mathrm{on}\partial \Omega ,$ is $w=0$.

From this and (10) we have

$$u_x+v_y=0,$$

in $\Omega$ with the boundary conditions

$$u_x+v_y=0,\mathrm{on}\partial \Omega .$$

The proof is complete. □

Remark 2. 

The process of obtaining such a system with a Laplace operator in P in the third equation is a standard and well known one.

The novelty here is the identification of the corrected related boundary conditions obtained through an appropriate solution of equation (10).

3. An approximate proximal approach

In this section we develop an approximate proximal numerical procedure for the model in question.

Such results are extensions of previous ones published in F.S. Botelho, [3] now for the Navier-Stokes system context.

More specifically, neglecting the gravity field, we solve the system of equations

$$\begin{array}{c}\left\{\begin{array}{cc}\nu {\nabla }^{2}u-u{\partial }_{x}u-v{\partial }_{y}u-{\partial }_{x}P=0,\hfill & \mathrm{in}\Omega ,\hfill \\ \\ \nu {\nabla }^{2}v-u{\partial }_{x}v-v{\partial }_{y}v-{\partial }_{y}P=0,\hfill & \mathrm{in}\Omega ,\hfill \\ \\ {\nabla }^{2}P+{\left({\partial }_{x}u\right)}^2+{\left({\partial }_{y}v\right)}^2+2\left({\partial }_{y}u\right)\left({\partial }_{x}v\right)=0,\hfill & \mathrm{in}\Omega .\hfill \end{array}\right.\end{array}$$

(11)

We present a software similar to those presented in [3], with $\nu =0.0177$, and with

$$\Omega =[0,1]\times [0,1]$$

with the boundary conditions

$$u=u_0=0.65y(1-y),v=v_0=0,P=p_0=0.15\mathrm{on}[0,y],\forall y\in [0,1],$$

$$u=v=P_y=0,\mathrm{on}[x,0]\mathrm{and}[x,1],\forall x\in [0,1],$$

$$u_x=v_x=0,\mathrm{and}{P}_x=0\mathrm{on}[1,y],\forall y\in [0,1].$$

The equation (11), in partial finite differences, stands for

$$\begin{array}{ccc}& & \nu \left(\frac{u_{n+1}-2u_n+u_{n-1}}{d^2}+\frac{{\partial }^2{u}_n}{\partial y^2}\right)-u_n\frac{(u_n-u_{n-1})}{d}-v_n\frac{\partial u_n}{\partial y}\hfill \\ & & -\frac{P_n-P_{n-1}}{d}=0,\hfill \end{array}$$

(12)

$$\begin{array}{ccc}& & \nu \left(\frac{v_{n+1}-2v_n+v_{n-1}}{d^2}+\frac{{\partial }^2{v}_n}{\partial y^2}\right)-u_n\frac{(v_n-v_{n-1})}{d}-v_n\frac{\partial v_n}{\partial y}\hfill \\ & & -\frac{\partial P_n}{\partial y}=0,\hfill \end{array}$$

(13)

$$\begin{array}{ccc}& & \left(\frac{P_{n+1}-2P_n+P_{n-1}}{d^2}+\frac{{\partial }^2{P}_n}{\partial y^2}\right)+(u_n-u_{n-1})\frac{(u_n-u_{n-1})}{d^2}+{\left(\frac{\partial v_n}{\partial y}\right)}^2\hfill \\ & & +2\frac{\partial u_n}{\partial y}\left(\frac{v_n-v_{n-1}}{d}\right)=0.\hfill \end{array}$$

(14)

After linearizing such a system about $U_0,V_0,P_0$ and introducing the proximal formulation, for an appropriate non-negative real constant K,we get

$$\begin{array}{ccc}& & \nu \left(\frac{u_{n+1}-2u_n+u_{n-1}}{d^2}+\frac{{\partial }^2{u}_n}{\partial y^2}\right)-{\left(U_0\right)}_n\frac{(u_n-{\left(U_0\right)}_{n-1})}{d}-{\left(V_0\right)}_n\frac{\partial u_n}{\partial y}\hfill \\ & & -\frac{{\left(P_0\right)}_n-{\left(P_0\right)}_{n-1}}{d}-K{u}_n+K{\left(U_0\right)}_n=0,\hfill \end{array}$$

(15)

$$\begin{array}{ccc}& & \nu \left(\frac{v_{n+1}-2v_n+v_{n-1}}{d^2}+\frac{{\partial }^2{v}_n}{\partial y^2}\right)-{\left(U_0\right)}_n\frac{(v_n-{\left(V_0\right)}_{n-1})}{d}-{\left(V_0\right)}_n\frac{\partial v_n}{\partial y}\hfill \\ & & -\frac{\partial {\left(P_0\right)}_n}{\partial y}-K{v}_n+K{\left(V_0\right)}_n=0,\hfill \end{array}$$

(16)

$$\begin{array}{ccc}& & \left(\frac{P_{n+1}-2P_n+P_{n-1}}{d^2}+\frac{{\partial }^2{P}_n}{\partial y^2}\right)+(u_{n+1}-{\left(U_0\right)}_n)\frac{({\left(u\right)}_{n+1}-{\left(U_0\right)}_n)}{d^2}+{\left(\frac{\partial {\left(V_0\right)}_n}{\partial y}\right)}^2\hfill \\ & & +2\frac{\partial {\left(U_0\right)}_n}{\partial y}\left(\frac{v_{n+1}-{\left(V_0\right)}_n}{d}\right)-K{P}_n+K{\left(P_0\right)}_n=0.\hfill \end{array}$$

(17)

At this point denoting $\nu =e_1,$ we define

$${\left(T_1\right)}_n=\left(-{\left(U_0\right)}_n\frac{(u_n-{\left(U_0\right)}_{n-1})}{d}-{\left(V_0\right)}_n\frac{\partial u_n}{\partial y}-\frac{{\left(P_0\right)}_n-{\left(P_0\right)}_{n-1}}{d}\right)\frac{d^2}{e_1}+\frac{{\partial }^2{u}_n}{\partial y^2}{d}^2,$$

$${\left(T_2\right)}_n=\left(-{\left(U_0\right)}_n\frac{(v_n-{\left(V_0\right)}_{n-1})}{d}-{\left(V_0\right)}_n\frac{\partial v_n}{\partial y}-\frac{\partial {\left(P_0\right)}_n}{\partial y}\right)\frac{d^2}{e_1}+\frac{{\partial }^2{v}_n}{\partial y^2}{d}^2,$$

and

$$\begin{array}{ccc}\hfill {\left(T_3\right)}_n& =& (u_{n+1}-{\left(U_0\right)}_n)\frac{({\left(u\right)}_{n+1}-{\left(U_0\right)}_n)}{d^2}{d}^2+{\left(\frac{\partial {\left(V_0\right)}_n}{\partial y}\right)}^2{d}^2\hfill \\ & & +2\frac{\partial {\left(U_0\right)}_n}{\partial y}\left(\frac{v_{n+1}-{\left(V_0\right)}_n}{d}\right)d^2+\frac{{\partial }^2{P}_n}{\partial y^2}{d}^2.\hfill \end{array}$$

(18)

Therefore, we may write

$$u_{n+1}-2u_n+u_{n-1}-K{u}_n\frac{d^2}{e_1}+{\left(T_1\right)}_n+{\left(f_1\right)}_n=0,$$

where

$${\left(f_1\right)}_n=K{\left(U_0\right)}_n\frac{d^2}{e_1},$$

$\forall n\in \{1,\cdots ,N-1\}.$

In particular for $n=1$, we obtain

$$u_2-2u_1+u_0-K{u}_1\frac{d^2}{e_1}+{\left(T_1\right)}_1+{\left(f_1\right)}_1=0,$$

so that

$$u_1=a_1{u}_2+b_1{u}_0+c_1{\left(T_1\right)}_1+{\left(h_1\right)}_1+{\left(E_r\right)}_1,$$

where

$$a_1={\left(2+K\frac{d^2}{e_1}\right)}^{-1},$$

$$b_1=a_1$$

$$c_1=a_1$$

$${\left(h_1\right)}_1=a_1{\left(f_1\right)}_1,$$

$${\left(E_r\right)}_1=0.$$

Similarly, for $n=2$ we get

$$u_3-2u_2+u_1-K{u}_2\frac{d^2}{e_2}+{\left(T_1\right)}_2+{\left(f_1\right)}_2=0,$$

so that

$$u_2=a_2{u}_3+b_2{u}_0+c_2{\left(T_1\right)}_2+{\left(h_1\right)}_2+{\left(E_r\right)}_2,$$

where

$$a_2={\left(2+K\frac{d^2}{e_1}-a_1\right)}^{-1},$$

$$b_2=a_2{b}_1$$

$$c_2=a_2(c_1+1)$$

$${\left(h_1\right)}_2=a_2({\left(h_1\right)}_1+{\left(f_1\right)}_2),$$

$${\left(E_1\right)}_2=a_2\left(c_1({\left(T_1\right)}_1-{\left(T_1\right)}_2)\right).$$

Reasoning inductively, having

$$u_{n-1}=a_{n-1}{u}_n+b_{n-1}{u}_0+c_{n-1}{\left(T_1\right)}_{n-1}+{\left(h_1\right)}_{n-1}+{\left(E_r\right)}_{n-1},$$

we obtain

$$u_n=a_n{u}_{n+1}+b_n{u}_0+c_n{\left(T_1\right)}_n+{\left(h_1\right)}_n+{\left(E_r\right)}_n,$$

(19)

where

$$a_n={\left(2+K\frac{d^2}{e_1}-a_{n-1}\right)}^{-1},$$

$$b_n=a_n{b}_{n-1}$$

$$c_n=a_n(c_{n-1}+1)$$

$${\left(h_1\right)}_n=a_n({\left(h_1\right)}_{n-1}+{\left(f_1\right)}_n),$$

$$\left(E_r\right)=a_n({\left(E_r\right)}_{n-1}+c_{n-1}({\left(T_1\right)}_{n-1}-{\left(T_1\right)}_n)),$$

$\forall n\in \{1,\cdots ,N-1\}.$

Observe now that $n=N-1$ we have $u_{N-1}=u_N,$ so that

$$\begin{array}{ccc}\hfill u_{N-1}& \approx & a_{N-1}{u}_{N-1}+b_{N-1}{u}_0+c_{N-1}{\left(T_1\right)}_{N-1}+{\left(h_1\right)}_{N-1}\hfill \\ & & a_{N-1}{u}_{N-1}+b_{N-1}{u}_0\hfill \\ & & +c_{N-1}\frac{{\partial }^2{u}_{N-1}}{\partial y^2}{d}^2\hfill \\ & & +c_{N-1}\left(-{\left(U_0\right)}_n\frac{(u_n-{\left(U_0\right)}_{n-1})}{d}-{\left(V_0\right)}_n\frac{\partial u_n}{\partial y}-\frac{{\left(P_0\right)}_n-{\left(P_0\right)}_{n-1}}{d}\right)\frac{d^2}{e_1}\hfill \\ & & +{\left(h_1\right)}_{N-1}\hfill \end{array}$$

(20)

This last equation is a second order ODE in $u_{N-1}$ which must be solved with the boundary conditions

$$u_{N-1}\left(0\right)=u_{N-1}\left(1\right)=0.$$

Summarizing we have obtained $u_{N-1}.$

Similarly, we may obtain $v_{N-1}$ and $P_{N-1}.$

Having $u_{N-1}$ we may obtain $u_{N-2}$ with $n=N-2$ in equation (19) (neglecting ${\left(E_r\right)}_{N-2}.$)

Similarly, we may obtain $v_{N-2}$ and $P_{N-2}.$

Having $u_{N-2}$ we may obtain $u_{N-3}$ with $n=N-3$ in equation (19) (neglecting ${\left(E_r\right)}_{N-3}.$)

Similarly, we may obtain $v_{N-3}$ and $P_{N-3}.$

And so on up to obtaining $u_1,v_1$ and $P_1$.

The next step is to replace $\{{\left(U_0\right)}_n,{\left(V_0\right)}_n,{\left(P_0\right)}_n\}$ by $\{u_n,v_n,P_n\}$ and repeat the process until an appropriate convergence criterion is satisfied.

Here we present a concerning software in MATLAB based in this last algorithm (with small changes and differences where we have set $K=5$ and $\nu =0.047$).

******************************

$clearall$

$m8=500;$

$d=1/m8;$

$m9=140;$

$d1=1/m9;$

$e1=0.05;$

$K=5.0;$

$m2=zeros(m9-1,m9-1);$

$fori=2:m9-2$

$m2(i,i)=-2.0;$

$m2(i,i+1)=1.0;$

$m2(i,i-1)=1.0;$

$end;$

$m2(1,1)=-1.0;$

$m2(1,2)=1.0;$

$m2(m9-1,m9-1)=-1.0;$

$m2(m9-1,m9-2)=1.0;$

$m22=zeros(m9-1,m9-1);$

$fori=2:m9-2$

$m22(i,i)=-2.0;$

$m22(i,i+1)=1.0;$

$m22(i,i-1)=1.0;$

$end;$

$m22(1,1)=-2.0;$

$m22(1,2)=1.0;$

$m22(m9-1,m9-1)=-2.0;$

$m22(m9-1,m9-2)=1.0;$

$m1a=zeros(m9-1,m9-1);$

$m1b=zeros(m9-1,m9-1);$

$fori=1:m9-2$

$m1a(i,i)=-1.0;$

$m1a(i,i+1)=1.0;$

$end;$

$m1a(m9-1,m9-1)=-1.0;$

$fori=2:m9-1$

$m1b(i,i)=1.0;$

$m1b(i,i-1)=-1.0;$

$end;$

$m1b(1,1)=1.0;$

$m1=(m1a+m1b)/2;$

$Id=eye(m9-1);$

$a\left(1\right)=1/(2+K\ast d^2/e1);$

$b\left(1\right)=1/(2+K\ast d^2/e1);$

$c\left(1\right)=1/(2+K\ast d^2/e1);$

$fori=2:m8-1$

$a\left(i\right)=1/(2-a(i-1)+K\ast d^2/e1);$

$b\left(i\right)=a\left(i\right)\ast b(i-1);$

$c\left(i\right)=\left(c\right(i-1)+1)\ast a\left(i\right);$

$end;$

$fori=1:m9-1$

$u5(i,1)=0.55\ast i\ast d1\ast (1-i\ast d1);$

$end;$

$uo=u5;$

$vo=zeros(m9-1,1);$

$po=0.15\ast ones(m9-1,1);$

$fori=1:m8-1$

$Uo(:,i)=0.25\ast ones(m9-1,1);$

$Vo(:,i)=0.05\ast ones(m9-1,1);$

$Po(:,i)=0.05\ast ones(m9-1,1);$

$end;$

$fork7=1:1$

$e1=e1\ast .94;$

$b14=1.0;$

$k1=1;$

$k1max=500;$

$\mathrm{while}(b14>{10}^{-3.5})\mathrm{and}(k1<500)$

$k1=k1+1;$

$a\left(1\right)=1/(2+K\ast d^2/e1);$

$b\left(1\right)=a\left(1\right);$

$c1(:,1)=a\left(1\right)\ast K\ast Uo(:,1)\ast d^2/e1;$

$c2(:,1)=a\left(1\right)\ast K\ast Vo(:,1)\ast d^2/e1;$

$c3(:,1)=a\left(1\right)\ast (K\ast Po(:,1)\ast d^2/e1);$

$fori=2:m8-1$

$a\left(i\right)=1/(2+K\ast d^2/e1-a(i-1));$

$b\left(i\right)=a\left(i\right)\ast \left(b\right(i-1\left)\right);$

$c1(:,i)=a\left(i\right)\ast (c1(:,i-1)+K\ast Uo(:,i)\ast d^2/e1);$

$c2(:,i)=a\left(i\right)\ast (c2(:,i-1)+K\ast Vo(:,i)\ast d^2/e1);$

$c3(:,i)=a\left(i\right)\ast (c3(:,i-1)+K\ast Po(:,i)\ast d^2/e1);$

$end;$

$i=1;$

$M50=(Id-a(m8-1)\ast Id-c(m8-1)\ast m22/d{1}^2\ast d^2);$

$z1=b(m8-1)\ast uo+c1(:,m8-i)$

$z1=z1+c(m8-1)\ast (-Vo(:,m8-i).\ast (m1\ast Uo(:,m8-i)))/d1\ast d^2/e1;$

$M60=(Id-a(m8-1)\ast Id-c(m8-1)\ast m22/d{1}^2\ast d^2);$

$z2=b(m8-1)\ast vo+c2(:,m8-i)$

$z2=z2+c(m8-1)\ast (-Vo(:,m8-i)).\ast (m1\ast Vo(:,m8-i))/d1\ast d^2/e1$;

$M70=(Id-a(m8-1)\ast Id-c(m8-1)\ast m2/d{1}^2\ast d^2);$$z3=b(m8-1)\ast po$

$z3=z3+c(m8-1)\ast ((m1/d1\ast Vo(:,m8-i)).\ast (m1/d1\ast Vo(:,m8-i))\ast d^2);$

$z3=z3+c3(:,m8-i)$;

$U(:,m8-1)=inv\left(M50\right)\ast z1;$

$V(:,m8-1)=inv\left(M60\right)\ast z2;$

$P(:,m8-1)=inv\left(M70\right)\ast z3;$

$fori=2:m8-1$

$M50=(Id-c(m8-i)\ast m22/d{1}^2\ast d^2);$

$z1=b(m8-i)\ast uo+a(m8-i)\ast Uo(:,m8-i+1)$

$z1=z1+c(m8-i)\ast (-Uo(:,m8-i).\ast (Uo(:,m8-i+1)-Uo(:,m8-i))\ast d/e1)$

$z1=z1-c(m8-i)\left(Po\right(:,m8-i+1)-Po(:,m8-i\left)\right)\ast d/e1);$

$z1=z1+c1(:,m8-i)+c(m8-i)\ast (-Vo(:,m8-i).\ast (m1\ast Uo(:,m8-i))/d1\ast d^2)/e1;$

$M60=(Id-c(m8-i)\ast m22/d{1}^2\ast d^2);$

$z2=b(m8-i)\ast vo+a(m8-i)\ast Vo(:,m8-i+1)$

$z2=z2+c(m8-i)\ast (-(Uo(:,m8-i).\ast \left(Vo\right(:,m8-i+1)-Vo(:,m8-i\left)\right)\ast d/e1)$

$z2=z2++c(m8-i)Vo(:,m8-i)).\ast (m1\ast Vo(:,m8-i)/d1\ast d^2)/e1)$

$z2=z2-c(m8-i)\ast (m1\ast Po(:,m8-i)/d1\ast d^2)/e1;$

$z2=z2+c2(:,m8-i);$

$M70=(Id-c(m8-i)\ast m2/d{1}^2\ast d^2);$

$z3=b(m8-i)\ast po+a(m8-i)\ast Po(:,m8-i+1)$

$z3=z3$

$+c(m8-i)\ast \left(\right(Uo(:,m8-i+1)-Uo(:,m8-i)).\ast (Uo(:,m8-i)-Uo(:,m8-i)\left)\right);$

$z3=z3+c(m8-i)(m1/d1\ast Vo(:,m8-i)).\ast (m1/d1\ast Vo(:,m8-i))\ast d^2);$

$z3=z3$

$+2\ast (m1/d1\ast Uo(:,m8-i\left)\right).\ast \left(Vo\right(:,m8-i+1)-Vo(:,m8-i\left)\right)\ast d+c3(:,m8-i);$

$U(:,m8-i)=inv\left(M50\right)\ast z1;$

$V(:,m8-i)=inv\left(M60\right)\ast z2;$

$P(:,m8-i)=inv\left(M70\right)\ast z3;$

$end;$

$b14=max\left(max\right(abs(U-Uo)\left)\right);$

$b14$

$Uo=U;$

$Vo=V;$

$Po=P;$

$k1$

$U(m9/2,10)$

$end;$

$k7$

$end;$

$fori=1:m9-1$

$y\left(i\right)=i\ast d1;$

$end;$

$fori=1:m8-1$

$x\left(i\right)=i\ast d;$

$end;$

$mesh(x,y,U);$

*********************************

For the field of velocities U, V and the pressure field P, please see Figure 1, Figure 2 and Figure 3, respectively.

4. A software in MATHEMATICA related to the previous algorithm

In this section we develop the solution for the Navier-Stokes system through the generalized method of lines, similarly as the results presented in [3], but now in a Navier-Stokes system context.

We present a software in MATHEMATICA for $N=10$ lines for the case in which

$$\begin{array}{c}\left\{\begin{array}{cc}\nu {\nabla }^{2}u-u{\partial }_{x}u-v{\partial }_{y}u-{\partial }_{x}P=0,\hfill & \mathrm{in}\Omega ,\hfill \\ \\ \nu {\nabla }^{2}v-u{\partial }_{x}v-v{\partial }_{y}v-{\partial }_{y}P=0,\hfill & \mathrm{in}\Omega ,\hfill \\ \\ {\nabla }^{2}P+{\left({\partial }_{x}u\right)}^2+{\left({\partial }_{y}v\right)}^2+2\left({\partial }_{y}u\right)\left({\partial }_{x}v\right)=0,\hfill & \mathrm{in}\Omega ,\hfill \end{array}\right.\end{array}$$

(21)

We consider it in polar coordinates, with $\nu =e_1=0.1$, and with

$$\Omega =\{(r,\theta )\in {\mathbb{R}}^2:1\le r\le 2,0\le \theta \le 2\pi \},$$

$$\partial {\Omega }_1=\{(1,\theta )\in {\mathbb{R}}^2:0\le \theta \le 2\pi \},$$

and

$$\partial {\Omega }_2=\{(2,\theta )\in {\mathbb{R}}^2:0\le \theta \le 2\pi \}.$$

The boundary conditions are

$$u=v=0,P=0.15\mathrm{on}\partial {\Omega }_1,$$

$$u=u_f\left[x\right],v=0,P=0.12\mathrm{on}\partial {\Omega }_2.$$

From now and on, x stands for $\theta$.

We remark some changes have been made, concerning the original conception, in order to make it suitable through the software MATHEMATICA for such a Navier-Stokes system.

We highlight, as $K>0$ is larger, the related approximation is of a better quality. However if $K>0$ is too much large, the converging process gets slower.

Here the concerning software.

************************************************

• $m8=10;$
  $Clear[t3,t4];$
  $d=1.0/m8;$
  $K=4.0;$
  $e1=0.1;$
  $Uoo\left[x_{-}\right]=0.0;$
  $Voo\left[x_{-}\right]=0.0;$
  $Poo\left[x_{-}\right]=0.15;$
• $For[i=1,i<m8+1,i++,$
  $uo\left[i\right]=0.05;$
  $vo\left[i\right]=0.05;$
  $Po\left[i\right]=0.05];$
• $For[k=1,k<80,k++,$ (here we have fixed the number of iterations)
  $Print\left[k\right];$
  $a\left[1\right]=1/(2.0+K\ast d^2/e1);$
  $b\left[1\right]=a\left[1\right];$
  $b11\left[1\right]=a\left[1\right];$
  $c1\left[1\right]=a\left[1\right]\ast (K\ast uo\left[1\right])\ast d^2/e1;$
  $c2\left[1\right]=a\left[1\right]\ast (K\ast vo\left[1\right])\ast d^2/e1;$
  $c3\left[1\right]=a\left[1\right]\ast (K\ast Po\left[1\right]+P1)\ast d^2/e1;$
• $For[i=2,i<m8,i++,$
  $a\left[i\right]=1/(2.0+K\ast d^2/e1-a[i-1]);$
  $b\left[i\right]=a\left[i\right]\ast \left(b\right[i-1]+1);$
  $b11\left[i\right]=a\left[i\right]\ast b11[i-1];$
  $c1\left[i\right]=a\left[i\right]\ast (c1[i-1]+(K\ast uo\left[i\right])\ast d^2/e1);$
  $c2\left[i\right]=a\left[i\right]\ast (c2[i-1]+(K\ast vo\left[i\right])\ast d^2/e1);$
  $c3\left[i\right]=a\left[i\right]\ast (c3[i-1]+(K\ast Po\left[i\right])\ast d^2/e1)];$
  $u\left[m8\right]=uf\left[x\right]\ast t3;v\left[m8\right]=vf\left[x\right]\ast t3;P\left[m8\right]=0.12;d1=1.0;$
• $For[i=1,i<m8,i++,$
  $Print\left[i\right];$
  $t[m8-i]=1.0+(m8-i)\ast d;$
  $Dxu=\left(uo\right[m8-i+1]-uo[m8-i\left]\right)/d\ast f1\left[x\right]\ast t4-D\left[uo\right[m8-i],x]\ast f2\left[x\right]/t[m8-i]\ast t4;$
  $Dyu=\left(uo\right[m8-i+1]-uo[m8-i\left]\right)/d\ast f2\left[x\right]\ast t4+D\left[uo\right[m8-i],x]\ast f1\left[x\right]/t[m8-i]\ast t4;$
  $Dxv=\left(vo\right[m8-i+1]-vo[m8-i\left]\right)/d\ast f1\left[x\right]\ast t4-D\left[vo\right[m8-i],x]\ast f2\left[x\right]/t[m8-i]\ast t4;$
  $Dyv=\left(vo\right[m8-i+1]-vo[m8-i\left]\right)/d\ast f2\left[x\right]\ast t4+D\left[vo\right[m8-i],x]\ast f1\left[x\right]/t[m8-i]\ast t4;$
  $DxP=\left(Po\right[m8-i+1]-Po[m8-i\left]\right)/d\ast f1\left[x\right]\ast t4-D\left[Po\right[m8-i],x]\ast f2\left[x\right]/t[m8-i]\ast t4;$
  $DyP=\left(Po\right[m8-i+1]-Po[m8-i\left]\right)/d\ast f2\left[x\right]\ast t4+D\left[Po\right[m8-i],x]\ast f1\left[x\right]/t[m8-i]\ast t4;$
  $T1=-(u[m8-i+1]\ast Dxu+v[m8-i+1]\ast Dyu+DxP)\ast d^2/e1$
  $+(uo[m8-i+1]-uo[m8-i])/d/t[m8-i]\ast d^2+D[uo[m8-i+1],\{x,2\}]/t{[m8-i]}^2\ast d^2;$
  $T2=-(u[m8-i+1]\ast Dxv+v[m8-i+1]\ast Dyv+DyP)\ast d^2/e1+(vo[m8-i+1]-vo[m8-i])/d/t[m8-i]\ast d^2$
  $+D[vo[m8-i+1],\{x,2\}]/t{[m8-i]}^2\ast d^2;$
  $T3=(Dx{u}^2+Dy{v}^2+2\ast Dyu\ast Dxv)\ast d^2$
  $+(Po[m8-i+1]-Po[m8-i])/d/t[m8-i]\ast d^2+D[Po[m8-i+1],\{x,2\}]/t{[m8-i]}^2\ast d^2;$
• $A1=a[m8-i]\ast u[m8-i+1]+b[m8-i]\ast T1+c1[m8-i];$
  $A1=Expand\left[A1\right];$
  $A1=Series[A1,\{t4,0,1\},\{t3,0,2\},\{uf\left[x\right],0,2\},\{u{f}^{\prime }\left[x\right],0,1\},\{u{f}^{\prime \prime }\left[x\right],0,1\},$
  $\{u{f}^{\prime \prime \prime }\left[x\right],0,0\},\{u{f}^{\prime \prime \prime \prime }\left[x\right],0,0\},\{vf\left[x\right],0,1\},\{v{f}^{\prime }\left[x\right],0,1\},\{v{f}^{\prime \prime }\left[x\right],0,1\},$
  $\{v{f}^{\prime \prime \prime }\left[x\right],0,0\},\{v{f}^{\prime \prime \prime \prime }\left[x\right],0,0\},$
  $\{f1\left[x\right],0,1\},\{f2\left[x\right],0,1\},\{f{1}^{\prime }\left[x\right],0,0\},\{f{2}^{\prime }\left[x\right],0,0\},\{f{1}^{\prime \prime }\left[x\right],0,0\},\{f{2}^{\prime \prime }\left[x\right],0,0\}];$
  $A1=Normal\left[A1\right];$
  $u[m8-i]=Expand\left[A1\right];$
• $A2=a[m8-i]\ast v[m8-i+1]+b[m8-i]\ast T2+c2[m8-i];$
  $A2=Expand\left[A2\right];$
  $A2=Series[A2,\{t4,0,1\},\{t3,0,2\},\{uf\left[x\right],0,1\},\{u{f}^{\prime }\left[x\right],0,1\},\{u{f}^{\prime \prime }\left[x\right],0,1\},$
  $\{u{f}^{\prime \prime \prime }\left[x\right],0,0\},\{u{f}^{\prime \prime \prime \prime }\left[x\right],0,0\},\{vf\left[x\right],0,2\},\{v{f}^{\prime }\left[x\right],0,1\},\{v{f}^{\prime \prime }\left[x\right],0,1\},$
  $\{v{f}^{\prime \prime \prime }\left[x\right],0,0\},\{v{f}^{\prime \prime \prime \prime }\left[x\right],0,0\},$
  $\{f1\left[x\right],0,1\},\{f2\left[x\right],0,1\},\{f{1}^{\prime }\left[x\right],0,0\},\{f{2}^{\prime }\left[x\right],0,0\},$
  $\{f{1}^{\prime \prime }\left[x\right],0,0\},\{f{2}^{\prime \prime }\left[x\right],0,0\}];$
  $A2=Normal\left[A2\right];$
  $v[m8-i]=Expand\left[A2\right];$
• $A3=a[m8-i]\ast P[m8-i+1]+b[m8-i]\ast T3+c3[m8-i]+b11[m8-i]\ast Poo\left[x\right];$
  $A3=Expand\left[A3\right];$
  $A3=Series[A3,\{t4,0,2\},\{t3,0,2\},\{uf\left[x\right],0,1\},\{u{f}^{\prime }\left[x\right],0,1\},\{u{f}^{\prime \prime }\left[x\right],0,0\},\{u{f}^{\prime \prime \prime }\left[x\right],0,0\},$
  $\{u{f}^{\prime \prime \prime \prime }\left[x\right],0,0\},\{vf\left[x\right],0,1\},\{v{f}^{\prime }\left[x\right],0,1\},\{v{f}^{\prime \prime }\left[x\right],0,0\},\{v{f}^{\prime \prime \prime }\left[x\right],0,0\},$
  $\{v{f}^{\prime \prime \prime \prime }\left[x\right],0,0\},\{f1\left[x\right],0,1\},\{f2\left[x\right],0,1\},$
  $\{f{1}^{\prime }\left[x\right],0,0\},\{f{2}^{\prime }\left[x\right],0,0\},\{f{1}^{\prime \prime }\left[x\right],0,0\},\{f{2}^{\prime \prime }\left[x\right],0,0\}];$
  $A3=Normal\left[A3\right];$
  $P[m8-i]=Expand\left[A3\right]];$
• $For[i=1,i<m8+1,i++,$
  $uo\left[i\right]=u\left[i\right];$
  $vo\left[i\right]=v\left[i\right];$
  $Po\left[i\right]=P\left[i\right]];d1=1.0;$
• $Print\left[Expand\right[U[m8/2]\left]\right]]$
• $For[i=1,i<m_8,i++,$
  $Print["u[",i,"]=",u[i\left]\right[x\left]\right]]$

Here we present the related line expressions obtained for the lines $n=1$, $n=5$ and n=9 of a total of $N=10$ lines.

• $Linen=1$

  $$\begin{array}{ccc}\hfill u\left[1\right]& =& 1.58658\ast {10}^{-9}+0.019216f_1\left[x\right]+3.68259\ast {10}^{-11}{f}_2\left[x\right]+0.132814u_f\left[x\right]\hfill \\ & & +2.28545\ast {10}^{-8}{f}_1\left[x\right]u_f\left[x\right]-1.69037\ast {10}^{-8}{f}_2\left[x\right]u_f\left[x\right]-0.263288f_1\left[x\right]u_f{\left[x\right]}^2\hfill \\ & & -6.02845\ast {10}^{-9}{f}_1\left[x\right]\left(u_f^{\prime }\right)\left[x\right]+6.02845\ast {10}^{-9}{f}_2\left[x\right]\left(u_f^{\prime }\right)\left[x\right]\hfill \\ & & +0.104284f_2\left[x\right]u_f\left[x\right]\left(u_f^{\prime }\right])\left[x\right]+0.0127544\left(u_f^{\prime \prime }\right)\left[x\right]\hfill \\ & & +9.39885\ast {10}^{-9}{f}_1\left[x\right]\left(u_f^{\prime \prime }\right)\left[x\right]-5.26303\ast {10}^{-9}{f}_2\left[x\right]\left(u_f^{\prime \prime }\right)\left[x\right]\hfill \\ & & -0.0340276f_1\left[x\right]u_f\left[x\right]\left(u_f^{\prime \prime }\right)\left[x\right]+0.0239544f_2\left[x\right]\left(u_f^{\prime }\right)\left[x\right]\left(u_f^{\prime \prime }\right)\left[x\right]\hfill \end{array}$$

  (22)
• $Linen=5$

  $$\begin{array}{ccc}\hfill u\left[5\right]& =& 4.25933\ast {10}^{-9}+0.0436523f_1\left[x\right]+9.88625\ast {10}^{-11}{f}_2\left[x\right]+0.572969u_f\left[x\right]\hfill \\ & & +6.87985\ast {10}^{-8}{f}_1\left[x\right]u_f\left[x\right]-4.40534\ast {10}^{-8}{f}_2\left[x\right]uf\left[x\right]-0.765222f_1\left[x\right]u_f{\left[x\right]}^2\hfill \\ & & -1.61319\ast {10}^{-8}{f}_1\left[x\right]\left(u_f^{\prime }\right)\left[x\right]+1.61319\ast {10}^{-8}{f}_2\left[x\right]\left(u_f^{\prime }\right)\left[x\right]\hfill \\ & & +0.363471f_2\left[x\right]u_f\left[x\right]\left(u_f^{\prime }\right)\left[x\right]+0.0333685\left(u_f^{\prime \prime }\right)\left[x\right]\hfill \\ & & +2.39576\ast {10}^{-8}{f}_1\left[x\right]\left(u_f^{\prime \prime }\right)\left[x\right]-1.27491\ast {10}^{-8}{f}_2\left[x\right]\left(u_f^{\prime \prime }\right)\left[x\right]\hfill \\ & & -0.0342544f_1\left[x\right]u_f\left[x\right]\left(u_f^{\prime \prime }\right)\left[x\right]+0.0509889f_2\left[x\right]\left(u_f^{\prime }\right)\left[x\right]\left(u_f^{\prime \prime }\right)\left[x\right]\hfill \end{array}$$

  (23)
• $Linen=9$

  $$\begin{array}{ccc}\hfill u\left[9\right]& =& 1.15848\ast {10}^{-9}+0.0136828f_1\left[x\right]+2.68892\ast {10}^{-11}{f}_2\left[x\right]+0.922534u_f\left[x\right]\hfill \\ & & +2.16498\ast {10}^{-8}{f}_1\left[x\right]u_f\left[x\right]-1.16065\ast {10}^{-8}{f}_2\left[x\right]u_f\left[x\right]-0.278966f_1\left[x\right]u_f{\left[x\right]}^2\hfill \\ & & -4.25263\ast {10}^{-9}{f}_1\left[x\right]\left(u{f}^{\prime }\right)\left[x\right]+4.25263\ast {10}^{-9}{f}_2\left[x\right]\left(u_f^{\prime }\right)\left[x\right]\hfill \\ & & +0.154642f_2\left[x\right]u_f\left[x\right]\left(u_f^{\prime }\right)\left[x\right]+0.0110114\left(u_f^{\prime \prime }\right)\left[x\right]\hfill \\ & & +6.13523\ast {10}^{-9}{f}_1\left[x\right]\left(u_f^{\prime \prime }\right)\left[x\right]-3.23081\ast {10}^{-9}{f}_2\left[x\right]\left(u_f^{\prime \prime }\right)\left[x\right]\hfill \\ & & +0.0146222f_1\left[x\right]u_f\left[x\right]\left(u_f^{\prime \prime }\right)\left[x\right]+0.0090088f_2\left[x\right]\left(u_f^{\prime }\right)\left[x\right]\left(u_f^{\prime \prime }\right)\left[x\right]\hfill \end{array}$$

  (24)

5. The software and numerical results for a more specific example

In this section we present numerical results for the same Navier-Stokes system and domain as in the previous one, but now with different boundary conditions.

In this example, we set $\nu =0.1$ and the boundary conditions are

$$u=v=0,P=0.15\mathrm{on}\partial {\Omega }_1,$$

$$u=-1.0sin\left[x\right],v=1.0cos\left[x\right],P=0.12\mathrm{on}\partial {\Omega }_2.$$

Here the concerning software:

************************************************

• $m8=10;$
  $Clear[t3,t4];$
  $d=1.0/m8;$
  $K=4.0;$
  $e1=0.1;$
  $Uoo\left[x_{-}\right]=0.0;$
  $Voo\left[x_{-}\right]=0.0;$
  $Poo\left[x_{-}\right]=0.15;$
• $For[i=1,i<m8+1,i++,$
  $uo\left[i\right]=0.05;$
  $vo\left[i\right]=0.05;$
  $Po\left[i\right]=0.05];$
  $f1\left[x_{-}\right]=Cos\left[x\right];$
  $f2\left[x_{-}\right]=Sin\left[x\right];$
• $For[k=1,k<80,k++,$ (here we have fixed the number of iterations)
  $Print\left[k\right];$
  $a\left[1\right]=1/(2.0+K\ast d^2/e1);$
  $b\left[1\right]=a\left[1\right];$
  $b11\left[1\right]=a\left[1\right];$
  $c1\left[1\right]=a\left[1\right]\ast (K\ast uo\left[1\right])\ast d^2/e1;$
  $c2\left[1\right]=a\left[1\right]\ast (K\ast vo\left[1\right])\ast d^2/e1;$
  $c3\left[1\right]=a\left[1\right]\ast (K\ast Po\left[1\right]+P1)\ast d^2/e1;$
• $For[i=2,i<m8,i++,$
  $a\left[i\right]=1/(2.0+K\ast d^2/e1-a[i-1]);$
  $b\left[i\right]=a\left[i\right]\ast \left(b\right[i-1]+1);$
  $b11\left[i\right]=a\left[i\right]\ast b11[i-1];$
  $c1\left[i\right]=a\left[i\right]\ast (c1[i-1]+(K\ast uo\left[i\right])\ast d^2/e1);$
  $c2\left[i\right]=a\left[i\right]\ast (c2[i-1]+(K\ast vo\left[i\right])\ast d^2/e1);$
  $c3\left[i\right]=a\left[i\right]\ast (c3[i-1]+(K\ast Po\left[i\right])\ast d^2/e1)];$
  $uf\left[x_{-}\right]=-1.0\ast Sin\left[x\right];$
  $vf\left[x\right]=1.0\ast Cos\left[x\right];$
  $u\left[m8\right]=uf\left[x\right]\ast t3;$
  $v\left[m8\right]=vf\left[x\right]\ast t3;$
  $P\left[m8\right]=0.12;$
• $For[i=1,i<m8,i++,$
  $Print\left[i\right];$
  $t[m8-i]=1.0+(m8-i)\ast d;$
  $Dxu=\left(uo\right[m8-i+1]-uo[m8-i\left]\right)/d\ast f1\left[x\right]\ast t4-D\left[uo\right[m8-i],x]\ast f2\left[x\right]/t[m8-i]\ast t4;$
  $Dyu=\left(uo\right[m8-i+1]-uo[m8-i\left]\right)/d\ast f2\left[x\right]\ast t4+D\left[uo\right[m8-i],x]\ast f1\left[x\right]/t[m8-i]\ast t4;$
  $Dxv=\left(vo\right[m8-i+1]-vo[m8-i\left]\right)/d\ast f1\left[x\right]\ast t4-D\left[vo\right[m8-i],x]\ast f2\left[x\right]/t[m8-i]\ast t4;$
  $Dyv=\left(vo\right[m8-i+1]-vo[m8-i\left]\right)/d\ast f2\left[x\right]\ast t4+D\left[vo\right[m8-i],x]\ast f1\left[x\right]/t[m8-i]\ast t4;$
  $DxP=\left(Po\right[m8-i+1]-Po[m8-i\left]\right)/d\ast f1\left[x\right]\ast t4-D\left[Po\right[m8-i],x]\ast f2\left[x\right]/t[m8-i]\ast t4;$
  $DyP=\left(Po\right[m8-i+1]-Po[m8-i\left]\right)/d\ast f2\left[x\right]\ast t4+D\left[Po\right[m8-i],x]\ast f1\left[x\right]/t[m8-i]\ast t4;$
  $T1=-(u[m8-i+1]\ast Dxu+v[m8-i+1]\ast Dyu+DxP)\ast d^2/e1$
  $+(uo[m8-i+1]-uo[m8-i])/d/t[m8-i]\ast d^2$
  $+D[uo[m8-i+1],\{x,2\}]/t{[m8-i]}^2\ast d^2;$
  $T2=-(u[m8-i+1]\ast Dxv+v[m8-i+1]\ast Dyv+DyP)\ast d^2/e1$
  $+(vo[m8-i+1]-vo[m8-i])/d/t[m8-i]\ast d^2+D[vo[m8-i+1],\{x,2\}]/t{[m8-i]}^2\ast d^2;$
  $T3=(Dx{u}^2+Dy{v}^2+2\ast Dyu\ast Dxv)\ast d^2$
  $+(Po[m8-i+1]-Po[m8-i])/d/t[m8-i]\ast d^2+D[Po[m8-i+1],\{x,2\}]/t{[m8-i]}^2\ast d^2;$
  $A1=a[m8-i]\ast u[m8-i+1]+b[m8-i]\ast T1+c1[m8-i];$
  $A1=Expand\left[A1\right];$
  $A1=Series[A1,\{t4,0,1\},\{t3,0,2\},\{Sin\left[x\right],0,2\},\{Cos\left[x\right],0,2\left\}\right];$
  $A1=Normal\left[A1\right];$
  $u[m8-i]=Expand\left[A1\right];$
  $A2=a[m8-i]\ast v[m8-i+1]+b[m8-i]\ast T2+c2[m8-i];$
  $A2=Expand\left[A2\right];$
  $A2=Series[A2,\{t4,0,1\},\{t3,0,2\},\{Sin\left[x\right],0,2\},\{Cos\left[x\right],0,2\left\}\right];$
  $A2=Normal\left[A2\right];$
  $v[m8-i]=Expand\left[A2\right];$
  $A3=a[m8-i]\ast P[m8-i+1]+b[m8-i]\ast T3+c3[m8-i]+b11[m8-i]\ast Poo\left[x\right];$
  $A3=Expand\left[A3\right];$$A3=Series[A3,\{t4,0,2\},\{t3,0,2\},\{Sin\left[x\right],0,2\},\{Cos\left[x\right],0,2\left\}\right];$
  $A3=Normal\left[A3\right];$
  $P[m8-i]=Expand\left[A3\right]];$
• $For[i=1,i<m8+1,i++,$
  $uo\left[i\right]=u\left[i\right];$
  $vo\left[i\right]=v\left[i\right];$
  $Po\left[i\right]=P\left[i\right]];$
  $Print\left[Expand\right[P[m8/2]\left]\right]]$
• $For[i=1,i<m_8,i++,$
  $Print["u[",i,"]=",u[i\left]\right[x\left]\right]]$

Here the corresponding line expressions for $N=10$ lines

• $Linen=1$

  $$\begin{array}{ccc}\hfill u\left[1\right]& =& 1.445\ast {10}^{-10}+0.0183921Cos\left[x\right]+1.01021\ast {10}^{-9}Cos{\left[x\right]}^2-0.120676Sin\left[x\right]\hfill \\ & & +3.62358\ast {10}^{-10}Cos\left[x\right]Sin\left[x\right]+1.37257\ast {10}^{-9}Sin{\left[x\right]}^2\hfill \\ & & +0.0620534Cos\left[x\right]Sin{\left[x\right]}^2\hfill \end{array}$$

  (25)
• $Linen=2$

  $$\begin{array}{ccc}\hfill u\left[2\right]& =& 2.60007\ast {10}^{-10}+0.0307976Cos\left[x\right]+1.81088\ast {10}^{-9}Cos{\left[x\right]}^2-0.233061Sin\left[x\right]\hfill \\ & & +6.53242\ast {10}^{-10}Cos\left[x\right]Sin\left[x\right]+2.46412\ast {10}^{-9}Sin{\left[x\right]}^2\hfill \\ & & +0.123121Cos\left[x\right]Sin{\left[x\right]}^2\hfill \end{array}$$

  (26)
• $Linen=3$

  $$\begin{array}{ccc}\hfill u\left[3\right]& =& 3.40796\ast {10}^{-10}+0.0384482Cos\left[x\right]+2.36167\ast {10}^{-9}Cos{\left[x\right]}^2-0.339657Sin\left[x\right]\hfill \\ & & +8.5759\ast {10}^{-10}Cos\left[x\right]Sin\left[x\right]+3.21926\ast {10}^{-9}Sin{\left[x\right]}^2\hfill \\ & & +0.180891Cos\left[x\right]Sin{\left[x\right]}^2\hfill \end{array}$$

  (27)
• $Linen=4$

  $$\begin{array}{ccc}\hfill u\left[4\right]& =& 3.83612\ast {10}^{-10}+0.0420843Cos\left[x\right]+2.64262\ast {10}^{-9}Cos{\left[x\right]}^2-0.441913Sin\left[x\right]\hfill \\ & & +9.66336\ast {10}^{-10}Cos\left[x\right]Sin\left[x\right]+3.60895\ast {10}^{-9}Sin{\left[x\right]}^2\hfill \\ & & +0.230559Cos\left[x\right]Sin{\left[x\right]}^2\hfill \end{array}$$

  (28)
• $Linen=5$

  $$\begin{array}{ccc}\hfill u\left[5\right]& =& 3.87923\ast {10}^{-10}+0.0421948Cos\left[x\right]+2.65457\ast {10}^{-9}Cos{\left[x\right]}^2-0.540729Sin\left[x\right]\hfill \\ & & +9.77606\ast {10}^{-10}Cos\left[x\right]Sin\left[x\right]+3.63217\ast {10}^{-9}Sin{\left[x\right]}^2\hfill \\ & & +0.266239Cos\left[x\right]Sin{\left[x\right]}^2\hfill \end{array}$$

  (29)
• $Linen=6$

  $$\begin{array}{ccc}\hfill u\left[6\right]& =& 3.56064\ast {10}^{-10}+0.0391334Cos\left[x\right]+2.419\ast {10}^{-9}Cos{\left[x\right]}^2-0.636718Sin\left[x\right]\hfill \\ & & +8.97185\ast {10}^{-10}Cos\left[x\right]Sin\left[x\right]+3.31618\ast {10}^{-9}Sin{\left[x\right]}^2\hfill \\ & & +0.281514Cos\left[x\right]Sin{\left[x\right]}^2\hfill \end{array}$$

  (30)
• $Linen=7$

  $$\begin{array}{ccc}\hfill u\left[7\right]& =& 2.93128\ast {10}^{-10}+0.033175Cos\left[x\right]+1.97614\ast {10}^{-9}Cos{\left[x\right]}^2-0.730328Sin\left[x\right]\hfill \\ & & +7.38127\ast {10}^{-10}Cos\left[x\right]Sin\left[x\right]+2.71426\ast {10}^{-9}Sin{\left[x\right]}^2\hfill \\ & & +0.269642Cos\left[x\right]Sin{\left[x\right]}^2\hfill \end{array}$$

  (31)
• $Linen=8$

  $$\begin{array}{ccc}\hfill u\left[8\right]& =& 2.0656\ast {10}^{-10}+0.0245445Cos\left[x\right]+1.38127\ast {10}^{-9}Cos{\left[x\right]}^2-0.821902Sin\left[x\right]\hfill \\ & & +5.19585\ast {10}^{-10}Cos\left[x\right]Sin\left[x\right]+1.90085\ast {10}^{-9}Sin{\left[x\right]}^2\hfill \\ & & +0.22362Cos\left[x\right]Sin{\left[x\right]}^2\hfill \end{array}$$

  (32)
• $Linen=9$

  $$\begin{array}{ccc}\hfill u\left[9\right]& =& 1.05509\ast {10}^{-10}+0.0134316Cos\left[x\right]+6.99554\ast {10}^{-10}Cos{\left[x\right]}^2-0.911718Sin\left[x\right]\hfill \\ & & +2.65019\ast {10}^{-10}Cos\left[x\right]Sin\left[x\right]+9.64573\ast {10}^{-10}Sin{\left[x\right]}^2\hfill \\ & & +0.13622Cos\left[x\right]Sin{\left[x\right]}^2\hfill \end{array}$$

  (33)

Here we present the related plots for the Lines $n=2$, $n=4$, $n=6$ and $n=8$ of a total of $N=10$ lines.

For each line we set $N=500$ nodes on the interval $[0,2\pi ]$, so that the units in x are $2\pi /500$, where again x stands for $\theta .$

For such lines, please see Figure 4, Figure 5, Figure 6 and Figure 7, respectively.

6. Numerical results through the original conception of the generalized method of lines for the Navier-Stokes system

In this section we develop the solution for the Navier-Stokes system through the generalized method of lines, as originally introduced in [4], with further developments in [5].

We present a software in MATHEMATICA for $N=10$ lines for the case in which

$$\begin{array}{c}\left\{\begin{array}{cc}\nu {\nabla }^{2}u-u{\partial }_{x}u-v{\partial }_{y}u-{\partial }_{x}P=0,\hfill & \mathrm{in}\Omega ,\hfill \\ \\ \nu {\nabla }^{2}v-u{\partial }_{x}v-v{\partial }_{y}v-{\partial }_{y}P=0,\hfill & \mathrm{in}\Omega ,\hfill \\ \\ {\nabla }^{2}P+{\left({\partial }_{x}u\right)}^2+{\left({\partial }_{y}v\right)}^2+2\left({\partial }_{y}u\right)\left({\partial }_{x}v\right)=0,\hfill & \mathrm{in}\Omega .\hfill \end{array}\right.\end{array}$$

(34)

Such a software refers to an algorithm presented in Chapter 27, in [2], in polar coordinates, with $\nu =1.0$, and with

$$\Omega =\{(r,\theta )\in {\mathbb{R}}^2:1\le r\le 2,0\le \theta \le 2\pi \},$$

$$\partial {\Omega }_1=\{(1,\theta )\in {\mathbb{R}}^2:0\le \theta \le 2\pi \},$$

and

$$\partial {\Omega }_2=\{(2,\theta )\in {\mathbb{R}}^2:0\le \theta \le 2\pi \}.$$

The boundary conditions are

$$u=v=0,P=0.15\mathrm{on}\partial {\Omega }_1,$$

$$u=-1.0sin\left(\theta \right),v=1.0cos\left(\theta \right),P=0.10\mathrm{on}\partial {\Omega }_2.$$

We remark some changes have been made, concerning the original conception, in order to make it suitable through the software MATHEMATICA for such a Navier-Stokes system.

We highlight the nature of this approximation is qualitative.

Here the concerning software in MATHEMATICA.

****************************************

• $m_8=10;$
• $Clear[z_1,z_2,z_3,u_1,u_2,P,b_1,b_2,b_3,a_1,a_2,a_3];$
• $Clear[P_f,t,a_{11},a_{12},a_{13},b_{11},b_{12},b_{13},t_3]$;
• $d=1.0/m8;$
• $e_1=1.0;$
• $a_1=0.0;$
• $a_2=0.0;$
• $a_3=0.15;$
• $For[i=1,i<m8,i++,$
  $Print\left[i\right];$
  $Clear[b_1,b_2,b_3,u_1,u_2,P];$
  $b_1\left[x_{-}\right]=u_1[i+1]\left[x\right];$
  $b_2\left[x_{-}\right]=u_2[i+1]\left[x\right];$
  $b_3\left[x_{-}\right]=P[i+1]\left[x\right];$
  $t\left[i\right]=1+i\ast d;$
  $du1x=Cos\left[x\right]\ast (b_1\left[x\right]-a_1)/d\ast t_3-1/t\left[i\right]\ast Sin\left[x\right]\ast D[b_1\left[x\right],x]\ast t_3;$
  $du1y=Sin\left[x\right]\ast (b_1\left[x\right]-a_1)/d\ast t_3+1/t\left[i\right]\ast Cos\left[x\right]\ast D[b_1\left[x\right],x]\ast t_3;$
  $du2x=Cos\left[x\right]\ast (b_2\left[x\right]-a_2)/d\ast t_3-1/t\left[i\right]\ast Sin\left[x\right]\ast D[b_2\left[x\right],x]\ast t_3;$
  $du2y=Sin\left[x\right]\ast (b_2\left[x\right]-a_2)/d\ast t_3+1/t\left[i\right]\ast Cos\left[x\right]\ast D[b_2\left[x\right],x]\ast t_3;$
  $dPx=Cos\left[x\right]\ast (b_3\left[x\right]-a_3)/d\ast t3-1/t\left[i\right]\ast Sin\left[x\right]\ast D[b_3\left[x\right],x]\ast t_3;$
  $dPy=Sin\left[x\right]\ast (b_3\left[x\right]-a_3)/d\ast t3+1/t\left[i\right]\ast Cos\left[x\right]\ast D[b_3\left[x\right],x]\ast t_3;$
• $For[k=1,k<6,k++,$ (in this example, we have fixed a relatively small number of iterations )
  $Print\left[k\right];$
  $z_1=(u_1[i+1]\left[x\right]+b_1\left[x\right]+a_1+1/t\left[i\right]\ast (b_1\left[x\right]-a_1)\ast d+1/t{\left[i\right]}^2\ast D[b_1\left[x\right],x,2]\ast d^2$
  $-(b_1\left[x\right]\ast du1x+b_2\left[x\right]\ast du1y)\ast d^2/e1-dPx\ast d^2/e_1)/3.0;$
  $z_2=(u_2[i+1]\left[x\right]+b_2\left[x\right]+a_2+1/t\left[i\right]\ast (b_2\left[x\right]-a_2)\ast d+1/t{\left[i\right]}^2\ast D[b_2\left[x\right],x,2]\ast d^2$
  $-(b_1\left[x\right]\ast du2x+b_2\left[x\right]\ast du2y)\ast d^2/e1-dPy\ast d^2/e_1)/3.0;$
  $z_3=(P[i+1]\left[x\right]+b_3\left[x\right]+a_3+1/t\left[i\right]\ast (b_3\left[x\right]-a_3)\ast d+1/t{\left[i\right]}^2\ast D[b_3\left[x\right],x,2]\ast d^2$
  $+(du1x\ast du1x+du2y\ast du2y+2.0\ast du1y\ast du2x)\ast d^2)/3.0;$
• $z_1=Series[z_1,\{u_1[i+1]\left[x\right],0,2\},\{u_1{[i+1]}^{\prime }\left[x\right],0,1\},\{u_1{[i+1]}^{\prime \prime }\left[x\right],0,1\},$
  $\{u_1{[i+1]}^{\prime \prime \prime }\left[x\right],0,0\},\{u_1{[i+1]}^{\prime \prime \prime \prime }\left[x\right],0,0\},\{u_2[i+1]\left[x\right],0,1\},$
  $\{u_2{[i+1]}^{\prime }\left[x\right],0,0\},\{u_2{[i+1]}^{\prime \prime }\left[x\right],0,0\},\{u_2{[i+1]}^{\prime \prime \prime }\left[x\right],0,0\},$
  $\{u_2{[i+1]}^{\prime \prime \prime \prime }\left[x\right],0,0\},\{P[i+1]\left[x\right],0,1\},\{P{[i+1]}^{\prime }\left[x\right],0,0\},$
  $\{P{[i+1]}^{\prime \prime }\left[x\right],0,0\},\{P{[i+1]}^{\prime \prime \prime }\left[x\right],0,0\},\{P{[i+1]}^{\prime \prime \prime \prime }\left[x\right],0,0\},$
  $\left\{Sin\right[x],0,1\},\left\{Cos\right[x],0,1\}];$
  $z_1=Normal\left[z_1\right];$
  $z_1=Expand\left[z_1\right];$
  $Print\left[z_1\right];$
• $z_2=Series[z_2,\{u_1[i+1]\left[x\right],0,1\},\{u_1{[i+1]}^{\prime }\left[x\right],0,1\},\{u_1{[i+1]}^{\prime \prime }\left[x\right],0,1\},$
  $\{u_1{[i+1]}^{\prime \prime \prime }\left[x\right],0,0\},\{u_1{[i+1]}^{\prime \prime \prime \prime }\left[x\right],0,0\},\{u_2[i+1]\left[x\right],0,2\},$
  $\{u_2{[i+1]}^{\prime }\left[x\right],0,0\},\{u_2{[i+1]}^{\prime \prime }\left[x\right],0,0\},\{u_2{[i+1]}^{\prime \prime \prime }\left[x\right],0,0\},$
  $\{u_2{[i+1]}^{\prime \prime \prime \prime }\left[x\right],0,0\},\{P[i+1]\left[x\right],0,1\},\{P{[i+1]}^{\prime }\left[x\right],0,0\},$
  $\{P{[i+1]}^{\prime \prime }\left[x\right],0,0\},\{P{[i+1]}^{\prime \prime \prime }\left[x\right],0,0\},\{P{[i+1]}^{\prime \prime \prime \prime }\left[x\right],0,0\},$
  $\left\{Sin\right[x],0,1\},\left\{Cos\right[x],0,1\}];$
  $z_2=Normal\left[z_2\right];$
  $z_2=Expand\left[z_2\right];$
  $Print\left[z_2\right];$
• $z_3=Series[z_3,\{u_1[i+1]\left[x\right],0,2\},\{u_1{[i+1]}^{\prime }\left[x\right],0,1\},\{u_1{[i+1]}^{\prime \prime }\left[x\right],0,1\},$
  $\{u_1{[i+1]}^{\prime \prime \prime }\left[x\right],0,0\},\{u_1{[i+1]}^{\prime \prime \prime \prime }\left[x\right],0,0\},\{u_2[i+1]\left[x\right],0,2\},$
  $\{u_2{[i+1]}^{\prime }\left[x\right],0,1\},\{u_2{[i+1]}^{\prime \prime }\left[x\right],0,0\},\{u_2{[i+1]}^{\prime \prime \prime }\left[x\right],0,0\},$
  $\{u_2{[i+1]}^{\prime \prime \prime \prime }\left[x\right],0,0\},\{P[i+1]\left[x\right],0,1\},\{P{[i+1]}^{\prime }\left[x\right],0,1\},$
  $\{P{[i+1]}^{\prime \prime }\left[x\right],0,0\},\{P{[i+1]}^{\prime \prime \prime }\left[x\right],0,0\},\{P{[i+1]}^{\prime \prime \prime \prime }\left[x\right],0,0\},$
  $\left\{Sin\right[x],0,1\},\left\{Cos\right[x],0,1\}];$
  $z_3=Normal\left[z_3\right];$
  $z_3=Expand\left[z_3\right];$
  $Print\left[z_3\right];$
• $b_1\left[x_{-}\right]=z_1;$
  $b_2\left[x_{-}\right]=z_2;$
  $b_3\left[x_{-}\right]=z_3;$
  $b_{11}=z_1;$
  $b_{12}=z_2;$
  $b_{13}=z_3;];$
• $a_{11}\left[i\right]=b_{11};$
  $a_{12}\left[i\right]=b_{12};$
  $a_{13}\left[i\right]=b_{13};$
  $Print\left[a_{11}\left[i\right]\right];$
  $Clear[b_1,b_2,b_3];$
  $u_1[i+1]\left[x_{-}\right]=b_1\left[x\right];$
  $u_2[i+1]\left[x_{-}\right]=b_2\left[x\right];$
  $P[i+1]\left[x_{-}\right]=b_3\left[x\right];$
  $a_1=Series[b_{11},\{t_3,0,1\};\{b_1\left[x\right],0,1\},\{b_2\left[x\right],0,1\},\{b_3\left[x\right],0,1\},$
  $\{b_1^{\prime }\left[x\right],0,0\},\{b_2^{\prime }\left[x\right],0,0\},\{b_3^{\prime }\left[x\right],0,0\},\{b_1^{\prime \prime }\left[x\right],0,0\},\{b_2^{\prime \prime }\left[x\right],0,0\},\{b_3^{\prime \prime }\left[x\right],0,0\}];$
  $a_1=Normal\left[a_1\right];$
  $a_1=Expand\left[a_1\right];$
  $a_2=Series[b_{12},\{t_3,0,1\};\{b_1\left[x\right],0,1\},\{b_2\left[x\right],0,1\},\{b_3\left[x\right],0,1\},$
  $\{b_1^{\prime }\left[x\right],0,0\},\{b_2^{\prime }\left[x\right],0,0\},\{b_3^{\prime }\left[x\right],0,0\},\{b_1^{\prime \prime }\left[x\right],0,0\},\{b_2^{\prime \prime }\left[x\right],0,0\},\{b_3^{\prime \prime }\left[x\right],0,0\}];$
  $a_2=Normal\left[a_2\right];$
  $a_2=Expand\left[a_2\right];$
  $a_3=Series[b_{13},\{t_3,0,1\};\{b_1\left[x\right],0,1\},\{b_2\left[x\right],0,1\},\{b_3\left[x\right],0,1\},$
  $\{b_1^{\prime }\left[x\right],0,0\},\{b_2^{\prime }\left[x\right],0,0\},\{b_3^{\prime }\left[x\right],0,0\},\{b_1^{\prime \prime }\left[x\right],0,0\},\{b_2^{\prime \prime }\left[x\right],0,0\},\{b_3^{\prime \prime }\left[x\right],0,0\}];$
  $a_3=Normal\left[a_3\right];$
  $a_3=Expand\left[a_3\right];$
• $b_1[x_{]}=-1.0\ast Sin\left[x\right];$
  $b_2[x_{]}=1.0\ast Cos\left[x\right];$
  $b_3[x_{]}=0.10;$
• $For[i=1,i<m8,i++,$
  $A_{11}=a_{11}[m8-i];$
  $A_{11}=Series[A_{11},\{Sin\left[x\right],0,2\},\{Cos\left[x\right],0,2\}];$
  $A_{11}=Normal\left[A_{11}\right];$
  $A_{11}=Expand\left[A_{11}\right];$
  $A_{12}=a_{12}[m_8-i];$
  $A_{12}=Series[A_{12},\{Sin\left[x\right],0,2\},\{Cos\left[x\right],0,2\}];$
  $A_{12}=Normal\left[A_{12}\right];$
  $A_{12}=Expand\left[A_{12}\right];$
  $A_{13}=a_{13}[m_8-i];$
  $A_{13}=Series[A_{13},\{Sin\left[x\right],0,2\},\{Cos\left[x\right],0,2\}];$
  $A_{13}=Normal\left[A_{13}\right];$
  $A_{13}=Expand\left[A_{13}\right];$
• $u_1[m8-i]\left[x_{-}\right]=A_{11};$
  $u_2[m8-i]\left[x_{-}\right]=A_{12};$
  $P[m8-i]\left[x_{-}\right]=Expand\left[A_{13}\right];$
• $t_3=1.0;$
• $Print["u_1[",m_8-i,"]=",A_{11}];$
  $Clear\left[t_3\right];$
  $b_1[x_{]}=A_{11};$
  $b_2[x_{]}=A_{12};$
  $b_3[x_{]}=A_{13};];$
• $t_3=1.0;$
• $For[i=1,i<m_8,i++,$
  $Print["u_1[",i,"]=",u_1\left[i\right]\left[x\right]]]$

***************************************************

Here the line expressions for the field of velocity $u=\left\{u_1\left[n\right]\left(x\right)\right\}$, where again we emphasize $N=10$ lines and $\nu =e_1=1.0$:

•  

  $$\begin{array}{ccc}\hfill u_1\left[1\right]\left(x\right)& =& 0.0044548Cos\left[x\right]-0.174091Sin\left[x\right]+0.00041254Cos{\left[x\right]}^{2}Sin\left[x\right]\hfill \\ & & +0.0260471Cos\left[x\right]Sin{\left[x\right]}^2-0.000188598Cos{\left[x\right]}^{2}Sin{\left[x\right]}^2\hfill \end{array}$$

  (35)
•  

  $$\begin{array}{ccc}\hfill u_1\left[2\right]\left(x\right)& =& 0.00680614Cos\left[x\right]-0.331937Sin\left[x\right]+0.000676383Cos{\left[x\right]}^{2}Sin\left[x\right]\hfill \\ & & +0.0501544Cos\left[x\right]Sin{\left[x\right]}^2-0.000176433Cos{\left[x\right]}^{2}Sin{\left[x\right]}^2\hfill \end{array}$$

  (36)
•  

  $$\begin{array}{ccc}\hfill u_1\left[3\right]\left(x\right)& =& 0.00775103Cos\left[x\right]-0.470361Sin\left[x\right]+0.000863068Cos{\left[x\right]}^{2}Sin\left[x\right]\hfill \\ & & +0.0682792Cos\left[x\right]Sin{\left[x\right]}^2-0.000121656Cos{\left[x\right]}^{2}Sin{\left[x\right]}^2\hfill \end{array}$$

  (37)
•  

  $$\begin{array}{ccc}\hfill u_1\left[4\right]\left(x\right)& =& 0.00771379Cos\left[x\right]-0.589227Sin\left[x\right]+0.000994973Cos{\left[x\right]}^{2}Sin\left[x\right]\hfill \\ & & +0.0781784Cos\left[x\right]Sin{\left[x\right]}^2-0.00006958Cos{\left[x\right]}^{2}Sin{\left[x\right]}^2\hfill \end{array}$$

  (38)
•  

  $$\begin{array}{ccc}\hfill u_1\left[5\right]\left(x\right)& =& 0.00701567Cos\left[x\right]-0.690152Sin\left[x\right]+0.00106158Cos{\left[x\right]}^{2}Sin\left[x\right]\hfill \\ & & +0.0796091Cos\left[x\right]Sin{\left[x\right]}^2-0.0000330485Cos{\left[x\right]}^{2}Sin{\left[x\right]}^2\hfill \end{array}$$

  (39)
•  

  $$\begin{array}{ccc}\hfill u_1\left[6\right]\left(x\right)& =& 0.00589597Cos\left[x\right]-0.775316Sin\left[x\right]+0.00104499Cos{\left[x\right]}^{2}Sin\left[x\right]\hfill \\ & & +0.0734277Cos\left[x\right]Sin{\left[x\right]}^2-0.0000121648Cos{\left[x\right]}^{2}Sin{\left[x\right]}^2\hfill \end{array}$$

  (40)
•  

  $$\begin{array}{ccc}\hfill u_1\left[7\right]\left(x\right)& =& 0.00452865Cos\left[x\right]-0.846947Sin\left[x\right]+0.000931782Cos{\left[x\right]}^{2}Sin\left[x\right]\hfill \\ & & +0.0609739Cos\left[x\right]Sin{\left[x\right]}^2-2.74137\ast {10}^{-6}Cos{\left[x\right]}^{2}Sin{\left[x\right]}^2\hfill \end{array}$$

  (41)
•  

  $$\begin{array}{ccc}\hfill u_1\left[8\right]\left(x\right)& =& 0.00303746Cos\left[x\right]-0.907103Sin\left[x\right]+0.000716865Cos{\left[x\right]}^{2}Sin\left[x\right]\hfill \\ & & +0.0437018Cos\left[x\right]Sin{\left[x\right]}^2\hfill \end{array}$$

  (42)
•  

  $$\begin{array}{ccc}\hfill u_1\left[9\right]\left(x\right)& =& 0.00150848Cos\left[x\right]-0.957599Sin\left[x\right]+0.000403216Cos{\left[x\right]}^{2}Sin\left[x\right]\hfill \\ & & +0.0229802Cos\left[x\right]Sin{\left[x\right]}^2\hfill \end{array}$$

  (43)

7. Conclusion

In this article, we develop solutions for examples concerning the two-dimensional, time-independent and incompressible Navier-Stokes system through the generalized method of lines. We also obtain the appropriate boundary conditions for an equivalent elliptic system. Finally, the extension of such results to ${\mathbb{R}}^3$, compressible and time dependent cases is planned for a future work.