In this work, approximate analytical solutions to the lid-driven square cavity flow problem, which satisfied two-dimensional unsteady incompressible Navier-Stokes equations, are presented using the kinetically reduced local Navier-Stokes equations. Reduced differential transform method and perturbation-iteration algorithm are applied to solve this problem. The convergence analysis was discussed for both methods. The numerical results of both methods are given at some Reynolds numbers and low Mach numbers, and compared with results of earlier studies in the review of the literatures. These two methods are easy and fast to implement, and the results are close to each other and other numerical results, so it can be said that these methods are useful in finding approximate analytical solutions to the unsteady incompressible flow problems at low Mach numbers.
Fluid flow is one of the most important engineering phenomena that have received widespread attention in theoretical and practical scientific research. Many of these studies focus on simulated mathematical models which represent these phenomena. Therefore, the equations of Navier-Stokes, which are the basic model for describing the movement of fluid, have received considerable attention from researchers to find their analytical and numerical solutions.
In this work, unsteady viscous incompressible flows characterized by two-dimensional Navier-Stokes equations are studied. The non-dimensional momentum and continuity equations have the following form
u t = − ( u u x + v u y + p x ) + 1 R e ( u x x + u y y ) , v t = − ( u v x + v v y + p y ) + 1 R e ( v x x + v y y ) , (1.1)
and
u x + v y = 0, (1.2)
where t is the physical time, u ( x , y , t ) and v ( x , y , t ) are the fluid velocity components, p ( x , y , t ) is the pressure, and Re is the Reynolds number. Since, the Navier-Stokes equations are nonlinear partial differential equations and there is no explicit equation for calculating pressure, these equations are difficult to solve, so many studies have suggested the alternative thermodynamic description of incompressible fluid flows. One of these alternative formulas is the kinetically reduced local Navier-Stokes (KRLNS) equations [
p = g + u 2 + v 2 2 , (1.3)
and the continuity equation by
g t = − 1 ( M a ) 2 ( u x + v y ) + 1 R e ( g x x + g y y ) , (1.4)
where Ma is the Mach number and g ( x , y , t ) is the grand potential. The time scale in INS equations is related to that of KRLNS equations; t K R L N S ( τ ) = M a × t N S . Then, the system of equations of KRLNS has the following form
u t = − ( 2 u u x + v v x + v u y + g x ) + 1 R e ( u x x + u y y ) , v t = − ( u v x + u u y + 2 v v y + g y ) + 1 R e ( v x x + v y y ) , g t = − 1 ( M a ) 2 ( u x + v y ) + 1 R e ( g x x + g y y ) . (1.5)
The KRlNS equations suggested in [
The lid-driven cavity problem refers to the flow in a box cavity with no-slip at the walls, one or more which move at constant speed. It has been used extensively as a benchmark case for the study of computational methods to solve Navier-Stokes equations, because the simplicity of its geometry and boundary conditions. Numerous literature studies have offered the solutions for this problem by using the different numerical methods in rectangular or square cavities. For example, in [
The main aim of this study is to obtain the approximate analytical solutions for two-dimensional lid-driven square cavity flow problem, since most of the research focused on the numerical solutions for this problem. Reduced differential transform method (RDTM) and perturbation-iteration algorithm (PIA) are used for this purpose for several reasons. The first reason is that both methods have not previously been applied to resolve this problem. Secondly, these methods can directly be applied to KRLNS equations. Moreover, these methods can reduce the size of the calculations and at the same time maintain the accuracy of the numerical solution.
We have organized this paper into seven sections, of which this introduction is the first. In Section 2 and 3, we describe the reduced differential transform method and perturbation-iteration algorithm, and applied them to KRLNS equations. We derived the condition of convergence for both methods (Section 4). We then present the approximate analytical solutions for two-dimensional lid-driven cavity flow, which are obtained by applying differential transform method and perturbation-iteration algorithm (Section 5). Next, we introduce the numerical results and compare these results with other works (Section 6). The last Section summarizes the major findings of this study.
The RDTM was first introduced by Keskin [
In this section, we give some properties of the (2 + 1)-dimensional RDTM [
U k ( X ) = 1 k ! [ ∂ k ∂ t k u ( X , t ) ] t = 0 , (2.1)
is the t-dimensional spectrum function of u ( X , t ) which is the transformed function. The reduced differential inverse transform of U k ( X ) is defined as
u ( X , t ) = ∑ k = 0 ∞ U k ( X ) t k , (2.2)
from Equation (2.1) and Equation (2.2), we can conclude that
u ( X , t ) = ∑ k = 0 ∞ 1 k ! [ ∂ k ∂ t k u ( X , t ) ] t = 0 t k . (2.3)
The fundamental mathematical operations performed by RDTM are readily obtained and listed in
| Functional form | Transformed form |
|---|---|
| w ( X , t ) = u ( X , t ) v ( X , t ) | W k ( X ) = U k ( X ) V k ( X ) |
| w ( X , t ) = α u ˙ ( X , t ) | W k ( X ) = α U ˙ k ( X ) , α is constant |
| w ( X , t ) = u ( X , t ) v ˙ ( X , t ) | W k ( X ) = ∑ i = 0 k U i ( X , t ) V ˙ k − i ( X , t ) |
| w ( x , t ) = ∂ r ∂ t r u ( X , t ) | W k ( X ) = ( k + 1 ) ⋯ ( k + r ) U ˙ k ( X ) = ( k + r ) ! k ! U ˙ k ( X ) |
| w ( X , t ) = ∂ r 1 + r 2 + ⋯ + r n ∂ x 1 r 1 ∂ x 2 r 2 ⋯ ∂ x n r n u ( X , t ) | W k ( x ) = ∂ r 1 + r 2 + ⋯ + r n ∂ x 1 r 1 ∂ x 2 r 2 ⋯ ∂ x n r n U k ( X ) |
In order to apply this method with KRLNS equations to find approximate analytical solutions for INS equations, we suppose that X = ( x , y ) , u = ( u , v ) and U k = ( U k , V k ) , where u ( X , t ) and v ( X , t ) are the fluid velocity components in the x and y directions, and U k ( X ) , V k ( X ) and G k ( X ) are t-dimensional spectrum functions of u ( X , t ) , v ( X , t ) and g ( X , t ) respectively. Then, we have
( k + 1 ) U k + 1 ( X ) = − ( 2 A k + B k + C k + ( G k ( X ) ) x − 1 R e ( ( U k ( X ) ) x x + ( U k ( X ) ) y y ) ) ,
( k + 1 ) V k + 1 ( X ) = − ( D k + E k + 2 F k + ( G k ( X ) ) y − 1 R e ( ( V k ( X ) ) x x + ( V k ( X ) ) y y ) ) ,
( k + 1 ) G k + 1 ( X ) = − ( 1 ( M a ) 2 ( ( U k ( X ) ) x + ( V k ( X ) ) y ) − 1 R e ( ( G k ( X ) ) x x + ( G k ( X ) ) y y ) ) , (2.4)
such that
A k = ∑ i = 0 k U i ( X ) ( U k − i ( X ) ) x , B k = ∑ i = 0 k V i ( X ) ( V k − i ( X ) ) x ,
C k = ∑ i = 0 k V i ( X ) ( U k − i ( X ) ) y , D k = ∑ i = 0 k U i ( X ) ( V k − i ( X ) ) x ,
E k = ∑ i = 0 k U i ( X ) ( U k − i ( X ) ) y , F k = ∑ i = 0 k V i ( X ) ( V k − i ( X ) ) y ,
where k = 0 , 1 , 2 , 3 , ⋯ , U 0 ( X ) = u ( X , 0 ) , V 0 ( X ) = v ( X , 0 ) and G 0 ( X ) = g ( X , 0 ) . Then the exact solution is obtained as follows:
u ( X , τ ) = lim n → ∞ u n ( X , τ ) ,
v ( X , τ ) = lim n → ∞ v n ( X , τ ) ,
g ( X , τ ) = lim n → ∞ g n ( X , τ ) , (2.5)
where
u n ( X , τ ) = ∑ k = 0 n U k ( X ) τ k , v n ( X , τ ) = ∑ k = 0 n V k ( X ) τ k , g n ( X , τ ) = ∑ k = 0 n G k ( X ) τ k .
This approach is referred to by (KRDTM) in this paper.
Perturbation methods are important analytical methods which have been used to construct approximate analytical solutions of algebraic equations, differential equations, and integro-differential equations. The main limitation of using the perturbation methods is to install a small auxiliary parameter in the equation. For this reason, the solutions of these methods are restricted by validity range of physical parameters, so many of perturbation techniques have been suggested by several authors. PIA is one of the techniques which was proposed by Pakdemirli and Boyac in [
In general, PIA is obtained by taking different numbers of terms in the perturbation expansions and different order of correction terms in the Taylor series expansions. Therefore, the perturbation-iteration algorithm is called PIA(m,n) where the m is the number of the correction terms in the perturbation expansion and n is the highest order derivative term in the Taylor series such that m should always be less than or equal to n.
To obtain approximate analytical solutions for two-dimensions Navier-Stokes equations, PIA (1, 1) will be applied to KRLNS equations and which will be referred to this article by (KPIA). Firstly, we write Equation (1.5) as follows:
F 1 ( u , v , u t , u x , u y , v x , g x , u x x , u y y , ϵ ) = u t + ϵ ( 2 u u x + v v x + v u y + g x − 1 R e ( u x x + u y y ) ) ,
F 2 ( u , v , v t , u y , v x , v y , g y , v x x , v y y , ϵ ) = v t + ϵ ( u v x + u u y + 2 v v y + g y − 1 R e ( v x x + v y y ) ) ,
F 3 ( g t , u x , v y , g x x , g y y , ϵ ) = g t + ϵ ( 1 ( M a ) 2 ( u x + v y ) − 1 R e ( g x x + g y y ) ) , (3.1)
where ϵ is a small perturbation parameter. Secondly, we define the following perturbation expansions with only one correction term:
u n + 1 = u n + ϵ ( u c ) n ,
v n + 1 = v n + ϵ ( v c ) n ,
g n + 1 = g n + ϵ ( g c ) n , (3.2)
where n represents the n_th iteration and u c , v c and g c are the correction terms in the perturbation expansion. Thirdly, by replacing (3.2) into (3.1) and writing in the Taylor series expansion for first order derivative terms about ϵ = 0 , yields
F 1 ( u n , v n , ( u n ) t , ( u n ) x , ( u n ) y , ( v n ) x , ( g n ) x , ( u n ) x x , ( u n ) y y , 0 ) + ϵ [ F 1 ϵ + F 1 u n + 1 ( u c ) n + F 1 v n + 1 ( v c ) n + F 1 ( u n + 1 ) t ( ( u c ) n ) t + F 1 ( u n + 1 ) x ( ( u c ) n ) x + F 1 ( u n + 1 ) y ( ( u c ) n ) y + F 1 ( v n + 1 ) x ( ( v c ) n ) x + F 1 ( g n + 1 ) x ( ( g c ) n ) x + F 1 ( u n + 1 ) x x ( ( u c ) n ) x x + F 1 ( u n + 1 ) y y ( ( u c ) n ) y y ] = 0 ,
F 2 ( u n , v n , ( v n ) t , ( u n ) y , ( v n ) x , ( v n ) y , ( g n ) y , ( v n ) x x , ( v n ) y y , 0 ) + ϵ [ F 2 ϵ + F 2 u n + 1 ( u c ) n + F 2 v n + 1 ( v c ) n + F 2 ( v n + 1 ) t ( ( v c ) n ) t + F 2 ( u n + 1 ) y ( ( u c ) n ) y + F 2 ( v n + 1 ) x ( ( v c ) n ) x + F 2 ( v n + 1 ) y ( ( v c ) n ) y + F 2 ( g n + 1 ) y ( ( g c ) n ) y + F 2 ( v n + 1 ) x x ( ( v c ) n ) x x + F 2 ( v n + 1 ) y y ( ( v c ) n ) y y ] = 0 ,
F 3 ( ( g n ) t , ( u n ) x , ( v n ) y , ( g n ) x x , ( g n ) y y , 0 ) + ϵ [ F 3 ϵ + F 3 ( g n + 1 ) t ( ( g c ) n ) t + F 3 ( u n + 1 ) x ( ( u c ) n ) x + F 3 ( v n + 1 ) y ( ( v c ) n ) y + F 3 ( g n + 1 ) x x ( ( g c ) n ) x x + F 3 ( g n + 1 ) y y ( ( g c ) n ) y y ] = 0. (3.3)
All derivatives in Equation (3.3) are evaluated at ϵ = 0 such that
F 1 ( u n , v n , ( u n ) t , ( u n ) x , ( u n ) y , ( v n ) x , ( g n ) x , ( u n ) x x , ( u n ) y y , 0 ) = ( u n ) t ,
F 2 ( u n , v n , ( v n ) t , ( u n ) y , ( v n ) x , ( v n ) y , ( g n ) y , ( v n ) x x , ( v n ) y y , 0 ) = ( v n ) t ,
F 3 ( ( g n ) t , ( u n ) x , ( v n ) y , ( g n ) x x , ( g n ) y y , 0 ) = ( g n ) t ,
F 1 ( u n + 1 ) t = F 1 ( v n + 1 ) t = F 3 ( g n + 1 ) t = 1 ,
F 1 u n + 1 = F 1 v n + 1 = F 1 ( u n + 1 ) x = F 1 ( u n + 1 ) y = F 1 ( v n + 1 ) x = F 1 ( u n + 1 ) x x = F 1 ( u n + 1 ) y y = 0 ,
F 2 u n + 1 = F 2 v n + 1 = F 2 ( v n + 1 ) x = F 2 ( v n + 1 ) y = F 2 ( u n + 1 ) y = F 2 ( v n + 1 ) x x = F 2 ( v n + 1 ) y y = 0 ,
F 1 ( g n + 1 ) x = F 2 ( g n + 1 ) y = F 3 ( u n + 1 ) x = F 3 ( v n + 1 ) y = F 3 ( g n + 1 ) x x = F 3 ( g n + 1 ) y y = 0 ,
F 1 ϵ = ( 2 u n ( u n ) x + v n ( v n ) x + v n ( u n ) y + ( g n ) x − 1 R e ( ( u n ) x x + ( u n ) y y ) ) ,
F 2 ϵ = ( u n ( v n ) x + u n ( u n ) y + 2 v n ( v n ) y + ( g n ) y − 1 R e ( ( v n ) x x + ( v n ) y y ) ) ,
F 3 ϵ = ( 1 ( M a ) 2 ( ( u n ) x + ( v n ) y ) − 1 R e ( ( g n ) x x + ( g n ) y y ) ) .
Finally, by substituting the above derivative in the formulas (3.3) and setting ϵ = 1 we obtain the following iteration equation formulas:
( ( u c ) n ) t = ( u n ) t − ( 2 u n ( u n ) x + v n ( v n ) x + v n ( u n ) y + ( g n ) x − 1 R e ( ( u n ) x x + ( u n ) y y ) ) ,
( ( v c ) n ) t = ( v n ) t − ( u n ( v n ) x + u n ( u n ) y + 2 v n ( v n ) y + ( g n ) y − 1 R e ( ( v n ) x x + ( v n ) y y ) ) ,
( ( g c ) n ) t = ( g n ) t − ( 1 ( M a ) 2 ( ( u n ) x + ( v n ) y ) − 1 R e ( ( g n ) x x + ( g n ) y y ) ) . (3.4)
The calculations start with initial condition u ( x , y ,0 ) , v ( x , y ,0 ) and g ( x , y ,0 ) where these values are used as estimate values for ( u c ) 0 , ( v c ) 0 and ( g c ) 0 in Equation (3.4), and then substitute the results of Equation (3.4) into Equation (3.2) to obtain u 1 , v 1 and g 1 which are the solutions at the first iteration. So we can get ( n + 1 ) iteration solutions by repeating this process and using the previous solution n as an initial guess.
We now study the convergence analysis of the approximate analytical solutions which are computed from the application KRDTM and KPIA.
Let us consider the Hilbert space H = L 2 ( ( a , b ) 2 × [ 0, T ] ) as defined by
u : H → ℝ with ∫ ( a , b ) 2 × [ 0, T ] u 2 ( X , t ) d X d t < ∞ ,
and the norm
‖ u ‖ 2 = ∫ ( a , b ) 2 × [ 0, T ] u 2 ( X , t ) d X d t ,
where X = ( x , y ) . Defined as
u = ( u , v , g ) : H 3 → ℝ 3 with ∫ ( a , b ) 2 × [ 0, T ] ( u 2 ( X , t ) + v 2 ( X , t ) + g 2 ( X , t ) ) d X d t < ∞ ,
such that ‖ u ‖ 2 = ‖ u ‖ 2 + ‖ v ‖ 2 + ‖ g ‖ 2 .
We consider the KRINS equation in the following form
L ( u ( X , τ ) ) = N ( u ( X , τ ) ) + R ( u ( X , τ ) ) , (4.1)
which is equivalent to the following formula
u ( X , τ ) = F ( u k ( X , τ ) ) , (4.2)
where L is the linear partial derivative with respect to τ , N is a nonlinear operator, R is a linear operator, and F is a general nonlinear operator involving both linear and nonlinear terms.
Case 1: According to KRDTM, formula (4.1) can be written in the following form
( k + 1 ) U k + 1 ( X ) = N ( U k ( X ) ) + R ( U k ( X ) ) ,
and the solutions
u ( X , τ ) = ∑ k = 0 ∞ U k ( X ) τ k = ∑ k = 0 ∞ B k , (4.3)
where B k = ( B 1 k , B 2 k , B 3 k ) . It is noted that the solutions by KRDTM is equivalent to determining the sequence
S 0 = U 0 ( X ) = B 0 ,
S 1 = U 0 ( X ) + U 1 ( X ) τ = B 0 + B 1 ,
S 2 = U 0 ( X ) + U 1 ( X ) τ + U 2 ( X ) τ 2 = B 0 + B 1 + B 2 ,
⋮
S n = ∑ k = 0 n U k ( X ) τ k = ∑ k = 0 n B k .
Case 2: To study the convergence of KPIA, we write the approximate solutions in different form. To do this, we define
B 0 = ( B 10 , B 20 , B 30 ) = ( u ( X , 0 ) , v ( X , 0 ) , g ( X , 0 ) ) = u ( X , 0 ) ,
B n + 1 = ( B 1 ( n + 1 ) , B 2 ( n + 1 ) , B 3 ( n + 1 ) ) = ( ( u c ) n ( X , τ ) , ( v c ) n ( X , τ ) , ( g c ) n ( X , τ ) ) = ( u c ) n ( X , τ ) ,
u 0 = B 0 = S 0 ,
u 1 = u 0 + ϵ ( u c ) 0 = B 0 + B 1 = S 1 ,
u 2 = u 1 + ϵ ( u c ) 1 = B 0 + B 1 + B 2 = S 2 ,
u 3 = u 2 + ϵ ( u c ) 2 = B 0 + B 1 + B 2 + B 3 = S 3 ,
⋮
u n = u n − 1 + ϵ ( u c ) n − 1 ≡ B 0 + B 1 + B 2 + B 3 + ⋯ + B n = ∑ k = 0 n B k = S n .
So the solutions, which are resulted from KPIA have the form
u ( X , τ ) = ∑ k = 0 ∞ U k ( X ) τ k = ∑ k = 0 ∞ B k . (4.4)
such that S n + 1 = F ( S n ) for both cases.
The sufficient condition for convergence of the series solution { S n } 0 ∞ is given in the following theorems.
Theorem 4.1. The series solution { S n = ( R n , S n , T n ) } 0 ∞ converges whenever there is γ such that 0 < γ < 1 , γ = γ 1 + γ 2 + γ 3 and ‖ B i ( k + 1 ) ‖ ⩽ γ i ‖ B i k ‖ .
Proof: Firstly, we show that { S n = ( R n , S n , T n ) } 0 ∞ is a Cauchy sequence in the Hilbert space H 3 . For this reason, we suppose that
‖ R n + 1 − R n ‖ = ‖ B 1 ( n + 1 ) ‖ ⩽ γ 1 ‖ B 1 n ‖ ⩽ γ 1 2 ‖ B 1 ( n − 1 ) ‖ ⩽ ⋯ ⩽ γ 1 n + 1 ‖ B 10 ‖ ,
‖ S n + 1 − S n ‖ = ‖ B 2 ( n + 1 ) ‖ ⩽ γ 2 ‖ B 2 n ‖ ⩽ γ 2 2 ‖ B 2 ( n − 1 ) ‖ ⩽ ⋯ ⩽ γ 2 n + 1 ‖ B 20 ‖ ,
‖ T n + 1 − T n ‖ = ‖ B 3 ( n + 1 ) ‖ ⩽ γ 3 ‖ B 3 n ‖ ⩽ γ 3 2 ‖ B 3 ( n − 1 ) ‖ ⩽ ⋯ ⩽ γ 3 n + 1 ‖ B 30 ‖ .
Then, by using the triangle inequality, we find that
‖ S n − S m ‖ = ‖ ( R n , S n , T n ) − ( R m , S m , T m ) ‖ = ‖ ( R n − R m , S n − S m , T n − T m ) ‖ ⩽ ‖ R n − R m ‖ + ‖ S n − S m ‖ + ‖ T n − T m ‖ ⩽ ‖ R n − R n − 1 ‖ + ‖ R n − 1 − R n − 2 ‖ + ⋯ + ‖ R m + 1 − R m ‖ + ‖ S n − S n − 1 ‖ + ‖ S n − 1 − S n − 2 ‖ + ⋯ + ‖ S m + 1 − S m ‖ + ‖ T n − T n − 1 ‖ + ‖ T n − 1 − T n − 2 ‖ + ⋯ + ‖ T m + 1 − T m ‖
⩽ ( γ 1 n + γ 1 n − 1 + ⋯ + γ 1 m + 1 ) ‖ B 10 ‖ + ( γ 2 n + γ 2 n − 1 + ⋯ + γ 2 m + 1 ) ‖ B 20 ‖ + ( γ 3 n + γ 3 n − 1 + ⋯ + γ 3 m + 1 ) ‖ B 30 ‖ ⩽ ( γ n + γ n − 1 + ⋯ + γ m + 1 ) ( ‖ B 10 ‖ + ‖ B 20 ‖ + ‖ B 30 ‖ ) = γ m + 1 ( γ n − m − 1 + γ n − m − 2 + ⋯ + 1 ) ( ‖ B 10 ‖ + ‖ B 20 ‖ + ‖ B 30 ‖ ) ⩽ γ m + 1 1 − γ ‖ B 0 ‖ ,
since ‖ B 0 ‖ < ∞ and 0 < γ < 1 , we then have lim n , m → ∞ ‖ S n − S m ‖ = 0 . Thus, we conclude that { S n } 0 ∞ is a Cauchy sequence in the Hilbert space H3, thus, the series solution { S n } 0 ∞ converges to some { S } ∈ H 3 .
Theorem 4.2. Let F = ( F 1 , F 2 , F 3 ) be a nonlinear operator satisfies Lipschitz condition from a Hilbert space H3 into H3 and u ( X , τ ) be the exact solution of INS equations. If the series solution { S n } 0 ∞ converges, then it is converged to u ( X , τ ) .
Proof: Let u 1 ( X , τ ) , u 2 ( X , τ ) , then we have
‖ F ( u 1 ) − F ( u 2 ) ‖ = ‖ ( F 1 ( u 1 ) , F 2 ( u 1 ) , F 3 ( u 1 ) ) − ( F 1 ( u 2 ) , F 2 ( u 2 ) , F 3 ( u 2 ) ) ‖ = ‖ ( F 1 ( u 1 ) − F 1 ( u 2 ) , F 2 ( u 1 ) − F 2 ( u 2 ) , F 3 ( u 1 ) − F 3 ( u 2 ) ) ‖ ⩽ ‖ F 1 ( u 1 ) − F 1 ( u 2 ) ‖ + ‖ F 2 ( u 1 ) − F 2 ( u 2 ) ‖ + ‖ F 3 ( u 1 ) − F 3 ( u 2 ) ‖ ⩽ α 1 ‖ u 1 − u 2 ‖ + α 2 ‖ u 1 − u 2 ‖ + α 3 ‖ u 1 − u 2 ‖ = ( α 1 + α 2 + α 3 ) ‖ u 1 − u 2 ‖ = α ‖ u 1 − u 2 ‖ .
Therefore, from the Banach fixed-point theorem, there is a unique solution of the problem (4.1). Now we have to prove that { S n } 0 ∞ converges to u ( X , τ )
u ( X , τ ) = F ( u ( X , τ ) ) = F ( ∑ k = 0 ∞ B k ) = F ( lim n → ∞ ∑ k = 0 n B k ) = lim n → ∞ F ( ∑ k = 0 n B k ) = lim n → ∞ F ( S n ) = lim n → ∞ S n + 1 = S .
Definition 4.1. For i = 1 , 2 , 3 and k ∈ ℕ ∪ { 0 } , we define
γ i k = ( ‖ B i ( k + 1 ) ‖ ‖ B i k ‖ , ‖ B i k ‖ ≠ 0, 0, ‖ B i k ‖ = 0.
then we can say that the series approximate solutions { S n } 0 ∞ converges to the exact solution u ( X , t ) when γ k = γ 1 k + γ 2 k + γ 3 k and 0 < γ k < 1 for all k ∈ ℕ ∪ { 0 } .
In this work we presented the recirculation viscous flow problem in a square cavity, that is called Burggraf Flow [
u ( x , y ) = 8 f ( x ) g ′ ( y ) , v ( x , y ) = − 8 f ′ ( x ) g ( y ) , p ( x , y ) = 8 R e ( F ( x ) g ‴ ( x ) + f ′ ( x ) g ′ ( x ) ) + 64 F 1 ( x ) ( g ( y ) g ″ ( y ) − ( g ′ ( y ) ) 2 ) , (5.1)
where
f ( x ) = x 4 − 2 x 3 + x 2 , g ( x ) = y 4 − y 2 ,
F ( x ) = ∫ f ( x ) d x , F 1 ( x ) = ∫ f ( x ) f ′ ( x ) d x ,
such that the stream function ψ and vorticity ω are defined as
ψ = 8 f ( x ) g ( y ) , such that ψ y = u , & ψ x = − v ,
ω = v x − u y = − 8 ( f ″ ( x ) g ( y ) + f ( x ) g ″ ( y ) ) .
The boundary conditions for the velocities u and v in this problem are of Dirichlet type, which are equal to zero everywhere except along the top surface where
u ( x ,1, t ) = 16 ( x 4 − 2 x 3 + x 2 ) .
To obtain the approximate analytical solutions of the unsteady lid-driven cavity flow problem, we consider the analytical solutions to this problem, which are given in (5.1) as initial conditions for u, v and p.
Then, by applying KRDTM with the initial conditions of this problem, we obtained the iterative solutions like the form (2.5), such that
U 1 ( x , y ) = 0 ,
U 2 ( x , y ) = − 192 ( R e ) 2 ( 12 x 2 + 2 y 2 − 12 x + 1 ) y + 256 5 ( R e ) 2 [ 3 ( 2 x − 9 ) ( 6 y 2 − 1 ) x 8 − 30 y 4 ( y 2 − 1 ) 2 + 4 ( 531 y 4 − 102 y 2 + 1 ) x 7 − 14 ( 531 y 4 − 156 y 2 + 10 ) x 6 + ( 2520 y 6 + 7344 y 4 − 2244 y 2 + 215 ) x 5 − 25 ( 252 y 6 − 9 y 4 + 9 y 2 + 5 ) x 4 + 5 ( 72 y 8 + 932 y 6 − 660 y 4 + 264 y 2 + 5 ) x 3 − 5 ( 108 y 6 + 138 y 4 − 201 y 2 + 111 ) x 2 y 2
+ 5 ( 48 y 6 − 62 y 4 + 19 y 2 + 11 ) x y 2 ] − 1024 x 2 ( x − 1 ) 2 y [ 6 ( 2 y 6 − 5 y 4 + 4 y 2 − 1 ) y 4 − 44 ( 2 y 2 − 1 ) y 4 ( y 2 − 1 ) 2 x + ( 36 y 6 − 18 y 4 + 8 y 2 − 1 ) ( x − 4 ) x 7 − 2 ( 90 y 8 − 236 y 6 + 105 y 4 − 37 y 2 + 3 ) x 6 + 2 ( 270 y 8 − 456 y 6 + 189 y 4 − 55 y 2 + 2 ) x 5
+ ( 144 y 10 − 954 y 8 + 1172 y 6 − 429 y 4 + 93 y 2 − 1 ) x 4 − 8 ( 36 y 8 − 126 y 6 + 124 y 4 − 39 y 2 + 5 ) x 3 y 2 + ( 232 y 8 − 634 y 6 + 544 y 4 − 149 y 2 + 7 ) x 2 y 2 ] ,
⋮
V 1 ( x , y ) = 32 R e [ 3 ( y 4 − y 2 + x 4 − 2 5 x 5 ) − 2 ( 12 y 2 − 1 ) x 3 + 6 ( 6 y 2 − 1 ) x 2 − 2 ( 3 y 4 + 3 y 2 − 1 ) x ] + 128 x 2 ( x − 1 ) 2 [ ( 6 y 4 − 2 y 2 + 1 ) ( x − 2 ) x 3 − ( 8 y 6 − 18 y 4 + 6 y 2 − 1 ) x 2 + 2 ( 2 y 4 − 3 y 2 + 1 ) ( 2 x − 1 ) y 2 ] ,
V 2 ( x , y ) = − 192 ( R e ) 2 ( 12 ( 2 x − 1 ) y 2 + 10 x 3 − 15 x 2 + x + 2 ) + 256 5 R e [ 10 y 7 + 15 ( 108 x 6 − 324 x 5 + 334 x 4 − 128 x 3 + 10 x 2 − 1 ) y 5 + ( 864 x 8 − 3456 x 7 + 4652 x 6 − 1860 x 5 − 710 x 4 + 480 x 3 + 30 x 2 + 5 ) y 3 − x 2 ( x − 1 ) y ( 132 x 5 − 396 x 4 + 130 x 3 + 400 x 2 − 315 x + 45 ) ]
− 2048 x 3 ( x − 1 ) 3 ( 2 x − 1 ) [ 4 ( 16 x 2 − 16 x + 9 ) y 10 − 2 ( 15 x 4 − 30 x 3 + 81 x 2 − 66 x + 37 ) y 8 + 2 ( 18 x 2 − 18 x + 23 ) ( x 2 − x + 1 ) y 6 − ( 7 x 4 − 14 x 3 + 21 x 2 − 14 x + 8 ) y 4 + x 2 ( x − 1 ) 2 y 2 ] ,
⋮
G 1 ( x , y ) = 192 ( R e ) 2 ( 4 x 2 + 2 y 2 − 4 x − 1 ) ( 2 x − 1 ) y
+ 128 R e [ − 3 ( 50 y 4 − 18 y 2 + 1 ) ( x − 4 ) x 7 − 6 ( 84 y 6 + 68 y 4 − 32 y 2 + 3 ) x 6 + 12 ( 126 y 6 − 73 y 4 + 15 y 2 + 1 ) x 5
− ( 120 y 8 + 1388 y 6 − 1320 y 4 + 372 y 2 + 3 ) x 4 + 16 ( 15 y 6 + 16 y 4 − 30 y 2 + 12 ) x 3 y 2 − 2 ( 78 y 6 − 98 y 4 + 21 y 2 + 15 ) x 2 y 2 + 2 y 4 ( y 2 − 1 ) 2 ( 18 x − 1 ) ,
G 2 ( x , y ) = 3456 ( R e ) 3 y ( 2 x − 1 ) + 256 ( R e ) 2 [ − 9 ( 50 y 2 − 3 ) ( x − 4 ) x 7 − 6 ( 980 y 4 + 78 y 2 − 9 ) x 6 + 18 ( 980 y 4 − 272 y 2 + 12 ) x 5 − 3 ( 1820 y 6 + 4490 y 4 − 1800 y 2 + 107 ) x 4 + 12 ( 910 y 6 − 205 y 4 − 45 y 2 + 13 ) x 3 − 6 ( 60 y 8 + 1058 y 6 − 905 y 4 + 207 y 2 + 4 ) x 2
+ 6 ( 60 y 6 + 148 y 4 − 210 y 2 + 66 ) y 2 x − 21 y 2 + 9 y 4 + 70 y 6 − 78 y 8 ] + 96 ( M a ) 2 R e ( 4 x 2 + 2 y 2 − 4 x − 1 ) y ( 2 x − 1 ) − 64 ( M a ) 2 x 2 ( x − 1 ) 2 [ ( 30 y 4 − 6 y 2 + 1 ) ( x − 2 ) x 3 − ( 56 y 6 − 90 y 4 + 18 y 2 − 1 ) x 2 + 2 ( 28 y 6 − 30 y 4 + 6 y 2 ) x − 2 ( 14 y 4 − 15 y 2 + 3 ) y 2 ] ,
⋮
To make a decision on the convergence of the KRDTM, we computed γ k as:
γ 10 = ‖ U 1 ( x , y ) τ ‖ ‖ U 0 ( x , y ) ‖ = 0 ,
γ 20 = ‖ V 1 ( x , y ) τ ‖ ‖ V 0 ( x , y ) ‖ = 714 ( 3227504 ( R e ) 2 − 12514788 R e + 1994117697 ) 14586 τ R e ,
γ 30 = ‖ G 1 ( x , y ) τ ‖ ‖ G 0 ( x , y ) ‖ = 60 ( 49405942 ( R e ) 2 + 1467358893 ) 1253680 ( R e ) 2 − 21900879 R e + 292485765 τ R e ,
γ 11 = ‖ U 2 ( x , y ) τ 2 ‖ ‖ U 1 ( x , y ) τ ‖ = 0,
γ 21 = ‖ V 2 ( x , y ) τ 2 ‖ ‖ V 1 ( x , y ) τ ‖ = τ [ 16422 ( 4861754265600 ( R e ) 4 + 2305887904320 ( R e ) 3 + 15827448430362608 ( R e ) 2
− 25967180162523600 R e + 3689889931130027625 ) ] 0.5 ÷ [ 260015 ( 3227504 ( R e ) 2 − 12514788 R e + 1994117697 ) 0.5 R e ] ,
γ 31 = ‖ G 2 ( x , y ) τ 2 ‖ ‖ G 1 ( x , y ) τ ‖ = τ [ 3 ( ( 2007409369056 ( R e ) 2 + 14901358071600 ) ( M a ) 4 + ( 1477832512 ( R e ) 2 − 165570645240 ) ( R e ) 2 ( M a ) 2 + ( 20792594 ( R e ) 2 + 4402076679 ) R e 4 ) ] 0.5 ÷ [ 6 ( 49405942 ( R e ) 2 + 1467358893 ) 0.5 R e ( M a ) 2 ] ,
⋮
such that γ 0 = γ 10 + γ 20 + γ 30 , γ 1 = γ 11 + γ 21 + γ 31 , ⋯ . For example, if M a = 0.1 , t = 0.1 , and R e = 1 such that τ = M a × t , for all x and y in domain [ 0,1 ] 2 , then
γ 0 = 0.9991283888 < 1 , γ 1 = 0.7958329986 < 1 , ⋯ ,
if M a = 0.1 , t = 0.01 , and R e = 10 then
γ 0 = 0.0129736262 < 1 , γ 1 = 0.2936820858 < 1 , ⋯
Thus, the iterative solutions (3.2) for this problem, which are obtained by using KPIA, have the following form
u 1 ( x , y , τ ) = 16 x 2 y ( 2 y 2 − 1 ) ( x − 1 ) 2 ,
u 2 ( x , y , τ ) = 16 x 2 y ( 2 y 2 − 1 ) ( x − 1 ) 2 + [ − 192 y ( R e ) 2 ( 12 x 2 − 12 x + 2 y 2 + 1 ) + 256 5 R e ( 3 ( 6 y 2 − 1 ) ( 2 x − 9 ) x 8 + 4 ( 531 y 4 − 102 y 2 + 1 ) x 7 − 14 ( 531 y 4 − 156 y 2 + 10 ) x 6 + ( 2520 y 6 + 7344 y 4 − 2244 y 2 + 215 ) x 5 − 25 ( 252 y 6 − 9 y 4 + 9 y 2 + 5 ) x 4 + 5 ( 72 y 8 + 932 y 6 − 660 y 4 + 264 y 2 + 5 ) x 3
− 5 ( 108 y 6 + 138 y 4 − 201 y 2 + 111 ) x 2 y 2 + 5 ( 48 y 6 − 62 y 4 + 19 y 2 + 11 ) x y 2 − 30 y 4 ( y 2 − 1 ) 2 ) − 1024 x 2 ( x − 1 ) 2 y ( ( 36 y 6 − 18 y 4 + 8 y 2 − 1 ) ( x − 4 ) x 7 − 2 ( 90 y 8 − 236 y 6 + 105 y 4 − 37 y 2 + 3 ) x 6 + 2 ( 270 y 8 − 456 y 6 + 189 y 4 − 55 y 2 + 2 ) x 5
+ ( 144 y 10 − 954 y 8 + 1172 y 6 − 429 y 4 + 93 y 2 − 1 ) x 4 − 4 ( 72 y 8 − 252 y 6 + 248 y 4 − 78 y 2 + 10 ) x 3 y 2
+ ( 232 y 8 − 634 y 6 + 544 y 4 − 149 y 2 + 7 ) x 2 y 2 − 44 ( 2 y 2 − 1 ) y 4 ( y 2 − 1 ) 2 x + 6 ( 2 y 6 − 5 y 4 + 4 y 2 − 1 ) y 4 ) ] τ 2 − 16384 [ ( − 1 R e ( 3 ( x − 2 ) x 3 + 3 ( 12 y 2 − 1 ) x 2 − 6 ( 6 y 2 − 1 ) x
+ 3 y 4 + 3 y 2 − 1 ) + 8 y ( ( 6 y 4 − 2 y 2 + 1 ) ( 2 x − 7 ) x 6 − 3 ( 4 y 6 − 24 y 4 + 8 y 2 − 3 ) x 5 + 5 ( 6 y 6 − 15 y 4 + 5 y 2 − 1 ) x 4 − ( 28 y 6 − 48 y 4 + 16 y 2 − 1 ) x 3 + ( 2 y 4 − 3 y 2 + 1 ) ( 6 x − 1 ) x y 2 ) ) × ( − 1 120 R e ( 3 ( 2 x − 5 ) x 4 + 10 ( 12 y 2 − 1 ) x 3
− 30 ( 6 y 2 − 1 ) x 2 + 10 ( 3 y 4 + 3 y 2 − 1 ) x − 15 y 2 ( y 2 − 1 ) ) + y 6 ( ( 6 y 4 − 2 y 2 + 1 ) ( x − 4 ) x 7 − 2 ( 4 y 6 − 24 y 4 + 8 y 2 − 3 ) x 6 + 4 ( 6 y 6 − 15 y 4 + 5 y 2 − 1 ) x 5 − ( 28 y 6 − 48 y 4 + 16 y 2 − 1 ) x 4 + 2 ( 2 y 4 − 3 y 2 + 1 ) ( 4 x − 1 ) x 2 y 2 ) ) ] τ 3 ,
⋮
v 1 ( x , y , τ ) = − 16 ( 2 x 3 − 3 x 2 + x ) ( y 4 − y 2 ) + [ 32 R e ( 3 ( y 4 − y 2 + x 4 − 2 x 5 5 ) − 2 ( 12 y 2 − 1 ) x 3 + 6 ( 6 y 2 − 1 ) x 2 − 2 ( 3 y 4 + 3 y 2 − 1 ) x ) + 128 x 2 ( x − 1 ) 2 y ( ( 6 y 4 − 2 y 2 + 1 ) ( x − 2 ) x 3 − ( 8 y 6 − 18 y 4 + 6 y 2 − 1 ) x 2 + 2 ( 2 y 4 − 3 y 2 + 1 ) y 2 ( 2 x − 1 ) ) ] τ ,
v 2 ( x , y , τ ) = − 16 ( 2 x 3 − 3 x 2 + x ) ( y 4 − y 2 ) + 128 [ y ( ( 6 y 4 − 2 y 2 + 1 ) ( x − 4 ) x 7 − 2 ( 4 y 6 − 24 y 4 + 8 y 2 − 3 ) x 6 + 2 ( 12 y 6 − 30 y 4 + 10 y 2 − 2 ) x 5 − ( 28 y 6 − 48 y 4 + 16 y 2 − 1 ) x 4 + 2 ( 2 y 4 − 3 y 2 + 1 ) ( 4 x − 1 ) x 2 y 2 ) − 1 20 R e ( 3 ( 2 x − 5 ) x 4 + 10 ( 12 y 2 − 1 ) x 3 − 30 ( 6 y 2 − 1 ) x 2 + 10 ( 3 y 4 + 3 y 2 − 1 ) x − 15 y 2 ( y 2 − 1 ) ) ] τ + [ − 192 ( R e ) 2 ( 2 x − 1 )
× ( 5 x 2 − 5 x + 12 y 2 − 2 ) + 256 y 5 R e ( 12 ( 72 y 2 − 11 ) ( x − 4 ) x 7 + 2 ( 810 y 4 + 2326 y 2 − 263 ) x 6 − 30 ( 162 y 4 + 62 y 2 + 9 ) x 5 + 5 ( 1002 y 4 − 142 y 2 + 143 ) x 4 − 120 ( 16 y 4 − 4 y 2 + 3 ) x 3 + 15 ( 10 y 4 + 2 y 2 + 3 ) x 2 + 5 ( 2 y 4 − 3 y 2 + 1 ) y 2 ) + 512 ( y 2 − 1 ) ( x − 1 ) 3 x 3 ( 2 x − 1 ) y 2 ( x 2 ( x − 1 ) 2
− 4 ( 16 x 2 − 16 x + 9 ) y 6 + 2 ( 15 x 2 − 15 x + 19 ) ( x 2 − x + 1 ) y 4 − 2 ( 3 x 2 − 3 x + 4 ) ( x 2 − x + 1 ) y 2 ) ] τ 2 + [ − 512 15 ( 20 y ( ( 6 y 4 − 2 y 2 + 1 ) ( x − 4 ) x 7 − 2 ( 4 y 6 − 24 y 4 + 8 y 2 − 3 ) x 6 + 2 ( 12 y 6 − 30 y 4 + 10 y 2 − 2 ) x 5 − ( 28 y 6 − 48 y 4 + 16 y 2 − 1 ) x 4 + 2 ( 2 y 4 − 3 y 2 + 1 ) ( 4 x − 1 ) x 2 y 2 ) − 1 R e ( 3 ( 2 x − 5 ) x 4
+ 10 ( 12 y 2 − 1 ) x 3 − 30 ( 6 y 2 − 1 ) x 2 + 10 ( 3 y 4 + 3 y 2 − 1 ) x − 15 y 2 ( y 2 − 1 ) ) ) × ( 2 ( 30 y 4 − 6 y 2 + 1 ) ( x − 4 ) x 7 − 4 ( 28 y 6 − 120 y 4 + 24 y 2 − 3 ) x 6 + 8 ( 42 y 6 − 75 y 4
+ 15 y 2 − 1 ) x 5 − 2 ( 196 y 6 − 240 y 4 + 48 y 2 − 1 ) x 4 + 16 ( 14 y 6 − 15 y 4 + 3 y 2 ) x 3 − 4 ( 14 y 6 − 15 y 4 + 3 y 2 ) x 2 − 3 y R e ( 4 ( 2 x − 3 ) x 2 + 2 ( 2 y 2 + 1 ) ) x − 2 y 2 + 1 ) ] τ 3 ,
⋮
g 1 ( x , y , τ ) = 32 x y R e [ ( 4 y 2 − 3 x + 6 x 2 5 ) x 2 − ( 2 y 2 − 1 ) ( 3 x − 1 ) ] − 64 ( x − 1 ) 2 x 2 y 2 [ ( 10 y 4 − 9 y 2 + 3 ) ( x − 2 ) x 3 + ( 8 y 6 − 6 y 4 − y 2 + 3 ) x 2 − 2 y 2 ( y 2 − 1 ) 2 ( 4 x − 1 ) ] − [ 192 ( R e ) 2 ( 2 x − 1 ) ( 4 x 2 + 2 y 2 − 4 x − 1 ) y
− 128 R e ( 3 ( 50 y 4 − 18 y 2 + 1 ) ( x − 4 ) x 7 + 6 ( 84 y 6 + 68 y 4 − 32 y 2 + 3 ) x 6 − 6 ( 252 y 6 − 146 y 4 + 30 y 2 + 2 ) x 5 + ( 120 y 8 + 1388 y 6 − 1320 y 4 + 372 y 2 + 3 ) x 4 − 16 ( 15 y 6 + 16 y 4 − 30 y 2 + 12 ) x 3 y 2 + 2 ( 78 y 6 − 98 y 4 + 21 y 2 + 15 ) x 2 y 2 − 2 ( 18 x − 1 ) y 4 ( y 2 − 1 ) 2 ) ] τ ,
g 2 ( x , y , τ ) = − 64 x 2 ( x − 1 ) 2 y 2 ( ( 10 y 4 − 9 y 2 + 3 ) ( x − 2 ) x 3 + ( 8 y 6 − 6 y 4 − y 2 + 3 ) x 2 − 2 ( y 2 − 1 ) 2 ( 4 x − 1 ) y 2 ) + 32 y 5 R e ( 3 ( 2 x − 5 ) x 4 + 20 x 3 y 2 − 5 ( 2 y 2 − 1 ) ( 3 x − 1 ) x ) + [ 192 y ( R e ) 2 ( 2 x − 1 ) ( 4 x 2 − 4 x + 2 y 2 − 1 ) − 128 R e ( 3 ( 50 y 4 − 18 y 2 + 1 ) ( x − 4 ) x 7 + 6 ( 84 y 6 + 68 y 4 − 32 y 2 + 3 ) x 6
− 12 ( 126 y 6 − 73 y 4 + 15 y 2 + 1 ) x 5 + ( 120 y 8 + 1388 y 6
− 1320 y 4 + 372 y 2 + 3 ) x 4 − 16 ( 15 y 6 + 16 y 4 − 30 y 2 + 12 ) x 3 y 2 + 2 ( 78 y 6 − 98 y 4 + 21 y 2 + 15 ) x 2 y 2 − 2 y 4 ( y 2 − 1 ) 2 ( 18 x − 1 ) ) ] τ + [ 3456 y ( R e ) 3 ( 2 x − 1 ) − 256 ( R e ) 2 ( 9 ( 50 y 2 − 3 ) ( x − 4 ) x 7 + 6 ( 980 y 4 + 78 y 2 − 9 ) x 6 − 72 ( 245 y 4 − 68 y 2 + 3 ) x 5
+ 3 ( 1820 y 6 + 4490 y 4 − 1800 y 2 + 107 ) x 4 − 12 ( 910 y 6 − 205 y 4 − 45 y 2 + 13 ) x 3 + 2 ( 180 y 8 + 3174 y 6 − 2715 y 4 + 621 y 2 + 12 ) x 2 − 12 ( 30 y 6 + 74 y 4 − 105 y 2 + 33 ) x y 2
+ ( 78 y 6 − 70 y 4 − 9 y 2 + 21 ) y 2 ) + 1 ( M a ) 2 ( 96 y R e ( 4 ( 2 x − 3 ) x 2 + 2 ( 2 y 2 + 1 ) x − 2 y 2 + 1 ) − 64 x 2 ( x − 1 ) 2 ( ( 30 y 4 − 6 y 2 + 1 ) ( x − 2 ) x 3 − ( 56 y 6 − 90 y 4 + 18 y 2 − 1 ) x 2 + 2 ( 14 y 4 − 15 y 2 + 3 ) ( 2 x − 1 ) y 2 ) ) ] τ 2 ,
⋮
To test the convergence of the approximate solutions, we calculated γ k as:
γ 10 = ‖ ( u c ) 1 ( x , y , τ ) ‖ ‖ u ( x , y , 0 ) ‖ = 0 , γ 11 = ‖ ( u c ) 2 ( x , y , τ ) ‖ ‖ ( u c ) 1 ( x , y , τ ) ‖ = 0 ,
γ 20 = ‖ ( v c ) 1 ( x , y , τ ) ‖ ‖ v ( x , y , 0 ) ‖ = 714 ( 3227504 ( R e ) 2 − 12514788 R e + 1994117697 ) 14586 τ R e ,
γ 30 = ‖ ( g c ) 1 ( x , y , τ ) ‖ ‖ g ( x , y , 0 ) ‖ = 60 ( 49405942 ( R e ) 2 + 1467358893 ) 1253680 ( R e ) 2 − 21900879 R e + 292485765 τ R e ,
γ 21 = ‖ ( v c ) 2 ( x , y , τ ) ‖ ‖ ( v c ) 1 ( x , y , τ ) ‖ = 2 534905 τ [ ( 252795448503308124160 τ 2 + 13571076548224512000 ) ( R e ) 4 − 1150099707715306782720 ( τ 2 + 20037330051 41150401552 τ − 369495 66021376 ) ( R e ) 3 + ( 71129205032874519404544 τ 2 + 4131885499121329297920 τ + 44180660411272032158160 ) ( R e ) 2
− 185898066637793079398400 ( τ 2 + 19609947 23628440 τ + 5360355 13747456 ) R e + 7961346208200723126497280 ( τ 2 + 34459425 275978504 τ + 182807249625 141300994048 ) ] 0.5 ÷ [ 3506302275 ( 3227504 ( R e ) 2 − 12514788 R e + 1994117697 ) 0.5 R e ] ,
γ 31 = ‖ ( g c ) 2 ( x , y , τ ) ‖ ‖ ( g c ) 1 ( x , y , τ ) ‖ = τ R e ( M a ) 2 [ ( ( 2007409369056 ( R e ) 2 + 14901358071600 ) ( M a ) 4 + ( 1477832512 ( R e ) 2 − 165570645240 ) ( R e ) 2 ( M a ) 2 + ( 20792594 ( R e ) 2 + 4402076679 ) ( R e ) 4 ) ÷ ( 12 ( 49405942 ( R e ) 2 + 1467358893 ) ) ] 0.5 ,
⋮
such that γ 0 = γ 10 + γ 20 + γ 30 , γ 1 = γ 11 + γ 21 + γ 31 , ⋯ . For example, if M a = 0.1 , t = 0.1 , and R e = 1 such that τ = M a × t , then
γ 0 = 0.9991283888 < 1 , γ 1 = 0.7959676332 < 1 , ⋯ ,
if M a = 0.1 , t = 0.01 , and R e = 10 then
γ 0 = 0.0129736262 < 1 , γ 1 = 0.2936820051 < 1 , ⋯
In this section, we introduce the numerical computations of velocity components u, v, vorticity function w and stream function ψ , which have been obtained by the application of KRDTM and KPIA. All calculations are run by Maple 2017 software with used various values of Reynolds numbers and Mach numbers in the domain [ 0,1 ] 2 .
In
| u ( 0.5, y ,0.1 ) | Re = 10 | Re = 100 | Re = 400 | Re = 1000 |
|---|---|---|---|---|
| 0.0625 | −0.0620116263 | −0.0620117178 | −0.0620117080 | −0.0620117057 |
| 0.125 | −0.1210934186 | −0.1210937382 | −0.1210937303 | −0.1210937281 |
| 0.1875 | −0.1743158597 | −0.1743163948 | −0.1743163886 | −0.1743163864 |
| 0.25 | −0.2187492697 | −0.2187500032 | −0.2187499985 | −0.2187499963 |
| 0.3125 | −0.2514639638 | −0.2514648747 | −0.2514648714 | −0.2514648693 |
| 0.375 | −0.2695302503 | −0.2695313138 | −0.2695313119 | −0.2695313098 |
| 0.4375 | −0.2700184296 | −0.2700196175 | −0.2700196172 | −0.2700196153 |
| 0.5 | −0.2499987981 | −0.2500000797 | −0.2500000812 | −0.2500000796 |
| 0.5625 | −0.2065416607 | −0.2065430027 | −0.2065430066 | −0.2065430054 |
| 0.625 | −0.1367173508 | −0.1367187180 | −0.1367187250 | −0.1367187244 |
| 0.6875 | −0.0375962574 | −0.0375976129 | −0.0375976240 | −0.0375976244 |
| 0.75 | 0.0937511524 | 0.0937498465 | 0.0937498298 | 0.0937498282 |
|---|---|---|---|---|
| 0.8125 | 0.2602543373 | 0.2602531196 | 0.2602530957 | 0.2602530924 |
| 0.875 | 0.4648427531 | 0.4648416627 | 0.4648416294 | 0.4648416238 |
| 0.9375 | 0.7104460577 | 0.7104451334 | 0.7104450882 | 0.7104450798 |
| v ( x ,0.5,0.1 ) | ||||
| 0.0625 | 0.1520352246 | 0.1535990206 | 0.1537294237 | 0.1537555060 |
| 0.125 | 0.2443698776 | 0.2458276287 | 0.2459491556 | 0.2459734619 |
| 0.1875 | 0.2840233785 | 0.2853353452 | 0.2854446967 | 0.2854665674 |
| 0.25 | 0.2797924904 | 0.2809293365 | 0.2810240781 | 0.2810430265 |
| 0.3125 | 0.2404575825 | 0.2413982264 | 0.2414766095 | 0.2414922860 |
| 0.375 | 0.1747890513 | 0.1755190079 | 0.1755798316 | 0.1755919962 |
| 0.4375 | 0.0915546641 | 0.0920648498 | 0.0921073612 | 0.0921158634 |
| 0.5 | −0.0004740000 | −0.0001880250 | −0.0001641937 | −0.0001594275 |
| 0.5625 | −0.0925173262 | −0.0924556955 | −0.0924505555 | −0.0924495275 |
| 0.625 | −0.1757864500 | −0.1759449258 | −0.1759581261 | −0.1759607661 |
| 0.6875 | −0.2414853287 | −0.2418548488 | −0.2418856383 | −0.2418917961 |
| 0.75 | −0.2808158432 | −0.2813816962 | −0.2814288551 | −0.2814382869 |
| 0.8125 | −0.2849851434 | −0.2857257912 | −0.2857875328 | −0.2857998815 |
| 0.875 | −0.2452139088 | −0.2460994800 | −0.2461733253 | −0.2461880953 |
| 0.9375 | −0.1527436594 | −0.1537341432 | −0.1538167704 | −0.1538332975 |
| u ( 0.5, y ,0.1 ) | Re = 10 | Re = 100 | Re = 400 | Re = 1000 |
|---|---|---|---|---|
| 0.0625 | −0.0620116267 | −0.0620117178 | −0.0620117080 | −0.0620117057 |
| 0.125 | −0.1210934190 | −0.1210937382 | −0.1210937303 | −0.1210937281 |
| 0.1875 | −0.1743158600 | −0.1743163948 | −0.1743163886 | −0.1743163864 |
| 0.25 | −0.2187492699 | −0.2187500032 | −0.2187499985 | −0.2187499963 |
| 0.3125 | −0.2514639639 | −0.2514648747 | −0.2514648714 | −0.2514648693 |
| 0.375 | −0.2695302503 | −0.2695313137 | −0.2695313119 | −0.2695313098 |
| 0.4375 | −0.2700184292 | −0.2700196175 | −0.2700196172 | −0.2700196153 |
| 0.5 | −0.2499987975 | −0.2500000797 | −0.2500000812 | −0.2500000796 |
| 0.5625 | −0.2065416599 | −0.2065430027 | −0.2065430066 | −0.2065430054 |
| 0.625 | −0.1367173502 | −0.1367187180 | −0.1367187250 | −0.1367187244 |
| 0.6875 | −0.0375962576 | −0.0375976130 | −0.0375976241 | −0.0375976244 |
| 0.75 | 0.0937511505 | 0.0937498462 | 0.0937498298 | 0.0937498282 |
| 0.8125 | 0.2602543328 | 0.2602531191 | 0.2602530956 | 0.2602530923 |
| 0.875 | 0.4648427455 | 0.4648416618 | 0.4648416292 | 0.4648416237 |
| 0.9375 | 0.7104460475 | 0.7104451322 | 0.7104450879 | 0.7104450797 |
| v ( x ,0.5,0.1 ) | ||||
| 0.0625 | 0.1520352173 | 0.1535990205 | 0.1537294237 | 0.1537555060 |
| 0.125 | 0.2443698698 | 0.2458276285 | 0.2459491556 | 0.2459734619 |
| 0.1875 | 0.2840233713 | 0.2853353451 | 0.2854446967 | 0.2854665674 |
| 0.25 | 0.2797924846 | 0.2809293364 | 0.2810240781 | 0.2810430265 |
|---|---|---|---|---|
| 0.3125 | 0.2404575786 | 0.2413982263 | 0.2414766095 | 0.2414922860 |
| 0.375 | 0.1747890492 | 0.1755190079 | 0.1755798316 | 0.1755919962 |
| 0.4375 | 0.0915546634 | 0.0920648498 | 0.0921073612 | 0.0921158634 |
| 0.5 | −0.0004739999 | −0.0001880249 | −0.0001641937 | −0.0001594275 |
| 0.5625 | −0.0925173258 | −0.0924556954 | −0.0924505555 | −0.0924495274 |
| 0.625 | −0.1757864500 | −0.1759449258 | −0.1759581261 | −0.1759607660 |
| 0.6875 | −0.2414853294 | −0.2418548488 | −0.2418856382 | −0.2418917961 |
| 0.75 | −0.2808158450 | −0.2813816962 | −0.2857875328 | −0.2814382869 |
| 0.8125 | −0.2849851463 | −0.2857257912 | −0.2461733253 | −0.2857998815 |
| 0.875 | −0.2452139127 | −0.2460994800 | −0.2461733253 | −0.2461880953 |
| 0.9375 | −0.1527436637 | −0.1537341433 | −0.1538167704 | −0.1538332975 |
| Ref. [ | KRDTM | KPIA | |
|---|---|---|---|
| ψ min | −0.125 | −0.1263030704 | −0.1262878352 |
| x ( ψ min ) | 0.5 | 0.5 | 0.5 |
| y ( ψ min ) | 0.70703 | 0.70703125 | 0.703125 |
| u min | −0.2721659 | −0.2720274424 | −0.2720273443 |
| y ( u min ) | 0.40869 | 0.41015625 | 0.40625 |
| v min | −0.2886756 | −0.365196882 | −0.3649970593 |
| x ( v min ) | 0.78857 | 0.78515625 | 0.78125 |
| v max | 0.2886756 | 0.3590485063 | 0.3588814265 |
| x ( v max ) | 0.21143 | 0.21484375 | 0.21875 |
| u ( 0.5 ; 0.0625 ) | −0.062011718741 | −0.0619894102 | −0.0619894551 |
| u ( 0.5 ; 0.125 ) | −0.121093749988 | −0.1210481233 | −0.1210481633 |
| u ( 0.5 ; 0.1875 ) | −0.174316406238 | −0.1742486127 | −0.1742486451 |
| u ( 0.5 ; 0.25 ) | −0.218749999990 | −0.2186617447 | −0.2186617667 |
| u ( 0.5 ; 0.3125 ) | −0.251464843745 | −0.2513583768 | −0.2513583858 |
| u ( 0.5 ; 0.375 ) | −0.2695312499997 | −0.2694093550 | −0.2694093482 |
| u ( 0.5 ; 0.4375 ) | −0.270019531254 | −0.2698855135 | −0.2698854884 |
| u ( 0.5 ; 0.5 ) | −0.250000000006 | −0.2498576781 | −0.2498576334 |
| u ( 0.5 ; 0.5625 ) | −0.206542968755 | −0.2063966790 | −0.2063966152 |
| u ( 0.5 ; 0.625 ) | −0.1367187500006 | −0.1365733705 | −0.1365732911 |
| u ( 0.5 ; 0.6875 ) | −0.037597656248 | −0.0374586584 | −0.0374585703 |
| u ( 0.5 ; 0.75 ) | 0.093749999998 | 0.0938764774 | 0.0938765646 |
| u ( 0.5 ; 0.8125 ) | 0.260253906243 | 0.2603609875 | 0.2603610630 |
| u ( 0.5 ; 0.875 ) | 0.46484374998 | 0.4649238234 | 0.4649238796 |
| u ( 0.5 ; 0.9375 ) | 0.710449218737 | 0.7104941432 | 0.7104941808 |
| v ( 0.0625 ; 0.5 ) | 0.153808593744 | 0.1365030994 | 0.1365023819 |
|---|---|---|---|
| v ( 0.125 ; 0.5 ) | 0.24609374999 | 0.2298505049 | 0.2298497694 |
| v ( 0.1875 ; 0.5 ) | 0.28564453123 | 0.2709291888 | 0.2709285329 |
| v ( 0.25 ; 0.5 ) | 0.28124999999 | 0.2684293744 | 0.2684288605 |
| v ( 0.3125 ; 0.5 ) | 0.241699218747 | 0.2310463824 | 0.2310460353 |
| v ( 0.375 ; 0.5 ) | 0.175781250002 | 0.1674821344 | 0.1674819447 |
| v ( 0.4375 ; 0.5 ) | 0.092285156254 | 0.0864478792 | 0.0864478112 |
| v ( 0.5 ; 0.5 ) | 2.3e−14 | −0.00333375 | −0.003333749 |
| v ( 0.5625 ; 0.5 ) | −0.092285156254 | −0.0931287051 | −0.0931286946 |
| v ( 0.625 ; 0.5 ) | −0.175781250002 | −0.1741943413 | −0.1741943774 |
| v ( 0.6875 ; 0.5 ) | −0.241699218746 | −0.2377853663 | −0.2377854934 |
| v ( 0.75 ; 0.5 ) | −0.281249999989 | −0.2751626592 | −0.2751629021 |
| v ( 0.8125 ; 0.5 ) | −0.28564453123 | −0.2776041425 | −0.2776044993 |
| v ( 0.875 ; 0.5 ) | −0.24609374999 | −0.2364163347 | −0.2364167719 |
| v ( 0.9375 ; 0.5 ) | −0.153808593744 | −0.1429446548 | −0.1429451057 |
| Grid size | Ref. [ | KRDTM | KPIA | |||
|---|---|---|---|---|---|---|
| ψ | ω | ψ | ω | ψ | ω | |
| R e = 10 | ||||||
| 21 × 21 | 3.230e−7 | 1.008e−5 | 2.0098e−8 | 8.1319e−7 | 5.8382e−8 | 1.5946e−6 |
| 41 × 41 | 2.347e−8 | 7.740e−7 | 3.4138e−8 | 9.7680e−7 | 7.2296e−8 | 1.8527e−6 |
| 81 × 81 | 1.559e−9 | 5.168e−8 | 4.4644e−8 | 1.0757e−7 | 9.2885e−8 | 2.0028e−6 |
| R e = 100 | ||||||
| 21 × 21 | 8.087e−5 | 4.081e−3 | 4.2498e−9 | 3.6230e−7 | 4.2494e−9 | 3.6230e−7 |
| 41 × 41 | 7.120e−6 | 2.508e−4 | 5.1135e−9 | 3.6230e−7 | 5.1130e−9 | 3.6230e−7 |
| 81 × 81 | 4.927e−7 | 1.717e−5 | 5.6559e−9 | 3.6372e−7 | 5.6555e−9 | 3.6372e−7 |
| R e = 1000 | ||||||
| 41 × 41 | 3.322e−4 | 1.449e−2 | 5.0673e−9 | 3.6448e−7 | 5.0669e−9 | 3.6449e−7 |
| 81 × 81 | 3.916e−5 | 1.659e−3 | 5.7419e−9 | 3.6448e−7 | 5.7419e−9 | 3.6449e−7 |
| 161 × 161 | 2.795e−6 | 1.469e−4 | 6.1146e−9 | 3.6448e−7 | 6.1146e−9 | 3.6449e−7 |
methods at R e = 1 , M a = 0.01 and t = 0.1 with the numerical results which have been evaluated by using the finite volume method and introduced by [
In this paper, we applied the reduced differential transform method and the perturbation-iteration algorithm on the kinetically reduced local Navier-Stokes equations to find approximate solutions to the problem of lid-driven square cavity flow. The calculations in this study show that KRDTM and KPIA are fast and
successful techniques and yield remarkably good results to solve unsteady viscous incompressible flow problems at low Mach numbers. Therefore, the application of KRDTM and KPIA could be expanded to include various and multi-dimensions of flow problems. In addition, these methods can be combined with other methods to increase the accuracy of solutions.
The authors declare no conflicts of interest regarding the publication of this paper.
Al-Saif, A.-S.J. and Harfash, A.J. (2018) A Comparison between the Reduced Differential Transform Method and Perturbation-Iteration Algorithm for Solving Two-Dimensional Unsteady Incompressible Navier-Stokes Equations. Journal of Applied Mathematics and Physics, 6, 2518-2543. https://doi.org/10.4236/jamp.2018.612211