Erratum to “Method of Analytical Resolution of the Navier-Stokes Equations” [Open Journal of Fluid Dynamics Vol.13 No.5, (2023) 226-231]

Abstract

The original online version of this article (Jean Luc Wendkouni Tougma (2023) Method of Analytical Resolution of the Navier-Stokes Equations. Volume 13, 226-231 https://doi.org/10.4236/ojfd.2023.135017 was published with one method missing in Results and Discussion section. The author wishes to add a new section as follows.

Keywords

Erratum

Share and Cite:

Tougma, J. (2025) Erratum to “Method of Analytical Resolution of the Navier-Stokes Equations” [Open Journal of Fluid Dynamics Vol.13 No.5, (2023) 226-231]. Open Journal of Fluid Dynamics, 15, 133-135. doi: 10.4236/ojfd.2025.153008.

1. Results and Discussion 2

Then Navier Stokes equations become ν=α the viscosity constant:

v t +( N10 ) n=1 + v 500n+2 n! 10 2 p=0 N10 pv=f+α 2 v 1 ρ P (1)

v t +( N10 ) n=1 + v 500n+2 n! ( N10 )( N9 ) 2× 10 2 v=f+α 2 v 1 ρ P (2)

Let v=h( t )g( r ) with r{ x,y,z } a coordinate:

1 h h t +( N10 ) n=1 + v 500n+1 n! ( N10 )( N9 ) 2× 10 2 = f v +α 1 g 2 g 1 vρ P (3)

with 1 h h t = K 2 ; K 2 = 1 τ c the temporal decay, we obtain:

h= h 0 exp( t τ c ) (4)

Let’s take ω 2 = 1 τ c + ( N10 )( N9 ) 2× 10 2 we get:

fd r 2 +α d 2 v 1 ρ Pd r 2 ( N10 ) n=1 + v 500n+2 n! d r 2 + ω 2 vd r 2 =0 (5)

f r 2 2 +α v 2 2 Pr ρ ( N10 ) n=1 + 1 n! v 500n+2 d r 2 + ω 2 vd r 2 =0 (6)

with g=exp( ar ) we have dv=avdr that implies dr= 1 a dv v , then we get:

a 2 f r 2 2 + a 2 α v 2 2 a 2 Pr ρ ( N10 ) n=1 + 1 n! v 500n d v 2 + ω 2 1 v d v 2 =0 (7)

a 2 f r 2 2 + a 2 α v 2 2 a 2 Pr ρ ( N10 ) n=1 + v 500n+2 n!( 500n+1 )( 500n+2 ) + ω 2 [ ln( v )1 ]v=0 (8)

a 2 f r 2 2 + a 2 α v 2 2 a 2 Pr ρ ( N10 ) n=1 + v 500n+2 n!( 500n+1 )( 500n+2 ) = ω 2 [ 1ln( v ) ]v (9)

a and τ c must numerically verify for v :

v( r{ x,y,z } )= a 2 ω 2 ( 1+ t τ c ar ) { f r 2 2 +α exp[ 2( ar t τ c ) ] 2 Pr ρ ( N10 ) n=1 + exp[ ( 500n+2 )( ar t τ c ) ] n!( 500n+1 )( 500n+2 ) a 2 } (10)

by continuity equation, a must also be verified:

Dρ Dt =av (11)

( t +v )ρ=av (12)

( t +v ) ρ v =a (13)

( t +v ) ρ v =a (14)

with λ=a , we get:

{ λ liquid =( t +v )ρ ρ t for a liquid ρ liquid v fluid λ gaz ρfor a gaz ρ gaz v gaz

v( r{ x,y,z } )= λ 2 ω 2 ( 1+ t τ c λr ) { f r 2 2 +α exp[ 2( λr t τ c ) ] 2 Pr ρ ( N10 ) n=1 + exp[ ( 500n+2 )( λr t τ c ) ] n!( 500n+1 )( 500n+2 ) λ 2 } (15)

Conflicts of Interest

The authors declare no conflicts of interest.

Copyright © 2025 by authors and Scientific Research Publishing Inc.

Creative Commons License

This work and the related PDF file are licensed under a Creative Commons Attribution 4.0 International License.