On Global Solution of Incompressible Navier-Stokes Equations - IAENG

Proceedings of the World Congress on Engineering 2014 Vol II, WCE 2014, July 2 - 4, 2014, London, U.K.

On Global Solution of Incompressible Navier-Stokes Equations Algirdas Antano Maknickas

Abstract—The fluid equations, named after Claude-Louis Navier and George Gabriel Stokes, describe the motion of fluid substances. These equations arise from applying Newton’s second law to fluid motion, together with the assumption that the stress in the fluid is the sum of a diffusing viscous term (proportional to the gradient of velocity) and a pressure term - hence describing viscous flow. Due to specific of NS equations they could be transformed to full/partial inhomogeneous parabolic differential equations: differential equations in respect of space variables and the full differential equation in respect of time variable and time dependent inhomogeneous part. Finally, orthogonal polynomials as the partial solutions of obtained Helmholtz equations were used for derivation of analytical solution of incompressible fluid equations in 1D, 2D and 3D space for rectangular boundary. Solution in 3D space for any shaped boundary is expressed in term of 3D global solution of 3D Helmholtz equation accordantly. Index Terms—Analytical solution, Incompressible fluid, Helmholtz equation, Navier-Stokes equations.

Aim of this article is to propose new approach for solution of incompressible fluid equations. The article has three basic parts: first part explains how to solve NS in one dimension, second part extend solution into two-dimensional space and, finally, third part summarize with three-dimensional space. II. PARABOLIC FORMULATION OF EQUATIONS Incompressible fluid equations are expressed as follow ∂v ρ + (v · ∇)v − µ∆v + ∇p = f, (1) ∂t ∂ρ + ∇ · (ρv) = 0, (2) ∂t where equation (2) for incompressible flow reduces to dρ dt = 0 or ρ = const due to ∇v = 0. Equations of fluid motion (1) could be expressed in full time derivative replacing covariant time derivative by

I. I NTRODUCTION In physics, the fluid equations, named after Claude-Louis Navier and George Gabriel Stokes, describe fluid substances motion. These equations arise from applying Newton’s second law to fluid motion, together with the assumption that the stress in the fluid is the sum of a diffusing viscous term (proportional to the gradient of velocity) and a pressure term - hence describing viscous flow. Equations were introduced in 1822 by the French engineer Claude Louis Marie Henri Navier [1] and successively re-obtained, by different arguments, by a several authors including Augustin-Louis Cauchy in 1823 [2], Simeon Denis Poisson in 1829, Adhemar Jean Claude Barre de Saint-Venant in 1837, and, finally, George Gabriel Stokes in 1845 [3]. Detailed and thorough analysis of the history of the fluid equations could be found in by Olivier Darrigol [4]. The invention of the digital computer led to many changes. John von Neumann, one of the CFD founding fathers, predicted already in 1946 that automatic computing machines’ would replace the analytic solution of simplified flow equations by a numerical’ solution of the full nonlinear flow equations for arbitrary geometries. Von Neumann suggested that this numerical approach would even make experimental fluid dynamics obsolete. Von Neumann’s prediction did not fully come true, in the sense that both analytic theoretical and experimental research still coexist with CFD. Crucial properties of CFD methods such as consistency, stability and convergence need mathematical study [5]. Manuscript received December 01, 2013; revised February 14, 2014. This work was partly supported by the Project of Scientific Groups (Lithuanian Council of Science), Nr. MIP-13204. A.A. Maknickas is with the Institute of Mechanical Sciences, Vilnius Gediminas Technical University, Saul˙etekio al. 11, LT-10223 Vilnius, Lithuania e-mail: ([email protected]).

ISBN: 978-988-19253-5-0 ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online)

d ∂ = + dt ∂t

(v · ∇).

(3)

So, we obtain dv 1 − a2 ∆v = (−∇p + f ). dt ρ

(4)

3 inhomogeneouspparabolic like equation for full time derivative, where a = µ/ρ. III. O NE DIMENSIONAL INHOMOGENEOUS SOLUTION Consider the initial-boundary value problem for v = v(x, t) dv 1 − a2 ∆v = (−∇p + f ) in Ω × (0, ∞), dt ρ v(x, 0) = v0 (x) x ∈ Ω, ∂v = 0 on ∂Ω × (0, ∞), ∂n

(5) (6) (7)

where p = p(x, t) and f = f (x, t), Ω ⊂ Rn , n the exterior unit normal at the smooth parts of ∂Ω, a2 a positive constant and v0 (x) a given function. So according to [6] equation (4), when x is normed to a = 1, could be rewritten as follow dv = vxx + Q(x, t), x ∈ Ω, t > 0. dt

(8)

We expand v and Q in the eigenfunctions sin (nπx) on mπx space Ω ∈ [0, L] where sin( nπx L ) and sin( L ) functions orthogonality could be applied. So, we obtain Q(x, t) =

∞ X n=1

qn (t) sin (

nπx ), L

(9)

WCE 2014

Proceedings of the World Congress on Engineering 2014 Vol II, WCE 2014, July 2 - 4, 2014, London, U.K. with qn (t)

1 I1 Z

=

I1

Z Q(x, t) sin (

nπx )dx, L

(10)

Ω

sin2 (

=

nπx )dx L

(11)

where p = p(x, y, t) and f = f (x, y, t), Ω ⊂ R2n , n the exterior unit normal at the smooth parts of ∂Ω, a2 a positive constant and v0x (x, y), v0y (x, y) a given function. So, when x and y are normed to a = 1, equation (4) could be rewritten as follow dv i i i = vxx + vyy + Qi (x, y, t), x, y ∈ Ω, t > 0. dt

Ω

and

∞ X

v(x, t) =

un (t) sin (

n=1

nπx ). L

(12)

A. Rectangular boundary We expand v and Q in the eigenfunctions exp ( jnπx Lx ) jmπy jnπx exp ( Ly ) on space Ω ∈ [0, Lx ] × [0, Ly ] where exp ( Lx )

Thus we get the inhomogeneous ODE nπ 2 un (t) = qn (t), u˙ n (t) + L whose solution is

(13)

un (t) = un (0) exp (−(nπ/L)2 t) Rt + qn (τ ) exp (−(nπ/L)2 (t − τ ))dτ,

0

(14)

1 I1

Z v0 (x) sin (

nπx )dx. L

(15)

i qmn (t)

Again, we substitute all obtained equations into (12), denote sin2 (x, y) = sin(πx)sin(πy) and have

I2

v(x, t) = ds(v0 (s)(

1 I1

nx 2 sin2 ( ns L , L ) exp (−(nπ/L) t)) +

Rt ∞ P n=1

1 I1

and

∂v(x, t) = 0. (17) ∂x This is equation of extreme for coordinate x. Solving this equation gives extreme point xex . Finally, solution of 1D incompressible Navier-Stokes equation is

2 i u˙ imn (t) + km,n uimn (t) = qmn (t), 2 2 nπ 2 km,n = L + mπ . Ly x

n=1

2

sin

(28)

2 uimn (t) = uimn (0) exp (−km,n t) t R i 2 (t − τ ))dτ, + qjmn (τ ) exp (−km,n

1 I1

nxex 2 sin2 ( ns L , L ) exp (−(nπ/L) t) +

(29)

where uimn (0) =

dτ Q(s, τ )

0 1 I1

(27)

0

Rt ∞ P

(26)

whose solution is

v(x, t) = n=1

nx my , ). Lx Ly

Thus we get the inhomogeneous ODE

∇·v =

ds(v0 (s)

uimn (t) exp2 (

m,n=1

Now we must apply continuity condition

Ω

∞ X

v i (x, y, t) =

dτ Q(s, τ )

nx 2 sin2 ( ns L , L ) exp (−(nπ/L) (t − τ ))). (16)

∞ P

(23)

ZZ 1 nx my = Qi (x, y, t) exp2 ( , )dΩ, (24) I2 Lx Ly ZZ Ω nx my 2 = (exp2 ( , )) dΩ (25) Lx Ly

0

×

nx my , ), Lx Ly

Ω

∞ P

n=1

Ω

×

i qmn (t) exp2 (

with

Ω

R

∞ X

Qi (x, y, t) =

m,n=1

un (0) =

R

0

jn πx jm πy ) functions orthogexp ( jmπy Ly ) and exp ( Lx ) exp ( Ly √ onality could be applied, where j = −1. Let’s denote dΩ = dxdy and exp2 (x, y) = exp (jπx + jπy). So, we obtain

0

where

(22)

nxex 2 ( ns L , L ) exp (−(nπ/L) (t

1 I2

ZZ

v0i (x, y) exp2 (

nx my , )dΩ. Lx Ly

(30)

Ω

− τ ))). (18)

Again, we substitute all obtained equations into (26) and have

If we investigate each point of fluid in moving coordinate system of this point, Galilean transform must by applied v(r0 − vt, t). IV. T WO DIMENSIONAL INHOMOGENEOUS SOLUTION Consider the initial-boundary value problem for v = v(x, y, t) i

dv 1 − a2 ∆v i = (−∇i p + fi ) in Ω × (0, ∞), (19) dt ρ v i (x, y, 0) = v0i (x, y) x, y ∈ Ω, (20) ∂v i = 0 on ∂Ω × (0, ∞), (21) ∂n


RR

v i (x, y, t) = 2 ds0 ds(v0i (s0 , s)S(s, s0 , x, y) exp (−km,n t) +

Ω

Rt

2 Qi (s0 , s, τ )S(s, s0 , x, y) exp (−km,n (t − τ )))dτ ),

0

S(s, s0 , x, y) = ∞ P m=1 n=1

1 I2

0

2 nx my exp2 ( Lnsx , ms Ly ) exp ( Lx , Ly ).

(31)

Now we must apply continuity condition ∇·v =

∂v x (x, y, t) ∂v y (x, y, t) + = 0. ∂x ∂y

(32)

WCE 2014

Proceedings of the World Congress on Engineering 2014 Vol II, WCE 2014, July 2 - 4, 2014, London, U.K. y x So, we obtain relation conditions between v0mnf and v0mnf , y x Qmnf and Qmnf x (v0nmf + Qxnmf )

nπ mπ y + (v0nmf + Qynmf ) = 0, Lx Ly

(33) V. T HREE DIMENSIONAL INHOMOGENEOUS SOLUTION

i where v0f , Qif are i v0nmf =

where m(n) notes dependence of m and n. If we investigate each point of fluid in moving coordinate system of this point, Galilean transform must by applied v(r0 − vt, t).

0

0 v0i (s0 , s) exp2 ( Lnsx , ms Ly )ds ds, Ω RR 0 Qinmf = I12 ds ds

1 I2

RR

(34)

Consider the initial-boundary value problem for v = v(x, y, z, t)

(35)

dv i 1 − a2 ∆v i = (−∇i p + fi ) in Ω × (0, ∞), (45) dt ρ v i (x, y, z, 0) = v0i (x, y, z) x, y, z ∈ Ω, (46) ∂v i = 0 on ∂Ω × (0, ∞), (47) ∂n

Ω

Rt 0 2 × Qi (s0 , s, τ ) exp2 ( Lnsx , ms Ly ) exp (kmn τ )dτ. 0

Finally, solutions of 2D incompressible Navier-Stokes equations are ∞ P x v x (x, y, t) = (v0mnf + Qxmnf ) m,n=1

jπmx(v y +Qy ) 2 × exp (− Ly (vx 0mnf+Qx mnf) + jπmy Ly − kmn t), mnf 0mnf ∞ P y v y (x, y, t) = (v0mnf + Qymnf ) m,n=1 x jπny(v0mnf +Qx mnf ) y Lx (v0mnf +Qy mnf )

× exp ( jπnx Lx −

2 − kmn t).

(36)

dv i i i i = vxx + vyy + vzz + Qi (x, y, z, t), dt

(37)

If we investigate each point of fluid in moving coordinate system of this point, Galilean transform must by applied v(r0 − vt, t).

(48)

where x, y, z ∈ Ω, t > 0. A. Rectangular boundary Let’s denote

B. Any shaped boundary For any shaped boundary ∂Ω, equation (23) could be replaced by ∞ X

Qi (x, y, t) =

where p = p(x, y, z, t) and f = f (x, y, z, t), Ω ⊂ R3n , n the exterior unit normal at the smooth parts of ∂Ω, a2 a positive constant and v0x (x, y, z), v0y (x, y, z), v0z (x, y, z) a given function. So, when x, y and z are normed to a = 1, equation (4) could be rewritten as follow

i mn mn qmn (t)H∂Ω (x)H∂Ω (y)

(38)

m,n=1

exp3 (x, y, z) = exp(jπx) exp(jπy) exp(jπz).

(49)

nx my pz , Ly , Lz ) We expand v and Q in the eigenfunctions exp3 ( L x on space Ω ∈ [0, Lx ] × [0, Ly ] × [0, Lz ] where jmπy jpπz jn0 πx jm0 πy exp ( jnπx Lx ) exp ( Ly ) exp ( Lz ), exp ( Lx ) exp ( Ly ) 0

exp ( jpLzπz ) functions orthogonality could be applied. Let’s denote dΩ = dxdydz. So, we obtain

and equation (26) by ∞ X

v i (x, y, t) =

mn mn uimn (t)H∂Ω (x)H∂Ω (y).

(39)

mn mn H∂Ω (x)H∂Ω (y)

where are partial solutions of Helmholtz 2D equation for given boundary ∂Ω and could be taken for example from [7]. So equation (31) transforms to ∞ P i v i (x, y, t) = (v0mnf + Qimnf ) mn mn 2 (y) exp (−kmn t), (40) ×H∂Ω (x)H∂Ω RR 1 i 0 mn mn 0 0 = I2mn v0 (s , s)H∂Ω (s)H∂Ω (s )ds ds, (41) Ω

Qimnf =

1

RR

I2mn

ds0 ds

Rt

where I2mn is expressed as follow ZZ mn mn I2mn = (H∂Ω (x)H∂Ω (y))2 dxdy.

i (t) = qmnp

n=1 m(n)n

(x)H∂Ω

I3 =

2 (y) exp (−km(n)n t),


RRR Ω RRR Ω

(50)

nx my pz Qi (Ω, t) exp3 ( L , Ly , Lz )dΩ, (51) x

nx my pz 2 (exp3 ( L , Ly , Lz )) dΩ x

(52)

and

(42)

∞ X

uimn (t) exp3 (

m,n,p=1

nx my pz , , ). (53) Lx Ly Lz

Thus we get the inhomogeneous ODE (43)

Applying of continuity condition ∇ · v = 0 gives similar to equations (33)(36)(37) relations between n and m. Finally, we obtain ∞ P i v i (x, y, t) = (v0m(n)nf + Qim(n)nf ) m(n)n

1 I3

v i (x, y, z, t) =

∂Ω

×H∂Ω

nx my pz , , ), Lx Ly Lz

with

dτ

0 Ω mn mn 0 2 ×Qi (s0 , s, τ )H∂Ω (s)H∂Ω (s ) exp (kmn τ )),

i qmn (t) exp3 (

m=1 n=1 p=1

m,n=1

i vmnf

∞ X

Qi (Ω, t) =

m,n=1

(44)

2 i u˙ imnp (t) + kmnp uimnp (t) = qmnp (t), 2 2 2 pπ nπ 2 + mπ + L , kmnp = L Ly x y

(54) (55)

whose solution is 2 uimnp (t) = uimnp (0) exp (−kmnp t) t R i 2 + qmnp (τ ) exp (−kmnp (t − τ ))dτ,

(56)

0

WCE 2014

Proceedings of the World Congress on Engineering 2014 Vol II, WCE 2014, July 2 - 4, 2014, London, U.K. where uimnp (0) = 1 I3

RRR Ω

mπy pπz v0i (x, y, z) exp3 ( nπx Lx , Ly , Lz )dΩ.

(57)

Again, we substitute all obtained equations into (53) and have v i (x, y, z, t) = RRR i 00 0 2 v0 (s , s , s)(S(Ω0 , Ω) exp (−kmnp t))dΩ0 + Ω

dΩ0

RRR

Rt

i

00

0

dτ

0

Ω 0

2 , Ω) exp (−kmnp (t

×Q (s , s , s, τ )(S(Ω − τ ))), (58) ∞ P 1 0 00 S(Ω0 , Ω) = I3 Smnp (s, s , s )Smnp (x, y, z), (59)

mnp where H∂Ω,k (x, y, z) are partial solutions of Helmholtz 3D equation for given boundary ∂Ω. and could be taken for example from [8].So equation (58) transforms to ∞ P i v i (x, y, z, t) = (v0mnpf + Qimnp ) m,n=1 mnp 2 ×H∂Ω,k (x, y, z) exp (−kmnp t), RRR mnp i i 00 0 v0mnpf = v0 (s , s , s)H∂Ω,k (s, s0 , s00 )dΩ0 , Ω RRR Rt Qimnp = dΩ0 dτ 0 Ω mnp 2 ×Qi (s00 , s0 , s, τ )H∂Ω,k (s, s0 , s00 ) exp (kmnp τ ).

(60)

∂v x (x,y,z,t) ∂x

+

∇·v = +

∂v y (x,y,z,t) ∂y

∂v z (x,y,z,t) ∂z

v i (x, y, z, t) =

∞ P

×H∂Ω,k = 0.

(61)

(71)

i (v0mnp(m,n)f + Qimnp(m,n) )

m,n=1 mnp(m,n)

Now we must apply continuity condition

(70)

Applying of continuity condition ∇ · v = 0 gives relations between n, m and p or p = p(m, n). Finally, we obtain

m,n,p=1

mπy pπz Smnp (x, y, z) = exp3 ( nπx Lx , Ly , Lz ).

(69)

2 (x, y, z) exp (−kmnp(m,n) t).

(72)

If we investigate each point of fluid in moving coordinate system of this point, Galilean transform must by applied v(r0 − vt, t).

So, we obtain relation conditions between n, m and p VI. C ONCLUSIONS

y nπ x (v0nmpf + Qxnmpf ) L + (v0nmpf + Qynmpf ) mπ Ly x pπ z +(v0nmpf + Qznmpf ) L = 0, z

(62)

i where v0f , Qif are i v0mnpf =

1 I3

v0i (s00 , s0 , s)Smnp (s, s0 , s00 )dΩ0 , (63) Ω RRR dΩ0 Qimnpf = I13

RRR

Ω

Rt

× dτ Qi (s00 , s0 , s, τ )Smnp (s, s0 , s00 ).

(64)

0

Finally, solutions of 3D incompressible Navier-Stokes are ∞ P

v i (x, y, z, t) =

R EFERENCES

i (v0mnpf + Qimnpf )

m,n=1

2 ×Smnp(m,n) (x, y, z) exp (−kmnp(m,n) t), 3 nx my p(m,n)z Smnp(m,n) (x, y, z) = exp ( Lx , Ly , Lz ),

(65) (66)

where integers p(m, n) must satisfy (62) relations for each m and n and could be expressed by substituting of p/Lz corresponding expression as the sum of two others n/Lx , m/Ly . If we investigate each point of fluid in moving coordinate system of this point, Galilean transform must by applied v(r0 − vt, t). B. Any shaped boundary For any shaped boundary ∂Ω, equation (50) could be replaced by Qi (x, y, z, t) =

∞ X

Due to the form of fluid equations they could be transformed into the full/partial inhomogeneous parabolic differential equations: differential equations in respect to space variables and full differential equations in respect to the time variable and inhomogeneous time dependent part. Finally, orthogonal polynomials as the partial solutions of obtained Helmholtz equations were used for derivation of analytical solution of velocities for incompressible fluid in 1D, 2D and 3D space for rectangular boundary. Solution in 3D space for any shaped boundary is expressed in term of 3D global solution of 3D Helmholtz equation accordantly.

mnp q i (t)H∂Ω,k (x, y, z)

(67)

mnp ui (t)H∂Ω,k (x, y, z),

(68)

[1] C. L. M. H. Navier, “M e´ moire sur les loisdu movement des fluides,” M´em. de l’ Acad. R. Sci. Paris, vol. 6, pp. 389–416, 1823. [2] A. L. Cauchy, “Recherches sur l’eqnilibre et le mouvement interieur des corps solides et fluides, elastiques ou non elastiques,” Bull. Philomat., vol. 1, pp. 9–13, 1823. [3] G. Stokes, “On the theories of the internal friction of fluids in motion,” Transactions of the Cambridge Philosophial Society, vol. 8, pp. 287– 305, 1845. [4] O. Darrigol, “Between hydrodynamics and elasticity theory: The first five births of the navier-stokes equation.” Arch. Hist. Exact Sci., vol. 56, pp. 95–150, 2002. [5] Centrum Wiskunde & Informatica, “History of theoretical fluid dynamics,” retrieved 11/29/2013. [6] B. Neta, Partial Differential Equations, MA 3132 Lecture Notes, Department of Mathematics Naval Postgraduate School, Monterey, California,. Online, 2012. [7] W. Read, “Analytical solutions for a helmholtz equation with dirichlet boundary conditions and arbitrary boundaries,” , Mathematical and Computer Modelling, vol. 24, no. 2, pp. 23–34, July 1996. [8] R. H. Hoppe, Computational Electromagnetics, Handout of the Course Held at the AIMS Muizenberg. Online, December 17–21 2007.

m,n,p=1

and v i (x, y, z, t) =

∞ X m,n,p=1


WCE 2014