On Global Solution of Incompressible Navier-Stokes equations

Transcription

1 On Global Solution of Incompressible Navier-Stokes equations Algirdas Antano Maknickas Institute of Mechanics, Vilnius Gediminas Technical University, Sauletekio al., Vilnius, ithuania December 3, 24 Abstract The fluid equations, named after Claude-ouis 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 D, 2D and 3D space for rectangular boundary. Solution in 2D and 3D space for any shaped boundary was expressed in term of 2D and 3D global solution of Helmholtz equation accordantly. Introduction In physics, the fluid equations, named after Claude-ouis 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 822 by the French engineer Claude ouis Marie Henri Navier [] and successively re-obtained, by different arguments, by a several authors including Augustin-ouis Cauchy in 823 [2], Simeon Denis Poisson in 829, Adhemar Jean Claude Barre de Saint-Venant in 837, and, finally, George Gabriel Stokes in 845 [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 946 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]. Aims of this article are 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 to two-dimensional space and, finally, third part summarize with three-dimensional space. 2 Parabolic formulation of equations Incompressible fluid equations are expressed as follow ( ) v ρ t + (v )v µ v+ p = f () ρ + (ρv) = (2) t

2 equation (2) for incompressible flow reduces to dρ dt = or ρ = const due to v =. Equations of fluid motion () could be expressed in full time derivative replacing covariant time derivative by So, we obtain d dt = + (v ) (3) t dv dt a2 v = ( p+f) (4) ρ 3 inhomogeneous parabolic like equation for full time derivative, a = µ/ρ. Tensor of the inner pressure of fluid for existing solution of velocities could be found by using of equations p i in = pe i +f i e i +µ i ( e i are eigenvectors of corresponding coordinate system. 3 v i e i ) (5) 3 Galilean transform invariance of Navier Stokes PDEs It is known fact that Galilean coordinate transform is invariance transform for Navier Stokes equations [6]. Definition 3.. A group G of transformations of v(r, t) is symmetry group of Navier Stokes equations if and only if g G,v a NS solution = gv a NS solution. Theorem 3.. A Galilean coordinate transformation group is symmetry group of Navier Stokes equations. Proof. Set Now we apply operator 3 onto ṽ Summation of left and right parts gives i= gu Gal v(r,t) = v(r ut,t)+u,u Rd ṽ(r,t) = v(r ut,t)+u (6) t ṽ(r,t) = t v(r ut,t) (u )v(r ut,t) (7) (ṽ )ṽ(r,t) = (v(r ut,t)+u) v(r ut,t) (8) t ṽ(r,t)+(ṽ )ṽ(r,t) = t v(r ut,t)+v(r ut,t) v(r ut,t) (9) On other hand, the transformations of spatial derivatives with respect to the coordinates are x = x x + x t t = x () Corollary 3.2. For not constant u the displacement of coordinates system of each fluid parcel moving with velocity u = v(r vdτ,t) transform full time derivative of Navier Stokes equations into partial time derivative Dv Dt = t v(r,t)+ t r r v(r,t) = t v(r = t v(r vdτ,t)+ t (r vdτ,t) = t v(r,t) vdτ) r v(r,t) So, NS equations could be solved at first in partial time derivative form or local coordinate system in agrangian point of view and than transformed into global coordinate system of Eulerean coordinates v(r,t) = ṽ(r,t)+ṽ(r ṽdτ,t) () 2

3 4 One dimensional inhomogeneous solution Consider the initial-boundary value problem for v = v(x, t) dv dt a2 v = ( p+f) in (, ) ρ (2) v(x,) = v (x) x (3) v = on (, ) n (4) p = p(x,t) and f = f(x,t), R n, n the exterior unit normal at the smooth parts of, a 2 a positive constant and v (x) a given function. So according to [7] equation (4), when x is scaled to a =, could be rewritten as follow dv dt = 2 v +Q(x,t), x, t > (5) x2 We expand v and Q in the eigenfunctions sin( nπx ) on space [,] sin(nπx ) and sin(mπx ) functions orthogonality could be applied. So, we obtain Q(x,t) = q n (t)sin( nπx ) (6) with and Thus we get the inhomogeneous ODE whose solution is n= q n (t) = I I = v(x,t) = Q(x,t)sin( nπx )dx (7) sin 2 ( nπx )dx = 2 n= (8) u n (t)sin( nπx ) (9) ( nπ ) 2un u n (t)+ (t) = q n (t), (2) u n (t) = u n ()e ( (nπ/)2t) + u n () = I q n (τ)e ( (nπ/)2 (t τ)) dτ (2) v (x)sin( nπx )dx (22) Again, we substitute all obtained equations into (9) and have v(x,t) = v (s)( sin( nπs I )sin(nπx t) )e( (nπ/)2 )ds + ds n= n= Q(s,τ)( sin( nπs I )sin(nπx 2 )e( (nπ/) (t τ)) )dτ (23) Now we must apply continuity condition v = xv(x,t) =. This is equation of extreme for coordinate x. Solving this equation gives extreme point x ex. Finally, solution of D incompressible Navier-Stokes equation is v(t) = v (s)( sin( nπs I )sin(nπx ex )sin( (nπ/)2 t))ds + ds n= n= Q(s,τ)( sin( nπs I )sin(nπx ex )sin( (nπ/)2 (t τ)))dτ (24) 3

4 If we investigate each point of fluid in moving coordinate system of this point, Galilean transform eq. () must by applied. 5 Two dimensional inhomogeneous solution Consider the initial-boundary value problem for v = v(x, y, t) dv i dt a2 v i = ρ ( ip+f i ) in (, ) (25) v i (x,y,) = v i (x,y) x,y (26) v i n = on (, ) (27) p = p(x,y,t) and f = f(x,y,t), R 2n, n the exterior unit normal at the smooth parts of, a 2 a positive constant and v(x,y),v x y (x,y) a given function. So, when x and y scale was determined to a =, equation (4) could be rewritten as follow 5. Rectangular boundary dv i dt = 2 v i x v i y 2 +Qi (x,y,t), x,y, t > (28) We will expand v and Q in base of orthogonal functions. At first, we must find vector potential Q o of Q so that Q = Q o. To any potential Q o, an arbitrary gradient field can be added to get another vector potential with the same curl every. For simplicity, the second component of Q o can be taken to be zero, since a gradient field can take care of that if needed. This means that some equations simplify Q o = Q x (29) Q o = Q 2 x 2 (3) These can be solved sequentially, namely Q o is determined using the first equation up to a function of x, while Q o is determined by the second equation, up to a function of x 2. Than we expand Q o as follow Now we could express vector Q as follow and Q x o = Q x = Q y = m,n= m,n= m,n= qmn x (t) = I mn I mn = q x mn(t)sin( nπx x y ) (3) q x mnp(t) x sin(nπx x y ) (32) q x mnp(t) y sin(nπx x y ) (33) ds ds 2 Q o (s,s 2,t)sin( nπs s s2 ) (34) d(sin( nπs s s2 )) 2 (35) The same way we will find vector v for t = by using equations (29) and (3). We must find vector potential v o of v so that v = v o. Than we expand v o as follow v x o = m,n= u x mn()sin( nπx x y ) (36) 4

5 Now we could express vector v for any t as follow v x = v y = Thus we get the inhomogeneous ODE whose solution is m,n= m,n= u x mn (t) x sin(nπx x y ) (37) u x mn (t) y sin(nπx x y ) (38) u x mn (t)+k2 mn ux mn (t) = qx mn (t), (39) ( ) 2 ( ) 2 nπ mπ kmn 2 = + (4) u x mn(t) = u x mn()e ( k2 mn t) + x y q x mn(τ)e ( k2 mn (t τ)) dτ (4) u x mn() = v I o(s x,s 2 )sin( nπs )ds ds 2 (42) mn s s2 Again, we substitute all obtained equations into (54) and have [i,j] [[x,y],[y,x]] v i (x,y,t) = ( ) i vo(s x,s 2 )( +( ) i ds ds 2 Qx o(s,s 2,τ)( m,n= m,n= I mn S(ns,ms 2 ) S(nx,my)e ( k2 mnp t) )ds ds 2 I mn S(ns,ms 2 ) S(nx,my)e ( k2 mnp (t τ)) )dτ (43) S(nx,mx 2 ) = sin( nπx x y ) (44) If we investigate each point of fluid in moving coordinate system of this point, Galilean transform eq. () must by applied. 5.2 Any shaped boundary For any shaped boundary we will use the similar equations (32), (33) Q i (x,x 2,t) = ( ) i m,n= i j and equations (37) and (38) for velocities v i (x,x 2,t) = ( ) i m,n= q mn(t) H,k (nx,mx 2 ) (45) u mn(t) H,k (nx,mx 2 ) (46) H,k (nx,my) are partial solutions of Helmholtz 2D equation for given boundary. and could be taken for example from [9]. The equations of inverse curl operator in any coordinate system could be obtained by solving equations (29) and (3). So, equation (43) transforms to v i (x,x 2,t) = ( ) i v mnf = Q mn = I mn = m,n= (v mnf +Q mn) I mn H,k (nx,mx 2 )e ( k2 mnp t) (47) v o(s,s 2 )H,k (ns,ms 2 )d (48) d Q o(s,s 2,τ)H,k (ns,ms 2 )e (k2 mn τ) dτ (49) d(h,k (ns,ms 2 )) 2 (5) 5

6 [i,j] [[x,x 2 ],[x 2,x ]] and denotes coordinate indexes. If we investigate each point of fluid in moving coordinate system of this point, Galilean transform eq. () must by applied. 6 Three dimensional inhomogeneous solution Consider the initial-boundary value problem for v = v(x,y,z,t) dv i dt a2 v i = ρ ( ip+f i ) in (, ) (5) v i (x,y,z,) = v i (x,y,z) x,y,z (52) v i n = on (, ) (53) p = p(x,y,z,t) and f = f(x,y,z,t), R 3n, n the exterior unit normal at the smooth parts of, a 2 a positive constant and v x(x,y,z),vy (x,y,z),vz (x,y,z) a given function. So, when x, y and z scale was determined to a =, equation (4) could be rewritten as follow 6. Rectangular boundary dv i dt = 2 v i x v i y v i z 2 +Qi (x,y,z,t), x,y,z, t >. (54) We will expand v and Q in base of orthogonal functions. At first, we must find vector potential Q o of Q so that Q = Q o. To any potential Q o, an arbitrary gradient field can be added to get another vector potential with the same curl every. For simplicity, the third component of Q o can be taken to be zero, since a gradient field can take care of that if needed. This means that some equations simplify Q2 o = Q x 3 (55) Q o = Q 2 x 3 (56) Q 2 o Q o = Q 3 x x 2 (57) These can be solved sequentially, namely Q 2 o is determined using the first equation up to a function of x and x 3, while Q o is determined by the second equation, up to a function of x 2 and x 3. The third equation can then be solved provided our solvability conditions holds. Than we expand Q o as follow Q i o = m,n,p= q z mnp =. Now we could express vector Q as follow and Q x = Q y = Q z = m,n,p= m,n,p= m,n,p= qmnp i (t) = q i mnp (t)sin(nπx x y )sin( pπz z ), i [,2] (58) (qmnp(t) y y qz mnp(t) z )sin(nπx )sin( pπz ) (59) x y z (qmnp(t) z z qx mnp(t) x )sin(nπx )sin( pπz ) (6) x y z (qmnp(t) x x qy mnp(t) y )sin(nπx )sin( pπz ) (6) x y z = ds ds 2 ds 3 Q i o (s,s 2,s 3,t)sin( nπs s s2 )sin( pπs 3 s3 ) (62) ds ds 2 ds 3 (sin( nπs s s2 )sin( pπs 3 s3 )) 2 (63) 6

7 The same way we will find vector v for t = by using equation (55), (56) and (57). Than we expand v o as follow v i o = m,n,p= Now we could express vector v for any t as follow v x = v y = v z = m,n,p= m,n,p= m,n,p= u z mnp (t) =. Thus we get the inhomogeneous ODE whose solution is u i mnp()sin( nπx x y )sin( pπz z ), i [,2] (64) (u y mnp(t) y uz mnp(t) z )sin(nπx )sin( pπz ) (65) x y z (u z mnp(t) z ux mnp(t) x )sin(nπx )sin( pπz ) (66) x y z (u x mnp(t) x uy mnp(t) y )sin(nπx )sin( pπz ) (67) x y z u i mnp(t)+kmnpu 2 i mnp(t) = qmnp(t), i (68) ( ) 2 ( ) 2 ( ) 2 nπ mπ pπ kmnp 2 = + + (69) u i mnp(t) = u i mnp()e ( k2 mnp t) + u i mnp() = Again, we substitute all obtained equations into (54) and have v i (x,y,z,t) = v j o (s,s 2,s 3 )( vo(s l,s 2,s 3 )( m,n,p= x y y q i mnp(τ)e ( k2 mnp (t τ)) dτ (7) v i o(s,s 2,s 3 )sin( nπs s s2 )sin( pπs 3 s3 )ds ds 2 ds 3 (7) m,n,p= + ds ds 2 ds 3 Qj o(s,s 2,s 3,τ)( ds ds 2 ds 3 Ql o(s,s 2,s 3,τ)( [i,j,k] [[x,y,z],[y,z,x],[z,x,y]] S(ns,ms 2,ps 3 ) S(nx,my,pz)e ( k2 mnp t) )ds ds 2 ds 3 S(ns,ms 2,s 3 ) x l S(nx,my,pz)e ( k2 mnp t) )ds ds 2 ds 3 m,n,p= m,n,p= S(ns,ms 2,ps 3 ) S(nx,my,pz)e ( k2 mnp (t τ)) )dτ S mnp (ns,ms 2,ps 3 ) x l S(nx,my,pz)e ( k2 mnp (t τ)) )dτ (72) S(nx,mx 2,px 3 ) = sin( nπx x y )sin( pπz z ) (73) If we investigate each point of fluid in moving coordinate system of this point, Galilean transform eq. () must by applied. 6.2 Any shaped boundary For any shaped boundary we will use the similar equations (59), (6), (6) ( Q i (x,x 2,x 3,t) = qmnp(t) j q x mnp(t) l ) H,k (nx,mx 2,px 3 ) (74) j x l m,n,p= [i,j,l] [[x,x2,x3],[x2,x3,x],[x3,x,x2]] and equations (65), (66) and (67) for velocities ( v i (x,x 2,x 3,t) = u j mnp(t) u l x mnp(t) ) H,k (nx,mx 2,px 3 ) (75) j x l m,n,p= 7

8 H,k (nx,my,pz) are partial solutions of Helmholtz 3D equation for given boundary. and could be taken for example from [9]. The equations of inverse curl operator in any coordinate system could be obtained from []. So, equation (72) transforms to v i (x,x 2,x 3,t) = vmnpf i = Q i mnp = = m,n,p= ( (vj mnpf +Qj mnp) (vl mnpf +Ql mnp ) )H,k (nx,mx 2,px 3 )e ( k2 mnp t) (76) x l v i o (s,s 2,s 3 )H,k (ns,ms 2,ps 3 )d (77) d Q i o(s,s 2,s 3,τ)H,k (ns,ms 2,ps 3 )e (k2 mnp τ) dτ (78) d(h,k (ns,ms 2,ps 3 )) 2 (79) [i,j,k] [[x,x 2,x 3 ],[x 2,x 3,x ],[x 3,x,x 2 ]] and denotes coordinate indexes. If we investigate each point of fluid in moving coordinate system of this point, Galilean transform eq. () must by applied. 7 Conclusions Due to the form of fluid equations they could be transformed into the full/partial inhomogeneous parabolic differential equations: partial differential equations in respect to space variables and full differential equations in respect to the time variable and inhomogeneous time dependent part. Velocity and outer forces density components were expressed in form of curl for obtaining solution satisfying continuity condition. Orthogonal polynomials as the partial solutions of obtained Helmholtz equations were used for derivation of analytical solution of velocities for incompressible fluid in D, 2D and 3D space for rectangular boundary. Solution in 2D and 3D space for any shaped boundary was expressed in term of 2D and 3D global solution of Helmholtz equation accordantly. Acknowledgement This work was partly supported by the Project of Scientific Groups (ithuanian Council of Science), Nr. MIP The author is very grateful all participants of seminar on 3 January, 24 and specially for prof. habil. dr. Konstantinas Pileckas for detailed discussions on completeness of obtained solutions. References [] Navier, C.. M. H. Mem acad. R. sci. paris, Vol. 6 pp , 823. [2] Cauchy, A.. Exercises de mathematique, p.83, Paris, 828. [3] Stokes. G. G. trans. Camb. Phil. Soc., vol 8, pp , 845. [4] Darrigol O., Between Hydrodynamics and Elasticity Theory: The First Five Births of the Navier-Stokes Equation. Arch. Hist. Exact Sci. 56, pp 95 5, 22. [5] Centrum Wiskunde and Informatica, History of theoretical fluid dynamics, retrieved /29/23 [6] Frisch, U. Turbulence: The egacy of A. N. Kolmogorov. Cambridge University Press, pp. 296, 995. [7] Beny Neta, Partial Differential Equations, MA 332 ecture Notes, Department of Mathematics Naval Postgraduate School, Monterey, California, p. 353, 22 [8] Read W.W. Analytical solutions for a helmholtz equation with dirichlet boundary conditions and arbitrary boundaries, Mathematical and Computer Modelling, V. 24, Nr 2, pp , July 996. [9] Hoppe Ronald H.W. Computational Electromagnetics, Handout of the Course Held at the AIMS Muizenberg, South Africa December 7 2, p., 27. 8

9 [] Sahoo S. Inverse Vector Operators, arxiv: v3, 8, 2. 9