On the Navier Stokes equations

Transcription

1 On the Navier Stokes equations Daniel Thoas Hayes April 26, 2018 The proble on the existence and soothness of the Navier Stokes equations is resolved. 1. Proble description The Navier Stokes equations are thought to govern the otion of a fluid in R 3, see [1]. Let u = u(x, t) R 3, p = p(x, t) R, and f = f(x, t) R 3 be the velocity, pressure, and given externally applied force respectively, each dependent on position x R 3 and tie t 0. The fluid is here assued to be incopressible with constant viscosity ν > 0 and to fill all of R 3. The Navier Stokes equations can then be written as with initial condition u + (u )u = ν 2 u p + f, (1) u = 0 (2) u(x, 0) = u 0 (3) where u 0 = u 0 (x) R 3. In these equations = ( x 1, x 2, x 3 ) is the gradient operator and 2 = 3 2 i=1 x 2 is the Laplacian operator. When ν = 0, equations (1), (2), (3) are called i the Euler equations. Solutions of (1), (2), (3) are to be found with u 0 (x + e j ) = u 0 (x), f(x + e j, t) = f(x, t) for 1 j 3 (4) where e 1 = (1, 0, 0), e 2 = (0, 1, 0), e 3 = (0, 0, 1). The initial condition u 0 is a given C divergence-free vector field on R 3 and α x β t f C αβγ(1 + t ) γ on R 3 [0, ) for any α, β, γ. (5) A solution of (1), (2), (3) would then be accepted to be physically reasonable if and u(x + e j, t) = u(x, t), p(x + e j, t) = p(x, t) on R 3 [0, ) for 1 j 3 (6) I provide a proof of the following stateent (D), see [2]. (D) Breakdown of Navier Stokes Solutions on R 3 /Z 3. u, p C (R 3 [0, )). (7) Take ν > 0. Then there exist a sooth, divergence-free vector field u 0 on R 3 and a sooth f on R 3 [0, ), satisfying (4), (5), for which there exist no solutions (u, p) of (1), (2), (3), (6), (7) on R 3 [0, ). 1

2 2. Proof of stateent (D) Herein I take f = 0. I seek an approxiation of the for u = p = 1 n L= 1 1 n 1 L= 1 l u L t l l e ıkl x t=0, (8) l p L t l l e ıkl x t=0 (9) to the solution of (1), (2), (3), (4), (5), (6) in light of Theore 1 and Theore 2 in the Appendix. Here u L = u L (t), p L = p L (t), ı = 1, k = 2π, and H L= H denotes the su over all L Z 3 with H L j H. Herein the sooth 1 divergence-free initial condition u 0 on R 3 is chosen to be u 0 = 1 L (L a L )δ e ıkl x L, 3 (10) L= 1 where δ i, j is the Kronecker delta defined by δ i, j = { 1, i = j 0, i j (11) and a L are constant vectors that are chosen such that u 0 R 3. ethod 1 Let u = p = n l u l t l t=0, (12) n 1 t l l p. (13) Substituting (12), (13) into (1) and equating like powers of t in accordance with Theore 1 yields l+1 u l+1 t=0 + l ( l u l t=0 ) u l t=0 =0 = ν 2 l u l p (14) where l =!(l )!. Substituting (12) into (2) and equating like powers of t in accordance with Theore 1 yields l u = 0. (15) 1 In this paper, sooth functions and C functions will both ean continuous functions whose derivatives and integrals are all continuous. 2

3 Applying to (14) and using the identities along with (15) gives 2 l+1 u l+1 t=0 = a = ( a) 2 a, (16) a = 0 (17) l ( l u l t=0 ) u l t=0 =0 Applying the inverse Laplacian 2 to (18) gives l+1 u l+1 t=0 = 2 l ( l u l t=0 ) u l t=0 =0 where Φ l ust satisfy the Laplace equation + ν 4 l u. (18) + ν 2 l u + Φ l (19) 2 Φ l = 0. (20) The required solution to (20) is Φ l = 0 in light of (4), (6). Equation (19) is then solved for l+1 u l+1 t=0 where l = 0, 1,..., n 1. Applying to (14) and noting (15) yields Applying 2 to (21) gives where 2 l p = l p = 2 l ( l u l t=0 ) u l t=0. (21) =0 l ( l u l t=0 ) u l t=0 + ψ l (22) =0 2 ψ l = 0. (23) Arbitrary constant ψ l R is the solution to (23) in light of (4), (6). Equation (22) is then solved for l p l t=0 where l = 0, 1,..., n 1. After truncating (12), (13) in their odes, expressions for (8), (9) fro ethod 1 are then known in ters of given functions. Note that for the Fourier series g = g e ıkl x L (24) L 0 where L 0 denotes the su over all L Z 3 with L 0, the 2 operator is defined herein as 2 g e ıkl x g e ıkl x L L = k 2 L 2. (25) L 0 L 0 3

4 ethod 2 Let u = p = 1 u e ıkl x L, (26) L= 1 1 L= 1 p L e ıkl x. (27) Substituting (26), (27) into (1) and equating like powers of e in accordance with Theore 2 yields u L + (u L ık)u = νk 2 L 2 u L ıklp L. (28) Substituting (26) into (2) and equating like powers of e in accordance with Theore 2 yields L u L = 0. (29) Applying L L to (28) and noting the vector identity along with (29) yields L 2 u L Equation (31) iplies u L a (b c) = (c a)b (b a)c (30) = L (L (u L ık)u ) νk 2 L 4 u L. (31) = ˆL ( ˆL (u L ık)u ) νk 2 L 2 u L (32) where the right hand side of (32) is 0 when L = 0 and ˆL = L/ L is the unit vector in the direction of L. Applying L to (28) and noting (29) gives ık L 2 p L = (u L ık)(u L) (33) iplying that where p 0 R is an arbitrary function of t. Let p L = (u L ˆL)(u ˆL) (34) u L = p L = n n 1 l u L t l, (35) l p L t l. (36) 4

5 Substituting (35) into (32) and equating like powers of t in accordance with Theore 1 yields l+1 u L l+1 t=0 = l =0 ˆL ( ˆL ( l u L l t=0 ık) u t=0) l νk 2 L 2 l u L. (37) Equation (37) is then solved for l+1 u L l+1 t=0 where l = 0, 1,..., n 1 and 1 L j 1. Substituting (35), (36) into (34) and equating like powers of t in accordance with Theore 1 yields l p L l t=0 = l =0 ( l u L l t=0 ˆL)( u t=0 ˆL) l. (38) Equation (38) is then solved for l p L where l = 0, 1,..., n 1 and 1 L j 1. Expressions for (8), (9) fro ethod 2 are then known in ters of given functions. At l = 0 in (37) it is found that u L t=0 = ˆL ( ˆL (u L t=0 ık)u t=0 ) νk 2 L 2 u L t=0. (39) In (39) with 1 L 2 3, u t=0 = 0 unless 2 = 3 and u L t=0 = 0 unless L 2 = 3. With L 2 = 3 and 2 = 3 the equation L 2 = 3 then iplies 2L = 3 which is not possible as an even nuber can not be equal to an odd nuber. Likewise, with L 2 = 1 and 2 = 3 the equation L 2 = 3 then iplies 2L = 1 which is not possible as an even nuber can not be equal to an odd nuber. With L 2 = 2 and 2 = 3 the equation L 2 = 3 then iplies L = 1 which is not possible as in this instance L {0, 2} when 1 L j 1, 1 j 1. Therefore u L t=0 = 3k 2 νu L t=0. (40) At O(t), I find that ethod 2 gives the sae result for (8) as given by ethod 1. At l = 1 in (37) it is found that 2 u L 2 t=0 = ˆL ( ˆL (( u L t=0 ık)u t=0 + (u L t=0 ık) u t=0 )) νk 2 L 2 u L t=0. (41) By a siilar arguent as that applied to (39) it is found in ethod 2 that In fact for l 0 it is found in ethod 2 that 2 u L 2 t=0 = 3k 2 ν u L t=0 = 9k 4 ν 2 u L t=0. (42) l+1 u L l+1 t=0 = ( 3k 2 ν) l+1 u L t=0. (43) 5

6 With ethod 1 for ν = 0, I find that u tt t=0 0 when truncated onto the odes with 1 L j 1. Therefore at O(t 2 ), the approxiation (8) found fro ethod 1 is different to the approxiation (8) found fro ethod 2. Because of this nonuniqueness at least one of the assuptions used was invalid. An exact solution Herein I denote u = (u, v, w) and x = (x, y, z). Let the initial condition be u 0 = (cos(k(x + y z)), cos(k(x y z)), cos(k(x + y z)) cos(k(x y z))) (44) which is consistent with (10). I used aple to find the aclaurin series of the solution (u, p) to (1), (2), (3), (4), (5), (6) using (44). The nonuniqueness of results found with ethod 1 and ethod 2 does occur when using (44). It appeared fro the aclaurin series of the solution (u, p) that v = cos(k(x y z))e νλt, (45) w = u cos(k(x y z))e νλt, (46) p = 0 (47) where λ = 3k 2. On substitution of (45), (46), (47) into (1), (2), (6), I found that u ust satisfy u + ( u y u z )eνλt cos(k(x y z)) ν 2 u = 0, (48) u x + u = 0, z (49) u(x + e j, t) = u(x, t), for 1 j 3. (50) For ν = 0, I used aple to find that the exact general solution of (48) is u = F(x, y + z, t cos(k(x y z)) y ) (51) cos(k(x y z)) where F is an arbitrary function. On atching (51) with (44) at t = 0, I then deduced that u = cos(2tk cos(k(x y z)) k(x + y z)). (52) The solution (52) also satisfies (49), (50). The resulting (u, p) was then verified to be an exact solution to (1), (2), (3), (4), (5), (6) for ν = 0. Integrating (52) with respect to t yields t sin(2tk cos(k(x y z)) k(x + y z)) u dt = (53) 2k cos(k(x y z)) which is undefined for soe values of x R 3 and t 0. For ν > 0, it is found that for the sall tie O(t) solution the equation (48) for u is u + ( u y u z )eνλt cos(k(x y z)) νλu = 0. (54) 6

7 Equation (54) iplies Then a change of variables yields Equation (49) becoes the initial condition (44) iplies (ue νλt ) + ( u y u ) cos(k(x y z)) = 0. (55) z τ = eνλt 1, νλ (56) u(x, t) = a(x, τ) τ (57) a τ + ( a y a ) cos(k(x y z)) = 0. (58) z and the spatially periodic boundary conditions (50) iply a x + a = 0, (59) z a(x, 0) = cos(k(x + y z)), (60) a(x + e j, τ) = a(x, τ) for 1 j 3. (61) Equations (58), (59), (60), (61) define an Euler proble. In light of this and (52), it is then clear that u = e νλt cos( 2k νλ (eνλt 1) cos(k(x y z)) k(x + y z)) (62) is valid for sall tie when ν > 0. Integrating (62) with respect to t yields t u dt = sin( 2k νλ (eνλt 1) cos(k(x y z)) k(x + y z)) 2k cos(k(x y z)) (63) which is undefined for soe values of x R 3 and t 0. Therefore stateent (D) is true. Appendix Theore 1 Providing that the aclaurin series Ă = n l A l t l n t=0 = l Ă l t l t=0 7 (64)

8 of the exact general solution to a Q th order partial differential equation Q A Q = Ψ (65) exists, it will solve the coefficients of t l for all l = 0, 1,..., n Q in (65) with A = Ă provided Ψ A=Ă is expandable in aclaurin series as Ψ A=Ă = l Ψ A=Ă t l (66) where n. Here all of the partial derivatives of A with respect to t are defined at t = 0. Proof of Theore 1 Since the aclaurin series of A exists and all of the partial derivatives of A with respect to t are defined at t = 0, one can integrate (65) Q ties with respect to t and then substitute the result into (64) to find Ă = n l Q Ψ l Q t l n t=0 = l Q Ψ dt A=Ă t l (67) where Q Ψ dt denotes the Qth integral of Ψ with respect to t. Substituting A = Ă into the residual r of (65) then gives r = n l Q Ψ A=Ă l Q t=0 t l Q (l Q)! l Ψ A=Ă t l (68) providing Ψ A=Ă is expanded in aclaurin series as in (66). Collecting like powers of t in (68) yields n Q l Ψ r = A=Ă t l l Ψ A=Ă t l (69) which shows that Theore 1 is true. Theore 2 Providing that the Fourier series à = N P(A, e ıkl x )e ıkl x = N of the exact general solution to a Q th order partial differential equation P(Ã, e ıkl x )e ıkl x (70) Q A Q = Ψ (71) 8

9 exists, it will solve the coefficients of e ıkl x for all N L j N in (71) with A = à provided Ψ A=à is expandable in Fourier series as Ψ A=à = L= P(Ψ A=Ã, e ıkl x )e ıkl x (72) where N. Here A is spatially periodic and sooth for all x R 3, k > 0 is a constant, and P(h, e ıkl x ) denotes the projection of h onto e ıkl x. Proof of Theore 2 Since the Fourier series of A exists where A is spatially periodic and sooth for all x R 3, one can integrate (71) Q ties with respect to t and then substitute the result into (70) to find à = N P( Ψ dt, e ıkl x )e ıkl x = Q N Substituting A = à into the residual r of (71) then gives r = Q Q N P( Q Ψ dt A=Ã, e ıkl x )e ıkl x P( Ψ dt, e ıkl x )e ıkl x A=Ã. (73) Q L= P(Ψ A=Ã, e ıkl x )e ıkl x (74) providing Ψ A=à is expanded in Fourier series as in (72). Equation (74) can be written as r = N P(Ψ A=Ã, e ıkl x )e ıkl x which shows that Theore 2 is true. References L= P(Ψ A=Ã, e ıkl x )e ıkl x (75) [1] Batchelor, G. K An introduction to fluid dynaics. Cabridge University Press: Cabridge. [2] Fefferan, C. L Existence and soothness of the Navier Stokes equation. Clay atheatics Institute: official proble description. 9