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Correct Expression of Material and Application to the Navier-Stokes Equation —– The solution existence condition of Navier-Stokes equation

Bo-Hua Sun1 1 School of Civil Engineering & Institute of Mechanics and Technology Xi’an University of Architecture and Technology, Xi’an 710055, China http://imt.xauat.edu.cn email: [email protected] (Dated: June 15, 2020) The material derivative is important in continuum physics. This paper shows that the expression d ∂ dt = ∂t + (v · ∇), used in most literature and textbooks, is incorrect. This article presents correct d(:) ∂ expression of the material derivative, namely dt = ∂t (:) + v · [∇(:)]. As an application, the form- solution of the Navier-Stokes equation is proposed. The form-solution reveals that the solution existence condition of the Navier-Stokes equation is that ”The Navier-Stokes equation has a solution if and only if the determinant of flow velocity is not zero, namely det(∇v) 6= 0.” Keywords: material derivative, velocity gradient, calculus, tensor determinant, Navier-Stokes equa- tions, solution existence condition

In continuum physics, there are two ways of describ- derivative as in Ref. 1. Therefore, to revive the great ing continuous media or flows, the Lagrangian descrip- influence of Landau in physics and fluid mechanics, we at- tion and the Eulerian description. In the Eulerian de- tempt to address this issue in this dedicated paper, where scription, the material derivative with respect to time we revisit the material derivative to show why the oper- d ∂ dv ∂v must be defined. For mass density ρ(x, t), flow velocity ators dt = ∂t + (v · ∇) and expression dt = ∂t + (v∇)v v(x, t) = vkek, and stress tensor σ = σijei ⊗ ej, the are incorrect, and derive a correct expression by using dρ ∂ρ ∂ρ standard tensor calculus [21–24]. material are given by: dt = ∂t + v1 ∂x1 + ∂ρ ∂ρ dv ∂v ∂v ∂v ∂v In Landau & Lifshitz,[1] the material derivative is given v2 ∂x2 + v3 ∂x3 ,, dt = ∂t + v1 ∂x1 + v2 ∂x2 + v3 ∂x3 , and dσ ∂σ ∂σ ∂σ ∂σ as: dt = ∂t + v1 ∂x1 + v2 ∂x2 + v3 ∂x3 , respectively, where t is time, x = xkek is position, ek is a base vector and ∂v ∂v ∂v ∂xk dx + dy + dz = (dr · grad)v. (1) vk = ∂t is the flow velocity. ∂x ∂y ∂z In most popular textbooks, handbooks and encyclope- dia, such as Refs. 1–12, the above material derivatives Although Landau’s fluid mechanics is well known world- dρ ∂ρ dv ∂v wide, the above expression is incorrect. are expressed as: dt = ∂t +(v ·∇)ρ, dt = ∂t +(v ·∇)v, and dσ = ∂σ + (v · ∇)v, where the gradient operator dt ∂t Proof: Since v · grad = v · ∇ = (viei) · ∂ ∇ = ek . ∂xk (∂jej) = vi,jei · ej = vi,jδij = divv, and To simplify the aforementioned equations further, a d- (v · grad)v = (v · ∇)v = (vj,iei · ej)(vkek) = d ∂ ifferential operator dt = ∂t + (v · ∇) is introduced. This δijvj,ivkek = vi,ivkek = (divv)v, where the operator is used by most fluid mechanics textbooks, in- ∂v1 ∂v2 ∂v3 divergence is divv = ∂x1 + ∂x2 + ∂x3 , so cluding well-known graduate textbooks such as Landau (divv)v = ( ∂v1 + ∂v2 + ∂v3 )v. Thus: & Lifshitz,[1, 2] Prandtl,[3, 4] Anderson,[5] Pope,[6] Cen- ∂x1 ∂x2 ∂x3 gel & Cimbala,[7], Kundu et al.[8], Woan [9], wikipedia ∂v ∂v ∂v (v · grad)v 6= v + v + v . (2) [10, 11], Britannica [12] and even mathematical paper 1 ∂x1 2 ∂x2 3 ∂x3 [13]. In the latest version of Landau & Lifshitz,[14] the material derivative of flow velocity is given in another In the latest version of Landau & Lifshitz,[14] for rea- dv ∂v form: dt = ∂t + (v∇)v. sons which are unclear, the material derivative expression Besides the books mentioned above, there is a large dv ∂v has been changed to dt = ∂t + (v∇)v; this is also incor- amount of literature in fluid mechanics taking the same rect. expressions as Refs. 1 and 14. This makes it difficult for readers to identify which expression for the materi- Proof: Since v∇ = (viei)(∂jej) = vi,jeiej, al derivative is correct, causing great confusion to both students and scholars. (v∇)v = (vj,ieiej)(vkek) = (vj,ieiej)(vkek), Although some authors, such as Lighthill,[15] which indicates that v∇ is a 3rd-order tensor Batchelor,[16] Frisch,[17], Nhan & Nam [18], Xie,[19] and rather than a vector. Thus: Zhao [20] have used the correct expression, it seems that ∂v ∂v ∂v most fluid mechanics textbooks and academic literature (v∇)v 6= v + v + v . (3) have adopted an incorrect expression for the material 1 ∂x1 2 ∂x2 3 ∂x3

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T ∂σ To obtain the correct formulation of the material (∇σ) , hence dσ = ∂t dt + (σ∇) · vdt, or dividing both derivative, we perform some basic tensor calculus [21, 22]. dσ ∂σ sides by dt leads to dt = ∂t +(σ∇)·v. Since (σ∇)·v = The operator is a vector differential operator, and de- v · (∇σ), ∂ fined by ∇ = ei . An important note concerning the ∂xi dσ ∂σ del operator ∇ is in order. Two types of are = + v · (∇σ). (6) used in continuum physics: left and right gradients. The dt ∂t left gradient is the usual gradient and the right gradient If an operator must be introduced for the material is the transpose of the forward gradient operator. derivative, it should be in the following form: To see the difference between the left and right of gradients, consider a vector function A = Ai(x)ei. d(:) ∂ ∂ = (:) + v · [∇(:)]. (7) The left gradient of a vector A is ∇ ⊗ A ≡ ej ⊗ ∂xj dt ∂t ∂Ai (Aiei) = ej ⊗ ei = Ai,jej ⊗ ei, or written as ∇A ≡ ∂xj As an application, the Navier-Stokes equations of in- ∂ ∂Ai ej (Aiei) = ejei = Ai,jejei. The right gradient compressible flow can be expressed as follows: ∂xj ∂xj ∂ ∂Ai of a vector A is A ⊗ ∇ ≡ (Aiei) ⊗ ej = ei ⊗ ∂xj ∂xj ∂v 1 2 ∂ + v · (∇v) = − ∇p + ν∇ v, (8) ej = Ai,jei ⊗ ej, or written as A∇ ≡ (Aiei)ej = ∂t ρ ∂xj ∂Ai ∂Ai eiej = Ai,jeiej, where Ai,j = [21–24]. ∇ · v = 0. (9) ∂xj xj The gradient of a scalar function is a vector, the diver- The Eq. 8 is momentums equation and Eq. 9 is mass gence of a vector-valued function is a scalar ∇ · A, and conservation equation. In which, v(x, t) is flow velocity the gradient of a vector-valued function is a second-order field, ρ is constant mass density, p(x, t) is flow pressure, tensor ∇A. Although the del operator has some of the ν is kinematical viscosity, t is time, x = xke is position properties of a vector, it does not have them all, because k coordinates, e is a base vector and v is flow velocity, it is an operator. For instance, ∇ · A is a scalar (called k ∇ = e ∂ is gradient operator, and ∇2 = ∇ · ∇. the divergence of A) whereas A·∇ is a scalar differential k ∂xk Applying the divergence operation to both sides of the operator, where A is a vector. Thus the del operator ∇ equation Eq.8 and use the mass conservation does not commute in this sense [21–24]. leads to a pressure equation: ∇2 ·(p1) = −ρ∇·(v·∇v) = It worth to emphasise that the right gradient of a vec- −ρ{(∇v)2 + [∇(v∇)] · v}, where 1 = e e is an identity tor A∇ is a more natural one, is often used in defining k k tensor and v∇ = (∇v)T . the gradient tensor, displacement gradien- The NavierõStokes existence and smoothness prob- t tensor, and velocity gradient tensor. It is clear that lem is an open problem in mathematics [25], regardless of A∇ = (∇A)T . numerous abstract studies that have been done by math- With the above-mentioned understanding of tensor ematicians. Now the question is that, is it possible to calculus, we material derivative expression for scalar, vec- propose a simple criteria on the solution existence of the tor and the 2nd order tensor as follows: Navier-Stokes equation without complicated mathemat- 1. Mass density ρ = ρ(x, t) is a scalar-valued function ics. of x and t and its differential is dρ = ∂ρ dt + ∂ρ · ∂x dt. ∂t ∂x ∂t In order to find some useful information from the ∂ρ ∂ρ For a scalar ∂x = ρ∇ = ∇ρ, so dρ = ∂t dt+(ρ∇)·vdt = Navier-Stokes equation, let’s have a look at the meaning ∂ρ ∂t dt + v · (∇ρ)dt, or, dividing both sides by dt, of v ·(∇v) in Eq. 1. This term is called convective accel- eration that is caused by the flow velocity gradient. It is dρ ∂ρ ∂ρ = + (ρ∇) · v = + v · (∇ρ). (4) obvious that the convective acceleration v · (∇v) is the dt ∂t ∂t central point of the Navier-Stokes equations. Without the convective acceleration, the solution existence would 2. Flow velocity v = v(x, t) is a vector-valued function ∂v ∂v ∂x not be a problem at all. The understanding on the con- of x and t and its differential is dv = ∂t dt + ∂x · ∂t dt. ∂v T vective term is quite important for the study on the so- For a vector ∂x = v ⊗ ∇ = v∇ = (∇v) . Strictly lution existence should be focus on v · (∇v). Therefore, speaking, the right gradient of v should be written as we attack the open problem from the v · (∇v). v ⊗ ∇; here, in order to agree as far as possible with Assuming the determinant of the velocity gradient is conventional presentation, we write v ⊗ ∇ = v∇ and, not zero, namely det ∇v 6= 0, the form-solution of the similarly for the left gradient, ∇⊗v = ∇v. Hence (v∇)· Navier-Stokes momentum equation in Eq.(8) can be ex- ∂v ∂ρ v = v · (∇v) and dv = ∂t dt + (v∇) · vdt = ∂t dt + v · pressed as follows (∇v)dt, or, dividing both sides by dt,   2 1 ∂v −1 dv ∂v ∂v v = ν∇ v − ∇p − · (∇v) , (10) = + (v∇) · v = + v · (∇v). (5) ρ ∂t dt ∂t ∂t equivalently 3. The 2nd order stress tensor σ = σ(x, t) is a tensor-   valued function of x and t and its differential is dσ = −T 2 1 ∂v ∂σ ∂σ ∂x ∂σ v = (∇v) · ν∇ v − ∇p − , (11) ∂t dt + ∂x · ∂t dt. For an arbitrary tensor ∂x = σ∇ 6= ρ ∂t Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 June 2020 doi:10.20944/preprints202003.0030.v4

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equivalently The great challenge to find solution of the Navier- Stokes equation are all from the existence of the velocity  p ∂v  v = ∇ · (ν∇v − 1) − · (∇v)−1, (12) gradient ∇v. The form solution in Eq.(18) reveals that ρ ∂t the difficulty of finding a solution for N-S equations is equivalently because of existence of the nonlinear term v · (∇v), or in other words, due to the existence of the velocity field  p ∂v  gradient ∇v. Accordingly, the solution of the Navier- v = (∇v)−T · ∇ · (ν∇v − 1) − . (13) ρ ∂t Stokes equation will be blowup as det ∇v tends to an infinitesimal, and has no solution when det ∇v = 0. The new formats of Naver-Stokes equations in Eqs. 10, The 2D N-S equations can be written as follows: 11, Eq.(12) and Eq.(13) have never been seen in litera- 2 2 ture. They are formulated for the first time by Bo-Hua ∂v1 ∂v1 ∂v1 ∂ v1 ∂ v1 1 ∂p + v1 + v2 = ν( 2 + 2 ) − , Sun [24]. Those form-solutions provide a rich information ∂t ∂x1 ∂x2 ∂x1 ∂x2 ρ ∂x1 on the solution existence. (19) Therefore we have a conjecture as follows: 2 2 ∂v2 ∂v2 ∂v2 ∂ v2 ∂ v2 1 ∂p + v1 + v2 = ν( 2 + 2 ) − . Conjecture 1 The 3D Navier-Stokes equation has a so- ∂t ∂x1 ∂x2 ∂x1 ∂x2 ρ ∂x1 lution if and only if the determinant of flow velocity gra- (20) dient is not zero, namely ∂vi where vi,j = . The above equations can be expressed ∂xj det(∇v) 6= 0. in matrix format Although we have successfully split the velocity field       ∂ v1 v1,1 v1,2 v1 v from the convective term v · (∇v), the calculation of + (21) ∂t v2 v2,1 v2,2 v2 the inverse of the velocity gradient (∇v)−1 is still great 1 ∂p !   challenge. ρ ∂x1 2 v1 = − 1 ∂p + ν∇ , (22) According to the Cayley-Hamilton theorem [22, 23, 26, v2 ρ ∂x2 27], for the 2nd order tensor ∇v, the following polyno- 2 2 2 ∂ ∂ v2 mial holds: where the 2D Laplace operator ∇ = 2 + 2 . ∂x1 ∂x2 3 2 The 2D velocity gradient is (∇v) − I1(∇v) + I2∇v − I31 = 0, (14)   1 2 2 v1,1 v1,2 where I1 = tr(∇v) = ∇ · v,I2 = 2 [(tr∇v) − tr(∇v) ] ∇v = , (23) v2,1 v2,2 and I3 = det(∇v). Hence, for the case of det(∇v) 6= 0, we have: its determinant is (∇v)2 − I ∇v + I 1 (∇v)−1 = 1 2 . (15) det(∇v) = v v − v v , (24) det(∇v) 1,1 2,2 1,2 2,1 and the inverse of the 2D velocity gradient is thus For incompressible flow, the divergence of velocity gra-   dient is zero, namely, I1 = tr(∇v) = ∇ · v = 0, thus −1 1 v2,2 −v1,2 1 2 2 1 2 (∇v) = I2 = 2 [(tr∇v) − tr(∇v) ] = − 2 tr(∇v) . Therefore, det(∇v) −v2,1 v1,1 the inverse of the velocity gradient for incompressible flow  v −v  takes a simpler form: 2,2 1,2 −v2,1 v1,1 = . (∇v)2 − 1 1tr(∇v)2 v v − v v (∇v)−1 = 2 . (16) 1,1 2,2 1,2 2,1 (25) det(∇v) Therefore, from Eq. 21, the 2D flow velocity is given by Therefore, the incompressible flow velocity field in   Eq.(12) is then reduced to the following form: v2,2 −v1,2        v1 −v2,1 v1,1 2 ∂ v1 1 p,1 h 2 1 ∂v i 2 1 2 = (ν∇ − ) − . ν∇ v − ρ ∇p − ∂t · [(∇v) − 2 1tr(∇v) ] v2 v1,1v2,2 − v1,2v2,1 ∂t v2 ρ p,2 v = , det(∇v) (26) (17) 2 For the 2D steady flow, Eq.26 is reduced to where ∇v = vj,ieiej,(∇v) = ∇v · ∇v = vj,kvk,ieiej, tr(∇v)2 = 1 :(∇v)2 = v v and det(∇v) =   i,k k,i v2,2 −v1,2 εijkv1,iv2,jv3,k, and εijk is permutation symbol.        v −v2,1 v1,1 v 1 p ∂v 1 = ν∇2 1 − ,1 . For steady flow, ∂t = 0, the Eq. 18 is reduced to v2 v1,1v2,2 − v1,2v2,1 v2 ρ p,2 h 2 1 i 2 1 2 (27) ν∇ v − ρ ∇p · [(∇v) − 2 1tr(∇v) ] v = , (18) det(∇v) Hence, we have a similar conjecture for 2D flow as follows. Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 15 June 2020 doi:10.20944/preprints202003.0030.v4

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Conjecture 2 The 2D Navier-Stokes equation has a so- The Eq.31 will be invalid as det(∇v) → 0. These might lution if and only if the determinant of flow velocity gra- be viewed as another interpretation of the conjectures on dient is not zero, namely det(∇v) 6= 0, or equivalently solution existence.

v1,1v2,2 − v1,2v2,1 6= 0. In conclusion, the incorrect expression of material d ∂ Concerning the geometrical and physical meaning of derivative operator dt = ∂t +(v ·∇) is mainly because of the conjectures, let’s try to give a basic interpretation. the miscalculation of v ·∇; now we know that ∇ is a gra- Assume that d2X is small surface area of a moving fluid dient operator rather than a vector, noting ei · ej = δij, element with volume d3X, they become to d2x and d3x so that: (v · ∇)v = [(eivi) · (ej∂j)]v = [vi,jei · ej]v = ∂v1 ∂v2 ∂v3 due to the velocity gradient det(∇v), respectively. Their [vi,jδij]v = vi,iv = ( ∂x1 + ∂x2 + ∂x3 )v, while v · (∇v) = relations are ∂ ∂ ∂ v1 ∂x1 +v2 ∂x2 +v3 ∂x3 , therefore (v·∇)v 6= v·(∇v). The correct material derivative operator should be defined as: 2 −T 2 d x = det(∇v)(∇v) · d X, (28) d(:) ∂ dt = ∂t (:) + v · [∇(:)]. and the volume induced by the velocity gradient ∇v is By taking into account of the importance of the con- d3x = det(∇v)d3X. (29) vective term v ·∇v, the conjectures on solution existence condition of Navier-Stokes equation have been proposed, From the relations in Eqs.28 and 29, both the surface which state that ”The Navier-Stokes equation has a solu- 2 3 area d x and volume d x induced by the velocity gradient tion if and only if the determinant of flow velocity gradi- ∇v will shrink to a point as det(∇v) → 0. If we image ent is not zero, namely det(∇v) 6= 0.” [24]. To be honest, 2 the finite surface area (d x 6= 0) as a window that flow this study on the solution existence is still in a very ba- 2 can go through, it means that, if d x = 0, the window is sic stage. From future perspective, the mathematician- closed and no flow can go through it. s should be invited for comprehensive investigation and Denoting the momentum flux density tensor in a vis- proof of the conjectures. cous fluid Π = pI + ρv ⊗ v − µ∇v, and µ dynamical viscosity, according to Landau and Lifshitz [14], the lo- cal form of the equation of motion of the viscous fluid is Availability of data: There are no data in this the- ∂ρv oretical article. ∂t + ∇ · Π = 0, which can be rewritten in integral form R  ∂ρv  3 ∂t + ∇ · Π d X = 0, and further simplified to Conflict of interests: The authors declare that they ∂ Z I have no known competing financial interests or personal ρvd3X + Π · d2X = 0, (30) ∂t relationships that could have appeared to influence the work reported in this paper by Green’s formula. Using Eqs.28 and 29, we can get an induced form of Bo-Hua Sun: Conceptualization, Methodology, For- Eq.30 by the velocity gradient ∇v as follows mulations, Formal analysis, Funding acquisition, Inves- ∂ Z ρvd3x I Π · (∇v)T · d2x tigation, Writing- Original draft preparation, Writing- + = 0. (31) Reviewing and Editing and all relevant works. ∂t det(∇v) det(∇v)

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