Correct Expression of the Material Derivative in Continuum Physics —The Solution Existence Condition for the Navier-Stokes Equation

Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 28 April 2020 doi:10.20944/preprints202003.0030.v3

Correct expression of the material derivative in continuum physics —the solution existence condition for the Navier-Stokes equation

Bohua 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: April 27, 2020) The material derivative is important in continuum physics. This Letter shows that the expression d ∂ dt = ∂t + (v · ∇), used in most literature and textbooks, is incorrect. The correct expression d(:) ∂ dt = ∂t (:) + v · [∇(:)] is formulated. The solution existence condition of Navier-Stokes equation has been proposed from its form-solution, the conclusion is that ”The Navier-Stokes equation has solution if and only if the determinant of ﬂow velocity gradient is not zero, namely det(∇v) 6= 0.”

the material derivative with respect to time must be In the latest version of Landau & Lifshitz,[13] the ma- defined. For mass density ρ(x, t), flow velocity v(x, t) = terial derivative of flow velocity is given in another form: vkek, and stress tensor σ = σijei ⊗ ej, the material dv ∂v derivatives are given by: = + (v∇)v. (8) dt ∂t dρ ∂ρ ∂ρ ∂xk ∂ρ ∂ρ ∂ρ ∂ρ = + = +v1 1 +v2 2 +v3 3 , (1) dt ∂t ∂xk ∂t ∂t ∂x ∂x ∂x Besides the books mentioned above, there is a large amount of literature in fluid mechanics taking the same expressions as Refs. 1 and 13. This makes it difficult dv ∂v ∂v ∂xk ∂v ∂v ∂v ∂v = + = + v1 1 + v2 2 + v3 3 , for readers to identify which expression for the materi- dt ∂t ∂xk ∂t ∂t ∂x ∂x ∂x (2) al derivative is correct, causing great confusion to both and students and scholars. Although some authors, such as Lighthill,[14] dσ ∂σ ∂σ ∂xk ∂σ ∂σ ∂σ ∂σ Batchelor,[15] Frisch,[16], Nhan & Nam [17], Xie,[18] and = + = + v1 1 + v2 2 + v3 3 , dt ∂t ∂xk ∂t ∂t ∂x ∂x ∂x Zhao [19] have used the correct expression, it seems that (3) most fluid mechanics textbooks and academic literature respectively, where t is time, x = xkek is position, ek is have adopted an incorrect expression for the material a base vector and v = ∂xk is the flow velocity. k ∂t derivative as in Ref. 1. Therefore, to revive the great In most textbooks, handbooks and encyclopedia, such influence of Landau in physics and fluid mechanics, we as Refs. 1–12, the above material derivatives are ex- attempt to address this issue in this dedicated paper, pressed as: where we revisit the material derivative to show why Eqs. dρ ∂ρ (7) and (8) are incorrect, and derive a correct expression = + (v · ∇)ρ, (4) dt ∂t using standard tensor calculus. In Landau & Lifshitz,[1] the material derivative is given dv ∂v as: = + (v · ∇)v, (5) dt ∂t ∂v ∂v ∂v dx + dy + dz = (dr · grad)v. (9) and ∂x ∂y ∂z dσ ∂σ = + (v · ∇)v, (6) Although Landau’s fluid mechanics is well known world- dt ∂t wide, the above expression is incorrect. ∂ where the gradient operator ∇ = ek . ∂xk Proof: To simplify Eqs. (1)–(3) further, a differential operator is introduced as follows: Since v · grad = v · ∇ = (viei) · (∂jej) = vi,jei · ej = vi,jδij = divv, and (v · grad)v = d ∂ = + (v · ∇). (7) (v · ∇)v = (vj,iei · ej)(vkek) = δijvj,ivkek = dt ∂t vi,ivkek = (divv)v, where the divergence is ∂v1 ∂v2 ∂v3 ∂v1 This operator is used by most fluid mechanics textbooks, divv = ∂x1 + ∂x2 + ∂x3 , so (divv)v = ( ∂x1 + ∂v2 ∂v3 including well-known graduate textbooks such as Landau ∂x2 + ∂x3 )v. Thus: & Lifshitz,[1, 2] Prandtl,[3, 4] Anderson,[5] Pope,[6] Cen- gel & Cimbala,[7], Kundu et al.[8], Woan [9], wikipedia ∂v ∂v ∂v (v · grad)v 6= v + v + v . (10) [10, 11] and britannica [12]. 1 ∂x1 2 ∂x2 3 ∂x3

∂v T In the latest version of Landau & Lifshitz,[13] for rea- For a vector ∂x = v ⊗ ∇ = v∇ = (∇v) . Strictly sons which are unclear, the material derivative expression speaking, the right gradient of v should be written as dv ∂v has been changed to dt = ∂t + (v∇)v; this is also incor- v ⊗ ∇; here, in order to agree as far as possible with rect. conventional presentation, we write v ⊗ ∇ = v∇ and, similarly for the left gradient, ∇⊗v = ∇v. Hence (v∇)· Proof: ∂v ∂ρ v = v · (∇v) and dv = ∂t dt + (v∇) · vdt = ∂t dt + v · Since v∇ = (viei)(∂jej) = vi,jeiej, (∇v)dt, or, dividing both sides by dt, dv ∂v ∂v (v∇)v = (vj,ieiej)(vkek) = (vj,ieiej)(vkek), = + (v∇) · v = + v · (∇v). (13) dt ∂t ∂t which indicates that v∇ is a 3rd-order tensor 3. The 2nd order stress tensor σ = σ(x, t) is a tensor- rather than a vector. Thus: valued function of x and t and its differential is dσ = ∂v ∂v ∂v ∂σ dt + ∂σ · ∂x dt. For an arbitrary tensor ∂σ = σ∇ 6= (v∇)v 6= v + v + v . (11) ∂t ∂x ∂t ∂x 1 ∂x1 2 ∂x2 3 ∂x3 T ∂σ (∇σ) , hence dσ = ∂t dt + (σ∇) · vdt, or dividing both dσ ∂σ sides by dt leads to dt = ∂t +(σ∇)·v. Since (σ∇)·v = To obtain the correct formulation of the material v · (∇σ), derivative, we perform some basic tensor calculus [20]. The del operator is a vector differential operator, and dσ ∂σ ∂ = + v · (∇σ). (14) defined by ∇ = ei . An important note concerning ∂xi dt ∂t the del operator ∇ is in order. Two types of gradients If an operator must be introduced for the material are used in continuum physics: left and right gradients. derivative, it should be in the following form: The left gradient is the usual gradient and the right gradient is the transpose of the forward gradient operator. d(:) ∂ = (:) + v · [∇(:)]. (15) To see the difference between the left and right of dt ∂t gradients, consider a vector function A = Ai(x)ei. ∂ For a scalar function f(x, t), this leads to df = ∂f + v · The left gradient of a vector A is ∇ ⊗ A ≡ ej ⊗ dt ∂t ∂xj ∂Ai (∇f). If f is considered as a distributed function of gas (Aiei) = ej ⊗ ei = Ai,jej ⊗ ei, or written as ∇A ≡ ∂xj molecules in their phase space, this derivative is as in Eq. ∂ ∂Ai ej (Aiei) = ejei = Ai,jejei. The right gradient ∂xj ∂xj (3.2) in Physical Kinetics by Lifshitz and Pitaevskii. [2] ∂ ∂Ai du of a vector A is A ⊗ ∇ ≡ (Aiei) ⊗ ej = ei ⊗ For a vector function u(x, t), Eq. (15) yields = ∂xj ∂xj dt ∂ ∂u + v · (∇u). For a tensor function A(x, t), Eq. (15) ej = Ai,jei ⊗ ej, or written as A∇ ≡ (Aiei)ej = ∂t ∂xj dA ∂A ∂Ai ∂Ai yields = + v · (∇A). eiej = Ai,jeiej, where Ai,j = . dt ∂t ∂xj xj The Navier-Stokes equations of incompressible flow can The gradient of a scalar function is a vector, the diver- be expressed as follows: the momentum equation: gence of a vector-valued function is a scalar ∇ · A, and the gradient of a vector-valued function is a second-order ∂v 1 + v · (∇v) = − ∇p + ν∇2v, (16) tensor ∇A. Although the del operator has some of the ∂t ρ properties of a vector, it does not have them all, because it is an operator. For instance, ∇ · A is a scalar (called and mass conservation equation: ∇ · v = 0. the divergence of A) whereas A·∇ is a scalar differential Applying the divergence operation to both sides of operator, where A is a vector. Thus the del operator ∇ the momentum equation Eq.16 and use the mass con- 2 does not commute in this sense. servation leads to a pressure equation: ∇ · (p1) = 2 It worth to emphasise that the right gradient of a vec- −ρ∇ · (v · ∇v) = −ρ{(∇v) + [∇(v∇)] · v}. In which, tor A∇ is a more natural one, is often used in defining v(x, t) is flow velocity field, ρ is constant mass density, the deformation gradient tensor, displacement gradien- p(x, t) is flow pressure, ν is kinematical viscosity, t is k t tensor, and velocity gradient tensor. It is clear that time, x = x ek is position coordinates, ek is a base vec- ∂ T A∇ = (∇A) . tor and v is flow velocity, ∇ = ek ∂xk is gradient operator T 1. Mass density ρ = ρ(x, t) is a scalar-valued function and 1 = ekek is an identity tensor, and v∇ = (∇v) . of x and t and its differential is dρ = ∂ρ dt + ∂ρ · ∂x dt. Both solution existence and smoothness of the Navier- ∂t ∂x ∂t Stokes equations are still an open problem in mathemat- For a scalar ∂ρ = ρ∇ = ∇ρ, so dρ = ∂ρ dt+(ρ∇)·vdt = ∂x ∂t ics [21], regardless of numerous abstract studies that have ∂ρ dt + v · (∇ρ)dt, or, dividing both sides by dt, ∂t been done by mathematicians. Now the question is that, dρ ∂ρ ∂ρ is it possible to propose an alternative and simple direct = + (ρ∇) · v = + v · (∇ρ). (12) approach to express the flow velocity field without using dt ∂t ∂t complicated mathematics. Even we can not find the final 2. Flow velocity v = v(x, t) is a vector-valued function solution of the velocity field, it is still plausible if we can ∂v ∂v ∂x of x and t and its differential is dv = ∂t dt + ∂x · ∂t dt. get rid of the nonlinear convective term v · (∇v). Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 28 April 2020 doi:10.20944/preprints202003.0030.v3

Assuming the determinant of the velocity gradient is gradient ∇v. Although the Eq.(17) and Eq.(22) can on- not zero, namely det ∇v 6= 0, the Navier-Stokes momen- ly give us a form-solution, but they still provide a rich tum equation in Eq.(16) can be rewritten as follows information. Both Eq.(17) and Eq.(22) reveal that the current difficulty of finding a solution is because of the p ∂v v = ∇ · (ν∇v − 1) − · (∇v)−1, (17) nonlinear term v · (∇v), or in other words, due to the ρ ∂t existence of the velocity field gradient ∇v. Accordingly, the solution of the Navier-Stokes equation will be blowup equivalently as det ∇v tends to an infinitesimal, and has no solution when det ∇v = 0. p ∂v v = (∇v)−T · ∇ · (ν∇v − 1) − . (18) The study might open an avenue for the understanding ρ ∂t of turbulence. It would be important to check the value The Eq.(17) and Eq.(18) can never be obtained, if the of det ∇v for the study of flow stability and turbulence d ∂ transition. wrong material derivative dt = ∂t + (v · ∇) is used in d ∂ the Navier-Stokes momentum equation as in the most of The mistake in the definition dt = ∂t +(v·∇) is mainly textbook such as [13]. the incorrect representation of v · ∇; ∇ is a gradient Therefore we have a statement as follows: operator rather than a vector, so that: ∂ ∂ ∂ The Navier-Stokes equation has solution if v · ∇ = (v1e1 + v2e2 + v3e3) · (e1 + e2 + e3 ) ∂x1 ∂x2 ∂x3 and only if the determinant of flow velocity ∂v ∂v ∂v ∂ ∂ ∂ gradient is not zero, namely det(∇v) 6= 0. = 1 + 2 + 3 6= v + v + v . ∂x1 ∂x2 ∂x3 1 ∂x1 2 ∂x2 3 ∂x3 Although we have successfully split the velocity field In conclusion, the material derivative operator is cor- v from the convective term v · (∇v), the calculation of d(:) ∂ rectly defined as: dt = ∂t (:) + v · [∇(:)]. The solu- the inverse of the velocity gradient (∇v)−1 is still great tion existence condition of Navier-Stokes equation has challenge. been proposed from its form-solution, the conclusion is According to the Cayley-Hamilton theorem [22–24], for that ”The Navier-Stokes equation has solution if and on- the 2nd order tensor ∇v, the following polynomial holds: ly if the determinant of flow velocity gradient is not zero, namely det(∇v) 6= 0.” 3 2 (∇v) − I1(∇v) + I2∇v − I31 = 0, (19)

1 2 2 where I1 = tr(∇v) = ∇ · v,I2 = 2 [(tr∇v) − tr(∇v) ] and I3 = det(∇v). Hence, for the case of det(∇v) 6= 0, we have: [1] L. D. Landau and E. M. Lifshitz, Fluid Mechanics (En- glish version) (1987). (∇v)2 − I ∇v + I 1 (∇v)−1 = 1 2 . (20) [2] E. M. Lifshitz and L. P. Pitaevskii, Physical Kinetics det(∇v) (Elsevier (Singapore) Ltd, 2008) [3] L. Prandtl, Führerdurch die Strömungslehre, Herbert For incompressible flow, the divergence of velocity gra- Oertel (ed.) (Springer Vieweg, 2012). dient is zero, namely, I1 = tr(∇v) = ∇ · v = 0, thus [4] L. Prandtl, Prandtl’s essentials of fluid mechanics, Her- 1 2 2 1 2 bert Oertel (ed.) (Springer-Verlag, 2004). I2 = 2 [(tr∇v) − tr(∇v) ] = − 2 tr(∇v) . Therefore, the inverse of the velocity gradient for incompressible flow [5] J. D. Anderson, Computational Fluid Dynamics–the ba- sics with applications (The McGraw-Hill Companies, In- takes a simpler form: c., 1995). [6] S. B. Pope, Turbulent flows (Cambridge University Press, (∇v)2 − 1 1tr(∇v)2 (∇v)−1 = 2 . (21) 2000). det(∇v) [7] Y. A. Cengel and J. M. Cimbala, Fluid Mechanics: Fun- damentals and Applications (McGraw-Hill Education, Therefore, the incompressible flow velocity field in Inc., 2010). Eq.(17) is then reduced to the following form: [8] P. K. Kundu, I. M. Cohen and D. R. Dowling, Fluid Mechanics (Elsevier, 2012). h i [9] G. Woan, The Cambridge Handbook of Physics Formu- ∇ · (ν∇v − p 1) − ∂v · [(∇v)2 − 1 1tr(∇v)2] ρ ∂t 2 las, Cambridge University Press (2000) v = , det(∇v) [10] https://en.wikipedia.org/wiki/Gauge covariant derivative (22) [11] https://en.wikipedia.org/wiki/Navier Stokes equations 2 [12] https://www.britannica.com/science/Navier-Stokes- where ∇v = vj,ieiej,(∇v) = ∇v · ∇v = vj,kvk,ieiej, 2 2 equation tr(∇v) = 1 :(∇v) = vi,kvk,i and det(∇v) = [13] L. D. Landau and E. M. Lifshitz, Fluid Mechanics (Rus- εijkv1,iv2,jv3,k, and εijk is permutation symbol. sian version) (2017). The great challenge to find solution of the Navier- [14] J. Lighthill, Waves in Fluids (Cambridge University Stokes equation are all from the existence of the velocity Press, 1978) Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 28 April 2020 doi:10.20944/preprints202003.0030.v3

[15] G. K. Batchelor, An introduction to fluid dynamics [20] J.N. Reddy, An Introduction to Continuum Mechanics, (Cambridge University Press, 2007) (Cambridge University Press NY, 2007). [16] U. Frisch, Turbulence (Cambridge University Press, [21] C.L. Fefferman, Existence and smoothness of the Navier- 1995) Stokes equation, The millennium prize problems, Clay [17] P.T. Nhan and M.D. Nam, Understanding viscoelastic- Math. Institute, pp. 57-67(2006). ity an introduction to rheology (Springer-Verlag Berlin [22] W.R. Hamilton, Lectures on Quaternions. Dublin(1853). Heidelberg,2013). [23] A. Cayley, A Memoir on the Theory of Matrices. Philos. [18] X. C. Xie, Modern Tensor Analysis and Its Applications Trans. 148.(1858). in Continuum Mechanics (Fudan University Press, 2014) [24] C. Truesdell, A first course in rational continuum me- [19] Y. P. Zhao, Lectures on Mechanics (Science Press, China, chanics. (Academic Press, New York, 1977). 2018)