Math 575-Lecture 13 In 1845, tokes extended Newton s original idea to find a constitutive law which relates the Cauchy stress tensor to the velocity gradient, and then derived a system of equations. The fluids satisfying such a constitutive law is called Newtonian fluids. The system of equations is called Navier-tokes equations. Navier derived this system of equations very much earlier but based on wrong assumptions. 1 Viscous Newtonian fluid and the Navier-tokes equations Recall the conservation of momentum ρ Du Dt = ρb + σ σ is symmetric by conservation of angular momentum. For ideal fluid, σ = pi. We now follow tokes and consider a particular viscous fluid. We decompose σ as the pressure and a term due to viscosity: σ = pi + τ (1.1) where τ is called the deviatoric stress. Assumptions: (1). τ is linear in u. (2). The fluid is isotropic, or more mathematically, τ is invariant under rotations: τ(u u U 1 ) = U σ( u) U 1 where U is a rotation matrix. (3). The fluid property is homogeneous in space. The second condition comes from the intuition that rigid rotation yields no diffusion of momentum. uch fluids are called Newtonian fluids. Theorem 1. Under Assumptions (1) and (2), there exist two scalars (may depend on the location x) µ > 0, ζ > 0 such that τ = 2µ[E 1 ( u)i] + ζ( u)i, 3 where E = 1 2 ( u + ut ) is the rate of strain tensor (which can be regarded as the time derivative of the strain tensor). 1
Proof. By the linearity, we have τ ij = T ijmn u n x m Let P be a rotation matrix so that P 1 mn = P T mn. T is a fourth order tensor we must determine. Now, we pick four vectors A, B, C, D and consider = T ijkl A i B j C k D l. By the isotropy, we have, Hence, implying P im (T mnkl C k D l )P T nj = T ijkl P km (C m D n )P T nl A i B j P im (T mnkl C k D l )P T nj = A i B j T ijkl P km (C m D n )P T nl T mnkl (P T A) m (P T B) j C k D l = T ijkl A i B j (P C) k (P D) l. This means that is unchanged if we rotate A, B, C, D at the same time. Hence, only depends on the cosine of the angles between them and their magnitudes. In other words, = α(a B)(C D) + β(a C)(B D) + γ(a D)(C B) Terms like (A B) (C D) can also be included but they bring nothing new. As a result, T ijkl = αδ ij δ kl + βδ ik δ jl + γδ il δ jk. Further, since τ is symmetric and thus T must be symmetric in i and j. Consequently, β = γ =: µ, and we denote α := λ. As a result, Letting ζ = λ + 2 3 µ finishes the proof. τ = λδ ij u + 2µE. µ and ζ are called the first and second coefficient of viscosity respectively. For liquids, usually ζ = 0. For viscous gases, ζ is usually positive. Remark 1. λ ui is also a scalar and one can combine it with the pressure term and get a new pressure. This is the difference between mechanical pressure and thermodynamic pressure. ee Batchelor 1967(introduction to fluid dynamics), P154. 2
If we assume (3), then µ, ζ = λ+ 2 3 µ are constants. Then, the conservation of momentum equation is reduced to the Navier-tokes equations: ρ Du Dt = p + (λ + µ) ( u) + µ u + ρb. If further we assume the fluid is an incompressible Newtonian fluid, we have the Navier-tokes equations for incompressible flows: { ρ(ut + u u) = p + µ u + f, u = 0 where f = ρb. For incompressible fluids with constant density, we can introduce called the kinematic viscosity. ν = µ/ρ 2 The Reynolds number and the physical significance Assume that in the problem we study, the typical length scale is L and velocity scale is U. The time scale associated with these two scales is T = L/U. (Note that T here may be regarded as the transport time scale. There could be other time scales in the problem, like the time for diffusion, or period for an object to oscillate. In those problems, one may want to use a different time scale from L/U.) Now, we define t = t/t, x = x/l, u = u/u so that t, x, u are dimensionless. The Navier-tokes equations are reduced to (go over this) ρu 2 L (u t + u x u ) = 1 L µu x p + L 2 x u. (2.1) If we make the importance of the diffusion term to be O(1) by dividing µu L 2, we have ρul µ (u t + u x u ) = L µu x p + x u (2.2) The importance of the inertia term is measured by the following number Re = ρul µ, (2.3) 3
called the Reynolds number. Read: The hydrodynamics of swimming microorganisms by Eric Lauga and Thomas. Powers. Life at Low Reynolds Number by Purcell The significance The time scale for advection is T = L/U. However, the time scale for diffusion is T 2 = ρl 2 /µ. Re = T 2 /T. This means if Re is large, the diffusion happens very slowly, while Re 0 implies that the diffusion happens quickly. The scale of the inertial term for ρu u is ρu 2 /L while the scale for µ u is µu/l 2. Hence, Re measures the relative importance between them. The scale for viscous stress is σ vis µu/l and then the force is of scale µul ( σ force/l 3 ). The invertial force is like a pressure in Bernoulli dynamics given by p ρu 2 and hence the force is ρu 2 L 2. The ratio is Re. Consider a typical microorganism, ρ 10 3 kg/m 3, µ 10 3 P a s, U 10µm/s and L 1µm. Then, Re 10 5. The inertial term is not important. For fish in water, Re 10 2. The Re number that a honeybee sees is like 1000, and the Re number that a Boeing 747 feels is around 10 9. Usually, when Re > 10 4, there will be turbulence flow appearing for which, the Navier- tokes equations are believed to be a sufficient model. 3 Boundary conditions: No-slip condition Mathematically, the Navier-tokes equations have second order spatial derivatives in the equation, so the no-penetration boundary condition in Euler equations will not be enough. Instead, we should specify the whole velocity. Physically, we have the following assumption: (No-slip condition) At a solid boundary, the relative motion of fluid to the boundary is zero. By this assumption, if the solid region does not move, then, we have the Dirichlet boundary condition u(x) = 0, x Ω. (3.1) The no-slip boundary condition turns out to be good for Newtonian fluids and it may fail for some rarefied gases or in small domains. 4
4 A local existence theorem from mathematics We now consider incompressible fluids with constant density. Assume all quantities are scaled so that all variables are dimensionless. We denote ν = 1/Re. The equations become u t + u u + p = ν u, u = 0 It is well-known that for 2D Navier-tokes equations, there is unique weak solution (by weak solution we mean in distribution sense, it satisfies the equation, or if you pairing with test function, you have equality) for any given initial data in some space. ince any strong solutions (strong solution means smooth functions that satisfy the equation pointwise) are weak solutions, then, the uniqueness of weak solutions solves the problem in 2D totally. For 3D, however, there are examples so that the weak solutions are not unique. Then, one turns to strong solutions and people hope to prove the existence of unique strong solutions. Regarding this issue, we have the following claim: Theorem 2. Given u 0 H m (R 3 ; R 3 ), with u 0 = 0, m 3. There is a T > 0 so that the initial value problem u t + u u + p = ν u, u = 0 has a unique solution with v C(0, T ; H m (R 3 ; R 3 )) C 1 (0, T ; H m 2 (R 3 ; R 3 )) and p C(0, T ; G H m 2 (R 3 ; R 3 )), where G = { p : p L 6, p L 2 } uch kind of results are called local existence because it asserts the existence on some time interval [0, T ]. If one wants to show the existence of solution on [0, ), then this is the global existence question. This is a very hard question and is open. (ome people are trying very hard to prove the global existence and some partial results are shown.) For the proof, see the supplementary material on the coursepage. From here on, we will assume that the Navier-tokes equations always have solutions in the problems we study. 5 viscous dissipation of energy Consider the kinetic energy where is a material volume of fluid. E = 1 2 ρ u 2 dv 5
Then, by the second convection theorem d dt E = 1 ρ D 2 Dt u 2 dv = ρu D Dt udv = u ( p + µ u + ρb)dv R t = (up)dv +µ u udv + ρb udv = u ( pn+µn u)d µ u 2 dv + ρb udv In the last step, we did the following: u j i 2 u j dv = i (u j i u j )dv i u j i u j dv = n i u j i u j i u j i u j dv This is some energy estimate that may be useful if u = 0 on, but it does not have enough physical meaning. Now, consider the fact 0 = µu ( u T )dv = µu i j i u j dv = µu i n j i u j d u i i u j dv Adding this to the equation, we have d dt E = u ( pn + n 2µE)d µ = i u j ( i u j + j u i )dv + ρb udv R t u (n σ)d 2µ u : EdV + ρb udv Note u : E = E : E and thuswe have d dt E = u (n σ)d 2µ E 2 dv + ρb udv The first term is the work done by hydrodynamics force; the second term represents the energy dissipated and therefore it is the dissipation rate; the last term means the work done by body force. 6