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A Summary of the Course Navier-Stokes Equations I A summary of the course Navier-Stokes equations I Sauli Lindberg University of Helsinki, Fall 2019 Sauli Lindberg Navier-Stokes equations I Fall 2019 1 / 34 The mathematical/physical basis We will follow an axiomatic approach where we use the following axioms. The continuum assumption. Mass is neither created nor destroyed. Newton’s second law: the rate of change of the momentum of a body equals the net force F on the body, F = d(mv)=dt. Energy is neither created nor destroyed. Extra assumptions are introduced and modified later as needed. Sauli Lindberg Navier-Stokes equations I Fall 2019 2 / 34 The main quantities of interest Let D ⊂ R3 be a domain (connected open set) filled with a fluid, and let −∞ < T1 < T2 < 1. We assume that the following quantities are smooth functions of (x; t) 2 D × [T1; T2]: velocity u(x; t) 2 R3, pressure p(x; t) 2 R, fluid density %(x; t) 2 (0; 1). Questions: What kinds of equations do u, p and % satisfy? How are those equations derived from the axioms of the previous slide? How do the solutions of the equations behave? Sauli Lindberg Navier-Stokes equations I Fall 2019 3 / 34 Conservation of mass I We consider volumes W ⊂ D ⊂ R3 of fluid where standard tools of vector analysis are available (and call them admissible volumes). Axiom 1 (Principle of conservation of mass) The rate of change of mass in W equals the net rate of mass crossing @W in the inward direction, i.e., d Z Z Z % dV = − % u · n dA = − div (% u)dV (T1 < t < T2): dt W @W W As a consequence, we get the continuity equation @% + div (% u) = 0: (1) @t Sauli Lindberg Navier-Stokes equations I Fall 2019 4 / 34 Eulerian and Lagrangian frames of reference Eulerian frame of reference: Consider the particle moving through x at time t 2 (T1; T2). Denote its velocity by u(x; t). This is the frame of reference of an observer fixed in space. Lagrangian frame of reference: Suppose a particle is at x 2 D at time t = T1. 3 Denote its position at time t 2 [T1; T2] by '(x; t) 2 R . Thus '(x; T1) = x. The velocity of the fluid particle at time t is @' (x; t) = u('(x; t); t): (2) @t We call ' the flow map. We assume furthermore: 1 ' = ('1;'2;'3) 2 C (D × [T1; T2]; D). x 7! '(x; t): D ! D is a bijection at every t 2 [T1; T2]. Sauli Lindberg Navier-Stokes equations I Fall 2019 5 / 34 Acceleration of fluid particles The acceleration of a fluid particle is @2' @ @u (x; t) = [u('(x; t); t)] = ··· = + u · ru ('(x; t); t): @t2 @t @t We denote the so-called material derivative of u by Du @u = + u · ru: Dt @t More generally: Definition 2 1 Let f 2 C (D × [T1; T2]). The material derivative of f is Df @f = + u · rf : Dt @t 1 3 Similarly, if v 2 C (D × [T1; T2]; R ), the material derivative of v is Dv @v = + u · rv: Dt @t Sauli Lindberg Navier-Stokes equations I Fall 2019 6 / 34 A geometric interpretation of the material derivative When x 2 D and T1 ≤ t1 < t2 ≤ T2, Z t2 @ Z t2 Df f ('(x; t2); t2) − f ('(x; t1); t1) = [f ('(x; t); t)] dt = ('(x; t); t) dt: t1 @t t1 Dt Thus Df =Dt measures the rate of change of f in the frame of reference of the moving fluid particle. Sauli Lindberg Navier-Stokes equations I Fall 2019 7 / 34 Volumes moving with the fluid I Given t 2 [T1; T2], we denote 't (x) = '(x; t); Wt = 't (W ) = f't (x): x 2 W g: We will study the following questions: R How does the mass % dV of Wt vary in time? Wt R How does the volume dV of Wt vary in time? Wt R R How does the average density % dV = dV in Wt vary in time? Wt Wt How does the momentum R % u dV vary in time? Wt How does the kinetic energy R % juj2 =2 dV vary in time? Wt Sauli Lindberg Navier-Stokes equations I Fall 2019 8 / 34 Volumes moving with the fluid II We consider the following general question: Question 3 1 R If f 2 C (D × [Tt ; T2]), what is the rate of change (d=dt) f dV ? Wt How about (d=dt) R % f dV ? Wt Theorem 4 (Transport theorem) 1 For every f 2 C (D × [T1; T2]), d Z Z @f f dV = + div (f u) dV : dt Wt Wt @t Corollary 5 (Transport theorem with density) 1 Let f 2 C (D × [T1; T2]). Then d Z Z Df % f dV = % dV : dt Wt Wt Dt The proofs proceed via a change of variables from Wt into W . Sauli Lindberg Navier-Stokes equations I Fall 2019 9 / 34 Change of variables in an integral The proofs are based on a change of variables. Denote 2@'1 @'1 @'1 3 6 @x1 @x2 @x3 7 6@' @' @' 7 6 2 2 2 7 J' = det D' = det 6 7 : 6 @x1 @x2 @x3 7 4@'3 @'3 @'3 5 @x1 @x2 @x3 Theorem 6 (Change of variables in an integral) R R 1 f (x; t)dV = f ('(x; t); t)J'(x; t)dV for every f 2 C (D × [T1; T2]). Wt W Theorem 7 (Euler’s identity) For all (x; t) 2 D × (T1; T2), @J' t (x) = [div u](' (x); t) J' (x): @t t t Sauli Lindberg Navier-Stokes equations I Fall 2019 10 / 34 Forces acting on a fluid I The forces acting on a volume of fluid are divided into two types: • Stresses, whereby the rest of the continuum acts on a volume of fluid W by forces across its surface @W . (E.g. pressure.) • Body forces (external forces) which generate a force per unit volume on the fluid. (E.g. gravity.) Definition 8 An ideal fluid is a fluid with the following property: the total force exerted on the fluid inside W by means of stress on @W at time t is Z Z − p(x; t)n(x)dA = − rp(x; t) dV : @W W Definition 9 If b(x; t) denotes the net body force per unit mass at (x; t), the total body force R on W is W %(x; t) b(x; t) dx. Sauli Lindberg Navier-Stokes equations I Fall 2019 11 / 34 Newton’s second law I We express Newton’s second law for ideal fluids: Axiom 10 (Law of balance of momentum) For an ideal fluid, the rate of change of momentum d Z Z Z % u dV = − rp dV + % b dV : (3) dt Wt Wt Wt Via the Transport theorem with density we obtain the following easier form: Theorem 11 Suppose a fluid is ideal. Then, in D × (T1; T2), Du % = −∇p + % b: (4) Dt Equation (4) is (a special case of) the Cauchy momentum equation or the law of balance of momentum. Sauli Lindberg Navier-Stokes equations I Fall 2019 12 / 34 Rates of change Under the assumptions of the course, we have (d=dt) R % dV = 0, Wt (d=dt) R dV = R div u dV , Wt Wt (d=dt)(R % dV = R dV ) = − R % dV R div u dV =(R dV )2, Wt Wt Wt Wt Wt (d=dt) R %u dV = − R rp dV + R % b dV (for ideal fluids), Wt Wt Wt (d=dt) R % juj2 =2 dV = R (−∇p + ρb) · u dV (for ideal fluids). Wt Wt In particular, divergence controls the rate of change of volume. This phenomenon is captured in the notion of incompressibility. Sauli Lindberg Navier-Stokes equations I Fall 2019 13 / 34 Incompressibility Definition 12 We call a flow incompressible if for every admissible W ⊂ D, Z volume of Wt = dV = constant in t: Wt Theorem 13 The following conditions are equivalent: (i)The flow is incompressible, (ii) div u = 0, (iii) J' = 1. (iv) D%=Dt = 0. Sauli Lindberg Navier-Stokes equations I Fall 2019 14 / 34 Homogeneous, incompressible Euler equations A fluid is said to be homogeneous if the density % is independent of the position x 2 D. Typically one sets % = 1. Definition 14 The homogeneous, incompressible Euler equations are given by @u + u · ru + rp = b; (5) @t div u = 0: (6) One typically sets one of the following boundary conditions: Impenetrability (of the container wall): u(x; t) · n(x) = 0 on @D if @D 6= ;. Periodic conditions: u(x + k; t) = u(x; t) for all x 2 R3 and k 2 Z3. Decay conditions: u(x; t) ! 0 in some sense when jxj ! 1 in the case D = R3. Sauli Lindberg Navier-Stokes equations I Fall 2019 15 / 34 Stationary solutions of Euler equations Suppose that @u=@t = 0 and b = −∇(gx3) is gravity, so that the Euler equations reduce to u · ru = −∇(p + gx3); (7) div u = 0: (8) Equation (7) is hard to analyse because the term u · ru is nonlinear in u. However, juj2 u · ru = r − u × (r × u): (9) 2 2 Thus, if div u = 0 and r × u = 0, then setting p = −g x3 − juj =2 gives a solution (u; p) of (7)–(8). The remark above (among other things) motivates us to study irrotational fields v (i.e. ones satisfying r × v = 0). Sauli Lindberg Navier-Stokes equations I Fall 2019 16 / 34 Solenoidal, irrotational vector fields Definition 15 If v 2 C 1(D) satisfies div v = 0, then v is said to be solenoidal.
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