Week 6 Notes, Math 865, Tanveer. Energy Methods for Euler and Navier-Stokes Equation We will consider this week basic energy estimates. These are estimates on the L 2 spatial norms of the solution u(x, t) and its higher deriviatives with respect to x. Like other PDE initial value problems, these estimates are most useful in establishing existence and uniqueness of solutions. For simplicity, we will first take Ω = R N, where N = 2 or 3. We will drop the boldfonted notation for vectors since by this time you are sufficiently familiar to recognize which quantities are vectors and scalars. The exposition of this topic follows Bertozzi & Majda (Text), though with some differences in notation... Kinetic Energy Dissipation: Consider the incompressible constant density Navier-Stokes equation (.) t u + u u = p + ν u + b (.2) u = Then the i-th component of (.) is given by (.3) t u i + u j xj u i = xi p + ν 2 x j u i + b i Multiplying (.3) by u i and summing over i and using (.2), we have on using Einstein convention of summation over repeated indices: ( ) (.4) t 2 u2 i + xj 2 u ju 2 ( ) i = xi (pu i ) + ν xj ui xj u i ν xj u i xj u i + b i u i Assuming that u ( u 2 + p) + u Du = o( x N+ ), as x, we obtain from (.4) after integration over R N the following identity: (.5) t 2 u(., t) 2 = ν Du 2 + (b, u) The physical kinetic energy in the fluid is (.6) E(t) = R N 2 u(x, t) 2 dx = 2 u(.t) 2, In the absence of any force, i.e. b =, we obtain from (.5) d (.7) dt E = ν Du 2 = ν Du(x, t) 2 dx = ǫ, x R N where ǫ is usually referred to as the rate of energy dissipation. For zero viscosity, i.e. Euler flow, kinetic energy E is conserved and E(t) = E(). In the presence of a force b, the term (b, u) is the work done by the force b per-unit time. So, (.5) is a physical statement that rate of change of Kinetic Energy is the rate of work done by the external force minus the Energy dissipated /lost due to viscous friction. This dissipated energy is actually converted to heat. For compressible fluid flow, we have to couple heat energy and thermodynamics with momentum equation to get a complete set of equations. But this is not the case for incompressible Navier- Stokes equation, which we are studying here, and we do not have to consider thermodynamics to solve for fluid flow field u.
2.2. Energy Estimate, Uniqueness and ν dependence of Smooth Solutions. Let u, w be two Navier-Stokes solution, corresponding to forcing b and c respectively. We assume b and c to be smooth as well and decaying sufficiently fast in x at. We denote the corresponding pressures by p and q. Then consider the difference v = u w. It is easy to check that v satisfies: (.8) t v +v v +w v +v w = P +ν v +f, where f = b c, P = p q The i-th component of the above equation may be written as (.9) t v i + v j xj v i + w j xj v i + v j xj w i = xi P + ν 2 x j v i + f i Multiplying above by v i and integrating we obtain that (.) t 2 v2 i + 2 [ x j (vj + w j )vi 2 ] ( ) +vi v j xj w i = xi (u i P)+ν xj vi xj v i ν( xj v i )( xj v i )+v i f i So integrating over R N with usual assumptions on decay of velocity and pressure fields at, we obtain by using (.) (v, f) v f, and (v, v w) Dw(., t) v 2, d (.2) dt 2 v 2 + ν Dv 2 Dw v 2 + v f So, in particular, (.3) d dt v w v + f Also, on integrating (.2) between t = to t = T, we obtain (.4) T 2 v(., T) 2 +ν Dv(., t) 2 dt T T 2 v(., ) 2 + Dw(., t) v(., t) 2 dt+ v(., t) f(., t) dt Using well-known Gronwall s inequality on (.3) and the definition of v, we obtain the following Lemma Lemma.. Let u and w be two smooth L 2 (R N ) solutions to the Navier-Stokes equation for t [, T] for the same viscosity ν, but different forcing b and c respectively. Then, (.5) { T } [ T sup t [,T] u(., t) w(., t) u(., ) w(., ) + b(., t) c(., t) dt exp Corollary.2. Uniqueness of smooth solutions Let u(., t) and w(., t) be two smooth L 2 (R N ) solutions to incompressible constant density Navier-Stokes equation for t [, T] with same initial data and forcing. Then, the solution is unique. Proof. This simply follows from Lemma??, since u(., ) w(., ) = and b c =. w(., t) dt ] Remark.3. The energy estimate (.5) does not explicitly depend on ν and is equally valid for ν =, i.e. for the Euler equation.
3 The energy estimate (.5) is also useful in estimating the difference between smooth Euler and Navier-Stokes solution with the same initial data and forcing. Let u [] be a smooth solution to the Euler equation, i.e. ν =, while u [ν] is a solution to Navier-Stokes equation with the same initial data and forcing. Then, we can obtain an equation for v = u [ν] u [] : (.6) t v + v v + u [] v + v u [] = P + ν v + f where f = ν u [], P = p q is the difference of pressure This is the same equation as for (.8), with w replaced by u [], and a different meaning of f. Therefore, the energy estimate (.5) in this case becomes (.7) ( ) { T } T sup u [ν] (., t) u [] (., t) ν t [,T] u [] (.t) dt exp Du [] (., t) dt Notice that (.4) with w replaced by u [], and with f = ν u [] gives rise to (.8) T T T ν Dv(., t) 2 dt Du [] (., t) v(., t) 2 dt + ν v(., t) u [] (., t) dt νtc(u [], T) Using estimate (.7) estimate, we obtain, (.9) T T } T Dv(., t) 2 {CT dt νtc Du [] (., t) dt + ν u [] (., t) dt νt 2 c 2 (u [], T) So, (.2) T ( /2 T Dv(., t) dt T /2 Dv(., t) dt) 2 ν /2 T 3/2 C 2 (u [], T) This implies the following proposition: Proposition.4. Comparison of smooth Euler and Navier-Stokes Solution Given the same initial data and forcing, then the difference v between smooth L 2 (R N ) Navier-Stokes and Euler solution over a common interval of existence [, T] satisfies (.7) and (.2). In particular for any fixed T, as ν, u [ν] (., t) u [] (., t), and Du [ν] (., t) Du [] (., t) uniformly for t [, T]..3. Kinetic Energy of 2-D flow. The Theorems in the last section hold for solutions to Navier-Stokes/Euler equation that decay sufficiently rapidly as x so that velocity u(., t) L 2. This is a reasonable physical assumption in R 3. For 2-D flow, this is not necessarily the case, unless the integral of vorticity in the flow is zero, as will be seen shortly. Suppose supp ω { x : x R 2, x < R } Applying 2-D Biot-Savart Law: (.2) u(x, t) = K(x y)ω(y, t)dy, where K(x) = 2π x 2 [ x 2, x ] We first note that (.22) y R
4 We note that ( (.23) x y 2 = x 2 2 y x ) x 2 + y 2 x 2 Now, if x 2R, then since y R, it follows that as x x y 2 = x 2 + O( x 3 ) So, from (.2), (.24) u(x, t) = K(x) ω(y, t) + O( x 3 ) y R 2 Since It follows that x R 2 ( + x ) l dx <, iff l > N Lemma.5. A 2-D incompressible flow with compact vorticity ω has finite energy iff (.25) ω(x)dx = R 2 Remark.6. Note that the vorticity ω(x, t) will satisfy (.25) for t >, if (.26) ω(x, )dx =, R 2 since integration of 2-D Navier-Stokes equation in the vorticity form gives d (.27) ω(x, t)dx = dt x R 2 Remark.7. The statement that finite energy is implied only iff (.26) is satisfied is not limited merely to flow with compact support. It is more generally true for ω L (R 2 ). When (.26) is violated, it is possible to decompose a solution to Navier-Stokes equation to such that a part of it is in L 2 (R 2 ) (hence finite energy), while the other part is generated by a radial distribution of vorticity whose integral is the same as the integral of initial vorticity over R 2. Consider an initial vorticity distribution ω (x) L (R 2 ). We chose any compact radial vorticity distribution ω ( x ) such that ω ( x )dx = ω (x)dx R 2 R 2 We determine radial vorticity solution ω( x, t) with initial value ω( x, ) = ω ( x ) to Navier-Stokes equation without forcing. We know from worked out problems two weeks back, that ω( x, t) satisfies 2-D heat equation with corresponding velocity (.28) ũ(x, t) = ( x 2, x ) x 2 x Therefore, we now consider the decomposition s ω (s, t)ds (.29) u(x, t) = ũ(x, t) + v(x, t)
5 Since ( v)(x, ) = ω (x) ω (x), it follows that (.3) ( v)(x, )dx = from construction of ω. From (.27) and the fact that heat solution preserves ω(x, t)dx, it follows that R 2 (.3) d dt R 2 ( v)(x, t)dx =, implying by above This implies that v(x, t) has finite energy. Thus, we have proved the following Lemma: x R 2 ( v)(x, t)dx = Lemma.8. Any smooth solution u(x, t) to 2-D Navier-Stokes equation with an initial L (R 2 ) vorticity can be decomposed into (.32) u(x, t) = v(x, t) + ũ(x, t) where v L 2 (R 2 ) and divergence free, while x (.33) ũ(x) = ( x 2, x ) x 2 s ω(s, t)ds for some smooth radial vorticity distribution ω( x, t) with an initial compact support..4. Energy Inequality for 2-D flow. Consider the radial-energy decomposition (.34) u(x, t) = ũ(x, t) + v(x, t) of solution to the Navier-Stokes equation where v L 2 (R 2 ). v satisfies (.35) t v + v v + ũ v + v ũ = p + ν v + F Consider two solutions to Navier-Stokes equation u, u 2 with radial decompositions: (.36) u = ũ + v, u 2 = ũ 2 + v 2 Then, if we denote (.37) w = v v 2, ũ ũ 2 = û, ˆF = F F 2, ˆp = p p 2, then w satisfies (.38) t w+v w+w v 2 +ũ w+û v 2 +v 2 û+w ũ = ˆp+ν w+ ˆF Then using the same integration by parts procedure as in the last section, we have (.39) d dt 2 w 2 + ν w 2 w { w ( v 2 + ũ ) + (ũ ũ 2 ) v 2 + ˆF } + ũ ũ 2 v 2 Using Gronwall s inequality, as in previous section, we end up with the following proposition
6 Proposition.9. 2-D Energy Estimate and Gradient Control Let u and u 2 be two smooth divergence free solutions to the Navier-Stokes equation with radial-energy decomposition u j (x, t) = v j (x, t) + ũ j (x, t) and with external forces F and F 2. Then we have the following estimates: (.4) [ T ] sup t [,T] v v 2 v (., ) v 2 (., ) + exp ( v 2 + ũ )dt (.4) ν T T [ (F F 2 )(., t) + ũ ũ 2 v 2 + ũ ũ 2 v 2 ] dt (v (., t) v 2 (., t)) 2 dt C(v 2, ũ, T) { [ T (u u 2 )(., ) 2 + ( F (., t) F 2 (., t) + ũ ũ 2 v 2 (., t) + ũ ũ 2 v 2 (., t) + ũ ũ 2 v (., t) )dt] 2} Exercise: Derive (.4) and (.4) and use it to prove the above proposition..5. Calculus Inequalties for Sobolev Spaces and Mollifiers. We have already introduced the Sobolev space H m (R N ) for integer m. We now extend it to H s (R N ) for any s R. In the Schwartz space S(R N ) of smooth functions with rapid decay at, we introduce the norm { } /2 (.42) u s = ( + k ) 2s û(k)dk R N where û(k) = F[u](k), i.e. the Fourier-Transform of u. The completion of S(R N ) with norem (.42) will be referred to as H s (R N ). You can check that for s = m, that this is equivalent to the original definition of H m. One of the most important Sobolev space property that we will use is the Sobolev inequality below: Lemma.. Sobolev embedding Theorem The space H s+k (R n ), for s > N/2, k Z + {} is continuously embedded in the space C k (R N ), and there exists a constant c > such that (.43) v C k c v s+k, for any v H s+k (R N ) Some other calculus inequalities in the following Lemma will be useful for our purposes: Lemma.. i. For all m Z + {}, there exists c > such that for all u, v L H m (R N ), α m uv m c { u D m v + D m u v } D α (uv) ud α v c { u D m v + D m } u v ii. For all s > N/2, H s (R N ) is a Banach algebra, i.e. there exits a constant c so that for all u, v H s (R N ), uv s c u s v s
7 We now introduce a mollifier. Suppose ρ = ρ( x ) Cc (RN ), i.e. infinitely smooth function with compact support. Suppose also ρ, and R ρdx =. N Then we define mollification of v, denoted by I ǫ v to be a function given by: ( ) x y (.44) (I ǫ v)(x) = ǫ R N ρ v(y)dy ǫ N Lemma.2. Properties of Mollifier Let I ǫ be the mollifier defined in (.44). Then I ǫ v C (R N ) and i. for all v C (R N ), I ǫ v v uniformly on any compact set Ω R N and I ǫ v v ii. Mollifiers commute with distribution derivatives D α I ǫ v = I ǫ D α v for any α m, v H m iii. For all u L p (R N ), v L q (R N ), /p + /q =, (I ǫ u)vdx = u(i ǫ v)dx R N R N iv. For all v H s (R N ), I ǫ v converges to v in H s and the rate of convergence in the H s norm is linear in ǫ, i.e. lim ǫ + I ǫ v v s = I ǫ v v s Cǫ v s v. For all v H m (R N ), k Z + {}, and ǫ >, I ǫ v m+k c mk ǫ k v m I ǫ D k v c k ǫ v N/2+k 2. More properties of Hodge Projection Lemma 2.. Any vector field v H m (R N ) for m Z {} has a unique orthogonal decomposition v = φ + w, where φ, w H m, w = We define w = Pv as the Hodge projection of v onto the divergence free vector field. Further, i. (Pv, φ) m = and Pv 2 m + φ 2 m = v 2 m. ii. P commutes with D α in H m for α m: PD α v = D α Pv. iii. PI ǫ v = I ǫ Pv iv. P is symmetric: (Pu, v) m = (u, Pv) m Proof. We only consider m =. Other cases follow simply by noting property ii: that P commutes with D. We further consider only v C c (R N ). This space is dense in H m and hence all the results will follow for more general v, except that derivatives have to be understood in the sense of a distribution. Define φ by solving φ = v, with φ as x Using Green s function for Laplacian, we know (2.45) φ(x) = y R N G(x y)( v)(y)dy,
8 where G(x) = log x for N = 2, G(x) = 2π 4π x for N = 3 Then it is clear that (2.46) φ = G(x y)( v)(y) vdy y R N Now, notice that as x, φ [ G](x) ( v)(y)dy + O( x N ) y R N From applying Green s theorem on the first term, and hence φ L 2 (R N ). Define φ = O( x N ) as x, Pv = w = v φ Clearly since v, φ L 2 (R N ), so is w = Pv. It is clear that w = v φ = So, Pv is divergence free, and from the decay rate of φ for large φ, it follows that w O( x N ) as x Now, property i. follows since (w, φ) = w j xj φdx = xj (w j φ) = lim R N R N R x =R φ(w n)dx = since for large x, φ = O(log x ) for N = 2 and φ = O( x N+2 ) for N = 3, while w = O( x N ). Also, v 2 = (w + φ, w + φ) = (w, w) + ( φ, φ) = Pv 2 + φ 2 because of the orthogonality property. Property ii. follows simply from the observation that D α Pv = D α w = D α v D α φ = D α v D α φ = PD α v, since (D α φ) = (D α v) Property iii. follows from the commuting property of I ǫ with ( defined in (2.45)) and with any differential operator, since I ǫ Pv = I ǫ v I ǫ v = (I ǫ v) (I ǫ v) = PI ǫ v Property iv follows for m = because (Pv, u) = ( u, v v ) = (u, v) + ( u, v ) = (u, v) + ( u, v ) = (u, v) ( ( u), v ) = ( u u, v ) = (v, Pu)