Math 575-Lecture 19. In this lecture, we continue to investigate the solutions of the Stokes equations.

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1 Math 575-Lecture 9 In this lecture, we continue to investigate the solutions of the Stokes equations. Energy balance Rewrite the equation to σ = f. We begin the energy estimate by dotting u in the Stokes equation and integrating: f udv = u ( σ)dv = u (n σ)ds + u : σdv (.) where we have used the simple relation above. Note that σ = pi + 2µE where E = 2 ( u + ut ). Using u = 0, we reduce the relation to u f S ds + f udv = 2µ E : EdV (.2) S where f S = n σ = σ n is the fce per unit area on the surface acting on the fluid. The obtained relation is nothing but the conservation of energy. The left hand side is the wk done by the surface fce and body fce and the right hand side is the dissipation of energy. If the body fce is conservative, the f term is zero and the energy dissipation is given by the surface fce only. 2 A variational fmulation We have seen that the energy dissipation is given by the following quantity Φ = 2µ E : EdV = µ ( u + u T ) : udv. We now take the minimum of this functional subject to u = 0 and the boundary condition u S = v S. The fmulation is to L = µ( u + u T ) : u λ udv with a fixed boundary condition. S

2 To take the variation, we require the variation δu S = 0 since the boundary conditions are imposed. We may take δu = ɛw. Then, 0 = dl(u + ɛw) ɛ=0 dɛ µ( w + w T ) : u + µ( u + u T ) : wdv λ wdv = 0 Integrating by parts, we have 2µ( u + u T ) : wdv λ wdv = 0 2µ u wdv + λ wdv = 0 Here w is no longer divergence free since we have used the Lagrange multiplier. Then the arbitrariness implies that (λ/2) µ u = 0. The derivative on λ recovers the incompressibility condition. We find that p = λ/2 plays the role of Lagrange multiplier f the incompressibility condition. 3 Some particular situations We now consider Stokes flows passing by cylinders spheres. 3. 2D: Stokes paradox Stokes paradox: Theem. There is no non-trivial, steady state solution to Stokes equations in R 2 in the region exteri to a disc. To illustrate, let us consider the unifm flow passing the cylinder with radius a. Taking the curl on the Stokes equations, we get ω = 0 2

3 Since ω = ψ where ψ is the 2D streamfunction, we have 2 ψ = 0. The streamfunction satisfies the biharmonic equation. As r, ψ Ur cos θ. The equation is linear in θ with constant coefficient, so there is no new modes generated, and we can just use the ansatz The Laplacian is = r ψ = f(r) sin θ. r (r r ) + 2 r 2 θ 2 f(r) = Ar 3 + Br ln r + Cr + Dr. Matching the boundary conditions, A = B = 0. The no-slip condition requires that C = D = 0. ( Note u = (ψẑ) = (ˆr r + ˆθ r θ + ẑ z ) (ψẑ)) F the reason, see the wds on P7-P8 in Childress. Roughly speaking, the reason is that in 2D outside the cylinder, Reu u is comparable to u no matter how small Re is. Outside a distance are, we should use the Navier-Stokes equations Oseen s equations to approximate. In 3D, there no such an issue D: Unifm Stokes flow past a sphere We use the spherical codinates (ρ, θ, φ) such that x = ρ sin θ cos φ, y = ρ sin θ sin φ, z = ρ cos θ. Introduce the Stokes stream function in Lecture 0 Ψ(ρ, θ), we can rewrite the velocity field as u = θψ ρ 2 sin θ ˆρ ρψ ρ sin θ ˆθ. By the Stokes equation: we find µ ω = 0. ω ρ 2 (ρ2 ρ ρ ) + ω ρ 2 (sin θ sin θ θ θ ) + 2 ω ρ 2 sin 2 θ φ 2 = 0. 3

4 As in Lecture 0, we find that ( ω = ˆφ ρ ( ρψ ρ sin θ ) + ρψ ρ 2 sin θ + ρ θ( ) θψ ρ 2 sin θ ) = ω ˆφ. ω ρ 2 (ρ2 ρ ρ ) + ρ 2 sin θ ω (sin θ θ θ ) ω ρ 2 sin 2 θ = 0. Recall the boundary conditions u Uẑ as x. In cylindrical, ψ(r, z) 2 Ur2, r 2 + z 2 Ψ 2 UR2 sin 2 θ, R. F completeness, one could try the separation of variables. Here, we may simply try the fm Ψ = sin 2 θf(ρ) (Note that there is sin θ in the equation so there could be new modes generated.) Consequently, ω = ((f /ρ) sin θ + sin θ ρ 2 f f ρ 3 (2 sin θ)) = sin θ(f /ρ 2f/ρ 3 ) = sin θh(ρ). Inserting this into the equation, we find ρ 2 (ρ2 h ) sin θ + h ρ 2 sin θ (sin θ cos θ) h ρ 2 sin θ = 0 ρ 2 (ρ2 h ) 2h ρ 2 = 0. This means that the assumption that Ψ has a single mode is reasonable. The equation f h is simplified as ρ 2 h + 2ρh 2h = 0 This is an Euler ODE and the general solution is of the fm ρ n. We find n = 2,. Hence, h(ρ) = Aρ + Bρ 2 4

5 Hence, ρ 2 f 2f = ρ 3 h = Aρ 4 + Bρ This is again Euler ODE with fcing term. The general fm f f is thus given by f(ρ) = C ρ + C 2 ρ + C 3 ρ 2 + C 4 ρ 4. Comparing with the limit as ρ, we must have C 4 = 0 and C 3 = 2U. To determine C, C 2, we use the boundary conditions at ρ = a. We have θ Ψ = 0, ρ Ψ = 0 at ρ = a. Hence, f(a) = f (a) = 0. The Stokes streamfunction is C = 4 Ua3, C 2 = 3 4 Ua. Ψ = 4 U(a3 ρ 3aρ + 2ρ 2 ) sin 2 θ. 5