Bounds on Surface Stress Driven Flows

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1 Bounds on Surface Stress Driven Flows George Hagstrom Charlie Doering January 1, 9 1 Introduction In the past fifteen years the background method of Constantin, Doering, and Hopf has been used to derive rigorous bounds on transport in fluid mechanics equations. The background method is based on the decomposition of the velocity or temperature field into a background field and a fluctuation field. The background field is specified so that the fluctuation field satisfies a partial differential equation with homogeneous boundary conditions. Many bounding problems have a variational formulation under the background method. The background method has been used to prove bounds on the mechanical dissipation rate in various driven turbulent flows. This work was started by Constantin and Doering [4]. They studied the mechanical dissipation in shear flows at high Reynolds number. Since then there has been a large number of papers about dissipation in turbulent shear flow with different boundary conditions and geometries. We extend this technique to flows driven by stress at the boundary, namely Marangoni convection and surface shear stress driven flow. In the case of surface shear stress driven flow we study the energy stability of the laminar flow solution as a function of the Grashoff number and prove an upper bound on the friction coefficient for high Reynolds number. Tang, Caulfield, and Young [6] first used the background method to prove bounds for this type of problem and we compare our results to theirs. The problem of Marangoni convection is related to that of stress driven shear flow. In the Marangoni case the surface stress is due to surface tension gradients. Pearson [5] developed the linear theory for Marangoni convection in 1958, and Davis [3] used variational methods to study the nonlinear stability problem in In the case of infinite Prandtl number we use nonvariational methods to improve the estimate of the critical Marangoni number for nonlinear stability of the conduction solution. We also use the background method to prove an upper bound on the Nusselt number in Marangoni convection. Surface Stress Driven Flow We consider the problem of flow in the two dimensional domain in Figure subject to periodic boundary conditions in the horizontal x direction, no slip conditions at z =, and 56

2 Figure 1: Domain for the surface stress driven flow. a fixed shear stress at z = h. The full system of equations is: u t + u u + p = ν u (1) u = () u z= = (3) u z z=h = (4) ν u x = τ (5) We take periodic boundary conditions in x, with a domain length L..1 Energy Stability Analysis in Two Dimensions The laminar flow solution of these equations is u x = τz ν. We perform the energy stability analysis of this solution by making the substitution u = U + τz ν ˆx. The resulting system of equations is: U t + U U + τz ν U x + U τ z ˆx + p = ν U ν (6) U = (7) U x z= = (8) U x z=h = (9) U z z=,1 = (1) 57

3 Take the dot product of the momentum equation with U and integrate over the domain to derive the energy expression: 1 d U τ + dt ν U xu z dxdz = ν U (11) If d U dt < for all U satisfying the perturbation s boundary conditions then the base solution is stable. This derivative is given by a quadratic form Q in U: τ Q = ν U + ν U xu z dxdz (1) We use variational methods to minimize Q, subject to the constraints U = and U = 1. The resulting Euler-Lagrange equations are: λu x = ν U x + q x + τ ν U z (13) λu z = ν U z + q + τ ν U x (14) U = (15) U = 1 (16) Here q is the Lagrange multiplier associated with incompressibility and λ is the Lagrange multiplier associated with the normalization condition. We non-dimensionalize the equations by making the substitutions x = x h and z = z τh h. Define the Grashoff number Gr = ν. Then if U (x, z ) = U(x, z) we have the following equations: λu x = U x + Gr U z + q x λu z = U z + Gr U x + q If the smallest eigenvalue λ min > then Q is positive definite and the base solution is stable. Since the system is two dimensional we introduce the stream function Ψ, which satisfies Ψ x = U z and Ψ = U x. Take linear combinations of the x and z derivatives of the resulting equations to eliminate the pressure. The Euler Lagrange equations become a fourth order equation: (17) (18) (19) λ Ψ = Ψ Gr Ψ x () Ψ x z=,1 = (1) Ψ z= = () Ψ z=1 = (3) (4) 58

4 The system is periodic in x so we write Ψ in terms of its Fourier series. Let k = πj/l be the wavenumber. We write Ψ = ˆΨ k k e ıkx. Then the equations become: ( ) ( ) ˆΨ λ k ˆΨ 4 ˆΨ = 4 k ˆΨ + k4 ˆΨ ıgrk ˆΨ (5) ˆΨ z=,1 = (6) ˆΨ z= = (7) ˆΨ z=1 = (8) We have supressed the k dependence of ˆΨ k. We search for the critical Grashoff number below which all of the eigenvalues λ are positive. This can be done with numerical methods. Each Fourier mode ˆΨ satisfies a fourth order ODE in z. We used finite differences to calculate the smallest eigenvalue corresponding to each k and determined the critical Grashoff number from this. After discretizing in z the differential equation becomes a generalized eigenvalue problem. We find that the critical Grashoff number is Gr = 14, at a value of k = Energy Stability for a Three Dimensional Stress Driven Flow Consider the following non-dimensionalized equations for a three dimensional stress driven flow: u t + u u + p = u (9) u = (3) u z z=,1 = (31) u x z= = (3) u y z= = (33) u x z=1 = Gr (34) u y z=1 = (35) The domain is periodic in x and y. A steady solution to this equation is u x = Grz. If we make the substitution U = u + îgrz we can perform the same analysis as in the two dimensional case. The variational formulation is different and so is the set of Euler-Lagrange equations: λu x = U x + Gr U z + q x λu y = U y + q y λu z = U z + Gr U x + q (36) (37) (38) 59

5 If we assume that the eigenfuction for the lowest eigenvalue is not a function of x, we can introduce the stream function Ψ defined by Ψ y = U z and Ψ = U y. The lowest frequency eigenmodes of shear driven flows tend to be Langmuir circulation flows that are independent of the streamwise direction. Our use of this assumption is justified by empirical observations within this field [6]. The Euler Lagrange equations become: λu x = Gr λ Ψ = q y + Ψ Ψ y U x (39) (4) λ Ψ y = q Ψ y + Gr U x (41) Taking the z derivative with respect to the second expression and the y derivative with respect to the first and third we can subtract the last two to eliminate the pressure: λ U x y = Gr Ψ y U x y λ Ψ = Ψ + Gr U x y We can write an ODE for the fourier modes of Ψ: λ ( ˆΨ k ˆΨ λıku x = Gr k ˆΨ ık ) = ( U x k U x ) ( 4 ˆΨ 4 k ˆΨ + k4 ˆΨ ) (4) (43) (44) + Gr ıku x (45) ˆΨ z=,1 = (46) ˆΨ z= = (47) ˆΨ z=1 = (48) U x z= = (49) U x z=1 = (5) This is an ODE and we can discretize it and convert it into a generalized eigenproblem. We look for the critical Grashoff number Gr where the system first has a negative eigenvalue for some k. We find that Gr c = 51.7 at k =.1. This is in agreement with work done by Tang, Caulfield, and Young [6]..3 Bounds on the Energy Dissipation in Two Dimensions Define to be the space time average and to be the horizontal and time average. Begin by taking the dot product of the momentum equation with u. After averaging we find that: ν u = τ hūx(h) (51) 6

6 Define the dissipation ɛ = ν u, Reynolds number of the flow to be Re = ūx(h)h ν, and the friction coefficient to be C f = ɛh ū x(h). For the steady state solution C 3 f (Re) = 1 Re. We want to determine a limit on how C f might scale with the Reynolds number by proving bounds on the mean horizontal velocity at the z = h. Introduce a background field horizontal velocity U(z) such that U satisfies the same boundary conditions as u. Consider the decomposition u = îu + ũ. The following are the dimensionless equations for ũ: ũ t + ũ ũ + p + U ũ x + îũ z du dz = ũ + îd U dz (5) Take the dot product with ũ and compute the space time average: = ũ du dz ũxũ d U z + ũ x dz ũ z= = (53) ũ z z=h = (54) ũ x z=h = (55) The time derivative term vanishes because ũ is bounded. If we integrate the last term by parts in the z direction the previous expression becomes: = ũ du dz ũxũ ũx du z + Gru x (1) (57) dz = ũ du dz ũxũ ũx du z + Gr(ū x (1) U(1)) (58) dz Substitute îu(z) + ũ(z) = ū(z) into u : 1 u = 1 ũ + 1 du + dz ũx du dz We take a linear combination of these expressions so as to eliminate the ũ x 1 u 1 = ũ + du dz ũxũ z + Gr(ū x (1) U(1)) + 1 Now we define the quadratic form Q U (ũ): 1 Q U (Ũ) = ũ + du dz ũxũ z du dz du dz term: (56) (59) (6) Finally we combine this definition with the identity u = Grū x (1) to produce: u x (1) = U(1) 1 du + Gr dz Gr Q (6) If we choose U so that Q is positive definite we arrive at the following bound for ū x (1): ū x (1) U(1) 1 du (63) Gr dz (61) 61

7 Define the background horizontal velocity profile to be linear in z in horizontal boundary layers near the top and bottom boundary and constant elsewhere: U(z) = Grz < z < δ 1 (64) = Grδ 1 δ 1 < z < 1 δ (65) = Gr(δ 1 + δ + z 1) 1 δ < z < 1 (66) Then U(1) = Gr(δ 1 +δ ) and ū x (1) Gr(δ 1 +δ ). We need to choose δ 1 and δ to maximize the sum while keeping Q positive definite. We drop the accents and refer to the fluctuations away from the background as u. Q = 1 δ1 1 u + Gr u x u z dzdx + Gr u x u z dxdz (67) 1 δ We start with the second term on the right hand side: δ1 δ1 u x u z dzdx = z ( ux u u ) z z + u x dz dxdz (68) u Now we use incompressibility to eliminate the u z u x = u x x x = ux x term. Then we find: δ1 δ1 ( z ) u x u z dzdx = u z x u z dz dz dxdz (69) δ1 z u x z u z δ 1,zdz dxdz (7) δ1 z u x δ 1,z u z δ 1,zdxdz (71) ( δ C u x δ 1 + C u x x δ 1 + C ) u z δ 1 (7) 6

8 ( δ1 Here δ1 = dxdz) 1/ ( 1 1/. and 1 δ = 1 δ dxdz) Similarly define ( ) δ1 1/. δ1,z = ( ) dz If we choose C = 1 we find δ 1 u xu z dzdx δ 1 16 u. We can perform an identical analysis at the top boundary layer, and set δ 1 = δ to prove that Q 1 u Gr δ 1 8 u δ 1 Gr δ 1 8 u 1 δ (73) ( ) 1 Gr δ 1 u (74) 8 This is positive if δ 1 Gr 1/. This means that ū x (1) 4Gr 1/. In terms of units ū x (h) = ū x (1) ν h 4τ 1/. The friction coefficient C f = τ ū(h) Bounds in Three Dimensions We can use much of the same algebra to derive bounds for the three dimensional case. The equations are the same except that there is a y component in the velocity field. As a ( result we have the same ū(h) U(1) 1 du ) Gr dz as long as Q. We pick the same background profile as in the two dimensional case, linear near each boundary and constant in the bulk. We begin with the second term in Q and find that: Z Z Z δ1 Z Z Z δ1 u xu zdxdydz Z Z δ 1 1 dx dy ux C δ 1,z + C (1 a) = δ 1 4 z ux δ 1,z uz δ 1,zdxdydz (75) Z Z dx dy uz δ 1,z + C Z Z «a dx dy ux x + uy y δ 1,y (76) 1 C ux δ 1 + C(1 a) uz δ 1 + Ca ux x δ 1 + Ca uy y δ 1 + Ca ux y δ 1 + Ca uy x δ 1 Choose a = 1 and C = to find that Gr δ 1 u xu z dxdydz δ 1 4 u δ 1. We make the transformation z 1 z to study the upper boundary layer: dx 3δ3/ π δ δ dy u x u z dz dx dy dz ( 1 ( u ) 1/ x π δ ) dz u z z 1 δ,z (78) ( 1 C u x + C u z δ + C u x x δ + C ) u y y δ (79) «(77) Combining these two terms we find that: 3/ Q 1 u Gr δ 1 4 u δ 1 + Gr 3δ π ([ ] δ1 Gr 4 + 3δ 3/ 1 π C ([ δ1 = Gr ] δ 3 π δ 1 ( 1 C u x + C u δ ) u x + δ 1 4 u u x + δ 1 4 u ) ) (8) (81) (8) 63

9 Figure : Stanton Diagram of the TCY bounds and the bounds proven here on C f In the last line we made the choice C = πδ 1 3 δ 3/. This shows that the ratio of the boundary layers δ δ 1 Gr 1/6. In order to produce a prefactor for the scaling of ū(1) we take δ =. Then δ 1 = Gr 1/ and ū(h) τ 1/. The corresponding friction coefficient is C f 1 =.354. The bound proved by Tang, Caulfield, and Young[6] is C f <.37, which is stronger than this bound. 3 Infinite Prandtl Number Marangoni Convection Consider the equations describing infinite prandtl number Marangoni convection: T t p = u (83) + u T = T (84) u z= = (85) T z= = (86) u z z=1 = (87) T z=1 = 1 (88) u x z=1 = Ma T x z=1 (89) 64

10 Take the divergence of the momentum equation: p = u (9) = u (91) = (9) If we take the Laplacian of the momentum equation we will find that u solves the biharmonic equation: u = p (93) = p (94) = (95) We will be interested in the z component of the velocity for the background method. If we take the Fourier transform in the z direction it satisfies an ordinary differential equation: d 4 u z dz 4 k d u x dz + k4 u x = (96) u z z= = (97) du z dz z= = (98) u z z=1 = (99) This has an exact solution: u k = Maθ k (1) u k Maθ k (1). sinh(k) sinh(k) cosh(k) k d u z dz z=1 = k Maθ(1) (1) (sinh(kz) kz cosh(kz) + (k coth(k) 1) z sinh(kz)) (11) Define f k (z) = We have plotted the modes in Figure 3. For large value of k the functions are concentrated near z = 1. For very large values of k the maximum value of f k goes to zero. For small values of k the maximum of f k is extremely small and the function is concentrated over the entire unit interval. For values of k near 3 and 4 the maximum is large and the concentration is also over a significant portion of the unit interval. Set up the background method by making the substitution T = τ(z)+θ, where τ() = and dτ dz (1) = 1. Then the field θ satisfies the equation: θ t + u dτ z dz + u θ = θ + d τ dz (1) θ(z = ) = (13) θ (1) = (14) If we multiply the θ evolution equation by θ and integrate over space we get the expression: 1 θ = dx dz ( θu z τ θτ ) θ (15) t 65

11 Figure 3: Plots of the functions f k. 3.1 Energy Stability Make the substitution τ = z in the previous equation. The solution τ = z will be stable when the quadratic form Q = θ u z θda is positive definite. Allow u k to stand for the kth Fourier coefficient of u z. Substitute the expression for the Fourier series of θ and u z into Q: Q = 1 ( dz R(θ k u k ) + ( dθ ) k k dz ) + k θk dz (16) Then use the Cauchy-Schwarz inequality: 1 dzr(θ k u k ) Maθ k (1) f k θ k (17) Ma dθ k dz θ k f k (18) Ma ( f k k dθ k dz + 1 ) k θ k (19) If for each k the quantity Ma k f k is less than 1 the form Q will be positive definite. Since we know the f k we can write this condition explicitly: Z " dz sinh(k) sinh(k) cosh(k) k (sinh(kz) kz cosh(kz) + (k coth(k) 1) z sinh(kz)) # 1 1/ A k Ma This allows us to calculate the critical Marangoni number to be Ma c > The maximum occurs at k =.4. This estimate is slightly better than the estimate published Davis in (11) 66

12 1969. Davis used variational methods and his bound is valid for all Prandtl numbers. The variational problem that he set up used natural boundary conditions to eliminate the terms that come from integrating by parts along the boundary. This weakening of the boundary condition is what makes it possible to improve the scaling using nonvariational methods. 3. Upper Bounds on the Nusselt Number We define the Nusselt number to be the ratio of heat transport to heat transport due to conduction, so that it measures the strength of convection. Conduction requires a temperature gradient, so one would expect that when convection is strong the fixed heat flux is maintained with a minimal temperature gradient. The result is that the temperature at the top becomes low. Therefore we define the Nusselt number as 1 T (z=1). If we multiply the time evolution equation for T by T and average we find that: 1 Nu = T (111) Take the energy evolution equation for θ and time average it: = dx dz(θu z τ θτ ) θ (11) Now use the expression θ = T τ dxdzτ θ to write: = dx dz(θu z τ θτ ) θ T + dxdzτ θ + τ (113) = dx dz(θu z τ ) T θ(1) θ + τ (114) = dx dz(θu z ) T (1) + τ(1) θ + τ + T (115) 1 Nu = dx dz(θu z τ ) θ + τ(1) + τ (116) If the functional Q = dx dz(θu z τ ) + θ is always positive we can prove an upper bound on the Nusselt number: 1 Nu τ(1) dzτ (117) We choose the derivative of the background profile τ to be equal to 1 in a layer of width δ 1 near the bottom and δ near the top. We choose τ to be constant in the bulk: τ(z) = z < z < δ 1 (118) = δ 1 δ 1 < z < 1 δ (119) = δ 1 z + 1 δ 1 δ < z < 1 (1) Then 1 Nu δ 1 + δ as long as dx dz(θu z ) + θ is positive. Write this in terms of the Fourier decomposition of θ and u z : k [ δ1 dzr(θ k u k ) δ dzr(θ k u k ) + dθ k dz + k θ k ] (11) 67

13 Figure 4: Background profile for Marangoni convection. If for all k the quantity δ 1 dzr(θ ku k ) δ dzr(θ k u k ) + dθ k dz + k θ k > the form Q will be positive definite. We begin the analysis with the lower boundary layer. We start by replacing u k with Maθ k (1)f k. Then we use the fundamental theorem of calculus: δ1 f k θ k (1)θ k dz f k θ k dz dθ k dz (1) δ1 z fk dθ k dz (13) Let F (z) = sup k f(z). Then we need to evaluate δ 1 zf (z)dz to determine the scaling. As z goes to zero each f k scales as c k z so there is reason to believe that F might have the same behavoir. We verified numerically that for extremely small z the value of k at which the maximum over f k is realized has a lower bound, implying that there is some c such that F (z) < cz. We show this in Figure 3.. This means we can choose δ 7/ 1 = O(Ma 1 ), or that Nu < O(Ma /7 ). We can calculate the prefactor numerically. We set δ =. We find that Nu <.84Ma /7. Boeck and Thess have used numerical methods to find solutions of the infinite Prandtl number Marangoni problem. They analyzed the scaling of the Nusselt number of their numerical solutions with respect to the Marangoni number and found that Nu =.446Ma.38. They also theorized that Nu Ma /9 [1]. We compare their data to our bound in Figure 3.. δ1 3.3 Heat Transport with finite Prandtl Number We consider the problem of Marangoni Convection in a system with finite Prandt number. The system is described by the equations: 68

14 Figure 5: Scaling of the functions f k for different values of k near z =. Figure 6: Plot of Nusselt and Marangoni number of numerical data compared to the rigorous upper bound and the critical Marangoni number 69

15 P 1 ( u + u u) = p + u t (14) u = (15) T + u T = T t (16) u(z = ) = (17) T (z = ) = 1 (18) u 3 (z = 1) = (19) T (z = 1) = 1 (13) u 1 = Ma T (z = 1) (131) x θ (z = 1) = (13) The Nusselt number of this flow is defined as Nu = T 1. Multiply the equation for (1) the temperature by T and average it over the volume. All the terms on the left hand side vanish, the first term because the norm of temperature is bounded and the advection term because there is no flux across the upper and lower boundaries. Therefore we find that: = T T dv (133) V = T + T T da (134) ΓV = T da T (135) z=1 This means that Nu = 1. Introduce a background profile τ(z). We choose τ to satisfy T a homogeneous Dirichlet condition at z = and the Neumann condition τ = 1 at z = 1. Then we can decompose the temperature T = τ(z) + θ(x, z, t) where θ = at z = and = at z = 1. For each τ there is an equation for θ: θ θ t + u θ + u τ z = θ + τ (136) Multiply this equation by θ and take the average over space and time. Again the first two terms on the left hand side vanish: Now substitute τ and θ for T into T = 1 Nu : = θ θ(1) θu z τ θ (137) θ τ dv (138) = θ θ(1) θu z τ + τ dv (139) 1 Nu = θ + τ + 7

16 In the last line we added two times the identity in the previous equation. In order to incorporate the Marangoni condition we need the momentum equation. We take the dot product of the momentum equation with u and calculate the space time average: = u u x + u x da (14) z=1 = u θ Ma u x da (141) z=1 x = u Ma (u θ x )dv (14) x Consider linear combinations of this expression and the equation for the Nusselt number: V Z Z 1 Nu = θ θ(1) θu zτ + τ dv + C( u θ + Ma V V (ux )dv ) (143) x Z Z = θ ( T (1) τ(1)) θu zτ + τ dv + C( u θ + Ma V V (ux )dv ) (144) x = τ(1) τ + Q(θ, τ, u) (145) In the second to last line we used the fact that θ = T τ. In the last line we used the fact that Nu = T 1 and defined the quadratic form Q: (1) Q(θ, τ, u) = θ + θu z τ + C( u + Ma (u θ x )) (146) x Suppose that Q > for all u, θ, for some given choice of τ. Then 1 Nu < τ(1) τ. The bound on the Nusselt number will only depend on the choice of τ. Therefore the task will be to choose a τ that maximizes the bound on 1 Nu subject to the constraint that Q is positive. The background τ only affects Q through the term θu z τ, which can be negative. Therefore it would make sense to make this term as small as possible. Define τ such that τ is zero everywhere except near the upper and lower boundaries. τ = z z < δ 1 (147) τ = δ 1 δ 1 < z < 1 δ (148) τ = δ 1 δ + 1 z 1 δ < z < 1 (149) With this choice of τ the expression θu z τ reduces to the inner product of u z and θ over regions of width δ 1 and δ near the boundaries. The τ also satisfies the boundary conditions. We treat the top and bottom layer separately, starting with the bottom layer: δ1 θu z dv δ1 δ1 ( z ) ( z θdz ) u zdz dxdz (15) z θ u z dxdz (151) = δ 1 θ u z (15) 71

17 It is more difficult to deal with the upper layer because θ is not zero on the upper boundary. As a result we have to use the Poincare inequality instead of the method we just used on the lower layer. We can choose δ = and not effect the scaling. The term involving the Marangoni number must also be controlled by the norms of the derivatives of u and θ. Ma ( ) ( θ ux θ u x = Ma x x u x x ) θ (153) u x θ x + u z θ (154) Here we have used the fact that u is incompressible. Now we have bound on Q in terms of first derivatives of u and θ. We are only interested in the scaling and do not make an effort here to derive an optimal bound. If we take C = 1 we find that Q is positive when the following equation for λ has no negative solutions: (1 λ) 4 ( δ4 4 + Maδ )(1 λ) = (155) This is satisfied as long as δ4 4 + Maδ < 1. This suggests that for large Ma δ Ma 1/. This suggests that Nu Ma 1/. The scaling Nu Ma 1/ is considerably weaker than the infinite Prandtl number case. On the other hand, simulations by Boeck and Thess do not indicate that the actual scaling is weaker for the finite Prandtl number case. The reason that this scaling is so much worse is because we were unable to take advantage of the relationship between the temperature and velocity fields. It was possible to achieve the /7th bound because we knew what the Fourier modes of the velocity field were. However the relationship is much more complicated in the finite Prandtl number case and we were unable to use it. In general, improvements of the bounds produced by the background method in Raleigh-Benard convection were achieved by using the relationship [], so it seems reasonable to expect that this bound could be improved by such considerations. 4 Conclusion We presented rigorous upper bounds for the friction coefficient in stress driven shear flow and also derived the critical Grashoff number for the stability of the laminar flow solution. Tang, Caulfield, and Young first used the background method to analyze this problem. We were able to prove an analytic bound on the friction coefficient using the full stress boundary condition. We improved the lower bound on the critical Marangoni number for nonlinear stability in infinite dimensional Prandtl number Marangoni convection. We also proved that Nu <.84Ma /7 for the same problem. In finite Prandtl number Marangoni convection we were unable to do any better than Nu Ma 1/. References [1] T. Boeck and A. Thess, Power-law scaling in benard-marangoni convection at large prandtl numbers, Phys. Rev. E, 64 (1). 7

18 [] F. O. C. Doering and M. Reznikoff, Bounds on vertical heat transport for infinite prandtl number rayleigh-benard convection, Flu. Mech., 56 (6), pp [3] S. Davis, Buoyancy surface tension instability by the method of energy, Flu. Mech., 39 (1969), pp [4] C. Doering and P. Constantin, Energy dissipation in shear driven turbulence, Phys. Rev. A, 69 (199), pp [5] J. Pearson, On convection cells induced by surface tension, Flu. Mech., 4 (1958), pp [6] C. C. W. Tang and W. Young, Bounds on dissipation in stress driven flow, Flu. Mech., 51 (4), p

On upper bounds for infinite Prandtl number convection with or without rotation

On upper bounds for infinite Prandtl number convection with or without rotation Charles R. Doering a and Peter Constantin b a Department of Mathematics, University of Michigan, Ann Arbor, MI 4819-119 b