Bounds on buoyancy driven flows with Navier-slip conditions on rough boundaries

We consider two-dimensional Rayleigh–Bénard convection with Navier-slip and fixed temperature boundary conditions at the two horizontal rough walls described by the height function h. We prove rigorous upper bounds on the Nusselt number Nu which capture the dependence on the curvature of the boundary κ and the (non-constant) friction coefficient α explicitly. If h∈W2,∞ and κ satisfies a smallness condition with respect to α, we find Nu≲Ra12+∥κ∥∞, where Ra is the Rayleigh number, which agrees with the predicted Spiegel–Kraichnan scaling when κ = 0. This bound is obtained via local regularity estimates in a small strip at the boundary. When h∈W3,∞ , the functions κ and α are sufficiently small in L∞ and the Prandtl number Pr is sufficiently large, we prove upper bounds using the background field method, which interpolate between Ra12 and Ra512 with non-trivial dependence on α and κ. These bounds agree with the result in Drivas et al (2022 Phil. Trans. R. Soc. A 380 20210025) for flat boundaries and constant friction coefficient. Furthermore, in the regime Pr⩾Ra57 , we improve the Ra12 -upper bound, showing Nu≲α,κRa37, where ≲α,κ hides an additional dependency of the implicit constant on α and κ.


Introduction
In this paper we deal with the Rayleigh-Bénard convection problem, modelled by the Boussinesq system for the velocity field u = (u1, u2) ∈ R 2 , the scalar temperature field T and the pressure p where h describes the height of the bottom boundary and e2 = (0, 1).The nondimensional numbers Ra and Pr are the Rayleigh and Prandtl number respectively.We assume periodicity of all variables in the e1 = (1, 0)-direction and impose where γ − = {(y1, y2) ∈ R 2 | 0 < y1 < Γ, y2 = h(y1)} and γ + = {(y1, y2) ∈ R 2 | 0 < y1 < Γ, y2 = 1 + h(y1)} are the bottom and top boundary respectively.Our goal is to derive physically meaningful upper bounds for the Nusselt number, a quantity that measures the average heat transport in the vertical direction.In the last thirty years, the problem of deriving scaling laws for the Nusselt number in turbulent convection between two horizontal plates has been highly investigated both in experiments and numerical studies (cf.[AGL09] and references therein).Our analysis wants to capture the role of geometry and boundary conditions in the scaling laws for the Nusselt number.Intentionally, we did not yet specify the boundary conditions for u at γ + and γ − .Before doing that, we recall some important results under no-slip and free-slip boundary conditions, which are the most studied for this problem.The no-slip boundary conditions in the case of flat boundaries (i.e.h = 0) read u = 0 at y2 = {0, 1} .
In this setting, Doering and Constantin in 1996 [DC96] rigorously proved the upper bound Nu Ra 1 2 in three-dimensions.From this seminal result, many works followed aiming at optimizing the Ra 1 2 -upper bound (see [Nob23], [FAW22] and references therein).In the infinite-Prandtl number setting, a series of works [CD99], [DOR06], [OS11] established the bound Nu Ra up to a logarithmic correction.A lower bound within the method proposed in [CD99] was proved in [NO17].In the finite Prandtl number case, Choffrut, Otto and the second author of this paper in [CNO16] improved the perturbative result of Wang [Wan08] showing Nu (Ra ln Ra) 1 3 for Pr (Ra ln Ra) 1 3 and the crossover to the bound Nu (Pr −1 Ra ln Ra) 1 2 for Pr (Ra ln Ra) 1 3 .The methods proposed by Doering and Constantin in the nineties to produce scaling laws for the Nusselt number in boundary-driven convection continues to be fruitfully employed in other (related) problems [Tob22], [DC01], [FNW20], [Ars+21] (see also [Gol16] and references therein).Recently, remarkable results have been produced concerning the saturation of the upper bounds [DT19], [Kum22].The question around the "ultimate scaling law" in the Rayleigh-Bénard convection problem was recently object of animated discussions [DTW19] and it remains unclear whether at large Rayleigh number the scaling Nu ∼ Ra 1 2 will prevail over the scaling Nu ∼ Ra 1 3 or whether another scaling arises.The no-slip boundary conditions are the most used in theoretical and numerical studies, but whether or not they represent the most "realistic" conditions, is subject to debate (see [Nob23] and references therein).In 2011 Doering and Whitehead [WD11] remarkably proved Nu Ra 5 12 in the two-dimensional Rayleigh-Bénard convection problem between flat horizontal boundaries, with free-slip boundary conditions, i.e. u2 = 0 and ∂2u1 = 0 at y2 = {0, 1} .This result rules out the Nu ∼ Ra 1 2 scaling in this setting.But the question about the optimality of this upper bound remains along with the question whether the scaling Nu ∼ Ra 5 12 carries a physical meaning.Motivated by this result, in [DNN22] Drivas, Nguyen and the second author of this paper considered the two-dimensional Rayleigh-Bénard convection problem with the following conditions on the horizontal plates: where Ls is the (constant) "slip-length".Under these assumptions the authors proved the interpolation bound Nu Ra Relatively few theoretical works have addressed the problem of the Nusselt number scaling in the case of rough boundaries in Rayleigh-Bénard convection: In [SW11] Shishkina and Wagner developed an analytical model to estimate the Nusselt number deviations caused by the wall roughness and in [WS15] the same authors performed direct numerical simulations.In [GD16] Goluskin and Doering considered the three-dimensional Rayleigh-Bénard problem between rough boundaries.In particular, under the assumption that the profile h (which may differ at the top and bottom boundaries) is a continuous and piecewise differentiable function of the horizontal coordinates and has squared-integrable gradients, the authors showed Nu C( ∇h 2 2 )Ra 1 2 .Here we want to cite also [Ker16] and [Ker18], where Kerswell derives an upper bound for the energy dissipation rate per unit mass for a pressure-driven flow in a rough channel.
The boundary conditions in (2) are the simplified version of the original Navier-slip boundary conditions [Nav23] τ Here we denoted with n and τ the outward unit normal and tangential vector to the boundary respectively, with uτ = u • τ the tangential velocity and with Duij = 1 2 (∂iuj + ∂jui) the strain tensor.The space-dependent function α ≥ 0 is the friction coefficient and measures the tendency of the fluid to slip on the boundary.It may vary along γ − and γ + .The problem is illustrated in Figure 1.We notice that the Navier-slip boundary conditions have been studied in a variety of problems in fluid mechanics.They have been considered in problems related to mixing [HW18] as well as rotating systems [DG17].In [AEG21] and [Ace+19] the authors studied the Stokes operator with Navier-slip boundary conditions and the convergence to no-slip boundary conditions as α → ∞.
In this paper we want to generalize the result in [DNN22], considering the original Navier-slip boundary conditions (3), hence allowing a non-constant (space dependent) friction coefficient.Furthermore, we allow a certain degree of roughness at the upper and lower plates.
Our first, more general result is the following Theorem 1.1.Let u and T solve system (NS)-(DF)-(AD), with boundary conditions (1) and pointwise almost everywhere on γ − ∪ γ + .Then there exists a constant C > 0 depending only on h ∞ and |Ω| such that for all Ra ≥ 1.
The constants are given by , where α and κ are the derivatives of α and κ along the boundary and C > 0 denotes a constant depending only on the size of the domain |Ω|, h ∞ and r.
First of all, we notice that our results in Theorem 1.1 and Theorem 1.2 also cover the case of flat-boundaries and generic friction coefficient α.In fact, all results (i.e. ( 5), ( 7), (8) and ( 9)) hold setting κ = 0.
and ν ≥ µ.This means that the best bound is achieved in the case 1 24 ≤ µ < 1 2 and Pr Ra 3 2 µ+ 3 4 .Notice that, while (7) holds only under the assumption |κ| ≤ α, the upper bound (8) holds under the weaker assumption |κ| ≤ 2α + especially when α and |κ| are very small.We observe that in the regime of small α and κ, say becomes independent of α and κ.Finally we remark that if the condition Ra −1 ≤ α for (8) is violated, then (5) yields a stricter bound.Interestingly (9) improves the upper bound (5) in the case of small α + κ ∞ and big Pr, i.e.Pr ≥ Ra 5 7 .While the interpolation bound (7) yields a better result if α ∼ Ra −µ , (9) provides the sharpest bound if α and κ are independent of Ra.Notice that, differently from the constant prefactor in (5), the constant C 3 7 also depends on α and κ.

Preliminaries
We start with introducing some notation and facts we will use in the whole paper.We will often use that the tangential vector to the boundary τ can be written as τ = n ⊥ = (−n2, n1), where n is outward unit normal and uτ = u • τ .The curvature of the boundaries κ is given by d dλ τ = κn, where λ is the parameterization of the boundaries by arc length in τ -direction.The problem is illustrated in Figure 1.
We will use the following notation to denote space-time averages Moreover, on the set Figure 2: Illustrations of γ(x 2 ), Ω(x 2 ) and Ω (z where 0 ≤ x2 ≤ 1, illustrated in Figure 2, we define the average In what follows we will consider 0 ≤ T0 ≤ 1, implying for all times t > 0, by the maximum principle for the temperature.
Next we define the Nusselt number.
Definition 2.1.The Nusselt number is defined as It also admits other equivalent representations, as stated in the following.
Proposition 2.1.The Nusselt number satisfies where n+ is the normal at the curve γ(x2) pointing in the same direction as n on γ + , illustrated in Figure 2.
Proof.Argument for (13): Testing the temperature equation with T , integrating by parts and using the the divergence-free condition and the boundary conditions for T , we obtain 1 2 Taking the long-time averages and using (11), the maximum principle for the temperature, T is universally bounded in time and we get Argument for (14): For 0 ≤ x2 ≤ 1, we define the sets Figure 3: Illustration of Ω (z) illustrated in Figure 2, and integrate the equation for T , obtaining where we used the incompressibility condition and that u • n = 0 at γ − .Taking the long-time average we get n In particular, from this identity we infer that the Nusselt number is independent of x2.

Integrate the previous equation in z between min h and 1 + max h and write
We analyze the three terms separately: Term C: We notice that this term can be simply rewritten as Term B: This integral has a sign, in fact Term A: We decompose the integral in A further: Notice that The term A1 instead will be estimated as follows where in the first inequality we used that n− • ∇T ≥ 0 at γ − .Putting all together, we obtain Taking the long-time average and observing that lim sup T dy dz dt = 0 by the maximum principle for T , we have

A-priori bounds
In this section we collect a-priori bounds on the energy and enstrophy of the solution u and derive pressure estimates that will be used to prove the main result in the Section 4.
3.1 A-priori estimate for the velocity Proposition 3.1 (Energy Balance).Strong solutions of (NS), (DF), (3) satisfy 1 2Pr Proof.The balance follows by testing the Navier Stokes equations with u, integrating by parts and observing that by the incompressibility and boundary conditions, and In this identity we used the algebraic identity the Navier slip boundary conditions to deduce proved in (84) in the Appendix.
Under a smallness assumption on κ, the second and third term on the left-hand side of (17) are positive and bounded from below by the H 1 -norm of u even though (2α + κ) might be negative on some parts of the boundary.This will be essential in what follows, especially in order to prove the energy decay and the main theorem.
Lemma 3.2.Assume α ≥ α almost everywhere on γ − ∪ γ + for some constant α > 0 and κ satisfies almost everywhere on γ − ∪ γ + .Then Proof.In this proof we use the notation which is the evaluation of the functions on the bottom or top boundary.Notice that, because of the symmetry of the domain, κ− = −κ+.
The idea of the proof is that if κ− is negative for some y1, then κ+ is positive and we can compensate by the fundamental theorem of calculus.
By the fundamental theorem of calculus, Young's and Hölder's inequality for all > 0, where |∂2u(y1, y2)| 2 dy2 and analogously Integrating ( 22) and ( 23) in y2 one gets where in the last inequality we have smuggled in the factor 1 holds for either i = + or i = −.Using (25), (24) turns into Integrating in y1 and choosing = 3 yields which implies the following bound for the H 1 -norm It is only left to show that (25) holds.In order to prove the claim we distinguish between two cases.
Remark 3.1.Note that this Lemma can be improved: In fact for every κ and α with κ∓ < , where we use the notation (21), it holds ∇u 2 2 + (2α + κ)u 2 τ > 0. This implies energy decay in (17).Nevertheless, in order to simplify the estimates and improve readability we will work with assumption (19) instead.This choice will have no effects in terms of optimality of the bounds for the Nusselt number.
We will use the energy balance and (20) to prove the following decay estimate for the energy of u: Lemma 3.3 (Energy Decay).Let the assumptions of Lemma 3.2 be satisfied.Then the energy of u is bounded by Proof.By the energy balance (17), Young's inequality and the maximum principle (11), we obtain 1 2Pr Plugging ( 20) into (27), we find 1 2Pr and choosing = 1 8 min{1, α} yields Applying Grönwall's inequality we obtain (26).
Taking the long-time average of the energy balance (17), using the fact that lim sup thanks to the uniform bound (26) one gets and observing that, by (15), we deduce the following Corollary 3.4.

A-priori estimate for the vorticity
We now introduce the two-dimensional vorticity ω = ∇ ⊥ • u, where ∇ ⊥ = (−∂2, ∂1).It is easy to see that ω satisfies the equation Notice that the boundary term is deduced from the following computation where we used that τ = (−n2, n1), the boundary conditions (3) and the identity proved in (83) in the Appendix.
Proposition 3.5.The following vorticity balance holds Proof.Testing the vorticity equation with ω yields 1 2Pr Using the incompressibility condition, it is easy to see that the first term on the right-hand side vanishes since u • n = 0 at ∂Ω.In order to analyze the second term, we first notice that where we used incompressibility in the first identity.Then, using the boundary conditions for the vorticity and temperature and (32), we have Plugging it into (31) yields 1 2Pr Let us observe that the term (α + κ)u • (u • ∇)u is, in general, non-zero as the parameters α and κ depend on the space variables.
We now want to relate the L 2 -norm of the vorticity with the L 2 -norm of the enstrophy.
holds, where the constant C only depends on p, |Ω| and h ∞.
Remark 3.2.Note that in the case of flat boundaries the estimates simplify to as proven in Lemma 7 in [DNN22]. Proof.
• Integrating by parts twice, we find where we used the identity = ∆u, due to incompressibility.Next notice that τiτj +ninj = δij.Therefore the second boundary term of the right-hand side of (33) can be rewritten as where in the last identity we used that τ The first term on the right-hand side of (33) cancels with the first term on the right-hand side of (34), implying on γ − ∪ γ + , which is proven in (84) in the Appendix, yields the claim.• Let φ be the stream function of u, i.e. ∇ ⊥ φ = u, then ∆φ = ω φ| γ ± = φ± with constants φ+ and φ− and without loss of generality set φ− = 0. We can calculate φ+ by In order to flatten the boundary we introduce the change of variables Here and in the rest of the paper C > 0 denotes a constant that possibly depends on h ∞, the size of the domain |Ω| and the Sobolev exponent and may change from line to line.Note that and for ξ(x) = χ(Ψ(x)) = χ(y) one has and analogous for the transformation in the other direction.Then where L φ = i,j ∂x i (ãi,j∂x j φ(x)) with ã1,1 = 1, ã1,2 = ã2,1 = −h and ã2,2 = 1 + (h ) 2 , φ(x) = φ(Ψ(x)) and f = ω − h 1 |Ω| Ω u1dy with ω(x) = ω(Ψ(x)).As this operator is elliptic we get for some constant C > 0 depending only on p, |Ω| and h ∞.Using Hölder's inequality and the estimates for the change of variables (36), (38) becomes Going back to the definition of φ, we find that (36), Hölder's inequality and (39) yield implying In order to estimate the L p -norm of u by the L q -norm use interpolation and Young's inequality to get and all > 0. Then choosing proving the claim.
• In order to prove the W 2,p bound notice that by (37) φ = ∂x 1 φ solves x 2 φ ∈ L p .Again using elliptic regularity and Hölder's inequality we find In order to estimate the missing term ∂ 3 Taking the norm in (43) we find Combining ( 42) and (44) Using Hölder's inequality we find which together with (39) yields where in the last inequality we used By the definitions and the change of variables estimate (36) and Hölder's inequality one gets where in the last inequality we used (45), ( 39) and (46).Finally using the W 1,r -bound for u, (41), and Young's inequality yields the claim.
The next result concerns a crucial L ∞ t L p x −bound for the vorticity.
Lemma 3.7.Let p ∈ (2, ∞) and assume that the conditions of Lemma 3.2 are satisfied.Then there exists a constant C depending only on p, |Ω| and h ∞ such that where Cα,κ = 1 + α + κ Proof.Fix an arbitrary time t > 0 and decompose the solution ω to (29) as Since the boundary and the initial values have a sign, i.e. ω+ ≤ 0 on γ + ∪ γ − , ω+,0 ≤ 0 in Ω and ω− ≥ 0 on γ + ∪ γ − , ω−,0 ≥ 0 in Ω, then, by the maximum principle, ω − ω+ = ω+ ≤ 0 and 0 ≤ ω− = ω − ω− yielding ω− ≤ ω ≤ ω+.In particular Hence, it remains to find upper bounds for |ω−| and |ω+|.By symmetry, it suffices to show an upper bound for ω+.We divide the proof in three steps: Step 1: Omitting the indices, we define Testing the equation with ω|ω| p−2 we obtain Using T ∞ = 1, Young's and Hölder's inequality, we estimate the second term of the right-hand side as By the Poincaré estimate applied to the second term of the right-hand side (remember that ω vanishes at the boundary by definition), we obtain where Cp denotes the Poincaré constant.Dividing through by ω p−2 p we obtain the inequality By the Grönwall inequality Step 2: We now turn to the estimate for Λ.We have which can be bounded using interpolation and Young's inequality by for p > 2 and arbitrary > 0, where θ = p 2(p−1) and x ) .
As Ω is bounded Hölder inequality and Lemma 3.7 yield that if ω0 ∈ L p for any p < ∞ then ω p is universally bounded in time.By trace Theorem and Lemma 3.6 Using Lemma 3.3 this is also universally bounded in time, therefore taking the long time average of the vorticity balance, see Lemma 3.5, we get the following.
Corollary 3.8.Assume that the conditions of Lemma 3.7 are satisfied and ω0 ∈ L p for some p > 2, then (48)

A-priori estimate for the pressure
The pressure satisfies The equation in the bulk is easy to obtain by applying the divergence to the Navier-Stokes equations, using incompressibility and writing compactly ∇ • ((u • ∇)u) = (∇u) T : ∇u.In order to track the pressure at the boundary we look at Navier-Stokes equations at where n is the normal at the boundary.It is clear that using the boundary condition for the vorticity in (29).Thanks to (84) in the Appendix we also have Hence at the boundary.Now, it is only left to observe that T = 0 at γ + and T = 1 at γ − .Proposition 3.9.For any r ∈ (2, ∞) there exists a constant C depending on |Ω|, r and h ∞ such that Proof.On one hand, integrating by parts and using the boundary conditions (for u and T ), we have On the other hand using the equation satisfied by the pressure ( 49) where we used the boundary conditions for T in the last identity.Combining these estimates one gets We estimate the right-hand side: By Hölder inequality with 1 r + 1 q + 1 2 = 1 and Sobolev embedding 1 p(∇u) T : where C depends on |Ω|, r and h ∞.For the temperature term we apply Hölder's inequality use that T ∞ ≤ 1 by the maximum principle (11) and for the second term we compute where in the last inequality we use the trace estimate.Finally, we estimate the first term: Similar to (51) for 1 r + 1 q + 1 2 = 1 and every r > 2 where C depends on |Ω|, r and h ∞.
1 Since Ω is bounded, for any 1 ≤ µ < ∞ choose s > max{2, µ} and q = s µ > 1, such that ns n+s = 2s 2+s ≤ 2s s = 2.By Hölder and Sobolev inequality we obtain Combining the estimates we find Using that the pressure p is only defined up to a constant so we choose p to have zero mean such that Poincaré yields p q ≤ C ∇p q which implies p 2 Finally dividing by p H 1 we conclude that there exists a constant C > 0 depending on Ω and r such that for any r > 2.

Upper bounds on the Nusselt number
Combining the a-priori estimates derived in the previous section we are now able to prove the Ra 1 2 bound, that was first derived for the flat, no slip case in 3 dimensions by Doering and Constantin [DC96].

Proof of Theorem 1.1
Let Ω δ be given by as illustrated in Figure 4. Taking the average in z ∈ (1 − δ, 1) in the representation of the Nusselt number (14) we find Nu = lim sup In order to estimate the first term on the right-hand side notice that by the fundamental theorem of calculus for (y1, y2) where ∇u L 2 (γ − ,γ + ) = ∇u(y1, •) L 2 (h(y 1 ),1+h(y 1 )) and we used the non-penetration boundary condition for u and that n+ is constant in y2-direction in the first inequality and Hölder's inequality in the second estimate.Analogously for the temperature and (y1, y2) ∈ Ω δ it holds as T = 0 on γ + .In order to estimate the second integral in (52) partial integration and the boundary condition T = 0 on γ + yields By the maximum principle (11) the temperature is bounded by T ∞ ≤ 1, so the first two terms on the right-hand side of (55) are bounded by a constant depending on h ∞ and |Ω|.In order to estimate the last term notice that , where h = ∂1h(y1) and h = ∂ 2 1 h(y1) as derived in ( 80) and (82) in the Appendix.Therefore |∇ • n+| = |κ|, which implies for the last term in (55).Combining these observations Plugging ( 53), ( 54) and (56) into (52) and using Hölder inequality, there exists a constant depending on h ∞ and |Ω| such that Nu ≤ lim sup By ( 12) and ( 28) we can substitute both gradients and get Balancing the terms by choosing δ = Nu − 1 2 Ra − 1 4 we get

Introduction of the background field method
In order to improve the bound of Theorem 1.1 we follow the "background field" strategy used [DNN22], which is based on [WD11].This approach consists of specifying a stationary background field for the temperature and show its "marginal stability" as we will explain in what follows.This will be achieved by applying the a-priori bounds derived in Section 3.
To this end we define the background profile for the temperature by for δ > 0 and the difference θ by This profile is illustrated in Figure 5.Note that η fulfills the boundary conditions of T , so θ vanishes on γ ± .Also since n− can be expressed by as derived in (80), its gradient is given by for almost every y ∈ Ω. Inserting this decomposition in the definition of the Nusselt number, we have Proposition 4.1.Let η and θ be defined by (57) and (58).Then The proof of this identity is essentially the same as the one for Proposition 7 in [DNN22] and we report it here just for convenience of the reader.
Proof.Plugging the definitions of η and θ into (AD) we have and, integrating this equation against θ we find 0 = 1 2 The third term on the right-hand side of (61) vanishes since u • n = 0 and u is divergence-free.
For the fourth term on the right-hand side of (61) we get where in the last equality we used that θ vanishes on the boundary by definition.Similarly Therefore taking the long time average of (61) and using that θ is universally bounded in time as both T and η fulfill 0 ≤ T, η ≤ 1 we find Using this identity in the definition of the Nusselt number (12) we find Next we define where the last identity is due to (48) and Using (60) we can rewrite the Nusselt number as where the quadratic form Q is defined as In this new representation a > 0 and 0 ≤ b < (1+max h−min h) −1 and notice that the balancing term M Ra2 , with M > 0, was introduced.The choice of the parameters a, b, M will follow from an optimization procedure at the end.We now want to prove that for a suitable choice of δ the form Q is non negative.In order to do so we need the following Lemma.
Proof.By (59) 2 We focus on the first term on the right-hand side.The second one can be treated similarly.By the fundamental theorem of calculus and Hölder's inequality for h(y1) ≤ y2 ≤ h(y1) + δ, where in the last inequality we used the boundary condition for u.Similarly as θ vanishes on the boundary.In order to estimate ∂2(n− • u) notice that by partial integration and the boundary condition for u 1+h(y 1 ) Therefore for every y1 there exists h(y1) ≤ ȳ2 ≤ 1 + h(y1) such that ∂2(u • n−)(y1, ȳ2) = 0. Applying the fundamental theorem of calculus again we find where in the last inequality we used Hölder's inequality, that n− is constant in e2 direction and |n−| = 1.Combing (65) with (66), ( 67) and (68) and using Young's inequality twice yields for some µ, ν > 0 that will be determined later and Taking the long time average and setting µ = 2δ With all these preparations at hand we are able to prove the main result in the next subsection.
As b ≤ 0 by (28) plugging in the definition of a, i.e. (4.2), yields (70) Next we estimate some of the terms individually.
• For the eighth term on the right-hand side of (70) we can shift the derivative onto u and α + κ as the boundary is periodic and get Using Hölder's inequality and Trace Theorem one gets where α and κ denotes the derivative of α and κ along the boundary.The pressure bound derived in Proposition 3.9 and Young's inequality imply (71) for all > 0, where C > 0 depends on |Ω|, r, h ∞ and and in the last inequality we used that κ ≤ 1.
• For the ninth term on the right-hand side of (70) Hölder's and Young's inequality yield • In order to estimate the tenth term on the right-hand side of (70) we first use Hölder's inequality and Trace Theorem to get Again Hölder's inequality with 1 r + 1 p + 1 2 = 1 and Sobolev Theorem as in the proof of Proposition 3.9 imply for all r > 2. Combining (72) and (73) and using Young's inequality and the assumption α for all > 0 where C > 0 depends on |Ω|, r, h ∞ and .
• In order to estimate the eleventh term on the right-hand side of (70) notice that by Trace Theorem and Young's inequality 2Ra In order to apply these estimates we first notice that by Lemma 3.6 The L p -norm of the vorticity and energy are bounded by Lemma 3.7 and Lemma 3.3 respectively, implying where we exploited (69).Using this bound for the W 1,r norm of u, the prefactors in the individual estimates are independent of time.Then taking the long time average of (71), ( 74) and (75) and plugging the bounds into (70) yields In order to estimate the second term on the right-hand side of (76) use Lemma (4.2) to get Next according to Lemma (3.2) the first two terms on the right-hand side can be estimated by the H 1 norm, i.e. Then and by Lemma 3.6 and the smallness conditions (69) where we used Young's inequality and that α ∞ ≤ 2 as κ ∞ ≤ 1 by (69) and α + κ ∞ ≤ 1. Setting = 1 6C 1 and using the smallness assumption α + κ 2 ∞ ≤ C = 1 2C 2 the first bracket is positive and we are left with Next we have to differentiate between the two conditions on κ.

• Case |κ| ≤ α
In order to estimate the vorticity term notice that by Lemma 3.6 and the condition |κ| ≤ α one has and taking the long time average (78) turns into From the second squared bracket on the right-hand side it becomes clear that a has to decay at least as fast as Ra − 3 2 for Q to be non negative.Setting a = a0Ra The assumption Pr ≥ 1 and since Ra − 1 2 < α where without loss of generality C ≥ 1.In order for the two squared brackets to be non-negative we choose Again applying (77) we find (79) which because of the second squared bracket on the right-hand side imposes the condition on a to decay at least as fast as Ra − 3 2 .We differentiate between two choices of a.
• Tensor product: • If not explicitly stated differently C will denote a positive constant, which might depend on the size of the domain |Ω| = Γ, h ∞ and potentially on the exponent of the Sobolev norm.
• n+ is the normal vector pointing upwards.

Figure 5 :
Figure 5: Illustrations of the background profile η.