Comment on Geometrical Control of Active Turbulence in Curved Topographies
aa r X i v : . [ c ond - m a t . s o f t ] A ug COMMENT ON “GEOMETRICAL CONTROL OFACTIVE TURBULENCE IN CURVEDTOPOGRAPHIES”
In a recent letter [1] Pearce et al. investigate theturbulent dynamics of a two-dimensional active nematicliquid crystal which is constrained to a curved surface.The underlying model combines an incompressible sur-face Navier-Stokes equation with friction and active forc-ing with a surface Landau-de Gennes model for nematicliquid crystals. To solve the surface Navier-Stokes equa-tion a vorticity-stream function approach is considered.The approach is based on a decomposition of the veloc-ity field in divergence- and curl-free parts. While ap-propriate on simply connected surfaces, this is not suf-ficient on non-simply connected surfaces, such as theconsidered torus. As a consequence of the topologyalso non-trivial harmonic parts, velocity fields which aredivergence- and curl-free, exist, which are not representedby the vorticity-stream function approach, see [2] for anexample. We here explain the underlying situation andprovide details and examples in the Supplementary In-formation (SI).Consider the surface Navier-Stokes equation with fric-tion and forcing terms, eq. (1a) of [1]. Without thefriction and forcing terms, with the exception of somespecial initial conditions for which the solution convergesto zero, any solution of the surface Navier-Stokes equa-tion on a torus converges to a stationary Killing vectorfield, which contains non-trivial harmonic parts, see [2, 3].Clearly, with the friction and without the forcing term,the velocity converges to zero. In contrast to the argu-mentation in [1] this means that the harmonic and theanti-harmonic part vanish identically over time. More-over, considering also the forcing term introduces addi-tional harmonic parts to the velocity. There is no reasonto assume that these solution components are negligible.In fact neglecting them changes the velocity qualitativelyin an nonphysical manner. In the SI we consider an ex-ample which takes friction and forcing via a constructedQ tensor in the sense of eq. (1a) of [1] into account anddemonstarte that the harmonic parts are not negligible.We further show that the argumentation following eq.(S43) in the SI of [1] is not correct.To solve the active nematic liquid crystal modelconsidered in [1] on general curved topographies thusrequires an approach which also handles the harmonicparts of the velocity field and directly acts on thevelocity and pressure variables. Numerical approachescan be found in [2] using discrete exterior calculus(DEC) and in [3, 4] using surface finite elements foreach component of an extended velocity field in theembedding space and a penalization of the normalcomponent. General approaches how to solve surfacevector- and tensor-valued partial differential equationscan be found in [5]. Ingo Nitschke, Sebastian ReutherInstitute of Scientific Computing, TU Dresden01062 Dresden, GermanyAxel VoigtInstitute of Scientific Computing, TU Dresden andDresden Center for Computational Materials Science(DCMS) andCenter for Systems Biology Dresden (CSBD) andCluster of Excellence Physics of Life (PoL)01062 Dresden, Germany [1] D. J. G. Pearce, P. W. Ellis, A. Fernandez-Nieves, andL. Giomi, Phys. Rev. Lett. , 168002 (2019).[2] I. Nitschke, S. Reuther, and A. Voigt, in
TransportProcesses at Fluidic Interfaces , edited by D. Bothe andA. Reusken (Springer, 2017) pp. 177–197.[3] S. Reuther and A. Voigt, Phys. Fluids , 012107 (2018).[4] I. Nitschke, S. Reuther, and A. Voigt,Phys. Rev. Fluids , 044002 (2019).[5] M. Nestler, I. Nitschke, and A. Voigt,J. Comput. Phys. , 48 (2019). SUPPLEMENTARY INFORMATION
We follow the notation of [1] but use a more com-pact vector notation for the differential operators, see[2]. Let Ker div S be the space of divergence free vectorfields on the surface S . This space can be decomposedw. r. t. the L inner product into parts with and withoutcurl-components by using the Hodge theorem, i. e.Ker div S = rot S C ∞ ( S ) ⊕ L H ( S ) , where the last considers the harmonic parts. Here, weexplicitly note that this decomposition does not hold inthe sense of the local inner product, but in the sense ofthe L inner product. In the following we restrict to thesurface of the torus with radii ratio ξ := a/b with ma-jor and minor radius a and b , respectively, and denotethe two basis vector fields by ∂ θ x and ∂ φ x as well as thetwo basis harmonic vector fields by Θ and Φ , as con-sidered in [1]. It is thereby noted that these harmonicvector fields are locally orthogonal to each other w. r. t.the local inner product. Thus, an orthogonal projectionof a vector fields into the space of harmonic vector fieldsΠ H : T S → H ( S ) can be considered using basic orthog-onalization techniques, i. e.Π H v = h v , Θ i L k Θ k L Θ + h v , Φ i L k Φ k L Φ ,where T S denotes the space of vector fields on S . Us-ing this, the anti-harmonic projection Π ⊥H : Ker div S → rot S C ∞ ( S ) is defined byΠ ⊥H v = v − Π H v .We recall eq. (1a) of [1] in vector notation, i. e. ρ D v Dt = η ( ∆ B v + K v ) − grad S p − ζ v + α div S Q (1)where v denotes the velocity, DDt the covariant materialtime derivative, ∆ B the Bochner Laplacian, grad S thesurface gradient, div S the surface diverence, K the Gaus-sian curvature, p the surface pressure, Q a Q-tensor, ρ thematerial density, η the dynamic viscosity, ζ the frictioncoefficient and α the strength of the external forcing.In the following we mainly focus on three different con-figurations of eq. (1). Firstly, no external contributionsare considered, i. e. ζ = 0 and α = 0. Secondly, we inves-tigate damped flow, i. e. ζ = 0 and α = 0, and show thatthere is a strong influence of the harmonic contributions.This is the same setup as in the Supplementary Infor-mation of [1], where it is argued that harmonic contribu-tions are negligible. Thirdly, the external contributionsare considered, i. e. ζ = 0 and α = 0. In this case wedemonstrate on an example that the harmonic parts donot vanish over time. Additionally, we show that the argumentation in theSupplementary Information of [1] is based on a wrongassumption and therefore contradicts with the followingargumentation in [1]. Flow with no external contributions
Let ζ = 0 and α = 0. The solution of eq. (1) convergesto a Killing vector field for t → ∞ (except for some spe-cial initial data with symmetric behavior which implies avanishing solution), see [2]. These vector fields exist onrotationally symmetric surfaces and are characterized bya vanishing deformation tensor. In the present case of atorus, the Killing vector field – denoted by v K – can bedetermined by v K = α K ∂ φ x with a constant α K and de-scribes a rigidly rotating torus. It can be easily verifiedthat Π H v K = 0 and thus, non-trivial Killing vector fieldscontain non-trivial harmonic parts, which cannot vanishasymptotically in general. Cancelling out the harmonicpart for a rigidly rotating torus would magically gener-ate internal friction and the velocity dissipates to zerowithout any physical reason. Damped flow
If we consider external friction, i. e. ζ = 0, and no ad-ditional forces, i. e. α = 0, then we indeed agree that theharmonic part of the solution vanishes exponentially. Butthe anti-harmonic part do the same and we do not see anyreasons to only highlight this behavior for the harmonicpart. In particular, it can be verified that the dampedKilling solution v K ( t ) = v K (0) exp ( − ζ/ρt ) solves eq. (1)with the Killing vector field from above as initial con-dition. Thus, the harmonic part Π H v K and the anti-harmonic part Π ⊥H v K vanish exponentially with the sameorder (in time). Figure 1 shows this behavior, where thekinetic energy of the full damped Killing vector field aswell as its harmonic and anti-harmonic part is shown.The harmonic contribution is even larger than the anti-harmonic part. Moreover, the ratio between these partscan be determined by r H := k Π H v K ( t ) k L k Π ⊥H v K ( t ) k L = ξ (cid:0) ξ − (cid:1) / ξ − ξ ( ξ − / ! / which only depends on the radii ratio ξ of the torus andis monotonically increasing for ξ >
1. Furthermore, ξ > ((7 + 2 · / ) / / ≈ .
619 yields r H > − − Time K i n e t i ce n e r g y fullharmonicanti-harmonic FIG. 1. Kinetic energy of the Killing solution as well as itsharmonic and anti-harmonic parts for a = 2, b = 1, ρ = 1, η = 0 . ζ = 0 . α = 0. The initial Killing vector field v K (0) is rescaled, such that its L -norm is normalized. Damped and forced flow
Consider the full equation (1) with ζ = 0 and α = 0.The force div S Q feeds harmonic parts into the solution,since div S Q has harmonic contributions itself in gen-eral. To highlight this behavior, we construct an ap-propriate Q-tensor. Generally, a Q-tensor can be ob-tained by Q = grad S V , where V ∈ H ( S ). This holds astr grad S V = div S V = 0 and h grad S V , ε i = − rot S V =0 with the Levi-Civita tensor ε . Thus, grad S V is a Q-tensor and we obtain div S Q = K V . For the latter iden-tity we used the Weizenb¨ock identity ∆ B v = ∆ dR v + K v with the Laplace-deRham operator ∆ dR , which can beobtained by ∆ dR v = − rot S rot S v − grad S div S v , see[2]. Consider the special choice Q = − b ( ξ −
1) grad S Φ .Thus, we obtain div S Q = − b ( ξ − K Φ and the re-sulting harmonic force is given by Π H div S Q = Φ . Thismeans we constantly feed the system with the harmonicfield α Φ such that the harmonic part of the solution can-not vanish. In this situation we expect a balance of inter-nal and external friction with the applied force, such thatthe solution converges to a stationary non-trivial vectorfield for t → ∞ . We further expect that the reachedsteady state solution does not have negligible harmoniccontributions. To demonstrate this we discretize eq. (1)in time by using an implicit Euler scheme and in space byusing a symmetric ansatz and finite differences in poloidaldirection. Figure 2 shows the kinetic energy of the fullsolution as well as the harmonic and anti-harmonic com-ponents. Indeed the harmonic contributions are not neg-ligible and are even larger than the anti-harmonic parts inthis simple example. The harmonic-anti-harmonic ratio r H | t =60 ≈ .
374 in the steady state regime also reflectsthis behavior.
Wrong assumption