# A Space-Time Adaptive Algorithm to Illustrate the Lack of Collision of a Rigid Disk Falling in an Incompressible Fluid

• Samuel Dubuis , Marco Picasso and Peter Wittwer

## Abstract

A space-time adaptive algorithm to solve the motion of a rigid disk in an incompressible Newtonian fluid is presented, which allows collision or quasi-collision processes to be computed with high accuracy. In particular, we recover the theoretical result proven in [M. Hillairet, Lack of collision between solid bodies in a 2D incompressible viscous flow, Comm. Partial Differential Equations 32 2007, 7–9, 1345–1371], that the disk will never touch the boundary of the domain in finite time. Anisotropic, continuous piecewise linear finite elements are used for the space discretization, the Euler scheme for the time discretization. The adaptive criteria are based on a posteriori error estimates for simpler problems.

MSC 2010: 65M50; 74F10

## A Derivation of the Penalty Method (2.1)

We present the derivation of the penalty formulation (2.1) used to approximate the motion of the rigid disk inside the cavity. We start from the physical model presented in [12]. We study the motion of a rigid disk of radius R inside a bounded, convex cavity Ω 2 filled with a fluid of constant viscosity and density. The fluid is governed by the incompressible Navier–Stokes equations while the dynamic of the rigid disk is ruled by the Newton law. Given a final time T ( 0 , + ] , we denote by ( 𝑿 ( t ) ) the disk centered at 𝑿 ( t ) , where t [ 0 , T ] . The equations of motions reads [12]: find ( 𝑿 , 𝐮 , p , 𝑽 , ω ) the solutions of

(A.1) { ρ ( 𝐮 t + ( 𝐮 ) 𝐮 ) - μ Δ 𝐮 + p = ρ 𝐠 , ( 𝐱 , t ) Ω ( 𝑿 ( t ) ) ¯ × ( 0 , T ] , div 𝐮 = 0 , ( 𝐱 , t ) Ω ( 𝑿 ( t ) ) ¯ × ( 0 , T ] , 𝐮 = 0 , ( 𝐱 , t ) Ω × ( 0 , T ] , 𝐮 ( 𝐱 , t ) = 𝑽 ( t ) + ω ( 𝐱 - 𝑿 ( t ) ) , ( 𝐱 , t ) ( 𝑿 ( t ) ) × ( 0 , T ] , 𝑿 ˙ ( t ) = 𝑽 ( t ) ,

together with

(A.2) { m 𝐠 - ( 𝑿 ( t ) ) ( 2 μ D ( 𝐮 ) - p I ) 𝒏 𝑑 σ = m 𝑽 ˙ , t ( 0 , T ] , ( 𝑿 ( t ) ) ρ 𝐠 ( 𝐱 - 𝑿 ( t ) ) - ( 𝑿 ( t ) ) ( 2 μ D ( 𝐮 ) - p I ) ( 𝐱 - 𝑿 ( t ) ) 𝒏 𝑑 σ = J ω ˙ , t ( 0 , T ] ,

where 𝐠 is gravitational acceleration, μ and ρ denote the viscosity and the density of the fluid, ρ the density of the disk, m and J its mass and inertia. In the above system, ( 𝐮 , p ) are the velocity and pressure fields of the fluid and ( 𝑽 , ω ) the translational and rotational speeds of ( 𝑿 ( t ) ) . Here above we note

( 𝐱 - 𝑿 ( t ) ) = ( - ( x 2 - X 2 ( t ) ) , x 1 - X 1 ( t ) ) 2 .

Note that since 𝐠 is applied at the center of mass, its angular momentum satisfies in particular

( 𝑿 ( t ) ) ρ 𝐠 ( 𝐱 - 𝑿 ( t ) ) = 0

which is also true for any constant vector 𝑽 inside the disk, i.e.,

( 𝑿 ( t ) ) 𝑽 ( 𝐱 - 𝑿 ( t ) ) = 0 .

To derive equations (2.1), the first step consists to establish a variational formulation that can easily be approximated with Galerkin methods, that is to say involving test functions that are defined on the whole cavity Ω. We follow the formulation introduced in [15]. Let us consider the problem at a given time t, and to lighten the presentation, we drop the dependence of 𝑿 upon the time. To extend the velocity field inside the disk, one may consider the space of rigid motion inside ( 𝑿 ) given by

H ( 𝑿 ) = { 𝐯 ( H 0 1 ( Ω ) ) 2 : there exists  ( 𝑽 , ω ) 2 ×  such that  𝐯 = 𝑽 + ω ( 𝐱 - 𝑿 )  a.e. in  ( 𝑿 ) }

that has the equivalent definition (see for instance [27])

H ( 𝑿 ) = { 𝐯 ( H 0 1 ( Ω ) ) 2 : D ( 𝐯 ) = 0  a.e. in  ( 𝑿 ) } .

We now multiply the momentum equation of (A.1) by a test function 𝐯 H ( 𝑿 ) and using the fact that for free divergence field

Δ 𝐮 = 2 div ( D ( 𝐮 ) ) ,

one has after integration by part

Ω ( 𝑿 ) ¯ ρ ( 𝐮 t + 𝐮 𝐮 ) 𝐯 + 2 μ Ω ( 𝑿 ) ¯ D ( 𝐮 ) : D ( 𝐯 ) - Ω ( 𝑿 ) ¯ p div 𝐯 - ( 𝑿 ) ( 2 μ D ( 𝐮 ) - p I ) 𝒏 𝐯 = Ω ( 𝑿 ) ¯ ρ 𝐠 𝐯 .

Using the Newton’s laws (A.2) and the fact that 𝐯 = 𝑽 v + ω v ( 𝐱 - 𝑿 ) in ( 𝑿 ) , one can write that

- ( 𝑿 ) ( 2 μ D ( 𝐮 ) - p I ) 𝒏 𝐯 = m 𝑽 ˙ 𝑽 v - m 𝐠 𝑽 v + J ω ˙ ω v - ( 𝑿 ) ρ 𝐠 ω v ( 𝐱 - 𝑿 ) .

Moreover, one has that

m 𝑽 ˙ 𝑽 v - m 𝐠 𝑽 v = B ( 𝑿 ) ρ 𝑽 ˙ 𝑽 v - B ( 𝑿 ) ρ 𝐠 𝑽 v ,

and

J ω ˙ ω v = B ( 𝑿 ) ρ ω ˙ ω v | 𝐱 - 𝑿 | 2 = B ( 𝑿 ) ρ ω ˙ ( 𝐱 - 𝑿 ) ω v ( 𝐱 - 𝑿 ) .

Thus since 𝐮 , 𝐯 H ( 𝑿 ) , we can prove that

- ( 𝑿 ) ( 2 μ D ( 𝐮 ) - p I ) 𝒏 𝐯 = m 𝑽 ˙ 𝑽 v + J ω ˙ ω v - m 𝐠 𝑽 v - ( 𝑿 ) ρ 𝐠 ω v ( 𝐱 - 𝑿 )
= B ( 𝑿 ) ρ ( 𝑽 ˙ 𝑽 v + ω ˙ ( 𝐱 - 𝑿 ) ω v ( 𝐱 - 𝑿 ) )
- B ( 𝑿 ) ρ 𝐠 𝑽 v - ( 𝑿 ) ρ 𝐠 ω v ( 𝐱 - 𝑿 )
= B ( 𝑿 ) ρ B ( 𝐮 t + ( 𝐮 ) 𝐮 ) 𝐯 - B ( 𝑿 ) ρ 𝐠 𝐯 ,

Ω ( 𝑿 ) ¯ ρ ( 𝐮 t + ( 𝐮 ) 𝐮 ) 𝐯 + B ( 𝑿 ) ρ B ( 𝐮 t + ( 𝐮 ) 𝐮 ) 𝐯 + 2 μ Ω ( 𝑿 ) ¯ D ( 𝐮 ) : D ( 𝐯 ) - Ω ( 𝑿 ) ¯ p div 𝐯
= Ω ( 𝑿 ) ¯ ρ 𝐠 𝐯 + B ( 𝑿 ) ρ 𝐠 𝐯 .

Finally, since D ( 𝐮 ) | ( 𝑿 ) = D ( 𝐯 ) | ( 𝑿 ) = 0 (in particular div 𝐮 = div 𝐯 = 0 inside the disk), it is possible to write the previous equation on the whole domain Ω, which leads to the variational formulation

Ω ρ 𝑿 ( 𝐮 t + 𝐮 𝐮 ) 𝐯 + 2 μ Ω D ( 𝐮 ) : D ( 𝐯 ) - Ω p div 𝐯 = Ω ρ 𝑿 𝐠 𝐯 for all  𝐯 H ( 𝑿 ) ,
Ω q div 𝐮 = 0 for all  q L 2 ( Ω ) ,

where we note

ρ 𝑿 = ρ χ Ω ( 𝑿 ) ¯ + ρ χ ( 𝑿 ) .

The constrains D ( 𝐮 ) B ( 𝑿 ) = D ( 𝐯 ) B ( 𝑿 ) = 0 can be approximated by a penalization of the viscous term. Thus the variational formulation can be written with test functions taken in H 0 1 ( Ω ) by considering (we keep the same notations for the unknowns to lighten the presentation)

Ω ρ 𝑿 ( 𝐮 t + ( 𝐮 ) 𝐮 ) 𝐯 + Ω 2 μ 𝑿 D ( 𝐮 ) : D ( 𝐯 ) - Ω p div 𝐯 = Ω ρ 𝑿 𝐠 𝐯 for all  𝐯 ( H 0 1 ( Ω ) ) 2 ,

where

μ 𝑿 = μ χ Ω ( 𝑿 ) ¯ + 1 ε χ ( 𝑿 ) , ε 1 .

Written in the strong form, the approximated variational formulation gives

ρ 𝑿 ( 𝐮 t + ( 𝐮 ) 𝐮 ) - div ( 2 μ 𝑿 D ( 𝐮 ) ) + p = ρ 𝑿 𝐠

with div 𝐮 = 0 . To obtain (2.1), we finally have to consider equations to compute the translational and rotational velocities inside the disk. In our approach, we assume that we are in a symmetric situation and that the disk does not spin. Thus it remains to have an equation to update only the position of the disk that is obtained by approximating the velocity inside the disk by taking its mean value

(A.3) 𝑿 ˙ = 1 π R 2 ( 𝑿 ) 𝐮 .

Observe that the last equation reduces to 𝑿 ˙ = 𝐮 in ( t ) if 𝐮 is constant in ( t ) , that is expected when ε 0 . Thus it is justified to use (A.3) to approximate the motion of the disk.

## B Justification of (2.8)

We now prove an a posteriori error estimate for the time discretization of a simplified problem. We simplify the first equation of (2.1) and consider the following problem:

(B.1) { - div ( 2 μ 𝑿 D ( 𝐮 𝑿 ) ) + p 𝑿 = ρ 𝑿 𝐠 in  Ω × ( 0 , T ) , div 𝐮 𝑿 = 0 in  Ω × ( 0 , T ) , 𝑿 ˙ ( t ) = 1 π R 2 ( 𝑿 ( t ) ) 𝐮 𝑿 ( x , t ) 𝑑 x , t ( 0 , T ) ,

where ρ 𝑿 and μ 𝑿 are still defined by (2.2). Here 𝐮 𝑿 is zero on the boundary Ω and 𝑿 ( 0 ) = 𝑿 0 .

We consider the forward Euler method. Given a integer N > 0 and a partition 0 = t 0 < t 1 < < t N = T of [ 0 , T ] , let τ n + 1 = t n + 1 - t n , τ = max n = 0 , 1 , 2 , , N - 1 τ n + 1 . For every n = 0 , 1 , 2 , , N - 1 , given 𝑿 n , we compute first ( 𝐮 𝑿 n , p 𝑿 n ) solution of

(B.2) { - div ( 2 μ 𝑿 n D ( 𝐮 𝑿 n ) ) + p 𝑿 n = ρ 𝑿 n 𝐠 in  Ω , div 𝐮 𝑿 n = 0 in  Ω , 𝐮 𝑿 n = 0 on  Ω ,

then 𝑿 n + 1 solution of

𝑿 n + 1 - 𝑿 n τ n + 1 = 1 π R 2 ( 𝑿 n ) 𝐮 𝑿 n 𝑑 x .

We assume that there exists a unique

( 𝑿 , 𝐮 𝑿 , p 𝑿 ) C 1 [ 0 , T ] × C ( [ 0 , T ] ; ( H 0 1 ( Ω ) ) 2 ) × L 2 ( ( 0 , T ) ; L 0 2 ( Ω ) )

solution of (B.1), that d ( ( 𝑿 ( t ) ; Ω ) > 0 for all t [ 0 , T ] , and that 𝑿 n , 𝐮 𝑿 n , p 𝑿 n converge to 𝑿 ( t n ) , 𝐮 𝑿 ( t n ) , p 𝑿 ( t n ) when τ goes to zero, and prove an a posteriori error estimate.

Observe that problem (B.1) reads: starting from 𝑿 ( 0 ) = 𝑿 0 , find 𝑿 ( t ) , for t > 0 , such that

𝑿 ˙ ( t ) = 𝑭 ( 𝑿 ( t ) ) ,

where, for any 𝒀 2 such that ( 𝒀 ) is compactly supported in Ω, 𝑭 ( 𝒀 ) is given by

𝑭 ( 𝒀 ) = 1 π R 2 ( 𝒀 ) 𝐮 𝒀 𝑑 x ,

and ( 𝐮 𝒀 , p 𝒀 ) ( H 0 1 ( Ω ) ) 2 × L 0 2 ( Ω ) is the weak solution of

{ - div ( 2 μ 𝒀 D ( 𝐮 𝒀 ) ) + p 𝒀 = ρ 𝒀 𝐠 , div 𝐮 𝒀 = 0 ,

where μ 𝒀 , ρ 𝒀 are defined as in (2.2). In order to prove an a posteriori error estimate, we need to prove that 𝑭 is Lipschitz.

## Lemma 1.

Let Y = ( Y 1 , Y 2 ) , Z = ( Z 1 , Z 2 ) Ω be such that the disks B ( Y ) , B ( Z ) are compactly supported in Ω. Let ( u 𝐘 , p 𝐘 ) , ( u 𝐙 , p 𝐙 ) ( H 0 1 ( Ω ) ) 2 × L 0 2 ( Ω ) be the weak solutions of

{ - div ( 2 μ 𝒀 D ( 𝐮 𝒀 ) ) + p 𝒀 = ρ 𝒀 𝐠 , div 𝐮 𝒀 = 0 ,
{ - div ( 2 μ 𝒁 D ( 𝐮 𝒁 ) ) + p 𝒁 = ρ 𝒁 𝐠 , div 𝐮 𝒁 = 0 ,

where μ 𝐘 , μ 𝐙 , ρ 𝐘 , ρ 𝐙 are defined as in (2.2). If | 𝐘 - 𝐙 | d ( B ( 𝐘 ) ; Ω ) is small enough, then there exists a constant C > 0 depending only on Ω such that

(B.3) | 1 π R 2 ( 𝒀 ) 𝐮 𝒀 𝑑 𝐱 - 1 π R 2 ( 𝒁 ) 𝐮 𝒁 𝑑 𝐱 | C L | 𝒀 - 𝒁 | d ( ( 𝒀 ) ; Ω ) ,

where

L = ρ | g | ε R μ 2 .

The key point to prove this lemma consists in mapping ( 𝒀 ) into ( 𝒁 ) as in [11]. For details we refer to [6, Appendix B]. The fact that estimate (B.3) blows up when d ( ( 𝑿 ) ; Ω ) approaches zero seems reasonable; it is due to the mapping from ( 𝒀 ) into ( 𝒁 ) which becomes singular. However, we do not know if the power one (on d) is sharp.

Our main result is then the following.

## Theorem 2.

Let ( 𝐗 ( t ) , u 𝐗 ( t ) ) , t [ 0 , T ] , be the solution of (B.1) and ( 𝐗 n , u 𝐗 n ) , n = 0 , 1 , N , the solution of (B.2). Let τ be the maximal time step, assume that for all n = 1 , 2 , 3 , , N , 𝐗 n converges to 𝐗 ( t n ) as τ goes to 0. Then, if τ is small enough, there exists a constant C > 0 depending only on Ω such that

| 𝑿 ( t N ) - 𝑿 N | C L exp ( 0 T C L d ( ( 𝑿 ( s ) ) ; Ω ) 𝑑 s ) n = 0 N - 1 t n t n + 1 1 d ( ( 𝑿 ( t ) ; Ω ) ) ( t - t n ) | 𝑿 n + 1 - 𝑿 n τ n + 1 | 𝑑 t .

## Proof.

Let 𝑿 τ be the piecewise linear reconstruction defined by

𝑿 τ ( t ) = 𝑿 n + ( t - t n ) 𝑿 n + 1 - 𝑿 n τ n + 1 , t [ t n , t n + 1 ] , n = 0 , 1 , 2 , , N - 1 .

Let e ( t ) = 𝑿 ( t ) - 𝑿 τ ( t ) , n 0 and t ( t n , t n + 1 ) . Using (B.1) and (B.2), we have

𝒆 ˙ ( t ) = 1 π R 2 ( 𝑿 ( t ) ) 𝐮 𝑿 ( t ) 𝑑 x - 1 π R 2 ( 𝑿 n ) 𝐮 𝑿 n 𝑑 x .

Taking the scalar product of the last equation with 𝒆 ( t ) , using the Cauchy–Schwarz inequality and Lemma 1 yields

1 2 d d t | 𝒆 ( t ) | 2 C L d ( ( 𝑿 ( t ) ) ; Ω ) | 𝑿 ( t ) - 𝑿 n | | 𝒆 ( t ) | .

Without loss of generality, we assume that 𝒆 ( t ) 0 , t ( t n , t n + 1 ) . Therefore, dividing by | 𝒆 ( t ) | , we get

d d t | 𝒆 ( t ) | = d d t | 𝒆 ( t ) | 2 2 | 𝒆 ( t ) | C L d ( ( 𝑿 ( t ) ) ; Ω ) | 𝑿 ( t ) - 𝑿 n |

with | 𝒆 ( t ) | = | 𝒆 ( t ) | 2 . Applying the triangle inequality, we obtain

d d t | 𝒆 ( t ) | C L d ( ( 𝑿 ( t ) ) ; Ω ) | 𝑿 τ - 𝑿 n | + C L d ( ( 𝑿 ( t ) ) ; Ω ) | 𝒆 ( t ) | ,

Multiplying the last inequality by

exp ( - 0 t C L d ( ( 𝑿 ( s ) ) ; Ω ) 𝑑 s ) ,

and integrating from t n to t n + 1

| 𝒆 ( t n + 1 ) | exp ( - 0 t n + 1 C L d ( ( 𝑿 ( t ) ) ; Ω ) 𝑑 t ) - | 𝒆 ( t n ) | exp ( - 0 t n C L d ( ( 𝑿 ( t ) ) ; Ω ) 𝑑 t )
C L t n t n + 1 ( | 𝑿 τ ( t ) - 𝑿 n | d ( ( 𝑿 ( t ) ; Ω ) ) ) exp ( - 0 t C L d ( ( 𝑿 ( s ) ) ; Ω ) 𝑑 s ) 𝑑 t .

Summing up over n leads to

| 𝒆 ( T ) | C L n = 0 N - 1 t n t n + 1 | 𝑿 τ ( t ) - 𝑿 n | d ( ( 𝑿 ( t ) ; Ω ) ) exp ( t T C L d ( ( 𝑿 ( s ) ) ; Ω ) 𝑑 s ) 𝑑 t ,

where we use that e ( 0 ) = 0 to get rid of the error at t = 0 . The desired estimate is then obtained by using the definition of 𝑿 τ . ∎

## Acknowledgements

Frédéric Alauzet is acknowledged for providing the Wolf-Interpol program corresponding to conservative interpolation [3]. Samuel Dubuis acknowledges Diane Guignard and Swarnendu Sil for advices and discussions concerning the derivation of the time error indicator. It should be noted that some numerical experiments with space adaptation only were already reported in [26].

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