Multi-Scale Behaviour in the Geo-Science I: The Onset of Convection

and Interfacial Instabilitiesby

Hans Mühlhaus

The Australian Computational Earth Systems Simulator

(ACcESS)

OverviewWhat is GeodynamicsMantle Convection, Spreading, Folding, Landscape Evolution, Earthquakes, Volcanoes

The onset of Natural Convection:Linear instability analysis

Numerical SimulationsNusselt Number

Remarks on Weakly Non-Linear AnalysisGalerkin methods

hot spot

subducting plate lithosphereasthenosphere

shield volcano

strato volcanotrench

trench

convergentplate boundary

convergentplate boundary

oceanicspreading ridge

divergentplate boundary

transformplate boundary

island arc

continentalcrust

oceaniccrust

Earth dynamics

TERRA MESH

BENEFITS

Earth dynamics

hot spot

subducting plate lithosphereasthenosphere

shield volcano

strato volcanotrench

trench

convergentplate boundary

convergentplate boundary

oceanicspreading ridge

divergentplate boundary

transformplate boundary

island arc

continentalcrust

oceaniccrust

22ndnd talk: Volcano modelling talk: Volcano modelling

Montserrat, West Indies

3rd talk: Particle Processes

–Spherical particles

–Selection of contact physics:

–Non-rotational and rotational dynamics

–Friction interactions

–Linear elastic interactions

–Bonded interactions

4th talk: modeling of geological folds

Director evolution

n : the director of the anisotropyW, Wn : spin and director spinD, D’: stretching and its deviatoric part

)( kikjkjkiijn

ij DDWW

ij nin j

jn

iji nWn

Governing Equations

))(()( Tpt

vvgvvv

g

0g

vdt

d

Navier Stokes Equations:

Heat Equation:

TkTt

Tcp

2)(

v

v(t,x) is the velocity vector, T is the temperature, kg/m2) is the density, Pas)is the viscosity, cp (10-3WK-1s/kg) is the specific heat at constant pressure and k (4Wm-1K-1) is the thermal conductivity

Governing Equations, cont.

))(1( 00 TTp

Temperature dependence of density:

Simplified convection model:

p ( 3 10-5K-1 ) is the thermal expansion

coefficient, T0 (288K) is the surface temperature

x1

x2

T=T1 , v2=0

T=T0 , v2=0

T,1 , v1=0T,1 , v1=0 H

L

Consider a square planet……

Governing Equations, cont.

,~xx H ,~

][

02

tk

cHt

t

p

,~

][vv

t

H

Nondimensionalisation:

)~

1)(( 2010 T

H

xTTTT

))((1

)1()(Pr

12

TpTxRat

vvgg

vvv

TTt

T 2)(

v

Raleigh Number: )1010()( 86

301

20

k

HTTgcRa p

Prandtl Number: )1025.0()/(

/Pr 18

0

0 pck

Relevant limit in

Geophysics:Pr

Insertion and dropping tildes…..

Governing Equations, cont.

2,2

1

x

v 1,2 v

timxexnT 121, sin),(

01,4 RaT TTTT t

21,1,2,2,1,, )(

Stream function

and

In this way we satisfy the incompressibility constraint div v=0 identically.Insertion into the velocity equations and the heat equation, assuming infinitePrandtl number gives:

and

Dropping nonlinear terms and insert the “Ansatz”

, gives:

RaTmnm 21,

2222 )( and 1,222 )( Tnm Thus

2222

32222

)(

)(

nm

nmRam

Marginal instability:

HLRa 22,4

27min1

4

min

For m=1 we obtain:

2/,

2/

)(

0

12

2

3222

L

Hn

L

Hm

m

nmRa

HL1

2

1

322

1

)4

(

))4

((

LH

LH

Ra

Finite Element Approximations…Ra=104, mesh: 128X128

The Nusselt Number

onlyconductionbyflux

heatfluxtotal

dVkT

dVkTTv

Nu

V

V

2,

2,2 )(

Definition:

It can be shown that for zero normal velocity b.c.’s and fixed top and bottom Temperature

dVRaV

NuV

PowerMechanicalessDimensionl

TT

))((:))((2

111 vvvv

The Nusselt NumberHint for derivation of Nu-Power relationship

))(()1( 22TpTxRa vve dV

V v(.)form

And apply Gauss’s Theorem. For the given b.c.’s it follows that

dVdVvTxRaV

TT

V ))((:))((

2

1)1( 22 vvvv

Finite Element Approximations…Ra=104, mesh: 128X128

Nusselt Number

Finite Element Approximations…Ra=105, mesh: 128X128

Finite Element Approximations…Ra=105, mesh: 128X128

Nusselt Number

Finite Element Approximations…Ra=106, mesh: 128X128

Finite Element Approximations…Ra=106, mesh: 128X128

Nusselt Number

Finite Element Approximations…Ra=107, mesh: 128X128

Finite Element Approximations…Ra=107, mesh: 128X128

Nusselt Number

Galerkin Method

212 2sin),(cossin),( xtmCmxxtmBT 12 cossin),( mxxtmA

0))((1

0

21

2

0

21,1,2,2,1,, dxdxTTTTT t

We consider the ansatz:

0)( 211,

1

0

2

0

dxdxTRa

Insert into:

and

We obtain:

abm

Ramaa

322

2

)(

and

3322

22

)(

2a

m

Rambb

Rayleigh-Taylor Instabilities

1x

1x

RT fingers evident in the Crab Nebula

Consider two fluids of different densities, the heaviest above the lightest. An horizontal interface separates the two fluids. This situation is unstable because of gravity. Effectively, if the interfaces modified then a pressure want of balance grows. Equilibrium can be found again tanks to surface tension that's why there is a competition between surface tension and gravity. Surface tension is stabilizing instead gravity is destabilizing for this configuration.

2x

Benchmark Problem

Method mesh γ0 (vrms)max reached at t=

van Keken coarse 0.01130 0.003045 212.14

van Keken fine 0.01164 0.003036 209.12

Particle-in-cell 1024 el 0.01102 0.003098 222

Particle-in-cell 4096 el 0.01244 0.003090 215

Level set 5175 el 0.01135 0.003116 215.06

Rayleigh-Taylor instability

benchmark

Linear Instability Analysis

1h

1h

00

2/1,

2/1,

2/1

kkt vSSSdt

d

Equilibrium to be satisfied in ground state at time t=t0 and at t0+dt:

Ground state:

0,

0,,0

222

221

xgxconstp

andxgxconstpvi

2h2x n

Continued Equilibrium:

0))((),(

v

gvvpS

Tp

Stokes equation:

Linear Instability Analysis

1h

0)( ,2/1

,2/1

,2/12/1

, jijjii uuP

02/1,

2/12/1, jjii uP

obtainwe

andwith ,, ttii pPvu

2h2x n

Or, considering the incompressibility constraint: 02/1

, jju1h

Linear Instability Analysis: Boundary and interface conditions

j,)(

1h

1h

We assume that the velocities are zero on top and bottom; on the sides we assume that v1=0 and i.e. symmetry boundary.On the interface the velocities as well as the natural boundary terms have to be continuous.Natural b.c.’s: replace

2h2x n

as well as its time derivative have to be continuous on the interface

By . The vector

in

0))(( ,2/1

,2/1

,2/12/1 jijjiij uuP

jn

jijjiiji nuuPt ))(( 2/1,

2/1,

2/12/1

Linear Instability Analysis: Boundary and interface conditions

0))(( 0221 dAnn

dt

djijjij

1h

1h

We assume that the velocities are zero on top and bottom; on the sides we assume that v1=0 and i.e. symmetry boundary.On the interface the velocities as well as the natural boundary terms have to be continuous.Natural b.c.’s

2h2x n

Or:

Result:

0)()(:

0)()(:

221212

2,221

2,21

2

21,2

22,1

211,2

12,1

11

gvPPuux

uuuux

Exercises

1. The normal component of the surface velocity of a 2D half plane reads . The half plane is occupied by a Stokes fluid with the viscosity The normal stress is obtained as . Show that .

2. Consider a Rayleigh-Taylor instability problem involving 2 infinite half-planes; i.e. . The normal velocity of the

Interface reads . Show that .

Hint: Use the relationship for a gravity free Half-plane and note that .

02 x12 cos kxVv

122 cos kxQ kVQ 2

12 cos kxVev tk

g

)(2

)(

21

21

1, 21 khkh

UVUUkQ tgt

,0

, ,2gVQQ t

gt

,0

,

Exercises (folding)

w

Exercise (folding) cont.

Exercise (folding) cont.

Excercise 4:

01,4 RaT TTTT t

21,1,2,2,1,, )(

Solve convection equations

and

using the perturbation expansion

up to terms of order .

Hint:

....22

1 ....22

1 TTT t2 22

0 RaRaRa

3

1. Insert into pde’s, collect coefficients of and 2. Individual coefficients must be equal to zero. Get

etc and T3 are a little bit harder to get since the pde’s contain inhomogeneous terms which areproportional to (

Perturbation solution for the weakly nonlinear problem

4. We require a solubility condition (Fredholm’s alternative) In the present case this just means that the coefficient of the resonant inhomogeneous term must vanish. The coefficient has the form of an ode which can be written as:

3222

22

, 2

1

)(aa

m

Rama

a

RaRacrit