Num Geophys

Numerical Methods in Geodynamics

Daniel Kohn

Lets consider the problem (not very physical) of a flow into a barrier of constanttemperature. What is the temperature field for the following situation:

T=0

x=0 x=1

T=1

constant speed U

Abbildung 1: Problem 1

Assume steady state:

UT

x=

2T

x2(1)

With the boundary conditions

at x = 0 T = 0

at x = L T = 1

we can find the analytical solution

T =exp(Pe x

L) 1

exp(Pe) 1 (2)

with the Peclet number defined as Pe := UL. Using a central difference dis-

cretization for the advective term we get:

UTk+1 Tk1

2h=

Tk+1 2Tk + Tk1h2

. (3)

Let = Uh2

the mesh Peclet number.

Exercise 1 Set up this problem in MATLAB.

2Writing eq. (3) as matrix equation we can implement the problem in MATLABby the following script:

% Numerical Methods in Geodynamics

% Exercise 1

%

% Daniel Koehn

% Kiel, the 23 rd of march 2004

clear all

close all

% define constants

alpha = 0.5; % mesh peclet number

N=6; % number of grid points

T0=0; % temperature at grid point number 0

TN=1; % temperature at grid point number N

L=10; % length of the rod

h=L./(N-1); % grid point spacing

kappa =1; % thermal diffusivity

U=(alpha.*2.*kappa)./h; % velocity

Pe=(U.*L)./kappa; % Peclet number

% central difference scheme

% -------------------------

% build matrix-equation

% cofficient matrix

e0 = -2.*ones(N-2,1); % elements on main diagonal

e1 = -(alpha-1).*ones(N-3,1); % 1st upper diagonal

e2 = (alpha+1).*ones(N-3,1); % 1st lower diagonal

S = diag(e0) + diag(e1,1) + diag(e2,-1)

% RHS vector

b = zeros(N-2,1);

% apply boundary conditions

b(1) = -(alpha+1).*T0;

b(N-2) = (alpha-1).*TN;

% plot RHS vector b

b

% matrix inversion

% ----------------

Ta=inv(S)*b;

3% add boundary points

Tc=[T0 Ta TN];

% plot temperature field

% ----------------------

% position vector

x=0:h:L;

% calculate exact solution

Tex=(exp(Pe.*(x./L))-1)./(exp(Pe)-1)

plot(x,Tc,--,x,Tex);

% axis and title

af=xlabel(x/m);

bf=ylabel(T/C);

df=title(Advection/Diffusion in 1D (central differences));

ef=legend(central differences,exact,2)

% fitting the fonts

set(gca,FontSize,15,FontName,Times);

set(af,FontSize,15,FontName,Times);

set(bf,FontSize,15,FontName,Times);

set(df,FontSize,15,FontName,Times);

set(ef,FontSize,14,FontName,Times);

% output as eps-file

print -deps exer_1_0p5.eps

For = 0.5, 1.0 and 2.5 you get the results as shown in figure 2. Note the strongoscillations of the solution, when > 1.

40 2 4 6 8 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x/m

T/C

Advection/Diffusion in 1D (central differences)central differencesexact

0 2 4 6 8 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x/m

T/C


0 2 4 6 8 100.5

0

0.5

1

x/m

T/C


Abbildung 2: Results using the central difference discretization for = 0.5 (top), = 1.0 (middle), = 2.5 (bottom). The numerical solutions are shown as dashedlines, the exact solution (2) as solid lines.

5Now what if we use upwind differences on the advection term ?

UT

x= U

Tk Tk1h

2T

x2=

Tk+1 2Tk + Tk1h2

(4)

In this case, the solution is formally less accurate but it never oscillates andconverges towards the correct solution.

Exercise 2 Program this in MATLAB


% Exercise 2

%

% Daniel Koehn


clear all

close all

% define constants

alpha = 1.0; % mesh peclet number


T0=0; % temperature at grid point number 0

TN=1; % temperature at grid point number N

L=10; % length of the rod

h=L./(N-1); % grid point spacing

kappa =1; % thermal diffusivity

U=(alpha.*2.*kappa)./h; % velocity

Pe=(U.*L)./kappa; % Peclet number

% upwind difference scheme

% -------------------------


% cofficient matrix

e0 = 2.*(alpha+1).*ones(N-2,1); % elements on main diagonal

e2 = -((2.*alpha)+1).*ones(N-3,1); % 1st lower diagonal

e1 = -ones(N-3,1); % 1st upper diagonal

S = diag(e0) + diag(e1,1) + diag(e2,-1)

% RHS vector

b = zeros(N-2,1);


b(1) = ((2.*alpha)+1).*T0;

b(N-2) = TN;

6% plot RHS vector b

b

% matrix inversion

% ----------------

Ta=inv(S)*b;

% add boundary points

Tc=[T0 Ta TN];

% plot temperature field

% ----------------------

% position vector

x=0:h:L;


Tex=(exp(Pe.*(x./L))-1)./(exp(Pe)-1)

plot(x,Tc,--,x,Tex);

% axis and title

af=xlabel(x/m);

bf=ylabel(T/C);

df=title(Advection/Diffusion in 1D (upwind differences));

ef=legend(upwind differences,exact,2)

% fitting the fonts







print -deps exer_2_1p0.eps

70 2 4 6 8 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x/m

T/C

Advection/Diffusion in 1D (upwind differences)upwind differencesexact

0 2 4 6 8 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x/m

T/C


0 2 4 6 8 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x/m

T/C


Abbildung 3: Results using the upwind difference discretization for = 0.5 (top), = 1.0 (middle), = 2.5 (bottom). The numerical solutions are shown as dashedlines, the exact solution (2) as solid lines.

8Exercise 3 Upwind differencing can also be viewed as adding artifical diffusionavis =

Uh2, so that the

effective diffusion is eff = +

uh2, i.e.:

UTk Tk1

h= U

Tk+1 Tk12h

Uh2

Tk+1 2Tk + Tk1h2

. (5)

With this artifical diffusion, show that the MATLAB solution for central diffe-rences gives the answer for upwind differences.

First we replace the thermal diffusivity in the mesh Peclet number by effin the MATLAB script for the central difference scheme. Then we can changethe artifical diffusion avis between 1 and 5. The results are plotted in figure 4.For avis = 1 we get a solution comparable with the original central differencescheme. With increasing avis the oscillations are more and more damped. Foravis = 4 we get nearly the analytical solution.

0 2 4 6 8 100.5

0

0.5

1

x/m

T/C

Advection/Diffusion in 1D (central differences+artifical viscosity)central differences+artifical viscosityexact

kav=1

kav=2

kav=3

kav=4

kav=5

Abbildung 4: The solutions of the central difference scheme for values of avis between1 and 5.

9Exercise 4 Compare the advection of a Gaussian temperature distribution using

central differences

upwind differences

The time evolution of the Gaussian temperature distribution can be computedwith the following MATLAB script:


% Exercise 4

%

% Daniel Koehn

% Kiel, the 24 th of march 2004

clear all

close all

% define constants

U=0.5; % velocity


L=150; % some length

dx=L./(N-1); % grid spacing

dt=dx./(200.*U) % Courant criteria

a=0.15; % half width of Gaussian

x0=15; % offset of Gaussian

tsamp= 20000.*dt; % take snapshot every tsamp

t1=0; % take first snapshot at time t1

t0=0;

t=t0; % start timer at t0

% set up initial condition (Gaussian)

x=0:dx:L;

Tm=exp(-(a.*(x-x0)).^2);

Ti=Tm;

for i=1:300000

% calculate temperature T at time step n

% from temperature Tm at time step n-1

% allocate memory for T

T=Tm;

% iteration

for k=2:N-1

T(k) = Tm(k)+dt.*((-U./(2.*dx)).*(Tm(k+1)-Tm(k-1)));

end

10

% schuffle vector elements

Tm=T;

if(t>t1)

plot(x,T);

hold on;

% axis and title

af=xlabel(x/m);

bf=ylabel(T/C);

df=title(Advection in 1D (forward differences));

% fitting the fonts





t1=t1+tsamp;

end

t=t+dt;

% boundary

Tm(1)=Tm(2);

Tm(N)=Tm(N-1);

end


print -deps exer_3_gauss_cent_diff.eps

The results are shown in figure 5. While the upwind discretization scheme pro-duces strong diffusion of temperature, the central difference scheme shows noone.

11

0 50 100 1500

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x/m

T/C

Advection in 1D (upwind differences)

0 50 100 1500.2

0

0.2

0.4

0.6

0.8

1

1.2

x/m

T/C

Advection in 1D (forward differences)

Abbildung 5: Time evolution of a Gaussian temperature distribution using an upwinddifference discretization scheme (top) and a central difference discretization scheme(bottom).

12

Exercise 5 Explore the 1D-problem from the last exercise in 2D using the ve-locity field

V = (u, v) = (y, x) (6)

2 1.5 1 0.5 0 0.5 1 1.5

2

1.5

1

0.5

0

0.5

1

1.5

x/m

y/m

Velocity field

The 2D-version of the MATLAB script from the last exercise using an upwinddiscretization scheme is listed below:


% 2D-Advection-Eq.

%

% Daniel Koehn

% Kiel, the 24 th of march 2004

clear all

close all

% define constants

N=100; % number of grid points in x-direction

M=100; % number of grid points in y-direction

L=4.0; % some length

13

dx=L./(N-1); % grid spacing

offx =2.5 % offset in x- and y-direction

% set up mesh

for j=1:M

for i=1:N

V(i,j)= i.*dx-offx; % velocity in y direction

U(i,j)=-(j.*dx-offx); % velocity in x direction

X(i,j)=i.*dx-offx; % x-coordinate

Y(i,j)=j.*dx-offx; %y-coordinate

end

end

% plot velocity field

%quiver(X,Y,U,V);

axis tight

grid off

hold off

% axis and title

af=xlabel(x/m);

bf=ylabel(y/m);

df=title(Velocity field);

% fitting the fonts






%print -deps exer_6_gauss_vel.eps

vmaxx = 2.5;

dt=dx./(4.*vmaxx) % Courant criteria

a=15; % half width of Gaussian

x0=-1.0; % offset of Gaussian in x-direction

y0=0.0; % offset of Gaussian in y-direction


for j=1:M

for i=1:N

xi=i.*dx-offx; % x-coordinate

yi=j.*dx-offx; % y-coordinate

Tm(i,j)=exp(-(a.*((xi-x0).^2+(yi-y0).^2)));

end

end

% introduce boundary conditions

Tm(1,:)=0;

Tm(:,1)=0;

Tm(N,:)=0;

Tm(:,N)=0;

14

for k=1:1




T=Tm;

% iteration (upwind differences)

for j=2:M-1

for i=2:N-1

if(U(i,j)>0)

dT1=(Tm(i,j)-Tm(i-1,j));

else

dT1=(Tm(i+1,j)-Tm(i,j));

end

if(V(i,j)>0)

dT2=(Tm(i,j)-Tm(i,j-1));

else

dT2=(Tm(i,j+1)-Tm(i,j));

end

T(i,j) = Tm(i,j)+dt.*(((-U(i,j)./dx).*dT1)+((-V(i,j)./dx).*dT2));

end

end


length(Tm)

length(T)

Tm=T;

end

imagesc(Tm);

axis equal;

axis tight;

hold on;

% axis and title

af=xlabel(x/GP);

bf=ylabel(y/GP);

df=title(Advection in 2D (upwind differences));

% fitting the fonts






print -deps exer_6_gauss_1.eps

The pictures in figure 6 are illustrating the evolution of the Gaussian in time. Note thestrong numerical diffusivity by using the upwind difference scheme.

15

x/GP

y/G

P


20 40 60 80 100

10

20

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40

50

60

70

80

90

100

x/GP

y/G

P


20 40 60 80 100

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x/GP

y/G

P


20 40 60 80 100

10

20

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50

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90

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x/GP

y/G

PAdvection in 2D (upwind differences)

20 40 60 80 100

10

20

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50

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90

100

x/GP

y/G

P


20 40 60 80 100

10

20

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x/GP

y/G

P


20 40 60 80 100

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x/GP

y/G

P


20 40 60 80 100

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90

100

x/GP

y/G

P


20 40 60 80 100

10

20

30

40

50

60

70

80

90

100

Abbildung 6: A Gaussian temperature distribution in the velocity field (6) after 1,100, 500, 1000, 2000, 5000, 10000 and 60000 timesteps (from top to bottom).

16

Exercise 6 Solve the 1-D heat conduction equation

T

t=

2T

x2, (7)

using a singular solution method.

1. Nearest neighbor criteria for t (Courant number).

2. Nearest 2-neighbors can give and receive heat.

3. For any t, superimposing analytical solutions. Compare with finite difference ap-proximation.

Using the parameters from the dike cooling problem in section 4-19 in Turcotte andSchubert: Geodynamics the following MATLAB script calculates the temperature as afunction of distance for t=1000 days using finite differences.


% Exercise 4a

%

% Daniel Koehn


clear all

close all

% define constants


L=20; % some length

dx=2.*L./(N-1); % grid spacing

x0=L; % offset of Gaussian

% numerical values from Turcotte and Schubert, Geodynamics

% Section 4-19, Section 4-21

Q=8.8e9; % heat

rho=2900; % density

c=1.2e3; % specific heat

kappa=5e-7; % thermal diffusivity

dt=dx./(100.*kappa) % size of time step

tmax=1000.*86400; % end of calculation after tmax days


t0=100.*86400; % start calculation after 8.64e6 s or 100 days

Tau=round((tmax-t0)./dt) % number of time steps

x=0:dx:2.*L;

Tm=(Q./(2.*rho.*c.*sqrt(pi.*kappa.*t0))).*exp((-(x-x0).^2)./(4.*kappa.*t0));

Ti=Tm;

T1000ex=(Q./(2.*rho.*c.*sqrt(pi.*kappa.*tmax))).*exp((-(x-x0).^2)./(4.*kappa.*tmax));

% set up boundary conditions

T0=Tm(1)

TN=Tm(N)

for i=1:Tau

17

% calculate temperature T at time step n+1

% from temperature Tm at time step n


T=Tm;

% iteration

for k=2:N-1

T(k) = Tm(k) + dt.*((kappa./(dx.^2)).*(Tm(k-1)+Tm(k+1)-2.*Tm(k)));

end


Tm=T;

% implement boundary

Tm(1)=T(2);

Tm(N)=T(N-1);

end

plot(x,Ti,x,T1000ex,x,T,--);

% axis and title

af=xlabel(x/m);

bf=ylabel(T-T_0/K);

df=title(Diffusion in 1D (forward differences));

% fitting the fonts






print -deps exer_4_heat_trans.eps

The analytical solution for this problem can be written as

T =Q

2cpit

exp

( (x x0)

2

4t

), (8)

where Q is the heat content, density, c specific heat and the thermal diffusivity. Acomparison between the analytical and the numerical solution is schown in figure 7. Asingular solution method is realized with the MATLAB script:


% Exercise 4b

%

% Daniel Koehn


%

% update: 26 th of march 2004 - fixed artifical diffusion bug

18

clear all

close all

% define constants


L=20; % some length

dx=2.*L./(N-1); % grid spacing

x0=L; % offset of Gaussian

% numerical values from Turcotte and Schubert, Geodynamics

% Section 4-19, Section 4-21

Q=8.8e9; % heat

rho=2900; % density

c=1.2e3; % specific heat

kappa=5e-7; % thermal diffusivity

dt=dx./(13.5*kappa) % size of time step

tmax=1000.*86400; % end of calculation after tmax days


t0=100.*86400; % start calculation after 8.64e6 s or 100 days

Tau=round((tmax-t0)./dt) % number of time steps

x=0:dx:2.*L; % coordinates of the grid points

xi=(dx./2):dx:2.*L; % coordinates of the integration points

Tm=(Q./(2.*rho.*c.*sqrt(pi.*kappa.*t0))).*exp((-(x-x0).^2)./(4.*kappa.*t0));

Ti=Tm; % calculate initial condition

T=Ti;

% exact solution after 1000 days

T1000ex=(Q./(2.*rho.*c.*sqrt(pi.*kappa.*tmax))).*exp((-(x-x0).^2)./(4.*kappa.*tmax));

% set up boundary conditions

T0=Tm(1)

TN=Tm(N)

t1=t0;

% calculate heat content in each cell

for k=1:N

Q0(k)=rho.*c.*Tm(k).*dx;

end

NS=N-3;

NE=4;

dir=-1;

for i=1:Tau

t=t1-t0;

% calculate change in heat content for each cell k

QSUM=0.0;

for k=NS:dir:NE

% calculate heat content in neighbor no. 1

Q1=Q0(k).*(erf(abs(xi(k-2)-x(k))./(2.*sqrt(kappa.*t)))-...

erf(abs(xi(k-1)-x(k))./(2.*sqrt(kappa.*t))))./2;

% calculate heat content in neighbor no. 2

19

Q2=Q0(k).*(erf(abs(xi(k-3)-x(k))./(2.*sqrt(kappa.*t)))-...

erf(abs(xi(k-2)-x(k))./(2.*sqrt(kappa.*t))))./2 ;

% heat content in neighbor no. 3 is symmetric to heat content in neighbor no. 1

Q3=Q1;

% heat content in neighbor no. 4 is symmetric to heat content in neighbor No. 2

Q4=Q2;

QSUM = QSUM + Q0(k);

% add heat content to neighbor cells

Q0(k-2)=Q0(k-2)+abs(Q2);

Q0(k-1)=Q0(k-1)+abs(Q1);

Q0(k+1)=Q0(k+1)+abs(Q3);

Q0(k+2)=Q0(k+2)+abs(Q4);

% subtract heat content from cell no. k

Q0(k)=Q0(k)-abs(Q1)-abs(Q2)-abs(Q3)-abs(Q4);

end

% implement boundary

Q0(1)=Q0(2);

Q0(2)=Q0(3);

Q0(3)=Q0(4);

Q0(N-2)=Q0(N-3);

Q0(N-1)=Q0(N-2);

Q0(N)=Q0(N-1);

for k=1:N

% calculate temperature in cell k

T(k)=Q0(k)./(rho.*c.*dx);

end

t1=t1+dt;

NSS=NE;

dir = -dir;

NE=NS;

NS=NSS;

end

QSUM

Tau

plot(x,Ti,x,T1000ex,x(4:N),T(4:N),--);

% axis and title

af=xlabel(x/m);

bf=ylabel(T-T_0/K);

df=title(Diffusion in 1D (singular solution));

% fitting the fonts





20


print -deps exer_4_heat_trans_sing_sol.eps

A comparison between the analytical and the numerical solution using the singularsolution method is shown in figure 8. Using a fixed number of grid points the error ofthe solution highly depends on the size of the time step, as can be seen by choosingdifferent time-step-sizes in the MATLAB script.

21

0 5 10 15 20 25 30 35 400

50

100

150

200

250

300

350

x/m

TT 0

/K

Diffusion in 1D (forward differences)

initial condition t=100 days

exact solution t=1000 days finite difference solution t=1000 days

Abbildung 7: Comparsion between the analytical (8) and the numerical solution usinga forward finite difference scheme.

0 5 10 15 20 25 30 35 400

50

100

150

200

250

300

350

x/m

TT 0

/K

Diffusion in 1D (singular solution)

initial condition t=100 days

exact solution t=1000 days

singular solution method t=1000 days

Abbildung 8: Comparsion between the analytical (8) and the numerical solution usingthe singular solution method.

22

Exercise 7 Form a 5 point central difference scheme for y in the form

y3 =1

h2(1y2 + 2y2 + 3y3 + 4y4 + 5y5) (9)

with equidistant sampling points. Estimate its O(h?) accuracy.

x

y

y1 y2y3

y4

y5

h h h h

Abbildung 9: The construction of the five-point FD-scheme.

Using Taylors theorem together with figure 9 to expand around x = 0 = x3 we get

y1 = y3 2hy3 +1

2!4h2y3

1

3!8h3y3 +

1

4!16h4y3 +O(h5y3 )

y2 = y3 hy3 +1

2!h2y3

1

3!h3y3 +

1

4!h4y3 +O(h5y3 )

y4 = y3 + hy3 +

1

2!h2y3 +

1

3!h3y3 +

1

4!h4y3 +O(h5y3 )

y5 = y3 + 2hy3 +

1

2!4h2y3 +

1

3!8h3y3 +

1

4!16h4y3 +O(h5y3 )

(10)

Using Eq. (10) Eq. (9) can be written as

y3 (x) =1

h2[(1 + 2 + 3 + 4 + 5)y3

+ (21 2 + 03 + 4 + 25)hy3+ (41 + 2 + 03 + 4 + 45)

h2

2!y3

+ (81 2 + 03 + 4 + 85)h3

3!y3

+ (161 + 2 + 03 + 4 + 165)h4

4!y3 ]

(11)

23

In order to make the rhs as exact in y3 as possible when h 0 we need to make asmany low order terms as possible identically zero:

1 + 2 + 3 + 4 + 5 = 0

21 2 + 03 + 4 + 25 = 041 + 2 + 03 + 4 + 45 = 2!

81 2 + 03 + 4 + 85 = 0161 + 2 + 03 + 4 + 165 = 0

(12)

This set of equations can be solved for the five unknowns i using the MATLAB-script

% 5-point difference scheme

%

% Daniel Koehn

% Raisdorf, the 18th of May 2004

clear all

close all

syms h;

% set up coefficient matrix

A=[1 1 1 1 1

-2 -1 0 1 2

4 1 0 1 4

-8 -1 0 1 8

16 1 0 1 16];

% set up rhs

b=[0 0 2 0 0];

b=b;

% invert matrix A

x=inv(A)*b;

x=sym(x)

The solution is 1 = 1/12, 2 = 4/3, 3 = 5/2, 4 = 4/3 and 5 = 1/12. So thefive point-difference-scheme can be written in the form:

yi =1

h2( 1

12yi2 +

4

3yi1 5

2yi +

4

3yi+1 1

12yi+2) (13)

It is accurate to O(h4y3 )-terms.

24

Exercise 8 Choose f(x) = sin(pix2). Solve the systems

2 11 2 1

......

...

1 2 11 1

u1

u2...

uN

=

h2f1

h2f2......

hfN

(14)

2 11 2 1

......

...

1 2 11 1

u1

u2...

uN

=

h2f1

h2f2......

h2 fN2

(15)

exactly for grids of h,h2,h4,h8and determine how the accuracy uFD ucont. is improved

as a function of h as h 0. Try 2 Norms, the 2-norm, and the -norm, and onlyconsider errors at the finite difference grid points. Also, take fi = f(xi).Try your super-duper 5-point difference scheme at the points 3 N 2. How muchaccurate is it ? (both within system (14) and (15)).

To find the exact solution ucont. to the problem we integrate

2u

x2= f(x) = sin

(pix

2

)(16)

twice and get

u

x=

2

picos

(pix

2

)+ c1

u =4

pi2sin

(pix

2

)+ c1x+ c2

(17)

where c1 and c2 denote constants of integration. We estimate them by applying theboundary conditions

u(0) = 0

u(1) = 0(18)

It follows that c1 = c2 = 0. We can conclude that the exact solution of (16) is

u =4

pi2sin

(pix

2

). (19)

The FD-Code for system (14) was realized in the following MATLAB-script:


% Exercise 9

% 1st order flux b.c.

25

%

% Daniel Koehn

% Kiel, the 18 th of may 2004

clear all

close all

% define constants


dh=1; % gridsize refineing factor

for j=1:4

h=0.1./dh % grid point spacing

N=(L./h)+1; % number of grid points

% 1st order scheme system

% -------------------------


% cofficient matrix

e0 = 2.*ones(N,1); % elements on main diagonal

e1 = -1.*ones(N-1,1); % 1st upper diagonal

e2 = -1.*ones(N-1,1); % 1st lower diagonal

S = diag(e0) + diag(e1,1) + diag(e2,-1);

S(N,N)=1;

S;

% RHS vector

b = ones(N-1,1);

x=0;

% build elements 1 - (N-1)

for i=1:N-1

b(i)=(h.^2).*sin(pi.*i.*h./2);

x=x+h;

end


% build element N

%b(N)=h.*sin(pi.*N.*h./2);

b(N)=0;

% plot RHS vector b

b;

% matrix inversion

% ----------------

u=inv(S)*b;

% plot solutions

% ----------------------

% position vector

26

xe=0:h:L;


uex=((4./(pi.^2)).*sin(pi.*xe./2));

% calculate 2-norm

du(dh)=0;

for i=1:N

dus=(u(i)-uex(i)).^2;

du(dh)=du(dh)+dus;

end

du(dh)=sqrt(du(dh));

% calculate infinity-norm

dui(dh)=0;

for i=1:N

dus=(u(i)-uex(i)).^100;

dui(dh)=dui(dh)+dus;

end

dui(dh)=(dui(dh)).^(1./100);

% plot solutions

plot(xe,u,--,xe,uex);

hold on;

% refine grid

dh=dh.*2;

end

% axis and title

af=xlabel(x);

bf=ylabel(u);

df=title(Poisson Equation with fixed and flux b.c. (1st order));

ef=legend(1st order b.c.,exact,2);

% fitting the fonts







print -deps 1st_order_scheme.eps

% output of 2-norm

du

% output of infinity-norm

dui

27

The code for the 2nd order scheme (15) is only a slightly modification of the 1st orderscheme code. The last element of the rhs vector has to be changed:

% apply flux boundary condition

% build element N

b(N)=(h.^2).*sin(pi.*N.*h./2)./2;

The results of calculations with grids h,h2,h4and h

8are shown in figure 10 and figure ??.

The MATLAB-script for the five-point scheme with 2nd order flux boundary schemelooks like this:


% Exercise 9a

% 5-point scheme

% 2nd order flux boundary scheme

%

% Daniel Koehn

% Kiel, the 18 th of may 2004

clear all

close all

% define constants


dh=1; % gridsize refineing factor

for j=1:4

h=0.1./dh % grid point spacing

N=(L./h)+1; % number of grid points

% 1st order scheme system

% -------------------------


% cofficient matrix

e0 = (5./2).*ones(N,1); % elements on main diagonal

e1 = (-4./3).*ones(N-1,1); % 1st upper diagonal

e2 = (-4./3).*ones(N-1,1); % 1st lower diagonal

e3 = (1./12).*ones(N-2,1); % 2nd upper diagonal

e4= (1./12).*ones(N-2,1); % 2nd lower diagonal

S = diag(e0) + diag(e1,1) + diag(e2,-1)+diag(e3,2)+diag(e4,-2);

% build 1st and 2nd equation

S(1,1)=2; S(1,2)=-1; S(1,3)=0;

S(2,1)=-1; S(2,2)=2; S(2,3)=-1; S(2,4)=0;

% build last 2 equation

S(N,N)=1; S(N,N-1)=-1; S(N,N-2)=0;

S(N-1,N)=-1; S(N-1,N-1)=2; S(N-1,N-2)=-1; S(N-1,N-3)=0;

S;

% RHS vector

b = ones(N-1,1);

28

x=0;

% build elements 1 - (N-1)

for i=1:N-1

b(i)=(h.^2).*sin(pi.*i.*h./2);

x=x+h;

end


% build element N

b(N)=(h.^2).*sin(pi.*N.*h./2)./2;

% plot RHS vector b

b;

% matrix inversion

% ----------------

u=inv(S)*b;

% plot solutions

% ----------------------

% position vector

xe=0:h:L;


uex=((4./(pi.^2)).*sin(pi.*xe./2));

% calculate 2-norm

du(dh)=0;

for i=1:N

dus=(u(i)-uex(i)).^2;

du(dh)=du(dh)+dus;

end

du(dh)=sqrt(du(dh));

% calculate infinity-norm

dui(dh)=0;

for i=1:N

dus=(u(i)-uex(i)).^100;

dui(dh)=dui(dh)+dus;

end

dui(dh)=(dui(dh)).^(1./100);

% plot solutions

plot(xe,u,--,xe,uex);

hold on;

% refine grid

dh=dh.*2;

end

29

% axis and title

af=xlabel(x);

bf=ylabel(u);

df=title(Poisson Equation with fixed and flux b.c. (2nd order));

ef=legend(2nd order b.c.,exact,2);

% fitting the fonts







print -deps 2nd_order_scheme.eps

% output of 2-norm

du

% output of infinity-norm

dui

The accuracy of the schemes was measured by the 2-norm S2 and the -norm S.The norms are defined as:

S2 =

( Ni=1

(uFD,i ucont.,i)2)1/2

S ( Ni=1

(uFD,i ucont.,i)100)1/100 (20)

The results of the accuracy measurements are collected in table 1 and are plotted infigure 11.

30

h h/2 h/4 h/8

three-point-scheme 1st order flux b.c.

S2 0.2236 0.1527 0.1060 0.0743

S 0.0752 0.0371 0.0185 0.0093

three-point-scheme 2nd order flux b.c.

S2 0.3223 0.2160 0.1484 0.1034

S 0.1066 0.0523 0.0260 0.0130

five-point-scheme 1st order flux b.c.

S2 0.2220 0.1521 0.1058 0.0742

S 0.0748 0.0370 0.0185 0.0093

five-point-scheme 2nd order flux b.c.

S2 0.3205 0.2153 0.1482 0.1033

S 0.1059 0.0521 0.0259 0.0130

Tabelle 1: S2 and S-norm for 1st and 2nd order flux b.c. for the three- andthe five-point scheme.

31

0 0.2 0.4 0.6 0.8 10

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

x

u

Poisson Equation with fixed and flux b.c. (1st order)1st order b.c.exact

Abbildung 10: Results of the FD-scheme with 1st order flux b.c. for grids h = 0.1, h2,h4

and h8, compared with the exact solution (19).

0 0.02 0.04 0.06 0.08 0.10

0.05

0.1

0.15

0.2

0.25

0.3

0.35

h

u

Comparison of 2 and norm for the different schemesS2 1stS

1st

S2 2ndS

2nd

S2 5p 1stS

5p 1st

S2 5p 2ndS

5p 2nd

Abbildung 11: Comparison between the 2- and the-norm for the different schemes.

32

Problem 1 Multigrid

Make a MATLAB finite difference programm that reproduces figure 18a-e. The startingcondition is shown in figure 18a

vj =1

2

[sin

(12jpi

N

)+ sin

(30jpi

N

)](21)

with N=48, with a weighted Jacobi smoother = 2/3.

The main part of the multigrid code is in the MATLAB-script multigrid.m:


% Multigrid

%

% Daniel Koehn

% Kiel, the 13th of april 2004

clear all

close all

% define constants


w=2./3; % damp factor

enorm = 1; % dummy normalization factor

% central difference scheme

% -------------------------


% coefficient matrix

e0 = 2.*ones(N-1,1); % elements on main diagonal

e1 = -1.*ones(N-2,1); % 1st upper diagonal

e2 = -1.*ones(N-2,1); % 1st lower diagonal

S = diag(e0) + diag(e1,1) + diag(e2,-1);

% RHS vector

b = zeros(N-1,1);

% boundary conditions are both zero

% v(0)=v(N)=0;

% define gridpoints

J=1:N-1;

% initial guess

for j=1:N-1

v(j) = (1./2).*(sin(12.*j.*pi./N)+sin(30.*j.*pi./N));

end

% save 1st guess v0

v0=v;

% plot inital guess with boundary points

33

plot([0 J N],[0 v0 0]);

axis tight;

% axis and title

af=xlabel(grid point);

bf=ylabel(v_j);

df=title(Initial guess );

% fitting the fonts






print -deps multigrid_a.eps;

% apply 1 weighted Jacobi iteration step

[vn]=jac(v,w,S,b);

% save solution v1n after 1 weighted Jacobi step

v1n = vn;

% plot solution v1n after 1 weighted Jacobi step

% with boundary points

figure

plot([0 J N],[0 v0 0],[0 J N],[0 v1n 0]);

axis tight;

% axis and title


bf=ylabel(v_j);

df=title(Solution after 1 fine grid relaxation);

% fitting the fonts






print -deps multigrid_b.eps;

% apply 2 more Jacobi iteration steps

for j=1:2

v = vn;

[vn]=jac(v,w,S,b);

end

% save solution v3n after 3 Jacobi iterations

v3n=vn;

% plot solution v3n after 3 weighted Jacobi steps

34

% with boundary points

figure

plot([0 J N],[0 v0 0],[0 J N],[0 v1n 0],[0 J N],[0 v3n 0]);

axis tight;

% axis and title


bf=ylabel(v_j);

df=title(Solution after 3 fine grid relaxations);

% fitting the fonts






print -deps multigrid_c.eps

% calculate residual

r = b - S * vn;

% full-weighted restriction

[J2h,I2h,v2h,S1,b1] = weight_res(N,vn,S,r);

% one Jacobi iteration step on coarse grid

v = v2h;

v=v;

[v2h]=jac(v,w,S1,b1);

% Correct fine grid solution (Prolongation)

[v1m] = prolong(N,I2h,v3n,v2h);

% plot sparseisity of matrix I2h

%figure

%spy(I2h);

figure

% plot solution v3n after 1 weighted Jacobi step

% on the coarse grid with boundary points

plot([0 J N],[0 v0 0],[0 J N],[0 v1n 0],[0 J N],[0 v3n 0],J,v1m);

axis tight;

% axis and title


bf=ylabel(v_j);

df=title(Solution on coarse grid after full-weighted restriction);

% fitting the fonts





35


print -deps multigrid_d.eps

% apply two more Jacobi steps on coarse grid

for i=1:2

v2hs = v2h;

[v2h]=jac(v2hs,w,S1,b1);

end

% Correct fine grid solution (Prolongation)

[v3m] = prolong(N,I2h,v3n,v2h);

% plot solution v2h after 3 weighted Jacobi steps

% on the coarse grid with boundary points

figure

plot([0 J N],[0 v0 0],[0 J N],[0 v1n 0],[0 J N],[0 v3n 0],J,v3m);

axis tight;

% axis and title


bf=ylabel(v_j);

df=title(Solution on coarse grid after 3 Jacobi iteration steps);

% fitting the fonts






print -deps multigrid_e.eps

The weighted jacobi iteration will be done with the function jac.m

function [vn] = jac(v,w,A,f)

% Weighted Jacobi iteration

%

% Daniel Koehn

% Raisdorf, the 13th of april 2004

% calculate residual r

r=f-A*v;

% calculate inverse of diagonal D of matrix A

D=diag(A);

D=1./D;

S = diag(D);

% Make one Jacobi iteration step

vn = v + (w * S * r);

The full-weighted restriction is realized in the function weight res.m

function [J2h,I2h,v2h,A2h,b1] = weight_res(N,vn,S,b)

36

% Full-Weighted Restriction

%

% Daniel Koehn


% full-weighted restriction

% build restriction matrix I2h

I2h=zeros((N./2)-1,N-1);

dj=0;

for i=1:((N/2)-1)

I2h(i,1+dj) = 1./4;

I2h(i,2+dj) = 2./4;

I2h(i,3+dj) = 1./4;

dj=dj+2;

end

% apply full-weighted restriction on vn

v2h = I2h*vn;

% define gridpoints on coarse grid

J2h=2:2:N-1;

% calculate Matrix A2h

A2h = I2h*S*(2.*I2h);

% apply full-weighted restriction on residual rh

b1=I2h*b;

The fine grid correction will be done with the function prolong.m:

function [vn1] = prolong(N,I2h,vn,v2h)

% Fine Grid Correction

%

% Daniel Koehn


% apply fine grid correction

vn1 = vn + (2.*I2h) * v2h;

37

0 5 10 15 20 25 30 35 40 450.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

grid point

vj

Initial guess

Abbildung 12: The initial guess vj = 12

[sin

(12jpiN

)+ sin

(30jpiN

)]with N=48 for a

coarse grid correction scheme applied with the weighted Jacobi iteration ( = 2/3)

0 5 10 15 20 25 30 35 40 450.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

grid point

vj

Solution after 1 fine grid relaxation

Abbildung 13: The approximation (and error) after one fine grid relaxation sweepsuperimposed upon the initial guess.

38

0 5 10 15 20 25 30 35 40 450.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

grid point

vj

Solution after 3 fine grid relaxations

Abbildung 14: The error after three fine grid relaxation sweeps superimposed uponprevious iterates.

0 5 10 15 20 25 30 35 40 450.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

grid point

vj

Solution on coarse grid after fullweighted restriction

Abbildung 15: The fine grid residual is transferred to the coarse grid by full weighting.This figure shows the error after one relaxation sweep on the coarse grid residualequation, superimposed upon previous approximationes.

39

0 5 10 15 20 25 30 35 40 450.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

grid point

vj

Solution on coarse grid after 3 Jacobi iteration steps

Abbildung 16: The error after three relaxation sweeps on the coarse grid residualequation superimposed upon previous approximations.

40

Problem 2 Temperature and melting of continental crust

Here we are solving 2 types of problems.

1. Heat conduction with a source term

T

t=

2T

z2+ q(z) (22)

2. Conduction-advection for a very simple case of advective flow

u = U0x

Lw = U0 z

L, (23)

where U0 is the stretching/compression velocity at the edge of the region of in-terest. For this particular velocity field, the temperature is constant with x, i.e.T=T(z), so we can again make a 1-D approximation to the problem

T

t U0 z

L

T

z=

2T

z2+ q(z). (24)

The boundary conditions are

T = 0 at z = 0

T = Tmantle = 1300C at z = L = 300 km

(25)

Moho

crust

mantle q=0

q=0z

T=0

at z=L, T=Tmantle

xLL

z=L=300 km

z=0

Abbildung 17: Model of the crust and the upper mantle.

Take L=300 km and assume that the initial Moho is 30 km deep (figure 17). Formulatea 1-D Finite difference algorithm for cases 1 and 2 in MATLAB.

41

1. Start with T = Tm everywhere except z = 0. Solve the problem of halfspacecoolingwith no crustal heat source and no avection. This has an error functionsolution given in Turcotte & Schubert 4-22 Ocean Plate Topography ([1]).

The solution to the halfspace cooling problemgiven in [1] has the form

T (z) = (T0 Tm)erf(

z

2t

)+Tm (26)

with T0 = 0 K, Tm = Tmantle = 1300C and = 1.0 106 m2/s. Using a central

difference scheme for the heat conduction term in eq. (22) and an upwind differencescheme for the time derivative we can write down the following MATLAB code to solvethe heat conduction problem.


% Crustal_melting_cond

%

% Daniel Koehn

% Kiel, the 22nd of may 2004

clear all

close all

% define constants


L=300.*1000; % depth of the model (km)

Lmoho = 30.*1000; % depth of the Moho (km)

kappa = 1e-6; % thermal diffusivity

rhoc=2700; % density of the crust

Tmantle = 1300 + 273; % temperature at lithosphere base

tend=1e7.*31536000; % stop calculation after tend Mio years

dz=L./(N-1); % grid spacing

dt=3e10; % time step

TN=round(tend./dt) % number of time steps

zi=0;

% setup inital temperature Tm, depth z and heat production q

for i=1:N

Tm(i) =Tmantle;

z(i)=zi;

% radioactive heat production in the crust

q(i) = 0.0;

% no radioactive heat production in the mantle

if(zi>=Lmoho)

q(i)=0.0;

end

zi=zi+dz;

end

% setup boundary conditions

Tm(1)=0;

Tm(N)=Tmantle;

42

t=0;

for i=1:TN




T=Tm;

% iteration

for k=2:N-1

T(k) = Tm(k)+dt.*((kappa./(dz.^2)).*((Tm(k-1)+Tm(k+1)-2.*Tm(k)))+q(k));

end


Tm=T;

t=t+dt;

end

% Calculate exact solution to halfspace cooling model

% given in Turcotte & Schubert 4-16 Cooling of the Oceanic Ltihosphere

Tex =((Tm(1)-Tm(N)).*erfc(z./(2.*sqrt(kappa.*t))))+Tm(N);

% plot solution

plot(T,z./1000,Tex,z./1000,--);

axis ij;

hold on

% axis and title

af=xlabel(T/K);

bf=ylabel(z/km);

df=title(Halfspace cooling model);

ef=legend(numerical solution,exact solution,2);

% fitting the fonts







print -deps halfspace_cooling.eps

43

Figure 18 compares the numerical solution calculated using the MATLAB code withthe exact solution eq.(26).

0 200 400 600 800 1000 1200 1400 1600

0

50

100

150

200

250

300T/K

z/km

Halfspace cooling model

numerical solutionexact solution

Abbildung 18: Comparison between the numerical solution and the exact solution eq.(26) for the

halfspace cooling problem after 10 Ma.

2. Solve the problem for a constant crustal heat production of H = 3 1010Wkg andc = 2700 kg/m

3, m = 3300 kg/m3 and = 1 mm2/s.

To calculate the crustal heat production rate q = Hcp we assume a specific heat cp 850 J/kg/K. Figure 19 shows the distribution of the heat production rate q in themodel. The MATLAB script crust melt cond sour.m includes heat production in thecrust.


% Crustal_melting_cond_sour

%

% Daniel Koehn

% Kiel, the 22nd of may 2004

clear all

close all

% define constants



44



H=3e-10; % rate of radioactive heat production in

% the crust

cp=850; % specific heat


Tmantle = 1300 + 273; % temperature at lithosphere base





zi=0;


for i=1:N

Tm(i) =Tmantle;

z(i)=zi;


q(i) = H./cp;

%q(i) = 0.0;


if(zi>=Lmoho)

q(i)=0.0;

end

zi=zi+dz;

end


Tm(1)=0;

Tm(N)=Tmantle;

t=0;

for i=1:TN




T=Tm;

% iteration

for k=2:N-1

T(k) = Tm(k)+dt.*((kappa./(dz.^2)).*((Tm(k-1)+Tm(k+1)-2.*Tm(k)))+q(k));

end


Tm=T;

t=t+dt;

end

% Calculate exact solution to halfspace cooling model

% given in Turcotte & Schubert 4-16 Cooling of the Oceanic Ltihosphere

Tex =((Tm(1)-Tm(N)).*erfc(z./(2.*sqrt(kappa.*t))))+Tm(N);

45

% plot solution

plot(T,z./1000,Tex,z./1000,--);

axis ij;

hold on

% axis and title

af=xlabel(T/K);

bf=ylabel(z/km);

df=title(Halfspace cooling model);

ef=legend(numerical solution,exact solution with q=0,2);

% fitting the fonts







print -deps halfspace_cooling_with_heat.eps

% plot distribution of radioactive elements

figure

plot(q,z./1000);

axis ij;

hold on

% axis and title

af=xlabel(q/K/s);

bf=ylabel(z/km);

df=title(q-Distribution);

% fitting the fonts






print -deps q-distribution.eps

46

The solution to the heat-conduction problem with sources is plotted in figure 20.

0 0.5 1 1.5 2 2.5 3 3.5 4x 1013

0

50

100

150

200

250

300q/K/s

z/km

qDistribution

Abbildung 19: Distribution of the heat production rate q in the model with constantcrustal heat production.

0 200 400 600 800 1000 1200 1400 1600

0

50

100

150

200

250

300T/K

z/km

Halfspace cooling model

numerical solutionexact solution with q=0

Abbildung 20: Comparison between the solution of the halfspace cooling model withheat source in the crust (solid line) and the exact solution with no heat sources (eq.(26)/dashed line) after 100 Ma.

47

3. Assume the crust thickness from 30 km to 60 km in 10 My. What happens toits temperature? (solve equation /problem 2). Does it increase /decrease withdepth ?

Using an upwind scheme for the advective term in eq. (24) we can formulate thefollowing MATLAB code


% Crustal_melting_cond_adv_sour

%

% Daniel Koehn

% Kiel, the 23rd of may 2004

clear all

close all

% define constants





H=3e-10; % rate of radioactive heat production in

% the crust

cp=850; % specific heat


Tmantle = 1300+273; % temperature at lithosphere base





U0=30000./tend; % convergence velocity [m/s]

zi=0;


for i=1:N

Tm(i) =Tmantle;

z(i)=zi;


q(i) = H./cp;


if(zi>=Lmoho)

q(i)=0.0;

end

zi=zi+dz;

end


Tm(1)=0;

Tm(N)=Tmantle;


T=Tm;

48

% 1. calculate temperature distribution after 10 Ma with crustal thickening

t=0;

% initial Moho depth

Lmoho0=Lmoho;

for i=1:TN



% depth of the Moho is linear increasing with time

Lmoho=U0.*t + Lmoho0;

zi=0;

% setup heat production rate q

for j=1:N


q(j) = H./cp;

z(j)=zi;


if(zi>=Lmoho)

q(j)=0.0;

end

zi=zi+dz;

end

% solve advection-conduction equation with sources by iteration

for k=2:N-1

T(k) = Tm(k)+dt.*((kappa./(dz.^2)).*((Tm(k-1)+Tm(k+1)-2.*Tm(k)))...

+q(k)+(U0.*(z(k)./L).*(Tm(k)-Tm(k-1))./dz));

end


Tm=T;

t=t+dt;

end

% plot solution

figure

plot(T,z./1000);

axis ij;

% axis and title

af=xlabel(T/K);

bf=ylabel(z/km);

df=title(crustal thickening);

ef=legend(geotherm after 10 Ma crustal thickening,1);

% fitting the fonts





49



print -deps halfspace_cooling_adv.eps

% plot distribution of radioactive elements

figure

plot(q,z./1000);

axis ij;

hold on

% axis and title

af=xlabel(q/K/s);

bf=ylabel(z/km);

df=title(q-Distribution);

% fitting the fonts






print -deps q-distribution_1.eps

50

The results of the calculation are plotted in figure 21 and figure 22. figure 22 showsthe geotherm after 10 Ma with heat sources in the crust and crustal thickening. Thetemperature increases with depth.

0 0.5 1 1.5 2 2.5 3 3.5 4x 1013

0

50

100

150

200

250

300q/K/s

z/km

qDistribution

Abbildung 21: Distribution of q after 10 Ma of crustal thickening.

0 200 400 600 800 1000 1200 1400 1600

0

50

100

150

200

250

300T/K

z/km

crustal thickening

geotherm after 10 Ma crustal thickening

Abbildung 22: The geotherm after 10 Ma of crustal thickening.

51

4. Assume the crust stays 60 km thick for 10 Ma and then thins to 30 km in another10 Ma. How does its temperature compare the original steady-state profile?

Using the same MATLAB-Code as in problem 3 we get the geotherms plotted in fi-gure 23. The geotherm after one cycle of 10 Ma crustal thickening, steady state for10 Ma and 10 Ma thining (dashed line) has a greater slope than the original steadystate profile (solid line), but a smaller slope than the geotherm after 10 Ma of crustalthickening (dotted line).

5. Assume the crust stays 60 km thick for 30 Ma and then thins in 10 Ma back to30 km. What are the differences between this and problem 4 ?

In figure 23 and figure 24 the problems 4 and 5 are compared. Cause the crust stays30 Ma in steady state the slope of the geotherm at the end of the simulation (dashedline) in problem 4 is greater, than in problem 5.

6. Assume the mantle beneath the crust is removed and replaced by 1300 C mantle(T = Tm). If no advection occurs (30 km crust and no deformation), what is themaximum increase in temperature in the crust at z=15 km depth ? (Before itgoes back to steady state). Say crust melts at 800 C. Where melts the lowercrust ?

At z=15 km depth the maximum temperature, after replacing the mantle beneath thecrust by 1300 C mantle, is Tmax 169 C (figure 25). By looking at figure 26 we canestimate, that the lower crust can melt at a depth of about 52 km.

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Abbildung 23: Geotherm after 10 Ma of crustal thickening (dotted line) and thegeotherm after another 10 Ma of steady state and 10 Ma of crustal thining (dashedline), compared with a 30 Ma steady state geotherm (solid line).

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Abbildung 24: Geotherm after 10 Ma of crustal thickening (dotted line) and thegeotherm after 30 Ma of steady state and 10 Ma of crustal thining (dashed line),compared with a 30 Ma steady state geotherm (solid line).

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Abbildung 25: Temperature evolution at a depth of z = 15 km after replacing themantle beneath the crust by 1300C mantle.

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Abbildung 26: Geotherm after 10 Ma of crustal thickening (dotted line). The geothermafter 30 Ma of steady state, 10 Ma of crustal thining and replacing the mantle beneaththe crust by 1300C mantle (dashed line).The geotherm after another 10 Ma of steadystate (solid line).

54 LITERATUR

Literatur

[1] Donald L. Turcotte and Gerald Schubert. Geodynamics. Cambridge UniversityPress, 2nd edition, 2002.

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