Analysis of Runge Kutta Method

7/24/2019 Analysis of Runge Kutta Method

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ENGR90024

Computational Fluid Dynamics

Lecture 09

Analysis of the Runge-Kutta method &

the Crank-Nicolson method


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Stability analysis of 2nd-order Runge-Kutta (page 39 of printedlecture notes):

l+1 =l +t

1

2k1 +

1

2k2

k1 = f(l, tl)

k2 = f(l +tk1, t

l +t)

to model problem

d

dt ==f()

Apply 2nd-order Runge-Kutta


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l+1 =l +t

1

2k1 +

1

2k2

k1 = f(

l, tl)

k2 = f(l +tk1, t

l +t)

dt ==f()

k1 = f(l, tl) =l f(l, tl) = l

l l

k2 =

f(l

+tk1, t

l

+t

)

tl

f( +tk1, t +t) =( +tk1)

l +tk1 l+tk1

f(l +tk1, tl +t) =(l +t(l))

= k1

= k2

tl+t


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l+1 =l +t

1

2k1 +

1

2k2

k2k1

l

(l

+t

(l))

=

=

l+1 =l +t12

l +12(l +tl)

l+1 =l +tl +1

2t

22l

l+1 = (1 +t +t22/2)l

where = 1 +t +t22/2l+1 = l = l1


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1 = Given

2 = 1

3=

2=

2

1

= 314= 3

.

.

. =

.

.

.

l+1 = l=

.

.

.

= l1

Prove that l+1 = l = l1

Hence, for stability

|| =|1 +t +t2

2

/2| 1


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|| =|1 +t +t22/2| 1

The stability boundary is can be found by solving

|1 +t +t22/2| = 1

A complex number

Need to find!!t where the above complex number has amagnitude of 1. Any complex number that has a magnitude 1 canbe represented by

Hence we want to find values of !!t that make

|1 +t +t22/2| = 1

1 +t +t22/2 =ei

1 +t +t22/2

ei = 0

|ei| = 1


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You can follow the steps above and show that the stability regionfor the 4th order Runge-Kutta method can be found by solving

t+

2t2

2

+

3t3

6

+

4t4

24

+ 1 ei

= 0

where

0 8 = 0

1 +t +t22/2 ei = 0

Note that0 2n

where n is the order of the polynomial.


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Example O09.1:Write MATLAB program to find stability

boundary for the 2nd-order Runge-Kutta method solving the

model problem

dt

==f()


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1 +t +

t

22

/2

e

i= 0

To be consistent with notation introduced previously, we let

1

t = p

Hence

ei = 0+p +p2/2

This is a 2nd order polynomial in terms of p. We can use Newton-Raphson polynomial to solve this equation. Note that

0 2n

where n is the order of the polynomial.

1 +p +p2/2 ei = 0

From the theory presented above, we need to solve


10/40

functionMPO09p1a()

clear all;

close all;

theta=0:0.1:4*pi;

%initial guess value of p

p=0;

forn=1:length(theta)

gp=1+p+p^2/2.0-exp(i*theta(n));

dgdp=1+p;

whileabs(gp)>1.0e-6

p=p-gp/dgdp

gp=1+p+p^2/2.0-exp(i*theta(n));

dgdp=1+p;

end

lamdt(n)=p;

end

plot(real(lamdt),imag(lamdt),'b-','linewidth',5);

xlabel('\lambda_{Re}\Delta t');

ylabel('\lambda_{Im}\Delta t');

axis([-4 2 -3 3]);

daspect([1 1 1]);

!4 !3 !2 !1 0 1 2!3

!2

!1

0

1

2

3

!Re"t

!Im

"

t

(p) = 1 +p+p2/2 ei = 0

dg

dp= 1 +p

= 0

=

= 2

= 3

= 4

pnew= pold g(pold)dg

dp(pold)

Newton Raphson


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Alternatively, we can get MATLAB to plot contours of

|| =|1 +t +t22/2|

We can do that using the contour()command in MATLAB

(4 4)( 4 4)


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functionMPO09p1b()

clear all;

close all;

relamdt=-4:0.1:4;

imlamdt=-4:0.1:4;

[x,y]=meshgrid(relamdt,imlamdt);

axis square;

lamdt=x+i*y;

sig=1+lamdt+lamdt.^2/2;contour_levels=[1 2 3 4];

contour(x,y,abs(sig),contour_levels,'linewidth',4);

xlabel('Re \lambda \Delta t','Fontsize',14);ylabel('Im \lambda \Delta t','Fontsize',14);

hold on;

plot([-4 4],[0 0],'k');

plot([0 0],[-4 4],'k');

title('contour |\sigma|','Fontsize',14);

Re !"t

Im!

"

t

contour |#|

!4 !3 !2 !1 0 1 2 3 4!4

!3

!2

!1

0

1

2

3

4

t= Re(t) + iIm(t)

= 1 +t +t22/2

|| = 4

|| = 3

|| = 2

|| = 1

(-4,-4) (4,-4)

(4,4)(-4,4)

||


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h+

2h2

2+

3h3

6+

4h4

24+ 1 e

I= 0

1 + h+2h2

2 e

I= 0

-2.79

2.83

-2.83

-2.0

4th order Runge Kutta

2nd order Runge Kutta

Euler


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h+

2h2

2+

3h3

6+

4h4

24+ 1 e

I= 0

1 + h+

2h2

2e

I

= 0

-2.79

Stability region for RK4 is bigger than RK2

In order for your code to be stable,you need to pick such that it falls withinthe stability region of your numerical scheme.

t

For example, if you want to solve

d

dt = 10

using Euler, the maximum value of !t youcan use is

tmax,Euler =2

10=

1

5= 0.2

If you were to use RK4, then the maximumvalue of!t you can use is

tmax,RK4 =2.79

10= 0.279


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0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

!

1!exp(!t)

Explicit Euler

0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

!

1!exp(!t)

2nd-orderRunge-Kutta

0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

!

1!exp(!t)

4th-orderRunge-Kutta

Back to this slide again: stability analysis predicts this behaviour

0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

!

1!exp(!t)

0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

!

1!exp(!t)

0 1 2 3 4 5 6 7 8!30

!20

!10

0

10

20

30

t

!

1!exp(!t)

0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

!

1!exp(!t)

0 1 2 3 4 5 6 7 8!30

!20

!10

0

10

20

30

t

!

1!exp(!t)

0 1 2 3 4 5 6 7 8!30

!20

!10

0

10

20

30

t

!

1!exp(!t)


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End of Example O09.1


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Runge-Kutta methods for aSystem of ODEs


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In general, a system of M ODEs can be written as

d1

dt =f1(1,2,3, . . . . . . , M, t)

d2dt

=f2(1,2,3, . . . . . . , M, t)

d3

dt =f3(1,2,3, . . . . . . , M, t)

... =...

dM

dt =fM(1,2,3, . . . . . . , M, t)

Runge-Kutta method can easily be extended to solve a system of Mequations


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d1

dt =f1(1, t)

RK-2

k1 = f(l

1, tl

)k2 = f(

l

1+tk1, t

l +t)

l+11 =l

1 +t

1

2k1 +

1

2k2


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d1

dt =f1(1, t) d1

dt =f1(1,2, t)

d2

dt =f2

(1,2, t

)

RK-2

RK-2

k1 = f(l

1, tl)

k2 = f(l

1+tk1, t

l +t)

l+11 =l

1 + t12k1 +

12k2

l+11 = l

1 +t

1

2k11 +

1

2k21

l+12 = l

2 +t

1

2k12 +

1

2k22

k11 = f1(l

1,l

2, tl

)k12 = f2(

l

1,l

2, tl)

k21 = f1l1+tk11,

l

2+tk12, t

l+t

k22 = f2

l

1 +tk11,l

2 +tk12, tl

+t

k1associated with

"1

k2associated with

"1

k2associated with

"2

k1associated with

"2


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d1

dt =f1(1, t)

RK-2

k1 = f(l

1, tl)

k2 = f(l

1+tk1, t

l +t)

l+11 =l

1 + t12k1 +

12k2

d1

dt =f1(1,2, t)

d2

dt =f2(1,2, t)

RK-2

l+11 = l

1 + t

1

2k11 +

1

2k21

l+12 = l

2 + t

1

2k12 +

1

2k22

k11 = f1(l

1,l

2, tl)

k12 = f2(l

1,l

2, tl)

k21 = f1l1+tk11,

l

2+tk12, t

l+t

k22 = f2l1+tk11,

l

2+tk12, t

l+t

d1

dt =f1(1,2,3, t)

d2

dt =f2(1,2,3, t)

d3

dt =f3(1,2,3, t)

l+11 = l

1 + t

1

2k11 +

1

2k21

l+12 = l2 + t12k12 + 1

2k22

l+13 =

l

3 + t

1

2k13 +

1

2k23

k11 = f1(l

1,l

2,l

3, tl)

k12 = f2(l

1,l

2,l

3, tl

)k13 = f3(

l

1,l

2,l

3, tl)

k21 = f1l1 +tk11,

l

2 +tk12,l

3 +tk13, tl+t

k22 = f2l1 +tk11,

l

2 +tk12,l

3 +tk13, tl+t

k23 = f3

l

1 +tk11,l

2 +tk12,l

3 +tk13, tl

+t

RK-2

k1associated with"1

k2associated with

"1

k2associated with

"3

k1associated with

"2

k1associated with

"3

k2associated with

"2


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Example O09.2:Write a MATLAB program to solve the

nonlinear ODE

with "(0)=0, (d"/dt)(0)=0, (d2"/dt2)(0)=0.332 using the 4th-order Runge-Kutta method. Plot d"/dt for 0


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d3

dt3 +

1

2

d2

dt2 = 0

Let

1 = , 2 = d/dt, 3 = d2/dt2

(0) = 0.

0 d

dt(0) = 0.0 d

2

dt2(0) = 0.332 0 t 25

(9.1)

sod1

dt =

d

dt = 2

d2

dt =

d2

dt2 = 3

and

1(0) = 0.

0 2(0) = 0.

0 3(0) = 0.

332

Equation (9.1) can be rewritten as

d3

dt

= 1

2

d

dt2 =

1

23

131


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functionMPO9p2()

clear all;

close all;

Delta_t=0.1;t=0:Delta_t:25;

phi=zeros(length(t),3);phi(1,:)=[0.0 0.0 0.332];

forn=1:length(t)-1

k1=f(phi(n,:));

k2=f(phi(n,:)+Delta_t*k1);

phi(n+1,:)=phi(n,:)+Delta_t*(k1/2+k2/2);

end

plot(t,phi(:,2),'ko-')hold on

xlabel('t');

ylabel('d\phi/dt');

functiondphidt=f(phi)

dphidt=[phi(2) phi(3) -0.5*phi(3)*phi(1)];

d3dt

=

1

231

d1

dt

d2

dt

= 2

= 3

1(0) = 0.0 2(0) = 0.0 3(0) = 0.332

t 5

d1

dt =f1(1,2,3, t)

d2

dt =f2(1,2,3, t)d3

dt =f3(1,2,3, t)

l+11 = l

1 + t

1

2k11 +

1

2k21

l+12 = l

2 + t

1

2k12 +

1

2k22

l+13 = l

3 + t

1

2k13 +

1

2k23

k11 = f1(l

1,l

2,l

3, tl)

k12 = f2(l

1,l

2,l

3, tl)

k13 = f3(l

1,

l

2,

l

3, t

l

)k21 = f1

l1+tk11,

l

2+tk12,

l

3+tk13, t

l+t

k22 = f2l1+tk11,

l

2+tk12,

l

3+tk13, t

l+t

k23 = f3l1+tk11,

l

2+tk12,

l

3+tk13, t

l+t

RK-2


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functionMPO9p2()

clear all;

close all;

Delta_t=0.1;

t=0:Delta_t:25;

phi=zeros(length(t),3);phi(1,:)=[0.0 0.0 0.332];

forn=1:length(t)-1

k1=f(phi(n,:));

k2=f(phi(n,:)+Delta_t*k1);

phi(n+1,:)=phi(n,:)+Delta_t*(k1/2+k2/2);

end

plot(t,phi(:,2),'ko-')

hold on

xlabel('t');

ylabel('d\phi/dt');

functiondphidt=f(phi)

dphidt=[phi(2) phi(3) -0.5*phi(3)*phi(1)];

0 5 10 15 20 250

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t

d!/dt


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27/40

CRANK-NICOLSON FORMULA

l+1 =l +t1

2

f(l, tl) +f(l+1, tl+1)

Apply Crank-Nicolson:

dt(t) =f(, t)

ODE to solve:


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GRAPHICAL COMPARISON

tl+1

t

tl

ll+1

!El+1

local

tl+1)

t

!

Explicit Euler:take slope at tl

tl+1

t

tl

l

l+1

! El+1local

tl+1)

t

!

Implicit Euler:take slope at tl+1

tl+1

t

tl

l

l+1

! El+1localtl+1)

t

!

Crank-Nicolson:take average of slopesat tl and tl+1

+1 = +tf( , t) +1 = +tf( +1, t +1) l+1 =l +t1

2

f(l, tl) +f(l+1, tl+1)

(t) Illustration of explicit Eulers method


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ddt

=f(l, tl)

(t)

t

t

l

t

l+1

Error

Predicted value of #l+1

True value of #l+1

Illustration of explicit Euler s method

l+1 =l +tf(l, tl)

(t)(tl+1)

(tl)Eulers formula

(t) Ill i f h i li i E l h d


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(t)

t

l

t +1

l

t


True value of #l+1

Error

+1 = +tf( +1, t +1)

Illustration of the implicit Euler method

Summary


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Error

(t)

tl

tl+1

l

t

Predicted value of !l+1

True value of !l+1

Error

+1 = +tf( +1 , t+1)

Illustration of the implicit Euler method

d

dt =f(l, tl)

(t)

t

tl

tl+1

Error

Predicted value of !l+1

True value of !l+1

Illustration of explicit Eulers method

l+1 =l +tf(l, tl)

(t)(tl+1

)

(tl)Eulers formula

Summary

(t)

t

t t+1

l

l+1 =l +t1

2

f(l, tl) +f(l+1, tl+1)

Illustration of Crank Nicolson method



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Example O09.3:

Using Crank-Nicolson method, solve

For 0


33/40

l+1 =l +t1

2

f(l, tl) +f(l+1, tl+1)

d

dt = 1 = f(, t)

Crank Nicolson formula

l+1 =l +t 12

f(l, tl) +f(l+1, tl+1)

d

dt = 1

1l 1

l+1

1 +

t

2

l+1 =

1

t

2

l + t

l+1 =

1

t

2

l +t

1 +

t

2

function MPO09p3()


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functionMPO09p3()

close allclear all

tmin=0.0;tmax=8.0;

[t1,phi1]=MyCrankNicolson([tmin tmax],0.0,2.0);

plot(t1,phi1,'ko-');hold onezplot(@(t)1-exp(-t),[0,8,0,2])xlabel('t');ylabel('\phi');legend('Crank-Nicolson','True');

function[t,phi]=MyCrankNicolson(tspan,phi0,Delta_t)t=tspan(1):Delta_t:tspan(2);

phi=zeros(length(t),1);

phi(1)=phi0;forn=1:length(t)-1

phi(n+1)=((1-Delta_t/2)*phi(n)+Delta_t)/(1+Delta_t/2);end

0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

1!

exp(!

t)

!

Euler

True

l+1 =

1

t

2

l +t

1 + t

2

Output

Crank-Nicolson


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0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

1!exp(!t)

!

Euler

True

!t=2.0

0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

!

EulerTrue!t=2.0

Compare implicit Euler, explicit Euler and Crank Nicolson method for !t=2.0

0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

1!exp(!t)

!

Euler

True

!t=2.0


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37/40

Example O09.4:Write a MATLAB program that uses the Crank-

Nicolson method to solve

for 0 < t < 10 with !(t=0)=1.

Applying the Crank-Nicolson method, we obtain

Analytical solution:

d

dt

= i2

(t) = ei2t

l+1 =l +t1

2

i2l + i2l+1

(1 it)l+1 = (1 + it)l

l+1 = 1 + it

1 itl


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functionMPO9p4()

% Solve dphi/dt = i*2*phi, phi(0) = 1, 0 < t

Analysis of Runge Kutta Method

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