numerical methods for partial differential equations caam 452 spring 2005 lecture 9 instructor: tim...

Numerical Methods for Partial Differential Equations

CAAM 452

Spring 2005

Lecture 9

Instructor: Tim Warburton

CAAM 452 Spring 2005

Today

• Change of plan – I will go through in detail how to solve homework 3

step by step.


Homework 3

Q1) Build a finite-difference solver for

Q1a) use your Cash-Karp Runge-Kutta time integrator from HW2 for time stepping

Q1b) use the 4th order central difference in space (periodic domain)

Q1c) perform a stability analysis for the time-stepping (based on visual inspection of the CK R-K stability region containing the imaginary axis)

Q1d) bound the spectral radius of the spatial operator

Q1e) choose a dt well in the stability region

Q1f) perform four runs with initial condition (use M=20,40,80,160) and compute maximum error at t=8

Q1g) estimate the accuracy order of the solution.

Q1h) extra credit: perform adaptive time-stepping to keep the local truncation error from time stepping bounded by a tolerance.

u uc

t x

24cos,0 , 0,2xu x e x


Q1a – Use Cash-Karp RK for Time-Stepping

• We will use a vector version of Cash-Karp:

• Where f is a vector valued function – in our case it will be a linear operator acting on a vector argument.

1

2 121

3 1 231 32

4 1 2 341 42 43

5 1 2 3 451 52 53 54

6 1 2 3 4 561 62 63 64 65

1 1 2 31 2 3 4

n

n

n

n

n

n

n n

k dtf u

k dtf u b k

k dtf u b k b k

k dtf u b k b k b k

k dtf u b k b k b k b k

k dtf u b k b k b k b k b k

u u c k c k c k c

4 5 65 6k c k c k


Where the b,c coefficients arehttp://www.library.cornell.edu/nr/bookcpdf/c16-2.pdf

c= 37/378 0 250/621 125/594 0 512/1771

0 0 0 0 0

1/5 0

3/40 9/40 0

3/10 -9/10 6/5 0

-11/54 5/2 -70/27 35/27 0

1631/55296 175/512 575/13824 44275/110592 253/4096 0

b

Provided in cashkarp.m on the web site.Original paper at:

http://portal.acm.org/citation.cfm?id=79507&coll=GUIDE&dl=GUIDE&CFID=38381912&CFTOKEN=60548034


Homework 3










u uc

t x

24cos,0 , 0,2xu x e x


Recall: 4th Order Central Difference Scheme

4 2 1 1 2

18 8

12 m m m mmv v v v v

dx

The fourth order central difference derivative acts on any vector and gives the following value for each entry of a result vector:

Where the increments on the indexing is done modulo M.Since we will use this operator repeatedly I made a function


Code

• Each time step consists of evaluating the 6 intermediate vectors k1,k2,..,k6

• And piecing them together.

• Notice – here I set up u at the start as a vector, and delta4 accepts a vector (and dx) as arguments and returns a vector.

• i.e. each of k1,k2,..,k6 is a vector.


Alternate Code

• I could also have used loops

• Note: u is a row vector of length M so the k’s are a matrix of size 6 x M


Homework 3










u uc

t x

24cos,0 , 0,2xu x e x


Cash-Karp

• The Cash-Karp Runge-Kutta integrator in our case is:

• Notice, we do not need the time component on the right hand side, because our finite-difference right hand side is independent of time.

1

2 121

3 1 231 32

4 1 2 341 42 43

5 1 2 3 451 52 53 54

6 1 2 3 4 561 62 63 64 65

1 1 2 3 4 5 61 2 3 4 5 6

n

n

n

n

n

n

n n

k dtf u

k dtf u b k

k dtf u b k b k

k dtf u b k b k b k

k dtf u b k b k b k b k

k dtf u b k b k b k b k b k

u u c k c k c k c k c k c k


Linear Stability Analysis

• We achieve the linear stability analysis by assuming f is linear in u: f u u

1

2 121

3 1 231 32

4 1 2 341 42 43

5 1 2 3 451 52 53 54

6 1 2 3 4 561 62 63 64 65

1 1 2 3 4 5 61 2 3 4 5 6

n

n

n

n

n

n

n n

k dtu

k dt u b k

k dt u b k b k

k dt u b k b k b k

k dt u b k b k b k b k

k dt u b k b k b k b k b k



cont

• We replace mu*dt with nu

1

2 121

3 1 231 32

4 1 2 341 42 43

5 1 2 3 451 52 53 54

6 1 2 3 4 561 62 63 64 65

1 1 2 3 4 5 61 2 3 4 5 6

n

n

n

n

n

n

n n

k u

k u b k

k u b k b k

k u b k b k b k

k u b k b k b k b k

k u b k b k b k b k b k



cont

• We wish to remove all the intermediate variables and express the scheme in a one-step form like:

• Where the multiplier function pi(nu) is a 6th order polynomial in nu.

• It is certainly possible to find all the coefficients of this polynomial by hand but a little messy.

• We can take a short cut – assuming Cash-Karp is actually a 5th order scheme as claimed we know that the first 6 terms of the polynomial multiplier must be the first 6 terms of the Taylor series for

1n nu u

e


cont

• So we know that:

• Where C is to be determined.

• However, looking at the definition of the stages we see that there is only one contribution to the 6th order term:

• So

1 2 3 4 5 61 1 1 1 11

1! 2! 3! 4! 5!C

66 65 54 43 32 21c b b b b b

1 2 3 4 5 66 65 54 43 32 21

1 1 1 1 11

1! 2! 3! 4! 5!c b b b b b


Stability Condition

• We need to plot the stability region, so we determine the margin of stability by finding nu such that

• The curve of points in the (root) complex plane with magnitude 1 can be parameterized in theta by

• So for each theta in [0,2pi) we seek the corresponding nu such that:

1 2 3 4 5 66 65 54 43 32 21

1 1 1 1 11 1

1! 2! 3! 4! 5!c b b b b b

ie

1 2 3 4 5 66 65 54 43 32 21

1 1 1 1 11

1! 2! 3! 4! 5!ic b b b b b e


Matlab roots Command

• The matlab roots command can be used to find roots of a polynomial.

• If the polynomial is say:

• Then construct a vector of coefficients (highest order first):

• And invoke: roots(a)• A vector of the roots (if any) of the polynomial will

be returned.

21 2 3x a a x a x

3 2 1, ,a a a a


Adapting RKabstab.m

• So I took the routine from the class web page and modified it slightly

• I hard coded it to use the multiplier polynomial for the Cash-Karp

• I changed the plotting to use a scatter plot


Absolute Stability for Cash-Karp

• This is a classic picture.

• What is interesting is that it does not convincingly cover the imaginary axis!.


On The Imaginary Axis

Here I used Matlab’s polyval to evaluate themultiplier polynomial for nu on the imaginary axis.Conclusion – the absolute stability region is justto the left of the imaginary axis (not a big issue here)


Homework 3










u uc

t x

24cos,0 , 0,2xu x e x


Q1d) Bound the spectral radius of the derivative operator

• All the eigenvalues of the 4th order central difference are on the imaginary axis, with values (recall from Lecture 6):

• So the gotcha is that there is no dt such that the eigenvalues*dt are all exactly inside the stability region.

• However, we will estimate the largest eigenvalue and make an assumption..

8sin sin 26m m m

icdx


Q1d) continued

2

2

-1

8sin sin 26

8cos 2cos 26

critical points for which satisfy:

0 8cos 2cos 2

=8cos 2 2cos 1

=-4cos 8cos 2

8 64 32 3. . cos 1

8 2

3. . = cos (1- ) 1.37222...

2 m

icdx

d icd dx

i e

ci e

dx

We can estimate the largestmagnitude eigenvalue directly.


Homework 3










u uc

t x

24cos,0 , 0,2xu x e x


Q1e)

• Since we are only integrating for 8 periods let’s choose a reasonable maximum |nu|<=1

• So for close to absolute (linear) stability

• Implies that we can choose:

• i.e. the time step restriction is:

• We can further reduce the right hand side of the inequality to reduce marginal growth.

1.37222...m

cdx

1.37222 1cdtdx

1.37222dx

dtc


Homework 3










u uc

t x

24cos,0 , 0,2xu x e x


Q1f: Test Runs

• I already set up this initial condition for the web demos:

• I modified the testrig.m file to call the centraldifference4CKRK54.m file

• It looks like reasonable 4th order convergence..

• We examine the error ratios to be more certain

T=4


Homework 3










u uc

t x

24cos,0 , 0,2xu x e x


Q1g) Estimate Order of Accuracy of Solution

M Max error at t=4 Order of accuracy

20 0.44353855151728

40 0.12852934806853 1.79

80 0.02040188007649 2.66

160 0.00134076263509 3.93

320 0.00008319425374 4.01

Cool – asymptotically it looks like a fourth order code (in space)


Homework 3










u uc

t x

24cos,0 , 0,2xu x e x


Q1h) Estimating the Local Truncation Error

• Given the k vectors we can also construct a lower order (4th order in time) approximation to the updated solution:

• Then we can ball park what the time step should be:

1 * 1 * 2 * 3 * 4 * 5 * 61 2 3 4 5 6

n nu u c k c k c k c k c k c k

0.24

5 4min

rk

est rk rk

tol udt dt

u u


Q1h) Code

Form alternative embeddedfourth order approximation

Estimate reasonable dt

Check to see if we shouldreduce time step.

Modify last time step to landon EndTime


Q1h) Testing

• I started with a fairly large time step and let the code adjust itself

M Max error at t=4 Order of accuracy

20 0.44353855318971

40 0.12852934719014 1.79

80 0.02040190480104 2.66

160 0.00134075190765 3.93

320 0.00008319462183 4.01

These are pretty much unchanged from before.


Q1h) Adaptive Time Stepping(started with dt=100*dx)

Q)

Why does it take moretime steps for theLower resolution runs?


Note on the Cash-Karp RK

• Check out:• http://www.math.sfu.ca/~cbm/mscthesis/cbm-mscthesis.ps.gz

• There is an interesting treatment of RK schemes, with the idea of parameterizing embedded RK schemes (n+1 stage, n’th order) by choosing the coefficient in the multiplier directly.

• Increasing this parameter a little will make sure that at least part of the imaginary axis is inside the absolute stability region.

http://www.math.sfu.ca/~cbm/mscthesis/cbm-mscthesis.ps.gz

numerical methods for partial differential equations caam 452 spring 2005 lecture 9 instructor: tim...

Documents

time step

adaptive time

th order central difference

function slide

result vector

vector argument

vector version of cash

q1a use cashkarp rk