1

Spring 2003

Prof. Tim Warburtontimwar@math.unm.edu

MA557/MA578/CS557Lecture 5a

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Homework 2 correctionQ4) Implement the numerical approximation of:

By:

2

3 0

( , 0) x

t x

x t e

1 1

11

0 1

00

Geometry:

14, 4,

1 1

Scheme:

1

3

Initial Condition :

, 02

Boundary Condition :

0

N i N

n n ni i i

i ii

N i ix x x x x

N N

dt

dx

x xt

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Homework 2 contQ4 cont)

For N=10,40,160,320,640,1280 run to t=10, with: dx = (8/(N-1)) dt = dx/6

Use

On the same graph, plot t on the horizontal axis and error on the vertical axis. The graph should consist of a sequence of 6 curves – one for each choice of dx.

Comment on the curves.

NOTE: For the purposes of this test we define error as:

1

1max ,

2n n i i

ii N

x xerror ndt

10 0 11n n n

N

dt dtu udx dx

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Lecture 5

• We will define stability for a numerical scheme and investigate stability for the upwind scheme.

• We will compare this scheme with a finite difference scheme.

• We will consider alternative ways to approximate the flux functions.

5

Basic Upwind Finite Volume Method

1

1

n ni i n n

i i

dxu u

dt

Approximate time derivative and look forsolution

dtudx

111n n n

i i i

Note we must supply a value for the left most average at each time step: 0n

Recall

1, ,i i i

dq dx uq x t uq x t

dt

Approximate fluxes with upwind flux

1i i i

dq dx uq uq

dt

n ni iq

6

Convergence• We have constructed a physically reasonable numerical scheme to

approximate the advection equation.

• However, we need to do some extra analysis to determine how good at approximating the true PDE the discrete scheme is.

• Let us suppose that the i’th subinterval cell average of the actual solution to the PDE at time T=n*dt is denoted by

1 1

1

1

1

1 1, ,

where q satisfies:

, , ,

i i

i i

i

i

x xni

i i x x

x

i i i

x

q q x ndt q x ndtx x dx

d dq x t dx dxq uq x t uq x t

dt dt

7

Error Equation• The goal is to estimate the difference of the exact

solution and the numerically obtained solution at some time T=n*dt.

• So we are interested in the error:

• For the given finite volume scheme dt and dx will be related in a fixed manner (i.e. dt = Cdx for some C, independent of dx).

• Suppose we let and then

the scheme is said to be of order s.

, nn n ni i i

TE q

dt

0dt n sE O dt

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Norms and Definitions

• We define the discrete p-norms:

• We say that the scheme is convergent at time T in the norm ||.|| if:

• It is said to be accurate of order s if:

1/ pip

ipi

E dx E

0lim 0n

dtndt T

E

as 0n sE O dt dt

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Local Truncation Error• Suppose at the beginning of a time step we actually have the exact

solution -- one question we can ask is how large is the error committed in the evaluation of the approximate solution at the end of the time step.

• i.e. choose

• Then

• We expand about xi,tn with Taylor series:

11

1 1

1

1:

n ni i

n n ni i i

n n ni i i

q

dt dtu q u q

dx dx

R qdt

11,n ni iq q

________2 2

31 2

________2 2

1 32

( )2

( )2

n ni i

n ni i

q dx qq q dx O dx

x x

q dt qq q dt O dt

t t

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Estimating Truncation Error

• Inserting the formulas for the expanded q’s:

11

________2 2

32

________2 2

32

1: 1

1

1( )

2

( )2

n n n ni i i i

ni

ni

ni

dt dtR u q u q q

dt dx dx

dtu q

dx

dt q dx qu q dx O dx

dt dx x x

q dt qq dt O dt

t t

11

________2 2

32

________2 2

32

1

1: ( )

2

( )2

ni

n ni i

ni

dtu q

dx

dt q dx qR u q dx O dx

dt dx x x

q dt qq dt O dt

t t

Estimating Truncation Error

• Removing canceling terms:

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Estimating Truncation Error

• Simplifying:________

2 23

2

________2 2

32

( )2

1:

( )2

ni

dt q dx qu dx O dx

dx x x

Rdt

q dt qdt O dt

t t

____ ____

____ ____2 2

2 22 2

:

( ) ( )2 2

ni

q qu

t x

R

dt q dx qO dt u O dx

t x

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Final Form

• Using the definition of q

____2

22

: 1 ( )2

ni

udx udt qR O dx

dx x

Using that:

____ ____2 2

22 2

q qu

t x

____ ____

____ ____2 2

2 22 2

:

( ) ( )2 2

ni

q qu

t x

R

dt q dx qO dt u O dx

t x

____ ____

2 22 2

2 2: ( ) ( )

2 2ni

dt q dx qR O dt u O dx

t x

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Interpretation of Consistency

So the truncation error is O(dx) under the assumption that dt/dx is a constant..

This essentially implies that the numerical solution diverges from the actual solution by an error of O(dx) every time step.

If we assume that the solution q is smooth enough then the truncation error converges to zero with decreasing dx. This property is known as consistency.

____2

22

: 1 ( )2

ni

udx udt qR O dx

dx x

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Error Equation• We define the error variable:

• We next define the numerical iterator N:

• Then:

• So the new error consists of the action of the numerical scheme on the previous error and the error commited in the approximation of the derivatives.

n n ni i iq E

1n nN

1 1

1

n n n n

n n n n n

n n n n

E N q E q

N q E N q N q q

N q E N q dtR

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Abstract Scheme• Without considering the specific construction of the

scheme suppose that the numerical N operator satisfies:

• i.e. N is a contraction operator in some norm then…

N P N Q P Q

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Estimating Error in Terms of Initial Error and Cumulative Truncation Error

1

1

1

1

1 1

1 0

1

triangle inequality

contraction property of

.....

n n n n n

n n n n n

n n n n n

n n n

n n n

m nn m

m

E N q E N q dtR

E N q E N q dt R

E q E q dt R N

E E dt R

E dt R dt R

E E dt R

by induction

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Error at a Time T (independent of dx,dt)

• If the method is consistent (and the actual solution is smooth enough) then:

• As dx -> 0 the initial error ->0 and consequently the numerical error at time T tends to zero with decreasing dx (and dt).

1 0

1

1 0

1,..max

m nn m

m

n m

m n

E E dt R

E E T R

1 0 nE E T O dx

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Specific Case: Stability and Consistency for the Upwind Finite Volume Scheme

• We already proved that the upwind FV scheme is consistent.

• We still need to prove stability of.

111n n n

i i i

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Stability in the Discrete 1-norm

11

11 1

11

1

11

1 1

11 1

1

01

0 1

1

1 triangle inequality

1 assuming 0 1

n n ni i i

Nn n

ii

Nn ni i

i

N Nn ni i

i i

Nn n

ii

n n

dx

dx

dx dx

dx dx

dx

So here’s the interesting story. In the case of a zero boundarycondition then we automatically observe that the operator is a contraction operator.

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Boundary Condition

• Suppose are two numerical solutions with

• Then:

• i.e. if we are spot on with the left boundary condition the N iterator is indeed a contraction.

, 0 0n n

1 10 01 1

1

n n n n n n

n n

dx

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Relaxation on Stability Condition

• Previous contraction condition on the numerical iterator N

• A less stringent condition is:

• Where alpha is a constant independent of dt as dt->0

N P N Q P Q

1N P N Q dt P Q

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Relaxation on Stability Condition

• In this case the stability analysis yields:

1

1

1

1 1

11 0

1

0

1,..

1

1 1

.....

1 1

max Ndt=T

n n n n n

n n n n n

n n n

n n n

m nn n mn m

m

T m

m n

E N q E N q dtR

E q E q dt R

E dt E dt R

dt dt E dt R dt R

E dt E dt dt R

e E T R

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Interpretation

• Relaxing the stability yields a possible exponential growth – but this growth is independent of T so if we reduce dt (and dx) then the error will decay to zero for fixed T.

1 0

1,..max , Ndt=Tn T m

m nE e E T R

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Next Lecture (6)

• Alternative flux formulations

• Alternative time stepping schemes