Numerical 2

7/30/2019 Numerical 2

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Numerical Methods

Marisa Villano, Tom Fagan,

Dave Fairburn, Chris Savino,

David Goldberg, Daniel Rave


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An Overview

The Method of Finite DifferencesError Approximations and DangersApproxmations to DiffusionsCrank Nicholson SchemeStability Criterion


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Finite Differences

Best known numerical method ofapproximation

Marisa Villano


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Finite Differences

Approximating the derivative with adifference quotient from the Taylor series

Function of One VariableChoose mesh size x

Then u j ~ u( j x)


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First Derivative Approximations

Backward difference: ( u j u j-1) / x

Forward difference: ( u j+1 u j) / x

Centered difference: ( u j+1 u j-1) / 2 x


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Taylor Expansionu( x + x) = u( x) + u ( x) x + 1/2 u ( x)( x)

+ 1/6 u ( x)( x) + O( x)

u( x x) = u( x) u ( x) x + 1/2 u ( x)( x)- 1/6 u ( x)( x) + O( x)

2

3

4

4

2

3


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Taylor Expansionu ( x) = u( x) u( x x) + O( x)

x u ( x) = u( x + x) u( x) + O( x)

x u ( x) = u( x + x) u( x x) + O( x)

2 x

2


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Second Derivative Approximation

Centered difference: ( u j+1 2u j + u j-1) / ( x)

Taylor Expansionu ( x) = u( x + x) 2u( x) + u( x x) + O( x)

( x)

2

2

2


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Function of Two Variables

u( j x, nt ) ~ u j Backward difference for t and x

( j x, nt ) ~ (u j u j ) / t

( j x, nt ) ~ (u j u j ) / x

n

n n-1

n-1

n

u t

u x


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Forward difference for t and x

( j x, nt ) ~ (u j u j ) / t

( j x, nt ) ~ (u j u j ) / x

n+1

n+1 n

nu t

u x


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Centered difference for t and x

( j x, nt ) ~ (u j u j ) / (2 t )

( j x, nt ) ~ (u j u j ) / (2 x)

n+1

n+1 n-1

n-1u t

u x


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Error Truncation Error: introduced in the solution by theapproximation of the derivative

Local Error: from each term of the equationGlobal Error: from the accumulation of localerror

Roundoff Error: introduced in the computation bythe finite number of digits used by the computer


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Example from 8.1

Consider u t = u xx u(x,0) = h(x)We will use the finite difference method toapproximate the solutionForward difference for u tCentered difference for u xxRe-write equation in terms of the finitedifference approximations


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Finite Difference Eqn.

u jn+1 - u jn = u n j+1 - 2u jn + u n j-1

t x( ) 2

Error: The local truncation error is O( t)

from the left hand side and is O( x) 2 from

the right hand side.


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Assumptions

Assume that we choose a small change inx, and that the denominator on both sides

of the equation are equal.We are now left with the scheme:

u jn+1 = u n j+1 - u n j + u n j-1Solving u with this scheme is now easy todo once we have the initial data.


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Initial Data

Let u(x,0) = h(x) = a step function withthe following properties:

h(x) = 0 for all j except for j = 5, soh j = 0 0 0 0 1 0 0 0 0 0 0 .

Initially, only a certain section, which is

at j = 5 is equal to the value of 1. j serves as the counter for the xvalues.


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How to solve?We know u 0 j = 1 at j = 5 and 0 at all other jinitially (given by superscript 0).We can plug into our scheme to solve for u 1

jat

all js. u1 j = u 0 j-1 - u 0 j + u 0 j+1u15 = -1; u 14 = 1; u 16 = 1

Now we can continue to increase the # ofiterations, n, and create a table


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Solution for 4 iterations4 1 -4 10 - 16 19 -16 10 -4 1 0

3 0 1 -3 6 -7 6 -3 1 0 0

2 0 0 1 -2 3 -2 1 0 0 0

1 0 0 0 1 -1 1 0 0 0 0

0 0 0 0 0 1 0 0 0 0 0

1 2 3 4 5 6 7 8 9 10

j values

n-valu

es


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Analysis of SolutionIs this solution viable?Maximum principle states that the solution must

be between 0 and 1 given our initial dataAt n = 4, our solution has already ballooned to u= 19!Clearly, there are cases when the finitedifference method can pose serious problems.


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Charting the ErrorAssume the solution is constant and equal to 0.5 (halfway betweenthe possible 0 and 1)


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Lessons Learned

While the finite difference method is easyand convenient to use in many cases,

there are some dangers associated withthe method.We will investigate why the assumption

that allowed us to simplify the schemecould have been a major contributor tothe large error.


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Approximations of Diffusions

Neumann Boundary Conditions

and the Crank-Nicolson Scheme

Chris Savino


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Approximations of DiffusionsErrors have accumulated from theapproximations of the derivatives using theprevious schemeThe problem is the choice of the mesh tto the mesh x Let s=

can solve scheme

2( )t

x

11 1( ) (1 2 )n n n n j j j ju s u u s u


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Neumann Boundary Conditions0 x l

Simplest Approximations are

(0, ) ( ) xu t g t ( , ) ( ) xu l t h t

1 0n n nu u g x

1n n j j nu u h x


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To get smallest error, we use centereddifferences for the derivatives on the boundaryIntroduce ghost points

Boundary Conditions become1

nu 1n

ju

1 1

2

n nnu u

g x1 1

2

n n j j nu u

h x


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Crank-Nicolson Scheme

Can avoid any restrictions on stabilityconditions

Unconditionally stable no matter what thevalue of s is.


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Centered Second Difference:

Pick a number theta between 0 and 1Theta scheme:

1 122

2( )( )

n n n j j j

n ju u u

u x

12 2 1(1 )( ) ( )

n n j j n n

j ju u

u ut


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We analyze the scheme by plugging in aseparated solution

Therefore

1 2(1 ) (1 cos )( )

1 2 (1 cos )

s k xk

s k x

( ) ( ( ))n ik x j n ju e k


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Must Check stability condition

If then

Therefore

is always true

( ) 1k

( ) 1k (1 2 )(1 cos ) 1 s k x

1 2 0


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If then there is no restriction on the sizeof s for stability to holdThe scheme is unconditionally stableWhen theta = it is called the Crank-NicolsonschemeIf theta < then the scheme is stable if

11

2

2

1( ) 2 4

t s

x


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Stability Criterion

Approximations of the diffusion

equation, u t=u xx

David Goldberg


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Stability Criterion

The method of finite differences gives ananswer, but it does not guarantee that thisanswer is meaningful.Values must be chosen appropriately, toensure that the results make sense andare applicable to real world scenarios.This condition, that values must satisfy inorder to be worthwhile, is called thestability criterion.


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Example

As per the book, take, for instance, thediffusion problem:

, 0 = = in 0, 2 in 2 ,

= for 00

= 0 at x=0, , that is0, = , = 0


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Example, continued

As can be easily shown, the graph of (x)looks like this.

0

0.2

0.4

0.6

0.8

1

1.2

1.41.6

1.8

0 0.5 1 1.5 2 2.5 3 3.5


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Example, continued

In attempting to use the method of finitedifferences, we are using a forwarddifference for u

t and a centered difference

for u xx.This means that

It is important to note here that thesuperscript n denotes a counter on the tvariable, and the subscript j denotes a

counter on the x variable.

+1

= +1 2 + 1

2


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Example, continued

In order to make the calculations a bitcleaner, we are introducing a variable, s ,which is defined by

Rearranging, we have

It would be nice if we could just plug in

values and get a valid result

2( )

t

x s

11 1

11 1

11 1

2

2

1 2

n n n n n j j j j j

n n n n n j j j j j

n n n n j j j j

u s u u u u

u s u s u u s u

u s u u s u


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Example, continued

However, putting in different values can lead to theresults being close to, or far from, that actual answer.For instance, letting x= /20, and letting s=5/11, we geta relatively nice result. Letting s=5/9 does not get sucha nice result.

So what, of significance, changes?

0

0.5

1

1.5

2

0 1 2 3 40

0.5

1

1.5

2

0 1 2 3 4


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Example, Continued

As it turns out, changing the value of s cansignificantly change the validity of the

solution. To see why, we return to ourequation. 1 1 1

1 11

...

1 2

Seperate variables

So, by combining like terms,

1 2

n n n n j j j j

n j j n

j jn

n j

u s u u s u

u XT u X T

X X T s s

T X


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Example, continuedSince the left hand side is a function of Tand the right hand side is a function of X,they must be equal to a constant.

1 11

11 0

1 1

...

1 2

and also

1 2

j jn

n j

nnn n n

n

j j

j

X X T s s

T X

T T T T T

T

X X s s

X


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Example, continued

This is a discrete version of an ODE,which when solved gives

0

...

1 2 ( )1 2 2 cos( )

Since, as discovered before,

if 1, T will grow without bound.

By above,

1 4 1

1So 1 4 1,

2

ik x ik x

nn

s s e e s s k x

T T

s

s s


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Example, finished

Thus, to achieve stability, . This iswhy setting s=5/9 didnt give a valid

result.It is to be noted that usually the necessarycriterion is that , but

that in this case it was irrelevant.So the stability criterion must be workedout before one can effectively use the

method of finite differences.

1 ( ) instead of 1O t

212

t

x


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Approximations of Diffusions

Example from 8.2

Daniel Rave


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Summary

Breif Review of Methods

Wide Applicability

Importance of Stability

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