Bvp Multiple Shooting

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ReviewMultiple Shooting

Richardson ExtrapolationSummary

BVPS – MULTIPLE SHOOTING TECHNIQUE

Dr. Johnson

School of Mathematics

Semester 1 2008

Dr. Johnson MATH65241

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OUTLINE

1 REVIEW

2 MULTIPLE SHOOTING

High Order ProblemsCorrecting the guess

3 RICHARDSON EXTRAPOLATION

Exploiting our knowledge of errors

4 SUMMARY


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OUTLINE

1 REVIEW

2 MULTIPLE SHOOTING




4 SUMMARY


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OUTLINE

1 REVIEW

2 MULTIPLE SHOOTING




4 SUMMARY


R i

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BOUNDARY VALUE PROBLEMS

Boundary value problems are commonplace in CFD andmathematics in general

Newton’s Shooting method:

gn+1 = gn −φ( gn)

φ′( gn)

Can use the Secant method to generate φ′,

or solve the augmented system to generate φ′ and useNewton’s method

Newton’s method is quadratic, so should converge in 5-15iterations


R i

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BOUNDARY VALUE PROBLEMS

Boundary value problems are commonplace in CFD andmathematics in general

Newton’s Shooting method:

gn+1 = gn −φ( gn)

φ′( gn)

Can use the Secant method to generate φ′,

or solve the augmented system to generate φ′ and useNewton’s method

Newton’s method is quadratic, so should converge in 5-15iterations


Review

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OUTLINE

1 REVIEW

2 MULTIPLE SHOOTING




4 SUMMARY


Review

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A FOURTH ORDER PROBLEM

Consider the problem:

d4 y

dx4

= y3 −dy

dx

2

,

with the boundary conditions

y(0) = 1, y′(0) = 0, y(1) = 2, and y′(1) = 1

We have two starting conditions at x = 0,

and two final conditions at x = 1.

Therefore we need to guess y′′′ and y(IV ) at x = 0


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INITIAL CONDITIONS

Let us define

Y1 = y, Y2 = y′

, Y3 = y′′

, and Y4 = y′′′

then our initial conditions will look like

Y1 = 1, Y2 = 0, Y3 = e, and Y4 = g

where e and g are guesses


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Multiple ShootingRichardson Extrapolation

Summary


INITIAL CONDITIONS

Let us define

Y1 = y, Y2 = y′

, Y3 = y′′

, and Y4 = y′′′

then our initial conditions will look like

Y1 = 1, Y2 = 0, Y3 = e, and Y4 = g

where e and g are guesses


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Summary


SATISFYING THE BOUNDARY CONDITIONS

We now have two conditions to satisfy, and hence twofunctions

φ1(e, g) =Y1(x = 1; e, g) − 2

φ2(e, g) =Y2(x = 1; e, g) − 1

We need to iterate on both e and g to ensure both φ1 = 0and φ2 = 0

To proceed we need to use a Taylor expansion for afunction of two variables

f (x + a, y + b) = f (x, y) + a∂ f

∂x+ b

∂ f

∂ y+ R1(x, y)


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Summary


SATISFYING THE BOUNDARY CONDITIONS

We now have two conditions to satisfy, and hence twofunctions

φ1(e, g) =Y1(x = 1; e, g) − 2

φ2(e, g) =Y2(x = 1; e, g) − 1

We need to iterate on both e and g to ensure both φ1 = 0and φ2 = 0

To proceed we need to use a Taylor expansion for afunction of two variables

f (x + a, y + b) = f (x, y) + a∂ f

∂x+ b

∂ f

∂ y+ R1(x, y)


Reviewl i l h i i h d bl

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Summary


OUTLINE

1 REVIEW

2 MULTIPLE SHOOTING




4 SUMMARY


ReviewM lti l Sh ti Hi h O d P bl

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Summary


TAKING A GUESS

Then let us perform a Taylor expansion around the guessese and ˆ g to get

φ1(e + de, ˆ g + dg) =φ1(e, ˆ g) + de

∂φ1

∂e + dg

∂φ1

∂ˆ g + R1(e, ˆ g)

φ2(e + de, ˆ g + dg) =φ2(e, ˆ g) + de∂φ2

∂e+ dg

∂φ2

∂ˆ g+ R1(e, ˆ g)


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Summary


TAKING A GUESS


φ1(e + de, ˆ g + dg) =φ1(e, ˆ g) + de

∂φ1

∂e + dg

∂φ1

∂ˆ g + R1(e, ˆ g)

φ2(e + de, ˆ g + dg) =φ2(e, ˆ g) + de∂φ2

∂e+ dg

∂φ2

∂ˆ g+ R1(e, ˆ g)

Set both φ(e + de, ˆ g + dg) = 0, and the remainder R1 = 0.


ReviewMultiple Shooting High Order Problems

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Summary


TAKING A GUESS


∂φ1

∂e de +

∂φ1

∂ˆ g dg = − φ1(e, ˆ g)

∂φ2

∂ede +

∂φ2

∂ˆ gdg = − φ2(e, ˆ g)

and we are left with simultaneous equations for de and dg.In matrix form this can be written as:

J( g )dg = −φ( g )


ReviewMultiple Shooting High Order Problems

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Summary


THE SOLUTION

Given the equations in matrix form,

J( g )dg = −φ( g )where g is a vector of guesses, φ is the vector of conditionsand J is the Jacobian matrix.

We can write the correction as

dg = − J−1( g )φ( g ).



l k l d f

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Summary


OUTLINE

1 REVIEW

2 MULTIPLE SHOOTING




4 SUMMARY



E l iti k l d f

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Summary


TRUNCATION ERRORS

Suppose that we use a method with truncation error of O(hm) to compute an approximation wi.

We can exploit the way in which our approximationconverges towards the solution

0.63

0.64

0.65

0.66

0.67

0.68

0.69

0.7

0.71

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07

w n

h

The most common technique is Richardson extrapolation.



E l iti k l d f

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p gRichardson Extrapolation

Summary


TRUNCATION ERRORS

Suppose that we use a method with truncation error of O(hm) to compute an approximation wi.

We can exploit the way in which our approximationconverges towards the solution

0.63

0.64

0.65

0.66

0.67

0.68

0.69

0.7

0.71

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07

w n

h

The most common technique is Richardson extrapolation.




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p gRichardson Extrapolation

Summary


TWO APPROXIMATIONS ARE BETTER THAN ONE

Suppose w(1)n is an approximation with step size h

and w(2)n with step size 2h

Then we can write

w(1)n = yi + E0hm + E1hm+1 + . . .

and

w(2)n = yi + E0(2h)m + E1(2h)m+1 + . . .




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Suppose w(1)n is an approximation with step size h

and w(2)n with step size 2h

Then we can write

w(1)n = yi + E0hm + E1hm+1 + . . .

and

w(2)n = yi + E0(2h)m + E1(2h)m+1 + . . .



h d lExploiting our knowledge of errors

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So eliminating the first term E0 we have

2mw(1)n − w

(2)n = (2m − 1) yi + O(hm+1)

thenw∗

n =2mw

(1)n − w

(2)n

2m − 1= yi + O(hm+1)

and the approximation w∗n is better than w

(1)n or w

(2)n .

For a fourth order method such as RK4 the above gives

w∗n =

16w(1)n − w

(2)n

15



Ri h d E t l tiExploiting our knowledge of errors

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p g g

EXAMPLE - EULER’S METHOD

Solve y′ = x + sin( y) with y(0) = 0 at x = 1

n h wn w∗n = 2w

(1)n − w

(2)n |w

(1)n − w

(2)n | ratio

16 0.0625 0.633632 0.03125 0.6709 0.7083 0.0373

64 0.01562 0.6902 0.7095 0.0192 1.9370128 0.00781 0.7000 0.7098 0.0097 1.9687256 0.00390 0.7049 0.7098 0.0049 1.9844512 0.00195 0.7074 0.7099 0.0024 1.9922

1024 0.00097 0.7086 0.7099 0.0012 1.9961

We get similar accuracy extrapolating with n = 16 andn = 32, as we do by taking n = 1024

Extrapolation will not always work this well! It dependson how smooth the errors are.



Richardson ExtrapolationExploiting our knowledge of errors

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p g g

MORE GENERAL RICHARDSON EXTRAPOLATION

Let A(h) and A

hk

be approximations to Y that depends

on the step size h for some k

A(h) = Y + a0hm + . . . , Ah

k

= Y + a0h

km

+ . . .

then we may write the general approximation A∗(h) as

A∗(h) = km

A h

k− A(h)

km − 1

We may perform further extrapolations on theextrapolated results to eliminate higher order errors



Richardson ExtrapolationExploiting our knowledge of errors

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MORE GENERAL RICHARDSON EXTRAPOLATION

Let A(h) and A

hk

be approximations to Y that depends

on the step size h for some k

A(h) = Y + a0hm + . . . , Ah

k

= Y + a0h

km

+ . . .

then we may write the general approximation A∗(h) as

A∗(h) = km

A h

k− A(h)

km − 1

We may perform further extrapolations on theextrapolated results to eliminate higher order errors



Richardson Extrapolation

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SUMMARY

If we have a vector of guesses, the correction can bewritten as:

dg = − J−1( g )φ( g )

We can use either the Secant method or solve theaugmented system to generate the Jacobian J.

If we know the order of a scheme, we can exploit it togenerate better approximations at low expense.

The general formula for Richardson extrapolation is:

A∗(h) =km A

hk

− A(h)

km − 1



Richardson Extrapolation

http://find/

http://goback/




SUMMARY

If we have a vector of guesses, the correction can bewritten as:

dg = − J−1( g )φ( g )

We can use either the Secant method or solve theaugmented system to generate the Jacobian J.

If we know the order of a scheme, we can exploit it togenerate better approximations at low expense.

The general formula for Richardson extrapolation is:

A∗(h) =km A

hk

− A(h)

km − 1


http://find/

http://goback/

Bvp Multiple Shooting

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