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Introduction to Simulation - Lecture 8

1-D Nonlinear Solution Methods

Jacob White

Thanks to Deepak Ramaswamy Jaime Peraire, Michal

Rewienski, and Karen Veroy

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Outline

Nonlinear Problems

Struts and Circuit Example Richardson and Linear Convergence

Simple Linear Example

Newtons Method Derivation of Newton

Quadratic Convergence

Examples

Global Convergence

Convergence Checks

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Nonlinear

problemsStrut Example

0 0( , )x y 1 1( , )x y

Given: x0, y0, x1, y1, WFind: x2, y2

2 2( , )x y

Load forceW

0yf W Need to Solve 0xf

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SMA-HPC 2003 MIT

Nonlinear Problems

Struts Example

Reminder: Strut Forces

1 1,x y

Y 0 00

1

1

2 2

1 1

c

x

y

L Lf EA L LL

xf f

L

yf f

L

L x y

L

yff

xf

(0,0)

X

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Nonlinear

problemsStrut Example

2 01 1

1

( )x ox xf L LL

2 2

1 2 0 2 0L x x y y

Load forceW

0 0( , )x y 1 1( , )x y

2 2

2 2 1 2 1L x x y y

2 12 2

2

( )x ox x

f L L

L

2 2( , )x y

1f2f

1 2 0x xf f 1 2 0y yf f W

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Nonlinear

problems

Strut Example

Why Nonlinear?

2 12

2( )o

y y

L LL

2 0

11

( ) 0o

y yL L W

L

The strut forces change

in both magnitude and

direction

Pull Hard on the

Struts

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Circuit ExampleNonlinear

problems

Need to Solve

0d rI I

+-

-

+

( 1) 0d

t

VV

d sI I e 10v

10

dV

1v 2v 1 010

r rI V

0srcv r

I I

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Solve IterativelyNonlinear

problems

Hard to find analytical solution for ( ) 0f x

Solve iterativelyguess at a solution

repeat for k = 0, 1, 2, .

until

Ask

Does the iteration converge to correct solution ?

How fast does the iteration converge?

0

0x x

1k kx W x 1 0kf x

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DefinitionRichardson

Iteration

Richardson Iteration Definition

1 ( )k k kx x f x

An iteration stationary point is a solution1k kx x

( ) 0k

f x * ( )kx x Solution

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Example 1Richardson

Iteration

( ) 0.7 10f x x

Start with 0 0x 1 0 0

2 1 1

2 2

3

3 3

4

( ) 10( ) 13

( ) 13.9( ) 14.17

x x f xx x f x

x x f xx x f x

5

6

7

8

14.2514.27

14.2814.28

xx

xx

Converged

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Example 1Richardson

Iteration( ) 0.7 10f x x

*kx x

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Example 2Richardson

Iteration

( ) 2 10f x x

Start with 0 0x

1 0 0

2 1 1

3 2 2

4 3 3

( ) 10( ) 40

( ) 130

( ) 400

x x f xx x f x

x x f x

x x f x

No convergence !

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Convergence

Setup

Richardson

Iteration

1

( )

k k k

x x f x

* * *

0

( )x x f x

Iteration Equation

Exact Solution

Computing Differences

1 * * *

( ) ( )

k k k

x x x x f x f x

Need to Estimate

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Convergence

Mean Value Theorem

Richardson

Iteration

( ) f vf v f y v yx

,v v y

v y

f

v

x

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Convergence

Use MVT

Richardson

Iteration

Iteration Equation

Computing Differences

1

( )

k k k

x x f x

Exact Solution

* * *

0

( )x x f x

1 * * *( ) ( )k k kx x x x f x f x

*1 k

f xx x

x

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Convergence

Richardson Theorem

Richardson

Iteration

*1 1 for all s.t.

f x

x x xx

If0 *x x And

Then 1 * *k kx x x x

Or 1 * 0 *lim lim 0k kk kx x x x

Linear Convergence

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Example 1Richardson

Iteration( ) 0.7 10f x x

*kx x

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ProblemsRichardson

Iteration

Convergence is only linear

x, f(x) not in the same units:x is a voltage,f(x) a current in circuits

x is a displacement,f(x) a force in struts Adding 2 different physical quantities

But a Simple Algorithm Just calculatef(x) and update

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Newtons method

Another approach

From the Taylor series about solution

* *0 ( ) ( ) ( ) ( )k k kdff x f x x x xdx

Define iteration

Do k = 0 to . 11

1

( ) ( )

( )

k k k k

k

dfx x x f x

dx

dfif x exists

dx

until convergence

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Graphically

Newtons Method

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ExampleNewtons Method

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ExampleNewtons Method

*kx x

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ConvergenceNewtons Method

2* * * 2

2

*

0 ( ) ( ) ( )( ) ( )( )

some [ , ]

k k k k

k

df d f f x f x x x x x x x

dx dx

x x x

Mean Value theorem

truncates Taylor series

But10 ( ) ( )( )k k k

dff x x x x

dx

k by Newtondefinition

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Contd.

21 * * 2

2( )( ) ( )( )k k k

df d f x x x x x x

dx d x

Subtracting

21 * 1 * 2

2( ) [ ( )] ( )( )k k k

df d f x x x x x x

dx d x

Dividing through

1 2

2Suppose ( ) ( ) for all

df d f x x L x

dx d x

21 * *then k kx x L x x

Convergence is quadratic if L is bounded

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Newtons Method Convergence

Example 1

21

2 21 * * *

* 2

2

1 *

*

( ) 2

( ) 1 0,

2 ( ) 1

2 ( ) 2 ( )

1( ) (

( 1

)2

)

k k

k k k k

k k k k k

k k

k

dfx x

dx

x x x x

x x x x x x x

f x x fin

x

d x x

or x x x xx

Convergence is quadratic

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Example 2

1 2

1 *

* *

1

2 *

( ) 2

2 ( 0) ( 0)1

0 0

( ) 0,

for 02

1( ) ( )

2

0

k k

k k k

k k k

k k

df x xdx

x x x

x x x x

or x x x

f x

x

x x

Convergence is linear

Note : not bounded

away from zero

1

dfdx

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Examples 1 , 2

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*0

*

if 1

then converges tok

L x x

x x

Proof * * *

1 0 0

* *

1 0

* * *

2 0 1

* 2 * 3 *

2 1 0

* 4 * 7 *

3 2 0

( )x x L x x x x

x x x x

x x L x x x x

or x x x x x x

x x x x x x

1 2

2Suppose ( ) ( ) for all

df d f x x L x

dx d x

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Theorem

2

2If is bounded ( bounded away from zero ; bounded)

then Newton's method is guaranteed to converge given a "close

enough" guess

df d f L

dx dx

Always converges ?

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SMA-HPC 2003 MIT


Example

Convergence Depends on a Good Initial Guess

X

f(x)

0

x

1x2

x

0

x

1x

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Convergence

Convergence Checks

Newtons Method

Need a "delta-x" check to avoid false convergence

X

f(x)

1kx kx *x

1 ak ff x

1 1

a r

k k k

x xx x x

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Convergence

Convergence Checks

Newtons Method

Also need an " " check to avoid false convergencef x

X

f(x)

1kx kx

*x

1

a

k

ff x

1 1

a r

k k k

x xx x x

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Summary

Nonlinear Problems

Struts and Circuit Example

Richardson and Linear Convergence

Simple Linear Example

1-D Newtons Method

Derivation of Newton

Quadratic Convergence

Examples

Global Convergence Convergence Checks

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