M. Dumbser 1 / 23 Analisi Numerica Università degli Studi di Trento Dipartimento dIngegneria Civile...

M. Dumbser

1 / 23

Analisi NumericaUniversità degli Studi di TrentoDipartimento d‘Ingegneria Civile ed AmbientaleDr.-Ing. Michael Dumbser

Lecture on Numerical Analysis

Dr.-Ing. Michael Dumbser

08 / 11 / 2007

M. Dumbser

2 / 23


Numerical Solution of Nonlinear Equations

0)( xf

Task: compute approximately the solution x of the nonlinear algebraic equation

If the problem is not in form (1), then it can always be cast into form (1) by putting all terms on the left hand side.

Examples:

(1)

2

2

2

)sin()exp()(

0)sin()exp(

)sin()exp(

xxxxf

xxx

xxx

)()()()(

)()()(

xexhxgxf

xexhxg

M. Dumbser

3 / 23


Numerical Solution of Nonlinear Equations

0)( xf

Task: compute approximately the solution x of the nonlinear algebraic equation

(1)

Two problems present even at the analytical level:

• For general functions f(x), there may be no solution at all. Example:

• For general functions f(x), there may be more than one solution. Example:

02)sin()( xxf

01)( 2 xxf 1,1 21 xx

M. Dumbser

4 / 23


x0 x2 x3 x1

x

f

x4

(I) Before we can start using this method, we must guess two points x0 and x1 where the function f(x) has opposite sign, i.e.

The Regula Falsi Method

0)()( 10 xfxf

(II) Now connect the two points (x0, f(x0)) and (x1, f(x1)) with a straight line and compute the intersection of this line with the x-axis, denoted by xi. The point xi is given by

10

1010

ff

fxxfxi

(III) Compute the sign of f(x2) and compare against the signs of f(x0) and f(x1). Then update the interval boundaries according to the result, i.e. set

0)()(if: 00 ii xfxfxx 0)()(if: 11 ii xfxfxx If |f(xi)| < tolerance then STOP (in Maple: break), otherwise go back to step (II).

M. Dumbser

5 / 23


x0 x2 x3 x1

x

f

x4

Advantages: - it is guaranteed that the method finds a solution of f(x)=0 because we already know that it

the solution is contained in the initial interval [x0;x1]. - The method is very robust and reliable

The Regula Falsi Method

Disadvantage: - it can not be applied if we do not have an initial guess of the interval [x0;x1].

M. Dumbser

6 / 23


x0x2 x1

x

f

x3

• Variant of the regula falsi method: we start from any two points x0 and x1

The Secant Method

• Advantage: no initial interval must be guessed

• Disadvantage: may not always converge towards a solution !

M. Dumbser

7 / 23


x0 x1

x

f

• Variant of the regula falsi method: we start from any two points x0 and x1

The Secant Method

• Advantage: no initial interval must be guessed


M. Dumbser

8 / 23


x0x1

x

f

• Differential form of the secant method. Instead of using the secant, Newton proposes to use the tangent!

Newton‘s Method

• Advantages: - no initial interval must be guessed - very fast


x2

M. Dumbser

9 / 23


x0x1

x

f

Newton‘s Method

x2

• Differential form of the secant method. Instead of using the secant, Newton proposes to use the tangent!

0)(

ii xxx

fxf

x

fxfx i

/)(

xxx ii 1

ixxx :

(1) If |f(xi)| < tolerance then STOP (in Maple/MATLAB: break) or(2) If | x | < tolerance then STOP

M. Dumbser

10 / 23


Exercise 1

Write a Maple worksheet and a MATLAB script that solve the nonlinear equation

using the regula falsi method. Choose an appropriate starting interval.

Exercise 2

Now write a MATLAB script that solves the problem of exercise 1 using the Newton method. The user should have the possibility to choose the starting guess, the maximum number of iterations as well as the tolerance.

0)sin()exp()( 2 xxxxf

M. Dumbser

11 / 23


Globally Convergent Newton Methods

From the previous Example 4 we find that even for strictly monotone functions f(x) Newton‘s method may not always converge from arbitrary starting points x0, but only from starting points that are sufficiently close to the solution of the problem. However, if the starting point is sufficiently close, the method converges. We therefore say the original Newton method is only locally convergent.

The local convergence behaviour of Newton‘s method and the possible failure to converge for initial guesses far away from the solution can be mainly explained by too large step sizes x:

M. Dumbser

12 / 23



As seen in the sketch of the previous slide, the absolute value of the function f(x) slightly increased during one Newton step. To avoid this problem, we therefore add the following requirement: The function |f(x)| must always decrease during one Newton step, i.e.:

)()( iii xfxxf

x

fxfx i

/)(

xxx ii 1

If this is not fulfilled, the step size is sucessively reduced until the criterion is satisfied. The globally convergent Newton method then reads as follows:

)()( iii xfxxf 1

10

WHILE

END

DO i=0,1,…

UNTIL tolxf i )(

Free parameter. Choose e.g. = 0.5

M. Dumbser

13 / 23



Exercise 5

Write a modified MATLAB script GlobalNewton.m to solve the nonlinear equation

with the globally convergent Newton method.

0)1arctan()( xxf

1) Use x0 = 2 as initial guess.

2) Use x0 = 2.5 as initial guess. What do you realize?

3) Use x0 = 106 as initial guess. Does the method converge?

M. Dumbser

14 / 23


Computation of the Roots of a Polynomial

We now apply Newton‘s method to a special case of nonlinear algebraic equation where thefunction f(x) is a polynomial of degree n:

The values x which satisfy eqn. (1) are called the roots of the polynomial pn(x).We suppose in the following that all the roots of the polynomial are distinct and real.

0)()(0

n

i

iin xaxpxf

Suppose x = is a root of pn(x), i.e.

then we have

If we knew all the roots i we would have the following expression of the polynomial:

)()()( 1 xpxxp nn

0)( np

))...()(()( 21 nn xxxxp

M. Dumbser

15 / 23



A simple and straightforward application of Newton‘s method now consists in finding a firstroot 1 (typically the largest one) of the polynomial pn and then compute the largest root 2 of the polynomial pn-1 obtained by dividing the polynomial pn by ( x – ):

11

)()(

x

xpxp n

n

After having found the root 2 of pn-1 we divide again

))((

)()()(

212

12

xx

xp

x

xpxp nn

n

and compute the next root 3 and so on, until all the n roots of pn have been found. In general, we have:

))...((

)()(

1 j

njn xx

xpxp

j

i ij

n

j

nxjn xxx

xp

xx

xpxp

x 111

1

))...((

)(

))...((

)()(

M. Dumbser

16 / 23



To express the polynomial pn-j(x) and its derivative, two simple MATLAB loops defining the auxiliary variables Dj and Sj are necessary:

j

njn D

xpxp

)()(

jj

n

j

nxjn S

D

xp

D

xpxp

x

)()(

)(

END

)(*

j]:[1k FOR

1

kjj

j

xDD

D

with

END

)/(1

j]:[1k FOR

0

kjj

j

xSS

S

jnnx

njnijn Spp

pp

xxp

x

fxfx

/)(/)(

We then obtain the so-called Newton-Maehly method:

M. Dumbser

17 / 23


Exercise 3

(1) Write a MATLAB script PolyRoot.m that finds all the roots of a general polynomial of degree n. The user should be able to specify the coefficients a of the polynomial. Then, two MATLAB function files, called Polynomial.m and DPolynomial.m compute the polynomial and its first derivative in point x using the coefficients a. Further user inputs required are tol, NMAX and x0.

(2) Apply your general MATLAB script to the particular polynomial

(3) Now try your MATLAB program on the polynomial

Hint: Use Maple to rewrite this polynomial in standard form

2

1

4

3x

4

73)( 234 xxxxp

12

0

)2()(j

jxxp

n

j

jjxaxp

0

)(

M. Dumbser

18 / 23


Efficient Evaluation of Polynomials

The standard format of defining a polynomial is computationally not the most efficient way toevaluate a polynomial:

...)( 43210 xxxxaxxxaxxaxaaxp

A computationally more efficient but mathematically fully equivalent way of writing the polynomialp(x) is the so called Horner scheme, which uses the distributive law:

...)( 44

33

2210 xaxaxaxaaxp

In this example, we have 4 additions and 10 multiplications

...)( 43210 axaxaxaxaxp

Now, we still have 4 additions but only 4 multiplications!

M. Dumbser

19 / 23



Exercise 4

Now try your original MATLAB script Newton.m to solve the nonlinear equation

0)1arctan()( xxf

1) Use x0 = 2 as initial guess.

2) Use x0 = 2.5 as initial guess. What do you realize?

M. Dumbser

20 / 23


Newton‘s Method for Systems of Equations

0)(

ii xxx

fxf

)(1ixfDx

xxx ii

1

ixxx

:

(1) If ||f(xi)|| < tolerance then STOP (in Maple/MATLAB: break) or(2) If || x || < tolerance then STOP

The extension of Newton‘s method to systems of nonlinear algebraic equations with several unknown variables is straightforward.

0)( xf

The problem is written in standard form in matrix-vector notation as follows:

Linearization (multi-dimensional Taylor series up to first order) leads to:

j

iij x

f

x

fDD

:

The matrix D is called the Jacobian matrix of the vector function f with respect to ist vector argument x and is the multi-dimensional equivalent to the first derivative of a scalar function.

This corresponds to solving the linear system )( ixfxD

i.e. the vector norm of f is the so-called function residual.

222

21 ...)( nfffxf

M. Dumbser

21 / 23


Exercise 6

Write a MATLAB script NewtonSys.m that solves the nonlinear system of equations

using the Newton method for systems. The user should have the possibility to choose the initial guess for the vector of unknowns, the maximum number of iterations as well as the tolerance. The vector function and its Jacobian matrix should be implemented in two separate .m files, called FuncSys.m and DFuncSys.m.

021)2sin()1exp()2,1( 221 xxxxxxf

021)2sin()1cos()2,1(2 xxxxxxf

M. Dumbser

22 / 23


Globally Convergent Newton‘s Method for Systems

The extension of the globally convergent Newton‘s method to systems of nonlinear algebraic equations with several unknown variables is done as in the scalar case:

)(

1

ixfx

fx

xxx kii

1

)()( iiki xfxxf

10

10

WHILE

kk 1

END

DO i=0,1,…

UNTIL tolxf i )(

0k

1kk

M. Dumbser

23 / 23


Exercise 7

We want to solve the following nonlinear system of equations:

(1) Use the classical Newton method for systems starting from [0,0]

(2) Use the classical Newton method for systems starting from [1000,1000]

(3) Can you find a starting point from which the classical Newton method converges?

(4) Construct a globally convergent Newton method for systems and start again from the points [0,0] and [1000,1000]. Does the global method even converge from the starting point [1e6,1e6] ?

0)arctan()1arctan(),( yxyxf

03

1)exp(),( 2 yxyxg

Date post:	26-Mar-2015
Category:	Documents
Upload:	jesus-silva
View:	216 times
Download:	0 times

M. Dumbser 1 / 23 Analisi Numerica Università degli Studi di Trento Dipartimento dIngegneria Civile...

Documents