CE 6011

Non linear equations

in MATLAB

Lecture 7

Riddhi Singh

Email: riddhi@iith.ac.in

1

Todays outline

Non-linear equations

System of non-linear equations

2

NON-LINEAR EQUATIONS

3

A non-linear system does not follow the rules of a

linear system: additivity and homogeneity

4

An equation, f(x)=K, is linear if it satisfies the principles above,

nonlinear otherwise

additivity : ( ) ( ) ( )

homogeneity : _______________________

superposition : _______________________

f x y f x f y

5xf(

x)

x

g(x)

Nonlinear systemLinear system

Examples of non linear equations

6

2

1/3

2

( ) 1

( ) sin( )

( ) 1

( ) log( ) 2

....

f x x x

f x x

f x x

f x x

Non linear equations can have one or more than

one (system of equations) unknowns

7

In case of more than one unknown, the solutions is a _______

2( ) 1

( ) sin( )

f x x x

f x x

1. ______ unknown solution is a ___________________

2. ______________ unknown: _______________________

2

1 1 2

2 1 2

( ) 1

( ) sin( ) cos(x )

f x x x

f x x

We need to find the value of x at which:

8

( ) f x

1. One unknown solution is a scalar

2. More than one unknown: system of non-linear equations

( ) if x

Also termed, ___________

Case I. Single unknown: All method are iterative

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1. _____________ identification

2. _____________ method

3. _____________ method

4. _____________ method

5. _____________ method

Case II. Multiple unknowns in a system of non-

linear equations

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Each method uses more information about f(x)

than the previous one

Interval identification

Bisection

False position

Newton-Raphson

Secant

Signs of f(x)

Signs of f(x)

Signs and values of f(x)

Values of f(x) and its

derivatives

Values of f(x), computes

derivatives using f(x)

Method Information used

Incr

eas

ing

com

ple

xity

NON-LINEAR EQUATIONS

WITH ONE UNKNOWN

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MATLAB has inbuilt solvers to solve any type of

equation. The key command is solve

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%define an unknown

>> syms x;

%solve a nonlinear equations in x

>>solve(x^3-3*x+1)

%get the output in numbers

>>eval(solve(x^3-3*x+1))

%define a system of equations:

>> syms x1 x2 x3

>> S = solve('x1+x2+x3=3','x1^2+x2^2+x3^2=5','exp(x1)+x1*x2-x1*x3=1')

>> Sx1 = S.x1;

>> Sx2 = S.x2;

>> Sx3 = S.x3;

Method I. Interval identification is based on the

Intermediate value theorem

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If a function f is continuous on the closed interval [a,b], and if f(a)yf(b)

or f(b)yf(a), then there exists a point c such that acb and f(c)=y

If a root of f(x) lies in the interval [a,b],

then at some point r,

f(r) =_____. Therefore, f(a)and f(b) are

of ____________ signs, or

xf(

x)

a br0

( ) 0 ( )

f a f b

Method I. Interval identification algorithm*

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1. Start with a guess (eg. xo=0) and interval width x2. Let x1=xo and x2=xo + x3. While f(x1)*f(x2)>0

new x1 = x2new x2 = x1 + x

4. Root lies in the interval [x1,x2]

*Limitations:

1. Cannot detect double roots

2. Cannot detect closely spaced roots if the chosen interval

is too large

Method 2. Bisection method is an extension of

the interval identification method

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Repeat the interval identification method for successively small intervals

If a root of f(x) lies in the interval

[a,b], then construct a new interval,

c=(a+b)/2. Reevaluate:

xf(

x)

a br0

c

c

( )* ( ),

( )* ( )

f a f c and

f b f c

Method 2. Bisection method algorithm*

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1. If the root lies in the interval [x1,x2]2. While error>tolerable error

c=(x1+x2)/2 error = (x2-x1)/2If f(x1)*f(c)>0

x1=cElse

x2=c

*Limitations:

1. Slow convergence than other methods that use more

information

Method 3. False position is an extension of the

interval identification method

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Instead of using the mid-point, this method uses the point where the secant

between f(a) and f(b) intersects the x-axis

The triangles formed by the secant

are similar, therefore:

xf(

x)

a b0

( ) ( )

b c c a

f b f a

a, f(a)

b, f(b)

c

Method 3. False position algorithm (faster than

bisection)

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1. If the root lies in the interval [x1,x2]2. While error>tolerable error

c=

If f(x1)*f(c)>0x1=c

Elsex2=c

( ) ( )

( ) ( )

af b bf a

f b f a

*Limitations:

1. May select the same endpoint- in that case will converge

slowly

Method 4. Newton-Raphson approximates the

curve with a tangent

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This method assumes that the function f is differentiable.

The curve at xo is approximated

using:

xy

01

( ) ( )( ) ( )

o o o

o

l x f x x x f x

x x

tangent

y=f(x)

xox1r

Method 4. Newton-Raphson algorithm

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1. Assume an initial estimate of the root, xo2. fx=f(xo), fp=f(xo)3. While precision>prescribed_precision

x=x- fx/fpfx= f(x)fp=f(x)precision= abs(fx/fp);

*Limitations:

1. Sensitive to the nature of the function may get trapped

if the tangent to the curve is parallel to the x-axis

2. Detects a multiple root (root of the function is also a

root of its derivative) only with linear convergence

Method 5. Secant method approximates the

derivative instead of estimating them analytically

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This method is the same as Newton-Raphson except that the derivatives

are approximated

The derivative at xo is approximated

as:

xy

0

0

1

1

1

( ) ( )( ) lim

( ) ( )( )

____________ ( )

oh

n nn

n n

n n n

f x h f xf x

h

f x f xf x

x x

x x f x

secant

y=f(x)

xn-1xn+1r xn

Method 5. Secant method algorithm

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1. Assume two intial intervals in which the root will lie, x1 and x22. fx1=f(x1), fx2=f(x2)3. While precision>prescribed_precision

x1 = x2fx1 = fx2fp = (fx2-fx1)/(x2-x1)x2 = x2-fx2/fpprecision = fx2/fp

*Limitations:

1. Slower than Newton-Raphson (although faster than bisection)

2. Can diverge instead of converging

SYSTEM OF NONLINEAR

EQUATIONS

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Goal: to identify the vector, x such that:

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1 1 2

2 1 2

1 2

( , ,..., ) 0

( , ,..., ) 0

.

.

.

( , ,..., ) 0

n

n

n n

f x x x

f x x x

f x x x

Matrix notation:

( ) 0F X

The ___________ : derivatives for vectors

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In case of an unknown vector, the derivative of the function can be

estimated w.r.t each component of the vector

Then, the derivative of a function, f, w.r.t the vector

X is given by the ________ vector:

1 2[ , ,..., ]nx x xX

Suppose, vector X is defined as:

1 1 1

1 2

,...,n

F F FJ

x x x

The Jacobian for a system of equations is:

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1 1 2

2 1 2

1 2

( , ,..., ) 0

( , ,..., ) 0

.

.

.

( , ,..., ) 0

n

n

n n

f x x x

f x x x

f x x x

System Jacobian matrix

1 1

1

2 2

1

1

. . .

. . .

. .

. .

. .

. . .

n

n

n n

n

f f

x x

f f

x x

J

f f

x x

Newton-Raphson is extended to nonlinear

systems by using the matrix notation*:

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1 1

1 1

1 1

1

[ '( )] ( )

[ ] ( )

( ) [ ] ( )

( ) ( )

( )

k k k k

k k k k

k k k k

k k k k

k k k

J

J

J

J H

X X F X F X

X X F X

X X F X

X X F X

F X

Solving the Jacobian linear systems yields the value of the

vector for the next time step

*Derivation is based on _____________ expansion.

Example: Solve the following linear system using

Newton-Raphson

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2 2 2

3

5

1x

x y z

x y z

e xy xz

Step 1. Find the Jacobian (analytically)

29

? ? ?

? ? ?

? ? ?

J

Step 2. Implement Newton-Raphson for vectors

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1. Assume an initial estimate of the root vector x=[xo, x1,, xn]

2. While precision>required_precisionEstimate, F=f(x)Estimate, J=Jacobian(x)Solve: JH=-F X = X+Hprecision= norm(H);

In the same manner, the secant method can be extended to a

system of non-linear equations

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Convergence of these methods generally depends on

the required precision and initial guesses

Interval identification

Bisection

False position

Newton-Raphson

Secant

---

Method Convergence formula

Incr

eas

ing

com

ple

xity

log( ) log(2 )

log 2

b an

2

1

( )1

2 ( )

nn n

n

fe e

f x

1 1

1 ( )

2 ( )n n n

f re e e

f r

---

Linear

Linear

Quadratic

Convergence type

Super linear

1

1

( ) ( )

2 ( ) ( )( )

nn

n n

a r f re

f r f r e a r