maths final ppt

G. H. Patel College of Engineering & Technology

Branch : Information Technology

Name Of Subject : Numerical Methods For Computer Engineering

Subject Code : 2710210

Name Of Unit : Ordinary differential equations

Topic : Euler’s method and Runge-Kutta methods

• Guided By : Prof. Krupal Parikh

• Submitted By : Mira Y Patel (140110723006)

Euler’s Method Derivation of Euler’s method :

At x = 0, we are given the value of y = y₀. Let us

call x = 0 as x₀. Now since we know the slope of y

with respect to x, that is, f(x,y) , then at x =x₀, the

slope is f(x₀,y₀). Both x₀ and y₀ are known from the

initial condition y(x₀) = y₀.


Φ

Step size, h

x

y

x0,y0

True value

y1,

Predictedvalue

00,, yyyxfdx

dy

Run

Rise

01

01

xx

yy

00 , yxf

010001 , xxyxfyy

hyxfy 000 ,

Slope

Figure 1 Graphical interpretation of the first step of Euler’s method


Φ

Step size

h

True Value

yi+1, Predicted value

yi

x

y

xi xi+1

hyxfyy iiii ,1

ii xxh 1

This formula is known as

Euler’s method and is

illustrated graphically in

Figure 2.

In some books, it is also

called the Euler-Cauchy

method.

Figure 2. General graphical interpretation of Euler’s method

Euler’s Method Modified Euler’s method :

A fundamental source of error in Euler’s method is that the derivative

at the beginning of the interval is assumed to apply across the entire

interval.

Two simple modifications are available to circumvent this shortcoming:

o Heun’s Method

o The Midpoint (or Improved Polygon) Method

Euler’s Method Modified Euler’s method (Heun’s method) :

To improve the estimate of the slope, determine two derivatives for the interval:

o At the initial point

o At the end point

The two derivatives are then averaged to obtain an improved estimate of the slope for the entire interval.

hyxfyxf

yy

hyxfyy

iiiiii

iiii

2

),(),(:

),( :0

111

01

Corrector

Predictor

Euler’s Method Modified Euler’s method (The Midpoint

Method):

Uses Euler’s method to predict a

value of y using the slope value

at the midpoint of the interval:

1/2 ( , )2i i i i

hy y f x y

hyxfyy iiii ),( 2/12/11

Euler’s Method Error analysis for Euler’s method :

o Numerical solutions of ODEs involves two types of error:

Truncation error

• Local truncation error

• Propagated truncation error

The sum of the two is the total or global truncation

error

Round-off errors (due to limited digits in representing

numbers in a computer)

Euler’s Method Error analysis for Euler’s method :o We can use Taylor series to quantify the local truncation error in

Euler’s method.

o The error is reduced by 4 times if the step size is halved O(h2).

o In real problems, the derivatives used in the Taylor series are not easy to obtain.

o If the solution to the differential equation is linear, the method will provide error free predictions (2nd derivative is zero for a straight line).

)(!

),(...

!

),(),(

!...

!

),(' Given

)('

)("'

11

21

21

2

2

nniin

iiiiii

nn

nii

iii

hOhn

yxfh

yxfhyxfyy

Rhn

yh

yhyyy

yxfy

)(!2

),(!2

),(

22

32

hOhyxf

E

Rhyxf

E

iia

iia

EULER Local Truncation ERROR

Euler’s Method How to write Ordinary Differential Equation :

How does one write a first order differential equation in the form of

yxfdx

dy,

Example

50,3.12 yeydx

dy x

is rewritten as

50,23.1 yyedx

dy x

In this case

yeyxf x 23.1,

Euler’s Method Application of Euler’s method :

A ball at 1200K is allowed to cool down in air at an ambient temperature of 300K. Assuming heat is lost only due to radiation, the differential equation for the temperature of the ball is given by

Kdt

d12000,1081102067.2 8412

Find the temperature at t = 480 seconds using Euler’s method. Assume a step size of h = 240 seconds.

Euler’s Method Application of Euler’s method : Solution step 1:

8412 1081102067.2

dt

d

8412 1081102067.2, tf

K

f

htf

htf iiii

09.106

2405579.41200

24010811200102067.21200

2401200,01200

,

,

8412

0001

1

1 is the approximate temperature at 240240001 httt

K09.106240 1

Euler’s Method Application of Euler’s method : Solution step 2:

For 09.106,240,1 11 ti

K

f

htf

32.110

240017595.009.106

240108109.106102067.209.106

24009.106,24009.106

,

8412

1112

2 is the approximate temperature at 48024024012 httt

K32.110480 2

Runge-Kutta Methods Runge-Kutta (RK) methods achieve the accuracy of a Taylor

series approach without requiring the calculation of higher derivatives.

),(

),,(

1

2211

1

ii

nn

iiii

yxfk

kakaka

hhyxyy

constants are s

FunctionIncrement

'a

),(

),(

),(

11,122,1111

22212133

11112

hkqhkqhkqyhpxfk

hkqhkqyhpxfk

hkqyhpxfk

nnnnninin

ii

ii

constants are s’ and s’ q p

Runge-Kutta Methods Fourth-Order Runge-Kutta Methods :

For ODEs that are a function of x alone, the classical fourth-order RK method is similar to Simpson’s 1/3 rule.

In addition, the fourth-order RK method is similar to the Heun approach in that multiple estimates of the slope are developed in order to come up with an improved average slope for the interval.

As depicted in Fig., each of the k’s represents a slope.

Graphical depiction of the slope estimates comprising the fourth-

order RK method.

Runge-Kutta Methods Fourth-Order Runge-Kutta Methods :

For 0)0(),,( yyyxfdx

dy

Runge-Kutta 4th order method is given by

hkkkkyy ii 43211 226

1

where

ii yxfk ,1

hkyhxfk ii 12 2

1,

2

1

hkyhxfk ii 23 2

1,

2

1

hkyhxfk ii 34 ,

Runge-Kutta Methods Application of Fourth-Order Runge-Kutta

Methods :A ball at 1200K is allowed to cool down in air at an ambient temperature of 300K. Assuming heat is lost only due to radiation, the differential equation for the temperature of the ball is given by

Kdt

d12000,1081102067.2 8412

Find the temperature at t = 480 seconds using Runge-Kutta 4th order method. Assume a step size of h = 240 seconds.

8412 1081102067.2

dt

d

8412 1081102067.2, tf

hkkkkii 43211 226

1


Methods : Solution step 1:

1200)0(,0,0 00 ti

5579.410811200102067.21200,0, 841201 ftfk o

38347.0108105.653102067.205.653,120

2405579.42

11200,240

2

10

2

1,

2

1

8412

1002

f

fhkhtfk

8954.310810.1154102067.20.1154,120

24038347.02

11200,240

2

10

2

1,

2

1

8412

2003

f

fhkhtfk

0069750.0108110.265102067.210.265,240

240984.31200,2400,8412

3004

f

fhkhtfk



K

hkkkk

65.675

2401848.26

11200

240069750.08954.3238347.025579.46

11200

226

1432101

1is the approximate temperature at

240240001 httt

K65.675240 1


Methods : Solution step 2:Kti 65.675,240,1 11

44199.0108165.675102067.265.675,240, 8412111 ftfk

31372.0108161.622102067.261.622,360

24044199.02

165.675,240

2

1240

2

1,

2

1

8412

1112

f

fhkhtfk

34775.0108100.638102067.200.638,360

24031372.02

165.675,240

2

1240

2

1,

2

1

8412

2113

f

fhkhtfk

25351.0108119.592102067.219.592,480

24034775.065.675,240240,8412

3114

f

fhkhtfk



K

hkkkk

91.594

2400184.26

165.675

24025351.034775.0231372.0244199.06

165.675

226

1432112

q2 is the approximate temperature at

48024024012 htt

K91.594480 2

Comparison of Euler's Method and Runge-Kutta Methods

0

200

400

600

800

1000

1200

1400

0 100 200 300 400 500

Time, t(sec)

Tem

pera

ture

,θ(K

)

Exact

4th order

Heun

Euler

Figure 3. Comparison of Euler's Method and Runge-Kutta Methods

References

http://mat.iitm.ac.in

http://en.m.wikipedia.org/wiki/Euler_method

http://en.m.wikipedia.org/wiki/Runge-Kutta_methods

http://www.classzone.com

http://mathforcollege.com

Certificate

I Mira Y Patel(140110723006) assure that material which I

prepared is not plagiarized material. I Prepared it under the

guidance of subject faculty. The references are also mentioned,

which are used for the ppt.

Sign

Mira Y Patel

THANK YOU

maths final ppt

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