Euler Method - Numerical Methods - Lecture Slides, Slides of Mathematical Methods for Numerical Analysis and Optimization

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Euler Method

Euler’s Method

Φ

Step size, h

x

y

x 0 ,y (^0)

True value

y 1 , Predicted value

f ( x , y ), y ( ) 0 y 0

dx

dy = =

Slope Run

Rise

1 0

1 0 x x

y y

−

−

= f ( x 0 , y 0 )

y 1 = y 0 + f ( x 0 , y 0 )( x 1 − x 0 )

= y (^) 0 + f ( x 0 , y 0 ) h

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

How to write Ordinary Differential

Equation

Example

+ 2 y = 1. 3 e −^ , y ( ) 0 = 5

dx

dy (^) x

is rewritten as

= 1. 3 e −^ − 2 y , y ( ) 0 = 5 dx

dy (^) x

In this case

f ( x y ) e y

x

−

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

f ( x y ) dx

dy = ,

Example

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

( ) ( ) K

dt

d

1. 2067 10 81 10 , 0 1200

12 4 8 = − × − × =

− θ θ

θ

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

h = 240 seconds.

Solution Cont

For i =^1 ,^ t 1 =^240 , θ 1 =^106.^09

( )

( )

( )

K

f

f t h

12 4 8

2 1 1 1

= + − × − ×

−

θ θ θ

Step 2:

θ (^2) is the approximate temperature at (^) t = t 2 = t 1 + h = 240 + 240 = 480

θ ( 480 ) ≈θ 2 = 110. 32 K

Solution Cont

The exact solution of the ordinary differential equation is given by the

solution of a non-linear equation as

1. 8519 tan ( 0. 00333 ) 0. 22067 10 2. 9282 300

1. 92593 ln

1 3 − =− × −

θ t

The solution to this nonlinear equation at t=480 seconds is

θ( 480 )= 647. 57 K

Step, h θ(480) E t |єt|%

480

240

120

60

30

−987.

Effect of step size

θ( 480 )= 647. 57 K

Table 1. Temperature at 480 seconds as a function of step size, h

(exact)

Comparison with exact results

0

500

1000

1500

0 100 200 300 400 500 Temperature, Tim e, t (sec)

Exact solution

h= h=

h=

θ(K)

Figure 4. Comparison of Euler’s method with exact solution for different step sizes

Errors in Euler’s Method

2 Et ∝ h

It can be seen that Euler’s method has large errors. This can be illustrated using

Taylor series.

( ) ( ) ( ) ... 3!

1 ,

3

3 2 1 ,

2

2

1 ,

• 1 =^ + + − + + − + i + − i + x y

i i x y

i i x y

i i x x dx

d y x x dx

d y x x dx

dy y y i i i i i i

( ) ( ) ''( , )( ) ...

3 1

2

yi + 1 = yi + f xi yi xi + 1 − xi + f xi yi xi + 1 − xi + f xi yi xi + − xi +

As you can see the first two terms of the Taylor series

y (^) i + 1 = yi + f ( xi , yi ) h

The true error in the approximation is given by

( ) ( ) ... 3!

,

2!

, (^2 )

′′

′ = h

f x y h

f x y E

i i i i t

are the Euler’s method.