Runge-Kutta 4th Order Method for Ordinary Differential ...

Chapter 08.04 Runge-Kutta 4th Order Method for Ordinary Differential Equations

After reading this chapter, you should be able to 1. develop Runge-Kutta 4th order method for solving ordinary differential equations,

2. find the effect size of step size has on the solution, 3. know the formulas for other versions of the Runge-Kutta 4th order method

What is the Runge-Kutta 4th order method?

Runge-Kutta 4th order method is a numerical technique used to solve ordinary differential equation of the form

dy dx

=

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

y0

So only first order ordinary differential equations can be solved by using the Runge-Kutta 4th

order method. In other sections, we have discussed how Euler and Runge-Kutta methods are

used to solve higher order ordinary differential equations or coupled (simultaneous)

differential equations.

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

Example 1

Rewrite

dy + 2 y = 1.3e-x , y(0) = 5

dx

in

dy dx

=

f (x, y),

y(0) =

y0

form.

08.04.1

08.04.2

Chapter 08.04

Solution

dy + 2 y = 1.3e-x , y(0) = 5

dx

dy = 1.3e-x - 2 y, y(0) = 5

dx In this case

f (x, y) = 1.3e-x - 2 y

Example 2

Rewrite

e y dy + x2 y 2 = 2sin(3x), y(0) = 5

dx

in

dy dx

=

f (x, y),

y(0) =

y0

form.

Solution

e y dy + x2 y 2 = 2sin(3x), y(0) = 5

dx

dy = 2sin(3x) - x2 y 2 , y(0) = 5

dx

ey

In this case

f (x, y) = 2sin(3x) - x2 y 2

ey

The Runge-Kutta 4th order method is based on the following

( ) yi+1 = yi + a1k1 + a2k2 + a3k3 + a4k4 h

(1)

where knowing the value of y = yi at xi , we can find the value of y = yi+1 at xi+1 , and

h = xi+1 - xi

Equation (1) is equated to the first five terms of Taylor series

( ) ( ) ( ) yi+1

=

yi

+

dy dx

xi , yi

xi+1 - xi

+

1 2!

d2y dx 2

xi , yi

xi+1 - xi

2

+

1 3!

d3y dx3

xi , yi

xi+1 - xi

3

(2)

( ) +

1 4!

d4y dx 4

xi , yi

xi+1 - xi

4

Knowing that

dy dx

=

f (x, y)

and

xi+1

- xi

=h

yi+1

=

yi

+

f

(xi ,

yi )h +

1 2!

f

' (xi ,

yi )h2

+

1 3!

f

( ) '' xi , yi h3

+

1 4!

f

(''' xi ,

yi )h4

(3)

Based on equating Equation (2) and Equation (3), one of the popular solutions used is

yi+1

=

yi

+

1 6

(k1

+

2k2

+

2k3

+

k4

)h

(4)

Runge-Kutta 4th Order Method

k1 = f (xi , yi )

k2

=

f

xi

+

1 2

h,

yi

+

1 2

k1h

k3

=

f

xi

+

1 2 h, yi

+

1 2

k2

h

k4 = f (xi + h, yi + k3h)

08.04.3

(5a)

(5b) (5c) (5d)

Example 3

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

( ) d = -2.2067 ?10-12 4 - 81?108 , (0) = 1200 K

dt where is in K and t in seconds. Find the temperature at t = 480 seconds using RungeKutta 4th order method. Assume a step size of h = 240 seconds. Solution

( ) d = -2.2067 ?10-12 4 - 81?108

dt

( ) f (t, ) = -2.2067 ?10-12 4 - 81?108

i+1

=

i

+

1 6

(k1

+

2k2

+

2k3

+

k4

)h

For i = 0 , t0 = 0 , 0 = 1200K

k1 = f (t0 ,0 ) = f (0,1200)

( ) = -2.2067 ?10-12 12004 - 81?108

= -4.5579

k2

=

f

t0

+

1 2

h,

0

+

1 2

k1h

= f 0 + 1 (240),1200 + 1 (- 4.5579)? 240

2

2

= f (120,653.05)

( ) = -2.2067 ?10-12 653.054 - 81?108

= -0.38347

k3

=

f

t

0

+

1 2

h,

0

+

1 2

k

2

h

= f 0 + 1 (240),1200 + 1 (- 0.38347)? 240

2

2

= f (120,1154.0)

08.04.4

Chapter 08.04

( ) = -2.2067 ?10-12 1154.04 - 81?108

= -3.8954

k4 = f (t0 + h,0 + k3h) = f (0 + 240,1200 + (- 3.894)? 240) = f (240,265.10)

( ) = -2.2067 ?10-12 265.104 - 81?108

= 0.0069750

1

=

0

+

1 6

(k1

+

2k2

+

2k3

+

k4 )h

= 1200 + 1 (- 4.5579 + 2(- 0.38347) + 2(- 3.8954) + (0.069750))240

6

= 1200 + (- 2.1848)? 240

= 675.65 K

1 is the approximate temperature at

t = t1

= t0 + h

= 0 + 240 = 240

1 = (240)

675.65 K

For i = 1,t1 = 240,1 = 675.65 K

k1 = f (t1,1 ) = f (240,675.65)

( ) = -2.2067 ?10-12 675.654 - 81?108

= -0.44199

k2

=

f

t1

+

1 2

h,1

+

1 2

k1h

= f 240 + 1 (240),675.65 + 1 (- 0.44199)240

2

2

= f (360,622.61)

( ) = -2.2067 ?10-12 622.614 - 81?108

= -0.31372

k3

=

f

t1

+

1 2

h,1

+

1 2

k

2h

= f 240 + 1 (240),675.65 + 1 (- 0.31372)? 240

2

2

= f (360,638.00)

( ) = -2.2067 ?10-12 638.004 - 81?108

Runge-Kutta 4th Order Method

08.04.5

= -0.34775

k4 = f (t1 + h,1 + k3h) = f (240 + 240,675.65 + (- 0.34775)? 240) = f (480,592.19)

( ) = 2.2067 ?10-12 592.194 - 81?108

= -0.25351

2

= 1

+

1 6

(k1

+ 2k2

+ 2k3

+

k4 )h

= 675.65 + 1 (- 0.44199 + 2(- 0.31372) + 2(- 0.34775) + (- 0.25351))? 240

6

= 675.65 + 1 (- 2.0184)? 240

6

= 594.91K

2 is the approximate temperature at t = t2 = t1 + h = 240 + 240 = 480

2 = (480)

594.91K

Figure 1 compares the exact solution with the numerical solution using the Runge-Kutta 4th order method with different step sizes.

1600

Temperature, (K)

1200 800 400

h=120

Exact h=240

0 0

-400

h=480

200

400

600

Time,t(sec)

Figure 1 Comparison of Runge-Kutta 4th order method with exact solution for different step sizes.

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