Numerical Differentiation - Kemdikbud

Numerical

Differentiation

Runge Kutta

Nur Laila Hamidah

Runge-Kutta 2 ^nd Order Method

Runge Kutta 2nd order method is given by

( ^{a k a k} ) ^h

y

y

_i+1

=

_i

+

_{1 1}

+

_{2 2}

where

(

^x_i ^y_i

)

f k₁ =

,

(

^{x p h y q k h}

)

f

k₂ = _i + ₁ , _i + _{11 1}

For

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

₀

dx

dy = =

Heun’s Method

x y

x_i x_i+1

y_i+1, predicted

y_i

Figure 1 Runge-Kutta 2nd order method (Heun’s method) (x h y k h)

f

Slope= _i + , _i + ₁

( ) ( )

f x_i h y_i k h f x_i y_i 

Slope

Average , ,

2 1

1 +

+ +

=

(^x_i ^y_i)

f Slope= ,

Heun’s method

2 1

1 = a

1 =1 p

11

= 1 q

resulting in

h k k

y

y_{i i} 



 



 +

+

+1 = 1 2

2 1 2

1

where

(

^x_i ^y_i

)

f k₁ = ,

(x h y k h)

f

k₂ = _i + , _i + ₁

Here a₂=1/2 is chosen

Midpoint Method

Here a₂ =1 is chosen, giving

1 =

0

a

2 1

1 = p

2 1

11 = q

resulting in h k y

y_i+1 = _i + ₂

where

(

x_i y_i

)

f k₁ = ,



 



 + +

= f x h y k h

k_{2 i i 1}

2 , 1

2 1

Ralston’s Method

Here

3 2

2 =

a is chosen, giving

3 1

1 = a

4 3

1 = p

4 3

11 = q

resulting in

h k k

y

y_{i i} 



 



 +

+

+1 = 1 2

3 2 3

1

where

(

^x_i ^y_i

)

f k₁ = ,



 



 + +

= f x h y k h

k_{2 i i 1}

4 , 3

4 3

How to write Ordinary Differential Equation

Example

( )

0 5 ,

3 . 1

2 = =

+ y e⁻ y dx

dy _x

is rewritten as

( )

^{0 5}

, 2 3

.

1 − =

= e⁻ y y dx

dy _x

In this case

( )

^{x y e y}

f , =1.3 ^−x −2

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

( )

^{x y}

dx f

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 = −2.206710₋¹² ⁴ −8110⁸ , 0 =1200

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

=

240

h _seconds.

(

^{4 8}

)

12 81 10

10 2067 .

2   − 

−

 = ₋

dt d

( )

^{t, =−2.206710−12}

(

^{4 −81108}

)

f

h k

i k

i 



 



 +

+

+1 = 1 2

2 1 2

 1



Solution

Step 1:

_{i 0,t 0, (0) 1200K}

0

0 = = =

=  

( )

( )

( )

5579 .

4

10 81 1200

10 2067 .

2

1200 ,

0 ,

8 4

12 0

1

−

=



−



−

=

=

=

−

f t f

k _o

( )

( )

( )

( )

( )

017595 .

0

10 81 09

. 106 10

2067 .

2

09 . 106 , 240

240 5579 .

4 1200

, 240 0

,

8 4

12 1 0 0

2

=



−



−

=

=

− + +

=

+ +

=

−

f f

h k h

t f

k 

( ) ( )

( )

K

h k k

16 . 655

240 2702 .

2 1200

240 017595

. 2 0 5579 1 .

2 4 1200 1

2 1 2

1

2 1

0 1

=

− +

=



 



 − +

+

=



 



 +

+

=



Solution Cont

Step 2: ⁱ =^1,t1 = ^t0 +^h = ⁰+ ²⁴⁰= ^240,1 = ^655.16K

( )

( )

( )

38869 .

0

10 81 16

. 655 10

2067 .

2

16 . 655 , 240

,

8 4

12 1

1 1

−

=



−



−

=

=

=

−

f t f

k 

( )

( )

( )

( )

( )

20206 .

0

10 81 87

. 561 10

2067 .

2

87 . 561 , 480

240 38869 .

0 16

. 655 , 240 240

,

8 4

12 1 1 1

2

−

=



−



−

=

=

− + +

=

+ +

=

−

f f

h k h

t f

k 

( ) ( )

( )

K

h k k

27 . 584

240 29538 .

0 16

. 655

240 20206

. 2 0

38869 1 .

2 0 16 1

. 655

2 1 2

1

2 1

1 2

=

− +

=



 



 − + −

+

=



 



 +

+

=



Solution Cont

The exact solution of the ordinary differential equation is given by the solution of a non-linear equation as

(

0.0033333

)

0.22067 10 2.9282 tan

8519 .

300 1 ln 300

92593 .

0 − ¹ = −  ³ −

+

− _{− −}

 t





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

57 . 647 )

480

( =



Comparison with exact results

Figure 2. Heun’s method results for different step sizes

-400 0 400 800 1200

0 100 200 300 400 500

Time, t(sec)

Temperature,θ(K)

Exact h=120

h=240 h=480

Effect of step size

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

Step size, h

(480) E_t |є_t|%

480 240 120 60 30

−393.87 584.27 651.35 649.91 648.21

1041.4 63.304

−3.7762

−2.3406

−0.63219

160.82 9.7756 0.58313 0.36145 0.097625 K

57 . 647 )

480

( =



^(exact)

Effects of step size on Heun’s Method

Figure 3. Effect of step size in Heun’s method

-400 -200 0 200 400 600 800

0 100 200 300 400 500

Step size, h

Temperature,θ(480)

Step size, h

 (480)

Euler Heun Midpoint Ralston 480

240 120 60 30

−987.84 110.32 546.77 614.97 632.77

−393.87 584.27 651.35 649.91 648.21

1208.4 976.87 690.20 654.85 649.02

449.78 690.01 667.71 652.25 648.61

Comparison of Euler and Runge- Kutta 2 ^nd Order Methods

Table 2. Comparison of Euler and the Runge-Kutta methods

K 57 . 647 )

480

( =



^(exact)

Comparison of Euler and Runge- Kutta 2 ^nd Order Methods

Table 2. Comparison of Euler and the Runge-Kutta methods

Step size,

h Euler Heun Midpoint Ralston

480 240 120 60 30

252.54 82.964 15.566 5.0352 2.2864

160.82 9.7756 0.58313 0.36145 0.097625

86.612 50.851 6.5823 1.1239 0.22353

30.544 6.5537 3.1092 0.72299 0.15940

K

57 . 647 )

480

(

=



_(exact)

t %



Comparison of Euler and

Runge-Kutta 2 ^nd Order Methods

Figure 4. Comparison of Euler and Runge Kutta 2^nd order methods with exact results.

500 600 700 800 900 1000 1100 1200

0 100 200 300 400 500 600

Time, t (sec)

Temperature,

Analytical

Ralston Midpoint

Euler

Heun

θ(K)

Runge-Kutta 4 ^th Order Method

where

(

^{k k k k}

)

^h

y

y_{i 1 i 1} 2 ₂ 2 _{3 4} 6

1 + + +

+

+ =

( ^x

_i

^y

_i

)

f k

₁

= ,



 



 + +

= f x h y k h

k_{2 i i 1}

2 , 1

2 1



 



 + +

= f x h y k h

k_{3 i i 2}

2 , 1

2 1

( ^{x h y k h} )

f

k

₄

=

_i

+ ,

_i

+

₃

For

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

0

dx

dy = =

Runge Kutta 4^th order method is given by

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 = −2.206710₋¹² ⁴ −8110⁸ , 0 =1200

Find the temperature at t = 480 seconds using Runge-Kutta 4^th order method.

=

240

h _seconds.

(

^{4 8}

)

12 81 10

10 2067 .

2   − 

−

 = ₋

dt d

( )

^{t, =−2.206710−12}

(

^{4 −81108}

)

f

Assume a step size of

(

^{k k k k}

)

^h

i

i 1 1 2 2 2 3 4

6

1 + + +

+

+ =



Solution

Step 1:

_{0, 0, (0) 1200}

0

0 = = =

= t  

i

(

0^,

) (

^0,1200

)

^{2.2067 10 12}

(

^{12004 81 108}

)

^4.5579

1 = f t = f =−  ⁻ −  =−

k _o

( ) ( )

(

^120,653.05

)

^{2.2067 10}

(

^{653.05 81 10}

)

^0.38347

240 5579 .

2 4 1200 1 ,

2 240 0 1

2 , 1

2 1

8 4

12 1

0 0

2

−

=



−



−

=

=



 



 + + −

 =



 



 + +

= f −

f h k h

t f

k 

( ) ( )

(

^120,1154.0

)

^{2.2067 10}

(

^{1154.0 81 10}

)

^3.8954

240 38347

. 2 0

1200 1 ,

2 240 0 1

2 , 1

2 1

8 4

12 2

0 0

3

−

=



−



=

=



 



 + + −

 =



 



 + +

= f −

f h

k h

t f

k 

( ) ( ( ) ( ) )

( ²⁴⁰

⁰

^{, 265 ,}

⁰

^{. 10} )

³

^{2 . 2067 0 10 240}

¹²

( ^{, 1200 265 . 10}

⁴

^{3 81 . 984 10 240}

⁸

) ^{0 . 0069750}

4

=



−



=

=

− + +

= +

+

= f −

f h

k h

t f

k



Solution Cont



1 is the approximate temperature at 240

240

0 0

1 = + = + =

= t t h t

( ) ²⁴⁰

^

^

₁ ⁼

^{675 . 65}

^K



( )

( ) ( ) ( )

( )

( )

K

h k k

k k

65 . 675

240 1848

. 6 2

1200 1

240 069750

. 0 8954

. 3 2 38347

. 0 2 5579 .

6 4 1200 1

2 6 2

1

4 3

2 1

0 1

=

− +

=

+

− +

− +

− +

=

+ +

+ +

= 



Solution Cont

Step 2: ^{i =1,t1 = 240,}^{1 = 675.65K}

(

1^, 1

) (

^240,675.65

)

^{2.2067 10 12}

(

^{675.654 81 108}

)

^0.44199

1 = f t = f =−  ⁻ −  =−

k 

( ) ( )

(

^360,622.61

)

^{2.2067 10}

(

^{622.61 81 10}

)

^0.31372

240 44199 .

2 0 65 1 . 675 , 2 240 240 1

2 , 1

2 1

8 4

12 1

1 1

2

−

=



−



−

=

=



 



 + + −

 =



 



 + +

= f −

f h k h

t f

k 

( ) ( )

(

^360,638.00

)

^{2.2067 10}

(

^{638.00 81 10}

)

^0.34775

240 31372 .

2 0 65 1 . 675 , 2 240 240 1

2 , 1

2 1

8 4

12 2

1 1

3

−

=



−



=

=



 



 + + −

 =



 



 + +

= f −

f h

k h

t f

k 

( ) ( ( ) ( ) )

(

^{4801 ,592, 1.19}

)

^{3 2.20672401024012}

(

^{592,675.19.654 81 010.347758}

)

^0240.25351

4

−

=



−



=

=

− + +

= +

+

= f −

f h k h

t f

k 

Solution Cont

₂ is the approximate temperature at 480

240

1 240

2 = t + h = + =

t

( )

^{480 }₂ ^{= 594.91K}



( )

( ) ( ) ( )

( )

( )

K

h k k

k k

91 . 594

240 0184

. 6 2

65 1 . 675

240 25351

. 0 34775

. 0 2 31372

. 0 2 44199 .

6 0 65 1 . 675

2 6 2

1

4 3

2 1

1 2

=

− +

=

− +

− +

− +

− +

=

+ +

+ +

= 



Solution Cont

The exact solution of the ordinary differential equation is given by the solution of a non-linear equation as

(

^0.00333

)

^{0.22067 10 2.9282}

tan 8519 .

300 1 ln 300

92593 .

0 − ¹  = −  ³ −

+



−

 _{− −}

t

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

57 . 647 )

480

( =



Comparison with exact results

Figure 1. Comparison of Runge-Kutta 4th order method with exact solution

-400 0 400 800 1200 1600

0 200 400 600

Time,t(sec)

Temperature, _h=120

Exact

h=240

h=480

θ(K)

Step size, h  (480) E

_t

|є

_t

|%

480 240 120 60 30

−90.278 594.91 646.16 647.54 647.57

737.85 52.660 1.4122 0.033626 0.00086900

113.94 8.1319 0.21807 0.0051926 0.00013419

Effect of step size

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

K 57 . 647 )

480

( =



(exact)

Effects of step size on Runge- Kutta 4 ^th Order Method

Figure 2. Effect of step size in Runge-Kutta 4th order method

-200 0 200 400 600 800

0 100 200 300 400 500

Step size, h

Temperature,θ(480)

Comparison of Euler and Runge-Kutta Methods

Figure 3. Comparison of Runge-Kutta methods of 1st, 2nd, and 4th order.

0 200 400 600 800 1000 1200 1400

0 100 200 300 400 500

Time, t(sec)

Temperature,θ(K)

Exact 4th order

Heun Euler