A textbook of Computer Based Numerical and Statiscal Techniques part 38 potx

7.7 MILNE’S PREDICTOR-CORRECTOR METHODPredictor-Corrector Methods Predictor-Corrector formulae are easily derived but require the previous evaluation of y and y = f x y at a certain nu

k4 =hf x( 0+h y, 0+k3) =0.2 0 0.2 1 0.0198( + )( − )=0.039208

1

6

y =y + k + k + k +k

= +1 10 2+ −( 0.02) (+ −2 0.0198) (+ −0.039208)

6

=1.0000 0.198013−

= 0.9801986 ≅ 0.9802

The exact value of y(0.2) is 0.9802.

Example 3 Solve the equation y′ = +(x y)with y 0 = 1 by Runge-Kutta rule from x = 0 to x = 0.4 with h = 0.1.

Sol Here f x y( ), = +x y h, =0.1, given y0 =1 when x0 =0

We have,

k1=hf x y( 0, 0) =0.1 0 1( + =) 0.1

2 0 , 0 1

k h

k =hf x + y + 

=0.1 0.05 1.05( + ) =0.11

3 0 , 0 2

k h

k =hf x + y + 

=0.1 0.05 1.055( + ) =0.1105

k4 =hf x( 0+h y, 0+k3) =0.1 0.1 1.1105( + ) =0.12105

1 ( 0.1 ) 0 ( 1 2 3 4)

1

6

x

y =y = =y + k + k + k +k

1 1(0.1 0.22 0.2210 0.12105)

6

Similarly for finding y2=y x( =0.2), we get

k1=hf x y( 1 1=0.1)( )0.1 +1.11034=0.121034

2 1 , 1 1

k h

k =hf x + y + 

=0.1 0.15 1.11034 0.660517[ + + ]=0.13208

3 1 , 1 2

k h

k =hf x + y + 

=0.1 0.15 1.11034 0.06604[ + + ]=0.13208

k4 =hf x( 1+h y, 1+k3) =0.1 0.20 1.11034 0.13263[ + + ]=0.14263

1

6

x

y =y = =y + k + k + k +k

1.11034 1 0.121034 2 0.13208 0.13263 0.14429( ) 1.2428

6

Similarly, for finding y3 =y x( =0.3), we get

k1 =hf x y( 2 2)=0.1 0.2( )+1.2428=0.14428

2 2 , 2 1

k h

k =hf x + y + 

=0.1 0.25 132428 0.07214[ + + ]=0.15649

3 2 , 2 2

k h

k =hf x + y + 

=0.1 0.25 1.2428 0.07824[ + + ]=0.15710

k4 =hf x( 2+h y, 2+k3) =0.1 0.30 1.2428 0.15710[ + + ]=0.16999

1

6

x

y =y = =y + k + K + K +k

=0.13997 Similarly, for finding y4 =y x( =0.4), we get

k1 =( )[0.1 0.3 1.3997+ ]=0.16997 ⇒ k1=0.16997

k2=( )[0.1 0.35 1.3997 0.08949+ + ]=0.18347 ⇒ k2 =0.18347

k3 =( )[0.1 0.35 1.3997 0.9170+ + ]=0.18414 ⇒ k3 =0.18414

k4 =( )[0.1 0.4 1.3997 0.18414+ + ]=0.19838 ⇒ k1=0.19838

∴ 4 1.3997 1 0.16997 2 0.18347 0.18414 0.19838( )

6

y4=1.5836 Ans

Example 4 Given dy y x

dx= − with y(0) = 2, find y(0.1) and y(0.2) correct to 4 decimal places.

Sol We have x0=0, y0=2, h=0.1

Then, we get

k1=hf x y( 0, 0) =0.1 2 0( − )=0.2

2 0 , 0 1

k h

k =hf x + y + 

=0.1 2 +0.22 − +0 0.12 =0.205

3 0 , 0 2

k h

k =hf x + y + 

=0.1 2 +0.2052 −0+0.12 =0.20525

k4=hf x( 0+h y, 0+k3) =0.1 2 0.20525 (0 0.1)[ + − + ]=0.210525 Therefore,

0 [ 1 2 3 4]

1

6

y=y + k + k + k +k

= +2 0.2051708=2.2051708

⇒ y( )0.1 =2.2052 Corect to 4 decimal places

For y( )0.2 , we have x0 =0.1, y0 =2.2052, we get

k1=hf x y( 0, 0) =0.1 2.2052 0.1( − )=0.21052

2 0 , 0 1

k h

k =hf x + y + 

=0.1 2.2052 +0.210522 −0.1+0.12 =0.216046

3 0 , 0 2

k h

k =hf x + y + 

=0.1 2.2052 +0.2160462 −0.1+0.12 =0.2163223

k4=hf x( 0+h y, 0+k3) =0.1 2.2052 0.2163223 + −(0.1 0.1+ )=0.22215223

1

6

y = + k + k + k +k

=2.2052 0.2162348+

=2.4214 Ans

Example 5 Solve dy 2xy2

dx= − with y(0) = 1 and h = 0.2 on the interval [0.1] using Runge-Kutta

fourth order method.

Sol As per given, we have x0=0, y0=1, h=0.2

k1=hf x y( 0, 0) ( )( )( )2

2 0.2 0 1 0

k h

k =hf x + y + 

( ) 0.2 ( )2

2

k h

k =hf x + y + 

( ) 0.2 ( )2

2 0.2 0.98 0.38416

2

k4 =hf x( 0+h y, 0+k3) = −22 0.2 0.2 0.961584( )( )( )2= −0.0739715 Hence,

( ) 0 [ 1 2 3 4]

1

6

y =y + k + k + k +k

1 1[0 0.08 0.076832 0.0739715] 6

=0.9615328 Now, we have

x1=0.2,y1 =0.9615328,h=0.2, we get

k1 =hf x y( 1, 1) ( )( )( )2

2 0.2 0.2 0.9615328 0.0739636

k h

k =hf x + y + 

= − ( )( )( )2 =

2 0.2 0.3 0.924551 0.1025754

k h

k =hf x + y + 

= – 2(0.2)(0.3)(0.9102451)2 = 0.0994255

k4 = hf(x1 + h, y1 + k3)

( )( )( )2

2 0.2 0.4 0.8621073 0.1189166

Thus,  ( )= + + + + 

1

6

0.9615328 1[ 0.0739636 0.2051508 0.1988510 0.1189166]

6

=0.8620525

Similarly, we can obtained

y( )0.6 =0.7352784

y( )0.8 =0.6097519

y( )1.0 =0.500073 Ans

Example 6 Solve dy yz x,

dx= + dz =xz y+ ;

dx Given that y(0) = 1, z(0) = –1 for y(0.1), z(0.1).

Sol Here, f x y z1( , , )=yz x+

= +

2( , , )

f x y z xz y

h=0.1,x0=0,y0=1,z0= −1

k1=hf x y z1( 0, 0, 0) (=h y z0 0+x0)= −0.1

l1=hf x y z2( 0, 0, 0) (=h x z0 0+y0)=0.1

2 1 0 , 0 1, 0 1

h

k =hf x + y + z + 

=hf1(0.05, 0.95, 0.95− )= −0.08525

2 2 0 , 0 1, 0 1

h

l =hf x + y + z + 

=hf2(0.05, 0.95, 0.95)− = −0.09025

3 1 0 , 0 2, 0 2

h

k =hf x + y + z + 

=hf1(0.05, 0.957375, 0.954875− )= −0.0864173

3 2 0 , 0 2, 0 2

h

l =hf x + y + z + 

=hf2(0.05, 0.957375, 0.954875− = −0.0864173

k4 = hf1(x0 + h,y0 + h3, z0 + l3) = –0.073048

l4 = hf2(x0 + h, y0 + h3, z0 + l3) = + 0.0822679

=1( 1+2 2+2 3+ 4)= −0.0860637

6

k k k k k

(1 2 3 4)

1

2 2 0.0907823 6

l= l + l + l +l =

y1=y( )0.1 =y0+ = −k 1 0.0860637=0.9139363

z1=z( )0.1 =z0+ = − +k 1 0.0907823= −0.9092176

7.7 MILNE’S PREDICTOR-CORRECTOR METHOD

Predictor-Corrector Methods

Predictor-Corrector formulae are easily derived but require the previous evaluation of y and

( )

y = f x y at a certain number of evenly spaced pivotal point (discrete points of x1 of

x-axis) in the neighbourhood of x0

In general the Predictor-Corrector methods are the methods which require the values of

y at x x n, n−1,x n−2 For computing the value of y at x n+1 A Predictor formula is used to

predict the value of y n+1 Now we discuss Milne’s method which is known as Predictor-Corrector methods

Milne’s Method

The method is a simple and eresonable accurate method of solving the ordinary first order differential equation numerically To solve the differential equation

dy y f x y( ),

dx= =′

by this method we first obtain the approximate value of y n+1 by Predictor formula and then improve the value of y n+1by means of a corrector formula Both these formulas can be derived from the Newton forward interpolation formula as follows:

From Newton’s formula, we have

( )= ( + )= ( )+ ∆ ( )+ ( − )∆ ( )

1

1 2

u u

f x f x uh f x u f x f x

( ) 3 0

( 1)( 2)

1 2 3

u u u

f x

where u x x0,

h

−

= or x=x0+uh

Putting y′ = f x( ) and y0′ = f x( )0 in the above formula, we get

y′=y′+ ∆ +u y′ − ∆ y′+ − − ∆ y′+ − − − ∆ y′

Intergrating (2) from x0 to x0+4h i.e from u=0 to u=4, we get

0 4 0

0

h

x

x

4 0

( 1)( 2)( 3)

] 24

(3 hdu = dx) which gives

0 0

4

x

x h

y + −y =h h′+ ∆ +y′ ∆ y′+ ∆ y′+ ∆ y′

[considering upto fourth differences only]

Using ∆ = −E 1, we get

y −y =h y′+ E− y′+ E− y′+ E− y′+ ∆ y′

⇒ [ ] 2

h

This is known as Milne’s predictor formula The corrector formula is obtained by integrating

(2) from x0 to x0 + 2h i.e., from u = 0 to u = 2.

x x h ydx

0

0 + 2

2

⇒ y2 – y0 = h y2 2 y 1 y y

3

1 90

0′ + 0′ + 2 0′ 4 0′

L

Using ∆ = −E 1, and simplifying we get

Expression (4) is called Milne’s corrector formula

The general forms of equations (3) and (4) are

4

3

h

3

h

4

3

n

h

and 1 1 [ 1 4 1]

3

h

In terms of f the Predictor formula is

1

4

3

n

h

and the corrector formula is

3

h

Example 7 Tabulate by Milne’s method the numerical solution of dy x y

dx= + with initial conditions

x 0 = 0, y 0 = 1, from x = 0.20 to x = 0.30.

Sol Here y′ = +x y

y′′= +1 y y′ ′′′, =y y′′ ′′′′, =y′′′ ′′′′′,y =y′′′′,

Hence, y0′ =x0+y0 = + =0 1 1

y0′′= + + = + =1 y0′ 1 1 2

y0′′′=y0′′ =2

y′′′′= y′′′′′=

Now taking h=0.05, we get

y1 =1.1026, y2=1.2104, y3 =1.3237

Using Milne’s Predictor formula, we get

4

3

h

y =y + y′−y′+ y′

4 0.05

1 2.2052 1.2104 2.6474 1.2428 3

4 0.2 1.2428 1.4428

Using corrector formula, we get

3

h

y =y + y′+ y′ +y′

1.1104 1.2104 5.2948 1.4428

3

which is the same as the predicted value

∴ y4=y0.20 =1.2428

and y4′ =1.4428

and putting n = 4, h = 0.05, we get

4

3

h

y =y + y′−y′+ y′

4(0.05) 1.0526 2.4208 1.3237 2.8856

3

5 5 5 0.25 1.3181 1.5681

y′ =x +y = + = Using Milne’s corrector formula, we get

5

3

h

y =y + y′+ y′ +y′ 

1.1737 1.3237 5.7712 1.5681

3

which is the same as the predicted value

∴ y5=y0.25=1.3181

and y5′ =1.5681

Again putting n=5, h=0.05 and using Milne’s predictor formula, we get

4

3

h

y =y + y′−y′ + y′

4 0.05 1.1104 2.6474 1.4428 3.1362

3

y′ =6 0.3 1.39972 1.6997+ =

which is corrected by

3

h

y =y + y′+ y′ +y′

1.2428 1.4428 6.2724 1.6997 1.3997

3

which is same as the predicted value.

∴ y6 =y0.30=1.3997

and y′ =6 1.6997

The result can be put in the tabular form

Example 8 Compute y(2) if y(x) is the solution of dy 1( )

= x + y

dx 2 using Milne Predictor-corrector method Given y 0 = 2, y 0.5 = 2.6336, y 1.0 = 3.595, y 1.5 = 4.968.( ) ( ) ( ) ( )

Sol Here, we have

x = y = 0 ( )

1

0 2 1 2

1 0.5, 1 2.636

1 0.5 2.636 1.568 2

2 1, 2 3.595

1

1 3.595 2.2975 2

3 1.5, 3 4.968

1 1.5 4.968 3.234 2

Using Predictor formula, we get

y4 = y0 + 4

3

h (2f1 – f2 + 2f3) = 2.0 + 4 0 5

3

× (2 × 1.568 – 2.2975 + 2 × 3.234) =6.871 and 4 4 4 ( )

1

2 6.871 4.4335 2

f =x +y = + =

Using corrector formula, we get

4 2 [ 2 4 3 4]

3

h

y =y + f + f +f

3.595 0.5(2.2975 4 3.234 4.4355)

3

=6.873166≈ 6.8732 Thus, corrected 4 1( 4 4) 1( )

2.0 6.8732 4.4366

Again, using Corrector formula, we get

4 2 ( 2 4 3 4)

3

h

y =y + f + f +f

3.595 0.5(2.2975 4 3.234 4.4366)

3

=6.87335 6.8734≈

Example 9 The differential equation dy dx = 1 + y 2 satisfies the following sets of values of x and y.

Sol Firstly, we calculate the following

f0= 1 + y02 = 1

f1= 1 + y12 = 1.0411

f2= 1 + y22 = 1.1787

f3= 1 + y32 = 1.4681 Using Predictor formula, we get

4= 0+4 (2 1− 2+2 3)

3

h

y y f f f

0 4 0.2 2 1.0411( ) 1.1787 2 1.4681( )

3

×

⇒ = + 2 = +( )2 =

4 1 4 1 1.0239 2.0480

Using Corrector formula, we have

4 2 [ 2 4 3 4]

3

h

y =y + f + f +f

0.4228 0.2(1.1787 4 1.4681 2.0480)

3

⇒ y( )0.8 =1.0294

The corrected value of

f4= +1 y24 =2.0597 Now, to find f( )1 we use predictor formula such that

y5 = y1 + 4

3

h (2f1 – f3 + 2f4)

= 0.2027 + 4 0 2

3

×

[2(1.1787) – 1.4681 + 2(2.0597)] = 1.5384 and f5 = 1 + y52 = 1 + (1.5384)2 = 3.3667

Finally using corrector formula, we get

5 3 [ 3 4 4 5]

3

h

y =y + f + f +f

0.2( ( ) )

0.6841 1.4681 4 2.0597 3.3667

3

=1.5556733 1.5557=