SCHOOL OF MATHEMATICAL SCIENCESUNIVERSITI SAINS MALAYSIA

MSG 389: ENGINEERING COMPUTATION IISemester 2, Academic Session 2013/2014

ASSIGNMENTGROUP: 4b (Adam-Bashfort)

LECTURER:

DR.SARATHA A/P SATHASIVAM

NAME

IC NUMBER

MATRIC NUMBER

1. MARINA AISHAH BINTI MOHAMAD 921027-11-5534 112556

2.MAZIATUL ATHIRAH LONG HUSSIN 921026-11-5938 112557

3.MARDHIYANA BINTI ABIDIN 921116-12-5216 112555

DATE OF SUBMISSION: 31st March 2014

2.

(b) Find an approximate value of y (1.5) for the initial value problem

y '=x2+ y2 , y (1 )=0

using the multistep method

y i+1 = y i+h

12¿)

with h=0.1 . Calculate the starting values using third order Taylor series

method with the same step length, h .

[40 marks]

SOLUTION 2(b)

Initial value problem,

y '=x2+ y2 , y (1 )=0

Using the multistep method,

y i+1 = y i+h

12¿)

y i+3= y i+2+h

12(23 f i+2−16 f i+ 1+5 f i)

i=0 : y3= y2+h

12(23 f 2−16 f 1+5 f 0)

For starting values, using third order Taylor series method :

y j+1= y j+h y j' + h

2

2y j' '+ h

3

6y j' ' '

We have : y '=x2+ y2, y ' '=2 x+2 y y' , y ' ' '=2+2 yy ' '+2¿

For h=0.1,

j=0 : x0=1 , y0=0 , y0'=1 , y0

' '=2 , y0' ' '=4

y1≅ y (1.1 )= y0+h y0' + h

2

2y0' '+ h

3

6y0' ' '

¿0+( 0.1 )1+ 0.12

2(2 )+ 0.13

6(4 )

¿0.11067

j=1: x1=1.1 , y1=0.11067 , y1'=1.22225 , y1

' '=2.47053 , y1' ' '=5.53462

y2≅ y (1.2 )= y1+h y1' + h

2

2y1' '+ h

3

6y1' ' '

¿0.11067+(0.1 )1.22225+ 0.12

2(2.47053 )+ 0.13

6(5.53462 )

¿0.24617

Then, use multistep method,

x0=1 , y0=0 , f 0=x02+ y0

2=1

x1=1.1 , y1=0.11067 , f 1=x12+ y1

2=1.22225

x2=1.2 , y2=0.24617 , f 2=x22+ y2

2=1.50060

For i=0 :

y3≅ y (1.3 )= y2+h

12 (23 f 2−16 f 1+5 f 0 )

¿ 0.24617+ 0.112

(23 (1.50060 )−16 (1.22225 )+5(1))

¿0.41249

For i=1 :

x3=1.3 , y3=0.41249 , f 3=x32+ y3

2=1.86015

y4≅ y (1.4 )= y3+h

12 (23 f 3−16 f 2+5 f 1 )

¿ 0.41249+ 0.112

(23 (1.86015 )−16 (1.50060 )+5(1.22225))

¿0.61987

For i=2:

x4=1.4 , y4=0.61987 , f 4=x42+ y4

2=2.34424

y5≅ y (1.5 )= y4+h

12 (23 f 4−16 f 3+5 f 2 )

¿ 0.61987+ 0.112

(23 (2.34424 )−16 (1.86015 )+5(1.50060))

¿0.88369

∴ y5≅ y (1.5 )=0.88369

SOLUTION 1(a)

( 3 1 −12 4 1

−1 2 5 )(UVW )=(411)withinitial value(X1

X2

X3)=(000)up¿6iterations .

3 X1+X2−X3=42 X1+4 X2+X3=1

−X1+2 X2+5 X3=1

X1=13 (4−X2+X3 )

X2=14 ( 1−2 X1−X 3 )

X3=15 (1+X1−2 X2 )

i) Jacobi Iteration Method

K=0 X1=0X2=0X3=0

K=1 X1=13 (4−(0 )+(0))=1.333333

X2=14 (1−2 (0 )−(0))=0.25

X3=15 (1+(0)−2(0))=0.2

K=2 X1=13 (4−(0.25 )+ (0.2 ) )=1.316667

X2=14 (1−2 (1.333333 )−(0.2))=−0.466667

X3=15 (1+(1.333333)−2(0.25))=0.366667

K=3 X1=13 (4−(−0.466667 )+(0.366667 ) )=1.611111

X2=14 (1−2 (1.316667 )−(0.366667))=−0.50

X3=15 (1+(1.316667)−2 (−0.466667))=0.276667

K=4 X1=13 (4−(−0.50 )+ (0.276667 ) )=1.592222

X2=14 (1−2 (1.611111)−(0.276667))=−0.624722

X3=15 (1+(1.611111)−2(−0.50))=0.722222

K=5 X1=13 (4−(−0.624722)+ ( 0.722222) )=1.782315

X2=14 (1−2 (1.592222 )−(0.722222))=−0.726667

X3=15 (1+(1.592222)−2(−0.624722))=0.806352

K=6 X1=13 (4−(−0.726667 )+(0.806352 ) )=1.844340

X2=14 (1−2 (1.782315 )−(0.806352))=−0.842746

X3=15 (1+(1.782315)−2(−0.726667))=0.847130

OR

k X1 X2 X3

0 0 0 01 1.333333 0.25 0.22 1.316667 -0.466667 0.3666673 1.611111 -0.50 0.2766674 1.592222 -0.624722 0.7222225 1.782315 -0.726667 0.806352

6 1.844340 -0.842746 0.847130

∴ X1≈2

X2≈−1

X3≈1

ii) Gauss-Seidel Method

K=0 X 1=0X2=0X3=0

K=1 X1=13 (4−(0 )+(0))=1.333333

X2=14 (1−2 (1.333333 )−(0))=−0.416667

X3=15 (1+(1.333333)−2(−0.416667))=0.633333

K=2 X1=13 (4−(−0.416667 )+(0.633333))=1.683333

X2=14 (1−2 (1.683333 )−(0.633333))=−0.75

X3=15 (1+(1.683333 )−2 (−0.75 ) )=0.836667

K=3 X1=13 (4−(−0.75 )+(0.836667))=1.862222

X2=14 (1−2 (1.862222 )−(0.836667))=−0.890278

X3=15 (1+ (1.862222 )−2 (−0.890278 ) )=0.928556

K=4 X1=13 (4−(−0.890278 )+(0.928556))=1.939611

X2=14 (1−2 (1.939611 )−(0.928556))=−0.951945

X3=15 (1+ (1.939611)−2 (−0.951945 ))=0.968700

K=5 X1=13 (4−(−0.951945 )+(0.968700))=1.973548

X2=14 (1−2 (1.973548 )−(0.968700))=−0.978949

X3=15 (1+ (1.973548 )−2 (−0.978949 ) )=0.986289

K=6 X1=13 (4−(−0.978949 )+(0.986289))=1.988413

X2=14 (1−2 (1.988413 )−(0.986289))=−0.990779

X3=15 (1+(1.988413 )−2 (−0.990779 ) )=0.993994

OR

k X1 X2 X3

0 0 0 01 1.333333 -0.416667 0.6333332 1.683333 -0.75 0.8366673 1.862222 -0.890278 0.9285564 1.939611 -0.951945 0.9687005 1.973548 -0.978949 0.9862896 1.988413 -0.990779 0.993994

∴ X1≈2

X2≈−1

X3≈1

SOLUTION 3(a)

(12 7 31 5 12 7 −11)(

X1

X2

X3)=( 2

−56 ) , initial value(X1

X2

X3)=(135)after3 iterations .

Gauss-Seidel Method :

12 X1+7 X2+3 X3=2X1+5 X2+X 3=−5

2 X1+7 X2−11 X3=6

X1=1

12 ( 2−7 X2−3 X3 )

X2=15 (−5−X1−X3 )

X3=−111 (6−2X 1−7 X2 )

K=0 X1=1X2=3X3=5

K=1 X1=1

12 (2−7(3)−3(5))=−2.833333

X2=15 (−5− (−2.833333 )−(5))=−1.433333

X3=−111 (6−2(−2.833333)−7(−1.433333))=−1.972727

K=2 X1=1

12 (2−7(−1.433333)−3 (−1.972727))=1.495959

X2=15 (−5− (1.495959 )−(−1.972727))=−0.904646

X3=−111 (6−2(1.495959)−7(−0.904646))=−0.849146

K=3 X1=1

12 (2−7(−0.904646)−3(−0.849146))=0.906663

X2=15 (−5− (0.906663 )−(−0.849146))=−1.011503

X3=−111 (6−2(0.906663)−7(−1.011503 ))=−1.024290

OR

k X1 X2 X3

0 1 3 51 -2.833333 -1.433333 -1.972727

2 1.495959 -0.904646 -0.8491463 0.906663 -1.011503 -1.024290

∴ X1≈1

X2≈−1

X3≈−1

Additional Questions.

Compute the solution of

y ' '−(0.1 ) (1− y2 ) y '+ y=0

with initial values y (0 )=1 , y ' (0 )=0 up to the third zero of y (t ). Use the Runge-Kutta(4th order) formulas for two first order equations.

Solution :

Convert into two system of first order differential equationLet y=z1∧ y’=z2

z2=z1’

y ”=z2’=(0.1)(1− y2) y ’ – y=(0.1)(1−z12)z2−z1

Therefore, we got

z '=f ( z )=[ z1 'z2 ' ]=[ z2

(0.1 ) (1−z12 ) z2−z1] , where z (0 )=[10]

The formula for Runge – Kutta 4th order method

zi+1 = zi + (h/6)(k1 + 2k2 + 2k3 + k4) where

k1 = f( ti , zi )k2 = f( ti + (h/2) , zi + (hk1)/2 )k3 = f( ti + (h/2) , zi + (hk2)/2 )k4 = f( ti + h , zi + hk3 )

For i=0, z0=[10]

k1=f (¿, [10 ])=[ 0(0.1 ) (1−12 ) (0 )−1]=[ 0

−1]

k 2=f (¿, [10 ]+ 0.12 [ 0

−1])=f (¿ ,[ 1−0.05])

¿ [ −0.05(0.1 ) (1−12 ) (−0.05 )−1]

¿ [−0.05−1 ]

k3=f (¿, [10]+ 0.12 [−0.05

−1 ])=f (¿, [0.9975−0.05 ])

¿ [ −0.05(0.1 ) (1−0.99752 ) (−0.05 )−0.9975]

¿ [ −0.05−0.997525]

k 4=f (¿ ,[10]+0.1[ −0.05−0.997525])=f (¿, [ 0.995

−0.099753])

¿ [ −0.099753(0.1 ) (1−0.9952 ) (−0.099753 )−0.995]

¿ [−0.099753−0.995100]

So,

z1=¿ [10] + 0.16

¿) = [ 0.995004−0.099835]

For i=1, z1=[ 0.995004−0.099835]

k1 ¿ f (__,[ 0.995004−0.099835]

¿ [ −0.099835(0.1 ) ( 1−0.9950042) (−0.099835 )−0.995004 ]

¿ [−0.099835−0.995104]

k 2=f (__,[ 0.995004−0.099835]+ 0.1

2 [[−0.099835−0.995104]]) = f (__,[ 0.990012

−0.149590])

¿ [ −0.149590(0.1 ) ( 1−0.9900122 ) (−0.149590 )−0.990012]

¿ [−0.149590−0.990309]

k3 ¿ f (__,[ 0.995004−0.099835]+0.1

2 [−0.149590−0.990309]) = f (__,[ 0.987525

−0.149350])

¿ [ −0.149350(0.1 ) ( 1−0.9875252 ) (−0.149350 )−0.987525]

¿ [−0.149350−0.987895]

k 4 ¿ f (__,[ 0.995004−0.099835]+0.1[−0.149350

−0.987895]) = f (__,[ 0.980069−0.198625])

¿ [ −0.198625(0.1 ) (1−0.9800692 ) (−0.198625 )−0.980069]

¿ [−0.198625−0.980853]

So,

z2 ¿ [ 0.995004−0.099835]+

0.16

¿)

¿ [ 0.980065−0.198708]

For i=2, z2=[ 0.980065−0.198708]

k1 ¿ f (__,[ 0.980065−0.198708]

¿ [ −0.198708(0.1 ) (1−0.9800652 ) (−0.198708 )−0.980065]

¿ [−0.198708−0.980849]

k 2 ¿ f (__,[ 0.980065−0.198708]+ 0.1

2 [−0.198708−0.980849]) = f (__,[ 0.970130

−0.247750])

¿ [ −0.247750(0.1 ) ( 1−0.9701302 ) (−0.247750 )−0.970130]

¿ [−0.247750−0.971588]

k3 ¿ f (__,[ 0.980065−0.198708]+0.1

2 [[−0.247750−0.971588]]) =f (__,[ 0.967678

−0.247287])

¿ [ −0.247287(0.1 ) ( 1−0.9676782 ) (−0.247287 )−0.967678]

¿ [−0.247287−0.969251]

k 4 ¿ f (__,[ 0.980065−0.198708]+0.1[[−0.247287

−0.969251]]) =f (__,[ 0.955336−0.295633])

¿ [ −0.295633(0.1 ) (1−0.9553362 ) (−0.295633 )−0.955336 ]

¿ [−0.295633−0.957918]

So,

z3=¿ [ 0.980065−0.198708]+

0.16

¿

+[−0.295633−0.957918])

¿ [ 0.955325−0.295715]

∴ z3=[ 0.955325−0.295715] .