numerical analysis solution

7/28/2019 Numerical Analysis Solution

1/14

NUMERICAL

ANALYSIS(SOLVED EXERCISES)

Olayinka G. Okeola


2/14

2

Question:Determine the values ofsuch that the b.v.p 0" uu ; 0)0( u ; 0)1( u has non-trivial

solution for 5n using canomical form of solution.

Solution

Recall Chebyshev polynomial

arcxTn cos()( )cos x ; 11 x

0n 10cos)( xTo

1n arcxT cos()(1 xx )cos

The recursive relation generated as follows:

12)( 22 xxT

xxxT 34)( 33

188)( 244 xxxT

xxxxT 52016)( 355

Given condition: 0)0( u ; ,0)1( u 10 x

Transforming )(xTn , 11 x into 10 x

x , are constants

0 (1)

1 (2)

21

Hence2

1

2

1 x

12 x

12 x


3/14

3

1)( xTo

12)(1 xxT

1881)12(2)( 222 xxxxT

14832)12(3)12(4)(233

3 xxxxxT

1321602561281)12(8)12(8)( 234244 xxxxxxxT

15040011201280512)12(5)12(20)12(16)( 2345355 xxxxxxxxxT

)()1(1)(

)()1()(1

)(

)()1()()(

)()1()(

)()1()}({

)()1()(

)1()(

)(

0

0)(

0

2

2

2

2

2

2

2

2

2

2

2

"

xQiixxQ

xQiixLQxQ

xQiixLQxQ

QxQiixLQ

LQxLQiixLQL

LQxLQiixLQ

xxiixLQ

xxLQ

xL

LU

Ux

UU

i

i

i

iii

iii

iii

iii

iii

ii

i

i

i


4/14

4

32

35

5

32

4

4

2

3

3

2

2

2

1

12020)(5

2412)(4

6)(3

2)(2

)(1

1)(0

xxxxQi

xxQi

xxxQi

xxQi

xxQi

xQi o

)()()()(

)()()(

)()()(

)()()(

)()()()()(

0

1

0

)1(

2

0

)(

1

1

0

)1(

2

0

)(

1

1

0

)1(

212

0

)(

11

121

xQcxcxxLU

xcxxcxxLU

xcxxTx

xcxxTx

xTxxTxxLU

LU

i

n

i

n

i

n

i

n

in

n

i

in

i

n

i

in

in

n

i

in

in

n

i

in

in

nnn

For U(0) = 0

)0()()(01

0

)1(

2

0

)(

1 i

n

i

n

i

n

i

n

i Qcxcx

(1)

For U(1) =0

)1()()(01

0

)1(

2

0

)(

1 i

n

i

n

i

n

i

n

i Qcxcx

(2)

0

0

)1()1(

)0()0(

2

1

1

0

)1(

0

)(

1

0

)1(

0

)(

n

i

i

n

i

n

i

i

n

i

n

i

i

n

i

n

i

i

n

i

QcQc

QcQc


5/14

5

For n=5

1)

32

32

5554

543

532

521

510

50

5

0

5

307208001

)0)(512(24)1280()0)(1120()2()400()0)(50(1)1(

)0()0()0()0()0()0()0(

QcQcQcQcQcQcQci

ii

2)

32

32

4

4

43

4

32

4

21

4

10

4

0

4

0

4

30723201

24)128()0)(256(

)2()160()0)(32(

1)1(

)0()0()0()0()0()0(

QcQcQcQcQcQci

ii

3)

32

23222

5554

543

532

521

510

50

5

0

5

921608001

201)512(

24121)1280(

61)1120(

21)400()1)(50(

1)1(

)1()1()1()1()1()1()1(

QcQcQcQcQcQcQci

ii

4)

32

3222

4

4

43

4

32

4

21

4

10

4

0

4

0

4

30723201

24121)128(

61)256(

21)160(

1)32(

1)1(

)1()1()1()1()1()1(

QcQcQcQcQcQci

ii

3232

3232

30723201921608001

30723201307208001

0

0

2

1

A


6/14

6

1474560091340867208)(

1887436801474560045670422402)(

01887436801474560045670422402

0188743601474560045675422402

03072320192160800130723201307208001

0

23'

234

234

65432

32323232

f

f

A

By Newton-Raphson method

)(

)('1

f

fnn

n 2 11.51573

3 11.06039

4 10.70213

5 10.42561

6 10.21848

7 10.07066

8 9.97386

9 9.92118

10 9.85577

9.9


7/14

7

Problem Statement1) Use the explicit method to solve for the temperature distribution of a long, thin rod with a

length lf 10cm and the following value: )./(49.0' CcmsCalk o , cmx 2 . At t = 0, the

temperature of the rod is zero and the boundary conditions are fixed for all times at

T(0)=1000C and T(10) =50

0C. Note that the rod is aluminum with )./(2174.0 CgCalc o

and 3/7.2 cmg . Compute results to t =0.2 and compare those in example 24.1

2) Use the Crank-Nocolson method to solve problem 1 above for cmx 5.2 .

Solutions1)

010437.0

/835.0)2174.07.2(

)49.0(

2.0

05.02

2

2

x

tk

scmk

t

stcm


8/14

8

)2(11

1 ji

ji

ji

ji

ji TTTTT

At t= 0.05s

0,1 ji 0437.1)100)0(20(0)2( 000

1

0

2

0

1

1

1 TTTTT

0,2 ji 0)0)0(20(0)2( 010

2

0

3

0

2

1

2 TTTTT

0,3 ji 0)0)0(20(0)2( 020

3

0

4

0

3

1

3 TTTTT

0,4 ji 52185.0)0)0(250(0)2( 03

04

05

04

14 TTTTT

At t = 0.1s

1,1 ji 04382.2)100)0875.2(20(0437.1)2( 101

1

1

2

1

1

2

1 TTTTT

1,2 ji 01089.0)0437.1)0(20(0)2( 111

2

1

3

1

2

2

2 TTTTT

1,3 ji 00544.0)0)0(252185.0(0)2( 121

3

1

4

1

3

2

3 TTTTT

1,4 ji 02191.1)0)52185.0(250(52185.0)2( 1314

15

14

24 TTTTT

At t = 0.15s

2,1 ji

04497.3)100)04382.2(201089.0(04382.2)2( 202

1

1

2

2

1

3

1 TTTTT

2,2 ji

03205.0)04382.2)01089.0(200544.0(01089.0)2( 212

2

2

3

2

2

3

2 TTTTT

2,3 ji

0161.0)01089.0)00544.0(202191.1(00544.0)2( 222

3

2

4

2

3

3

3 TTTTT

2,4 ji

52248.1)00544.0)02191.1(250(02191.1)2( 2324

25

24

34

TTTTT


9/14

9

At t = 0.2s

3,1 ji

02544.4)100)04497.3(203205.0(04497.3)2( 303

1

3

2

3

1

4

1 TTTTT

3,2 ji06333.0)04497.3)03205.0(20161.0(03205.0)2( 31

3

2

3

3

3

2

4

2 TTTTT

3,3 ji

03198.0)03205.0)00161.0(252248.1(0161.0)2( 323

3

3

4

3

3

4

3 TTTTT

3,4 ji 01272.2)0161.0)52248.1(250(52248.1)2( 33

34

35

34

44 TTTTT

(2)

Using Crank-Nicolson to solve the problem 1 with cmx 5.2

01336.0

/835.0

2.0

1.0

5.2

2

2

tk

scmk

st

st

cmx

First interior node

)()1(2)()1(2 102101

21

1 llllll tfTTtfTT

Last interior node

)()1(2)()1(2 111111

1

l

m

l

m

l

m

l

m

l

m

l

m tfTTtfTT

Inner nodes

l

i

l

i

l

i

l

i

l

i

l

i TTTTTT 111

1

11

1 )1(2)1(2


10/14

1 0

At t = 0.1s

0,1 li

672.201336.002672.2

)100(00)100(01336.002672.2

)()1(2)()1(2

12

11

12

11

10

02

01

00

12

11

TT

TT

tfTTtfTT

(1)

0,2 li001336.002672.201336.0

)1(2)1(2

1

3

1

2

1

1

0

3

0

2

0

1

1

3

1

2

1

1

TTT

TTTTTT (2)

0,3 li001336.002672.201336.0

)1(2)1(2

1

4

1

3

1

2

0

4

0

3

0

2

1

4

1

3

1

2

TTT

TTTTTT (3)

0,4 li336.102672.201336.0

)'()1(2)'()1(2

1

4

1

3

5

00

35

1

4

1

3

TT

tfTTtfTT (4)


11/14

1 1

336.1

0

0

672.2

02672.201336.0

01336.002672.201336.0

01336.002672.201336.0

01336.002672.2

14

13

12

11

T

T

T

T

Using Gauss elimination method[1] [1]

See Appendix

65917.0

00428.0

0087.0

3183.1

1

4

1

3

1

2

1

1

T

T

T

T

At t = 0.2s

1,1 li

2733.5336.100012.06014.2336.101336.002672.2

)100()0087.0()3183.1(97328.1)100(02672.2

)()1(2)()1(2

22

21

22

21

20

12

11

10

22

21

TT

TT

tfTTtfTT

(1)

1,2 li

0005.001336.002672.201336.0

0006.001717.001761.001336.002672.201336.0

)1(2)1(2

23

22

21

23

22

21

13

12

11

23

22

21

TTT

TTT

TTTTTT

(2)

1,3 li

01713.001336.002672.201336.0

00881.00084.000012.001336.002672.201336.0

)1(2)1(2

2

4

2

3

2

2

2

4

2

3

2

2

1

4

1

3

1

2

2

4

2

3

2

2

TTT

TTT

TTTTTT

(3)

1,4 li36182.1)50(00881.001701.0)50(02672.201336.0

)()1(2)'()1(2

2

4

2

3

2

5

1

4

1

35

2

4

2

3

TT

tfTTtfTT(4)


12/14

1 2

36182.1

01713.0

0005.0

2733.5

02672.201336.000

01336.002672.201336.00

001336.002672.201336.0

0001336.002672.2

24

23

22

21

T

T

T

T

Using Gauss elimination method[2] [2]

See Appendix

67526.0

50645.0

02074.0

60214.2

1

4

1

3

1

2

1

1

T

T

T

T

Appendix [1]

336.1

0

0

672.2

02672.201336.000

01336.002672.201336.00

001336.002672.201336.0

0001336.002672.2

1

4

1

3

1

2

1

1

T

T

T

T

02672.201336.0

21m which gives

336.1

0

01761.0

672.2

02672.201336.000

01336.002672.201336.00

001336.002672.20

0001336.002672.2

14

13

12

11

T

T

T

T

3

32 1059222.60266.2

01336.0 m which gives

336.1

000116.0

01761.0

672.2

02672.201336.000

01336.002672.200

001336.002663.20

0001336.002672.2

14

13

12

11

T

T

T

T


13/14

1 3

3243 mm which gives

3359.1000116.0

01761.0

672.2

02663.200001336.002663.200

001336.002663.20

0001336.002672.2

1

4

1

3

1

2

1

1

TT

T

T

Solving by back substitution give

3183.1672.201336.002663.2

008666.001761.001336.002663.2

004288.0000116.0)65917.0(01336.002663.2

05917.002663.2

3359.1

1

1

1

2

1

1

1

2

1

3

1

2

1

3

1

3

1

4

TTT

TTT

TT

T

Appendix [2]

36182.1

01713.0

0005.0

2733.5

02672.201336.000

01336.002672.201336.00

001336.002672.201336.0

0001336.002672.2

24

23

22

21

T

T

T

T

02672.201336.0

21m which gives

36182.1

01713.0

03526.0

2733.5

02672.201336.000

01336.002672.201336.00

001336.002663.20

0001336.002672.2

24

23

22

21

T

T

T

T

3

32 1059222.60266.201336.0 m which gives


14/14

1 4

36182.1

01736.1

03526.0

2733.5

02672.201336.000

01336.002672.200

001336.002663.20

0001336.002672.2

24

23

22

21

T

T

T

T

3243 mm which gives

36852.1

01736.1

03526.0

2733.5

02663.2000

01336.002672.200

001336.002663.20

0001336.002672.2

24

23

22

21

T

T

T

T

Solving by back substitution give

60214.22733.501336.002663.2

02074.003526.001336.002663.2

50645.001736.1)67526.0(01336.002663.2

67526.002663.2

36852.1

2

1

2

2

2

1

2

2

2

3

2

2

2

3

2

3

2

4

TTT

TTT

TT

T

numerical analysis solution

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