NUMERICAL ANALYSIS
Olayinka G. Okeola
Determine the values of such that the b.v.p u" u 0 ; u (0) 0 ; u (1) 0 has non-trivial solution for n 5 using canomical form of solution. Solution Recall Chebyshev polynomial T n ( x) cos(arc cos x) ; 1 x 1
n 0 T o ( x ) cos 0 1 n 1 T 1 ( x) cos(arc cos x) x The recursive relation generated as follows:
T 2 ( x ) 2 x 2 1 T 3 ( x ) 4 x 3 3 x
T 4 ( x ) 8 x 4 8 x 2 1 T 5 ( x ) 16 x 5 20 x 3 5 x
Given condition: u (0) 0 ; u (1) 0, 0 x 1
Transforming T n ( x) , 1 x 1 into 0 x 1 x λ, γ are constants 0
(1)
1
(2)
1
2 Hence
1 2
x
1 2
2 x 1 x 2 1
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T o ( x) 1
T 1 ( x) 2 x 1 T 2 ( x ) 2(2 x 1) 2 1 8 x 2 8 x 1 T 3 ( x ) 4( 2 x 1) 3 3( 2 x 1) 32 x 3 48 x 2 1
T 4 ( x) 8(2 x 1) 4 8(2 x 1) 2 1 128 x 4 256 x 3 160 x 2 32 x 1 T 5 ( x ) 16( 2 x 1) 5 20( 2 x 1) 3 5(2 x 1) 512 x 5 1280 x 4 1120 x 3 400 x 2 50 x 1
U U 0 "
2 ( 2 )U 0 x LU 0 2 L 2 x i LQi ( x) x LQi ( x) i(i 1) x
i 2
x i LQi ( x) i(i 1) LQi 2 ( x) LQi L{ LQi ( x)} i(i 1) LQi 2 ( x) LQi LQi ( x) i(i 1)Qi 2 ( x ) Qi Qi ( x) LQi ( x) i(i 1)Qi 2 ( x) Qi ( x ) Qi ( x )
1 1
LQi ( x) i (i 1)Qi 2 ( x)
x
i
i (i 1)Qi 2 ( x)
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1
i 0 Qo ( x ) i 1 Q1 ( x)
x
i 2 Q2 ( x ) i 3 Q3 ( x ) i 4 Q4 ( x ) i 5 Q5 ( x)
x 2 x
2
3
6 x
12
x 4 x 5
2 2 2
20 x 3
2
24 3
120 x 3
LU 0 LU n ( x) 1 ( x )T n ( x ) 2 ( x)T n 1 ( x) 1 ( x )T n ( x )
n
( x)c 1
( n)
x i
i
i 0 n 1
( x)c
2 ( x )T n 1 ( x)
2
( n 1) i
x i
i 0
LU n ( x)
n
( x)c 1
x
(n) i
i
i 0
n 1
( x)c 2
( n 1) i
x i
i 0
n 1 n (n) ( n 1) LU n ( x) 1 ( x)ci 2 ( x )c i Qi ( x) i 0 i 0
For U(0) = 0
n
n 1
Qi (0)
(1)
n 1 n ( n) ( n 1) 0 1 ( x )ci 2 ( x)ci Qi (1) i 0 i 0
(2)
0
1 ( x )ci
( n)
i 0
2 ( x)ci i 0
( n 1)
For U(1) =0
n ( n ) c i Qi (0) i n0 ( n) c i Qi (1) i 0
1 0 i 0 n 1 0 ( n 1) ci Qi (1) 2 i 0 n 1
c
( n 1)
i
Qi (0)
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For n=5 5
5
5
5
5
5
5
5
ci Qi (0) c0 Q0 (0) c1 Q1 (0) c2 Q2 (0) c3 Q3 (0) c4 Q4 (0) c5 Q5 (0)
i 0
1
1) ( 1)
1
(50)(0) (400)
( 2) 2
(1120)(0) (1280)
800 2
24 3
(512)(0)
30720 3
4
c Q (0) c Q (0) c Q (0) c Q (0) c Q (0) c Q (0) 4 i
4 0
i
0
4 1
4 2
1
4 3
2
4 4
3
4
i 0
1
2) (1)
(32)(0) (160)
1 320
5
2
5
( 2) 2
(256)(0) (128)
24 3
3072 3 5
5
5
5
5
5
ci Qi (1) c0 Q0 (1) c1 Q1 (1) c2 Q2 (1) c3 Q3 (1) c4 Q4 (1) c5 Q5 (1)
i 0
3) ( 1)
1
1 2 1 6 1 12 24 1 20 120 (50)( 1 ) (400) (1120) (1280) (512) 2 2 2 3 2 3 1
800 2
92160 3
4
c Q (1) c Q (1) c Q (1) c Q (1) c Q (1) c Q (1) 4 i
4 0
i
0
4 1
1
4 2
2
4 3
3
4 4
4
i 0
4) (1)
1
1
(32)
320 2
1 2 1 6 1 12 24 (160) 2 (256) 2 (128) 2 3 1
3072 3
1 800 30720 1 320 3072 2 3 0 2 3 1 0 1 800 92160 1 320 3072 2 2 3 2 3
A
5 e g a P
A 0
1 800 30720 1 320 3072 1 800 92160 1 320 3072 3 2 3 0 2 3 2 3 2
2 2
2240 3
456754 4
14745600 5
18874360 6
0
2 4 2240 3 456704 2 14745600 188743680 0 f ( ) 2 4 2240 3 456704 2 14745600 188743680 f ' ( ) 8 3 6720 2 913408 14745600 By Newton-Raphson method n 1 n
n 2 3 4 5 6 7 8 9 10
f ( ) f ' ( )
11.51573 11.06039 10.70213 10.42561 10.21848 10.07066 9.97386 9.92118 9.85577
9.9
6 e g a P
1) Use the explicit method to solve solve for the temperature distribution distribution of a long, thin rod rod with a length lf 10cm and the following value: k ' 0 .49 Cal /( s .cm o C ) , x 2cm . At t = 0, the temperature of the rod is zero and an d the boundary conditions are fixed for all times at T(0)=100 C and T(10) =50 C. Note that the rod is aluminum with c 0.2174Cal /( g .o C ) 0
0
and 2.7 g / cm 3 . 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 x 2.5cm .
1)
2cm t 0.05 s t 0.2 k
(0.49) k t
x 2
(2.7 0.2174)
0.835cm 2 / s
0.010437
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T i
j 1
T i j (T i j1 2T i j T i j 1 )
At t= 0.05s i 1, j 0
T 11 T 10 (T 20 2T 10 T 00 ) 0 (0 2(0) 100) 1.0437
i 2, j 0
T 21 T 20 (T 30 2T 20 T 10 ) 0 (0 2(0) 0) 0
i 3, j 0
T 31 T 30 (T 40 2T 30 T 20 ) 0 (0 2(0) 0) 0
i 4, j 0
1 0 0 0 0 T 4 T 4 (T 5 2T 4 T 3 ) 0 (50 2(0) 0) 0.52185
At t = 0.1s i 1, j 1
T 12 T 11 (T 21 2T 11 T 01 ) 1.0437 (0 2( 2.0875) 100) 2.04382
i 2, j 1
T 22 T 21 (T 31 2T 21 T 11 ) 0 (0 2(0) 1.0437 ) 0.01089
i 3, j 1
T 32 T 31 (T 41 2T 31 T 21 ) 0 (0.52185 2(0) 0) 0.00544
i 4, j 1
T 42 T 41 (T 51 2T 41 T 31 ) 0.52185 (50 2(0.52185) 0) 1.02191
At t = 0.15s i 1, j 2 T 13 T 12 (T 21 2T 12 T 02 ) 2.04382 (0.01089 2( 2.04382) 100) 3.04497 i 2, j 2 T 23 T 22 (T 32 2T 22 T 12 ) 0.01089 (0.00544 2(0.01089) 2.04382) 0.03205 i 3, j 2 T 33 T 32 (T 42 2T 32 T 22 ) 0.00544 (1.02191 2(0.00544) 0.01089) 0.0161 i 4, j 2 3 2 2 2 2 T 4 T 4 (T 5 2T 4 T 3 ) 1.02191 (50 2(1.02191) 0.00544 ) 1.52248
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At t = 0.2s i 1, j 3 T 14 T 13 (T 23 2T 13 T 03 ) 3.04497 (0.03205 2(3.04497 ) 100) 4.02544 i 2, j 3 T 24 T 23 (T 33 2T 23 T 13 ) 0.03205 (0.0161 2(0.03205) 3.04497 ) 0.06333 i 3, j 3 T 34 T 33 (T 43 2T 33 T 23 ) 0.0161 (1.52248 2(0.00161) 0.03205) 0.03198 i 4, j 3
4 3 3 3 3 T 4 T 4 (T 5 2T 4 T 3 ) 1.52248 (50 2(1.52248 ) 0.0161) 2.01272
(2) Using Crank-Nicolson to solve the problem 1 with x 2.5cm
x 2.5cm t 0.1 s t 0.2 s k 0.835cm 2 / s
k t
2
0.01336
First interior node l 1
2(1 )T 1
T 2l 1 f 0 (t l ) 2(1 )T 1l T 2l f 0 (t l 1 )
Last interior node
T ml 11 2(1 )T ml 1 f m 1 (t l ) 2(1 )T ml 1 T ml f m 1 (t l 1 ) Inner nodes
T i l 11 2(1 )T i l 1 T i l 11 T i l 1 2(1 )T i l T i l 1
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At t = 0.1s 2(1 )T 11 T 21 f 0 (t 0 ) 2(1 )T 10 T 20 f 0 (t 1 ) 1 1 i 1, l 0 2.02672T 1 0.01336T 2 (100) 0 0 (100)
(1)
2.02672T 11 0.01336T 21 2.672
i 2, l 0
i 3, l 0
i 4, l 0
T 11 2(1 )T 21 T 31 T 10 2(1 )T 20 T 30 0.01336T 11 2.02672T 21 0.01336T 31 0
T 21 2(1 )T 31 T 41 T 20 2(1 )T 30 T 40 0.01336T 21 2.02672T 31 0.01336T 41 0
T 31 2(1 )T 41 f 5 (t ' ) 2(1 )T 30 T 0 f 5 (t ' ) 0.01336T 31 2.02672T 41 1.336
(2)
(3)
(4) 0 1 e g a P
2.02672 0.01336 T 11 2.672 0.01336 2.02672 0.01336 1 0 T 2 1 0 0.01336 2.02672 0.01336 T 3 1 0.01336 2.02672 T 4 1.336
[1]
[1]
Using Gauss elimination method
See Appendix
T 11 1.3183 T 21 0.0087 T 31 0.00428 T 41 0.65917
At t = 0.2s 2(1 )T 12 T 22 f 0 (t 1 ) 2(1 )T 11 T 21 f 0 (t 2 ) 2 2 i 1, l 1 2.02672T 1 T 2 (100) 1.97328(1.3183) ( 0.0087) (100)
(1)
2.02672T 12 0.01336T 22 1.336 2.6014 0.00012 1.336 5.2733
T 12 2(1 )T 22 T 32 T 11 2(1 )T 21 T 31 i 2, l 1
0.01336T 12 2.02672T 22 0.01336T 32 0.01761 0.01717 0.0006
(2)
0.01336T 12 2.02672T 22 0.01336T 32 0.0005 T 22 2(1 )T 32 T 42 T 21 2(1 )T 31 T 41 i 3, l 1
0.01336T 22 2.02672T 32 0.01336T 42 0.00012 0.0084 0.00881
(3)
0.01336T 22 2.02672T 32 0.01336T 42 0.01713
i 4, l 1
T 32 2(1 )T 42 f 5 (t ' ) 2(1 )T 31 T 41 f 5 (t 2 ) 0.01336T 2.02672T (50) 0.01701 0.00881 (50) 1.36182 2 3
2 4
(4)
1 1 e g a P
0 0 2.02672 0.01336 T 12 5.2733 0.01336 2.02672 0.01336 2 0.0005 0 T 2 2 0.01713 0 0.01336 2.02672 0.01336 T 3 2 0.01336 2.02672 T 4 1.36182 0 0
[2]
Using Gauss elimination method
[2]
See Appendix
T 11 2.60214 T 21 0.02074 T 31 0.50645 T 41 0.67526
Appendix [1] 0 0 2.02672 0.01336 T 11 2.672 0.01336 2.02672 0.01336 T 1 0 0 2 0 0.01336 2.02672 0.01336 T 31 0 1 0 0 0 . 01336 2 . 02672 T 4 1.336
m 21 0.01336
2.02672
which gives
0 0 2.02672 0.01336 T 11 2.672 0 1 0.01761 2.02672 0.01336 0 T 2 0 0.01336 2.02672 0.01336 T 31 0 0.01336 2.02672 T 41 1.336 0 0 m32 0.01336
2.0266
6.59222 10 3 which gives
0 0 2.02672 0.01336 T 11 2.672 0 1 0.01761 2.02663 0.01336 0 T 2 1 0 0.000116 0 2.02672 0.01336 T 3 1 0.01336 2.02672 T 4 1.336 0 0
2 1 e g a P
m 43 m32 which gives 0 0 2.02672 0.01336 T 11 2.672 0 1 0.01761 2.02663 0.01336 0 T 2 1 0 0 2.02663 0.01336 T 3 0.000116 0 0 2.02663 T 41 1.3359 0
Solving by back substitution give T 41 1.3359
2.02663
0.05917
2.02663T 31 0.01336(0.65917) 0.000116 T 31 0.004288 2.02663T 21 0.01336T 31 0.01761 T 21 0.008666 2.02663T 11 0.01336T 21 2.672 T 11 1.3183
Appendix [2] 0 0 2.02672 0.01336 T 12 5.2733 0.01336 2.02672 0.01336 2 0.0005 0 T 2 2 0.01336 2.02672 0.01336 T 3 0 0.01713 0 0 0.01336 2.02672 T 42 1.36182
m 21 0.01336
2.02672
which gives
0 0 2.02672 0.01336 T 12 5.2733 0 2 0.03526 2.02663 0.01336 0 T 2 0 0.01336 2.02672 0.01336 T 32 0.01713 2 0 0 0 . 01336 2 . 02672 T 4 1.36182 m32 0.01336
2.0266
6.59222 10 3 which gives
3 1 e g a P
0 0 2.02672 0.01336 T 12 5.2733 0 2 0.03526 2.02663 0.01336 0 T 2 2 0 1.01736 0 2.02672 0.01336 T 3 2 0.01336 2.02672 T 4 1.36182 0 0
m 43 m32 which gives 0 0 2.02672 0.01336 T 12 5.2733 0 2 0.03526 2.02663 0.01336 0 T 2 2 0 1.01736 0 2.02672 0.01336 T 3 2 0 0 2.02663 T 4 1.36852 0 Solving by back substitution give T 42 1.36852
2.02663
0.67526
2.02663T 32 0.01336(0.67526) 1.01736 T 32 0.50645 2.02663T 22 0.01336T 32 0.03526 T 22 0.02074 2.02663T 12 0.01336T 22 5.2733 T 12 2.60214
4 1 e g a P
