Numerical Analysis Solution

NUMERICALANALYSIS (SOLVED EXERCISES) Olayinka G. Okeola Question: 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 Tn ( x)  cos(arc cos x) ;  1  x  1 n  0 To ( x)  cos 0  1 n  1 T1 ( x)  cos(arc cos x )  x The recursive relation generated as follows: T2 ( x)  2 x 2  1 T3 ( x )  4 x 3  3 x T4 ( x)  8 x 4  8 x 2  1 T5 ( x )  16 x 5  20 x 3  5 x Given condition: u (0)  0 ; u (1)  0, 0  x  1 Transforming Tn (x) ,  1  x  1 into 0  x  1   x   λ, γ are constants 0     (1) 1   (2) 1   2 Hence   1 1 x 2 2 Page x  2  1 2 2  x  1 To ( x)  1 T1 ( x)  2 x  1 T2 ( x)  2(2 x  1) 2  1  8 x 2  8 x  1 T3 ( x )  4( 2 x  1) 3  3( 2 x  1)  32 x 3  48 x 2  1 T4 ( x)  8(2 x  1) 4  8(2 x  1) 2  1  128 x 4  256 x 3  160 x 2  32 x  1 T5 ( 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   )U  0 x 2 LU  0 ( 2  x 2 LQi ( x)  x i L 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 Page 3 Qi ( x)  LQi ( x)  i(i  1)Qi 2 ( x) 1 Qi ( x)  LQi ( x)  i (i  1)Qi 2 ( x)  1 Qi ( x)  x i  i (i  1)Qi  2 ( x)  1  x i  1  Q1 ( x)   x2 2  i  2  Q2 ( x )   2 x 3 6x  i  3  Q3 ( x)   2 x 4 12 24   i  4  Q4 ( x )    2 3 x 5 20 x 3 120 x  2  3 i  5  Q5 ( x)     i  0  Qo ( x)  LU  0 LU n ( x)   1 ( x)Tn ( x)   2 ( x)Tn1 ( x) n  1 ( x)Tn ( x)   1 ( x)ci ( n ) x i i 0 n 1  2 ( x)Tn 1 ( x)   2 ( x)ci ( n 1) x i i 0 n n 1 LU n ( x)   1 ( x)ci x i   2 ( x)ci (n) i 0 ( n 1) xi i 0  (n) ( n 1)  LU n ( x)    1 ( x)ci   2 ( x)ci Qi ( x) i0  i 0  n 1 n For U(0) = 0 n 1  n ( n) ( n 1)  0     1 ( x )c i    2 ( x )c i Qi (0) i 0  i 0  (1) For U(1) =0 n 1  n (n) ( n 1)  0     1 ( x )c i    2 ( x ) c i Qi (1) i 0  i 0   Qi (0)  i 0   1    0  n 1    0  ( n 1) ci Qi (1)  2     i 0  n 1 ( n 1) i 4 c Page  n (n)   c i Qi (0)   i n0  (n)   c i Qi (1)  i 0 (2) For n=5 5 5 5 5 5 5 5 5  ci Qi (0)  c0 Q0 (0)  c1 Q1 (0)  c 2 Q2 (0)  c3 Q3 (0)  c 4 Q4 (0)  c5 Q5 (0) i 0 1)  (1)  4 1 (2) 24  (50)(0)  (400)  (1120)(0)  (1280)  (512)(0)  2 3 1 800 30720    2 3  c Q (0)  c Q (0)  c Q (0)  c Q (0)  c Q (0)  c Q (0) i 0 4 i i 4 0 4 1 0 4 2 1 4 3 2 4 4 3 4 1 (2) 24  (32)(0)  (160) 2  (256)(0)  (128) 3    1 320 3072   2  3    2)  (1) 5 5 5 5 5 5 5 5  ci Qi (1)  c0 Q0 (1)  c1 Q1 (1)  c 2 Q2 (1)  c3 Q3 (1)  c 4 Q4 (1)  c5 Q5 (1) i 0 3)  (1)  1 2  1 6   1 12 24   1 20 120  1   (1120)    (1280)    (512)    (50)( 1 )  (400)     2 2 2 3              2 3  1 800 92160    2 3 4  c Q (1)  c Q (1)  c Q (1)  c Q (1)  c Q (1)  c Q (1) i 0 4 i i 4 0 0 4 1 1 4 2 2 4 3 3 4 4 4 1 1 1 2  1 6   1 12 24   (32)  (160)  2   (256)  2   (128)  2  3              1 320 3072   2  3    4)  (1) 5 A 1 320 3072    3     0  2   1  1 320 3072   2   0    3   2   Page  1 800 30720   2  3     1  800  92160  3   2 A 0  1 800 30720  1 320 3072   1 800 92160  1 320 3072   3   2   3 0   2      3   2     3   2      2 2240 456754 14745600 18874360  2  3    0   4 5 6  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 Page 6   9.9 Problem Statement 1) Use the explicit method to solve for the temperature distribution of a long, thin 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 the boundary conditions are fixed for all times at T(0)=1000C and T(10) =500C. Note that the rod is aluminum with c  0.2174Cal /( g .o C ) 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 . Solutions 1) Page 7 x  2cm t  0.05s t  0.2 k  (0.49)  0.835cm 2 / s (2.7  0.2174) kt   2  0.010437 x Ti j 1  Ti j   (Ti j 1  2Ti j  Ti j 1 ) At t= 0.05s i  1, j  0 T11  T10   (T20  2T10  T00 )  0   (0  2(0)  100)  1.0437 i  2, j  0 T21  T20   (T30  2T20  T10 )  0   (0  2(0)  0)  0 i  3, j  0 T31  T30   (T40  2T30  T20 )  0   (0  2(0)  0)  0 i  4, j  0 T41  T40   (T50  2T40  T30 )  0   (50  2(0)  0)  0.52185 At t = 0.1s i  1, j  1 T12  T11   (T21  2T11  T01 )  1.0437   (0  2( 2.0875)  100)  2.04382 i  2, j  1 T22  T21   (T31  2T21  T11 )  0   (0  2(0)  1.0437)  0.01089 i  3, j  1 T32  T31   (T41  2T31  T21 )  0   (0.52185  2(0)  0)  0.00544 i  4, j  1 T42  T41   (T51  2T41  T31 )  0.52185   (50  2(0.52185)  0)  1.02191 At t = 0.15s i  1, j  2 T13  T12   (T21  2T12  T02 )  2.04382   (0.01089  2( 2.04382)  100)  3.04497 i  2, j  2 T23  T22   (T32  2T22  T12 )  0.01089   (0.00544  2(0.01089)  2.04382)  0.03205 i  3, j  2 T33  T32   (T42  2T32  T22 )  0.00544   (1.02191  2(0.00544)  0.01089)  0.0161 Page T43  T42   (T52  2T42  T32 )  1.02191   (50  2(1.02191)  0.00544 )  1.52248 8 i  4, j  2 At t = 0.2s i  1, j  3 T14  T13   (T23  2T13  T03 )  3.04497   (0.03205  2(3.04497)  100)  4.02544 i  2, j  3 T24  T23   (T33  2T23  T13 )  0.03205   (0.0161  2(0.03205)  3.04497)  0.06333 i  3, j  3 T34  T33   (T43  2T33  T23 )  0.0161   (1.52248  2(0.00161)  0.03205)  0.03198 i  4, j  3 T44  T43   (T53  2T43  T33 )  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  0.1s t  0.2s k  0.835cm 2 / s kt   2  0.01336 x First interior node 2(1   )T1l 1  T2l 1  f 0 (t l )  2(1   )T1l  T2l   f 0 (t l 1 ) Last interior node  Tml 11  2(1   )Tml 1  f m 1 (t l )  2(1   )Tml 1  Tml  f m 1 (t l 1 ) Inner nodes Page 9  Ti l 11  2(1   )Ti l 1  Ti l 11  Ti l1  2(1   )Ti l  Ti l1 At t = 0.1s 2(1   )T11  T21  f 0 (t 0 )  2(1   )T10  T20  f 0 (t 1 ) i  1, l  0 2.02672T11  0.01336T21   (100)  0  0   (100) (1) 2.02672T11  0.01336T21  2.672 i  4, l  0  T21  2(1   )T31  T41  T20  2(1   )T30  T40  0.01336T21  2.02672T31  0.01336T41  0  T31  2(1   )T41  f 5 (t ' )  2(1   )T30  T 0  f 5 (t ' )  0.01336T31  2.02672T41  1.336 (2) (3) (4) 10 i  3, l  0  0.01336T11  2.02672T21  0.01336T31  0 Page i  2, l  0  T11  2(1   )T21  T31  T10  2(1   )T20  T30    2.02672  0.01336  T11  2.672  0.01336 2.02672  0.01336  1  0     T2        0.01336 2.02672  0.01336 T31   0          0.01336 2.02672  T41  1.336   Using Gauss elimination method[1] [1] See Appendix T11  1.3183 T21  0.0087 T31  0.00428 T41  0.65917 At t = 0.2s 2(1   )T12  T22  f 0 (t 1 )  2(1   )T11  T21  f 0 (t 2 ) i  1, l  1 2.02672T12  T22   (100)  1.97328(1.3183)   (0.0087)   (100) (1) 2.02672T12  0.01336T22  1.336  2.6014  0.00012  1.336  5.2733  T12  2(1   )T22  T32  T11  2(1   )T21  T31 i  2, l  1  0.01336T12  2.02672T22  0.01336T32  0.01761  0.01717  0.0006 (2)  0.01336T12  2.02672T22  0.01336T32  0.0005  T22  2(1   )T32  T42  T21  2(1   )T31  T41 i  3, l  1  0.01336T22  2.02672T32  0.01336T42  0.00012  0.0084  0.00881 (3)  T32  2(1   )T42  f 5 (t ' )  2(1   )T31  T41  f 5 (t 2 )  0.01336T32  2.02672T42   (50)  0.01701  0.00881   (50)  1.36182 (4) Page i  4, l  1 11  0.01336T22  2.02672T32  0.01336T42  0.01713 0 0  2.02672  0.01336  T12   5.2733   0.01336 2.02672  0.01336   2   0.0005  0   T2       0.01336 2.02672  0.01336 T32  0.01713 0       0.01336 2.02672  T42  1.36182  0 0  Using Gauss elimination method[2] [2] See Appendix T11  2.60214 T21  0.02074 T31  0.50645 T41  0.67526 Appendix [1] 0 0  2.02672  0.01336  T11  2.672  0.01336 2.02672  0.01336  1   0  0   T2       0.01336 2.02672  0.01336 T31   0  0       0.01336 2.02672  T41  1.336  0 0  m21   0.01336 2.02672 which gives 0 0 2.02672  0.01336  T11   2.672   0   1   0.01761 2.02672  0.01336 0   T2      0   0.01336 2.02672  0.01336 T31   0   1   0  0.01336 2.02672  T4   1.336   0  6.59222  10 3 which gives 0 0 2.02672  0.01336  T11   2.672   0   1    0.01761  2 . 02663  0 . 01336 0   T2      0 0 2.02672  0.01336 T31   0.000116      0  0.01336 2.02672  T41   1.336   0 12 2.0266 Page m32   0.01336 m43  m32 which gives 0 0 2.02672  0.01336  T11   2.672   0   1    0.01761  2.02663  0.01336 0   T2      0 0 2.02663  0.01336 T31   0.000116      0 0 2.02663  T41   1.3359   0 Solving by back substitution give T41  1.3359  0.05917 2.02663 2.02663T31  0.01336(0.65917)  0.000116  T31  0.004288 2.02663T21  0.01336T31  0.01761  T21  0.008666 2.02663T11  0.01336T21  2.672  T11  1.3183 Appendix [2] 0 0  2.02672  0.01336  T12   5.2733   0.01336 2.02672  0.01336   2   0.0005  0   T2      0  0.01336 2.02672  0.01336 T32  0.01713      0 0  0.01336 2.02672  T42  1.36182   m21   0.01336 2.02672 which gives 2.0266  6.59222  10 3 which gives Page m32   0.01336 13 0 0 2.02672  0.01336  T12   5.2733   0   2  0.03526 2 . 02663  0 . 01336 0   T2      0  0.01336 2.02672  0.01336 T32  0.01713      0  0.01336 2.02672  T42  1.36182   0 0 0 2.02672  0.01336  T12   5.2733   0   2  0.03526 2.02663  0.01336 0   T2      0 0 2.02672  0.01336 T32  1.01736        0.01336 2.02672  T42  1.36182  0  0 m43  m32 which gives 0 0 2.02672  0.01336  T12   5.2733   0   2  0.03526 2 . 02663  0 . 01336 0   T2      0 0 2.02672  0.01336 T32  1.01736       0 0 2.02663  T42  1.36852   0 Solving by back substitution give T42  1.36852  0.67526 2.02663 2.02663T32  0.01336(0.67526)  1.01736  T32  0.50645 2.02663T22  0.01336T32  0.03526  T22  0.02074 Page 14 2.02663T12  0.01336T22  5.2733  T12  2.60214