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1 LU LU LU SOR ( ) SOR n x 1,, x n a 11 x 1 + a 1n x n = b 1 a n1 x 1 + a nn x n = b n

2 4 2 x 1 x 1 = b 1 a 12 x 2 a 1n x n a 11 n 1 x 2,, x n x n x n x n 1 x n x n n n Ax = b a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2 a n1 a n2 a nn b n 2 1 a 21 /a 11 a 2i = a 2i a 1i a 21 a 11, b 2 = b 2 b 1 a 21 a 11 i = 1,, n a 11 a 12 a 1n b 1 0 a 22 a 2n b a 32/a 22 0 a n2 a nn b n a i3 = a 3i a 2i b 3 = b 3 b 2 a 32 a 22 a 32 a 22, i = 2,, n a 11 a 12 a 1n b 1 0 a 22 a 2n b a 3n b a nn b n

3 4 3 A 0 a 11 a 12 a 1n b 1 0 a 22 a 2n b a (n 1) nn b (n 1) n 412 x n = b (n 1) n /a (n 1) n,n x n 1 = (b (n 2) n 1 a (n 2) n 1,n x n)/a (n 2) n 1,n 1 i n a (n 1) in /a (n 1) nn n 1 i ā b1 0 ā 22 0 b ā nn bn x i = b i /ā ii, i = 1, n 413 a (k 1) kk 0 0 a (k 1) lk 0 l 0 0 a (k 1) lk, l = k + 1,, n 0 A 414

4 4 4 1 // gauss elimination partial pivotting 2 #include "pchh" 3 #include <iostream> 4 #include <mathh> 5 #define N 4 6 int 7 main(int ac, char av[]) 8 { 9 double a[n][n+1] = {{10, 10, 30, 40, 10}, 10 {20, 10, 50, 10, 50}, 11 {30, 60, 20, 10, 80}, 12 {20, 20, 20, 30, 20}}; 13 int i, j, k, l, pivot; 14 double x[n]; 15 double p, q, t; for(i=0; i < N; i++) { 18 t = 0; 19 pivot = i; 20 for(k = i; k < N; k++) { 21 if(fabs(a[k][i]) > t) { 22 t = fabs(a[k][i]); 23 pivot = k; 24 } 25 } 26 if(pivot!= i) { 27 for(j=i; j < N+1; j++) { 28 t = a[i][j]; 29 a[i][j] = a[pivot][j]; 30 a[pivot][j] = t; 31 } 32 } 33 p = a[i][i]; 34 for(j = i+1; j < N; j++) { 35 q = a[j][i]/p; 36 for(k = i+1; k < N+1; k++) 37 a[j][k] = a[j][k] q a[i][k]; 38 a[j][i] = 0; 39 } 40 } 41 for(i = N 1; i >=0; i ) { 42 x[i] = a[i][n]; 43 for(j = N 1; j>i; j ) { 44 x[i] = x[i] a[i][j] x[j]; 45 } 46 x[i] = x[i]/a[i][i]; 47 } 48 printf("array a is\n"); 49 for(i=0; i < N; i++) { 50 for(j=0; j < N+1; j++) { 51 printf("%105f, ", a[i][j]); 52 } 53 printf("\n"); 54 } 55 printf("solution x is \n"); 56 for(i=0; i < N; i++) { 57 printf("x%d = %2517e\n", i+1, x[i]); 58 } 59 return 0; 60 }

5 x + y 3z 4w = 1 2x + y + 5z + w = 5 2x + y + (5 + ϵ)z + w = 8 2x + 2y + 2z 3w = 2 x = 2 y = ϵ 51 5ϵ ϵ = z = 3 ϵ w = 4ϵ 24 5ϵ x = 20, y = , z = 30000, w = ϵ = x = 20, y = , z = , w = ϵ = array a is , , , , , , , , , , , , , , , , , , , , solution x is x1 = e+00 x2 = e+05 x3 = e+04 x4 = e+04 ϵ = array a is , , , , , , , , , , , , , , , , , , , ,

6 4 6 solution x is x1 = e+00 x2 = e+05 x3 = e+04 x4 = e+04 ϵ = array a is , , , , , , , , , , , , , , , , , , , , solution x is x1 = e+00 x2 = e+07 x3 = e+06 x4 = e+06 ϵ = array a is , , , , , , , , , , , , , , , , , , , , solution x is x1 = e+00 x2 = e+07 x3 = e+06 x4 = e+06 ϵ = array a is , , , , , , , , , , , , , , , , , , , , index is 2, 3, 0, 1, solution x is x1 = e+00 x2 = e+07 x3 = e+06 x4 = e+06

7 4 7 ϵ i i i 1 1 p = a[i][i]; 2 for(j = i+1; j < N+1; j++) 3 a[i][j] = a[i][j]/p; // a {i,i} 1 4 a[i][i] = 1; 5 for(j = i+1; j < N; j++) { 6 q = a[j][i]; 7 for(k = i+1; k < N+1; k++) 8 a[j][k] = a[j][k] q a[i][k]; 9 a[j][i] = 0; 10 } 1 p = a[i][i]; 2 for(j = i+1; j < N+1; j++) 3 a[i][j] = a[i][j]/p; // a {i,i} 1 4 a[i][i] = 1; 5 for(j = 0; j < N; j++) { // i+1n 1 6 if (j == i) continue; // i 7 q = a[j][i]; 8 for(k = i+1; k < N+1; k++) 9 a[j][k] = a[j][k] q a[i][k]; 10 a[j][i] = 0; 11 } A a[i][n] i i + 1

8 LU Ax = b j i a j,i /a i,i A 1 a ji /a ii 1 i i i M i = a i+1,i /a i,i 1 a n,i /a i,i 1 i j A = a 11 a 12 a 1n a 21 a 22 a 2n a n1 a n2 a nn 2 a 11 a 12 a 1n 0 a A 22 a 2n = 0 a n2 a nn M 1 = m m n1 0 1 A A = M 1 A m i1 = a i1 /a 11 A (k) = M k M 2 M 1 A M k = m k+1,k 1 0 m nk 0 1

9 4 9 m jk = a (k 1) jk /a (k 1) kk n 1 A (n 1) = â 11 â 12 â 1n 0 â 22 â 2n 0 0 â 3n 0 0 â nn = U U L = M 1 n 1 M 1 2 M 1 1 A = M 1 n 1 M 1 n 2 M 1 1 U = LU L M k M k = m k+1,k 1 0 m n,k 0 1 L = 1 m k,1 1 m k+1,1 m k+1,k 1 m n,1 m n,k m n,k+1 1 U D 1 U â m â 22 0 A = LDU = m n1 m n â nn 1 â 12 â 1n 0 1 â 2n LDU â ij = â ij/â ii A (a ij = a ji ) L T = U A = LDL T A LU Ax = LUx = b Ux = y

10 4 10 Ly = b Ux = y L U y 1 = b 1 y 2 = b 2 m 2,1 y 1 y n = b n (m n,1 y m n,n 1 y n 1 ) x n = y n /â n,n x n 1 = (y n 1 â n 1,n y n )/â n 1,n 1 x 1 = (y 1 â 1,n y n â 1,n 1 y n 1 â 1,2 y 2 )/â 1,1 431 LU 1 // lu decomposition 2 #include "pchh" 3 #include <iostream> 4 #include <mathh> 5 #define N 4 6 int 7 main(int ac, char av[]) 8 { 9 double a[n][n] = { 10 {10, 10, 30, 40}, 11 {20, 10, 50, 10}, 12 {30, 60, 20, 10}, 13 {20, 20, 20, 30} }; 14 int i, j, k, l; 15 double b[n][n]; 16 double c[n][n]; 17 double p, q, t; 18 double s; 19 printf("array A is\n"); 20 for (i = 0; i < N; i++) { 21 for (j = 0; j < N; j++) { 22 printf("%105f, ", a[i][j]); 23 } 24 printf("\n"); 25 } 26 for (i = 0; i < N; i++) { 27 for (j = 0; j < N; j++) { 28 b[i][j] = 0; 29 }

11 b[i][i] = 1; 31 } for (i = 0; i < N; i++) { 34 for (j = i + 1; j < N; j++) { 35 q = a[j][i] / a[i][i]; 36 for (k = i + 1; k < N; k++) 37 a[j][k] = a[j][k] q a[i][k]; 38 a[j][i] = 0; 39 b[j][i] = q; 40 } 41 } 42 printf("array L is\n"); 43 for (i = 0; i < N; i++) { 44 for (j = 0; j < N; j++) { 45 printf("%105f, ", b[i][j]); 46 } 47 printf("\n"); 48 } 49 printf("array U is\n"); 50 for (i = 0; i < N; i++) { 51 for (j = 0; j < N; j++) { 52 printf("%105f, ", a[i][j]); 53 } 54 printf("\n"); 55 } 56 for (i = 0; i < N; i++) { 57 for (j = 0; j < N; j++) { 58 s = 0; 59 for (k = 0; k < N; k++) { 60 s = s + b[i][k] a[k][j]; 61 } 62 c[i][j] = s; 63 } 64 } printf("array L*U is\n"); 67 for (i = 0; i < N; i++) { 68 for (j = 0; j < N; j++) { 69 printf("%105f, ", c[i][j]); 70 } 71 printf("\n"); 72 } 73 return 0; 74 } 432 L b a b L 1 U b a

12 LU i j P ij P ij = i j A i j P ij P ij A (i j P ij AP ij ) j i a j,i /a i,i A 1 i a ji /a ii 1 j i i i a i+1,i /a i,i 1 a n,i /a i,i 1 LU U L A A A A = LU U = , L = A A = = U

13 A 0 0 n 0 0 Ax (k) b 0, k {x (k), k = 1, 2, } A F E D D = a a a nn, E = A = D + E + F a a n1 a n2 0, F = 0 a 12 a 1n 0 0 a 2n = a 1 11 {b 1 (a 12 x (k) 2 + a 13 x (k) a 1n x (k) 2 = a 1 22 {b 2 (a 21 x (k) 1 + a 23 x (k) a 2n x (k) j = a 1 jj {b j (a j1 x (k) a jj 1 x (k) j 1 + a jj+1x (k) j a jnx (k) n = a 1 nn{b n (a n1 x (k) 1 + a n2 x (k) a nn 1 x (k) n 1 )} = D 1 {b (E + F)x (k) } = D 1 (E + F)x (k) + D 1 b

14 x j x j 1 = a 1 11 {b 1 (a 12 x (k) 2 + a 13 x (k) a 1n x (k) 2 = a 1 22 {b 2 (a a 23 x (k) a 2n x (k) j = a 1 jj {b j (a j a jj 1 j 1 + a jj+1 x (k) j a jnx (k) n = a 1 nn{b n (a n1 1 + a n a nn 1 n 1 )} = D 1 {b E Fx (k) } = (D + E) 1 Fx (k) + (D + E) 1 b 443 SOR ( ) j j x (k) j 1 ω SOR ξ (k+1) 1 = a 1 11 {b 1 (a 12 x (k) 2 + a 13 x (k) a 1n x (k) 1 = x (k) 1 + ω(ξ (k+1) 1 x (k) 1 ) ξ (k+1) 2 = a 1 22 {b 2 (a a 23 x (k) a 2n x (k) 2 = x (k) 2 + ω(ξ (k+1) 2 x (k) 2 ) ξ (k+1) j = a 1 jj {b j (a j a jj 1 j 1 + a jj+1 x (k) j a jnx (k) j = x (k) j + ω(ξ (k+1) j x (k) j ) ξ (k+1) n = a 1 nn{b n (a n1 1 + a n a nn 1 n 1 )} n = x (k) n + ω(ξ n (k+1) x (k) n ) ξ (k+1) = D 1 {b E Fx (k) } = x (k) + ω(ξ (k+1) x (k) ) ξ = (I + ωd 1 E) 1 {(1 ω)i ωd 1 F}x (k) + ω(d + ωe) 1 b

15 = Mx (k) + b M < 1 x (k) () M M Mx M = sup x 0 x M // Jacobi method 2 #include "pchh" 3 #include <iostream> 4 #include <mathh> 5 #define N 4 6 int 7 main(int ac, char av[]) 8 { 9 double a[n][n] = { 10 {50, 10, 30, 40}, 11 {10, 50, 10, 10}, 12 {30, 60, 80, 10}, 13 {20, 20, 20, 90} }; 14 double b[n] = { 10, 50, 80, 20 }; int i, j, k, n; 17 double x[n] = { 10, 10, 10, 10 }, y[n]; 18 double e, p; 19 e = 00; 20 for (i = 0; i < N; i++) { 21 p = 0; 22 for (j = 0; j < N; j++) 23 p = p + a[i][j] x[j]; 24 p = p b[i]; 25 e = e + fabs(p); 26 } 27 n = 0; 28 while (e > 10e 5) { 29 n++; 30 printf("%d: err = %f\n", n, e); 31 for (i = 0; i < N; i++) { 32 y[i] = b[i]; 33 for (j = 0; j < i; j++) 34 y[i] = y[i] a[i][j] x[j]; 35 for (j = i + 1; j < N; j++) 36 y[i] = y[i] a[i][j] x[j]; 37 y[i] = y[i] / a[i][i]; 38 } 39 for (i = 0; i < N; i++)

16 x[i] = y[i]; e = 00; 43 for (i = 0; i < N; i++) { 44 p = 0; 45 for (j = 0; j < N; j++) 46 p = p + a[i][j] x[j]; 47 p = p b[i]; 48 e = e + fabs(p); 49 } 50 } 51 for (i = 0; i < N; i++) { 52 printf("x%d = %2517e\n", i + 1, x[i]); 53 } 54 return 0; 55 } 446 while 1 while (e > 10e 5) { 2 n++; 3 printf("%d: err = %f\n", n, e); 4 for (i = 0; i < N; i++) { 5 x[i] = b[i]; 6 for (j = 0; j < i; j++) 7 x[i] = x[i] a[i][j] x[j]; 8 for (j = i + 1; j < N; j++) 9 x[i] = x[i] a[i][j] x[j]; 10 x[i] = x[i] / a[i][i]; 11 } 447 SOR while

17 #define omega while (e > 10e 5) { 3 n++; 4 printf("%d: err = %f\n", n, e); 5 for (i = 0; i < N; i++) { 6 xi = b[i]; 7 for (j = 0; j < i; j++) 8 xi = xi a[i][j] x[j]; 9 for (j = i + 1; j < N; j++) 10 xi = xi a[i][j] x[j]; 11 xi = xi / a[i][i]; 12 x[i] = x[i] + omega (xi x[i]); 13 } SOR SOR ω = SOR ω x + 3y + 4z = 10 2x + y + 5z = 7 6x + 5y + z = A = LU 43 3x + y + z = 0 x + 3y + z = 4 x + y + 3z = 6 SOR omega SOR omega = 1

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