Resolución Numérica de Sistemas de Ecuaciones no Lineales
Se resuelve el sistema lineal J HP3L DP = -F HP3L:
jjjijjjjj 3.00000 7.65687 µ 10-6 -4.08422 µ 10-10 zzzzzzzyz.jjjjjjjij Dx1 zzzzzzzyz
1.00000 -16.2045 0.866026 {k Dx2 {
-0.499993 20.0000 Dx3
k -0.0000279286
DP = jjjjjjjij Dx1 zzzzzyzzz = jjjjijjjjj -2.54571 µ 10-7 zzzzzyzzzz
k Dx2 { k -0.0000279251 {
Dx3 -7.30415 µ 10-7
El siguiente punto de la iteración es:
P4 = P3 + DP
P4 = jjjjjjjji 0.5 zzzzzyzzz + jjijjjjjjj -2.54571 µ 10-7 zzzzzzzzyz = jjjjjjjij 0.5 zzzzzyzzz
k 0.000027929 { k -0.0000279251 { k 3.90671 µ 10-9 {
-0.523598 -7.30415 µ 10-7 -0.523599
Iteración i = 4.
P4 = jijjjjjjjjjj x1H4L zzzzzzzzzyzz = jjjjjjjij 0.500000 zzzyzzzzz
k x2H4L { k 3.90671 µ 10-9 {
x3H4L -0.523599
F HP4L = jjijjjjjjjj 1.06894 µ 10-10 zzzyzzzzzzz
k -6.31645 µ 10-8 {
9.03668 µ 10-11
jjjjjijjjj 3.00000 1.07105 µ 10-9 -7.99135 µ 10-18 zzzzzzzyzz
k 1.00000 -16.2000 0.866025
J HP4L = -3.90671 -0.500000
µ 10-9 20.0000 {
Se resuelve el sistema lineal J HP4L DP = -F HP4L:
jjjjijjjjj 3.00000 µ 10-9 1.07105 µ 10-9 -7.99135 µ 10-18 zzzzzyzzzz.ijjjjjjjj Dx1 zzzzzzzzy
k 1.00000 -16.2000 0.866025 {k Dx2 {
-3.90671 -0.500000 20.0000 Dx3
DP = jjjjjjijj Dx1 zzzzzzzzy = jjjjjjjijjj -3.56313 µ 10-11 zzzzzyzzzzz
k Dx2 { k -3.90671 µ 10-9 {
Dx3 -1.02186 µ 10-10
El siguiente punto de la iteración es:
P5 = P4 + DP
P5 = jjjjjjjij 0.5 yzzzzzzzz + jjjijjjjjjj -3.56313 µ 10-11 zzzzyzzzzzz = jjjjjjijj 0.5 zzzzzzyzz
k 3.90671 µ 10-9 { k -3.90671 µ 10-9 { k 8.71767 µ 10-17 {
-0.523599 -1.02186 µ 10-10 -0.523599
89
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Tabla de datos.
i Pi DP = jjjjijjjj Dx1 zzzzyzzzz Pi+1 = Pi + DP ¥Pi+1 - Pi∞¶
k Dx2 { 0.42296
Dx3
0 jjjjjjjji 0.5 zzzyzzzzz jjjjjjjij -0.000132437 zyzzzzzzz jjjjjjijj 0.499868 yzzzzzzzz
k 0.1 { k -0.0756424 { k 0.0243576 {
-0.1 -0.42296 -0.52296
1 jjijjjjjj 0.499868 zzzzzzzyz ijjjjjjjj 0.000154114 zzzzzyzzz jjjjjjjij 0.500022 zzzzzzzyz 0.0219684
k 0.0243576 { k -0.0219684 { k 0.00238921 {
-0.52296 -0.000576215 -0.523536
2 jjjjjjjij 0.500022 zzzzyzzzz jjjijjjjj -0.0000214227 zzzzzzzyz ijjjjjjjj 0.5 zzzzzzzyz 0.00236128
k 0.00238921 { k -0.00236128 { k 0.000027929 {
-0.523536 -0.0000621228 -0.523598
3 jijjjjjjj 0.5 zzzzzzyzz jjjjjjjjji -2.54571 µ 10-7 zzyzzzzzzz jjijjjjjj 0.5 zzzyzzzzz 0.0000279251
k 0.000027929 { k -0.0000279251 { k 3.90671 µ 10-9 {
-0.523598 -7.30415 µ 10-7 -0.523599
4 jjjjjjijj 0.5 µ 10-9 zzyzzzzzz jjjjjjjjijj -3.56313 µ 10-11 zzzzzzyzzzz jjjjjjjij 0.5 zzzyzzzzz 3.90671 µ 10-9
3.90671 k -3.90671 µ 10-9 { k 8.71767 µ 10-17 {
-1.02186 µ 10-10 -0.523599
k -0.523599 {
La solución aproximada del sistema es:
P5 = jjjijjjjj 0.500000 zzyzzzzzz
k 8.71767 µ 10-17 {
-0.523599
à Problema 17. Mediante el método de Newton para sistemas no lineales calcular las
aproximaciones P1 y P2 , partiendo del punto inicial P0 para los sistemas no lineales
siguientes:
a) f1Hx1, x2L = 4 x12 - 20 x1 + 1 ê 4 x22 + 8 = 0,
f2Hx1, x2L = 1 ê 2 x1 x22 + 2 x1 - 5 x2 + 8 = 0,
P0 = IxH10L, xH20LMT = H0, 0LT .
b) f1Hx1, x2L = senH4 p x1 x2L - 2 x2 - x1 = 0,
f2Hx1, x2L = HH4 p - 1L ê H4 pLL He2 x1 - eL + 4 e x22 - 2 e x1 = 0,
P0 = Ix1H0L, x2H0LMT = H0, 0LT .
c) f1Hx1, x2, x3L = 3 x1 - cosHx2 x3L - 1 ê 2 = 0,
f2Hx1, x2,x3L = 4 x12 - 625 x22 + 2 x2 - 1 = 0,
f3Hx1, x2, x3L = e-x1 x2 + 20 x3 + H10 p - 3L ê 3 = 0,
P0 = Ix1H0L, xH20L, x3H0LMT = H0, 0, 0LT .
d) f1Hx1, x2, x3L = x21 + x2 - 37 = 0,
f2Hx1, x2, x3L = x1 - x22 - 5 = 0,
90
Resolución Numérica de Sistemas de Ecuaciones no Lineales
f3Hx1, x2, x3L = x1 + x2 + x3 - 3 = 0,
P0 = IxH10L, xH20L, x3H0LMT = H0, 0, 0LT
Solución
a)
Clear@ecuaciones, p, m, dD;
ecuaciones = 84 x21 − 20 x1 + x22 ê 4 + 8, x1 x22 ê 2 + 2 x1 − 5 x2 + 8 <;
p = 80.0, 0.0<;
m = 2;
d = 10.−6;
newtonSistemasNoLineal@ecuaciones, p, m, dD;
Método de Newton-Raphson para sistemas de ecuaciones no lineales.
fi Hx1 , x2L = jjjjjij 4 x12 - 20 x1 + ÅxÅ4Å2Å2ÅÅ + 8 zzyzzzz = jij 0 zyz
k Å21ÅÅÅ x1 x22 - 5 x2 + 2 x1 + { k 0 {
8
P0 = jjjji x1H0L yzzzz = jij 0. zyz
k x2H0L { k 0. {
La función vectorial y la matriz jacobiana son:
F Hx1, x2L = jjjjjij 4 x12 - 20 x1 + ÅxÅ4Å2Å2ÅÅ + 8 zzzzzyz
k Å12ÅÅÅ x1 x22 - 5 x2 + 2 x1 + {
8
J Hx1, x2L = ijjjjj 8 x1 - 20 ÅxÅ2Å2ÅÅ x2 - 5 zzzzyz
k ÅxÅ2Å2Å2ÅÅ + 2 x1 {
Iteración i = 0.
P0 = ijjjj x1H0L yzzzz = jij 0 yzz
k x2H0L { k 0 {
F HP0L = ijj 8.00000 yzz
k 8.00000 {
J HP0L = jij -20.0000 0 zzy
k 2.00000 -5.00000 {
91
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Se resuelve el sistema lineal J HP0L DP = -F HP0L:
jji -20.0000 0 yzz.jji Dx1 yzz
k 2.00000 -5.00000 {k Dx2 {
DP = jij Dx1 yzz = ijj 0.4 yzz
k Dx2 { k 1.76 {
El siguiente punto de la iteración es:
P1 = P0 + DP
P1 = jji 0. yzz + jij 0.4 zzy = ijj 0.4 yzz
k 0. { k 1.76 { k 1.76 {
Iteración i = 1.
P1 = ijjjj x1H1L zyzzz = jji 0.400000 zzy
k x2H1L { k 1.76000 {
F HP1L = ijj 1.41440 zzy
k 0.619520 {
J HP1L = ijj -16.8000 0.880000 zzy
k 3.54880 -4.29600 {
Se resuelve el sistema lineal J HP1L DP = -F HP1L:
jji -16.8000 0.880000 yzz.jji Dx1 zzy
k 3.54880 -4.29600 {k Dx2 {
DP = ijj Dx1 zyz = ijj 0.0958936 zyz
k Dx2 { k 0.223423 {
El siguiente punto de la iteración es:
P2 = P1 + DP
P2 = ijj 0.4 zyz + ijj 0.0958936 yzz = ijj 0.495894 yzz
k 1.76 { k 0.223423 { k 1.98342 {
Tabla de datos.
i Pi DP = ijj Dx1 zzy Pi+1 = Pi + DP ¥Pi+1 - Pi∞¶
k Dx2 {
0 ijj 0. yzz jij 0.4 yzz jji 0.4 yzz 1.76
k 0. { k 1.76 { k 1.76 {
1 jji 0.4 zzy jij 0.0958936 zyz jji 0.495894 zzy 0.223423
k 1.76 { k 0.223423 { k 1.98342 {
92
Resolución Numérica de Sistemas de Ecuaciones no Lineales
La solución aproximada del sistema es:
P2 = jji 0.495894 zzy
k 1.98342 {
Solución
b)
Clear@ecuaciones, p, m, dD;
ecuaciones = 8 Sin@4 ∗ Pi ∗ x1 ∗ x2D − 2 x2 − x1,
HH4 Pi − 1L ê H4 PiLL HExp@2 x1D − EL + 4 E Hx2L2 − 2 E ∗ x1<;
p = 80.0, 0.0<;
m = 2;
d = 10.−6;
newtonSistemasNoLineal@ecuaciones, p, m, dD;
Método de Newton-Raphson para sistemas de ecuaciones no lineales.
fi Hx1 , x2L = ijjjj sinH4 p x1 x2L - x1 - 2 x2 zzyzz = jji 0 zyz
k 4 ‰ x22 - 2 ‰ x1 + ÅHÅÅ-ÅʼnÅÅÅ+ÅʼnÅÅÅ2ÅÅxÅÅ41ÅÅLÅpÅHÅ-ÅÅÅ1ÅÅÅ+ÅÅÅ4Å ÅÅpÅÅLÅÅ { k 0 {
P0 = jjijj x1H0L yzzzz = jij 0. zyz
k x2H0L { k 0. {
La función vectorial y la matriz jacobiana son:
F Hx1, x2L = ijjjj sinH4 p x1 x2L - x1 - 2 x2 yzzzz
k 4 ‰ x22 - 2 ‰ x1 + ÅHÅÅ-ÅʼnÅÅÅ+ÅʼnÅÅÅ2ÅÅxÅÅ41ÅÅLÅpÅHÅ-ÅÅÅ1ÅÅÅ+ÅÅÅ4ÅÅÅpÅÅLÅÅ {
jjjji 4p cosH4 p x1 x2L x2 - 1 4 p cosH4 p x1 x2L x1 - 2 zzzyz
k -2 ‰ + ʼnÅÅ2ÅÅÅxÅ1ÅÅÅHÅ2Å-ÅÅÅ1pÅÅ+ÅÅÅ4ÅÅÅÅpÅÅLÅ 8 ‰ x2 {
J Hx1, x2L =
Iteración i = 0.
P0 = jjjji x1H0L zzzyz = jji 0 zyz
k x2H0L { k 0 {
F HP0L = ijj 0 yzz
k -1.58155 {
J HP0L = jji -1.00000 -2.00000 zzy
k -3.59572 0{
93
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Se resuelve el sistema lineal J HP0L DP = -F HP0L:
jji -1.00000 -2.00000 yzz.jji Dx1 zzy
k -3.59572 0 {k Dx2 {
DP = ijj Dx1 yzz = ijj -0.439841 zyz
k Dx2 { k 0.219921 {
El siguiente punto de la iteración es:
P1 = P0 + DP
P1 = ijj 0. yzz + jij -0.439841 zyz = jji -0.439841 zzy
k 0. { k 0.219921 { k 0.219921 {
Iteración i = 1.
P1 = ijjjj x1H1L yzzzz = jji -0.439841 zzy
k x2H1L { k 0.219921 {
F HP1L = ijj -0.937560 yzz
k 0.797033 {
J HP1L = jij -0.0387518 -3.92250 yzz
k -4.67277 4.78245 {
Se resuelve el sistema lineal J HP1L DP = -F HP1L:
jji -0.0387518 -3.92250 yzz.jji Dx1 zzy
k -4.67277 4.78245 {k Dx2 {
DP = jji Dx1 zzy = ijj -0.0733204 zyz
k Dx2 { k -0.238297 {
El siguiente punto de la iteración es:
P2 = P1 + DP
P2 = jji -0.439841 zzy + jji -0.0733204 zyz = jji -0.513162 zzy
k 0.219921 { k -0.238297 { k -0.0183762 {
Tabla de datos.
i Pi DP = jij Dx1 zyz Pi+1 = Pi + DP ¥Pi+1 - Pi∞¶
k Dx2 {
0 jji 0. yzz jji -0.439841 yzz ijj -0.439841 yzz 0.439841
k 0. { k 0.219921 { k 0.219921 {
1 jji -0.439841 zzy ijj -0.0733204 zzy jji -0.513162 zyz 0.238297
k 0.219921 { k -0.238297 { k -0.0183762 {
94
Resolución Numérica de Sistemas de Ecuaciones no Lineales
La solución aproximada del sistema es:
P2 = jij -0.513162 zzy
k -0.0183762 {
Solución
c)
Clear@ecuaciones, p, m, dD;
ecuaciones = 83 x1 − Cos@x2 ∗ x3D − 1 ê 2, 4 Hx1L2 − 625 Hx2L2 + 2 x2 − 1,
Exp@−x1 ∗ x2D + 20 x3 + H10 Pi − 3L ê 3<;
p = 80.0, 0.0, 0.0<;
m = 2;
d = 10.−6;
newtonSistemasNoLineal@ecuaciones, p, m, dD;
Método de Newton-Raphson para sistemas de ecuaciones no lineales.
fi Hx1 , x2, x3L = jjjjjjjjjjji -cosHx2 x3L + 3 x1 - Å21ÅÅÅ zyzzzzzzzzzz = jjjjjjjji 0 zzyzzzzzz
k 4 x12 - 625 x22 + 2 x2 - 1 { k 0 {
20 x3 + ‰-x1 x2 + Å13ÅÅÅ H-3 + 10 pL 0
P0 = jijjjjjjjjjj x1H0L zzzzzzzzzzzy = jjjjjjjji 0. zzzzzzyzz
k x2H0L { k 0. {
x3H0L 0.
La función vectorial y la matriz jacobiana son:
F Hx1, x2, x3L = jjijjjjjjjjj -cosHx2 x3L + 3 x1 - Å21ÅÅÅ zzzzzzzzzzzy
k 4 x12 - 625 x22 + 2 x2 - 1 {
20 x3 + ‰-x1 x2 + Å31ÅÅÅ H-3 + 10 pL
jjjjijjjj 3 sinHx2 x3L x3 sinHx2 x3L x2 zzzzzzyzz
k 0 {
J Hx1, x2, x3L = 8 x1 2 - 1250 x2 20
-‰-x1 -‰-x1 x2 x1
x2 x2
95
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Iteración i = 0.
P0 = jjjjjjjijjjj x1H0L zzzzyzzzzzzz = jjjijjjjj 0 zzyzzzzzz
k x2H0L { k 0 {
x3H0L 0
F HP0L = jjjjjjjij -1.50000 zzzzyzzzz
k -1.00000 {
10.4720
J HP0L = jjjjjjjji 3.00000 0 0 zzzyzzzzz
k 0 2.00000 0 {
0 0 20.0000
Se resuelve el sistema lineal J HP0L DP = -F HP0L:
ijjjjjjjj 3.00000 0 0 zzzzyzzzz.ijjjjjjjj Dx1 zzyzzzzzz
k 0 2.00000 0 {k Dx2 {
0 0 20.0000 Dx3
DP = jjjjjjjji Dx1 zzzyzzzzz = jjjijjjjj 0.5 zzzzzzyzz
k Dx2 { k 0.5 {
Dx3 -0.523599
El siguiente punto de la iteración es:
P1 = P0 + DP
P1 = jjjjjjjij 0. zzzzzyzzz + jjijjjjjj 0.5 zzzzzzyzz = jjjjjjijj 0.5 zzzzyzzzz
k 0. { k 0.5 { k 0.5 {
0. -0.523599 -0.523599
Iteración i = 1.
P1 = jijjjjjjjjjj x1H1L zzzzzzzzyzzz = jjijjjjjj 0.500000 zyzzzzzzz
k x2H1L { k 0.500000 {
x3H1L -0.523599
F HP1L = jjijjjjjj 0.0340742 yzzzzzzzz
k -155.250 {
-0.221199
J HP1L = jjjjjjijj 3.00000 0.135517 -0.129410 yzzzzzzzz
k 4.00000 -623.000 0 {
-0.389400 -0.389400 20.0000
96
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Se resuelve el sistema lineal J HP1L DP = -F HP1L:
jjjjjjjji 3.00000 0.135517 -0.129410 {zzyzzzzzz.ijjjjkjjjj Dx1 zyzzzzzzz
k 4.00000 -623.000 0 Dx2 {
-0.389400 -0.389400 20.0000 Dx3
DP = jjjjjjijj Dx1 zzzzzzyzz = jjjjjijjj 0.000166687 zzzzzzyzz
k Dx2 { k -0.249196 {
Dx3 0.00621135
El siguiente punto de la iteración es:
P2 = P1 + DP
P2 = jjjijjjjj 0.5 zzyzzzzzz + jjjjjjijj 0.000166687 zzyzzzzzz = jjjjjjjij 0.500167 zyzzzzzzz
k 0.5 { k -0.249196 { k 0.250804 {
-0.523599 0.00621135 -0.517387
Tabla de datos.
i Pi DP = jjjjjijjj Dx1 zyzzzzzzz Pi+1 = Pi + DP ¥Pi+1 - Pi∞¶
k Dx2 {
Dx3
0 jjjjjjjji 0. yzzzzzzzz jjjjjjjji 0.5 zzzzzzyzz jjjjjjjij 0.5 zyzzzzzzz 0.523599
k 0. { k 0.5 { k 0.5 {
0. -0.523599 -0.523599
1 jijjjjjjj 0.5 zzzzzyzzz jjjjjijjj 0.000166687 zzzzzzzzy ijjjjjjjj 0.500167 yzzzzzzzz 0.249196
k 0.5 { k -0.249196 { k 0.250804 {
-0.523599 0.00621135 -0.517387
La solución aproximada del sistema es:
P2 = jjjjjjjij 0.500167 zzzzzzzyz
k 0.250804 {
-0.517387
Solución
d)
97
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Clear@ecuaciones, p, m, dD;
ecuaciones = 8x21 + x2 − 37, x1 − x22 − 5, x1 + x2 + x3 − 3<;
p = 80.0, 0.0, 0.0<;
m = 2;
d = 10.−6;
newtonSistemasNoLineal@ecuaciones, p, m, dD;
Método de Newton-Raphson para sistemas de ecuaciones no lineales.
fi Hx1 , x2, x3L = jjjjjijjjj x12 + x2 - 37 zzzzzzzzyz = jjjijjjjj 0 zzzyzzzzz
k -x22 + x1 - 5 { k 0 {
x1 + x2 + x3 - 3 0
P0 = jjjjjjjjjijj x1H0L zzzzyzzzzzzz = jjjjjjjji 0. yzzzzzzzz
k x2H0L { k 0. {
x3H0L 0.
La función vectorial y la matriz jacobiana son:
F Hx1, x2, x3L = jjjjjijjjj x12 + x2 - 37 zzzzzzzzyz
k -x22 + x1 - 5 {
x1 + x2 + x3 - 3
jjijjjjjj 2 x1 1 0 zyzzzzzzz
k 1 -2 x2 0 {
J Hx1, x2, x3L = 1 1 1
Iteración i = 0.
P0 = jjjjjjjjjjji x1H0L zzzzzzzzzzyz = jjjijjjjj 0 zyzzzzzzz
k x2H0L { k 0 {
x3H0L 0
F HP0L = jjjjjjjij -37.0000 zzzzzzyzz
k -5.00000 {
-3.00000
J HP0L = jjijjjjjj 0 1.00000 0 zyzzzzzzz
k 1.00000 0 0 {
1.00000 1.00000 1.00000
98
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Se resuelve el sistema lineal J HP0L DP = -F HP0L:
jjjjjjjij 0 1.00000 0 zyzzzzzzz{.jjikjjjjjj Dx1 zzzzzzzzy
k 1.00000 0 0 Dx2 {
1.00000 1.00000 1.00000 Dx3
DP = jjjjijjjj Dx1 zzzzzzyzz = jijjjjjjj 5. zzzyzzzzz
k Dx2 { k 37. {
Dx3 -39.
El siguiente punto de la iteración es:
P1 = P0 + DP
P1 = jjjjjjijj 0. zzzyzzzzz + jjijjjjjj 5. zzzyzzzzz = jjjjjjjij 5. zzzzzzzyz
k 0. { k 37. { k 37. {
0. -39. -39.
Iteración i = 1.
P1 = jjjjjjjjjjij x1H1L zzzzzzzzzyzz = jjjjjjijj 5.00000 zzyzzzzzz
k x2H1L { k 37.0000 {
x3H1L -39.0000
F HP1L = jjjjjjjij 25.0000 yzzzzzzzz
k -1369.00 {
0
J HP1L = jjjjjijjj 10.0000 1.00000 0 zzzzzyzzz
k 1.00000 -74.0000 0 {
1.00000 1.00000 1.00000
Se resuelve el sistema lineal J HP1L DP = -F HP1L:
jjjjjjjji 10.0000 1.00000 0 zzzzzzyzz.jjjijjjjj Dx1 zzzzzzyzz
k 1.00000 -74.0000 0 {k Dx2 {
1.00000 1.00000 1.00000 Dx3
DP = ijjjjjjjj Dx1 yzzzzzzzz = ijjjjjjjj -0.649123 yzzzzzzzz
k Dx2 { k -18.5088 {
Dx3 19.1579
El siguiente punto de la iteración es:
P2 = P1 + DP
P2 = jijjjjjjj 5. zyzzzzzzz + jjjjjijjj -0.649123 zzzzzzzzy = jijjjjjjj 4.35088 zzzzzyzzz
k 37. { k -18.5088 { k 18.4912 {
-39. 19.1579 -19.8421
Tabla de datos.
99
Resolución Numérica de Sistemas de Ecuaciones no Lineales
i Pi DP = jjjjjjijj Dx1 zyzzzzzzz Pi+1 = Pi + DP ¥Pi+1 - Pi∞¶
k Dx2 {
Dx3
0 jjjijjjjj 0. zzzzzzzzy jjjijjjjj 5. zzzzzzyzz jjijjjjjj 5. zzzzzzzyz 39.
k 0. { k 37. { k 37. {
0. -39. -39.
1 jjjjjijjj 5. zzzzzyzzz jijjjjjjj -0.649123 zzzzyzzzz ijjjjjjjj 4.35088 yzzzzzzzz 19.1579
k 37. { k -18.5088 { k 18.4912 {
-39. 19.1579 -19.8421
La solución aproximada del sistema es:
P2 = jijjjjjjj 4.35088 zzzzzzzzy
k 18.4912 {
-19.8421
à Problema 18. Dado el siguiente problema no lineal
f1Hx1, x2L = x1 H1 - x1L + 4 x2 = 12,
f2Hx1, x2L = Hx1 - 2L2 + H2 x2 - 3L2 = 25.
a) Representar gráficamente las curvas f1 y f2.
b) Calcular la solución aproximada del sistema empleando el método de Newton
comenzando en los puntos
P0 = Ix1H0L, x2H0LMT = H2, 3LT ,
P0 = Ix1H0L, xH20LMT = H-2, -2LT
e iterando hasta que ∞Pi+1 - Pi¥¶ § 5 µ 10-2.
Solución
a)
<< Graphics`ImplicitPlot`;
Clear@ecuaciones, p, m, d, f, g, g1, g2, gD;
f = x H1 − xL + 4 y − 12;
g = Hx − 2L2 + H2 y − 3L2 − 25;
Print@"Representación gráfica de la solución",
"\n\t f1Hx, yL = ", f, "\n\t f2Hx, yL = ", gD;
g1 = ImplicitPlot @f == 0, 8x, −3, 7<,
PlotStyle −> 88Thickness @0.010D, RGBColor @0, 0, 1D<<,
AxesLabel −> 8"X", "Y"<, DisplayFunction −> Identity D;
g2 = ImplicitPlot @g == 0, 8x, −3, 7<,
PlotStyle −> 88Thickness @0.010D, RGBColor @1, 0, 0D<<,
D
100
Resolución Numérica de Sistemas de Ecuaciones no Lineales
AxesLabel −> 8"X", "Y"<, DisplayFunction −> Identity D;
g = Show @g1, g2, AxesLabel −> 8"X", "Y"<,
AspectRatio −> Automatic, DisplayFunction −> $DisplayFunctionD;
Representación gráfica de la solución
f1Hx, yL = H1 - xL x + 4 y - 12
f2Hx, yL = Hx - 2L2 + H2 y - 3L2 - 25
Y
12 X
10 246
8
6
4
2
-2
Solución
b)
Clear@ecuaciones, p, m, dD;
ecuaciones = 8x1 H1 − x1L + 4 x2 − 12, Hx1 − 2L2 + H2 x2 − 3L2 − 25<;
p = 82.0, 3.0<;
m = 10;
d = 5 ∗ 10.−2;
newtonSistemasNoLineal@ecuaciones, p, m, dD;
p = 8−2.0, 2.0<;
m = 10;
d = 5 ∗ 10.−2;
newtonSistemasNoLineal@ecuaciones, p, m, dD;
Método de Newton-Raphson para sistemas de ecuaciones no lineales.
fi Hx1 , x2L = jji H1 - x1L x1 + 4 x2 - 12 25 yzz = jji 0 zzy
k Hx1 - 2L2 + H2 x2 - 3L2 - { k 0 {
P0 = ijjjj x1H0L zzyzz = jij 2. zyz
k x2H0L { k 3. {
101
Resolución Numérica de Sistemas de Ecuaciones no Lineales
La función vectorial y la matriz jacobiana son:
F Hx1, x2L = ijj H1 - x1L x1 + 4 x2 - 12 25 yzz
k Hx1 - 2L2 + H2 x2 - 3L2 - {
J Hx1, x2L = jji 1 - 2 x1 4 H2 x2 - 3L zzy
k 2 Hx1 - 2L 4 {
Iteración i = 0.
P0 = ijjjj x1H0L zyzzz = jji 2.00000 yzz
k x2H0L { k 3.00000 {
F HP0L = ijj -2.00000 yzz
k -16.0000 {
J HP0L = jji -3.00000 4.00000 yzz
k 0 12.0000 {
Se resuelve el sistema lineal J HP0L DP = -F HP0L:
jji -3.00000 4.00000 yzz.jji Dx1 zyz
k 0 12.0000 {k Dx2 {
DP = ijj Dx1 zzy = ijj 1.11111 yzz
k Dx2 { k 1.33333 {
El siguiente punto de la iteración es:
P1 = P0 + DP
P1 = jji 2. zyz + ijj 1.11111 yzz = jij 3.11111 zzy
k 3. { k 1.33333 { k 4.33333 {
Iteración i = 1.
P1 = jjjji x1H1L yzzzz = ijj 3.11111 zzy
k x2H1L { k 4.33333 {
F HP1L = ijj -1.23457 zzy
k 8.34568 {
J HP1L = jji -5.22222 4.00000 zzy
k 2.22222 22.6667 {
Se resuelve el sistema lineal J HP1L DP = -F HP1L:
jji -5.22222 4.00000 yzz.jji Dx1 zyz
k 2.22222 22.6667 {k Dx2 {
DP = jij Dx1 zyz = jij -0.482214 zyz
k Dx2 { k -0.320916 {
102
Resolución Numérica de Sistemas de Ecuaciones no Lineales
El siguiente punto de la iteración es:
P2 = P1 + DP
P2 = jji 3.11111 zyz + jji -0.482214 zzy = ijj 2.6289 yzz
k 4.33333 { k -0.320916 { k 4.01242 {
Iteración i = 2.
P2 = jjjji x1H2L zyzzz = jji 2.62890 zzy
k x2H2L { k 4.01242 {
F HP2L = ijj -0.232531 yzz
k 0.644479 {
J HP2L = jji -4.25779 4.00000 yzz
k 1.25779 20.0993 {
Se resuelve el sistema lineal J HP2L DP = -F HP2L:
jji -4.25779 4.00000 yzz.jji Dx1 yzz
k 1.25779 20.0993 {k Dx2 {
DP = jji Dx1 yzz = jji -0.0800312 zyz
k Dx2 { k -0.0270564 {
El siguiente punto de la iteración es:
P3 = P2 + DP
P3 = jji 2.6289 yzz + jij -0.0800312 yzz = jij 2.54887 zyz
k 4.01242 { k -0.0270564 { k 3.98536 {
Iteración i = 3.
P3 = ijjjj x1H3L yzzzz = ijj 2.54887 zzy
k x2H3L { k 3.98536 {
F HP3L = ijj -0.00640500 yzz
k 0.00933319 {
J HP3L = jij -4.09773 4.00000 zzy
k 1.09773 19.8829 {
Se resuelve el sistema lineal J HP3L DP = -F HP3L:
jji -4.09773 4.00000 yzz.jji Dx1 zzy
k 1.09773 19.8829 {k Dx2 {
DP = jji Dx1 zyz = jji -0.00191791 yzz
k Dx2 { k -0.000363521 {
103
Resolución Numérica de Sistemas de Ecuaciones no Lineales
El siguiente punto de la iteración es:
P4 = P3 + DP
P4 = jji 2.54887 zzy + ijj -0.00191791 zzy = jji 2.54695 zzy
k 3.98536 { k -0.000363521 { k 3.985 {
Tabla de datos.
i Pi DP = jij Dx1 zyz Pi+1 = Pi + DP ¥Pi+1 - Pi∞¶
k Dx2 {
0 jji 2. yzz jij 1.11111 zzy jij 3.11111 zyz 1.33333
k 3. { k 1.33333 { k 4.33333 {
1 jji 3.11111 zzy ijj -0.482214 zzy jij 2.6289 zzy 0.482214
k 4.33333 { k -0.320916 { k 4.01242 {
2 ijj 2.6289 zzy jij -0.0800312 zzy jji 2.54887 yzz 0.0800312
k 4.01242 { k -0.0270564 { k 3.98536 {
3 ijj 2.54887 zzy jij -0.00191791 zzy ijj 2.54695 zzy 0.00191791
k 3.98536 { k -0.000363521 { k 3.985 {
La solución aproximada del sistema es:
P4 = ijj 2.54695 zyz
k 3.98500 {
Método de Newton-Raphson para sistemas de ecuaciones no lineales.
fi Hx1 , x2L = jji H1 - x1L x1 + 4 x2 - 12 25 yzz = jij 0 zzy
k Hx1 - 2L2 + H2 x2 - 3L2 - { k 0 {
P0 = jjjji x1H0L zyzzz = ijj -2. zzy
k x2H0L { k 2. {
La función vectorial y la matriz jacobiana son:
F Hx1, x2L = ijj H1 - x1L x1 + 4 x2 - 12 25 yzz
k Hx1 - 2L2 + H2 x2 - 3L2 - {
J Hx1, x2L = jji 1 - 2 x1 4 H2 x2 - 3L yzz
k 2 Hx1 - 2L 4 {
104
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Iteración i = 0.
P0 = ijjjj x1H0L yzzzz = ijj -2.00000 yzz
k x2H0L { k 2.00000 {
F HP0L = ijj -10.0000 yzz
k -8.00000 {
J HP0L = jij 5.00000 4.00000 yzz
k -8.00000 4.00000 {
Se resuelve el sistema lineal J HP0L DP = -F HP0L:
jji 5.00000 4.00000 zzy.jji Dx1 yzz
k -8.00000 4.00000 {k Dx2 {
DP = jji Dx1 yzz = ijj 0.153846 zzy
k Dx2 { k 2.30769 {
El siguiente punto de la iteración es:
P1 = P0 + DP
P1 = ijj -2. yzz + ijj 0.153846 yzz = jij -1.84615 zzy
k 2. { k 2.30769 { k 4.30769 {
Iteración i = 1.
P1 = jjjij x1H1L zzzzy = jij -1.84615 zzy
k x2H1L { k 4.30769 {
F HP1L = ijj -0.0236686 yzz
k 21.3254 {
J HP1L = jij 4.69231 4.00000 zzy
k -7.69231 22.4615 {
Se resuelve el sistema lineal J HP1L DP = -F HP1L:
jji 4.69231 4.00000 zzy.jji Dx1 zyz
k -7.69231 22.4615 {k Dx2 {
DP = jji Dx1 zzy = ijj 0.63036 yzz
k Dx2 { k -0.733544 {
El siguiente punto de la iteración es:
P2 = P1 + DP
P2 = jji -1.84615 zyz + ijj 0.63036 zzy = ijj -1.21579 zzy
k 4.30769 { k -0.733544 { k 3.57415 {
105
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Iteración i = 2.
P2 = jjjji x1H2L zzzzy = jij -1.21579 zzy
k x2H2L { k 3.57415 {
F HP2L = ijj -0.397354 yzz
k 2.54970 {
J HP2L = jij 3.43159 4.00000 zyz
k -6.43159 16.5932 {
Se resuelve el sistema lineal J HP2L DP = -F HP2L:
jji 3.43159 4.00000 zzy.jji Dx1 yzz
k -6.43159 16.5932 {k Dx2 {
DP = ijj Dx1 zzy = jji 0.203129 zzy
k Dx2 { k -0.0749256 {
El siguiente punto de la iteración es:
P3 = P2 + DP
P3 = ijj -1.21579 yzz + ijj 0.203129 zyz = jij -1.01266 zyz
k 3.57415 { k -0.0749256 { k 3.49922 {
Iteración i = 3.
P3 = jijjj x1H3L zzzyz = jji -1.01266 zyz
k x2H3L { k 3.49922 {
F HP3L = ijj -0.0412615 yzz
k 0.0637169 {
J HP3L = jji 3.02533 4.00000 yzz
k -6.02533 15.9938 {
Se resuelve el sistema lineal J HP3L DP = -F HP3L:
jji 3.02533 4.00000 zzy.jji Dx1 zzy
k -6.02533 15.9938 {k Dx2 {
DP = jji Dx1 yzz = jij 0.01262 yzz
k Dx2 { k 0.000770471 {
El siguiente punto de la iteración es:
P4 = P3 + DP
P4 = jji -1.01266 zzy + ijj 0.01262 zyz = jji -1.00004 zzy
k 3.49922 { k 0.000770471 { k 3.49999 {
Tabla de datos.
106
Resolución Numérica de Sistemas de Ecuaciones no Lineales
i Pi DP = jji Dx1 zyz Pi+1 = Pi + DP ¥Pi+1 - Pi∞¶
k Dx2 {
0 ijj -2. zyz jij 0.153846 zyz jji -1.84615 yzz 2.30769
k 2. { k 2.30769 { k 4.30769 {
1 jji -1.84615 yzz jji 0.63036 zzy jji -1.21579 zyz 0.733544
k 4.30769 { k -0.733544 { k 3.57415 {
2 jji -1.21579 zzy jji 0.203129 zzy jji -1.01266 zzy 0.203129
k 3.57415 { k -0.0749256 { k 3.49922 {
3 jji -1.01266 zyz jji 0.01262 yzz jji -1.00004 zzy 0.01262
k 3.49922 { k 0.000770471 { k 3.49999 {
La solución aproximada del sistema es:
P4 = jji -1.00004 zzy
k 3.49999 {
à Problema 19. Aproximar las soluciones de los siguientes sistemas no lineales,
empleando el método de Newton con la aproximacióin inicial dada, iterando hasta que
»» Pi+1 - Pi »»¶ § 10-6.
a) f1Hx1, x2L = 3 x21 - x22 = 0, f2Hx1, x2L = 3 x1 x22 - x31 - 1 = 0,
P0 = IxH10L, xH20LMT = H1, 1LT y P0 = IxH10L, xH20LMT = H1, -1LT .
b) f1Hx1, x2L = lnHx21 + x22L - senHx1 x2L - Hln 2 + ln pL,
f2Hx1, x2L = eHx1-x2L + cosHx1 x2L = 0,
P0 = IxH10L, x2H0LMT = H2, 2LT .
c) f1Hx1, x2, x3L = x31 + x21 x2 - x1 x3 + 6 = 0,
f2Hx1, x2, x3L = ex1 + ex2 - x3 = 0,
f3Hx1, x2, x3L = x22 - 2 x1 x3 = 4,
P0 = IxH10L, xH20L, x3H0LMT = H-1, -2, 1LT .
d) f1Hx1, x2, x3L = 6 x1 - 2 cosHx2 x3L - 1 = 0,
f2Hx1, x2, x3L = 9 x2 + "#x##12##+####s##e##n##H#x##3##L##+####1##.#0###6## + 0.9 = 0,
f3Hx1, x2, x3L = 60 x3 + 3 e-x1 x2 + 10 p - 3 = 0
P0 = Ix1H0L, xH20L, xH30LMT = H0, 0, 0LT
Solución
a)
<< Graphics`ImplicitPlot`;
Clear@f, g, g1, g2, ecuaciones, p, m, dD;
f = 3 x2 − y2;
g = 3 x ∗ y2 − x3 − 1;
107
Resolución Numérica de Sistemas de Ecuaciones no Lineales
g = 3 x ∗ y2 − x3 − 1;
Print@"Representación gráfica de la solución",
"\n\t f1Hx, yL = ", f, "\n\t f2Hx, yL = ", g D;
g1 = ImplicitPlot @f == 0, 8x, −2, 2<,
PlotStyle −> 88Thickness @0.010D, RGBColor @0, 0, 1D<<,
AxesLabel −> 8"X", "Y"<, DisplayFunction −> Identity D;
g2 = ImplicitPlot @g == 0, 8x, −2, 2<,
PlotStyle −> 88Thickness @0.010D, RGBColor @1, 0, 0D<<,
AxesLabel −> 8"X", "Y"<, DisplayFunction −> Identity D;
g = Show @g1, g2, AxesLabel −> 8"X", "Y"<,
AspectRatio −> Automatic, DisplayFunction −> $DisplayFunctionD;
ecuaciones = 83 x12 − x22, 3 x1 x22 − x31 − 1<;
p = 81.0, 1.0<;
m = 10;
d = 10.−6;
newtonSistemasNoLineal@ecuaciones, p, m, dD;
p = 81.0, −1.0<;
m = 10;
d = 10.−6;
newtonSistemasNoLineal@ecuaciones, p, m, dD;
Representación gráfica de la solución
f1Hx, yL = 3 x2 - y2
f2Hx, yL = -x3 + 3 y2 x - 1
Y
4
2
-2 -1 X
12
-2
-4
108
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Método de Newton-Raphson para sistemas de ecuaciones no lineales.
fi Hx1 , x2L = ijjjj 3 x12 - x22 x1 - 1 yzzzz = jji 0 zzy
k -x13 + 3 x22 { k 0 {
P0 = ijjjj x1H0L zyzzz = jij 1. zzy
k x2H0L { k 1. {
La función vectorial y la matriz jacobiana son:
F Hx1, x2L = ijjjj 3 x12 - x22 x1 - 1 yzzzz
k -x13 + 3 x22 {
J Hx1, x2L = jij 6 x1 - 3 x12 -2 x2 zzy
k 3 x22 6 x1 x2 {
Iteración i = 0.
P0 = jijjj x1H0L zyzzz = ijj 1.00000 zzy
k x2H0L { k 1.00000 {
F HP0L = ijj 2.00000 yzz
k 1.00000 {
J HP0L = jij 6.00000 -2.00000 zzy
k 0 6.00000 {
Se resuelve el sistema lineal J HP0L DP = -F HP0L:
jji 6.00000 -2.00000 yzz.jji Dx1 zzy
k 0 6.00000 {k Dx2 {
DP = ijj Dx1 zzy = jji -0.388889 yzz
k Dx2 { k -0.166667 {
El siguiente punto de la iteración es:
P1 = P0 + DP
P1 = ijj 1. zyz + jji -0.388889 zyz = ijj 0.611111 zzy
k 1. { k -0.166667 { k 0.833333 {
Iteración i = 1.
P1 = jijjj x1H1L zyzzz = jij 0.611111 yzz
k x2H1L { k 0.833333 {
F HP1L = ijj 0.425926 yzz
k 0.0449246 {
J HP1L = jji 3.66667 -1.66667 zzy
k 0.962963 3.05556 {
109
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Se resuelve el sistema lineal J HP1L DP = -F HP1L:
jji 3.66667 -1.66667 yzz.jji Dx1 zzy
k 0.962963 3.05556 {k Dx2 {
DP = ijj Dx1 zyz = ijj -0.107452 zyz
k Dx2 { k 0.0191611 {
El siguiente punto de la iteración es:
P2 = P1 + DP
P2 = ijj 0.611111 zyz + ijj -0.107452 yzz = jji 0.503659 zzy
k 0.833333 { k 0.0191611 { k 0.852494 {
Iteración i = 2.
P2 = jjjji x1H2L yzzzz = jji 0.503659 zzy
k x2H2L { k 0.852494 {
F HP2L = ijj 0.0342707 yzz
k -0.0296667 {
J HP2L = ijj 3.02195 -1.70499 yzz
k 1.41922 2.57620 {
Se resuelve el sistema lineal J HP2L DP = -F HP2L:
jji 3.02195 -1.70499 yzz.jji Dx1 zyz
k 1.41922 2.57620 {k Dx2 {
DP = jji Dx1 zyz = jji -0.00369496 yzz
k Dx2 { k 0.0135512 {
El siguiente punto de la iteración es:
P3 = P2 + DP
P3 = jji 0.503659 zzy + jji -0.00369496 zyz = jij 0.499964 yzz
k 0.852494 { k 0.0135512 { k 0.866046 {
Iteración i = 3.
P3 = jjjij x1H3L zyzzz = jji 0.499964 yzz
k x2H3L { k 0.866046 {
F HP3L = ijj - 0.000142677 zyz
k - 1.25763 µ 10-6 {
J HP3L = jji 2.99978 -1.73209 yzz
k 1.50021 2.59795 {
110
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Se resuelve el sistema lineal J HP3L DP = -F HP3L:
jji 2.99978 -1.73209 yzz.jji Dx1 yzz
k 1.50021 2.59795 {k Dx2 {
DP = jij Dx1 zyz = ijj 0.0000358789 zzy
k Dx2 { k -0.0000202346 {
El siguiente punto de la iteración es:
P4 = P3 + DP
P4 = ijj 0.499964 zzy + jji 0.0000358789 zzy = ijj 0.5 yzz
k 0.866046 { k -0.0000202346 { k 0.866025 {
Iteración i = 4.
P4 = jjijj x1H4L zyzzz = jij 0.500000 zyz
k x2H4L { k 0.866025 {
F HP4L = ijjjj 3.45246 µ 10-9 zzzyz
k -5.08917 µ 10-9 {
J HP4L = jij 3.00000 -1.73205 zyz
k 1.50000 2.59808 {
Se resuelve el sistema lineal J HP4L DP = -F HP4L:
jji 3.00000 -1.73205 yzz.jji Dx1 yzz
k 1.50000 2.59808 {k Dx2 {
DP = jji Dx1 yzz = jijjj -1.49197 µ 10-11 zyzzz
k Dx2 { k 1.96744 µ 10-9 {
El siguiente punto de la iteración es:
P5 = P4 + DP
P5 = ijj 0.5 zzy + jjjij -1.49197 µ 10-11 zzzzy = jji 0.5 zzy
k 0.866025 { k 1.96744 µ 10-9 { k 0.866025 {
Tabla de datos.
i Pi DP = ijj Dx1 zyz Pi+1 = Pi + DP ¥Pi+1 - Pi∞¶
k Dx2 { 0.388889
0.107452
0 ijj 1. yzz jij -0.388889 yzz jji 0.611111 zzy
k 1. { k -0.166667 { k 0.833333 {
1 ijj 0.611111 yzz jji -0.107452 zzy ijj 0.503659 zzy
k 0.833333 { k 0.0191611 { k 0.852494 {
111
Resolución Numérica de Sistemas de Ecuaciones no Lineales
2 jij 0.503659 zzy jji -0.00369496 zzy jij 0.499964 zzy 0.0135512
k 0.852494 { k 0.0135512 { k 0.866046 { 0.0000358789
1.96744 µ 10-9
3 ijj 0.499964 zzy ijj 0.0000358789 yzz jji 0.5 zzy
k 0.866046 { k -0.0000202346 { k 0.866025 {
4 jji 0.5 yzz jijjj -1.49197 µ 10-11 zyzzz ijj 0.5 zzy
k 0.866025 { k 1.96744 µ 10-9 { k 0.866025 {
La solución aproximada del sistema es:
P5 = jji 0.500000 zzy
k 0.866025 {
Método de Newton-Raphson para sistemas de ecuaciones no lineales.
fi Hx1 , x2L = ijjjj 3 x12 - x22 x1 - 1 yzzzz = jji 0 zzy
k -x13 + 3 x22 { k 0 {
P0 = ijjjj x1H0L zzzzy = jji 1. yzz
k x2H0L { k -1. {
La función vectorial y la matriz jacobiana son:
F Hx1, x2L = ijjjj 3 x12 - x22 x1 - 1 yzzzz
k -x13 + 3 x22 {
J Hx1, x2L = jji 6 x1 - 3 x12 -2 x2 yzz
k 3 x22 6 x1 x2 {
Iteración i = 0.
P0 = ijjjj x1H0L zyzzz = jji 1.00000 zyz
k x2H0L { k -1.00000 {
F HP0L = ijj 2.00000 yzz
k 1.00000 {
J HP0L = jij 6.00000 2.00000 yzz
k 0 -6.00000 {
Se resuelve el sistema lineal J HP0L DP = -F HP0L:
jji 6.00000 2.00000 yzz.jji Dx1 zyz
k 0 -6.00000 {k Dx2 {
DP = jij Dx1 zzy = jji -0.388889 zzy
k Dx2 { k 0.166667 {
112
Resolución Numérica de Sistemas de Ecuaciones no Lineales
El siguiente punto de la iteración es:
P1 = P0 + DP
P1 = jji 1. yzz + ijj -0.388889 zzy = jji 0.611111 zyz
k -1. { k 0.166667 { k -0.833333 {
Iteración i = 1.
P1 = jijjj x1H1L zzzyz = jji 0.611111 yzz
k x2H1L { k -0.833333 {
F HP1L = ijj 0.425926 yzz
k 0.0449246 {
J HP1L = jij 3.66667 1.66667 zzy
k 0.962963 -3.05556 {
Se resuelve el sistema lineal J HP1L DP = -F HP1L:
jji 3.66667 1.66667 yzz.jji Dx1 zzy
k 0.962963 -3.05556 {k Dx2 {
DP = jij Dx1 zyz = jji -0.107452 zzy
k Dx2 { k -0.0191611 {
El siguiente punto de la iteración es:
P2 = P1 + DP
P2 = jji 0.611111 yzz + ijj -0.107452 yzz = jji 0.503659 zyz
k -0.833333 { k -0.0191611 { k -0.852494 {
Iteración i = 2.
P2 = jjjij x1H2L yzzzz = jji 0.503659 zyz
k x2H2L { k -0.852494 {
F HP2L = ijj 0.0342707 yzz
k -0.0296667 {
J HP2L = ijj 3.02195 1.70499 yzz
k 1.41922 -2.57620 {
Se resuelve el sistema lineal J HP2L DP = -F HP2L:
jji 3.02195 1.70499 yzz.jji Dx1 zzy
k 1.41922 -2.57620 {k Dx2 {
DP = jji Dx1 zzy = ijj -0.00369496 yzz
k Dx2 { k -0.0135512 {
113
Resolución Numérica de Sistemas de Ecuaciones no Lineales
El siguiente punto de la iteración es:
P3 = P2 + DP
P3 = ijj 0.503659 yzz + jij -0.00369496 zyz = ijj 0.499964 yzz
k -0.852494 { k -0.0135512 { k -0.866046 {
Iteración i = 3.
P3 = ijjjj x1H3L zzyzz = jij 0.499964 zzy
k x2H3L { k -0.866046 {
F HP3L = ijj - 0.000142677 zzy
k - 1.25763 µ 10-6 {
J HP3L = jji 2.99978 1.73209 zzy
k 1.50021 -2.59795 {
Se resuelve el sistema lineal J HP3L DP = -F HP3L:
jji 2.99978 1.73209 yzz.jji Dx1 yzz
k 1.50021 -2.59795 {k Dx2 {
DP = jji Dx1 zzy = jji 0.0000358789 zzy
k Dx2 { k 0.0000202346 {
El siguiente punto de la iteración es:
P4 = P3 + DP
P4 = jij 0.499964 zyz + ijj 0.0000358789 zyz = jij 0.5 yzz
k -0.866046 { k 0.0000202346 { k -0.866025 {
Iteración i = 4.
P4 = jjjij x1H4L zzzzy = jij 0.500000 yzz
k x2H4L { k -0.866025 {
F HP4L = ijjjj 3.45246 µ 10-9 zzzzy
k -5.08917 µ 10-9 {
J HP4L = ijj 3.00000 1.73205 yzz
k 1.50000 -2.59808 {
Se resuelve el sistema lineal J HP4L DP = -F HP4L:
jji 3.00000 1.73205 yzz.jji Dx1 zyz
k 1.50000 -2.59808 {k Dx2 {
DP = jji Dx1 zzy = jijjj -1.49197 µ 10-11 zyzzz
k Dx2 { k -1.96744 µ 10-9 {
114
Resolución Numérica de Sistemas de Ecuaciones no Lineales
El siguiente punto de la iteración es:
P5 = P4 + DP
P5 = jji 0.5 zzy + ijjjj -1.49197 µ 10-11 zyzzz = jji 0.5 yzz
k -0.866025 { k -1.96744 µ 10-9 { k -0.866025 {
Tabla de datos.
i Pi DP = jji Dx1 zyz Pi+1 = Pi + DP ¥Pi+1 - Pi∞¶
k Dx2 {
0 jji 1. zzy ijj -0.388889 zyz jji 0.611111 zzy 0.388889
k -1. { k 0.166667 { k -0.833333 {
1 ijj 0.611111 zyz jji -0.107452 zyz jji 0.503659 zyz 0.107452
k -0.833333 { k -0.0191611 { k -0.852494 {
2 ijj 0.503659 zzy jji -0.00369496 zyz jji 0.499964 zyz 0.0135512
k -0.852494 { k -0.0135512 { k -0.866046 {
3 jji 0.499964 yzz jji 0.0000358789 yzz ijj 0.5 yzz 0.0000358789
k -0.866046 { k 0.0000202346 { k -0.866025 {
4 jji 0.5 zyz jjjij -1.49197 µ 10-11 zzzyz jji 0.5 zzy 1.96744 µ 10-9
k -0.866025 { k -1.96744 µ 10-9 { k -0.866025 {
La solución aproximada del sistema es:
P5 = ijj 0.500000 zzy
k -0.866025 {
Solución
b)
Clear@ecuaciones, p, d, m, dD;
ecuaciones = 8 Log@x21 + x22D − Sin@x1 ∗ x2D − HLog @2D + Log @PiDL,
Exp@x1 − x2D + Cos@x1 ∗ x2D<;
p = 82.0, 2.0<;
m = 10;
d = 10.−6;
newtonSistemasNoLineal@ecuaciones, p, m, dD;
115
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Método de Newton-Raphson para sistemas de ecuaciones no lineales.
fi Hx1 , x2L = jji logHx12 + x22L - sinHx1 x2L - logHpL - logH2L zyz = jji 0 zyz
k cosHx1 x2L + ‰x1 -x2 { k 0 {
P0 = jjjij x1H0L zzzzy = ijj 2. zyz
k x2H0L { k 2. {
La función vectorial y la matriz jacobiana son:
F Hx1, x2L = jji logHx12 + x22L - sinHx1 x2L - logHpL - logH2L zyz
k cosHx1 x2L + ‰x1 -x2 {
J Hx1, x2L = jjjij ÅxÅÅ21Å2ÅÅ+ÅxÅÅxÅ1Å22ÅÅÅ - cosHx1 x2L x2 ÅxÅÅ21Å2ÅÅ+ÅxÅÅxÅ2Å22ÅÅÅ - cosHx1 x2L x1 yzzzz
k ‰x1-x2 - sinHx1 x2L x2 -sinHx1 x2L x1 - ‰x1-x2 {
Iteración i = 0.
P0 = jjijj x1H0L zzzzy = jji 2.00000 zyz
k x2H0L { k 2.00000 {
F HP0L = ijj 0.998367 zzy
k 0.346356 {
J HP0L = ijj 1.80729 1.80729 zyz
k 2.51360 0.513605 {
Se resuelve el sistema lineal J HP0L DP = -F HP0L:
jji 1.80729 1.80729 yzz.jji Dx1 zyz
k 2.51360 0.513605 {k Dx2 {
DP = jij Dx1 yzz = jij -0.0313174 zzy
k Dx2 { k -0.521094 {
El siguiente punto de la iteración es:
P1 = P0 + DP
P1 = ijj 2. zzy + jji -0.0313174 yzz = jji 1.96868 yzz
k 2. { k -0.521094 { k 1.47891 {
Iteración i = 1.
P1 = jjjij x1H1L zzzzy = jij 1.96868 yzz
k x2H1L { k 1.47891 {
F HP1L = ijj -0.263765 yzz
k 0.658308 {
J HP1L = ijj 2.08935 2.40465 zyz
k 1.29466 -2.08095 {
116
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Se resuelve el sistema lineal J HP1L DP = -F HP1L:
jji 2.08935 2.40465 yzz.jji Dx1 zzy
k 1.29466 -2.08095 {k Dx2 {
DP = ijj Dx1 yzz = jij -0.138603 zzy
k Dx2 { k 0.230118 {
El siguiente punto de la iteración es:
P2 = P1 + DP
P2 = jji 1.96868 zzy + ijj -0.138603 yzz = ijj 1.83008 yzz
k 1.47891 { k 0.230118 { k 1.70902 {
Iteración i = 2.
P2 = jijjj x1H2L zzzyz = jji 1.83008 zzy
k x2H2L { k 1.70902 {
F HP2L = ijj -0.0160496 yzz
k 0.128786 {
J HP2L = ijj 2.29262 2.37505 yzz
k 1.10486 -1.15420 {
Se resuelve el sistema lineal J HP2L DP = -F HP2L:
jji 2.29262 2.37505 yzz.jji Dx1 zzy
k 1.10486 -1.15420 {k Dx2 {
DP = jji Dx1 zyz = ijj -0.0545226 zzy
k Dx2 { k 0.0593879 {
El siguiente punto de la iteración es:
P3 = P2 + DP
P3 = jji 1.83008 zyz + jji -0.0545226 yzz = jji 1.77556 yzz
k 1.70902 { k 0.0593879 { k 1.76841 {
Iteración i = 3.
P3 = jijjj x1H3L zzyzz = jji 1.77556 yzz
k x2H3L { k 1.76841 {
F HP3L = ijj -0.00220154 yzz
k 0.00717276 {
J HP3L = jji 2.33388 2.33875 yzz
k 1.00421 -1.01015 {
117
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Se resuelve el sistema lineal J HP3L DP = -F HP3L:
jji 2.33388 2.33875 yzz.jji Dx1 zzy
k 1.00421 -1.01015 {k Dx2 {
DP = jji Dx1 zzy = jji -0.003092 yzz
k Dx2 { k 0.00402689 {
El siguiente punto de la iteración es:
P4 = P3 + DP
P4 = ijj 1.77556 zzy + ijj -0.003092 zyz = jij 1.77247 zzy
k 1.76841 { k 0.00402689 { k 1.77244 {
Iteración i = 4.
P4 = jijjj x1H4L zzyzz = jji 1.77247 yzz
k x2H4L { k 1.77244 {
F HP4L = ijj -8.48201 µ 10-6 zzy
k 0.0000268752 {
J HP4L = jji 2.33663 2.33665 zyz
k 1.00002 -1.00004 {
Se resuelve el sistema lineal J HP4L DP = -F HP4L:
jji 2.33663 2.33665 yzz.jji Dx1 zzy
k 1.00002 -1.00004 {k Dx2 {
DP = jji Dx1 yzz = jji -0.0000116223 zzy
k Dx2 { k 0.0000152522 {
El siguiente punto de la iteración es:
P5 = P4 + DP
P5 = jji 1.77247 zzy + ijj -0.0000116223 zyz = jji 1.77245 yzz
k 1.77244 { k 0.0000152522 { k 1.77245 {
Iteración i = 5.
P5 = ijjjj x1H5L zyzzz = jij 1.77245 zzy
k x2H5L { k 1.77245 {
F HP5L = ijjjj -1.20840 µ 10-10 zyzzz
k 3.81824 µ 10-10 {
J HP5L = jji 2.33664 2.33664 zzy
k 1.00000 -1.00000 {
118
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Se resuelve el sistema lineal J HP5L DP = -F HP5L:
jji 2.33664 2.33664 yzz.jji Dx1 zzy
k 1.00000 -1.00000 {k Dx2 {
DP = ijj Dx1 zzy = jijjj -1.65054 µ 10-10 zzzzy
k Dx2 { k 2.16769 µ 10-10 {
El siguiente punto de la iteración es:
P6 = P5 + DP
P6 = jji 1.77245 yzz + jjijj -1.65054 µ 10-10 zzyzz = jji 1.77245 zzy
k 1.77245 { k 2.16769 µ 10-10 { k 1.77245 {
Tabla de datos.
i Pi DP = jji Dx1 yzz Pi+1 = Pi + DP ¥Pi+1 - Pi∞¶
k Dx2 {
0 ijj 2. yzz ijj -0.0313174 zzy jji 1.96868 zyz 0.521094
k 2. { k -0.521094 { k 1.47891 {
1 jji 1.96868 zzy jji -0.138603 zzy jji 1.83008 zyz 0.230118
k 1.47891 { k 0.230118 { k 1.70902 {
2 ijj 1.83008 zzy jji -0.0545226 yzz jji 1.77556 zzy 0.0593879
k 1.70902 { k 0.0593879 { k 1.76841 {
3 jji 1.77556 zyz ijj -0.003092 zzy jji 1.77247 zzy 0.00402689
k 1.76841 { k 0.00402689 { k 1.77244 {
4 jji 1.77247 zzy ijj -0.0000116223 zzy jji 1.77245 yzz 0.0000152522
k 1.77244 { k 0.0000152522 { k 1.77245 {
5 jji 1.77245 zzy jijjj -1.65054 µ 10-10 zzzyz ijj 1.77245 zzy 2.16769 µ 10-10
k 1.77245 { k 2.16769 µ 10-10 { k 1.77245 {
La solución aproximada del sistema es:
P6 = jji 1.77245 zzy
k 1.77245 {
Solución
c)
Clear@ecuaciones, p, m, dD;
ecuaciones = 8 x31 + x12 x2 − x1 x3 + 6,
Exp@x1D + Exp@x2D − x3, x22 − 2 x1 x3 − 4<;
p = 8−1.0, −2.0, 1.0<;
119
Resolución Numérica de Sistemas de Ecuaciones no Lineales
m = 10;
d = 10.−6;
newtonSistemasNoLineal@ecuaciones, p, m, dD;
Método de Newton-Raphson para sistemas de ecuaciones no lineales.
ijjjjjjjjj x13 + x2 x12 - x3 x1 + 6 zzzzzyzzzz jjjijjjjj 0 zyzzzzzzz
k -x3 + ‰x1 + ‰x2 { k 0 {
fi Hx1 , x2, x3L = = 0
x22 - 2 x1 x3 - 4
P0 = ijjjjjjjjjjj x1H0L zzzzzzzyzzzz = jjjijjjjj -1. zzzyzzzzz
k x2H0L { k -2. {
x3H0L 1.
La función vectorial y la matriz jacobiana son:
jijjjjjjjj x13 + x2 x12 - x3 x1 + 6 yzzzzzzzzz
k -x3 + ‰x1 + ‰x2 {
F Hx1, x2, x3L =
x22 - 2 x1 x3 - 4
jjjjjjjij 3 x12 + 2 x2 x1 - x3 x12 -x1 zzzzzzzzy
‰x1 ‰x2 -1
J Hx1, x2, x3L =
k -2 x3 2 x2 -2 x1 {
Iteración i = 0.
P0 = jjjjjjjjjjij x1H0L zzzzzyzzzzzz = jjijjjjjj -1.00000 zzzzzzzzy
k x2H0L { k -2.00000 {
x3H0L 1.00000
F HP0L = jjjjjjjij 4.00000 yzzzzzzzz
k -0.496785 {
2.00000
J HP0L = ijjjjjjjj 6.00000 1.00000 1.00000 zzzyzzzzz
k 0.367879 0.135335 -1.00000 {
-2.00000 -4.00000 2.00000
Se resuelve el sistema lineal J HP0L DP = -F HP0L:
jjjjjjijj 6.00000 1.00000 1.00000 zzzy{zzzzz.jjjjjjjkji Dx1 zzzzzzzzy
k 0.367879 0.135335 -1.00000 Dx2 {
-2.00000 -4.00000 2.00000 Dx3
DP = jjjijjjjj Dx1 zzyzzzzzz = jjjijjjjj -0.636738 zzzyzzzzz
k Dx2 { k 0.485723 {
Dx3 -0.665293
120
Resolución Numérica de Sistemas de Ecuaciones no Lineales
El siguiente punto de la iteración es:
P1 = P0 + DP
P1 = ijjjjjjjj -1. zzyzzzzzz + ijjjjjjjj -0.636738 zzyzzzzzz = ijjjjjjjj -1.63674 zzzzzzyzz
k -2. { k 0.485723 { k -1.51428 {
1. -0.665293 0.334707
Iteración i = 1.
P1 = jjjjjjjjjjji x1H1L zzzzzzyzzzzz = jjjjjjjij -1.63674 zzzzzzyzz
k x2H1L { k -1.51428 {
x3H1L 0.334707
F HP1L = jjjjjjijj -1.89347 zzzzyzzzz
k 0.0798737 {
-0.611308
J HP1L = ijjjjjjjj 12.6590 2.67891 1.63674 yzzzzzzzz
k 0.194614 0.219967 -1.00000 {
-0.669414 -3.02855 3.27348
Se resuelve el sistema lineal J HP1L DP = -F HP1L:
jjjijjjjj 12.6590 2.67891 1.63674 yzzzzzzzz.jjijjjjjj Dx1 zzzzyzzzz
k 0.194614 0.219967 -1.00000 {k Dx2 {
-0.669414 -3.02855 3.27348 Dx3
DP = jjjjjjijj Dx1 zzyzzzzzz = jjjjjjjji 0.171881 zzzzzzyzz
k Dx2 { k -0.153955 {
Dx3 0.0794592
El siguiente punto de la iteración es:
P2 = P1 + DP
P2 = ijjjjjjjj -1.63674 zyzzzzzzz + ijjjjjjjj 0.171881 zzzzzzzyz = jjjjjjjji -1.46486 zzzzzzzzy
k -1.51428 { k -0.153955 { k -1.66823 {
0.334707 0.0794592 0.414166
121
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Iteración i = 2.
P2 = jjjjjjjjjjji x1H2L zzzzzzzzyzzz = ijjjjjjjj -1.46486 zyzzzzzzz
k x2H2L { k -1.66823 {
x3H2L 0.414166
F HP2L = jjjjjjijj -0.116306 zyzzzzzzz
k 0.00552485 {
-0.00361305
J HP2L = jjjjjijjj 10.9107 2.14581 1.46486 zzzzzzzyz
k 0.231111 0.188580 -1.00000 {
-0.828333 -3.33646 2.92971
Se resuelve el sistema lineal J HP2L DP = -F HP2L:
jjjjjjjij 10.9107 2.14581 1.46486 zzzzzzzzy.jjjjjjjji Dx1 zzzzzzzyz
k 0.231111 0.188580 -1.00000 {k Dx2 {
-0.828333 -3.33646 2.92971 Dx3
DP = ijjjjjjjj Dx1 zzyzzzzzz = jijjjjjjj 0.00874949 zzzzzzzzy
k Dx2 { k 0.00404093 {
Dx3 0.00830899
El siguiente punto de la iteración es:
P3 = P2 + DP
P3 = jjjjjjjji -1.46486 zzzzzyzzz + jjjjjijjj 0.00874949 zzzyzzzzz = jjijjjjjj -1.45611 zzzyzzzzz
k -1.66823 { k 0.00404093 { k -1.66419 {
0.414166 0.00830899 0.422475
Iteración i = 3.
P3 = jjjjjjjjjijj x1H3L yzzzzzzzzzzz = ijjjjjjjj -1.45611 zzzzzzzzy
k x2H3L { k -1.66419 {
x3H3L 0.422475
F HP3L = jjjjjjjji -0.000639433 yzzzzzzzz
k 0.0000104138 {
-0.000129070
J HP3L = jjjjjijjj 10.7848 2.12025 1.45611 zzzzzzyzz
k 0.233142 0.189344 -1.00000 {
-0.844951 -3.32838 2.91221
122
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Se resuelve el sistema lineal J HP3L DP = -F HP3L:
ijjjjjjjj 10.7848 2.12025 1.45611 yzzzzzzzz{.jjjjkjjjij Dx1 zzzzzzzzy
k 0.233142 0.189344 -1.00000 Dx2 {
-0.844951 -3.32838 2.91221 Dx3
DP = jjjjjjjji Dx1 zzzzzzzyz = jjjjjijjj 0.0000646082 zzzzzyzzz
k Dx2 { k -0.0000394195 {
Dx3 0.0000180128
El siguiente punto de la iteración es:
P4 = P3 + DP
P4 = jjjjjjjji -1.45611 zzzzyzzzz + ijjjjjjjj 0.0000646082 zzzzzzzzy = jjjjjjjji -1.45604 zzyzzzzzz
k -1.66419 { k -0.0000394195 { k -1.66423 {
0.422475 0.0000180128 0.422493
Iteración i = 4.
P4 = jjjjjjjijjjj x1H4L zzzzyzzzzzzz = jjjjjjijj -1.45604 zzzzzzzyz
k x2H4L { k -1.66423 {
x3H4L 0.422493
F HP4L = jjjjijjjjjj -1.89278 µ 10-8 yzzzzzzzzzz
k 6.33712 µ 10-10 {
-7.73657 µ 10-10
J HP4L = jjjjjijjj 10.7841 2.12006 1.45604 zzzzzzzzy
k 0.233157 0.189336 -1.00000 {
-0.844987 -3.32846 2.91209
Se resuelve el sistema lineal J HP4L DP = -F HP4L:
jjjjijjjj 10.7841 2.12006 1.45604 zzzzyzzzz.ijjjjjjjj Dx1 zyzzzzzzz
k 0.233157 0.189336 -1.00000 {k Dx2 {
-0.844987 -3.32846 2.91209 Dx3
DP = ijjjjjjjj Dx1 zyzzzzzzz = jjjjjjjijjj 1.55555 µ 10-9 zzzzzzzzyzz
k Dx2 { k 2.92941 µ 10-10 {
Dx3 1.05186 µ 10-9
El siguiente punto de la iteración es:
P5 = P4 + DP
P5 = ijjjjjjjj -1.45604 yzzzzzzzz + jjijjjjjjjj 1.55555 µ 10-9 zzyzzzzzzzz = jjjjjjjji -1.45604 zzzyzzzzz
k -1.66423 { k 2.92941 µ 10-10 { k -1.66423 {
0.422493 1.05186 µ 10-9 0.422493
123
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Tabla de datos.
i Pi DP = jjijjjjjj Dx1 zzzzyzzzz Pi+1 = Pi + DP ¥Pi+1 - Pi∞¶
k Dx2 {
Dx3
0 jjjijjjjj -1. zzzzzyzzz jijjjjjjj -0.636738 zzzzzzzyz jjijjjjjj -1.63674 zzzzzzyzz 0.665293
k -2. { k 0.485723 { k -1.51428 {
1. -0.665293 0.334707
1 jjjjjjijj -1.63674 zzzyzzzzz jjjjijjjj 0.171881 zyzzzzzzz jjjjjjjji -1.46486 zzzzyzzzz 0.171881
k -1.51428 { k -0.153955 { k -1.66823 {
0.334707 0.0794592 0.414166
2 ijjjjjjjj -1.46486 yzzzzzzzz jjjjijjjj 0.00874949 zzzzzzzyz jjjijjjjj -1.45611 zzzzzzzzy 0.00874949
k -1.66823 { k 0.00404093 { k -1.66419 {
0.414166 0.00830899 0.422475
3 jjjjjjjji -1.45611 zzzzzzzzy jjjjjijjj 0.0000646082 zzzyzzzzz jijjjjjjj -1.45604 zzzzzzzyz 0.0000646082
k -1.66419 { k -0.0000394195 { k -1.66423 {
0.422475 0.0000180128 0.422493
4 jijjjjjjj -1.45604 zzzyzzzzz jjjjjjjjijj 1.55555 µ 10-9 zyzzzzzzzzz jijjjjjjj -1.45604 zzzzzzzyz 1.55555 µ 10-9
k -1.66423 { k 2.92941 µ 10-10 { k -1.66423 {
0.422493 1.05186 µ 10-9 0.422493
La solución aproximada del sistema es:
P5 = jjjjjjjji -1.45604 zzzzzyzzz
k -1.66423 {
0.422493
Solución
d)
Clear@f, g, g1, g2, ecuaciones, p, m, dD;
ecuaciones = 9 6 x1 − 2 Cos@x2 x3D − 1, 9 x2 + "#x##21###+###S##i###n##@##x###3#D####+###1##.###0##6### + 0.9,
60 x3 + 3 Exp@−x1 x2D + 10 Pi − 3=;
p = 80., 0., 0.<;
m = 10;
d = 10.−6;
newtonSistemasNoLineal@ecuaciones, p, m, dD;
124
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Método de Newton-Raphson para sistemas de ecuaciones no lineales.
fi Hx1 , x2, x3L = jjjjjjjjjji -2 cosHx2 x3L + 6 x1 - 1 + 0.9 zzzzzzzzzzy = jjjjjjjji 0 zzzzyzzzz
k { k 0 {
9 x2 + "#x##12###+####s##i##n###H#x##3##L###+###1###.##0###6## 0
60 x3 + 3 ‰-x1 x2 + 10 p - 3
P0 = jjjjjjjijjjj x1H0L zzzyzzzzzzzz = jjjjjijjj 0. yzzzzzzzz
k x2H0L { k 0. {
x3H0L 0.
La función vectorial y la matriz jacobiana son:
F Hx1, x2, x3L = jjjjjjjijjj -2 cosHx2 x3L + 6 x1 - 1 + 0.9 zzzzzyzzzzz
k {
9 x2 + "#x##12###+####s##i##n###H#x##3##L###+###1###.##0###6##
60 x3 + 3 ‰-x1 x2 + 10 p - 3
J Hx1, x2, x3L = ijjjjjjjjjjj 6 2 sinHx2 x3L x3 2 sinHx2 x3L x2 zzzzzzzzzzzy
k 9 Å2ÅÅÅÅ"ÅÅÅÅ#xÅ#Å#12Å#Å#Å#+Å#cÅ#sÅ#Åo#Å#iÅ#snÅ#Å#HÅ#Hx#ÅxÅ#Å3#3Å#Å#LÅL#Å#+Å#Å##1ÅÅ#Å#.Å#Å#0Å#Å#6Å#Å#Å#ÅÅ
Å"ÅÅÅÅ#xÅ#Å#Å12#Å#Å#+Å#Å#Ås#Å#Å#iÅ#Ån#Å#xÅ#HÅ#Å1x#Å#Å3#Å#ÅL#Å#Å+#Å#Å#1Å#Å#Å.#Å#0Å#ÅÅ#6#Å#Å#ÅÅ
-3 ‰-x1 x2 x2 -3 ‰-x1 x2 x1 60 {
Iteración i = 0.
P0 = jjjjjjjjijjj x1H0L zzzzyzzzzzzz = jjjjjjjij 0 zzyzzzzzz
k x2H0L { k 0 {
x3H0L 0
F HP0L = jjjjjjjij -3.00000 yzzzzzzzz
k 1.92956 {
31.4159
J HP0L = ijjjjjjjj 6.00000 0 0 zzzzzzzyz
k 0 9.00000 0.485643 {
0 0 60.0000
Se resuelve el sistema lineal J HP0L DP = -F HP0L:
jjjjijjjj 6.00000 0 0 yzzzzzzzz.jjjjijjjj Dx1 zzzyzzzzz
k 0 9.00000 0.485643 {k Dx2 {
0 0 60.0000 Dx3
DP = jjjjijjjj Dx1 zzzzzzzzy = jjjjjijjj 0.5 yzzzzzzzz
k Dx2 { k -0.186142 {
Dx3 -0.523599
125
Resolución Numérica de Sistemas de Ecuaciones no Lineales
El siguiente punto de la iteración es:
P1 = P0 + DP
P1 = jjjijjjjj 0. zzzzzzzyz + jjjjjjjij 0.5 zzyzzzzzz = jijjjjjjj 0.5 zzzzzzyzz
k 0. { k -0.186142 { k -0.186142 {
0. -0.523599 -0.523599
Iteración i = 1.
P1 = jjjjjjjjjjij x1H1L zzyzzzzzzzzz = jjjjijjjj 0.500000 zzzzzzzyz
k x2H1L { k -0.186142 {
x3H1L -0.523599
F HP1L = jjjjjjijj 0.00949169 zzzyzzzzz
k 0.124719 {
0.292620
J HP1L = jjjjijjjj 6.00000 -0.101902 -0.0362269 zzzzzzzzy
k 0.555556 9.00000 0.481125 {
0.612896 -1.64631 60.0000
Se resuelve el sistema lineal J HP1L DP = -F HP1L:
jjjjjjjij 6.00000 -0.101902 -0.0362269 zzzzzzzyz.jjjjjijjj Dx1 yzzzzzzzz
k 0.555556 9.00000 0.481125 {k Dx2 {
0.612896 -1.64631 60.0000 Dx3
DP = jjjjjjjij Dx1 zzzzzzyzz = jjjjijjjj -0.00184219 yzzzzzzzz
k Dx2 { k -0.0134645 {
Dx3 -0.00522762
El siguiente punto de la iteración es:
P2 = P1 + DP
P2 = jjijjjjjj 0.5 yzzzzzzzz + ijjjjjjjj -0.00184219 zzzzyzzzz = jjjjjjjji 0.498158 zzyzzzzzz
k -0.186142 { k -0.0134645 { k -0.199607 {
-0.523599 -0.00522762 -0.528826
126
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Iteración i = 2.
P2 = ijjjjjjjjjjj x1H2L zzzzzzzzzyzz = jjjjijjjj 0.498158 yzzzzzzzz
k x2H2L { k -0.199607 {
x3H2L -0.528826
F HP2L = jijjjjjjj 0.0000788744 zzyzzzzzz
k -1.26903 µ 10-6 {
-0.0000148414
J HP2L = jjjjjjijj 6.00000 -0.111436 -0.0420617 zzzzzyzzz
k 0.555694 9.00000 0.481561 {
0.661426 -1.65072 60.0000
Se resuelve el sistema lineal J HP2L DP = -F HP2L:
jjjjjijjj 6.00000 -0.111436 -0.0420617 zzzzzzyzz.jjjjjjijj Dx1 zyzzzzzzz
k 0.555694 9.00000 0.481561 {k Dx2 {
0.661426 -1.65072 60.0000 Dx3
DP = jjjijjjjj Dx1 zzzzzzzyz = jijjjjjjjj -0.0000131256 zzzzzyzzzz
k Dx2 { k 9.2908 µ 10-7 {
Dx3 4.1761 µ 10-7
El siguiente punto de la iteración es:
P3 = P2 + DP
P3 = jjjjjjjji 0.498158 zzzzzzyzz + jjjijjjjjj -0.0000131256 zzzzzyzzzz = jjjijjjjj 0.498145 yzzzzzzzz
k -0.199607 { k 9.2908 µ 10-7 { k -0.199606 {
-0.528826 4.1761 µ 10-7 -0.528826
Iteración i = 3.
P3 = jjjjjjijjjjj x1H3L yzzzzzzzzzzz = jjjjjjjji 0.498145 zzzyzzzzz
k x2H3L { k -0.199606 {
x3H3L -0.528826
F HP3L = ijjjjjjjjjj 4.09894 µ 10-13 zzyzzzzzzzz
k 6.80560 µ 10-11 {
5.61542 µ 10-11
J HP3L = jjjjijjjj 6.00000 -0.111435 -0.0420613 yzzzzzzzz
k 0.555684 9.00000 0.481565 {
0.661421 -1.65067 60.0000
127
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Se resuelve el sistema lineal J HP3L DP = -F HP3L:
jijjjjjjj 6.00000 -0.111435 -0.0420613 zzzzzyzzz.jjjjjjjij Dx1 zzzzyzzzz
k 0.555684 9.00000 0.481565 {k Dx2 {
0.661421 -1.65067 60.0000 Dx3
DP = jjjjjjjji Dx1 zyzzzzzzz = jjjjjjjjjij -2.15366 µ 10-13 zzzzzzzzzzy
k Dx2 { k -7.48751 µ 10-12 {
Dx3 -1.13952 µ 10-12
El siguiente punto de la iteración es:
P4 = P3 + DP
P4 = jjjjjjjij 0.498145 zzzzzzzyz + ijjjjjjjjjj -2.15366 µ 10-13 yzzzzzzzzzz = jijjjjjjj 0.498145 zzzzyzzzz
k -0.199606 { k -7.48751 µ 10-12 { k -0.199606 {
-0.528826 -1.13952 µ 10-12 -0.528826
Tabla de datos.
i Pi DP = jjjjjjjji Dx1 yzzzzzzzz Pi+1 = Pi + DP ¥Pi+1 - Pi∞¶
k Dx2 {
Dx3
0 jjjjjjjij 0. zzyzzzzzz jjjjijjjj 0.5 zzzzzyzzz jjjjjijjj 0.5 zzzzzzzzy 0.523599
k 0. { k -0.186142 { k -0.186142 {
0. -0.523599 -0.523599
1 jjjjjjjji 0.5 zyzzzzzzz ijjjjjjjj -0.00184219 zzyzzzzzz jjjjjijjj 0.498158 zyzzzzzzz 0.0134645
k -0.186142 { k -0.0134645 { k -0.199607 {
-0.523599 -0.00522762 -0.528826
2 jijjjjjjj 0.498158 zzzzyzzzz jijjjjjjjj -0.0000131256 zzyzzzzzzz jjjjjjjji 0.498145 zzzzzzzyz 0.0000131256
k -0.199607 { k 9.2908 µ 10-7 { k -0.199606 {
-0.528826 4.1761 µ 10-7 -0.528826
3 jjjjjjjij 0.498145 zzzzzzzyz jjjjjjjjjij -2.15366 µ 10-13 zzyzzzzzzzz ijjjjjjjj 0.498145 zzzzzzzzy 7.48751 µ 10-12
k -0.199606 { k -7.48751 µ 10-12 { k -0.199606 {
-0.528826 -1.13952 µ 10-12 -0.528826
La solución aproximada del sistema es:
P4 = jjijjjjjj 0.498145 zzzzyzzzz
k -0.199606 {
-0.528826
128
Resolución Numérica de Sistemas de Ecuaciones no Lineales
à Problema 20. El sistema de ecuaciones no lineal :
f1Hx1, x2, x3, x4L = 4 x1 - x2 + x3 = x1 x4,
f2Hx1, x2, x3, x4L = -x1 + 3 x2 - 2 x3 = x2 x4,
f3Hx1, x2, x3, x4L = x1 - 2 x2 + 3 x3 =x3x4,
f4Hx1, x2, x3, x4L = x21 + x22 + x23 = 1
tiene vaias soluciones. Aplíquese el método de Newton para aproximarlas tomando como
puntos inicial el punto P0, y aplicando el método hasta que »» Pi+1 - Pi »»¶ § 10-5.
a) P0 = IxH10L, x2H0L, x3H0L, xH40LMT = H0, 1, 1, 1LT ,
b) P0 = IxH10L, x2H0L, xH30L, x4H0LMT = H1, -1, 1, 1LT .
c) P0 = Ix1H0L, xH20L, xH30L, xH40LMT = H1, 0, 0, 1LT .
Solución
a)
Clear@ecuaciones, p, m, dD;
ecuaciones = 8 4 x1 − x2 + x3 − x1 x4, −x1 + 3 x2 − 2 x3 − x2 x4,
x1 − 2 x2 + 3 x3 − x3 x4, x21 + x22 + x32 − 1 <;
p = 80.0, 1.0, 1.0, 1.0<;
m = 10;
d = 10.−5;
newtonSistemasNoLineal@ecuaciones, p, m, dD;
Método de Newton-Raphson para sistemas de ecuaciones no lineales.
fi Hx1 , x2, x3, x4L = jjjjjjjjjjjjjji -x4 x1 + 4 x1 - x2 + x3 zzzzzzzzzzzzyzz = jjjjjjjjjijjjj 0 zzzzzzzzzzzzyz
k -x1 + 3 x2 - 2 x3 - x2 x4 { k 0 {
x1 - 2 x2 + 3 x3 - x3 x4 0
x12 + x22 + x32 - 1 0
P0 = ijjjjjjjjjjjjjjjjjj x1H0L zzzzzzzzzzzzyzzzzzz = ijjjjjjjjjjjjj 0. yzzzzzzzzzzzzz
k x2H0L { k 1. {
x3H0L 1.
x4H0L 1.
129
Resolución Numérica de Sistemas de Ecuaciones no Lineales
La función vectorial y la matriz jacobiana son:
F Hx1, x2, x3, x4L = jjjjjjjjijjjjjj -x4 x1 + 4 x1 - x2 + x3 zzzzzzzzzzyzzzz
k -x1 + 3 x2 - 2 x3 - x2 x4 {
x1 - 2 x2 + 3 x3 - x3 x4
x12 + x22 + x32 - 1
J Hx1, x2, x3, x4L = ijjjjjjjjjjjjj 4 - x4 -1 1 -x1 yzzzzzzzzzzzzz
k -1 3 - x4 -2 -x2 {
1 -2 3 - x4 -x3
2 x1 2 x2 2 x3 0
Iteración i = 0.
P0 = jjjjjjjjjjjjjjijjjj x1H0L zzzzzzzzzyzzzzzzzzz = jjjjjjjjjjjijj 0 zzyzzzzzzzzzzz
k x2H0L { k 1.00000 {
x3H0L 1.00000
x4H0L 1.00000
F HP0L = jjjjijjjjjjjjj 0 zzzzzzzzzzzzzy
k 0 {
0
1.00000
J HP0L = ijjjjjjjjjjjjj 3.00000 -1.00000 1.00000 0 zzzzzzzzzzzzzy
k -1.00000 2.00000 -2.00000 -1.00000 {
1.00000 -2.00000 2.00000 -1.00000
0 2.00000 2.00000 0
Se resuelve el sistema lineal J HP0L DP = -F HP0L:
jjijjjjjjjjjjj 3.00000 -1.00000 1.00000 0 zzzzzyzzzzzzzz.jjjjjjjjjjjjjij Dx1 zzzzzzyzzzzzzzz
-1.00000 2.00000 -2.00000 -1.00000 {k Dx2 {
1.00000 -2.00000 2.00000 -1.00000
2.00000 2.00000 0 Dx3
k0 Dx4
DP = jjjjjjjjjjjjjji Dx1 zzyzzzzzzzzzzzz = ijjjjjjjjjjjjj 0. yzzzzzzzzzzzzz
k Dx2 { k -0.25 {
-0.25
Dx3 0.
Dx4
130
Resolución Numérica de Sistemas de Ecuaciones no Lineales
El siguiente punto de la iteración es:
P1 = P0 + DP
P1 = ijjjjjjjjjjjjj 0. zzzzzzzzzzzzzy + jjjjjjjjjjjjji 0. yzzzzzzzzzzzzz = ijjjjjjjjjjjjj 0. yzzzzzzzzzzzzz
k 1. { k -0.25 { k 0.75 {
1. -0.25 0.75
1. 0. 1.
Iteración i = 1.
P1 = jjjjjjjjjjjjijjjjjj x1H1L yzzzzzzzzzzzzzzzzzz = jjjjjjjjjjijjj 0 zyzzzzzzzzzzzz
k x2H1L { k 0.750000 {
x3H1L 0.750000
x4H1L 1.00000
F HP1L = jjjjjjjjjjjjji 0 zzzzzzzzzzzzzy
k 0 {
0
0.125000
J HP1L = ijjjjjjjjjjjjj 3.00000 -1.00000 1.00000 0 zzzzzzzzzzzzzy
k -1.00000 2.00000 -2.00000 -0.750000
1.00000 -2.00000 2.00000 -0.750000
0 1.50000 1.50000
0{
Se resuelve el sistema lineal J HP1L DP = -F HP1L:
jjjjjjjjjijjjj 3.00000 -1.00000 1.00000 0 zzyzzzzzzzzzzz.jjjjijjjjjjjjjj Dx1 zzzzzzzyzzzzzzz
k -1.00000 2.00000 -2.00000 -0.750000 {k Dx2 {
1.00000 -2.00000 2.00000 -0.750000
0 1.50000 1.50000 0 Dx3
Dx4
DP = ijjjjjjjjjjjjjj Dx1 zzzzyzzzzzzzzzz = ijjjjjjjjjjjjj 0. yzzzzzzzzzzzzz
k Dx2 { -0.0416667 {
-0.0416667
Dx3
Dx4 k 0.
El siguiente punto de la iteración es:
P2 = P1 + DP
P2 = ijjjjjjjjjjjjj 0. yzzzzzzzzzzzzz + ijjjjjjjjjjjjj 0. yzzzzzzzzzzzzz = ijjjjjjjjjjjjj 0. zzzzzzzzzzzzzy
k 0.75 { k -0.0416667 { k 0.708333 {
0.75 -0.0416667 0.708333
1. 0. 1.
131
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Iteración i = 2.
P2 = jjjjjijjjjjjjjjjjjj x1H2L zzzzzyzzzzzzzzzzzzz = jjjjjjjjjjijjj 0 zyzzzzzzzzzzzz
k x2H2L { k 0.708333 {
x3H2L 0.708333
x4H2L 1.00000
F HP2L = jjjijjjjjjjjjjjj 0 zzzzzzzzzzzzzzyz
k -1.11022 µ 10-16 {
-1.11022 µ 10-16
0.00347222
J HP2L = ijjjjjjjjjjjjj 3.00000 -1.00000 1.00000 0 zzzzzzzzzzzzzy
k -1.00000 2.00000 -2.00000 -0.708333 {
1.00000 -2.00000 2.00000 -0.708333
0 1.41667 1.41667 0
Se resuelve el sistema lineal J HP2L DP = -F HP2L:
jijjjjjjjjjjjj 3.00000 -1.00000 1.00000 0 zzzzzzzzzzzzzy.jjjjjjjjjjjjjji Dx1 zyzzzzzzzzzzzzz
k -1.00000 2.00000 -2.00000 -0.708333 {k Dx2 {
1.00000 -2.00000 2.00000 -0.708333
0 1.41667 1.41667 0 Dx3
Dx4
DP = ijjjjjjjjjjjjjj Dx1 zzzzzzzzzyzzzzz = ijjjjjjjjjjjjjj 0. zzzzzzzzyzzzzzz
k Dx2 { k -0.00122549 {
-0.00122549
Dx3 -1.56737 µ 10-16
Dx4
El siguiente punto de la iteración es:
P3 = P2 + DP
P3 = ijjjjjjjjjjjjj 0. zzzzzzzzzzzzzy + jjjjjjjjjjjjjji 0. zzzzzzzzzzzzzzy = ijjjjjjjjjjjjj 0. zzzzzzzzzzzzzy
k 0.708333 { k -0.00122549 { k 0.707108 {
0.708333 -0.00122549 0.707108
1. -1.56737 µ 10-16 1.
132
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Iteración i = 3.
P3 = jjjjjjjjjjjjjjjijjj x1H3L zzzzzzzzzzzzzzzyzzz = ijjjjjjjjjjjjj 0 yzzzzzzzzzzzzz
k x2H3L { k 0.707108 {
x3H3L 0.707108
x4H3L 1.0000
F HP3L = jjjjjjjjijjjjjjjj 0 µ 10-16 yzzzzzzzzzzzzzzzz
k 2.22045 µ 10-16 {
2.22045 µ 10-6
3.00365
J HP3L = ijjjjjjjjjjjjj 3.00000 -1.00000 1.00000 0 zzzzzzzzzzzzzy
-1.00000 2.00000 -2.00000 -0.707108
1.00000 -2.00000 2.00000 -0.707108
1.41422 1.41422
k0 0{
Se resuelve el sistema lineal J HP3L DP = -F HP3L:
jjjjjijjjjjjjj 3.00000 -1.00000 1.00000 0 zzzzzzyzzzzzzz.jjjjjjjjjjjjjji Dx1 zzzyzzzzzzzzzzz
-1.00000 2.00000 -2.00000 -0.707108 {k Dx2 {
1.00000 -2.00000 2.00000 -0.707108
1.41422 1.41422 0 Dx3
k0 Dx4
DP = ijjjjjjjjjjjjjj Dx1 zzzzzzzzzzzzzyz = ijjjjjjjjjjjjjjjj 0. µ 10-6 zzyzzzzzzzzzzzzzz
k Dx2 { k -1.06195 µ 10-6 {
-1.06195 10-16
Dx3 3.14018 µ
Dx4
El siguiente punto de la iteración es:
P4 = P3 + DP
P4 = ijjjjjjjjjjjjj 0. zzzzzzzzzzzzzy + jjjjjjjjjjjjjjjji 0. zzzzzzyzzzzzzzzzz = ijjjjjjjjjjjjj 0. zzzzzzzzzzzzzy
k 0.707108 { k -1.06195 µ 10-6 { k 0.707107 {
0.707108 -1.06195 µ 10-6 0.707107
1. 3.14018 µ 10-16 1.
Tabla de datos.
133
Resolución Numérica de Sistemas de Ecuaciones no Lineales
i Pi DP = jjjjjjjjjjjjijj Dx1 zzzzzzzyzzzzzzz Pi+1 = Pi + DP ¥Pi+1 - Pi∞¶
k Dx2 {
Dx3
Dx4
0 jjjjjijjjjjjjj 0. zzzzyzzzzzzzzz jjjjjjjjjijjjj 0. yzzzzzzzzzzzzz ijjjjjjjjjjjjj 0. zzzzzzzzyzzzzz 0.25
k 1. { -0.25 0.75
1. -0.25 0.75
1.
k 0. { k 1. {
1 jjijjjjjjjjjjj 0. zzyzzzzzzzzzzz jjjjjijjjjjjjj 0. zzzzzzzzzzzzzy jjijjjjjjjjjjj 0. zzzzzzzzzzzzzy 0.0416667
k 0.75 { k -0.0416667 { 0.708333
0.75 -0.0416667 0.708333
1. 0.
k 1. {
2 jijjjjjjjjjjjj 0. zzzzzzzzzzzyzz jjjjjijjjjjjjjj 0. zzzzzzzzzzzzzyz ijjjjjjjjjjjjj 0. zzyzzzzzzzzzzz 0.00122549
0.708333 k -0.00122549 { k 0.707108 {
0.708333 -0.00122549 0.707108
-1.56737 µ 10-16 1.
k 1. {
3 jjjjjjjjjjjjji 0. zzzzzzzzzzzzzy jjjjjjjjjjjjjijjj 0. yzzzzzzzzzzzzzzzz jjjjjjjjjjjjji 0. zzzzzzzzzzzzzy 1.06195 µ 10-6
k 0.707108 { k -1.06195 µ 10-6 { k 0.707107 {
0.707108 -1.06195 µ 10-6 0.707107
1. 3.14018 µ 10-16 1.
La solución aproximada del sistema es:
P4 = ijjjjjjjjjjjjj 0 zzzzzzzzzzzzzy
0.707107
0.707107
k 1.00000 {
Solución
b)
Clear@ecuaciones, p, m, dD;
ecuaciones = 8 4 x1 − x2 + x3 − x1 x4, −x1 + 3 x2 − 2 x3 − x2 x4,
x1 − 2 x2 + 3 x3 − x3 x4, x12 + x22 + x32 − 1 <;
p = 81.0, −1.0, 1.0, 1.0<;
m = 10;
d = 10.−5;
newtonSistemasNoLineal@ecuaciones, p, m, dD;
134
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Método de Newton-Raphson para sistemas de ecuaciones no lineales.
fi Hx1 , x2, x3, x4L = jjjjjjjjjjjjjji -x4 x1 + 4 x1 - x2 + x3 zzzzzzzzyzzzzzz = jjjijjjjjjjjjj 0 zzzzzzzzzzzzyz
k -x1 + 3 x2 - 2 x3 - x2 x4 { k 0 {
x1 - 2 x2 + 3 x3 - x3 x4 0
x12 + x22 + x32 - 1 0
P0 = ijjjjjjjjjjjjjjjjjj x1H0L zzzzzzzzzyzzzzzzzzz = ijjjjjjjjjjjjj 1. zzzzzzzzzzzzzy
k x2H0L { k -1. {
x3H0L 1.
x4H0L 1.
La función vectorial y la matriz jacobiana son:
F Hx1, x2, x3, x4L = jjjjjjjjijjjjjj -x4 x1 + 4 x1 - x2 + x3 zzzzzzzzzzyzzzz
k -x1 + 3 x2 - 2 x3 - x2 x4 {
x1 - 2 x2 + 3 x3 - x3 x4
x12 + x22 + x32 - 1
J Hx1, x2, x3, x4L = ijjjjjjjjjjjjj 4 - x4 -1 1 -x1 yzzzzzzzzzzzzz
k -1 3 - x4 -2 -x2
1 -2 3 - x4 -x3
2 x1 2 x2 2 x3
0{
Iteración i = 0.
P0 = jjjjjjjjjjjjjjjjijj x1H0L zzzzzzzzzzzzzzzzzyz = jjjjjjjjjjjijj 1.00000 zzzzzzzyzzzzzz
k x2H0L { k -1.00000 {
x3H0L 1.00000
x4H0L 1.00000
F HP0L = jjjijjjjjjjjjj 5.00000 zzzzzzzyzzzzzz
-5.00000
5.00000
k 2.00000 {
J HP0L = ijjjjjjjjjjjjj 3.00000 -1.00000 1.00000 -1.00000 zzzzzzzzzzzzzy
-1.00000 2.00000 -2.00000 1.00000 {
1.00000 -2.00000 2.00000 -1.00000
-2.00000 2.00000 0
k 2.00000
135
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Se resuelve el sistema lineal J HP0L DP = -F HP0L:
jjjjjjjjjjjjji 3.00000 -1.00000 1.00000 -1.00000 zzzz{yzzzzzzzzz.jjijjjjjjjjjjjjk Dx1 zzzzzzzzzzzyzzz
k -1.00000 2.00000 -2.00000 1.00000 Dx2 {
1.00000 -2.00000 2.00000 -1.00000
2.00000 -2.00000 2.00000 0 Dx3
Dx4
DP = jjjjjjjjjjjjjji Dx1 zzzzzzzzzyzzzzz = ijjjjjjjjjjjjj -0.333333 yzzzzzzzzzzzzz
k Dx2 { 0.333333
-0.333333
Dx3
Dx4 k 3.33333 {
El siguiente punto de la iteración es:
P1 = P0 + DP
P1 = ijjjjjjjjjjjjj 1. yzzzzzzzzzzzzz + ijjjjjjjjjjjjj -0.333333 yzzzzzzzzzzzzz = ijjjjjjjjjjjjj 0.666667 yzzzzzzzzzzzzz
k -1. { k 0.333333 { k -0.666667 {
1. -0.333333 0.666667
1. 3.33333 4.33333
Iteración i = 1.
P1 = jjjjjjjjjjjjijjjjjj x1H1L zzzzzzzyzzzzzzzzzzz = jjjjijjjjjjjjj 0.666667 zyzzzzzzzzzzzz
k x2H1L { k -0.666667 {
x3H1L 0.666667
x4H1L 4.33333
F HP1L = jjjjjjijjjjjjj 1.11111 zzzzzzyzzzzzzz
-1.11111 {
1.11111
k 0.333333
J HP1L = ijjjjjjjjjjjjj -0.333333 -1.00000 1.00000 -0.666667 zzzzzzzzzzzzzy
-1.00000 -1.33333 -2.00000 0.666667 {
1.00000 -2.00000 -1.33333 -0.666667
-1.33333 1.33333 0
k 1.33333
136
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Se resuelve el sistema lineal J HP1L DP = -F HP1L:
jjjjjjjjjjjjij -0.333333 -1.00000 1.00000 -0.666667 zz{zzzzzzzzzyzz.ikjjjjjjjjjjjjjj Dx1 zzzzzzzzzzzzyzz
-1.00000 -1.33333 -2.00000 0.666667 Dx2 {
1.00000 -2.00000 -1.33333 -0.666667
-1.33333 1.33333 0 Dx3
k 1.33333 Dx4
DP = jjjjjjjjjjjjjji Dx1 zzzzzzyzzzzzzzz = ijjjjjjjjjjjjj -0.0833333 yzzzzzzzzzzzzz
k Dx2 { k 0.0833333 {
-0.0833333
Dx3 1.45833
Dx4
El siguiente punto de la iteración es:
P2 = P1 + DP
P2 = ijjjjjjjjjjjjj 0.666667 yzzzzzzzzzzzzz + ijjjjjjjjjjjjj -0.0833333 yzzzzzzzzzzzzz = ijjjjjjjjjjjjj 0.583333 yzzzzzzzzzzzzz
k -0.666667 { k 0.0833333 { k -0.583333 {
0.666667 -0.0833333 0.583333
4.33333 1.45833 5.79167
Iteración i = 2.
P2 = jjjjjjjjjjjjjjjjjji x1H2L zzzzyzzzzzzzzzzzzzz = jjijjjjjjjjjjj 0.583333 zzzzyzzzzzzzzz
k x2H2L { k -0.583333 {
x3H2L 0.583333
x4H2L 5.79167
F HP2L = jjjjijjjjjjjjj 0.121528 zzzyzzzzzzzzzz
k -0.121528 {
0.121528
0.0208333
J HP2L = ijjjjjjjjjjjjj -1.79167 -1.00000 1.00000 -0.583333 zzzzzzzzzzzzzy
-1.00000 -2.79167 -2.00000 0.583333 {
1.00000 -2.00000 -2.79167 -0.583333
-1.16667 1.16667 0
k 1.16667
137
Resolución Numérica de Sistemas de Ecuaciones no Lineales
Se resuelve el sistema lineal J HP2L DP = -F HP2L:
jjjjjjjjjjjjij -1.79167 -1.00000 1.00000 -0.583333 zz{zzzzzzzzzzyz.jjjjjjjikjjjjjjj Dx1 zzzzzyzzzzzzzzz
-1.00000 -2.79167 -2.00000 0.583333 Dx2 {
1.00000 -2.00000 -2.79167 -0.583333
-1.16667 1.16667 0 Dx3
k 1.16667 Dx4
DP = ijjjjjjjjjjjjjj Dx1 zzzzyzzzzzzzzzz = ijjjjjjjjjjjjj -0.00595238 yzzzzzzzzzzzzz
k Dx2 { k 0.00595238 {
-0.00595238
Dx3 0.206207
Dx4
El siguiente punto de la iteración es:
P3 = P2 + DP
P3 = ijjjjjjjjjjjjj 0.583333 yzzzzzzzzzzzzz + ijjjjjjjjjjjjj -0.00595238 yzzzzzzzzzzzzz = ijjjjjjjjjjjjj 0.577381 yzzzzzzzzzzzzz
-0.583333 0.00595238 -0.577381
0.583333 -0.00595238 0.577381
k 5.79167 { k 0.206207 { k 5.99787 {
Iteración i = 3.
P3 = jjjjjijjjjjjjjjjjjj x1H3L zzzzzzyzzzzzzzzzzzz = ijjjjjjjjjjjjj 0.577381 zzzyzzzzzzzzzz
k x2H3L { k -0.577381 {
x3H3L 0.577381
x4H3L 5.99787
F HP3L = jijjjjjjjjjjjj 0.00122743 zzzzzzyzzzzzzz
k -0.00122743 {
0.00122743
0.000106293
J HP3L = ijjjjjjjjjjjjj -1.99787 -1.00000 1.00000 -0.577381 zzzzzzzzzzzzzy
k -1.00000 -2.99787 -2.00000 0.577381
1.00000 -2.00000 -2.99787 -0.577381
1.15476 -1.15476 1.15476
0{
138
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Resolución numérica de sistemas de ecuaciones no lineales
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