Elementos Finitos en Ecuaciones Diferenciales

Ordinarias

E. A. P. de Computación Cientí�ca

Facultad de Ciencias Matemáticas

Universidad Nacional Mayor de San Marcos

15 de septiembre de 2010

Alumno: Diaz Avalos Josué

Ejercicio

1. Se tiene el modelo:

−((x2 + 1)y′)′+ (x2 + 2x+ 1)y = 2ex, 0 ≤ x ≤ 4

y(0) = y(4) = 0

Tomando siete funciones de base lineales en los puntos de [0, 4] igual-mente espaciados, calcular la matriz de ensamblaje y el vector de carga,averiguar el valor de la de�exión(solución) en cada nodo, y la función desolución proximada.

La forma general es:

−(a(x)y′)′ + b(x)y′+ c(x)y = f(x)

a(x) = x2 + 1, b(x) ≡ 0, c(x) = x2 + 2x+ 1, f(x) = 2ex

7 funciones base8 intevalos iguales9 nodos igualmente espaciados

h =4− 0

8=

1

2xi = ih, i = 1, . . . , 8

−(a(x)y′)′ + c(x)y = f(x)

1

−(a(x)y′)′v + c(x)yv = f(x)v

−´ 40(a(x)y

′)′vdx+

´ 40c(x)yvdx =

´ 40f(x)vdx

´ 40a(x)y

′v′dx+

´ 40c(x)yvdx =

´ 40f(x)vdx

u '∑7

j=1 αjφj ; v = φi, i = 1, . . . , 7∑7i=1 αi

´ 40(a(x)φ

′

iφ′

j + c(x)φiφj)dx =´ 40f(x)φidx

La matriz de ensamblaje:

Para j = i, donde i = 1, . . . , 7

Aii =´ 40(a(x)(φ

′

i)2 + c(x)(φi)

2dx

=´ xi+1

xi−1a(x)(

1

h)2dx+

´ xi

xi−1c(x)(

1

h)2(x− xi−1)

2dx

+´ xi+1

xic(x)(

1

h)2(x− xi+1)

2dx

= (1

h)2[´ xi+1

xi−1a(x)dx+

´ xi+1

xi−1c(x)x2dx− 2xi−1

´ xi

xi−1c(x)xdx

−2xi+1

´ xi+1

xic(x)xdx+ x2i−1

´ xi

xi−1c(x)dx+ x2i+1

´ xi+1

xic(x)dx

Necesitamos la siguientes integrales:

´a(x)dx =

´(x2 + 1)dx =

x3

3+ x

´c(x)x2dx =

´(x2 + 2x+ 1)x2dx =

x5

5+x4

2+x3

3´c(x)xdx =

´(x2 + 2x+ 1)xdx =

x4

4+

2x3

3+x2

2´c(x)dx =

´(x2 + 2x+ 1)dx =

x3

3+ x2 + x

Reemplazando:

A11 = 4[(x3

3+ x) |10 +(

x5

5+x4

2+x3

3) |10 −2(0)(

x4

4+

2x3

3+x2

2) |1/20

−2(1)(x4

4+

2x3

3+x2

2) |11/2 +(0)2(

x3

3+x2+x) |1/20 +(1)2(

x3

3+x2+x) |11/2]

= 4[(1,3333) + (1,0333)− 2(1)(1,1927) + (1)(1,5417)] = 6,0917

A22 = 4[(x3

3+ x) |3/21/2 +(

x5

5+x4

2+x3

3) |3/21/2 −2(

1

2)(x4

4+

2x3

3+x2

2) |11/2

2

−2(32)(x4

4+2x3

3+x2

2) |3/21 +(

1

2)2(

x3

3+x2+x) |11/2 +(

3

2)2(

x3

3+x2+x) |3/21 ]

= 4[(2,0833) + (5,0958) − 2(1

2)(1,1927) − 2(3)(3,2240) + (

1

4)(1,5417) +

(9

4)(2,5417)] = 9,6750

A33 = 4[(x3

3+ x) |21 +(

x5

5+x4

2+x3

3) |21 −2(1)(

x4

4+

2x3

3+x2

2) |3/21

−2(2)(x4

4+

2x3

3+x2

2) |23/2 +(1)2(

x3

3+x2+x) |3/21 +(2)2(

x3

3+x2+x) |23/2]

= 4[(3,3333)+ (16,0333)− 2(3,2240)− 4(6,6927)+ (2,5417)+4(3,7917)] =15,4252

A44 = 4[(x3

3+ x) |5/23/2 +(

x5

5+x4

2+x3

3) |5/23/2 −2(

3

2)(x4

4+

2x3

3+x2

2) |23/2

−2(52)(x4

4+2x3

3+x2

2) |5/22 +(

3

2)2(

x3

3+x2+x) |23/2 +(

5

2)2(

x3

3+x2+x) |5/22 ]

= 4[(5,0833)+(39,0958)−3

2(6,6927)−5

2(11,9740)+(

9

4)(3,7917)+(

25

4)(5,2917)] =

23,3417

A55 = 4[(x3

3+ x) |32 +(

x5

5+x4

2+x3

3) |32 −2(2)(

x4

4+

2x3

3+x2

2) |5/22

−2(3)(x4

4+

2x3

3+x2

2) |35/2 +(2)2(

x3

3+x2+x) |5/22 +(3)2(

x3

3+x2+x) |35/2]

= 4[(7,3333)+(81,0333)−4(11,974)−6(19,4427)+4(5,2917)+9(7,0417)] =33,426

A66 = 4[(x3

3+ x) |7/25/2 +(

x5

5+x4

2+x3

3) |7/25/2 −2(

5

2)(x4

4+

2x3

3+x2

2) |35/2

−2(72)(x4

4+2x3

3+x2

2) |7/23 +(

5

2)2(

x3

3+x2+x) |35/2 +(

7

2)2(

x3

3+x2+x) |7/23 ]

= 4[(10,0833) + (150,0958) − 5(19,4427) − 7(29,474) + (25

4)(7,0417) +

(49

4)(9,0417)] = 45,6762

A77 = 4[(x3

3+ x) |43 +(

x5

5+x4

2+x3

3) |43 −2(3)(

x4

4+

2x3

3+x2

2) |7/23

−2(4)(x4

4+

2x3

3+x2

2) |47/2 +(3)2(

x3

3+x2+x) |7/23 +(4)2(

x3

3+x2+x) |47/2]

= 4[(13,3333)+(256,0333)−6(29,474)−8(42,4427)+9(9,0417)+16(11,2917)] =60,094

3

Ahora para j = i+ 1 i = 1, . . . , 6

Ai i+1 =´ 40(a(x)φ

′

iφ′

i+1 + c(x)φiφi+1)dx

=´ xi+1

xia(x)(− 1

h)(1

h)dx+

´ xi+1

xic(x)(− 1

h)(x− xi+1)(

1

h)(x− xi)dx

= (1

h)2[−

´ xi+1

xia(x)dx−

´ xi+1

xic(x)x2dx+(xi+xi+1)

´ xi+1

xic(x)xdx−xixi+1

´ xi+1

xic(x)dx]

Necesitamos la siguientes integrales:

´a(x)dx =

´(x2 + 1)dx =

x3

3+ x

´c(x)x2dx =

´(x2 + 2x+ 1)x2dx =

x5

5+x4

2+x3

3´c(x)xdx =

´(x2 + 2x+ 1)xdx =

x4

4+

2x3

3+x2

2´c(x)dx =

´(x2 + 2x+ 1)dx =

x3

3+ x2 + x

Reemplazando:

A12 = 4[−(x3

3+x) |11/2 −(

x5

5+x4

2+x3

3) |11/2 +(

1

2+1)(

x4

4+

2x3

3+x2

2) |11/2

−(12)(1)(

x3

3+ x2 + x) |11/2]

= 4[−(0,7917)− (0,9542) + (1,5)(1,1927)− (0,5)(1,5417)] = −2,9108

A23 = 4[−(x3

3+x) |3/21 −(

x5

5+x4

2+x3

3) |3/21 +(1+

3

2)(x4

4+

2x3

3+x2

2) |3/21

−(32)(1)(

x3

3+ x2 + x) |3/21 ]

= 4[−(1,2917)− (4,1416) + (2,5)(3,224)− (1,5)(2,5417)] = −4,7434

A34 = 4[−(x3

3+x) |23/2 −(

x5

5+x4

2+x3

3) |23/2 +(

3

2+2)(

x4

4+

2x3

3+x2

2) |23/2

−(32)(2)(

x3

3+ x2 + x) |23/2]

= 4[−(2,0417)− (11,8917) + (3,5)(6,6927)− (3)(3,7917)] = −7,536

A45 = 4[−(x3

3+x) |5/22 −(

x5

5+x4

2+x3

3) |5/22 +(

5

2+2)(

x4

4+

2x3

3+x2

2) |5/22

−(52)(2)(

x3

3+ x2 + x) |5/22 ]

4[−(3,0417)− (27,2042) + (4,5)(11,974)− (5)(5,2917)] = −11,2856

A56 = 4[−(x3

3+x) |35/2 −(

x5

5+x4

2+x3

3) |35/2 +(

5

2+3)(

x4

4+

2x3

3+x2

2) |35/2

4

−(52)(3)(

x3

3+ x2 + x) |35/2]

4[−(4,2917)− (53,8292) + (5,5)(19,4427)− (7,5)(7,0417)] = −15,9952

A67 = 4[−(x3

3+x) |7/23 −(

x5

5+x4

2+x3

3) |7/23 +(

7

2+3)(

x4

4+

2x3

3+x2

2) |7/23

−(72)(3)(

x3

3+ x2 + x) |7/23 ]

4[−(5,7917)− (96,2667) + (6,5)(29,474)− (10,5)(9,0417)] = −21,661

Ahora el vector de carga:

Fi =´ 40f(x)φidx =

´ xi

xi−1f(x)(

1

h)(x−xi−1)dx−

´ xi+1

xif(x)(

1

h)(x−xi+1)dx

= (1

h)[´ xi

xi−1f(x)xdx−xi−1

´ xi

xi−1f(x)dx−

´ xi+1

xif(x)xdx+xi+1

´ xi+1

xif(x)dx]

Necesitamos las siguientes integrales:

´f(x)dx =

´2exdx =´

f(x)xdx =´2exxdx = 2ex(x− 1)

Reemplazando:

Fi = (1

h)[2ex(x−1) |xi

xi−1−xi−12e

x |xixi−1−2ex(x−1) |xi+1

xi +xi+12ex |xi+1

xi ]

= (2

h)[ex(x− 1) |xi

xi−1−xi−1e

x |xixi−1−ex(x− 1) |xi+1

xi +xi+1ex |xi+1

xi ]

F1 = 4[ex(x− 1) |1/20 −(0)ex |1/20 −ex(x− 1) |11/2 +(1)ex |11/2]

4[(0,1756)− (0)(0,6487)− (0,8244) + (1)(1,0696)] = 1,6848

F2 = 4[ex(x− 1) |11/2 −(1

2)ex |11/2 −e

x(x− 1) |3/21 +(3

2)ex |3/21 ]

4[(0,8244)− (0,5)(1,0696)− (2,2408) + (1,5)(1,7634)] = 2,7756

F3 = 4[ex(x− 1) |3/21 −(1)ex |3/21 −ex(x− 1) |23/2 +(2)ex |23/2]

4[(2,2408)− (1)(1,7634)− (5,1482) + (2)(2,9074)] = 4,5760

F4 = 4[ex(x− 1) |23/2 −(3

2)ex |23/2 −e

x(x− 1) |5/22 +(5

2)ex |5/22 ]

4[(5,1482)− (1,5)(2,9074)− (10,8847) + (2,5)(4,7934)] = 7,5436

F5 = 4[ex(x− 1) |5/22 −(2)ex |5/22 −ex(x− 1) |35/2 +(3)ex |35/2]

4[(10,8847)− (2)(4,7934)− (21,8973) + (3)(7,903)] = 12,4384

F6 = 4[ex(x− 1) |35/2 −(5

2)ex |35/2 −e

x(x− 1) |7/23 +(7

2)ex |7/23 ]

5

4[(21,8973)− (2,5)(7,903)− (42,6176) + (3,5)(13,03)] = 20,5088

F7 = 4[ex(x− 1) |7/23 −(3)ex |7/23 −ex(x− 1) |47/2 +(4)ex |47/2]

4[(42,6176)− (3)(13,03)− (81,0058) + (4)(21,4827)] = 33,8104

Como Aij =´ 40(a(x)φ

′

iφ′

j + c(x)φiφj)dx = Aji

A =

6,0917 −2,9108 0 0 0 0 0−2,9108 9,6750 −4,7434 0 0 0 0

0 −4,7434 15,4252 −7,536 0 0 00 0 −7,536 23,3417 −11,2856 0 00 0 0 −11,2856 33,426 −15,9952 00 0 0 0 −15,9952 45,676 −21,6610 0 0 0 0 −21,661 60,094

F =

1,68482,77564,57607,543612,438420,508833,8104

Luego tenemos el siguiente sistema:

A

α1

α2

α3

α4

α5

α6

α7

= F

Resolviendo el sistema:

α1

α2

α3

α4

α5

α6

α7

=

0,88561,27461,47111,60171,6621,56541,1269

Entonces el valor de la de�exión en cada nodo seria:

6

Figura 1: Grá�ca de la solución aproximada.

xi y aprox.

0 0

0.5 0.8856

1 1.2746

1.5 1.4711

2 1.6017

2.5 1.662

3 1.5654

3.5 1.1269

4 0

La función solución aproximada es:

y(x) w 0,8856φ1(x)+1,2746φ2(x)+1,4711φ3(x)+1,6017φ4(x)+1,662φ5(x)+1,5654φ6(x) + 1,1269φ7(x)

7