Autovalores do Laplaciano - Departamento de Matemática - UFMG

Rodney Josué Biezuner 59
u(xi − ∆x, yj + ∆y)
= u(xi, yj) +
+ 1
2 ∂ u
2!
+ 1
3!
+ 1
4!
+ 1
5!
− ∂u
∂x (xi, yj)∆x + ∂u
∂y (xi,
yj)∆y
∂x 2 (xi, yj)∆x 2 − 2 ∂2 u
∂x∂y (xi, yj)∆x∆y + ∂2 u
∂y 2 (xi, yj)∆y 2
− ∂3 u
∂ 4 u
∂x 3 (xi, yj)∆x 3 + 3 ∂3 u
∂x 2 ∂y (xi, yj)∆x 2 ∆y − 3 ∂3 u
∂x∂y 2 (xi, yj)∆x∆y 2 + ∂3 u
∂y3 (xi, yj)∆y 3
∂x 4 (xi, yj)∆x 4 − 4 ∂4 u
∂x 3 ∂y (xi, yj)∆x 3 ∆y + 6 ∂4 u
∂x∂y 3 (xi, yj)∆x 2 ∆y 2 − 4 ∂3 u
∂x∂y 3 (xi, yj)∆x∆y 3 + ∂4 u
∂y 4 (xi, yj)∆y 4
− ∂5 u
∂x 5 (xi, yj)∆x 5 + 5 ∂5 u
∂x 4 ∂y (xi, yj)∆x 4 ∆y − 10 ∂5 u
−5 ∂5 u
∂x∂y 4 (xi, yj)∆x∆y 4 + ∂5 u
∂y 5 (xi, yj)∆y 5
+ O ∆x 6 , ∆y 6 .
Substituindo estas expressões na fórmula acima, obtemos:
−∆dud = (c1 + c2 + c3 + c4 + c5 + c6 + c7 + c8 + c9) u (xi, yj)
+ ∆x (−c1 + c3 − c4 + c6 − c7 + c9) ∂u
∂x (xi, yj)
∂x3∂y 2 (xi, yj)∆x 3 ∆y 2 + 10 ∂5u ∂x∂y4 (xi, yj)∆x 2 ∆y 3
+ ∆y (−c1 − c2 − c3 + c7 + c8 + c9) ∂u
∂y (xi, yj)
+ ∆x 2
1
2 c1 + 1
2 c3 + 1
2 c4 + 1
2 c6 + 1
2 c7 + 1
2 c9
2 ∂ u
∂x2 (xi, yj)
+ ∆x∆y (c1 − c3 − c7 + c9) ∂2u ∂x∂y (xi, yj)
+ ∆y 2
1
2 c1 + 1
2 c2 + 1
2 c3 + 1
2 c7 + 1
2 c8 + 1
2 c9
2 ∂ u
∂y2 (xi, yj)
+ ∆x 3
− 1
6 c1 + 1
6 c3 − 1
6 c4 + 1
6 c6 − 1
6 c7 + 1
6 c9
3 ∂ u
∂x3 (xi, yj)
+ ∆x 2
∆y − 1
2 c1 − 1
2 c3 + 1
2 c7 + 1
2 c9
3 ∂ u
∂x2∂y (xi, yj)
+ ∆x∆y 2
− 1
2 c1 + 1
2 c3 − 1
2 c7 + 1
2 c9
3 ∂ u
∂x∂y2 (xi, yj)
+ ∆y 3
− 1
6 c1 − 1
6 c2 − 1
6 c3 + 1
6 c7 + 1
6 c8 + 1
6 c9
3 ∂ u
∂y3 (xi, yj)
+ ∆x 4
1
24 c1 + 1
24 c3 + 1
24 c4 + 1
24 c6 + 1
24 c7 + 1
24 c9
4 ∂ u
∂x4 (xi, yj)
+ ∆x 3
1
∆y
6 c1 − 1
6 c3 − 1
6 c7 + 1
6 c9
4 ∂ u
∂x3∂y (xi, yj)
+ ∆x 2 ∆y 2
1
4 c1 + 1
4 c3 + 1
4 c7 + 1
4 c9
4 ∂ u
∂x2∂y 2 (xi, yj)
+ ∆x∆y 3
1
6 c1 − 1
6 c3 − 1
6 c7 + 1
6 c9
4 ∂ u
∂x∂y3 (xi, yj)
+ ∆y 4
1
24 c1 + 1
24 c2 + 1
24 c3 + 1
24 c7 + 1
24 c8 + 1
24 c9
4 ∂ u
∂y4 (xi, yj)