10/03/10 00:03:16 ljW7qcS4
>>432
x で1階偏微分すると、
exp(-(x^2+y^2+z^2))∂/∂x = exp(-(x^2+y^2+z^2)) (-2x)
そして x で2階偏微分すると、
exp(-(x^2+y^2+z^2))∂^2/∂x^2
= (exp(-(x^2+y^2+z^2)))’ (-2x) + exp(-(x^2+y^2+z^2)) ((-2x))’
= exp(-(x^2+y^2+z^2)) (-2x)^2 + (-2) exp(-(x^2+y^2+z^2))
=( (-2) + (-2x)^2) exp(-(x^2+y^2+z^2))
=(4x^2 - 2) exp(-(x^2+y^2+z^2))
= 2(2x^2 - 1) exp(-(x^2+y^2+z^2)) (1,1)
式(1,1)より、
exp(-(x^2+y^2+z^2))∂^2/∂y^2 = 2(2y^2 - 1) exp(-(x^2+y^2+z^2)) (1,2)
exp(-(x^2+y^2+z^2))∂^2/∂z^2 = 2(2z^2 - 1) exp(-(x^2+y^2+z^2)) (1,3)
式(1,1),(1,2),(1,3)より、( n(r) = exp(-(x^2+y^2+z^2)) ※ r はベクトル(x^2 + y^2 + z^2))
△exp(-(x^2+y^2+z^2)) = n(r)∂^2/∂x^2 + n(r)∂^2/∂y^2 + n(r)∂^2/∂z^2
= (2(2x^2 - 1) + 2(2y^2 - 1) + 2(2z^2 - 1)) exp(-(x^2+y^2+z^2))
= 2((2x^2 + 2y^2 + 2z^2) -1-1-1) exp(-(x^2+y^2+z^2))
= 2(2(x^2 + y^2 + x^2) - 3) exp(-(x^2+y^2+z^2))
= 2(2r - 3) exp(-r)
になると思うのですが・・・