设函数u=f(r),r=√(x^2+y^2+z^2),则э^2u/эx^2+э^2u/эy^2+э^2u/эz^2=
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设函数u=f(r),r=√(x^2+y^2+z^2),则э^2u/эx^2+э^2u/эy^2+э^2u/эz^2=
э^2u/эx^2代表u对x的二阶偏导数
э^2u/эx^2代表u对x的二阶偏导数
эu/эx=f'(r)*эr/эx
=f'(r)*x/r
э^2u/эx^2=f''(r)*(x/r)^2+f'(r)*(r-x*x/r)/r^2
=f''(r)*(x/r)^2+f'(r)*(r^2-x^2)/r^3
同理
э^2u/эy^2=f''(r)*(y/r)^2+f'(r)*(r^2-y^2)/r^3
э^2u/эz^2=f''(r)*(z/r)^2+f'(r)*(r^2-z^2)/r^3
所以
э^2u/эx^2+э^2u/эy^2+э^2u/эz^2
=f''(r)*(x/r)^2+f'(r)*(r^2-x^2)/r^3+f''(r)*(y/r)^2+f'(r)*(r^2-y^2)/r^3+f''(r)*(z/r)^2+f'(r)*(r^2-z^2)/r^3
=f'(r)*(x^2+y^2+z^2)/r^2+f''(r)*[3r^2-(x^2+y^2+z^2)]/r^3
=f'(r)+2f'(r)/
=f'(r)*x/r
э^2u/эx^2=f''(r)*(x/r)^2+f'(r)*(r-x*x/r)/r^2
=f''(r)*(x/r)^2+f'(r)*(r^2-x^2)/r^3
同理
э^2u/эy^2=f''(r)*(y/r)^2+f'(r)*(r^2-y^2)/r^3
э^2u/эz^2=f''(r)*(z/r)^2+f'(r)*(r^2-z^2)/r^3
所以
э^2u/эx^2+э^2u/эy^2+э^2u/эz^2
=f''(r)*(x/r)^2+f'(r)*(r^2-x^2)/r^3+f''(r)*(y/r)^2+f'(r)*(r^2-y^2)/r^3+f''(r)*(z/r)^2+f'(r)*(r^2-z^2)/r^3
=f'(r)*(x^2+y^2+z^2)/r^2+f''(r)*[3r^2-(x^2+y^2+z^2)]/r^3
=f'(r)+2f'(r)/
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