设r=√(x^2+y^2+z^2)证明∂^2/∂x^2+∂^2r/∂r^2+∂^2r/∂z^2=2/r

Question

设 r=√(x^2+y^2+z^2 ) 证明 ∂^2/∂x^2+∂^2r/∂r^2+∂^2r/∂z^2=2/r

Answer 1

显然∂r/∂x= x /√(x^2+y^2+z^2 ) = x/r,
而∂²r/∂x²= ∂(x/r) / ∂x
= (r -x*∂r/∂x) /r^2 = (r- x^2 /r) /r^2 = 1/r - x^2 /r^3
同理可以得到
∂²r/∂y²= 1/r -y^2/r^3
∂²r/∂z²= 1/r -z^2/r^3
所以
∂²r/∂x²+∂²r/∂y²+∂²r/∂z²
= 1/r - x^2 /r^3 + 1/r - y^2 /r^3 +1/r - z^2 /r^3
= 3/r - (x^2+y^2 +z^2)/r^3
而r=√(x^2+y^2+z^2 ),即(x^2+y^2+z^2)=r^2,
故
∂²r/∂x²+∂²r/∂y²+∂²r/∂z²
=3/r - (x^2+y^2 +z^2)/r^3
=3/r - r/r^3
=2/r
问题得到了证明