Vector Differentiation

Question

Prove that (grad)^2 f(r) = D^2/dr^2 + (2/r) Df/Dr

Where D^2/dr^2 refers to the second partial derivative of f , w.r.t. r and

Df/ Dr refers to partial derivative of f, w.r.t r

vector r = xi + yj + zk and f(r) is twice differentiable

Answer 1

Definition of grad^2 : (grad)^2 f = f_xx+f_yy+f_zz where the subscripts mean partial derivatives.

For example, f_x=∂f/∂x.

Note that r²=x²+y²+z², so taking partial derivatives of both sides ...

2rr_x=2x => r_x=x/r and likewise r_y=y/r and r_z=z/r

So, taking the first partial of f(r) wrt x ....

f_x = f_rr_x = (x/r) f'(r) where ∂f/∂r = f_r = f'(r) since f only depends on r.

f_xx = ∂/∂x [ (x/r) f'(r) ] = (1/r)f' + x ∂/∂x [ (1/r) f'(r) ] by Product Rule

= (1/r)f' + x [(1/r)f']_rr_x by Chain Rule

= (1/r)f' + x [ (1/r)f'' - (1/r²)f' ] (x/r) again by Product rule on the ∂/∂r

⇒ f_xx = (1/r -x²/r³)f' + (x²/r²)f''

and f_yy = (1/r -y²/r³)f' + (y²/r²)f''

and f_zz = (1/r -z²/r³)f' + (z²/r²)f''

adding these yields and terms we will see only depend on r:

So f_xx+f_yy+f_zz = (3/r - (x²+y²+z²)/r³)f' + ((x²+y²+z²)/r²)f''

= (3/r - 1/r) f' + f'' since x²+y²+z² = r²

= (2/r)f' + f''

= (2/r) ∂f/∂r + ∂²f/∂r² Your result, written in more standard form.