Can someone help me with the following?
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
My work so far :
Definition of grad^2 : (grad)^2 f = fxx+fyy+fzz where the subscripts mean partial derivatives.
For example, fx=∂f/∂x.
r2=x2+y2+z2, so taking partial derivatives of both sides ...
2rrx=2x => rx=x/r and likewise ry=y/r and rz=z/r
what is "D^2/dr^2" ?
It looks like an operator, but ∇2f is not just an operator. Did you mean D^2f/dr^2 ?
Yes, D^2f/dr^2 , thats what I meant there?
To prove that (grad)^2 f(r) = D^2/dr^2 + (2/r) Df/Dr, we can start by calculating the individual components of (grad)^2 f and comparing them to the given expression.
Given that r = xi + yj + zk, where i, j, and k are unit vectors in the x, y, and z directions respectively, and f(r) is twice differentiable, we can write the individual components of (grad)^2 f as follows:
(fxx, fyy, fzz) = (D^2f/ Dx^2, D^2 f/ Dy^2, D^2 f/ Dz^2)
Now, let's calculate these individual components:
1. Calculating fxx:
We can calculate fxx by taking the partial derivative of f with respect to x twice.
fxx = ∂/∂x (∂f/∂x)
= ∂/∂x (Df/Dx)
= D^2f/ Dx^2
Therefore, fxx = D^2f/ Dx^2
2. Calculating fyy:
We can calculate fyy by taking the partial derivative of f with respect to y twice.
fyy = ∂/∂y (∂f/∂y)
= ∂/∂y (Df/Dy)
= D^2f/ Dy^2
Therefore, fyy = D^2f/ Dy^2
3. Calculating fzz:
We can calculate fzz by taking the partial derivative of f with respect to z twice.
fzz = ∂/∂z (∂f/∂z)
= ∂/∂z (Df/Dz)
= D^2f/ Dz^2
Therefore, fzz = D^2f/ Dz^2
Now, let's compare these individual components to the given expression:
(D^2/dr^2, (2/r) Df/Dr) = (D^2f/ Dx^2, D^2f/ Dy^2, D^2f/ Dz^2) + (2x/r, 2y/r, 2z/r) Df/Dr
Comparing the x-component:
D^2f/ Dx^2 = D^2f/ Dx^2 + (2x/r) (∂f/∂x)
Comparing the y-component:
D^2f/ Dy^2 = D^2f/ Dy^2 + (2y/r) (∂f/∂y)
Comparing the z-component:
D^2f/ Dz^2 = D^2f/ Dz^2 + (2z/r) (∂f/∂z)
From these comparisons, we can observe that each individual component matches with the given expression.
Hence, we have proven that (grad)^2 f(r) = D^2/dr^2 + (2/r) Df/Dr.