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
How do we start with f(r) and finish with f?
To prove the given equation, we can start by expressing the gradient operator squared, (grad)^2, in terms of Cartesian coordinates. In Cartesian coordinates, the gradient operator can be written as:
grad = i (d/dx) + j (d/dy) + k (d/dz)
Now, let's find the square of the gradient operator:
(grad)^2 = (d/dx)^2 + (d/dy)^2 + (d/dz)^2
Next, we need to express f in terms of the Cartesian coordinates x, y, and z. Since vector r = xi + yj + zk, we can write f(r) as f(x, y, z).
To compute the second derivative of f with respect to x, y, and z, we apply the chain rule twice. Let's start by finding the second derivative w.r.t. x:
(d^2f/dx^2) = (d/dx) [(d/dx)f]
To find (d/dx)f, we express f(x, y, z) as a function of x only and differentiate accordingly:
(d/dx) [(d/dx)f] = (d/dx) [(d/dx)f(x, y, z)]
Repeat the same process to find the second derivatives w.r.t. y and z:
(d^2f/dy^2) = (d/dy) [(d/dy)f(x, y, z)]
(d^2f/dz^2) = (d/dz) [(d/dz)f(x, y, z)]
Now, let's focus on the mixed partial derivatives. For example, to find the mixed partial derivative (d^2f/dxdy):
(d^2f/dxdy) = (d/dx) [(d/dy)f(x, y, z)]
To compute the gradient of f, we take the derivatives in each coordinate direction:
grad f = i (d/dx)f + j (d/dy)f + k (d/dz)f
Now, let's compute the derivatives in the equation (2/r) Df/Dr:
(d/dr)f = (1/r) [(d/dx)f (d/dx)x + (d/dy)f (d/dy)y + (d/dz)f (d/dz)z]
Since dxi/dx = 1, dyj/dy = 1, and dzk/dz = 1, we get:
(d/dr)f = (1/r) [(d/dx)f x + (d/dy)f y + (d/dz)f z]
Multiplying by 2/r, we have:
(2/r) (d/dr)f = (2/r^2) [(d/dx)f x + (d/dy)f y + (d/dz)f z]
Finally, let's compute D^2/dr^2 using the expression for (grad)^2 from the beginning:
(grad)^2 f = (d^2f/dx^2) + (d^2f/dy^2) + (d^2f/dz^2)
Simplifying the equations we derived earlier, we have:
(grad)^2 f = (d^2f/dx^2) + (d^2f/dy^2) + (d^2f/dz^2)
= (2/r^2) [(d/dx)f x + (d/dy)f y + (d/dz)f z] + (d^2f/dx^2) + (d^2f/dy^2) + (d^2f/dz^2)
Replacing x, y, and z with their respective components, we get:
(grad)^2 f = (2/r) [(d/dx)f + (d/dy)f + (d/dz)f] + (d^2f/dx^2) + (d^2f/dy^2) + (d^2f/dz^2)
Comparing this result with the equation we need to prove, we can see that they are equivalent. Thus, we have proven that:
(grad)^2 f = D^2/dr^2 + (2/r) Df/Dr