Let $\mathbf{R} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$, and $r = |\mathbf{R}|$. Find $f(r)$ such that $\operatorname{div}[\operatorname{grad}f(r)] = 0$. Let $ \mathbf{R} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} $, and $ r = |\mathbf{R}| $. Find $ f(r) $ such that $ \operatorname{div}[\operatorname{grad}f(r)] = 0 $.
Below is my work done:
$ \nabla f = \frac{\partial f}{\partial x}\mathbf{i} + \frac{\partial f}{\partial y}\mathbf{j} + \frac{\partial f}{\partial z}\mathbf{k}. $
Since $ r = |\mathbf{R}| = \sqrt{x^2 + y^2 + z^2} $, we have $ r^2 = x^2 + y^2 + z^2 $, and differentiating both sides with respect to $x$, $y$, and $z$ gives:
$ 2r \frac{\partial r}{\partial x} = 2x, $ $ 2r \frac{\partial r}{\partial y} = 2y, $ $ 2r \frac{\partial r}{\partial z} = 2z. $
Dividing each equation by $2r$ gives:
$ \frac{\partial r}{\partial x} = \frac{x}{r}, $ $ \frac{\partial r}{\partial y} = \frac{y}{r}, $ $ \frac{\partial r}{\partial z} = \frac{z}{r}. $
Now, substitute these partial derivatives into the expression for the gradient:
$ \nabla f = \frac{\partial f}{\partial x}\mathbf{i} + \frac{\partial f}{\partial y}\mathbf{j} + \frac{\partial f}{\partial z}\mathbf{k} = f'(r) \left(\frac{x}{r}\mathbf{i} + \frac{y}{r}\mathbf{j} + \frac{z}{r}\mathbf{k}\right). $
Now, find the divergence of this gradient:
$ \operatorname{div}(\nabla f) = \frac{\partial}{\partial x}\left(f'(r)\frac{x}{r}\right) + \frac{\partial}{\partial y}\left(f'(r)\frac{y}{r}\right) + \frac{\partial}{\partial z}\left(f'(r)\frac{z}{r}\right). $
It is a complicated expression I don't know how to deal with.