# Thread: Is this function harmonic?

1. ## Is this function harmonic?[SOLVED]

$u=\frac{1}{r}$ and $r=\sqrt{x^2+y^2}$ Calculate the Laplacian L(u) and give your anser in terms of r. Use dr/dx=d/x etc.

$\frac{du}{dx}=\frac{du}{dr}=\frac{-1}{r^2}\frac{dr}{dx}=\frac{-x}{r^3}$

now I'm stuck calculating the second order derivative

2. Originally Posted by superdude
$u=\frac{1}{r}$ and $r=\sqrt{x^2+y^2}$ Calculate the Laplacian L(u) and give your anser in terms of r. Use dr/dx=d/x etc.

$\frac{du}{dx}=\frac{du}{dr}=\frac{-1}{r^2}\frac{dr}{dx}=\frac{-x}{r^3}$

now I'm stuck calculating the second order derivative
If $u_x = r_x u_r$ then $
u_{xx} = r_x^2 u_{rr} + r_{xx} u_r
$

3. Originally Posted by Danny
If $u_x = r_x u_r$ then $
u_{xx} = r_x^2 u_{rr} + r_{xx} u_r
$
could you please elaborate? where does this come from? Does $r_x^2$ mean square r before taking it's derivative with respect to x?

This is what I've got but I know it's wrong:
$\frac{\partial^2 u}{\partial x^2}=\frac{\partial}{\partial x}(\frac{-x}{r^3}) = \frac{3x}{r^45}\times\frac{x}{r}=\frac{3x^2}{r^5}$

4. Originally Posted by Danny
If $u_x = r_x u_r$ then $
u_{xx} = r_x^2 u_{rr} + r_{xx} u_r
$
If $u_x = r_x u_r$ then we have the operator $\frac{\partial}{\partial x} = r_x \frac{\partial}{\partial x}$

Therefore $
\frac{\partial}{\partial x} u_x = \frac{\partial}{\partial x} \left( r_x u_r\right) = \frac{\partial}{\partial x} \left(r_x \right)u_r + r_x \frac{\partial}{\partial x} \left( u_r\right)
$

$
= r_{xx}u_r + r_x r_x \frac{\partial}{\partial r} \left( u_r\right) \left(\text{from the operator above}\right)$

$
= \left(r_x\right)^2 u_{rr} + r_{xx} u_r.
$

5. got it