Is this function harmonic?

**superdude** · October 7th 2009, 04:51 PM

$u=\frac{1}{r}$ 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

**Danny** · October 8th 2009, 08:37 AM

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 $

**superdude** · October 8th 2009, 09:43 AM

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:
$\frac{\partial^2 u}{\partial x^2}=\frac{\partial}{\partial x}(\frac{-x}{r^3}) = \frac{3x}{r^45}\times\frac{x}{r}=\frac{3x^2}{r^5}$

**Danny** · October 8th 2009, 10:19 AM

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. $

**superdude** · October 8th 2009, 04:19 PM

got it

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