Does the function u match the differential equation

**patzer** · February 4th 2013, 06:17 AM

Problem statement: If $u=\frac{1}{r}$ where $r=\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}$ , for $r\ne0$ prove that $\bigtriangleup u \equiv \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2}=0$

This is my approach: $\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\frac{dr}{dx}=-\frac{x-a}{r^2\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}$
$\frac{\partial^2 u}{\partial x^2}=\frac{\partial(\frac{\partial u}{\partial x})}{\partial x}=\frac{\partial(\frac{\partial u}{\partial x})}{\partial r}\frac{dr}{dx}=\frac{2(x-a)^2}{r^3[(x-a)^2+(y-b)^2+(z-c)^2]}$

$\frac{\partial^2 u}{\partial y^2}=\frac{\partial(\frac{\partial u}{\partial y})}{\partial y}=\frac{\partial(\frac{\partial u}{\partial y})}{\partial r}\frac{dr}{dy}=\frac{2(y-b)^2}{r^3[(x-a)^2+(y-b)^2+(z-c)^2]}$

$\frac{\partial^2 u}{\partial z^2}=\frac{\partial(\frac{\partial u}{\partial z})}{\partial z}=\frac{\partial(\frac{\partial u}{\partial z})}{\partial r}\frac{dr}{dz}=\frac{2(z-c)^2}{r^3[(x-a)^2+(y-b)^2+(z-c)^2]}$

So $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2}=\frac{2(x-a)^2+2(y-b)^2+2(z-c)^2}{r^3[(x-a)^2+(y-b)^2+(z-c)^2]}=\frac{2}{r}\ne0$

It seems I made an major error somewhere or I have conceptual problems.
Help me!

Thank you

**johng** · February 4th 2013, 01:19 PM

Yes, you made an error in the calculation of the 2nd partials. Here's the situation: you have a function r(x,y,z) and a function u=f(r) (in your case 1/r). So you want the partials of u w.r.t. x , y and z. The first partial u_x=f'(r(x,y,z) times r_x is used correctly. But the second partial is:

$u_{xx}=f^{\prime\prime}(r(x,y,z)r_x\cdot r_x+f^{\prime}(r(x,y,z)r_{xx}$ , using the product rule.

If you do this (replacing occurrences of r) you get:

$u_{xx}={3(x-a)^2\over r^5}-{1\over r^3}$

Similarly for the other 2nd partials. It is then easy to see that Laplace's equation is satisfied.

**patzer** · February 4th 2013, 01:26 PM

Thank you John

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