Can anyone help me proofing this? This is not homework.

  • Thread starter yungman
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In summary, the conversation is discussing the derivation of Green's function, where the goal is to prove the equation \frac{x}{r} \frac{\partial u}{\partial x} + \frac{y}{r} \frac{\partial u}{\partial y} +\frac{z}{r} \frac{\partial u}{\partial z} \;=\; \frac{\partial u}{\partial r} using the chain rule. However, the person is unsure how to proceed and needs guidance on how to find \frac{\partial x}{\partial r}.
  • #1
yungman
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This is part of a bigger derivation of Green's function. The book just jump to the ending and I cannot understand how to do this. I need someone to give me guidence how to proof this:

[tex] \frac{x}{r} \frac{\partial u}{\partial x} + \frac{y}{r} \frac{\partial u}{\partial y} +\frac{z}{r} \frac{\partial u}{\partial z} \;=\; \frac{\partial u}{\partial r} [/tex]

Where

[tex] r=\sqrt{x^2+y^2+z^2}[/tex]

This is not a homework problem.

Thanks
 
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  • #2
Use the chain rule:

[tex]\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\cdots[/tex]
 
  • #3
cristo said:
Use the chain rule:

[tex]\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\cdots[/tex]

I thought of that, but what is [tex]\frac{\partial x}{\partial r}[/tex]

I know

[tex]\frac{\partial r}{\partial x}=\frac{x}{r}[/tex]

That's exactly where I got stuck!
 

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