Divergence in spherical coordinates

[billiards]

I am stuck on this problem.

Use these equations:

\textbf{v}(\textbf{r}) = f(r)\textbf{r}

\frac{\partial r}{\partial x} = \frac{x}{r}

And the chain rule for differentiation, show that:

(\nabla\cdot\textbf{v}) = 2f(r) + r\frac{df}{dr}
(cylindrical coordinates)

Any help greatly appreciated, I will post my progres so far in a following post.

Cheers

 

[billiards]

So far I have tried to crack this by subbing in

\textbf{v}(\textbf{r}) = f(r)\textbf{r}

in the bottom equation I get:

(\nabla\cdot f(r)\textbf{r}) = 2f(r) + r\frac{df}{dr}

But I think that if we write out the left hand side of the equation and separate out the differentials we get this:

(\nabla\cdot f(r)\textbf{r}) = r_{x}\frac{\partial f(r)}{\partial x} + f(r)\frac{\partial r_{x}}{\partial x} + r_{y}\frac{\partial f(r)}{\partial y} + f(r)\frac{\partial r_{y}}{\partial y}

Which I think reduces further to:

(\nabla\cdot f(r)\textbf{r}) = 2f(r) + r_{x}\frac{\partial f(r)}{\partial x} + r_{y}\frac{\partial f(r)}{\partial y} = 2f(r) + r\frac{df}{dr}

Which IF that is right (and it's a big IF) basically means that I'm left to prove that:

r\frac{df}{dr} = r_{x}\frac{\partial f(r)}{\partial x} + r_{y}\frac{\partial f(r)}{\partial y}

That's where I'm stuck. I have a feeling that the total derivative is needed next and I am thinking along those lines, I still haven't used the chain rule yet so I think that will come into play in the total derivative context. It could well be that I've strayed off and am heading nowhere with this, which is why I need a bit of guidance.

Thanks

 

[LCKurtz]

I assume you know, or can prove, that for scalar f and vector V in rectangular coordinates,

\nabla\cdot f\vec V = \nabla f \cdot \vec V + f(\nabla \cdot\vec V)

Also, I assume the standard notation

\vec r = \langle x, y \rangle,\ r = |\vec r|

Then by the first identity

\nabla\cdot f(r)\vec r = \nabla f(r) \cdot \vec r + f(r)(\nabla \cdot \vec r)

The only place you need the chain rule is to calculate the first term on the right.

[Edit] Corrected typos.

 

[billiards]

I assume you know, or can prove, that for scalar f and vector V in rectangular coordinates,

\nabla\cdot f\vec V = \nabla f \cdot \vec V + f(\nabla \cdot\vec V)

Also, I assume the standard notation

\vec r = \langle x, y \rangle,\ r = |\vec r|

Then by the first identity

\nabla\cdot f(r)\vec r = \nabla f(r) \cdot \vec r + f(r)(\nabla \cdot \vec r)

The only place you need the chain rule is to calculate the first term on the right.

[Edit] Corrected typos.

Thank you for your reply.

If I understand you correctly then I have more or less followed your reasoning thus far.

I am stuck on the bit you mention in your last line, I am struggling to calculate the first term on the right.

The way I see it is that

\nabla f(r) \cdot \vec r = r_{x}\frac{\partial f(r)}{\partial x} + r_{y}\frac{\partial f(r)}{\partial y}

Is that correct?

That's my last line in my earlier post. How do I invoke the chain rule here to get this into the form I am after?

Thanks

 

[LCKurtz]

The way I see it is that

\nabla f(r) \cdot \vec r = r_{x}\frac{\partial f(r)}{\partial x} + r_{y}\frac{\partial f(r)}{\partial y}

Is that correct?

Thanks

When you differentiate f(r) with respect to, for example, x, you get f'(r)r_x. And do the dot product last.

 

[billiards]

Which IF that is right (and it's a big IF) basically means that I'm left to prove that:

r\frac{df}{dr} = r_{x}\frac{\partial f(r)}{\partial x} + r_{y}\frac{\partial f(r)}{\partial y}

That's where I'm stuck.

Okay, I think I've got it now.

Use the chain rule:

\frac{\partial f(r)}{\partial x} = \frac{df}{dr}\times \frac{\partial r}{\partial x}

And the result:

\frac{\partial r}{\partial x} = \frac{x}{r}

To get:

r\frac{df}{dr} = r_{x}\frac{df}{dr}\frac{x}{r} + r_{y}\frac{df}{dr}\frac{y}{r}

Now r_x=x. Use this result as well as the fact that in cylindrical coordinates r² = x²+y²,and the term on the right reduces to the term on the left. Is this OK?

Cheers

 

[LCKurtz]

Yes, I think you have it figured out now.

 

[billiards]

Yes, I think you have it figured out now.

Ahhh good. For some reason that one had me scratching my head for a while, but it seems quite simple now.

Cheers