- #1

kostoglotov

- 234

- 6

**I have completed the exercise, but I did something weird in one step to make it work, and I'd like to know more about what I did...or if what I did was at all valid.**

1. Homework Statement

1. Homework Statement

Show that Laplace's equation

[tex]\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}+\frac{\partial^2u}{\partial z^2}=0[/tex]

when converted to cylindrical coordinates gives

[tex]\frac{\partial^2u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial^2u}{\partial \theta^2}+\frac{\partial^2u}{\partial z^2}=0[/tex]

## Homework Equations

obviously

[tex]x = r\cos{\theta}[/tex]

[tex]y = r\sin{\theta}[/tex]

[tex]z = z[/tex]

## The Attempt at a Solution

It was a long exercise, so I'll skip ahead to just before the step I'm asking about

[tex]r^2\frac{\partial^2u}{\partial r^2}+\frac{\partial^2u}{\partial \theta^2}+\frac{\partial^2u}{\partial z^2} =

\frac{\partial^2u}{\partial x^2}(r^2\cos^2{\theta}+r^2\sin^2{\theta})+\frac{\partial^2u}{\partial y^2}(r^2\cos^2{\theta}+r^2\sin^2{\theta})+\frac{\partial u}{\partial x}(-r\cos{\theta})+\frac{\partial u}{\partial y}(-r\sin{\theta})+\frac{\partial^2u}{\partial z^2}[/tex]

[tex]r^2\frac{\partial^2u}{\partial r^2}+\frac{\partial^2u}{\partial \theta^2} =

r^2\left(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}\right)+\frac{\partial u}{\partial x}(-r\cos{\theta})+\frac{\partial u}{\partial y}(-r\sin{\theta})[/tex]

Now, I got rid of the second partial of u to z since it just takes up space, and is not interesting right now. What I am interested in is the last two terms of the RHS. By the chain rule,

[tex]\frac{\partial u}{\partial x}(-r\cos{\theta})+\frac{\partial u}{\partial y}(-r\sin{\theta}) = \frac{\partial u}{\partial r}\frac{\partial r}{\partial x}(-r\cos{\theta})+\frac{\partial u}{\partial r}\frac{\partial r}{\partial y}(-r\sin{\theta})[/tex]

[tex]\frac{\partial r}{\partial x} = \frac{1}{\cos{\theta}} \ and \ \frac{\partial r}{\partial y} = \frac{1}{\sin{\theta}}[/tex]

So

[tex]\frac{\partial u}{\partial x}(-r\cos{\theta})+\frac{\partial u}{\partial y}(-r\sin{\theta}) = -r\left(\frac{\partial u}{\partial r}+\frac{\partial u}{\partial r}\right)[/tex]

Now...here's where I need more insight, I have an inkling that the following is

**NOT**right

[tex]\frac{\partial u}{\partial x}(-r\cos{\theta})+\frac{\partial u}{\partial y}(-r\sin{\theta}) = -2r\frac{\partial u}{\partial r}[/tex]

My intuition that this was wrong came from a few places. First off, it did not look like it would give the solution I was being directed to find...but that isn't sufficient for real understanding.

My understanding is that when using partial derivatives, especially when using them with the chain rule, one is finding rates of change in given directions, thus it's kind of like a vector (or basis vector - I haven't studied linear algebra yet) situation...and one cannot simply add basis vectors to each other...so I opted for this

[tex]\frac{\partial u}{\partial x}(-r\cos{\theta})+\frac{\partial u}{\partial y}(-r\sin{\theta}) = -r\frac{\partial u}{\partial r}[/tex]

and of course, by reinserting the second partial for u on z and rearranging one can show the very thing the problem asked us to show.

However, it bothers me that I have

[tex]-r\left(\frac{\partial u}{\partial r}+\frac{\partial u}{\partial r}\right) = -r\frac{\partial u}{\partial r}[/tex]

What is going on here?

Like I said, I've completed the exercise, but I feel like my insight is lacking.