Show arctan(y/x) satisfies Laplace's equation

1. Jan 10, 2010

alex3

1. The problem statement, all variables and given/known data
Show that arctan(y/x) satisfies Laplace's equation.

2. Relevant equations
Laplace's equation:
$$\frac{\partial^{2}f}{\partial x^{2}}+\frac{\partial^{2}f}{\partial y^{2}}=0$$

3. The attempt at a solution
We haven't really done this is class too thoroughly, I've looked a http://www.wolframalpha.com/input/?i=second+derivative+of+arctan(y/x)" and it looks pretty messy, and I get the feeling I missing something. I have done partial derivatives, and we've done an exercise in class where we've transformed cartesian to polar co-ords, but I can't see the connection.

Thanks for any help.

Last edited by a moderator: Apr 24, 2017
2. Jan 10, 2010

Altabeh

It is never messy. Actually its second derivative with respect to x is $$-2\,x{y}^{-3} \left( 1+{\frac {{x}^{2}}{{y}^{2}}} \right) ^{-2}$$. So I think you better put all your energy into finding its derivative wrt y. There you'll be left with some addition\subtraction(s) to put an end to this challenge!

AB

Last edited by a moderator: Apr 24, 2017
3. Jan 10, 2010

alex3

Aha, I've noticed I made a mistake and made a reference to arctan(x/y) rather than arctan(y/x), which is the problem. Does this change things?

Am I right in saying then that I need to find those double partial derivatives and then find the sum (and hope I get zero)? That's 'all' there is to the question?

4. Jan 10, 2010

Altabeh

It makes no difference. In both cases you'll hit the answer 'zero'. That online integrator probably assumes y as a function of x and then this makes the first and second derivatives of f look enormously immense as the derivatives of y wrt x are to be involved as well.

If y is a function of x, I have a really simple solution for that which also applies to this one, too. And you talked about it in your first post... :)

Last edited: Jan 10, 2010
5. Jan 10, 2010

alex3

I'm having problems with the second derivative, I'm using the product rule like so

$$\frac{\partial}{\partial x} \left( \frac{y}{x^{2}} \times \frac{1}{\frac{y^{2}}{x^{2}} + 1}\right)$$

And getting a pretty nasty result. I don't mean to question the help, but are you sure that's the second derivative?

6. Jan 10, 2010

Altabeh

The derivative of arctanu(x) is $$u'(x)/(1+u^2)$$. From this rule, you must get for u=x/y, $$[arctanu(x/y)]'= (1/y)/(1+(x/y)^2)$$. This is totally different from what you have given us and of course yours sounds so cumbersome to go on with and above all else it's incorrect. Taking another derivative of f wrt x simply gives $$-(2x/y^2)(1/y)/[1+(x/y)^2]^2$$. So isn't this the same as the one I gave earlier!? Check out your calculations again, but carefully, to find out where you went wrong with differentiation.

Another way to solve this problem is to use the spherical coordinates in place of Cartesian coordinates: Take $$\theta=arctan(x/y)$$ and then use the the chain rule to replace d/dx and d/dy by $$d/d\theta$$. This will make calculations way easier than when you keep going with the traditional method.

AB

7. Jan 10, 2010

alex3

u=y/x, rather than x/y. I know you said this makes no difference, but the differential is different, surely? The first will become

$$-\frac{x^{-2}y}{1+\frac{y^{2}}{x^{2}}}$$

Won't it? Apologies if I'm being frustrating, I really don't mean to be!

8. Jan 10, 2010

vela

Staff Emeritus
That's right so far. It looks like it gets messy, but it's really not that bad.

By the way, your second derivative won't match what Altabeh got because you're working on different terms since Altabeh started with arctan(x/y) and you started with arctan(y/x).

Last edited: Jan 10, 2010
9. Jan 11, 2010

Count Iblis

One line proof:

arctan(y/x) = Im[Log(z)] and we know that the real and imaginary parts of analytic functions satisfy the Laplace equation.

10. Jan 11, 2010

alex3

I'm afraid I'm not familiar with analytic functions, I've only very briefly touched on imaginary/complex numbers, and my knowledge of Laplace's equation extends only as far as the form it takes, rather than general rules about its satisfaction.

It certainly looks much nicer than the way I have to do it though! I managed to work it all out in the end mind; thanks to everyone who helped!