# I 2D Laplacian in polar coordinates

1. May 1, 2017

### Chan Pok Fung

The 2D Laplacian in polar coordinates has the form of
$$\frac{1}{r}(ru_r)_r +\frac{1}{r^2}u_{\theta \theta} =0$$
By separation of variables, we can write the $\theta$ part as
$$\Theta'' (\theta) = \lambda \Theta (\theta)$$

Now, the book said because we need to satisfy the condition $\Theta (0)=\Theta (2\pi)$, $\lambda = n^2$ with n =1,2,3...

I tried this in detailed steps.
$\Theta (\theta) = Acos\sqrt{\lambda}\theta + Bcos\sqrt{\lambda}\theta$
With the BC,
$B = Acos(\sqrt{\lambda}2\pi) + Bsin(\sqrt{\lambda}2\pi)$
This is just an equality but not an identity. Why can't I conclude that
$B = \frac{cos(\sqrt{\lambda}2\pi)}{1-sin(\sqrt{\lambda}2\pi)} A$
instead of imposing constraint on $\lambda$

Thanks!

2. May 1, 2017

### Orodruin

Staff Emeritus
Because you also need to satisfy $\Theta'(0) = \Theta'(2\pi)$. For general lambda that gives you a system of equations that lead to trivial solutions only. Only for a discrete set of lambda do you get non-trivial solutions.

Note: You are missing the $\lambda=0$ solution.

3. May 2, 2017

### Chan Pok Fung

Thank you! Is it also true that $\Theta^{(n)}(0)=\Theta^{(n)}(2\pi)$?

4. May 2, 2017

### Orodruin

Staff Emeritus
That follows from the differential equation if you have the other two.