Laplace's equation after change of variables 
Show that if $u(r, \theta)$ is dependent on $r$ alone, Laplace's
  equation becomes $$u_{rr} + \frac{1}{r}u_r=0.$$

My first reaction is to replace $r=x$ and $\theta=y$, but obviously it does not work. Then I recall $x=r\cos \theta$ and $y=r\sin \theta$. Then I obtain the following: $$v(r, \theta) = u(r\cos \theta, r\sin \theta).$$
Then I start to differentiate it, but what is $u_r$ and $u_{rr}$? Can anyone give me some hints to move on?
 A: The Laplace operator in $2$D is 
$$\nabla^2 \equiv \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$$
To express this in polar-coordinates we need to express the $x,y$-derivatives in terms of $r,\theta$-derivatives. Since $(x,y)=(r\cos\theta,r\sin\theta)$ the chain-rule gives us that
$$\frac{\partial}{\partial x} = \frac{\partial r}{\partial x} \frac{\partial}{\partial r} + \frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta}$$
Now $r=\sqrt{x^2+y^2} \implies \frac{\partial r}{\partial x} = \frac{x}{r} = \cos\theta$ and $\frac{y}{x}=\tan(\theta) \implies \frac{\partial\theta}{\partial x} = -\frac{y}{r^2} = -\frac{\sin\theta}{r}$ so
$$\frac{\partial}{\partial x} = \cos\theta \frac{\partial}{\partial r}  - \frac{\sin\theta}{r}\frac{\partial}{\partial \theta}$$
From this we can compute $\frac{\partial^2}{\partial x^2}$ as follows
$$\frac{\partial^2}{\partial x^2} = \left[\cos\theta \frac{\partial}{\partial r}  - \frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\right]\left[\cos\theta \frac{\partial}{\partial r}  - \frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\right]$$
To compute this (see this answer for more info) remember that a derivative acts on everything that is behind it so for example
$$\cos\theta \frac{\partial}{\partial r} \cos\theta \frac{\partial}{\partial r} \equiv \cos\theta \frac{\partial}{\partial r}\left(\cos\theta \frac{\partial}{\partial r}\right)$$
You also need to do the same thing as above to compute $\frac{\partial^2}{\partial y^2}$. Once you have done this you will have an expression for  $\nabla^2 $ in polar coordinates. Then you can let this act on $u$, i.e. consider $\nabla^2 u$, and use that $u = u(r)$ so $\frac{\partial u}{\partial \theta} = 0$ to get the desired result.
A: You have
$$ \frac{\partial f}{\partial r} = \frac{\partial x}{\partial r} \frac{\partial f}{\partial x} + \frac{\partial y}{\partial r} \frac{\partial f}{\partial y} = f_x \cos{\theta} + f_y \sin{\theta}. $$
Then differentiating again (and using that $\partial \theta/\partial r=0$),
$$ f_{rr} = \cos{\theta} (\cos{\theta} \partial_x+\sin{\theta} \partial_y)f_x + \sin{\theta} (\cos{\theta} \partial_x+\sin{\theta} \partial_y)f_y \\
f_{xx} \cos^2{\theta} + f_{yy}\sin^2{\theta} + 2f_{xy} \sin{\theta}\cos{\theta} $$
Similarly, we can show that
$$ \partial_{\theta} = -r\sin{\theta} \partial_x + r\cos{\theta}\partial_y, $$
so
$$ f_{\theta\theta} = \partial_{\theta}(-r\sin{\theta} f_x + r\cos{\theta}f_y) = r^2[(-\cos{\theta} f_x -\sin{\theta} f_y ) +\sin^2{\theta} f_{xx}+\cos^2{\theta}f_yy -2\sin{\theta}\cos{\theta} f_{xy}] $$
Thus
$$ f_{rr} + \frac{1}{r^2}f_{\theta\theta} + \cos{\theta} f_x + \sin{\theta} f_y = f_{xx}+f_{yy} $$
But the last terms on the left are just $r^{-1}\partial_r $, so
$$ f_{xx} + f_{yy} = f_{rr}+\frac{1}{r}f_r + \frac{1}{r^2}f_{\theta\theta}. $$

That's a bit of a mess, though. Instead, we can use $u(r,\theta)=f(r)$, and $r=\sqrt{x^2+y^2}$, and compute
$$ \Delta f = \frac{\partial^2}{\partial x^2} f(\sqrt{x^2+y^2})+\frac{\partial^2}{\partial y^2} f(\sqrt{x^2+y^2}) \\
= \frac{\partial}{\partial x} \left( \frac{x}{\sqrt{x^2+y^2}} f'(\sqrt{x^2+y^2}) \right) + \frac{\partial}{\partial y} \left( \frac{y}{\sqrt{x^2+y^2}} f'(\sqrt{x^2+y^2}) \right) \\
= \frac{y^2}{(x^2+y^2)^{3/2}}f'(\sqrt{x^2+y^2}) + \left(\frac{x}{\sqrt{x^2+y^2}}\right)^2 f''(\sqrt{x^2+y^2}) + \frac{x^2}{(x^2+y^2)^{3/2}}f'(\sqrt{x^2+y^2}) + \left(\frac{y}{\sqrt{x^2+y^2}}\right)^2 f''(\sqrt{x^2+y^2}) \\
= f''(\sqrt{x^2+y^2})+\frac{1}{\sqrt{x^2+y^2}}f'(\sqrt{x^2+y^2}) = u_{rr}+\frac{1}{r}u_r. $$
