Showing an equation satisfies laplace equation I'm completely lost with the laplace equation I've searched different explanations of it on google and on this website and nothing is helping explain it. The question I was given is: 
Show that the function 
$$\ f(x, y) = log(\sqrt{x^2 + y^2}) $$ 
  Satisfies a Laplace equation of the form
$$\frac{ ∂^2f}{∂x^2} + \frac {∂^2f}{∂y^2}= 0 $$
I'm just not too sure what to do even an example using a different equation in 2D would be a massive help. Thanks
 A: I think it's just a matter of computing the two partials, adding them up, and simplifying to demonstrate that it simplifies to $0$. In fact, you only should need to compute one of them and use that again, swapping $x$ and $y$, by symmetry.
To start, note you can write $f(x,y) = \frac12\log(x^2+y^2)$, so $$\frac{\partial f}{\partial x} = \frac12\frac{2x}{x^2+y^2}=\frac{x}{x^2+y^2}$$
and then
$$\frac{\partial^2 f}{\partial x^2} = \frac{\partial }{\partial x}\left(\frac{\partial f}{\partial x}\right) =\frac{(x^2+y^2)\cdot 1 -(x)\cdot2x}{(x^2+y^2)^2} = \frac{-x^2+y^2}{(x^2+y^2)^2}$$
Can you proceed from here?
A: This has been answered already adequately, but there's more can be added. 
Note the function $f(x,y)=log(\sqrt{x^2+y^2})$ satisfies the Laplace equation $\Delta u=0$, in $\mathbb{R}^2$ It turns out this function is actually the fundamental solution to the Laplace equation (up to a constant). For $n=2$ the fundamental solution is $f(|x|)=(2\pi)^{-1}ln|x|$ where $|x|=\sqrt{x^2+y^2}$. The derivation is long, but it takes advantage that solutions to the Laplace equation have a radial nature. 
