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
On
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.
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?