<s>[INST] General forms of Harmonic functions I wish to find the harmonic functions, $u(x,y)$, of the following form $$u(x,y) = \phi(\frac{x}{y})$$ and $$u(x,y) = \phi(\frac{x^2+y^2}{x^2})$$
where $\phi$ is a certain real-valued unknown function. How do I go about finding this harmonic function if I am given no specific function to work with?
I know that a function is harmonic if it has continuous second partials and that it satisfies the Laplace equation in it's domain. 
So, i'm looking for a function $\phi(\frac{x}{y})$ which satisfies that property.
 [/INST] $u(x,y):=\phi({x\over y}) $ where  $y\not=0$. 
$u_x(x,y)=\phi'({x\over y}){1\over y}$
$u_{xx}(x,y)=\phi''({x\over y}){1\over y^2}$
Similar computation gives : 
$u_y(x,y)=\phi'({x\over y}){x}$
$u_{yy}(x,y)=\phi''({x\over y}){x^2}$
$u_{xx}(x,y)+u_{yy}(x,y)=\phi''({x\over y})({x^2}+{1\over y^2})=0$
Therefore we can choose $\phi$ such that $\phi''(t)=0 $  for all $t\in \mathbb{R}$.
Therefore $\phi(t)=at+b$.
$u(x,y):=\phi({{x^2+y^2}\over x^2})$ where  $x\not=0$. 
$u_x(x,y)=\phi'({{x^2+y^2}\over x^2}){y^2 ({-2\over x^3})}$
$u_{xx}(x,y)=-2y^2[{1\over x^3}\phi''({{x^2+y^2}\over x^2}){y^2 {-2\over x^3}}+\phi'({{x^2+y^2}\over x^2}){-3\over x^4}]$$={4y^4\over x^6}\phi''({{x^2+y^2}\over x^2})+{6y^2\over x^4}\phi'({{x^2+y^2}\over x^2})$
Similar computation gives : 
$u_y(x,y)$$=\phi'({{x^2+y^2}\over x^2}){2y\over x^2}$

$u_{yy}(x,y)={2\over x^2}[\phi'({{x^2+y^2}\over x^2})+\phi''({{x^2+y^2}\over x^2}){2y^2\over x^2}]$$={2\over x^2}\phi'({{x^2+y^2}\over x^2})+\phi''({{x^2+y^2}\over x^2}){4y^2\over x^4}$Let $t={{x^2+y^2}\over x^2}$.
$u_{xx}(x,y)+u_{yy}(x,y)=({4y^4\over x^6}+{4y^2\over x^4})\phi''({{x^2+y^2}\over x^2})+({6y^2\over x^4}+{2\over x^2})\phi'({{x^2+y^2}\over x^2})$$={4y^2\over x^4}({{x^2+y^2}\over x^2})\phi''({{x^2+y^2}\over x^2})+{2\over x^2}({2y^2\over x^2}+{{x^2+y^2}\over x^2})\phi'({{x^2+y^2}\over x^2})$$={4\over x^2}(t-1)t\phi''(t)+({4\over x^2}(t-1)+{2\over x^2}t)\phi'(t)$$={4}(t-1)t\phi''(t)+(6t-4)\phi'(t)$$=(2t-2)t\phi''(t)+(3t-2)\phi'(t)$$=0$

Wolfram Alpha gives me that $\phi(t)=c+log[{{1-\sqrt(1-t)}\over{1+\sqrt(1-t)}}]$. 
</s>