Is it a harmonic function or not? I am trying to resolve a question whether a certain function is harmonic or not. If yes, I should find its harmonic conjugate.
The function is $u = \frac{x}{x^2+y^2}$.
I found that it is a harmonic function by using Laplace equation, but I am not sure. How can I find its harmonic conjugate, please help anyone.
 A: To the function to be harmonic its Laplacian should be zero
$$
\Delta f = f_{xx}(x,y)+f_{yy}(x,y) = 0
$$
Just check it
$$
f_{xx} = \left(\frac {x}{x^2+y^2}\right)_{xx} = \left [\frac {-x^2+y^2}{\left( x^2+y^2\right)^2}\right]_x = \frac {2x(x^2-3y^2)}{\left( x^2+y^2\right)^3} \\
f_{yy} = \left(\frac {x}{x^2+y^2}\right)_{yy} = \left [-\frac {2xy}{\left ( x^2+y^2\right)^3} \right ]_y = -\frac {2x(x^2-3y^2)}{\left( x^2+y^2\right)^3}
$$
which means $\Delta f = 0$ hence harmonic.
A: You can do this without calculation. Recall that a holomorphic function is harmonic.
The function
$$
f:z\longmapsto \frac{1}{z}
$$
is holomorphic, so it is harmonic.
Hence its real part
$$
u:z=x+iy\longmapsto \mbox{Re}f(z)=\mbox{Re}\frac{\bar{z}}{|z|^2}=\frac{\mbox{Re} \bar{z}}{|z|^2}=\frac{x}{x^2+y^2}
$$
is harmonic.
And the harmonic conjugate of $u$ is
$$
v:z=x+iy\longmapsto \mbox{Im} f(z) =\frac{\mbox{Im} \bar{z}}{|z|^2}=-\frac{y}{x^2+y^2}.
$$
A: If a function $f(z)=u(x,y)+i v(x,y)$, with $z=x+iy$ and $u,v:\mathbb R^2\to\mathbb R$, is holomorphic, it is also harmonic. So, if you can find a $v$, such that the Cauchy-Riemann equations hold, i.e.:
\begin{align}
\frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y} \\
\frac{\partial u}{\partial y} &= -\frac{\partial v}{\partial x},
\end{align}
you've found your harmonic conjugate. 
If you plug in your $u$ into these equation, you get two differential equations that you can solve to get your $v$ up to some (arbitrary) constant of integration.
Edit: The equations you get in this case are:
\begin{align}
\frac{\partial v}{\partial y} &= \frac{y^2-x^2}{(x^2+y^2)^2} \\
\frac{\partial v}{\partial x} &= \frac{2xy}{(x^2+y^2)^2}
\end{align}
Each of these you can integrate separately, and you get something like $v(x,y)=v_1(x,y) + f(x)$ from the first, $v(x,y)=v_2(x,y) + g(y)$ from the second equation. The functions $f$ and $g$ are "constants" of integration, which you need to match with corresponding terms in $v_2$ and $v_1$, respectively. 
Edit 2: There is also a way to do it without using the CR equations. I knew I read it some time ago, but it took me quite a while to find it again: William T. Shaw "Recovering Holomorphic Functions from Their Real or Imaginary Parts without the Cauchy--Riemann Equations"   SIAM Rev., 46(4), 717–728. 
You can recover $f$ by 
$$f(z)=2u\left( \frac{z+\bar a}{2},\frac{z-\bar a}{2i}\right)-\overline{f(a)}$$
where $a$ is an arbitrary point, such that $f$ is holomorphic in a neighborhood of $a$, and you need to extend $u$ here is the extension from the given $u$ to $\mathbb C\times\mathbb C$. Why that works, you would need to read up in the paper I cited. (Or maybe search for Milne-Thomson method, I think it's more or less the same).
