In complex analysis , if $f(z)= u(x,y) + \iota v(x,y) $ is analytic,then v is called harmonic conjugate of u. If I try to find the harmonic conjugate of $u = x^2$ , I'm not getting any valid function v.
Then what are the conditions for existence of harmonic function?
Which condition is violated , when I take the above $u$?
A function $u$ (defined on an open connected set $\Omega\subseteq\Bbb C$) has a harmonic conjugate $v$ iff the function $f=u+iv$ is holomorphic on $\Omega.$ Consequently, if $u$ has a harmonic conjugate, then both are harmonic real-valued functions on $\Omega,$ meaning that they are twice continuously differentiable and satisfies the LaPlace equations $$\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 v}{\partial x^2}+\frac{\partial^2 v}{\partial y^2}=0.$$
On the other hand, let's suppose that $u$ is a harmonic function on $\Omega.$ We'd like to show that there exists $f=u+iv$ such that $u$ and $v$ satisfy the Cauchy-Riemann equations $$\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}$$ and $$\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}.$$ Now the LaPlace equations allow us to use line integrals to define $v$ locally as $$v=\int_C\left(-\frac{\partial u}{\partial y}\,dx+\frac{\partial u}{\partial x}\,dy\right).$$ By "locally," I mean that, given any point $z_0\in\Omega,$ there is some $r>0$ such that $z\in\Omega$ whenever $|z-z_0|<r,$ and for every such $z,$ we can define $v$ at $z$ by a line integral from $z_0$ to $z$ in this way. We may not be able to do better, because otherwise we can't guarantee independence of path, as we may have singularities throwing us off. However, what this does accomplish is proving that $v$ exists, though is only unique up to a constant term, so we've really only proved that a family of such functions $v$ exist, but that we don't necessarily have a canonical way to choose one.