i have a problem regarding holomorphy of a complex valued functions.
there are a lot of applications where the real part or imaginary part of a function is given and you have to find the function which is holomorphic.
ok...now my problem:
we know that for a real function u(x,y) that is harmonic on a simply connected domain, D, exist a complex valued function, f, holomorphic on D and Re(f)=u(x,y);
what happens when my function is not defined on a simply connected domain, such as:
$$u(x,y)=\frac{x}{x^2+y^2}+x$$
i found guys that solved a lot of problems like that but they say nothing about the domain of holomorphy!!
thanks ;)
If the region $D$ is not simply connected, the statement can fail. The canonical example is the radial function $$u(x, y) := \tfrac{1}{2}\log(x^2 + y^2)$$ on its maximal domain, $\Bbb C - \{ 0 \}$.
If $u$ had a harmonic conjugate $v$, by the Cauchy-Riemann equations it would satisfy \begin{align} \frac{\partial v}{\partial y} &= \phantom{-}\frac{\partial u}{\partial x} = \phantom{-}\frac{y}{x^2 + y^2}\\ \frac{\partial v}{\partial x} &= -\frac{\partial u}{\partial y} = -\frac{x}{x^2 + y^2} . \end{align} But we can conclude from this that no such conjugate exists: If $S^1$ is the unit circle (oriented anticlockwise), evaluating $\int_{S^1} dv$ by expanding $dv = \frac{\partial v}{\partial x} dx + \frac{\partial v}{\partial y} dy$, substituting the above formulas, and, e.g., parameterizing, gives $\int_{S^1} dv = 2 \pi$. But since $S^1$ is compact, the Fundamental Theorem of Calculus gives $\int_{S^1} dv = 0$, a contradiction.
On the other hand, if we consider some simply connected subset $D' \subset \Bbb C - \{ 0 \}$, then $u\vert_{D'}$ does have a harmonic conjugate, and any such harmonic conjugate has the form $v(x, y) = \arg(x + iy) + v_0$ for some constant $v_0$, where $\arg$ is a choice of branch of the argument on $D'$.
For example, the occurrence of the expression $x^2 + y^2$ in the equations for the partial derivatives of $v$ suggest changing to polar coordinates, for which the equations become $\frac{\partial v}{\partial r} = 0$, $\frac{\partial v}{\partial \theta} = 1$, giving $v = \theta + \theta_0$. Converting back to rectangular coordinates on a suitable domain (say, $D' = \{x > 0\})$ and choosing a convenient $\theta_0$ gives $v(x, y) = \arctan \frac{y}{x}$.