Complex function of complexvariable

Ruun · #1 $Old$ November 3rd, 2009, 11:24 AM

If I am given the function $u(x,y)$ wich could be the real part of an analytic complex function of complex variable, and $z=x+iy$ , how can I compute a function $f(z)=u(x,y)+iv(x,y)$ ? I have to apply the Cauchy Riemann conditions ( $u_{x}=v_{y}$ and $u_{y}=-v_{x}$ ) and integrate both equations?

Thanks in advance

shawsend · #2 $Old$ November 3rd, 2009, 12:56 PM

Assume $u(x,y)$ is harmonic, then it has a harmonic conjugate $v(x,y)$ such that $f(x,y)=u(x,y)+i v(x,y)$ is analytic and satisfies the partials:

$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\quad \quad \frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$

so yes, you can partially integrate them like:

$\int \partial v=\int \frac{\partial u}{\partial x}\partial y$

and after the integration with respect to y with x held constant, we obtain something like:

$v(x,y)=\int \frac{\partial u}{\partial x}\partial y+\phi(x)$

where the constant of integration is now a function of x right? Now we use the second CR equations to solve for $\phi(x)$ by writing:

$\frac{\partial u}{\partial y}=-\frac{\partial}{\partial x}\left\{\int \frac{\partial u}{\partial x}\partial y+\phi(x)\right\}$

Now, solve for $\phi'(x)$ and integrate.
Do it manually a few times first to get the hang of it then explain how that's the same as:

$v(x,y)=\int\frac{\partial u}{\partial x}\partial y-\int\left(\frac{\partial}{\partial x}\int \frac{\partial u}{\partial x}\partial y+\frac{\partial u}{\partial y}\right)dx$

where all the integrations are without the arbitrary function, and it looks intimidating at first sight but becomes a piece of cake after you solve a few of them. For example $u(x,y)=y^3-3x^2y$ , then:

$\int \frac{\partial u}{\partial x}\partial y=\int (-6xy)\partial y=-3xy^2$

Now take the derivative of that with respect to x:

$\frac{\partial}{\partial x}\left(\int \frac{\partial u}{\partial x}\partial y\right)=\frac{\partial}{\partial x}(-3xy^2)=-3y^2$

and continue working the partials and integrals until you get the expression for v(x,y). Then the most general harmonic conjugate is v(x,y)+k.

Ruun · #3 $Old$ November 3rd, 2009, 02:46 PM

Yes, by "integrating the equations" I mean what you have done, taking in count that function $\phi (x)$ with the partial respect to $y$ , not just integrating both.

Thanks