0
$\begingroup$

I am stuck with a math problem that I thought should be straightforward. Maybe I'm missing something here and you can help me.

The key idea is that I have to integrate this function over a volume:

$ I = \int_V \frac{\partial u}{\partial x} dV $

where the volume V is in the x-y plane. I thought of using complex analysis since I do not have a closed form for u, but instead I have an expression for F(z) with

$ F(z) = u + iv \quad \text{with} \quad z=x+iy$

I therefore thought of using complex analysis.

I'll explain my first approach, which I believe was incorrect: First, I thought of the integral "I" as the real part of the whole thing:

$ I = Re\left(\int_V \frac{d F}{d z} dV \right)$

and then use the divergence theorem to cast this not as a volume integral, but as a contour integral.

As you can see already, the approach is wrong in many levels: mostly because dF/dz does not correspond to the divergence (does it?).

I am sure there should be an easy way of converting this integral into a contour integral, but I have been unsuccessful finding it.

Anyone has dealt with this type of problem before?

Many thanks in advance!

Ignacio

Edit:

Ok, after some research I found out that, by defining J as

$J = \displaystyle \frac{i}{2} \int_V \frac{dF}{dz} dz \wedge dz^* $

it is shown that I = Re(J), since $dz\wedge dz^* = -2idxdy$

I believe that J can be calculated using Stokes theorem... work in progress...

$\endgroup$
1
$\begingroup$

If you are still interested now--

This is the correct formula you can use: $$\int_S \frac{\partial F(z,\bar{z})}{\partial z}dz d\bar{z}=i\oint_{\partial S}F(z,\bar{z})d\bar{z},$$ where $\bar{z}=z^*=x-iy$, $dzd\bar{z}=2dxdy$, $S$ is the region we are integrating over, and $\partial S$ is the boundary of $S$ (assuming the contour to be anti-clockwise). Here I have reserved the possibility that $F$ might not be meromorphic, i.e. it may depend explicitly on $\bar{z}$.

$\mathbf{Proof}$: \begin{eqnarray} \int_S \frac{\partial F(z,\bar{z})}{\partial z}dz d\bar{z}=\int_{S}2\partial_z F dxdy=\int_S (\partial_x-i\partial_y)F dx dy=\oint_{\partial S}(F dy+i Fdx)=i\oint_{\partial S} F d\bar{z}, \end{eqnarray} where in the third step I have used Green's formula $\int_S (\partial_x M+\partial_y L)dx dy=\oint_{\partial S}(Mdy-Ldx)$.

Now returning to your question, we have to calculate $\oint_{\partial S} F(z)d\bar{z}$ or its complex conjugate $\oint_{\partial S} F^*(\bar{z})dz$. Because $F^*(\bar{z})$ is not analytic, the integral would depend on the shape of the contour--not just the topology.

Let's see an example here: Take $S$ to be a circular region of radius $R$ centered on $z=0$ and suppose $F(z)$ can be Taylor expanded in $S$. We have $$\oint_{\partial S} F^*(\bar{z}) dz=\oint_{\partial S}\sum_{n\geq 0} \frac{\bar{z}^n}{n!}F^{(n)}(0)^* dz=\oint_{\partial S}\sum_{n\geq 0} \frac{R^{2n}}{z^n n!}F^{(n)}(0)^* dz=2\pi i R^2 F'(0)^*.$$ Here is a specific application $$\int_S 2\cos z dS=\int_S \partial_z \sin z dz d\bar{z}=i\oint_{\partial S} \sin z d\bar{z}=i [2\pi i R^2 \cos(0)]^*=2\pi R^2.$$ Basically with circular geometry, all functions $F(z,\bar{z})$ can be integrated this way, as long as it can be Taylor expanded.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.