# Riemann Boundary Value Problem

1. Dec 21, 2006

### marcusl

Edit: Am having a lot of trouble with the equations!
Hope this (“Analysis”) is the right forum for my question. I am studying the Plemelj-Sokhotskii (P-S) formulas and solutions to Riemann’s boundary value problem, using the book Boundary Value Problems by Gakhov. On p. 91 the homogeneous Riemann problem is stated as finding two functions, analytic on either side of a simple closed curve L, which satisfy at points t on the curve the relation
$$\Phi^+(t) = G(t) \Phi^-(t)$$
Here + indicates the interior of L and – the exterior.
From here on I get confused. The next paragraph considers a particular type of this problem where the function undergoes a jump across L,
$$\Phi^+(t) - \Phi^-(t) = \varphi(t)$$
where $$\varphi(t)$$ is a function defined only on L. This is the P-S formula having solution
$$\Phi(z) = {\frac{1}{2\pi i}} \int_L {\frac{\varphi(\tau)}{\tau-z}}{d\tau}$$.
Why list it with the homogeneous problem if it has a separate solution? Is there some relation that I'm not seeing?

In the next subsection Gakhov comes back to the homogeneous Riemann problem and on p. 93 considers the case where the index or winding number $$\chi$$ of G(t) is positive. He notes that
$$t^{-\chi}G(t)$$
has zero index and may be written as
$$t^{-\chi}G(t) = \frac{e^{\Gamma^+(t)}}{e^{\Gamma^-(t)}}$$
where
$$\Gamma(z) = {\frac{1}{2\pi i}} \int_L {\frac{ln[\tau^{-\chi}G(\tau)]}{\tau-z}}{d\tau}$$
Where does this come from? I’m not seeing the equivalence (unless this is just a Cauchy integral, but then why go to the trouble of writing it like this? And wouldn't this require that G be analytic throughout both domains that are separated by L?)

Last edited: Dec 21, 2006