Consider the following Riemann-Hilbert problem as given on Page 9 of this paper: $$\Phi^+(t)- \Phi^-(t) = 2u(t)$$ $$\Phi^+(t)+ \Phi^-(t) = \frac{P}{\pi i } \int_{t_1}^{t_2} \frac{d\zeta}{\zeta} u(\zeta) \frac{\zeta +t}{\zeta - t} = E(t - t^{-1})$$ where $E\in \mathbb{R}$, $P$ denotes the principal value of the integral, and $\Phi^{\pm}(t)$ denote the values of the function $\Phi$ as the point $t$ on the counterclockwise arc $(t_1, t_2)$ is approached from the left and the right respectively. The function $u(t)$ takes the boundary values $u(t_1) = u(t_2) = 0$. We have an additional constraint here as well:

$$\Phi(\infty) = -1.$$

  1. The author then proceeds to write down the solution of this problem which I am confused by. The first confusion is the claim is that "the function $h(z) = \left[ (z-t_1)(z-t_2)\right]^{1/2}$ solves the problem $h^+ + h^- =0$ upto an entire function." I am not able to see the same.

The author then writes the ansatz for the function $\Phi(z) = h(z)H(z)$. Using the Plemelj formula and calculating the residues, the author then obtains the expression for $H(z)$ as:

$$H(z) = \frac{E}{2} \left[ \frac{z-z^{-1}}{\sqrt{(z-t_1)(z-t_2)}}+1 + \frac{1}{z}\right] $$

  1. Is there an argument why the minus sign in the first term's numerator of the above equation gets converted into a plus sign in the following equation for $\Phi(z)$?

$$\Phi(z) = \frac{E}{2} \left( z+ \frac{1}{z}\right) + \frac{E}{2} \left(\frac{1}{z} + 1\right) \sqrt{(z-t_1)(z-t_2)}$$

  1. Using the above equation the author substitutes $\Phi(\infty) = -1$ to obtain:

$$\frac{1-\cos \alpha}{2} = \frac{1}{E}$$ This follows if I set $z = e^{i\alpha}$. However it does not seem reasonable to do so when $z=\infty$. Is there some other way to see this?


Your Answer

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

Browse other questions tagged or ask your own question.