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