For large $q$ limit, we have general solution $$e^{g(\tau)}=\frac{c^2}{\mathcal{J}^2}\frac{1}{\left[\sin\left(c(|\tau|+\tau_0)\right)\right]^2},\tag{2.17}$$ $$\mathcal{J}\equiv\frac{\sqrt{q}J}{2^{\frac{q-1}{2}}},\tag{2.16}$$ Given the boundary condition $$g(0)=g(\beta)=0,\tag{p.12}$$ equations (2.18) and (2.19) in this paper https://arxiv.org/abs/1604.07818 directly give the answer with another variable $v$: $$e^g(\tau)=\left(\frac{\cos\frac{\pi v}{2}}{\cos\left[\pi v(\frac{1}{2}-\frac{|\tau|}{\beta})\right]}\right)^2,\tag{2.18}$$ $$\beta\mathcal{J}=\frac{\pi v}{\cos\frac{\pi v}{2}}.\tag{2.19}$$ My question is how to derive the above expression with new variable $v$?
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A purely technical derivation. From $g(0) = g(\beta) = 0$ follows $$ \sin(c\tau_0) = \pm\sin(c(\beta+\tau_0)) \tag{1} $$ and $$ \beta\cal{J} = \frac{c\beta}{|\sin(c\tau_0)|}.\tag{2} $$ I do not know why, probably there should be some kind of physics here, but the authors of the article chose the following specific solution to equation (1): $$ c(\beta +\tau_0) = \pi - c\tau_0\quad \rightarrow \quad \tau_0 = \frac{\pi}{2c}-\frac\beta2. $$ Hence $$\sin(c(|\tau|+\tau_0)) = \cos\left(c\beta\left(\frac12-\frac{|\tau|}\beta\right)\right)\quad\mbox{and}\quad \sin(c\tau_0) = \cos\left(\frac{c\beta}2\right). $$ By substituting $c\beta = \pi v$ and assuming $\sin(c\tau_0) >0$, the reasons for which I also do not know, we get the required formulas. I hope this helps and makes some sense.
