= Solution
For $\gamma=2$, $\alpha_1=(q/\beta)^{1/2}\to0$. The jump cubic is dominated by $\alpha_2^3\sim3q^2/\alpha_1$, since the other two polynomial terms contain powers of the much smaller $\alpha_1$. Hence
$$
\alpha_2\sim3^{1/3}q^{1/2}\beta^{1/6},\qquad\alpha_2^2\sim3^{2/3}q\beta^{1/3}.
$$
In the downstream integral, $\alpha\geq\alpha_2\to\infty$ and $q=O(1)$, so
$$
\frac{\alpha^4-q^2}{\alpha^3-(q/\beta)\alpha}=\alpha\frac{1-q^2/\alpha^4}{1-q/(\beta\alpha^2)}\sim\alpha.
$$
Its relative corrections are uniformly small over this interval. Integration therefore gives
$$
\frac{G}{c_0^2}(L-x_s)\simeq\frac12(\alpha_3^2-\alpha_2^2),\qquad
\boxed{\frac{2g\sin\theta}{c_0^2}(L-x_s)\simeq\alpha_3^2-3^{2/3}q\beta^{1/3}}.
$$
The stated sufficient condition $0<\alpha_3^2-3^{2/3}q\beta^{1/3}<GL/c_0^2$ places the jump in the downstream half of the tube, $L/2<x_s<L$, to this order. The full condition for $0<x_s<L$ is the weaker upper bound $2GL/c_0^2$, because the displayed length relation contains a factor two. The stricter printed bound is sufficient rather than necessary. This large-$\beta$ branch also requires the downstream prescribed area to scale at least as $\beta^{1/6}$; keeping $\alpha_3$ fixed while sending $\beta$ to infinity would make its right-hand side negative.
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