 Solution

For $\gamma =2$, ${\alpha }_1=(q/\beta {)}^{1/2}\to 0$. The jump cubic is dominated by ${\alpha }_2^3\sim 3q^2/{\alpha }_1$, since the other two polynomial terms contain powers of the much smaller ${\alpha }_1$. HenceIn the downstream integral, $\alpha \ge {\alpha }_2\to \infty$ and $q=O(1)$, soIts relative corrections are uniformly small over this interval. Integration therefore givesThe 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.