For fixed $0<\mathcal M_0<1$, eliminate $\alpha$ by
$$
\alpha=\frac{x^{\nu/2}}{2\sqrt{\mathcal M_0}}.
$$
The remaining equation is
$$
\nu x+2\sqrt{\mathcal M_0}\,x^{-\nu/2}
=\nu+\frac32+\frac{\mathcal M_0^2}{2}.
$$
Its left side diverges at both ends and has one minimum, at
$$
x=\mathcal M_0^{1/(\nu+2)}.
$$
For $0<\mathcal M_0<1$ this minimum lies strictly below the right side, so there are exactly two positive values of $x$, and hence exactly two values of $\alpha$. The branches merge at the marginal point $(\alpha,\mathcal M_0)=(1/2,1)$.

On the very hot branch, $\alpha\to\infty$ and $\mathcal M_0=O(\alpha^{-2})$. Dropping $1/\alpha$ and $\mathcal M_0^2$ in the relation from part b gives
$$
\boxed{
\mathcal M_0\sim\frac1{4\alpha^2}
\left(\frac{2}{5-3\gamma}\right)^\nu},
\qquad
\nu=\frac{5-3\gamma}{2(\gamma-1)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 314 / 3 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-314/3/d/solution