Differentiating the equilibrium velocity gives
$$
2\Omega+\frac{du_y}{dx}
=\frac\Omega2+\frac{c_s^2}{2\Omega}\frac{d^2\sigma}{dx^2}.
$$
On the other hand, the definition of <vortensity> and $\Sigma=\Sigma_0e^\sigma$ give
$$
2\Omega+\frac{du_y}{dx}=\zeta\Sigma_0e^\sigma.
$$
Multiplying by $2/\Omega$, and introducing the <dimensionless variables>
$$
\xi=\frac\Omega{c_s}x,
\qquad
q=\frac{2\Sigma_0}{\Omega}\zeta,
$$
yields
$$
\boxed{\frac{d^2\sigma}{d\xi^2}-qe^\sigma+1=0}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 321 / 3 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-321/3/b/ii/solution