With the <dimensionless variables> $x=r/r_s$ and $y=u/c_s$, <mass conservation> and part (c) give
$$
\frac\rho{\rho_0}=\frac{e^{3/2}}{x^2y}.
$$
Substitution into the <Bernoulli function>, together with $GM/r=2c_s^2/x$, eliminates every dimensional parameter and gives the <transcendental equation>
$$
\boxed{\frac{y^2}{2}-\ln y-2\ln x+\frac32-\frac2x=0}.
$$
Equivalently,
$$
\boxed{y^2-\ln(y^2)=\frac4x+4\ln x-3},
$$
or
$$
\boxed{y^2e^{-y^2}=x^{-4}e^{3-4/x}}.
$$
The <transonic branch> passes through $(x,y)=(1,1)$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 314 / 2 / d / Solution

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
$$
\boxed{\frac{d^2\sigma}{d\xi^2}-qe^\sigma+1=0}.
$$

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

A dimensionless variable is unchanged by a consistent change of physical units. Scaling a variable by a characteristic quantity with the same dimensions produces a dimensionless ratio.


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 Incoming links: Dimensionless variable