Reduction of Order
Starting from the derived ODEs :
\[(Cf'')' + ff'' + \beta_H(\tau - f'^2) = 0\]
\[\left(\frac{C}{\mathrm{Pr}}\tau'\right)' + f\tau'
+ (\gamma-1)M_e^2\!\left[Cf''^2 - \beta_H\tau f'\right] = 0\]
where \(C = \rho\mu/(\rho_e\mu_e)\) is the Chapman–Rubesin factor, \(f' = u/u_e\) the
dimensionless velocity, \(\tau = T/T_e\) the dimensionless temperature, and all primes
denote differentiation with respect to the wall-normal similarity variable \(\eta\) .
Chapman–Rubesin factor
For a perfect gas \(\rho/\rho_e = 1/\tau\) , so the Chapman–Rubesin factor simplifies to:
\[C = \frac{\rho\mu}{\rho_e\mu_e} = \frac{\mu}{\mu_e\tau}\]
Differentiating with respect to \(\eta\) via the quotient rule:
\[C' = \frac{d}{d\eta}\!\left(\frac{\mu}{\mu_e\tau}\right)
= \frac{1}{\mu_e}\frac{\mu'\tau - \mu\tau'}{\tau^2}\]
Expanding the compound derivatives
Applying the product rule to \((Cf'')'\) and \(\bigl(C\tau'/\mathrm{Pr}\bigr)'\) :
\[(Cf'')' = C'f'' + Cf'''\]
\[\left(\frac{C}{\mathrm{Pr}}\tau'\right)' = \frac{C'}{\mathrm{Pr}}\tau' + \frac{C}{\mathrm{Pr}}\tau''\]
Substituting into the two ODEs:
\[C'f'' + Cf''' + ff'' + \beta_H(\tau - f'^2) = 0\]
\[\frac{C'}{\mathrm{Pr}}\tau' + \frac{C}{\mathrm{Pr}}\tau'' + f\tau'
+ (\gamma-1)M_e^2\!\left[Cf''^2 - \beta_H\tau f'\right] = 0\]
Substituting C and C'
Replacing \(C = \mu/(\mu_e\tau)\) and \(C' = (\mu'\tau - \mu\tau')/(\mu_e\tau^2)\) :
Momentum:
\[\frac{\mu'\tau - \mu\tau'}{\mu_e\tau^2}\,f''
+ \frac{\mu}{\mu_e\tau}\,f'''
+ ff'' + \beta_H(\tau - f'^2) = 0\]
Energy:
\[\frac{(\mu'\tau - \mu\tau')}{\mathrm{Pr}\,\mu_e\tau^2}\,\tau'
+ \frac{\mu}{\mathrm{Pr}\,\mu_e\tau}\,\tau''
+ f\tau'
+ (\gamma-1)M_e^2\!\left[\frac{\mu}{\mu_e\tau}\,f''^2 - \beta_H\tau f'\right] = 0\]
Isolating the highest derivatives
Momentum
Multiply by \(\mu_e\) :
\[\frac{\mu'\tau - \mu\tau'}{\tau^2}\,f''
+ \frac{\mu}{\tau}\,f'''
+ \mu_e ff'' + \mu_e\beta_H(\tau - f'^2) = 0\]
Multiply by \(\tau/\mu\) :
\[\frac{\mu'\tau - \mu\tau'}{\mu\tau}\,f''
+ f'''
+ \frac{\mu_e\tau}{\mu}\,ff''
+ \frac{\mu_e\tau}{\mu}\,\beta_H(\tau - f'^2) = 0\]
Noting that \(\mu_e\tau/\mu = 1/C\) and \((\mu'\tau - \mu\tau')/(\mu\tau) = \mu'/\mu - \tau'/\tau\) , and rearranging:
\[\boxed{f''' = \frac{\tau'}{\tau}\,f'' - \frac{\mu'}{\mu}\,f'' - \frac{ff''}{C} - \frac{\beta_H(\tau - f'^2)}{C}}\]
Energy
Multiply by \(\mu_e\) :
\[\frac{(\mu'\tau - \mu\tau')}{\mathrm{Pr}\,\tau^2}\,\tau'
+ \frac{\mu}{\mathrm{Pr}\,\tau}\,\tau''
+ \mu_e f\tau'
+ (\gamma-1)M_e^2\!\left[\frac{\mu}{\tau}\,f''^2 - \beta_H\mu_e\tau f'\right] = 0\]
Multiply by \(\mathrm{Pr}\,\tau/\mu\) , using \(\mu_e\tau/\mu = 1/C\) and \(\mu_e\tau^2/\mu = \tau/C\) :
\[\frac{(\mu'\tau - \mu\tau')}{\mu\tau}\,\tau'
+ \tau''
+ \frac{\mathrm{Pr}\,f\tau'}{C}
+ \mathrm{Pr}(\gamma-1)M_e^2\!\left[f''^2 - \frac{\beta_H\tau f'}{C}\right] = 0\]
Applying \((\mu'\tau - \mu\tau')/(\mu\tau) = \mu'/\mu - \tau'/\tau\) and rearranging:
\[\boxed{\tau'' = \frac{\tau'^2}{\tau} - \frac{\mu'\tau'}{\mu}
- \frac{\mathrm{Pr}\,f\tau'}{C}
- \mathrm{Pr}(\gamma-1)M_e^2\!\left[f''^2 - \frac{\beta_H\tau f'}{C}\right]}\]
Viscosity derivatives
The \(\mu'/\mu\) term requires the derivative of viscosity with respect to \(\eta\) .
By the chain rule:
\[\mu' = \frac{d\mu}{d\eta} = \frac{d\mu}{dT}\frac{dT}{d\eta}\]
Since \(T = T_e\tau(\eta)\) :
\[\frac{dT}{d\eta} = T_e\frac{d\tau}{d\eta} = T_e\tau'\]
Therefore:
\[\mu' = \frac{d\mu}{dT}\,T_e\tau'\]
Substituting into the expressions above:
\[f''' = \frac{\tau'}{\tau}\,f''
- \frac{T_e}{\mu}\frac{d\mu}{dT}\,\tau'f''
- \frac{ff''}{C}
- \frac{\beta_H(\tau - f'^2)}{C}\]
\[\tau'' = \frac{\tau'^2}{\tau}
- \frac{T_e}{\mu}\frac{d\mu}{dT}\,\tau'^2
- \frac{\mathrm{Pr}\,f\tau'}{C}
- \mathrm{Pr}(\gamma-1)M_e^2\!\left[f''^2 - \frac{\beta_H\tau f'}{C}\right]\]
State variables
The third-order equation in \(f\) and the second-order equation in \(\tau\) yield a
fifth-order system. Define:
\[y_1 = f, \qquad y_2 = f', \qquad y_3 = f'', \qquad y_4 = \tau, \qquad y_5 = \tau'\]
First-order ODE system
Using the state variables and the expressions above, with \(T = T_e y_4\) and
\(C = \mu(T)/(\mu_e y_4)\) :
\[y_1' = y_2\]
\[y_2' = y_3\]
\[y_3' = \frac{y_5}{y_4}\,y_3
- \frac{T_e}{\mu}\frac{d\mu}{dT}\,y_5 y_3
- \frac{y_1 y_3 + \beta_H(y_4 - y_2^2)}{C}\]
\[y_4' = y_5\]
\[y_5' = \frac{y_5^2}{y_4}
- \frac{T_e}{\mu}\frac{d\mu}{dT}\,y_5^2
- \frac{\mathrm{Pr}\,y_1 y_5}{C}
- \mathrm{Pr}(\gamma-1)M_e^2\!\left[y_3^2 - \frac{\beta_H y_4 y_2}{C}\right]\]
Boundary conditions
\[\eta = 0:\quad f = 0,\quad f' = 0,\quad
\tau = \tau_w\ \text{(isothermal)}\]
\[\eta \to \infty:\quad f' \to 1,\quad \tau \to 1\]
For an adiabatic wall, replace \(\tau = \tau_w\) at \(\eta = 0\) with \(\tau' = 0\) at \(\eta = 0\) .