Derivation (Illingworth-Stewartson)
Note
This derivation uses the Illingworth-Stewartson (IS) transformation.
For the equivalent derivation using the Levy-Lees transformation,
see Derivation (Levy-Lees) .
Starting from the steady 2D compressible BL equations :
\[
\frac{\partial(\rho u)}{\partial x} + \frac{\partial(\rho v)}{\partial y} = 0
\]
\[
\rho\!\left(u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y}\right)
= -\frac{dp}{dx} + \frac{\partial}{\partial y}\!\left(\mu\frac{\partial u}{\partial y}\right)
\]
\[
\rho c_p\!\left(u\frac{\partial T}{\partial x} + v\frac{\partial T}{\partial y}\right)
= u\frac{dp}{dx} + \frac{\partial}{\partial y}\!\left(k\frac{\partial T}{\partial y}\right)
+ \mu\!\left(\frac{\partial u}{\partial y}\right)^{\!2}
\]
Similarity Ansatz
Edge velocity power law :
\[u_e(x) = C x^m\]
Dimensionless stream function :
\[\psi = \sqrt{2\xi}\,f(\eta)\]
Outer flow
The outer inviscid flow satisfies the Euler x-momentum equation (see
2D BL equations ):
\[-\frac{dp}{dx} = \rho_e u_e \frac{du_e}{dx}\]
For the power-law edge velocity \(u_e = Cx^m\) :
\[
\frac{du_e}{dx} = \frac{d}{dx}\left(Cx^{m}\right)
= m C x^{m-1} = \frac{m}{x} \underbrace{C x^m}_{u_e} = \frac{m u_e}{x}
\]
so the pressure gradient becomes:
\[-\frac{dp}{dx} = \rho_e u_e^2\frac{m}{x}\]
Definitions
Transformed coordinates (IS physical-to-transformed map). The IS transformation references freestream constants \(\rho_\infty\) , \(\mu_\infty\) , so \(\xi\) requires no assumption on the streamwise variation of edge quantities:
\[\bar{x} = \int_0^x \frac{\rho_e \mu_e}{\rho_\infty \mu_\infty}\,dx', \qquad
\bar{y} = \frac{1}{\rho_\infty}\int_0^y \rho\,dy'\]
In the transformed space \((\bar{x}, \bar{y})\) , the continuity equation takes the
incompressible form (unit density). The stream function \(\psi\) is therefore defined
without a density weighting:
\[\frac{\partial\psi}{\partial \bar{y}} = u, \qquad
\frac{\partial\psi}{\partial \bar{x}} = -\bar{v}\]
The similarity coordinates follow from \(\bar{x}\) and \(\bar{y}\) :
\[\xi = \int_0^{\bar{x}} u_e\,d\bar{x}' = \int_0^x \frac{\rho_e \mu_e u_e}{\rho_\infty\mu_\infty}\,dx',
\qquad
\eta = \frac{u_e}{\sqrt{2\xi}}\,\bar{y} = \frac{u_e}{\rho_\infty\sqrt{2\xi}}\int_0^y\rho\,dy'\]
The dimensionless stream function is:
\[\psi = \sqrt{2\xi}\,f(\eta), \qquad f' = \frac{u}{u_e}\]
Hartree parameter :
\[\beta_H = \frac{2m}{m+1}\]
Chapman-Rubesin factor (IS, referenced to freestream):
\[C = \frac{\rho\mu}{\rho_\infty\mu_\infty}\]
Temperature ratio :
\[\tau = \frac{T}{T_e}\]
Ideal gas law :
From the ideal gas law \(p = \rho R T\) , the density is \(\rho = p/(RT)\) , and at the boundary
layer edge \(\rho_e = p_e/(RT_e)\) . From the
y-momentum boundary layer equation
\[
\frac{\partial p}{\partial y} = 0 \rightarrow p = p_e
\]
Therefore:
\[
\frac{\rho}{\rho_e}
= \frac{p/(RT)}{p_e/(RT_e)}
= \frac{T_e}{T}
= \frac{1}{\tau}
\]
Edge Mach number :
\[
M_e = \frac{u_e}{\sqrt{\gamma R T_e}}
\]
With \(c_p = \gamma R/(\gamma-1)\) it follows that
\[
\frac{u_e^2}{c_p T_e} = (\gamma-1)M_e^2
\]
Prandtl number :
\[\mathrm{Pr} = \frac{\mu c_p}{k}\]
so that \(k = \mu c_p/\mathrm{Pr}\) .
Partial derivatives of the similarity coordinates
By the Leibniz integral rule:
\[\frac{\partial\xi}{\partial x} = \frac{\rho_e\mu_e u_e}{\rho_\infty\mu_\infty}, \qquad
\frac{\partial\eta}{\partial y} = \frac{\rho u_e}{\rho_\infty\sqrt{2\xi}}\]
For the power-law \(u_e = Cx^m\) with \(\rho_e\mu_e = \text{const}\) (uniform freestream, \(\rho_e = \rho_\infty\) , \(\mu_e = \mu_\infty\) ):
\[
\xi = \frac{\rho_\infty\mu_\infty}{\rho_\infty\mu_\infty}\int_0^x Cx'^m\,dx'
= \frac{\overbrace{Cx^m}^{u_e}\,x}{m+1}
= \frac{\rho_\infty\mu_\infty u_e x}{m+1}
\]
so that \(m/x = \beta_H\rho_\infty\mu_\infty u_e/(2\xi)\) . The \(\partial\eta/\partial x\) term cancels in the convective and energy terms.
\[\frac{\partial F}{\partial y}\bigg|_x
= \frac{\partial F}{\partial \xi}\bigg|_\eta \underbrace{\frac{\partial \xi}{\partial y}\bigg|_x}_{=\,0}
+ \frac{\partial F}{\partial \eta}\bigg|_\xi \frac{\partial \eta}{\partial y}\bigg|_x
= \frac{\rho u_e}{\rho_\infty\sqrt{2\xi}}\frac{\partial F}{\partial \eta}\bigg|_\xi\]
\[\frac{\partial F}{\partial x}\bigg|_y
= \frac{\partial F}{\partial \xi}\bigg|_\eta \frac{\partial \xi}{\partial x}\bigg|_y
+ \frac{\partial F}{\partial \eta}\bigg|_\xi \frac{\partial \eta}{\partial x}\bigg|_y\]
Streamwise velocity
From \(\partial\psi/\partial\bar{y} = u\) it follows that \(u = u_e f'(\eta)\) . The wall-normal derivative:
\[
\frac{\partial \overbrace{u}^{u_e f'(\eta)}}{\partial y}
= \frac{\partial \left(u_e f'(\eta)\right)}{\partial\eta} \overbrace{\frac{\partial\eta}{\partial y}}^{\frac{\rho u_e}{\rho_\infty\sqrt{2\xi}}}
= \frac{\rho u_e}{\rho_\infty\sqrt{2\xi}}\frac{\partial(u_e f'(\eta))}{\partial\eta}
= \frac{\rho u_e^2}{\rho_\infty\sqrt{2\xi}}\,f''(\eta)
\]
The streamwise derivative:
\[
\frac{\partial \overbrace{u}^{u_e f'(\eta)}}{\partial x}\bigg|_y
= \overbrace{\frac{du_e}{dx}}^{mu_e/x} f'(\eta)
+ u_e\,f''(\eta)\,\frac{\partial\eta}{\partial x}\bigg|_y
= \frac{m u_e}{x}\,f'(\eta) + u_e\,f''(\eta)\,\frac{\partial\eta}{\partial x}\bigg|_y
\]
Continuity
Satisfied identically by construction of \(\psi\) .
x-momentum
Using the expressions derived above:
\(\rho u\frac{\partial u}{\partial x}\) term.
\[\rho u\frac{\partial u}{\partial x} = \rho u_e f' \left(\frac{m u_e}{x}f' + u_e f'' \frac{\partial\eta}{\partial x}\right)
= \frac{\rho u_e^2}{x}\!\left(m f'^2 + x f'' f'\frac{\partial\eta}{\partial x}\right)\]
\(\rho v \frac{\partial u}{\partial y}\) term.\(\partial\psi/\partial\bar{x} = -\bar{v}\) , applying the product rule to \(\psi = \sqrt{2\xi}\,f(\eta)\) with \(\rho_e = \rho_\infty\) , \(\mu_e = \mu_\infty\) :
\[
\rho v = -\frac{\rho_\infty\mu_\infty u_e}{\sqrt{2\xi}}\,f - \sqrt{2\xi}\,\rho_\infty\,f'\,\frac{\partial\eta}{\partial x}\bigg|_y
\]
Multiplying by \(\partial u/\partial y = \rho u_e^2 f''/(\rho_\infty\sqrt{2\xi})\) :
\[
\rho v\,\frac{\partial u}{\partial y}
= \left(-\frac{\rho_\infty\mu_\infty u_e}{\sqrt{2\xi}}\,f - \sqrt{2\xi}\,\rho_\infty\,f'\,\frac{\partial\eta}{\partial x}\bigg|_y\right)
\frac{\rho u_e^2 f''}{\rho_\infty\sqrt{2\xi}}
= -\frac{\rho\mu_\infty u_e^3}{\rho_\infty \cdot 2\xi}\,ff''
- \rho u_e^2\,f'f''\,\frac{\partial\eta}{\partial x}\bigg|_y
\]
Combined convective term. Adding both terms, the \(\partial\eta/\partial x\) pieces cancel:
\[
\rho u\,\frac{\partial u}{\partial x} + \rho v\,\frac{\partial u}{\partial y}
= \frac{m\rho u_e^2}{x}\,f'^2
+ \cancel{\rho u_e^2\,f'f''\,\frac{\partial\eta}{\partial x}}
- \frac{\rho\mu_\infty u_e^3}{\rho_\infty \cdot 2\xi}\,ff''
- \cancel{\rho u_e^2\,f'f''\,\frac{\partial\eta}{\partial x}}
= \frac{m\rho u_e^2}{x}\,f'^2 - \frac{\rho\mu_\infty u_e^3}{\rho_\infty \cdot 2\xi}\,ff''
\]
Recast \(\frac{m}{x}\) using \(\xi = \rho_\infty\mu_\infty u_e x/(m+1)\) :
\[
\frac{m}{\underbrace{x}_{\frac{\xi(m+1)}{\rho_\infty\mu_\infty u_e}}}
= \frac{m\rho_\infty\mu_\infty u_e}{\xi(m+1)}
= \overbrace{\frac{m}{(m+1)}}^{\beta_H/2} \frac{\rho_\infty\mu_\infty u_e}{\xi}
= \frac{\beta_H\rho_\infty\mu_\infty u_e}{2\xi}
\]
Substituting:
\[
\rho u\,\frac{\partial u}{\partial x} + \rho v\,\frac{\partial u}{\partial y}
= \overbrace{\frac{m}{x}}^{\frac{\beta_H\rho_\infty\mu_\infty u_e}{2\xi}} \rho u_e^2 f'^2
- \frac{\rho\mu_\infty u_e^3}{\rho_\infty \cdot 2\xi}\,ff''
= \frac{\beta_H\rho\mu_\infty u_e^3}{\rho_\infty \cdot 2\xi} f'^2 - \frac{\rho\mu_\infty u_e^3}{\rho_\infty \cdot 2\xi}\,ff''
\]
\[\boxed{
\rho u\,\frac{\partial u}{\partial x} + \rho v\,\frac{\partial u}{\partial y}
= \frac{\rho\mu_\infty u_e^3}{\rho_\infty \cdot 2\xi}\!\left(\beta_H f'^2 - ff''\right)
}\]
Pressure term.
\[
-\frac{dp}{dx}
= \rho_e u_e^2\frac{m}{x}
= \rho_e u_e^2 \frac{\beta_H \rho_\infty\mu_\infty u_e}{2\xi}
= \overbrace{\rho_e}^{\rho\tau} \frac{\mu_\infty u_e^3}{\rho_\infty \cdot 2\xi}\beta_H
= \frac{\rho\mu_\infty u_e^3}{\rho_\infty \cdot 2\xi}\,\tau\beta_H
\]
\[\boxed{-\frac{dp}{dx} = \frac{\rho\mu_\infty u_e^3}{\rho_\infty \cdot 2\xi}\,\tau\beta_H}\]
Viscous term.
\[
\mu\frac{\partial \overbrace{u}^{u_e f'}}{\partial y}
= \overbrace{\mu}^{C\rho_\infty\mu_\infty/\rho} \cdot \overbrace{\frac{\partial u}{\partial y}}^{\rho u_e^2 f''/(\rho_\infty\sqrt{2\xi})}
= \frac{C\mu_\infty u_e^2}{\sqrt{2\xi}}\,f''
\]
Applying the outer \(\partial/\partial y = (\rho u_e/(\rho_\infty\sqrt{2\xi}))\,\partial/\partial\eta\) :
\[
\frac{\partial}{\partial y}\!\left(\mu\frac{\partial u}{\partial y}\right)
= \overbrace{\frac{\partial}{\partial y}}^{\frac{\rho u_e}{\rho_\infty\sqrt{2\xi}}\partial/\partial\eta}
\left(\overbrace{\mu\frac{\partial u}{\partial y}}^{C\mu_\infty u_e^2 f''/\sqrt{2\xi}}\right)
= \frac{\rho u_e}{\rho_\infty\sqrt{2\xi}}\cdot\frac{\mu_\infty u_e^2}{\sqrt{2\xi}}\,(Cf'')'
\]
\[\boxed{\frac{\partial}{\partial y}\!\left(\mu\frac{\partial u}{\partial y}\right)
= \frac{\rho\mu_\infty u_e^3}{\rho_\infty \cdot 2\xi}\,(Cf'')'}\]
Assembly.
The x-momentum equation becomes:
\[
\cancel{\frac{\rho\mu_\infty u_e^3}{\rho_\infty \cdot 2\xi}}\!\left(\beta_H f'^2 - ff''\right)
= \cancel{\frac{\rho\mu_\infty u_e^3}{\rho_\infty \cdot 2\xi}}\beta_H\tau
+ \cancel{\frac{\rho\mu_\infty u_e^3}{\rho_\infty \cdot 2\xi}}\,(Cf'')'
\]
Dividing through by \(\rho\mu_\infty u_e^3/(\rho_\infty \cdot 2\xi)\) and rearranging:
\[(Cf'')' + ff'' + \beta_H(\tau - f'^2) = 0\]
Energy
The energy equation is:
\[
\rho c_p\!\left(u\frac{\partial T}{\partial x} + v\frac{\partial T}{\partial y}\right)
= u\frac{dp}{dx}
+ \frac{\partial}{\partial y}\!\left(k\frac{\partial T}{\partial y}\right)
+ \mu\!\left(\frac{\partial u}{\partial y}\right)^{\!2}
\]
Note \(T = T_e\,\tau(\eta) \rightarrow\) function of \(\eta\) only
\[
\frac{\partial \overbrace{T}^{T_e\,\tau(\eta)}}{\partial y}
= T_e\,\overbrace{\frac{\partial\tau}{\partial\eta}}^{\tau'}\,
\overbrace{\frac{\partial\eta}{\partial y}}^{\rho u_e/(\rho_\infty\sqrt{2\xi})}
= \frac{\rho u_e T_e}{\rho_\infty\sqrt{2\xi}}\,\tau'
\]
\[
\frac{\partial \overbrace{T}^{T_e\,\tau(\eta)}}{\partial x}\bigg|_y
= T_e\,\overbrace{\frac{\partial\tau}{\partial\eta}}^{\tau'}\,\frac{\partial\eta}{\partial x}\bigg|_y
= T_e\,\tau'\,\frac{\partial\eta}{\partial x}\bigg|_y
\]
Convective term. Streamwise and wall-normal contributions combined (the \(\partial\eta/\partial x\) cross terms cancel):
\[
\rho c_p\!\left(u\frac{\partial T}{\partial x} + v\frac{\partial T}{\partial y}\right)
= \cancel{\rho c_p u_e T_e f'\tau'\frac{\partial\eta}{\partial x}}
- \frac{\rho\mu_\infty u_e^2 c_p T_e}{\rho_\infty \cdot 2\xi}\,f\tau'
- \cancel{\rho c_p u_e T_e f'\tau'\frac{\partial\eta}{\partial x}}
\]
\[\boxed{\rho c_p\!\left(u\frac{\partial T}{\partial x} + v\frac{\partial T}{\partial y}\right)
= -\frac{\rho\mu_\infty u_e^2 c_p T_e}{\rho_\infty \cdot 2\xi}\,f\tau'}\]
Diffusion term.
\[
k\frac{\partial T}{\partial y}
= \overbrace{k}^{C\rho_\infty\mu_\infty c_p/(\rho\,\mathrm{Pr})} \cdot
\overbrace{\frac{\partial T}{\partial y}}^{\rho u_e T_e\tau'/(\rho_\infty\sqrt{2\xi})}
= \frac{C\mu_\infty u_e c_p T_e}{\mathrm{Pr}\sqrt{2\xi}}\,\tau'
\]
Applying the outer \(\partial/\partial y = (\rho u_e/(\rho_\infty\sqrt{2\xi}))\,\partial/\partial\eta\) :
\[
\frac{\partial}{\partial y}\!\left(k\frac{\partial T}{\partial y}\right)
= \overbrace{\frac{\partial}{\partial y}}^{\frac{\rho u_e}{\rho_\infty\sqrt{2\xi}}\partial/\partial\eta}
\left(\overbrace{k\frac{\partial T}{\partial y}}^{C\mu_\infty u_e c_p T_e\tau'/(\mathrm{Pr}\sqrt{2\xi})}\right)
= \frac{\rho u_e}{\rho_\infty\sqrt{2\xi}}\cdot\frac{\mu_\infty u_e c_p T_e}{\mathrm{Pr}\sqrt{2\xi}}\,\left(\frac{C}{\mathrm{Pr}}\tau'\right)'
\]
\[\boxed{\frac{\partial}{\partial y}\!\left(k\frac{\partial T}{\partial y}\right)
= \frac{\rho\mu_\infty u_e^2 c_p T_e}{\rho_\infty \cdot 2\xi}\left(\frac{C}{\mathrm{Pr}}\tau'\right)'}\]
Pressure work term.
\[
u\frac{dp}{dx}
= \overbrace{u_e f'}^{u}\cdot\left(-\rho_e u_e^2\overbrace{\frac{m}{x}}^{\beta_H\rho_\infty\mu_\infty u_e/(2\xi)}\right)
= -\frac{\beta_H\rho_e\mu_\infty u_e^4 f'}{\rho_\infty \cdot 2\xi}
= -\frac{\rho\mu_\infty u_e^2 c_p T_e}{\rho_\infty \cdot 2\xi}\cdot
\overbrace{\frac{\beta_H\tau u_e^2 f'}{c_p T_e}}^{(\gamma-1)M_e^2\beta_H\tau f'}
\]
\[\boxed{u\frac{dp}{dx}
= -\frac{\rho\mu_\infty u_e^2 c_p T_e}{\rho_\infty \cdot 2\xi}\,(\gamma-1)M_e^2\,\beta_H\tau f'}\]
Dissipation term.
\[
\mu\!\left(\frac{\partial u}{\partial y}\right)^{\!2}
= \overbrace{\frac{C\rho_\infty\mu_\infty}{\rho}}^{\mu}
\left(\overbrace{\frac{\rho u_e^2 f''}{\rho_\infty\sqrt{2\xi}}}^{\partial u/\partial y}\right)^{\!2}
= \frac{C\rho\mu_\infty u_e^4 f''^2}{\rho_\infty\cdot 2\xi}
= \frac{\rho\mu_\infty u_e^2 c_p T_e}{\rho_\infty \cdot 2\xi}\cdot
\overbrace{\frac{C u_e^2 f''^2}{c_p T_e}}^{(\gamma-1)M_e^2\,Cf''^2}
\]
\[\boxed{\mu\!\left(\frac{\partial u}{\partial y}\right)^{\!2}
= \frac{\rho\mu_\infty u_e^2 c_p T_e}{\rho_\infty \cdot 2\xi}\,(\gamma-1)M_e^2\,Cf''^2}\]
Assembly. Every term carries \(\rho\mu_\infty u_e^2 c_p T_e/(\rho_\infty \cdot 2\xi)\) :
\[
-\cancel{\frac{\rho\mu_\infty u_e^2 c_p T_e}{\rho_\infty \cdot 2\xi}}\,f\tau'
= -\cancel{\frac{\rho\mu_\infty u_e^2 c_p T_e}{\rho_\infty \cdot 2\xi}}\,(\gamma-1)M_e^2\beta_H\tau f'
+ \cancel{\frac{\rho\mu_\infty u_e^2 c_p T_e}{\rho_\infty \cdot 2\xi}}\left(\frac{C}{\mathrm{Pr}}\tau'\right)'
+ \cancel{\frac{\rho\mu_\infty u_e^2 c_p T_e}{\rho_\infty \cdot 2\xi}}\,(\gamma-1)M_e^2 Cf''^2
\]
Dividing through and rearranging:
\[\left(\frac{C}{\mathrm{Pr}}\tau'\right)' + f\tau'
+ (\gamma-1)M_e^2\!\left[Cf''^2 - \beta_H\tau f'\right] = 0\]