Suggested derivation of boundary layer equations
The boundary layer equations (Eq. (4-32) in [Whi06] are derived by multiplying the Navier-Stokes equations by \(L/U^2\) and then eliminating all terms proportional to \(1/Re\) and \(1/Re^2\). This is justified for flows with high Reynolds number. The NS equations are non-dimensionalized using
\[\begin{split}
\begin{aligned}
\overline{x} = \frac{x}{L} && \overline{u} = \frac{u}{U} && \overline{v} = \frac{v}{U}\sqrt{Re} \\
\overline{y} = \frac{y}{L}\sqrt{Re} && \overline{t} = \frac{t U}{L} && \overline{p} = \frac{p}{\rho U^2}
\end{aligned}
\end{split}\]
Term by term the momentum equation in the streamwise (\(x\)) direction is multiplied by \(L/U^2\)
\[
\frac{L}{U^2} \frac{\partial u}{\partial t} = \frac{\partial (u/U)}{\partial (t U / L)} = \frac{\partial \overline{u}}{\partial \overline{t}}
\]
\[
\frac{L}{U^2} u\frac{\partial u}{\partial x} = \frac{u}{U} \frac{\partial (u/U)}{\partial (x/L)} = \overline{u} \frac{\partial \overline{u}}{\partial \overline{x}}
\]
\[
\frac{L}{U^2} v\frac{\partial u}{\partial y} = \frac{v}{U}\sqrt{Re} \frac{\partial (u/U)}{\partial (\frac{y}{L}\sqrt{Re})} = \overline{v} \frac{\partial \overline{u}}{\partial \overline{y}}
\]
\[
\frac{L}{U^2} \nu\frac{\partial^2 u}{\partial x \partial x} = \frac{\nu}{UL} \frac{\partial^2 (u/U)}{\partial (x/L) \partial (x/L)} = \frac{1}{Re} \frac{\partial^2 \overline{u}}{\partial \overline{x}^2}
\]
\[
\frac{L}{U^2} \nu \frac{\partial^2 u}{\partial y \partial y} = \frac{\partial^2 (u/U)}{\partial (\frac{y}{L} \sqrt{Re}) \partial (\frac{y}{L} \sqrt{Re})} = \frac{\partial^2 \overline{u}}{\partial \overline{y}^2}
\]
\[
\frac{L}{U^2} \frac{1}{\rho}\frac{\partial p}{\partial x} = \frac{\partial \frac{p}{\rho U^2}}{\partial x/L} = \frac{\partial \overline{p}}{\partial \overline{x}}
\]
The second derivative of \(\overline{u}\) in the \(x\)-direction is neglected and we end up with
\[
\frac{\partial \overline{u}}{\partial \overline{t}} + \overline{u} \frac{\partial \overline{u}}{\partial \overline{x}} +\overline{v} \frac{\partial \overline{u}}{\partial \overline{y}} = \frac{\partial^2 \overline{u}}{\partial \overline{y}^2} - \frac{\partial \overline{p}}{\partial \overline{x}}
\]
In a similar manner it is trivial to show that the continuity equation is unchanged through normalization
\[
\frac{\partial \overline{u}}{\partial \overline{x}} + \frac{\partial \overline{v}}{\partial \overline{y}} = 0.
\]
whereas the momentum equation in the wall normal direction is found from
\[
\frac{L}{U^2} \frac{\partial v}{\partial t} = \frac{1}{\sqrt{Re}} \frac{\partial \overline{v}}{\partial \overline{t}},
\]
\[
\frac{L}{U^2} u\frac{\partial v}{\partial x} = \frac{\overline{u}}{\sqrt{Re}} \frac{\partial \overline{v}}{\partial \overline{x}},
\]
\[
\frac{L}{U^2} v\frac{\partial v}{\partial y} = \frac{\overline{v}}{\sqrt{Re}} \frac{\partial \overline{v}}{\partial \overline{y}},
\]
\[
\frac{L}{U^2} \nu\frac{\partial^2 v}{\partial x \partial x} = \frac{1}{Re^{\frac{3}{2}}} \frac{\partial^2 \overline{v}}{\partial \overline{x}^2},
\]
\[
\frac{L}{U^2} \nu\frac{\partial^2 v}{\partial y \partial y} = \frac{1}{\sqrt{Re}} \frac{\partial^2 \overline{v}}{\partial \overline{y}^2},
\]
\[
\frac{L}{U^2} \frac{1}{\rho}\frac{\partial p}{\partial y} = \sqrt{Re} \frac{\partial \frac{p}{\rho U^2}}{\partial (\frac{y}{L} \sqrt{Re})} = \sqrt{Re} \frac{\partial \overline{p}}{\partial \overline{y}}.
\]
Dividing all terms by \(\sqrt{Re}\) the only term left is \(\partial \overline{p} / \partial \overline{y}\) and thus naturally from the momentum equation in the wall normal direction we obtain
\[
\frac{\partial \overline{p}}{\partial \overline{y}} = 0.
\]
Note that there is an error in White Eq. (4-32), where the temperature is also accounted for. In Eq. (4-32) the temperature term should be divided by \(\sqrt{Re}\).
