Suggested derivation of Blasius equationΒΆ
The Blasius solution is derived from the boundary layer equations using a similarity variable
\[
\eta(x, y) = y \sqrt{\frac{U}{2 \nu x}}.
\]
The pressure gradient is zero and the stream function and the velocity components are assumed given at a constant \(x\)-value as
\[
\psi = \sqrt{2 \nu U x} f(\eta), \quad u = U f^{\prime}, \quad v = \sqrt{\frac{\nu U}{2 x}} \left(\eta f^{\prime} - f \right).
\]
The derivatives of \(\eta\) are
\begin{align*} \frac{\partial \eta}{\partial x} &= -\frac{y}{2}\sqrt{\frac{U}{2 \nu}} x^{-\frac{3}{2}} = -\frac{\eta}{2 x}, \ \frac{\partial \eta}{\partial y} &= \sqrt{\frac{U}{2 \nu x}}. \end{align*}
These are then used in the partial derivatives of the velocity components
\[
\frac{\partial u}{\partial x} = \frac{\partial u}{\partial \eta} \frac{\partial \eta}{\partial x} = -U f^{\prime\prime}\frac{\eta}{2x}
\]
\[
\frac{\partial u}{\partial y} = \frac{\partial u}{\partial \eta} \frac{\partial \eta}{\partial y} = U f^{\prime\prime}\sqrt{\frac{U}{2 \nu x}}
\]
\[
\frac{\partial^2 u}{\partial y^2} = U f^{\prime\prime\prime} \frac{U}{2 \nu x}
\]
Inserting into the momentum equation
\[
U f^{\prime} (-1)U f^{\prime\prime}\frac{\eta}{2x} + \sqrt{\frac{\nu U}{2 x}} \left(\eta f^{\prime} - f \right) U f^{\prime\prime}\sqrt{\frac{U}{2 \nu x}} = \nu U f^{\prime\prime\prime} \frac{U}{2 \nu x}
\]
Multiply by \(2x/U^2\) and rearrange
\[\begin{split}
\begin{aligned}
-f^{\prime} f^{\prime\prime} \eta + \left(\eta f^{\prime} - f \right) f^{\prime\prime} = f^{\prime\prime\prime}, \\
f^{\prime\prime\prime} + f f^{\prime\prime} = 0
\end{aligned}
\end{split}\]