# Boundary layer momentum equation

How does the author get that formula(d alpha formula) using the momentum equation . The book here is Mechanics and thermodynamics of propulsion by Philip Hill and Carl Peterson.

We have $$\frac{dP}{\rho} = -UdU$$ and $$\frac{dP}{\rho} = -udu$$. These equations can be re-organised as $$\frac{-dP}{\rho U} = dU, ~\frac{-dP}{\rho u} = du$$ Given, $$\alpha = \frac{u}{U}$$. Differentiate this relation on both sides, we get $$d\alpha = \frac{du}{U} - \frac{u}{U^2}dU$$ Now, substitute the expressions for $$du$$ and $$dU$$ in the above, we have $$\begin{eqnarray*} d\alpha &= &\frac{1}{U}\frac{-dP}{\rho u} - \frac{u}{U^2}\frac{-dP}{\rho U}\\ &= &-\frac{1}{U^2}\frac{U}{u}\frac{dP}{\rho} + \frac{1}{U^2} \frac{u}{U}\frac{dP}{\rho} \\ &= &-\frac{1}{U^2}\frac{1}{\alpha}\frac{dP}{\rho} + \frac{1}{U^2} \alpha\frac{dP}{\rho} \\ &= & \frac{dP}{\rho U^2}\left(\alpha - \frac{1}{\alpha}\right)\\ &= & \frac{dP}{\rho U^2}\left(\frac{\alpha^2 - 1}{\alpha}\right) \end{eqnarray*}$$