L'equazione di Blasius

Sostituendo le espressioni trovate nella equazione principale:

$$\ensuremath{\frac{\partial \Psi}{\partial y}}\ \ensuremath{\frac{... ...{\partial^2 \Psi}{\partial y^2}}\ = \ \nu\ \frac{\partial^3\Psi}{\partial y^3}$$

$$U \ \ensuremath{\frac{d\,f}{d\,\eta}} \left( - \frac{1}{2 x} \ U ... ... \ \sqrt{\frac{U}{\nu \ x}}= \nu \ \frac{U^2}{\nu \ x} \ \frac{d^3f}{d\eta^3}$$

$$- \frac{U^2}{2 \ x} \ \ensuremath{\frac{d\,f}{d\,\eta}} \ \ensure... ...ac{d^2\, f}{d\, \eta^2}} = \ \nu \ \frac{U^2}{\nu \ x} \ \frac{d^3f}{d\eta^3}$$

$$- \ \frac{U^2}{2 \ x} \ f \ \ensuremath{\frac{d^2\, f}{d\, \eta^2}} = \ \nu \ \frac{U^2}{\nu \ x} \ \frac{d^3f}{d\eta^3}$$

$$- \ \frac{1}{2} \ f \ \ensuremath{\frac{d^2\, f}{d\, \eta^2}} \ = \ \frac{d^3f}{d\eta^3}$$

Si ottiene finalmente l'equazione di Blasius:

$$2 \ \frac{d^3f}{d\eta^3} \ + \ f \ \ensuremath{\frac{d^2\, f}{d\, \eta^2}} \ = \ 0$$

$$\boxed{ 2 \ f^{'''} \ + \ f \ f^{''} \ = \ 0 }$$ (7.2)

con le condizioni al contorno:

   c.c.$$\left\{ \begin{array}{lclcll} \displaystyle f(\eta=0) & = & ... ...ow 1}&=& 1 & \mbox{Velocit\\lq a indisturbata} \end{array}\right.$$