Derivata sostanziale o totale di una funzione vettoriale

Ci si propone di trovare una espressione della derivata totale lagrangiana di una funzione vettoriale.

$$\ensuremath{\frac{d\,\vec{V}}{d\,t}}\ =\ \cdots$$

Per semplicità si può procedere con una componente:

$$\ensuremath{\frac{d\,u}{d\,t}} = \ensuremath{\frac{\partial u}{\partial t}}+\ensuremath{\frac{\par... ...d\,t}}+\ensuremath{\frac{\partial u}{\partial z}}\ensuremath{\frac{d\,z}{d\,t}}$$

$$= \ensuremath{\frac{\partial u}{\partial t}}+\ensuremath{\frac{\par... ...ath{\frac{\partial u}{\partial y}}v+\ensuremath{\frac{\partial u}{\partial z}}w$$

$$= \ensuremath{\frac{\partial u}{\partial t}}+\vec{V}\nabla u\footnotemark$$   ^2.7

In forma vettoriale, considerando analoghe relazioni per le rimanenti componenti:

$$\ensuremath{\frac{d\,\vec{V}}{d\,t}}=\ensuremath{\frac{\partial \vec{V}}{\partial t}}+(\vec{V}\nabla)\vec{V}$$ (2.14)

in cui l'operatore:

$$(\vec{V}\nabla)(\cdots)=\ensuremath{\frac{\partial (\cdots)}{\par... ...ial (\cdots)}{\partial y}}v+\ensuremath{\frac{\partial (\cdots)}{\partial z}}w$$

o anche:

$$(\vec{V}\nabla)(\cdots)=\vec{V} \$$   grad$$\ \vec{V}$$

Una ulterire manipolazione permette di mettere in risalto le componenti di rotazione del moto.

$$\ensuremath{\frac{d\,u}{d\,t}} = \ensuremath{\frac{\partial u}{\partial t}}+\ensuremath{\frac{\par... ...ath{\frac{\partial u}{\partial y}}v+\ensuremath{\frac{\partial u}{\partial z}}w$$

$$= \ensuremath{\frac{\partial u}{\partial t}}+\ensuremath{\frac{\par... ...rac{\partial u}{\partial y}}- \ensuremath{\frac{\partial v}{\partial x}}\right)$$

$$= \ensuremath{\frac{\partial u}{\partial t}}+\ensuremath{\frac{\par... ...rac{\partial u}{\partial y}}- \ensuremath{\frac{\partial v}{\partial x}}\right)$$

$$= \ensuremath{\frac{\partial u}{\partial t}}+\ensuremath{\frac{\par... ...rac{\partial u}{\partial y}}- \ensuremath{\frac{\partial v}{\partial x}}\right)$$

$$\ensuremath{\frac{d\,u}{d\,t}} = \ensuremath{\frac{\partial u}{\partial t}}+\frac{1}{2}\ensuremath... ... v}{\partial x}}- \ensuremath{\frac{\partial u}{\partial y}}\right)}^{\omega_3}$$

$$\ensuremath{\frac{d\,v}{d\,t}} = \ensuremath{\frac{\partial v}{\partial t}}+\frac{1}{2}\ensuremath... ... v}{\partial x}}- \ensuremath{\frac{\partial u}{\partial y}}\right)}^{\omega_3}$$

$$\ensuremath{\frac{d\,w}{d\,t}} = \ensuremath{\frac{\partial w}{\partial t}}+\frac{1}{2}\ensuremath... ...w}{\partial x}}- \ensuremath{\frac{\partial u}{\partial z}}\right)}^{-\omega_2}$$

Al secondo membro delle equazioni di sopra si riconosce la presenza di termini del rotore del vettore $\vec{V}$ :

$$\left\vert \begin{array}{ccc} \vec{\imath} & \vec{\jmath} & \... ...rac{\partial }{\partial z}}\\ u & v & w \end{array}\right\vert = \vec{\imath}\left( \ensuremath{\frac{\partial w}{\partial y}}-\en... ...frac{\partial v}{\partial x}}-\ensuremath{\frac{\partial u}{\partial y}}\right)$$

$$= \vec{\imath}\ \omega_1\ + \vec{\jmath}\ \omega_2+ \vec{k}\ \omega_3$$

$$= \nabla \wedge \vec{V}$$

$$= \vec{\omega}$$

$$\ensuremath{\frac{d\,u}{d\,t}} = \ensuremath{\frac{\partial u}{\partial t}}+\frac{1}{2}\ensuremath{\frac{\partial (u^2+v^2+w^2)}{\partial x}} +\ w\ \omega_2 \ -\ v\ \omega_3$$

$$\ensuremath{\frac{d\,v}{d\,t}} = \ensuremath{\frac{\partial v}{\partial t}}+\frac{1}{2}\ensuremath{\frac{\partial (u^2+v^2+w^2)}{\partial y}}-\ w\ \omega_1\ +\ u\ \omega_3$$

$$\ensuremath{\frac{d\,w}{d\,t}} = \ensuremath{\frac{\partial w}{\partial t}}+\frac{1}{2}\ensuremath{\frac{\partial (u^2+v^2+w^2)}{\partial z}}+ v\ \omega_1 \ -\ u\ \omega_2$$

A questo punto sono evidenti al secondo membro delle equazioni sopra i termini del prodotto vettoriale $\vec{\omega}\wedge\vec{V}$ :

$$\vec{\omega}\wedge\vec{V} = \left\vert \begin{array}{ccc} \vec{\imath} & \vec{\jmath} & \... ...mega_1& \omega_2 & \omega_3\\ u & v & w \end{array}\right\vert$$

$$= \vec{\imath}(\omega_2\ w\ - \omega_3\ v)\ -\ \vec{\jmath}(\omega_1\ w\ - \omega_3\ u)\ +\ \vec{k}(\omega_1\ v\ - \omega_2\ u)\$$

Si può riscrivere:

$$\ensuremath{\frac{d\,u}{d\,t}} = \ensuremath{\frac{\partial u}{\partial t}}+\frac{1}{2}\ensuremath{\frac{\partial (V^2)}{\partial x}} +(\vec{\omega}\wedge\vec{V})_x$$

$$\ensuremath{\frac{d\,v}{d\,t}} = \ensuremath{\frac{\partial v}{\partial t}}+\frac{1}{2}\ensuremath{\frac{\partial (V^2)}{\partial y}}-\ (\vec{\omega}\wedge\vec{V})_y$$

$$\ensuremath{\frac{d\,w}{d\,t}} = \ensuremath{\frac{\partial w}{\partial t}}+\frac{1}{2}\ensuremath{\frac{\partial (V^2)}{\partial z}}+ (\vec{\omega}\wedge\vec{V})_z$$

e in definitiva:

$$\ensuremath{\frac{d\,\vec{V}}{d\,t}}\ =\ \ensuremath{\frac{\parti... ...al t}}\ +\ \frac{1}{2} \nabla(\vec{V}\ \vec{V})\ + \ \vec{\omega}\wedge \vec{V}$$ (2.15)

La Eq. (2.15) permette allora la scrittura della Eq. (2.13).