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Derivata sostanziale o totale di una funzione vettoriale

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

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

Per semplicità si può procedere con una componente:

$\displaystyle \ensuremath{\frac{d\,u}{d\,t}}$ $\displaystyle =$ $\displaystyle \ensuremath{\frac{\partial u}{\partial t}}+\ensuremath{\frac{\par...
...d\,t}}+\ensuremath{\frac{\partial u}{\partial z}}\ensuremath{\frac{d\,z}{d\,t}}$  
  $\displaystyle =$ $\displaystyle \ensuremath{\frac{\partial u}{\partial t}}+\ensuremath{\frac{\par...
...ath{\frac{\partial u}{\partial y}}v+\ensuremath{\frac{\partial u}{\partial z}}w$  
  $\displaystyle =$ $\displaystyle \ensuremath{\frac{\partial u}{\partial t}}+\vec{V}\nabla u\footnotemark $   2.7

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

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

in cui l'operatore:

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

o anche:

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

Una ulterire manipolazione permette di mettere in risalto le componenti di rotazione del moto.
$\displaystyle \ensuremath{\frac{d\,u}{d\,t}}$ $\displaystyle =$ $\displaystyle \ensuremath{\frac{\partial u}{\partial t}}+\ensuremath{\frac{\par...
...ath{\frac{\partial u}{\partial y}}v+\ensuremath{\frac{\partial u}{\partial z}}w$  
  $\displaystyle =$ $\displaystyle \ensuremath{\frac{\partial u}{\partial t}}+\ensuremath{\frac{\par...
...rac{\partial u}{\partial y}}- \ensuremath{\frac{\partial v}{\partial x}}\right)$  
  $\displaystyle =$ $\displaystyle \ensuremath{\frac{\partial u}{\partial t}}+\ensuremath{\frac{\par...
...rac{\partial u}{\partial y}}- \ensuremath{\frac{\partial v}{\partial x}}\right)$  
  $\displaystyle =$ $\displaystyle \ensuremath{\frac{\partial u}{\partial t}}+\ensuremath{\frac{\par...
...rac{\partial u}{\partial y}}- \ensuremath{\frac{\partial v}{\partial x}}\right)$  


$\displaystyle \ensuremath{\frac{d\,u}{d\,t}}$ $\displaystyle =$ $\displaystyle \ensuremath{\frac{\partial u}{\partial t}}+\frac{1}{2}\ensuremath...
... v}{\partial x}}- \ensuremath{\frac{\partial u}{\partial y}}\right)}^{\omega_3}$  
$\displaystyle \ensuremath{\frac{d\,v}{d\,t}}$ $\displaystyle =$ $\displaystyle \ensuremath{\frac{\partial v}{\partial t}}+\frac{1}{2}\ensuremath...
... v}{\partial x}}- \ensuremath{\frac{\partial u}{\partial y}}\right)}^{\omega_3}$  
$\displaystyle \ensuremath{\frac{d\,w}{d\,t}}$ $\displaystyle =$ $\displaystyle \ensuremath{\frac{\partial w}{\partial t}}+\frac{1}{2}\ensuremath...
...w}{\partial x}}- \ensuremath{\frac{\partial u}{\partial z}}\right)}^{-\omega_2}$  

Al secondo membro delle equazioni di sopra si riconosce la presenza di termini del rotore del vettore $ \vec{V}$ :
\begin{displaymath}\left\vert
\begin{array}{ccc}
\vec{\imath} & \vec{\jmath} & \...
...rac{\partial }{\partial z}}\\
u & v & w
\end{array}\right\vert\end{displaymath} $\displaystyle =$ $\displaystyle \vec{\imath}\left( \ensuremath{\frac{\partial w}{\partial y}}-\en...
...frac{\partial v}{\partial x}}-\ensuremath{\frac{\partial u}{\partial y}}\right)$  
  $\displaystyle =$ $\displaystyle \vec{\imath}\ \omega_1\ + \vec{\jmath}\ \omega_2+ \vec{k}\ \omega_3$  
  $\displaystyle =$ $\displaystyle \nabla \wedge \vec{V}$  
  $\displaystyle =$ $\displaystyle \vec{\omega}$  


$\displaystyle \ensuremath{\frac{d\,u}{d\,t}}$ $\displaystyle =$ $\displaystyle \ensuremath{\frac{\partial u}{\partial t}}+\frac{1}{2}\ensuremath{\frac{\partial (u^2+v^2+w^2)}{\partial x}}
+\ w\ \omega_2 \ -\ v\ \omega_3$  
$\displaystyle \ensuremath{\frac{d\,v}{d\,t}}$ $\displaystyle =$ $\displaystyle \ensuremath{\frac{\partial v}{\partial t}}+\frac{1}{2}\ensuremath{\frac{\partial (u^2+v^2+w^2)}{\partial y}}-\
w\ \omega_1\ +\ u\ \omega_3$  
$\displaystyle \ensuremath{\frac{d\,w}{d\,t}}$ $\displaystyle =$ $\displaystyle \ensuremath{\frac{\partial w}{\partial t}}+\frac{1}{2}\ensuremath{\frac{\partial (u^2+v^2+w^2)}{\partial z}}+
v\ \omega_1 \ -\ u\ \omega_2$  

A questo punto sono evidenti al secondo membro delle equazioni sopra i termini del prodotto vettoriale $ \vec{\omega}\wedge\vec{V}$ :
$\displaystyle \vec{\omega}\wedge\vec{V}$ $\displaystyle =$ \begin{displaymath}\left\vert
\begin{array}{ccc}
\vec{\imath} & \vec{\jmath} & \...
...mega_1& \omega_2 & \omega_3\\
u & v & w
\end{array}\right\vert\end{displaymath}  
  $\displaystyle =$ $\displaystyle \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:
$\displaystyle \ensuremath{\frac{d\,u}{d\,t}}$ $\displaystyle =$ $\displaystyle \ensuremath{\frac{\partial u}{\partial t}}+\frac{1}{2}\ensuremath{\frac{\partial (V^2)}{\partial x}}
+(\vec{\omega}\wedge\vec{V})_x$  
$\displaystyle \ensuremath{\frac{d\,v}{d\,t}}$ $\displaystyle =$ $\displaystyle \ensuremath{\frac{\partial v}{\partial t}}+\frac{1}{2}\ensuremath{\frac{\partial (V^2)}{\partial y}}-\
(\vec{\omega}\wedge\vec{V})_y$  
$\displaystyle \ensuremath{\frac{d\,w}{d\,t}}$ $\displaystyle =$ $\displaystyle \ensuremath{\frac{\partial w}{\partial t}}+\frac{1}{2}\ensuremath{\frac{\partial (V^2)}{\partial z}}+
(\vec{\omega}\wedge\vec{V})_z$  

e in definitiva:

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

La Eq. (2.15) permette allora la scrittura della Eq. (2.13).
next up previous contents
Next: Cinematica dei fluidi: relazioni Up: Forma differenziale lagrangiana delle Previous: Forma differenziale lagrangiana delle   Indice
2009-01-26