Se vuoi quel tipo di comportamento puoi modificare il tuo metodo che hai scritto all’inizio.
Prova così:
`
\begin{equation}
\begin{array}{r}
\forall t \in (0,T] \quad \mathrm{trovare} \quad
(\stackrel{\circ}{\Ub}\!\!(t),P(t)) \in \Vb_\zerob \times Q \,
:\\[4mm]
\newsavebox{\tempbox}
\sbox{\tempbox}{$\dfrac{d}{dt}(\rho\stackrel{\circ}{\Ub}\!\!(t),\vb)+c^{\rho}(\stackrel{\circ}{\Ub}\!\!(t),
\stackrel{\circ}{\Ub}\!\!(t),\vb)+c^{\rho}(\stackrel{\circ}{\Ub}\!\!(t),\Rb_{\Ub_D},\vb)$}
\left\{\begin{array}{rl}
\usebox{\tempbox}\\[4mm]
+c^{\rho}(\Rb_{\Ub_D},\stackrel{\circ}{\Ub}\!\!(t),\vb)+\dfrac{1}{\mathrm{Re}}a^{\mu}(\stackrel{\circ}{\Ub}\!\!(t),\vb)+b(\vb,P(t))\\[4mm]
=\mathrm{G}\mathcal{F}^{\rho}(\vb)+\mathcal{F}^{R_1}(\vb) & \forall \vb \in \Vb_\zerob,\\[4mm]
\makebox[\wd\tempbox]{$b(\stackrel{\circ}{\Ub}\!\!(t),q)=\mathcal{F}^{R_2}(q)$\hfill} & \forall q \in Q,\\[4mm]
\makebox[\wd\tempbox]{$\stackrel{\circ}{\Ub}\!\!(0) = \Ub_0,$\hfill}&
\end{array} \right. \end{array}\label{cap2eq12}
\end{equation}
`
Ciao
Andrea