Re: Formule con allineamenti particolari

#16504
DMW
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    A volte le cose sono più semplici di quanto si pensi.:D 😀
    Direi che così ti puoi accontentare senza porsi troppi problemi (non c’è nemmeno bisogno del pacchetto mathtools):
    `
    \begin{gather}
    \makebox[.8\linewidth]{\hfill$\forall t \in \mathopen{(}0,T\mathclose{]} \quad \mathrm{trovare} \quad
    (\stackrel{\circ}{\Ub}\!\!(t),P(t)) \in \Vb_\zerob \times Q \,
    :$}\nonumber\\[2mm]
    \begin{cases}
    \dsfrac{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)\\[2mm]
    \qquad\qquad+c^{\rho}(\Rb_{\Ub_D},\stackrel{\circ}{\Ub}\!\!(t),\vb)+\dsfrac{1}{\mathrm{Re}}a^{\mu}
    (\stackrel{\circ}{\Ub}\!\!(t),\vb)+b(\vb,P(t))\\[4mm]
    \hspace{4cm}=\mathrm{G}\mathcal{F}^{\rho}(\vb)+\mathcal{F}^{R_1}(\vb)\quad \forall \vb \in \Vb_\zerob,\\[2mm]
    b(\stackrel{\circ}{\Ub}\!\!(t),q)=\mathcal{F}^{R_2}(q)\qquad\forall q\in Q,\\[2mm]
    \stackrel{\circ}{\Ub}\!\!(0)=\Ub_0,
    \end{cases}
    \end{gather}
    \end{gather}
    `

    Ciao

    Andrea

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