Avete qualche idea su come sistemarlo?
No, se non ci invii un esempio minimale compilabile.
Ho provato a compilare il tuo esempio:
`\documentclass{article}
\usepackage[latin1]{inputenc}
\usepackage[italian]{babel}
%\usepackage{textcomp}
\usepackage[T1]{fontenc}
\usepackage{type1ec}%per i font cmsuper
\usepackage[pdftex]{graphicx}
\usepackage{epstopdf}
\usepackage{fancyhdr}
\usepackage{color}
\usepackage{hyperref}
\usepackage{makeidx}
\usepackage{amssymb}
\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{theorem}
\usepackage{lscape}
\usepackage{glossary}
\usepackage{array}
\usepackage{xtab}
\usepackage{eufrak} % Per fare la F particolare!
\usepackage{tocloft}
\newcommand{\Vb}{{\mib V}}
\newcommand{\Ub}{{\mib U}}
\newcommand{\vb}{{\mib v}}
\newcommand{\zerob}{{\mib 0}}
\newcommand{\Rb}{\mib R}
\begin{document}
\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]
\left\{ \begin{array}{rl} \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)\\[4mm]
+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]
= \mathrm{G}\mathcal{F}^{\rho}(\vb)+\mathcal{F}^{R_1}(\vb) & \forall \vb \in \Vb_\zerob,\\[4mm]
b(\stackrel{\circ}{\Ub}\!\!(t),q) = \mathcal{F}^{R_2}(q) & \forall q \in Q,\\[4mm]
\stackrel{\circ}{\Ub}\!\!(0) = \Ub_0, &
\end{array} \right. \end{array}\label{cap2eq12}
\end{equation}
\end{document}`
ma c’è un errore:
`! Undefined control sequence.
\Ub ->{\mib
U}
l.36 (\stackrel{\circ}{\Ub}
\!\!(t),P(t)) \in \Vb_\zerob \times Q \,
?
`
L.