\nonumber \\
&{}& - m\left [ a^{m}_{j}(t)
+\overline{a^{(-m)}_{j}}(t) \right ] \Lambda^{ml}_{jp}
\overline{d^{ml}_{p}}(t) \biggr \}. \label{eq:rate-change-L-two}\end{aligned}$$ Define $a^m_j(t)=u^m_j(t)+{\rm i}v^m_j(t)$ with $u^m_j(t)$ and $v^m_j(t)$ being real functions of time. According to identity (\[eq:a-versus-d\]), one may further simplify equation (\[eq:rate-change-L-two\]) to $$\begin{aligned}
{ d {\cal L}\over  d t} &=&
\sum_{m=-\infty}^{\infty} L_m(t),
\label{eq:rate-change-L-three} \\
L_m(t) &=& -\frac {m}{2}\sum_{j=0}^{\infty}
D_j(m) \nonumber \\
&{}& \times \left [ u^m_j(t)v^{-m}_j(t)+u^{-m}_j(t)v^m_j(t) \right ].
\label{eq:rate-change-L-four}\end{aligned}$$ The share of the $m$th mode from $ d {\cal L}/ d t$ is thus determined by $L_m(t)$. As one could anticipate for an isolated stellar disk, $ d {\cal L}/ d t$ vanishes and the total angular momentum remains constant because the terms $L_m(t)$ and $L_{-m}(t)$ cancel each other in (\[eq:rate-change-L-three\]), and $L_0(t)$ is annulled by the factor $m$ in (\[eq:rate-change-L-four\]).

Trial and Weighting Functions {#sec::trial-and-test-fncs}
-----------------------------

Choosing the trial functions $\Phi^{ml}_j(\textbf{\textit{J}})$ is the most delicate step in the reduction of the CBE to a system of ODEs. One possible way is to set $\partial f_1/\partial t=0$ in (\[eq:CBE-perturbed\]) and solve the first order equation $$\left [f_1,{\cal H}_0 \right ]+\left [f_0,{\cal H}_1 \right ]=0,
\label{eq:CBE-linear-equilibrium}$$ for $f_1$. Substituting (\[eq:expansion-f1\]) and (\[eq:expansion-V1\]) in (\[eq:CBE-linear-equilibrium\]) gives $$\begin{aligned}
d^{ml}_j\Phi^{ml}_j(\textbf{\textit{J}}) &=& b^{ml}_j
\varrho^{ml}_{0}(\textbf{\textit{J}}) \Psi^{ml}_j(\textbf{\textit{J}}),
\label{eq:trial-functions-one} \\
\varrho^{ml}_{0}(\textbf{\textit{J}}) &=& 