action space. There will be two types of such integrals (instead of three) if one uses the trial functions defined in (\[eq:trial-functions-two\]). Let me introduce the auxiliary integral $${\cal I}^{ml}_{jk}=
\int  d \textbf{\textit{J}} \left ( l{\partial f_0\over \partial J_R}+
m{\partial f_0\over \partial J_\phi} \right )
\Psi^{ml}_{j}(\textbf{\textit{J}})\Psi^{ml}_{k}(\textbf{\textit{J}}),$$ and apply the trial functions $\Phi^{ml}_{j}=\varrho^{ml}_0\Psi^{ml}_{j}$ in (\[eq:A-tensor\]). The elements of $\textbf{\textit{M}}$ and $\textbf{\textit{C}}$ are thus computed from $$\begin{aligned}
M_{pq} &=& \delta_{l,l'}\Lambda^{ml}_{jj'}, \label{eq:bar-M-matrix} \\
C_{pq} &=& \delta_{l,l'}{\cal I}^{ml}_{jj'}-
\sum_{k=0}^{j_{\rm max}} \left [ {4\pi^2\over D_{k}(m)}\right ]
{\cal I}^{ml}_{jk}\Lambda^{ml'}_{kj'}. \label{eq:bar-A-matrix}\end{aligned}$$ Both $\Lambda^{ml}_{jk}$ and ${\cal I}^{ml}_{jk}$ consist of boundary integrals when the unperturbed stellar disk is unidirectional with the DF $f_0(\textbf{\textit{J}})=H(J_\phi)f^P_0(\textbf{\textit{J}})$. Here $H$ is the Heaviside function. The boundary terms are $$\begin{aligned}
\tilde \Lambda^{ml}_{jk} &=& \int_{0}^{\infty} d J_R\left [
{mf^P_0(\textbf{\textit{J}})
\Psi^{ml}_{j}(\textbf{\textit{J}})\Psi^{ml}_{k}(\textbf{\textit{J}}) \over
l\Omega_R(\textbf{\textit{J}})+m\Omega_{\phi}(\textbf{\textit{J}}) }
\right ]_{J_{\phi}=0}, \\
%
{\tilde {\cal I}}^{ml}_{jk} &=& \int_{0}^{\infty} d J_R\left [
mf^P_0(\textbf{\textit{J}})
\Psi^{ml}_{j}(\textbf{\textit{J}})\Psi^{ml}_{k}(\textbf{\textit{J}}) \right
]_{J_{\phi}=0}.\end{aligned}$$ Dynamics of modes with different azimuthal wavenumbers are decoupled in the linear regime and the matrix $\textbf{\textit{A}}$ is an odd function of the wavenumber $m$. i.e., $\textbf{\textit{A}}(-m)=-\textbf{\textit{A}}(m)$. An immediate result of this property is $a^{-m}_j(t)=\overline {a^m_j}(t)$. Consequently, $L_m(t)$ becomes equal to zero for all $|m|\ge 0$ and each mode individually conserves the total angular momentum.

The present 