\[fig:fig1\]), as required by the boundary conditions Eq. (\[boundcond\]). For this reason we only consider the case of $n$ odd in the sequel.

According to the general procedure outlined in Sec.\[sec3\], the final step in the construction of an unconditionally stable algorithm is to decompose $H$. Guided by previous work on Schrödinger and diffusion problems [@DeRaedt87; @DeRaedt94; @Michielsen98; @DeRaedt00], we split $H$ into two parts, i.e. $H=H_{1}+H_{2}$, where $$\begin{aligned}
H_{1} &=&
\mathop{{\sum}'}_{i=1}^{n}
\beta_{i+1,i}\left(
\be^{{\phantom{T}}}_{i}\be^T_{i+1}-
\be^{{\phantom{T}}}_{i+1}\be^T_{i}
\right),
\label{h1}
\\
H_{2} &=&
\mathop{{\sum}'}_{i=1}^{n}
\beta_{i+1,i+2}\left(
\be^{{\phantom{T}}}_{i+1}\be^T_{i+2}-
\be^{{\phantom{T}}}_{i+2}\be^T_{i+1}
\right),
\label{h2}\end{aligned}$$ i.e. we divide the lattice into odd and even numbered cells. In matrix notation we have $$H_{1} =
\left( \begin{array}{cccccccc}
0         & \beta_{2,1} &       &        &        &           &        & \\
-\beta_{2,1} & 0      & 0       &        &        &           &        & \\
        & 0      & 0       & \beta_{4,3} &        &           &        & \\
        &        & -\beta_{4,3} & 0      & 0      &           &        & \\
        &        &         & 0           & \ddots & 0         &        & \\
        &        &         &        & 0  & 0         & \beta_{n-1,n-2} & \\
        &        &         &        &    & -\beta_{n-1,n-2}   & 0      & 0 \\
        &        &         &        &    &                    & 0      & 0
\end{array} \right)
\label{h1m}$$ and $$H_{2} =
\left( \begin{array}{cccccccc}
0 