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Discrete memory function of the VACF within the AR model
Starting from the memory function equation of the VACF (Eq. 4.72), the first step towards a numerical computation of the memory function consists in discretizing Eq. 4.72
\begin{displaymath}
\frac{VACF(n+1)-VACF(n)}{\Delta t} = -\sum_{k=0}^{n-1}
\Delta t \xi(n-k)\psi(k),
\end{displaymath} (4.79)

Eq. (4.79) is now subjected to a one-sided z-transform. Using that
\begin{displaymath}
Z_{>}\left\{f(n+1)-f(n)\right\} =
zF_{>}(z) - zf(0)
\end{displaymath} (4.80)

for any discrete function $f(n)$ whose one-sided z-transform exists, one obtains from (4.79)
\begin{displaymath}
\Xi_{>}(z) = \frac{1}{\Delta t^2} \left(\frac{z}{\mathrm{VACF}_{>}(z)} + 1 -
z\right),
\end{displaymath} (4.81)

using that $VACF(0)=1$. The one-sided z-transform of an arbitrary discrete function $f(n)$ is defined as $F_{>}(z) = \sum_{n=0}^{\infty} f(n) z^{-n}$. Here it has been used that the one-sided z-transform of the discrete convolution integral is just the product $\Xi_{>}(z)\mathrm{VACF}_{>}(z)$. Inserting the definition of the one-sided z-transform for $\Xi_{>}(z)$ and $\mathrm{VACF}_{>}(z)$, this equation can be rewritten as
\begin{displaymath}
\sum_{j=0}^\infty  \xi(j)   z^{-j} = \frac{1}{\Delta t^2}...
...ight]   z^{-j}}
{ \sum_{j=0}^\infty   VACF(j)   z^{-j} }.
\end{displaymath} (4.82)

Note that the term proportional to z cancels out. The time dependent memory function is, in principle, obtained by comparing the coefficients of the series on the lhs and the rhs of Eq. (4.82). To construct a numerical method, we replace the series by polynomials of order N, where $t=N\Delta t$ defines the time window for the memory function to be computed. After this first step a polynomial division is performed on the rhs of Eq. (4.82), and after a subsequent multiplication of both sides with $z^{-N}$ one obtains the time dependent memory function, $\xi(j)$, by comparison of coefficients,
$\displaystyle \frac{z^{-N}}{\Delta t^2} \frac{\sum_{j=0}^N   \left[VACF(j) -
VACF(j+1)\right]   z^{N-j}} { \sum_{j=0}^N   VACF(j)  
z^{-j}}$ $\textstyle =$ $\displaystyle z^{-N} \left(\sum_{j=0}^N   c_j   z^{N-j} + R \right)$  
  $\textstyle =$ $\displaystyle \sum_{j=0}^N \xi(j)   z^{-j}.$ (4.83)

Within nMOLDYN $VACF(n)$ is replaced by the autocorrelation function calculated in the framework of the autoregressive model, $VACF^{(AR)}(n)$, as in Eqs. 4.75 and 4.76. The coefficients $c_j$ are obtained by polynomial division and $R$ is a rest which does not contain information on the memory function within the time interval $t\in [0,N\Delta t]$. The discrete memory function is therefore given by $\xi(j) = c_{N-j}$.

A remark concerning the discretization scheme (4.79) is in place here. The discrete convolution sum is effectively a first order approximation of the convolution integral. More sophisticated approximations could be used, but they would lead to less convenient expressions upon z-transformation. Correspondingly, we have chosen a first order approximation for the differentiation on the left-hand side of (4.72). In this way the first order (integro-)differential equation (4.72) is transformed into the first order difference equation (4.79).

However, this simple discretization scheme together with the use of the one-sided z-transform leads to a significant error in $\xi(0)$. It is clear from Eq. (4.72) that due to the symmetry of the autocorrelation function ( $\psi(t)=\psi(-t)$), the derivative $d\psi/dt$ should vanish at $t=0.$ However, in the discretized version it is approximated by a forward difference that is always negative. A higher-order calculation shows that the estimate for $\xi(0)$ that results from the procedure described above should be doubled.


next up previous contents
Next: MSD within the AR Up: Theory and implementation Previous: Density of states within   Contents
pellegrini eric 2009-10-06