next up previous
次へ: solve directly 上へ: the case of 戻る: the case of

solve iteratively


\begin{displaymath}
\frac{d^2}{dz^2} V_H^0(z) = -4 \pi \rho^0 (z)
\end{displaymath} (210)

where \( V_H^0(z)=V_H(G_\parallel =0,z) \) and \( \rho^0 (z) = \rho(G_\parallel =0,z) \). The above equation is equivalent to
\begin{displaymath}
\left(\frac{d^2}{dz^2} -K^2 \right) V_H^0(z) = -4 \pi \rho^0 (z)
-K^2 V_H^0(z)
\end{displaymath} (211)

with arbitrary $K$. Using the Green function $G(K,z,z')$ with the boundary condition \( G(K,z=z_0,z')=0 \) and \( G(K,z=z_{l+1},z')=0 \) (same as eq.(201))
$\displaystyle V_H^0(z)$ $\textstyle =$ $\displaystyle \int_{z_0}^{z_{l+1}}dz' G(K,z,z')
\left( 4 \pi \rho^0 (z') + K^2 ...
...right)
- \left[
V_{H}(z') \partial_{z'} G(K,z,z') \right]^{z'=z_{l+1}}_{z'=z_0}$ (212)

It seems that eq.(213) is especially stable to solve $ V_H^0(z)$ iteratively. Nara-san says that a few of iterations are enough to converge $V_H^0$.



kino 平成18年4月17日