solve iteratively

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

. Using the Green function

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

iteratively. Nara-san says that a few of iterations are enough to converge

kino 平成18年4月17日