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次へ: , another (easier) derivation 上へ: the case of 戻る: solve iteratively

solve directly

An alternative way to solve eq. (211) (Kobayashi's method)
$\displaystyle \frac{d}{dz} V_H^0(z)$ $\textstyle =$ $\displaystyle -4 \pi \int_{z_0}^{z} dz' \rho^0 (z') + a$ (213)
$\displaystyle V_H^0(z)$ $\textstyle =$ $\displaystyle -4 \pi \int_{z_0}^{z} dz'' \int_{z_0}^{z''} dz' \rho^0 (z') +
a (z-z_0) + b$ (214)
  $\textstyle =$ $\displaystyle \triangle _2(z) + a (z-z_0) + b$ (215)

where $a$ and $b$ are some constants. They can be determined from the he boundary conditions
$\displaystyle V_H^0(z_0)$ $\textstyle =$ $\displaystyle \triangle _2(z_0) + b = b$ (216)
$\displaystyle V_H^0(z_{l+1})$ $\textstyle =$ $\displaystyle \triangle _2(z_{l+1}) + a(z_{l+1}-z_0) + b$ (217)

Nara-san says that when the $z$-mesh is fine enough, the latter method with a simple integration scheme to evaluate $\triangle _2$ gives almost the same precision as the former method.



kino 平成18年4月17日