solve directly

An alternative way to solve eq. (211) (Kobayashi's method)

$$\frac{d}{dz} V_H^0(z) = -4 \pi \int_{z_0}^{z} dz' \rho^0 (z') + a$$ (213)

$$V_H^0(z) = -4 \pi \int_{z_0}^{z} dz'' \int_{z_0}^{z''} dz' \rho^0 (z') + a (z-z_0) + b$$ (214)

$$= \triangle _2(z) + a (z-z_0) + b$$ (215)

where $a$ and $b$ are some constants. They can be determined from the he boundary conditions

$$V_H^0(z_0) = \triangle _2(z_0) + b = b$$ (216)

$$V_H^0(z_{l+1}) = \triangle _2(z_{l+1}) + a(z_{l+1}-z_0) + b$$ (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.