add boundary condition without any charge

How to solve

$$\nabla^2 V_H(r) = -4 \pi \rho(r )$$ (215)

when $V_H(x,y,z=z_0)$ and $V_H(x,y,z=z_{l+1})$ are given.

split $V_H(r) = \phi(r) + \psi(r)$,

$$\nabla^2 \phi(r) = -4 \pi \rho(r )$$ (216)

$$\nabla^2 \psi(r) = 0$$ (217)

$$\nabla^2 ( \phi(r) + \psi(r) ) = -4 \pi \rho(r )$$ (218)

where $\phi(r)$ is the solution of the Poission equation with charge $\rho(r)$ without any boundary condition in the region from $z=z_1$ to $z=z_l$. $\psi(r)$ is the solution of the Poission equation without charge and with the boundary condition $V_H(x,y,z=z_0)$ and $V_H(x,y,z=z_{l+1})$ in the region from $z=z_0$ to $z=z_{l+1}$.

$\phi(G)$, the Fourier transform of $\phi(r)$, can be solved in the usual way for periodic boundary condition as

$$\phi(G) = 4\pi \rho(G)/G^2$$ (219)

$\psi(G_\parallel,z)$ can be solved analytically using the formula of the previous chapter with the condition $\rho=0$. Surely eq.(207) has such a form. $\psi(G_\parallel=0,z)$ can be also solved using the formula of the previous chapter with the condition $\rho^0=0$. The solution, eq(218), is simply,

$$\psi(r)= (\mu_R - \mu_L) /( z_{l+1} - z_0 )$$ (220)