# Solving Poisson-Boltzmann equation for flat surface without excess salt

I'm trying to understand the solution of Poisson-Boltzmann (PB) equation as I read it in a review. The surface is uniformly charged and flat. I am only considering the $x$ direction.

The PB equation under these conditions is $$\frac{\partial ^2 \psi}{\partial x^2} = -\frac{4 \pi}{\epsilon \epsilon_0} C_0 z q \, {\text e}^{-z q \psi/kT}$$ I set $$a = -\frac{4 \pi C_0 z q}{\epsilon \epsilon_0}; \qquad b = \frac{zq}{kT}.$$ Also, I have a boundary condition at $x=0$ $$\left . \frac{\partial \psi}{\partial x} \right |_{x=0} = - \frac{4 \pi}{\epsilon \epsilon_0} \sigma$$ where $\sigma$ is the surface charge density.

Just to be clear, $\epsilon$ and $\epsilon_0$ are the solvent and vacuum dielectric constants, $kT$ is the thermal energy, $z$ is the ion valency and q is the proton charge. Finally, $C_0$ is a reference concentration at which $\psi = 0$.

The solution, as published in the review paper (for monovalent ions) is $$\psi(x) = \frac{2 T}{q} \log(x+l_{gc}) + \psi_0$$ here $l_{gc} = \frac{q}{2 \pi \left | \sigma \right | \ell}$ is the Gouy-Chapman length and $\ell= \frac{q^2}{\epsilon \epsilon_0 kT}$ is the Bjerrum length. I tried to derive this result myself: $$\int d\psi \int {\text e}^{b \psi} d\psi = \int dx \int adx$$ $$\int \frac{ {\text e}^{b \psi} d\psi}{b} = \int \left ( ax + c \right ) dx$$ $$\frac { {\text e}^{b \psi} }{b^2} = \frac{ax^2}{2} + cx + d$$ I want to use my boundary condition: $$\psi = \log \left [ \left ( \frac{ax^2}{2} + cx + d \right )b^2\right ]/b$$ $$\left . \frac{\partial \psi}{\partial x} \right |_{x = 0} = \left . \frac{ax + c}{ax^2 + cx + d} \frac{1}{b} \right |_{x=0} = \frac{c}{db}$$ I can only find the ratio between $c$ and $d$ to be $\frac{16 \pi^2}{\epsilon^2 \epsilon_0^2 \sigma}$ and I can't really do anything else with this.

This is supposed to be a fundamental result for this system but I can't find any explanation to this result or derivation. I can find the solution for a system with excess salt thought.

So, how do I solve this system/where is the flow in my solution?

Edit:
The review I am referring to is David Andelman's chapter in the book structure and dynamics of membranes. The chapter title is: Electrostatic properties of membranes: The Poisson-Boltzmann equation

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You seem to have mixed your $x$s' and $z$s'. You say "..a boundary condition at $z=0$.." and then you say "...$z$ is the ion valency...". Can you clean up? Also can you please point to the review...? – Vijay Murthy May 1 '12 at 18:33
@VijayMurthy: Thank you for your remark. Nevertheless, the solution I have posted, and my approach are derived under the right notations. – Yotam May 2 '12 at 6:49
It still seems messed up..! $x$ seems to have disappeared from the solution $\psi$. Sorry. But could you please cite the review? – Vijay Murthy May 2 '12 at 8:53
OK. I give up..! The notation is not clear. $z$ is sometimes used as the ion valency and sometimes used as a space variable. And no reference to the review even after two requests. – Vijay Murthy May 2 '12 at 10:25
@VijayMurthy Sorry I didn't post the source of this. I wasn't in my office (where I have the book) until now. Reading my question I have noticed that I have switched my notepad while typing in. Sorry – Yotam May 2 '12 at 10:57

We want to solve the Poisson-Boltzmann (PB) equation for the case of a uniformly charged flat membrane of surface charge density $\sigma$ located at $z=0$. Also we want to consider the case of monovalent ions in the solution with no added electrolyte. We thus want to solve $$\psi''(z) = -\frac{4\pi e n_0}{\varepsilon_w} e^{-e\psi(z)/T} = -\alpha e^{-\beta \psi(z)}$$ in the half-space $z>0$ where $\alpha= 4\pi e n_0/\varepsilon_w$ and $\beta=e/T$ with the boundary condition $$\psi'(z) \big\vert_{z=0} = - \frac{4\pi}{\varepsilon_w} \sigma \,\, > 0.$$ Notice that this boundary condition implicitly assumes that the surface charge density is negative $\sigma<0$. This also means that the potential $\psi(z)$ will be an increasing function of $z$.
To solve non-linear equations of the type $$\psi''(z) = -\alpha e^{-\beta \psi(z)}$$ there is a general trick as outlined here. Using this trick, we arrive at $$\psi'(z) = \pm \sqrt{\frac{2\alpha}{\beta} e^{-\beta\psi(z)} + C}.$$
Now for some physical input. Since the potential increases with $z$, and we require that the electric field $\psi'(z)$ must vanish for $z\to\infty$, we get $C=0$. This also indicates that we must take the positive root for $\psi'(z)$ above. With this, we get $$\psi'(z) = \sqrt{\frac{2\alpha}{\beta}} e^{-\beta\psi(z)/2}$$ which is easily solved to give $$\psi(z) = \frac{2}{\beta} \ln(z+b) + \psi_0 = \frac{2T}{e} \ln(z+b) + \psi_0$$ where $$\psi_0 = \frac{1}{\beta} \ln \frac{2\alpha}{\beta} = \frac{T}{e} \ln \frac{8 \pi T n_0}{\varepsilon_w}.$$ By imposing the boundary condition, you will easily find the Gouy–Chapman length $$b=\frac{\varepsilon_w T}{2\pi |\sigma| e}.$$