The full one-dimensional case

In many practical cases we can use the low-potential-assumption and it leads to realistic results. In addition, it is a simple equation and dependencies like the one on the salt concentration can easily be seen. In some cases, however, we have high potentials and we cannot linearize the Poisson-Boltzmann equation. Now we treat the general solution of the onedimensional Poisson-Boltzmann equation and drop the assumption of low potentials. It is convenient to solve the equation with the dimensionless potential y = ertp/kBT. Please do not mix this up with the spacial coordinate y In this section we use the symbol y for the 

Using sinhy = 1/2 (ey — e y). To solve the differential equation we multiply both sides by 2 dy/dx  

Ci is an integration constant. It is determined by the boundary conditions. At large distances the dimensionless potential y and its derivative dy/dx are zero. Since coshy = 1 for y = 0 this constant is Ci = -2k2. It follows that 

In front of the square root there is a minus sign because y has to decrease for a positive potential with increasing distance, i.e. y 0 = dy/dx 0. Now we remember the mathematical 

2 In (tanh ) = —2kx + 2C2 C2 is another integration constant. Written explicitly we get 

By using the dimensionless surface potential yo = y(x — 0) = etpo/keT we can determine the integration constant 

Both expressions are identical, which is easily seen when remembering the definition of the tanh function  

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