Boundary Conditions involving exponential functions of nodal unknowns

I am fairly new to Computational Engineering and I have mainly been exposed to using the Finite Difference Method to produce Linear Systems and solve them using Iterative Methods.

I am trying to solve a problem from the field of Electrochemistry, involving the Laplace equation and boundary conditions that link the unknown potential to the current density resulting from the electrochemical reactions at the boundaries (electrodes) , i.e. $$-\sigma \frac{\partial \phi}{\partial y} = \alpha e^{\beta\phi}$$ from which I derive that $$\phi_{i,j-1} = \phi_{i,j+1} - \frac{2\,Δy\,\alpha}{\sigma}e^{\beta\phi_{i,j}}$$ which provides the value of the potential at a fictitious node below the bottom boundary of a unit square computational domain.

The equation that I have arrived to using central differences on a 5-point stencil is the following: $$\frac{1}{Δx^2}\phi_{i-1,j} -2 \bigg(\frac{1}{Δx^2} + \frac{1}{Δy^2}\bigg)\phi_{i,j} - \frac{2\, \alpha}{Δy\, \sigma}e^{\beta \phi_{i,j}} + \frac{2}{Δy^2}\phi_{i,j+1} + \frac{1}{Δx^2}\phi_{i+1,j} = 0$$ but I am clueless on how to handle it, as this appears to lead to neither a linear system nor to a nonlinear system that I can solve using Newton's method.

Any suggestions from someone who has a wide perspective of the available options?

• At a quick glance, seems like something that could be handled by Newton's method if you had a good enough guess. Why do you doubt you can use that here? – spektr Mar 4 at 14:49
• I thought that Newton's method applies to nonlinear system where the values of the coefficient matrix are functions of the nodal unknowns, but this does not seem to be true here; the exponential term is not a coefficient for any nodal value. Now I'm thinking that maybe the answer is simple, and I can just move the exponential term to the right hand side and refresh its value based on the new solution after every iteration. – Gouyoku Mar 5 at 7:02
• You can use Newton’s method to handle any nonlinear systems of equations that are sufficiently smooth, which yours are. There should be no issue using it here. – spektr Mar 5 at 13:37