Solving Laplace equation with nonlinear diffusity/conductivity

chriss85 · October 20, 2014, 07:40

I want to solve a Laplace equation, div(sigma * grad(phi)) = 0, where sigma depends nonlinearly on the solution of phi. What is the best approach for doing something like this? If I solve the equation iteratively and update sigma each time based on the current solution, what kind of conditions are required to guarantee convergence and unique solutions?
Is it possible (and beneficial) to perform this kind of iteration directly in the matrix solver?

marupio · October 20, 2014, 09:19

I'd use Newton-Raphson linearization. Semi-implicit. Take a phi to the power of one out of your sigma equation and put it into the matrix, and the rest goes into the source term as a derivative, I believe. Use relaxation and iterate in an inner-loop.

chriss85 · October 21, 2014, 04:38

Thanks for your reply!

For completeness:
Newton-Raphson is an iterative procedure for finding roots of a function. In this case this means that x_(n+1)=x_n + f(x_n) / f'(x_n) with x_n = phi_n and f(phi) = laplace(sigma(phi), phi) and semi implicit in this case means to update sigma and phi_n in an alternating sequence.

What do you mean with " Take a phi to the power of one out of your sigma equation and put it into the matrix, and the rest goes into the source term as a derivative, I believe"?

marupio · October 21, 2014, 14:05

I just reread my last message, and now I think I should go back and reread all the code I was working on at the time. Does not make much sense.

You want to use Newton-Raphson linearization. That uses information from one point (phi0, sigma0, and its derivative) to guess the likely sigma value at another point (phi1).

sigma*(phi1) = sigma(phi0) + derivativeOfSigmaWrtPhi(phi0) * (phi1 - phi0)

When you put this relation into the matrix equation, the diagonal and source are changed according to:

Ap -= derivativeOfSigmaWrtPhi(phi0)
Bp -= derivativeOfSigmaWrtPhi(phi0) * phi0

This is from memory... my reference material is not available to me at the moment.

chriss85 · October 22, 2014, 04:27

Why would I put this equation in a matrix and not calculate it directly?
From what I understand right now this is an extrapolation on the behaviour of sigma while phi changes.

Some more information maybe: This is for electric currents, and in this case sigma is the electrical conductivity, which is a function of the current density here, so sigma(j)=sigma(-grad(phi)).

October 20, 2014, 07:40	Solving Laplace equation with nonlinear diffusity/conductivity	#1
chriss85 Senior Member Join Date: Oct 2013 Posts: 397 Rep Power: 19	I want to solve a Laplace equation, div(sigma * grad(phi)) = 0, where sigma depends nonlinearly on the solution of phi. What is the best approach for doing something like this? If I solve the equation iteratively and update sigma each time based on the current solution, what kind of conditions are required to guarantee convergence and unique solutions? Is it possible (and beneficial) to perform this kind of iteration directly in the matrix solver?

October 20, 2014, 09:19		#2
marupio Senior Member David Gaden Join Date: Apr 2009 Location: Winnipeg, Canada Posts: 437 Rep Power: 22	I'd use Newton-Raphson linearization. Semi-implicit. Take a phi to the power of one out of your sigma equation and put it into the matrix, and the rest goes into the source term as a derivative, I believe. Use relaxation and iterate in an inner-loop. mm.abdollahzadeh likes this. __________________ ~~~ Follow me on twitter @DavidGaden

October 21, 2014, 14:05		#4
marupio Senior Member David Gaden Join Date: Apr 2009 Location: Winnipeg, Canada Posts: 437 Rep Power: 22	I just reread my last message, and now I think I should go back and reread all the code I was working on at the time. Does not make much sense. You want to use Newton-Raphson linearization. That uses information from one point (phi0, sigma0, and its derivative) to guess the likely sigma value at another point (phi1). sigma(phi1) = sigma(phi0) + derivativeOfSigmaWrtPhi(phi0) (phi1 - phi0) When you put this relation into the matrix equation, the diagonal and source are changed according to: Ap -= derivativeOfSigmaWrtPhi(phi0) Bp -= derivativeOfSigmaWrtPhi(phi0) * phi0 This is from memory... my reference material is not available to me at the moment. __________________ ~~~ Follow me on twitter @DavidGaden

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
High Courant Number @ icoFoam	Artex85	OpenFOAM Running, Solving & CFD	11	February 16, 2017 14:40
Help for the small implementation in turbulence model	shipman	OpenFOAM Programming & Development	25	March 19, 2014 11:08
rhoSimplecFoam Mach0.8 no pressure values	CFDnewbie147	OpenFOAM Running, Solving & CFD	16	November 23, 2013 06:58
calculation stops after few time steps	sivakumar	OpenFOAM Running, Solving & CFD	7	March 17, 2013 07:37
Orifice Plate with a fully developed flow - Problems with convergence	jonmec	OpenFOAM Running, Solving & CFD	3	July 28, 2011 06:24

October 21, 2014, 04:38		#3
chriss85 Senior Member Join Date: Oct 2013 Posts: 397 Rep Power: 19	Thanks for your reply! For completeness: Newton-Raphson is an iterative procedure for finding roots of a function. In this case this means that x_(n+1)=x_n + f(x_n) / f'(x_n) with x_n = phi_n and f(phi) = laplace(sigma(phi), phi) and semi implicit in this case means to update sigma and phi_n in an alternating sequence. What do you mean with " Take a phi to the power of one out of your sigma equation and put it into the matrix, and the rest goes into the source term as a derivative, I believe"?

October 22, 2014, 04:27		#5
chriss85 Senior Member Join Date: Oct 2013 Posts: 397 Rep Power: 19	Why would I put this equation in a matrix and not calculate it directly? From what I understand right now this is an extrapolation on the behaviour of sigma while phi changes. Some more information maybe: This is for electric currents, and in this case sigma is the electrical conductivity, which is a function of the current density here, so sigma(j)=sigma(-grad(phi)).