Linearity of solution dependent property

Linear Algebra
1

Hi, I am working on a problem with solution dependent property, for instance k(u)\frac{du}{dt}=\frac{d^2u}{dx^2}. I am wondering how to implement it. I saw some posts giving almost the same solution: define u=Function(V), and then define k as an expression such as k=Expression('u+1',u=u,degree=2). I am confused by this implementation, since it is treating each time step as a non-linear problem (linear problem defines TrialFunction first). It looks to me that Expression freezes k(u) at each time step based on the solution of u in the previous step, thus we are solving a linear problem at each time step. Why in all examples (e.g. Material property depending on solution) we do not define u=TrialFunction(V)?

I tried u=TrialFunction(V) and k=Expression('u+1',u=u,degree=2). It reports error, so it seems Expression does not input TrialFunction. Is it designed on purpose?

2

Perhaps look at the nonlinear Poisson demo as a guide.

3

Thanks for your reply. I have read the demo. The demo expresses k(u)=u**2 directly without an Expression. I can understand that the demo is a non-linear case since u**2 is computed in Jacobian. I will need a conditional expression (e.g.,u>1?u+1:u+2) so I do not think I can use a simple form.

4

Why not? Your conditional is piecewise gateaux differentiable.

If you don’t want to use Newton’s method you could look into Picard iteration. E.g. the Navier Stokes demo.

5

Thanks again for your reply. How to express the conditional expression in the variational form?

6

Check the documentation.

E.g.

from dolfin import *

mesh = UnitSquareMesh(32, 32)
V = FunctionSpace(mesh,"DG",0)
x = SpatialCoordinate(mesh)
f0 = conditional(lt(x[0], 0.5), 1.0, 0.0)

image