Issue with gradient inner product

1

Hi,

I am trying to implement a weighted version of the inner product (u * grad u * v),
where u = (u1, u2, u3) is a three-dimensional TrialFunction and v is a 3D TestFunction. Since it should be weighted (w1, w2, w3 constants), I am trying to use
w1* inner(u, grad(u1)) * v1 * dx + w2* inner(u, grad(u_prev2)) * v2 * dx +
w3*inner(u, grad(u_prev3)) * v3 * dx,
which gives me the “ValueError: too many values to unpack (expected 2)” error.
The dimensions seem to be ok to me, e.g. u is three-dimensional, u1 is 1D, thus grad(u1) is 3D.

Could someone maybe give me a hint?

Thank you so much!
Nora

2

Hi,
inner(u, grad(u1))*v1*dx makes no sense if u=(u1,u2,u3) is a TrialFunction. What you are certainly looking for is to solve a nonlinear problem with u being a Function. Checkout the nonlinear demos
https://fenicsproject.org/docs/dolfin/2019.1.0/python/demos/nonlinear-poisson/demo_nonlinear-poisson.py.html
http://home.simula.no/~hpl/homepage/fenics-tutorial/release-1.0-nonabla/webm/nonlinear.html

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3

Thank you so much for your answer. I am trying to understand when to categorise the unknown function as “Function” or “TrialFunction”. TrialFunction is for linear problems and Function is used for nonlinear ones? So if I am looking for a p-u pair with the problem being nonlinear in u and linear in p, one unknown function classifies as Function and the other as TrialFunction? Thank you again.

4

Not exactly, TrialFunctions are used to define bilinear forms a(u,v) with u,v \in V being respectively the trial and test functions. a(u,v) is represented as a matrix after assembly. Linear or non-linear forms F(v) are defined using TestFunctions but may also depend on the unknown Function w\in V in the case of non-linear problems, in this case F(v) = (G(w),v). When assembled, they are represented by a vector which may depend on the Function w.

FEniCS can solve a linear problem in two different ways:

  • bilinear/linear forms: a(u,v)=L(v) for all v\in V with w the solution
  • residual form: F(v) = a(w, v)-L(v) = 0 for all v\in V, in this form no TrialFunction appears
    In the second case, FEniCS will treat it as a nonlinear problem F(v) and will use a Newton method for finding a solution, this is done by computing the Jacobian (derivative of F with respect to w) along a TrialFunction direction u: J(u,v) = derivative(F, w, u), the Newton system is then in the bilinear/linear form with J being the bilinear form

For you u-p problem, the formulation will depend if you want to solve it in a monolithic fashion (both u and p at the same time) or in a staggered fashion (solve p at fixed u then solve for u at fixed p, then iterate possibly). In the first case, you will have to formulate your coupled problem as a non-linear one (both u and p as Functions, then compute the Jacobian with respect to the (u,p) pair). In the second case, you can define the linear p-problem in any of the two above-mentioned ways with a fixed u Function and then formulate the u-problem as a non-linear problem with a fixed p Function.
Hope this helps.

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