How To Find A Scalar Field From Its Gradient

variational formulation
1

Hello everyone!

I am a novice at Fenics, but after staring at the variational form for a bit was able to successfully solve a set of PDEs for a scalar field. Now I would like to again use Fenics to solve for a separate scalar field from its gradient which I know at every point within the domain. This seems really elementary but I am struggling to figure it out. I understand how to utilize UserExpressions to specify my gradient at every point in the domain, but I need help with putting it into variational form and solving for it.

Any help would be appreciated,

Garrett

2

Hi,
It would be helpful to provide some context. Simply based on what you say, consider the following field

u_e = x^2 y

to be recovered from it’s gradient

\nabla u_e = \begin{bmatrix}2xy\\ x^2\end{bmatrix} 
from dolfin import *
mesh = UnitSquareMesh(10, 10)
V = FunctionSpace(mesh, "CG", 3)
u = Function(V)
v = TestFunction(V)
du = TrialFunction(V)
x = SpatialCoordinate(mesh)
gradu = as_vector([2*x[0]*x[1], x[0]*x[0]])

solve(inner(grad(du), grad(v))*dx == inner(gradu, grad(v))*dx, u, [DirichletBC(V, Constant(0), "near(x[0], 0) && on_boundary")])

# Exact solution
ue = Function(V)
ue.interpolate(Expression("x[0]*x[0]*x[1]", degree=3))

# Test the solution 
max(abs(u.vector()[:] - ue.vector()[:]))  # 9.703349235223868e-14

As long as you can write what you want to achieve in a variational form, you should be able to implement it like above.

3 Likes
3

Hello bhaveshshrimali,

That answer was just what I was looking for! Thank you so much!

Garrett