Behavior of the norm function

Linear Algebra
1

Hello

I’m trying to understand the behavior of the norm function in dolfin.

Given a mixed vector function space, I compute a vector function, say U.

If I want to find its norm, what is the difference between norm(U) and norm(U.vector())? Both give me different results and I’m trying to understand why.

Thank you!

2

It sounds like you may be confusing two different notions of “vector” here. The functions in a VectorFunctionSpace are vector-valued, in the sense that they evaluate to a vector at each point in the domain. However, functions in finite element function spaces can be represented as vectors of coefficients scaling each basis function (e.g., a list of nodal values in a Lagrange finite element space). U.vector() refers to the latter notion of “vector”.

The function norm is overloaded to have different behaviors when given a function or a vector as its argument. If the norm of a Function is computed, it returns some Sobolev norm, defaulting to L^2, but optionally specified by a second string argument, as illustrated below:

from dolfin import *
mesh = UnitIntervalMesh(10)
V = VectorFunctionSpace(mesh,"CG",1,dim=2)
x = SpatialCoordinate(mesh)
u = project(as_vector((sin(pi*x[0]),cos(pi*x[0]))),V)
print(norm(u))
print(sqrt(assemble(inner(u,u)*dx)))
print(norm(u,"H1"))
print(sqrt(assemble((inner(u,u) + inner(grad(u),grad(u)))*dx)))

If a vector is passed, norm computes some algebraic norm of the vector, defaulting, if I recall, to the Euclidean (or \ell^2) norm, i.e., the square root of the sum of the squares of the components.

1 Like
3

Thanks for your reply!

I understand what you said but I guess its still a bit hazy for me. Say, I have a function in a VectorFunctionSpace, which as you said would be vector-valued at each point. Let’s say this function is a velocity vector, U.

Suppose I want to normalize the velocity vector by its own norm, am I right in thinking that I will need to use “norm(U.vector())” as the dividing factor ?

I’m asking because the norm of a vector field could have multiple definitions (even the L2 norm) based on local L2 norms, Frobenius norms etc.

4

You must distinguish between the local norm of your physical vectorial field and its global norm when considering it as a vector of degrees of freedom. I think you are referring to the former so U_normalized = U/sqrt(dot(U, U)) is a normalized vectorial field.
The Froebenius norm is a norm on matrices, I don’t see how the L2 norm can have different definitions…

5

Gotcha, that makes sense. Thanks!