Daniel,
Hi all -- I'm writing a library that involves solving a nonlinear elliptic
PDE, which I'll write as f(u) = 0. There is an exact solution for this PDE for
a certain simplified geometry. To test everything, I check that my numerical
solution is tolerably close to the analytic solution, with both bilinear and
biquadratic elements, and on a mesh that has been adaptively refined in part
of the domain. This all works fine.
The PDE can be derived from an action principle, i.e. there's a convex
functional P such that f = dP. I've decided to reimplement some of my code to
explicitly use the action principle, so I wrote a routine to calculate P. The
action is analytically computable for the exact solution I'm already using to
test the code, so this is already one unit test. However, I decided to take it
a step further, and check that the nonlinear differential operator f is the
derivative of P, and that its linearization df is the Hessian operator of P.
To check that, I take some vector field u, another vector field v, and check
that
P(u + h * v) = P(u) + h*f(u) . v + h^2 * (df(u)v) . v / 2 + O(h^3)
> [...]
I agree it makes not very much sense, but that often points to a
misunderstanding rather than a bug. So let me poke:
* In the formula above, P(.) is a functional, I assume, i.e., it takes a
function and returns a number, right?
* If so, what exactly does
f(u) . v
actually mean? How do you compute this?
* Same for the second derivatives?
This test passes in the simple geometry with both bilinear and biquadratic
elements, but *not* for an adaptively refined mesh. The errors in both the
local linear approximation to the action functional and the local quadratic
approximation do not go to 0 as the increment h is reduced.
I assume you mean by "error" the size of the second and third term?
In addition, the
value of the action is roughly the same for each h for both the adaptively
refined and uniform meshes.
I don't think I understand this statement.
The only other clue I have is that the error of
the numerical solution against the analytic solution is actually somewhat
worse on the adaptively-refined mesh than for the uniform mesh, but before I
assumed this just had something to do with the linear solver.
How do you measure "worse"? As the error as a function of the number of
unknowns?
Best
W.
PS: I do like this approach to testing. It shows great maturity in designing
code and testing numerical methods to think this way. Well done!
--
------------------------------------------------------------------------
Wolfgang Bangerth email: bange...@colostate.edu
www: http://www.math.colostate.edu/~bangerth/
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