Solve DGL with non-constant (discrete) variable?

Question

Solve DGL with non-constant (discrete) variable?

stefmech opened this issue 2 years ago · comments

Hi Mikael,

i would like to solve a DGL of this form:
$u_{xx} -q(x)u = -f(x)$ .

$q(x)$ is only defined by discrete values, so it cannot be incorparated as a expression. This is maybe a stupid question, but in the documentation I couldn't find a proper solution for such a problem. Is it possible to solve this kind of equation in shenfun with non-constant coefficients?

Best regards,
Steffen

Mikael Mortensen · Answer 1 · Tue Jan 18 2022 17:57:03 GMT+0800 (China Standard Time)

Hi
You would be the first to implement something like this in shenfun, but it is not necessarily very difficult. You could probably use something like

$q(x) = \sum_{i=0}^{M} \delta(x-x_i)q_i$

and the inner product would then be

$(qu, v)_{\omega} = \sum_{i=0}^{M} (\delta(x-x_i)q_iu, v)_{\omega}=\sum_{i=0}^{M} q_i u(x_i)v(x_i)\omega (x_i)$

With test function $\phi_k$ and trial $\psi_j$ we get

$\sum_{i=0}^{M} q_i u(x_i)v(x_i)\omega (x_i) = \sum_{i=0}^M \sum_{j=0}^{N-1} q_i \psi_j(x_i) \phi_k(x_i) \omega (x_i) \hat{u}_j$

So you get a dense matrix for each i. Perhaps not very clever after all?

stefmech · Answer 2 · Wed Jan 19 2022 06:42:52 GMT+0800 (China Standard Time)

Thank you for reply! Sorry for the not clearly formulated question. I mean, the non-constant variable is defined on the underlying grid N.

Would be the inner product not something like this?
$(qu, v)_\omega = \sum_{j=0}^{N-1} q_j \psi_j \phi_k \omega \hat{\u}_j$ .

Today, I saw that there's a scale functionality for an Expression, for example. What's about this? This could probably work for the above inner product, right?

Best regards

Mikael Mortensen · Answer 3 · Wed Jan 19 2022 16:22:10 GMT+0800 (China Standard Time)

If q is defined on underlying grid, then there is nothing special to it. You can get it in spectral space simply by a forward transform. However, you do have something that needs to be computed like a nonlinear term. So I think you should compute it with something like

u = u_hat.backward() # get u on grid
qu = q*u # assuming u and q are now on the same grid in physical space
inner(v, qu)

The scale functionality can only be used for a constant scaling of the entire expression. The scale can not be a function of x.

stefmech · Answer 4 · Thu Jan 20 2022 03:51:21 GMT+0800 (China Standard Time)

First of all, i would like to say thank you for your help! I saw your example "Demo - Lid driven cavity", where the nonlinear rhs was handled it in a similiar way as you mentioned above. However, I think the problem in your approach is , that the $inner(v,qu)$ would result in a linear form, but not in a bilinear form. Or should I use an conjugate gradient solver which can use the matrix in the form "A*x"?
So would it be necessary to add a new case to the "inner"-function, e.g. setting up a bilinear form for the specific case, that there are two expressions consisting of a function and a testfunction?

I add an small example:

from shenfun import FunctionSpace, Function, grad, \
BlockMatrix, extract_bc_matrices, TestFunction, TrialFunction, inner, Array
import numpy as np

l = 100.
h = l/2
domain_x = (0, l)

N = 18
T = FunctionSpace(N, family='C', bc=(0,2), domain=domain_x)

u = TrialFunction(T)
v = TestFunction(T)

q = 100 # Is q is constant it can be simply multiplied

A = inner(q*grad(u), grad(v))

u_hat = Function(T)

# get boundary matrices
bc_mats = extract_bc_matrices([A])

# BlockMatrix for inhomogeneous part
BM = BlockMatrix(bc_mats)

# inhomogeneous part of solution
uh_hat = Function(T).set_boundary_dofs()

# additional part to be passed to the right hand side
b = Function(T)
# negative because added to right hand side
b = BM.matvec(-uh_hat, b)

# homogeneous part of solution
u_hat = A[0].solve(b)

u_ = u_hat.backward(kind='uniform')

print('Done')

Mikael Mortensen · Answer 5 · Thu Jan 20 2022 15:35:17 GMT+0800 (China Standard Time)

Hi
You are right. Any nonlinear term or similar with multiplication of two functions need to use a linear form and go to the right hand side. So you would need to solve it iteratively. This is not just a shenfun problem, it is because when you multiply two spectral functions you need to compute a convolution and the order of the discretization goes naturally from N to 2N.

$u \cdot q=(\sum_{i=0}^{N-1} T_i \hat{u}_i) \cdot (\sum_{i=0}^{N-1} T_i \hat{q}_i)=\sum_{i=0}^{2N-1} T_i \hat{c}_i$

Dealiasing and pseudospectral techniques are usually the preferred way around this. Multiply u and q in real space (usually with 3/2N quadrature points) and then forward the outcome. There is no way I am aware of that can compute this easily in a bilinear form. In fact, it would take a trilinear form.