Periodic Boundary Conditions

[J Harr]

Hi,

I would like to solve a periodic BC problem (with real values on the LHS) using Dedalus. However a "Factor is exactly singular" error keeps popping.

I am aware of other threads dealing with this problem but since I don't have BCs I don't see how to relate it to my situation.

As I am very new to Dedalus and spectral methods I am sure the error is right under my nose but I can't seem to find it.

Here is my code:

Lx, Ly = (28, 28)
x_basis = de.Fourier('x', 152, interval=(0, Lx))
y_basis = de.Fourier('y', 152, interval=(0, Ly))
domain = de.Domain([x_basis, y_basis], grid_dtype=np.float64)

def func():
       return one

Gfunc=de.operators.GeneralFunction(domain, 'g', func)
de.operators.parseables["f"] = Gfunc

problem = de.LBVP(domain, variables=['u'])
problem.substitutions["uxx"] = "dx(dx(u))"
problem.substitutions["uyy"] = "dy(dy(u))"
problem.substitutions["uxxxx"] = "dx(dx(dx(dx(u))))"
problem.substitutions["uyyyy"] = "dy(dy(dy(dy(u))))"
problem.add_equation("uxx + uyy + uxxxx + uyyyy = f ")

solver = problem.build_solver()
solver.solve()

Where "one" is a numpy.ndarray of shape (28,28) composed of ones and zeros (describing a shape).

Do you have an idea where the error comes from ?

Kind regards,

Josquin

[Daniel Lecoanet]

Hi Josquin,

The horizontally-averaged mode has dx(u) = dy(u) = 0. For that mode, the equation becomes 0*u = average(f). You can fix this by telling Dedalus to not solve that equation for the horizontally-averaged mode:

problem.add_equation(“uxx + uyy + uxxxx + uyyyy = f”, condition = “(nx !=0) or (ny != 0)”)

problem.add_equation(“u = 0”, condition = “(nx == 0) and (ny == 0)”)

Here I set the horizontal average of u to zero — maybe you want it to be non-zero!

Daniel

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[J Harr]

Hi Daniel,

Many thanks for the fast reply, everything works fine now !

I am not sure about the impact of setting the horizontal average to zero but I guess this last equation is necessary for Dedalus ?

If I get rid of it I get a ValueError: Pencil [0, 0] has 0 equations for 1 variables.

If I understand well I can set the horizontal average to n by changing in the last equation "u=n", I think this is fine for my problem.

Kind regards,

Josquin

[Daniel Lecoanet]

Hi Josquin,

Your equation does not set the horizontal average of u. If u = u_0(x, y) is a solution to the equation, so would u_0(x, y) + a + b x + c y, for a, b, c constants. By periodicity, b = c = 0 (assuming the average of f is zero). But you can shift everything by a constant a. This is a degree of freedom that must be specified in order to completely determine u. I think “u=n” is close to what you’re looking for. I think that might set the integral of u to n, not the average. You can experiment to see which it is.

Daniel

To view this discussion on the web visit https://groups.google.com/d/msgid/dedalus-users/65fa8031-8e85-4761-977f-e42a4eb18bf9n%40googlegroups.com.

[J Harr]

Hi Daniel,

Ok, I get it now.

I've tried changing the constant in "u=n", we can see it shifts everything.

Thanks for the help !

Josquin