How to do a convolution

[nathan magnan]

Hello,

I am trying to solve an EVP, but one of the internal equations requires a convolution. Mathematically, it looks like this:

variable_2 (x) = integral[ dx' Kernel(x, x') variable_1(x') ]

This convolution can be conceived as an operator acting on variable_1, but I don't know how to write it in Dedalus.

As a side note, if omega is my eigenvalue, is omega^2 allowed to appear in one of the internal equations, or would that break linearity?

Thanks for your help !

Nathan

[nathan magnan]

Hello again,

I've tried to build a work-around that looks like this:

# i) build the trial functions as fields
Trial_functions = []

for i in range(Nx):
    name = 'trial_function_' + str(i)
    trial_function = dist.Field(name = name, bases = basis)
    trial_function['c'][i] = 1
   
    Trial_functions.append(trial_function)

# ii) build x = cst slices of the kernel as fields

def kernel(x_outer, x_inner):
    return # WHATEVER MY KERNEL IS

K_slices = []

for i in range(Nx):
    x_outer = x[i]
    name = 'K_slice_' + str(i)
    K_slice = dist.Field(name = name, bases = basis)

    for j in range(Nr):
        x_inner = r[j]
        K_slice['g'][j] = kernel(x_outer, x_inner)

    K_slices.append(K_slice)

# iii) build the image of the trial functions via my convolution operator. The output should be fields

Images_of_trial_functions = []

for i in range(Nx):
    trial_function = Trial_functions[i]
   
    name = 'image_of_trial_function_' + str(i)
    image_of_trial_function = dist.Field(name = name, bases = basis)

    for j in range(Nx):
        K_slice = K_slices[j]

        image_of_trial_function['g'][j] = (d3.Integrate(K_slice * trial_function, 'x').evaluate())['g'][0]
        K_slice.change_scales(1)
        trial_function.change_scales(1)
        image_of_trial_function.change_scales(1)
   
    Images_of_trial_functions.append(image_of_trial_function)

## iv) build the operator

variable_2 = 0 * variable_1

for i in range(Nx):
    trial_function = Trial_functions[i]
    image_of_trial_function = Images_of_trial_functions[i]

    coefficient_of_Sigma = d3.Integrate(trial_function * variable_1, 'x') # I DON'T KNOW IF THIS IS EQUIVALENT TO variable_1['c'][i]
    variable_2 += coefficient_of_Sigma * image_of_trial_function

My first issue concerns the penultimate line.

I'm not sure that the i-th element in variable_1['c'] is always given by integral(variable_1 * trial_function_i).

I'm afraid that this only works for some choices of trial functions. For instance, the trial function may form a basis that is not orthonormal, or they might be orthonormal but the scalar product may not be the direct integral, there could be a complex conjugation hidden in there.

Ideally, I would like an operator that gives me the i-th coefficient in the decomposition of variable_1. A bit like d3.Interpolate gives me the k-th coefficient in the grid representation of variable_1. Is such a thing available ? If not, any idea for a workaround?

My second issue is that I'm worried about the numerical efficiency of my method. How catastrophicly slow is it gonna be?

Thanks for your help !

Nathan

[Daniel Lecoanet]

Hi Nathan,

If you’re doing a periodic problem, you may be able to use that a convolution in physical space is a product in wavenumber space. Then you would want to make a GeneralFunction operator which multiplies by the Fourier transform of your kernel in coefficient space.

I haven’t thought about convolutions in bounded domains.

Daniel

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[Nathan Magnan]

Hi Daniel,

Thanks for the idea!
Unfortunately, I have Dirichlet boundary conditions on both sides.

Best regards,
Nathan

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