Comments (4)
maroba
commented on September 15, 2024
Hi,
I'm not sure if you mean stencil or matrix representation. Anyways, let's look at both.
First of all, yes it is
x, y = np.linspace(0, 1, 100), np.linspace(0, 1, 100)
or simply
x = y = np.linspace(0, 1, 100)
For the stencil, consider the simplified version with only a 5x5 grid of grid spacing 1:
laplace_2d = FinDiff(0, 1, 2) + FinDiff(1, 1, 2)
stencil = laplace_2d.stencil((5, 5))
print(stencil) # as of version v0.9.1 (just released) it is just stencil an no longer stencil.data
which yields
{('L', 'L'): {(0, 0): 4.0, (0, 1): -5.0, (0, 2): 4.0, (0, 3): -1.0, (1, 0): -5.0, (2, 0): 4.0, (3, 0): -1.0},
('L', 'C'): {(0, -1): 1.0, (0, 1): 1.0, (1, 0): -5.0, (2, 0): 4.0, (3, 0): -1.0},
('L', 'H'): {(0, -3): -1.0, (0, -2): 4.0, (0, -1): -5.0, (0, 0): 4.0, (1, 0): -5.0,
(2, 0): 4.0, (3, 0): -1.0},
('C', 'L'): {(-1, 0): 1.0, (0, 1): -5.0, (0, 2): 4.0, (0, 3): -1.0, (1, 0): 1.0},
('C', 'C'): {(-1, 0): 1.0, (0, -1): 1.0, (0, 0): -4.0, (0, 1): 1.0, (1, 0): 1.0},
('C', 'H'): {(-1, 0): 1.0, (0, -3): -1.0, (0, -2): 4.0, (0, -1): -5.0, (0, 0): 8.881784197001252e-16,
(1, 0): 1.0},
('H', 'L'): {(-3, 0): -1.0, (-2, 0): 4.0, (-1, 0): -5.0, (0, 0): 4.0, (0, 1): -5.0,
(0, 2): 4.0, (0, 3): -1.0},
('H', 'C'): {(-3, 0): -1.0, (-2, 0): 4.0, (-1, 0): -5.0, (0, -1): 1.0, (0, 0): 8.881784197001252e-16,
(0, 1): 1.0},
('H', 'H'): {(-3, 0): -1.0, (-2, 0): 4.0, (-1, 0): -5.0, (0, -3): -1.0, (0, -2): 4.0,
(0, -1): -5.0, (0, 0): 4.0}}
which looks ok. In the interior of the grid, the (C,C) stencil applies, near the boundary one of the others. L means low index boundary region, H means high index boundary region. The stencil in your case looks strange because it contains 1 / grid_spacing **2 terms, which is just 1 in my example.
If you mean matrix, then this is how you can get it:
laplace_2d = FinDiff(0, dx, 2) + FinDiff(1, dy, 2)
mat = laplace_2d.matrix(u.shape) # returns a scipy sparse matrix
# apply matrix representation to u:
mat.dot(u.reshape(-1)).reshape(u.shape)
Regarding the Helmholtz operator in 1D, the following works:
from findiff import *
dx = 1
H = Identity() - FinDiff(0, dx, 2)
print(H.matrix((10,)).toarray())
which gives
[[-1. 5. -4. 1. 0. 0. 0. 0. 0. 0.]
[-1. 3. -1. 0. 0. 0. 0. 0. 0. 0.]
[ 0. -1. 3. -1. 0. 0. 0. 0. 0. 0.]
[ 0. 0. -1. 3. -1. 0. 0. 0. 0. 0.]
[ 0. 0. 0. -1. 3. -1. 0. 0. 0. 0.]
[ 0. 0. 0. 0. -1. 3. -1. 0. 0. 0.]
[ 0. 0. 0. 0. 0. -1. 3. -1. 0. 0.]
[ 0. 0. 0. 0. 0. 0. -1. 3. -1. 0.]
[ 0. 0. 0. 0. 0. 0. 0. -1. 3. -1.]
[ 0. 0. 0. 0. 0. 0. 1. -4. 5. -1.]]
which looks correct. For actual computation you need a reasonable grid spacing, of course.
The stencil method for Helmholtz example really gives an error. I will fix it in the next days
as version 0.9.2.
Edit: What part is a feature request?
from findiff.
maroba
commented on September 15, 2024
Ok, I just released v0.9.2 with the necessary fix.
from findiff.
sotakao
commented on September 15, 2024
Thank you very much! I just checked that I can create the stencil for the Helmholtz operator with v0.9.2.
Have a nice day.
from findiff.
maroba
commented on September 15, 2024
Thanks for checking. Have a nice day, too!
from findiff.
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