Kuramoto Sivashinky 1D PDE #9
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Simple 1D Chaotic test problem that is significantly better understood than Lorenz '96
elswit
reviewed
Sep 7, 2020
elswit
reviewed
Sep 7, 2020
Note: the jacobian is dense for all non-trivial inputs. for the initial condition it might seem like the jacobian is sparse, but this is a red herring. The jacobian is dense, and the way in which it is computed in this code is as efficient as I dare make it.
src/+otp/+kuramotosivashinsky/jac.m
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| function j = jac(~, u, k, k2, k4) | |||
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| j = -diag(k2) - diag(k4) - diag(k)*ifft(fft(diag(ifft(u))).').'; | |||
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diag(k) can be replaced with .* to scale rows
| end | ||
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| methods | ||
| function soly = solution2real(soly) |
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To be consistent with Pendulum problem, let's rename to something like convert2grid and not bother supporting solution struct.
| % right boundary point | ||
| x = linspace(h, L, N).'; | ||
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| u0 = cos(x/16).*(1+sin(x/16)); |
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Replace hardcoded 16 so i.c. is periodic for different L. It might be easier to have x in the range 0 to 2 pi.
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If I have x in 0 to 2 pi, then I would have to calculate h differently. I will change u0, to be a function of L and use cospi and sinpi. I think that is simpler.
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Simple 1D Chaotic test problem that is significantly better understood than Lorenz '96