Dear Daniel,
This is the current script referred to in the previous email, where I have
removed the phi.set value for phi.
Were can one find the function that results as the solution to the given PDE in
FiPY? For isntance, if this solution was phi = sec²(x) can FiPY print it out?
Thanks
Sergio
dphi/dt + d^2phi/dx² + |phi|² phi = 0
#!/usr/bin/env python
# testing a non-complex variant of the NLSE
import numpy
import cmath as math
from fipy import *
from fipy import numerix
nx = 50
dx = 1. / float(nx)
mesh = Grid1D(nx=nx,dx=dx)
X = mesh.cellCenters[0]
phi = CellVariable(mesh=mesh, name="Solution")
#phi.setValue(0.5-0.5*numerix.exp((1j*X)))
vi = Viewer(vars=phi,datamin=0.0, datamax=1.0)
vi.plot()
raw_input("Initialization ...")
phi.constrain(1., mesh.facesLeft)
phi.constrain(0., mesh.facesRight)
phi_sq = CellVariable(mesh=mesh)
phi_sq.setValue( phi*phi )
#We now represent the equation, where u = phi, du/dt + d^2u/dx^2 + u^2 = 0 in
fipy commands. du/dx : is a convection term with a unit scalar coefficient,
i.e. <SpecificConvectionTerm>(coeff=(1.,), var=u) ,
# du/dt : is a transient term that one can treat as before, i.e.
TransientTerm(var=u). one can add a second order derivative as
ExplicitDiffusionTerm(coeff=D)
eq = TransientTerm(coeff=1., var=phi) + ExponentialConvectionTerm(coeff=(1.,),
var=phi) + abs((phi_sq))*(phi) == 0.0
dt = 0.01
steps = 100
for step in range(steps):
eq.sweep(dt=dt)
#
phi_sq.setValue( phi * phi )
#
vi.plot()
phiAnalytical = CellVariable(name="Analytical value",
mesh=mesh)
vi = Viewer(vars=(phi, phiAnalytical))
vi.plot()
raw_input("Press <return> ...")
Sergio Manzetti
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From: "Daniel Wheeler" <[email protected]>
To: "fipy" <[email protected]>
Sent: Monday, May 22, 2017 6:05:25 PM
Subject: Re: Complex conjugates in FiPY
On Sat, May 20, 2017 at 5:02 AM, Sergio Manzetti
<[email protected]> wrote:
>
> Dear Daniel, I am wondering if you can clarify a small. thing.
>
> In the given script, phi is set as e^ix, and the numerical simulation treats
> the given PDE. Is phi tested for wether it is a result of the given PDE in
> this script ? Or does the script do something else?
I'm not quite sure what you're asking, but I'm sure that the script
below does not work as you intend it to work.
> #!/usr/bin/env python
> # testing a non-complex variant of the NLSE
>
> import numpy
> import cmath as math
> from fipy import *
> from fipy import numerix
>
> nx = 50
> dx = 1. / float(nx)
>
> mesh = Grid1D(nx=nx,dx=dx)
> X = mesh.cellCenters[0]
>
> phi = CellVariable(mesh=mesh, name="Solution")
> phi.setValue(0.5-0.5*numerix.exp((1j*X)))
At this point your script is broken, "phi.value.imag" is all zero
while "(0.5-0.5*numerix.exp((1j*X))).value.imag" is non-zero. The type
of the CellVariable is wrong initially and the type doesn't change
when the value is reset.
--
Daniel Wheeler
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