examples.levelSet.advection.mesh1D¶

Solve the distance function equation in one dimension and then advect it.

This example first solves the distance function equation in one dimension:

$\abs{\nabla \phi} = 1$

with $\phi = 0$ at $x = L / 5$ .

The variable is then advected with,

$\frac{ \partial \phi } { \partial t} + \vec{u} \cdot \nabla \phi = 0$

The scheme used in the FirstOrderAdvectionTerm preserves the var as a distance function.

The solution to this problem will be demonstrated in the following script. Firstly, setup the parameters.

>>> from fipy import CellVariable, Grid1D, DistanceVariable, TransientTerm, FirstOrderAdvectionTerm, AdvectionTerm, Viewer
>>> from fipy.tools import numerix, serialComm

>>> velocity = 1.
>>> dx = 1.
>>> nx = 10
>>> timeStepDuration = 1.
>>> steps = 2
>>> L = nx * dx
>>> interfacePosition = L / 5.

Construct the mesh.

>>> mesh = Grid1D(dx=dx, nx=nx, communicator=serialComm)

Construct a distanceVariable object.

>>> var = DistanceVariable(name='level set variable',
...                        mesh=mesh,
...                        value=-1.,
...                        hasOld=1)
>>> var.setValue(1., where=mesh.cellCenters[0] > interfacePosition)
>>> var.calcDistanceFunction() 

The advectionEquation is constructed.

>>> advEqn = TransientTerm() + FirstOrderAdvectionTerm(velocity)

The problem can then be solved by executing a serious of time steps.

>>> from builtins import range
>>> if __name__ == '__main__':
...     viewer = Viewer(vars=var, datamin=-10., datamax=10.)
...     viewer.plot()
...     for step in range(steps):
...         var.updateOld()
...         advEqn.solve(var, dt=timeStepDuration)
...         viewer.plot()

The result can be tested with the following code:

>>> from builtins import range
>>> for step in range(steps):
...     var.updateOld()
...     advEqn.solve(var, dt=timeStepDuration)
>>> x = mesh.cellCenters[0]
>>> distanceTravelled = timeStepDuration * steps * velocity
>>> answer = x - interfacePosition - timeStepDuration * steps * velocity
>>> answer = numerix.where(x < distanceTravelled,
...                        x[0] - interfacePosition, answer)
>>> print(var.allclose(answer)) 
1

Last updated on Jun 27, 2023. Created using Sphinx 6.2.1.