# Re: sources errors in advection/diffusion problems, and one solution

No worries -- I had to do it to figure out the problem in my more complex
domain and equation... The issue which surprised me was that the value the
variable was initialized to had an effect on the steady solution.

Jamie

On Fri, Sep 16, 2016 at 8:14 AM, Guyer, Jonathan E. Dr. (Fed) <
jonathan.gu...@nist.gov> wrote:

> James -
>
> I think Daniel will have more insight into why this is happening and if
> there is anything to be done about it besides artificial relaxation, but I
> just want to say how much I appreciate your putting this together. This is
> a very lucid illustration.
>
> - Jon
>
> > On Sep 15, 2016, at 5:13 PM, James Pringle <jprin...@unh.edu> wrote:
> >
> > Dear FiPy users --
> >
> >    This is a simple example of how and why fipy may fail to solve a
> >    problem can reduce the error. I also found something that was a
> >    surprise -- the "initial" condition of a steady problem can affect
> >    the solution for some solvers.
> >
> >    At the end are two interesting questions for those who want to
> >    understand what FiPy is actually doing.... I admit to being a bit
> >    lost
> >
> >    The equation I am solving is
> >
> >         \Del\dot (D\del psi + u*psi) =0
> >
> >    Where the diffusion D is 1, and u is a vector (1,0) -- so there is
> >    only a flow of speed -1 in the x direction.  I solve this equation
> >    on a 10 by 10 grid. The boundary conditions are no normal gradient
> >    on the y=0 and y=10 boundary:
> >
> >         psi_y =0 at y=0 and y=10
> >
> >    For the x boundary, I impose a value of x=1 on the inflow boundary at
> x=10
> >    (this is a little tricky -- the way the equation is written, u is
> >    the negative of velocity).
> >
> >         psi=1 at x=10
> >
> >    and a no-normal-gradient condition at x=0.
> >
> >         psi_x=0 at x=0
> >
> >    since all of the domain and boundary is symmetrical with respect to
> >    y, we can assume psi_y=0 is zero everywhere. This reduces the
> equation to
> >
> >         psi_xx + psi_x =0
> >
> >    The general solution to this equation is
> >
> >         psi=C1+C2*exp(-x)
> >
> >    Where C1 and C2 are constants. For these boundary conditions, C1=1
> >    and C2=0, so psi=1 everywhere.
> >
> >    Now run the code SquareGrid_HomemadeDelaunay and look at figure(3)
> >    -- this is the plot of psi versus x, and you can see that it does
> >    not match the true solution of psi=1 everywhere! Instead, it
> >    appears to be decaying exponential. The blue line is actually just
> >    (1+exp(-x)). What is going on? It appears that small errors in the
> >    boundary condition at x=0 are allowing C2 to not be exactly 0, and
> >    this error is this exponential mode. The error is the artificial
> >    exiting of a correct mode of the interior equation, albeit one that
> >    should not be excited by these BC's.
> >
> >    Interestingly, the size of this error depends on the value the psi
> >    is initially set to. If the line
> >
> >        psi=fipy.CellVariable(name='psi',mesh=mesh,value=0.0)
> >
> >    is changed so psi is initially 1, the error goes away entirely; if
> >    it is set to some other value, you get different errors. I do not
> >    know if this is a bug, or just numerical error in a well programmed
> >    solver.
> >
> >    Now if you run SquareGrid_HomemadeDelaunay_transient  which
> implements
> >
> >          psi_t = \Del\dot (D\del psi + u*psi)
> >
> >    you can see that the error in the numerical solution is advected
> >    out of the x=0 boundary, and the solution approaches the true
> >    solution of psi=1 rapidly.
> >
> >    The interesting question is if the formulation of the boundary
> >    condition at x=0 could be altered to less excite the spurious mode?
> >
> >    Also, why does the "initial" condition have any effect on the
> >
> >    Cheers,
> >    Jamie
> >
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