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comparison of discretization scheme

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Old   December 19, 2017, 11:26
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Martin Hegedus
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When you add dT/dt into the equation you get

(Fe - Fw +1/delta_t + Fn - Fs) * Tp = Fw*TW - Fe*TE + Fs*TS - Fn*TN

If you made delta_t equal to infinity, you recover your equation (which is Newtonian iteration). Unfortunately, as you noticed, the determinant of that is zero.

As a general statement for central differencing one should add dissipation and march forward in time with a reasonable time step, i.e. large but not very large. Non-linear equations can have multiple solutions and marching forward in time (in a reasonable way) helps insure that a solution is found that one can actually get to.
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Old   December 21, 2017, 10:34
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Quote:
Originally Posted by Martin Hegedus View Post
When you add dT/dt into the equation you get

(Fe - Fw +1/delta_t + Fn - Fs) * Tp = Fw*TW - Fe*TE + Fs*TS - Fn*TN

If you made delta_t equal to infinity, you recover your equation (which is Newtonian iteration). Unfortunately, as you noticed, the determinant of that is zero.

As a general statement for central differencing one should add dissipation and march forward in time with a reasonable time step, i.e. large but not very large. Non-linear equations can have multiple solutions and marching forward in time (in a reasonable way) helps insure that a solution is found that one can actually get to.
if we write your eqn:
(Fe - Fw +1/delta_t + Fn - Fs) * Tp = Fw*TW - Fe*TE + Fs*TS - Fn*TN

in this form:

ap*Tp=aE*TE + aW*TW + aN*TN + aS*TS + b

we can find a negetive coeff for aE and aN for posetive dirrection of flow:
aE = - Fe <0 , aN = - Fn <0
so certainly or solution will be diverged!

I want to conclude that we never can solve pure advection eqn with central scheme.

is it correct?

thanks
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Old   December 21, 2017, 11:41
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Central schemes are the border between upwind and downwind. I'm not sure of what they are capable of without dissipation.
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Old   December 21, 2017, 16:27
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Overall, it will depend on your boundary conditions. For example lets look at the determinant of the left hand matrix for Newtonian iteration on a 1-D case with five nodes. Also, lets fix the temperature at the left and the flow moves from left to right.

The LHS matrix for the case where the right hand side temperature (along with left hand side) is fixed is

1 0 0 0 0
-1 0 1 0 0
0 -1 0 1 0
0 0 -1 0 1
0 0 0 0 1

the determinant for that is 0. However, if you had four points, the determinant would be 1. So, that is not robust.

Now lets use first order upwind at the last point

1 0 0 0 0
-1 0 1 0 0
0 -1 0 1 0
0 0 -1 0 1
0 0 0 0-1 1

The determinant of that is 1.

Not sure how well that example transfers over to 2-D, but it's a start and should be straight forward to test.
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