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Questions about fvsolution, Why OF set relTol to control convergence?

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Old   March 14, 2016, 05:31
Smile Questions about fvsolution, Why OF set relTol to control convergence?
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Ping Chang
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Hi everyone,
I am a newer to OF, I am confusing that why OF set relTol to control convergence .In fvsolution, tolerance is already exist to control the convergence of the equation.

And I found the value of reTol are varies in many cases. Could someone tell me the reason of this?

Thanks a lot.
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Old   May 10, 2016, 04:51
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Good question (I was wondering exactly the same )
any reply ?
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Old   May 10, 2016, 05:50
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Nene Lee
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Firstly, the 'relTol' entry is as an aternative method to control the solving of linear algebraic equations. To avoid the tolerance getting too small, even under the VSMALL, the 'relTol' is used.
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Old   May 10, 2016, 10:21
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Mahdi Hosseinali
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As the documents says:

Before solving an equation for a particular field, the initial residual is evaluated based on the current values of the field. After each solver iteration the residual is re-evaluated. The solver stops if either of the following conditions are reached:

the residual falls below the solver tolerance, tolerance;
the ratio of current to initial residuals falls below the solver relative tolerance, relTol;
the number of iterations exceeds a maximum number of iterations, maxIter;
The solver tolerance should represent the level at which the residual is small enough that the solution can be deemed sufficiently accurate. The solver relative tolerance limits the relative improvement from initial to final solution. In transient simulations, it is usual to set the solver relative tolerance to 0 to force the solution to converge to the solver tolerance in each time step. The tolerances, tolerance and relTol must be specified in the dictionaries for all solvers; maxIter is optional.
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Old   July 2, 2016, 07:23
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Amelie Rohlfs
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Hello,

are there reasons why I would want to control the difference between the initial and the final residual with relTol?
Are there any disadvantages when the step from initial to final is very big?

Greetings, Amelie
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Old   July 3, 2016, 13:54
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Simply because you do not need to push convergence to an absolute tolerance value during each inner iteration loop. Consider a transient simulation with several outer iteration loops in each time step. Inside the inner iteration loop, you perform several iterations (aka sweeps) and achieve an absolute tolerance, let's say 10e-6. You finish the outer loop and recalculate the coefficient matrix to start another outer loop. Now again you have to iterate till the absolute tolerance achieves 10e-6. This is, however, usually not necessary to do in every single loop since it may take a lot of iterations to achieve the absolute tolerance. The subsequent update of the coefficient matrix increases the residual value significantly anyway (initial residual) and kind of "ruins" the efforts to push the tolerance that low. Therefore, you are just happy to push the residual down by a factor, let's say, 0.01 (=relTol), and go to the next outer loop. In OF, you usually set the relative tolerance to 0 in the last outer iteration loop of the time step (UFinal, pFinal and so on), which means that you enforce the absolute residual only at the end of the time step. To sum it up, it is convenient to use relTol to avoid superfluous iterations and so to save some computational time.
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Old   July 4, 2016, 04:54
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Thank you for that explanation, I got it
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