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Math Operator Question
Posted 25 giu 2012, 10:16 GMT-4 Parameters, Variables, & Functions Version 4.2a 1 Reply
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Hey guys,
Just had a question about math operators. I'm a little new to comsol.
Let's say I'm working with a heat source term in nonisothermal flow. On my Q, I defined Q as a function in the variables section. But, what if I want Q to be the calculated value of the function I already defined for only certain values.
What I have now:
Q = Qdefined*(Qdefined>0)
which I'm taking to mean, if the calculated Q is greater than 0, use that value. But what exactly does the *Qdefined>0 mean? Does it mean that it's 0 for all other values?
If not, what mathematical operator do I use so that if Q is NOT a positive number calculated by the function (i.e negative or 0), it gives 0 as the answer.
Thanks!
Just had a question about math operators. I'm a little new to comsol.
Let's say I'm working with a heat source term in nonisothermal flow. On my Q, I defined Q as a function in the variables section. But, what if I want Q to be the calculated value of the function I already defined for only certain values.
What I have now:
Q = Qdefined*(Qdefined>0)
which I'm taking to mean, if the calculated Q is greater than 0, use that value. But what exactly does the *Qdefined>0 mean? Does it mean that it's 0 for all other values?
If not, what mathematical operator do I use so that if Q is NOT a positive number calculated by the function (i.e negative or 0), it gives 0 as the answer.
Thanks!
1 Reply Last Post 25 giu 2012, 10:21 GMT-4
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Posted:
1 decade ago
Hi
the (Qdefined>0) is a Bolena operator, giving 0 if not true and 1 if true, so if true Q=Qdefined, ifnot Q=0 in your case
Note that such hard "dirac type transition can make life tough for the solvers, often it's worth to consider smoothing the transition with a "step()" operator or whatever applies (but simetimes its OK, all model dependent)
--
Good luck
Ivar
the (Qdefined>0) is a Bolena operator, giving 0 if not true and 1 if true, so if true Q=Qdefined, ifnot Q=0 in your case
Note that such hard "dirac type transition can make life tough for the solvers, often it's worth to consider smoothing the transition with a "step()" operator or whatever applies (but simetimes its OK, all model dependent)
--
Good luck
Ivar