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Differentiation operator gives zero everywhere

Posted Jan 11, 2012, 6:48 p.m. EST Parameters, Variables, & Functions, Results & Visualization Version 4.2 2 Replies

Anna d'Entremont
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I'm having trouble using the differentiation operator when plotting some results from my simulation and for defining variables.

Every time I try to use a derivative of the form d(u,t) or d(u,x), the result is zero for all times and locations. For instance, plotting Vx gives non-zero results, but d(V,x) gives zero results for the same solution, same location and same times.

What I primarily need is the time derivative of the electric field es.Ex. I think I should be able to use d(es.Ex,t), but that gives me zero as well.
2 Replies Last Post Jan 12, 2012, 2:30 a.m. EST
Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)
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Posted: 1 decade ago Jan 12, 2012, 1:48 a.m. EST
Hi

What about looking into the equations a bit ? ;)

With ES physics you solve for V, and your E field x component is defined internally in COMSOL as Ex = -Vx
Turn on the equation view (Options Preferences Show Equation View) and do not get confused with the poor notation Vx is in fact defined as d(V,x) but Ex is the "X" component of "E"

So if you are asking for d(E,x) you are looking for -Vxx the second derivative of the dependent variable "V".

but the derivatives are extracted from the "shape" functions you use and by default COMSOL mostly uses second order shape functions, and the 2nd derivative of a 2nd order polynomial is a constant

to get a better value, try using a third forth of fifth order shape function. (turn on the discretisation tab with Options Preferences Show Discretisation)

Concerning the time derivatives, you need to have a time dependent study

Try it out on a models like this one (4.2a)


--
Good luck
Ivar
Hi What about looking into the equations a bit ? ;) With ES physics you solve for V, and your E field x component is defined internally in COMSOL as Ex = -Vx Turn on the equation view (Options Preferences Show Equation View) and do not get confused with the poor notation Vx is in fact defined as d(V,x) but Ex is the "X" component of "E" So if you are asking for d(E,x) you are looking for -Vxx the second derivative of the dependent variable "V". but the derivatives are extracted from the "shape" functions you use and by default COMSOL mostly uses second order shape functions, and the 2nd derivative of a 2nd order polynomial is a constant to get a better value, try using a third forth of fifth order shape function. (turn on the discretisation tab with Options Preferences Show Discretisation) Concerning the time derivatives, you need to have a time dependent study Try it out on a models like this one (4.2a) -- Good luck Ivar

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Anna d'Entremont
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Posted: 1 decade ago Jan 12, 2012, 2:30 a.m. EST
Hi Ivar,
I'm currently using a fifth-order shape function and a time-dependent study, so the derivatives I want should be available. The current problem is that I get zero for any derivative of the form d(u,t) or d(u,x) that I try to plot. For instance, I can plot "Vx", so I know that this derivative is non-zero, but if I try to plot it as d(V,x), I get zero.

Unfortunately, I couldn't open the model you attached, because it was created in a later version of COMSOL than I have. I'm using COMSOL 4.2--I guess you must be using 4.2a?

Thanks.
Hi Ivar, I'm currently using a fifth-order shape function and a time-dependent study, so the derivatives I want should be available. The current problem is that I get zero for any derivative of the form d(u,t) or d(u,x) that I try to plot. For instance, I can plot "Vx", so I know that this derivative is non-zero, but if I try to plot it as d(V,x), I get zero. Unfortunately, I couldn't open the model you attached, because it was created in a later version of COMSOL than I have. I'm using COMSOL 4.2--I guess you must be using 4.2a? Thanks.

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