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How to input the square of a differential?

Posted 18.01.2011, 11:28 GMT-5 Version 4.0 3 Replies

Xiaoyi Yu
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comsol version 4.0 or higher
How to express (?u)^2, ? represents differential here, u is a dependent variable, u(x, y)
Is this right: (ux+uy)^2 ?
3 Replies Last Post 19.01.2011, 01:48 GMT-5
Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)
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Posted: 1 decade ago 18.01.2011, 14:40 GMT-5
Hi

I do not believe so, if your vector field "_u_(x,y,z)" (_in bold_) is in fact the vector field of scalar values (u,v,w) for 3D then what do you really mean by du ?

In COMSOL notation

(ux+uy)^2 = (du/dx+dudy)^2

I agree I find too that using "u" for the general vector field and for the 1st part of the vector is confusing, normally they were using bold case for the full vector field and normal case for u,v,w. tehn if you look up the doc, you will see that ux=du/dx, uy=du/dy vx=dv/dx and wt=dw/dt, just as uxx=d^2u/dx^2 vxy=d^2u/dx/dy
This means als that one need to take some precautions when inventing variable names with COMSOL and not to redefine internal ones, the results might be strange ;)

--
Good luck
Ivar
Hi I do not believe so, if your vector field "_u_(x,y,z)" (_in bold_) is in fact the vector field of scalar values (u,v,w) for 3D then what do you really mean by du ? In COMSOL notation (ux+uy)^2 = (du/dx+dudy)^2 I agree I find too that using "u" for the general vector field and for the 1st part of the vector is confusing, normally they were using bold case for the full vector field and normal case for u,v,w. tehn if you look up the doc, you will see that ux=du/dx, uy=du/dy vx=dv/dx and wt=dw/dt, just as uxx=d^2u/dx^2 vxy=d^2u/dx/dy This means als that one need to take some precautions when inventing variable names with COMSOL and not to redefine internal ones, the results might be strange ;) -- Good luck Ivar
Xiaoyi Yu
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Posted: 1 decade ago 18.01.2011, 21:50 GMT-5

Hi

I do not believe so, if your vector field "_u_(x,y,z)" (_in bold_) is in fact the vector field of scalar values (u,v,w) for 3D then what do you really mean by du ?

In COMSOL notation

(ux+uy)^2 = (du/dx+dudy)^2

I agree I find too that using "u" for the general vector field and for the 1st part of the vector is confusing, normally they were using bold case for the full vector field and normal case for u,v,w. tehn if you look up the doc, you will see that ux=du/dx, uy=du/dy vx=dv/dx and wt=dw/dt, just as uxx=d^2u/dx^2 vxy=d^2u/dx/dy
This means als that one need to take some precautions when inventing variable names with COMSOL and not to redefine internal ones, the results might be strange ;)

--
Good luck
Ivar


Sorry, maybe I didn't say what I want to say clearly.
the "?" refer to the vector differential operator (gradient), ?u=(ux, uy) or ?u=(du/dx, du/dy)
and I want to know how to express (?u)^2 in comsol.
(?u)^2=(ux)^2+(uy)^2, or (?u)^2=(ux+uy)^2, which is right in comsol?
[QUOTE] Hi I do not believe so, if your vector field "_u_(x,y,z)" (_in bold_) is in fact the vector field of scalar values (u,v,w) for 3D then what do you really mean by du ? In COMSOL notation (ux+uy)^2 = (du/dx+dudy)^2 I agree I find too that using "u" for the general vector field and for the 1st part of the vector is confusing, normally they were using bold case for the full vector field and normal case for u,v,w. tehn if you look up the doc, you will see that ux=du/dx, uy=du/dy vx=dv/dx and wt=dw/dt, just as uxx=d^2u/dx^2 vxy=d^2u/dx/dy This means als that one need to take some precautions when inventing variable names with COMSOL and not to redefine internal ones, the results might be strange ;) -- Good luck Ivar [/QUOTE] Sorry, maybe I didn't say what I want to say clearly. the "?" refer to the vector differential operator (gradient), ?u=(ux, uy) or ?u=(du/dx, du/dy) and I want to know how to express (?u)^2 in comsol. (?u)^2=(ux)^2+(uy)^2, or (?u)^2=(ux+uy)^2, which is right in comsol?
Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)
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Posted: 1 decade ago 19.01.2011, 01:48 GMT-5
Hi

for me that is pure maths, no ?

the norm of a gradient is the sum of squares of the components (and not the square of a sum)

or have I missed something ?

--
Good luck
Ivar
Hi for me that is pure maths, no ? the norm of a gradient is the sum of squares of the components (and not the square of a sum) or have I missed something ? -- Good luck Ivar

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