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Integration from 0 to y in 2D
Posted 12 août 2009, 04:13 UTC−4 Version 3.5a 6 Replies
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I have a function f(x,y) defined on a 2D subdomain x=0..a and y=0..b and I want to calculate at each point (x,y) the Integral I2:
I2(x,y)=int_0^y f(x,y') dy'
In a 1d model, this can be done using an Integration coupling variable and the dest operator :
for example I1(x) = int_0^x f(x') dx' is done by writing as integrand : f*(x<dest(x))
But in 2d if I write f*dest(y) it performs int_0^a int_0^y f(x,y') dy' dx and I don't want my expression to be integrated over x !
I've tried a projection from my 2d domain on a 1d domain x=0..a and then extrude it again on a 2d domain x=0..a y=0..b but it calculates the integral from 0 to b : I3(x,y)=int_0^b f(x,y') dy'
Does any of you have an idea ?
I know that i could define subdomain but I want the value of the integral at every point of the 2d domain so it makes a lot of subdomain to create...
I've attached my model with the post.
Thanks a lot. Best Regards
Thomas Doki-Thonon
Attachments:
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Gamma = 0 0
F = uy-f(x,y)
The solution to the PDE, u(x,y) is then the integral that you desire, i.e. u(x,y)=I2(x,y)=int_0^y f(x,y') dy'
To do this with coupling variables you would have to extrude the solution to a 3D prism and then project back to the 2D domain which is tricky and computationally more expensive.
/ Dan
COMSOL Development
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Thanks for your answer,
I've tried both methods:
PDE Method
I'm confusing about the PDE Methods. In the attached file i tried to integrate a simple function f(x,y)=y but what i obtain disagree with the analytical integral (plotted in Geom 4)
About the projection/extrusion method :
I think that I've understood what you meant :
1) Extruding 2D surface onto the first parallel face of a 3D Prism
2) Projecting 3D Prism along a direction Perpendicular to the second parallel face of the Prism back on the 2D Surface. Since the extruded Surface are "cut" by the obliqued plane. When I project I get the integral from 0 to y.
Am I right ? Because I didn't manage to create the extrusion and projection.
I think the first method could be perfect for my problem, do see in my file where my mistake is ?
Thanks a ton for your help and answers.
Attachments:
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/ Dan
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It works, thanks a lot !
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Have you ever tried 3D. My application can be simplified as:
? (? u)=exp(int(-a*z'),z'=0 to z)
note: alpha is dependent on u, so, I can not use an analytical form for the right part of the equation.
To this, assume u1, u2
? (? u1)=u2
du2/dz=-a*u2
As one test step to solve the above problem, I tried to solve the second equation alone by assuming the right part is a known function:
du2/dz=f(x,y,z)
The geometry is a cylinder. The top boundary is set to be R=1-u2=0. other boundaries with G=0, R=0.
However, it is quit slow and says "can not find the solution" sometimes.
Is there any special way to do that in 3D?
Thanks in advance
Thanks,
Jiang
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I want to use I2(x,b), which is defined as I2(x,b)=int_0^b f(x,y') dy', as a coefficient in a PDE. This PDE has a dependent variable,say 'u'. But, f(x,y) is itself a function of 'u'.
Anybody has an idea how to do it?
Regards,
Tanmay