Writing Integral with Bessel Function

Madeinspace · April 30, 2020, 10:49

Hi,

Bessel function seem to have been implemented in OpenFOAM library. As mentioned in the programmer's guide, Type 1 Bessel function of zeroth order in OF is j0(S) and first order is j1(S). Then how do I write to calculate this bessel function. Can someone please help?

All the notations are known but I don't know how to write it in OF.

Thanks in advance!

CFD_10 · April 30, 2020, 12:10

Try the following:

Code:

Foam::jn(0, s) 
Foam::jn(1, s) 

// or
Foam::j0(s)
Foam::j1(s)

Madeinspace · May 4, 2020, 13:36

Hi CFD_10,

Thanks. Do you know what is lambda in the equation? All the other notations are process parameters but lambda must come from Bessel function. It is not mentioned in the literature what is lambda. I am completely new to this equation.

Thanks again!

CFD_10 · May 4, 2020, 15:30

Quote:

Originally Posted by Madeinspace

Hi CFD_10,

Thanks. Do you know what is lambda in the equation? All the other notations are process parameters but lambda must come from Bessel function. It is not mentioned in the literature what is lambda. I am completely new to this equation.

Thanks again!

What is the reference of that equation?

Madeinspace · May 4, 2020, 16:25

Initially developed by the authors in reference
https://link.springer.com/content/pd...BF02815302.pdf

It has since been used by several authors like
Eq. 32 in https://www.sciencedirect.com/scienc...17931019319295
Eq. 3a in https://www.researchgate.net/publica...ng_arc_welding

Regards,

CFD_10 · May 5, 2020, 07:16

Quote:

Originally Posted by Madeinspace

Initially developed by the authors in reference
https://link.springer.com/content/pd...BF02815302.pdf

It has since been used by several authors like
Eq. 32 in https://www.sciencedirect.com/scienc...17931019319295
Eq. 3a in https://www.researchgate.net/publica...ng_arc_welding

Regards,

My guess is that $\lambda$ is an eigenvalue of the Laplacian operator. In this case, it looks like it has dimension of 1/length.

But I think you don't have to care about it since it is just an integration variable.

Madeinspace · May 5, 2020, 10:06

Quote:

Originally Posted by CFD_10

My guess is that $\lambda$ is an eigenvalue of the Laplacian operator. In this case, it looks like it has dimension of 1/length.

But I think you don't have to care about it since it is just an integration variable.

Hi CFD_10,
Appereciate your help. Yes, It should have the dimension of 1/length so that the exponential term is dimensionless. So I defined a dimensionedScalar variable named lambda and gave it value of one with dimension (1/length). It compiles and runs however the computed values are way off. How do I compute the integration from 0 to infinity. Does Foam::J0 take care of that part?

Code:

volScalarField Jz = I_A/(2*pi)*lambda*Foam::j0(lambda*r_local)*Foam::exp(-Foam::sqr(lambda*SigmaQ)/2)*(Foam::sinh(lambda*(L0-cellCentre.component(vector::Y))))/(Foam::sinh(lambda*L0));

regards,

CFD_10 · May 5, 2020, 14:09

You can use your favorite numerical integration method to compute that integral (Simpson, Romberg, Gauss quadratures, etc.).

You have:

$J_{ze} = \int_0^\infty{f\left(\lambda; \cdots\right) \mathrm d\lambda}$

You can change the limits of the integral to make it easy (over a finite interval):

$\lambda = {t\over 1-t}$
$J_{ze} = \int_0^1{f\left({t\over 1-t}; \cdots\right){1\over (1-t)^2} \mathrm dt}$

I think you can declare Jze as volScalarField and loop over all the cells, and compute the integral for each cell.

That's my opinion take it with a grain of salt.

Madeinspace · June 24, 2020, 12:32

Thank You CFD_10. I was able to do it by computing the integral using Trapezoid rule and looping over the entire cell in the domain.

Regards,

Schmidtgen · May 4, 2022, 05:37

Quote:

Originally Posted by Madeinspace

Thank You CFD_10. I was able to do it by computing the integral using Trapezoid rule and looping over the entire cell in the domain.

Regards,

How many intervals did you use to integrate with the Trapezoid rule?

April 30, 2020, 12:10		#2
CFD_10 Member CFD USER Join Date: May 2019 Posts: 40 Rep Power: 6	Try the following: Code: Foam::jn(0, s) Foam::jn(1, s) // or Foam::j0(s) Foam::j1(s) Madeinspace likes this. Last edited by CFD_10; April 30, 2020 at 15:34.

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May 4, 2020, 13:36		#3
Madeinspace Member James Join Date: Jan 2014 Posts: 38 Rep Power: 12	Hi CFD_10, Thanks. Do you know what is lambda in the equation? All the other notations are process parameters but lambda must come from Bessel function. It is not mentioned in the literature what is lambda. I am completely new to this equation. Thanks again!

May 4, 2020, 16:25		#5
Madeinspace Member James Join Date: Jan 2014 Posts: 38 Rep Power: 12	Initially developed by the authors in reference https://link.springer.com/content/pd...BF02815302.pdf It has since been used by several authors like Eq. 32 in https://www.sciencedirect.com/scienc...17931019319295 Eq. 3a in https://www.researchgate.net/publica...ng_arc_welding Regards,

May 5, 2020, 14:09		#8
CFD_10 Member CFD USER Join Date: May 2019 Posts: 40 Rep Power: 6	You can use your favorite numerical integration method to compute that integral (Simpson, Romberg, Gauss quadratures, etc.). You have: $J_{ze} = \int_0^\infty{f\left(\lambda; \cdots\right) \mathrm d\lambda}$ You can change the limits of the integral to make it easy (over a finite interval): $\lambda = {t\over 1-t}$ $J_{ze} = \int_0^1{f\left({t\over 1-t}; \cdots\right){1\over (1-t)^2} \mathrm dt}$ I think you can declare Jze as volScalarField and loop over all the cells, and compute the integral for each cell. That's my opinion take it with a grain of salt.

June 24, 2020, 12:32		#9
Madeinspace Member James Join Date: Jan 2014 Posts: 38 Rep Power: 12	Thank You CFD_10. I was able to do it by computing the integral using Trapezoid rule and looping over the entire cell in the domain. Regards,