Can Bessel Functions be Simplified for Numerical Integration?

In summary, the conversation discusses the use of Bessel functions and Hankel transforms in the analytical computation of S(k) and V(R). The speaker has successfully obtained an analytic solution for S(k) using a known Hankel transform, but still needs to find a Hankel transform for the computation of V(R). They are currently unable to do so and suggest avoiding the truncated law of distribution or resorting to numerical computation.
  • #1
Astaroth.
10
0
Hello, need your help.
11.png

J_1 and J_0 - Bessel function
Necessary to solve analytically or to be able to simplify the numerical solution.
 
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  • #2
The analytic solution for S(k) is given in the joint page.
It was obtained thanks to a known Hankel transform.
Then, for the analytic computation of V(R), another Hankel transform is needed, which I have'nt available. This will require more research. Sorry, I have not enough time now. I will look again within a few days.
 

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  • #3
Thank you. I have been modeling the rotation of galaxies.
The journal «Freeman, K. C. Astrophysical Journal, vol. 160, p.811» http://articles.adsabs.harvard.edu//full/1970ApJ...160..811F/0000813.000.html described, but I need for different density profiles [itex]\sigma[/itex].
[itex] \sigma (r)) = \sigma_{0} \exp(-\frac{r}{L}) [/itex]
and
[itex] \sigma (r) = \sigma (Rt)*(1+(Rt-r)/SCL)= \sigma_{0} * \exp(-\frac{Rt}{L})*(1+(Rt-r)/SCL) [/itex]
Where Rt - radius of truncation. SCL - scale truncation
 
  • #4
Finally, it was not so arduous than I thought first.
The result is expressed in terms of Bessel functions (in attachment)
 

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  • #5
Thank you very much.
And if the [itex] \sigma (r) = \sigma_ {0} * \ exp (- \frac {Rt} {L}) * (1 + (Rt-r) / SCL) [/itex]

The result will differ greatly?
 
  • #6
Astaroth. said:
And if the [itex] \sigma (r) = \sigma_ {0} * \ exp (- \frac {Rt} {L}) * (1 + (Rt-r) / SCL) [/itex]
The result will differ greatly?

In the function sigma(r), if you replace the exopential function (which tends to zero when r tends to infinity) by a linear function which tends to -infinity when r tends to infinity), indeed, the result will differ greatly.
So greatly that the integral (from 0 to infinity) is no longer convergent.
 
  • #7
JJacquelin said:
In the function sigma(r), if you replace the exopential function (which tends to zero when r tends to infinity) by a linear function which tends to -infinity when r tends to infinity), indeed, the result will differ greatly.
So greatly that the integral (from 0 to infinity) is no longer convergent.

What do I do?

The exponential law of density distribution [itex] \sigma (r)) = \sigma_{0} \exp(-\frac{r}{L}) [/itex]
I want to apply a linear truncation for [itex] \sigma (r) [/itex]
[itex] \sigma (r) =
\left\{\begin{matrix}\sigma (r),
&r < Rt \\ \sigma (Rt)(1+(Rt-r)/SCL),
& Rt < r < Rt+SCL\\ 0,
& r > Rt+SCL
\end{matrix}\right.[/itex]
 
Last edited:
  • #8
Astaroth. said:
What do I do?

The exponential law of density distribution [itex] \sigma (r)) = \sigma_{0} \exp(-\frac{r}{L}) [/itex]
I want to apply a linear truncation for [itex] \sigma (r) [/itex]
https://www.physicsforums.com/attachments/41402

OK, I understand : the linear function is truncated, which is very different regarding the Hankel integral.
Bur I cannot read the attachment : "Invalid Attachment specified"
 
  • #9
[itex] \sigma (r) =
\left\{\begin{matrix}\sigma (r),
&r < Rt \\ \sigma (Rt)(1+(Rt-r)/SCL),
& Rt < r < Rt+SCL\\ 0,
& r > Rt+SCL
\end{matrix}\right. [/itex]
 
  • #10
Well, I am afraid, this will be more complicated than with the previous sigma(r).
At first sight, solving analytically the integal S(k) seems possible.
Then, about the other integral V(R), I don't know yel. Now, I have not enough time available.
 
  • #11
Only you can help me.
 
  • #12
After more study in the case of the truncated law for sigma(r), I come to a conclusion :
Using the background provided by the Hankel transforms is no longer possible. We have to split the integral in two, each one on a limited range of integration. The background is less extended than in case of integration from zero to infinity.
Nevertheless, the integral S(k) can be analytically solved. In fact, it can be expressed in terms of a series of hypergeometric functions (this is of few interest for numerical calculus, since computing a lot of hypergeometric functions is not avantageous compare to a direct numerical integration).
Then, the other integral V(R) becommes very complicated, involving the product of hypergeometric functions and the Bessel function. I don't think that it could be analytically solved in present state of knowledge.
As a conclusion I suggest, if possible, to avoid the trucaded law of distribution : In is much simpler to apply the analytical solution found in case of the untruncated law.
If the truncated law is essential, my opinion is that there in no other way than the numerical computation of the integrals.
Sorry, that all I can do to help you concerning the particular case of truncated law of distribution.
 
  • #13
Thank you very much.
 

FAQ: Can Bessel Functions be Simplified for Numerical Integration?

What is the purpose of calculating integrals?

The purpose of calculating integrals is to find the exact area under a curve or the volume of a solid, which can represent real-life quantities such as displacement, velocity, and acceleration.

What is the difference between definite and indefinite integrals?

A definite integral has specific limits of integration, and its value represents the precise area or volume being calculated. An indefinite integral does not have limits and is represented by a general function, which can be used to find the value of a definite integral.

How do you solve an integral?

To solve an integral, you must first determine the integrand (the function being integrated) and the limits of integration. You can then use various techniques, such as substitution, integration by parts, or trigonometric identities, to simplify the integral and find its exact value.

What is the relationship between derivatives and integrals?

Derivatives and integrals are inverse operations. This means that the derivative of a function represents its rate of change, while the integral of a function represents its accumulated value. Additionally, the fundamental theorem of calculus states that the derivative of an integral is equal to the original function.

Can you use technology to calculate integrals?

Yes, technology such as graphing calculators and computer software can be used to calculate integrals. These tools use numerical methods, such as the trapezoidal rule or Simpson's rule, to approximate the value of an integral. However, it is still important to understand the concepts and techniques behind integration to ensure accurate results.

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