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This is the basic form of the problem i am solving. enter image description here

The main difference with the integral on the photo and my example is that i have my integral defined from b to c instead of 0 to a, and Bessel function definition is written in a simpler way.

FullSimplify[
  Integrate[
   r BesselJ[0, Subscript[s, n] r] BesselJ[0, Subscript[s, m] r], {r, 
    b, c}]] // TraditionalForm

The code is really basic, but i can't define that sn and sm are orthogonal in mathematica, even though they are in reality.

How does this change in integral borders and Bessel function definition affect integral result?

P.S.: This is the equation used to numerically define values of ss (or s). enter image description here

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  • $\begingroup$ What exactly are the s-terms? Perhaps you might try to tell Mathematica in some way what these are, as it will not know otherwise. Also In general, it might also be better to use indexed terms rather than subscripted terms. That is, something like s[n]. $\endgroup$ May 25 at 13:18
  • $\begingroup$ Related: mathematica.stackexchange.com/q/253675/4999 $\endgroup$
    – Michael E2
    May 25 at 13:38
  • $\begingroup$ Have you seen BesselJZero[]? You don't seem to use it to get the alphas. $\endgroup$
    – Michael E2
    May 25 at 13:42
  • $\begingroup$ The s-terms are numerical solutions (zeroes) of my equation - it is a combination of J0 and Y0 Bessel functions. I know their values, but that is not the relevant part of this question. Why i would like to know if the change of integral bounds from [0,a] to [b,c] affects basic form of the solution as it is shown on the picture. $\endgroup$
    – U.Grammy
    May 25 at 13:52

2 Answers 2

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The OP's integral is missing information about the subscripted variables (which I avoid) and the change of coordinates leading to the interval from b to c, which information is necessary to evaluate the integral. Here's the integral from the quote:

res = Integrate[
  r BesselJ[0, BesselJZero[0, n] r/a] BesselJ[0, 
    BesselJZero[0, m] r/a], {r, 0, a}, 
  Assumptions -> {m, n} ∈ PositiveIntegers]

FullSimplify yields a generically true result:

FullSimplify[res, {m, n} ∈ PositiveIntegers]
(*  0  *)

The exception is clear from the denominator; one can also use FullSimplify[FunctionSingularities[res, {m, n}], {m, n} ∈ PositiveIntegers] programmatically in theory, but there does not seem to be enough information about BesselJ encoded in Mathematica to reach the conclusion m == n.

Limit[res, m -> n, Assumptions -> n ∈ Integers] // FullSimplify
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  • $\begingroup$ Thank you it is very useful. But one dilemma still remains - does change of integral bounds affect the style/type of solution? I have always only seen solution for integral from 0 to a, but what about from a to b? Numerical integration is not a problem in this case, i want to understand the background of the problem, which type of solution appers when. $\endgroup$
    – U.Grammy
    May 25 at 14:08
  • $\begingroup$ If you look at my post, at the end I added a picture of equation from which i numerically determine values of s or zeroes, so I can't help myself a lot with BesselJZero. $\endgroup$
    – U.Grammy
    May 25 at 14:10
  • 1
    $\begingroup$ @U.Grammy Changing the bounds without changing the integrand should change the result in general, unless the integrand is symmetric with respect to the change (e.g. periodicity in trigonometric functions). If you do a substitution (change of coordinates), then the integrand and interval change and of course the result stays the same. $\endgroup$
    – Michael E2
    May 25 at 14:11
  • $\begingroup$ @U.Grammy If you use numerically determined values, then Integrate probably won't solve your problem. It is an exact solver and is not as good with inexact (floating-point) input. NIntegrate is better, but all parameters such as n and m must be given specific values. $\endgroup$
    – Michael E2
    May 25 at 14:15
  • $\begingroup$ Yes, i understand that side of the problem. What i would like to know if I use integral bounds [a,b] instead of [0,a] how does it change the result? Is it the same as the first picture in question, but (b^2-a^2) instead of a^2 or is the result in the form as your final equation (with both J0 and J1 in the solution). $\endgroup$
    – U.Grammy
    May 25 at 14:19
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This is really a mathematics question rather than a Mathematica question, but from Abramowitz and Stegun section 11.4 we have

$\int_a^b t C\left[\nu ,\lambda _m t\right] C\left[\nu ,\lambda _n t\right] \, dt=0$ for $m\neq n$

and for m = n, a>b, the indefinite integral is

$\frac{1}{2} t^2 \left\{\left(1-\frac{\nu ^2}{\lambda _n^2 t^2}\right) C\left[\nu ,\lambda _n t\right]{}^2+C'\left(\nu ,\lambda _n t\right){}^2\right\}$

Apply the limits on $t$ at $a$ and $b$ to get the definite integral.

In the above $C[\nu ,z]=A J_{\nu }(z)+B Y_{\nu }(z)$

And $\lambda _n$ is a real root of $h_1 \lambda C[\nu +1,\lambda b]-h_2 C[\nu ,\lambda b]=0$

and you must have k's that fit $k_1 \lambda _n C\left[\nu +1,\lambda _n a\right]-k_2 C\left[\nu ,\lambda _n a\right]=0$.

Since you don't tell us exactly what problem you are trying to solve, it is hard to know whether your problem can be make to fit these parameters. You mention your solution is a combination of J0 and Y0 bessel functions, but are only integrating the J's, which makes no sense to me. It almost looks like you are trying to solve a bounded circular reservoir problem.

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