Bessel Functions

  • #1
joshmccraney
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Hi PF!

I was wondering if anyone could shed some light on my understanding of arriving at the coefficients of Bessel Equations? Namely, why do we use the indicial equation to determine coefficients?

As an example, if we have to solve $$s^2 \alpha'' + 2 s \alpha ' - \frac{1}{4} \gamma^2 s^2 \alpha = 0$$ we gues a solution as $$\alpha(s) = \sum_{k=0} a_k s^{n+k}$$

Plugging this guess into the above ode yields $$\sum_{k=0} a_k (n+k)(n+k+1) s^{n+k} - \frac{1}{4} \gamma^2 \sum_{k=2} a_{k-2} s^{n+k} = 0$$

Now for solving this, I know we let ##k=0 \implies n(n+1) = 0## but what about when ##k=1##? Can someone help me know why we look only at the indicial equation? I can provide more work if you need it.

Thanks
 

Answers and Replies

  • #2
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Its not about looking at a particular k. The point is, the two sums don't have a common starting point so we take out the k=0 and k=1 parts of the first sum. Now we can have one sum and some other terms. Those other terms are called the indicial equation. Here the indicial equation is [itex] a_0 n(n+1) s^n+a_1 (n+1)(n+2) s^{n+1} =0 [/itex] and because different powers of s are independent, we'll have [itex] n(n+1)=0 [/itex] and [itex] (n+1)(n+2)=0 [/itex]. But these two equations are satisfied at the same time only when [itex] n=-1 [/itex].
 
Last edited:
  • #3
joshmccraney
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How do they tell us the same thing? One implies ##n=0,-1## and the other says ##n=-1,-2##.
 
  • #4
joshmccraney
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Thanks for the fast response!
 
  • #5
2,792
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Yeah, that was a mistake. I corrected it. Both should be satisfied at the same time, so [itex]n=-1[/itex].
 
  • #6
joshmccraney
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But my work leads me to believe we need both ##n=-1## and ##n=0## regardless of the fact that ##n=0## does not solve the second equation. Also, this is what leads to a pair of solutions. If we only use ##n=-1## then we only have one equation (I think of this as one Bessel Equation). To get the other we need the other piece, namely ##n=-1##.
 
  • #7
2,792
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But my work leads me to believe we need both ##n=-1## and ##n=0## regardless of the fact that ##n=0## does not solve the second equation. Also, this is what leads to a pair of solutions. If we only use ##n=-1## then we only have one equation (I think of this as one Bessel Equation). To get the other we need the other piece, namely ##n=-1##.
If the other equation is not satisfied, the series is not a solution to the ODE! You want a solution, right?

Also I can ask, why [itex] n=0 [/itex] and not [itex] n=-2 [/itex]?

The second solution can be found by the following procedure:
Consider two functions u and w satisfying your differential equation.
[itex]
s^2 u''+2 s u'-\frac 1 4 g^2 s^2 u=0 \\
s^2 w''+2 s w'-\frac 1 4 g^2 s^2 w=0
[/itex]
Now I multiply first by w and second by u and subtract the second from the first to get:
[itex]
s^2(w u''-u w'')+2s(wu'-uw')=0 \Rightarrow \frac{d}{ds}[s^2(wu'-uw')]=0 \\ \Rightarrow s^2(wu'-uw')=C_1 \Rightarrow \frac{wu'-uw'}{w^2}=\frac{C_1}{s^2 w^2} \Rightarrow d(\frac{u}{w})=\frac{C_1}{s^2 w^2} ds \\ \Rightarrow u=C_2 w+C_1 w \int \frac{ds}{s^2 w^2}
[/itex]
Clearly u is linearly independent of w and so is the second solution we were seeking. So you can build a solution using [itex] n=-1[/itex] and then use it and the above result to build the second solution. I should say the above result is usually hard to use and so its just for proving that given a solution, there always exist a linearly independent solution from it. So there should be other ways for finding that second solution which you should find in texts about this particular ODE.
 
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  • #8
joshmccraney
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Are you serious!?! That's crazy! Like I follow you all the way but I would never dream of concluding the way you did. That's brilliant! But why is this hard to use (or do you mean generally it is difficult, but here given this ODE we got lucky)?

And I have no idea why ##n=0,-1## works and not ##n=-2##. To be honest, I haven't tried ##n=-2##; perhaps it's a linear combination of ##n=0,-1##? Your thoughts?

Thanks for posting (and if you could, do you mind responding to both of my questions above?
 
  • #9
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I was missing something that showed up in the calculations I did.
At first, you have only one choice:[itex] n=-1 [/itex].
The recurrence relation will be [itex] a_k=\frac{\gamma^2}{4k(k-1)} a_{k-2} [/itex].
Starting from k=0, you'll get a series with even powers of s and starting from k=1, you'll get a series with odd powers of s and these two series are obviously linearly independent and so you have two solutions with two unknowns([itex] a_0 [/itex] and [itex] a_1 [/itex]), as expected.
 
  • #10
joshmccraney
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You mean starting with ##k \geq 2##? Thanks a ton for the help!
 
  • #11
2,792
594
Yes!
 

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