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How do I show the above relation with Sturm-Liouville theory (assume the usual boundary conditions for the identity)? Here is what I have tried: if we start with $$ \big(xJ_a'(\ell x) \big)'+\left(\ell^2-\frac{a^2}{x}\right)J_a(\ell x)=0 $$ and $$ \big(xJ_a'(\ell' x) \big)'+\left(\ell'^2-\frac{a^2}{x}\right)J_a(\ell' x)=0 $$ and multiply the top expression by $xJ_a(\ell' x)$ and bottom by $xJ_a(\ell' x)$ and subtract the two equations, $$ x\big(xJ_a'(\ell x) \big)'-x\big(xJ_a'(\ell' x) \big)'+x(\ell^2-\ell'^2)J_a(\ell x)J_a(\ell' x)=0. $$ Then by rearrangement and integration over $[0, a]$ my answer is wrong. To be explicit, I get: $$ \int_0^axJ_m(\ell x)J_m(\ell'x)=\frac{a\big(J_m'(\ell a)J_m(\ell' a)-J_m'(\ell' a)J_m(\ell a)\big)}{\ell^2-\ell'^2} $$ which is just barely wrong--- the correct result is $$ \int_0^axJ_m(\ell x)J_m(\ell'x)=\frac{a\big(\ell J_m'(\ell a)J_m(\ell' a)-\ell'J_m'(\ell' a)J_m(\ell a)\big)}{\ell^2-\ell'^2} $$ which is zero (by the usual boundary conditions)

What am I doing wrong? Also-- is the reason we choose to multiply by $x\,\times$ the bessel function because of the "weight function" present in the Sturm–Liouville form of the Bessel function? Is this to ensure that our functions vanish at certain points?

Thank you!

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  • $\begingroup$ @AlapanDas Yes I changed $q$ to $a$. However changing $a$ to $m$ is unnecessary. I just used it as a dummy variable. $\endgroup$
    – user161698
    Commented Jan 1, 2021 at 16:18
  • $\begingroup$ @AlapanDas when you say "that" what do you mean? $\endgroup$
    – user161698
    Commented Jan 1, 2021 at 16:33
  • $\begingroup$ Ahhh sorry that is a typo. $\endgroup$
    – user161698
    Commented Jan 1, 2021 at 16:35
  • $\begingroup$ The Bessel equation is $(xJ'_a(lx))'+(l^2x-\frac{q^2}{x})J_a(lx)=0$. $\endgroup$
    – Alapan Das
    Commented Jan 1, 2021 at 16:47
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    $\begingroup$ Eigenfunctions of a self-adjoint operator are orthogonal, so if you set it up correctly, you don't need any calculation. $\endgroup$
    – Christian Remling
    Commented Jan 1, 2021 at 16:50

2 Answers 2

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For, $bl$ and $bl'$ being distinct zeroes of Bessel function , we have
$(xJ_a(lx)')'+(l^2x-q^2/X)J_a(lx)=0$ and

$(xJ_a(l'x)')'+(l'^2x-q^2/X)J_a(l'x)=0$ .

Multiplying first with $J_a(l'x)$ and second with $J_a(l'x)$ and then subsratcting second from first ,

$-[J_a(l'x)(xJ_a(l'x))'-J_a(lx)(xJ'_a(l'x))']=x(l^2-l'^2)xJ_a(lx)J_a(l'x)$

Now, $J_a(l'x)(xJ'_a(l'x))'-J_a(lx)(xJ'_a(l'x))'=\frac{d}{dx}(xJ_a(l'x)J'_a(lx)-xJ_a(lx)J'_a(l'x)$

Hence, $\int_{0}^{b} xJ_a(lx)J_a(l'x)=\frac{bJ_a(l'b)J'_a(lb)-bJ_a(lb)J'_a(l'b)}{l^2-l'^2}$.

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  • $\begingroup$ I get the same thing, but according to math.usm.edu/lambers/mat415/lecture15.pdf, it is wrong. Thanks for the answer. $\endgroup$
    – user161698
    Commented Jan 1, 2021 at 16:54
  • $\begingroup$ See, what I have meant by $J'_a(lx)$ say, $A(x)$ is $\frac{d J_a(lx)}{dx}$. But what is meant there by $J'_a(lx)$ say, $B(x)$ is $\frac{d J_a'(t)}{dt}$ at $t=lx$. And $B(x)=lA(x)$. That's why those terms are appearing. $\endgroup$
    – Alapan Das
    Commented Jan 1, 2021 at 17:01
  • $\begingroup$ So they are appearing from the boundary conditions? $\endgroup$
    – user161698
    Commented Jan 1, 2021 at 17:04
  • $\begingroup$ Or is this just the chain rule? $\endgroup$
    – user161698
    Commented Jan 1, 2021 at 17:10
  • $\begingroup$ Yes, it's just due to chain rule. $\endgroup$
    – Alapan Das
    Commented Jan 1, 2021 at 17:17
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If $l \ne m$, then $$ \int_{0}^{b}\!x{{\rm J}_{a}\left(lx\right)}{{\rm J}_{a}\left(mx\right) }\,{\rm d}x={\frac {b \big( {{\rm J}_{a+1}\left(lb\right)}\;{{\rm J}_{a }\left(mb\right)}\;l-{{\rm J}_{a+1}\left(mb\right)}\;{{\rm J}_{a}\left(lb \right)}\;m \big) }{{l}^{2}-{m}^{2}}} $$

Assume $l,m$ are zeros of $J_a$ and $b=1$. Then we get the value $0$.

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  • $\begingroup$ Thanks for the answer Gerald. I should have been more specific when I said "for the usual conditions." This identity follows from assuming $\ell$ and $\ell'$ are the zeros of the Bessel function. $\endgroup$
    – user161698
    Commented Jan 1, 2021 at 16:31

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