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Let $n\geq 0$ be an integer and let $J_n=J_n(r)$ denote the usual Bessel function (of the first kind) of order $n$ i.e. one of the solutions to Bessel's differential equation

$$r^2\frac{d^2y}{dr^2}+r\frac{dy}{dr}+(r^2-n^2)y=0.$$

I am interested in the quantities

$$\beta_n:=\int_0^\infty J_n(r)^2 J_0(r)^2 dr$$ which arise naturally in the study of certain classes of oscillatory integrals. In particular, I would like to explore the convexity of the sequence $\{\beta_n\}$. My question is the following: is it true that

$$\beta_n<\frac{\beta_{n-1}+\beta_{n+1}}{2}$$ for every $n\geq 1$? This inequality seems to be numerically verifiable (Mathematica) for small values of $n$. To prove it, I was trying to use the so-called ''Neumann's formula'' (see e.g. Watson's treatise (1966), p.32)

$$J_n(r)^2=\frac{1}{\pi}\int_0^\pi J_{2n}(2r\sin\theta)d\theta,$$ together with the well-known recurrence

$$J_{n-1}(r)-J_{n+1}(r)=2 \frac{d J_n}{dr}(r)$$ and some integration by parts. Unfortunately I wasn't able to make this approach work.

Ideas or/and references are welcome, thank you very much in advance.

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2 Answers 2

I fully suspect the answer to your question is yes. The reason is as follows. \begin{eqnarray} \beta_{n-1} + \beta_{n+1} &=& \int [J_{n+1}^2 + J_{n-1}^2] J_0^2 \nonumber \\ &=& \int [(J_{n+1} + J_{n-1})^2 - 2 J_{n+1} J_{n-1}] J_0^2 \nonumber \\ &=& \int [\frac{4 n^2}{r^2} J_n^2 - 2 J_{n+1} J_{n-1}] J_0^2 \end{eqnarray} where we we use the recurrence relation $2 \frac{n}{2} J_n = J_{n-1} + J_{n+1}$ to get the last inequality. Notice that since $$\int 2J_{n+1} J_{n-1}J_0^2 \leq \beta_{n+1} + \beta_{n-1}$$ we have that $$ \beta_n + \int (\frac{n^2}{r^2} - 1) J_n^2 J_0^2 \leq \frac{\beta_{n-1} + \beta_{n+1}}{2} $$ You now just need that one equation to be non-negative to get the answer you want. This quantity should be non-negative by the simple fact that $\frac{n^2}{r^2} - 1$ is positive near $r=0$ and this is where the Bessel functions have most of their mass. Further, $\frac{n^2}{r^2} - 1$ is negative where the Bessel function is small. I'm not sure how to get this rigorously though.

I hope this helps.

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Thank you! Your heuristics are insightful but I think you're missing one point: the function $J_n(r)$ starts oscillating only around $r\sim n$, until then it looks like a convex function with absolute value very close to 0. And in fact I just checked (numerically) that the integrals $\int (n^2/r^2-1)J_n^2 J_0^2$ are negative for small values of $n$. I also suspect that my original question should have an affirmative answer, but something rather subtle seems to be going on... Any further thoughts? Thanks again. –  user17240 Apr 12 at 15:23
    
That's interesting. I didn't think the estimate would have been that damaging. The only place where we estimated was to replace $2ab$ with $a^2 + b^2$. If we don't estimate, then we get $2ab = a^2 + b^2 - (a - b)^2$. Setting $a = J_{n+1}$ and $b=J_{n-1}$ and using the recurrence relation $2 J_n' = J_{n-1} - J_{n+1}$ we get $$\beta_n + \int\{(\frac{n^2}{r^2} - 1)J_n^2 + J_n'^2\}J_0^2 = \frac{\beta_{n-1} + \beta_{n+1}}{2}$$ It seems interesting that one gets that $(\frac{n^2}{r^2} - 1)J_n^2 + J_n'^2$ turns out to be non-negative in an averaged sense. I'll keep thinking about this. –  k3thomps Apr 12 at 20:19
    
I don't think a 2 is missing. I re-did the calculation and what I had above looks correct. –  k3thomps May 21 at 18:03
    
@Wurlitzer and then you have to divide by 4. –  k3thomps May 22 at 12:43

This is not an answer, just a comment following k3thomps computations in remarks, the question is equivalent to, it seems: why is $$ \int_0^\infty \left(\left(\frac{n^2}{x^2}-1\right) J_n^2 +(J_n^\prime)^2\right)J_0^2\geq 0 \,? $$ Below is an alternative formulation of the same question (I can't make it work, but here it is anyway). We have, by definition, $$J_n^{\prime\prime}+\frac{1}{x}J_n -\left(\frac{n^2}{x^2}-1\right) J_n=0$$ Integrating against $J_nJ_0^2$, $n\geq1$, we obtain \begin{eqnarray*} \int_0^\infty \left(\left(\frac{n^2}{x^2}-1\right) J_n^2 + (J_n^\prime)^2\right) J_0^2 &=& \frac{1}{2}\int_0^\infty -(J_n^2)^\prime(J_0^2)^\prime +\frac{1}{x}(J_n^2)^\prime J_0^2\\ &=& \frac{1}{2}\int_0^\infty -(J_n^2)^\prime(J_0^2)^\prime -\frac{1}{x}(J_0^2)^\prime J_n^2+\frac{1}{x^2} J_n^2 J_0^2 \\ &=& \frac{1}{2}\int_0^\infty -(J_n^2)^\prime(J_0^2)^\prime +\frac{1}{x}(J_0^2)^\prime J_n^2\\ &+&\frac{1}{2}\int_0^\infty \frac{1}{x^2} J_n^2 J_0^2 - \frac{2}{x}(J_0^2)^\prime J_n^2\\ &=&\int_0^\infty \left( \frac{1-2x^2}{2x^2} J_0^2 +\frac{x-1}{x}(J_0^2)^\prime \right) J_n^2. \end{eqnarray*}

This is therefore puzzling, it this calculation is correct, because the first term isn't positive.

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