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I know that the following differential equation: $$x^2\frac{d^2y(x)}{dx^2}+x\frac{dy(x)}{dx}+(x^2-\alpha^2)y(x)$$ has the solution: $$y(x)=C_1\cdot J_\alpha(x)+C_2\cdot Y_\alpha(x)$$ In my case, the differential equation is of the same form, but the parameter $\alpha$ is a random variable having a Gaussian distribution with zero mean and variance $\sigma$. I have problems to find the distribution of the solution $y(x)$. Can someone give me a hint? Thanks.

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A proper thing would be to look for a solution into a series form, as routinely done in such a case. Coefficients of the series are random variables depending on $\alpha$. – Jon Jun 15 '12 at 10:55
Thanks. So you mean there is no way to give an analytical expression of $y(\alpha,x)$ – Riccardo.Alestra Jun 15 '12 at 11:12
I do not know as I have not done an explicit computation. But the series you get, after averages, can have some simple form. – Jon Jun 15 '12 at 11:49
I have found a simpler route. I hope this helps. – Jon Jun 15 '12 at 12:12
up vote 1 down vote accepted

The simplest approach is to use the following integral formula for a Bessel function (I just consider $J_\alpha$ but for $Y_\alpha$ the argument is very similar) $$ J_\alpha(x)=\frac{1}{2\pi}\int_{-\pi}^\pi e^{-i\alpha\tau+ix\sin\tau}d\tau. $$ From this we get the following correlators $$ \langle J_\alpha(x)\rangle=\frac{1}{2\pi}\int_{-\pi}^\pi e^{-\sigma_\alpha^2\tau^2+ix\sin\tau}d\tau $$ being $\sigma_\alpha$ the variance. Similarly, you will have $$ \langle J_\alpha(x)J_\alpha(y)\rangle=\frac{1}{2\pi}\int_{-\pi}^\pi d\tau\int_{-\pi}^\pi d\tau' e^{-\sigma_\alpha^2(\tau+\tau')^2+ix\sin\tau'+iy\sin\tau} $$ and so on. Note that this integrals cam be evaluated very easily in the limit $x,\ y\rightarrow\infty$ otherwise they have not a closed form.

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