Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.