I am trying to find the eigenfunctions of the following operator:

$$\mathcal{L}f=(-\gamma x+\frac{\mu}{x})f_x+\mu f_{xx}$$

I know that I must somehow use Laguerre polynomials, the solutions to the following ODE. $L_n(x)$ solves $$xy''+(1-x)y'+ny=0$$.

Here is the current ODE I have which should be solved $$\mu {\phi}_{xx}+(-\gamma x + \frac{\mu}{x})\phi_x -\lambda \phi =0$$

That operator is the generator of the following stochastic differential equation $$dA_t=(-\gamma x + \frac{\mu}{x})dt+\sqrt{2\mu}dW_t$$ where $W_t$ is Brownian motion. I am about 90% sure this is true, but this $may$ be something I'm doing wrong. In an example in the notes for the section in which I find this problem, the Professor finds the eigenfunctions of $$\mathcal{L}f=-\alpha x f_x + \frac{1}{2}\sigma^2f_{xx}$$ by solving the ODE $$\phi '' - \frac{2\alpha x}{\sigma^2} \phi ' + \frac{2\lambda }{\sigma^2}\phi=0$$ via substitution with some $y$ to get $$\phi '' - y\phi ' +\frac{\lambda}{\alpha}\phi =0 $$ which is then solved with Hermite polynomials. The reason I include this is that I believe some clever substitution may be necessary in my problem as well, but I cannot figure it out! I greatly appreciate any help.

EDIT: Based on the question, I am almost certain that the eigenvalues are $\lambda = -2\gamma n$ for positive integers $n$. Maybe this can help. EDIT2: For clarification, the generator of a process $dX_t=b(X_t,t)dt+\sigma(X_t,t)dW_t$ is found thus: $$\mathcal{L}f=b(x,t)\cdot \nabla f(x)+\frac{1}{2}a(x,t):\nabla^2 f(x)$$ where $a(x,t)=\sigma \sigma^T$. This is how I found the generator, which I'm 90% sure is right.

  • $\begingroup$ Should there be $-\gamma x $ rather $-\gamma$ in your first equation? $\endgroup$ – Chinny84 May 11 '15 at 6:18
  • $\begingroup$ Yes, thank you ^_^ $\endgroup$ – John Ryan May 11 '15 at 6:29
  • $\begingroup$ Are you totally sure that you require Laguerre polys? $\endgroup$ – Chinny84 May 11 '15 at 6:37
  • $\begingroup$ Here is the question quoted directly: Find the eigenfunctions of the generator of this process. Write these in terms of the Laguerre polynomials. $\endgroup$ – John Ryan May 11 '15 at 6:41

Let $x=t^a$ then Where $$ \dfrac{dx}{dt} = at^{a-1} $$ $$ \dfrac{d}{dx} = \dfrac{dt}{dx}\dfrac{d}{dt} = \frac{1}{a}t^{-a+1}\dfrac{d}{dt}\\ \dfrac{d^2}{dx^2} = \frac{1}{a}t^{-a+1}\dfrac{d}{dt}\frac{1}{a}t^{a-1}\dfrac{d}{dt}= \frac{1}{a^2}t^{2(1-a)}\dfrac{d^2}{dt^2} + \frac{1-a}{a^2}t^{1-2a}\dfrac{d}{dt} $$ Thus we get $$ \frac{\mu}{a^2}t^{2(1-a)}\dfrac{d^2y}{dt^2} + \frac{\mu(1-a)}{a^2}t^{1-2a}\dfrac{dy}{dt} -\gamma t^a\frac{1}{a}t^{-a+1}\dfrac{dy}{dt} +\frac{\mu}{t^a}\frac{1}{a}t^{-a+1}\dfrac{dy}{dt}-\lambda y = 0 $$ This leads to

$$ \frac{\mu}{a^2}t^{2(1-a)}\dfrac{d^2y}{dt^2} + \frac{\mu(1-a)}{a^2}t^{1-2a}\dfrac{dy}{dt} -\gamma t\frac{1}{a}\dfrac{dy}{dt} +\mu \frac{1}{a}t^{-2a+1}\dfrac{dy}{dt}-\lambda y = 0 $$ Let $a =1/2$ Then we get $$ \frac{\mu}{a^2}t\dfrac{d^2y}{dt^2} + \frac{\mu(1-a)}{a^2}\dfrac{dy}{dt} -\gamma t\frac{1}{a}\dfrac{dy}{dt} +\mu \frac{1}{a}\dfrac{dy}{dt}-\lambda y = 0 $$ Which sort of looks like the Lagurre poly equation, up to normalisation. (But check the calculations). Subbing in $a$ we find $$ t\ddot{y} +\left(1-\frac{\gamma }{2\mu}t\right) \dot{y} -\frac{\lambda}{4\mu}y =0. $$ Then using the transformation $t\rightarrow \beta s$ We find $$ s\dfrac{d^2y}{ds^2} +\left(1- s\right)\dfrac{dy}{ds} - \frac{\lambda}{2\gamma}y = 0 $$ Where $\beta = \frac{2\mu}{\gamma}$ the final statement is fixing $$ n = -\frac{\lambda}{2\gamma} \implies \lambda = -2\gamma n. $$

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    $\begingroup$ WOW that is impressive. Thank you so much for the detailed explanation! If I may ask, by what intuition did you start with $x=t^a$? Is there a name for this trick? Also, for anyone who may stumble upon this, the 3rd to last equation should be $$ty''+(1-\frac{\gamma}{2\mu}t)y'-\frac{\lambda}{4\mu}y=0$$ I never would have come up with this, but it is so cool to me. Clearly I need to read an ODE textbook. $\endgroup$ – John Ryan May 11 '15 at 11:08
  • $\begingroup$ Nice catch :). You can edit if you want :). As for the sub, to be honest it was a calculated guess, which when you go through the steps you either become more confident that the ansatz is correct or not. In this case it worked with a suitable $a$. But not to claim a deep insight, but I saw that if I multiply the $x$ through in the original equation I saw you had an $x^2$ so I thought I could get rid of that with an $t^{1/2}$ sub, but wanted the result to be distiller from the generic sub. $\endgroup$ – Chinny84 May 11 '15 at 11:24
  • $\begingroup$ Also, it shows how you can do this for in general. This will not always work (clearly) but it is my thought process :). $\endgroup$ – Chinny84 May 11 '15 at 11:25

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