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Consider the following symmetric second order diffusion operator, defined, for $\phi \in \mathcal{C}^{2,1}_c\left(\mathbb{R}\times \mathbb{R}_+\right)$, by: $$L\phi := \lambda_1 \partial_{R_1}(R_1 \phi) - \lambda_2 \partial_{R_2}((\sigma^2 - R_2)\phi) + \frac{1}{2}\lambda_1^2 \partial^2_{R_1, R_1}(\sigma^2\phi)$$ Where $\lambda_1, \lambda_2 > 0$ and $\sigma: (R_1, R_2) \in \mathbb{R}\times \mathbb{R}_+ \mapsto \beta_0 + \beta_1 R_1 + \beta_2 \sqrt{R_2}$, s.t $\beta_0 >0$, $\beta_1 < 0$ and $\beta_2 \in (0, 1]$.

I'm trying to find an expansion to the solutions of the Kolmogorov PDE defined by $L$, i.e: $$\partial_t u(t, R) = L u(t, R), \quad (t, R) \in (0, T) \times \mathbb{R}\times \mathbb{R}_+ $$

The expansion is defined by : $$\forall (t, R) \in [0, T] \times \mathbb{R}\times \mathbb{R}_+: \ u(t, R) = \sum_{k = 0}^{+\infty}\alpha_k e^{\kappa_k t} \Psi_k(R)$$

where $(\Psi_k)_{k\in \mathbb{N}}$ is the family of eigenvectors of $L$ associated with eigenvalues $(\kappa_k)_{k\in \mathbb{N}}$ and $(\alpha_k) \in \mathbb{R}^{\mathbb{N}}$.

Does anyone know how can we determine such eigenvalues/vectors ?

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  • $\begingroup$ usually the approach goes by separation of variables i.e. finding which ODE equations are satisfied by $e^{kR_{1}}f(R_{2})$ and solving those separately. What equations do you get, are they hard to solve? $\endgroup$ May 27, 2023 at 2:56
  • $\begingroup$ @ThomasKojar Yes I've tried, but I don't think that this class of functions is in the solutions set of this PDE. We know that the $j$-th eigenfunction satisfies the following PDE: $(L - \kappa_j \text{Id})\Psi_j = 0$. Suppose that $\Psi_j$ can be written in the form that u've proposed, then $f$ will satisfy the following ODE: $\lambda_2 \left(\sigma^2 - R_2\right)f' = (\lambda_1\kappa_j R_1 + \frac{1}{2}\lambda_1^2 \kappa_j^2 \sigma^2 - \kappa_j)f$. We can solve this ODE in an integral form, but notice that $f$ still depends on $R_1$!. But how can this help me determine the eigenvalues ? $\endgroup$
    – Greyearl
    May 27, 2023 at 10:45
  • $\begingroup$ Another approach on separability is the one suggested in the answer here: math.stackexchange.com/questions/1649303/…. First he studies the 1d-interval version and from that extend to the rectangle version. In your case you will then have to take limit in order to obtain the full upper half plane. $\endgroup$ May 27, 2023 at 15:52
  • $\begingroup$ @ThomasKojar Thanks for the link, it's an interesting read but still, it doesn't help me answer my question. I think I didn't formulate my question well. We're interested in computing the point spectrum of this particular operator (eigenvalues). I think the difficulty stems from the form of the operator, so one idea is to introduce some change of variable so that it has a more friendly form so we can compute the spectrum and the associated eigenfunctions using separability as you have mentioned. $\endgroup$
    – Greyearl
    May 27, 2023 at 20:58
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    $\begingroup$ I was more thinking about separately solving $L_{R_{1}}\psi_{k}=\lambda_{k} \psi_{k}$ and $L_{R_{2}}\phi_{k}=\nu_{k} \phi_{k}$ and then testing whether $\psi_{k}(R_{1})\phi_{k}(R_{2})$ generate a basis for the eigenfunctions of $(L_{R_{1}}+L_{R_{2}})f=\rho f$. $\endgroup$ May 28, 2023 at 15:10

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