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Let us suppose that I have some differential operator $D_{x,y}$ acting on functions of two variables $(x,y)$. I want to solve one eigenvalue equation

$$D_{x,y}f_{ij}(x,y)=\lambda_{i,j} f_{ij}(x,y).$$

I know that the $\lambda_{ij}$ and I know that $f_{ij}$ admit a power series expansion

$$f_{ij}(x,y)=\sum_{m,n=0}^\infty a_{mn}x^{\alpha+m}y^{\beta+n}.$$

I need to find the coefficients $a_{mn}$, which everyone in the literature says its very easy to do using the differential equation with a computer. I want to know how to do that using Mathematica, because it feels to me that it should be possible. Even by hand it is possible, it is just that it is very tedious and I wanted to use Mathematica for this.

What is the way to use Mathematica in order to find the coefficients of a power series expansion of a function that solves some differential equation? How can I find the $a_{mn}$, $\alpha$ and $\beta$, given $D_{x,y}$ and $\lambda_{ij}$?

user1620696
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  • Without a test example it's hard to know exactly how to tackle this question. I would encourage you to code up a system of ODEs showing the type of system you're looking for and [edit] your question to include it. In particular, you might want to clarify whether the $\lambda_{i,j}$ are functions of $x$ and $y$ or not, because if they're constants (and your system is linear, which it appears to be) then I'm pretty sure the solution is regular everywhere and you won't need the coefficients $\alpha$ and $\beta$. – Michael Seifert Jun 23 '23 at 17:41
  • Right, I didn't want to write the specific differential operator to try to make the question more useful to a broader audience. The differential operator in question is the conformal Casimir $$D_{x,y}=x^2(1-x)\partial_x^2+y^2(1-y)\partial_y^2-(a+b+1)(x^2\partial_x+y^2\partial_y)-ab(x+y)+\epsilon \dfrac{xy}{x-y}\left((1-x)\partial_x-(1-y)\partial_y\right).$$ Here $a,b,\epsilon$ are just numbers. Moreover, $\lambda_{ij}$ are eigenvalues, so they are also just numbers the specific solution depends on. – user1620696 Jun 23 '23 at 17:44
  • You'll be more likely to get help if you could give the problem in code (and TeX). It makes it easier for others copy/paste and test ideas. (The generic answer is use AsymptoticDSolveValue, but I don't think it works on PDEs. Another possibility: If you equate and truncate two series, you can solve for the coefficients with SolveAlways. Or equate the CoefficientList[] of each. – Michael E2 Jun 23 '23 at 17:56
  • Here's a pre AsymptoticDSolveValue approach for simple ODEs: https://mathematica.stackexchange.com/a/96696/4999. I avoided SolveAlways because SolveAlways can choke on a large system – Michael E2 Jun 23 '23 at 18:02
  • @MichaelE2 thanks, I'm going to look at that. Indeed I think truncating the series would be fine already. While finding a closed form expression for the coefficients $a_{mn}$ as functions of $(m,n)$ would be optimal, just having Mathematica to evaluate the first few coefficients would also work. – user1620696 Jun 23 '23 at 18:11

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