ODEs-in-Stokes-Groupoids

A Project written from pure python to symbolically solve ODEs on the Stokes groupoid.

Instructor: Professor Marco Gualtieri

Introduction

Restate the problem that we want to solve:

$$z^k \frac{d\psi}{dz} = A(z)\psi$$, where $\psi$ is a vector-valued function and $A(z)$ as a given matrix-valued function is holomorphic.

At $k = 0$, it is the regular case, we just work on C and find the solution easily.

At $k = 1$, it is the regular singular case

At $k \geq 2$, it is the irregular singular case

Why singular: if we try to solve by putting it into a vector field, we need to divide both sides by $z^k$, in which case, the vector field has a pole of order $k$, causing the flow to be not defined.

Instead of working away from singular point, we consider it on Stok_k($\mathbb{C}$), the Stokes groupoid, where we have a universal solution $\Phi$

In the case where we could find a fundamental solution easily (e.g. when $A$ is constant)

$$\Psi = \begin{pmatrix} \psi_1 & ... & \psi_n \end{pmatrix}$$

on a simply connected domain $\Omega$, then $\Phi(u, z) = \Psi(t(u, z)) \Psi^{-1}(s(u,z))$, which is defined on $\Omega$ while could be analytically continued to all of Sto_k.

In the case where we can't find it easily, we use Formal Gauge transform (call it $F$) to simplify the system into a diagonal version, and solve the simplified system. Since we know that the Formal Gauge transforms actually form a group acting on Matrix-valued functions, then we transform the solution back by the transform of $F^{-1}$. Then, $\hat{\Phi}(u, z) = \hat{\Psi}(t(u, z)) \hat{\Psi}^{-1}(s(u,z))$ is the formal universal solution on Sto_k, while since $\Phi$ (the actual universal solution) is unique, then this means that $\hat{\Phi}$ is the power series of $\Phi$, so it must be convergent since $\Phi$ is entire.

To do the Gauge transformation, the core idea is to transfer it to the case where $z^k \frac{d\tilde{\psi}}{dz} = (D_0+zD_1+...z^{k-1}D_{k-1})\tilde{\psi}$, where $D_i$ are diagonal, which can be easily solve by entries on both sides.

In order to solve this problem, we first need to make a change of variables $\psi' = F \psi$ (a change of coordinate), whereas $F = 1+zF_1 + z^2F_2 + ...$

Then, let $B(z) = \frac{A(z)}{z}$, then after this change of coordinate, the original system becomes $\frac{d\psi'}{dz} = (FBF^{-1} + \frac{dF}{dz}F^{-1})\psi'$

In this step, we let $F = (I+zH_1)(I+z^2H_2)...$ where $H_p = {ad}^{-1}_{A_0}(A_p^{OD})$ if $k > 1$

and $({ad}_{A_0} - p)^{-1}(A_p^{OD})$ if $k=1$

After this transform, the ODE becomes $z^k \frac{d\psi'}{dz} = (diagonal(A_0) + diagonal(A_1)z + ...) \psi'$

After doing this, we make the second Gauge Transform to transform it to the finite case,

which is the transform $K = \exp(-\int{D_k+D_{k+1}z+...})$

Finally, the simplified system is $z^k \frac{d\psi''}{dz} = (diagonal(A_0) + diagonal(A_1)z + ...+ diagonal(A_{k-1})z^{k-1}) \psi''$

and the overall Gauge transform is $(KF)$[...], and the inverse is $(KF)^{-1}$[...], so $\psi = (KF)^{-1}\psi''$

After we get the fundamental solution $\Psi$, we can construct $\Phi$ on the Sto_k in a similar way.

Dependency

sympy 1.13.2

Get Started

The Notebook of Constant_Case.ipynb contains many examples where $A$ is constant.

The General_Case.ipynb contains many examples where $A$ is not constant and we solve in the way that stated above.

One could view the examples by opening the ipynb files in github directly.

Reference

Gualtieri, Marco, Songhao Li, and Brent Pym. "The stokes groupoids." Journal für die reine und angewandte Mathematik (Crelles Journal) 2018.739 (2018): 81-119.