Asymptotic Solutions of Linear ODEs. 2


### Ordinary points of differential equation

#### Theory

See Definition 3.1.1.

We consider complex-analytic solutions to the equation $$u^{(n)}(z)+ p_{n-1} (z)u^{(n-1)}(z)+\ldots+p_1 (z)u'(z)+p_0(z)u(z)=0 \label{eq-3.2.1}$$ where $p_k(z)$ are complex-analytic functions at $z_0\in \bC$.

Assume for a sake of simplicity notations that $z_0=0$ (we can always reach it by the change of variables $\zeta=z-z_0$). Then $$p_k(z)= \sum_{l\ge 0} p_{kl}z^l \label{eq-3.2.2}$$ and we are looking for solution $$u(z)=\sum_{l\ge 0} u_l z^l. \label{eq-3.2.3}$$ Plugging into (\ref{eq-3.2.1}) we get \begin{equation*} \sum_{l\ge n} u_l \frac{l!}{(l-n)!} z^{l-n} + \sum_{0\le k\le n-1}\ \sum_{m\ge 0}\ \sum_{l\ge k} p_{km}u_{l-k} \frac{l!}{(l-k)!} z^{l-k+m}=0 \end{equation*} or changing index in summation ($l:=l+n$ in the first term, $l:=l+k-m$ in the other terms) we get \begin{equation*} \sum_{l\ge 0} u_{l+n} \frac{(l+n)!}{l!} z^{l} + \sum_{0\le k\le n-1}\ \sum_{m\ge 0}\ \sum_{l\ge m} p_{km}u_{l-k-m} \frac{(l+k-m)!}{(l-m)!} z^{l}=0 \end{equation*} or equivalently \begin{equation*} u_{l+n} \frac{(l+n)!}{l!} + \sum_{0\le k\le n-1}\ \sum_{m\le l} p_{km}u_{l+k-m} \frac{(l+k-m)!}{(l-m)!} =0 \end{equation*} and finally $$u_{l+n} = -\frac{l!}{(l+n)!} \sum_{0\le k\le n-1}\ \sum_{m\le l} p_{km}u_{l +k-m} \frac{(l+k-m)!}{(l-m)!} \label{eq-3.2.4}$$ which defines $u_l$ for $l=n,n+1,\ldots$ recurrently as long as $u_0,\ldots, u_{n-l}$ could be chosen arbitrarily.

Theorem 1. If series (\ref{eq-3.2.2}) converge as $|z|< R$ and $u_l$ for $l=n,n+1,\ldots$ are defined by (\ref{eq-3.2.4}) then series (\ref{eq-3.2.3}) converges as $|z|< R$.

Corollary 1. If series (\ref{eq-3.2.2}) converge for all $z$ and $u_l$ for $l=n,n+1,\ldots$ are defined by (\ref{eq-3.2.4}) then series (\ref{eq-3.2.3}) converges for all $z$.

Remark 1. Such functions are called entire.

#### Proof of Theorem 1 (optional)

We sketch the proof. We can rewrite equation as the first order system $$U'(z)=\Lambda (z) U(z) \label{eq-3.2.5}$$ where $U=(u\ u'\ \ldots \ u^{(n-1)})^T$, $^T$ means a transposed matrix and $\Lambda(z)$ is analytic as $|z|< R$.

Definition 1. Let us consider two series: $\sum a_k z^k$ and $\sum_k b_kz^k$. We say that $\sum_k b_kz^k$ dominates $\sum_k a_kz^k$, $\sum_k b_kz^k\ggg \sum_k b_kz^k$ if $|a_k|\le b_k$ for all $k$. Attention! Norm is only in the left! For vector- or matrix- valued functions it should be for each component.

One can prove easily that if $U$ is solution of (\ref{eq-3.2.5}) and $V$ is solution of the similar system $$V'(z)=\Lambda_1 (z) V(z) \label{eq-3.2.6}$$ with matrix $\Lambda_1(z)$ which dominates $\Lambda(z)$ and if $|U_j (0)|\le V_j(0)$ then $U(z)\lll V(z)$.

But if series (\ref{eq-3.2.2}) converges as $|z|< R$ then $\Lambda(z)\lll M E(r-z)^{-1}$ for any $r< R$ and $M=M_r$. Here $E$ is a matrix with all elements $1$. If we add all equations in system (\ref{eq-3.2.6}) we get a single equation $$v'=Mn(r-z)^{-1} v \label{eq-3.2.7}$$ with $v=V_1+\ldots +V_n$. Solution of this equation is $v(z)=v(0) + Mn (\ln r- \ln (r-z))$ which is analytic as $|z|< r$ and therefore it Taylor decomposition at $z=0$ converges in the disk $\{z:\,|z|< r\}$. Since $v(z)\ggg u(z)$ Taylor decomposition of $u(z)$ at $z=0$ converges in the same disk.