Asymptotic Solutions of Linear ODEs. 3


### Regular Singular Points of Differential Equation

#### Incidical Equation

See Definition 3.1.2 for regular singular points.

Again we can assume that $z_0=0$. Let us multiply our $$u^{(n)}(z)+ p_{n-1} (z)u^{(n-1)}(z)+\ldots+p_1 (z)u'(z)+p_0(z)u(z)=0 \label{eq-3.3.1}$$ by $z^n$: $$Lu:=z^n u^{(n)}(z)+q_{n-1} (z) z^{n-1} u^{(n-1)}(z)+\ldots+ q_1 (z) zu'(z)+q_0(z)u(z)=0 \label{eq-3.3.2}$$ and according to Definition 3.1.2 $$q_k(z) := z^{n-k}p_k(z)\qquad k=0,1,\ldots, n-1 \label{eq-3.3.3}$$ are analytic functions at $0$. Let $$\bar{q}_k := q_k(0)\qquad k=0,1,\ldots, n-1 \label{eq-3.3.4}$$ and $$\bar{L}u:=z^n u^{(n)}(z)+\bar{q}_{n-1} z^{n-1} u^{(n-1)}(z)+\ldots+ \bar{q}_1 zu'(z)+\bar{q}_0(z)u(z) \label{eq-3.3.5}$$ and $$L':=L-\bar{L}. \label{eq-3.3.6}$$ Observe that $$\bar{L}z^\alpha:=I(\alpha)z^\alpha,\qquad I(\alpha)=\sum_{0\le k \le n} \alpha (\alpha-1)\cdots (\alpha -k+1)\bar{q}_k. \label{eq-3.3.7}$$ Therefore if $L=\bar{L}$ (the case of the Euler equation) $u=z^\alpha$ satisfies $Lu=0$ if and only if $I(\alpha)=0$.

Definition 1. We call $I(\alpha)$ incidical polynomial and its roots incidical exponents.

#### Simple case

On the other hand, $L'z^\alpha$ contains only $z^{\alpha+1},z^\alpha+2,\ldots$. Let $\alpha$ be an incidical exponent. Then $Lz^\alpha$ contains only $z^{\alpha+1},z^\alpha+2,\ldots$: \begin{equation*} Lz^\alpha = \sum _{l\ge 1} c_l(\alpha) z^{\alpha+l}. \end{equation*} To compensate $z^{\alpha+1}$ we add to $z^\alpha$ correction $u_1z^{\alpha+1}$; then \begin{equation*} Lz^\alpha = I(\alpha+1) u_1 z^{\alpha+1} + \sum _{l\ge 1} c_l(\alpha) z^{\alpha+l} + \sum _{l\ge 1} c_l(\alpha+l+1) u_1 z^{\alpha+l+1} \end{equation*} and the coefficient at $z^{\alpha+1}$ is $I(\alpha+1)u_1+c_1(\alpha)$. Choosing \begin{equation*} u_1= - \frac{1}{I(\alpha)}c_1(\alpha) \end{equation*} we can get rid off $z^{\alpha+1}$. In a similar manner we can get rid of $z^{\alpha+l}$ for all $l=1,2,\ldots$ provided $$I(\alpha+k)\ne 0 \qquad\forall l=1,2,\ldots \label{eq-3.3.8}$$

It is a condition of Subsection 3.1.3. So we proved the first statement here

Theorem 1. Let $I(\alpha=0)$ and (\ref{eq-3.3.8}) be fulfilled. Then

1. There exists a formal solution of $Lu=0$ $$u(z)=\sum_{l\ge 0} u_l (z-z_0)^{\alpha +l}\qquad\text{with }\ u_0=1. \label{eq-3.3.9}$$
2. If $q_k(z)$ are analytic in the disk $\{z:\, |z-z_0|< R \}$ then (\ref{eq-3.3.9}) converges in the same disk (may be without center).

Proof of Statement (b). It could be done in the same manner as of Theorem 3.2.1. We skip it.

Assume now that $\alpha$ is an incidical exponent of multiplicity $r\ge 2$ which means that $$I(\alpha)=I'(\alpha)=\ldots =I^{(r-1)}(\alpha)=0, \qquad I^{(r)}(\alpha)\ne 0. \label{eq-3.3.10}$$ Then not only $u=z^\alpha$ satisfies $\bar{L}u=0$, but also $u=z^\alpha (\log z)^j$ for all $j=1,\ldots,r-1$ (but not for $j=r$).

But what about $L' z^{\alpha} (\log z)^j$? It contains all powers of logarithm $\le j$: \begin{equation*} L' z^\alpha (\log z)^j = \sum _{l\ge 1}\sum_{k\le j}c_{lk}(\alpha) z^{\alpha+l} (\log z)^k. \end{equation*} Also \begin{equation*} \bar{L} z^{\alpha+l}(\log z)^k = I(\alpha+l) z^{\alpha+l}(\log z)^k + \sum_{m\le k-1} d_{km} z^{\alpha+l} (\log z)^m. \end{equation*} Therefore in the same manner as above we can compensate term with $z^{\alpha+l}(\log z)^j$, then term with $z^{\alpha+l}(\log z)^{j-1}$ and so on up to term with $z^{\alpha+l}$ and then move to $l+1$ instead of $l$.

So we proved the first statement here

Theorem 2. Let (\ref{eq-3.3.10}) and (\ref{eq-3.3.8}) be fulfilled. Then for all $j=0,1,\ldots, r-1$

1. There exists a formal solution of $Lu=0$ $$u(z)=\sum_{l\ge 0}\sum_{k\le j} u_{lk} (z-z_0)^{\alpha +l}(\log (z-z_0))^k \qquad\text{with }\ u_{0j}=1. \label{eq-3.3.11}$$
2. If $q_k(z)$ are analytic in the disk $\{z:\, |z-z_0|< R \}$ then $$\sum_{l\ge 0} u_{lk} (z-z_0)^{\alpha +l} \label{eq-3.3.12}$$ converge in the same disk (may be without center).

Proof of Statement (b). It could be done in the same manner as of Theorem 3.2.1. We skip it.

#### General case

Now we get rid off assumption (\ref{eq-3.3.8}). At some moment we need to compensate term with $z^{\alpha+l}(\log z)^m$ (but we allow new terms with increased $l' > l$ instead of $l$ and $m'$ instead of $m$ and also terms with $m' < m$.

To do this we used $u_{lm}z^{\alpha+l}(\log z)^m$ but it worked only as $I(\alpha+l)\ne 0$. Now we assume that $$I(\alpha+l)=I'(\alpha+l)=\ldots =I^{(r'-1)}(\alpha+l)=0, \qquad I^{(r')}(\alpha+l)\ne 0 \label{eq-3.3.13}$$ with $r'\ge 1$.

One can prove that \begin{multline} \bar{L} z^{\alpha+l}(\log z)^{m'}= \frac{m'!}{r'!(m'-r')!} I^{(r')}(\alpha+l)z^{\alpha+l}(\log z)^{m'-r'}+\\ \sum_{k< m'-r'} d_{m'k}(\alpha+l) z^{\alpha+l}(\log z)^{k}\qquad \label{eq-3.3.14} \end{multline} and we can do the compensation at the expense of raising $m$ to $m'=m+r'$.

So we proved the first statement here

Theorem 3. Let (\ref{eq-3.3.10}) be fulfilled (with $r\ge 1$). Then for all $j=0,1,\ldots, r-1$

1. There exists a formal solution of $Lu=0$ $$u(z)=\sum_{l\ge 0}\sum_{k} u_{lk} (z-z_0)^{\alpha +l}(\log (z-z_0))^k \qquad\text{with }\ u_{0j}=1 \label{eq-3.3.15}$$ where $u_{lk}=0$ unless $k$ does not exceed $j$ plus the total multiplicity of incidical exponents $\alpha+1,\ldots,\alpha+l$ 1
2. Statement (b) of Theorem 2 holds.

Proof of Statement (b). It could be done in the same manner as of Theorem 3.2.1. We skip it.

1. if $\beta$ is not an incidical exponent its multiplicity is $0$.