\documentclass[twoside]{article} \usepackage{amssymb} % font used for R in Real numbers \pagestyle{myheadings} \markboth{\hfil Analytic solutions of n-th order differential equations \hfil EJDE--2002/12} {EJDE--2002/12\hfil Brian Haile \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol. {\bf 2002}(2002), No. 12, pp. 1--14. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \vspace{\bigskipamount} \\ % Analytic solutions of n-th order differential equations at a singular point % \thanks{ {\em Mathematics Subject Classifications:} 30A99, 34A30, 34M35, 46E15. \hfil\break\indent {\em Key words:} linear differential equation, regular singular point, analytic solution. \hfil\break\indent \copyright 2002 Southwest Texas State University. \hfil\break\indent Submitted July 28, 2001. Published February 4, 2002.} } \date{} % \author{Brian Haile} \maketitle \begin{abstract} Necessary and sufficient conditions are be given for the existence of analytic solutions of the nonhomogeneous n-th order differential equation at a singular point. Let $L$ be a linear differential operator with coefficients analytic at zero. If $L^*$ denotes the operator conjugate to $L$, then we will show that the dimension of the kernel of $L$ is equal to the dimension of the kernel of $L^*$. Certain representation theorems from functional analysis will be used to describe the space of linear functionals that contain the kernel of $L^*$. These results will be used to derive a form of the Fredholm Alternative that will establish a link between the solvability of $Ly = g$ at a singular point and the kernel of $L^*$. The relationship between the roots of the indicial equation associated with $Ly=0$ and the kernel of $L^*$ will allow us to show that the kernel of $L^*$ is spanned by a set of polynomials. \end{abstract} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{example}{Example}[section] \renewcommand{\theequation}{\thesection.\arabic{equation}} \catcode@=11 \@addtoreset{equation}{section} \catcode@=12 \section{Introduction} In 1969, Harris, Sibuya and Weinberg \cite{h3} proved a theorem, based on certain techniques of functional analysis, that contains as corollaries the existence theorems of O. Perron and F. Lettenmeyer. Harris, Sibuya and Weinberg were able to show, under certain conditions, that there exists a polynomial $f(z)$ such that the linear differential system $z^D\frac{dy}{dz} - A(z)y = f(z),$ where $A(z)$ is analytic at $z=0$, and $D=diag\{d_1,\dots, d_n\}$ with nonnegative integers $d_i$, has a solution analytic at $z=0$. Two years later, H.L. Turrittin \cite{t2} posed the following problem: Given the equation $\frac{dW}{dz} = \sum_{j=0}^{\infty}\,z^{-j}A_j\,W + \sum_{j=1}^{\infty}\,z^{-j}B_j$ where both series converge if $|z|\ge z_0 > 0$, with a formal solution $W(z) = \sum_{j=0}^{\infty}\,z^{-j}C_j,$ what are the necessary and sufficient conditions that the formal solution converges? At the same time Turrittin speculated that the afore mentioned paper of Harris, Sibuya and Weinberg might lead to an answer to this question. L.J. Grimm and L.M. Hall answered the question posed by Turrittin in \cite{g2}. Although their initial work was inspired by the Harris, Sibuya and Weinberg paper, Grimm and Hall actually employed the theory of normally solvable and Noetherian operators, much as Korobe\u inik did, to derive an analogue of the Fredholm alternative for certain systems of functional differential operators. Then they used A.E. Taylor's representation of Banach spaces of analytic functions \cite{t1} to obtain conditions for the solvability of certain nonhomogeneous equations they had studied previously in \cite{g3}. The primary difficulty encountered when applying the results of \cite{g2} comes when attempting to describe the cokernel (see \cite{g2}, \cite{h1}, \cite{h2}) of the operator. In this paper we will completely describe the cokernel of the operator $L$, defined by $$\label{eq1} Ly(z) = z^ny^{(n)}(z)+z^{n-1}a_1(z)y^{(n-1)}(z)+\cdots+a_n(z)y(z)$$ where $a_1(z),\dots, a_n(z)$ are analytic at the point $z=0$. We will produce solvability conditions for the nonhomogeneous Bessel equation, that are independent of the order, but match those of \cite{h2} if it is an equation of integer order. First, we will define a collection of Banach spaces, each containing functions analytic at the origin, and such that $L$ is a continuous linear operator on this collection. We will show that the dimension of the kernel of $L$ is equal to the dimension of the cokernel of $L$. Then we will use Taylor's representation of Banach spaces of analytic functions \cite{t1} to essentially set up a relationship between the roots of the indicial equation (from the Frobenius method) and the necessary form of the nonhomogeneous term $g$ that allows us to find a solution to $Ly=g$ analytic at the origin. \section{Definitions and Preliminaries} Let $B_1$ and $B_2$ be Banach spaces and let $T: B_1 \to B_2$ be a continuous linear operator with domain $B_1$. Denote the conjugate of $T$ by $T^\ast$. \paragraph{Definition} The operator $T$ is normally solvable if $T(B_1)$ is closed in $B_2$. \smallskip The following lemma, due to Korobe\u inik \cite{j1}, gives another characterization of normal solvability. The annihilator of the set $W$ will be denoted by $W^{\,\perp}$. \begin{lemma} \label{Lm2.1} Let $B_1$ and $B_2$ be Banach spaces, and let $T:B_1 \to B_2$ be a continuous linear operator. Then the following two properties are equivalent. \begin{enumerate} \item[(a)] $T$ is normally solvable. \item[(b)] The equation $Ty = u$, $u \in B_2$, is solvable in $B_1$ if and only if $u \in (\ker T^*)^\perp$. \end{enumerate} \end{lemma} \paragraph{Definition} % Defn2.2 The d-characteristic of $T$ is the ordered pair $(\alpha(T),\beta(T))$, where $\alpha(T) = \dim(\ker T)$ and $\beta(T) = \dim(B_2/T(B_1)$. $\beta(T)$ is called the defect number of $T$. \paragraph{Definition} % Defn2.3 The index of the operator $T$, denoted by $ind(T)$, is the number $\beta(T) - \alpha(T)$. \paragraph{Definition} % Defn2.4 The operator $T$ is called a Noetherian operator if $T$ is normally solvable and if both $\alpha(T)$ and $\beta(T)$ are finite. \smallskip \paragraph{Remark} If $\beta(T) < \infty$, then $\beta(T) = \dim(\mathop{\rm coker} T) = \dim(\ker T^*)$. \smallskip The following theorem is due to I.C. Gohberg and M.G. Kre\u in \cite{g1}. \begin{theorem} \label{Thm2.1} Let $T: B_1 \to B_2$ be a continuous Noetherian operator, and let $T_1: B_1 \to B_2$ be a compact operator. Then $T_2 = T + T_1$ is a Noetherian operator from $B_1$ into $B_2$ with $\mathop{\rm ind} (T_2) = \mathop{\rm ind}(T)$. \end{theorem} For an arbitrary Noetherian operator the following theorem holds. \begin{theorem}[Fredholm Alternative] \label{Thm2.2} If $B_1$ and $B_2$ are Banach spaces and $T:B_1\to B_2$ is a Noetherian operator with conjugate $T^*$, then \begin{enumerate} \item[(i)] The equation $T\,y = u$ has a solution (in $B_1$) if and only if $u\in (\ker T^*)^\perp$. \item[(ii)] The equation $T^*\,f = g$ has a solution (in $B_2^*$) if and only if $g\in (\ker T^{**})^\perp$. \end{enumerate} \end{theorem} The proof of this theorem can be found in Korobe\u inik \cite[Lemma 2.1]{j1}. \section{Properties of the Operator} Let $G = \{z: |z|<1\}$ and define $A_p$ to be the Banach space of functions $\upsilon(z)$ analytic in $G$ and $p$ times continuously differentiable on $\overline{G}$, with norm $\|\upsilon(z)\|_p = \{\max |\upsilon^{(i)}(z)|,\; 0 \leq i \leq p,\; |z| = 1\}.$ Consider the operator $L:A_n \to A_0$, defined in (\ref{eq1}), where the $a_i$ are now in $A_0$. In order to apply Theorem \ref{Thm2.1} to the operator $L$, we need the following two results. A proof for each of these two lemmas can be easily obtained by following the techniques used by Korobe\u inik \cite{j1}. \begin{lemma} \label{Lem3.1} The operator $l_0:A_n \to A_0$ defined by $l_0y(z) \equiv z^ny^{(n)}(z)$ is Noetherian with $\alpha(l_0) = \beta(l_0)$. \end{lemma} \begin{lemma} \label{Lm3.2} The operator $l_i:A_n \to A_0$, $1 \leq i \leq n$, defined by \\ $l_iy(z) = z^{n-i}a_i(z)y^{(n-i)}(z)$, $a_i \in A_0$, is compact. \end{lemma} Now we are able to apply Theorem \ref{Thm2.1} to the operator $L$ to obtain the following theorem. \begin{theorem} \label{Thm3.1} $L$ is a Noetherian operator with $\alpha(L) = \beta(L)$. \end{theorem} \section{Description of the Banach Spaces} Let $\tilde{A}$ be the space of all functions that are analytic in $G$. If $f \in \tilde{A}$, $f(z)$ can be expanded in a power series in $z$, $$f(z) = \sum_{k=0}^{\infty} f_kz^k , \quad \mbox{convergent in } G.$$ \paragraph{Definition} % Defn4.1 If $f$ and $g$ are elements of $\tilde{A}$ such that $$f(z) = \sum_{k=0}^{\infty} f_kz^k \quad \mbox{and}\quad g(z) = \sum_{k=0}^{\infty} g_kz^k\,,$$ we define the Hadamard product of $f$ and $g$ by $$B(f,g;z) = \sum_{k=0}^{\infty} f_kg_kz^k \,,\quad z \in G.$$ The series $B(f,g;z)$, thus defined, is an element of $\tilde{A}$ and defines a continuous linear functional of $f$ over $A_0$. \paragraph{Definition} % Defn4.2 Define $A_0^0$ as the class of all $F \in \tilde{A}$ such that $$\lim_{r \to 1^-}B(u,F;r)$$ exists for each $u \in A_0$. \begin{theorem} \label{Thm4.1} $A_0^0$ is a Banach space if the norm is defined by $$\|F\| = \lim_{r \to 1^-} \sup_{\|u\|_0=1} |B(u,F;r)|,\quad u \in A_0.$$ \end{theorem} Let $A_0^*$ denote the space of all linear functionals defined on $A_0$. Recall that, as $u$ varies over $A_0$, $B(u,F;r)$ defines a linear functional with norm $\sup_{\|u\|_0=1} |B(u,F;r)|,\;\; u \in A_0$. Then, if $F \in A_0^0$, $$\label{eq2} \gamma(u) = \lim_{r \to 1} B(u,F;r), \;\; u \in A_0$$ defines an element $\gamma$ of $A_0^*$. Define the mapping $\Lambda: A_0^0 \to A_0^*$ by $\Lambda(F) = \gamma$. Let $\phi \in A_0^*$ and define the function $$H(z) = \sum_{k=0}^{\infty} \phi(z^k)z^k,\quad z \in G.$$ \begin{theorem} \label{Thm4.2} $H(z) \in A_0^0$. \end{theorem} Now, we may also define the mapping $\Gamma: A_0^* \to A_0^0$ by $\Gamma(\phi) = H$. \begin{theorem} \label{Thm4.3} The mapping $\Lambda$ is an isometric isomorphism and $\Lambda^{-1} = \Gamma$. \end{theorem} Therefore, as a result of theorems \ref{Thm4.2} and \ref{Thm4.3}, each linear functional in $A_0^*$ can be uniquely represented (or determined) by an element in $A_0^0$. We restate this result in the following theorem. \begin{theorem} \label{Thm4.4} Every linear functional $\gamma \in A_0^*$ is representable in the form $$\label{eq3} \gamma(u) = \lim_{r \to 1}\frac{1}{2\pi}\int_{0}^{2\pi}u(\rho e^{i\theta}) F \big(\frac{r}{\rho}e^{-i\theta} \big)d \theta,\quad u \in A_0,$$ where $r < \rho < 1$ and $F \in A_0^0$. $F$ uniquely determines and is uniquely determined by $\gamma$, and $\|\gamma\| = \|F\|$. \end{theorem} \section{Description of the Cokernel} Using the results of Theorem \ref{Thm4.4}, we can apply the Fredholm Alternative to obtain the following theorem which gives necessary and sufficient conditions for the solvability of $Ly = g$, $g \in A_0$. \begin{theorem} \label{Thm5.1} For $g \in A_0$, the equation $Ly = g$ has a solution in $A_n$ if and only if $$\label{eq4} \lim_{r \to 1} B(g,f;r) = 0$$ for all $f \in A_0^0$ such that $$\label{eq5} \lim_{r \to 1} B(Ly,f;r) = 0$$ for all $y \in A_n$. \end{theorem} \paragraph{Proof.} Define the space ${\cal F}(L)$ by ${\cal F}(L) = \Gamma(\ker(L^*))$. Hence ${\cal F}(L)$ is a subspace of $A_0^0$ and can be represented by $$\label{eq6} {\cal F}(L) = \{f \in A_0^0: \mbox{f satisfies (\ref{eq5}) for all }y \in A_n \}.$$ Now, from Theorem \ref{Thm2.2}, the equation $Ly = g$ has a solution in $A_n$ if and only if $u$ is in $ker(L^*)^{\perp}$, and $u$ is in $ker(L^*)^{\perp}$ if and only if $\gamma(u) = 0$ for all $\gamma \in ker(L^*)$. But by Theorem \ref{Thm4.4} this means $$\gamma(u) = \lim_{r \to 1} B(g,f;r) = 0$$ for every $f \in \Gamma(\ker(L^*)) = {\cal F}(L)$. \hfill$\diamondsuit$ Equation (\ref{eq5}) characterizes the cokernel of $L$ $(\ker(L^*))$ and equation (\ref{eq4}) characterizes the annihilator of the cokernel of $L$ $(\ker(L^*)^{\perp})$. Next, we will use theorems \ref{Thm3.1} and \ref{Thm5.1} to describe completely the structure of the cokernel of $L$. We will do this by developing a rather intriguing relationship between the coefficients of a Frobenius series solution for $Ly = 0$ and the coefficients of a series for any member of $A_0^0$ that satisfies (\ref{eq5}). %----- Definition 5.1 ----- \paragraph{Definition} The factorial function'' $r^{(k)}$ is defined as follows, according to the value of $k$: \begin{enumerate} \item[(a)] if $k=1,2,3,\dots$, then $r^{(k)}=r(r-1)(r-2) \cdots (r-k+1)$, \item[(b)] if $k=0$, then $r^{(0)}=1$, \item[(c)] if $k=-1,-2,-3,\dots$, then $r^{(k)}=\frac{1}{(r+1)(r+2)\cdots (r-k)}$, \item[(d)] if $k$ is not an integer, then $r^{(k)} = \frac{\Gamma(r+1)}{\Gamma(r-k+1)}$. \end{enumerate}\smallskip If we seek a Frobenius series solution for $Ly = 0$ of the form \newline $w(z) = \sum_{k=0}^{\infty}c_kz^{k+r}$, the indicial equation will be $$F(r) = r^{(n)}+r^{(n-1)}a_{1_0}+ \cdots +r^{(2)}a_{(n-2)_0}+r^{(1)}a_{(n-1)_0}+a_{n_0} = 0.$$ The following theorem is an immediate consequence of theorem \ref{Thm3.1} and \ref{Thm5.1}. \begin{theorem} \label{Thm5.2} If the equation $F(r) = 0$ does not have a nonnegative integer root, then the equation $Ly = g$ has a solution in $A_n$ for any $g \in A_0$. \end{theorem} \paragraph{Proof.} We know that $dim(\ker(L^*)) = 0$ from Theorem \ref{Thm3.1}. So $f \equiv 0$ is the only solution to (\ref{eq5}) and (\ref{eq4}) is satisfied for any $g \in A_0$. Therefore, by Theorem \ref{Thm5.1}, equation $Ly = g$ has a solution in $A_n$ for any\\ $g \in A_0$. \hfill $\diamondsuit$ \smallskip Now assume that the indicial equation, $F(r) = 0$, has at least one nonnegative integer root. We will show that the cokernel of $L$ is always spanned by a set of polynomials and therefore the existence of a solution to $Ly = g$ in $A_n$ depends only on a finite number of the coefficients in the series expansion of $g(z)$. First, we need to rewrite equation (\ref{eq5}). Let $$f(z)= \sum_{k=0}^{\infty}f_kz^k \quad\mbox{and}\quad a_i(z)= \sum_{k=0}^{\infty}a_{i_k}z^k, \quad i=1,\dots,n.$$ Then another form of (\ref{eq5}) is: $$\label{eq7} \begin{array}{c} \vspace{.15cm} F(0)f_0 + \sum_{k=1}^\infty a_{n_k}\,f_k = 0 \\ \vspace{.15cm} F(1)f_1 + \sum_{k=1}^\infty \left[a_{(n-1)_k}+a_{n_k}\right]f_{k+1} = 0\\ \vspace{.15cm} F(2)f_2 + \sum_{k=1}^\infty \left[2a_{(n-2)_k}+2a_{(n-1)_k}+a_{n_k}\right]f_{k+2} = 0\\ F(3)f_3 + \sum_{k=1}^\infty \left[6a_{(n-3)_k}+6a_{(n-2)_k}+3a_{(n-1)_k}+a_{n_k}\right]f_{k+3} = 0\\ \vdots \\ F(n)f_n + \sum_{k=1}^\infty \left[n^{(n-1)}a_{1_k}+n^{(n-2)}a_{2_k}+\cdots +na_{(n-1)_k}+a_{n_k}\right]f_{k+n}=0\\ \vdots \\ F(r)f_r + \sum_{k=1}^\infty \left[r^{(n-1)}a_{1_k}+r^{(n-2)}a_{2_k}+\cdots +ra_{(n-1)_k}+a_{n_k}\right]f_{k+r}=0\\ \vdots \end{array}$$ \begin{theorem} \label{Thm5.3} If the equation $F(r) = 0$ has exactly one nonnegative integer root, $r$, then the cokernel of $L$ is spanned by a polynomial of degree $r$. \end{theorem} \paragraph{Proof.} Since there is only one nonnegative integer root, $r$, we may reduce (\ref{eq7}) to $$\label{eq8} \begin{array}{c} \vspace{.15cm} F(0)f_0 + \sum_{k=1}^r a_{n_k}\,f_k = 0 \\ \vspace{.15cm} F(1)f_1 + \sum_{k=1}^{r-1} \left[a_{(n-1)_k}+a_{n_k}\right]f_{k+1} = 0\\ \vspace{.15cm} F(2)f_2 + \sum_{k=1}^{r-2} \left[2a_{(n-2)_k}+2a_{(n-1)_k}+a_{n_k}\right]f_{k+2} = 0\\ F(3)f_3 + \sum_{k=1}^{r-3} \left[6a_{(n-3)_k}+6a_{(n-2)_k}+3a_{(n-1)_k}+a_{n_k}\right]f_{k+3} = 0\\ \vdots \\ F(r-1)f_{r-1} + \left[(r-1)^{(n-1)}a_{1_1} + \cdots + (r-1)^{(1)}a_{(n-1)_1}+a_{n_1}\right]f_r = 0\\ F(r)f_r = 0 \end{array}$$ by setting $f_k = 0$ for $k > r$. Back substitution for $f_k,\,\, k < r$, yields a polynomial of degree $r$ in the cokernel of $L$. We also know that $dim(\ker(L)) = 1$, which implies, from Theorem \ref{Thm3.1}, that $\dim(\ker(L^*)) = 1$. Therefore the cokernel of $L$ is spanned by a polynomial of degree $r$. \hfill $\diamondsuit$\smallskip We now assume that the indicial equation for $Ly = 0$ has $q > 1$ nonnegative integer roots, $r_1,\, \dots,\, r_q$. Assume $r_1 > r_2 > \cdots > r_q$ and let $N = r_1 - r_q$. Under these assumptions we are guaranteed, following the standard Frobenius series solution technique, at least one solution in $A_n$. If there is to be more than one linearly independent solution in $A_n$, it can be determined by using the smaller root, $r_q$, of the indicial equation and calculating the first $N + 1$ coefficients of the Frobenius series. If $c_0,\,c_1,\dots,c_N$ denote these coefficients and $H=(h_{i,j})$ is the upper-triangular matrix where $$h_{i,j}= \left\{ \begin{array}{ll} 0, & \mbox{if i>j}\\ F(r_1-i+1), & \mbox{if i=j}\\ (r_1-j+1)^{(n-1)}a_{1_{j-i}} + \cdots \\ +(r_1-j+1)^{(1)}a_{(n-1)_{j-i}}+a_{n_{j-i}}, & \mbox{if i r_1, then the number of linearly independent solutions to (\ref{eq10}) coincides with the number of linearly independent polynomials in the cokernel of L. The following lemma is the basis of a very lucrative relationship between the number of linearly independent solutions in A_n of Ly = 0 and the number of linearly independent polynomials in the cokernel of L. \begin{lemma} \label{Lma5.1} Let B be an n \times n matrix and B^t be its backward transpose, then \\ B^t \sim B. \end{lemma} \paragraph{Proof.} Denote the n \times n backward identity by P, i.e.,$$P = \left[\begin{array}{ccccc} 0 & \cdots & 0 & 0 & 1 \\ 0 & \cdots & 0 & 1 & 0 \\ \vdots & &\cdots & & \vdots \\ 0 & 1 & 0 &\cdots & 0 \\ 1 & 0 & 0 &\cdots & 0 \end{array}\right].$$Then B^t = PB^TP where B^T is the usual transpose of B. It is known that B^T \sim B which implies that there exists a nonsingular matrix R such that B^T = RBR^{-1}. So, if we let Q = PR then Q is a nonsingular matrix such that B^t = QBQ^{-1}. \hfill \diamondsuit\smallskip Now, from Lemma \ref{Lma5.1}, we know that H \sim H^t. In other words, there exists a nonsingular matrix, Q, such that QH = H^tQ. Therefore, equations (\ref{eq9}) and (\ref{eq10}) will have the same number of linearly independent solutions directly related to each other by the matrix Q. We state this result in the following lemma. \begin{lemma} \label{Lma5.2} If Q is a nonsingular matrix such that QH = H^tQ and [c_N \cdots c_0]^T is a solution of (\ref{eq9}) then Q[c_N \cdots c_0]^T is a solution of (\ref{eq10}). \end{lemma} Together, lemmas \ref{Lma5.1} and \ref{Lma5.2} describe the unique relationship between the kernel and cokernel of the operator L. Thus, we can completely describe the structure of the cokernel of L when F(r) = 0 has two or more nonnegative integer roots. These results are presented in the following theorem. \begin{theorem} \label{Thm5.4} Assume the indicial equation, F(r) = 0, has q (q \leq n) nonnegative integer roots, r_1,\, \dots,\,r_q, with r_1 > r_2 > \, \cdots > r_q. Let N_i = r_i - r_{i+1} and \\ \eta(p) = \sum_{i=1}^{p} N_i, p \in \{1, \dots, q-1\}. Then, if Ly = 0 has m + 1 (m+1 \leq q) linearly independent solutions in A_n, there exist integers p_1 < p_2 < \cdots < p_m \leq q-1 such that the cokernel of L is spanned by the functions f_0(z) \cup \{f_j(z)\}_{j=1}^{m} where f_0(z) is a polynomial of degree r_q and f_j(z) is a polynomial of degree r_q + \eta(p_j). \end{theorem} \paragraph{Proof.} Following the Frobenius method for solving Ly = 0 we find a solution in A_n associated with the largest root, r_1. Then we check for a solution in A_n associated with each of the smaller roots. With the hypotheses above we will get solutions associated with the following m + 1 roots,$$\{r_1, r_1 - \eta(p_1), r_1 - \eta(p_2), \dots, r_1 - \eta(p_m) \},$$where p_1, p_2, \dots, p_m are integers such that 1 \leq p_1 < p_2 < \cdots < p_m \leq q-1, and these will be linearly independent. Thus (\ref{eq9}) will have m + 1 linearly independent solutions, where each of the coefficients$$\{c_N, c_{N-\eta(p_1)}, c_{N-\eta(p_2)}, \dots, c_{N-\eta(p_m)} \}$$are arbitrary. From lemmas \ref{Lma5.1} and \ref{Lma5.2}, we will get m+1 linearly independent solutions to (\ref{eq10}), where each of the coefficients$$\{f_{r_q}, f_{r_q+\eta(p_1)}, f_{r_q+\eta(p_2)}, \dots, f_{r_q+\eta(p_m)} \}$$are arbitrary. Therefore, each of the polynomials $$\label{eq11} \begin{array}{c} f_0(z) = f_{0_0} + f_{0_1}z + \cdots + f_{0_{r_q}}z^{r_q} \vspace{.1cm} \\ f_1(z) = f_{1_0} + f_{1_1}z + \cdots + f_{1_{r_q+\eta(p_1)}}z^{r_q+\eta(p_1)}\vspace{.1cm}\\ f_2(z) = f_{2_0} + f_{2_1}z + \cdots + f_{2_{r_q+\eta(p_2)}}z^{r_q+\eta(p_2)} \\ \vdots \\ f_m(z) = f_{m_0} + f_{m_1}z + \cdots + f_{m_{r_q+\eta(p_m)}}z^{r_q+\eta(p_m)} \end{array}$$ is an element of \ker(L^*). But, from Theorem \ref{Thm3.1} we know that$$\dim(\ker L^*) = \dim(\ker L) = m + 1.$$Hence, the cokernel of L is spanned by the polynomials in (\ref{eq11}). \hfill \diamondsuit \smallskip Theorems \ref{Thm5.1}, \ref{Thm5.2}, \ref{Thm5.3}, and \ref{Thm5.4} describe completely the cokernel of the operator L, which is either zero dimensional or is spanned by polynomials. Therefore, if the nonhomogeneous term in Ly = g is represented by the series $$\label{eq12} g(z) = \sum_{k=0}^{\infty} g_kz^k,$$ the solvability in A_n of Ly = g is determined by a finite number of the coefficients in (\ref{eq12}). \section{Examples} In this section we will illustrate some of the preceding results. Consider the nonhomogeneous Bessel equation of order \nu. %----- Example 6.1 ----- \begin{example} \rm Let L_1:A_2 \to A_0 be defined by$$L_1\,y(z) = z^2\,y'' (z) + zy' (z) + (z^2-\nu^2)y(z).$$\end{example} If we use the Frobenius method to solve L_1y = 0 we obtain the indicial function F(r) = (r - \nu)(r + \nu), so \nu and -\nu are the roots of the indicial equation. Without loss of generality assume that \nu \geq 0. If \nu is not an integer, F(r) = 0 will not have a nonnegative integer root. Thus, by Theorem \ref{Thm5.2}, Ly = g will have a solution in A_2 for any function g \in A_0. If \nu is an integer, then F(r) = 0 has exactly one nonnegative integer root, \nu. Thus, by Theorem \ref{Thm5.3}, the cokernel of L_1 is spanned by a polynomial of degree \nu. Following the proof of Theorem \ref{Thm5.3} we are able to produce this polynomial. First, we derive the finite system of equations$$\begin{array}{c} F(0)f_0 + \sum_{k=1}^{\nu} a_{n_k}\,f_k = 0 \vspace{.1cm} \\ \vspace{.1cm} F(1)f_1 + \sum_{k=1}^{\nu -1} \left[a_{(n-1)_k}+a_{n_k}\right]f_{k+1} = 0\\\vspace{.1cm} F(2)f_2 + \sum_{k=1}^{\nu -2} \left[2a_{(n-2)_k}+2a_{(n-1)_k}+a_{n_k}\right]f_{k+2} = 0\\ F(3)f_3 + \sum_{k=1}^{\nu -3} \left[6a_{(n-3)_k}+6a_{(n-2)_k}+3a_{(n-1)_k}+a_{n_k}\right]f_{k+3} = 0\\ \vdots \\ F(\nu -1)f_{\nu -1} + \left[(\nu -1)^{(n-1)}a_{1_1} + \cdots + (\nu -1)^{(1)}a_{(n-1)_1}+a_{n_1}\right]f_{\nu} = 0\\ F(\nu)f_{\nu} = 0, \end{array} $$similar to those in (\ref{eq8}); then, back substitution yields the polynomial$$f(z) = z^{\nu} + \sum_{m=1}^{\lfloor \frac{\nu}{2} \rfloor} \frac{(\nu - m -1)!}{2^{2m}\, m!(\nu - 1)!} z^{\nu -2m},$$which spans the cokernel of L_1. So, by Theorem \ref{Thm5.1}, L_1y = g will have a solution in A_2 if and only if the coefficients in the series expansion of g(z) = \sum_{k=0}^\infty g_kz^k satisfy$$g_{\nu} + \sum_{m=1}^{\lfloor \frac{\nu}{2} \rfloor} \frac{(\nu - m -1)!}{2^{2m}\, m!(\nu - 1)!} g_{\nu -2m} = 0. $$%----- Remark 6.1 ----- \paragraph{Remark} This example was given previously by Hall \cite{h2}. However, by applying the techniques developed in this article, we have obtained solvability conditions for L_1y=g, the nonhomogeneous Bessel equation of order \nu, independent of the value of \nu. We obtained this condition, which matches that of \cite{h2} if \nu is an integer, without converting the equation to a system. \smallskip %----- Example 6.2 ----- \begin{example} \rm Let L_2:A_2 \to A_0 be defined by$$L_2\,y(z) = z^2\,y''(z) - 2z(z-2)y'(z) + 2(2-3z)y(z).$$\end{example} Here, F(r) = 0 has roots r_1 = 4, r_2 = 1 and N = r_1 - r_2 = 3. We will use Theorem \ref{Thm5.4} to describe the cokernel of the operator L_2. First, we obtain the coefficient matrices of (\ref{eq9}) and (\ref{eq10}).$$H = \left[\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & -2 & -2 & 0 \\ 0 & 0 & -2 & -4 \\ 0 & 0 & 0 & 0 \end{array}\right] \quad and \quad H^t = \left[\begin{array}{cccc} 0 & -4 & 0 & 0 \\ 0 & -2 & -2 & 0 \\ 0 & 0 & -2 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right].$$Then, from Lemma \ref{Lma5.1}, we know that H \sim H^t and so there exists a nonsingular matrix$$Q = \left[\begin{array}{cccc} 1 & -2 & 2 & 8 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \end{array}\right]$$such that QH = H^tQ. Now, \left[\begin{array}{cccc} 1 & 2 & -2 & 1 \end{array}\right]^T is a solution to H\vec(c) = \vec{0} which implies, by Lemma \ref{Lma5.2}, that Q\left[\begin{array}{cccc} 1 & 2 & -2 & 1 \end{array}\right]^T = \left[\begin{array}{cccc} 1 & 0 & 0 & 1 \end{array}\right]^T is a solution to \\ H^t\left[\begin{array}{cccc} f_1 & f_2 & f_3 & f_4 \end{array}\right]^T = \vec{0}. Thus, by setting f_k = 0, k > 4, in (\ref{eq7}) we can use back substitution to get \displaystyle{f_0 = {3 \over 2}f_1}. So, we get two linearly independent polynomials, f_1(z) = z^4 and \\ f_2(z) = z + {3 \over 2} that span the cokernel of L_2. Thus, by Theorem \ref{Thm5.1}, L_2y = g will have a solution in A_2 if and only if the coefficients in the series expansion of g(z) satisfy both g_4 = 0 and g_1 = -{3 \over 2}g_0. %----- Remark 6.2 ----- \paragraph{Remark} Back substitution is necessary to find the f_k, k < r_q=1, since the equation H^t\left[\begin{array}{cccc} f_1 & f_2 & f_3 & f_4 \end{array}\right]^T = \vec{0} is derived from the first through the fourth equations of (\ref{eq7}). \smallskip %----- Example 6.3 ----- \begin{example} \rm Let L_3:A_4 \to A_0 be defined by$$L_3\,y(z) = z^4\,y^{(4)}(z) + z^3a_1(z)y^{(3)}(z) + z^2a_2(z)y''(z) + za_3(z)y' + a_4(z)y(z),$$where a_1(z) = z^4-4, a_2(z) = 12z^2-\frac{1}{2}z+6, a_3(z) = -3z^3+13z^2-z-1, and a_4(z) = 6z^4+z^2-11z+1. \end{example} In this case, F(r) = 0 has roots r_1=5, r_2=4, r_3=1, and r_4=0. So, by Theorem \ref{Thm5.4}, we know that the cokernel of L_3 is spanned by the function f_0(z) = 1 and, possibly as many as three more polynomials, f_j(z), of degree \eta(p_j), 1 \leq p_j \leq 3. To apply Theorem \ref{Thm5.4}, i.e., determine the polynomials in the cokernel of L_3, it is enough to find the matrix H^t of (\ref{eq10}) and solve H^t\left[\begin{array}{cccccc} f_0&f_1&f_2&f_3&f_4&f_5 \end{array}\right]^T = \vec{0}.$$H^t = \left[\begin{array}{cccccc} 0 & -11 & 1 & 0 & 6 & 0 \\ 0 & 0 & -12 & 14 & -3 & 6 \\ 0 & 0 & 12 & -14 & 3 & -6 \\ 0 & 0 & 0 & 12 & -17 & 112 \\ 0 & 0 & 0 & 0 & 0 & -21 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right] Thus, $\left[\begin{array}{cccccc} f_0&f_1&f_2&f_3&f_4&f_5 \end{array}\right]^T = \left[\begin{array}{cccccc} \alpha&\frac{533}{11}\beta&101\beta&102\beta&72\beta&0 \end{array}\right]^T$ is a solution to $H^t\left[\begin{array}{cccccc} f_0&f_1&f_2&f_3&f_4&f_5 \end{array}\right]^T = \vec{0}$, where $\alpha$ and $\beta$ are arbitrary constants. Therefore, by setting $f_k = 0$, $k > 5$, we get two linearly independent polynomials, $f_0(z) = 1$ and $f_1(z) = \frac{533}{11}z + 101z^2 + 102z^3 + 72z^4$ that span the cokernel of $L_3$. So there exists only one integer, $p_1=2$, such that the cokernel of $L_3$ is spanned by $f_0(z)$, a polynomial of degree $r_4=0$, and $f_1(z)$, a polynomial of degree $r_4+ \eta(p_1)=0+(1+3)$. So, by Theorem \ref{Thm5.1}, $L_3y = g$ will have a solution in $A_4$ if and only if the coefficients in the series expansions of $g(z)$ satisfy the conditions $g_0 = 0$ and $\frac{533}{11}g_1 + 101g_2 + 102g_3 + 72g_4 = 0$. \paragraph{Acknowledgments.} I wish to express my gratitude to Professor Leon Hall for his guidance. This paper stems from work done under his direction as part of my dissertation at the University of Missouri at Rolla. \begin{thebibliography}{00} \frenchspacing \bibitem{g1} I. C. Gohberg and M. G. Kre\u in, {\em The basic propositions on defect numbers, root numbers, and indices of linear operators}, Amer. Math. Soc. Transl. {\bf 2} (1960), 185-264. \bibitem{g2} L. J. Grimm and L. M. Hall, {\em An alternative theorem for singular differential systems}, J. Differential Equations {\bf 18} (1975), 411-422. \bibitem{g3} L. J. Grimm and L. M. Hall, {\em Holomorphic solutions of functional differential systems near singular points}, Amer. Math. Soc. {\bf 42} (1974), 167 - 170. \bibitem{h1} L. M. Hall, {\em A characterization of the cokernel of a singular Fredholm differential operator}, J. Differential Equations {\bf 24} (1977), 1 - 7. \bibitem{h2} L. M. Hall, {\em Regular singular differential equations whose conjugate equation has polynomial solutions}, SIAM J. Math. Anal. {\bf 8} (1977), 778-784. \bibitem{h3} W. A. Harris, Jr., Y. Sibuya, and L. Weinberg, {\em Holomorphic solutions of linear differential systems at singular points}, Arch. Rational Mech. Anal. {\bf 35} (1969), 245 - 248. \bibitem{j1} Ju. F. Korobe\u inik, {\em Normal solvability of linear differential equations in the complex plane}, Math USSR Izvestija {\bf 6} (1972), 445-466. \bibitem{t1} A. E. Taylor, {\em Banach spaces of functions analytic in the unit circle, I, II}, Studia Math. {\bf 11} (1950), 145 - 170; Studia Math. {\bf 12} (1951), 25 - 50. \bibitem{t2} H. L. Turrittin, {\em My mathematical expectations}, Symposium on Ordinary Differential Equations, Lecture Notes in Mathematics {\bf 312}, Springer-Verlag (1973), 3 - 22. \end{thebibliography} \noindent{\sc Brian Haile}\\ Department of Mathematics and Statistics\\ Northwest Missouri state university \\ 800 University Drive\\ Maryville, MO 64468 USA\\ e-mail: bhaile@mail.nwmissouri.edu \end{document}