Series Methods I: Ordinary Points

William O. Bray

Department of Mathematics & Statistics

Spring 2002

Motivation

Consider the simple differential equation
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Our task is to approximate a solution with a polynomial, say
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Substituting into the differential equation and rearranging terms we obtain
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The first $n$ terms can be made to disappear if we assume the recursion formula
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The fact that we can do nothing about the last term merely reflects that we are trying to construct an approximation. The recursion formula leads to:
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and our approximation is then
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Of course, we know that the solution to the differential equation is $y(x)=a_{0}e^{x},$ where $a_{0}$ is an arbitrary constant. The reader should note that the above approximation is the $n^{th}$ partial partial sum of the Maclaurin series for $e^{x}.$ This example gives an indication of the method of power series as well as the form for the solution.

A second motivation for the use of power series techniques lies in the realm of partial differential equations and a fundamental method for solving them, separation of variables. A simple example comes from the linear model of a clamped perfectly elastic vibrating string of length $\pi .$ The initial boundary value problem for this model has the following form.
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Here, $x$ is the spacial variable, $t$ represents time, $u=u(x,t)$ represents the vertical displacement of the string from the horizontal, $BC$ represents the boundary conditions corresponding to the assumption that the string is clamped, and $IC$ are the initial conditions prescribing the initial mechanical state of the string. The structure of the partial differential equation suggests seeking a solution satisfing the boundary conditions of the form $u(x,t)=X(x)T(t).$ Substituting in the equation and simplifing leads to two ordinary differential equations.
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With alittle work we arrive at separated solutions of the following form: for $n=1,2,\ldots $
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Note the vibratory nature of these solutions. For fixed $t,$ $\sin x$ represents the fundamental wave form for the string, $\sin nx$ represents the harmonics. Complete description of the motion of the string can be obtained by superposition. So what has all this to do with power series? Well, a two dimensional analog to the above problem would be the vibrating drum head. Formulating the mechanics correctly and applying separation of variables leads to what is known as Bessel's equation:
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Here $r$ is the radial variable (polar coordinates) and $R(r)$ represents the radial part of the separated solution. Understanding the nature of the solutions of Bessel's equation is significant from a physical perspective: the solutions play the same role as $\sin nx$ did for the vibrating string. The bottom line here is that power series techniques give a powerful approach to the study of solutions of Bessel's equation and many others arising in mathematical physics.

The third and final motivation for the use of power series techniques lies in the realm of numerical approximations to solutions. On one hand, the derivation of error estimates for numerical schemes usually is based on some version of Taylor's theorem mentioned below. On the other hand, fast variable step numerical methods often have power series techniques built in for the purpose of estimating error and controling step size.

Taylor Series

A fundamental result from calculus is the mean value theorem:
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where $c$ is a number between $x$ and $x_{0}.$ Rewritten, this says that the derivative achieves the average value of the function between $x$ and $x_{0}, $ i.e.,
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There is a higher order derivative version of the mean value theorem, known as Taylor's formula with remainder:
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This result assumes $f^{(l)}(s)$ is continuous on an interval containing $x$ and $x_{0}$ for $l=0\ldots k,$ and $f^{(k+1)}$ exists on this interval.

We are interested here in letting $k$ tend to infinity, leading to the Taylor series of $f$ about the point $x_{0}$:
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The validity of this last formula rests on the assumption that $f$ is differentiable to all orders and in showing that the remainder term tends to zero as $k$ tend to infinity for $c$ in a suitable interval. A few well known series are the following.
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It is an important fact (in the case of functions of a real variable) that the assumption that $f$ be differentiable to all orders is not sufficient for $f$ to have a Taylor series or that this assumption does not control the interval where the series is convergent. Regarding the second point, consider the second entry in the first row above. The function $(1+x^{2})^{-1}$ is infinitely differentiable for all $x,$ yet its series expansion converges only on the interval given. This point can be circumvented and explained if we view this function as a function of a complex variable $z$:
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Now the function is differentiable for all $z$ save two points $z=\pm i$; note that these points are one unit away from the origin in the complex plane. Rather than turn our attention to complex analysis, we make the following definition.

Definition

A function $f$ is said to be analytic at the point $x_{0}$ if there is a power series expansion of $f$ about $x_{0}$ which converges to $f(x)$ in an interval MATH where $R>0,$ i.e.,
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At first look, this definition may seem ambiguous: we do not imply how the coefficients are related to $f$ nor do we assume that $f$ is infinitely differentiable. A fact from calculus comes to the rescue here: power series may be differentiated term by term in their interval of convergence and the resulting series has the same interval of convergence and is the derivative of the original series. In terms of the above formula we get:
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This tells us that an analytic function is infinitely differentiable near $x_{0}.$ Furthermore, the above derived series leads to the formula
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i.e., the power series assumed in the definition of an analytic function must be the Taylor series of that function at $x_{0}.$

Examples in Differential Equations

Example (1)

Consider the equation
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Recalling the beginning of this set of notes, we would assume a solution of the form
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Substituting into the differential equation, differentiating term by term, and rearranging terms leads to the following.
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It follows that $a_{k-1}-ka_{k}=0$ for all $k,$ i.e., we have the recursion formula
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Applying this successively, we can determine all the coefficients provided $a_{0}$ is known. In other words our solution takes the form
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Example (2)

Consider the second order equation
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Again we will assume a series solution of the form
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It follows that
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Substituting the series into the differential equation we obtain
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Our recursion formula now takes the form
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In order to apply this formula we note the following sucessions:
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In words, from $a_{0}$ all even order coefficients can be calculated, and from $a_{1}$ all odd order coefficients can be calculated. The pattern for the coefficients leads to:
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Returning to the original form of the series expansion for the solution and breaking into the even and odd powers of $x$ leads to the following.
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Please note that the to series on the right hand side are the Maclaurin series for $\cos x$ and $\sin x,$ respectively. In other words power series techniques have led to what we already know, the general solution of the equation has the form
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Example (3)

Consider the variable coefficient equation
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Our solution form is
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Substututing the series into the differential equation we obtain
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or
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This can be rewritten as follows:
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We have the following recursion formulas:
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As in the preceding example, from $a_{0}$ we can compute all even order coefficients and from $a_{1}$ we can compute all odd order coefficients. In the latter case, note that $a_{5}=0$ and so all odd coefficients terminate after the third. We have a solution of the form:
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Exercise (1)

For each of the following, find series solutions of the form
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(a) $y^{\prime }+y,$ $x_{0}=0;$ (b) MATH $x_{0}=0;$ (c) MATH $x_{0}=-1.$

Exercise (2)

Find the first six terms of the power series solution to the following initial value problem
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Graph your approximate solution against that obtained by a numerical solver.

Exercise (3)

Consider the equation
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Here $\mu $ is a fixed parameter. Assuming a power series solution expanded about the point $x_{0}=0,$ find the recursion formula for the coefficients. Show that if $\mu =n$ is a non-negative integer, we always have a polynomial of degree $n$ as one solution to this equation. (HINT: reflect on the last example above.)

Ordinary Points of Differential Equations

The general problem which we face concerns finding power series expansions for solutions of a second order differential equation of the form
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We know from theory to expect two linearly independant solutions. From our preceding examples we know that both solutions will have power series expansions built from a recursion formula for the coefficients. In order to make precise when we can expect to have a power series expansion we make the following definition.

Definition

We say that $x_{0}$ is an ordinary point for the differential equation if the functions $p$ and $q$ are analytic at $x=x_{0}.$ In other words, both functions have power series expansions of the form
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convergent in an interval about $x_{0},$ i.e., in an interval of the form MATH for some positive $R.$

The common radius of convergence in the above definition actually tells us where we can expect power series solutions to our differential equation. The fundamental result is as follows.

Theorem

Suppose that $x_{0}$ is an ordinary point for the differential equation
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Let $R>0$ be such that the power series expansions for the functions $p$ and $q$ about $x_{0}$ converge in the interval MATH Then the differential equation has general solution of the form
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where $y_{1}$ and $y_{2}$ have power series expansions about $x_{0}$ of radius of convergence at least $R.$

The proof of this result actully lies in the realm of complex analysis. The scheme is to assume a series expansion of the form
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substitute the known power series expansions for $p$ and $q,$ and after considerable manipulation obtain a recursion formula for the coefficients $a_{m}.$ Careful estimates based on this formula then leads to the conclusion of the theorem. We content ourselves with two examples.

Example (1)

Consider Legendre's equation
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Here the functions $p$ and $q$ are given by
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The point $x_{0}=0$ is an ordinary point for this differential equation. Indeed,
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the series valid in MATH These were obtained by substituting $x^{2}$ for $x$ in the geometric series
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Consequently, there are two linearly independant solutions with power series expansions at least in the interval MATH NOTE: The recursion formula for the coefficients was obtained in Exercise 3 of the preceding set of notes, it takes the form
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In the preceding example both $p$ and $q$ are rational functions. There is a simple method in this case for determining the radius of convergence as follows. Think of $x$ as a complex variable and suppose $p$ has the form
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where $r$ and $s$ are polynomials. About any point $x_{0}$ where $s(x_{0})\neq 0$ the function $p$ will have a power series expansion. Its radius of convergence is the distance from $x_{0}$ to the closest zero of $s(x)$ in the complex plane. (This fact can be demonstrated by writing $s(x)$ as a product of (complex) factors, computing the partial fraction expansion, and applying the geometric sum formula to each term). For example, the function
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has a power series expansion about $x_{0}=0$ with radius of convergence $\sqrt{2}$ and about $x_{0}=1$ with radius of convergence $\sqrt{3}.$

Example (2)

Consider the differential equation
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Any point $x_{0}$ such that $1+2x_{0}^{2}\neq 0$ is an ordinary point. Choosing $x_{0}=0,$ our theorem tells us that the radius of convergence of power series solutions would be at least $1/\sqrt{2}.$

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