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Math 555: Differential Equations

5.3 Series Solutions Near an
Ordinary Point, Part II

5.3.1 Alternate Approach to Finding the Power Series

We showed last section that it is possible to find solutions to

$$ P(x)y'' + Q(x)y' + R(x)y = 0, $$
where $P$, $Q$, and $R$ are polynomials, near an ordinary point $x_0$. The procedure for that section involved assuming that the solution $y = \phi(x)$ is a generic power series

$$ \phi (x) = \sum_{n=0}^\infty a_n (x-x_0)^n $$
convergent for $\vert x-x_0 \vert\lt\rho$ for some positive $\rho$, and then finding the recurrence relation of the coefficients $a_n$. Here, we will study an alternative to that approach that uses Taylor's theorem and the fact that $x_0$ is an ordinary point.

At an ordinary point $x_0$, $\phi$ is differentiable, so the Taylor series representation

$$ \phi (x) = \sum_{n=0}^\infty \dfrac{\phi^{(n)}(x_0)}{n!} (x-x_0)^n $$
for $\phi$ gives us the expression for each coefficient $a_n$. We can move the factorial to obtain

$$ \phi^{(n)}(x_0) = n! a_n. $$
This is useful because we can perform a manipulation of the ODE that will allow us to determine these values directly.

Plugging $\phi$ into the ODE and rewriting the equation in standard form (which is allowed, since $P(x_0)\neq 0$),

$$ \phi''(x) + p(x)\phi'(x) + q(x)\phi(x) = 0. $$
This equation may be solved for $\phi''$. In addition, by recalling that $\phi(x_0) = a_0$ and $\phi'(x_0) = a_1$, we may write down

$$ \phi''(x_0) = -p(x_0)\phi'(x_0) - q(x_0)\phi(x_0) = -p(x_0)a_1 - q(x_0)a_0. $$
Using the Taylor series formula from above, we have that

$$ 2!a_2 = \phi''(x_0) = -p(x_0)a_1 - q(x_0)a_0. $$
By taking advantage of the fact that $\phi(x)$ is an analytic function (it has a convergent power series) at $x_0$, we may differentiate it as many times as we like. Taking these derivatives allow us to compute terms for the sequence $\phi^{(n)}(x_0)$ one at a time. To get the third derivative, we take the formula for $\phi''$, differentiate it, then evaluate it at $x = x_0$:

$$ \begin{aligned} \phi'''(x_0) &= 3!a_3 = -\left.\left( p(x)\phi'(x) + q(x)\phi(x)\right)'\right\vert_{x = x_0} \\ \\ &= -\left.\left( p'(x)\phi'(x) + p(x)\phi''(x) + q'(x)\phi(x) + q(x)\phi'(x) \right)\right\vert_{x=x_0} \\ \\ &= -2!p(x_0)a_2 - (p'(x_0) + q(x_0))a_1 - q'(x_0)a_0. \end{aligned}$$
At this point, we can plug in the value we have for $a_2$ to get the entire expression in terms of $a_0$ and $a_1$.

To find more coefficients, we repeat this process.

On a theoretical note, this process will work whenever the functions $p$ and $q$ are analytic at $x_0$. In fact, this gives us a more precise definition for ordinary points: $x_0$ is an ordinary point if $p$ and $q$ are analytic, otherwise $x_0$ is a singular point .

Example 5.3.1

Let $y = \phi(x)$ be a solution of the initial value problem

$$ y'' + x^2 y' + (\sin x)y = 0,\qquad y(0) = a_0,\ \ y'(0) = a_1. $$
Determine $\phi''(x_0)$, $\phi'''(x_0)$, and $\phi^{(4)}(x_0)$.

Solution

Finding $\phi''(0)$ is simple, we just evaluate the ODE at $x = 0$:

$$ \phi''(0) + 0^2 \phi'(0) + (\sin 0)\phi(0) = 0, $$
hence $\phi''(0) = 0$.

For $\phi'''(0)$, we take the ODE and differentiate the whole expression with respect to $x$:

$$ \begin{aligned} \Big(\phi''(x) + x^2 \phi'(x) + (\sin x)\phi(x)\Big)' &= 0 \\ \\ \phi'''(x) + 2x\phi'(x) + x^2 \phi''(x) + (\cos x)\phi(x) + (\sin x)\phi'(x) &= 0 \end{aligned} $$
Then set $x = 0$,

$$ \begin{aligned} \phi'''(0) + 2(0)\phi'(0) + 0^2 \phi''(0) + (\cos 0)\phi(0) + (\sin 0)\phi'(0) &= 0 \\ \\ \phi'''(0) + \phi(0) &= 0 \\ \\ \phi'''(0) &= -a_0 \end{aligned} $$
Differentiating once more to find $\phi^{(4)}(0)$,

$$ \begin{aligned} \Big(\phi'''(x) + 2x\phi'(x) + x^2 \phi''(x) + (\cos x)\phi(x) + (\sin x)\phi'(x)\Big)' &= 0 \\ \\ \phi^{(4)}(x) + 2\phi'(x) + 2x\phi''(x) + 2x\phi''(x) + x^2 \phi'''(x)\qquad \\ - (\sin x)\phi(x) + (\cos x)\phi'(x) + (\cos x)\phi'(x) + (\sin x)\phi''(x) & = 0 \\ \\ \phi^{(4)}(0) + 2\phi'(0) + 2(0)\phi''(0) + 2(0)\phi''(0) + 0^2 \phi'''(0)\qquad \\ - (\sin 0)\phi(0) + (\cos 0)\phi'(0) + (\cos 0)\phi'(0) + (\sin 0)\phi''(0) & = 0 \\ \\ \phi^{(4)}(0) + 2a_1 + (\cos 0)a_1 + (\cos 0)a_1 & = 0 \\ \\ \phi^{(4)}(0) &= -4a_1. \end{aligned} $$
Collecting our results:

$$ \phi''(0) = 0,\quad \phi'''(0) = -a_0,\quad \phi^{(4)}(0) = -4a_1. $$
Following the procedure, work this exercise out for yourself.

Exercise 5.3.1

Let $y = \phi(x)$ be a solution of the initial value problem

$$ y'' + xy' + y = 0,\qquad y(0) = 1,\ \ y'(0) = 0. $$
Determine $\phi''(x_0)$, $\phi'''(x_0)$, and $\phi^{(4)}(x_0)$.

Check Your Work
$$ \phi''(0) = -1,\quad \phi'''(0) = 0,\quad \phi^{(4)}(0) = 3. $$

Video Solution

5.3.2 Radii of Convergence

Several times during our discussion of power series, radii and intervals of convergence have been mentioned as the restriction on the validity of our solutions. It is always important to know when the power series we are using are convergent to have the best possible understanding of when our solutions are useful. Typically, the radius of convergence for a power series is determined by using the ratio test, but that tool can be inconvenient in cases where we do not already have a general term for the power series.

Thankfully, there is a very useful theorem from Fuchs which allows us to take a different approach.

Theorem 5.3.1

If $x_0$ is an ordinary point of the differential equation

$$ P(x)y'' + Q(x)y' + R(x)y = 0, $$
so we have $p(x) = Q(x)/P(x)$ and $q(x) = R(x)/P(x)$ are analytic at $x_0$, then the general solution of the ODE is

$$ y = \sum_{n=0}^\infty a_n (x-x_0)^n = a_0 y_1(x) + a_1 y_2(x), $$
where $a_0$ and $a_1$ are arbitrary values and $y_1$ and $y_2$ are two power series solutions analytic at $x_0$. These solutions form a fundamental set of solutions, and the radius of convergence for each of the series solutions $y_1$ and $y_2$ is at least as large as the minimum of the radii of convergence for $p$ and $q$.

This means that we need only concern ourselves with the radii of convergence for $p$ and $q$ to be able to have a minimal value for the radius of convergence to the solution to the ODE.

The simplest approach to finding the radius of convergence for a rational function like $p(x) = Q(x)/P(x)$ involves a small bit of complex analysis. It is a theorem that for the power series expansion of a complex valued function about a point $x_0$ so that $P(x_0)\neq 0$ that the radius of convergence is the distance from $x_0$ to the nearest zero of $P(x)$. That is the distance from $x_0$ to a root of $P(x)$ in the complex plane.

Let's do a couple of examples to illustrate this idea.

Example 5.3.2

What is the radius of convergence of the Taylor series for $\frac{1}{1+x^2}$ about $x = 0$?

Solution

We know that

$$ \dfrac{1}{1+x^2} = \sum_{n=0}^\infty (-1)^n x^{2n}, $$
which by the ratio test will give us $\rho = 1$. However, we would like to use the geometric approach. The series is expanded about $x_0 = 0$, and the zeros of the denominator are $x = \pm i$. If we graph this on the complex plane, taking the horizontal axis to be real and the vertical axis to be imaginary: Radius of convergence schematic for 1/(1+x^2)

Here, we have drawn the largest circle centered at $x = 0$ possible without intersecting the roots of $P(x)$, $x = \pm i$. The radius of this circle is the radius of convergence $\rho$.

Example 5.3.3

What is the radius of convergence of the Taylor series for $\dfrac{1}{x^2 - 3x + 4}$ about $x = 0$? About $x = \frac{3}{2}$?

Solution

The equation

$$ x^2 - 3x + 4 = 0 $$
has solutions $x = \frac{1}{2}\left(3 \pm i\sqrt{7}\right)$. The distance from $0$ to $\frac{1}{2}\left(3 \pm i\sqrt{7}\right)$ in the complex plane may be determined using the Pythagorean theorem

$$ \rho = \sqrt{\left(\frac{3}{2} - 0\right)^2 + \left(\frac{\sqrt{7}}{2} - 0\right)^2} = 2. $$
In the case that $x = \frac{3}{2}$, the distance to the complex root is just $\rho = \frac{\sqrt{7}}{2}$. Both radii of convergence are depicted visually here: Radius of convergence schematic for 1/(x^2-3x+4) with x = 0 and x = 3/2

Exercise 5.3.2

Determine a minimum for the radius of convergence for the series solution to

$$ (4 + x^2)y'' - xy' + y = 0 $$
about $x = 0$ and about $x = 1$.

Check Your Work
$$ x_0 = 0\ \ \Rightarrow\ \ \rho = 2,\quad x_0 = 1\ \ \Rightarrow\ \ \rho = \sqrt{5} $$
Radius of convergence schematic for solutions (4 + x^2)y'' - xy' + y = 0 with x = 0 and x = 1

Hints to Find Solution
Begin by putting the equation into standard form
$$\begin{align*} \\ y'' - \underbrace{\dfrac{x}{4+x^2}}_{p(x)} y' + \underbrace{\dfrac{1}{4+x^2}}_{q(x)} y = 0 \end{align*}$$
and look for the smaller radius of convergence for the power series representations of $p(x)$ and $q(x)$. Note that these representations are virtually identical (since the denominators have the same roots), so we need only worry about one of them. From here, follow the technique of Example 5.3.3.

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