We continue our study of the method of Frobenius for finding series solutions of linear second order differential equations, extending to the case where the indicial equation has a repeated real root.

The Method of Frobenius II

In this section we discuss a method for finding two linearly independent Frobenius solutions of a homogeneous linear second order equation near a regular singular point in the case where the indicial equation has a repeated real root. As in the preceding section, we consider equations that can be written as

where . We assume that the indicial equation has a repeated real root . In this case Theorem thmtype:7.5.3 implies that (eq:7.6.1) has one solution of the form but does not provide a second solution such that is a fundamental set of solutions. The following extension of Theorem thmtype:7.5.2 provides a way to find a second solution.

Theorem thmtype:7.5.2 implies (eq:7.6.4). Differentiating formally with respect to in (eq:7.6.3) yields
which proves (eq:7.6.5).

To prove that satisfies (eq:7.6.6), we view in (eq:7.6.2) as a function of two variables, where the prime indicates partial differentiation with respect to ; thus, With this notation we can use (eq:7.6.2) to rewrite (eq:7.6.4) as

Differentiating both sides of (eq:7.6.7) with respect to yields By changing the order of differentiation in the first two terms on the left we can rewrite this as or which is equivalent to (eq:7.6.6).

Since is a repeated root of , the indicial polynomial can be factored as so which is nonzero if . Therefore the assumptions of Theorem thmtype:7.6.1 hold with , and (eq:7.6.4) implies that . Since it follows that , so (eq:7.6.6) implies that This proves that and are both solutions of . We leave the proof that is a fundamental set as an exercise.

Since the recurrence formula (eq:7.6.11) involves three terms, it’s not possible to obtain a simple explicit formula for the coefficients in the Frobenius solutions of (eq:7.6.9). However, as we saw in the preceding sections, the recurrence formula for involves only two terms if either or in (eq:7.6.1). In this case, it’s often possible to find explicit formulas for the coefficients. The next two examples illustrate this.

If the solution of reduces to a finite sum, then there’s a difficulty in using logarithmic differentiation to obtain the coefficients in the second solution. The next example illustrates this difficulty and shows how to overcome it.

Text Source

Trench, William F., ”Elementary Differential Equations” (2013). Faculty Authored and Edited Books & CDs. 8. (CC-BY-NC-SA)