Ch 5.7: Series Solutions Near a Regular Singular Point, Part II

Transcription

1 Ch 5.7: Series Solutions Near a Regular Singular Point, Part II! Recall from Section 5.6 (Part I): The point x 0 = 0 is a regular singular point of with and corresponding Euler Equation! We assume solutions have the form

2 Substitute Derivatives into ODE! Taking derivatives, we have! Substituting these derivatives into the differential equation, we obtain

3 Multiplying Series

4 Combining Terms in ODE! Our equation then becomes

5 Rewriting ODE! Define F(r) by! We can then rewrite our equation in more compact form:

6 Indicial Equation! Thus our equation is! Since a 0 0, we must have! This indicial equation is the same one obtained when seeking solutions y = x r to the corresponding Euler Equation.! Note that F(r) is quadratic in r, and hence has two roots, r 1 and r 2. If r 1 and r 2 are real, then assume r 1 r 2.! These roots are called the exponents at the singularity, and they determine behavior of solution near singular point.

7 Recurrence Relation! From our equation, the recurrence relation is! This recurrence relation shows that in general, a n depends on r and the previous coefficients a 0, a 1,, a n-1.! Note that we must have r = r 1 or r = r 2.

8 Recurrence Relation & First Solution! With the recurrence relation we can compute a 1,, a n-1 in terms of a 0, p m and q m, provided F(r + 1), F(r + 2),, F(r + n), are not zero.! Recall r = r 1 or r = r 2, and these are the only roots of F(r).! Since r 1 r 2, we have r 1 + n r 1 and r 1 + n r 2 for n 1.! Thus F(r 1 + n) 0 for n 1, and at least one solution exists: where the notation a n (r 1 ) indicates that a n has been determined using r = r 1.

9 Recurrence Relation & Second Solution! Now consider r = r 2. Using the recurrence relation we compute a 1,, a n-1 in terms of a 0, p m and q m, provided F(r 2 + 1), F(r 2 + 2),, F(r 2 + n), are not zero.! If r 2 r 1, and r 2 - r 1 n for n 1, then r 2 + n r 1 for n 1.! Thus F(r 2 + n) 0 for n 1, and a second solution exists: where the notation a n (r 2 ) indicates that a n has been determined using r = r 2.

10 Convergence of Solutions! If the restrictions on r 2 are satisfied, we have two solutions where a 0 =1 and x > 0. The series converge for x < ρ, and define analytic functions within their radii of convergence.! It follows that any singular behavior of solutions y 1 and y 2 is due to the factors x r1 and x r2.! To obtain solutions for x < 0, it can be shown that we need only replace x r1 and x r2 by x r1 and x r2 in y 1 and y 2 above.! If r 1 and r 2 are complex, then r 1 r 2 and r 2 - r 1 n for n 1, and real-valued series solutions can be found.

11 Example 1: Singular Points (1 of 5)! Find all regular singular points, determine indicial equation and exponents of singularity for each regular singular point. Then discuss nature of solutions near singular points.! Solution: The equation can be rewritten as! The singular points are x = 0 and x = -1.! Then x = 0 is a regular singular point, since

12 Example 1: Indicial Equation, x = 0 (2 of 5)! The corresponding indicial equation is given by or! The exponents at the singularity for x = 0 are found by solving indicial equation:! Thus r 1 = 0 and r 2 = -1/2, for the regular singular point x = 0.

13 Example 1: Series Solutions, x = 0 (3 of 5)! The solutions corresponding to x = 0 have the form! The coefficients a n (0) and a n (-1/2) are determined by the corresponding recurrence relation.! Both series converge for x < ρ, where ρ is the smaller radius of convergence for the series representations about x = 0 for! The smallest ρ can be is 1, which is the distance between the two singular points x = 0 and x = -1.! Note y 1 is bounded as x 0, whereas y 2 unbounded as x 0.

14 Example 1: Indicial Equation, x = -1 (4 of 5)! Next, x = -1 is a regular singular point, since and! The indicial equation is given by and hence the exponents at the singularity for x = -1 are! Note that r 1 and r 2 differ by a positive integer.

15 Example 1: Series Solutions, x = -1 (5 of 5)! The first solution corresponding to x = -1 has the form! This series converges for x < ρ, where ρ is the smaller radius of convergence for the series representations about x = -1 for! The smallest ρ can be is 1. Note y 1 is bounded as x -1.! Since the roots r 1 = 2 and r 2 = 0 differ by a positive integer, there may or may not be a second solution of the form

16 Equal Roots! Recall that the general indicial equation is given by! In the case of equal roots, F(r) simplifies to! It can be shown (see text) that the solutions are given by where the b n (r 1 ) are found by substituting y 2 into the ODE and solving, as usual. Alternatively, as shown in text,

17 Roots Differing by an Integer! If roots of indicial equation differ by a positive integer, i.e., r 1 r 2 = N, it can be shown that the ODE solns are given by where the c n (r 1 ) are found by substituting y 2 into the differential equation and solving, as usual. Alternatively, and! See Theorem for a summary of results in this section.