GENERALIZED POWER SERIES OR THE METHOD OF FROBENIUS

11. GENERALIZED POWER SERIES OR THE METHOD OF FROBENIUS

It may happen that the solution of a differential equation is not a power series

n=0 a n x but may either (a) contain some negative powers of x, for example,

or (b) have a fractional power of x as a factor, for example,

Both these cases (and others—see Section 21) are covered by a series of the form

where s is a number to be found to fit the problem; it may be either positive or negative and it may be a fraction. (In fact, it may even be complex, but we shall

not consider this case.) Since a 0 x s is to be the first term of the series, we assume that a 0 is not zero. The series (11.1) is called a generalized power series. We shall consider some differential equations which can be solved by assuming a series of the form (11.1); this way of solving differential equations is called the method of Frobenius .

Example 1. As an illustration of this method we solve the equation (11.2)

x 2 y ′′ + 4xy ′ + (x 2 + 2)y = 0.

From (11.1) we have

y=a x s 0 +a 1 x s+1 +a 2 x s+2

a +···= n+s n x ,

n=0

y ′ = sa 0 x s−1 + (s + 1)a 1 x s + (s + 2)a s+1 2 x +···

(n + s)a n x n+s−1 ,

n=0

y ′′ = s(s − 1)a 0 x s−2 + (s + 1)sa 1 x s−1 + (s + 2)(s + 1)a 2 x s +···

(n + s)(n + s − 1)a n x n+s−2 .

n=0

We substitute (11.3) into (11.2) and set up a table of powers of x as we did for the Legendre equation:

x n+s x 2 y ′′ s(s − 1)a 0 (s + 1)sa 1 (s + 2)(s + 1)a 2 (n + s)(n + s − 1)a n

4xy ′ 4sa 0 4(s + 1)a 1 2(s + 2)a 2 4(n + s)a n x 2 y

a 0 a n−2 2y

2a 0 2a 1 2a 2 2a n

586 Series Solutions of Differential Equations Chapter 12

The total coefficient of each power of x must be zero. From the coefficient of x s we

get (s 2 + 3s + 2)a 0 = 0, or since a 0

s 2 + 3s + 2 = 0.

This equation for s is called the indicial equation. We solve it and find

s = −2,

s = −1.

From here on we solve two separate problems, one when s = −2, and another when s = −1; a linear combination of the two solutions so obtained is then the general solution just as A sin x + B cos x is the general solution of y ′′ + y = 0.

Example 2. For s = −1, the coefficient of x s+1 in the table gives a 1 = 0. From the x s+2 column on, we can use the general formula given by the last column. Notice, however, that the first two columns in the table do not contain the a n−2 term, so you must be careful about using the general term at first (Problems 13 and 14). From the general column with s = −1, we have

a n [(n − 1)(n + 2) + 2] = −a n−2

Since a 1 = 0, this gives all odd a’s equal to zero. For even a’s:

Then one solution of (11.2) is

(11.6) x 3 x 5 a 0 sin x

=a 0 x −2 x−

x 2 The other solution, when s = −2, will be left to Problem 1.

PROBLEMS, SECTION 11

1. Finish the solution of equation (11.2) when s = −2. Write your solution in closed form as in (11.6). To avoid confusion with the a n values we found when s = −1, you may want to call the coefficients in your series a ′ n or b n ; however, this is not essential as long as you realize that there are two separate problems, one when s = −1 and one when s = −2, and each series has its own coefficients.

Solve the following differential equations by the method of Frobenius (generalized power series). Remember that the point of doing these problems is to learn about the method (which we will use later), not just to find a solution. You may recognize some series [as we did in (11.6)] or you can check your series by expanding a computer answer.

2. x 2 y ′′ + xy ′ − 9y = 0 3. x 2 y ′′ + 2xy ′ − 6y = 0 4. x 2 y ′′ − 6y = 0

5. 2xy ′′ +y ′ + 2y = 0 6. 3xy ′′ + (3x + 1)y ′ +y=0

− (x 2 + 2)y = 0 8. x 2 y ′′ + 2x 2 y ′ − 2y = 0

7. x 2 y ′′

9. xy ′′ −y ′ + 9x 5 y=0 10. 2xy ′′ −y ′ + 2y = 0

11. 36x 2 y ′′

+ (5 − 9x 2 )y = 0

Section 12 Bessel’s Equation 587

12. 3xy ′′ − 2(3x − 1)y ′ + (3x − 2)y = 0 Consider each of the following problems as illustrations showing that, in a power series

solution, we must be cautious about using the general recursion relation between the coefficients for the first few terms of the series.

13. Solve y ′′ +y ′ /x 2 = 0 by power series to find the relation

If, without thinking carefully, we test the series P ∞ n=0 a n x n for convergence by the ratio test, we find

(Show this.)

n→∞

|a n x n

Thus we might conclude that the series diverges and that there is no power series solution of this equation. Show why this is wrong, and that the power series solution is y = const. 14. ′′ Solve y = −y by the Frobenius method. You should find that the roots of the indicial equation are s = 0 and s = 1. The value s = 0 leads to the solutions cos x P ∞

and sin x as you would expect. For s = 1, call the series y = n=0 b n x n+1 , and find the relation

b b n+2 n =−

(n + 3)(n + 2)

Show that the b 0 series obtained from this relation is just sin x, but that the b 1 series is not a solution of the differential equation. What is wrong?