The Method of Frobenius

The Method of Frobenius Department of Mathematics IIT Guwahati

RA/RKS

MA-102 (2016)

The Method of Frobenius

If either p(x) or q(x) in y 00 + p(x)y 0 + q(x)y = 0 is not analytic near x0 , power series solutions valid near x0 may or may not exist. Example: Try to find a power series solution of x 2 y 00 − y 0 − y = 0 about the point x0 = 0. Assume that a solution y (x) =

∞ X

an x n

n=0

exists. RA/RKS

MA-102 (2016)

(1)

The Method of Frobenius

Substituting this series in (1), we obtain the recursion formula an+1 =

n2 − n − 1 an . n+1

The ratio test shows that this power series converges only for x = 0. Thus, there is no power series solution valid in any open interval about x0 = 0. This is because (1) has a singular point at x = 0. The method of Frobenius is a useful method to treat such equations.

RA/RKS

MA-102 (2016)

The Method of Frobenius

Cauchy-Euler equations revisited Recall that a second order homogeneous Cauchy-Euler equation has the form ax 2 y 00 (x) + bxy 0 (x) + cy (x) = 0, x > 0,

(2)

where a(6= 0), b, c are real constants. Writing (2) in the standard form as y 00 + p(x)y 0 + q(x)y = 0, where p(x) =

c b , q(x) = 2 . ax ax

Note that x = 0 is a singular point for (2). We seek solutions of the form y (x) = x r and then try to determine the values for r . RA/RKS

MA-102 (2016)

The Method of Frobenius

Set L(y )(x) := ax 2 y 00 (x) + bxy 0 (x) + cy (x) and w (r , x) := x r . Now L(w )(x) = ax 2 r (r − 1)x r −2 + bxrx r −1 + cx r = {ar 2 + (b − a)r + c}x r . Thus, w = x r is a solution ⇐⇒ r satisfies ar 2 + (b − a)r + c = 0.

(3)

The equation (3) is known as the auxiliary or indicial equation for (2).

RA/RKS

MA-102 (2016)

The Method of Frobenius

Case I: When (3) has two distinct roots r1 , r2 . Then L(w )(x) = a(r − r1 )(r − r2 )x r . The two linearly independent solutions are y1 (x) = w (r1 , x) = x r1 , y2 (x) = w (r2 , x) = x r2 for x > 0. Case II: When r1 = α + iβ, r2 = α − iβ. Then x α+iβ = e (α+iβ) ln x = e α ln x cos(β ln x) + i e α ln x sin(β ln x) = x α cos(β ln x) + i x α sin(β ln x). Thus, two linearly independent real-valued solutions are y1 (x) = x α cos(β ln x), y2 (x) = x α sin(β ln x).

RA/RKS

MA-102 (2016)

The Method of Frobenius

Case III: When r1 = r2 = r0 is a repeated roots. Then L(w )(x) = a(r − r0 )2 x r . Setting r = r0 yields the solution y1 (x) = w (r0 , x) = x r0 , x > 0. To find the second linearly independent solution, we note that ∂ {L(w )(x)}|r =r0 = {a(r − r0 )x r ln x + 2a(r − r0 )x r }|r =r0 = 0. ∂r   ∂ Since ∂r∂ L(w ) = L ∂w ( ∂r and L commute), we obtain ∂r   ∂w L = 0. ∂r r =r0

RA/RKS

MA-102 (2016)

The Method of Frobenius

A second linearly independent solution is ∂ r ∂w (r0 , x) = (x ) = x r0 ln x, x > 0. y2 (x) = ∂r ∂r r =r0 Example: Find a general solution to 4x 2 y 00 (x) + y (x) = 0, x > 0. Note that L(w )(x) = (4r 2 − 4r + 1)x r . The indicial equation has repeated roots r0 = 1/2. Thus, the general solution is √ √ y (x) = c1 x + c2 x ln x, x > 0.

RA/RKS

MA-102 (2016)

The Method of Frobenius

The Method of Frobenius To motivate the procedure, recall the Cauchy-Euler equation in the standard form y 00 (x) + p(x)y 0 (x) + q(x)y (x) = 0,

(4)

where p(x) =

p0 q0 , q(x) = 2 with p0 = b/a q0 = c/a. x x

The indicial equation is of the form r (r − 1) + p0 r + q0 = 0.

(5)

If r = r1 is a root of (5), then w (r1 , x) = x r1 is a solution to (4). RA/RKS

MA-102 (2016)

The Method of Frobenius

Observe that xp(x) and x 2 q(x) are analytic functions. We have ∞ X xp(x) = p0 + p1 x + p2 x + · · · = pn x n , (6) n=0 2

2

x q(x) = q0 + q1 x + q2 x + · · · =

∞ X

qn x n

(7)

n=0

in some neighborhood of x = 0. Then, it follows that lim xp(x) = p0 and lim x 2 q(x) = q0 .

x→0

x→0

Therefore, it is reasonable to expect that the solutions to (2) will behave (for near 0) like the solutions to the Cauchy-Euler equation x 2 y 00 (x) + p0 xy 0 (x) + q0 y (x) = 0. RA/RKS

MA-102 (2016)

The Method of Frobenius

When p(x) and q(x) satisfy (6) and (7), we say that the singular point x = 0 is regular. This observation leads to the following definition. Definition: A singular point x0 of y 00 (x) + p(x)y 0 (x) + q(x)y (x) = 0 is said to be a regular singular point if both (x − x0 )p(x) and (x − x0 )2 q(x) are analytic at x0 . Otherwise x0 is called an irregular singular point. Example: Classify the singular points of the equation (x 2 − 1)2 y 00 (x) + (x + 1)y 0 (x) − y (x) = 0. The singular points are 1 and −1. Note that x = 1 is an irregular singular point and x = −1 is a regular singular point. RA/RKS

MA-102 (2016)

The Method of Frobenius

Series solutions about a regular singular point Assume that x = 0 is a regular singular point for y 00 (x) + p(x)y 0 (x) + q(x)y (x) = 0 so that p(x) =

∞ X

pn x

n−1

, q(x) =

n=0

∞ X

qn x n−2 .

n=0

In the method of Frobenius, we seek solutions of the form w (r , x) = x r

∞ X

an x n =

n=0

∞ X

an x n+r , x > 0.

n=0

Assume that a0 6= 0. We now determine r and an , n ≥ 1. RA/RKS

MA-102 (2016)

The Method of Frobenius

Differentiating w (r , x) with respect to x, we have ∞ X w (r , x) = (n + r )an x n+r −1 , 0

n=0

w 00 (r , x) = 0

∞ X

(n + r )(n + r − 1)an x n+r −2 .

n=0 00

Substituting w , w , w , p(x) and q(x) into (4), we obtain ∞ X

(n + r )(n + r − 1)an x n+r −2

n=0

+ +

∞ X n=0 ∞ X

! pn x n−1 ! qn x n−2

n=0

∞ X n=0 ∞ X

! (n + r )an x n+r −1 ! an x n+r

n=0 RA/RKS

MA-102 (2016)

= 0.

The Method of Frobenius

Group like powers of x, starting with the lowest power, x n−2 . We find that [r (r − 1) + p0 r + q0 ]a0 x r −2 + [(r + 1)ra1 + (r + 1)p0 a1 +p1 ra0 + q0 a1 + q1 a0 ]x r −1 + · · · = 0. Considering the first term, x r −2 , we obtain {r (r − 1) + p0 r + q0 }a0 = 0. Since a0 6= 0, we obtain the indicial equation. Definition: If x0 is a regular singular point of y 00 + p(x)y 0 + q(x)y = 0, then the indicial equation for this point is r (r − 1) + p0 r + q0 = 0, where p0 := lim (x − x0 )p(x), q0 := lim (x − x0 )2 q(x). x→x0

x→x0

RA/RKS

MA-102 (2016)

The Method of Frobenius

Example: Find the indicial equation at the the singularity x = −1 of (x 2 − 1)2 y 00 (x) + (x + 1)y 0 (x) − y (x) = 0. Here x = −1 is a regular singular point. We find that 1 p0 = lim (x + 1)p(x) = lim (x − 1)−2 = , x→−1 x→−1 4 1 q0 = lim (x + 1)2 q(x) = lim [−(x − 1)−2 ] = − . x→−1 x→−1 4 Thus, the indicial equation is given by 1 1 r (r − 1) + r − = 0. 4 4

RA/RKS

MA-102 (2016)

The Method of Frobenius

The method of Frobenius To derive a series solution about the singular point x0 of a2 (x)y 00 (x) + a1 (x)y 0 (x) + a0 (x)y (x) = 0, x > x0 .

(8)

Set p(x) = a1 (x)/a2 (x), q(x) = a0 (x)/a2 (x). If both (x − x0 )p(x) and (x − x0 )2 q(x) are analytic at x0 , then x0 is a regular singular point and the following steps apply. Step 1: Seek solution of the form w (r , x) =

∞ X

an (x − x0 )n+r .

n=0

Using termwise differentiation and substitute w (r , x) into (8) to obtain an equation of the form A0 (x − x0 )r +J + A1 (x − x0 )r +J+1 + · · · = 0. RA/RKS

MA-102 (2016)

The Method of Frobenius

Step 2: Set A0 = A1 = A2 = · · · = 0. (Notice that A0 = 0 is a constant multiple of the indicial equation r (r − 1) + p0 r + q0 = 0). Step 3: Use the system of equations A0 = 0, A1 = 0, . . . , Ak = 0 to find a recurrence relation involving ak and a0 , a1 , . . ., ak−1 . Step 4: Take r = r1 , the larger root of the indicial equation, and use the relation obtained in Step 3 to determine a1 , a2 , . . . recursively in terms of a0 and r1 . Step 5: A series expansion of a solution to (8) is w (r1 , x) = (x − x0 )

r1

∞ X

an (x − x0 )n , x > x0 ,

n=0

where a0 is arbitrary and an ’s are defined in terms of a0 and r1 . RA/RKS

MA-102 (2016)

The Method of Frobenius

Theorem: Let x0 be a regular singular point for y 00 + p(x)y 0 (x) + q(x)y (x) = 0 and let r1 and r2 be the roots of the associated indicial equation, where r1 ≥ r2 or Re r1 ≥ Re r2 . Case a: If r1 − r2 is not an integer, then there exist two linearly independent solutions of the form y1 (x) = y2 (x) =

∞ X n=0 ∞ X

an (x − x0 )n+r1 ; a0 6= 0, bn (x − x0 )n+r2 , b0 6= 0.

n=0

RA/RKS

MA-102 (2016)

The Method of Frobenius

Case b: If r1 = r2 , then there exist two linearly independent solutions of the form y1 (x) =

∞ X

an (x − x0 )n+r1 ; a0 6= 0,

n=0

y2 (x) = y1 (x) ln(x − x0 ) +

∞ X

bn (x − x0 )n+r1 .

n=1

Case c: If r1 − r2 is a positive integer, then there exist two linearly independent solutions of the form y1 (x) =

∞ X

an (x − x0 )n+r1 ; a0 6= 0,

n=0

y2 (x) = Cy1 (x) ln(x − x0 ) +

∞ X

bn (x − x0 )n+r2 , b0 6= 0,

n=0

where C is a constant that could be zero. RA/RKS

MA-102 (2016)