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The Method of Frobenius

R. C. Daileda

Trinity University

Partial Differential Equations

April 7, 2015

Motivating example

Failure of the power series method

Consider the ODE 2xy ′′^ + y ′^ + y = 0. In standard form this is

y

′′

2 x

y

′

2 x

y = 0 ⇒ p(x) = q(x) =

2 x

, g (x) = 0.

In exercise A.4.25 you showed that 1/x is analytic at any a > 0,

with radius R = a. Hence:

Every solution of 2xy ′′^ + y ′^ + y = 0 is analytic at a > 0

with radius R ≥ a (i.e. given by a PS for 0 < x < 2 a).

However, since p, q, g are continuous for x > 0, general theory guarantees that:

Every solution of 2xy ′′^ + y ′^ + y = 0 is defined for all x > 0.

Question: Can we find series solutions defined for all x > 0?

What now?

To find a second independent solution, we instead assume

y = x

r

∑^ ∞

n=

anx

n

PS with R> 0

∑^ ∞

n=

anx

n+r (a 0 6 = 0)

for some r ∈ R to be determined. Since

y ′^ =

∑^ ∞

n=

(n + r )anxn+r^ −^1 , y ′′^ =

∑^ ∞

n=

(n + r )(n + r − 1)anxn+r^ −^2 ,

plugging into the ODE gives

2 x

∑^ ∞

n=

(n+r )(n+r −1)anxn+r^ −^2 +

∑^ ∞

n=

(n+r )anxn+r^ −^1 +

∑^ ∞

n=

anxn+r^ = 0.

Distributing the 2x and setting m = n − 1 in the first two series

yields

∑^ ∞

m=− 1

2(m + r + 1)(m + r )am+1xm+r^ +

∑^ ∞

m=− 1

(m + 1 + r )am+1xm+r

∑^ ∞

n=

anxn+r^ = 0

or, replacing m with n

(2r (r − 1) + r )a 0 x

r − 1 ︸ ︷︷ ︸ n=− 1

∑^ ∞

n=

((n + r + 1) (2(n + r ) + 1) an+1 + an) x

n+r = 0.

This requires the coefficients on each power of x to equal zero.

Taking a 0 = 1 in the second we eventually find that

an =

(−1)n 2 n

(2n + 1)!

⇒ y 2 = x^1 /^2

∑^ ∞

n=

(−1)n 2 n

(2n + 1)!

xn^ =

sin

2 x

This gives the second (linearly independent) solution to the ODE,

and we have the general solution

y = c 1 y 1 + c 2 y 2 = c 1 cos

2 x

• c 2 ′ sin

2 x

(x > 0).

Remarks:

The fact that both series yielded familiar functions is simply a

coincidence, and should not be expected in general.

One could also have obtained y 2 from y 1 (or vice-verse) using

a technique called reduction of order.

Method of Frobenius - First Solution

When will the preceding technique work at an “extraordinary” point? Here’s a partial answer:

Theorem

Suppose that at least one of p(x) or q(x) is not analytic at x = 0,

but that both of xp(x) and x^2 q(x) are. If

lim x→ 0

xp(x) = p 0 and lim x→ 0

x^2 q(x) = q 0 ,

then there is a solution to y ′′^ + p(x)y ′^ + q(x)y = 0 (x > 0 ) of the form

y = xr

∑^ ∞

n=

anxn^ (a 0 6 = 0),

where r is a root of the indicial equation r 2 + (p 0 − 1)r + q 0 = 0.

Example

Find the general solution to x^2 y ′′^ + xy ′^ + (x − 2)y = 0.

In standard form this ODE has

p(x) =

x

and q(x) =

x − 2

x^2

neither of which is analytic at x = 0. However, both

xp(x) = 1 and x^2 q(x) = x − 2

are analytic at x = 0, so we have a regular singularity with

p 0 = lim x→ 0

xp(x) = 1 and q 0 = lim x→ 0

x^2 q(x) = − 2.

The indicial equation is

r 2 + (1 − 1)r − 2 = 0 ⇒ r = ±

Applying the method of Frobenius, we set

y = x

r

∑^ ∞

n=

anx

n

∑^ ∞

n=

anx

n+r (a 0 6 = 0)

and substitute into the ODE, obtaining

(r 2 − 2)a 0 xr^ +

∑^ ∞

n=

(n + r )^2 − 2

an + an− 1

xn+r^ = 0.

Hence we must have r 2 − 2 = 0 (which we already knew) and

an =

−an− 1

(n + r )^2 − 2

−an− 1

n(n + 2r )

for n ≥ 1.

Taking a 0 = 1 one readily sees that

an =

(−1)n

n!(1 + 2r )(2 + 2r )(3 + 2r ) · · · (n + 2r )

Method of Frobenius - Second Solution

What do we do if the indicial roots differ by an integer?

Theorem

Suppose that x = 0 is a regular singular point of

y ′′^ + p(x)y ′^ + q(x)y = 0, and that the roots of the indicial equation are r 1 and r 2 , with r 1 − r 2 ∈ N 0.

If r 1 = r 2 = r , the second solution has the form

y 2 = y 1 ln x + x

r

∑^ ∞

n=

bnx

n .

If r 1 > r 2 (so that y 1 uses r 1 ), the second solution has the form

y 2 = ky 1 ln x + xr^2

∑^ ∞

n=

bnxn^ (b 0 6 = 0).

Example

Find the general solution to xy ′′^ + (1 − x)y ′^ + 2y = 0, x > 0.

In standard form we have

p(x) =

1 − x

x

and q(x) =

x

which are non-analytic at x = 0, and

xp(x) = 1 − x and x

2 q(x) = 2x,

which are. This makes x = 0 a regular singularity with

p 0 = lim x→ 0

1 − x = 1 and lim x→ 0

2 x = 0,

and indicial equation

r 2 + (1 − 1)r + 0 = 0 ⇒ r = 0.

According to the theorem, a second independent solution has the

form

y 2 = y 1 ln x + x

0

∑^ ∞

n=

bnx

n

w

and we need to solve for the bn. The product rule gives us

y 2 ′ = y 1 ′ ln x +

y 1

x

• w ′,

y

′′ 2 =^ y^

′′ 1 ln^ x^ +

2 y 1 ′

x

y 1

x^2

• w ′′,

and plugging these into xy 2 ′′ + (1 − x)y 2 ′ + 2y 2 = 0 we obtain

( xy

′′ 1 + (1^ −^ x)y^

′ 1 + 2y 1

=

ln x − y 1 + 2y

′ 1 +^ xw^

′′

• (1 − x)w

′

• 2w = 0,

xw

′′

• (1 − x)w

′

• 2w = − 2 y

′ 1 +^ y^1.

We now plug y 1 = 1 − 2 x + x^2 /2 and w =

n=1 bnx

n (^) into this

equation to obtain a recurrence for the bn:

b 1 +

∑^ ∞

n=

(n + 1)^2 bn+1 − (n − 2)bn

xn^ = 5 − 4 x +

x^2

2

Hence

b 1 = 5, 4 b 2 + b 1 = − 4 , 9 b 3 =

and

bn+1 =

(n − 2)bn

(n + 1)^2

⇒ bn =

36 b 3

n(n − 1)(n − 2)n!

for n ≥ 3.

Thus, since b 3 = 1/18,

y 2 =

1 − 2 x +

x^2

2

y 1

ln x + 5x −

x^2 + 2

∑^ ∞

n=

xn

n(n − 1)(n − 2)n! ︸ ︷︷ ︸ w