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ACM95b/100b Lecture Notes
Caltech 2004
The Method of Frobenius
Consider the equation x^2 y′′^ + xp(x)y′^ + q(x)y = 0, (1)
where x = 0 is a regular singular point. Then p(x) and q(x) are analytic at the origin and have convergent power series expansions
p(x) =
∑^ ∞
k=
pkxk, q(x) =
∑^ ∞
k=
qkxk, |x| < ρ (2)
for some ρ > 0. Let r 1 , r 2 (R(r 1 ) ≥ R(r 2 )) be the roots of the indicial equation
F (r) = r(r − 1) + p 0 r + q 0 = 0. (3)
Depending on the nature of the roots, there are three forms for the two linearly independent solutions on the intervals 0 < |x| < ρ. The power series that appear in these solutions are convergent at least in the interval |x| < ρ. (Proof: Coddington)
Case 1: Distinct roots not differing by an integer (r 1 − r 2 6 = N, N ∈ Z)
The two solutions have the form
y 1 (x) = xr^1
∑^ ∞
k=
ak(r 1 )xk^ (4)
y 2 (x) = xr^2
∑^ ∞
k=
bk(r 2 )xk^ (5)
where ak(r 1 ) and bk(r 2 ) are determined by substitution of (4) or (5) into equation (1) to obtain the corre- sponding recurrence relation.
Case 2: Repeated root (r 1 = r 2 )
The first solution y 1 (x) has form (4) and the second solution has the form
y 2 (x) = y 1 (x) logx + xr^1
∑^ ∞
k=
bk(r 1 )xk. (6)
Note that the term k = 0 is ommitted as it would just give a multiple of y 1 (x).
Case 3: Roots differing by an integer (r 1 − r 2 = N, N ∈ Z+)
The first solution y 1 (x) has form (4) and the second solution has the form
y 2 (x) = c y 1 (x) logx + xr^2
∑^ ∞
k=
bk(r 2 )xk. (7)
where c may turn out to be zero. The constant bN (r 2 ) is arbitrary and may be set to zero. This is evident by writing
xr^2
∑^ ∞
k=
bk(r 2 )xk^ = b 0 xr^2 + ... + bN − 1 xr^2 +N^ −^1 + xr^1 (bN + bN +1x + bN +2x^2 + ...) ︸ ︷︷ ︸ form of y 1 (x)
so we see that bN (r 2 ) plays the same role as a 0 (r 1 ) and merely adds multiples of y 1 (x) to y 2 (x).
Example: Case 1
Consider 4 xy′′^ + 2y′^ + y = 0 (9)
so x = 0 is a regular singular point with p(x) = 12 and q(x) = x 4. The power series in y 1 and y 2 will converge for |x| < ∞ since p and q have convergent power series in this interval. By (3), the indicial equation is
r(r − 1) +
r = 0 ⇒ r^2 −
r = 0 (10)
so r 1 = 12 and r 2 = 0 (Note: p 0 = 12 , q 0 = 0). Substituting y = xr^
k=0 akx k (^) into (9) and shifting the
indices of the first two series so all terms are of form xk+r^ we get
∑^ ∞
k=− 1
(k + r + 1)(k + r)ak+1xk+r^ + 2
∑^ ∞
k=− 1
(k + r + 1)ak+1xk+r^ +
∑^ ∞
k=
akxk+r^ = 0. (11)
All coefficients of powers xk+r^ must equate to zero to obtain a solution. The lowest power is xr−^1 for k = − 1 and this yields
4 r(r − 1) + 2r = 0 ⇒ r^2 −
r = 0 (12)
which is just the indicial equation as expected. For k ≥ 0, we obtain
4(k + r + 1)(k + r)ak+1 + 2(k + r + 1)ak+1 + ak = 0 (13)
corresponding to the recurrence relation
ak+1 =
−ak (2k + 2r + 2)(2k + 2r + 1)
, k = 0, 1 , 2 ... (14)
First Solution: To find y 1 apply (14) with r = r 1 = 12 to get the recurrence relation
ak+1 =
−ak (2k + 3)(2k + 2)
, k = 0, 1 , 2 ... (15)
Then a 1 = −a 0 3 · 2
, a 2 = −a 1 5 · 4
, a 3 = −a 2 7 · 6
so a 1 = − a 0 3!
, a 2 = a 0 5!
, a 3 = − a 0 7!
Since a 0 is arbitrary, let a 0 = 1 so
ak(r 1 ) =
(−1)k (2k + 1)!
, k = 0, 1 , 2 ... (18)
and
y 1 (x) = x^1 /^2
∑^ ∞
k=
(−1)k (2k + 1)!
xk. (19)
Second Solution: To find y 2 , just apply (14) with r = r 2 = 0 to get the recurrence relation
bk+1 = −bk (2k + 2)(2k + 1)
Letting the arbitrary constant b 0 = 1, then
bk(r 2 ) =
(−1)k (2k)!
Now we know L[y 1 ] = 0 so this gives
L
[ ∞
k=
bkxk
]
= − 2 y 1 ′ (34)
or in detail after appropriate index shifts to the first and second series
b 1 +
∑^ ∞
k=
[(k + 1)kbk+1 + (k + 1)bk+1 − bk] xk^ = − 2 − x −
x^2 6
Equating coefficients gives
b 1 = − 2 4 b 2 − b 1 = − 1 9 b 3 − b 2 = −
so
b 1 = − 2 , b 2 = −
, b 3 = −
The second linearly independent solution is then
y 2 (x) = y 1 (x) log x +
[
− 2 x −
x^2 −
x^3 − ...
]
Example: Case 3 (log term required)
Consider Ly ≡ xy′′^ + y = 0 (38)
with p(x) = 0 and q(x) = x and a regular singular point at x = 0. The power series in y 1 and y 2 will converge for |x| < ∞ since p and q have convergent power series in this interval. The indicial equation is given by
r(r − 1) = 0 (39)
so r 1 = 1 and r 2 = 0. First solution: Substituting y = xr^
k=0 akx k (^) into (38) results in
∑^ ∞
k=
(r + k)(r + k − 1)akxr+k−^1 +
∑^ ∞
k=
akxr+k^ = 0. (40)
Shifting indices in the second series and regrouping terms gives
r(r − 1)a 0 xr−^1 +
∑^ ∞
k=
[(r + k)(r + k − 1)ak + ak− 1 ]xr+k−^1 = 0. (41)
Setting the coefficient of xr−^1 to zero we recover the indicial equation with r 1 = 1 and r 2 = 0. Setting all the other coefficients to zero gives the recurrence relation
ak = −ak− 1 (r + k)(r + k − 1)
, k ≥ 1. (42)
With r = r 1 this gives
ak =
−ak− 1 (k + 1)k
ak− 2 (k + 1)k^2 (k − 1)
(−1)ka 0 (k + 1)(k!)^2
Setting the arbitrary constant a 0 (r 1 ) = 1, the first solution is then
y 1 (x) = x
∑^ ∞
k=
(−1)k (k + 1)(k!)^2
xk^ = x −
x^2 2
x^3 12
x^4 144
Second solution: First, let’s see how we run into trouble if we fail to include the log term in the second solution. The recurrence relation (42) with r = r 2 = 0 becomes (with bk replacing ak since we are now using r = r 2 )
bk =
−bk− 1 k(k − 1) , k ≥ 1. (45)
This formula fails for k = 1. As was anticipated, for roots of the form r 2 − r 1 = N with N ∈ Z+^ it may not be possible to determine bN if the log term is ommitted from y 2 (in our case N = 1). For the second solution consider substituting
y = c y 1 (x) log x + x^0
∑^ ∞
k=
bkxk^ (46)
into (38) so
xy′′^ =
[
cy 1 x
- 2cy 1 ′ + cxy′′ 1 log x + x
d^2 dx^2
∑^ ∞
k=
bkxk
]
We then obtain
cL[y 1 ] log x + 2cy 1 ′ −
cy 1 x
+ L
[ ∞
k=
bkxk
]
Now we know L[y 1 ] = 0 so this gives
L
[ ∞
k=
bkxk
]
= − 2 cy′ 1 +
cy 1 x
Expanding the left hand side gives
b 0 + (2b 2 + b 1 )x + (6b 3 + b 2 )x^2 + (12b 4 + b 3 )x^3 + (20b 5 + b 4 )x^4 + ... (50)
and expanding the right hand side gives
−c +
cx −
cx^2 +
cx^3 −
cx^4 ± ... (51)
Equating coefficients gives the system of equations
b 0 = −c 2 b 2 + b 1 =
c
6 b 3 + b 2 = −
c
12 b 4 + b 3 =
c .. .
Now b 0 (r 2 ) is an arbitrary constant and c = −b 0. Notice that b 1 can also be chosen arbitrarily. This is because it is the coefficient of xr^1 = x^1 = x in the series
xr^2
∑^ ∞
k=
bkxk^ = b 0 + xr^1 (b 1 + b 2 x + b 3 x^2 + ...) ︸ ︷︷ ︸ form of y 1 (x)
If the roots are of the form r 2 − r 1 = N with N ∈ Z+^ then the log term is generally required to enable the calculation of bN (r 2 ). However, in the present case, we see from (58) that the coefficient of xr+1^ will vanish for r 2 = − 1 /2 regardless of the value of b 1 (where b 1 replaces a 1 since we are now using r = r 2 ). Hence, the log term is unnecessary in y 2 and c = 0. Let’s suppose we did not notice this ahead of time and attempted to substitute the general form (64) into (54). Then notice
y′^ =
cy 1 x
d dx
[
x−^1 /^2
∑^ ∞
k=
bkxk
]
and
y′′^ = −
cy 1 x^2
cy′ 1 x
d^2 dx^2
[
x−^1 /^2
∑^ ∞
k=
bkxk
]
We then obtain
cL[y 1 ] log x + 2cxy′ 1 + L
[
x−^1 /^2
∑^ ∞
k=
bkxk
]
Now we know L[y 1 ] = 0 so this gives
L
[
x−^1 /^2
∑^ ∞
k=
bkxk
]
= − 2 cxy′ 1. (68)
Expanding the left hand side (for convenience we may use (58) and (59) with bk replacing ak since we are using r = r 2 ) gives 0 · b 0 x−^1 /^2 + 0 · b 1 x^1 /^2 + (2b 2 + b 0 )x^3 /^2 + (6b 3 + b 1 )x^5 /^2 + ... (69)
and expanding the right hand side gives
−cx^1 /^2 +
cx^5 /^2 −
cx^9 /^2 ± ... (70)
Equating coefficients give the system of equations
0 · b 1 = −c 2 b 2 + b 0 = 0
6 b 3 + b 1 =
c 12 b 4 + b 2 = 0 .. .
So b 0 (r 2 ) and b 1 (r 2 ) are both arbitrary as expected and c = 0. The other coefficients are then given by
b 2 k =
(−1)kb 0 (2k)!
, b 2 k+1 =
(−1)kb 1 (2k + 1)!
, k = 1, 2 , ... (71)
Hence the second solution has the form
y 2 (x) = x−^1 /^2
[
b 0
∑^ ∞
k=
(−1)k (2k)!
x^2 k^ + b 1
∑^ ∞
k=
(−1)k (2k + 1)!
x^2 k+
]
Note that as expected, b 1 just introduces a multiple of y 1 (x) so we may choose b 1 = 0. Setting the arbitrary constant b 0 = 1, the second solution finally becomes
y 2 (x) = x−^1 /^2
∑^ ∞
k=
(−1)k (2k)!
x^2 k. (73)
