Physics 451
Fall 2004 Homework Assignment #13 — Solutions
Textbook problems: Ch. 8: 8.4.1, 8.4.3, 8.5.6, 8.5.11, 8.5.14, 8.5.17 Chapter 8 8.4.1 Show that Legendre’s equation has regular singularities at x = −1, 1, and ∞. Legendre’s equation may be written as y 00 −
2x 0 l(l + 1) y + y=0 1 − x2 1 − x2
so that P (x) = −
2x 2x = , 2 1−x (x − 1)(x + 1)
Q(x) =
l(l + 1) l(l + 1) =− 2 1−x (x − 1)(x + 1)
Written in this fashion, we see that both P (x) and Q(x) have simple poles at x = 1 and x = −1. This is sufficient to indicate that these two points are regular singular points. For the point at ∞, we make the substitution x = 1/z. As worked out in the text, we end up with 2z − P (z −1 ) 2z + 2z −1 /(1 − z −2 ) 2 2 2z Pe(z) = = = + = 2 2 2 2 z z z z(z − 1) z −1 and Q(z −1 ) l(l + 1)/(1 − z −2 ) l(l + 1) e Q(z) = = = 2 2 4 4 z z z (z − 1) e as z → 0, we see that Pe is regular, while Q e Examining the behavior of Pe and Q e has a double pole. Because of the double pole in Q, Legendre’s equation also has a regular singularity at x = ∞. 8.4.3 Show that the substitution x→
1−x , 2
a = −l,
b = l + 1,
c=1
converts the hypergeometric equation into Legendre’s equation.
Making the above substitution (along with dx → − 21 dx which implies y 0 → (−2)y 0 and y 00 → (−2)2 y 00 ) into the Hypergeometric equation, we find x(x − 1)y 00 + [(1 + a + b)x − c]y 0 + aby = 0 1−x 1−x 1−x 2 00 − 1 (−2) y + (1 − l + (l + 1)) − 1 (−2)y 0 ⇒ 2 2 2 − l(l + 1)y = 0 ⇒
−(1 − x2 )y 00 + 2xy 0 − l(l + 1)y = 0
Changing an overall sign yields Legendre’s equation (1 − x2 )y 00 − 2xy 0 + l(l + 1)y = 0 This indicates that Legendre’s equation is in fact a special case of the more general Hypergeometric equation. 8.5.6 Develop series solutions for Hermite’s differential equation a) y 00 − 2xy 0 + 2αy = 0 Since x = 0 is a regular point, we develop a simply Taylor series solution y=
∞ X
an xn ,
y0 =
n=0
∞ X
nan xn−1 ,
y 00 =
n=0
∞ X
n(n − 1)an xn−2
n=0
Substituting this in to Hermite’s equation, we find ∞ X
[n(n − 1)an xn−2 − 2nan xn + 2αan xn ] = 0
n=0
⇒
∞ X
[(n + 2)(n + 1)an+2 + 2(α − n)an ]xn = 0
n=0
To obtain the second line, we had made the substitution n → n + 2 in the first term of the series so that we could collect identical powers of xn . Since this series vanishes for all values of x, each coefficient must vanish. This yields the recursion relation 2(n − α) an+2 = an (1) (n + 2)(n + 1) which determines all higher an ’s, given a0 and a1 as a starting point. In fact, we obtain two series, one for n even and one for n odd. For n even, we set a0 = 1 and find a0 = 1,
a2 =
2(−α) , 2!
a4 =
2(2 − α) 22 (−α)(2 − α) a2 = , 4·3 4!
etc.
This gives the even solution yeven = 1 + 2(−α)
x4 x6 x2 + 22 (−α)(2 − α) + 23 (−α)(2 − α)(4 − α) + · · · (2) 2! 4! 6!
For n odd, we set a1 = 1 and find a1 = 1,
a3 =
2(1 − α) , 3!
a5 =
2(3 − α) 22 (1 − α)(3 − α) a3 = , 5·4 5!
etc.
This results in the odd solution yodd = x + 2(1 − α)
x5 x7 x3 + 22 (1 − α)(3 − α) + 23 (1 − α)(3 − α)(5 − α) + · · · (3) 3! 5! 7!
Note that, and an ordinary point, we did not have to solve the indicial equation. However, if we had chosen to do so, we would have found k = 0 or k = 1, yielding the even and odd solutions, respectively. b) Show that both series solutions are convergent for all x, the ratio of successive coefficients behaving, for large index, like the corresponding ratio in the expansion of exp(2x2 ). To test for convergence, all we need is to use the ratio test an xn (n + 2)(n + 1) n = lim = lim =∞ n→∞ an+2 xn+2 n→∞ 2(n − α)x2 n→∞ 2x2 lim
(4)
Since this is larger than 1, the series converges for all values of x. Note that the ratio an /an+2 was directly obtained from the recursion relation (1), and this result is valid for both yeven and yodd . Furthermore, if we compared this with exp(2x2 ), we would see that the n-th term in the Taylor series of the exponential is bn = (2x2 )n /n!, which leads to a ratio bn−1 n = 2 bn 2x in direct correspondence with that of (4). Hence the solutions to Hermite’s equations are (generically) asymptotic to exp(2x2 ). c) Show that by appropriate choice of α the series solutions may be cut off and converted to finite polynomials. Examination of the series solutions (2) and (3) indicates that yeven terminates for α = 0, 2, 4, . . . and yodd terminates for α = 1, 3, 5, . . .. This means the for α a non-negative integer either yeven or yodd (depending on α being even or odd) terminates, yielding a finite ‘Hermite polynomial’.
8.5.11 Obtain two series solutions of the confluent hypergeometric equation xy 00 + (c − x)y 0 − ay = 0 Test your solutions for convergence. We first observe that this equation has a regular singular point at x = 0 and an irregular one at x = ∞. We would like to develop a series solution around the regular singular point at x = 0. Thus we start with the indicial equation y 00 +
c−x 0 a y − y=0 x x
⇒
p0 = c,
q0 = 0
and k(k − 1) + p0 k + q0 = 0
⇒
k(k − 1) + ck = 0
⇒
k(k + c − 1) = 0
This shows that the indices at x = 0 are k1 = 0 and k2 = 1 − c. We start with k1 = 0. Since the index vanishes, we attempt an ordinary Taylor series solution y=
∞ X
an xn ,
y0 =
∞ X
nan xn−1 ,
y 00 =
n(n − 1)an xn−2
n=0
n=0
n=0
∞ X
Substituting this into the confluent hypergeometric equation, we obtain ∞ X
[n(n − 1)an xn−1 + ncan xn−1 − nan xn − aan xn ] = 0
n=0
Making the substition n → n + 1 in the first two terms and simplifying gives ∞ X
[(n + 1)(c + n)an+1 − (a + n)an ]xn = 0
n=0
Therefore we have a recursion relation of the form an+1 =
a+n an (n + 1)(c + n)
Setting a0 = 1, the first few terms in the series becomes a , c
a+1 a(a + 1) a1 = , 2(c + 1) 2!c(c + 1) a+2 a(a + 1)(a + 2) a3 = a2 = 3(c + 2) 3!c(c + 1)(c + 2)
a0 = 1,
a1 =
a2 =
(5)
This indicates that a(a + 1) x2 a(a + 1)(a + 2) x3 a + + ··· y =1+ x+ c c(c + 1) 2! c(c + 1)(c + 2) 3! ∞ X (a)n xn = (c)n n! n=0
(6)
where the notation (a)n is given by (a)n = a(a + 1)(a + 2) · · · (a + n − 2)(a + n − 1) =
Γ(a + n) Γ(a)
(7)
This is the ‘regular’ solution of the confluent hypergeometric equation. We now test this series for convergence using the ratio test. Given the recursion relation (5), we find an xn (n + 1)(c + n) n = lim = lim =∞ n+1 n→∞ an+1 x n→∞ n→∞ x (a + n)x lim
Therefore this series converges for all values of x, unless c is a non-positive integer, in which case the denominators in (6) will eventually all blow up. Turning next to k2 = 1 − c, we seek a series solution of the form y = x1−c
∞ X
y 0 = x−c
an xn ,
n=0
y 00 = x−1−c
∞ X
(n + 1 − c)an xn ,
n=0 ∞ X
(n + 1 − c)(n − c)an xn
n=0
Substituting this into the confluent hypergeometric equation, we find 1−c
x
∞ X
[(n+1−c)(n−c)an xn−1 +c(n+1−c)an xn−1 −(n+1−c)an xn −aan xn ] = 0
n=0
Performing the shift n → n + 1 in the first two terms and simplifying, we obtain 1−c
x
∞ X
[(n + 2 − c)(n + 1)an+1 − (n + 1 + a − c)an ]xn = 0
n=0
which yields the recursion relation an+1 =
n+1+a−c an (n + 2 − c)(n + 1)
Supposing that a0 = 1, the first few terms in this series are given by 1+a−c 2+a−c (1 + a − c)(2 + a − c) , a2 = a1 = , 2−c 2(3 − c) 2!(2 − c)(3 − c) 3+a−c (1 + a − c)(2 + a − c)(3 + a − c) a3 = a2 = 3(4 − c) 3!(2 − c)(3 − c)(4 − c)
a0 = 1,
a1 =
Following the notation of (7), we may write the series solution as 1−c
ynew = x
∞ X (1 + a − c)n xn (2 − c)n n! n=0
(8)
This series is rather similar to the standard one (6). In fact, the solution of (6) may be converted into ynew by making the substitions a → a + 1 − c and c → 2 − c and multiplying y by the prefactor x1−c . [Why this works may be seen by making the substitutions directly into the confluent hypergeometric equation itself.] As a result, by the same ratio test argument as before, ynew converges for all values of x, except when c = 2, 3, 4, . . . where the denominators in (8) would eventually all blow up. To summarize, for non-integer values of c, the two solutions (6) and (8) form a complete linearly independent set. For c = 1, both (6) and (8) are precisely the same, and we have found only one solution. For other integer values of c, only one of (6) or (8) makes sense (and the other one blows up because of a bad denominator). So in fact for all integer c, we have only obtained one solution by the series method, and the second solution would be of the ‘irregular’ form (which is not fun at all). 8.5.14 To a good approximation, the interaction of two nucleons may be described by a mesonic potential Ae−ax V = x attractive for A negative. Develop a series solution of the resultant Schr¨ odinger wave equation h2 d 2 ψ ¯ + (E − V )ψ = 0 2m dx2 We begin by substituting the explicit potential in the Schr¨ odinger equation 2mE d2 ψ 2mAe−ax + − ψ=0 dx2 h2 ¯ h2 x ¯ As in the text, it would be convenient to define E=
2mE , h2 ¯
A=
2mA h2 ¯
In this case, we want to solve the second order equation e−ax ψ + E −A x 00
ψ=0
(9)
which has a regular singular point at x = 0 and an irregular one at x = ∞. We now develop a series solution around x = 0. Noting that e−ax Q(x) = x
P (x) = 0,
⇒
p0 = 0,
q0 = 0
the indicial equation is trivial, k(k − 1) = 0. Since we have k1 = 1 and k2 = 0, we look for the k1 = 1 series (the larger index one always ‘works’). Here we have to worry that e−ax is non-polynomial. As a result, we will not be able to obtain a simple recursion relation. We thus content ourselves with just working out a few terms in the series. Normalizing the first term in the series to be x, we take y 0 = 1+2a2 x+3a3 x2 +· · · ,
y = x+a2 x2 +a3 x3 +· · · ,
y 00 = 2a2 +6a3 x+· · ·
Substitution into (9) gives 2a2 + 6a3 x + · · · + (Ex − Ae−ax )(1 + a2 x + a3 x2 + · · ·) = 0 Since we have used a series for the wavefunction ψ(x), we ought to also expand the exponential as a series, e−ax = 1 − ax + 21 a2 x2 − · · ·. Keeping appropriate powers of x, we find 0 = 2a2 + 6a3 x + · · · + (Ex − A(1 − ax + · · ·))(1 + a2 x + · · ·) = 2a2 + 6a3 x + · · · + (−A + (aA + E)x + · · ·)(1 + a2 x + · · ·) = 2a2 + 6a3 x + · · · + (−A) + (aA + E − a2 A)x + · · · = (2a2 − A) + (6a3 + aA + E − a2 A)x + · · · Setting the coefficients to zero gives a2 = 12 A,
a3 = 16 (a2 A − E − aA) = 16 ( 12 A2 − E − aA)
The series solution is the of the form ψ = x + 21 Ax2 + 16 ( 12 A2 − E − aA)x3 + · · · 8.5.17 The modified Bessel function I0 (x) satisfies the differential equation x2
d2 d I0 (x) + x I0 (x) − x2 I0 (x) = 0 2 dx dx
From Exercise 7.4.4 the leading term in an asymptotic expansion is found to be ex I0 (x) ∼ √ 2πx Assume a series of the form ex I0 (x) ∼ √ (1 + b1 x−1 + b2 x−2 + · · ·) 2πx Determine the coefficients b1 and b2 The (modified) Bessel equation has a regular singular point at x = 0 and an irregular one at x = ∞. Here we are asked to develop an asymptotic expansion around x = ∞. Although this is an irregular one (witness the essential singularity ex ), we are given the form of the series. As a result, all we have to do is to take derivatives and insert the expressions into the differential equation. To make it easier to obtain the derivatives, we write 1 3 5 7 ex I0 (x) ∼ √ (x− 2 + b1 x− 2 + b2 x− 2 + b3 x− 2 + · · ·) 2π The derivative d/dx acts either on the ex factor or the series in the parentheses. The resulting first derivative is 1 3 5 7 ex I00 (x) ∼ √ (x− 2 + (b1 − 12 )x− 2 + (b2 − 23 b1 )x− 2 + (b3 − 52 b2 )x− 2 + · · ·) 2π Taking one more derivative yields 1 3 5 ex − 72 I000 (x) ∼ √ (x− 2 + (b1 − 1)x− 2 + (b2 − 3b1 + 43 )x− 2 + (b3 − 5b2 + 15 + · · ·) 4 b1 )x 2π Substituting the above into the modified Bessel equation and collecting like powers of x, we find 3 1 1 ex − 32 0 ∼ √ (x 2 + (b1 − 1)x 2 + (b2 − 3b1 + 43 )x− 2 + (b3 − 5b2 + 15 + ··· 4 b1 )x 2π 3
− x2
+ x2
1
+ (b1 − 12 )x− 2
1
− b2 x− 2
− b1 x 2
3
1
+ (b2 − 23 b1 )x− 2 + · · ·
1
3
− b3 x− 2 − · · ·)
1 3 ex ∼ √ ((−2b1 + 14 )x− 2 + (−4b2 + 49 b1 )x− 2 + · · ·) 2π Setting the coefficients to zero gives
b1 = 18 ,
b2 =
9 16 b1
=
9 128
so that the asymptotic series develops as ex 9 I0 (x) ∼ √ (1 + 18 x−1 + 128 x−2 + · · ·) 2πx Note that, in order to find b1 and b2 , we needed to keep track of the b3 coefficient, even though it dropped out in the end.
