SOLUTIONS TO PROBLEMS 15 (ODD NUMBERS)
15.1
(a) In the standard form (15.1), the equation is
d2y(x)
dy(x)
+ p(x)
+ q(x)y(x) = 0 ,
2
dx
dx
with
p(x) =
−3x 2
(1 − x 2 )
and
q(x) =
(1 − x)
1
=
.
2
(1 − x ) (1 + x)
For any point x 0 ≠ ±1 ,
lim p(x)
x→x 0
and
lim q(x)
x→x 0
are both finite, and so a solution of the form
∞
y(x) = ∑ an (x − x 0 )n
n=0
is possible. But if x 0 = ±1 , lim p(x) is not finite, and if x 0 = −1 , neither is lim q(x) .
x→x 0
x→x 0
However for x 0 = ±1 ,
⎛ −3x 2 ⎞⎟
⎟=±3 ,
lim(x − x 0 )p(x) = lim (x  1)⎜⎜⎜
2⎟
x→x 0
x→±1
⎜⎝1 − x ⎟⎠
2
and
⎛ 1 ⎞⎟
⎟= 0.
lim(x − x 0 )2q(x) = lim (x  1)2 ⎜⎜⎜
x→x 0
x→±1
⎜⎝1 + x ⎟⎟⎠
Therefore, in both these cases, a solution of the Frobenius type
∞
∞
n=0
n=0
y(x) = x c ∑ bn (x − x 0 )n = x c ∑ bn (x + 1)n
is possible.
(b) Using the notations above, p(x) = 1 x 3 and q(x) = 1 x , so x 0 = 0 is an essential
singularity and no power solution is possible. But x 0 ≠ 0 is a regular point, and so
a power series solution exists.
15.3
Setting z = x −1 , the equation becomes
(1 − z 2 )
d2y
dy
− 3z
−y = 0 .
2
dz
dz
Since z = 0 is a regular point, we substitute
S15.1
∞
y = ∑ an z n ,
n=0
which gives
∞
∑ ⎡⎢⎣(1 − z
n=0
2
)(n + 1)(n + 2)an+2 z n − 3z(n + 1)an+1 z n −an z n ⎤⎥ = 0
⎦
and equating powers of z n gives
(n + 1)(n + 2)an+2 − n(n −1)an − 3nan −an = 0 ,
so that
⎛ n + 1 ⎞⎟
⎟a .
an+2 = ⎜⎜⎜
⎜⎝ n + 2 ⎟⎟⎠ n
Hence
a 2 = 12 a 0 , a 4 = 43 a 2 = 83 a 0 ,
a 3 = 23 a1 , a5 = 45 a 3 =
8
15
a1 ,
and the solution is
⎡
⎤
(x −1)2 3(x −1)4
y(x) = a 0 ⎢⎢1 +
+
+⎥⎥
2
8
⎢⎣
⎥⎦
3
⎡
⎤
2(x −1)
8(x −1)5
+ a1 ⎢⎢(x −1) +
+
+⎥⎥ ,
3
15
⎢⎣
⎥⎦
where a 0 and a1 are constants.
15.5
With y 2 = y1u , we have
y 2′ =
u′ u
u ′′ 2u ′ 2u
− 2 and y 2′′ =
− 2 + 3 .
x x
x
x
x
Substituting into the differential equation gives
xu ′′ + u ′ = 0 ,
which is a first-order equation in u ′ . So, setting u ′(x) = h(x) , we have
h = 1 x ⇒ u(x) = C ln x + D and hence y 2(x) =
1
(C ln x + D ) ,
x
where C and D are constants. The general solution is therefore
y(x) = (A + B ln x ) x ,
S15.2
where A and B are constants
15.7
2
From yn (x) = e −x /2H n (x) , we have
d2yn
dx
= H n′′(x)e −x
2
2
2
− 2xH n′ (x)e −x
2
2
2
+ (x 2 −1)H n (x)e −x 2,
so that (1) becomes
⎡ d2H (x)
⎤ 2
dH n (x)
n
⎢
⎥ e −x 2 = 0 .
−
2x
+
2nH
(x)
n
⎢ dx 2
⎥
dx
⎢⎣
⎥⎦
The term in brackets is just the left-hand side of Hermite’s equation (15.13), and
vanishes if H n (x) is a Hermite polynomial.
From (1) we have
d2yn
dx
2
− x 2yn + (2n + 1)yn = 0
and
d2ym
dx
2
− x 2ym + (2m + 1)ym = 0 .
Multiplying the first of these equations by ym and the second by yn , and then taking
the difference gives
⎛ d2y
d2ym ⎞⎟⎟
⎜⎜y
n
−y
⎟ = 2(m −n)ynym .
⎜⎜ m
n
2
dx 2 ⎟⎟⎠
⎝ dx
Then integrating from −∞ to ∞ , using integration by parts on the left-hand side,
gives
∞
⌠ ⎛⎜ d2yn
d2ym ⎞⎟⎟
⎮ ⎜⎜ym
−y
⎟ dx
n
⎮ ⎜ dx 2
dx 2 ⎟⎟⎠
⌡⎝
−∞
∞
∞
⎡ dy
dym ⎤⎥
n
⎢
= ⎢ym
−yn
− ∫ (ym′ yn′ − yn′ym′ ) = 0,
dx ⎥⎥⎦ −∞ −∞
⎢⎣ dx
2
because y(x) = e −x /2H n (x) → 0 . Hence, for m ≠ n ,
∞
∫
−∞
∞
ym (x)yn (x)dx =
∫H
2
m
(x)H n (x)e −x dx = 0,
−∞
as required.
S15.3
15.9
Substituting y = e mx gives the characteristic equation
m 2 + 2m + λ = 0 ,
with solutions
m± = −1 ± 1 −λ .
(1) If λ < 1 , then m± are real, so that the general solution is
y(x) = Ae
m+x
+ Be
m−x
,
and the boundary conditions require A = B = 0 , i.e. there are no non-trivial solution.
(2) if λ = 1 , the general solution is
y(x) = (Ax + B)e −x
and again there is no non-trivial solution satisfying the boundary conditions.
(3) If λ > 1 , the general solution is
y(x) = (A coskx + B sin kx) ,
where k = λ −1 > 0. The boundary condition y(0) = 0 gives A = 0 , while y(1) = 0
gives either B = 0 , or sin k = 0 , i.e.
k = λ −1 = nπ , ,
n = ±1, ± 2, …
Hence λ −1 = n 2π 2 , so that the eigenvalues are
λ = λn = 1 + n 2π 2
and the corresponding eigenfunctions are therefore
yn (x) = An sin(nπx).
*15.11 Substituting (15.35) for Ql (x) into the left and right-hand sides of (1) and comparing
the coefficients of powers of x one sees that the recurrence relation is satisfied.
Equation (1) is identical with the recurrence relation (15.39a) for Legendre
polynomials. Hence, since from Example 15.8,
Q0 (x) =
1 ⎛⎜1 + x ⎞⎟ P0 (x) ⎛⎜1 + x ⎞⎟
⎟⎟ =
⎟,
ln ⎜
ln ⎜⎜
⎜⎝ 1 − x ⎟⎟⎠
2 ⎜⎜⎝ 1 − x ⎟⎠
2
and
S15.4
Q1(x) =
P (x) ⎛1 + x ⎞⎟
x ⎛⎜1 + x ⎞⎟
⎟⎟ −1 = 1 ln ⎜⎜
⎟
ln ⎜⎜
⎜⎜⎝ 1 − x ⎟⎟⎠ −1,
2 ⎜⎝ 1 − x ⎟⎠
2
this implies
Ql (x) =
⎛1 + x ⎞⎟
⎟ −q (x) ,
ln ⎜⎜⎜
⎜⎝ 1 − x ⎟⎟⎠ l
2
Pl (x)
as required, where q 0 (x) = 0 , q1(x) = 1 , and
lql (x) = (2l −1)xql−1(x)−(l −1)ql−2(x) ,
so that
q 2(x) = 32 xq1(x)− 12 q 0 (x) = 32 x ,
q 3(x) = 53 xq 2(x)− 23 q1(x) = 52 x 2 − 23 ,
q 4 (x) = 74 xq 3(x)− 43 q 2(x) =
35
8
55
x 3 − 24
x.
*15.13 Differentiating the generating function partially with respect to h gives
∞
∂G(x,h)
(x −h)
=
=
lPl (x)h l−1 .
∑
∂h
(1 − 2xh + h 2 )3/2
l=0
Using the expression for G, this becomes
(1 − 2xh + h 2 )∑ lPl (x)h l−1 = (x − h)∑ Pl (x)h l .
l
l
Equating the coefficients of h l on each side gives
(l + 1)Pl+1 − 2lxPl + (l −1)Pl−1 + Pl−1 − xPl = 0 ,
and hence
(2l + 1)xPl = (l + 1)Pl+1 + lPl−1
as required.
*15.15 Since Pk (−x) = (−1)k Pk (x) , we have
P2n+1(0) = 0 and P2n′ (0) = 0.
Using this, the recurrence relation (15.39a) gives
2nP2n (0) = −(2n −1)P2n−2(0)
so that
S15.5
P2n (0) = −
(2n −1)
(2n −1)(2n − 3)
P2n−2(0) =
P2n−4 (0)
2n
2n(2n − 2)
==
(−1)n (2n −1)!!
(−1)n (2n −1)!!
P0 (0) =
,
(2n)!!
(2n)!!
as required, since P0 (0) = 1 . Similarly, from the recurrence relation (15.39b), we
obtain
n
(−1) (2n + 1)!!
P2n+1′(0) = (2n + 1)P2n (0) =
,
(2n)!!
using the previous result for P2n (0) .
*15.17 From (15.33) and Rodriques’ formula, one obtains
1
1
n
(2n + 1)
(2n + 1) 1
n
n d
cn =
x
P
(x)dx
=
x
(x 2 −1)n dx.
∫
∫
n
n
n
2
2
2 n ! −1 dx
−1
The proceeding as in Example 15.10, but with l = n , one obtains
1
1
dn
2
n
n
2
n
x
∫ dx n−1 (x −1) dx = (−1) n ! ∫ (x −1) dx
−1
−1
n
1
= n ! ∫ (1 + x)n (1 − x)n dx .
−1
Sbstituting x = 2t −1, dx = 2dt , 1 + x = 2t , 1 − x = 2(1 −t) , one obtains
1
∫
−1
1
(1 + x)n (1 − x)n dx = 22n+1 ∫ t n (1 −t)n dt = 22n+1
0
n !n !
(2n + 1)!
using the given integral. Hence
cn =
(2n + 1) 1
n !n !
2n n !n !
2n+1
n
!2
=
.
2
(2n + 1)!
(2n)!
2n n !
*15.19 These relations follow directly from the recurrence relations (15.71). In particular, at
a maximum of J n (x) , the derivative J n′ (x) = 0 , so that (15.71d) gives
J n−1(x) = J n+1(x),
while if J n (x) = 0 , (15.71c) gives
J n−1(x) = −J n+1(x).
S15.6
15.21
(a) From (15.68),
N ν (x) ≡
J ν (x)cos(νπ)−J −v (x)
sin(νπ)
,
where for integer m, this is interpreted as
⎡ J (x)cos(νπ)−J (x) ⎤
−ν
⎥.
N m (x) = lim ⎢⎢ ν
⎥
ν→m
sin(νπ)
⎢⎣
⎥⎦
For ν = 0 , the numerator and denominator both vanish, but by l’Hôpital’s rule,
⎡ J ′ (x)cos(νπ)− π sin(νπ)J (x)− J ′ (x) ⎤
ν
−ν
⎥,
N 0 (x) = lim ⎢⎢ ν
⎥
ν→0
π cos(νπ)
⎢⎣
⎥⎦
(2)
where the derivatives are with respect to ν . From (15.65) and (15.62a),
ν
ν+2
⎛x ⎞
1 ⎛⎜ x ⎞⎟
J ν (x) =
⎜⎜ ⎟⎟ +O ⎜⎜⎜ ⎟⎟⎟
⎜⎝ 2 ⎟⎠
νΓ(ν) ⎜⎝ 2 ⎟⎠
,
which using (1) is
ν
ν+2
⎛x ⎞
⎛x ⎞
J ν (x) = ⎡⎢1 + γν +O(ν 2 )⎤⎥ ⎜⎜⎜ ⎟⎟⎟ +O ⎜⎜⎜ ⎟⎟⎟
⎣
⎦ ⎜⎝ 2 ⎟⎠
⎜⎝ 2 ⎟⎠
,
(3a)
and similarly
−ν
2−ν
⎛x ⎞
⎛x ⎞
J −ν (x) = ⎡⎢1 − γν +O(ν 2 )⎤⎥ ⎜⎜⎜ ⎟⎟⎟ +O ⎜⎜⎜ ⎟⎟⎟
⎣
⎦ ⎜⎝ 2 ⎟⎠
⎜⎝ 2 ⎟⎠
.
(3b)
To find the derivatives of these functions, we set y = (x 2)ν , so that ln y = ν ln(x 2)
and d ln y dν = (1 y)dy dν = ln(x 2) . Hence
ν
ν
⎛x ⎞ ⎛x ⎞
d ⎛⎜ x ⎞⎟
⎜⎜ ⎟⎟ = ⎜⎜⎜ ⎟⎟⎟ ln ⎜⎜⎜ ⎟⎟⎟ ,
⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠
dx ⎜⎝ 2 ⎟⎠
(4a)
and similarly
−ν
d ⎛⎜ x ⎞⎟
⎜ ⎟⎟
dx ⎜⎜⎝ 2 ⎟⎠
−ν
⎛x ⎞
⎛x ⎞
= −⎜⎜⎜ ⎟⎟⎟ ln ⎜⎜⎜ ⎟⎟⎟ .
⎜⎝ 2 ⎟⎠
⎜⎝ 2 ⎟⎠
(4b)
The derivatives of the Bessel functions are then
ν
ν
⎛x ⎞ ⎛x ⎞
⎛x ⎞
J ν′ (x) = (1 + γν)⎜⎜⎜ ⎟⎟⎟ ln ⎜⎜⎜ ⎟⎟⎟ + γ ⎜⎜⎜ ⎟⎟⎟ ,
⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠
⎜⎝ 2 ⎟⎠
and
−ν
−ν
⎛x ⎞
⎛x ⎞
⎛x ⎞
′ (x) = −(1 − γν)⎜⎜ ⎟⎟⎟ ln ⎜⎜ ⎟⎟⎟ − γ ⎜⎜ ⎟⎟⎟ .
J −ν
⎜⎜⎝ 2 ⎟⎠
⎜⎜⎝ 2 ⎟⎠
⎜⎜⎝ 2 ⎟⎠
S15.7
So finally, using these expressions in (2) and taking the limit ν → 0 , gives
N 0 (x) =
2⎡
ln x − ln 2 + γ ⎤⎥ +O(x 2 ).
⎢
⎣
⎦
π
(b) As in part (a) above, from (15.69),
N 1(x) ≡ lim
J ν (x)cos(νπ)−J −v (x)
sin(νπ)
ν→1
,
which may be written
⎡ J (x)cos(νπ)−J (x) ⎤
−1−ε
⎥
N 1(x) = lim ⎢⎢ 1+ε
⎥
ε→0
sin(νπ)
⎢⎣
⎥⎦
where ν = 1 + ε . Both numerator and denominator vanish in the limit ε → 0 , but
using l’Hôpital’s rule, we have
⎡ J ′ (x)cos(νπ)− π sin(νπ)J (x)− J ′ (x) ⎤
1+ε
−1−ε
⎥ ,
N 1(x) = lim ⎢⎢ 1+ε
⎥
ε→0
π
cos(νπ)
⎢⎣
⎥⎦
(5)
where the derivatives are now with respect to ε . Proceeding as in part (a), and using
(15.65) and (15.62a), we have
1+ε
⎛ x ⎞⎟
1
⎜⎜ ⎟
J 1+ε (x) =
⎟
ε(1 + ε)Γ(ε) ⎜⎜⎝ 2 ⎟⎠
3+ε
⎛x ⎞
+O ⎜⎜⎜ ⎟⎟⎟
⎜⎝ 2 ⎟⎠
1+ε
⎛x ⎞
= ⎡⎢1 + (γ −1)ε +O(ε2 )⎤⎥ ⎜⎜⎜ ⎟⎟⎟
⎣
⎦ ⎜⎝ 2 ⎟⎠
3+ε
⎛x ⎞
+O ⎜⎜⎜ ⎟⎟⎟
⎜⎝ 2 ⎟⎠
,
where we have used the expansion (1) for εΓ(ε) . Similarly,
−1−ε
1 ⎛⎜ x ⎞⎟
J −1−ε (x) =
⎜ ⎟⎟
Γ(−ε) ⎜⎜⎝ 2 ⎟⎠
1−ε
⎛ x ⎞⎟
1
⎜⎜ ⎟
−
⎟
(−ε)Γ(−ε) ⎜⎜⎝ 2 ⎟⎠
−1−ε
⎛x ⎞
= ⎡⎢−ε +O(ε2 )⎤⎥ ⎜⎜⎜ ⎟⎟⎟
⎣
⎦ ⎜⎝ 2 ⎟⎠
3−ε
⎛x ⎞
+O ⎜⎜⎜ ⎟⎟⎟
⎜⎝ 2 ⎟⎠
1−ε
⎛x ⎞
− ⎡⎢1 − γε +O(ε2 )⎤⎥ ⎜⎜⎜ ⎟⎟⎟
⎣
⎦ ⎜⎝ 2 ⎟⎠
Hence
1+ε
⎛x ⎞
′ (x) = (γ −1)⎜⎜ ⎟⎟⎟
J 1+ε
⎜⎜⎝ 2 ⎟⎠
1+ε
⎛x ⎞
+ ⎜⎜⎜ ⎟⎟⎟
⎜⎝ 2 ⎟⎠
⎛x ⎞
ln ⎜⎜⎜ ⎟⎟⎟,
⎜⎝ 2 ⎟⎠
and
−1−ε
⎛x ⎞
′ (x) = −⎜⎜ ⎟⎟⎟
J −1−ε
⎜⎜⎝ 2 ⎟⎠
1−ε
⎛x ⎞
+ γ ⎜⎜⎜ ⎟⎟⎟
⎜⎝ 2 ⎟⎠
S15.8
1−ε
⎛x ⎞
+ ⎜⎜⎜ ⎟⎟⎟
⎜⎝ 2 ⎟⎠
3−ε
⎛x ⎞
+O ⎜⎜⎜ ⎟⎟⎟
⎜⎝ 2 ⎟⎠
⎛x ⎞
ln ⎜⎜⎜ ⎟⎟⎟ ,
⎜⎝ 2 ⎟⎠
.
where we have neglected terms which obviously vanish in the limit ε → 0 .
Substituting into (5) gives
N 1(x) =
⎛x ⎞
⎛ x ⎞ ⎛ x ⎞⎤
1 ⎡⎢ ⎛⎜ 2 ⎞⎟
⎟⎟ + (2γ −1)⎜⎜ ⎟⎟⎟ + 2 ⎜⎜ ⎟⎟⎟ ln ⎜⎜ ⎟⎟⎟⎥ +O(x 2 ),
−
⎜
⎜⎜⎝ 2 ⎟⎠
⎜⎜⎝ 2 ⎟⎠ ⎜⎜⎝ 2 ⎟⎠⎥
π ⎢⎢⎣ ⎜⎜⎝ x ⎟⎠
⎥⎦
as required.
S15.9