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Unformatted text preview: 74 Chapter 4 Partial Diﬀerential Equations in Polar and Cylindrical Coordinates Solutions to Exercises 4.8
1. (a) Using the series deﬁnition of the Bessel function, (7), Section 4.7, we have
d −p
[x Jp (x)] =
dx ∞ d
dx =
k =0
∞ = x
2 2k (−1)k
dx
p k !Γ(k + p + 1) dx 2
2 2k k =0 ∞ 2p (k −
m=0 (−1)m
+ p + 2) −x−p (−1)m
m!Γ(m + p + 2)
m=0 (−1)k 2k
1
+ p + 1) 2 2p k!Γ(k x
2 2k −1 2k −1 2m+1 x
2 ∞ = k =0 (−1)
x
− 1)!Γ(k + p + 1) 2 2p m!Γ(m ∞ = k k =0
∞ = (−1)k
+ p + 1) 2p k!Γ(k (set m = k − 1)
x
2 2m+p+1 = −x−p Jp+1 (x). To prove (7), use (1):
dp
[x Jp (x)] = xp Jp−1 (x)
dx ⇒ xp Jp−1 (x) dx = xp Jp (x) + C. Now replace p by p + 1 and get
xp+1 Jp (x) dx = xp+1 Jp+1 (x) + C,
which is (7). Similarly, starting with (2),
d −p
[x Jp (x)] = −x−p Jp+1 (x)
dx ⇒−
⇒ x−pJp+1 (x) dx = x−p Jp (x) + C
x−p Jp+1 (x) dx = −x−p Jp (x) + C. Now replace p by p − 1 and get
x−p+1 Jp (x) dx = −x−p+1 Jp−1(x) + C,
which is (8).
(b) To prove (4), carry out the diﬀerentiation in (2) to obtain
x−p Jp (x) − px−p−1Jp (x) = −x−p Jp+1 (x) ⇒ xJp (x) − pJp (x) = −xJp+1 (x), upon multiplying through by xp+1 . To prove (5), add (3) and (4) and then divide
by x to obtain
Jp−1 (x) − Jp+1 (x) = 2Jp (x).
To prove (6), subtract (4) from (3) then divide by x.
5. J1(x) dx = −J0 (x) + C , by (8) with p = 1. 9.
J3 (x) dx = x2[x−2J3 (x)] dx
x2 = u, x−2J3(x) dx = dv, 2x dx = du, v = −x−2 J2 (x) = −J2 (x) + 2 x−1 J2(x) dx = −J2 (x) − 2x−1 J1(x) + C 2
2
J1(x) − J1(x) + C (use (6) with p = 1)
x
x
4
= J0 (x) − J1(x) + C.
x
= J0 ( x ) − ...
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.
 Fall '11
 StuartChalk

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