Math 334 Assignment 9 Solutions

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Math 334 Assignment 8 — Solutions 3

(a) Setting µ = 1/2 into the expression for Jµ gives

J 1 2

(x) = ∞ ∑

n=0

(−1)nx 1

2 +2n

2 1

2 +2n n! Γ( 12 + n+ 1)

=

2

x

∞ ∑

n=0

(−1)nx1+2n

21+2n n! Γ( 12 + n+ 1) .

Using the result of question 1(c) with n replaced by n+ 1 we have

22n+1 n! Γ(n+ 1 + 1

2 ) = (2n+ 1)!

√ π.

Thus we get

J 1 2

(x) =

2

x

∞ ∑

n=0

(−1)nx1+2n

21+2n n! Γ( 12 + n+ 1) =

2

x

∞ ∑

n=0

(−1)nx2n+1

(2n+ 1)! √ π =

2

πx sinx.

(b) To obtain the result for J −

1

2

(x), we use the formula d

dx (xµJµ(x)) = x

µJµ−1(x) with µ = 1/2:

J −

1

2

(x) = x− 1

2

d

dx (x

1

2J 1 2

(x)) = x− 1

2

d

dx (

2

π sinx) =

2

πx cosx.

(c) To obtain the result for J 3 2

(x), we use the formula 2µ

x Jµ(x) = Jµ−1(x) + Jµ+1(x) with µ = 1/2:

J 3 2

(x) = 1

x J 1

2

(x)− J −

1

2

(x) =

2

πx

(

sinx

x − cosx

)

.

4. Show that

∫ ℓ

0

xJ2n( αnm ℓ

x) dx = ℓ2

2 J2n+1(αnm), where αnm is the m

th zero of Jn (i.e. Jn(αnm) = 0).

Hint:

• Show that Bessel’s equation can be written as d

dx (xu′)2 = (n2 − x2)

d

dx (u2), and integrate both

sides.

• The formula obtained above by integration must hold for all solutions of Bessel’s equation, so set u = Jn.

• Use the formula from question ?? to get the result.

Solution

Multiply Bessel’s equation of order n by 2u′ to get 2x2u′′u′ + 2x(u′)2 + 2(x2 − n2)uu′ = 0, which can be rearranged to the form

d

dx [(x2(u′)2] = (n2 − x2)

d

dx (u2).

Integrate both sides to get

x2u′ 2 (x) =

(n2 − x2)(u2)′ dx = (n2 − x2)u2(x) + 2

xu2(x) dx

Thus, every solution u of Bessel’s equation of order n (including u = Jn) must satisfy ∫

xu2(x) dx = 1

2

[

x2u′2(x)− (n2 − x2)u2(x) ]

. (1)

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