Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 3 potx

Since e1zt−1/tvt,has non-zero residues for vt = tn−1 and vt = t−n−1, choosing any simple, positive, closed contour about the originwill give us a non-trivial solution of Bessel’s equatio

.

Derive W [Iν, I−ν] from the value of W [Jν, J−ν] Derive W [Iν, Kν] from the value of W [Iν, I−ν].Hint 34.9

Hint 34.10

Hint 34.11

Hint 34.12

Hint 34.13

Hint 34.14

z214



t + 1t

2

+ z12



t − 1t

+ 1

2

+ x

2t +1t

+ 2

We apply integration by parts to move derivatives from the kernel to v(t).

C

e12 z(t−1/t) t2v00(t) + 3tv(t) + (1 − n2)v(t)
dt = 0

In order that the integral vanish, v(t) must be a solution of the differential equation

t2v00+ 3tv + 1 − n2
v = 0

This is an Euler equation with the solutions {tn−1, t−n−1} for non-zero n and {t−1, t−1log t} for n = 0

Consider the case of non-zero n Since

e1z(t−1/t) t2− 3t
v(t) − tv0

(t)

is single-valued and analytic for t 6= 0 for the functions v(t) = tn−1 and v(t) = t−n−1, the boundary term will vanish if

C is any closed contour that that does not pass through the origin Note that the integrand in our solution,

e12 z(t−1/t)v(t),

is analytic and single-valued except at the origin and infinity where it has essential singularities Consider a simpleclosed contour that does not enclose the origin The integral along such a path would vanish and give us y(z) = 0.This is not an interesting solution Since

e1z(t−1/t)v(t),has non-zero residues for v(t) = tn−1 and v(t) = t−n−1, choosing any simple, positive, closed contour about the originwill give us a non-trivial solution of Bessel’s equation These solutions are

Choosing v(t) = t−1log t would make both the boundary terms and the integrand multi-valued We do not pursue thepossibility of a solution of this form.

The solution y1(t) and y2(t) are not linearly independent To demonstrate this we make the change of variables

t → −1/t in the integral representation of y1(t)

C

(−1/t)n−1e1z(−1/t+t) −1

t2 dt

=Z

C

(−1)nt−n−1e12 z(t−1/t) dt

= (−1)ny2(t)Thus we see that a solution of Bessel’s equation for integer n is

y(t) =

Z

C

t−n−1e12 z(t−1/t) dt

where C is any simple, closed contour about the origin

Therefore, the Bessel function of the first kind and order n,

Jn(z) = 1

ı2πI

t − 1t

4



t − 1t



t + 1t



t − 1t

Now we can evaluate the desired sum.

2

+z2



t − 1t

+ z2 −z

2



t − 1t

Solution 34.5

W [Jν, Yν] =

Jν Jνcot(νπ) − J−νcsc(νπ)

Jν0 Jν0 cot(νπ) − J−ν0 csc(νπ)

= cot(νπ)

Iν I−ν

Iν0 I−ν0

=

ı−νJν(ız) ıνJ−ν(ız)

ı−νıJν0(ız) ıνıJ−ν0 (ız)

... 2πzSince the Wronskian does not vanish identically, the functions are independent for all values of ν.Solution 34 .6

Iν(z) = ı−νJν(ız)

W [Iν,