Page 563 - Advanced_Engineering_Mathematics o'neil

15.3 Bessel Functions 543

Then
w
(3)
y (x) + xy (x) = 0.
EI
Let u = y to write the second order differential equation

w

u + xu = 0.
EI
Compare this with equation (15.8), putting
w
2 2 2 2 2
2a − 1 = 0,a − ν c = 0,2c − 3 = 1, and b c = .
EI
Thus choose

1 3 1 2 w
a = ,c = ,ν = ,b = .
2 2 3 3 EI
The general solution for u(x) is

√ 2 w √ 2 w
3/2 3/2
u(x) = y (x) = c 1 xJ 1/3 x + c 2 xJ −1/3 x .
3 EI 3 EI
Since there is no bending moment at the top of the rod, y (0)=0. A routine differentiation shows

that this forces c 1 = 0. Now

√ 2 w
3/2
y (x) = c 2 xJ −1/3 x .
3 EI

Because the lower end of the rod is clamped vertically, then y (L) = 0, so

√ 2 w
L 3/2 = 0.
c 2 LJ −1/3
3 EI
We must have c 2 = 0 to have a nontrivial solution. The critical length L C is the smallest positive
value of L such that

2 w 3/2
J −1/3 L = 0.
3 EI
We find from a table or from MAPLE that the smallest positive number α such that J −1/3 (α) = 0
is approximately 1.8663. Therefore
2 w
L 3/2 ≈ 1.8663.
3 EI c
Solve for L C :
1/3
EI

L C ≈ 1.9863 .
w
This is the critical length.

15.3.6 Modified Bessel Functions
Some applications use modified Bessel functions. We will show how these are obtained. Begin
with the general solution

y(x) = c 1 J 0 (kx) + c 2 Y 0 (kx)
of the zero-order Bessel equation
1
2
y + y + k y = 0.
x

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October 14, 2010 15:20 THM/NEIL Page-543 27410_15_ch15_p505-562