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Bessel’s Equation
There are many special functions which arise as solutions to differential equations (Hermite, Legendre, Chebyshev, etc.). Here we will look at how one important class of functions, Bessel Functions, arise through a series solution to a differential equation. Bessel Functions of the First Kind are particularly important in the study of partial differential equations, and arise in the study of vibrating circular drumheads, heat equations, and many other areas where cylindrical symmetry is present.
Bessel’s equation is a differential equation of the form
x^2 y′′^ + xy′^ + (x^2 − v^2 )y = 0,
where v is a real number.
The point x = 0 is a singular point, since p(x) =
x
and q(x) = 1 −
v^2 x^2
are not analytic at x = 0. Recall that analytic
means there exists a Taylor series of the function about the point with nonzero radius of convergence.
Further, since xp(x) = 1 and x^2 q(x) = x^2 − v^2 are analytic at x = 0, the point x = 0 is a regular singular point.
Therefore, we can try to determine at least one series solution of Bessel’s equation which has the form y =
n=0 anx
r+n.
Bessel Functions of the First Kind
Differentiating y =
n=0 anx
r+n (^) and substituting into Bessel’s equation, we find
y =
∑^ ∞
n=
anxr+n, y′^ =
∑^ ∞
n=
(r + n)anxr+n−^1 , y′′^ =
∑^ ∞
n=
(r + n)(r + n − 1)anxr+n−^2 ,
x^2
n=
(r + n)(r + n − 1)anxr+n−^2
n=
(r + n)anxr+n−^1
n=
anxr+n
∑^ ∞
n=
(r + n)(r + n − 1)anxr+n^ +
∑^ ∞
n=
(r + n)anxr+n^ +
∑^ ∞
n=
anxr+n+2^ −
∑^ ∞
n=
v^2 anxr+n^ = 0
∑^ ∞
n=
(r + n)(r + n − 1)anxr+n^ +
∑^ ∞
n=
(r + n)anxr+n^ +
∑^ ∞
n=
an− 2 xr+n^ −
∑^ ∞
n=
v^2 anxr+n^ = 0 ( (^) ∞ ∑
n=
(r + n)(r + n − 1)anxn^ +
∑^ ∞
n=
(r + n)anxn^ +
∑^ ∞
n=
an− 2 xn^ −
∑^ ∞
n=
v^2 anxn
xr^ = 0
∑^ ∞
n=
(r + n)(r + n − 1)anxn^ +
∑^ ∞
n=
(r + n)anxn^ +
∑^ ∞
n=
an− 2 xn^ −
∑^ ∞
n=
v^2 anxn^ = 0 [ r(r − 1)a 0 + ra 0 − v^2 a 0
]
x^0 +
[
(r + 1)ra 1 + (r + 1)a 1 − v^2 a 1
]
x^1 + ∑^ ∞
n=
(r + n)(r + n − 1)anxn^ +
∑^ ∞
n=
(r + n)anxn^ +
∑^ ∞
n=
an− 2 xn^ −
∑^ ∞
n=
v^2 anxn^ = 0
[
r(r − 1)a 0 + ra 0 − v^2 a 0
]
x^0 +
[
(r + 1)ra 1 + (r + 1)a 1 − v^2 a 1
]
x^1 +
∑^ ∞
n=
[
(r + n)^2 an + an− 2 − v^2 an
]
xn^ = 0
We want this to be true for all values of x. Therefore, the coefficients of powers of x must be zero. This leads to the equations:
r(r − 1)a 0 + ra 0 − v^2 a 0 = 0 (1) (r + 1)ra 1 + (r + 1)a 1 − v^2 a 1 = 0 (2) (r + n)^2 an + an− 2 − v^2 an = 0 , n = 2, 3 , 4 ,... (3)
These equations are the recursion and indicial equation. We can choose any one of them as the indicial equation; let’s choose the indicial equation to be Eq. (1),
r(r − 1)a 0 + ra 0 − v^2 a 0 = 0.
If we assume a 0 6 = 0, the solutions to the indicial equation are r = ±v. We can get one solution to Bessel’s equation if we choose to work with the larger root from the indicial equation, so let’s assume r = v > 0 and proceed.
The recursions equations (Eq. (2) and (3)) become
(2v + 1)a 1 = 0 (4) an = −
an− 2 n(n + 2v)
, n = 2, 3 , 4 ,... (5)
Equation (4) tells us that a 1 = 0, unless v = − 1 /2, in which case the equation is satisfied regardless of the value of a 1. Since we have assumed that v > 0, this situation does not arise, and we may safely assume a 1 = 0.
We can now use Eq (5) to generate the first few an, and then try to determine the pattern.
a 0 = unspecified, does not equal zero a 1 = 0 a 2 = −
a 0 2(2 + 2v) a 3 = 0 a 4 = −
a 2 4(4 + 2v)
a 0 2 · 4(2 + 2v)(4 + 2v)
a 0 22 ·^2 2!(1 + v)(2 + v) a 5 = 0 a 6 = − a 4 6(6 + 2v)
a 0 2 · 4 · 6(2 + 2v)(4 + 2v)(6 + 2v)
a 0 22 ·^3 3!(1 + v)(2 + v)(3 + v)
This is enough for us to recognize the pattern. First, let’s work with (1 + v)(2 + v) · · · (m + v). If v is an integer, we observe that
(1 + v)(2 + v) · · · (m + v)
1 · 2 · 3 · · · (v − 1)(v) 1 · 2 · 3 · · · (v − 1)(v)(1 + v)(2 + v) · · · (m + v)
v! (v + m)!
Bessel Functions of the Second Kind
A second solution to Bessel’s equation can be found using reduction of order, since we now know a first solution. Here we use the formula we derived for reduction of order earlier, where we identify p(x) = 1/x from Bessel’s equation.
y2(x) = y 1 (x)
exp
p(x) dx
y^21 (x)
dx
= Jv (x)
exp
(1/x) dx
J v^2 (x)
dx
= Jv (x)
xJ v^2 (x)
dx
= π 2
Yv (x)
where the last integral is found from a book of integrals [1] or a more in-depth study of Bessel Functions than we have time for.
Notice that this procedure found a solution which is a constant multiple of Yv (x), so we can drop the constant and take as the second solution y 2 (x) = Yv (x).
Here is a plot of Yv (x), for v = 0, 1 , 2 , 3 , 4 , 5.
Notice that limx→ 0 Yv (x) = −∞ for integer values of v. This is a logarithmic singularity that we sometimes see with second solutions to differential equations about a regular singular point. This excludes, for physical reasons, Yv (x) from solutions in many applications.
The derivation of the second solution using other techniques can be a difficult task, but realize that although our solution looks simple, we had an integral that we worked out using a table of integrals, so to understand this integral we would likely have to put in an equal amount of work.
References
[1] Gradshteyn, I. S.; Ryzhik, I. M. Table of Integrals, Series, and Products, 6th ed., Academic Press, New York, 2000. Page 664 #6.539.
