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11.4 Bessel Functions

We have already seen several cases of Bessel’s equation of order p≧0 $p ≧ 0$ ,

x 2 y'' + x y' + (x 2 - p 2) y = 0.

$x^{2} y ″ + x y ′ + (x^{2} - p^{2}) y = 0.$ (1)

Its solutions are now called Bessel functions of order p. Such functions first appeared in the 1730s in the work of Daniel Bernoulli and Euler on the oscillations of a vertically suspended chain. The equation itself appears in a 1764 article by Euler on the vibrations of a circular drumhead, and Fourier used Bessel functions in his classical treatise on heat (1822). But their general properties were first studied systematically in an 1824 memoir by the German astronomer and mathematician Friedrich W. Bessel (1784–1846), who was investigating the motion of planets. The standard source of information on Bessel functions is G. N. Watson’s A Treatise on the Theory of Bessel Functions, 2nd ed. (Cambridge: Cambridge University Press, 1944). Its 36 pages of references, which cover only the period up to 1922, give some idea of the vast literature of this subject.

Bessel’s equation in (1) has indicial equation r2−p2=0, $r^{2} - p^{2} = 0,$ with roots r=±p. $r = \pm p .$ If we substitute y=∑cmxm+r $y = \sum c_{m} x^{m + r}$ in Eq. (1), we find in the usual manner that c1=0 $c_{1} = 0$ and that

[(m + r) 2 - p 2] c m + c m - 2 = 0

$[{(m + r)}^{2} - p^{2}] c_{m} + c_{m - 2} = 0$ (2)

for m≧2. $m ≧ 2 .$ The verification of Eq. (2) is left to the reader (Problem 6).

The Case r=p > 0

If we use r=p $r = p$ and write am $a_{m}$ in place of cm, $c_{m},$ then Eq. (2) yields the recursion formula

a m = - a m - 2 m ( 2 p + m ) .

$a_{m} = - \frac{a_{m - 2}}{m (2 p + m)} .$ (3)

Because a1=0, $a_{1} = 0,$ it follows that am=0 $a_{m} = 0$ for all odd values of m. The first few even coefficients are

a 2 = - a 0 2 ( 2 p + 2 ) = - a 0 2 2 ( p + 1 ), a 4 = - a 2 4 ( 2 p + 4 ) = a 0 2 4 \cdot 2 ( p + 1 ) ( p + 2 ), a 6 = - a 4 6 ( 2 p + 6 ) = - a 0 2 6 \cdot 2 \cdot 3 ( p + 1 ) ( p + 2 ) ( p + 3 ) .

$\begin{array}{l} a_{2} = - \frac{a_{0}}{2 (2 p + 2)} = - \frac{a_{0}}{2^{2} (p + 1)}, \\ a_{4} = - \frac{a_{2}}{4 (2 p + 4)} = \frac{a_{0}}{2^{4} \cdot 2 (p + 1) (p + 2)}, \\ a_{6} = - \frac{a_{4}}{6 (2 p + 6)} = - \frac{a_{0}}{2^{6} \cdot 2 \cdot 3 (p + 1) (p + 2) (p + 3)} . \end{array}$

The general pattern is

a 2 m = ( - 1 ) m a 0 2 2 m m ! ( p + 1 ) ( p + 2 ) \dots ( p + m ),

$a_{2 m} = \frac{{(- 1)}^{m} a_{0}}{2^{2 m} m! (p + 1) (p + 2) \dots (p + m)},$

so with the larger root r=p $r = p$ we get the solution

y 1 (x) = a 0 \sum m = 0 \infty ( - 1 ) m x 2 m + p 2 2 m m ! ( p + 1 ) ( p + 2 ) \dots ( p + m ) .

$y_{1} (x) = a_{0} \sum_{m = 0}^{\infty} \frac{{(- 1)}^{m} x^{2 m + p}}{2^{2 m} m! (p + 1) (p + 2) \dots (p + m)} .$ (4)

If p=0 $p = 0$ this is the only Frobenius series solution; with a0=1 $a_{0} = 1$ as well, it is the function J0(x) $J_{0} (x)$ we have seen before.

The Case r = -p < 0

If we use r=−p $r = - p$ and write bm $b_{m}$ in place of cm, $c_{m},$ Eq. (2) takes the form

m (m - 2 p) b m + b m - 2 = 0

$m (m - 2 p) b_{m} + b_{m - 2} = 0$ (5)

for m≧2, $m ≧ 2,$ whereas b1=0. $b_{1} = 0 .$ We see that there is a potential difficulty if it happens that 2p is a positive integer—that is, if p is either a positive integer or an odd positive integral multiple of 12. $\frac{1}{2} .$ For then when m=2p, $m = 2 p,$ Eq. (5) is simply 0⋅bm+bm−2=0. $0 \cdot b_{m} + b_{m - 2} = 0 .$ Thus if bm−2≠0, $b_{m - 2} \neq 0,$ then no value of bm $b_{m}$ can satisfy this equation.

But if p is an odd positive integral multiple of 12, $\frac{1}{2},$ we can circumvent this difficulty. For suppose that p=k/2, $p = k / 2,$ where k is an odd positive integer. Then we need only choose bm=0 $b_{m} = 0$ for all odd values of m. The crucial step is the kth step,

k (k - k) b k + b k - 2 = 0;

$k (k - k) b_{k} + b_{k - 2} = 0;$

and this equation will hold because bk=bk−2=0 $b_{k} = b_{k - 2} = 0$ .

Hence if p is not a positive integer, we take bm=0 $b_{m} = 0$ for m odd and define the coefficients of even subscript in terms of b0 $b_{0}$ by means of the recursion formula

b m = - b m - 2 m ( m - 2 p ), m ≧ 2.

$b_{m} = - \frac{b_{m - 2}}{m (m - 2 p)}, m ≧ 2.$ (6)

In comparing (6) with (3), we see that (6) will lead to the same result as that in (4), except with p replaced with −p. $- p .$ Thus in this case we obtain the second solution

y 2 (x) = b 0 \sum m = 0 \infty ( - 1 ) m x 2 m - p 2 2 m m ! ( - p + 1 ) ( - p + 2 ) \dots ( - p + m ) .

$y_{2} (x) = b_{0} \sum_{m = 0}^{\infty} \frac{{(- 1)}^{m} x^{2 m - p}}{2^{2 m} m! (- p + 1) (- p + 2) \dots (- p + m)} .$ (7)

The series in (4) and (7) converge for all x>0 $x > 0$ because x=0 $x = 0$ is the only singular point of Bessel’s equation. If p>0, $p > 0,$ then the leading term in y1 $y_{1}$ is a0xp, $a_{0} x^{p},$ whereas the leading term in y2 $y_{2}$ is b0x−p. $b_{0} x^{- p} .$ Hence y1(0)=0, $y_{1} (0) = 0,$ but y2(x)→±∞ $y_{2} (x) \to \pm \infty$ as x→0, $x \to 0,$ so it is clear that y1 $y_{1}$ and y2 $y_{2}$ are linearly independent solutions of Bessel’s equation of order p>0 $p > 0$ .

The Gamma Function

The formulas in (4) and (7) can be simplified by use of the gamma function Γ(x), $Γ (x),$ which (as in Section 10.1) is defined for x>0 $x > 0$ by

Γ (x) = \int \infty 0 e - t t x - 1 d t .

$Γ (x) = \int_{0}^{\infty} e^{- t} t^{x - 1} d t .$ (8)

It is not difficult to show that this improper integral converges for each x>0. $x > 0 .$ The gamma function is a generalization for x>0 $x > 0$ of the factorial function n!, which is defined only if n is a nonnegative integer. To see the way in which Γ(x) $Γ (x)$ is a generalization of n!, we note first that

Γ (1) = \int \infty 0 e - t d t = lim b \to \infty [- e - t] b 0 = 1.

$Γ (1) = \int_{0}^{\infty} e^{- t} d t = \lim_{b \to \infty} {[- e^{- t}]}_{0}^{b} = 1.$ (9)

Then we integrate by parts with u=tx $u = t^{x}$ and dv=e−t dt $d v = e^{- t} d t$ :

Γ (x + 1) = lim b \to \infty \int b 0 e - t t x d t = = lim b \to \infty ([- e - t t x] b 0 + \int b 0 x e - t t x - 1 d t) x (lim b \to \infty \int b 0 e - t t x - 1 d t);

$\begin{array}{rcl} Γ (x + 1) = \lim_{b \to \infty} \int_{0}^{b} e^{- t} t^{x} d t & = & \lim_{b \to \infty} ({[- e^{- t} t^{x}]}_{0}^{b} + \int_{0}^{b} x e^{- t} t^{x - 1} d t) \\ = & x (\lim_{b \to \infty} \int_{0}^{b} e^{- t} t^{x - 1} d t); \end{array}$

that is,

Γ (x + 1) = x Γ (x) .

$Γ (x + 1) = x Γ (x) .$ (10)

This is the most important property of the gamma function.

If we combine Eqs. (9) and (10), we see that

Γ (2) = 1 \cdot Γ (1) = 1!, Γ (3) = 2 \cdot Γ (2) = 2!, Γ (4) = 3 \cdot Γ (3) = 3!,

$Γ (2) = 1 \cdot Γ (1) = 1!, Γ (3) = 2 \cdot Γ (2) = 2!, Γ (4) = 3 \cdot Γ (3) = 3!,$

and in general that

Γ (n + 1) = n! for n ≧ 0 an integer .

$Γ (n + 1) = n! for n ≧ 0 an integer .$ (11)

An important special value of the gamma function is

Γ (1 2) = \int \infty 0 e - t t - 1/2 d t = 2 \int \infty 0 e - u 2 d u = π - - \sqrt,

$Γ (\frac{1}{2}) = \int_{0}^{\infty} e^{- t} t^{- 1/2} d t = 2 \int_{0}^{\infty} e^{- u^{2}} d u = \sqrt{π},$ (12)

where we have substituted u2 $u^{2}$ for t in the first integral; the fact that

\int \infty 0 e - u 2 d u = π - - \sqrt 2

$\int_{0}^{\infty} e^{- u^{2}} d u = \frac{\sqrt{π}}{2}$

is known, but is far from obvious. (See, for instance, Example 5 in Section 13.4 of Edwards and Penney, Calculus: Early Transcendentals, 7th edition, Hoboken, NJ: Pearson, 2008.)

Although Γ(x) $Γ (x)$ is defined in (8) only for x>0, $x > 0,$ we can use the recursion formula in (10) to define Γ(x) $Γ (x)$ whenever x is neither zero nor a negative integer. If −1<x<0, $- 1 < x < 0,$ then

Γ (x) = Γ ( x + 1 ) x;

$Γ (x) = \frac{Γ (x + 1)}{x};$

the right-hand side is defined because 0<x+1<1. $0 < x + 1 < 1 .$ The same formula may then be used to extend the definition of Γ(x) $Γ (x)$ to the open interval (−2,−1), $(- 2, - 1),$ then to the open interval (−3,−2), $(- 3, - 2),$ and so on. The graph of the gamma function thus extended is shown in Fig. 11.4.1. The student who would like to pursue this fascinating topic further should consult Artin’s The Gamma Function (New York: Holt, Rinehart and Winston, 1964). In only 39 pages, this is one of the finest expositions in the entire literature of mathematics.

FIGURE 11.4.1.

The graph of the extended gamma function.

Bessel Functions of the First Kind

If we choose a0=1/[2pΓ(p+1)] $a_{0} = 1 / [2^{p} Γ (p + 1)]$ in (4), where p>0, $p > 0,$ and note that

Γ (p + m + 1) = (p + m) (p + m - 1) \dots (p + 2) (p + 1) Γ (p + 1)

$Γ (p + m + 1) = (p + m) (p + m - 1) \dots (p + 2) (p + 1) Γ (p + 1)$

by repeated application of Eq. (10), we can write the Bessel function of the first kind of order p very concisely with the aid of the gamma function:

J p (x) = \sum m = 0 \infty ( - 1 ) m m ! Γ ( p + m + 1 ) (x 2) 2 m + p .

$J_{p} (x) = \sum_{m = 0}^{\infty} \frac{{(- 1)}^{m}}{m! Γ (p + m + 1)} {(\frac{x}{2})}^{2 m + p} .$ (13)

Similarly, if p>0 $p > 0$ is not an integer, we choose b0=1/[2−pΓ(−p+1)] $b_{0} = 1 / [2^{- p} Γ (- p + 1)]$ in (7) to obtain the linearly independent second solution

J - p (x) = \sum m = 0 \infty ( - 1 ) m m ! Γ ( - p + m + 1 ) (x 2) 2 m - p

$J_{- p} (x) = \sum_{m = 0}^{\infty} \frac{{(- 1)}^{m}}{m! Γ (- p + m + 1)} {(\frac{x}{2})}^{2 m - p}$ (14)

of Bessel’s equation of order p. If p is not an integer, we have the general solution

y (x) = c 1 J p (x) + c 2 J - p (x)

$y (x) = c_{1} J_{p} (x) + c_{2} J_{- p} (x)$ (15)

for x>0 $x > 0$ ; xp $x^{p}$ must be replaced with |x|p ${| x |}^{p}$ in Eqs. (13) through (15) to get the correct solutions for x<0 $x < 0$ .

If p=n, $p = n,$ a nonnegative integer, then Eq. (13) gives

J n (x) = \sum m = 0 \infty ( - 1 ) m m ! ( m + n ) ! (x 2) 2 m + n

$J_{n} (x) = \sum_{m = 0}^{\infty} \frac{{(- 1)}^{m}}{m! (m + n)!} {(\frac{x}{2})}^{2 m + n}$ (16)

for the Bessel functions of the first kind of integral order. Thus

J 0 (x) = \sum m = 0 \infty ( - 1 ) m x 2 m 2 2 m ( m ! ) 2 = 1 - x 2 2 2 + x 4 2 2 \cdot 4 2 - x 6 2 2 \cdot 4 2 \cdot 6 2 + \dots

$J_{0} (x) = \sum_{m = 0}^{\infty} \frac{{(- 1)}^{m} x^{2 m}}{2^{2 m} {(m!)}^{2}} = 1 - \frac{x^{2}}{2^{2}} + \frac{x^{4}}{2^{2} \cdot 4^{2}} - \frac{x^{6}}{2^{2} \cdot 4^{2} \cdot 6^{2}} + \dots$ (17)

and

J 1 (x) = \sum m = 0 \infty ( - 1 ) m 2 2 m + 1 2 2 m + 1 m ! ( m + 1 ) ! = x 2 - 1 2 ! (x 2) 3 + 1 2 ! \cdot 3 ! (x 2) 5 - \dots .

$J_{1} (x) = \sum_{m = 0}^{\infty} \frac{{(- 1)}^{m} 2^{2 m + 1}}{2^{2 m + 1} m! (m + 1)!} = \frac{x}{2} - \frac{1}{2!} {(\frac{x}{2})}^{3} + \frac{1}{2! \cdot 3!} {(\frac{x}{2})}^{5} - \dots .$ (18)

The graphs of J0(x) $J_{0} (x)$ and J1(x) $J_{1} (x)$ are shown in Fig. 11.4.2. In a general way they resemble damped cosine and sine oscillations, respectively. Indeed, if you examine the series in (17), you can see part of the reason why J0(x) $J_{0} (x)$ and cos x might be similar—only minor changes in the denominators in (17) are needed to produce the Taylor series for cos x. As suggested by Fig. 11.4.2, the zeros of the functions J0(x) $J_{0} (x)$ and J1(x) $J_{1} (x)$ are interlaced—between any two consecutive zeros of J0(x) $J_{0} (x)$ there is precisely one zero of J1(x), $J_{1} (x),$ and vice versa. The first four zeros of J0(x) $J_{0} (x)$ are approximately 2.4048, 5.5201, 8.6537, and 11.7915. For n large, the nth zero of J0(x) $J_{0} (x)$ is approximately (n−14)π $(n - \frac{1}{4}) π$ ; the nth zero of J1(x) $J_{1} (x)$ is approximately (n+14)π. $(n + \frac{1}{4}) π .$ Thus the interval between consecutive zeros of either J0(x) $J_{0} (x)$ or J1(x) $J_{1} (x)$ is approximately π $π$ —another similarity with cos x and sin x. You can see the way the accuracy of these approximations increases with increasing n by rounding the entries in the table in Fig. 11.4.3 to two decimal places.

It turns out that Jp(x) $J_{p} (x)$ is an elementary function if the order p is half an odd integer. For instance, on substitution of p=12 $p = \frac{1}{2}$ in Eqs. (13) and (14), respectively, the results can be recognized (Problem 2) as

J 1/2 (x) = 2 π x - - - \sqrt sin x and J - 1/2 (x) = 2 π x - - - \sqrt cos x .

$J_{1/2} (x) = \sqrt{\frac{2}{π x}} \sin x and J_{- 1/2} (x) = \sqrt{\frac{2}{π x}} \cos x .$ (19)

FIGURE 11.4.2.

The graphs of the Bessel functions J0(x) $J_{0} (x)$ and J1(x) $J_{1} (x)$ .

FIGURE 11.4.3.

Zeros of J0(x) $J_{0} (x)$ and J1(x) $J_{1} (x)$ .

`n`	`n`th zero of J0(x) $J_{0} (x)$	(n−14)π $(n - \frac{1}{4}) π$	`n`th zero of J1(x) $J_{1} (x)$	(n+14)π $(n + \frac{1}{4}) π$
1	2.4048	2.3562	3.8317	3.9270
2	5.5201	5.4978	7.0156	7.0686
3	8.6537	8.6394	10.1735	10.2102
4	11.7915	11.7810	13.3237	13.3518
5	14.9309	14.9226	16.4706	16.4934

Bessel Functions of the Second Kind

If n is not an integer, then Bessel’s equation of order n has no second Frobenius series solution independent of Jn(x). $J_{n} (x) .$ For Bessel’s equation of order 0, Example 4 in Section 8.4 of Edwards and Penney, Differential Equations and Boundary Value Problems: Computing and Modeling (5th edition, Hoboken, NJ: Pearson, 2014) gives the second solution

Y 0 (x) = 2 π [(γ + ln x 2) J 0 (x) + \sum m = 1 \infty ( - 1 ) m + 1 H m ( m ! ) 2 (x 2) 2 m],

$Y_{0} (x) = \frac{2}{π} [(γ + \ln \frac{x}{2}) J_{0} (x) + \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m + 1} H_{m}}{{(m!)}^{2}} {(\frac{x}{2})}^{2 m}],$ (20)

where

H m = 1 + 1 2 + 1 3 + \dots + 1 m

$H_{m} = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{m}$

denotes the mth partial sum of the harmonic series ∑n=1∞1n $\sum_{n = 1}^{\infty} \frac{1}{n}$ and

γ = lim n \to \infty (H n - ln n) \approx 0.57722

$γ = \lim_{n \to \infty} (H_{n} - \ln n) \approx 0.57722$

is Euler’s constant. The logarithmic term in (20) implies that this Bessel function Y0(x) $Y_{0} (x)$ of the second kind is not a Frobenius series, and it is typical of the case in which a second Frobenius series solution does not exist. It also implies that Y0(x)→−∞ $Y_{0} (x) \to - \infty$ as x→0 $x \to 0$ (Fig. 11.4.4), so Y0(x) $Y_{0} (x)$ is not continuous at x=0. $x = 0 .$ These properties are shared by the general Bessel function Yn(x) $Y_{n} (x)$ of the second kind (with n a positive integer), which is defined by a complicated generalization of the formula in (20).

A general solution of Bessel’s equation of integral order n is given by

y (x) = c 1 J n (x) + c 2 Y n (x) .

$y (x) = c_{1} J_{n} (x) + c_{2} Y_{n} (x) .$ (21)

But if y(x) is continuous at x=0, $x = 0,$ the fact that Yn(x)→−∞ $Y_{n} (x) \to - \infty$ as x→0 $x \to 0$ implies that c2=0. $c_{2} = 0 .$ It follows that any continuous solution of Bessel’s equation of integral order n must be a constant multiple of the Bessel function Jn(x) $J_{n} (x)$ of the first kind. Numerous physical applications of this fact—to heat flow in circular plates or cylinders and to vibrations of circular membranes, for instance—are discussed in Section 10.4 of the reference cited previously.

Figure 11.4.5 illustrates the fact that for n>1 $n > 1$ the graphs of Jn(x) $J_{n} (x)$ and Yn(x) $Y_{n} (x)$ look generally like those of J1(x) $J_{1} (x)$ and Y1(x). $Y_{1} (x) .$ In particular, Jn(0)=0 $J_{n} (0) = 0$ while Yn(x)→−∞ $Y_{n} (x) \to - \infty$ as x→0+, $x \to 0^{+},$ and both functions undergo damped oscillation as x→+∞ $x \to + \infty$ .

FIGURE 11.4.4.

The graphs of the Bessel functions Y0(x) $Y_{0} (x)$ and Y1(x) $Y_{1} (x)$ of the second kind.

FIGURE 11.4.5.

The graphs of the Bessel functions J2(x) $J_{2} (x)$ and Y2(x) $Y_{2} (x)$ .

Bessel Function Identities

Bessel functions are analogous to trigonometric functions in that they satisfy a large number of standard identities of frequent utility, especially in the evaluation of integrals involving Bessel functions. Differentiation of

J p (x) = \sum m = 0 \infty ( - 1 ) m m ! Γ ( p + m + 1 ) (x 2) 2 m + p

$J_{p} (x) = \sum_{m = 0}^{\infty} \frac{{(- 1)}^{m}}{m! Γ (p + m + 1)} {(\frac{x}{2})}^{2 m + p}$ (13)

in the case that p is a nonnegative integer gives

d d x [x p J p (x)] = = = d d x \sum m = 0 \infty ( - 1 ) m x 2 m + 2 p 2 2 m + p m ! ( p + m ) ! \sum m = 0 \infty ( - 1 ) m x 2 m + 2 p - 1 2 2 m + p - 1 m ! ( p + m - 1 ) ! x p \sum m = 0 \infty ( - 1 ) m x 2 m + p - 1 2 2 m + p - 1 m ! ( p + m - 1 ) !,

$\begin{array}{rcl} \frac{d}{d x} [x^{p} J_{p} (x)] & = & \frac{d}{d x} \sum_{m = 0}^{\infty} \frac{{(- 1)}^{m} x^{2 m + 2 p}}{2^{2 m + p} m! (p + m)!} \\ = & \sum_{m = 0}^{\infty} \frac{{(- 1)}^{m} x^{2 m + 2 p - 1}}{2^{2 m + p - 1} m! (p + m - 1)!} \\ = & x^{p} \sum_{m = 0}^{\infty} \frac{{(- 1)}^{m} x^{2 m + p - 1}}{2^{2 m + p - 1} m! (p + m - 1)!}, \end{array}$

and thus we have shown that

d d x [x p J p (x)] = x p J p - 1 (x) .

$\frac{d}{d x} [x^{p} J_{p} (x)] = x^{p} J_{p - 1} (x) .$ (22)

Similarly,

d d x [x - p J p (x)] = - x - p J p + 1 (x) .

$\frac{d}{d x} [x^{- p} J_{p} (x)] = - x^{- p} J_{p + 1} (x) .$ (23)

If we carry out the differentiations in Eqs. (22) and (23) and then divide the resulting identities by xp $x^{p}$ and x−p, $x^{- p},$ respectively, we obtain (Problem 8) the identities

J' p (x) = J p - 1 (x) - p x J p (x)

$J_{p}^{′} (x) = J_{p - 1} (x) - \frac{p}{x} J_{p} (x)$ (24)

and

J' p (x) = p x J p (x) - J p + 1 (x) .

$J_{p}^{′} (x) = \frac{p}{x} J_{p} (x) - J_{p + 1} (x) .$ (25)

Thus we may express the derivatives of Bessel functions in terms of Bessel functions themselves. Subtraction of Eq. (25) from Eq. (24) gives the recursion formula

J p + 1 (x) = 2 p x J p (x) - J p - 1 (x),

$J_{p + 1} (x) = \frac{2 p}{x} J_{p} (x) - J_{p - 1} (x),$ (26)

which can be used to express Bessel functions of higher order in terms of Bessel functions of lower orders. In the form

J p - 1 (x) = 2 p x J p (x) - J p + 1 (x),

$J_{p - 1} (x) = \frac{2 p}{x} J_{p} (x) - J_{p + 1} (x),$ (27)

it can be used to express Bessel functions of large negative order in terms of Bessel functions of numerically smaller negative orders.

The identities in Eqs. (22) through (27) hold wherever they are meaningful—that is, whenever no Bessel functions of negative integral order appear. In particular, they hold for all nonintegral values of p.

Example 1

With p=0, $p = 0,$ Eq. (22) gives

\int x J 0 (x) d x = x J 1 (x) + C .

$\int x J_{0} (x) d x = x J_{1} (x) + C .$

Similarly, with p=0, $p = 0,$ Eq. (23) gives

\int J 1 (x) d x = - J 0 (x) + C .

$\int J_{1} (x) d x = - J_{0} (x) + C .$

Example 2

Using first p=2 $p = 2$ and then p=1 $p = 1$ in Eq. (26), we get

J 3 (x) = 4 x J 2 (x) - J 1 (x) = 4 3 [2 x J 1 (x) - J 0 (x)] - J 1 (x),

$J_{3} (x) = \frac{4}{x} J_{2} (x) - J_{1} (x) = \frac{4}{3} [\frac{2}{x} J_{1} (x) - J_{0} (x)] - J_{1} (x),$

so that

J 3 (x) = - 4 x J 0 (x) + (8 x 2 - 1) J 1 (x) .

$J_{3} (x) = - \frac{4}{x} J_{0} (x) + (\frac{8}{x^{2}} - 1) J_{1} (x) .$

With similar manipulations every Bessel function of positive integral order can be expressed in terms of J0(x) $J_{0} (x)$ and J1(x) $J_{1} (x)$ .

Example 3

To antidifferentiate xJ2(x), $x J_{2} (x),$ we first note that

\int x - 1 J 2 (x) d x = - x - 1 J 1 (x) + C

$\int x^{- 1} J_{2} (x) d x = - x^{- 1} J_{1} (x) + C$

by Eq. (23) with p=1. $p = 1 .$ We therefore write

\int x J 2 (x) d x = \int x 2 [x - 1 J 2 (x)] d x

$\int x J_{2} (x) d x = \int x^{2} [x^{- 1} J_{2} (x)] d x$

and integrate by parts with

u d u = = x 2, 2 x d x, and d v v = = x - 1 J 2 (x) d x, - x - 1 J 1 (x) .

$\begin{array}{rcllrcl} u & = & x^{2}, & d v & = & x^{- 1} J_{2} (x) d x, \\ d u & = & 2 x d x, & and & v & = & - x^{- 1} J_{1} (x) . \end{array}$

This gives

\int x J 2 (x) d x = - x J 1 (x) + 2 \int J 1 (x) d x = - x J 1 (x) - 2 J 0 (x) + C,

$\int x J_{2} (x) d x = - x J_{1} (x) + 2 \int J_{1} (x) d x = - x J_{1} (x) - 2 J_{0} (x) + C,$

with the aid of the second result of Example 1.

Applications of Bessel Functions

The importance of Bessel functions stems not only from the frequent appearance of Bessel’s equation in applications, but also from the fact that the solutions of many other second-order linear differential equations can be expressed in terms of Bessel functions. To see how this comes about, we begin with Bessel’s equation of order p in the form

$z^{2} \frac{d^{2} w}{d z^{2}} + z \frac{d w}{d z} + (z^{2} - p^{2}) w = 0$ (28)

and substitute

$w = x^{- α} y, z = k x^{β} .$ (29)

Then a routine (but lengthy) transformation of Eq. (28) yields

$x^{2} y^{″} + (1 - 2 α) x y^{′} + (α^{2} - β^{2} p^{2} + β^{2} k^{2} x^{2 β}) y = 0;$

that is,

$x^{2} y^{″} + A x y^{′} + (B + C x^{q}) y = 0,$ (30)

where the constants A, B, C, and q are given by

$A = 1 - 2 α, B = α^{2} - β^{2} p^{2}, C = β^{2} k^{2}, and q = 2 β .$ (31)

It is a simple matter to solve the equations in (31) for

$\begin{array}{rcllrcl} α & = & \frac{1 - A}{2}, & β & = & \frac{q}{2}, \\ k & = & \frac{2 \sqrt{C}}{q}, & and & p & = & \frac{\sqrt{{(1 - A)}^{2} - 4 B}}{q} . \end{array}$ (32)

Under the assumption that the square roots in (32) are real, it follows that the general solution of Eq. (30) is

$y (x) = x^{α} w (z) = x^{α} w (k x^{β}),$

where

$w (z) = c_{1} J_{p} (z) + c_{2} J_{- p} (z)$

(assuming that p is not an integer) is the general solution of the Bessel equation in (28). This establishes the following result.

Example 4

Solve the equation

$4 x^{2} y ″ + 8 x y ′ + (x^{4} - 3) y = 0.$ (34)

Solution

To compare Eq. (34) with Eq. (30), we rewrite the former as

$x^{2} y ″ + 2 x y ′ + (- \frac{3}{4} + \frac{1}{4} x^{4}) y = 0$

and see that $A = 2, B = - \frac{3}{4}, C = \frac{1}{4},$ and $q = 4.$ Then the equations in (32) give $α = - \frac{1}{2}, β = 2, k = \frac{1}{4},$ and $p = \frac{1}{2} .$ Thus the general solution in (33) of Eq. (34) is

$y (x) = x^{- 1/2} [c_{1} J_{1/2} (\frac{1}{4} x^{2}) + c_{2} J_{- 1/2} (\frac{1}{4} x^{2})] .$

If we recall from Eq. (19) that

$J_{1/2} (z) = \sqrt{\frac{2}{π z}} \sin z and J_{- 1/2} (z) = \sqrt{\frac{2}{π z}} \cos z,$

we see that a general solution of Eq. (34) can be written in the elementary form

$y (x) = x^{- 3/2} (A \cos \frac{x^{2}}{4} + B \sin \frac{x^{2}}{4}) .$

Example 5

Solve the Airy equation

$y ″ + 9 x y = 0.$ (35)

Solution

First we rewrite the given equation in the form

$x^{2} y ″ + 9 x^{3} y = 0.$

This is the special case of Eq. (30) with $A = B = 0, C = 9,$ and $q = 3.$ It follows from the equations in (32) that $α = \frac{1}{2}, β = \frac{3}{2}, k = 2,$ and $p = \frac{1}{3} .$ Thus the general solution of Eq. (35) is

$y (x) = x^{1/2} [c_{1} J_{1/3} (2 x^{3/2}) + c_{2} J_{- 1/3} (2 x^{3/2})] .$

11.4 Problems

Differentiate termwise the series for $J_{0} (x)$ to show directly that $J_{0}^{′} (x) = - J_{1} (x)$ (another analogy with the cosine and sine functions).
1. Deduce from Eqs. (10) and (12) that
  
  $Γ (n + \frac{1}{2}) = \frac{1 \cdot 3 \cdot 5 \cdot (2 n - 1)}{2^{n}} \sqrt{π} .$
2. Use the result of part (a) to verify the formulas in Eq. (19) for $J_{1/2} (x)$ and $J_{- 1/2} (x)$ .
1. Suppose that m is a positive integer. Show that
  
  $Γ (m + \frac{2}{3}) = \frac{2 \cdot 5 \cdot 8 \dots (3 m - 1)}{3^{m}} Γ (\frac{2}{3}) .$
2. Conclude from part (a) and Eq. (13) that
  
  $\begin{array}{l} J_{- 1/3} (x) = \\ \frac{{(x / 2)}^{1/3}}{Γ (\frac{2}{3})} (1 + \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} 3^{m} x^{2 m}}{2^{2 m} m! \cdot 2 \cdot 5 \dots (3 m - 1)}) . \end{array}$
Apply Eqs. (19), (26), and (27) to show that

$J_{3/2} (x) = \sqrt{\frac{2}{π x^{3}}} (\sin x - x \cos x)$

and

$J_{- 3/2} (x) = - \sqrt{\frac{2}{π x^{3}}} (\cos x + x \sin x) .$
Express $J_{4} (x)$ in terms of $J_{0} (x)$ and $J_{1} (x)$ .
Derive the recursion formula in Eq. (2) for Bessel’s equation.
Verify the identity in (23) by termwise differentiation.
Deduce the identities in Eqs. (24) and (25) from those in Eqs. (22) and (23).
Use the relation $Γ (x + 1) = x Γ (x)$ to deduce from Eqs. (13) and (14) that if p is not a negative integer, then

$\begin{array}{l} J_{p} (x) = \\ \frac{{(x / 2)}^{p}}{Γ (p + 1)} [1 + \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m} {(x / 2)}^{2 m}}{m! (p + 1) (p + 2) \dots (p + m)}] . \end{array}$

This form is more convenient for the computation of $J_{p} (x)$ because only the single value $Γ (p + 1)$ of the gamma function is required.
Use the series of Problem 9 to find $y (0) = \lim_{x \to 0} y (x)$ if

$y (x) = x^{2} [\frac{J_{5/2} (x) + J_{- 5/2} (x)}{J_{1/2} (x) + J_{- 1/2} (x)}] .$

Any integral of the form $\int x^{m} J_{n} (x) d x$ can be evaluated in terms of Bessel functions and the indefinite integral $\int J_{0} (x) d x .$ The latter integral cannot be simplified further, but the function $\int_{0}^{x} J_{0} (t) d t$ is tabulated in Table 11.1 of Abramowitz and Stegun. Use the identities in Eqs. (22) and (23) to evaluate the integrals in Problems 11 through 18.

$\int x^{2} J_{0} (x) d x$
$\int x^{3} J_{0} (x) d x$
$\int x^{4} J_{0} (x) d x$
$\int x J_{1} (x) d x$
$\int x^{2} J_{1} (x) d x$
$\int x^{3} J_{1} (x) d x$
$\int x^{4} J_{1} (x) d x$
$\int J_{2} (x) d x$

In Problems 19 through 30, express the general solution of the given differential equation in terms of Bessel functions.

$x^{2} y ″ - x y ′ + (1 + x^{2}) y = 0$
$x y ″ + 3 y ′ + x y = 0$
$x y ″ - y ′ + 36 x^{3} y = 0$
$x^{2} y ″ - 5 x y ′ + (8 + x) y = 0$
$36 x^{2} y ″ + 60 x y ′ + (9 x^{3} - 5) y = 0$
$16 x^{2} y ″ + 24 x y ′ + (1 + 144 x^{3}) y = 0$
$x^{2} y ″ + 3 x y ′ + (1 + x^{2}) y = 0$
$4 x^{2} y ″ - 12 x y ′ + (15 + 16 x) y = 0$
$16 x^{2} y ″ - (5 - 144 x^{3}) y = 0$
$2 x^{2} y ″ - 3 x y ′ - 2 (14 - x^{5}) y = 0$
$y ″ + x^{4} y = 0$
$y ″ + 4 x^{3} y = 0$
Apply Theorem 1 to show that the general solution of

$x y ″ + 2 y ′ + x y = 0$

is $y (x) = x^{- 1} (A \cos x + B \sin x)$ .
Verify that the substitutions in (2) in Bessel’s equation [Eq. (1)] yield Eq. (3).
1. Show that the substitution
  
  $y = - \frac{1}{u} \frac{d u}{d x}$
  
  transforms the Riccati equation $d y / d x = x^{2} + y^{2}$ into $u ″ + x^{2} u = 0$ .
2. Show that the general solution of $d y / d x = x^{2} + y^{2}$ is
  
  $y (x) = x \frac{J_{3/4} (\frac{1}{2} x^{2}) - c J_{- 3/4} (\frac{1}{2} x^{2})}{c J_{1/4} (\frac{1}{2} x^{2}) + J_{- 1/4} (\frac{1}{2} x^{2})} .$
  
  Suggestion: Apply the identities in Eqs. (22) and (23).
1. Substitute the series of Problem 9 in the result of Problem 33 to show that the solution of the initial value problem
  
  $\frac{d y}{d x} = x^{2} + y^{2}, y (0) = 0$
  
  is
  
  $y (x) = x \frac{J_{3/4} (\frac{1}{2} x^{2})}{J_{- 1/4} (\frac{1}{2} x^{2})} .$
2. Deduce similarly that the solution of the initial value problem
  
  $\frac{d y}{d x} = x^{2} + y^{2}, y (0) = 1$
  
  is
  
  $y (x) = x \frac{2 Γ (\frac{3}{4}) J_{3/4} (\frac{1}{2} x^{2}) + Γ (\frac{1}{4}) J_{- 3/4} (\frac{1}{2} x^{2})}{2 Γ (\frac{3}{4}) J_{- 1/4} (\frac{1}{2} x^{2}) - Γ (\frac{1}{4}) J_{1/4} (\frac{1}{2} x^{2})} .$
  
  Some solution curves of the equation $d y / d x = x^{2} + y^{2}$ are shown in Fig. 11.4.6. The location of the asymptotes where $y (x) \to + \infty$ can be found by using Newton’s method to find the zeros of the denominators in the formulas for the solutions as listed here.
  
  FIGURE 11.4.6.
  
  Solution curves of $\frac{d y}{d x} = x^{2} + y^{2} .$

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Table of Contents for 11.4 Bessel Functions

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Table of Contents for
11.4 Bessel Functions