6.3 Bessel's Equation and Bessel Functions

Whenever a problem has circular or cylindrical symmetry, one particular differential equation appears. Separating variables in the wave, heat or Laplace equation in cylindrical coordinates always leads to it: for the vibrations of a drum skin, the temperature in a long cylinder, the fields inside a circular waveguide or a coaxial cable, the light in an optical fibre, the flow in a pipe. It is Bessel's equation, and its solutions, the Bessel functions, are the "sines and cosines of cylindrical problems".

In this section we apply the Frobenius method of Section 6.2 to Bessel's equation. Along the way we need one new tool, the gamma function, which extends the factorial to non-integer arguments. We finish with the properties engineers use every day: recurrence identities, zeros and the large-$x$ behaviour.

The order $\nu$ is not the order of the differential equation (which is always $2$); it is just the name of the parameter. In applications $\nu$ is usually an integer (from the angular dependence $\cos n\theta$) or a half-integer (from spherical problems).

Frobenius analysis

The equation is already in the standard form $x^2{y}^{\prime \prime }+xP(x)y^{\prime }+Q(x)y=0$ with

Both are polynomials, so $x=0$ is a regular singular point, there are no other singular points, and every series we find converges for all $x$ ($R=\infty$).

Indicial equation. $p_0=1$ and $q_0=-{\nu }^2$, so

The difference of the roots is $r_1-r_2=2\nu$. All three cases of Theorem 6.2.1 can occur, depending on $\nu$.

Recurrence. Substitute $y={\sum }_{n\ge 0}{a}_n{x}^{n+r}$. The terms $x^2{y}^{\prime \prime }+x{y}^{\prime }-{\nu }^{2}y$ give $\sum [(n+r)(n+r-1)+(n+r)-{\nu }^2]a_n{x}^{n+r}=\sum [(n+r{)}^2-{\nu }^2]a_n{x}^{n+r}$. The term $x^{2}y$ gives $\sum a_n{x}^{n+r+2}$, which after the shift $n\to n-2$ becomes ${\sum }_{n\ge 2}{a}_{n-2}{x}^{n+r}$. Collecting powers:

The larger root $r=\nu$. Then $(n+\nu {)}^2-{\nu }^2=n(n+2\nu )$, and the equations become

Since $1+2\nu \ne 0$, $a_1=0$, and then the recurrence (step 2) makes every odd coefficient zero. For even $n=2k$ the recurrence reads $2k(2k+2\nu )a_{2k}=-a_{2k-2}$, i.e.

Unwinding it step by step:

The product $(\nu +1)(\nu +2)\cdots (\nu +k)$ looks like a factorial, but $\nu$ need not be an integer. The gamma function lets us write it compactly.

The gamma function

The integral converges for every $x>0$: near $t=0$ the integrand behaves like $t^{x-1}$, which is integrable when $x>0$, and for large $t$ the factor $e^{-t}$ wins against any power.

Proof. (1) $\Gamma (1)={\int }_0^{\infty }{e}^{-t}dt=1$.

(2) Integrate by parts with $u=t^x$, $dv=e^{-t}dt$, so $du=x{t}^{x-1}dt$, $v=-e^{-t}$:

The boundary term vanishes at $\infty$ because $e^{-t}$ decays faster than any power grows, and at $0$ because $x>0$.

(3) Apply (2) repeatedly: $\Gamma (n+1)=n\Gamma (n)=n(n-1)\Gamma (n-1)=\cdots =n!\Gamma (1)=n!$.

(4) Substitute $t=u^2$, $dt=2udu$: $\Gamma \left(\frac{1}{2}\right)={\int }_0^{\infty }{t}^{-1/2}{e}^{-t}dt={\int }_0^{\infty }{u}^{-1}{e}^{-u^2}2udu=2{\int }_0^{\infty }{e}^{-u^2}du=\sqrt{\pi }$, using the Gaussian integral ${\int }_0^{\infty }{e}^{-u^2}du=\frac{\sqrt{\pi }}{2}$. $■$

So $\Gamma$ is a smooth function that interpolates the factorials: $\Gamma (n+1)=n!$. Property (2) also gives us the product we need. Applying it $k$ times,

so

Negative arguments. Rewrite property (2) as $\Gamma (x)=\frac{\Gamma (x+1)}{x}$. The right side makes sense for $-1<x<0$ (because then $x+1>0$), so we use it to define $\Gamma$ there; then for $-2<x<-1$, and so on. This extends $\Gamma$ to all real $x$ except $0,-1,-2,\dots$, where $|\Gamma (x)|\to \infty$. At these points we set $\frac{1}{\Gamma (x)}=0$, which is the natural limiting value and is exactly what the Bessel series below need.

Bessel functions of the first kind

Return to the coefficients $a_{2k}$. Using the product formula,

Any $a_0\ne 0$ gives a solution. The standard choice, which makes the formula as clean as possible, is

Then $a_{2k}{x}^{2k+\nu }=\frac{(-1{)}^k}{k!\Gamma (k+\nu +1)}\frac{x^{2k+\nu }}{2^{2k+\nu }}$, and we obtain:

For integer order $\nu =n$ we have $\Gamma (k+n+1)=(k+n)!$. The two most used cases are

The ratio of consecutive terms is $\frac{(x/2{)}^2}{(k+1)(k+\nu +1)}\to 0$, so the series converge for all $x$, and they converge fast for moderate $x$.

The second solution

Non-integer order: $J_{-\nu }$

Now try the smaller root $r=-\nu$. The computation is identical with $\nu$ replaced by $-\nu$, as long as no division by zero occurs, and gives

Why independent? As $x\to 0^{+}$, $J_{\nu }(x)\to 0$, while the first term of $J_{-\nu }$ is $\frac{(x/2{)}^{-\nu }}{\Gamma (1-\nu )}$, and $\Gamma (1-\nu )$ is finite and nonzero because $1-\nu$ is not $0,-1,-2,\dots$ So $J_{-\nu }$ is unbounded near $0$ and cannot be a constant multiple of $J_{\nu }$.

Integer order: $J_{-n}$ is not new

If $\nu =n$ is a nonnegative integer, the formula for $J_{-n}$ still makes sense with the convention $1/\Gamma (m)=0$ for $m=0,-1,-2,\dots$ But look at what happens. For $k=0,1,\dots ,n-1$ the number $k-n+1$ is $\le 0$, so those terms vanish. The series really starts at $k=n$. Put $k=m+n$:

So $J_{-n}=(-1{)}^n{J}_n$: just a multiple of $J_n$, not a second independent solution. This matches Theorem 6.2.1: for $\nu =0$ we are in Case 2, and for $\nu =n\ge 1$ in Case 3 with a logarithm that really is needed. The second solution must contain $J_n(x)\ln x$.

Bessel functions of the second kind

Instead of building the logarithmic solution for each $n$ separately, it is customary to use one particular combination that works for every order.

For non-integer $\nu$, $Y_{\nu }$ is a combination of $J_{\nu }$ and $J_{-\nu }$, so it is a solution. The point of the definition is that the limit $\nu \to n$ exists and is still a solution, independent of $J_n$; it contains the logarithm predicted by Frobenius. For example,

where $H_k=1+\frac{1}{2}+\cdots +\frac{1}{k}$ and $\gamma ={\lim }_{k\to \infty }(H_k-\ln k)\approx 0.5772$ is Euler's constant. Note the term $J_0(x)\ln x$, exactly the Case 2 form.

To see the Frobenius machinery at work on Bessel's equation, use the solver with $P=1$ and $Q=x^2-{\nu }^2$. The first preset is order $\nu =\frac{1}{3}$ (Case 1, two Frobenius series); the second is order $\nu =1$ (Case 3, where the solver will find that the logarithm is needed). Try $\nu =\frac{1}{2}$, i.e. $Q=x^2-1/4$, to watch Case 3 without a logarithm.

Half-integer order: elementary functions

Bessel functions are generally not elementary, with one remarkable exception: orders $\pm \frac{1}{2},\pm \frac{3}{2},\dots$ For $\nu =\frac{1}{2}$ we need $\Gamma \left(k+\frac{3}{2}\right)$. From $\Gamma \left(\frac{3}{2}\right)=\frac{1}{2}\sqrt{\pi }$ and repeated use of $\Gamma (x+1)=x\Gamma (x)$:

(The last step multiplies top and bottom by the even numbers $2\cdot 4\cdots (2k)=2^{k}k!$.) Substituting into the series:

In the same way, using $\Gamma \left(k+\frac{1}{2}\right)=\frac{(2k)!}{2^{2k}k!}\sqrt{\pi }$,

Recurrence identities

Bessel functions of neighbouring orders are tied together by simple identities. They let you compute derivatives and integrals, and build higher orders from lower ones.

Proof of (a). Multiply the series by $x^{\nu }$:

Differentiate term by term; the power $x^{2k+2\nu }$ brings down $2(k+\nu )$, and $\frac{k+\nu }{\Gamma (k+\nu +1)}=\frac{1}{\Gamma (k+\nu )}$ by property (2) of $\Gamma$:

Identity (b) is proved the same way. For (c) and (d), expand the derivatives in (a) and (b) with the product rule and divide by $x^{\pm \nu }$:

Subtracting gives (c); adding gives (d). $■$

Two special cases are worth remembering: with $\nu =0$ in (b), $J_0^{\prime }(x)=-J_1(x)$; with $\nu =1$ in (a), $\frac{d}{dx}[x{J}_1(x)]=x{J}_0(x)$, so $\int x{J}_0(x)dx=x{J}_1(x)+C$.

Zeros and large-x behaviour

Each $J_{\nu }$ and $Y_{\nu }$ has infinitely many positive zeros. The first few zeros of $J_0$ are

(the first zeros of $J_1$ and $J_2$ are $3.8317$ and $5.1356$). These numbers are tabulated in every engineering handbook and built into every scientific library. For large $x$ the Bessel functions behave like decaying cosines:

For $\nu =\pm \frac{1}{2}$ these "approximations" are exact, as we have just seen. For $J_0$ the formula predicts zeros near $\left(m-\frac{1}{4}\right)\pi$:

The agreement improves quickly, and the spacing between consecutive zeros approaches $\pi$.

Equations that reduce to Bessel's equation

The parametric form. In applications the equation usually appears as

Substitute $t=\lambda x$. By the chain rule $x\frac{dy}{dx}=t\frac{dy}{dt}$ and $x^2\frac{d^{2}y}{d{x}^2}=t^2\frac{d^{2}y}{d{t}^2}$, and ${\lambda }^2{x}^2=t^2$, so the equation becomes Bessel's equation in $t$. Therefore

Hidden Bessel equations. Many other equations become Bessel's equation after a substitution of the form $y=x^{a}u$, a change of the independent variable, or both. A standard result (which you can verify by direct substitution) is that

has the general solution $y=x^a[c_1{J}_{\nu }(b{x}^c)+c_2{Y}_{\nu }(b{x}^c)]$. In simple cases it is easier to do the substitution by hand.

A short note on Legendre's equation

Bessel's equation is the signature of cylindrical problems. Its partner for spherical problems (potential around a sphere, radiation patterns of antennas, gravitational fields) is Legendre's equation

where $x=\cos \theta$ is the cosine of the polar angle. Here $x=0$ is an ordinary point and the singular points are $x=\pm 1$, so Section 6.1 applies and the series converge at least for $|x|<1$. Substituting $\sum a_j{x}^j$ gives the recurrence

When $n$ is a nonnegative integer, the factor $n-j$ vanishes at $j=n$, so one of the two series stops: it is a polynomial. Scaled so that its value at $x=1$ is $1$, it is the Legendre polynomial $P_n(x)$: