Bessel functions

The Bessel function is canonical solution to Bessel's differential equation ${\displaystyle x^{2}y''+xy'+(x^{2}-\nu ^{2})y=0,\,\nu \in \mathbb {C} ~.}$ Solutions were first introduced by Daniel Bernoulli, but later generalized by Friedrich Bessel. The most common and most important case of the Bessel function is when ${\displaystyle \nu \in \mathbb {Z} ~,}$ which is called the order of the Bessel function.

Bessel functions arise when the method of separation of variables is applied to the Laplace or Helmholtz equation in cylindrical or spherical coordinates. They are very important for many problems dealing with physical phenomena, like wave or heat propagation.

Consider the Bessel equation:

${\displaystyle x^{2}y''+xy'+(x^{2}-\nu ^{2})y=0}$

${\displaystyle \Leftrightarrow y''+\underbrace {\left({\frac {1}{x}}\right)} _{p(x)}y'+\underbrace {\left(1-{\frac {\nu ^{2}}{x^{2}}}\right)} _{q(x)}y=0}$

We're seeking solutions near ${\displaystyle x_{0}=0~.}$ Since:

${\displaystyle {\begin{aligned}xp(x)&=1\\x^{2}q(x)&=x^{2}-\nu ^{2}\end{aligned}}}$

are power series in x, ${\displaystyle x_{0}=0}$ is a regular singular point of the Bessel equation. This allows Frobenius's method to be applied.

We are seeking solutions of the form:

${\displaystyle y(x)=\sum _{n=0}^{\infty }C_{n}x^{n+r},\,x>0,C_{n}\neq 0}$

Differentiating yields:

${\displaystyle {\begin{aligned}y'(x)&=\sum _{n=0}^{\infty }(n+r)C_{n}x^{n+r-1}\\y''(x)&=\sum _{n=0}^{\infty }(n+r-1)(n+r)C_{n}x^{n+r-2}\end{aligned}}}$

Conditions for ${\displaystyle C_{n}}$ must be found. Substituting our expressions back into the Bessel equation:

${\displaystyle {\begin{aligned}0&=x^{2}y''+xy'+(x^{2}-\nu ^{2})y\\&=\sum _{n=0}^{\infty }(n+r-1)(n+r)C_{n}x^{n+r}+\sum _{n=0}^{\infty }(n+r)C_{n}x^{n+r}+\sum _{n=0}^{\infty }C_{n}x^{n+r+2}-\sum _{n=0}^{\infty }\nu ^{2}C_{n}x^{n+r}\end{aligned}}}$

A substitution must be made in indices:${\displaystyle m=n+2~.}$ This yields:

${\displaystyle {\begin{aligned}0&=\sum _{n=0}^{\infty }\left[(n+r-1)(n+r)+(n+r)-\nu ^{2}\right]C_{n}x^{n+r}+\sum _{m=2}^{\infty }C_{m-2}x^{m+r}\\&=\sum _{n=0}^{\infty }\left[(n+r)^{2}-\nu ^{2}\right]C_{n}x^{n+r}+\sum _{n=2}^{\infty }C_{n-2}x^{n+r}\\&=(r^{2}-\nu ^{2})C_{0}x^{r}+[(r+1)^{2}-\nu ^{2}]C_{1}x^{r+1}+\sum _{n=2}^{\infty }\left\{[(n+r)^{2}-\nu ^{2}]C_{n}+C_{n-2}\right\}x^{n+r}\end{aligned}}}$

Dividing the equation above by ${\displaystyle x^{r}~(x>0)~}$ yields:

${\displaystyle 0=(r^{2}-\nu ^{2})C_{0}+[(r+1)^{2}-\nu ^{2}]C_{1}x+\sum _{n=2}^{\infty }\left\{[(n+r)^{2}-\nu ^{2}]C_{n}+C_{n-2}\right\}x^{n}}$

By the "Identity Theorem" (which states that xⁿ is linearly independent), it follows that:

${\displaystyle {\begin{aligned}&(r^{2}-\nu ^{2})C_{0}=0\\&[(r+1)^{2}-\nu ^{2}]C_{1}=0\\&[(n+r)^{2}-\nu ^{2}]C_{n}+C_{n-2}=0,\,n=2,3,4,\cdots \end{aligned}}}$

By assumption, ${\displaystyle C_{0}\neq 0~,}$ so we define a function:

${\displaystyle h(r):=r^{2}-\nu ^{2}=0\quad {\text{(indicial equation)}}}$

The possible values for ${\displaystyle r=\pm \nu ~.}$ Let ${\displaystyle r_{1}:=\nu ,\,r_{2}:=-\nu ~}$ and for convenience, let ${\displaystyle \nu >0~.}$ We obtain the following recurrence relations for ${\displaystyle C_{n}}$:

${\displaystyle {\begin{cases}C_{0}\neq 0\quad {\text{(arbitrarily defined)}}\\C_{1}=0\quad {\text{(follows from }}[(r+1)^{2}-\nu ^{2}]C_{1}=0{\text{)}}\\\underbrace {[(n+r)^{2}-\nu ^{2}]} _{h(n+r)}C_{n}=-C_{n-2},\,n=2,3,4,\cdots \end{cases}}}$

To get a solution to the Bessel equation, choose ${\displaystyle r_{1}=\nu ~,\,\nu \neq 0~.}$ Thus, ${\displaystyle h(n+r)=h(n+\nu )\neq 0,\,n=2,3,4,\cdots ~.}$ We can now solve for ${\displaystyle C_{n}}$:

${\displaystyle C_{n}=-{\frac {C_{n-2}}{(n+\nu )^{2}-\nu ^{2}}}=-{\frac {C_{n-2}}{n^{2}+2n\nu }}}$

We end up with the recursion:

${\displaystyle {\begin{cases}C_{0}\neq 0\\C_{1}=0\\C_{n}=-{\frac {C_{n-2}}{n(n+2\nu )}},\,n=2,3,4,\cdots \end{cases}}}$

Since the recursion has depth 2 and ${\displaystyle C_{1}=0}$, it follows that:

${\displaystyle {\begin{cases}C_{0}\neq 0\\C_{2n+1}=0,n=0,1,2,\cdots \\C_{2n}=-{\frac {C_{2n-2}}{2n(2n+2\nu )}}=-{\frac {C_{2n-2}}{2^{2}n(n+\nu )}},\,n=1,2,3,\cdots \end{cases}}}$

Because of the recursion, we get the following set of terms:

${\displaystyle {\begin{aligned}&C_{0}\neq 0\\&C_{2}=-{\frac {C_{0}}{2^{2}\cdot 1\cdot (1+\nu )}}\\&C_{4}=C_{2\cdot 2}=-{\frac {C_{2}}{2^{2}\cdot 2\cdot (2+\nu )}}={\frac {(-1)^{2}C_{0}}{2^{4}\cdot 1\cdot 2\cdot (1+\nu )(2+\nu )}}={\frac {(-1)^{2}C_{0}}{2^{4}\cdot 2!\cdot (1+\nu )(2+\nu )}}\\&C_{6}=C_{2\cdot 3}=-{\frac {C_{4}}{2^{2}\cdot 3\cdot (3+\nu )}}={\frac {(-1)^{3}C_{0}}{2^{6}\cdot 3!\cdot (1+\nu )(2+\nu )(3+\nu )}}\\&\vdots \\&C_{2n}={\frac {(-1)^{n}C_{0}}{2^{2n}\cdot n!\cdot (1+\nu )(2+\nu )\cdots (n+\nu )}},\,n=1,2,3,\cdots \end{aligned}}}$

In order to simplify the expansion of y, we normalize ${\displaystyle C_{0}}$ and choose:

${\displaystyle C_{0}:={\frac {1}{2^{\nu }\Gamma (1+\nu )}}}$

This simplifies our general term to:

${\displaystyle C_{2n}={\frac {(-1)^{n}}{2^{2n+\nu }\cdot n!\cdot \Gamma (n+1+\nu )}},\,n=0,1,2,\cdots }$

The first solution to the Bessel equation can be written like this:

${\displaystyle J_{\nu }(x)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!\cdot \Gamma (n+1+\nu )}}\left({\frac {x}{2}}\right)^{2n+\nu }}$

For the case that ${\displaystyle \nu \not \in \mathbb {N} ,\nu >0}$, we can define a second solution to the Bessel function. In this case, ${\displaystyle n+1-\nu \in \mathbb {R} \backslash (-\mathbb {N} \cup \{0\})}$ and therefore ${\displaystyle \Gamma (n+1-\nu )}$ is defined. Consider:

${\displaystyle J_{-\nu }(x):=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!\cdot \Gamma (n+1-\nu )}}\left({\frac {x}{2}}\right)^{2n-\nu }}$

Some theorems for this new function:

• ${\displaystyle J_{-\nu }(x)~}$ solves the Bessel equation.
• ${\displaystyle J_{\nu }(x),J_{-\nu }(x)~}$ are linearly independent.

These theorems are proved easily, but will not be shown here.

A second function that is defined on ${\displaystyle \nu \in \mathbb {R} ,\nu \not \in \mathbb {Z} }$ takes the form:

${\displaystyle Y_{\nu }(x):={\frac {\cos(\nu \pi )J_{\nu }(x)-J_{-\nu }(x)}{\sin(\nu \pi )}}}$

Deriving this result is fairly difficult and will not be shown here. This function, called the Bessel function of the second kind of order ${\displaystyle \nu }$, is linearly independent from ${\displaystyle J_{\nu }(x)}$.

A third type of function (complex-valued) for ${\displaystyle \nu \in \mathbb {R} ,\,\nu \not \in \mathbb {Z} ,}$ are:

${\displaystyle H_{\nu }^{(1)}:=J_{\nu }(x)+i\,Y_{\nu }(x)}$

${\displaystyle H_{\nu }^{(2)}:=J_{\nu }(x)-i\,Y_{\nu }(x)}$

and are called the Bessel functions of the 3rd kind or Hankel functions of order ${\displaystyle \nu }$. The Hankel functions ${\displaystyle H_{\nu }^{(1)},H_{\nu }^{(2)}}$ are linearly independent.

For all ${\displaystyle \nu \in \mathbb {R} ,\lambda \in \mathbb {R} ,}$ the complete solution of the Bessel equation:

${\displaystyle x^{2}y''+xy'+(\lambda ^{2}x^{2}-\nu ^{2})y=0}$

can be written as:

${\displaystyle y(x)=C_{1}J_{\nu }(\lambda x)+C_{2}Y_{\nu }(\lambda x)}$

or:

${\displaystyle y(x)=C_{1}H_{\nu }^{(1)}(\lambda x)+C_{2}H_{\nu }^{(2)}(\lambda x)~.}$

If ${\displaystyle \nu \in \mathbb {R} \backslash \mathbb {Z} ~,}$ then:

${\displaystyle y(x)=C_{1}J_{\nu }(x)+C_{2}J_{-\nu }(x)~.}$

Moreover:

• ${\displaystyle J_{\nu },J_{-\nu },Y_{\nu }}$ have countably many zeroes.
• If ${\displaystyle v\geq 0}$, then ${\displaystyle J_{\nu }(\lambda x)}$ is finite for all ${\displaystyle x\in \mathbb {R} }$, ${\displaystyle J_{-\nu }(\lambda x)}$ and ${\displaystyle Y_{\nu }(\lambda x)}$ are unbounded in the neighborhood of 0.

Here are some identities for the Bessel function. They can be deduced with reasonable effort.