Bessel's Differential Equation - textbook queries:

In order to ask this question I must first give some background information as written in my text book:

Given Bessel's Differential equation:

$$x^2y^{\prime\prime}+xy^{\prime}+(x^2-p^2)y=0$$ or $$x(xy^{\prime})^{\prime}+(x^2-p^2)y=0\tag{1}$$

where $p$ is a constant, but not necessarily an integer.

We find a generalized power series for $(1)$ by writing only the general terms in the series for $y$ and the derivatives we need in $(1)$, we have $$y=\sum_{n=0}^\infty a_nx^{n+s}$$ $$y^{\prime}=\sum_{n=0}^\infty a_n(n+s)x^{n+s-1}$$ $$xy^{\prime}=\sum_{n=0}^\infty a_n(n+s)x^{n+s}$$ $$(xy^{\prime})^{\prime}=\sum_{n=0}^\infty a_n(n+s)^2x^{n+s-1}$$ $$x(xy^{\prime})^{\prime}=\sum_{n=0}^\infty a_n(n+s)^2x^{n+s}$$ We now substitute the above into $(1)$ and tabulate the coefficients of powers of $x$:

\begin{array}{|c|l:l|}\hline & x^s & x^{s+1} & x^{s+2} & \cdots & x^{s+n} \\ \hline x(xy^{\prime})^{\prime} & s^2a_0 & (1+s)^2a_1 & (2+s)^2a_2 & & (n+s)^2a_n \\ x^2y & & & a_0 & & a_{n-2} \\ -p^2y & -p^2a_0 & -p^2a_1 & -p^2a_2 & & -p^2a_n \\ \hdashline \hline \end{array} The coefficient of $x^s$ gives the indical equation and the values of $s$: $s^2-p^2=0\implies s=\pm p$. The coefficient of $x^{s+1}$ gives $a_1=0$. The coefficient of $x^{s+2}$ gives $a_2$ in terms of $a_0$, etc. But we may as well write the general formula from the last column at this point. We get:$$\left[(n+s)^2-p^2\right]a_n+a_{n-2}=0$$ or $$a_n=-\frac{a_{n-2}}{(n+s)^2-p^2}\tag{2}$$ First we shall find the coefficients for the case $s=p$. From $(2)$ we have $$a_n=-\frac{a_{n-2}}{(n+p)^2-p^2}=-\frac{a_{n-2}}{(n^2+2np)}=-\frac{a_{n-2}}{n(n+2p)}\tag{3}$$ Since $a_1=0$, all odd $a$'s are zero. For even $a$'s it is convenient to replace $n$ by $2n$; then from $(3)$ we have $$a_{2n}=-\frac{a_{2n-2}}{2n(2n+2p)}=-\frac{a_{n-2}}{2^2n(n+p)}\tag{4}$$ The formulas for the coefficients can be simplified by the use of the $\Gamma$ function (Gamma function) notation by recalling that $\Gamma(p+1)=p\Gamma(p)$ for any $p$ so $$\Gamma(p+2)=(p+1)\Gamma(p+1)$$ $$\Gamma(p+3)=(p+2)\Gamma(p+2)=(p+2)(p+1)\Gamma(p+1)$$ $$\Gamma(p+4)=(p+3)\Gamma(p+3)=(p+3)(p+2)(p+1)\Gamma(p+1)$$ and so on. Then from $(4)$ we find $$a_2=-\frac{a_0}{2^2(1+p)}=-\frac{a_0\Gamma(1+p)}{2^2\Gamma(2+p)}$$ $$a_4=-\frac{a_2}{2^3(2+p)}=\frac{a_0}{2!\cdot2^4(1+p)(2+p)}=\frac{a_0\Gamma(1+p)}{2!\cdot2^4\Gamma(3+p)}$$ $$a_6=-\frac{a_4}{3!\cdot2(3+p)}=-\frac{a_0}{3!\cdot2^6(1+p)(2+p)(3+p)}=-\frac{a_0\Gamma(1+p)}{3!\cdot2^6\Gamma(4+p)}$$ and so on. Then the series solution (for the $s=p$ case) is $$\begin{align}&y=a_0x^p\Gamma(1+p)\left[\frac{1}{0!\cdot\Gamma(1+p)}-\frac{1}{1!\cdot\Gamma(2+p)}\left(\frac{x}{2}\right)^2+\frac{1}{2!\cdot\Gamma(3+p)}\left(\frac{x}{2}\right)^4-\frac{1}{3!\cdot\Gamma(4+p)}\left(\frac{x}{2}\right)^6+\cdots\right] \\&= a_02^p\left(\frac{x}{2}\right)^p\Gamma(1+p)\left[\frac{1}{\Gamma(1)\Gamma(1+p)}-\frac{1}{\Gamma(2)\Gamma(2+p)}\left(\frac{x}{2}\right)^2+\frac{1}{\Gamma(3)\Gamma(3+p)}\left(\frac{x}{2}\right)^4-\frac{1}{\Gamma(4)\Gamma(4+p)}\left(\frac{x}{2}\right)^6+\cdots\right]\end{align}$$ Where we have inserted $\Gamma(1)$ and $\Gamma(2)$ (which are both equal to $1$) in the first $2$ terms and written $x^p=2^p\left(\dfrac{x}{2}\right)^p$ to make the series appear more systematic. If we take $$a_0=\frac{1}{2^p\Gamma(1+p)}=\frac{1}{2^pp!}$$ then $y$ is called the Bessel function of the first kind of order $p$, and written $J_p(x)$ Then $$J_p(x)=\frac{1}{\Gamma(1)\Gamma(1+p)}\left(\frac{x}{2}\right)^p-\frac{1}{\Gamma(2)\Gamma(2+p)}\left(\frac{x}{2}\right)^{2+p}+\frac{1}{\Gamma(3)\Gamma(3+p)}\left(\frac{x}{2}\right)^{4+p}-\frac{1}{\Gamma(4)\Gamma(4+p)}\left(\frac{x}{2}\right)^{6+p}+\cdots$$ or $$\fbox{$J_p(x)=\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1+p)}\left(\frac{x}{2}\right)^{2n+p}$}\tag{5}$$ We have found just one of the two solutions of Bessel's equation, that is, the one when $s=p$; we must next find the solution when $s=-p$. It is unnecessary to go through all the details again; we can just replace $p$ by $-p$ in $(5)$. In fact, the solution when $s=-p$ is usually written $J_{-p}(x)$. So from $(5)$ we have $$\fbox{$J_{-p}(x)=\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1-p)}\left(\frac{x}{2}\right)^{2n-p}$}\tag{6}$$

I understand all of the above.

My question is regarding the following extract:

If $p$ is not an integer $J_p(x)$ is a series starting with $x^p$ and $J_{-p}(x)$ is a series starting with $x^{-p}$. Then $J_p(x)$ and $J_{-p}(x)$ are two independent solutions and a linear combination of them is a general solution. $\bbox[#AFF]{\text{But if }p\text{ is}\text{ an integer, then the first few terms in }{J_{-p}(x)}\text{ will be zero, as in the denominator }\,\Gamma(n-p+1)\text{ is }\Gamma \text{ of a negative number, which is infinite.}}$ $\bbox[#AFA]{\text{You can show that }J_{-p}(x)\text{ starts with the term }x^p\text{ for integer }p\text{ just as } J_{p}(x)\text{ does,}}$$\bbox[yellow]{\text{ and that }J_{-p}(x)=(-1)^pJ_p(x)\text{ for integer }p}$.

1. For the blue highlighted part it says the "first few terms in $J_{-p}(x)$ will be zero"; But how do they know this? This can only be true by my logic iff $p\gt 2$. Yet we are given no restriction on $p$ other than $p \gt 0$; Why is it only the first few (two) terms that are zero?
2. For the green highlighted part; by my logic $J_{-p}(x)$ always starts with $x^{-p}$ and not $x^p$. What am I missing here?
3. Lastly for the yellow part: I'm sorry but I have simply no idea how to show that $J_{-p}(x)=(-1)^pJ_p(x)$, and much to my annoyance this is the most important part of this post.

I have already checked the errata list for this book and I can tell you that none of the above queries of mine are due to book errors.

If anyone is able to provide me with some hints/advice on addressing some/all/any of the $3$ queries above I will be most grateful.

I have only just started to learn about Bessel's equation, so my knowledge is somewhat limited (apologies).

Kindest Regards.

Solution 1:

We consider the Bessel functions \begin{align*} J_p(x)&=\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1+p)}\left(\frac{x}{2}\right)^{2n+p}\\ J_{-p}(x)&=\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1-p)}\left(\frac{x}{2}\right)^{2n-p}\\ \end{align*}

Blue question:

Due to the symmetry of $J_p(x)$ and $J_{-p}(x)$ the author seems silently to assume that $p>0$ (or it is stated somewhere else). So, let's keep this in mind.

The point is that the reciprocal of the Gammafunction \begin{align*} \frac{1}{\Gamma(n+1-p)}\tag{1} \end{align*} is an entire function with $\color{blue}{\text{simple zeros for}}$ $\color{blue}{n \lt p}$ assuming $p$ being a positive integer.

$$ $$

Yellow question:

It's easier to answer the yellow question first, since the green follows from it.

We obtain assuming an integer $p>0$ \begin{align*} J_{-p}(x)&=\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1-p)}\left(\frac{x}{2}\right)^{2n-p}\\ &=\sum_{n=p}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1-p)}\left(\frac{x}{2}\right)^{2n-p}\tag{2}\\ &=\sum_{n=0}^\infty\frac{(-1)^{n+p}}{\Gamma(n+p+1)\Gamma(n+1)}\left(\frac{x}{2}\right)^{2n+p}\tag{3}\\ &=(-1)^p\sum_{n=0}^\infty\frac{(-1)^{n}}{\Gamma(n+1)\Gamma(n+p+1)}\left(\frac{x}{2}\right)^{2n+p}\\ &=(-1)^pJ_p(x) \end{align*} and the claim follows.

Comment:

• In (2) we start the index with $n=p$ because summands with $n<p$ are zero due to (1)

• In (3) we shift the index to start from zero again

Green question:

When we look at the section before we see that \begin{align*} J_{-p}(x)=(-1)^pJ_p(x)=\frac{(-1)^p}{p!}\left(\frac{x}{2}\right)^p-\cdots \end{align*} and this representation is addressed when the author says that $J_{-p}(x)$ starts with $x^p$.

Comment:

Note that the terms with $0\leq n <p$ in $J_{-p}(x)$ vanish due to (1). So, according to (2) we obtain \begin{align*} J_{-p}(x)&=\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1-p)}\left(\frac{x}{2}\right)^{2n-p}\\ &=\sum_{n=0}^{p-1}0+\sum_{n=p}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1-p)}\left(\frac{x}{2}\right)^{2n-p}\\ &=\frac{(-1)^p}{p!}\left(\frac{x}{2}\right)^p\color{red}{\Large\text{-}} \frac{(-1)^p}{(p+1)!}\left(\frac{x}{2}\right)^{p+1}+\cdots \end{align*} on the other hand we have \begin{align*} J_p(x)&=\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1+p)}\left(\frac{x}{2}\right)^{2n+p}\\ &=\frac{1}{p!}\left(\frac{x}{2}\right)^p- \frac{1}{(p+1)!}\left(\frac{x}{2}\right)^{p+1}+\cdots \end{align*}