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Mathematical Methods of Physics: For Undergraduate

Sunil Karna

Section 4.8 Examples D

Bessels’s Function.

Example 4.8.1.

Prove that \(J_{-n}(x)=(-1)^{n}J_{n}(x)\) where \(n\) is a positive integer.
Solution.
We have
\begin{equation*} J_{-n}(x)=\sum\limits_{r=0}^{\infty}\frac{(-1)^{r}}{r!(-n+r)!}\left(\frac{x}{2}\right)^{-n+2r} \end{equation*}
\begin{equation*} =\sum\limits_{r=0}^{n-1}\frac{(-1)^{r}}{r!(-n+r)!}\left(\frac{x}{2}\right)^{-n+2r}+\sum\limits_{r=n}^{\infty}\frac{(-1)^{r}}{r!(-n+r)!}\left(\frac{x}{2}\right)^{-n+2r} \end{equation*}
\begin{equation*} =0+\sum\limits_{r=n}^{\infty}\frac{(-1)^{r}}{r!(-n+r)!}\left(\frac{x}{2}\right)^{-n+2r} \end{equation*}
Since, \(!(-ve \quad integer) = \infty\text{.}\) Hence, for \(r=0,1,2,\cdots\text{,}\) we have
\begin{equation*} \frac{1}{(-n+r)!}=0 \end{equation*}
putting \(r=n+s\text{,}\) where \(s\) is a positive integer, we have
\begin{equation*} J_{-n}(x)=\sum\limits_{s=0}^{\infty}\frac{(-1)^{n+s}}{s!(n+s)!}\left(\frac{x}{2}\right)^{n+2s} \end{equation*}
\begin{equation*} =(-1)^{n}\sum\limits_{s=0}^{\infty}\frac{(-1)^{s}}{s!(n+s)!}\left(\frac{x}{2}\right)^{n+2s} =(-1)^{n}J_{n}(x) \end{equation*}

Example 4.8.2.

Prove that
  1. \begin{equation*} J_{\frac{1}{2}}(x)=\sqrt{\left(\frac{2}{\pi x}\right)}\sin x \end{equation*}
  2. \begin{equation*} J_{-\frac{1}{2}}(x)=\sqrt{\left(\frac{2}{\pi x}\right)}\cos x \end{equation*}
Solution.
We know that
\begin{equation*} J_{-n}(x)=\sum\limits_{r=0}^{\infty}\frac{(-1)^{r}}{r!(n+r)!}\left(\frac{x}{2}\right)^{n+2r} \end{equation*}
\begin{equation*} = \frac{x^{n}}{2^{n}} \left[\frac{1}{n!}-\frac{x^{2}}{2^{2}(n+1)!}+\frac{x^{4}}{2!(n+2)!2^{4}}-\cdots\right] \end{equation*}
\begin{equation} = \frac{x^{n}}{2^{n}n!}\left[1-\frac{x^{2}}{2\cdot 2(n+1)}+\frac{x^{4}}{2\cdot 4\cdot 2^{2}(n+1)(n+2)}-\cdots\right]\tag{4.8.1} \end{equation}
  1. putting \(n=\frac{1}{2}\) in eqn. (4.8.1), we get -
    \begin{equation*} J_{\frac{1}{2}}(x)=\frac{x^{\frac{1}{2}}}{2^{\frac{1}{2}}\frac{1}{2}!}\left[1-\frac{x^{2}}{2\cdot 2(\frac{1}{2}+1)}\right. \end{equation*}
    \begin{equation*} \left. +\frac{x^{4}}{2\cdot 4\cdot 2^{2}(\frac{1}{2}+1)(\frac{1}{2}+2)}-\cdots\right] \end{equation*}
    \begin{equation*} =\frac{\sqrt{x}}{\sqrt{2}\Gamma{(\frac{1}{2}+1})}\left[1-\frac{x^{2}}{3!}+\frac{x^{4}}{5!}-\cdots\right] \end{equation*}
    \begin{equation*} =\frac{\sqrt{x}}{\sqrt{2}\cdot\frac{1}{2}\Gamma{(\frac{1}{2})}}\cdot\frac{1}{x}\left[x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\cdots\right] \end{equation*}
    since, \(\left[n!=\Gamma(n+1)=n\Gamma n\right]\text{,}\) the Gamma function
    \begin{equation*} \therefore J_{\frac{1}{2}}(x)=\frac{1}{\sqrt{2x}\cdot \frac{\sqrt{\pi}}{2}}\sin x=\sqrt{\left(\frac{2}{\pi x}\right)}\sin x \end{equation*}
    \([\because \Gamma(\frac{1}{2})=\sqrt{\pi}\) and \((\frac{1}{2})!=\frac{\sqrt{\pi}}{2}] \text{.}\)
  2. Againg putting \(n=-\frac{1}{2}\) in equation (4.8.1), we get -
    \begin{equation*} J_{-\frac{1}{2}}(x)=\frac{x^{-\frac{1}{2}}}{2^{-\frac{1}{2}}\frac{1}{2}!}\left[1-\frac{x^{2}}{2\cdot 2(-\frac{1}{2}+1)}\right. \end{equation*}
    \begin{equation*} \left.+\frac{x^{4}}{2\cdot 4\cdot 2^{2}(-\frac{1}{2}+1)(-\frac{1}{2}+2)}-\cdots\right] \end{equation*}
    \begin{equation*} =\frac{\sqrt{2}}{\sqrt{x}\Gamma{(-\frac{1}{2}+1})}\left[1-\frac{x^{2}}{2!}+\frac{x^{4}}{2\cdot3\cdot4}-\cdots\right] \end{equation*}
    \begin{equation*} =\frac{\sqrt{2}}{\sqrt{x}\Gamma{(\frac{1}{2})}}\left[1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots\right] \end{equation*}
    \begin{equation*} =\sqrt{\left(\frac{2}{\pi x}\right)}\cos x \end{equation*}
    [since \(\Gamma(\frac{1}{2})=\sqrt{\pi}\)].

Example 4.8.3.

Find the values of \(J_{\pm\frac{3}{2}}(x)\text{.}\)
Solution.
We have the recurrence relation (4) from Subsubsection 4.7.1.3,
\begin{equation} 2nJ_{n}=x(J_{n-1}+J_{n+1}) \tag{4.8.2} \end{equation}
putting \(n=\frac{1}{2}\text{,}\) we get -
\begin{equation*} J_{\frac{1}{2}}=x\left(J_{-\frac{1}{2}}+J_{\frac{3}{2}}\right) \end{equation*}
or,
\begin{equation*} J_{\frac{3}{2}}=\frac{1}{x}\left(J_{\frac{1}{2}}\right)-J_{-\frac{1}{2}} \end{equation*}
\begin{equation} =\frac{1}{x}\sqrt{\frac{2}{\pi x}}\sin x- \sqrt{\frac{2}{\pi x}}\cos x = \sqrt{\frac{2}{\pi x}}\left[\frac{\sin x}{x}-\cos x\right] \tag{4.8.3} \end{equation}
Similarly on putting \(n=-\frac{1}{2}\) in eqn. (4.8.2), we get -
\begin{equation} J_{-\frac{3}{2}}=\sqrt{\frac{2}{\pi x}}\left[\frac{-\cos x}{x}-\sin x\right] \tag{4.8.4} \end{equation}
We can also prove it by putting \(n=\pm\frac{3}{2}\) in the eqn. (4.8.1) of previous problem 2.

Example 4.8.4.

Prove that \(J'_{2}(x)=(1-\frac{4}{x^{2}})J_{1}(x)+\frac{2}{x}J_{o}(x)\text{.}\)
Solution.
By recurrnce relation 2 from Subsubsection 4.7.1.3,
\begin{equation} xJ'_{n}(x)=-nJ_{n}(x)+xJ_{n-1}(x)\tag{4.8.5} \end{equation}
putting \(n=2\) in eqn. (4.8.5), we have
\begin{equation*} xJ'_{2}=-2J_{2}+xJ_{1} \end{equation*}
or,
\begin{equation} J'_{2}=-\frac{2}{x}J_{2}+J_{1} \tag{4.8.6} \end{equation}
By recurrnce relation 4 from Subsubsection 4.7.1.3,
\begin{equation*} 2nJ_{n}=x(J_{n-1}+J_{n+1}) \end{equation*}
putting \(n=1\text{,}\) we have
\begin{equation*} 2\cdot1\cdot J_{1}=x(J_{o}+J_{2}) \end{equation*}
\begin{equation} J_{2}=\frac{2J_{1}}{x}-J_{o} \tag{4.8.7} \end{equation}
putting the value of \(J_{2}\) in eqn. (4.8.6), we get -
\begin{equation*} J'_{2}=-\frac{2}{x}\left(\frac{2}{x}J_{1}-J_{o}\right)+J_{1}=-\frac{4}{x^{2}}J_{1}+\frac{2}{x}J_{o}+J_{1} \end{equation*}
\begin{equation*} = (1-\frac{4}{x^{2}})J_{1}(x)+\frac{2}{x}J_{o}(x) \end{equation*}

Example 4.8.5.

Using recurrence relations, show that \(4J''_{n}(x)=J_{n-2}(x)-2J_{n}(x)+J_{n+2}(x)\text{.}\)
Solution.
By the recurrnce relation 3 from Subsubsection 4.7.1.3,
\begin{equation} 2J'_{n}=J_{n-1}-J_{n+1}\tag{4.8.8} \end{equation}
differentiating, we get-
\begin{equation} 2J''_{n}=J'_{n-1}-J'_{n+1} \tag{4.8.9} \end{equation}
replacing \(n\) by \(n-1\)in eqn. (4.8.8) and \(n\) by \(n+1\) in eqn. (4.8.9), we get -
\begin{equation*} 2J'_{n-1}=J_{n-2}-J_{n} \end{equation*}
or,
\begin{equation} J'_{n-1}=\frac{1}{2}J_{n-2}-\frac{J_{n}}{2} \tag{4.8.10} \end{equation}
and
\begin{equation*} 2J'_{n+1}=J_{n}-J_{n+2} \end{equation*}
or,
\begin{equation} J'_{n+1}=\frac{1}{2}J_{n}-\frac{1}{2}J_{n+2}\tag{4.8.11} \end{equation}
putting the values of \(J'_{n-1} \) and \(J'_{n+1}\) in eqn. (4.8.9), we get -
\begin{equation*} 2J''_{n}=\frac{1}{2}\left[J_{n-2}-J_{n}\right]-\frac{1}{2}\left[J_{n}-J_{n+2}\right] \end{equation*}
or,
\begin{equation*} 4J''_{n}=J_{n-2}-2J_{n}+J_{n+2}. \end{equation*}

Example 4.8.6.

Prove that \(J''_{1}=\left(\frac{2}{x^{2}}-1\right)J_{1}-\frac{1}{x}J_{o}\text{.}\)
Solution.
By recurrence relation 2 from Subsubsection 4.7.1.3,
\begin{equation} xJ'_{n}=-nJ_{n}+xJ_{n-1}\tag{4.8.12} \end{equation}
Differentiating, we get -
\begin{equation*} xJ''_{n}+J'_{n}=-nJ'_{n}+xJ'_{n-1}+J_{n-1} \end{equation*}
Putting, \(n=1\text{,}\)
\begin{equation*} xJ''_{1}+J'_{1}=-J'_{1}+xJ'_{o}+J_{o} \end{equation*}
or,
\begin{equation} J''_{1}=-\frac{2}{x}J'_{1}+J'_{o}+\frac{1}{x}J_{o} \tag{4.8.13} \end{equation}
By recurrence relation 1,
\begin{equation*} xJ'_{n}=nJ_{n}-xJ_{n+1} \end{equation*}
putting \(n=0\)
\begin{equation*} xJ'_{o}=-xJ_{1} \end{equation*}
or,
\begin{equation*} J'_{o}=-J_{1} \end{equation*}
putting the value of \(J'_{o} \) in eqn. (4.8.13), we get -
\begin{equation*} J''_{1}=-\frac{2}{x}J'_{1}-J_{1}+\frac{1}{x}J_{o} \end{equation*}
putting \(n=1\text{,}\) in eqn. (4.8.12), we get -
\begin{equation*} xJ'_{1}=-J_{1}+xJ_{o} \end{equation*}
or,
\begin{equation} J'_{1}=-\frac{1}{x}J_{1}+J_{o} \tag{4.8.14} \end{equation}
putting \(J'_{1}\) in eqn. (4.8.14), we have
\begin{equation*} J''_{1}=-\frac{2}{x}\left(-\frac{J_{1}}{x}+J_{o}\right)-J_{1}+\frac{1}{x}J_{o} =\frac{2}{x^{2}}J_{1}-J_{1}+\frac{J_{o}}{x}(1-2) \end{equation*}
\begin{equation*} =\left(\frac{2}{x^{2}}-1\right)J_{1}-\frac{J_{o}}{x} \end{equation*}

Example 4.8.7.

Prove that \(\frac{\,d}{\,dx}\left[x^{n}J_{n}(x)\right]=x^{n}J_{n-1}(x)\text{.}\)
Solution.
\begin{equation*} x^{n}J_{n}(x)=x^{n}\sum\limits_{r=0}^{\infty}\frac{(-1)^{r}}{r!(n+r)!}\left(\frac{x}{2}\right)^{n+2r} \end{equation*}
\begin{equation*} =\sum\limits_{r=0}^{\infty}\frac{(-1)^{r}x^{2n+2r}}{r!(n+r)!\cdot 2^{n+2r}} \end{equation*}
\begin{equation*} \therefore \frac{d}{dx}\left[x^{n}J_{n}(x)\right]=\sum\limits_{r=0}^{\infty}\frac{(-1)^{r}(2n+2r)x^{2n+2r-1}}{r!(n+r)!\cdot 2^{n+2r}} \end{equation*}
\begin{equation*} =x^{n}\sum\limits_{r=0}^{\infty}\frac{(-1)^{r}(n+r)x^{n+2r-1}}{r!(n+r)!(n+r-1)\cdot 2^{n+2r-1}} \end{equation*}
\begin{equation*} =x^{n}\sum\limits_{r=0}^{\infty}\frac{(-1)^{r}}{r!(n-1+r)!}\left(\frac{x}{2}\right)^{\overline{n-1}+2r} \end{equation*}
\begin{equation*} = x^{n}J_{n-1}(x) \end{equation*}

Example 4.8.8.

Prove that
\begin{equation*} x\sin x=2\left[2^{2}J_{2}-4^{2}J_{4} +6^{2}J_{6}-\cdots\right] \end{equation*}
and
\begin{equation*} x\cos x=2\left[1^{2}J_{1}-3^{2}J_{3} +5^{2}J_{5}-\cdots\right] \end{equation*}
Solution.
We know that -
\begin{equation} \cos(x\sin\theta)=J_{o}+2J_{2}\cos2\theta+2J_{4}\cos4\theta+\cdots \tag{4.8.15} \end{equation}
and
\begin{equation} \sin(x\sin\theta)=2J_{1}\sin\theta+2J_{3}\sin3\theta+2J_{5}\sin5\theta+\cdots \tag{4.8.16} \end{equation}
Differentiating eqn. (4.8.15) w. r. t. \(\theta\text{,}\) we get -
\begin{equation} [-\sin(x\sin\theta)]x\cos\theta=0-4J_{2}\sin2\theta-8J_{4}\sin4\theta+\cdots \tag{4.8.17} \end{equation}
again, differentiating eqn. (4.8.17) w. r. t. \('\theta'\text{,}\) we get -
\begin{equation*} [-\sin(x\sin\theta)](-x\sin\theta)+[-\cos(x\sin\theta)](x\cos\theta)(x\cos\theta) \end{equation*}
\begin{equation} =-8J_{2}\cos2\theta-32J_{4}\cos4\theta+\cdots \tag{4.8.18} \end{equation}
putting \(\theta=\frac{\pi}{2}\) in eqn. (4.8.18), we get -
\begin{equation*} x\sin x=8J_{2}-32J_{4}+\cdots =2[2^{2}J_{2}-4^{2}J_{4}+\cdots] \end{equation*}
similarly, differentiating eqn. (4.8.16) twice and putting \(\theta=\frac{\pi}{2}\text{,}\) we get -
\begin{equation*} x\cos x=2[1^{2}J_{1}-3^{2}J_{3}+\cdots] \end{equation*}
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