Chapter 8.6, Problem 33E

(a)

To determine

To Prove: The Bessel function $J_0$ satisfies the differential function.

Explanation of Solution

Given:

The given equation differential equation is,

  $x^2{J^{″}}_0\left(x\right)+x{J}^{\prime }\left(x\right)+x^2{J}_0\left(x\right)=0$

Calculation:

Consider the Bessel function of order 0 is,

  $\begin{array}{c}{J}_0\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n{x}^{2n}}{2^{2n}{\left(n!\right)}^2}}\\ {J^{\prime }}_0\left(x\right)={\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n}n2x^{2n-1}}{2^{2n}{\left(n!\right)}^2}}\\ {J^{\prime }}_0\left(x\right)={\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n\left(2n\right)\left(2n-1\right)2x^{2n-2}}{2^{2n}{\left(n!\right)}^2}}\end{array}$

Then, the given differential equation is written as,

  $x^2{J^{″}}_0\left(x\right)+x{J}^{\prime }\left(x\right)+x^2{J}_0\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n{x}^{2n}}{2^{2n}{\left(n!\right)}^2}}+{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n}2n{x}^{2n-1}}{2^{2n}{\left(n!\right)}^2}}+{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n\left(2n\right)\left(2n-1\right)2x^{2n-2}}{2^{2n}{\left(n!\right)}^2}}$

Substitute $\left(n-1\right)$ for $n$ in the above equation as,

  $\begin{array}{c}{x}^2{J^{″}}_0\left(x\right)+x{J}^{\prime }\left(x\right)+x^2{J}_0\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n{x}^{2n}}{2^{2n}{\left(n!\right)}^2}}+{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n}2n{x}^{2n}}{2^{2n}{\left(n!\right)}^2}}+{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n\left(2n\right)\left(2n-1\right)2x^{2n}}{2^{2n}{\left(n!\right)}^2}}\\ ={\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^n\left[\frac{\left(2n\right)\left(2n-1\right)}{2^{2n}{\left(n!\right)}^2}+\frac{2n}{2^{2n}{\left(n!\right)}^2}+\frac{1}{{\left[2^{2n-2}\left(n-1\right)!\right]}^2}\right]}{x}^{2n}\\ =0\end{array}$

Hence, it is proved that $x^2{J^{″}}_0\left(x\right)+x{J}^{\prime }\left(x\right)+x^2{J}_0\left(x\right)=0$ .

(b)

To determine

To Find: The evaluated form of the given interval correct to three decimal places.

Answer to Problem 33E

The value of the integral is $0.920$ .

Explanation of Solution

Given:

The given integral is ${\displaystyle {\int }_0^1{J}_0}\left(x\right)dx$ .

Calculation:

Consider the integral as.

  $\begin{array}{c}{\displaystyle \int J_0\left(x\right)d\left(x\right)}={\displaystyle \int {\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n{x}^{2n}}{2^{2n}{\left(n!\right)}^2}}}\\ ={\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n{x}^{2n+1}}{2^{2n}{\left(n!\right)}^2\left(2n+1\right)}}+c\end{array}$

Then,

  $\begin{array}{c}{\displaystyle {\int }_0^1{J}_0}\left(x\right)dx=\left[x-\frac{x^3}{3\cdot \left(4\right)}+\frac{x^5}{5\cdot \left(64\right)}-\frac{x^7}{7\cdot \left(2304\right)}\right]\\ =\left[1-\frac{1}{12}+\frac{1}{320}-\frac{1}{16128}+\mathrm{.....}\right]\end{array}$

Since, the value of $\frac{1}{16128}$ is 0.000062 this error is neglected an the value is,

  $\begin{array}{c}{\displaystyle {\int }_0^1{J}_0}\left(x\right)dx=1-\frac{1}{12}+\frac{1}{320}\\ =0.920\end{array}$