Question

To show the following equation shown in the problem

$\mathbf{2}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}{\mathit{J}}_{\mathbf{a}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{J}}_{\mathbf{a}\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$.

Short answer

The answer is,$2\frac{\mathrm{d}}{\mathrm{dx}}{J}_{\mathrm{a}}\left(x\right)=J_{\mathrm{a}-1}\left(x\right)$ .

Step-by-step solution

 Step 1: Concept of Bessel Function:

The Bessel functions of the first kind${\mathbf{J}}_{\mathbf{n}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ are defined as the solutions to the Bessel differential equation.

${\mathit{x}}^{\mathbf{2}}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\mathit{x}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{+}\left(x^2-n^2\right)\mathit{y}\mathbf{=}\mathbf{0}$

Solve the equation 2ddxJa(x)=Ja-1(x)-Ja+1(x)  about the Bessel Function:

The given equation for the following equation is shown as,

$2\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{J}}_{\mathrm{a}}\left(\mathrm{x}\right)={\mathrm{J}}_{\mathrm{a}-1}\left(\mathrm{x}\right)-{\mathrm{J}}_{\mathrm{a}+1}\left(\mathrm{x}\right)$

By the use of the following proposition about the zeros of Bessel functions solve the equation as follows:

$$\begin{gathered} \frac{d}{dx}{J}_{\alpha }\left(x\right)=\frac{d}{dx}\left(x^a\left(x^{-\alpha }{J}_{\alpha }\right(x\left)\right)\right) \\ =\alpha x^{\alpha -1}\left(J_{\alpha }\left(x\right)\right)+x^{\alpha }\frac{d}{dx}\left(x^{\alpha }{J}_{\alpha }\left(x\right)\right) \\ =-\alpha x^{\alpha -1}{J}_a\left(x\right)+x^{-\alpha }\left(x^{\alpha }{J}_{\alpha -1}\left(x\right)\right) \\ =-\alpha x^{\alpha -1}{J}_{\alpha }\left(x\right)+J_{a-1}\left(x\right) \end{gathered}$$

Calculation of the equation ddx(xαx-αJα(x) :

By adding the two expressions as follows:

$$\begin{gathered} \frac{d}{dx}{J}_a\left(x\right)=\frac{d}{dx}\left(x^{\alpha }\left(x^{-\alpha }{J}_{\alpha }\right)\left(x\right)\right) \\ =\alpha x^{\alpha -1}\left(J_{\alpha }\left(x\right)\right)+x^{\alpha }\frac{d}{dx}\left(x^{\alpha }{J}_{\alpha }\left(x\right)\right) \\ =\left(J_{\alpha -1}\left(x\right)-J_{\alpha +1}\left(x\right)\right)/2 \\ 2\frac{d}{dx}{J}_{\alpha }\left(x\right)=J_{a-1}\left(x\right)-J_{a+1}\left(x\right) \end{gathered}$$