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Assume that $J_\alpha(x)$ is defined as below: (Actually, It's one of the solutions of Bessel's ODE) $J_\alpha(x)=\sum_{n=0}^{\infty} \frac{(-1)^n}{n! \Gamma(n+\alpha+1)}(\frac{x}{2})^{2n+\alpha}$

Such that $\Gamma(x)=\int_0^{+\infty} t^{x-1} e^{-t}dt$.

We know that (i) & (ii) hold.

(i) $\frac{d}{dx}(x^\alpha J_\alpha(x))=x^\alpha J_{\alpha-1}(x)$
(ii) $\frac{d}{dx}(x^{-\alpha} J_\alpha(x))=-x^{-\alpha}J_{\alpha+1}(x)$

Question:

Using (i) and (ii), Prove that:

a) $J'_\alpha(x)=\frac{1}{2}(J_{\alpha-1}(x)-J_{\alpha+1}(x))$
b) $\frac{2\alpha}{x}J_{\alpha}(x)=J_{\alpha-1}(x)+J_{\alpha+1}(x)$
c) $\frac{d}{dx}(xJ_\alpha(x)J_{\alpha+1}(x))=x(J_\alpha^2(x)-J_{\alpha+1}^2(x))$

My try:

a) First, I calculated the derivative of the LHS of (i) and then multiplied the whole equation by $x^{-\alpha}$. I did the same thing for (ii) and then multiplied it by $x^\alpha$. The summation of these two equations gives the desired result.

b) I solved it similar to a, with a tiny difference. (Multiplying by $-x^\alpha) $

I'm stuck on proving (c). I wrote a lot but it didn't work out...

Any idea?

Arman Malekzadeh
  • 4,108

2 Answers2

4

Here is an easy method using only the two identities you are given at the start of the question.

Notice that: $$\color{red}{x^{\alpha+1}J_{\alpha+1}(x)}\cdot \color{blue}{x^{-\alpha} J_\alpha(x)}=xJ_{\alpha}(x)J_{\alpha+1}(x)$$ Therefore, we can apply the product rule with $u=x^{\alpha+1}J_{\alpha+1}(x)$ and $v=x^{-\alpha}J_{\alpha}(x)$: $$\begin{align}(uv)'&=u'v+uv'\\&=(x^{\alpha+1}J_{\alpha+1}(x))'\cdot x^{-\alpha}J_{\alpha}(x)+x^{\alpha+1}J_{\alpha+1}(x)\cdot (x^{-\alpha}J_{\alpha}(x))' \end{align}$$ Substituting (i) and (ii) and simplifying gives the required identity: $$x^{\alpha+1}J_{\alpha}(x)\cdot x^{-\alpha}J_{\alpha}(x)+x^{\alpha+1}J_{\alpha+1}(x)\cdot -x^{-\alpha}J_{\alpha+1}(x)=x(J_{\alpha}^2(x)-J_{\alpha+1}^2(x))$$

projectilemotion
  • 15,428
1

$$\frac{\text{d}}{\text{d} x}\left(x\text{J}_{\alpha}(x)\text{J}_{\alpha+1}(x)\right)=$$

Use the product rule, $\frac{\text{d}}{\text{d}x}(uv)=v\frac{\text{d}u}{\text{d}x}+u\frac{\text{d}v}{\text{d}x}$, where $u=x$ and $v=\text{J}_{\alpha}(x)$:

$$\text{J}_{\alpha}(x)\text{J}_{1+\alpha}(x)\left[\frac{\text{d}}{\text{d}x}(x)\right]+x\left[\frac{\text{d}}{\text{d}x}\left(\text{J}_{\alpha}(x)\text{J}_{1+\alpha}(x)\right)\right]=$$ $$\text{J}_{\alpha}(x)\text{J}_{1+\alpha}(x)\left[1\right]+x\left[\frac{\text{d}}{\text{d}x}\left(\text{J}_{\alpha}(x)\text{J}_{1+\alpha}(x)\right)\right]=$$ $$\text{J}_{\alpha}(x)\text{J}_{1+\alpha}(x)+x\left[\frac{\text{d}}{\text{d}x}\left(\text{J}_{\alpha}(x)\text{J}_{1+\alpha}(x)\right)\right]=$$

Use the product rule, $\frac{\text{d}}{\text{d}x}(uv)=v\frac{\text{d}u}{\text{d}x}+u\frac{\text{d}v}{\text{d}x}$, where $u=\text{J}_{\alpha}(x)$ and $v=\text{J}_{\alpha+1}(x)$:

$$\text{J}_{\alpha}(x)\text{J}_{1+\alpha}(x)+x\left[\text{J}_{1+\alpha}(x)\cdot\frac{\text{J}_{\alpha-1}-\text{J}_{1+\alpha}(x)}{2}+\text{J}_{\alpha}(x)\left[\frac{\text{d}}{\text{d}x}\left(\text{J}_{1+\alpha}(x)\right)\right]\right]=$$ $$\text{J}_{\alpha}(x)\text{J}_{1+\alpha}(x)+x\left[\text{J}_{1+\alpha}(x)\cdot\frac{\text{J}_{\alpha-1}-\text{J}_{1+\alpha}(x)}{2}+\text{J}_{\alpha}(x)\left[\frac{1}{2}\left(\text{J}_{\alpha}(x)-\text{J}_{2+\alpha}(x)\right)\right]\right]=$$ $$x\left[\text{J}_{\alpha}^2(x)-\text{J}_{1+\alpha}^2(x)\right]$$

Jan Eerland
  • 28,671