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Bessel function of order 1

Alexmahone

Active member
Jan 26, 2012
268
Prove that $\displaystyle J_1(x)=\frac{1}{\pi}\int_0^\pi\cos(\theta-x\sin\theta)d\theta$ by showing that the right-hand side satisfies Bessel's equation of order 1 and that the derivative has the value $J_1'(0)$ when $x=0$. Explain why this constitutes a proof.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Prove that $\displaystyle J_1(x)=\frac{1}{\pi}\int_0^\pi\cos(\theta-x\sin\theta)d\theta$ by showing that the right-hand side satisfies Bessel's equation of order 1 and that the derivative has the value $J_1'(0)$ when $x=0$. Explain why this constitutes a proof.
Hi Alexmahone, :)

Let \(\displaystyle f(x)=\frac{1}{\pi}\int_0^\pi\cos(\theta-x\sin\theta)\,d\theta\). By the Leibniz integral rule we get,

\[f'(x)=\frac{1}{\pi}\int_0^\pi\sin\theta\sin(\theta-x\sin\theta)\,d\theta\]

\[f''(x)=-\frac{1}{\pi}\int_0^\pi\sin^{2}\theta\cos(\theta-x\sin\theta)\,d\theta\]

Substituting these in the left hand side of the Bessel's equation of order one we get,

\begin{eqnarray}

x^2 \frac{d^2 f(x)}{dx^2} + x \frac{df(x)}{dx} + (x^2-1)f(x)&=&-\frac{x^2}{\pi}\int_0^\pi\sin^{2}\theta\cos(\theta-x\sin\theta)\,d\theta+\frac{x}{\pi}\int_0^\pi\sin \theta\sin(\theta-x\sin\theta)\,d\theta\\

&~&+\frac{(x^2-1)}{\pi}\int_0^\pi\cos(\theta-x\sin\theta)\,d\theta\\

&=&\frac{x^2}{\pi}\int_0^\pi\cos^{2}\theta\cos( \theta-x\sin\theta)\,d\theta+\frac{x}{\pi}\int_0^\pi\sin \theta\sin(\theta-x\sin\theta)\,d\theta\\

&~&-\frac{1}{\pi}\int_0^\pi\cos(\theta-x\sin\theta)\,d\theta\\

\end{eqnarray}

Using integration by parts on \(\displaystyle\int_0^\pi\sin \theta\sin(\theta-x\sin\theta)\,d\theta\) we get,

\begin{eqnarray}

x^2 \frac{d^2 f(x)}{dx^2} + x \frac{df(x)}{dx} + (x^2-1)f(x)&=&\frac{x}{\pi}\int_0^\pi\cos \theta\cos(\theta-x\sin\theta)\,d\theta-\frac{1}{\pi}\int_0^\pi\cos(\theta-x\sin\theta)\,d\theta\\

&=&\frac{1}{\pi}\int_0^\pi(x\cos\theta-1)\cos(\theta-x\sin\theta)\,d\theta\\

&=&-\frac{1}{\pi}\int_0^\pi\cos(\theta-x\sin\theta)\,d(\theta-x\sin\theta)\\

&=&-\left.\frac{1}{\pi}\sin(\theta-x\sin\theta)\right|_{\theta=0}^{\theta=\pi}\\

&=&0

\end{eqnarray}

\[\therefore x^2 \frac{d^2 f(x)}{dx^2} + x \frac{df(x)}{dx} + (x^2-1)f(x)=0\]

Also, \(\displaystyle f'(0)=\frac{1}{\pi}\int_0^\pi\sin^{2}\theta\,d \theta=\frac{1}{2}\)

If we consider the Taylor expansion of the Bessel function of the first kind, differentiation of term by term is possible since it is a power series. Therefore we get,

\[J'_{1}(0)=\frac{1}{2}\]

Hence,

\[J'_{1}(0)=f'(0)=\frac{1}{2}~~~~~~~~~~~(1)\]

The Bessel's equation should have two linearly independent solutions. If the order of the Bessel equation is an integer then the the two solutions are: Bessel function of the first kind and Bessel function of the second kind. (Read this.)

In our case the order is 1 and hence the function \(f(x)\) should be linearly dependent to either the Bessel function of the first kind\((J_{1})\) or the Bessel function of second kind\((Y_{1})\). But the Bessel function of second kind is not continuous at \(x=0\) whereas \(f(0)\) is finite. Therefore the only possibility is that \(f(x)\) should be linearly dependent to the Bessel function of the first kind. That is,

\[f(x)=\beta J_{1}(x)\mbox{ where }\beta\in\Re\]

Now it can be easily shown that, \(f(0)=J_{1}(0)=0\). Therefore \(\beta\) could not be found out by substituting \(x=0\). However if we differentiate the above equation and substitute zero,

\[f'(0)=\beta J'_{1}(0)\]

By (1),

\[\beta=1\]

\[\therefore J_{1}(x)=f(x)=\frac{1}{\pi}\int_0^\pi\cos(\theta-x\sin\theta)\,d\theta\]

Kind Regards,
Sudharaka.
 
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