# Problem of the Week #9 - May 28th, 2012

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#### Chris L T521

##### Well-known member
Staff member
Thanks to those who participated in last week's POTW. I was glad to see the large turnout!

Here's the problem for this week; I've decided to revisit the topic of Bessel functions.

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Problem: The Bessel function of the first kind of order zero is defined by the following Taylor series:

$J_0(t) = \sum_{n=0}^{\infty}\frac{(-1)^nt^{2n}}{(n!)^22^{2n}}$

Let $\mathcal{L}\{f(t)\}=\displaystyle\int_0^{\infty}e^{-st}f(t)\,dt$ denote the Laplace transform of $f(t)$. Assuming that we can compute Laplace transforms term by term, show that

$\mathcal{L}\{J_0(t)\} = \frac{1}{\sqrt{s^2+1}},\qquad s>1$

and

$\mathcal{L}\{J_0(\sqrt{t})\}=\frac{e^{-1/4s}}{s},\qquad s>0.$

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#### Chris L T521

##### Well-known member
Staff member
This problem was correctly answered by Sudharaka. Here's his solution.

$J_0(t) = \sum_{n=0}^{\infty}\frac{(-1)^nt^{2n}}{(n!)^22^{2n}}$

$\Rightarrow\mathcal{L}\{J_{0}(t)\}=\int_0^{ \infty}e^{-st}\sum_{n=0}^{\infty}\frac{(-1)^nt^{2n}}{(n!)^22^{2n}}\,dt$

Since the series representation of the Bessel function is a power series we can integrate term by term,

${L}\{J_{0}(t)\}=\sum_{n=0}^{\infty}\left(\frac{(-1)^n}{(n!)^22^{2n}}\int_0^{ \infty}e ^{-st}t^{2n}\,dt\right)$

Since, $$\displaystyle\mathcal{L}\{t^{2n}\}=\int_0^{ \infty}e ^{-st}t^{2n}\,dt=\frac{(2n)!}{s^{2n+1}}\mbox{ for }s>0$$

${L}\{J_{0}(t)\}=\sum_{n=0}^{\infty}\left(\frac{(-1)^n}{(n!)^22^{2n}}\frac{(2n)!}{s^{2n+1}}\right)~~~~~~~~~~(1)$

Now we shall show by mathematical induction, $$\displaystyle\frac{(-1)^n(2n)!}{(n!)^22^{2n}}=\binom{-\frac{1}{2}}{n}\mbox{ for each }n\in\mathbb{N}=\mathbb{Z}\cup\{0\}$$

It is clear that the above statement holds for $$n=0$$. Suppose the statement is true for $$n=p\in\mathbb{N}$$. That is,

$\frac{(-1)^p(2p)!}{(p!)^22^{2p}}=\binom{-\frac{1}{2}}{p}$

Consider, $$\displaystyle\frac{(-1)^{p+1}(2(p+1))!}{((p+1)!)^22^{2(p+1)}}$$

\begin{eqnarray}

\frac{(-1)^{p+1}(2(p+1))!}{((p+1)!)^22^{2(p+1)}}&=&\frac{(-1)^p(2p)!}{(p!)^22^{2p}}\left(\frac{(-1)(2p+2)(2p+1)}{2^2(p+1)}\right)\\

&=&\binom{-\frac{1}{2}}{p}\left(\frac{-p-\frac{1}{2}}{p+1}\right)\\

&=&\binom{-\frac{1}{2}}{p+1}

\end{eqnarray}

Hence by mathematical induction, $$\displaystyle\frac{(-1)^n(2n)!}{(n!)^22^{2n}}=\binom{-\frac{1}{2}}{n}\mbox{ for each }n\in\mathbb{N}=\mathbb{Z}\cup\{0\}~~~~~~~~~~~~(2)$$

By (1) and (2),

\begin{eqnarray}

{L}\{J_{0}(t)\}&=&\sum_{n=0}^{\infty}\binom{-\frac{1}{2}}{n}\frac{1}{s^{2n+1}}\\

&=&\frac{1}{s}\sum_{n=0}^{\infty}\binom{-\frac{1}{2}}{n}\left(\frac{1}{s^2}\right)^n\\

\end{eqnarray}

$$\displaystyle\sum_{n=0}^{\infty}\binom{-\frac{1}{2}}{n}\left(\frac{1}{s^2}\right)^n$$ is the Taylor series expansion of $$\displaystyle\left(1+\frac{1}{s^2}\right)^{-\frac{1}{2}}\mbox{ for }\frac{1}{s^2}<1\Rightarrow s>1\mbox{ (Since s>0)}$$

$\therefore{L}\{J_{0}(t)\}=\frac{1}{s}\left(1+\frac{1}{s^2}\right)^{-\frac{1}{2}}=\frac{1}{\sqrt{s^2+1}}\mbox{ for }s\geq 1$

Now we shall show that, $$\displaystyle\mathcal{L}\{J_0(\sqrt{t})\}=\frac{e^{-1/4s}}{s}\mbox{ where }s>0\,.$$

$J_0(\sqrt{t}) = \sum_{n=0}^{\infty}\frac{(-1)^nt^{n}}{(n!)^22^{2n}}$

$\Rightarrow\mathcal{L}\{J_{0}(\sqrt{t})\}=\int_0^{ \infty}e^{-st}\sum_{n=0}^{\infty}\frac{(-1)^nt^{n}}{(n!)^22^{2n}}\,dt$

Since the series representation of the Bessel function is a power series we can integrate term by term,

${L}\{J_{0}(t)\}=\sum_{n=0}^{\infty}\left(\frac{(-1)^n}{(n!)^22^{2n}}\int_0^{ \infty}e ^{-st}t^{n}\,dt\right)$

Since, $$\displaystyle\mathcal{L}\{t^{n}\}=\int_0^{ \infty}e ^{-st}t^{n}\,dt=\frac{n!}{s^{n+1}}\mbox{ for }s>0$$

\begin{eqnarray}

{L}\{J_{0}(t)\}&=&\sum_{n=0}^{\infty}\left(\frac{(-1)^n}{(n!)^22^{2n}}\frac{n!}{s^{n+1}}\right)\\

&=&\frac{1}{s}\sum_{n=0}^{\infty}\frac{\left(-\frac{1}{4s}\right)^n}{n!}\\

\end{eqnarray}

Since, $$\displaystyle e^{-\frac{1}{4s}}=\sum_{n=0}^{\infty}\frac{\left(-\frac{1}{4s}\right)^n}{n!}$$

$\mathcal{L}\{J_0(\sqrt{t})\}=\frac{e^{-1/4s}}{s}\mbox{ for }s>0$

Q.E.D

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