Laplace Transform of Bessel Function of the First Kind

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Theorem

Let $J_n$ denote the Bessel function of the first kind of order $n$.


Then the Laplace transform of $J_n$ is given as:

$\laptrans {\map {J_n} {a t} } = \dfrac {\paren {\sqrt {s^2 + a^2} - s}^n} {a^n \sqrt {s^2 + a^2} }$


Proof 1

From Series Expansion of Bessel Function of the First Kind:

\(\ds \map {J_n} {a t}\) \(=\) \(\ds \dfrac {\paren {a t}^n} {2^n \, \map \Gamma {n + 1} } \paren {1 - \dfrac {\paren {a t}^2} {2 \paren {2 n + 2} } + \dfrac {\paren {a t}^4} {2 \times 4 \paren {2 n + 2} \paren {2 n + 4} } - \cdots}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {n + k + 1} } \paren {\dfrac {a t} 2}^{n + 2 k}\)


Hence:

\(\ds \laptrans {\map {J_n} {a t} }\) \(=\) \(\ds \laptrans {\sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {n + k + 1} } \paren {\dfrac {a t} 2}^{n + 2 k} }\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k a^{n + 2 k} } {2^{n + 2 k} k! \, \map \Gamma {n + k + 1} } \laptrans {t^{n + 2 k} }\) Linear Combination of Laplace Transforms
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k a^{n + 2 k} } {2^{n + 2 k} k! \, \map \Gamma {n + k + 1} } \dfrac {\map \Gamma {n + 2 k + 1} } {s^{n + 2 k + 1} }\) Laplace Transform of Power
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {\dfrac a 2}^{n + 2 k} \dfrac {\map \Gamma {n + 2 k + 1} } {k \, \map \Gamma k \, \map \Gamma {n + k + 1} } \dfrac 1 {s^{n + 2 k + 1} }\) rearrangement, Gamma Function Extends Factorial
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {\dfrac a 2}^{n + 2 k} \dfrac 1 {k \, \map \Beta {k, n + k + 1} } \dfrac 1 {s^{n + 2 k + 1} }\) Definition 3 of Beta Function



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Proof 2

\(\ds \laptrans {t^2 \frac {\d^2 x} {\d t^2} + t \frac {\d x} {\d t} + \paren {t^2 - \alpha^2} x} s\) \(=\) \(\ds 0\) Laplace Transform of Bessel's Equation
\(\ds \frac {\d^2} {\d s^2} \laptrans {x''} - \frac \d {\d s} \laptrans {x'} + \frac {\d^2} {\d s^2} \laptrans x - \alpha^2 \laptrans x\) \(=\) \(\ds 0\) setting the initial conditions as $\map x 0 = 1, \ \map {x'} 0 = 0$
\(\ds \paren {s^2 + 1} \LL'' \set x + 3 s \LL' \set x + \paren {1 - \alpha^2} \laptrans x\) \(=\) \(\ds 0\) Make the following change of variable $u = \sqrt {s^2 + 1} \laptrans x$
\(\ds u'' \sqrt {s^2 + 1} + \dfrac s {\sqrt {s^2 + 1} } u'\) \(=\) \(\ds \frac {\alpha^2 u} {\sqrt {s^2 + 1} }\)
\(\ds \paren {u' \sqrt {s^2 + 1} }'\) \(=\) \(\ds \frac {\alpha^2 u} {\sqrt {s^2 + 1} }\) multiplying both sides by $u' \sqrt {s^2 + 1}$
\(\ds \frac 1 2 \paren { \paren {u' \sqrt {s^2 + 1} }^2}'\) \(=\) \(\ds \frac 1 2 \alpha^2 \paren {u^2}'\)
\(\ds u' \sqrt {s^2 + 1}\) \(=\) \(\ds -\alpha u\) Constant of Integration removed by the Final Value Theorem of Laplace Transform
\(\ds \int \frac 1 u \rd u\) \(=\) \(\ds \int -\frac \alpha {\sqrt {s^2 + 1} } \rd s\)
\(\ds \map \ln u\) \(=\) \(\ds \alpha \map \ln {\sqrt {s^2 + 1} - s}\)
\(\ds u\) \(=\) \(\ds \paren {\sqrt {s^2 + 1} - s}^\alpha\) reverting the substitution
\(\ds \laptrans x\) \(=\) \(\ds \dfrac {\paren {\sqrt {s^2 + 1} - s}^\alpha} {\sqrt {s^2 + 1} }\)


Hence:

\(\ds \map {\laptrans {\map {J_\alpha} t} } s\) \(=\) \(\ds \dfrac {\paren {\sqrt {s^2 + 1} - s}^\alpha} {\sqrt {s^2 + 1} }\)

$\blacksquare$


Sources

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