Laplace Transform of Bessel Function of the First Kind/Proof 2
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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
| \(\ds \laptrans {t^2 \frac {\d^2 x} {\d t^2} + t \frac {\d x} {\d t} + \paren {t^2 - \alpha^2} x} s\) | \(=\) | \(\ds 0\) | 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:
| \(\ds \map {\laptrans {\map {J_\alpha} t} } s\) | \(=\) | \(\ds \dfrac {\paren {\sqrt {s^2 + 1} - s}^\alpha} {\sqrt {s^2 + 1} }\) |
$\blacksquare$