\section{Laplace Transform of Bessel Function of the First Kind}
Tags: Laplace Transforms of Bessel Functions, Laplace Transform of Bessel Function of the First Kind

\begin{theorem}
Let $J_n$ denote the [[Definition:Bessel Function of the First Kind|Bessel function of the first kind]] of [[Definition:Order of Bessel Function|order]] $n$.
Then the [[Definition:Laplace Transform|Laplace transform]] of $J_n$ is given as:
:$\laptrans {\map {J_n} {a t} } = \dfrac {\paren {\sqrt {s^2 + a^2} - s}^n} {a^n \sqrt {s^2 + a^2} }$
\end{theorem}

\begin{proof}
From [[Bessel Function of the First Kind of Order Zero]]:
{{begin-eqn}}
{{eqn | l = \map {J_0} t
      | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac t 2}^{2 k}
      | c = 
}}
{{eqn | r = 1 - \dfrac {t^2} {2^2} + \dfrac {t^4} {2^2 \times 4^2} - \dfrac {t^6} {2^2 \times 4^2 \times 6^2} + \dotsb
      | c = 
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \laptrans {\map {J_0} t}
      | r = \laptrans {\sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac t 2}^{2 k} }
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {2^{2 k} \paren {k!}^2} \laptrans {t^{2 k} }
      | c = [[Linear Combination of Laplace Transforms]]
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {2^{2 k} \paren {k!}^2} \dfrac {\paren {2 k}!} {s^{2 k + 1} }
      | c = [[Laplace Transform of Positive Integer Power]]
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {\dfrac 1 2}^{2 k} \dbinom {2 k} k \dfrac 1 {s^{2 k + 1} }
      | c = {{Defof|Binomial Coefficient}}: $\dbinom {2 k} k = \dfrac {\paren {2 k}!} {\paren {k!}^2}$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \dfrac 1 {\sqrt {s^2 + 1} }
      | r = \dfrac 1 {\sqrt {s^2} \sqrt {1 + \paren {1 / s}^2} }
      | c = 
}}
{{eqn | r = \dfrac 1 s \paren {1 + \paren {\dfrac 1 s}^2}^{-1/2}
      | c = {{Defof|Rational Power}}
}}
{{eqn | r = \dfrac 1 s \sum_{k \mathop = 0}^\infty \dbinom {-1/2} k \paren {\dfrac 1 s}^{2 k}
      | c = [[General Binomial Theorem]]
}}
{{eqn | r = \dfrac 1 s \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {4^k} \dbinom {2 k} k \paren {\dfrac 1 s}^{2 k}
      | c = [[Binomial Coefficient of Minus Half]]
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \dfrac 1 {2^{2 k} } \dbinom {2 k} k \paren {\dfrac 1 {s^{2 k + 1} } }
      | c = rearranging, and bringing $\dfrac 1 s$ inside the [[Definition:Summation|summation]]
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {\dfrac 1 2}^{2 k} \dbinom {2 k} k \dfrac 1 {s^{2 k + 1} }
      | c = further rearrangement
}}
{{end-eqn}}
The two expressions match, and the result follows.
{{qed}}
\end{proof}

\begin{proof}
From [[Bessel Function of the First Kind of Order Zero]]:
{{begin-eqn}}
{{eqn | l = \map {J_0} t
      | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac t 2}^{2 k}
      | c = 
}}
{{eqn | r = 1 - \dfrac {t^2} {2^2} + \dfrac {t^4} {2^2 \times 4^2} - \dfrac {t^6} {2^2 \times 4^2 \times 6^2} + \dotsb
      | c = 
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \laptrans {\map {J_0} t}
      | r = \dfrac 1 s - \dfrac 1 {2^2} \dfrac {2!} {s^3} + \dfrac 1 {2^2 4^2} \dfrac {4!} {s^5} - \dfrac 1 {2^2 4^2 6^2} \dfrac {6!} {s^7} + \dotsb
      | c = [[Laplace Transform of Positive Integer Power]]
}}
{{eqn | r = \dfrac 1 s \paren {1 - \dfrac 1 2 \paren {\dfrac 1 {s^2} } + \dfrac {1 \times 3} {2 \times 4} \paren {\dfrac 1 {s^4} } - \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} \paren {\dfrac 1 {s^6} } + \dotsb}
      | c = simplifying
}}
{{eqn | r = \dfrac 1 s \paren {\paren {1 - \dfrac 1 {s^2} } ^{-1/2} }
      | c = [[General Binomial Theorem]]
}}
{{eqn | r = \dfrac 1 {\sqrt {s^2 + 1} }
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
\end{proof}

\begin{proof}
By definition of [[Definition:Bessel Function of the First Kind|Bessel function of the first kind]], $\map {J_0} t$ satisfies [[Definition:Bessel's Equation|Bessel's equation]]:
{{begin-eqn}}
{{eqn | l = t^2 \, \map {\dfrac {\d^2} {\d t^2} } {\map {J_0} t} + t \, \map {\dfrac \d {\d t} } {\map {J_0} t} + \paren {t^2 - 0^2} {\map {J_0} t}
      | r = 0
      | c = 
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = t \, {\map {J_0''} t} + \map {J_0'} t + t \, \map {J_0} t
      | r = 0
      | c = 
}}
{{end-eqn}}
From [[Laplace Transform of Derivative]]:
:$\laptrans {\map {J_0'} t} = s \laptrans {\map {J_0} t} - \map {J_0} 0$
and from [[Laplace Transform of Second Derivative]]:
:$\laptrans {\map {J_0''} t} = s^2 \laptrans {\map {J_0} t} - s \, \map {J_0} 0 - \map {J_0'} 0$
From [[Bessel Function of the First Kind of Order Zero|Bessel Function of the First Kind of Order $0$]]:
:$\map {J_0} t = 1 - \dfrac {t^2} {2^2} + \dfrac {t^4} {2^2 \times 4^2} - \dfrac {t^6} {2^2 \times 4^2 \times 6^2} + \dotsb$
from which it follows immediately that:
:$\map {J_0} 0 = 1$
From [[Derivative of Bessel Function of the First Kind of Order 0|Derivative of Bessel Function of the First Kind of Order $0$]] we have:
{{ProofWanted|Derivative of Bessel Function of the First Kind of Order 0}}
from which it follows that:
:$\map {J'_0} 0 = 0$
Then from [[Derivative of Laplace Transform]]:
{{begin-eqn}}
{{eqn | l = \laptrans {t \, \map {J_0''} t}
      | r = -\map {\dfrac \d {\d s} } {\map {J_0''} t}
      | c = 
}}
{{eqn | r = -\map {\dfrac \d {\d s} } {s^2 \, \map {J_0} t - s \, \map {J_0} 0 - \map {J_0'} 0}
      | c = 
}}
{{eqn | r = -\map {\dfrac \d {\d s} } {s^2 \, \map {J_0} t - s}
      | c = $\map {J_0} 0 = 1$ and $\map {J'_0} 0 = 0$
}}
{{end-eqn}}
and:
:$\laptrans {t \map {J_0} t} = -\map {\dfrac \d {\d s} } {\map {J_0} t}$
Thus by taking the [[Definition:Laplace Transform|Laplace transform]] of $(1)$ we have:
Let $y = \laptrans {\map {J_0} t}$.
{{begin-eqn}}
{{eqn | l = \laptrans {t \, {\map {J_0''} t} + \map {J_0'} t + t \, \map {J_0} t}
      | r = \laptrans 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -\map {\dfrac \d {\d s} } {s^2 \, \laptrans {\map {J_0} t} - s} + \paren {s \laptrans {\map {J_0} t} - \map {J_0} 0} - \map {\dfrac \d {\d s} } {\map {J_0} t}
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -\map {\dfrac \d {\d s} } {s^2 y - s} + \paren {s y - 1} - \dfrac {\d y} {\d s}
      | r = 0
      | c = setting $y = \laptrans {\map {J_0} t}$ and noting that $\map {J_0} 0 = 1$ from above
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\d y} {\d s}
      | r = -\dfrac {s y} {s^2 + 1}
      | c = rearranging and simplifying
}}
{{eqn | ll= \leadsto
      | l = \int \dfrac {\d y} y
      | r = -\int \dfrac {s \rd s} {s^2 + 1}
      | c = [[Separation of Variables]]
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \dfrac c {\sqrt {s^2 + 1} }
      | c = for some [[Definition:Constant|constant]] $c$
}}
{{end-eqn}}
Now we have that:
{{begin-eqn}}
{{eqn | l = \lim_{s \mathop \to \infty} s \, \map y s
      | r = \lim_{s \mathop \to \infty} \dfrac {c s} {\sqrt {s^2 + 1} }
      | c = 
}}
{{eqn | r = c
      | c = 
}}
{{end-eqn}}
and:
:$\displaystyle \lim_{t \mathop \to 0} \map {J_0} t = 1$
Thus by the [[Initial Value Theorem of Laplace Transform]]:
:$c = 1$
and so:
:$\laptrans {t \, \map {J_0} t} = \dfrac 1 {\sqrt {s^2 + 1} }$
{{qed}}
\end{proof}

\begin{proof}
{{begin-eqn}}
{{eqn | l = \laptrans {t^2 \frac {\d^2 x} {\d t^2} + t \frac {\d x} {\d t} + (t^2-\alpha^2)x} s 
      | r = 0 
      | c = [[Definition:Laplace Transform|Laplace Transform]] of [[Definition:Bessel's Equation|Bessel's Equation]]
}}
{{eqn | l = \frac {\d^2} {\d s^2}\mathcal{L}[x''] - \frac \d {\d s} \mathcal{L}[x'] + \frac {\d^2} {\d s^2} \mathcal{L}[x] - \alpha^2 \mathcal{L} [x]
      | r = 0
      | c = setting the initial conditions as $x(0)=1,\ x'(0)=0$
}}
{{eqn | l = (s^2+1) \mathcal{L}''[x] + 3 s \mathcal {L}'[x] + (1 - \alpha^2) \mathcal{L}[x]
      | r = 0
      | c = Make the following change of variable $u = \sqrt {s^2 + 1} \mathcal {L} [x]$
}}
{{eqn | l = u'' \sqrt {s^2 + 1} + \dfrac s {\sqrt {s^2 + 1} } u'
      | r = \frac {\alpha^2 u} {\sqrt {s^2 + 1} }
      | c = 
}}
{{eqn | l = (u' \sqrt {s^2 + 1} )'
      | r = \frac {\alpha^2 u} {\sqrt {s^2 + 1} }
      | c = multiplying both sides by $u' \sqrt {s^2 + 1}$
}}
{{eqn | l = \frac 1 2 ( (u' \sqrt {s^2 + 1} )^2)'
      | r = \frac 1 2 \alpha^2 (u^2)'
      | c = 
}}
{{eqn | l = u' \sqrt {s^2 + 1}
      | r = -\alpha u
      | c = [[Definition:Arbitrary Constant|Constant of Integration]] removed by the [[Final Value Theorem of Laplace Transform]]
}}
{{eqn | l = \int \frac 1 u \rd u
      | r = \int -\frac \alpha {\sqrt {s^2 + 1} } \rd s
      | c = 
}}
{{eqn | l = \map \ln u
      | r = \alpha \ln (\sqrt {s^2 + 1} - s)
      | c = 
}}
{{eqn | l = u
      | r = (\sqrt {s^2 + 1} - s)^\alpha
      | c = reverting the substitution
}}
{{eqn | l = \mathcal{L} [x]
      | r = \dfrac {\paren {\sqrt {s^2 + 1} - s}^\alpha} {\sqrt {s^2 + 1} }
      | c = 
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \mathcal{L}[J_\alpha (t)](s)
      | r = \dfrac {\paren {\sqrt {s^2 + 1} - s}^\alpha} {\sqrt {s^2 + 1} }
      | c = 
}}
{{end-eqn}}
{{qed}}
\end{proof}


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