Laplace Transform of Bessel Function of the First Kind of Order Zero/Proof 3
Theorem
Let $J_0$ denote the Bessel function of the first kind of order $0$.
Then the Laplace transform of $J_0$ is given as:
- $\laptrans {\map {J_0} t} = \dfrac 1 {\sqrt {s^2 + 1} }$
Proof
By definition of Bessel function of the first kind, $\map {J_0} t$ satisfies Bessel's equation:
\(\ds t^2 \, \map {\dfrac {\d^2} {\d t^2} } {\map {J_0} t} + t \, \map {\dfrac \d {\d t} } {\map {J_0} t} + \paren {t^2 - 0^2} {\map {J_0} t}\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds t \, {\map {J_0} t} + \map {J_0'} t + t \, \map {J_0} t\) | \(=\) | \(\ds 0\) |
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 $0$:
- $\map {J_0} t = 1 - \dfrac {t^2} {2^2} + \dfrac {t^4} {2^2 \times 4^2} - \dfrac {t^6} {2^2 \times 4^2 \times 6^2} + \dotsb$
from which it follows immediately that:
- $\map {J_0} 0 = 1$
From Derivative of Bessel Function of the First Kind of Order $0$ we have:
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from which it follows that:
- $\map {J'_0} 0 = 0$
Then from Derivative of Laplace Transform:
\(\ds \laptrans {t \, \map {J_0} t}\) | \(=\) | \(\ds -\map {\dfrac \d {\d s} } {\map {J_0} t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\map {\dfrac \d {\d s} } {s^2 \, \map {J_0} t - s \, \map {J_0} 0 - \map {J_0'} 0}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\map {\dfrac \d {\d s} } {s^2 \, \map {J_0} t - s}\) | $\map {J_0} 0 = 1$ and $\map {J'_0} 0 = 0$ |
and:
- $\laptrans {t \map {J_0} t} = -\map {\dfrac \d {\d s} } {\map {J_0} t}$
Thus by taking the Laplace transform of $(1)$ we have:
Let $y = \laptrans {\map {J_0} t}$.
\(\ds \laptrans {t \, {\map {J_0} t} + \map {J_0'} t + t \, \map {J_0} t}\) | \(=\) | \(\ds \laptrans 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds -\map {\dfrac \d {\d s} } {s^2 \, \laptrans {\map {J_0} t} - s} + \paren {s \laptrans {\map {J_0} t} - \map {J_0} 0} - \map {\dfrac \d {\d s} } {\map {J_0} t}\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds -\map {\dfrac \d {\d s} } {s^2 y - s} + \paren {s y - 1} - \dfrac {\d y} {\d s}\) | \(=\) | \(\ds 0\) | setting $y = \laptrans {\map {J_0} t}$ and noting that $\map {J_0} 0 = 1$ from above | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\d y} {\d s}\) | \(=\) | \(\ds -\dfrac {s y} {s^2 + 1}\) | rearranging and simplifying | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \dfrac {\d y} y\) | \(=\) | \(\ds -\int \dfrac {s \rd s} {s^2 + 1}\) | Solution to Separable Differential Equation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \dfrac c {\sqrt {s^2 + 1} }\) | for some constant $c$ |
Now we have that:
\(\ds \lim_{s \mathop \to \infty} s \, \map y s\) | \(=\) | \(\ds \lim_{s \mathop \to \infty} \dfrac {c s} {\sqrt {s^2 + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds c\) |
and:
- $\ds \lim_{t \mathop \to 0} \map {J_0} t = 1$
Thus by the Initial Value Theorem of Laplace Transform:
- $c = 1$
and so:
- $\laptrans {t \, \map {J_0} t} = \dfrac 1 {\sqrt {s^2 + 1} }$
$\blacksquare$
Sources
- 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Solved Problems: Bessel Functions: $34 \ \text{(a)}$