Laplace Transform of Bessel Function of the First Kind of Order Zero

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:

$\laptrans {\map {J_0} t} = \dfrac 1 {\sqrt {s^2 + 1} }$

Corollary

$\laptrans {\map {J_0} {a t} } = \dfrac 1 {\sqrt {s^2 + a^2} }$

Proof 1

From Bessel Function of the First Kind of Order Zero:

\(\ds \map {J_0} t\)

\(=\)

\(\ds \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac t 2}^{2 k}\)

\(\ds \)

\(=\)

\(\ds 1 - \dfrac {t^2} {2^2} + \dfrac {t^4} {2^2 \times 4^2} - \dfrac {t^6} {2^2 \times 4^2 \times 6^2} + \dotsb\)

Hence:

\(\ds \laptrans {\map {J_0} t}\)

\(=\)

\(\ds \laptrans {\sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac t 2}^{2 k} }\)

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {2^{2 k} \paren {k!}^2} \laptrans {t^{2 k} }\)

Linear Combination of Laplace Transforms

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {2^{2 k} \paren {k!}^2} \dfrac {\paren {2 k}!} {s^{2 k + 1} }\)

Laplace Transform of Positive Integer Power

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {\dfrac 1 2}^{2 k} \dbinom {2 k} k \dfrac 1 {s^{2 k + 1} }\)

Definition of Binomial Coefficient: $\dbinom {2 k} k = \dfrac {\paren {2 k}!} {\paren {k!}^2}$

Then:

\(\ds \dfrac 1 {\sqrt {s^2 + 1} }\)

\(=\)

\(\ds \dfrac 1 {\sqrt {s^2} \sqrt {1 + \paren {1 / s}^2} }\)

\(\ds \)

\(=\)

\(\ds \dfrac 1 s \paren {1 + \paren {\dfrac 1 s}^2}^{-1/2}\)

Definition of Rational Power

\(\ds \)

\(=\)

\(\ds \dfrac 1 s \sum_{k \mathop = 0}^\infty \dbinom {-1/2} k \paren {\dfrac 1 s}^{2 k}\)

General Binomial Theorem

\(\ds \)

\(=\)

\(\ds \dfrac 1 s \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {4^k} \dbinom {2 k} k \paren {\dfrac 1 s}^{2 k}\)

Binomial Coefficient of Minus Half

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \dfrac 1 {2^{2 k} } \dbinom {2 k} k \paren {\dfrac 1 {s^{2 k + 1} } }\)

rearranging, and bringing $\dfrac 1 s$ inside the summation

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {\dfrac 1 2}^{2 k} \dbinom {2 k} k \dfrac 1 {s^{2 k + 1} }\)

further rearrangement

The two expressions match, and the result follows.

$\blacksquare$

Proof 2

From Bessel Function of the First Kind of Order Zero:

\(\ds \map {J_0} t\)

\(=\)

\(\ds \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac t 2}^{2 k}\)

\(\ds \)

\(=\)

\(\ds 1 - \dfrac {t^2} {2^2} + \dfrac {t^4} {2^2 \times 4^2} - \dfrac {t^6} {2^2 \times 4^2 \times 6^2} + \dotsb\)

Hence:

\(\ds \laptrans {\map {J_0} t}\)

\(=\)

\(\ds \dfrac 1 s - \dfrac 1 {2^2} \dfrac {2!} {s^3} + \dfrac 1 {2^2 4^2} \dfrac {4!} {s^5} - \dfrac 1 {2^2 4^2 6^2} \dfrac {6!} {s^7} + \dotsb\)

Laplace Transform of Positive Integer Power

\(\ds \)

\(=\)

\(\ds \dfrac 1 s \paren {1 - \dfrac 1 2 \paren {\dfrac 1 {s^2} } + \dfrac {1 \times 3} {2 \times 4} \paren {\dfrac 1 {s^4} } - \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} \paren {\dfrac 1 {s^6} } + \dotsb}\)

simplifying

\(\ds \)

\(=\)

\(\ds \dfrac 1 s \paren {\paren {1 - \dfrac 1 {s^2} } ^{-1/2} }\)

General Binomial Theorem

\(\ds \)

\(=\)

\(\ds \dfrac 1 {\sqrt {s^2 + 1} }\)

simplifying

$\blacksquare$

Proof 3

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:

$\laptrans {t \, \map {J_0} t} = \dfrac 1 {\sqrt {s^2 + 1} }$

$\blacksquare$