Bessel functions

\[ t^2 y'' (t) + t\,y' (t) + \left( t^2 - n^2 \right) y (t) =0 \]

\[ J_n^L (\lambda ) = {\cal L}\left[ J_n (t) \right] (\lambda ) = \int_0^{\infty} e^{-\lambda \,t} J_n (t) \,{\text d}t \]

\[ {\cal L} \left[ t\, y'' (t) \right] + {\cal L} \left[ y' (t) \right] + {\cal L} \left[ t\,y (t) \right] = 0 . \]

\begin{align*} {\cal L} \left[ t\, y'' (t) \right] &= \int_0^{\infty} t\, y'' (t) \, e^{-\lambda \,t} {\text d} t = \left( t\, y' (t) \, e^{-\lambda \,t} \right)_{t=0}^{\infty} - \left[ y (t) \, \frac{\text d}{{\text d}t} \left( t\, e^{-\lambda \,t} \right) \right]_{t=0}^{\infty} + \int_0^{\infty} y(t) \, \frac{{\text d}^2}{{\text d}t} \left( t\, e^{-\lambda \,t} \right) ; \\ {\cal L} \left[ t\,y (t) \right] &= \int_0^{\infty} t\,y (t) \, e^{-\lambda \,t} {\text d} t = - \frac{\text d}{{\text d}\lambda}\, \int_0^{\infty} y (t) \,e^{-\lambda \,t} {\text d} t = - \frac{\text d}{{\text d}\lambda}\,{\cal L} \left[ y (t) \right] = - \frac{\text d}{{\text d}\lambda}\,y^L (\lambda ). \end{align*} The former requires more manipulations: Therefore, we get This leads to the differential equation of the first order for the Laplace transform of y(t) because we can set y(0) = 0. The general solution of the latter is known to be where C is a n arbitrary constant that we set C = 1. Expanding the Laplace transform of the Bessel function into infinite series using binomial theorem \( \left( 1 + z \right)^{\alpha} = \sum_{k\ge 0} \binom{\alpha}{k} z^k , \) we get Then term-by-term integration yields This power series converges for all t because the signs of successive terms alternate and the general term turns to zero. Its Laplace transform is represented by uniformly convergent integral Taking the derivative, we have From the Bessel equation \( t\, y'' (t) + y' (t) + t\, y (t) =0 , \) we derive Upon defining \( J_1 (t) = -J'_0 (t) , \) ,/span> we can show that this function is a solution of the Bessel equation of order 1: span class="math">\( t^2\, y'' (t) + t\,y' (t) + \left( t^2 -1 \right) y (t) =0 . \) Its Laplace transform becomes

For arbitrary positive integer n, we obtain the Laplace transform of J_n from the corresponding transformation of Bessel's differential equation:

We have Therefore, application of the Laplace transform to the Bessel equation gives the differential equation for the Laplace transform of the Bessel function: Making substitution \( \phi (\lambda ) = \left( 1 + \lambda^2 \right)^{1/2} J_n^L (\lambda ) , \) we obtain Then for new dependent variable, we get the differential equation which upon multiplication by an integrating factor \( \phi' \left( 1 + \lambda^2 \right)^{1/2} , \) results in the exact equation Integration yields where C is a constant of integration. Setting C = 0, we have Checking with knowing formula for J₁, we have to choose minus sign and get with integral Now we obtain explicit expression for the Laplace transform of Bessel's function of first kind (upon setting C = 1): For large λ, we have \( \displaystyle {\cal L} \left[ J_n (t) \right] (\lambda ) \sim \frac{1}{2^n \lambda^{n+1}} + O\left( \lambda^{-n-3} \right) , \]

\[ J_n (t) \sim \frac{t^n}{2^n n!} + O \left( t^{n+2} \right) . \]

\[ \int_0^{\infty} J_n (t) \,{\text d}t = 1 \qquad \mbox{and} \qquad \int_0^{\infty} t^{-1} J_n (t) \,{\text d}t = \frac{1}{n} \qquad (n=0,1,2,\ldots ). \]

\[ \frac{1}{\left( 1 + \lambda^2 \right)^{1/2} - \lambda} + \left( 1 + \lambda^2 \right)^{1/2} - \lambda = 2 \left( 1 + \lambda^2 \right)^{1/2} , \]

\[ J_{n-1}^L (\lambda ) + J_{n+1}^L (\lambda ) = 2n {\cal L} \left[ J_n (t)/t \right] \]

\[ t \left[ J_{n+1} (t) + J_{n-1} (t) \right] = 2n\, J_n (t) . \]

\[ Similarly, the identity yields because \( J_n (0) =0 \ (n=1,2,\ldots ). \)

Eliminating J_n+1(t) from the above recurrence relations, we obtain

On the other hand, elimination of J_n-1(t) yields Application of the above relation m times gives Putting n = 0 in the latter shows References