# Laplace convolution with the Bessel function

The Question:

(i) Find the Laplace Transform of the Bessel Function $J_0(x)$

(ii) Hence, show that if $f(x)$ satisfies the differential equation

$$f''(x)+f(x)=J_0(x) \qquad f(0)=f'(0)=0$$

then $f$ is given by

$$f(x) = \int_0^x tJ_0(t)dt$$

(i) I found $\overline{J_0}(p) = \dfrac{1}{\sqrt{1+p^2}}$ which is definitely correct

(ii) I Laplace Transformed both sides of the equation and applied the boundary conditions to get

$$\bar f(p) = \frac{\overline{J_0}(p)}{1+p^2} = \big(1+p^2\big)^{-3/2}$$

It is not immediately clear how to Laplace Invert $(1+p^2)^{-3/2}$ and it probably won't give the required integral in terms of $J_0$ anyway.

Instead, I know that the Laplace Transform of $\sin(x)$ is precisely $(1+p^2)^{-1}$, i.e. we have

$$\bar f(p) = \overline {J_0} (p) \overline{\sin}(p)$$

So by the convolution theorem,

$$f(x) = \big(J_0 * \sin \big)(x) = \int_0^x \sin (x-t) J_0(t)dt$$

and I don't see how in the world you can make the $\sin(x-t)$ a $t$ as required.

Have I done something wrong, or is there a Bessel function identity I am unaware of?

• Ping me When answered. +1. – mick Jun 2 '18 at 23:47
• *ping ping ping* – glowstonetrees Jun 3 '18 at 12:20

The calculations are correct. In addition, you mat recognize that \begin{equation} \frac{1}{\left( p^2+1 \right)^{3/2}}=-\frac{1}{p}\frac{d}{dp}\frac{1}{\sqrt{p^2+1}} \end{equation} Then, using successively the frequency domain derivative and the time domain integration formulae for the Laplace transform, you find the expected result. In fact, you derived the general result \begin{equation} \int_0^x \sin (x-t) J_0(t)\,dt= \int_0^x tJ_0(t)\,dt \end{equation} Moreover, using the properties of the Bessel functions, it can be shown that $f(x)=xJ_1(x)$.