Fourier-Bessel series with different argument

Question

Trying to solve the PDE associated to a free hangin chain $$u_{tt}=g(xu_{xx}+u_x),$$ with the boundary conditions $|u(0,t)|<\infty$, $u(L,t)=0$, $u(x,0)=a(L-x)^2=f(x)$ and $u_t(x,0)=0$ i came up with the following situation, $$f(x)=\sum_{n=1}^\infty A_n J_0(2\lambda_n \sqrt{x}),$$ where $\lambda_n=\frac{\gamma_n}{2\sqrt{L}},$ and the $\gamma_n$ are the zeros of $J_0(z)$. I know about the Fourier-Bessel series and that the $A_n$ must be the coefficents of such an expansion for the function $f(x)$, but the argument $2\lambda_n \sqrt{x}$ is troubling me, how can i adapt the orthogonalit relationships of the Bessel functions to this case? Any help will be very appreciated!

Answer 1

Start with orthogonality property of $J_0(\gamma_k x)$

$$ \int\limits_{0}^{1} xJ_0(\gamma_k x)J_0(\gamma_n x)dx = \cases{ 0, & $k\neq n$,\\[2ex] \frac{1}{2}J_1^2(\gamma_k), & $k = n$. } $$

Then use following change of variables

$$ x = \sqrt\frac{\xi}{L}, \quad dx = \frac{1}{2\sqrt{L}}\frac{d\xi}{\sqrt{\xi}}, \quad \xi = Lx^2. $$

$$ \int\limits_{0}^{1} xJ_0(\gamma_k x)J_0(\gamma_n x)dx = \frac{1}{2L}\int\limits_{0}^{L} J_0\left(\gamma_k\sqrt\frac{\xi}{L}\right) J_0\left(\gamma_n\sqrt\frac{\xi}{L}\right)d\xi = \\ \frac{1}{2L}\int\limits_{0}^{L} J_0\left(2\lambda_k\sqrt{\xi}\right) J_0\left(2\lambda_n\sqrt{\xi}\right)d\xi. $$ And the result is

$$ \int\limits_{0}^{L} J_0(2\lambda_k \sqrt{\xi})J_0(2\lambda_n \sqrt{\xi})d\xi = \cases{ 0, & $k\neq n$,\\[2ex] LJ_1^2(2\sqrt{L}\lambda_n), & $k = n$. } $$

So the formula for $A_n$ is

$$ A_n = \frac{1}{LJ_1^2(2\sqrt{L}\lambda_n)}\int\limits_{0}^{L}f(\xi)J_0(2\lambda_n\sqrt{\xi})d\xi. $$