Consider the following set of equations $$ \sum_{k=-\infty}^{\infty} \left[ \psi_k(\lambda) \left( \lambda I_k'(\lambda R) \cos k\phi \cos \phi + \frac{k}{R} I_k(\lambda R) \sin k\phi \sin \phi \right) + f_1 (\lambda,k) \cos k \phi \right] = 0 \, , \\ \sum_{k=-\infty}^{\infty} \left[ \psi_k(\lambda) \left( \lambda I_k'(\lambda R) \cos k\phi \sin \phi - \frac{k}{R} I_k(\lambda R) \sin k\phi \cos \phi \right) + f_2 (\lambda,k) \sin k \phi \right] = 0 \, , $$ where $I_k$ is the modified Bessel function and $\psi_k(\lambda)$ is a real function of the variable $\lambda$. Here $f_1$ and $f_2$ are two real functions and $R>0$. Prime denotes derivative with respect to the argument. My goal is to rearrange and equate the Fourier coefficients in order to get a recurrence equation (i.e. the summation sign should disappear.) It would be great if someone could provide with useful ideas that help. Thanks. Best
r
the idea is quite simple. All what you need is to to apply the following trigonometric relations $$ \cos k\phi \cos \phi = \frac{1}{2} \left( \cos((k-1)\phi) + \cos((k+1)\phi) \right) \, , \\ \sin k\phi \sin \phi = \frac{1}{2} \left( \cos((k-1)\phi) - \cos((k+1)\phi) \right) \, , $$ and $$ \cos k\phi \sin \phi = \frac{1}{2} \left( \sin((k+1)\phi) - \sin((k-1)\phi) \right) \, , \\ \sin k\phi \cos \phi = \frac{1}{2} \left( \sin((k+1)\phi) + \sin((k-1)\phi) \right) \, . $$
Then try to express the first equation based on $\cos k\phi$ and the second equation on $\sin k\phi$. I.e. make the change of summation index $k$ accordingly to obtain the following equations $$ \frac{\lambda}{2} I_k (\lambda R) \psi_{k+1} (\lambda) + \frac{\lambda}{2} I_k (\lambda R) \psi_{k-1} (\lambda) + f_1 (\lambda, k) = 0 \, , \\ -\frac{\lambda}{2} I_k (\lambda R) \psi_{k+1} (\lambda) + \frac{\lambda}{2} I_k (\lambda R) \psi_{k-1} (\lambda) + f_2 (\lambda, k) = 0 \, , $$ which leads immediately to $$ \lambda I_{k+1}(\lambda R) \psi_{k} = -(f_1(\lambda,k) + f_2(\lambda,k)) \, , \\ \lambda I_{k-1}(\lambda R) \psi_k = -(f_1(\lambda,k) - f_2(\lambda,k)) \, . $$
Cheers