Solving Homogeneous System $R(r)=A_m\mathcal{J}_m(kr)+B_m\mathcal{Y}_m(kr)$

[Dustinsfl]

$R(r) = A_m\mathcal{J}_m(kr) + B_m\mathcal{Y}_m(kr)$
$$
\begin{pmatrix}
\mathcal{J}_m(ka) & \mathcal{Y}_m(ka)\\
\mathcal{J}_m(kb) & \mathcal{Y}_m(kb)
\end{pmatrix}
\begin{pmatrix}
A_{m}\\
B_{m}
\end{pmatrix} =
\begin{pmatrix}
0\\
0
\end{pmatrix}
$$
In order for our system to have a non-trivial solution, we require that the determinant be 0.
Therefore,
$$
\mathcal{J}_m(ka)\mathcal{Y}_m(kb) - \mathcal{J}_m(kb)\mathcal{Y}_m(ka) = 0
$$
which is our eigenvalue equation.

How do I get to this now:
$$
R(r) = A_m\left[\mathcal{J}_m(kr)-\frac{\mathcal{J}_m(ka)}{\mathcal{Y}_m(ka)} \mathcal{Y}_m(kr)\right]
$$

 

[jvicens]

To get to this expression, you can use the fact that the determinant of the matrix in the given equation is equal to 0. This means that either $\mathcal{J}_m(ka) = 0$ or $\mathcal{Y}_m(kb) = 0$.

Assuming $\mathcal{J}_m(ka) = 0$, we can substitute this into the eigenvalue equation to get:
$$
\mathcal{J}_m(ka)\mathcal{Y}_m(kb) - \mathcal{J}_m(kb)\mathcal{Y}_m(ka) = 0 \\
0 - \mathcal{J}_m(kb)\mathcal{Y}_m(ka) = 0 \\
\mathcal{Y}_m(ka) = 0
$$
Thus, $\mathcal{Y}_m(ka)$ must be equal to 0 for the determinant to be 0. This means that we can write our solution as:
$$
R(r) = A_m\mathcal{J}_m(kr) + B_m\mathcal{Y}_m(kr) = A_m\mathcal{J}_m(kr)
$$
since $\mathcal{Y}_m(kr)$ is 0.

Next, we can use the fact that $\mathcal{J}_m(ka) = 0$ to simplify our expression further. We can write $\mathcal{J}_m(kr)$ as:
$$
\mathcal{J}_m(kr) = \mathcal{J}_m(kr) - \mathcal{J}_m(ka) + \mathcal{J}_m(ka)
$$
We can then substitute this into our expression for $R(r)$ to get:
$$
R(r) = A_m\left[\mathcal{J}_m(kr) - \mathcal{J}_m(ka) + \mathcal{J}_m(ka)\right] = A_m\left[\mathcal{J}_m(kr) - \mathcal{J}_m(ka)\right]
$$
Finally, we can use the fact that $\mathcal{Y}_m(ka) = 0$ to simplify further and get the desired expression:
$$
R(r) = A_m\left[\mathcal{