we claim that the resolvent is given by $$\label{G0uv}
G_{nm}=-\dfrac{u_{n}\,v_{m}\Theta(m-n)+u_{m}\,v_{n}\Theta(n-m)}{W_{n}}.$$ Here $\Theta(n)$ is the unit-step function and $W_{n}=u_{n}v_{n-1}-u_{n-1}v_{n}$ is the Wronskian. Using Eqs. (\[un\], \[vn\]) it is straight forward to see that the Wronskian is independent of $n$. One can also check that $$G_{n+1m}+G_{n-1m}+z\,G_{nm}=\delta_{nm}.$$ We now set $u_{n}=v_{N+1-n}$ to match the initial value problem (\[un\], \[vn\]) to the boundary value problem (\[ODE\]). We find $$\label{Gun}
G_{11}=-\dfrac{u_{N}}{u_{N+1}}.$$ The ODE (\[un\]) is satisfied by the Chebyshev polynomial of the second kind, $U_{m}(-z/2)$, defined as $$U_{m}(x)=\dfrac{\sin[(m+1)\cos^{-1}(x)]}{\sin[\cos^{-1}(x)]},$$ for $U_{0}(x)=1$ and $U_{1}(x)=2x$. Since we have fixed $u_{1}=1$, therefore $u_{n}=U_{n-1}$, thus we can write Eq. (\[Gun\]) as $$\label{GF}
G_{11}=-\dfrac{U_{N-1}(-z/2)}{U_{N}(-z/2)}=-\dfrac{\sin[N\,k]}{\sin[(N+1)k]}.$$ Here the last equality follows from the energy dispersion relation. Using Eq. (\[GF\]) in Eq. (\[charcpol\]), we end up with an algebraic equation $$\label{fincharpol}
F(z)
=0=
1-\eta\lambda\dfrac{U_{N-1}(-z/2)}{U_{N}(-z/2)}.$$

![Repeated on the same pattern of Fig. \[ansatz-exact\] but for parametric resonances.[]{data-label="ansatz-parametric"}](ansatz-parametric.eps){width="75.00000%"}

Zeros of $F(z)$ are the roots of a polynomial of order $N$. For exact resonances Eq. (\[fincharpol\]) can be easily transformed into $$\label{Az}
[\mathsf{a}(z)]^{(2N+1)}=
\dfrac{\mathsf{a}(z)^{-1}-\eta\,\mathsf{a}(z)}{1-\eta},$$ where $$\label{Aexp}
\mathsf{a}(z)=-\exp[ik(z)].$$ In order to solve Eq. (\[Az\]), we propose an ansatz assuming that opening of the system at one end causes $\mathcal{O}(N^{-1})$ complex corrections to the $k_{\alpha}$’s. Let $$\label{ansatz}
 \tilde{k}_{\alpha}=k_{\alpha}+\dfrac{\Phi_{\alpha}}{N},$$ where 