# Mode distribution and Phase constant for nonlinear optical fiber

While going through the derivation of Nonlinear Schrodingers equation from the Helmholtz equation I came across the following two pdes,

$\nabla_\perp^2 F + \left[ \epsilon\left(\omega\right) k_0^2 - \tilde{\beta}^2 \right] F = 0$

$2i\beta_0 \frac{\partial\tilde{A}}{\partial z} + \left(\tilde{\beta_0}^2 - \beta_0^2\right) = 0$

where $F$ and $\tilde{A}$ is part of the Fourier transform of the electric field $E$, centered at $\omega_0$, given by

$\tilde{E}\left(\mathbf{r},\omega - \omega_0\right) = F\left(x,y\right) \tilde{A} \left(z,\omega - \omega_0 \right) \mathrm{exp}\left(i \beta_0 z \right)$.

Here the nonlinearity comes from the $\epsilon\left(\omega \right)$ term, which is given by

$\epsilon = \left(n + \Delta n \right)^2 \approx n^2 + 2n \Delta n$

and $\Delta n$ is given by

$\Delta n = \bar{n}_2 \left|E \right|^2 + \frac{i\bar{\alpha}}{2 k_0}$

Here $n$ is the refractive index and $\Delta n$ is small perturbation to $n$, $\bar{n}_2$ is the nonlinear component of the electric field and $\tilde{\alpha}$ is the absorption constant.

What my doubt is that how you solve the pde for $F$. In text(Nonlinear Fiber optics by Govind Agrawal) they are solving it using the first order perturation theory. I tried reading up perturbation theory but I was not able to apply it for solving this pde.

First if I ignore the perturbation, the solution for pde fo $F$ is the normal fiber mode distributions. When the effects of $\Delta n$ is included, the text says that the mode distribution $F(x,y)$ remains unchanged and only the eigen values of $\tilde{\beta}$ changes.

How can I get solution for $F$ and $\beta(\omega)$ ?