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I try to solve numerically the following PDE for $E(r, z)$ with a cylindrical symmetrie (i. e. $E(r, z) = E(-r, z)$).

$\frac{\partial E}{\partial z} = \frac{i}{2k} \Delta E + \mathcal{N}(E)$

Where $\Delta$ is the Laplace operator in transversal direction and $k$ a real number. I want to use the Crank-Nicolson scheme to treat the Laplace operator and the Adams-Bashforth scheme to treat the nonlinearity ($\mathcal N(E)$). $r$ is defined by $r_j = 0 + j\Delta r$ for $j = 0 \dots N$. Since the Laplace in cylinder coordinates is given by: $\Delta E = \frac{\partial^2E}{\partial r^2} + \frac{1}{r}\frac{\partial E}{\partial r} $ one gets the following representation of the Laplace operator:

$\Delta E_j^n = E^n_{j-1} -2 E^n_j + E^n_{j+1} + \frac{1}{2j}(E^n_{j+1} - E^n_{j-1})$

The given PDE equation is therefore given by the following, where the nonlinearity is treated by the Adams-Bashforth scheme.

$E_j^{n+1} - E_j^n = i\delta (\Delta_j E^{n+1}_j + \Delta_j E^{n}_j) + (3/2 \mathcal N^n_j - 1/2 N^{n-1}_{j})$

where $\delta = \frac{\Delta z}{4 k \Delta r^2}$. From this $E^{n+1}_j $ can be expressed the following way:

$E^{n+1}_j = L_-^{-1} [L_+E^n_j + 3/2\mathcal N^n_j - 1/2\mathcal N^{n-1}_j]$ with the following matrixes $L_-$ and $L_+$.

$ L_{\pm} = \left( \begin{array}{rrrr} 1\mp2i\delta & \pm i\delta v_0 \\ \pm i\delta u_1 & 1\mp2i\delta & \pm i\delta v_1\\ & & & & \\ & & & & \\ & & & \pm i\delta u_N & 1\mp2i\delta\\ \end{array}\right) $

Where $u_j = 1 - 1/(2j)$ and $v_j = 1 + 1/(2j)$.

For my problem the following boundary conditions are given: $ E(r = r_{max}, z) = 0 $ and $\frac{\partial E}{\partial r} |_{r=0}$. The first one for $r=r_{max}$ I can easily incorporate by changing the last row of $ L_{\pm} $.

The other one ($\frac{\partial E}{\partial r} |_{r=0}$) gives me trouble. I somehow have to change the entries of the first row of $L_{\pm}$, but I don't know how. I got some working boundary conditions for the case the nonlinearity is 0, but they break as soon as I add a nonlinearity.

I am greatful for any help.

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