Numerical solution of sine-Gordon using RK4

I would like to ask you for help with the program.

I am trying to write a program to solve the sine-Gordon equation (with disipation) using the RK4 method.

The equation is:

The initial conditions are

,

And the boundary conditions are

,

.

I wrote this equation as a system of equations:

,

.

However, I stuck on preparing the code. I don't know if I should prepare a separate matrix for the derivative (in the code ddxx_u)? I can't figure out how to run this. May I ask you for help with this code?

clc;  
clear all;

v=0.5;
x0=0;
a=0.01;

dx=0.1;               
x = -20:dx:20; 
dt=0.1;
t = 0:dt:10;

u = zeros(length(t),length(x)); 
du = zeros(length(t),length(x)); 

%initial condition

u(1,:) = 4*atan(exp((x-x0)/sqrt(1-v^2)));
du(1,:) = -(2*v/sqrt(1-v^2))*sech((x-x0)/sqrt(1-v^2));

%ddxx_u?


F_xyz = @(du) du; 
G_xyz = @(ddxx_u,u,du) ddxx_u-sin(u)-a*du;

for i=1:(length(t))
    for j=1:(length(x))
        
        %boundary conditions
        if (j==1)
            u(i,j)=0;
        elseif (j==length(x))
            u(i,j)=2*pi;
        end
        
        %ddxx_u?

        k_1 = F_xyz(du(i,j));
        L_1 = G_xyz(ddxx_u(i,j),u(i,j),du(i,j));

        k_2 = F_xyz(du(i,j)+0.5*dt*k_1);
        L_2 = G_xyz(ddxx_u(i,j)+0.5*dt,u(i,j)+0.5*dt*k_1,du(i,j)+0.5*dt*L_1);


        k_3 = F_xyz(du(i,j)+0.5*dt*k_2);
        L_3 = G_xyz(ddxx_u(i,j)+0.5*dt,u(i,j)+0.5*dt*k_2,du(i,j)+0.5*dt*L_2);

        k_4 = F_xyz(du(i,j)+0.5*dt);
        L_4 = G_xyz(ddxx_u(i,j)+0.5*dt,u(i,j)+0.5*dt,du(i,j)+0.5*dt);

        u(i,j+1) = u(i,j) + (1/6)*(k_1+2*k_2+2*k_3+k_4)*dt;
        du(i,j+1) = du(i,j) + (1/6)*(k_1+2*k_2+2*k_3+k_4)*dt;
    end 
end

[Torsten]

This is a partial differential equation that depends on t and x as independent variables.

Look up "method-of-lines" to see how to solve it.