Reputation: 21
I am solving Nonlinear Schrodinger equation by split-step Fourier method: i df/dz+1/2* d^2f/dX^2+|f|^2*f=0
using an initial condition: f=q*exp(-(X/X0)^24).
But I have to use the condition that q=1 for |x|<1, otherwise, q=0. So I write the following subroutine (excerpt of the code involving transverse variable) for transverse variable x:
fs=120;
N_fx=2^11; %number of points in frequency domain
dX=1/fs;
N_X=N_fx;
X=(-(N_X-1)/2:(N_X-1)/2)*dX;
X0=1;
Xn=length(X);
for m=1:Xn
Xnn=Xn/8;
pp=m;
if pp>3*Xnn && pp<5*Xnn
q=1.0;
f=q*exp(-(X/X0).^24);
else
f=0;
end
end
But it seems that 'f' is getting wrong and it is a 1 by 2048 vector with all entries zero. I'm not getting the expected result. If my initial condition is only f=q*exp(-(X/X0).^24), q=1, it is straightforward, but with the above condition (q=1 for |x|<1, otherwise, q=0) what I need to do? Any help will be much appreciated. Thanks in advance.
Upvotes: 1
Views: 130
Reputation: 1572
A MWE, this has 0 < [f(867) : f(1162)] <= 1
:
fs=120;
N_fx=2^11; %number of points in frequency domain
dX=1/fs;
N_X=N_fx;
X=(-(N_X-1)/2:(N_X-1)/2)*dX;
X0=1;
Xn=length(X);
f = zeros(1,Xn); % new: preallocate f size and initialise it to 0
for m=1:Xn
Xnn=Xn/8;
if m>3*Xnn && m<5*Xnn
%if abs(X(m)) < 1 %alternative to line above
q=1.0;
% error was here below: you overwrote a 1x1 f at each iteration
f(m)=q*exp(-(X(m)/X0).^24);
else
f(m)=0;
end
end
Upvotes: 1