Hi verybody, I wish I have plot a figure using a part of data avaible.

I am solving a nonlinear equation with a MATLAB code using the split-step fourier method. I wish I plot a 3D figure where the spatial extent of the grid is [-150 150], but the solution is computed on the spatial extent [-256000 255000] but I am unable. Here is my code: clear all; close all; clc;

g1D=0.; chi=1; epsi=0.2; omega=0.001; alp = 3/1280; epsiz=1;

V10=1; sigma = 1.47; L1 = 10*sigma; % external trapping potential parameters

beta2 = 1; gamma = 1; epsi = 0.50000e-012; P0 = 1;

L= 41; % Longueur de la fibre (dns NLSE duree max de propagation Lstep = 0.05e3; % pas arbitraire

fmax=0.001; % frequence maxi observable dx = 1/fmax; % pas temporelle n = 1; T0 = 0; while T0 < 500*dx % d?nition de la fenetre temporelle n = n+1; nt= 2^n; % nombre de points sur l'axe des temps

   T0 = dx*(2^n);
end

df = 1/T0; % pas en fr?ence dw = 2*pi*df; % pas en pulsation n1 = nt/2; % definition de l'axe des temps n2 = n1-1; N1 = n1 + n2 + 1; x = (-n1:n2)*1*dx; % axe des espaces dans NLSE, pts espacés de dx de -n1*dx à (n1-1)*dx

% dénition de l'axe des temps

w = (-n1:n2)*dw; %definition de l'axe des pulsations, dns NLSE, w= nbre d'onde f = w/(2*pi); %definition de l'axe des pulsations w2 = w.*w;

%%%% CHAMP INITIAL dénition de l'onde continue initiale: solution stationnaire mu = 2.8; c0 = 0; K = 1/12; delta = g1D^2-4*chi*(V10-mu); u0 = sqrt((-g1D+sqrt(delta))/(2*chi)); Vg = u0*sqrt(g1D+2*chi*u0^2);

b1 = 3/4+chi*u0^4/Vg^2; b2 = 1/(8*Vg); c11 = -6*b2*K/b1; d11 = 0; beta1 = -4*b2*K^3; X = x-Vg*0; eta1 = c0+c11*tanh(K*X-beta1*0); eta1X = c11*(1-tanh(K*X-beta1*0).^2)*K; U0 = u0*(1+eta1X/(2*Vg)).*exp(1i*(-mu*0+eta1))+10^-5*(rand(1,N1)-0.5);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% External field

Pot=V10*(exp(-(x+L1/2).^2/sigma^2)+exp(-(x-L1/2).^2/sigma^2)); % est donné pour tous les xi car x est 1 vect %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

V0 = fftshift(ifft(U0)); % Transformee de Fourier la transformee inverse est utilise coe transf directe est 1 vect qui fourier trans en tous les xi % V0p = fftshift(ifft(U0p)); % V est un vect de dim dim(x) % V0pp = fftshift(ifft(U0pp));

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

t = 0; % a t =0 on va initialiser U, t est le temps dns NLSE U = U0; % initialisation de U

V = V0; % initialisation de V %%%% PAS D'INTEGRATION SUIVANT Z (dns NLSE Z est plutot t)

L_NL = 1/(gamma*P0); % longueur nonlin?re % in O.F L_D = T0^2/abs(beta2); % longueur dispersion % in O.F Lmin = min([L_NL L_D Lstep]); dt = Lmin/500; %(pas temporel dns NLSEà % pas suivant z in F.O1

npt = round(L/dt); G = []; % gain numerically GdB = []; % gain numerically ZZ = []; PP0 = [];

DISP =exp(-1i*0.5*dt*w.^2); %terme dans l'espace spectral (integrale de la part dispersive)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

     TabU = []; TabUmax = []; %initialisation des grandeurs à représenter
     TabV = []; 
     TabG = [];  %NUMerical checking of gain 
     Tabt = []; Tabx = [];
     nplot=50; npt1=npt/nplot; %nplot est le nbre de courbes npt1 est l'espace entre les courbes

for k = 1:nplot

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

    for I = 1:npt1

    P = (abs(U)).^2;          %P module de psi au carré

    SPM=Pot+P*g1D+chi*P.^2; %  /lamda terme de nonlinéarité + potentiel(espace reel des x)
    U = U.*exp(-1i*dt*SPM); %soltion de la part nonlin, produit solution initial+terme de 
    V = fftshift(ifft(U));  % ifft(U) est la FFT (passage dns l'esp spectrale de la solution 
    V = V.*DISP; % produit des solutions des part nonlin et lin dns l'esp spectral  
    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

    U = fft(fftshift(V)); % fft est la IFFT (solution num dns l'esp reel)
    Umax=max(U);

    t=t+dt;
    end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

     TabU = [TabU; (abs(U)).^2]; % dans TabU on met valeur absolue de U solution espace reel 
     TabV = [TabV; abs(V)];  % solution espace spectral
     TabUmax = [TabUmax; abs(Umax)];    
     Tabt = [Tabt; ones(1,nt)*t]; % sauvergarde du temps
     Tabx=[Tabx; x.*ones(1,nt)]; % alp = half_spatial_extension/(nt*dx)
end

figure(1) mesh(Tabx,Tabt,TabU) grid off colorbar view(0,90) xlabel('x', 'fontsize', 30) % , 'fontsize', 30 ylabel('t') zlabel('|\Psi(x,t)|^2', 'fontsize', 30)

figure(2) plot(Tabx,TabU(10,:), 'red', 'LineStyle', ':', 'LineWidth',6); xlabel('x', 'fontsize', 30) ylabel('|\Psi(x,t=10)|^2', 'fontsize', 30)

One should notice that the code gives a figure (the one of figure(1)) on the spatial extent [-256000 255000], but the solution width is too tiny because its width is 12. That's why I need to represent on the spatial extent [-150 150] in order to have a correct view of the solution. Thank you for your help.

[Image Analyst]

Either read this : formatting code

or else just attach your m-file with the paper clip icon. The post is unreadable as is.