Matlab Not enough input arguments [closed] - matlab

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A nonlinear code
% This produces a black waterfall plot of the solution of
% u_t + 6u^2u_x + u_xxx - (b(x,t)u)_x = 0 on [-pi,pi]
% and returns the the time vector to use for efft.m
% The code is a simple modification of Trefethen's example p27.m
% The initial data is given by a double soliton with parameters
% a1, a2 (centers) and c1, c2 (velocities).
% The arguments of mkdvb.m are time t, parameters given in a vector form
% [a1,a2,c1,c2], a function bb = b(x,t), and the log_2 of the number of
% x grid points (the default value is 8).
% It return the time vector T, the position vector X, and the matrix of
% solution values at times given by T.
%
% Here is an example
%
% T = mkdvB(#(x,t) 100*cos(x-10^3*t).^2-50*sin(2*x+10^3*t), 0.05, [1,-1,4,-11],9)
%
% The time vector T can then be entered into effdyn.m code to compare the
% solution to the effective dynamics:
%
% [T,Y]=effdyn(#(x,t) 100*cos(x-10^3*t).^2-50*sin(2*x+10^3*t), T, [1,-1,4,-11])
%
function [T,X,U] = mkdvb2(fun,tmax,ZZ,N)
if (nargin < 2 )
tmax=0.065;
end
if (nargin < 3)
ZZ = [-1,-2,-6,9];
end
if (nargin < 4)
N = 8;
end
N = 2^N;
% Set up grid and two-soliton initial data:
dt = .4/N^2; x = (2*pi/N)*(-N/2:N/2-1)';
drawnow, set(gcf,'renderer','zbuffer')
u=sqrt(2)*qu(x,ZZ);
v = fft(u); k = [0:N/2-1 0 -N/2+1:-1]'; ik3 = 1i*k.^3;
% Solve PDE and plot results:
nplt = floor((tmax/25)/dt); nmax = round(tmax/dt);
udata = u; tdata = 0; h = waitbar(0,'please wait...');
for n = 1:nmax %#ok<ALIGN>
t = n*dt; g = -1i*dt*k;
qz=fun(x,t);
E = exp(dt*ik3/2); E2 = E.^2;
a = g.*fft(real( ifft( v ) ).^3 - ifft( v ).*qz);
b = g.*fft(real( ifft(E.*(v+a/2)) ).^3 - ifft(E.*(v+a/2)).*qz);
c = g.*fft(real( ifft(E.*v + b/2) ).^3 - ifft(E.*v + b/2).*qz);
d = g.*fft(real( ifft(E2.*v+E.*c) ).^3 - ifft(E2.*v+E.*c).*qz);
v = E2.*v + (E2.*a + 2*E.*(b+c) + d)/6;
if mod(n,nplt) == 0
u = real(ifft(v)); waitbar(n/nmax)
udata = [udata u]; tdata = [tdata t];
end
end
udata = udata/sqrt(2);
hold on
[~,P] = size(tdata);
z1=0;
z2=0;
for j=1:P
plot3(x,tdata(j)+0*x, udata(:,j) );
z1=max(z1,max(udata(:,j)));
z2=min(z1,min(udata(:,j)));
end
T = tdata;
U = udata;
X = x;
view(0,60)
xlabel x, ylabel t, axis([-pi pi 0 tmax z2 z1]), grid off
close(h)
function qu = qu(x,ZZ)
k1 = ZZ(3)*(x - ZZ(1));
k2 = ZZ(4)*(x - ZZ(2));
ga1 = exp(-k1);
ga2 = - exp(-k2);
m11=(1+ga1.^2)./(2*ZZ(3));
m22=(1+ga2.^2)./(2*ZZ(4));
m12=(1+ga1.*ga2)./(ZZ(3)+ZZ(4));
M=m11.*m22 - m12.^2;
M1=ga1.*(m12 - m22) + ga2.*(m12-m11);
qu = M1./M;

Answer 1

The code is a bit messy, but fortunately this kind of problem is very easy to troubleshoot.
Save your code in a function or script
turn on dbstop if error
See where the error occurs
You should now be on a line like
[y1, y2] = f(x1,x2,x3)
Try to evaluate all input arguments, at least 1 of them will not exist (yet?).