Question

非线性代码

    % 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

代码有点乱，但幸运的是这种问题很容易排除故障。

将代码保存在函数或脚本中
开启dbstop if error
查看错误发生的位置

你现在应该像

[y1, y2] = f(x1,x2,x3)

尝试评估所有输入参数，其中至少有一个不存在（还有？）。

Matlab没有足够的输入参数

1 个答案: