Error using vertcat; Dimensions of arrays being concatenated are not consistent.

I have this kind of error at line 64. The last plot. Someone could help me to fix it?

clc; clear all; close all;
% parametri inziali
N=30;
L=1;
x=linspace(0,L,N+1); 
h=x(2)-x(1);
dx=x(2)-x(1);
% condizione iniziale
phi0=sin(2*pi*x./L);
% parametri
k=1;
b=5;
beta=1.5;
dt=(beta*h*h)/k;
Nt=100;
Tf=Nt*dt;
t=linspace(0,Tf,Nt+1);
% plot([x,L],[phi0,phi0(1)]);
% costruisco i pesi
i=1;
xc=x(i);
wdd=PesiDer(x(1:4),xc,2);
w=PesiDer(x(1:4),xc,0);
switch i
    case 1
        v2=zeros(1,N);
        v2(1)=wdd(1); v2(2)=wdd(2); v3(3)=wdd(3); v2(4)=wdd(4);

      case 2
          v2=zeros(1,N);
          v2(1)=wdd(2); v2(2)=wdd(3); v3(3)=wdd(4); v2(end)=wdd(1);

      case 3
          v2=zeros(1,N);
          v2(1)=wdd(3); v2(2)=wdd(4); v3(end-1)=wdd(1); v2(end)=wdd(2);

      case 4
          v2=zeros(1,N);
          v2(1)=wdd(4); v2(end-2)=wdd(); v3(end-1)=wdd(2); v2(end)=wdd(3);

end
% matrice spaziale
v=zeros(1,N);
v(1)=1;
I=gallery('circul',v);
A=-k*gallery('circul',N,v2)+beta*gallery('circul',N,v);
% matrice temporale
dt=(beta*h*h)/k;
Nt=100;
Tf=Nt*dt;
B=I-.5*A*dt;
C=I+.5*A*dt;
T=B\C;
phi=phi0(1:end-1)';
% soluzione analitica
phia=zeros(Nt,N);
for ite=1:Nt
    for isp=1:N
        phia=exp((-(b+k)*(2*pi/L)^2)*t(ite))*sin(x(isp));
    end
end
for it=2:Nt+1
    phin=T*phi; %soluzione numerica
    plot([x;x(end)+dx],[phi0;phi0(1)],'.-r',[x;x(end)+dx],[phi;phi(1)],'g',[x;x(end)+dx],[phia(it,:);phia(it,1)],'y');
end

[Adam Danz]

Please format all of your code by editing it and using the {} button.

[Adam]

Just use

doc size

on the command line with a breakpoint at the line in question and you will instantly see the size of the two things you are attempting to concatenate, e.g.

size( phia(it,:) )

[Guillaume]

I've formatted the code for you. It would be useful if you told us which is line 64.

Answer 1

[Guillaume]

What is the intent of

[phia(it,:);phia(it,1)]

in your plot line.

phia(it, :) implies that phia has more than one column since it asks for all columns of row it. phia(it, 1) has of course only one column. Therefore vertically concatenating something with more than one column with something with only one column is always going to fail.

The only way the above can work is if phia has only one column, i.e. is a vector, in which case 2d indexing is misleading and the above could be simply written as

[phia(it); phia(it)]  %not likely to be what you meant.