Index exceeds the number of array elements. Index must not exceed 1.

Question

nune pratyusha 2022-5-24

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1726425-index-exceeds-the-number-of-array-elements-index-must-not-exceed-1

评论： nune pratyusha 2022-5-25

FDE12.m

function run_LE_FO_a(ne,ext_fcn,t_start,h_norm,t_end,x_start,h,q,a_min,a_max,n);
figure();
hold on;
ne=4;
ext_fcn=@LE_RF_a;
t_start=0;h_norm=0.02;t_end=1;
x_start=[0 0 1 1];h=0.2;q=1;
a_max=-11;
a_min=-12.2;
n=800;
p_step=(a_max-a_min)/n
a=a_min;
% for  p1=1.1:0.1:1.3;
while a<=a_max
    lp=FO_Lyapunov_a(ne,ext_fcn,0,0.2,1,[0 0 1 1],0.2,1,a)
    a=a+p_step;
    plot(a,lp,'.')
    drawnow();
    % end
end
function LE=FO_Lyapunov_a(ne,ext_fcn,t_start,h_norm,t_end,x_start,h,q,a);
% Memory allocation
x=zeros(ne*(ne+1),1); 
x0=x;
% y0=x_start
c=zeros(ne,1);
gsc=c; zn=c;
n_it = round((t_end-t_start)/h_norm);
% Initial values
x(1:ne)=x_start;
i=1;
while i<=ne
    x((ne+1)*i)=1.0;
    i=i+1;
end
t=t_start;
% Main loop
it=1;
while it<=n_it
    % Solutuion of extended ODE system of FO using FDE12 routine
    [T,Y] = FDE12(q,ext_fcn,t,t+h_norm,[0 0 1 1],h,a); 
    t=t+h_norm;
    Y=transpose(Y);
    x=Y(size(Y,1),:); %solution at t+h_norm
    i=1;
    while i<=ne
        j=1;
        while j<=ne;
            x0(ne*i+j)=x(ne*j+i);
            j=j+1;
        end;
        i=i+1;
    end;
    %       construct new orthonormal basis by gram-schmidt
    zn(1)=0.0;
    j=1;
    while j<=ne
        zn(1)=zn(1)+x0(ne*j+1)^2; 
        j=j+1;
    end;
    zn(1)=sqrt(zn(1));
    j=1;
    while j<=ne
        x0(ne*j+1)=x0(ne*j+1)/zn(1); 
        j=j+1;
    end
    j=2;
    while j<=ne
        k=1;
        while k<=j-1
            gsc(k)=0.0;
            l=1;
            while l<=ne;
                gsc(k)=gsc(k)+x0(ne*l+j)*x0(ne*l+k);
                l=l+1;
            end
            k=k+1;
        end
        k=1;
        while k<=ne
            l=1;
            while l<=j-1
                x0(ne*k+j)=x0(ne*k+j)-gsc(l)*x0(ne*k+l);
                l=l+1;
            end
            k=k+1;
        end;
        zn(j)=0.0;
        k=1;
        while k<=ne
            zn(j)=zn(j)+x0(ne*k+j)^2;
            k=k+1;
        end
        zn(j)=sqrt(zn(j));
        k=1;
        while k<=ne
            x0(ne*k+j)=x0(ne*k+j)/zn(j);
            k=k+1;
        end
        j=j+1;
    end
    %       update running vector magnitudes
    k=1;
    while k<=ne;
        c(k)=c(k)+log(zn(k));
        k=k+1;
    end;
    %       normalize exponent        
    k=1;
    while k<=ne
        LE(k)=c(k)/(t-t_start);
        k=k+1;
    end
    i=1;
    while i<=ne
           j=1;
            while j<=ne;
                x(ne*j+i)=x0(ne*i+j);
                j=j+1;
            end
            i=i+1;
        end;
        x=transpose(x);
        it=it+1;
end
function f=LE_RF_a(t,x,a)
%p is the parameter
f=zeros(16,1);
b=0.4;c=11;d=6;e=130;f1=10;
x=X(1); y=X(2); z=X(3);w=X(4);
Y= [X(5), X(9), X(13),  X(17);
    X(6), X(10), X(14), X(18);
    X(7), X(11), X(15), X(19);
    X(8), X(12),  X(16), X(20) ];
f(1)=a*y+(0.2+0.2*(abs(w))).*z;
f(2)=b*((w.^2)-13).*z-c*y;
f(3)=-d*x-e*y-f1*z;
f(4)=(z.^2)-w.^2;%Linearized system
 Jac=[0 a 0.2+0.2*abs(w) 0.2*w.*z/abs(w);
     0 -c b*((w.^2)-13) b*((w*2)-13).*z;
     -d -e -f1 0;
     0 0 2*z -2*w];
  
f(5:20)=Jac*Y;

error:>> run_LE_FO_a

p_step =

1.499999999999999e-03

Index exceeds the number of array elements. Index must not exceed 1.

Error in LE_RF_p1 (line 4)

X= [x(4) x(7) x(10);

Error in FDE12>f_vectorfield (line 300)

f = feval(Probl.fdefun,t,y,Probl.param) ;

Error in FDE12 (line 114)

f_temp = f_vectorfield(t0,y0(:,1),Probl) ;

Error in FO_Lyapunov_a (line 21)

[T,Y] = FDE12(q,ext_fcn,t,t+h_norm,[0 0 1 1],h,a);

Error in run_LE_FO_a (line 16)

lp=FO_Lyapunov_a(ne,ext_fcn,0,0.2,1,[0 0 1 1],0.2,1,a)

>>

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Answer 1

Cris LaPierre 2022-5-24

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1726425-index-exceeds-the-number-of-array-elements-index-must-not-exceed-1#answer_971040

在 MATLAB Online 中打开

See the full error message:

Index exceeds the number of array elements. Index must not exceed 1.

Error in LE_RF_p1 (line 4)

X= [x(4) x(7) x(10);

Apparently x is a scalar, not a vector, meaning it only has a single value. There error is because your code is requesting the 4th, 7th and 10th values, which don't exist.

x = 3;
% works
x(1)
ans = 3
% Your error
x(4)
Index exceeds the number of array elements. Index must not exceed 1.

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nune pratyusha 2022-5-25

编辑：Cris LaPierre 2022-5-25

在 MATLAB Online 中打开

I changed x,y & z as follows after that also showing same error

function f=LE_RF_a(t,x,a)
%p is the parameter
% x=1:20;
f=zeros(16,1);
b=0.4;c=11;d=6;e=130;f1=10;
X= [x(5) x(9) x(13)  x(17);
    x(6) x(10) x(14) x(18);
    x(7) x(11) x(15) x(19);
    x(8) x(12)  x(16) x(20) ];
f(1)=a*x(2)+(0.2+0.2*(abs(x(4)))).*x(3);
f(2)=b*((x(4).^2)-13).*x(3)-c*x(2);
f(3)=-d*x(1)-e*x(2)-f1*x(3);
f(4)=(x(3).^2)-x(4).^2;
Jac=[0 a 0.2+0.2*abs(x(4)) 0.2*x(4).*x(3)/abs(x(3));
    0 -c b*((x(4).^2)-13) b*((x(4)*2)-13).*x(3);
    -d -e -f1 0;
    0 0 2*x(3) -2*x(4)];
f(5:20)=Jac*X;

error:

>> run_LE_FO_a

p_step =

1.499999999999999e-03

Index exceeds the number of array elements. Index must not exceed 1.

Error in LE_RF_a (line 6)

X= [x(5) x(9) x(13) x(17);

Error in FDE12>f_vectorfield (line 300)

f = feval(Probl.fdefun,t,y,Probl.param) ;

Error in FDE12 (line 114)

f_temp = f_vectorfield(t0,y0(:,1),Probl) ;

Error in FO_Lyapunov_a (line 23)

[T,Y] = FDE12(q,ext_fcn,t,t+h_norm,[0 0 1 1],h,a);

Error in run_LE_FO_a (line 16)

lp=FO_Lyapunov_a(4,ext_fcn,0,0.2,1,[0 0 1 1],0.2,1,a)

>>

nune pratyusha 2022-5-25

编辑：Cris LaPierre 2022-5-25

在 MATLAB Online 中打开

FDE12.m

function f=LE_RF_a(t,x,a)
%p is the parameter
% x=1:20;
f=zeros(16,1);
b=0.4;c=11;d=6;e=130;f1=10;
X= [x(5) x(9) x(13)  x(17);
    x(6) x(10) x(14) x(18);
    x(7) x(11) x(15) x(19);
    x(8) x(12)  x(16) x(20) ];
f(1)=a*x(2)+(0.2+0.2*(abs(x(4)))).*x(3);
f(2)=b*((x(4).^2)-13).*x(3)-c*x(2);
f(3)=-d*x(1)-e*x(2)-f1*x(3);
f(4)=(x(3).^2)-x(4).^2;
Jac=[0 a 0.2+0.2*abs(x(4)) 0.2*x(4).*x(3)/abs(x(3));
    0 -c b*((x(4).^2)-13) b*((x(4)*2)-13).*x(3);
    -d -e -f1 0;
    0 0 2*x(3) -2*x(4)];
f(5:20)=Jac*X;
function LE=FO_Lyapunov_a(ne,ext_fcn,t_start,h_norm,t_end,x_start,h,q,a);
% Memory allocation
ne=4;
x=zeros(ne*(ne+1),1); 
x0=x;
q=1;
% y0=x_start
c=zeros(ne,1);
gsc=c; zn=c;
n_it = round((t_end-t_start)/h_norm);
% Initial values
x(1:ne)=x_start;
i=1;
while i<=ne
    x((ne+1)*i)=1.0;
    i=i+1;
end
t=t_start;
% Main loop
it=1;
while it<=n_it
    % Solutuion of extended ODE system of FO using FDE12 routine
    [T,Y] = FDE12(q,ext_fcn,t,t+h_norm,[0 0 1 1],h,a); 
    t=t+h_norm;
    Y=transpose(Y);
    x=Y(size(Y,1),:); %solution at t+h_norm
    i=1;
    while i<=ne
        j=1;
        while j<=ne;
            x0(ne*i+j)=x(ne*j+i);
            j=j+1;
        end;
        i=i+1;
    end;
    %       construct new orthonormal basis by gram-schmidt
    zn(1)=0.0;
    j=1;
    while j<=ne
        zn(1)=zn(1)+x0(ne*j+1)^2; 
        j=j+1;
    end;
    zn(1)=sqrt(zn(1));
    j=1;
    while j<=ne
        x0(ne*j+1)=x0(ne*j+1)/zn(1); 
        j=j+1;
    end
    j=2;
    while j<=ne
        k=1;
        while k<=j-1
            gsc(k)=0.0;
            l=1;
            while l<=ne;
                gsc(k)=gsc(k)+x0(ne*l+j)*x0(ne*l+k);
                l=l+1;
            end
            k=k+1;
        end
        k=1;
        while k<=ne
            l=1;
            while l<=j-1
                x0(ne*k+j)=x0(ne*k+j)-gsc(l)*x0(ne*k+l);
                l=l+1;
            end
            k=k+1;
        end;
        zn(j)=0.0;
        k=1;
        while k<=ne
            zn(j)=zn(j)+x0(ne*k+j)^2;
            k=k+1;
        end
        zn(j)=sqrt(zn(j));
        k=1;
        while k<=ne
            x0(ne*k+j)=x0(ne*k+j)/zn(j);
            k=k+1;
        end
        j=j+1;
    end
    %       update running vector magnitudes
    k=1;
    while k<=ne;
        c(k)=c(k)+log(zn(k));
        k=k+1;
    end;
    %       normalize exponent        
    k=1;
    while k<=ne
        LE(k)=c(k)/(t-t_start);
        k=k+1;
    end
    i=1;
    while i<=ne
        j=1;
        while j<=ne;
            x(ne*j+i)=x0(ne*i+j);
            j=j+1;
        end
        i=i+1;
    end;
    x=transpose(x);
    it=it+1;
end
function run_LE_FO_a(ne,ext_fcn,t_start,h_norm,t_end,x_start,h,q,a_min,a_max,n);
figure();
hold on;
ne=4;
ext_fcn=@LE_RF_a;
t_start=0;h_norm=0.02;t_end=1;
x_start=[0 0 1 1];h=0.2;q=1;
a_max=-11;
a_min=-12.2;
n=800;
p_step=(a_max-a_min)/n
a=a_min;
% for  p1=1.1:0.1:1.3;
while a<=a_max
    lp=FO_Lyapunov_a(4,ext_fcn,0,0.2,1,[0 0 1 1],0.2,1,a)
    a=a+p_step;
    plot(a,lp,'.')
    drawnow();
    % end
end

Cris LaPierre 2022-5-25

编辑：Cris LaPierre 2022-5-25

在 MATLAB Online 中打开

FDE12.m

The x input to LE_RF_a is a single value, not a vector. The error is happening in the following line of code in FDE12

% Check compatibility size of the problem with size of the vector field

f_temp = f_vectorfield(t0,y0(:,1),Probl) ;

This in turn calls the following function

f = feval(Probl.fdefun,t,y,Probl.param)

Here, y corresponds to the input 'x' in f=LE_RF_a(t,x,a). You are passing in a single value (y0(:,1) where y0 = [0 0 1 1]). Even if you didn't index y0, it only has 4 values. Your code in LE_RF_a is written assuming x has at least 20 elements.

The values of y0 are hard coded into your call to FDE12 inside FO_Lyapunov_a:

% Solutuion of extended ODE system of FO using FDE12 routine

[T,Y] = FDE12(q,ext_fcn,t,t+h_norm,[0 0 1 1],h,a);

Here's an executable version of your code with the error for others who may want to help.

ne=4;
ext_fcn=@LE_RF_a;
t_start=0;h_norm=0.02;t_end=1;
x_start=[0 0 1 1];h=0.2;q=1;
a_max=-11;
a_min=-12.2;
n=800;
p_step=(a_max-a_min)/n
p_step = 0.0015
a=a_min;
% for  p1=1.1:0.1:1.3;
while a<=a_max
    lp=FO_Lyapunov_a(4,ext_fcn,0,0.2,1,[0 0 1 1],0.2,1,a)
    a=a+p_step;
    plot(a,lp,'.')
    drawnow();
    % end
end
Index exceeds the number of array elements. Index must not exceed 1.

Error in solution>LE_RF_a (line 24)
X= [x(5) x(9) x(13)  x(17);

Error in FDE12>f_vectorfield (line 300)
    f = feval(Probl.fdefun,t,y,Probl.param) ;

Error in FDE12 (line 114)
f_temp = f_vectorfield(t0,y0(:,1),Probl) ;

Error in solution>FO_Lyapunov_a (line 61)
    [T,Y] = FDE12(q,ext_fcn,t,t+h_norm,[0 0 1 1],h,a); 
function f=LE_RF_a(t,x,a)
%p is the parameter
% x=1:20;
f=zeros(16,1);
b=0.4;c=11;d=6;e=130;f1=10;
X= [x(5) x(9) x(13)  x(17);
    x(6) x(10) x(14) x(18);
    x(7) x(11) x(15) x(19);
    x(8) x(12)  x(16) x(20) ];
f(1)=a*x(2)+(0.2+0.2*(abs(x(4)))).*x(3);
f(2)=b*((x(4).^2)-13).*x(3)-c*x(2);
f(3)=-d*x(1)-e*x(2)-f1*x(3);
f(4)=(x(3).^2)-x(4).^2;
Jac=[0 a 0.2+0.2*abs(x(4)) 0.2*x(4).*x(3)/abs(x(3));
    0 -c b*((x(4).^2)-13) b*((x(4)*2)-13).*x(3);
    -d -e -f1 0;
    0 0 2*x(3) -2*x(4)];
f(5:20)=Jac*X;
end
function LE=FO_Lyapunov_a(ne,ext_fcn,t_start,h_norm,t_end,x_start,h,q,a);
% Memory allocation
ne=4;
x=zeros(ne*(ne+1),1); 
x0=x;
q=1;
% y0=x_start
c=zeros(ne,1);
gsc=c; zn=c;
n_it = round((t_end-t_start)/h_norm);
% Initial values
x(1:ne)=x_start;
i=1;
while i<=ne
    x((ne+1)*i)=1.0;
    i=i+1;
end
t=t_start;
% Main loop
it=1;
while it<=n_it
    % Solutuion of extended ODE system of FO using FDE12 routine
    [T,Y] = FDE12(q,ext_fcn,t,t+h_norm,[0 0 1 1],h,a); 
    t=t+h_norm;
    Y=transpose(Y);
    x=Y(size(Y,1),:); %solution at t+h_norm
    i=1;
    while i<=ne
        j=1;
        while j<=ne;
            x0(ne*i+j)=x(ne*j+i);
            j=j+1;
        end;
        i=i+1;
    end;
    %       construct new orthonormal basis by gram-schmidt
    zn(1)=0.0;
    j=1;
    while j<=ne
        zn(1)=zn(1)+x0(ne*j+1)^2; 
        j=j+1;
    end;
    zn(1)=sqrt(zn(1));
    j=1;
    while j<=ne
        x0(ne*j+1)=x0(ne*j+1)/zn(1); 
        j=j+1;
    end
    j=2;
    while j<=ne
        k=1;
        while k<=j-1
            gsc(k)=0.0;
            l=1;
            while l<=ne;
                gsc(k)=gsc(k)+x0(ne*l+j)*x0(ne*l+k);
                l=l+1;
            end
            k=k+1;
        end
        k=1;
        while k<=ne
            l=1;
            while l<=j-1
                x0(ne*k+j)=x0(ne*k+j)-gsc(l)*x0(ne*k+l);
                l=l+1;
            end
            k=k+1;
        end;
        zn(j)=0.0;
        k=1;
        while k<=ne
            zn(j)=zn(j)+x0(ne*k+j)^2;
            k=k+1;
        end
        zn(j)=sqrt(zn(j));
        k=1;
        while k<=ne
            x0(ne*k+j)=x0(ne*k+j)/zn(j);
            k=k+1;
        end
        j=j+1;
    end
    %       update running vector magnitudes
    k=1;
    while k<=ne;
        c(k)=c(k)+log(zn(k));
        k=k+1;
    end;
    %       normalize exponent        
    k=1;
    while k<=ne
        LE(k)=c(k)/(t-t_start);
        k=k+1;
    end
    i=1;
    while i<=ne
        j=1;
        while j<=ne;
            x(ne*j+i)=x0(ne*i+j);
            j=j+1;
        end
        i=i+1;
    end;
    x=transpose(x);
    it=it+1;
end
end

nune pratyusha 2022-5-25

function run_LE_FO_p1(ne,ext_fcn,t_start,h_norm,t_end,x_start,h,q,p_min,p_max,n);

figure();

hold on;

ne=3;

ext_fcn=@LE_RF_p1;

t_start=0;h_norm=0.02;t_end=1;

x_start=[0.1 0.1 0.1];h=0.2;q=1;

p_max=1.3;

p_min=1.1;

n=800;

p_step=(p_max-p_min)/n

p1=p_min;

% for p1=1.1:0.1:1.3;

while p1<=p_max

lp=FO_Lyapunov_p1(ne,ext_fcn,0,0.2,1,[0.1 0.1 0.1],0.2,1,p1)

p1=p1+p_step;

plot(p1,lp,'.')

drawnow();

% end

end

function LE=FO_Lyapunov_p1(ne,ext_fcn,t_start,h_norm,t_end,x_start,h,q,p1);

% Memory allocation

x=zeros(ne*(ne+1),1);

x0=x;

c=zeros(ne,1);

gsc=c; zn=c;

n_it = round((t_end-t_start)/h_norm);

% Initial values

x(1:ne)=x_start;

i=1;

while i<=ne

x((ne+1)*i)=1.0;

i=i+1;

end

t=t_start;

% Main loop

it=1;

while it<=n_it

% Solutuion of extended ODE system of FO using FDE12 routine

[T,Y] = FDE12(q,ext_fcn,t,t+h_norm,x,h,p1);

t=t+h_norm;

Y=transpose(Y);

x=Y(size(Y,1),:); %solution at t+h_norm

i=1;

while i<=ne

j=1;

while j<=ne;

x0(ne*i+j)=x(ne*j+i);

j=j+1;

end;

i=i+1;

end;

% construct new orthonormal basis by gram-schmidt

zn(1)=0.0;

j=1;

while j<=ne

zn(1)=zn(1)+x0(ne*j+1)^2;

j=j+1;

end;

zn(1)=sqrt(zn(1));

j=1;

while j<=ne

x0(ne*j+1)=x0(ne*j+1)/zn(1);

j=j+1;

end

j=2;

while j<=ne

k=1;

while k<=j-1

gsc(k)=0.0;

l=1;

while l<=ne;

gsc(k)=gsc(k)+x0(ne*l+j)*x0(ne*l+k);

l=l+1;

end

k=k+1;

end

k=1;

while k<=ne

l=1;

while l<=j-1

x0(ne*k+j)=x0(ne*k+j)-gsc(l)*x0(ne*k+l);

l=l+1;

end

k=k+1;

end;

zn(j)=0.0;

k=1;

while k<=ne

zn(j)=zn(j)+x0(ne*k+j)^2;

k=k+1;

end

zn(j)=sqrt(zn(j));

k=1;

while k<=ne

x0(ne*k+j)=x0(ne*k+j)/zn(j);

k=k+1;

end

j=j+1;

end

% update running vector magnitudes

k=1;

while k<=ne;

c(k)=c(k)+log(zn(k));

k=k+1;

end;

% normalize exponent

k=1;

while k<=ne

LE(k)=c(k)/(t-t_start);

k=k+1;

end

i=1;

while i<=ne

j=1;

while j<=ne;

x(ne*j+i)=x0(ne*i+j);

j=j+1;

end

i=i+1;

end;

x=transpose(x);

it=it+1;

end

function f=LE_RF_p1(t,x,p1)

%p is the parameter

f=zeros(9,1);

X= [x(4) x(7) x(10);

x(5) x(8) x(11);

x(6) x(9) x(12)];

%RF equations

f(1)=x(2).*(x(3)-1+x(1).*x(1))+0.1*x(1);

f(2)=x(1).*(3*x(3)+1-x(1).*x(1))+0.1*x(2);

f(3)=-2.*x(3).*(p1+x(1).*x(2));

%Jacobian matrix

J=[2*x(1).*x(2)+0.1, x(1).*x(1)+x(3)-1, x(2);

-3*x(1).*x(1)+3*x(3)+1,0.1,3*x(1);

-2*x(2).*x(3),-2*x(1).*x(3),-2*(x(1).*x(2)+p1)];

f(4:12)=J*X; % To be modified if ne>3

In this code i put above 4 dimensional equations and i am getting error

Cris LaPierre 2022-5-25

编辑：Cris LaPierre 2022-5-25

I have already explained this error in a previous comment.

To summarize, in the code you shared for your 3D equation, you create the x input dynamically and pass that into FDE12.

ne=3;

...

x=zeros(ne*(ne+1),1); % 12x1

...

[T,Y] = FDE12(q,ext_fcn,t,t+h_norm,x,h,p1);

in the code you shared for your 4D equation, you also dynamically create x, but then hardcode your input as a 1x4 vector.

ne=4;

x=zeros(ne*(ne+1),1); % 20x1

...

[T,Y] = FDE12(q,ext_fcn,t,t+h_norm,[0 0 1 1],h,a);

Use the same syntax in both cases to obtain the same behavior

nune pratyusha 2022-5-25

it's working thank you so much

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