# I gave the initial condition correctly still the program not working.

2 views (last 30 days)
SAHIL SAHOO on 11 Oct 2022
Answered: Walter Roberson on 11 Oct 2022
ti = 0;
tf = 70E-8;
tspan=[ti tf];
k = (0.62).*10^(-5);
% y0= [(10e-6)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1));
% (10e-6)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1));
% (10e-6)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1));
% (10e-6)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1));
% (10e-6)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1));
% ((-3.14).*rand(5,1) + (3.14).*rand(5,1))];
y0 = [ 0.00001; 0.00001; 0.00001; 0.00001; 0.00001;
0.00001; 0.00001; 0.00001; 0.00001; 0.00001; 2.5669; 2.0482; 2.0454; -0.7968; 0.2303];
yita_mn = [
0 1 0 0 1;
1 0 1 0 0;
0 1 0 1 0;
0 0 1 0 1;
1 0 0 1 0;
]*(k);
N = 5;
tp = 1E-12;
[T,Y]= ode45(@(t,y) rate_eq(t,y,yita_mn,N),tspan./tp,y0);
Index exceeds the number of array elements. Index must not exceed 15.

Error in solution>rate_eq (line 81)
dy(n16) = -a.*(Gt(n2)-Gt(n1)) + (k).*(y(j2)./y(j5)).*cos(y(n16)) - (k).*(y( j5)./y(j2)).*cos(y(n16)) + (k).*(y(j8)./y(j5)).*cos(y(n17)) - (k).*(y(j19)./y(j2)).*cos(y(n20));

Error in solution (line 24)
[T,Y]= ode45(@(t,y) rate_eq(t,y,yita_mn,N),tspan./tp,y0);

Error in odearguments (line 92)
f0 = ode(t0,y0,args{:}); % ODE15I sets args{1} to yp0.

Error in ode45 (line 107)
odearguments(odeIsFuncHandle,odeTreatAsMFile, solver_name, ode, tspan, y0, options, varargin);
figure(1)
plot(T./t,(Y(:,16)),'linewidth',0.8);
hold on
for m = 16:20
plot(T./t,(Y(:,m)),'linewidth',0.8);
end
hold off
grid on
xlabel("time")
ylabel("phase difference")
set(gca,'fontname','times New Roman','fontsize',18,'linewidth',1.8);
function dy = rate_eq(t,y,yita_mn,N,o)
dy = zeros(4*N,1);
dGdt = zeros(N,1);
dAdt = zeros(N,1);
dOdt = zeros(N,1);
P = 0.5;
a = 1;
T = 2E3;
Gt = y(1:3:3*N-2);
At = y(2:3:3*N-1);
Ot = y(3:3:3*N-0);
k = (0.62).*10^(-5);
for i = 1:N
dGdt(i) = (P - Gt(i) - (1 + 2.*Gt(i)).*(At(i))^2)./T ;
dAdt(i) = (Gt(i).*(At(i)));
dOdt(i) = -a.*(Gt(i));
for j = 1:N
dAdt(i) = dAdt(i)+yita_mn(i,j).*(At(j))*sin(Ot(j)-Ot(i));
dOdt(i) = dOdt(i)+yita_mn(i,j).*((At(j)/At(i)))*cos(Ot(j)-Ot(i));
end
end
dy(1:3:3*N-2) = dGdt;
dy(2:3:3*N-1) = dAdt;
dy(3:3:3*N-0) = dOdt;
n1 = (1:5)';
n2 = circshift(n1,-1);
n16 = n1 + 15;
n17 = circshift(n16,-1);
n20 = circshift(n16,1);
j2 = 3*(1:5)-1;
j5 = circshift(j2,-1);
j8 = circshift(j2,-2);
j19 = circshift(j2,1);
dy(n16) = -a.*(Gt(n2)-Gt(n1)) + (k).*(y(j2)./y(j5)).*cos(y(n16)) - (k).*(y( j5)./y(j2)).*cos(y(n16)) + (k).*(y(j8)./y(j5)).*cos(y(n17)) - (k).*(y(j19)./y(j2)).*cos(y(n20));
end
##### 0 CommentsShow -1 older commentsHide -1 older comments

Sign in to comment.

### Accepted Answer

Walter Roberson on 11 Oct 2022
y0 = [ 0.00001; 0.00001; 0.00001; 0.00001; 0.00001;
0.00001; 0.00001; 0.00001; 0.00001; 0.00001; 2.5669; 2.0482; 2.0454; -0.7968; 0.2303];
That is 15 initial values.
for m = 16:20
plot(T./t,(Y(:,m)),'linewidth',0.8);
end
But you are trying to plot assuming 20 results. The only way to get 20 results is to have 20 or more initial values.
##### 0 CommentsShow -1 older commentsHide -1 older comments

Sign in to comment.

### More Answers (1)

Benjamin Thompson on 11 Oct 2022
circshift returns a vector of the same length as its input. So, j2, j5, j8, and j19 are vectors and not scalar values as the line having the failure seems to expect. You can use breakpoints in your script in MATLAB to investigate further and debug the problems.
##### 0 CommentsShow -1 older commentsHide -1 older comments

Sign in to comment.

### Categories

Find more on Data Type Identification in Help Center and File Exchange

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by