I gave the initial condition correctly still the program not working.
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SAHIL SAHOO
am 11 Okt. 2022
Beantwortet: Walter Roberson
am 11 Okt. 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);
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
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Akzeptierte Antwort
Walter Roberson
am 11 Okt. 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.
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Weitere Antworten (1)
Benjamin Thompson
am 11 Okt. 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.
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