how to rid of this warning and get correct solution?

ti = 0;
tf = 1E-7;
tspan=[ti tf];
k = 5E-3;
h = 10E-2;
y0= [(h)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1));
     (h)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1));
     (h)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1)); 
     (h)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1)); 
     (h)*rand(2,1); ((-3.14).*rand(1,1) + (3.14).*rand(1,1));
      ((-3.14).*rand(5,1) + (3.14).*rand(5,1))];

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);

Warning: Failure at t=4.032134e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.136868e-13) at time t.

figure(1)
plot(T,(Y(:,16)),'linewidth',0.8);
hold on 
for m = 16:20
    plot(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 = 1.25;
a = 5;
T = 2000;
Gt = y(1:3:3*N-2);
At = y(2:3:3*N-1);
Ot = y(3:3:3*N-0);
k = 5E-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

Answer 1

[Gokul Nath S J]

Dear Sahil Sahoo,

I have tried running your code on my machine. By setting the relative and absolute error values to 1e-2, the warning messages can be avoided.

options = odeset('RelTol',1e-2,'AbsTol',1e-2);

The code should be included before calling ode45 (line 22)

However, at the same time, it is essential to check the validity of the eventual answer.

Also, I have found similar cases that might be helpful. Refer to the link below for the same.

Unable to meet integration tolerances without reducing the step size below the smallest value allowed - MATLAB Answers - MATLAB Central (mathworks.com)