Plotting ODE solutions for a single time point

I have plotted the solution to four odes against time.

I now want to plot results for a single time point. Such as for when t=5, i want to plot c_{j}(5) against j. Is there a short command that can do this using the code i have written

function f = beckerdorin(t,C)
a=0.5;
b=1.5;
sigma=0.5;
f(1,1)= b*(C(4)+ C(3)+ 2*C(2))-a*C(1)*(2*C(1)+C(2)+C(3));
f(2,1)= a*(C(1))^2 - b*C(2) - a*C(1)*C(2) + b*C(3)-2*sigma*(a*C(2)^2-b*C(4));
f(3,1)= a*C(2)*C(1)-b*C(3)-a*C(3)*C(1)+b*C(4);
f(4,1)= a*C(3)*C(1)-b*C(4)+sigma*(a*C(2)^2-b*C(4));
end

Plotting

clf
[tv,c] = ode45('beckerdorin',[0,3],[10,0,0,0]);
figure(2)
plot(tv,c(:,1),'r');hold on
plot(tv,c(:,2),'b');hold on
plot(tv,c(:,3),'y');hold on
plot(tv,c(:,4),'g');
legend('C(1)','C(2)','C(3)','C(4)')
xlabel('Time t')
ylabel('c_{j}(t)')

Answer 1

[Star Strider]

It requires that you slightly change your ode45 call, then use the solution structure with the odextend (link) function:

sln = ode45(@beckerdorin,[0,3],[10,0,0,0])
tv = sln.x';
c = sln.y';
figure(2)
plot(tv,c(:,1),'r');hold on
plot(tv,c(:,2),'b');hold on
plot(tv,c(:,3),'y');hold on
plot(tv,c(:,4),'g');
legend('C(1)','C(2)','C(3)','C(4)')
xlabel('Time t')
ylabel('c_{j}(t)')

slnxtd = odextend(sln, @beckerdorin, 5);                % Extend Integration To ‘t=5’
c_5 = slnxtd.y(:,end)                                   % Values Of ‘c’ At ‘t=5’

c_5 =
          2.02359968542774
           1.3649852291002
         0.920727887972237
         0.621061548113791

[Aishah Malek]

Thankyou, but what if i want to change my j value ,i get the same results when i try another value for j, and also if i want to plot different j values on the same plot.

[Star Strider]

My pleasure.

What are you referring to? I do not see ‘j’ defined anywhere in your code. I have no idea what it is. It only appears as part of a character vector in your ylabel call.

You asked how to evaluate your ‘beckerdorin’ function at ‘t=5’, and I provided a way to do that.

I leave the rest to you.