How to Find the period of a periodic function

How do I find the period of this function?

*Note y(1) is equivalent to x and y(2) is equivalent to y in this set of equations.

My code:

sym vars

a=1; b=1; c=0.1; d=0.1;

RHS=@(t,y)([a*y(1)-b*y(1)*y(2); c*y(1)*y(2)-d*y(2)]);

[t,y]=ode45(RHS, [0 100], [5,0.2]);

figure

plot(t, y(:,1))

xlabel('t (in years)'); ylabel('Number of Prey (in millions)');

title('Number of Prey for 100 Year Period');

legend('x vs t');

set(gca, 'fontsize', 16);

Answer 1

[Star Strider]

Use the Signal Processing Toolbox findpeaks function:

a=1; b=1; c=0.1; d=0.1;
 
RHS=@(t,y)([a*y(1)-b*y(1)*y(2); c*y(1)*y(2)-d*y(2)]);
[t,y]=ode45(RHS, [0 100], [5,0.2]);
figure
plot(t, y(:,1))
xlabel('t (in years)'); ylabel('Number of Prey (in millions)');
title('Number of Prey for 100 Year Period');
legend('x vs t');
set(gca, 'fontsize', 16);

[pks,locs] = findpeaks(y(:,1));                                     % Peaks & Locations
mean_period = mean(diff(t(locs)));                                  % Period
text(min(xlim)+0.1*diff(xlim), min(ylim)+0.95*diff(ylim), sprintf('Mean Period = %.2f time units', mean_period))

.

[Nasir Holliday]

Thank you so much!! But could you explain the last line in the code starting with text(min(...

What does that do because I when I run the code, I do not get any results

[Star Strider]

My pleasure!

The text call simply writes that line on the plot. It is not otherwise necessary for the code. (I do not understand the reason it did not print the results. It worked correctly for me in R2020a.)

[Nasir Holliday]

Oh! I didn't run it with the rest of the code so that's why! Thank you!

[Star Strider]

As always, my pleasure!