You could quickly generalize the equilibrium points to be: x = 5 + sqrt(5) * sqrt(5 - 2*h) or x = 5 - sqrt(5) *sqrt(5 - 2*h) -- such that x is contained in the real numbers. You can use the quadratic formula. (I prefer to just complete the square). Now, in order to find the bifurcation point, just find the value of h which sets the second term of your x value equal to zero. In this case, it just so happens to be 2.5. So: x = 5 +/- sqrt(5)*sqrt(5 - 2 * 2.5) = 5 +/- sqrt(5)*sqrt(0) = 5 +/- 0 = 5 We can now conclude that there is but one equilibria. If we are on a value of x which provides only one equilibria, this is where the local minimum of our quadratic equation is located. This point is equivalent to our bifurcation point. After determining the bifurcation point, plug in higher and lower values for x and see if your derivative is positive or negative. This will qualitatively tell you how the equation is behaving before and after the bifurcation point.
Hello @admin, can you explain the figure 3.6.3 a, b and 3.6.4 a and b of Nonlinear Dynamics and Chaos by Steven Strogatz. It is related to the imperfect bifurcation.
Hi, I have a question to this tutorial, for C=0, you said that z has two stable points (at +2 and -2), however when I solve the differential equation in MATLAB "dz=4z-z^3", then gives me the solution z(t), however z can never reach the value -2, z(t) just oscillate between 0 and 2, then how -2 it is a stable point?, thanks
I'm not sure how you solved the differential equation dz/dt=4z-z^3 in MATLAB, but if you see that it oscillates between 0 and 2, then something must have gone wrong. If, for example, you type dz/dt=4z-z^3 in Wolfram Alpha, you will see that the solutions it shows (which have positive initial condition) go smoothly to 2. If the initial conditions were negative, then the solution would go smoothly to -2, showing that -2 is a stable point. I don't think I can put a link here, but if you go to the Math Insight web site (link above) and search for "solve differential equation graphical," you can find a page with a applet where you can type "4x-x^3" (as the variable there is x) and explore what the solutions look like.
+Jordan Jones Exactly at the bifurcation point, when c=c*, there will be 2 equilibria at the locations you mentioned. I don't think I would describe the as "critical points," though.
If you were to plot z(t) with an initial condition z(0)=0.1, then you would find that z(t) increases away from 0 (toward 2). On the other hand, if you plotted z(t) with a initial condition z(0)=-0.1, then z(t) would decrease away from 0 (toward -2). Since z(t) will always move away from 0, no matter how close to 0 you start, the equilibrium z(t)=0 is unstable.
+Pedro Hernandez Do you mean how one sketches the graph by hand? I determined where 4z-z^3=0, which is true for z=-2, 0, and 2. Since the z^3 is negative, I knew that for large value of z, the function should go negative, and for small values is should go positive. That was enough to sketch the general form, that it starts positive, becomes negative at -2, becomes positive at 0, and finally becomes negative again at 2.
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Here 10 years later, great video! Best I've seen so far..
Very Good work. I am in my first steps in Bifurcation and You put me in the true way. Thank you
You could quickly generalize the equilibrium points to be: x = 5 + sqrt(5) * sqrt(5 - 2*h) or x = 5 - sqrt(5) *sqrt(5 - 2*h) -- such that x is contained in the real numbers. You can use the quadratic formula. (I prefer to just complete the square).
Now, in order to find the bifurcation point, just find the value of h which sets the second term of your x value equal to zero.
In this case, it just so happens to be 2.5. So: x = 5 +/- sqrt(5)*sqrt(5 - 2 * 2.5) = 5 +/- sqrt(5)*sqrt(0) = 5 +/- 0 = 5
We can now conclude that there is but one equilibria. If we are on a value of x which provides only one equilibria, this is where the local minimum of our quadratic equation is located. This point is equivalent to our bifurcation point.
After determining the bifurcation point, plug in higher and lower values for x and see if your derivative is positive or negative. This will qualitatively tell you how the equation is behaving before and after the bifurcation point.
the best explanation of bifurcation so far. Thanks!
Thanks a lot, by watching your video I was able to solve the problem of imperfect bifurcation.
Thank you so much!
Could you please give an example with delay diffirential equation (dde)?
Thank you very much for posting such a nice video!
Hello @admin, can you explain the figure 3.6.3 a, b and 3.6.4 a and b of Nonlinear Dynamics and Chaos by Steven Strogatz. It is related to the imperfect bifurcation.
Great! do you know how to put those changes in a function?
Hi, I have a question to this tutorial, for C=0, you said that z has two stable points (at +2 and -2), however when I solve the differential equation in MATLAB "dz=4z-z^3", then gives me the solution z(t), however z can never reach the value -2, z(t) just oscillate between 0 and 2, then how -2 it is a stable point?, thanks
I'm not sure how you solved the differential equation dz/dt=4z-z^3 in MATLAB, but if you see that it oscillates between 0 and 2, then something must have gone wrong. If, for example, you type dz/dt=4z-z^3 in Wolfram Alpha, you will see that the solutions it shows (which have positive initial condition) go smoothly to 2. If the initial conditions were negative, then the solution would go smoothly to -2, showing that -2 is a stable point.
I don't think I can put a link here, but if you go to the Math Insight web site (link above) and search for "solve differential equation graphical," you can find a page with a applet where you can type "4x-x^3" (as the variable there is x) and explore what the solutions look like.
when c=c*, won't there then be 2 critical points? one at around z=-1 and then the one all the way to the right?
+Jordan Jones Exactly at the bifurcation point, when c=c*, there will be 2 equilibria at the locations you mentioned. I don't think I would describe the as "critical points," though.
thank you soooo much wanting to find a simple example and I found one thank youuuuu
Great videos, thanks for sharing your knowledge!
Very good. Bravo!
If I plot z(t), how would be the instability at point 0??
If you were to plot z(t) with an initial condition z(0)=0.1, then you would find that z(t) increases away from 0 (toward 2). On the other hand, if you plotted z(t) with a initial condition z(0)=-0.1, then z(t) would decrease away from 0 (toward -2). Since z(t) will always move away from 0, no matter how close to 0 you start, the equilibrium z(t)=0 is unstable.
same argument is valid for z=2, -2 than how they become the stable point.
Than you for this video, Sir !
how do u do to graph the equation at the graph at the beginning?
+Pedro Hernandez Do you mean how one sketches the graph by hand? I determined where 4z-z^3=0, which is true for z=-2, 0, and 2. Since the z^3 is negative, I knew that for large value of z, the function should go negative, and for small values is should go positive. That was enough to sketch the general form, that it starts positive, becomes negative at -2, becomes positive at 0, and finally becomes negative again at 2.
this is great!
THANKS BRO