00:01
So in this problem, we're given a differential equation, x double dot is equal to x minus lambda, x squared minus lambda.
00:11
I'm asked to investigate the equilibrium points of the parameter dependent system.
00:17
So in a differential equation system, there are equilibrium points correspond to the case where the differential is equal to zero, which in this case nicely gives us the polynomial expression x minus lambda times x squared minus lambda equals zero and so we have equilibrium points which i'm going to know as x starred at lambda one at the square root of lambda and x3 at minus the square root of lambda because each of these terms has to be zero.
00:58
Now to evaluate the stability of a differential equation what we do is we generally, so we'll take for example let's say that the differential equation is written as x double dot is equal to f of x.
01:12
What we need to do is we need to make a taylor expansion of this function f of x and truncate to the first derivative and we look at the sign of that derivative to determine what's happening locally to the function.
01:27
So what we do is we calculate df by dx, which in this case is d by dx of x minus lambda times x squared minus lambda, which we can use the product rule.
01:44
So differentiating this first term just gives us x so we have x squared minus lambda plus x minus lambda multiplied by this differentiator with respect to x which gives us 2x and so we can expand this out and simplify things so we have x squared minus lambda plus plus 2x squared minus 2 lambda x, which gives us 3x squared minus 2 lambda x minus lambda.
02:39
So now what we do is we evaluate this derivative out to the different equilibrium points and look at the sine of the function to determine the stability.
02:50
So let's consider x1 star, which is equal to lambda.
02:55
So we have the f, the x evaluated at x1 star is equal to 3 lambda squared minus 2 lambda squared minus lambda, which gives us lambda squared minus lambda, which is just just can be written as lambda times lambda minus 1.
03:24
So the conditions we consider are if the f by dx is positive, then the system is unstable, because locally the function is then moving away from the point.
03:43
So we consider that an unstable point.
03:47
And if, sorry, yes, an unstable point because the function is moving away from it...