3.1 By drawing the phase diagram, determine the nature of...

3.1 By drawing the phase diagram, determine the nature of stability of fixed points of the following dynamical system. Also write down the invariant sets of the system: dx/dt = (x - 1)(x - 2)(x - 3) 3.2 Consider the linear dynamical system dx/dt = -x - 4y dy/dt = x - y This system has its only fixed point at (0; 0). a) Explain why the above system is autonomous. b) Find the nullclines of this system and use these to draw a simple phase diagram. Indicate the direction of trajectories in each region of the plane. Based on your phase diagram, comment on the type of stability you suspect the fixed point at the origin might have. c) Confirm and/or expand on your findings in b) by analysing the stability of the fixed point at the origin using the trace/determinant method.

Transcript

00:01 In this question, we need to draw the phase diagram and then we have to determine the nature of stability of fixed point of the following dynamical system which is defined by dx over dt is equals to x minus 1 times x minus 2 times x minus 3.

00:21 Okay, so let us see how we are going to do this.

00:24 So, we can multiply this, so this becomes x minus 1, if you multiply this x square minus 5x plus 6.

00:33 Now, further we can simplify this as x cube minus 6x square plus 11x minus 6.

00:41 This is the dynamical system that we are having.

00:45 So, we have to find out first the fixed point.

00:49 So, fixed points, they are given by dx over dt is equals to 0.

00:57 That means x minus 1, x minus 2, x minus 3 equals to 0, which implies x value as 1, 2 and 3.

01:12 Correct.

01:14 Now, stability, it can be characterized, can be characterized, it is characterized by graph, i graph as well as the function f of x.

01:41 Okay, so what i will do, i will assume my function to be this x cube minus 6x square plus 11x minus 6.

01:56 Now, consider the derivative of this function, which is f dash of x, 2x minus 12x plus 11.

02:06 We will try to find the value at all the fixed points.

02:09 So, f dash of 1.

02:11 So, it will be 3 minus 12 plus 11.

02:15 So, this will be 14 minus 12, which is equals to 2.

02:21 Okay, so which is obviously greater than 0, this will implies that x equals to 1, is unstable.

02:31 Okay, so this point is unstable.

02:35 Next, we will try to find the value at 2, so it will be 3 times 4 minus 12 times 2 plus 11.

02:46 Solving this, we get this to be equals to minus 1, which is less than 0.

02:53 So, this implies x is stable, which x, x is equals to 2.

03:05 Next, what we can do is f dash of 3...

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