 Welcome everyone to lecture number 10 on non-linear dynamical systems. We will continue with the Bendixian criteria and the Poincare Bendixian criteria. In particular, we will see important examples. One is the Lotka-Volterra predator pre-model and also the van der Poel oscillator. Let us start with the Lotka-Volterra predator pre-model. This is studying how the population of two species vary. As a function of time, these two species are classified into prey and hunter. So, there is one species that is a prey, another species that is a hunter and we will study the model of this prey and hunter species. Of course, we are studying a simplified model. Let xh be the hunter specimen in the model and let xp be the prey specimen in the model. So, what does this equation say? x dot is equal to x dot h equal to minus x of h plus some quantity that depends on both xh and xp and x dot p equal to xp minus xh xp. So, the first term in each equation is how the particular species would evolve if there were no other species. So, the first system, first equation says that if there were no prey, that is if xp were equal to 0, then xh would just decrease as a function of time. It would decrease exponentially because there is no food. So, left to itself, the hunter species would just decrease. But for each interaction between xh and xp, the hunter eats the prey and hence this extra, the next term, the second term in this right hand side is causing an increase in the hunter population. So, the hunter population decreases because of its own population and it increases because of its interaction with xp. So, the rate of increase is proportional to both xp and xh population. It is bilinear in the two, it is equal to the product. That is the increasing, increase causing term. On the other hand, the prey itself is just going to multiply, it is going to increase exponentially when left to itself if there had been no hunter species and interaction with the hunter species causes xp to decrease. So, quantities xh and xp are all positive and whether the increase or decrease depends on its own population and also population of the other species. So, this is a reasonable model for how dynamics of two species that interact with each other evolves as a function of time. Of course, we have simplified most importantly in the sense that more generally there would be some constants x dot h would be equal to minus a times xh plus b times xh xp and x dot p is equal to c times xp. The rate of increase is proportional to some c times xp in general and the decrease, the interaction causes a decrease with this multiplication with d. So, this is how one could study a general model, but one can consider that we are choosing a different unit for xh and xp so that these constants become equal to 1. Also, there is some normalization that has been done so that we are studying this model. Of course, this itself is a simplification. This model is also simplification because there might be some higher order derivatives. We have seen already how the population of just one species can vary with resource availability, with the ability to reproduce depending on the interaction between species. All that has been ignored. We have assumed this first order dynamics with respect to itself and just the product, the interaction is just the product of the two species population. So, the question that we can ask for this particular model is what are the equilibrium points? What is the nature of the equilibrium point of the linearized system? Are there periodic orbits? These are the questions that we will ask. So, let us go back to this particular model and we will find the equilibrium points for this system. So, d by dt of xh and xp is equal to minus xh plus xp times xh and this is xp minus xp times xh. So, this is our f. So, equilibrium points are those values of xh and xp where f1 of xh comma xp equal to 0 and also f2 of xh comma xp equal to 0. So, what do we get by equating xh minus xh plus xp xh equal to 0 and xp minus xp xh equal to 0. For a particular value of xh and xp to be an equilibrium point, these two equations have to be satisfied. So, let us see what are the values for which these equations are satisfied. First equation says xh equal to 0 or xp equal to 1. Second equation says xp equal to 0 or xh equal to 1. So, this gives us how many pairs of equilibrium point. So, an equilibrium point has an xp and xh coordinate. So, let us see what all possibilities are there for equilibrium point. So, if both equations have to be satisfied, then one can have 0 comma 0. This is nothing but xp equal to 0 and xh equal to 0. So, the first component in this is xh, specific value. Second is the xp population value. So, both equal to 0 is one equilibrium point. That is what we get from here and both equal to 1, 1 comma 1, which means xh equal to 1 and xp equal to 1. This is another value for the equilibrium point. You see, notice that if xh is equal to 0, you cannot have xp equal to 1 because for both equations, this is equation 1 and this is equation 2. This is 1, this is 2. Equation 1 says that any one of these two possibilities, equation 2 says any one of these two possibilities. And when we combine them, we get that these two equilibrium points. These two points, these two values of xp and xh are situations where the population species does not change as a function of time. So, this is where xh, this is xp. So, one equilibrium point is here, another equilibrium point is here. This is the equilibrium point 1 comma 1, this is equilibrium point 0 comma 0. As I said, the first component denotes xh value. Let us see what happens if xp is always equal to 0. xp equal to 0 means this is the hunter population. So, our dynamical equation system says that if xp is equal to 0, which means that the second term is always equal to 0. And if you put xp equal to 0 here, then xh is just decreasing. That is why we have drawn these arrows. And if xh were equal to 0, so this is sitting on the xp axis, then xp just goes on increasing. This is how the arrows look. But more generally, it is a combination of the two. So, for example, let us take what happens at 0.5, 0.5. Let us draw the arrow at this particular point which corresponds to 0.5 and 0.5. So, at xh equal to half and xp equal to half, we get x dot h equal to, so this is just substituting 0.5 in place of these two. So, we get minus 0.5 plus 0.25, which is equal to minus 0.25. And xp dot is just 0.5 minus 0.25, which is equal to 0.25. So, this is the vector whose xh component is negative, but xp component is positive. So, this is an arrow that looks like this, so that it is xh component. The horizontal component is xh. It is decreasing, but xp component is increasing. So, like this we can draw arrows for all the points. One can check and this is how we get. Let me draw a bigger figure. Only the first quadrant is reasonable because the populations do not become negative. So, this is a point 1 comma 1. This is 0 comma 0. As I said, xh population is going to decrease if xp is equal to 0. xp equal to 0 corresponds to this xh axis and xp axis corresponds to xh equal to 0. So, I am sorry. So, xh xp, when left to itself, the prey population is going to increase. That is why the arrow should all be in the direction of increasing xp. So, the correct figure should be and this is a point 1 comma 1 and we already checked that at intermediate points, at this point it is like this. If it is a little higher, let us verify this that this is how it looks. So, this itself is an equilibrium point. If it happens to be at the point 1 comma 1, if the hunter population is equal to 1 unit and the prey population is also equal to 1 unit, then it remains constant. But for small perturbations about that point, the arrows I have drawn like this, but this requires verification. So, let us take a sample point. This particular point has xp coordinate equal to 1, but xh coordinates slightly more than 1. So, for example, let us consider the point xh comma xp equal to 1.1 comma 1. Let us see what happens for this particular point. For this particular point, we have drawn the arrow like this, but let us check whether it indeed is like this. So, xh dot xp dot equal to we are evaluating at the point xh comma xp equal to 1.1 and 1. So, this is minus 1.1 plus 1 times 1.1. So, this is equal to 1.1 and xp population rate of change of the prey population is equal to 1 minus 1 into 1.1. So, this turns out to be equal to xh dot xp dot is equal to the top component is 0 and lower value is minus 0.1. This is what happens when xh is slightly more than 1, slightly more than equilibrium point, but xp is equal to the equilibrium point value that is equal to 1. So, when we do this, then we are speaking of this point here. For this point, we are getting that xh rate of change is equal to 0. So, the horizontal component is equal to 0 and the vertical component is equal to minus 0.1. That is why it is vertically downwards. So, similarly one can check for each of these four points. What is the property of this point? It is xp population. The prey population is slightly more than 1, but xh population, the hunter population is equal to 1. For each of these four points, one can verify and see that the arrows are indeed like this, suggesting that there is a periodic orbit around this point. So, there are periodic orbits close to this, but this point on the other hand looks like a saddle point. So, let us verify this by linearizing the system at each of these two equilibrium points. So, let us go back to the dynamical system. So, xh dot xp dot minus xh plus xp xh xp minus xp xh. So, del f by del x. So, this is equal to f1 of x, f2 of x equals this. So, the first row, the first function here is called as f1 of x. The second function here is f2 of x. Del f by del x is equal to a 2 by 2 matrix. The entry here is derivative of this with respect to xh. That is equal to minus 1 plus xp. The entry here is derivative of this with respect to the second component of x, that is xp. So, this is equal to xh. The entry that comes here is derivative of f2 with respect to xh. Here, we get minus xp and the entry that comes here is the derivative of this with respect to xp, the second component of the state. For that, we get 1 minus xh. So, as expected, this is a matrix. This is a 2 by 2 matrix, which depends on xp and xh. So, we are going to evaluate this matrix at the equilibrium point. So, del f by del x evaluated at the equilibrium point 0 comma 0. This is one of the equilibrium points. For this particular equilibrium point, we get minus 1 0 0 1 and del f by del x evaluated at the other equilibrium point 1 comma 1. We get equal to, by putting xp and xh both equal to 1, we get 0 1 minus 1 0. So, we have these two A matrices, one A matrix for the equilibrium point 0 comma 0 and the other A matrix for the equilibrium point 1 comma 1. It is not difficult to see the eigenvalues of these two matrices. So, equilibrium point 0 comma 0 has eigenvalues for a diagonal matrix. The eigenvalues are nothing but the diagonal entries. Equivalent point 0 comma 0 has eigenvalues 1 and minus 1. So, we already saw that this is an example of a saddle point and the equilibrium point 1 comma 1 has eigenvalues. What are the eigenvalues of the matrix, which matrix of this matrix? Eigenvalues of this matrix are plus minus j. As we noted in one of the first few lectures that the eigenvalue of such a matrix, if beta is not equal to 0, then the eigenvalue of this matrix are equal to alpha plus minus j beta. The eigenvalues of such a matrix are complex. Precisely, what complex values are the eigenvalues? Alpha plus minus j beta. The diagonal entries are the real elements, real part and the off diagonal entries with opposite signs correspond to the imaginary part of the eigenvalue. So, these are the eigenvalues even when beta is equal to 0. So, for this particular equilibrium point, we have this special case and so the eigenvalues are plus minus j, which we know corresponds to a center. The equilibrium point is what we called a center. So, a center is one that has periodic orbits and we already saw that for this particular plot, indeed this particular equilibrium point, equilibrium point has periodic orbits and this is a saddle point. So, the linearized system is a center, which is nothing but a continuum of periodic orbits, very close by different initial conditions correspond to different periodic orbits. They all correspond to periodic orbits and different periodic orbits. Is that the same for the non-linear system also? This is the topic that we will see in detail today. So, please note that we have investigated the Lotka-Volterra predator prey model for convenience. The predator we have called as Hunter, so that we can use a subscript H and the prey we continue to call PXP. The simplified model shows two equilibrium points. One equilibrium point, the linearized system is a saddle point and the other equilibrium point of the Lotka-Volterra predator prey model correspond to 1 comma 1 corresponds to a center after linearizing. So, what is important is that this particular equilibrium point is a center. We already saw that these arrows are suggesting like this. If it were a linear system, then when we go close, this is a periodic orbit, when we go close to this and another initial condition it may or may not be a periodic may or may not be a different periodic orbit. The linearized system says so, but it need not mean for the original non-linear system also. For example, if these are two different initial conditions, they correspond to the same periodic orbit, but different initial conditions like this may correspond to different periodic orbits or might converge to the same periodic orbit. This is the subject that we will see in detail today. So, if all these initial conditions correspond to different periodic orbits, then we will like to say that there is a continuum of periodic orbits. These periodic orbits are not isolated, but very close to each periodic orbit. There is another periodic orbit in a very close vicinity. Suppose this is a periodic orbit, the initial conditions starting from here correspond to periodic orbits also. In that sense, there is a continuum of periodic orbits. So, it is a very well-known important fact that for the particular Lotka-Volterra model that we have taken, let us go back here. For this particular Lotka-Volterra predatory model, for constants a, b, c, d, we have two equilibrium points, 0, 0 and 1, 1. When you assume a, b, c, d equal to 1, but for a different point, when a, b, c, d are some positive constants, possibly not equal to 1, there are two equilibrium points. While the 0, 0 is a saddle point, the other equilibrium point is a center and moreover for the non-linear system for this Lotka-Volterra predatory model, there is a continuum of periodic orbits. This particular fact for this particular model for any of these two models is a very important fact and one can modify this model suitably so that we have isolated periodic orbits. So, today we are going to see a different example where there are indeed isolated periodic orbits. So, let us use Poincaré-Bendixson criteria and the Bendixson criteria to check if there are periodic orbits. Let us see the Bendixson criteria. What does this Bendixson criteria say? We will evaluate this particular quantity and check whether it is, whether this is identically equal to 0 or not. If it is not identically equal to 0, only then we can go ahead and apply the Bendixson criteria. So, let us evaluate this particular quantity for our example. For our example, f1 of x was equal to minus xh plus xp times xh and f2 of x is equal to xp minus xp times xh. So, this f2 we could also call as fp and this is equal to fh. fh denotes the rate of change of xh and fp denotes the rate of change of xp. So, let us evaluate del fh by del xh plus del fp by del xp. When we evaluate this, we get derivative of this with respect to xh is equal to minus 1 plus xp plus derivative of this with respect to xp. We get this equal to 1 minus xh. So, this is equal to xp minus xh. So, is this identically equal to 0? No, it is not identically equal to 0. So, that is why we can go ahead and apply the Bendixson criteria. Let us now apply it and see xp minus xh sign of this quantity. If the sign does not change over a region, the Bendixson criteria says that if the sign of this particular quantity does not change on a region, then there are no periodic orbits contained inside that region. So, when is xp minus xh equal to 0, it is along this line. So, everywhere to the right of this line, this is xh, this is xp. To the right of this line, this quantity is negative and above this line or to the left of this line, this quantity is positive. So, the Bendixson criteria says that there cannot be a periodic orbit contained to the right of this line nor can there be a periodic orbit to the top of this line. It does not. So, this is equilibrium 0.1 comma 1. The Bendixson criteria does not rule out such a periodic orbit. That does not lie entirely in this region nor does it lie entirely in this region. So, this is an important property to note that the Bendixson criteria only says that can such a periodic orbit exist inside this region? No, this is not possible. Can a periodic orbit lie entirely in this region where the sign of this is all positive? That is also not possible. However, this particular periodic orbit could exist. So, Bendixson criteria is only a sufficient condition for non-existence of a periodic orbit lying entirely inside a region. Let us now check what the Bendixson criteria says for a linear system. x dot is equal to Ax for which the equilibrium point is 0 comma 0. So, Bendixson criteria is applicable when for the planar case that is when x has two components at each time instant, x of t has two components x1 and x2. So, suppose A was equal to may be we see a slide about this. So, for the Lutka-Voltaire pre-interpre model, we have already seen this. Before we see another example, let us see this particular case periodic orbit for a A that looks that is of this form. So, our A we have already assumed in the it is of this form and now we will do del f1 by del x1 plus del f2 by del x2. Notice that these two terms are nothing but the diagonal entries of this matrix A. So, for this particular A, the diagonal entries are both 0. So, they add up to 0 also. So, they are identically equal to 0. No matter which x1, x2 you check, this is going to be equal to 0. This particular quantity is expected to be independent of x1, x2 for linear systems. Why? For linear systems, for linear time invariant systems, these four entries are all independent of x and hence you differentiate f1 and f2, f1 with respect to x1, f2 with respect to x2 which is nothing but just picking up these entries, picking the values at these two positions and they are going to be independent of x. So, for this particular A, we get this identically equal to 0. So, do we say that Bendixson criteria is not applicable or do we say that there are no periodic orbits? Of course, we know that for this particular A, the eigenvalues of A are equal to plus minus square root of 2 times A. So, if A is positive, then the eigenvalues are plus minus 2 times, minus of 2 times A. We will just verify this. So, what is S i minus A? S i minus A is equal to, so determinant of S i minus A is equal to S square plus 2 A. So, eigenvalues of the A matrix are nothing but roots of the determinant. The roots are square root of minus 2 A plus minus. So, if A is positive, A greater than 0, then complex, purely imaginary in fact. If the eigenvalues are purely imaginary, then we know for a linear system, there are periodic orbits. And if A is less than 0, then eigenvalues are real. One of them is greater than 0, other is less than 0. Why? Because if A is negative, this quantity itself under the square root sign is positive. So, we can take the square root and one is positive, one is negative. So, for this case, eigenvalues are here and for this case, eigenvalues are here and here. How far from the origin depends on the value of A of course, but whether they are depending on whether A is positive or negative affects whether the roots are purely imaginary or real. So, we know that for this case, the equilibrium point is a center and there are periodic orbits. While for this case, the equilibrium point is a saddle and there are no periodic orbits. So, the important case when this is identically equal to 0, that particular case could correspond to either there are periodic orbits or there are no periodic orbits. This is just to see that the Bendixon criteria is unable to say anything when this is identically equal to 0. That is precisely the reason that Bendixon criteria assume that this is not identically equal to 0 and then you start looking at whether the sign changes or not. So, let us take a case where for x dot is equal to Ax. Let us check what is del f1 by del x1 plus del f2 by del x2. What is the value of this? We will check that this is equal to 4 by calculation explicitly. So, x dot is equal to Ax means 2x1 plus 3x2, that is the meaning of A acting on x and the second row of A will be used to multiply with x to get minus 3x1 plus 2x2. So, when we do this, then we differentiate the first component of x dot with respect to x1 and we get this equal to 2. We are picking up just this entry and the second component of x dot that is f2 of x with respect to x2, we are doing this. So, notice that the derivative of this with respect to x1 is just this component, this first one by one entry and the derivative of this with respect to x2 is just this entry. That is the reason that I said that doing this particular to evaluate this quantity is nothing but to add the diagonal entries for a linear system, for a linear time invariant system. So, we get this equal to 4, this is greater than 0 and it is independent of x1 x2. For linear systems we expect that this will not depend on x1 x2 and it is indeed independent of x1 x2. Since it is greater than 0 for all x1 x2, we get that no periodic orbits, no periodic orbits in R2, in the entire state space, in the entire plane there are no periodic orbits. So, for linear systems we can check that as long as the diagonal entries do not add up to 0, as long as the diagonal entries do not add up to 0, this quantity will not be identically 0 and then we can see that periodic orbits are ruled out. When would periodic orbits be possible if the diagonal entries add up to 0? If the diagonal entries add up to 0, we cannot say that the periodic orbits exist because the Bendixon criteria is silent for that case. It does not say anything when the diagonal entries add up to 0 identically. We only saw that it is possible that there are periodic orbits, it is also possible that periodic orbits do not exist when the diagonal entries add up to 0. So, this is already the complication for linear systems. So, for the Lotka-Volterra-Pilipri model to show that there are periodic orbits is a difficult thing and it is an important research topic after which it has been concluded that there is a continuum of periodic orbits for the particular model that we studied. Now, let us take some other examples of dynamical system and check whether there are periodic orbits or not. So, consider this example. So, in which x1 dot equal to x2 plus x1 times x2 square, x2 dot is equal to minus x2 plus x2 times x1 square. So, we differentiate the first in order to use Bendixon criteria. This particular quantity that we were to evaluate is nothing but divergence of f. The divergence of f is also denoted as dot product of this operator with f. So, when we evaluate this, we get x2 square here and so there is something wrong here. This x2 should have been x1. So, please note that there is a small mistake here. If we have x1 here, then we get this. So, now we have that this is always positive. After you substitute x1 here, we get that this is always positive. So, if the Lotka-Volterra-Pilipri model, this is very similar to the Lotka-Volterra-Pilipri model. After we have x1 here, this is very similar to the Lotka-Volterra-Pilipri model. If it had depended on not just the product of x1, x2 but some higher order power of one of the species, then it is possible to show that the Bendixon criteria says that there are no periodic orbits. Why? Because the divergence of f is always positive except at the equilibrium point. So, hence by Bendixon criteria, there are no periodic orbits. Now, consider this example. This example also has we should have a modification here. We should have x1 here in place of in the second equation x2 dot equal to minus x2 minus x2 times x1 square. So, here we see that this changes sign. This only means that inside the region where it has the same sign, there there are no periodic orbit contained inside that region. This is an example that we have already seen. Now, we will study an important case where we have an epsilon here along the diagonal entries. So, for this particular A in which we have epsilon along the diagonal, what can we say about the equilibrium point? So, when epsilon is greater than 0, then the equilibrium point 0, 0 is unstable focus. This is something that we have already seen. What happens when epsilon is less than 0? When epsilon is less than 0, diagonal values of this matrix are epsilon plus minus 2j. So, epsilon greater than 0 implies unstable. Unstable, node or focus, it is unstable focus because the imaginary part is non-zero. And for epsilon less than 0, we have a stable focus. So, this is what we will say that if this epsilon, this is our epsilon value and this is let us say radius, distance of the point from the origin. So, if epsilon is some positive quantity, epsilon itself is not dependent on x1, x2. So, it is just the same positive number. Positive means unstable focus and if it is some same fixed negative number, then it is stable. So, this is unstable and this is a stable focus. So, how about modifying this epsilon as a function of radius? So, that we have trajectories that converge to a periodic orbit. So, this is what we will see in detail. So, what if epsilon changes its sign? Depends on the distance from the origin and changes its sign. So, consider this differential equation in which along the diagonal we have put 25 minus x1 square minus x2 square along the diagonal and off diagonal term we keep constant, does not depend with change with radius. So, this is nothing but writing it in this form in which along the diagonal we have some function that depends on the radius, depends on the distance from the origin. Now, we consider the case when r is equal to 5. For that case, we have this epsilon of r is equal to 0. We can consider the case when r is less than 5. For r less than 5, the diagonal entries are positive and when r is greater than 5, the diagonal entries are negative or both negative. We cannot speak of eigenvalues of this, the eigenvalues of this matrix itself depends on x1, x2. We speak of eigenvalues of only constant matrices. So, it appears like if we make this radius, if we make this diagonal entry depend on radius, then we will have trajectories either coming towards the origin or going away from the origin, depending on whether we are inside a particular circle, whether we are inside the circle of radius 5 or outside and on the circle itself, we are going to be remaining on the circle. So, let us check. This is a circle of radius 5. So, when x1 is equal to 0, so this is radius 5 circle. So, when x1 is equal to 0 and x2 is equal to 5, the time x1 dot, so x1 equal to 0, sorry, the orientations, we start again. How do we get this orientation? We expect that for radius equal to 5, equal to 5, we have a periodic orbit. Why is it that we have a periodic orbit? You put r equal to 5 and you see that the matrix A, it looks just as if, so check, check that x1 of t, x2 of t equal to 5 times. So, along a circle, x1, x2 are like cos and sine of a function of what frequency? T. Now, just cos t and sine t. Why? Because omega is equal to 1. For this particular A, the solutions are x1, x2 are equal to cos and sine to the sine of the quantities and of radius 5. And why cos t, sine t? In general, it would have been cos omega t, sine omega t and then omega is equal to 1 because of diagonal entries are equal to 1. Now, we are going to decide why is it clockwise and not anticlockwise. That we can check by taking some sample points. So, consider this point. This is x1 component equal to 0 and x2 equal to 5. So, consider the point 0, 5 at 0, 5. When this acts on 0, 5, when A acts on this matrix, then we get that this is equal to 5 common 0. So, the x1 component is increasing at this particular point. That is why it is along this direction. So, by using the same argument, we can decide where cos of t should come, where sine of t should come and whether there should be a negative sign to one of these. Now, the focus of this particular topic is to see what happens for radius larger than 5, for radius larger than 5 and for radius smaller than 5. So, for radius larger than 5, there is the off diagonal term indeed cause some rotation, but the diagonal entries cause a decrease in the radius. That is why it is coming inwards. So, we have these arrows coming inwards and for the circles inside the circle of radius 5, that is for circles of radius less than 5, there is some rotation causing cause because of the off diagonal terms, but the diagonal entries themselves are positive, which is causing this radius to grow as a function of time. This is an important property that we will exploit to say that all initial conditions except the equilibrium point 0, 0 are all converging to this special periodic orbit, which periodic orbit, the periodic orbit of radius equal to 5. So, let us check. Let us use Poincare-Bendixson criteria and check that there indeed exists a periodic orbit. We are not able to say that this is a center kind of arguments because we can use that only for a linear system by linearizing at an equilibrium point. So, take m to be equal to the set, set of all x1, x2 such that x1 square plus x2 square lies in the interval 4 to 6, closed interval. So, what is our set m? Let me write again. Set m is a set of all x1, x2 points such that x1 square plus x2 square is less than or equal to 6 and x1 square plus x2 square is greater than or equal to 4. In other words, this is a circle, supposed to be a circle. This is another circle. This is the circle of radius 4. This is the circle of radius 6. All the points in this ring, these are all the points whose distance from the origin is greater than or equal to 4 and less than or equal to 6 also, this and this. So, we will now check that this particular set m is positively invariant and has no equilibrium points inside it and it is compact. The compactness is satisfied because it is a compact set and it is a closed set because these inequalities are non-strict inequalities. They are not strict inequalities but non-strict inequalities because of the fact this is a closed set and it is a bounded set because we see that all the points are at most distance 6 away from the origin hence it is a bounded set. So, in order to use Poincare-Bendixson criteria, we are going to check that this set m is a positively invariant set also. So, when would Poincare-Bendixson criteria be applicable, the set m should be a closed and bounded set, should be positively invariant and there should be no equilibrium points inside it or at most one equilibrium point which is either an unstable node or an unstable focus. So, let us check what is the property of this particular m. We show that m is positively invariant. So, how can we show that this set m is positively invariant? So, what we will do is we will take the circle of radius 6 and we will check that this circle of radius 6, the outward, the vector that is perpendicular to this boundary is outward like this. This is a vector. We will check what is the inner product of this vector with the vector field. If at every point along the boundary, this vector field is directed inwards, it means that all the points or trajectories are coming inwards. What are these vectors? This is a unit vector perpendicular to the boundary and directed outside the region. The region is directed, the region is inside this as far as the boundary 6 is concerned, as far as this boundary of radius 6, circle of radius 6 is concerned, this is a vector that is directed outwards. So, let us check whether what is this vector, it is nothing but x1, x2 vector. It is inner product with f of x at that point. f of x at each point is again a vector of dimension 2. We will check whether this inner product is positive or negative. If this inner product is negative on circle of radius 6, it means that along this circle, all trajectories are going inwards. Why is it inwards? Because this is a vector outward and this is the f of x. If this particular angle is, this dot product being negative means that the angle between the two vectors is an obtuse angle. Why? Because what is the dot product of W dot product with V? This is equal to W norm times V norm times cos of the angle between W and V. So, if this quantity is negative, it means these two quantities cannot be negative. So, this cos theta is negative and cos theta is negative only for theta beyond 90 degrees which means that this angle between these two vectors is greater than 90 degrees and since this vector is a vector that is directed outside the boundary, this angle being obtuse means the f is directed inwards. So, let us check whether this quantity is negative. So, x1, x2 times f of x. So, f of x is what we can see from this particular thing. f1 of x is first row times x1, x2. So, this is nothing but x1 times 25 minus x1 square minus x2 square plus x2. This is f1 of x and f2 of x is minus x1 plus x2 times 25 minus x1 square minus x2 square. So, when we do the dot product of this, that is nothing but this row vector times this column vector. When we evaluate this, let us see what we get. So, x1 square times 25 minus x1 square minus x2 square plus x1, x2. This is just this quantity here that I have written is just x1 times the first component here plus x2 times minus x1 plus x2 square times 25 minus x1 square minus x2 square. So, x1 x2 minus x1 x2, these both cancel. So, we get x1 square plus x2 square in common 25 minus x1 square minus x2 square. Now, we note that we are going to evaluate this along the circle of radius 6. So, we get 6 square times 25 minus 36. So, this is some quantity that is less than 0. That is all we needed. So, this proves that along the circle of radius 6, the arrow is directed inwards. Now, let us check what happens along the inner circle. This is a circle of radius 4. And along this boundary, this is a unit normal and the f itself. So, there are two vectors at each point along the boundary. One vector is the direction of f of x at that point. And another vector is the direction of the unit normal. And in this case, this vector says that it is directed inwards. Why? Because m lies to the outside of this region, outside of this circle. m was the set of all points of radius greater than or equal to 4 and less than or equal to 6. Since we are taking a circle of radius 4, the region is to the inside. So, this vector is directed inwards of the region. So, at each point, what is this vector? It is this x1, x2 again. And the direction of f at that point is f1x f2 of x. Now, because this vector is a vector directed towards inside the region along the boundary, this particular quantity being greater than 0 means that the angle is acute angle. The angle between the two vectors is less than 90 degrees. And then it would mean that the trajectories are all coming inwards into the region. So, there are all these trajectories that are coming into the region as far as the boundary is concerned. As far as which boundary is concerned, the circle of radius 4. So, by the same argument, all we have to do is we have to now substitute x1 square plus x2 square is equal to 4 square and not 6 square. And this quantity becomes 4 square when we are looking at the circle of radius 4. And this quantity becomes 25 minus 16, which is now positive. So, this is greater than 0 on circle of radius 4. So, this proves that the set M is positively invariant. How have we shown that this set is positively invariant? We have said that this set M has boundary consisting of two circles. So, this is one boundary of the circle 4. The other boundary is circle of radius 6 and the region is like this. This is the region M. All along the outer circle, the trajectories are coming inwards into the region. This is what we checked because the angle was obtuse there. All along the inner boundary, also the trajectories are coming inwards. So, check that as long as the region M is defined as set of all points where the radius of the x1 square plus x2 square is greater than or equal to say 4.9 and less than or equal to 5.1. It will still be positively invariant. That is the only property that we used. So, such an M will be positively invariant and hence it will contain a periodic orbit. Is there an equilibrium point inside this region? That is another thing we are supposed to check before we use the Poincare-Bendixson criteria. This is the last thing we will check before today's lecture ends. So, let us put f1 of x equal to 0 and f2 of x equal to 0. So, f1 of x we had evaluated and we had got that equal to, so what is shown here is f1 of x and this is f2 of x. We have to substitute both equal to 0 and find the values of x1, x2 such that both these functions are equal to 0. So, those x1, x2 values will comprise the equilibrium points. So, one can check that the only equilibrium point for this is x1, x2 equal to 0, 0. In other words, if M is a set of all points of distance greater than or equal to 0, if the radius is strictly greater than 0, then M will not have any equilibrium points. That is why this implies that x1, x2 equal to 0, 0. x1, x2 equal to 0, 0 is the only equilibrium point. The equilibrium point itself is stable or unstable. One can check, let this be as a homework that this equilibrium point is unstable. Why? Because at this equilibrium point, we ensured that the diagonal entries of this particular matrix, of which matrix, let us go back to the slide of this particular matrix for x1 equal to 0 and x2 equal to 0. This particular matrix has diagonal entries positive and hence the equilibrium point is an unstable focus. So, this allows us to use Poincare-Bendixson criteria since the region that we have considered, set of all points of radius greater than 4 and less than 6, greater than or equal to 4 and less than or equal to 6 is compact, is positively invariant and has no equilibrium points and hence there is a periodic orbit. By making this set M smaller, smaller and smaller such that it just contains the circle of radius 5. So, we can take the region of M to be set of all points of distance slightly less than 5 and slightly more than 5 and it will still the same argument will hold and there will be a periodic orbit. This shows that there is no continuum of periodic orbits here. There is an isolated periodic orbit for this particular example. We will consider modifying this example and obtain the van der Paul oscillator as a special case. Thank you.