Hello everyone, I am Sridham Jogade and I welcome you all to the lecture number 11 of non-linear dynamical systems. In today's lecture, we will look into the Bendixson and Poincare Bendixson criteria. The application of it, we will consider one or two examples of it. Next, we will consider a van der Poel oscillator. We will study van der Poel oscillator, which is a non-linear oscillator. Then, we will take the example of a van der Poel oscillator, which is a RLC circuit, L-C tank connected to a active circuit. During the lecture, we will refer to the following six figures. First, we will consider the Bendixson criteria. Here are the state equations of an example. Here, we can see in the matrix the function 25 minus x 1 square minus x 2 square. This function is dependent on both x 1 and x 2. Let us say the radius is R. Then, we can say that x 1 square plus x 2 square is equal to R square. So, the function 25 minus x 1 square minus x 2 square can be written as 25 minus R square. So, the function is dependent on R. Let us represent that function as epsilon R, a function dependent on R. So, the state equations reduce to the following form, which are shown in the slide.  This results out to be 2 into 25 minus 2 R square. The root of this equation is R equal to phi by root 2, which is approximately equal to 3.536. So, at R equal to 3.536, we will have this expression value to be 0. Now, let us try to apply Bendixon criteria to the following example. For this, we will consider two regions. The first region we will consider bounded by R, R which is bounded by 3.53. So, R is strictly less than 3.53.  Now, the second region we will consider for R strictly greater than 3.54.  So, we can see in both the regions, the sign of the expression does not change. In the first region, the sign of the expression remains positive and in the second it remains negative. So, let us try to apply a Bendixon criteria. Let us consider the first region where R is strictly less than 3.53. So, there is no sign change. We can say that by Bendixon criteria, no periodic orbit exist in the region. Now, consider the second region where R is strictly greater than 3.53. Now, the question arises whether we can apply Bendixon criteria to this? The answer is no. We cannot conclude in this case because Bendixon criteria requires simply connected region. Now, what is simply connected region? A simply connected region is a region which has no holes or we can define it in other sense. If we take that region and if we take a simply closed curve in that region and shrunk it to the point, then it should also remain within the region. During this shrinking, every curvature and up to the point should remain in the region. So, that is how we define simply connected region. So, for R greater than 3.54, we cannot conclude whether there are periodic orbits or not or we cannot apply Bendixon criteria since this region is not simply connected region. So, we conclude here that Bendixon criteria is not applicable for region R greater than 3.54. Next, we will consider an example of point care Bendixon criteria. Here, we take a system of the form x dot equal to A x where A is equal to 0, 2 minus 2 epsilon. The diagonal elements are 0 and epsilon. Now, in this case as epsilon will vary, the behavior of the system will change. So, we will consider three cases. In the first case, we will consider epsilon as greater than 0. In the second case, we will consider epsilon equal to 0 and in the third case, we will consider epsilon less than 0. Now, let us take that example previous one. In the previous one, we had both the diagonal elements as 25 minus x 1 square minus x 2 square. Now, in this example, we have one diagonal element as 0 and the second as 25 minus x 1 square minus x 2 square. So, we will try to analyze the behavior of the system and for that we will convert the coordinates into polar coordinates. So, we will have a clearer picture. So, we can convert it polar coordinate into the following form. So, x 1 will become R cos theta and x 2 will become R sin theta where R is the radius and theta is the angle. So, x 1 square plus x 2 square we get as equal to R square. Differentiating it with respect to time, we will get x 1 into x 1 dot plus x 2 into x 2 dot equal to R into R dot. Now, we have the expressions for x 1 and x 2 and x 1 dot and x 2 dot from the state equations. So, we can substitute in this expression and we will get the following expression at R dot equal to expression shown on the slide. So, we will cancel out the common factors and rearrange it and the equation will reduce to the final form R dot equal to 25 minus R square into R into sin square theta. So, except theta equal to 0 or theta equal to 180 degrees, sin square theta is always positive. So, let us consider one figure. In this figure, we have plotted 25 minus R square into R into sin square theta versus R. In the first case, we have considered R greater than 0. So, we can find that for R equal to phi, the expression reduces to 0. So, we can see that for R equal to phi, R dot is equal to 0. So, when R dot is equal to 0, we can say that the circle with radius phi is a periodic orbit. Since the radius is not changing, now we will consider the second case where sin square theta equal to 0 means theta is equal to either what 0 or 180 degrees. So, in that case, we will get x 2 equal to 0. So, the x 2 equal to 0 is say is along the x 1 axis. So, if we substitute these values in the state equation, then our state equation reduced to the following form where x 1 dot equal to 0 and x 2 dot equal to minus x. In this case, we have a point on the x 1 axis which is the initial condition. Then according to the state equations, we will have x 1 dot equal to 0 and x 2 dot equal to minus x 1. So, for x 1 to be x 1, x 1 is positive. We will have a vector pointing in downward direction and it will be perpendicular to the x 1 axis. Similarly, when x 1 is negative, the direction vector of the vector will be pointing outwards and it will be perpendicular to the x 1 axis. So, the magnitude of this vector will depend on the value of x 1. So, we can see that for x 1 not equal to 0, the vector is always non-zero. So, from the state equation, we can see that for the equilibrium point, we need x 1 dot to be 0 and x 2 dot to be 0. So, in this case, the only equilibrium point we can see is when x 1 dot is equal to 0. So, the only equilibrium point is x 1 equal to 0 and x 2 equal to 0. So, we will go back to the previous figure where we had drawn plotted the r dot versus r. In this case, we can see that for r equal to phi, if there is a disturbance or a perturbation, then the trajectories are approaching towards r equal to phi when r is greater than phi or r is less than phi. So, the point at r equal to phi on the periodic orbit, we will conclude that it is stable. We can also say that the limit cycle is isolated. The meaning of isolated is if we consider a small region around r equal to phi, then we will have no periodic orbits. So, in that region, r equal to phi is the only periodic orbit existing. We have already proved that how it is stable since for disturbance, all the trajectories are pointing towards r equal to phi. So, it is a stable limit cycle. Next we will consider van der pol oscillator. Van der pol oscillator is a non-linear oscillator. If we consider the same system form as x dot equal to x, where a is equal to 0, 1, minus 1 and epsilon into 1 minus x 1 square. So, epsilon here is a positive constant. Here we can see that one diagonal element is dependent on x 1 square. So, it is not the diagonal element is dependent only on x 1 square. It is not dependent on x 2 square. So, we cannot conclude that it depends on radius. It does not depend on radius. It depends only on x 1 square. This system we will call it as van der pol oscillator and van der pol oscillator is a special case of Lienard's equation. Next we will consider Lienard's equation. As we previously mentioned that van der pol oscillator is a special case, where Lienard's equation defines the generalised case for the non-linear oscillators. Let us consider two functions f and g, which are continuously differentiable. Let us consider that f is an even function that is f of minus x is equal to f of x and g is a odd function. So, that g of minus x is equal to minus g x. So, the second order differential equation of the form x double dot plus f of x into x dot plus g of x is equal to 0. This equation is called Lienard's equation. This is the generalised form of the equation for non-linear oscillators. In general non-linear oscillators are considered for the modeling of the physical oscillators. We have defined the general Lienard's equation for the non-linear oscillators. We will have to next look into the stability of the oscillations for the non-linear oscillators. For that we will first define a function capital f of x which is equal to integral of small f of x. Then for a Lienard's system, if we consider the following conditions like g of x is greater than 0 for all x greater than 0, then capital f of x tends to infinity as x tends to infinity and for some p capital f of x satisfies that it is negative for the range when x is between 0 to p and f of x also satisfies that it is positive and monotonic for x greater than p that is when x is greater than p the f of x is monotonically increasing. If these conditions are satisfied then we can say that the Lienard's system is having a unique and stable limit cycle and this is what is called Lienard's theorem. Lienard's theorem gives the conditions for the stability of oscillations for non-linear oscillators. For the reference you can see the book titled Non-linear Oscillations by Nicholas Mianorski with the respective addition. Next, let us consider f of x is equal to minus epsilon into 1 minus x square where epsilon is a scalar and it is strictly greater than 0 and g of x equal to x. Then the system is called Van der Poel oscillator. Previously we consider a Lienard's equation where it was generalized case. Now we are defining Van der Poel oscillator in that equation where f of x and g of x are defined as I said before. So the differential equation the Lienard's equation gets transformed to the form x double dot minus epsilon into 1 minus x square into x dot plus x equal to 0. Let us now investigate the stability of oscillation. We had Lienard's theorem which gives us the condition for the stability of oscillation. For Van der Poel oscillator we can specifically investigate for its stability like if epsilon is much greater than 0 then our oscillations for the Van der Poel oscillator are very stable. Now as epsilon goes on decreasing the relative stability of the oscillations goes on decreasing. When epsilon is equal to 0 we can see that the equation is transformed to x double dot plus x equal to 0. So the oscillator no longer remains non-linear. It turns into a linear oscillator and for the third case where epsilon is less than 0 you will have unstable oscillations. So we can see that the Van der Poel oscillator will have stable oscillations only if epsilon is greater than 0. So we can also conclude that for epsilon greater than 0 f of x and g of x will satisfy the Lienard's condition given for the stability. Let us look into the existence of a closed orbit for the Van der Poel oscillator. We will consider the same equation where v double dot plus epsilon hv h of v into v dot plus v is equal to 0 where v can be a voltage across the element in the given circuit. Now h of v here is equal to minus 1 plus v square. Now for analyzing the behavior let us choose state variables as x1 equal to v and x2 equal to v dot plus epsilon into capital H of v. Now here we will define capital H of v as a d by dv of capital H of v is equal to small v and capital H of v at v equal to 0 is equal to 0. So therefore if we differentiate the equations of x1 and x2 we will get the following state equations where x1 dot is equal to x2 minus epsilon h of x1 and x2 dot equal to minus x1. So we can see here if we put x1 equal to 0 and x2 equal to 0 it has a unique equilibrium point and since capital H is equal to 0 only at v equal to 0 it is the only equilibrium point. So origin is the only equilibrium point for this Van der Poel oscillator. Next we will consider one figure. Let us look at the state plane where x1 and x2 are the axis. So we will divide this plane into four regions with the help of the curves given as follows x1 dot is equal to x2 minus epsilon H of capital H of x1. So it is the this curve where x2 is equal to epsilon into capital H of x1 and the second curve is x2 dot equal to minus x1 equal to 0 which is the x2 axis. So we will next look into how this curve divides the plane into four regions. We can say that each curve divides or separates x1 xi dot greater than 0 from xi dot less than 0. Like for example let us consider the first curve where x2 is equal to epsilon capital H of x1. So above this curve in this region we can have x1 dot greater than 0 since x2 is greater than epsilon capital H of x1. So x1 dot is greater than 0 in this whole region and below this region we have x1 dot which is less than 0. Now let us consider the second curve which is the x2 axis. Now to the right side of the x2 axis we have x2 dot less than 0 since x2 dot is equal to minus x1. So this to the right side of x2 axis the x1 is positive so x2 dot will be negative. So x2 dot is less than 0 to the right of half of the plane and to the left half of the plane x2 dot is positive. So now we will consider the four regions. Now in the first region we have x1 dot greater than 0 and x2 dot less than 0. In the second region we have x1 dot less than 0 and x2 dot less than 0. In the third region we have x1 dot less than 0 and x2 dot greater than 0 and the fourth region is x1 dot greater than 0 and x2 dot greater than 0. So as we have seen here the two curves are dividing the state plane in the four regions. Now we will see how these four regions will be helpful to us in finding the existence of the periodic orbit. So we will consider another figure. So let us take the initial condition on the x2 axis so that x1 is equal to 0 and x2 is equal to minus k. These are the initial conditions we have taken and here k is greater than 0. Let us name the point as a. If we draw the trajectory according to the directions given so here is the trajectory which will be intersecting x1 axis to at point c and x2 axis again at point e. Let us say that the coordinates for the e point is 0 and minus alpha of k where alpha is positive. So alpha is greater than 0. The reason I have taken alpha as a function of k because alpha depends on k. Now if we change the initial condition or if we change the value of k then we will get a different alpha. So the value of alpha is actually dependent on k. So the intersection at point e or at the x2 axis again so it is dependent on the value of k. So we can say that alpha is a function of k. Now if we take k as large enough then we can prove that alpha k is less than k. So that it is same as saying if we start out with the initial condition and around the orbit if we consider 180 degrees curvature so trajectory will come closer to the periodic orbit since k is greater than alpha of k. Now let us look why it is like that. In the first slide we consider that x1 dot and x2 dot the state equations. In that we can see they are the function of capital H of x1 and x. So both these capital H of x1 and x are odd functions. So we can say that if x1 and x2 are the solutions to the van der poel oscillator then minus of x1 t and minus of x1 x2 t are also the solutions. Now as stated before the reason for this is H is an odd function. Now let us consider that if the trajectory completes 360 degrees then if alpha k is less than k the trajectory will become more closer to the periodic orbit. Let us consider the function v of x equal to x1 square plus x2 square upon 2. Now v of x is equivalent to the total energy in a LC circuit in a LC tank circuit. So if we differentiate v dot v of x with respect to time then we will get is equal to minus epsilon into x1 into capital H of x1. We can verify this by substituting the values of x1 dot and x2 dot from the state equations. Now suppose we consider the curvature of H. So this is the curvature of H of x1. So it has a positive root p when x is greater than 0 and a negative root x1 is less than 0. So when x1 is greater than p or v dot x is less than 0 since for x1 greater than p H of x1 is positive and x1 is also positive. So from the expression we can see that v dot x is negative. If we consider the reason for x1 to be between 0 and p then we will get that v dot x is greater than 0. So we can conclude that the v dot x is changing along the x1 axis. Now let us take delta k as a change in energy when we intersect the curvature to the x2 axis. So delta k is the change of energy from point A to point E. So we can define delta k as vE minus vA where vE is the energy at point E and vA is the energy at point A. So it is equal to the integral along the curvature AE of v dot x with respect to time. We can divide the following curve into three curves A B where B is the point just above the x1 equal to p. So at point B along the curvature x1 is equal to p point C is the intersection of the trajectory with the x1 axis. Point D is another point where x1 is equal to p on the curvature and DE is the remaining curvature. So the whole curvature from A to E is divided into three curves A B, BCD and DE. So delta k can be represented as a change of energy from A to B, B to D and D to E. Let us represent it in the form of delta 1 k, delta 2 k and delta 3 k. Where delta 1 k is the change of energy from A to B, delta 2 k is the change of energy from B to D and delta 3 k is the change of energy from D to E along the curve. Now let us take the case where delta 1 k is greater than 0. We can say that delta 1 k is greater than 0 because x1 here is greater than 0 and H of x1 as we can see is negative. So delta 1 k is greater than 0. So we can say that the change of energy from A to B is positive. In the second case we can say that delta 3 k which is the change of energy along the curve from D to E is positive. It is similar to delta 1 k. In this case also x1 is greater than 0 and H of x1 is less than 0. H of x1 is less than 0 because x1 is restricted to the point P as we can see B and D are the points along the x1 equal to P axis. So we will have delta 1 k greater than 0 and delta 3 k always greater than 0. Now let us consider the change of energy along the curve BCD which is denoted by delta 2 k. Here we can say the change of energy along the BCD is less than 0. We can give the reason because x1 is greater than 0 in along the curve and H of x1 as we can see x1 is greater than P. So capital H of x1 will be always greater than 0. So delta 2 k along the curve BCD will be less than 0. Now as we see as I increase the initial conditions as I increase the k the curvature will expand and delta 2 k that is the change of energy along BCD will go on decreasing and we can also say that as limit x tends to minus infinity the change of energy along BCD that is delta 2 k will go to minus infinity. In the other context if we look at the delta 1 k and delta 2 k expressions that is the change of energy along AB and DE then for large k they will not grow as much as delta 2 k. So as k will go on increasing the delta 2 k will grow much faster than delta 1 k and delta 3 k. Since delta 2 k is negative so the net summation of delta 1 k delta 2 k and delta 3 k will be negative. Hence for a large value of k we will have delta k less than 0. Since delta k is less than 0 we can say that along the curvature from A to E the energy at point E is less than energy at point A that is the energy along the curve is decreasing. Since the energy is decreasing along the curve we can say that alpha k is less than k. So the trajectory will move closer to the periodic orbit or it will approach the periodic orbit. Now due to the symmetry of the in the solutions as we stated before if x 1 t is a solution and x 2 t is a solution then minus x 1 t and minus x 2 t are also solutions. So the next 180 degrees will be similar to that. Let us consider point care bendixon criteria again. For applying point care bendixon criteria we need a compact positively invariant set M such that either M has a no equilibrium point or it can have at most one equilibrium point such that after linearization if we consider the Eigen values they will be in open right half plane. So the equilibrium point if it is there in the M region it will be unstable. So here we can choose an region we consider the curve A, B, C, D, E before. Now let us consider another curve where the initial condition is 0 minus k. So the curvature will be F, G, H, I and back to A. So if we consider the whole region if we connect close this region then we can say that this region is a positively invariant set. Now this curve is also contained in the van der pol oscillation or it is also solutions with due to symmetry as stated before that if a portion x 1 t and x 2 t is in the solution then a portion minus x 1 t and minus x 2 t will also in the solution. So if we consider the whole region it will be a positively invariant region that is for initial condition in this region the trajectory will remain in the region and it will approach to a periodic. So next we will consider a Jacobian matrix which will be formed after linearizing the system. So the A is the Jacobian matrix which is defined as dou F by dou x at x equal to 0 since we have the equilibrium point at x equal to 0. So we will get the matrix as shown in the slide which is minus epsilon into small h of 0 1 minus 1 and 0. So the characteristic polynomial of A will be given as S square plus epsilon into h of 0 S plus 1. Now from this equation we can say that the product of Eigen values is 1 and sum is equal to minus epsilon h of 0 because the roots of this characteristic equation are the Eigen values of the system. Now since we define before epsilon is greater than 0 and h of 0 as negative then both of the Eigen values we will have as having positive real parts. So we can say since the Eigen values are having positive real part the equilibrium point is unstable. So now the conditions we concluded for this system of Van der Poel oscillator are that we have a invariant set M then we have a equilibrium point inside that which is unstable. So we can apply the Poincare-Bendixson criteria here and by Poincare-Bendixson criteria we can say that there is a closed orbit in M. Now for the example of the Van der Poel oscillator we will consider an RLC circuit where the R is an active resistive element. Now let us consider this figure. In this figure this is an RLC parallel circuit this is the capacitance C which is greater than 0, L inductance which is greater than 0 and we have connected in parallel an active resistance element. This is having VI characteristics as I equal to capital H of V where we had defined capital H of V before. Now so we can say that the capital H of V satisfies the following conditions capital H of 0 is equal to 0 then capital H dash that is derivative of capital H with respect to V satisfies capital H dash of 0 is less than 0 and capital H of V tends to infinity as V tends to infinity. Now here we have a point saying that capital H of V is similar to capital H of A F in Liener's equation. We can see that the conditions that are satisfied by capital H of V capital F of X are same. So we can say that H and F are both odd functions. Now let us go back to the circuit again. Here we consider a point D and we will apply KCL here. IC is the current to the capacitor, IL is the current to the inductor and IR is the current for the active resistance element. So at D we will apply KCL so that we will get the summation of all the three currents is equal to 0 then in the differential form we will get the following expression where we have H of V coming into the picture. Now we will use a transformation we will define a element tau equal to T upon root L c and substitute in the following equation. So we will get a normalized time variable equation. It is a second order differential equation and we can see that it is in the form of a van der pol oscillator where we have epsilon equal to root of L c. Here we can see that root of L c is epsilon is greater than 0. Now next we will consider the coefficient of V dot which determines the damping of the system. So the coefficient of V dot is H of V into root L c and this determines the damping of the voltage. The damping here is a non-linear damping. We can practically implement active resistance element in the form of tunnel diodes. So it will act as a negative resistance for some value of V and a positive resistance for other values of V. Now we will define capital H of V as minus V plus V cube by 3 and small H of V into root L c is equal to V square minus 1 into root L c. So let us analyze the system. Consider V much greater than 1. So we have much greater than 1 the damping coefficient is positive. So the V can say the damping is positive and since the damping is positive energy is dissipated in the active resistance element. The transaction of energy is from L c circuit to active resistive element and it is getting dissipated in the resistive element. The resistive element in this case will be positive. The value of the resistor will be positive and we can say that since the energy is getting dissipated V r into I r is strictly greater than 0. Now in the second case we will consider V value of V voltage is much less than 1. So in that case the damping will be negative as the damping constant coefficient V is negative and the energy will be fed into the L c tank circuit. So the transactional energy is from active resistance element to the L c circuit. So we can say that for the active resistance element we have V r into I r less than 0 and here that since the damping is negative the resistance is actually an active element. It is acting as an active element or we can also say the resistance is negative in this case. Let us look at the behavior of the system for given an initial condition. Now that initial condition may be a voltage across the capacitor an initial voltage or it may be an initial current of the inductor. Now in this case the trajectories will remain bounded. Now how can we say that as we looked before for V greater than 1 and V less than 1 we have damping positive and negative. So we can say the trajectories are remaining bounded. When damping is negative positive the trajectories are approaching towards the orbit and for the when damping is negative the trajectories also in that case also the trajectories are approaching the orbit. So we can say that the trajectories are remaining bounded. Now the second point we can say that trajectories encircle the origin. Now for this case we will consider when for an initial given condition as I said before the initial voltage the damping is positive and negative depending on V is remaining in V greater than 1 region or V less than 1 region. So the value of V and I are repeatedly changing the sign. So in that case we can say that since V and I are repeatedly changing the sign so the trajectories are actually encircling the origin. Now we have concluded the two points that the trajectory remain bounded and they are also encircling the origin. The next we will consider after sufficient time that is for an given initial condition if we have a sufficient time we can say that the trajectories are almost periodic. Since for a given initial condition the trajectories are approaching the periodic orbit so after much sufficient time we can say that they are almost periodic. They cannot be periodic because the two trajectories cannot intersect so they will be almost periodic. Next we will consider oscillation along the periodic orbit. When we have an oscillation along the periodic orbit the active resistor feeds the energy into LC circuit for some time that is when the active resistor is negative the damping is negative it feeds the energy into LC circuit and it also absorbs the energy in the LC from the LC circuit when it is positive. In that case damping is positive. So we can say that during the periodic orbit active resistance element is feeding energy and also absorbing energy for some time. Now let us see how can we say that the periodic orbit is stable or not. Now suppose along periodic orbit energy fed by the resistance is equal to the energy absorbed by the resistance then we can say that periodic orbit is stable. We can say this because when the energy fed is equal to the energy absorbed net energy extend is equal to 0 so the periodic orbit in that case will be a stable one. So we will have a stable oscillation for van der pol oscillator. Further we will see some animations about van der pol oscillator and also about the Lutka-Volterra predator pre-model. Thank you.