 Welcome back again. So, in today's class, we will be discussing periodic orbits, especially their existence. So, these are also referred to as closed orbits. So, the existence of periodic orbits is an important part of qualitative theory and quite useful with respect to many physical and other applications. I am not going to prove any result in this discussion, mainly because the proofs are quite involved and they also require new concepts and tools. I will try to explain the results through some examples and through some pictures. The discussion is restricted to 2 D systems and with regard to 2 D systems, there are fairly general results. Those things we are going to discuss. The important theorem with regard to 2 D autonomous system is the celebrated Poincare Bendixson theorem, which will be coming more or less at the end of this hour. So, since we are discussing only 2 D systems, so let me change a notation little bit and write our system as x dot equal to f of x y autonomous. So, there is no t dependence on the right hand functions, so g x y. So, x and y are the unknown functions of t and this is our 2 D system. So, we would like to discuss the existence of periodic orbits for the system 1. So, what are the necessary conditions? What are the sufficient conditions? There are only sufficient conditions. Necessary conditions are too hard to come by. There are some that we will discuss them. So, recall again from the general lemmas we have learnt about autonomous systems. So, let me recall that. So, in this 2 D setup if x t, so let me write that y t is a solution of 1 satisfying x of t 0 plus t is equal to x t 0 and y of t 0 plus t equal to y t 0 for some t 0 and t positive, then x t plus t equal to x t, y of t plus t is equal to y t for all t. So, thus the solution is periodic. So, this we have already seen in the general context. So, in particular this 2 D system. So, our first result let me state it as theorem 1. So, any periodic orbit, so orbit that is related to a periodic solution that is periodic orbit, any periodic orbit necessarily surrounds at least one and equilibrium point of. So, let me just what does this mean. So, this suppose c is a periodic orbit of 1. So, in the phase space, so it will be like this. So, that will be. So, described by a periodic solution of the system 1. So, the interior of c, so that is what is meant by this surrounding. So, there is at least one equilibrium. So, this is. So, theorem 1 is in some sense a negative result. So, theorem 1 implies that no equilibrium points, no periodic orbit. So, in order to have periodic orbits necessarily the system 1 has to have some equilibrium points. So, let me just briefly describe how this theorem 1 is proved. So, that is where the new tools, new concepts, new ideas come in and they are bit advanced. So, just let me restrict myself to a discussion of that. So, let me just discuss that. So, to prove theorem 1 the concept an index, this also called Poincare index. So, let me just, index is introduced. So, let me just briefly discuss what does that mean. So, this is again very specific to 2 D. So, this even this concept of index. So, you consider any closed curve, not necessarily an orbit or curve in the phase space, in the phase plane, where in R 2. So, that is. So, let me describe how the index is defined. So, this is, suppose this. So, at every point on the curve you just compute this f x y g x y, this vector. So, this is in R 2 now. So, that is the reason why you call system 1 as a vector field. So, vector vector field suppose this vector is non-zero vector at all points of C. Then it defines a definite direction. So, let me just draw here. So, this is. So, this say x 0 y 0 a specific thing and this is f of x 0 y 0 g of x 0 y 0. So, we say that this vector field is non-vanishing on this closed curve C. So, then at all points on the curve, this vector has a unique direction. So, we can measure its angle from some fixed direction, say x axis. So, we can just keep it here x axis. So, we measure the angle of this vector with respect to this axis and then you start moving along the curve. So, you suggest move along this thing. So, they may just depending on f. So, you do that. So, you have draw all the vectors along this and then you come back. Then you come back and again you measure the angle at the point we started. So, measure the angle at a starting point x 0 y 0 on C. Then make a round. So, you keep on measuring the angles at all points make a counter clockwise round clockwise movement along C return to return to that starting point again and then measure the angle again. So, the difference will be 2 pi k. So, k is an integer. So, it could be positive it could be negative or it could even be 0. So, that is what you do. So, this k is called. So, take some simple examples of closed curve like circles ellipses and you take a vector field which is non-vanishing on that closed curve and then try to measure this angle. So, there is an analytic formula for this angle measurement I will describe that in a minute. So, first the definition. So, this k is called the index of C. So, here the vector field is fixed that f and g that is coming from our system 1. So, that is why you are not otherwise that is also part of the definition, but we have fixed that vector field f g. So, f and g are assumed to be smooth enough. So, the analytical formula for this k is analytical description. So, k is equal to 1 by 2 pi. So, this is a contour integral over C. So, f d g minus g d f this is just by taking the angle angle is in terms of arc tan of g by f and then you work out and this will be the definition and since we are dividing by f square plus g square you see that assumption is necessary that this f the vector field is non-vanishing on C otherwise we will have trouble there. So, this is a line integral. So, when you have a parametric representation of this closed curve C then this d g and d f will be expressed in that thing. So, it will be a one dimensional integral. So, I am not going to the details. So, one important observation about this index and that is very much used in proof of theorem 1 and this is the key observation. So, you start with C again whatever C is you determine its index. So, if you deform this thing this closed curve smoothly to another closed curve C tilde. So, this smooth deformation so there are again very precise description of this thing in terms of mappings. So, again I am not going to do that this is the thing. The only requirement is this deformation should happen without crossing without passing through any without that is important any equilibrium points of system 1. So, when we deform this C into C tilde. So, nowhere we should come across the equilibrium points of 1. So, that is that we should keep in mind then one proves this is the important observation the index of C is same as index of C tilde. So, that is one important observation. So, in particular if a closed curve is C is such that in the interior of that C if there are no equilibrium points of 1 then I can keep on deforming deforming and make this length of this closed curve very small and eventually going to 0. So, what we get is so thus if the closed curve C does not contain any equilibrium points of 1 in its interior that is important in its interior. So, I can keep on doing this deformation we see that index of C is 0. So, whereas one can show. So, whereas one can show there are many results in computation of indices. So, it is one of them if C is a periodic orbit of 1 then of the system 1 then its index is. So, these are deep results they are not at all trivial results and if you compare the previous remark and this result you will see that theorem 1 is both. So, the next result. So, let me state it as theorem 2. So, this proves this implies theorem 1. Theorem 2 this is called Bendikson's criterion again in some sense it is a negative result if in any region I will write script R of the phase plane this quantity divergence of the vector field del f by del x plus del g by del y f and g are smooth functions. So, this is in 2 d this is referred to a divergence of the vector field f g has definite sign. So, that is either it is greater than 0 or in R at all points in R. So, either it is positive or negative then one cannot have periodic orbits in that region the system 1 cannot have periodic orbits. So, this is a simple consequence of Green's theorem. So, let me just describe that. So, Green's theorem in 2 d. So, it is something like integration by parts. So, there are of course, some certain smoothness assumptions on the curve and the domain I am not explicitly saying them. So, if again in just in terms of the f and g let me state that. So, if so let better write this c this is d if c is a smooth closed curve d as its interior as in the figure then this integral d of the divergence del f by del x plus del g by del y del g by del y. So, this is a double integral d x d y and this is given by line integral c f d y minus c d. So, when you parametrize the curve c. So, d y and d x will be expressed in that terms of parametrization. So, this will be a one dimensional integral that is line integral contour integral line integral. So, now if this region does have a periodic orbit then I can take that as c if this given region has a closed orbit to get a contradiction. And when c is a periodic orbit you see that using the system 1 the right hand side is 0 if c is a periodic orbit. And whereas, by hypothesis the left hand side is non 0 since this del f the divergent del f by del x del g by del y has a definite sign. So, the left hand side will be non 0 left hand l h s will be non 0 and that contradiction proves that the region will not have any periodic orbits. So, now we go to the celebrated Poincare Appendixian theorem. So, let R be a region in the phase plane. So, that means it is bounded and closed and bounded not containing any equilibrium states. So, let me state that not containing suppose an orbit. So, let me write it since we are only two dimensions. So, let me just write explicitly x t y t. So, that is a solution of system 1 enters the region for example, it can start in region R region R or it can even come from somewhere else enter the region R at time t equal to t 0 and stays in R. So, this is the assumption on the orbit. So, either it can start in the region R at sometime t equal to t 0 and then never leaves that region R for all future times or it can enter the region R at t equal to t 0 and then it does not leave the region for all future times. So, this is the hypothesis on the orbit then the conclusion this orbit itself is a periodic orbit if not. So, there are only two alternatives or it approaches a periodic orbit as t tends to be. So, thus under the hypothesis one will always have. So, this is to some extent a satisfactory result. So, it under this assumptions on the orbit it give the existence of periodic solutions, but however if you look at the hypothesis it is somewhat difficult to verify in a general situation. This can be done in particular situations, but to say that an orbit entering the region R will stay there for all time that requires again some analysis and it is not easy to verify in a general context. So, there are some simplified theorems will state one of them which are specific to second order equations. So, this remember this is valid for first order systems and this is more general than that, but whereas in that other theorem of Lenard we are going to state that the verification directly in terms of the functions involved in the equation and not in terms of the orbits of the equation. So, that is there is one difference. So, let me just briefly discuss the hypothesis, the contents and the conclusions of the Poincare-Pendixson theorem. So, if you compare theorem one compare with theorem one. Theorem one says if there are no equilibrium points then there are no periodic orbits whereas Poincare-Pendixson theorem requires that compact region not to have any equilibrium points. So, how should that look like that compact region in the hypothesis of Poincare-Pendixson theorem? How should it look like? So, it should look like typically an annulus type. So, this is something like that there is another one here and there is some let me call it x 0 y 0. So, x 0 y 0 is an equilibrium point of one system one and this is R. So, this is a typical example of R a compact closed and bounded of R in Poincare-Pendixson theorem. So, the requirement there is an orbit of one it enters may be a time t equal to 0 here t equal to t 0, but once it enters it will stay there. So, you can just visualize that it may be approaching a periodic orbit. So, this state that it stays in that R forces that orbit to approach a periodic orbit. We will see some examples some simple examples we will see. So, examples. So, these illustrate the Poincare-Pendixson theorem. So, this first one let me just take that x dot equal to minus y minus y plus x 1 minus x square minus y square y dot is equal to x plus y into 1 minus x square minus y square. So, because of the presence of x square plus y square it is convenient to express this equations in polar coordinates. So, usual polar coordinates so introduce and this are helpful in many situations in 2 D systems introduce polar coordinates. So, in instead of variables x y we have R and theta R cos theta and now would like to have differential equations for R and theta from the given equations from action y and in general you see that. So, compute this thing this is not at all difficult. So, this you can also check simple algebra R R dot is equal to x x dot plus y y and you also for theta you have that R square theta dot is equal to x y dot minus y x dot this is a more general set if you introduce this polar coordinates x equal to R cos theta y equal to R sin theta. So, then we will have differential equations for R and theta given by this and now if you substitute the given expressions for x dot and y dot. So, this implies so I have this and let me write that thing and you can easily check that R dot R into 1 minus R square. So, you already see that it is coming from that x square plus y square expressions and theta dot equal to 1. So, in this setup you see the R and theta variable they are decoupled. So, R does not depend on theta and theta does not depend on R. So, we can separately solve for them whereas, the original system x and y are coupled very much. So, we can solve that thing, but before that so let us see the applicability of Poincare Bendixson theorem. So, if you consider this annulus consider the annulus. So, this is half less than or equal to R. So, that is the annulus and here 0 0 is the only equilibrium point that is you can check 0 0 is the only equilibrium point. So, this R does not contain that because we are away from the R and now if you look at the equation for R you see that R dot equal to R 1 minus R square and in the on the internal spheres. So, at R equal to half. So, R dot is positive and at R equal to 2 R dot is negative. So, that means if an orbit in R try to approach the boundary of inner sphere it will be again pushed back in the region because R dot is positive. So, as soon as it comes here it has to again go back in the region and when the orbit tries to leave the upper the outer surface of the outer sphere outer circle then it will be pushed in because R dot is negative. So, something comes here again it has to be pushed out and here if it comes here it has to be pushed inside this annulus. So, therefore, an orbit which enters or which starts there enters starts in R will remain in R for all future times. So, this is what I was telling. So, in this particular case we are able to prove that the orbit never leaves the region R once it enters there or when it starts there. So, this but in general it may be difficult. So, Poincare-Bendixian theorem implies existence of periodic R in R it will be there. So, in fact in this case we can explicitly construct that. So, again recall we have this thing R dot equal to R into 1 minus R square and theta dot equal to 1. So, explicitly we have this. So, solve R is equal to 1 plus c e to the minus 2 t minus half theta is equal to you can just write t plus t 0 t 0 is some orbit constant c is also a constant. So, therefore, we have x t is equal to R cos t plus t 0 and y t R sin t plus t 0 and t 0 and this constant c are fixed by the initial condition. So, suppose c is equal to 0 then R is identically 1 from that equation. So, this in this case this implies the and certainly that is periodic solution. So, R equal to 1 is a periodic orbit. So, we are explicitly finding in this case that R equal to 1 is a periodic orbit and it is that orbit is obtained by taking s t is equal to cos of t plus t 0 and y t equal to sin of t plus t 0. So, let me just draw that. So, this is 1. So, you put the arrow appropriately in which directions counter clockwise clockwise direction. So, let me not do that thing and when c is negative just look at here if c is negative this quantity is some c there if c is negative then this is less than 1 and we are taking in the denominator. So, R will be bigger than 1. So, this if c is less than 1 less than 0 R will be bigger than 1. So, the orbit starts outside this circle of radius 1 and eventually approaches let me use different color it eventually approaches or it can do it counter clockwise also. So, if c is. So, this is c negative. So, for every c negative you get such an orbit and when c is positive. So, this is more than 1. So, in the denominator so then R will be less than 1. So, c positive R will be less than 1. So, from inside it will be approaching either clockwise or counter clockwise it will not go on that thing. So, just it will approach that that is the c equal to c positive in this picture. So, you can try a similar example. So, let me just say that. So, example 2 x dot is equal to 4 x plus 4 y minus x square plus y square y dot is equal to minus 4 x plus 4 y minus y same thing. So, and again you introduce the polar coordinates R theta it is similar to the previous example, but there is some subtle difference I would like you to work it out and see what the difference is. So, here we have R dot is equal to R 4 minus R square R square. So, again and theta dot is different theta dot is minus 4 I suppose and this makes a difference work it out from the previous example. So, again R equal to 2 is a periodic R. So, now you work in the region say 1. So, now next I will describe this Lenard's theorem theorem 4 this theorem is very well motivated by the Van der Poel equation. So, in fact the hypothesis are well suited to that Van der Poel equation. So, this applies to second order equations. So, f double dot plus f x f dot plus g. So, f and g are smooth functions. So, f c 1 functions satisfy g is odd that is g of minus x is minus g of x. So, necessarily g of 0 is 0 and g of x is positive for x positive 2 f is even. So, that is f of minus x is equal to f of x and if I integrate now little f that will be an odd function and the odd function f of x capital f of x which is integral 0 to x f s d s. So, you integrate this even function. So, the requirement on capital F has a unique positive root a positive root f x is positive and x bigger than a. And approach is infinity as x tends to a. So, as I said earlier now the hypothesis is purely in terms of the coefficients of the equation. So, no orbits are involved here it is purely in terms of the coefficients in the equation. So, somewhat easier to verify than the Poincare Bendixson theorem. So, here p here is the picture of f. So, this is f since it is odd it is 0 here and we require another 0 here. So, it is negative here and I did not say that and f x is negative or 0 x less than. So, this is f and g is that is positive that conclusion there exist a unique. So, there is more unique period and here in this case you can easily check that. So, 0 0 is the only equilibrium point in this case there is more to it. So, any other orbit will spirally approach that periodic orbit. So, the example as I said it is motivated by the van der Poel equation. So, recall this f double dot plus mu x square minus 1 f dot. So, here this is our f x in the Lenard's theorem and this is. So, capital f of x will be just mu x cube by 3 minus x. So, you see that there is a positive root namely root 3 and that will remain positive after that. So, the Lenard's theorem implies you existence of unique periodic study. So, that thing will come to an end of this discussion on periodic orbits.