 So, recall in the previous class, we are discussing two important theorems. One is Poincare-Bendixson theorem and another one is Lenard's theorem. So, both give sufficient conditions for the existence of periodic orbits in 2D. Let me stress that again. These are very 2D specific theorems and this is applicable for 2D systems of first order equations and this one, second order equation f double dot plus f x f dot plus g x equal to 0. Under certain conditions on f and g, which we stated in the previous class. So, Lenard's theorem proved that there exist unique periodic orbit surrounding the only equilibrium point 0 0 and all other non-trivial orbits. So, here the trivial orbit is only the equilibrium solution. All other non-trivial orbits tend to this periodic orbit as t tends to infinity. So, that is an important conclusion of this Lenard's theorem. Today, we will discuss another class of equations where the phase plane analysis can be done somewhat easily and we obtain a complete phase portrait of 2D systems and these are termed as conservative equations. We will shortly see why this name conservative equation and these are again second order equations of the form x double dot plus d by dx of v x equal to g. So, this function v is assumed to be smooth. The function v is called the potential function. So, the terminology again comes from classical mechanics. So, consider this quantity. So, if x is a solution of that conservative second order equation, consider this quantity f naught t square plus v x. So, let me call this as e t e of t. So, x t is a solution of that equation. So, this represents kinetic energy and this represents potential energy. So, this is the total energy at time t, total energy of the particle. So, if we differentiate this e with respect to t, what we get is x dot t x double dot t plus using chain rule. So, we have this d by dx of v of x t and now x dot is common and if you look at the other terms x double dot t plus d by dx of v x of t and that is 0 by the equation. So, what we are seeing by this simple calculation that total energy is conserved. So, it is the same for all t. So, therefore, e t is identically equal to e a constant. So, this is the reason why that second order equation is called a conservative equation. So, and the position x t and velocity x dot t, these are called as phases of the particle. Again terminology comes from mechanics. So, when you do analysis involving phases that is phase plane analysis and when we describe the position and velocity for all time and that is described as phase portrait. So, that is phase phase plane analysis. These are the terminology coming from this. It is plane because there is only position and velocity. So, it is two dimensional phase plane analysis phase portrait refer to this analysis and description of all the orbits and that is what we are going to do for some simple examples. So, again the key equation is the following. So, this is the key equation. This half x dot t square plus v of x of t is equal to e and this is remember a constant. And this term is always non-negative. So, this v has to be less than or equal to e. So, for different levels of this we will see how this phase portrait will change. And so, once we are given this potential function and this different energy levels e we take and that will restrict the values of x and then we use this key equation to analyze the behavior of x t and x dot t. That is the position and velocity of the particle. So, we will explain with through simple examples how this is done. The first example is pendulum equation. So, we are not adding any force. So, this is unforced and there is no damping. So, this is undamped. So, this is the equation x double dot plus k sin x equal to j. So, k is some fixed cost, k is a positive cost. So, this can be written as a conservative equation. So, take there are many choices and I will explain why this particular choice. So, this you take v x is equal to k minus. So, you can always add a constant to the potential function without changing the equation and that constant can add it in such a way to achieve some simplification for the potential function. So, here you see that since it is 1 minus cos x. So, I added this constant. Otherwise, it is just minus k cos x. We can also take that one. I have added k here just to make this potential function non-negative for all x. So, then the equation becomes x double dot plus d by d x of v x equal to 0 and therefore, the key equation in this situation is. So, half x dot t square plus v x. So, remember this is constant and our v is given by this. So, remember this equation. So, that is what we are going to do now. So, let me draw this potential function as a function of x and let us see how it looks like. So, this is the function v of x, v of x is given here. So, remember this is v of x and the graph of v of x suggests that we take these different energy levels e equal to 0 and then e between 0 and 2 k and then e equal to 2 k and then finally, e bigger than 2 k. This graph itself suggests. So, let us take one by one. So, just observe carefully. So, let us start with e equal to 0. You know that again the key equation you remember that half of x dot t square plus v of x t equal to e and now, we are taking e equal to 0 and v will be 0 only at these points. These are all even multiples of pi and x dot will be 0 because of that conservative key equation. So, here we obtain all equilibrium points n pi 0 and n integer. So, all even multiples of pi we obtain at this energy level important to remember that e equal to 0 we are taking now. So, x dot is 0 there and we obtain these equilibrium points of the system. So, what about 0 less next level? The next level 0 e less than 2 k and now, if you see observe the graph of v the values of x will be restricted to these portions. So, these are symmetric intervals around each of these equilibrium points 2 n pi 0. They are around that and now, x values restricted only to that interval. So, you can explicitly calculate what this point is that is not difficult. So, it is strictly in the interval here minus pi pi and similarly, it will be in different intervals. So, since all are have similar structure. So, let us calculate let us concentrate around this interval z 0 the equilibrium point 0 and corresponding to that values of x we obtain periodic orbits which are exhibited here and arrow is this. So, we obtain the periodic orbits when 0 e is less than 2 k. So, obtain let me write here periodic orbits around each equilibrium point 2 n pi 0. So, here I have just shown it around 0 0 and same thing happens at all other points and next when you go to this energy level e equal to 2 k that is that level and here we do not have any restriction on x just to the point that see now, in this case first we obtain we obtain equilibrium points 2 n plus 1 pi 0. So, n again any restriction integer which we did not obtain in the energy level 0 now with energy level equal to 2 k we obtain all these are the equilibrium points odd multiples of 2 pi and now you see start an orbit at a point different from this equilibrium point and this is what we get this is the I am just this is the one actually there are 2 orbits both approaching 2 different equilibrium points the upper one is as t increases the arrow is marked like that and the bottom one like this because in the upper portion x dot is positive. So, x is increasing and in the bottom orbit x dot is negative. So, it is approaching it going to minus pi as t tends to infinity. So, we obtain addition to this thing we obtain 2 orbits and together they are referred to as. So, here I highlighted it. So, homo clinic is not there let me it is a separate x as it is going to separate 2 kinds of orbits and the final one e bigger than 2 k and there is absolutely no restriction on the values of x and from the key equation you also see that in this case e bigger than 2 k let me write that you also see that from the key equation x dot t square is always bigger than or equal to there is half let me write that half e minus 2 k and now e is bigger than 2 k. So, this is always positive. So, the x dot t is always bounded away from 0. So, either it is always positive or again always negative. So, that is why you get 2 orbits. So, one here is since x dot is positive. So, x is going in that direction and the bottom one is in this direction. So, in this case for the undamped unforced pendulum equation we obtain a complete phase portrait of this orbits several orbits we see that. So, there are period again let me repeat it. So, for there are 4 energy levels that to be considered in this case. So, e equal to 0 we obtain all equilibrium points which are even multiples of pi and 0 and for the energy level 0 let me write that e less than e less than 2 k we obtain periodic orbits around each of the equilibrium points 2 and pi 0 and for e equal to 2 k we obtain the equilibrium points which are odd multiples of pi and in addition to that we also obtain 2 orbits and there around each this is important. So, the same picture happens at all other points. So, that is important. It is not just restricted to this 0 the same thing you repeat it to other points other 2 n multiples of e 1 multiples of pi and for e bigger than 2 k we obtain unbounded solutions and either the solution is always increasing or always decreasing and as shown in the picture second example. So, this is again unforced undamped Duffing equation. So, earlier we have discussed about this thing and now we are taking the undamped Duffing's equation and this is again second order equation given by. So, there is x dot term is missing that is the damp one. So, we are taking delta equal to 0. So, here we take looking at the equation we take v x. So, here we are not going to add any constant. So, just to keep it like at x square by 2 plus x 4 by 2. So, with this potential function the given equation is written as x dot plus d by dx of v x equal to 0. So, therefore, we have the key equation again x dot t square. So, any solution will satisfy this this is a card. So, this is the before drawing the phase portrait. So, let me just spend a few minutes on this potential function. So, that the explanation of phase portrait will be very clear. So, here the v x is minus x square by 2 plus x 4 by 4. So, this is symmetric in x v of x is equal to v of minus x and quickly draw I am showing this mode in the next picture. So, this is just v here. So, this is a 0 here and then if you solve this thing you have two more 0s minus root 2 and plus root 2 and this is negative here 0 here. It is symmetric very smooth because it is a polynomial just you can do that yourself. So, this minimum occurs. So, this is level minus quarter ok. So, this occurs at plus 1 and this is minus 1 and recall that in this case we have 0 0 plus r minus 1 0 are the equilibrium points and you may recall the linear analysis we have done at this stage to understand the behavior of orbits very clearly. This we have already done it. So, this 0 0 is always unstable and plus r minus 1 0 in this undamped case they are stable, but not asymptotically stable. So, you can recall that thing and the graph of we suggest that the energy levels we are to consider is equal to minus quarter and then we have this minus quarter less than e less than 0 and then we have equal to 0 and e. So, let me just for reference. So, this is the equal to 0 level and this one in anywhere in between. So, there are many. So, this is minus quarter less than e less than 0 and then we will have this x axis is also energy level 0 and the last one. So, anything above e. So, again in the like in the pendulum case we also have 4 energy levels and at e equal to 0 you only get. So, let me again see it here. So, this level we obtain only equilibrium solutions equilibrium points plus r minus 1 0 and when energy level is between minus quarter and 0 the values of x is restricted to this portion it is called potential well in physics language. So, around each of this equilibrium points. So, they are around this plus r minus 1 0 and again we obtain periodic solutions around that and then we have to find out then at energy level 0 we do obtain this equilibrium point 0 0 and then when we start the orbit at a different point we do obtain again separatrix and when e is bigger than 0 again you see this value of x is restricted to this interval this interval. So, it is restricted to that. So, again we obtain periodic solutions and these are shown in the next picture. So, that that is what I am going to show you. So, just remember this diagram it is more elaborately done here. So, what we saw in the previous picture same thing is done here and say at energy level e equal to minus quarter we obtain only the equilibrium points plus r minus 1 0 and then when the energy level is between minus quarter and 0 we obtain this periodic orbits I am just showing one around each of this equilibrium points plus r minus 1 0. So, this is minus 1 0 and that is this is minus 1 0 and that is plus 1 0 and when energy level is exactly 0 apart from this equilibrium point 0 0 we do obtain orbits one on the right side and another one. So, this is single orbit let me just explain that it is not periodic orbit that you should keep in mind because this is an equilibrium point. So, another one and the nature of these orbits clearly indicate that the origin is unstable and when again e is greater than 0 as I said. So, it is restricted to x is restricted to this interval and we obtain this it is going in this direction. So, that is the periodic orbit. So, we have in the case of undamped unforced duffing equation we do have again several different kinds of orbits at one energy level we get only the equilibrium points and then we get periodic orbits then we get separate ricks and then again periodic orbits. So, in the in the in the previous example in the final thing we got only unbounded solutions, but here at all levels we get only periodic orbits. So, that is a big difference. So, let me just make few remarks about this duffing equation again. So, again here let me just concentrate on this level energy level 0. So, we obtain let me write that thing let us go to the next one. So, duffing equation again so, duffing equation let me just call make a remark about that. So, we have again half f dot t square plus this let me write this x square by 2 plus everything x t x 4 t by 4 equal to 0. So, energy level 0 is equal to 0. I am considering that particular energy level want to make a remark on that. So, first we obtain the equilibrium point 0 0. So, this let me just draw that. So, this 0 0. So, if you start the orbit at a different point then this one then we got. So, this let me draw that. So, for example, if I take x dot positive. So, we obtain this. So, this will go up to that root 2 root 2 0 and similarly we obtain the other one here. They are supposed to be symmetric and the remark I want to make is following. So, these are. So, this is orbit suppose it starts here. So, it will make an entire round here and come back to the origin it can come here only at infinity and you can see that as t goes to minus infinity also it goes to 0. So, this is equilibrium point. So, remember that. So, and similarly here if an orbit starts here it starts here it goes all the way around and again eventually goes to 0 as t goes to infinity and here also it goes to 0 as t goes to minus infinity. So, let me just draw this. So, both these orbits have the spatial property. So, these orbits so let me now write these two orbits tend to the equilibrium point 0 0 both as t goes to infinity as t tends to infinity and t tends to minus infinity. So, this is very special here and such orbits are called when an orbit tends to an equilibrium point both as t goes to infinity as well as t goes to minus infinity. See in the previous case that does not happen you can just see that in pendulum case that does not happen as t goes to infinity it goes to one equilibrium point and t goes to minus infinity it goes to another equilibrium point. It does not go to the same equilibrium point and such orbits are called homoclinic orbits. So, in this case the Duffing equation exhibits the existence of there are two homoclinic orbits and they corresponding to energy level 0 and again when you go e positive. So, this these are also now act as separate x. So, these orbits separate a two kinds of periodic orbits one inside this. So, we have here several we saw that thing there are several periodic orbits here and here around and above that we have this another periodic orbits. So, let me give different color. So, this is you saw that right. So, this corresponds to e positive and this these two orbits corresponds to minus quarter less than e less than e less than x. So, for every e there we have periodic orbit. So, that is fine. So, they are not difficult, but you have to just work out patiently in order to see the phase portrait of the orbits. So, with that thing we come to end of this discussion on qualitative theory of differential equations. So, before ending let me make few comments. So, this qualitative theory of differential equations is an very important and quite deep subject. What we have done in the last few lectures? We have been able to scratch the surface of this vast and important subject. So, if you understand this properly you will be able to read what is the difference more advanced topics and there are excellent textbooks providing that material. So, we have that we have included in the references. So, you will be able to understand much more advanced topics once you grasp this elementary knowledge. Thank you.