 So, welcome to control of nonlinear dynamical systems, I think we started a rather interesting lecture last time, may be a little bit complicated, but we started discussing the Lassalle invariance principle. This is for scenarios like I said, where you have several asymptotically stable systems with a you know nice enough Lyapunov candidate right, but when you take the v dot that the directional derivative it turns out that it is only negative semi definite yeah and this is a sort of an issue yeah and because this only by your typically Lyapunov theorem this only gives you stability right, this is not enough obviously yeah you know that in most cases you know that these systems are asymptotically stable just by looking at the behavior in real life yeah just like this pendulum, simple pendulum okay. Then of the second motivation was systems with limit cycle behavior like the van der Paul oscillator I mean you have the linear oscillator of course then van der Paul oscillator is like a nonlinear oscillator. So, they have this limit cycle type of behavior which is you know also something you want to sort of capture or encapsulate in your stability or whatever notion you want to call it yeah okay good. So, in order to talk about the Lassal invariance we defined few things that is the invariant set, limit points and limit sets I hope all of you are clear on these definitions if you still have some confusion talk to me now or later anything is okay okay. And then we went on to state the Lassal invariance principle alright. So, I will restate it not stated but I will sort of explain what is going on the Lassal invariance principle basically constructs a bunch of sets okay. So, you basically start with the domain yeah and I if you remember I told you that this is the BR type of a set that you were working with last time where everything holds all the derivatives are negative and all the nice things happen yeah. Inside that set is where all the complications begin in Lassal invariance alright you have to construct a omega which is a invariant and compact set okay it has two properties invariance and compactness compactness was just closed and boundedness in reals. So, you need a closed and bounded set which is invariant okay which essentially means that any trajectory that starts inside this set remains inside this set and because it is a closed and bounded set it means that you have all your trajectories will remain bounded yeah if you start inside the omega set your trajectories remain bounded inside the omega set because it is not like some kind of elongated cylinder or some funny set like that no it's a closed and compact set yeah sorry closed and bounded set okay. So, it cannot just escape anywhere. So, already by constructing such an omega we have in a sense said something very nice about the system. So, remember that yeah so although we are not assuming Lyapunov candidates here we are still making some assumptions okay and then in this set we require V to be positive semi definite and V dot to be negative semi definite okay in this set we don't care what happens beyond it yeah we are saying we are we are restricting our entire analysis to this invariant set if you cannot for your system find such an invariant set you cannot apply LaSalle invariance yeah. So, be very careful I am not just constructing these sets for the fancy of it okay if you cannot construct this omega set you cannot apply LaSalle invariance okay and so within this set you have these two nice properties holding okay and once you have this set omega the set E is then constructed by taking the set V dot equal to 0 okay set of the states where V dot is exactly equal to 0 obviously we have said that V dot is less than equal to 0. So, equal to 0 is also part of omega set right therefore the set E is completely inside the omega set that is evident okay and once we have the set E we construct what is called the largest invariant set inside E this is the set M and the claim of the theorem or the principle is that if you start your initial conditions inside the omega set you are guaranteed to converge to the set M okay which is again a positive limit set and it is the largest invariant set okay notice we didn't say anything about the compactness of M or for that matter we didn't say anything about the compactness of E either. So, this is something you need to ask yourself I mean do you think E is a compact set or can you say anything about the compactness or closed and boundedness of the set E okay let me start simpler is the set E bounded yes I mean it is even evident by the picture but you should not again ever give me proof by picture yeah please do not do that the set E is bounded just because it is contained inside omega and omega is bounded so E and M both acquire those properties both are bounded sets what about closed is E a closed set is E a closed set you've done this argument many times yes how are you going to so you think it's closed you're doing proof by picture aren't you because I drew this line you think inside and outside those ideas work if you know the spaces here I'm not giving you any particular shapes and I mean I'm making something but M E could be very well something very complex looking do you think E will always be just a you know closed curve not necessarily right you could always play with the system just like I constructed all these funny dynamical system I could construct it so that E becomes a disc or something like that or maybe you choose the V very badly then also is possible so is E closed we talked about two ways of testing whether a set is closed one was this complimenting which you learned there was another way right which is what we've been using a lot to test whether its set is open or closed how do you do that we've mentioned this a few times no we never used this in this class if you remember I mean fine I mean maybe once the same thing what she said contains all its limit point means supremum is also a limit point actually but that's not the test we never used in this class we use something else to construct open sets and closed sets specifically in the proof of stability we kept constructing open sets on this side and that side how do we do that any other way of testing any other way of constructing open sets forget proving open sets how do I construct if I give you open set in one space how do I construct open set in another space thank you very much same with closed right inverse of closed is closed under continuous function inverse of open is open under continuous function excellent so in this case what can I say what is the set E set E is what it's written here V dot of yes I'm hearing something but not what is set E and why is it so complicated it's defined here what is set E okay you folks are not used to this notation please get used to this this is not difficult at all you will because this notation will show up in your exams from from this E definition set E is obvious what is set E no no no what is no nothing complicated at all from here I have defined E can I write E in shorthand in any smart way you just said it's a function continuous function is there a function involved here or not wow okay disappointed to say the least okay why what's so complicated why what's E you know I won't go ahead without this sensor there will be complete silence then what is set E I have defined it can it be written as something else any short form any shorthand what is the set E yes it's a function of what set E is the image of the function wow okay no unfortunately no so is it can I write it in another way or no is this it you are just saying words I need the math I can write one thing in 20 different ways in math and we've done this so many times you use these notations all the time what are we doing what did we do here what is it that we did here can you second somebody tell me what is it that we were doing here how did I construct this E is a different E by the way but still how was this constructed why why forget inverse image of open set why did I construct it exactly like this what was the reason to construct it like this what was the origin of this idea I mean you know because I am not going to spoon feed you here I am telling you the proof and giving you step by step so it seems easy if I give you a simple proof in the exam you will not be able to do it one step if you follow it like this yeah so what so what so we wanted a bound of what or norm x we wanted a bound of norm x is it I don't think what what did we want to bound on okay so so so we wanted the V to be upper bounded by alpha epsilon 1 square okay V of X to be bounded by alpha epsilon 1 square okay then how did I go from here to here what is the logic V upper bounded by alpha epsilon 1 square so why this and so you are saying the what is where does the set E lie by the way set E is in which space it's an RN fine I will tell you it's an RN so it's in the X space space where your states are sitting okay so V of X is I wanted to be less than alpha epsilon 1 square so from there I went to this one I just constructed minus alpha epsilon no don't do this and you do this in the exam also you have to construct is the same logic that I am asking you you are saying you are thinking it's obvious because I am telling you the steps when I tell you the steps everything seems obvious in the exam there are no steps right you will not be given steps you have to start from the beginning so why did I construct an inverse because I knew V X had to be less than alpha epsilon 1 square alright then X has to be V inverse of whatever alpha 0 to alpha epsilon 1 square or minus alpha epsilon 1 square to observe okay great what's what what am I doing different here where is this guy this guy isn't E also a set in the space of X okay excellent now can I write E in another short form he said E is image of a function I mean it's wrong but not at least it's a direction is E the image of a function is E the image of a function okay what is it a pre-image of thank you what is E then huh V dot inverse what what what V dot inverse of what thank you V dot inverse of 0 by the way is it just V dot inverse of 0 or is it enough is this enough this is not enough I can tell you why yeah these are things you have to be very used to you don't state is there is a prerequisite unfortunately but I should probably analysis is a little bit of a prerequisite here so X is in omega X is not an RN not the entire space if X was an RN this is okay but X is not an RN alright so what how do I fix this see in this definition here yes yes in words good in in math yeah thank you very much intersection omega this is precisely what it is okay please be little bit comfortable with this this will be a bit of trouble for you otherwise we will be you will have to do this okay now I am claiming E is closed why why is E closed so these are basically standard set operations I am not doing any analysis I am just writing a set in for example whatever you are saying in words these things have to be captured in math why do I have to capture it in math if you don't do that you can't claim anything about closed and open just by saying these words you can't prove that it is closed or an open set but as soon as I wrote it like this I can claim something is it closed why function yes yes function is continuous so a little bit more a little bit more sure function is continuous v dot is continuous okay because I said v is c1 which means it's a continuously differentiable function so the derivative is continuous so v dot is continuous great now what omega is closed okay all right sure but but what about this is this a closed set why inverse image of a closed set because you single I said this last time in fact write these few things down write these down in your notebooks what is it inverse image of closed set is closed under continuous function inverse image of open set is open under continuous function okay closed set contains its supremum and all limit points okay then finite element finite element sets that is finite sets with finite elements are always closed these are some key facts write this memorize this memorize this no need to prove for anything memorize it nobody is asking to prove it is not an analysis course inverse of open open inverse of closed closed under continuous function finite sets always closed closed sets always contain their supremum infimums and all limit points supremum infimums are basically limit points okay supremum and infimums are also limit points okay okay good so this is E is closed okay I found that E is closed can I say something about m m m it will lead you to more facts that you have to copy down what about m m is bounded we already said m inherit the property of omega is it closed see m is already invariant right we already said that m that is what we defined it at it is the largest invariant so let us forget about that it is bounded closed is the only a property that is left is m a closed set so m is a positive limit set I already mentioned it here okay are limit sets closed sets anybody know limit sets are what I defined it right it is a set of limit points so it is the points to which the sequences converge now now what is a closed set a closed set is a set which contains all its limit points I just said that okay it is one of the points I hope you noted it down okay now what I am asking is is the limit set itself a closed set which means which means does the limit set contain all its limit points almost sounds like I am doing a rap song but I am not does the limit set contain all its limit points okay even if you do not know the answer is yes okay because a limit set will have no limit points other than the elements of the set itself okay think of the circle here this is a limit set we discussed this right in fact this is a too simple a limit set wherever you start it will just making the circle so forever it is there okay so if you see there are no other limit points other than this okay remember the limit points depend on the initial condition so do not get confused by the other circles right so all of this is corresponding to an initial condition okay so I apologize sorry so this the limit set the limit points of the limit set are the limit set itself okay note this limit points of a limit set are the same set okay so if you call omega or you can just write in shorthand omega if you can if you remember from our notes limit points of omega are the set omega cannot be beyond omega okay which means omega is a sorry not omega m m m m sorry m m was the limit set not omega m limit points of m or any limit set are the same set itself cannot be beyond that set okay it is you can check with any sequence if you are even if you not able to prove it you can check any sequences half one half one sequence the limit set is one and a half one comma half right it can have no other limit points because anyway it is too discrete point so it is a closed set therefore m is a closed set limit set is a closed set okay any limit set is a closed set okay these are facts that I want you to memorize we will not have the luxury to prove it right now okay good so we have a bunch of very nice sets I hope you are now convinced that all our sets are super nice except for the D domain set inside that we had omega which we constructed which nobody gave it to go to us which was compact and invariant inside that we found in set E which is also compact and invariant now because it has closed bounded sorry it is only compact the set E is only compact not invariant and then inside that we found the largest invariant set therefore m is also compact and invariant so m has exactly the same properties as omega okay so you start from a similar type of set and you end in a similar type of set okay so Lassalle invariance is that is why geometers love Lassalle invariance principle because it is very mathematical okay but this is also very useful okay so just keep this in mind that all these sets have very very nice properties and the ability to apply the Lassalle invariance relies on you having this omega set if you do not cannot apply okay we will look at the proof but before that I remember we were doing the example so we will restart the example okay this was the pendulum correct and what did we do we took just the energy as the V function okay we do not care we only need semi-definiteness and it is okay it is in fact if you take any x1 x2 it is semi it is not going to be negative because of 1 minus cos x1 is lower bounded at 0 same this guy so it is semi-definite if you want positive definiteness then you have to fix x1 to a small range minus pi to pi but we do not care to have positive definiteness okay we are semi-definiteness is enough so we choose a larger range of x1 why why do we choose a choose the larger range of x1 do you remember for the pendulum to have more than one equilibrium because we want to show the power of Lassalle invariance so we were capturing this equilibrium and this equilibrium also by x1 minus 2 pi to 2 pi but x2 was free to be within R and this was our domain D okay now to construct omega we did some interesting manipulation okay how did we construct omega we constructed omega using V itself okay so we did somehow the reverse thing first we chose a V and use that to construct omega how did we do that we said that so first we did this V dot we saw that it is less than equal to 0 okay wherever we take it does not matter it is definitely negative semi definite the domain of x1 x2 everywhere in D in fact anywhere in D this is negative semi definite so I am not concerned about the set omega anymore so this V will help me construct omega so what do I know V is non-increasing okay so if I start the V at some value it will stay below that value okay so I am going to construct the set omega such that V remains below this value okay and how did we do that I think we did we wrote it we write it I think I erased it if you remember we construct the set omega as I wrote this by the way x1 x2 in D such that V less than equal to some constants okay and this is the same as or this is same as saying V t0 less than equal to this constant I am more than okay because V dot is negative semi definite in all of D okay and to do this I just wrote this guy so I want this quantity to be less than equal to C okay so how do I do that I take the smallest value of this which is 0 right and then I have x2 is bounded by square root 2C okay this we calculated last time so therefore I get my omega as this set okay I get my omega as this set okay just note that it is written here this is from how I get this the only thing is I will not make them open ended I will make this closed okay I have made this closed just to ensure omega is a closed set if I take the open bracket then it is an open set that I do not want I want omega to be closed so this is what I take my omegas okay this is fine I mean in fact I can take my domain also domain has no restriction it can be open or closed okay so I take the domain also with a closed bracket not the open bracket okay so so so this is my omega I have a nice omega now okay now I am ready to apply the lasal invariance forget this yeah so how do I apply lasal invariance so yeah so I construct the E set what is the E set the set where V dot is exactly 0 right so that is just requires me to have the x2 term to be 0 okay I do not care about the x1 term okay so this is the set okay again I will make this the closed bracket not open bracket okay yes why are we taking the maximum value we are taking the minimum value yeah because it is just an inequality requirement look at this I want vx less than equal to c we have vx less than equal to c so this is I want x2 square by 2 plus k1 minus cosine x1 less than equal to c okay you can look at it in a couple of ways first is I am not interested in bounding of x1 because x1 is already bounded the way I want okay so this is only going to make a positive contribution right it is only going to make the left side larger so if or else you can write if you may I will write this do you believe me huh so effectively I made this 0 only you know see if I if I took the maximum what will I get I will get this as less than equal to x2 square by 2 plus 2k but that is not helpful for me this is also an upper bound this is also an upper bound I cannot compare the two although no I do not require that I do not require that I only need vx to be less than equal to c why the maximum value okay it may not reach the maximum value understand yeah so so this will not be whenever you are comparing inequalities this is pretty standard you have to be able to write it like this once I write it like this I can compare this and this because it is a natural inequality going from left to right but if I write it like this I cannot compare this there is no guarantee which one is larger this is standard in inequality I know you are thinking max of v has to be but no it does not have to be because the max of v may never be achieved yeah yeah we are taking but it may not hit all the sides we do not care we only need this this much because if you do this you are significantly restricting x2 significantly which we do not need which we do not need which we do not need okay we come up with the best estimate a largest possible estimate of x2 okay all right