 Now, we have now constructed the set E ok. Now what? This is the slightly complicated part everybody has trouble with using this part of the La Salle invariance is to find the set M ok and M has to be invariant and inside E and it is the largest invariant set ok. Can't be anything cannot be any invariant set larger than this ok. Now, how do we do this is how do we usually do this? I mean because I do not know how to find the largest and smallest right I mean it is not like an optimization problem that I find largest and smallest because you can see that these will be very complicated sets yeah this kind of optimization you are truly impossible to do. Basically you are you will not find an engine which will solve this optimization problem if I actually thought of finding the largest invariant set and all because you will have to initialize you will have to take all initial conditions and try to do something ok not possible not possible. So, what I will do is these or what everybody does not what I will do yeah what everybody does is assume E is invariant ok. So, we actually assume that E is itself an invariant set ok. So, this sort of helps us continue forward. What does it mean if E is invariant ok? What do you need for E to be invariant? It means if I start inside E I must remain inside E, but look at the interesting thing what is the set E is this guy ok alright it is this guy. Now, the first term is a continuous could be a continuous quantity right it can take any value between minus 2 pi to 2 pi notice this guy can take any value between minus 2 pi to 2 pi. So, I cannot really conclude too much about this, but what we use is the fact that the second term that is x 2 has to be 0 is fixed. So, for me to be invariant and to remain inside E I need x 2 to be identically 0 for all E in this case yeah I need x 2 to be identically 0 and what does this mean if something is 0 forever or some constant forever and all these are continuous right x 1 x 2 is continuous because I have a differential equation right. So, obviously continuous nice properties. So, what does it mean if x 2 is 0 for all time or any constant for all time no not really right that would mean all states are 0 only one state being 0 is not just like you know that alpha comma 0 is not an equilibrium right. I hope you understand that alpha any x 1 is not the equilibrium otherwise it would be this 0 velocity 0 velocity 0 velocity all this will be equilibrium. So, this is not an equilibrium no what can I say if any quantity is if any function is the constant function what do we know about the function this is a function of time right I mean although I write it like this it is actually a function of time then x 2 dot is identically 0 yeah x 2 dot has to be identically that is only a function can be a constant function alright which means what what is x 2 dot anybody remember this minus k sin x 1 minus c x 2 this has to be identically 0 all of this to make e invariant all of this is to make the set e invariant without this happening set e cannot be made invariant ok. Now, I already know that this guy is already going to 0 right and the left hand side is already 0. So, what do I know what do I conclude this also has to converge to 0 not converge to 0 whatever has to be 0 has to be 0 and what do I know sin x 1 is equal to 0 at what x 1 equal to what n pi right. So, 0 pi I will not write 2 pi because after that it is repetitions right I will just write 0 and pi and you see 0 and pi are both in our set omega right both in our set omega. So, we started by assuming that e is the invariant set, but to ensure that e is in fact invariant I required sin x 1 to be 0 which gave me exactly 2 values not all of e not all of e ok everything else is the same right 0 and pi all other angles are the same you can write them if you want, but yeah you can write 0 and pi I know 2 pi is also there, but it is all the same. So, I am not including them. So, what do I have I have that the invariant set is what m is equal to what now what is the set m now 0 0 and pi 0 ok alright. Just by starting with e being invariant I have concluded that only these 2 points constitute the invariant set ok. So, m is the invariant set the largest invariant set you cannot have anything more why because I started with e being invariant which is the largest set I had to work with from there I could only get out that these 2 points give invariant and what are these points these are exactly the this and this right. So, what does Lassal invariance say because I have satisfied all the requirements of the Lassal invariance I constructed a compact invariant omega right I had a v and a v positive semi definite v dot negative semi definite in omega in fact it is in d, but in omega also ok. And so and then I constructed the e and the m sets. So, I have satisfied all the requirements of the Lassal invariance principle. So, Lassal invariance principle says that if I start in this set omega not anywhere in the domain if I start in this set omega then I will converge to this ok. So, interestingly what I have been able to prove is still not if you say everything because although there is no restriction on the angle right it says if you start in omega ok. So, I can start at any angle because minus 2 pi to 2 pi close set includes all the angles no problem ok. And it says I will converge to either this point or this point it does not specify which point you can converge here or here no problem. But the velocity is restricted there is a restriction in the velocity although again in reality you know what I can start at any velocity theoretically and I will converge to an equilibrium one of the two equilibrium for a pendulum ok. In fact, I will converge to this equilibrium only I will never converge to this equilibrium if the velocity is anything but 0 I will converge to this equilibrium only ok. So, in fact any velocity is allowed in reality, but I have only been able to prove this much ok. So, there is a caveat ok I mean even Lassalle invariance does not let us prove what we know in reality. But this is more than enough because I can choose any c which will give me velocity 1 e power 6 I will choose a c accordingly and tell you yeah this will work yeah because all I have to do c c was nothing c is just a artifact of our imagination ok. So, c is nothing we created it if you give me 1 e to the power 6 1 e to the power 10 velocity for whatever reason you want to put a rocket thruster on the pendulum, but you will I can still guarantee that you will converge to the equilibrium by choosing a large enough c ok. But yeah in theory once you fix a c your velocities are bounded yeah not unbounded velocity unbounded velocities are not ok which is obviously fair enough for all intents and purposes ok. So, do you all understand how to use the Lassalle invariance I have also as the TS to try some examples I hope they try something other than the pendulum anyway with Lassalle invariance application of the general Lassalle invariance with multiple equilibrium ok. So, it is important that we follow this. Now, I also want to state if you have no other no questions on this I want to state the stability versions of these theorems ok. So, Lassalle invariance was obviously by Lassalle yeah, but then the stability versions of these theorems where almost parallely developed by Krasovsky-Barbersian ok. So, therefore, I like to call it Krasovsky-Barbersian-Lassalle theorems some books call it Krasovsky-Lassalle some call it Barbersian-Lassalle but whatever it is better to put all the names. So, these are very close to Lassalle theorems, but they only deal with stability of the zero equilibrium ok. Here you have a general multi in multi stability what you can call multi stability was multiple points you can reach a limit set yeah. Here you specialize to reaching the origin only yeah. And how do we do that? We no longer start with V which is only positive semi definite we actually start with a proper Lyapunov candidate ok. This is now a proper Lyapunov candidate positive definite in C1 and we just have semi-definiteness on the V dot such Lyapunov candidates are called non-strictly Lyapunov function ok. For all x in D notice for all x in D not in any omega yeah. You see that the omega is carefully missing here there is no omega here. Why because I know that I can construct omega just by this fact we just did the example and then if I define the S similarly whatever we called E here is the S here just because it is stability theorems I am using a different notation that is all. If you define the S as such and if x equal to 0 is the only invariant trajectory in S this is just wording that some texts use but this is the same as saying the set x equal to 0 is the largest invariant set inside S then x equal to 0 asymptotically stable ok. On the other end then you can also state a global version which says that if now you do not have a domain you have all of Rn ok again negative semi definite in all of Rn and you have the same things happening then you have global SM dot extensibility ok that is it. If the domain goes away then you have global SM dot extensibility ok. Notice all these results LaSalle invariance, Barbashin Krasovsky LaSalle all these are only for time invariant systems or autonomous systems ok. For time varying systems the results are significantly more complicated ok I mean not easy to apply these results ok because the notions of limit points and limit sets itself becomes very messy ok alright. So these are the stability versions of the theorems alright now if you go back to our example how would I do the stability version I mean in fact this is actually pretty straight forward yeah what will I do I will no longer take minus 2 pi to 2 pi I will just take minus pi 2 pi because I only want the bottom equilibrium included in the set now because I am looking to do stability of 0 equilibrium. So I will not include the top equilibrium in my set D ok. So my set D will now contain only minus pi 2 pi ok anyway this is something I will ask you to complete yeah but the rest of the steps are exactly the same you still have a V ok notice that as soon as I choose x 1 and minus pi 2 pi this is positive definite this we discussed right earlier in the example we did this example ok. If x 1 is in the minus pi 2 pi range not minus 2 pi to 2 pi range in the minus pi to pi range this is a positive definite function right I hope you all agree yes because this function will not be 0 anywhere but at 0 0 the problem with the larger ranges 2 pi is included then it is not positive definite but if 2 pi is not included so minus pi to pi then 0 is the only point that is x 1 equal to 0 is the only point where this can become 0 yeah x 1 x 2 equal to 0 nowhere else ok. So the only difference is in the domain ok. So it will become minus pi to pi cross r after that it is very straight forward V is positive definite V dot is negative semi definite in the domain ok and after that the same analysis goes through it this analysis is not changing the set E or in this case the set S whatever you can call it S but this is going to be exactly the same right because V dot is exactly the same so this set is the same if this set is the same the set M is the same without this case because minus pi to pi does not contain pi at all minus pi to pi open set remember we took for this domain is minus pi comma pi ok for applying the Bar-Baschine-Krasovsky-Lasall we took we take this. So this is not even part of this set ok so M contains only the origin M contains only the origin ok all this so nothing changed in the analysis I just shortened my or reduce the size of my domain and just by reducing the size of my domain instead of applying the general Lasall principle I am applying the Bar-Baschine-Krasovsky-Lasall why because by shrinking the domain I removed this equilibrium minus pi to pi does not contain this gate yeah because I excluded pi ok that is all so by shrinking so what am I doing by saying minus pi to pi it is starting from here and here so it is like it goes from here all the way to slightly here that is it minus pi to pi is just everything but that vertical upward vertical line ok so I have essentially taken everything but that top vertical line and so I have skipped the second equilibrium once I skip the second equilibrium there is only one equilibrium ok and remember I always told you when you have one equilibrium only then you can talk about you know stability or global stability multiple equilibria then you have to think Lasall type ideas. So that is all I shrunk the domain so that I have one equilibrium in this domain and then applying Bar-Baschine-Krasovsky-Lasall is very easy because the set E remains the same and the set M contains only the 0 equilibrium and that is proof that 0 is asymptotically stable not global because of this restriction in the domain is not global but whatever asymptotically stable more than enough ok and this is in fact sounds more general I did not have to do all this omega construction and all right it gave me asymptotic stability. So basically what it is saying is if you start anywhere but at this equilibrium anywhere but this equilibrium so you can start anywhere arbitrarily close to it but away from that equilibrium then you will fall here which is what is the reality also right you can verify that very easily in experiments all right is that clear ok. The spring mass damper example is also very similar and much easier this is a spring mass damper is a just a linear version of the pendulum, linearized pendulum if you may. This is what the dynamics looks like x 1 dot is x 2 x 2 dot is minus k 1 x 1 minus k 2 x 2 the constants k 1 and k 2 depend on yeah spring mass coefficients all right and what do I do I take V as the energy right this is the what is this term potential energy potential energy is the spring energy energy stored in the spring this guy and this is the kinetic energy term ok spring energy potential energy kinetic energy if you take the derivative and of course you can see that this is all nice and radially unbounded in fact V is radially unbounded I hope that is clear to you and this is yeah in fact in all of R n the domain you do not even have to worry about the domain here sorry all of R 2 ok this is V is valid in all of R 2 V dot is negative semi definite in all of R 2 and V is radially unbounded in all of R 2 ok. So, once you have that it is again the same kind of S construction right because sorry again E has been used it does not matter. So, what is the set E now the set E is just V dot equal to 0 which is the same amazing comes out to be exactly the same you just need x 1 comma 0 all sets of the form x 1 comma 0 ok. Now, I know 0 has to remain you have to the second variable of x 2 has to remain at 0 for all time. Therefore, x 2 dot has to remain at 0 for all time which means that minus k 1 x 2 minus k 2 x 2 has to be 0 for all time same logic, but x 2 is already 0. So, x 1 also has to be 0 for all time. So, what is the largest invariant set in E x 2 was already 0 now I also want x 1 to be 0. So, 0 0 is the only invariant set inside this ok from every other point you will move ok. So, they are not invariant set ok you cannot find any other invariant set. So, this is the largest invariant set inside E ok. So, this is in fact, your whatever M set if you know yeah and. So, any trajectory which starts anywhere in fact, will converge to the 0 0 equilibrium ok. Again it is a linear system you could have very well computed the Eigen values and obviously, found out that they were negative and you would have concluded exponential stability. This is just giving you an alternate way, but this is just a way of using this theorem that is all I mean you have already seen that I can use it for non-linear systems also. In fact, you can you use it Lassal invariance like I said is a method of choice for typically for geometers because like I said because they like to use the energy as the Lyapunov candidate and whenever you use the energy of the as a Lyapunov candidate for these conservative type systems that is with no external forces invariably your V dot will be 0 right because energy is conserved right it will not be less than equal to anything will be exactly 0 and from there if you want to conclude anything you have to use Lassal invariance ok you have no choice but to use Lassal invariance ok. So, Lassal invariance is a pretty very very strong method yeah. I am not sure if you will have time, but there is also what adaptive control folks use is the notion of Barbalat's lemma is a different way of doing this Barbashin Krasov Lassal ok which is more I would say you might find it easier to use and it can be used for time varying systems also ok here you cannot yeah you cannot use these for time varying system these results in this form. But Barbalat's lemmas also cannot be used for multiple equilibria case it is so the general Lassal invariance principle is a very powerful result yeah it is a very very strong result you cannot get multiple equilibria type results from any typically any other method ok not easy not easy you have Poincaré-Benedictson theorem and all that but not easy they have all sorts of interesting assumptions which you may not satisfy yeah so Lassal invariance actually a very very powerful tool that way yeah.