 stage 2 now we go to the general class k function nothing much changes you already have this stuff so this is going to help you anyway but now all this alpha epsilon square cannot happen I hope that is evident because alpha epsilon 1 square came because I assumed alpha norm x square as my class k function that cannot happen so I restate everything I have v tx greater than equal to phi norm of x all right for all t greater than equal to 0 for all x and b r and I also have v dot to be semi definite again under these assumptions so I do I write the same kind of statement this is the same kind of statement that I wrote that there is a lower bound phi norm of x which is this basically I am rewriting this here and then I have an upper bound this guy this is just coming from my semi-definiteness condition this guy is just from the semi-definiteness condition now what do I claim or what do I want earlier I wanted alpha epsilon 1 square on the right hand side now I say I just have on the right hand side less than phi epsilon 1 yeah I just use this function itself so I have phi epsilon 1 now remember these functions phi have beautiful properties they are 0 at 0 they are continuous they are strictly agree we are monotonic therefore if if you have phi norm of x to be less than phi epsilon 1 which is what you do then norm x has to be less than epsilon 1 I hope this is sort of evidently 0 at 0 monotone increasing function so if the function value is less than phi epsilon 1 then the argument has to be less than epsilon there is no two ways about it so this is rather neat sort of a result okay and this is also in this case there is invertibility in play yeah because the both sides are real numbers argument of phi is a real number what it outputs is a real number okay and phi invertibility is guaranteed by what monotonic the monotone function invertibility is guaranteed done so if phi x is less than phi epsilon 1 norm of x is less than epsilon 1 okay so if I can prove this that this happens just like before instead of alpha epsilon 1 square I just have phi epsilon 1 then I am done okay so not too different so what do I want for this to happen so I just have to write it in slightly more complicated language that is all here I had minus alpha epsilon square epsilon 1 square alpha epsilon square here I have to write it as this way this is what we need for this to happen yeah this condition is rewritten as this I hope you understand just by taking the just this condition is rewritten as this way okay just by taking phi inverse on the left hand side so is just saying that because I have to write everything as open sets in some sense like open sets I mean these are these are relative open sets so things seem complicated but this is what I want I want phi inverse of 0 to V t 0 x 0 to lie within 0 epsilon 1 okay all right so here this notion of open inverse open and all seems a bit murky here but don't worry it is not murky yeah what happens we are talking about relative open sets so don't I am not going to explain it don't worry about that open issue is not giving I mean taking a beating here all right so we know that phi is increasing and continuous right so I have drawn a sort of a picture here right so I know that if you look at this picture here V t 0 x is continuous because I am again fixing the t 0 right so it's a function of x therefore this notation so V t 0 x is continuous it is 0 at 0 here and phi epsilon 1 is greater than 0 here right therefore there has to exist some norm x bound right some x bound such that V t 0 x lies within this range okay just by continuity of V all right just by continuity of V I can get this okay all right now yeah so that's what I say here I say here in this more mathematical language that we choose delta such that soup of norm x less than delta V t 0 x is less than equal to this guy exactly the same statement as before here it was alpha epsilon 1 square now I have just written the phi epsilon 1 nothing has changed exactly the same argument okay okay so all right so I will go to the aside later on but if this happens if x 0 is less than delta then I do have from this condition that V t 0 x 0 is less than phi epsilon 1 and if this happens right I know that V t x is less than equal to V t 0 x 0 right because because delta is less than equal to r and norm x 0 is less than equal to delta okay so which means that norm x 0 is also less than equal to r so I am in the good place where all my negative semi-definiteness etc hold okay so if delta is less than equal to r that's what I've assumed then norm x 0 is also less than equal to r okay which means I am in a good place huh so V t x is less than equal to V t 0 x 0 okay so this also holds and once this holds you have of course that phi norm the first statement here I have just repeated that here phi norm of x less than equal to V t x less than equal to V t 0 x 0 less than phi epsilon 1 the only purpose of this statement was to sort of tell you that your initial condition is within the r ball okay and if the initial condition is within the r ball you have some space to go again the r ball is also an open set so if your initial condition is within a delta ball within the r ball there is some more space to go so you are within the r ball your analysis is going on within the r ball your trajectories are still within the r ball so if you start within the delta ball then this negative semi-definiteness will hold because your trajectories are within the r ball if you started in the delta ball so therefore this negative semi-definiteness holds and therefore you just add these two pieces from the beginning that's all okay these these things are of course also holding because here in the r ball so once you have this you have norm x is less than epsilon 1 which is less than epsilon and again less than r okay so continues to hold so the only thing that I did not prove is this guy which I am saying not exactly prove but I sort of indicated to you that this is again going to happen by continuity okay because continuity will give me some bound on x and once I get some bound on x I will get a bound on norm x it can be conservative or whatever it doesn't matter I will get about the aside that I want to sort of say here is that is this particular sequence of things okay so delta has to be upper bounded by epsilon 1 which is upper bounded by r this is evident by the choice of epsilon 1 itself and epsilon 1 is defined so that this happens so this is not complicated of course but I am claiming that this has to be the case okay this has remember when we defined stability we already said delta is less than epsilon or less than equal to epsilon but that was in the definition of stability here we are trying to prove stability so this delta that we are getting is not from the stability definition this we are getting from here okay so it is important for us to sort of prove that delta is going to be less than r because if delta is not less than r then this cannot be claimed I hope this is clear if delta is not less than r this is not true anymore because your initial condition may already be outside the r ball then negative semi-definiteness doesn't hold so that is not guaranteed somehow evident here just by looking at this that whatever delta you get will it be less than r or not is not evident just by looking at this okay so that's what I have just tried to prove very quickly nothing too complicated yeah so what I am saying is let's assume for contradiction that delta is greater than epsilon 1 okay and if delta is greater than epsilon 1 then I know by my positive definiteness that this happens is just the positive definiteness statement which means that by monotonicity of phi I know this is why this is true when this happens I am assuming that delta is greater than epsilon 1 so there exists some norm of x between epsilon 1 and delta correct delta is strictly greater than epsilon 1 so there is some value in between so I can choose a norm of x is in between that value yeah because delta is greater than epsilon 1 so there exists some norm of x in between norm of x is just a number just a number okay so there exists some number in between now if if norm of x is in between this epsilon 1 and delta then if norm of x is greater than epsilon 1 I know that phi norm of x is greater than phi epsilon 1 right by monotonicity of the phi so whatever I just proved I have proved that v t0 x is greater than equal to phi epsilon 1 for some norm of x okay and this norm of x is still within the initial condition bound because the initial condition bound was less than delta right so this norm of x is still satisfying this initial condition bound and within this bound I have now proved that v t0 is greater than phi epsilon 1 okay but this is a contradiction right this is a contradiction I chose my delta such that this upper bound holds okay see if so there is a problem with my assumption there is a problem with my assumption okay so the assumption is invalid this is one of the ways of proving results in mathematics by contradiction all right so that is what we have done so just I have assumed that there is a contradiction that is delta is actually greater than epsilon 1 and then I show that something goes wrong here all right which it can't I am not allowing it to my delta is chosen in this way okay so it's important for us to sort of ensure that delta is less than r so that this satisfies and if this satisfies then I have these two additional things and I am done all right I am done with the proof so in the general file case also I can do this the only thing is it's not very evident that this e and so on and so forth how the picture looks and how the open sets look but it's still the proof goes along in exactly the same lines you you basically have this sort of a see in this when there was alpha I had a 1 by alpha here instead of phi inverse I just had a 1 by alpha here all right here just I have a fine okay so so basically I had something so here I had a alpha epsilon 1 square right so so that was the idea not not too complicated here now I know I have written it in this form I am wondering if I can write it somehow in this form also to construct the set e can anybody of you suggest how this set e will look in this case so so here the equivalent was just alpha epsilon 1 square here that was the only difference here there was an alpha epsilon 1 square here so how do I define the set e can anybody tell me it is something v t0 inverse of minus phi epsilon 1 to phi epsilon 1 okay that's it same deal and I know I wrote it in a different way like this but you have to worry this is it is just this set okay and remember phi epsilon 1 is just a number so therefore this is an open set right just by previous notion right inverse of open set under continuous function is open so this is also an open set and this is an open set means I just have this picture again some e yeah we saw how to get the equation right somehow you have an equation it could be an ellipsoid it could be some funny shape doesn't matter important thing to remember is that origin is contained in this why is origin contained in this because origin is in this set and inverse under this function of origin is origin by definition therefore origin is so in fact I can even say something like this 0 belongs to me okay origin is contained in this so origin is in e and e is an open set same same deal make these things then I get a delta so it is evident now that I got the delta right I mean it is not evident maybe from this whatever but this expression but it is evident from this picture not constructive don't expect constructive thing in general non-linear function cases but it is a delta fair enough okay all right questions comments is too complicated this is more or less I mean well the lasali invariance prove is a little bit more complicated little bit more I mean that's it's the geometry guys so that stuff is always but but this is fairly straightforward actually yeah I am trying to wonder if exponential stability proof goes simpler or doesn't particularly think yeah I am not sure there will be any particular advantage there either because there you have to use the same order of magnitude idea to get a exponential decay see here in all these proofs again as is expected until now we have just talked about stability but you know there is no rate of convergence even when we look at asymptotic stability next you will see that there is no particular rate of convergence notion as such so you can't expect any rate of convergence anyway so the exercise is to complete uniform stability how do you think you will go about it so you so I hope it's evident again again I hope it's evident that here I'm taking a vt0 inverse here I'm taking a vt0 inverse of this guy to find a delta right so so the e is e is somehow I mean dependent on t0 right and of course e0 and epsilon right because epsilon is right here inside this so e is a function of t0 epsilon yeah so if I want to get rid of this t0 that's what I will need to prove uniform stability because if not then then if e depends on these then delta has to depend on this the set e is depending on this so the only way for me to get rid of t0 dependence is to get rid of t0 dependence in e it doesn't depend on t0 or initial time then I have arbitrary choice of delta which is independent of t0 so how do you think I would be able to remove that t0 dependence here this is fine what about this t0 dependence here how do you to prove uniform stability how do you prove anything when I ask you to prove something anyway you should when you do that you will hopefully be a little bit more comfortable with these see some of you have already done proofs but mathematical proofs are you understand it they are different require a little bit of a different mindset but eventually whenever we prove anything what how does it work I give you some statement generally I am just not talking about this I am just saying there are some statements or assumption and based on those you get a result right so what are the assumptions in this case for uniform stability yes what is the uniform stability Lyapunov theorem all right so that's what I'm assuming should be useful to you because otherwise instability proof we used all the other features of V and V dot right the only thing that we did not have and use was decrescence so obviously I need to use that no otherwise I can't just prove an additional property for free without assuming something so you can think what you can do with decrescence all right okay I think that's it we stop here