 So, let us look at another example ok. So, I took the same Lyapunov function as you can see that was my aim as I can keep the same Lyapunov function and come up with a different example that was the plan alright great. What is this and we already know this is continuous positive definite and radial and bounded and all the nice things not decrescent. Now v dot as usual is x 1 x 1 dot 1 plus t x 2 x 2 dot and x 2 squared by 2 here I have just use the chain rule or the product rule, product rule not the chain rule the product rule I always get confused between the product this is not the chain rule it is the product rule ok. Chain rule is for chain of functions and then as always I substitute the dynamics this remains the same yeah here 1 plus t x 2 is as it is then I have substituted for x 2 dot right here and because of this wicked choice of system that I have made the first term will cancel this right if you see 1 plus t 1 plus t cancels out and then the first term will cancel this ok alright that was the plan yeah and then the second term again by my wicked choice of system will cancel this term alright that is exactly how I chose it this guy multiplied by this is minus x 2 squared by 2 and this is plus x 2 squared by 2 this was what was remaining last time so I made sure I cancelled it yeah ok is that clear notice this is all that was remaining here so I just introduce the term here to cancel this ok cancelled now I have v dot to be exactly 0 ok v dot to be exactly 0 yeah very ridiculous looking system but it is stable at least alright it is stable it is not v is not decrescent so can't claim uniform stability again I am saying not uniformly stable but it is more appropriate to say can't claim uniform stability with this v ok if you don't get a property from a v does not mean that the system does not have the property that you have to conclude in different ways I am most certain that this is not uniformly stable but but it is not obvious just by choosing only upon our function and doing one v dot and claim no that's not enough yeah in fact if you do 100 even then you can't yeah it's just a deficiency in your choice of v and not in you know your you know in terms of the system itself yeah ok great any questions alright so it would be very illustrative to see the difference between these two systems ok if you look at this guy and if you look at this guy ok forget how I modified it and whatever did whatever just to cancel terms this was just a modified harmonic oscillator right because if this term was not there or for small time just like we discussed it's a pure harmonic oscillator and this system is actually a damped harmonic oscillator it's a modified damped harmonic oscillator ok why damped because of this term this term is acting as a damping right if you all of you folks who do PID control know that the derivative term is the all connected so derivative term is the damping term yeah so this is the derivative term this is the proportional term this is the derivative term there is no integral term here but honestly speaking in typical nonlinear control design you don't get a integral term it would be rather unusual yeah but there is a PID control if you think about it yeah but with a time varying modification yeah so this is a modified damped harmonic oscillator ok if time was small or if this term did not exist or it was one then this system is actually what you mean I mean in Lyapunov terms what is this system in terms of Lyapunov stability terms uniform what what do you know about this system about a damped harmonic oscillator what is the example of a damped harmonic oscillator spring mass damper alright ok see if I leave a spring mass damper from any arbitrary initial condition what happens come to rest at origin unless you are springer poor quality and they start stretching and all yeah which is the case in general but yeah they come to rest at the origin so what does it mean what does it mean about stability property of the system asymptotic stability it is asymptotic stability yeah in fact uniform because of no time dependence yeah so if this was not there this system is in fact uniform asymptotic stable alright so very strong property in fact it's exponentially stable why why do I claim it's exponentially stable yes yeah I am looking at you why am I claiming it's exponentially stable without doing any analysis all I talked about was asymptotic stability by giving an example of leaving a mass in a spring mass damper why is it exponentially stable yes we discussed this for a linear system asymptotic stability exponential stability is the same we proved it we proved it ok well we didn't prove it did you prove it exactly yeah we proved the stability one but from that you can see that the exponential stability one is also basically any convergence for a linear system is always exponential a linear system will never converge at any other rate ok than exponent linear time invariant system ok so but all as soon as I introduce this time dependence I drastically reduce what I got I am now only stable not even uniformly stable I am quite certain it's not uniformly stable ok so significant drop in what I can do now it's very simple to see why things go wrong as time becomes large this stabilizing term the so-called PD controller if you put a PD controller on a system and after a certain time you stop the controller control you make the control 0 ok then the system is not going to be stable anymore right ok here it works in the absence of the time dependence it works because it's perpetually acting as soon as you introduce this time the effect of the proportional derivative term is dropping at a linear rate ok so which means that your proportional derivative term which is what is stabilizing the system is dying ok so the proportional term gives you stability the derivative term schemes you asymptotic stability yeah so these terms are the contribution of these terms are dying as time increases and therefore the system will not be stable anymore so you lose all these properties ok so at best what you get is stability in fact you are lucky that you at least get stability ok alright any questions yeah if you think about the control gains this is the gain right 1 over 1 plus t right k p k d whatever if you think about it the gains are going down right if you ever design a PD controller a PID control with your gains going down you will see the system will not work it's obvious ok alright great so let's go to this argai yeah when the time dependence is removed I already said that this is exponentially stable ok because for linear systems asymptotic and exponential is the same in fact in this case you can even solve this system right this is not difficult to get a solution for the solution will have exponential decay term so it's exponentially stable but if I wanted to do the very very hard work it turns out to be very hard work of proving exponential stability via Lyapunov functions and then it's very complicated that's what I've sort of illustrated here now but then sometimes you do need to do this kind of an exercise because your system may not be linear always ok so I cannot choose something as simple as x 1 square plus x 2 square by 2 as my Lyapunov function that does not work I can promise you it will not work it will not work so I have to choose something like this ok yeah later on we will go into the motivation of choosing something like this and why something like this might make sense and so on when we do design yeah so I am not telling you anything about why and how I choose it ok I am just choosing it yeah so this is a little bit complicated but the idea it's it's basically based on the notion of back stepping ok but we are not we have not done back stepping yet so don't worry about it this is just think about this as just an example alright alright so what is this half k x 1 plus x 2 squared plus 1 over 2 alpha x x 1 squared so it's not just a combination of 2 terms I have basically introduced some random constants also like this k and alpha are positive constant ok I know that this is positive definiteness and definite in fact radially unbounded do you understand why why do you think this is positive definite and radially unbounded see all I need to verify for positive definiteness is what that for non-zero values of the state it is strictly positive so what about do you think this is k and alpha are also positive so is it that easy ok yeah but if you remember we had discussed terms v terms like x 1 plus x 2 whole square and I said it is not positive definite right because this is also 0 when k x 1 equal to x 2 is minus k x 1 right so there is a problem right because you guys immediately said yeah everything is positive this is positive then what does it do these are not good arguments no no no no can't be so vague no no no don't do that will be more precise you tell me a little bit will be this is where I am I may not ask you to design these things but you have to be very precise but why this is positive definite we have already discussed this wait I am going to write this again this now what are you saying both terms are positive you mean non-negative then ok so when when is the when are the 0 which means that individual terms are 0 when x 2 is minus k x 1 and this is x 1 equal to 0 and this is an and condition which implies x 2 is also 0 so the only place where this so things change because I added this x 1 square if I did not add this did not have this then this is not positive definite anymore because it is 0 on a line and that is not positive definite ok so only because I added this so there are now two conditions need to be satisfied this has to be 0 and this has to be 0 because they can't cancel each other these are all non-negative terms therefore this and this both have to hold which means this happens ok so be very precise when I ask you do not just say positive positive positive ok otherwise why did we do all these definitions there was a very good reason ok of doing these definitions carefully so please be careful when you state that something is positive definite or not all right fine sounds good that is what I have said so anyway the k is missing here it does not matter I will put a k ok all right great we take the derivatives yeah just like we are doing pretty straight forward right lot of bookkeeping in this case it look complicated yeah but eventually I am not doing anything very fancy I am just taking derivatives and substituting for the derivatives so here I will get k x 1 plus x 2 times k x 1 plus x 2 derivative I will get x 1 and x 1 derivative and the half will go away this half will go away ok that is it all right now I substitute for the dynamics here k x 1 dot is k x 2 x 2 dot is minus x 1 minus x 2 similarly this is x 1 x 2 all right now we have to do a lot of manipulation yeah k x 1 plus x 2 k x 2 minus x 1 minus x 2 and then I write this as x 1 times k x 1 plus x 2 minus k x 1 square by alpha ok so this is just equal to this do you believe me yes I have just added the k x 1 term and why did I do this again these are manipulations that you will have to do these are the kind of manipulations you will have to do in any Lyapunov analysis so better be comfortable if you are not asking why did I do this why do you think I did this see in the harmonic oscillator case this guy was getting cancelled directly ok now I know I cannot cancel it directly I have a problem so what do I do they do the next best thing I try to club it with the term I have already and I see that here I have a k x 1 plus x 2 term so I try to write it as k x 1 plus x 2 so what did I do I took the x 2 I wrote it as k x 1 plus x 2 minus k x 1 so I got from this one term these two terms now I know that this guy can be clubbed with this term ok alright I cannot cancel so I try to club them together this is the only few manipulations you can do ok there is one or two more which we will get too soon but right now remember this is the only thing I can cancel or I can club terms so I cannot cancel so I try to club it with the term I already have so this is the only term I have it does not make sense to club it with this is way more complicated anyway why would I do that I will just club it I will just introduce this term here ok and the cool thing that happens when I do this this is again a product of the back stepping method which we have not discussed by introducing the k x 1 and a minus k x I actually got a negative term in x 1 yeah both k and alpha are positive so this term is actually a nice negative term it is a good term it is a helpful term ok now this guy gets combined with this and what do I have k x 1 plus x 2 k x 2 minus x 1 minus x 2 and x 1 by alpha right this is coming from here alright make sense ok this first term is coming just by the clubbing of the terms alright and I already have a good term here which I keep as is yeah I love the good terms as soon as I get negative quadratic terms I keep them as it is never touched them they are what will help me eventually so I never touch them this gets carried on until the end of the analysis now I have only two variables actually x 1 and x 2 yeah so I just club all the terms in terms of x 1 and x 2 ok alright so I get something like minus 1 over 1 minus 1 by alpha minus 1 minus k and all this mess but I get some terms ok so now what do I do I take this 1 minus k common I pull it out yeah these are all variables that I choose that is why I kept these handles or knobs ok so that I can play with them this is for me to gives me freedom to play with this yeah otherwise I will not be able to choose a good V yeah when I was starting to choose a V I did not have any idea what is going to be k and alpha but one once you conclude the analysis you will see that the choices of k and alpha will become obvious alright so I have I have deliberately written everything as negative terms why because V dot needs to be negative definite so writing it as positive terms is ridiculous makes no sense so I have written everything as negative terms how will I get V dot to be negative definite I want this term and this term to look identical this should also look like k x 1 plus x 2 yeah I am doing this carefully just follow the steps ok these are things we do often huh so I take 1 minus minus 1 minus k common outside now the first term is the same the second term has 1 minus 1 over alpha divided by 1 minus k plus x 2 and this is of course my favorite term remains as it is now what will I say I will say that I want my k to be equal to this guy yeah and if it is suppose it is yeah just to wit it is equal to k then these two terms together become k x 1 plus x 2 whole square right and this is some nice negative term this is nice negative term I have negative definite V dot ok now all I need is this has to be equal to k ok and of course I want k to be less than 1 that is the other requirement otherwise k has to be strictly less than 1 all good it is my choice it is just a V right it is system function I use for analysis it is not changing anything alright now you just have to see if this is feasible or not you just have to check the feasibility of this guy huh so k is between 0 and 1 alpha is positive I want to check the feasibility of this this will give me a quadratic equation in k ok which is going to solve this I have to choose one of them yeah I can choose any one of them right they will both satisfy so if I choose k as this guy I am fine yeah I can choose actually any one of them apparently alright so I think I took an example or what let us see yeah so first I am trying to ensure that this because that is what is going to give me the feasibility right I want to check that the discriminant is going to be positive or not so that I get a real outcome here so that is all I need to do I get alpha less than 4 by 3 ok because this is a linear requirement yeah so alpha is positive but less than 4 by 3 anything less than 4 by 3 is good ok now if I assume that this is equal to half some value less than 4 by 3 then I get whatever then I get alpha is 8 by 7 which is fine no problem yeah and I can choose k as any one of actually it has to be less than 1 so I will choose the it is preferable to choose the let us see negative one oops I cannot erase this can I I will just choose the negative sign right yeah I will just choose the negative sign because whatever appropriate value of alpha I choose less than 4 by 3 you see that the quantity inside is going to be less than 1 because this 1 minus some quantity less than 1 it is something less than 1 so it is square root of that is also less than 1 so it is better to choose the 1 minus because if I choose the 1 plus 1 plus will also work for some time because I have a divided by 2 and all that but this will be an easier choice yeah this is guaranteed to work yeah so basically I have given you a choice of a k and a choice of an alpha yeah so alpha is exactly this and and so on and so forth I mean whatever alpha is exactly this in if you want something like this and k comes from here with this choice of alpha alright so with this very very complicated construction I have proven exponential stability of this system which is very easy yeah see in this case it does take a lot of work because the first the because exponential stability requires the same order of magnitude functions and so on and so forth alright so if you look at this guy and you look at the V dot these are exactly same looking functions right they are the same order of magnitude functions they both have the same quadratic terms k x1 plus x2 square and x1 square yeah so I actually got same order of magnitude functions yeah and so by my Lyapunov theorem it is exponentially stable yeah so this is not too easy to do for nonlinear systems in general okay