 So, that is what I was saying that this result is very powerful, because it really gives a easy check. If I have k vector fields in the distribution, I just have to check their combinations, yeah. So, k c 2, when you are done, ok, check that their combinations belong to the distribution and you are good to go, ok, proof, of course, whenever we do something we have to prove. It is an if and only if, so you have to prove both sides. But suppose if you that you have involutivity, if you have involutivity, then this is obvious, right. Because involutivity means that if f 1 to f k belong to delta, then their lee bracket have to be in delta. So, this is obvious by in you know that if you are starting with involutivity, obviously this will happen, no problem. The bigger question is if you start with this that only their combinations are in delta, then can you claim that the distribution is involutive as per this definition, ok. How do we go about it? We take some two vector fields, ok, g 1 and g 2 in delta. Now, the important thing is because g 1 and g 2 are in delta and that happens at all points, they you should be able to write them as a linear combination of the f i's is the only way g 1 and g 2 can belong to delta, right, because they have to be the linear combination of the f i's, right. And of course, these f phi ij's are smooth functions, yeah. Smoothness is coming out of the fact that f i fj's are also smooth, yeah, and g 1 g 1 g 2 are also smooth, ok. Therefore, this coefficients have to be smooth, ok, so it is everything is smooth, so coefficients also have to be smooth, ok. For simplicity, if you assume k equal to 2 that is you have only two vector fields here f 1 and f 2 forming the distribution just for so that the proof looks easy, yeah. Then what do you have? You have that the lee bracket of g 1 g 2 is actually this guy, lee bracket of this, because I have written g 1 as phi 1 1 f 1 plus phi 1 2 f 2 and g 2 as phi 2 1 f 1 plus phi 2 2 f 2, right, because g 1 g 2 belong to the distribution, right. Now, I want to compute this lee bracket, ok. Now, there is a problem with this computation, ok. So, I want you to actually this is another exercise, I need you to complete this computation. This is computation is pretty straight forward, it is just doing this with some coefficients phi 1 i f i plus phi 2 j f j, remember phi 1 i and phi 2 j are depending on x and they are just their functions, they are not just, yeah, they are not just constants or anything, ok, alright. So, I want you to prove this computation anyway, but let us look at the lee bracket of g 1 and g 2 as per this expression, ok. It is known that the lee brackets are distributed, I have not talked about this, yeah, but it is easy to prove that they distribute that is I can break this open. How? In all these four terms, yeah, this works like a nice distributive bracket, ok. So, from g 1 g 2 I will get phi 1 1, phi 2 1, f 1 f 1 first term, then I will get phi 1 1, phi 2 2, f 1 f 2, second phi 1 2, phi 2 1, f 2 f 1 and phi 1 2, phi 2 2, f 2 f 2. Now, it should be obvious to you that the lee bracket of a vector field with itself is 0, just, yeah, 0, ok. So, these two are gone, ok, these two are gone. So, what am I left with? I am left with these two, ok. And these are actually the same term, I can actually flip the sign. So, also this is another thing that is true, lee bracket is anti-symmetric. So, f 1 f 2 lee bracket is negative of f 2 f 1 lee bracket, yeah. So, I can flip the sign. So, what am I left with? I am left with something rather simple, right? It is, yeah, this is my g 1 g 2, ok, ok. Now, whatever I assumed that f 1 f 2 belongs to delta, yes, yes, which means what? g 1 g 2 also belongs to delta. Why? Because it is just a scalar multiple of f 1 f 2, yeah, no. What do I have to prove for invulnerability? Take arbitrary two vectors, vector fields in the distribution and prove that the lee bracket is in the distribution, yes. What does it mean to be in the distribution? They are their linear combinations of the forming vector fields, ok. I have assumed that there are only two vector fields, forming vector fields, yeah, or generator vector fields, which is f 1 and f 2. So, g 1 and g 2 if they are in delta means g 1 can be written as phi 1 1 f 1 plus phi 1 2 f 2 and g 2 can be written as phi 2 1 f 1 and phi 2 2 f 2, ok. Notice phi 1 phi 1 1 phi 1 2 phi 2 2 are functions of x, but scalars, they are not changing the vector direction. This is in the direction of f 1, this is in the direction of f 2. It is like multiplying a scalar and a vector. And just that here, it is not just scalars, it is a scalar function. Similarly, it is not just vectors, it is a vector field, that is it, ok. So, phi 1 1 f 1 is in the direction of f 1, phi 1 2 f 2 is in the direction of f 2. Similarly, phi 2 1 f 1 is in the direction of f 1 and phi 2 2 f 2 is in the direction of f 2, ok. The important thing to remember about lee bracket is that they distribute nicely, that is what I have used. I have just broken open the bracket just like you break open any product, works exactly like this, yeah. So, I get 4 terms. Now, out of these 2 terms, I have same, you know, f 1 comma f 1 and f 2 comma f 2, ok. Just because the lee bracket is anti-symmetric, this structure, this is 0. If you put, if g and f are the same, this is actually the 0, then, you know, becomes a 0 vector field. So, this guy is 0, this guy is 0. I am left with f 1 f 2 and f 2 f 1. And so, that is negative. So, f 1 f 2 is negative of f 2 f 1 again by anti-symmetry. So, all I have left is this entire guy, ok. And this is just a scalar, yeah. I mean scalar function, but still a scalar, not changing the direction of the vector, yeah. And I have already assumed that f 1 f 2 belongs to the distribution. So, I am done, g 1 g 2 belongs to the distribution now. Now, if I had assumed instead of 2 f, there were many more k generating f's. Yeah, I will say 3. What would happen? I will get 3, 5 1 1, 5 1 2, 5 1 3, 5 2 1, 5 2 2, 5 3 3. But I will still get combinations of these and repetitions, combinations, repetitions. And the combinations are already assumed to be in the distribution. Repetitions are dying. So, therefore, I will always remain in that, ok. So, straightforward actually, yeah, you do not have to worry too much. Yes, good point. I was also wondering, one should always ask oneself, are we using all the assumptions? A non-singular distribution, if delta is a non-singular distribution, then involutivity, if and only if, ok, ok, ok, ok. You know what, so what you are asking is that where do we use the non-singularity of the distribution, all right. Now, you know, this is where I think the problem will happen, this place. I think this equality is no longer writable, is what I am wondering or am I wrong. I think this, this equality is where I will have, start having trouble. Yeah, I have to think about it more carefully, but I believe this equality you will not be able to write anymore. Because if your distribution is changing rank, the span is changing rank. Now, the span definitely has to hold the vector, because we are assuming that g i is in delta. Obviously, the span has to hold the vector, but my feeling is you will not be able to do it with smooth functions here, yeah. If you suddenly want to jump from one point p to another point p prime, you go from a three-dimensional distribution to a two-dimensional distribution, all right, because it is singular, then this phi will undergo a rather drastic change. I do not think you will be able to retain smooth phi's anymore. Again, I have to think about this more carefully, but this is where I think things will go wrong. This assignment will not lead to smooth functions phi, but that is a good point. I will try to hunt it up. That is also the next question I was asking myself, all right. Where are we using the smoothness of the functions phi? See, when I look at this without a smooth phi, this is poorly defined. Fine, I wrote this as this split and all this, all right. But if I try to actually compute this guy by this formula, you can see I am starting to take partials of the phi's and here. So, this is poorly defined. So, even writing like this is not okay. Because, I mean, eventually all the bracket operations are, is there all derivatives of some kind or the other? So, I am frequently taking derivatives. I will definitely mess up. I mean, this is not a good, I mean not a well-defined object anymore. So, it almost certainly, I believe what I am saying is right that non-singularity will result in these phi's becoming non-smooth. Because, you cannot just jump rank and expect that everything will turn out to remain smooth. This will definitely create some trouble. You are suddenly projecting to a plane. Say you are in three-dimension, suddenly you are projecting to a plane. Or in five dimensions to a four-dimensional hyperplane or a three-dimensional hyperplane that will not retain good values of the phi. I mean, yeah, I mean, as of now whatever I am saying is imagination. It is nicer to if we can construct, if we can construct and sort of evaluate. But I believe that is what we will go down. I will check anyway, if I can find some fun examples. So, here I am just giving some example of this normal form business. This is again going back to the previous topic. But anyway, let us look at this. So, this is sort of the system. And I want to put this in normal form. You can see it is already non-linear and messy. I am using y as x2. I am using y output as x2. I want to get this to normal form. How do I do this? I simply start taking derivatives, first to get the relative degree. Because anyway, the y and its derivatives become my states, my linear states. So, the first state is x2 itself. Then I take a derivative and I get to x2 dot, which is x3. And then I take another derivative y double dot, which is x3 dot, which gives me the control. So, this is a relative degree what system? What is the relative degree of the system? 2. Yeah, it is 2. I have also written it here. I took two derivatives of the output. Whatever output I was given, I took two derivatives, I reached the control. So, obviously, my first two states become this guy. Yeah, that is y and its derivative. Those become my first two states or the last two states. Then I need to find the phi state, which will make it a diffeomorphism. Suppose I choose this guy, x1. This is a problem because the derivative of x1 contains the control. So, I do not like it so much because it is not going to give me the normal form. Now, what do I need for the normal form? If you remember, I need l, yeah, I have actually written it here. I need l, g, phi equal to 0. So, that there is no control term appearing. So, what is l, g, phi? l, g, phi is basically partial of phi with multiplied by g. What is g? g is basically whatever is multiplying the control. So, that is 2 plus x3, 1 plus x3 is whatever 2 plus x3 square, 1 plus x3 square is 0 and 1, this guy. So, if I compute this product, I get this. I can actually break this open to get 1 plus x3 square this del phi del x1 del phi del x3. Now, I want this to be 0 and I want to use this to motivate my new state. So, I have of course, chosen it like this. Pretty scary looking actually. How do you think I came with this? So, this is my phi. This is what I chose as phi. How do you think I came up with this? I did some guess work. So, if you look at this guy, this looks like tan inverse derivative of tan inverse x3. This looks like derivative of tan inverse x3. So, if I keep x1 as linear, I get del phi del x1 as 1 and then I get del phi del x3 as 1 over 1 plus x3 square. Yeah, 1 over 1 plus x3 square and then this guy is again just multiplying del phi del x1, which is 1. So, these two will cancel out and I am left with my apologies. Did I get this right? Yeah, yeah and this and part of this will cancel out. This is basically del phi del x3 is going to give me minus 1 minus 1 over 1 plus x3 square. So, I did a little bit of guess work. How would you do it? If I wanted to guess a phi starting from this equation, is there any other smarter way? I did just bunch of guess work. I just looked at what I can cancel. Is there any nicer way if can that you can think of to sort of arrive at this? No, because individually I cannot say that individually these are 0 or something like that. It is not a quadratic or anything. So, I cannot say that individually these are 0 or some such. Right? One thing that is obvious to me is that I do not have the this entire equation as only x1 and x3. So, I do not need the phi to depend on x2. That much is obvious to me. My phi does not have to depend on x2 because it is playing no role in this entire equation. So, phi is only a function of x1 and x3 is all I know. Yeah, I think to me it seems this is what works fine. Okay. And anyway, if I choose this funny looking coordinate, I get z1 dot as this, z2 dot as z3 and z3 dot as this z1 z3 plus u. Oh no, it would not work. x1 minus tan inverse x3 is it? So, so del phi del x1 will be 1 and del phi del x3 will be minus 1 over 1 plus x3 squared. So, this will be 1, 1 over 1 plus x3 squared and this will be 1 over 1 plus x3 squared. So, 1 remains no. So, I am trying to cancel the 1 also. That is why I put the minus x3 here. I must have tried with that only, but then I added the minus x3 because I have to cancel the 1 also. Okay. That is it. It is just a little bit of hit and trial. Okay. But this is what you get as the dynamics. You still get nice enough zero dynamics by the way if you see. Yeah, because if the linear part goes to zero, this guy is zero and this is again an exponentially stable system. Okay. So, again started with something complicated and very non-linear, not evident how I would design a control based on any method that I know. Yeah, but I still end up with something rather nice. Okay. Oh, this one. I guess it is already in your notes. Why is this exercise then? Well, I see. Apparently, it is not exercise. I have already solved it. So, it is already in your notes that I have uploaded. So, I should probably just not say this is an exercise and solve it. Yeah, because you can see that it is here. I mean, I have actually done it. Anyway, you can take a look at this. Yeah. You can take a look at this in your leisure time. This whatever expression you get here. Yeah. I believe this expression has a problem. This expression has a problem. So, anyway, so I got to this expression. I am not sure why I am saying this is specialized to k equal to 2. It does not seem like this is k equal to 2 is necessary here. Yeah, it seems it is specialized to k equal to 2. Yeah, because otherwise there will be summations and stuff. That is all. Alright. Okay. Anyway, this is something you can just look at it on your own. Alright. So, anyway, next time we will continue with our discussion on Frobenius theorem. Alright. Okay. Alright. We will stop there. Thank you.