 Мы продолжаем, и теперь мы начинаем более серьезно изучать конкурс-мэп. И я бы хотел бы писать это в таком виде, что мы можем реально оперировать. И в центре, это всегда лучше делать с линией объектом, объекта, потом с точками на манифултах. Я хочу, чтобы мы получили нашу систему, F U of T Q. И, конечно же, мы можем даже предоставить третий точек, И с ассоциацией Q of t, это Ft, как вы сказали, но, в первую очередь, первую очередь, я хотел показать, что это как-то ровная мета, в которой ты видишь. От паспорта контролов, контролл, функции в манифултах. И я хотел бы сделать даже более генерал, и на этом уровне это не важно. На этом уровне, ты просто переметриваешь вектор-филд, и мы можем считать, что мэп, это рестриция следующего мэпа. Если мы возьмем любые нон-автономы вектор-филд, Xt, это временная варианта вектор-филд. И считаем, что q dot equals Xt of q. Такая спецификация в этом мэпе, мы возьмем, не все, нон-автономы вектор-филд, это специфика, переметриваешь вектор-филд. И уже на этом уровне, я хотел бы показать, что если мы возьмем нон-автономы вектор-филд, то нон-автономы вектор-филд, это просто нон-автономы вектор-филд, Qt будет в секунде 0, 1. Norm Xt, Norm Ck, это Norm Ck. Поэтому, когда мы рассматриваем все древиции патриоты с оптодиками, это на компактах, на компактах, окей? Мы должны выживаться на компактах, но я не буду. Если вы рассматриваете такие enormы и рассматриваете мэп x dot в qt, мы оцениваем, что qt0 фиксируется. Тогда я буду показать, что будет смузное название нормы в стопологии. Мы можем тоже поставить хип, какие-то интегральные нормы, и это будет смузное. В этой стопологии. Я просто нашел тейлерыкспансион в этом мэпе, и это будет использовано. Мы всегда изучаем асимптотики, асимптотики для Certain family, Certain perturbation of giving control, и это очень важная информация. Чтобы сделать это, я буду изучать, qt билонствует сам нон-линио-объект для нас, даже если это домашний рен, это нон-линио-эквация. Но у нас есть классический трик, дуэл-апрочка, как считать это нон-линио-эквация. Конечно, нон-линио-оперейт-эквация это помогает большое в правильной экспенсии. Мы просто возьмем смузное название, асимптотики, асимптотики, асимптотики, асимптотики, асимптотики, асимптотика, асимптотика. too much. Тыllcard, oh, oh, oh, oh, oh, Мы можем уйти в интегральную эквейсию, и потом мы можем уйти. И это тоторикоордионет. По-моему, это фиксировано. Это A of Q0, плюс интеграль Q dot. Интеграль Q dot. Но мы знаем, что это интеграль Q dot. Это дифференциал. Это будет D of A Q of tau. И X of tau. И снова D of A X of tau. Поговорим так. Вы можете написать так. Мы знаем, что это ничего иначе, но экшен объекта филд. Это один дифференциал дифференциал. Вы можете написать так. X of tau. Аплитенция A. И это тоторикоордионет. Итак, потом мы можем делать тоторикоордионет. Теперь мы посмотрим на другую функцию. У нас есть другая функция. Поговорим так. У нас есть другая функция. X of tau A. Это только другая функция. Вы можете написать B of tau. И мы можем использовать абсолютно один дифференциал. Этот дифференциал является очень важным для всех функций. Вместе с T мы можем взять тоор. Вместе с функцией A мы можем использовать эту функцию B. У нас есть следующая функция. Аплитенция A of T equals A of Q0 plus tau T x tau A computed at Q0. Plus double integral. Это второй дифференциал, потому что теперь нужно использовать. Поговорим так. Это будет двумя дифференциалами. T2 smaller than T1 smaller than T2 smaller than T1 smaller than T. У нас есть X T2. У нас есть X T1. У нас есть A and everything computed at Q of T2. Of course we can keep. Вот мы делаем. Таким образом, мы делаем степь по степь. Тейлор экспенсион. У нас есть map. У нас есть map. У нас есть zero. У нас есть zero in the space of non-autonomous vector field. Если X is filled with zero, We stop here. Otherwise we can write this guy. This is linear with respect to X. Non-autonomous field form a linear space. This is linear with respect to field. This is just a point. So it's a linear with respect to field. There are no trajectories anymore. Hidden non-linearity. And then we continue. We can continue. Keep calculating. And put here. This term is linear. This term is quadratic, but not really quadratic. Because Q of T was hidden. Q of T depend on X. It does not depend on X, but Q of T depends on X. It's our map somehow. But we again now can repeat the same trick. Working with this function instead of original function A. And applying our equation to this function. And at the moment T2. And then we have here again Q zero. And so on and so on. We have the whole expression. And this is a Taylor expression. Each time we will retain the term. And it's not so hard to estimate it. To see that it is indeed has order. Has order. If we have a K guys here. It has order K in some norm. So we are working here actually. Not in the Banach space to be fair. But in the Frische space. Norm of the result. CK norm of the result. May be estimated in Cn. Norm of the source. Of the vector field. In the norm of the. Of the remainder for this expression. As always needs more derivative. So we have indeed each time. Here we have linear term. Here we have quadratic term. Then we have. Order. Order n. Torm. XTn. XT1a. Just in Q zero. So first differentiate that evaluated Q zero. We can write like that Q zero. DT1, DTn. And plus it. So on plus remainder that we should know how to write. But each time remainder. Has more and more derivative of X. Because here we apply high and high order operators. So this. And so we. If we treat this. Order n. As a derivative of order n. Then it is derivative. Not in the Banach space. It's fixed number. But in the Friches space. So remainder. If you would like to estimate norm of this. K norm of this. Then it will be. It will be. K remainder. It will be half of order. N1. But it's norm. N1. K plus n. Because we have n derivatives. N minus 1. But does not matter. Because X is differentiated only. N minus 1 times. So this sense. It is a Taylor expansion. And this sense. So. The norm is not the same. But even always different. OK. If you would like to estimate this. Yes, of course. N plus 1. N plus 1. N plus 1. N plus 1. For the moment I do not care about that. For the moment I do not care about that. What? What my first term is. I simply take a of q of t. I simply take a of q of t. And write it as a of q of 0. Plus integral of differential. Plus integral of differential. That's all. That's all. Newton-Leibniz formula. So. But. Actually I do not care about that. Because. Because. What I really. This expansion. Quite convenient again. Because it's universal. I always insisted. Depend on particular control problem. But if we have particular control problem. We have a composition of two maps. One map associated to u. f of u is just point wise maps. And then we use this map. And if f already fix. We have a normal. Smooth map. In integral. In some integral norm for u. In integral norm for u. And. If we have f u here. f u. Of toy. f u. Of toy. We consider composition. Then this is. Because f i are always the same. We differentiate our vector field. But. They are fixed. And they give you some constant only. Plus o. Of. Norm u. Say. Say. L-infinite norm, if you want. But if we depend linearly. It's very general. If we depend linearly on u. Then we can take any LP norm. Whatever you want. u here will be L-infinite. And n plus 1. So it's a Taylor expansion. Quite natural. And can be interpreted. Well. Even in the Banach scale. If we work not with all vector fields. But with. With vector fields. Which depend on finite dimensional parameter. So. We have such an expansion. But this is expansion only. Only. By the way. For the moment. In the. In the. Zero. It's no problem to get expansion everywhere. Again. With simple tricks. First of all. I would like to even generalize it. I would like because no. It's convenient. To think about it even in a more general setting. Do not fix q zero. I don't want to fix q zero anymore. Okay. And because from any. Because it's a well post. Well post. Coshi problem. Why I have to fix q zero. So. If I have a u. U. Now. My idea to. Consider expansion. Expansion. Let me consider. Okay. I would like to rewrite it. I can rewrite it. And the following way. Let p t. Is a. Familiar of defymorphism. That send q of zero. Into q of t. With. For all q zero. For all q zero. Just belong to him. I don't want to consider anymore. One individual trajectory. But the whole flow. Generated by. By. By the system. By non-autonomous system. And that very similar equation. It's not a bad idea. To think about this non-linear transformation. Again. It's a linear operator. On the space of function. Quite natural. If. Vector field is just derivation. In certain direction. Defymorphism exchange of variables. Okay. I write p a. A is again p t a. By definition. By definition is just. Composition of A and p. A and p. Composition of A and p. Composition of A and p. Then when I. Compute p at certain p t. At certain p t. At certain point it will be exactly q of t. And then I can rewrite this equation. Even without any a. As a universal. Okay. I can write like that. That. p t applied to a. A of q of t is p t applied to a. By definition. It's just change of variables. P a of q. I can write like that. A of p of q. P a was a function. P was change of variables. Transformation and composition is another function. Okay. So I can write this equation. As a. As a a. Because p zero is just identity. Plus. Plus x. This expansion x t a. Plus. Plus. And so on and so on. Erase q zero. Work with functions. Work with functions. And I can even write a remainder. SID. The same integral of the same type. Plus one. Plus one. And here we would. We would have. We would have p. P. N plus 1. Composition of operators. X t n. Because. Apply p to some function. Function is obtained by. Consecutive. Deferential. Of a. And then we make change of variables. We compute this guy at q of t. Not at q of t. Zero. It can be estimated also. Of order n plus 1. Of order n plus 1. And at this moment we can of course. Erase. Erase a. Because it's a linear expression. Now it's totally linear expression. Because we're just linear instead of nonlinear. Linear transformation becomes linear operators. Linear transformation becomes linear operators. As operator equation on the space of functions. It can be written like that. And we can use. Such a presentation. In order to. To write everything. How to use it. Write everything. Expansion not only. On. Zero. But. At any point. Reduce actually calculation. At any point to the calculation of zero. Actually. Of course the first idea. We come in mind. As we compute derivative another point. Take a difference. Between trajectory. So now I would like to. To compute. The following thing. The following thing. I would like to. To compute. What happens if I consider the map. We need some. Some. Some symbol to do it. 5. That send. That send x. Into. Into. The corresponding. Transformation pt. So it's the same map. Almost the same sft. As this one. But we consider. Not one initial condition. But all initial conditions. Instead of point on the trajectory. И initial points are transformed. Ok. So if we consider such a map. We already know expansion. Then we know that. Phi of x. Such a. Expansion. Phi t of x. Such an expansion. This we already know. At zero. But we would like to have it. Not at zero. Phi t. x. y t. As a series. Where we have linear quadratic and so on. Term. With depend on t. You can put here epsilon. If you compute. And then write. Expansion is in epsilon. In epsilon. If you consider. Directional derivatives here. Апсайлон. Just to think about. y t. Without. The last. The last. Because. Each time. We obtain. If we do not want to have. To have a remainder term. We obtain q of. This is what we did. Without remainder term. Так. Если мы делаем еще один. Мы получим. Ку зиру. Но в конце ремейна. У нас есть q of t. p t. p t. some function b. Call this guy b. p t. p t. p t. p t. Just b p t. Okay. Maybe the confusion is that p. p is considered in two different ways. Here p is a linear operator. p b. Better not to put this circle here. Let us do like that. Let us do like that. This p has different. Here p is a linear operator. I don't want to long in this direction. Introduce all the notation. Otherwise it will be. I never arrived to interesting stuff. To interesting stuff. This is just to show. That map is smooth. It is map of functional spaces. It is map of functional spaces. In principle, we can compute its derivative. That is what I want to say. So here it is a composition of the map. That is p t of b. p t of b. p t of b. p t of b. p is at a point. p t is a point of p t. p t is b of p t. p t of p t is b of p t. p t of p t. p t of p t. So this is a b of p t. p t is b of p t. By definition. So what is written here. It is just formal notation. Very convenient for calculating. Because you can keep calculating. Okay. У нас есть много фольта без координата. Это невозможно считать F plus Y minus phi T of X. Но, окей, мы не можем это сделать в координате. Но даже в координате это невозможно. Это сложно получить резонабольный результат. Потому что если мы дифференшируем без респекта T, то это не деференширует эквейсию. Это самое сложнее, чем мы дифференшируем. Самые сложные дифференшируют. Мы не можем это сделать в координате. Но мы всегда можем это сделать. Это почему я делаю без респекта T. Это невозможно. Это невозможно, что я делаю без респекта T, но все флори. Потому что флори в форме групп. Мы можем считать флори без респекта T. И мы контролируем, потому что они не работают без респекта T. Флори начинают быть важными, поэтому я беру... Мы хотим считать флори. Мы считаем флори без респекта T. И все равно мы делаем без респекта T. И это дефеморфизм. Это не пункт на монифу, но трансформация. И, по-моему, это хорошая идея, чтобы он считал как оператор. Это оператор, который отправил А в... Потому что у нас есть линия. Вы видите, что у нас будет линия-эквейшн. Но это линия-эквейшн. Мой эквейшн может быть написан. Эквейшн для индивидио-транжекты, это первая одопедия. Это характеристика. ОДЕ это характеристика для первой одопедии. Если мы пишем А от P of T, P of Q0, то QT, D A over DT, это XT A of P of Q0. Мы можем пишать так. First X, then we compute everything in point PT. So, in operator sense, we write PT A, D over DT equals PT XA, just by differing. First X, then we make change of variables in XT A. Это линия-эквейшн. Мы можем сказать, что это линия-эквейшн. Это просто variable. Мы пишем генеральную эквейшн, но не только форму, потому что, если мы объясним принцип, мы можем контролировать нормы. Мы получим линия-эквейшн. Мы считаем, что это линия-эквейшн. Это всегда удобно. Это A to P T of X plus Y. Это то, что я имею в виду. Это линия-эквейшн. Теперь, сделаем фоллен. Девайте PT of X plus Y. Девайте. Если я делаю PT, то это будет комфортно. Чтобы немного simplifier. PT is just phi of XT. Я всегда использую такую линию. ПТ is phi of XT. Я делаю PT of X minus Y. Это очень легко сделать эквейшн для PT of X minus Y. Потому что это оперативная эквейшн для PT of X minus Y. Это что-то, которое имеет 0 идентики. Мы в группе, а не 0 идентики, а 0 идентики. Если у нас просто PT, PT minus Y. Но мы можем... Но хорошая вещь о такой эквейшн. Если мы считаем, как это является PT, это еще solution to OD of the same type. Мы можем... Это solution of OD is vector field is not Y. But when Y is 0, it is linear with respect to Y. And we can write our expansion at 0. We reduce expansion at any point to the expansion of 0. Because of that many, many arguments, many observations I will do later in 0. Because you more or less automatically can do it. Just change your vector field. You have to... It's very natural. The CT satisfies the following equations. Let us write CT dot. First of all, we can easily write... Let us write expression for PT minus Y. How to do it? It's quite simple. So we can write PT minus Y. PT is identity. It's more or less standard calculations when you do with linear equations. And all differentiations are legal, by the way. In the space of the operator space differential are legal. Leibniz rule is working. That's no problem. So let us... Let us do it. I would like to... For inverse, fortunately PT is invertible. Because we have a well-posed question problem. PT minus Y. I write like that for me. It's more convenient. Let us differentiate this. It's for all T. We differentiate this identity. We obtain PTX. Differentiate this identity. PTX, okay. It is here. PT minus Y. Minus Y. And plus PT. And here we have D over DT of PT minus Y, okay. And this is... Sum is zero. Sum is zero. It's correct or not? What is written here? Yes. So then we can write it like that. D over DT. This is invertible, so we can cancel it. We can write D over DT PT minus Y equals to... Equals to the following guy. The X PT minus Y, okay. X PT minus Y. Minus, of course minus. S operator. S operator, but very clear which operator. Here. This is... And I will use this formula in order to find differential equation for T. I will see the derivative of T. It is some vector field. Not Y anymore, but some vector field. Y somehow changed with respect to YP by X. So we can... It's just simple exercises. I will not write details. I will not write details. You differentiate this. Okay, let us write. It's so simple. Let me instead of D over DT write a dot of X plus Y. PT minus Y plus... And then I differentiate this guy. PT dot X plus Y. And I have to differentiate this one. So this guy will be minus X PT minus Y. Okay. So this is CT... No, I'm sorry. I'm sorry. No, no, no. X plus Y PT minus Y. Okay, that's fine. X plus Y come linearly. And what else? And this is equal to CT dot. CT dot. So I differentiate this identity. It's equal to CT dot. What happens? I'm sorry. This I know. This CT dot X plus Y is actually according to our... This is a differential equation with a vector field X plus Y. This vector field X plus Y. So this is PT X plus Y. Plus Y. And minus PT... I'm sorry. X PT minus Y. Because when I differentiate this guy, I have only X. Okay. So I take this one. This is a solution of equation corresponding to X plus Y. When we differentiate it, you obtain the same PT. But X plus Y. Plus Y multiplied by X. And then this one. So this is CT dot. CT dot. We see that a term with X, as we expected. Because of course when X is zero, the result might be... When Y is zero, the result might be identity. The term with X disappears. And we obtain that C dot equals... Equals the following things. I write shortly. C dot equals C. The same guy. PT YT. PT minus Y. Now look at this term. Look at this term. These are quite common standard calculations for linear system. But I repeat it because we are formally... We started with non-linear, but now we rewrite this linear system. Okay? So it's this equation of the same type. We have family of defymorphins. Flow derivative is family of defymorphins. Treated as substitution of variables. And then... And this is vector field. This is vector field. You can check it. This is just nothing else that PT minus one star YT. This is a good exercise for you to check, but you can imagine. Because P is the element of the space of linear operators. P is a change of variables in the space. If you change variables in the linear operator, you have a similar matrix. Similar operator, okay? And that is what is here. You have a similar operator. And similar is again a vector field. It's a just change of what is here. Just change of variables in the vector field. You have to be careful with one minus one, but it is like that. That you can check, write, remember all the definitions. So C dot is going to be ODE. Right-hand side is linear with respect to Y. So we have C dot equals C. And here we have another vector field. Not Y, but Y is changed by X. Change of variables by X. Very natural. So if you divide it, very natural. And you can write expansion for that. Because the result is... And here you have a P T star minus one Y. P tau. Write it without arguments, because it's for all... Or you want to write it for T or tau, as you want. So what we have now? Phi of X plus Y is P. Now we would like to find expansion with respect to Y. Как мы сделали? Вновь вернувшись. Это C T, P T. P T не зависит от Y, только от X. Так что мы получим C T, P T, P T. Итак, что у нас? Это не зависит от Y. И здесь мы компьют... P T — это изображение X для нашей map. И мы компьют expansion. Это, конечно, начнет с P T. Мы перетерпим X. И мы получим expansion. Старфорс P T. Y is zero. Но expansion of this we already know. Expansion of this we know. Right hand side is linear with respect to our perturbation. We already know. So in our setting for non-linear equation, we cannot take a difference of solutions, but we can multiply. So we have a group. You think about it as a huge group, group of defymorphisms, or flows, group of flows even. And of course, we may translate things to union just dividing by boundary. So it can be computed also here. So it is indeed smooth. And in principle, you can write expansion. Let us now do it. Think a little bit about it. Come back in order to not make more complicated notation. X, Y, P — too many. Let us come back to the expansion in zero. Because expansion at any point is obtained like that. I repeat that P is just a change of variables. So you take expansion of C and make a change of variables by P. Very natural. Just another variable. It is a trick, which people all the time use working with non-linear ODE's and PDE's. If you want to have a good celestial mechanics since the 19th century, or maybe the 19th, if you would like to have a good asymptotic, controlled asymptotic expansion, you do it by change of variables. Not by taking difference, but always by changing variables. To kill low order terms, to kill low order terms, you change variables. So it is essentially what we do here, abstractly and formally, and to show that it is indeed the only way that can work well. Otherwise you cannot control anything. Okay, so now let me consider this expansion. This expansion identity plus. Or even we may evaluate it. Okay, identity plus Pt of X. Equal this. You can erase where F was specification, but for the moment I still do not need specification to explain, to understand some formulas. Pn, Xt1, dt1, dtn. So I write something, which does not have an operator. It's okay, but it's just a series of operators of more and more high order. So in order to compute something, I have applied it to some function. I compute some variable. But that's what I assume, what I would like to explain. No first term, I'm sorry. You do not correct me. X dot is identity plus integral. X2, d2 plus and so on. So we have some formula algebraic equation. Anyway, first term. So we have a flow. And maybe we would like to compute this flow at some particular point. Then we have a trajectory. And some strange series of operators. Of course, we may compute this operator also in some particular point. It means that you apply it to function at this point. We can do it formally. But it does not give you expansion of Q of t. This is Q of t. Like that, it does not give you expansion of Q of t. It gives you first term. It's okay that we already discussed it. First term is just average velocity. It's where we go, really, if it is not zero. Now one more tool in all the story. Pure algebraic. Since we have succeeded to write such an algebraic. Pure algebraic tool is like that. Assume that this guy is zero. This guy is zero. Then I claim that what is written here is equal. That you can even write. Look, please. This integral t2 t1 smaller than t x2 xt1 dt1 dt2 you can write it writing symmetric and anti-symmetric part. You can present it this way. It is going to be product square of average guy and plus integral just symmetric and anti-symmetric part of x2 t2 x2 t1 At Q0, here since it is vector field not just operator, commutator I can write it like that. So it's operator of second order. That's true, but if this one computed at Q0. If this one is zero at Q0 if average is zero, then this disappears. We know that if principle term is zero that second order term must be vector field. And we explicitly compute this vector field. Okay? And actually we can continue like that, but applying tool from the algebra. You expect that if some n-1 terms are zero that n-th term is a velocity of your trajectory and it should be vector field some combination of bracket. This is a fuel algebraic expression. And using some standard tool from the algebra, you can show actually it's an absolute universal property. If even without algebraically even without assumption that previous guy has zero you can present it as a bracket and quite specific concrete brackets white concrete brackets. So vector field simply substitute all products by bracket and divided by y no n. This is a projection on whole polynomials of non-commutative variables. We consider them justice getting more and more general but in a good way. So at every moment we can go down applied to some function compute instead of getting FU It's a formalism but you immediately compute things for your not just abstract it's the same story. Simply as we do in mathematics economy something you're thinking because when you need just algebraic manipulation think about forget about all other information all the information but for algebraic we forget about that think about them as a non-commutative variables. So this is of course what I do it's not the same but plus plus some polynomial on low order terms. As example I show it for second order terms but it's always like this. You can write the product any word with non-commutative variables for us it's important that this guy do not commute for different times otherwise they do not produce some anything. Polynomial of low order terms low order terms. So it sits algebraic property from the algebra and ok but if a previous term if it's just we do not specify q0 у нас but now if we specify q0 take one individual trajectory such that this low order terms vanish at this q0 then principle term has a specific bracket term. I already showed in the first lecture that example we set but it is always like that principle term is vector field more often with this algebra we already have no expression and this is a good deal and with that trick we can do it not only at zero but we can do it perturbing non-zero vector field also so that is what I wanted to say plus something smaller of high order something of high order now I erase and look more carefully on the bracket I formulate the theorem so with this calculations we can see what kind of controls what kind of non-autonomous vector fields allow us to go in the direction of some bracket we have to select controls such that these terms are zero and actually it's sufficient to consider bracket terms of zero because it's induction if first is zero then second is written as bracket if also second is zero then third is written as bracket and so on we can try to understand important set of vector fields almost specifically controls which make it vanishing all the guys up to order on it starts to be operationally deep, complicated and so on but this give us some specific algorithmic even way to go in certain direction of brackets always approximately we go very slow because it's high order x is small and expansion works when x is small that this is x of order epsilon that this is epsilon n we go slow or we go slow but at approximately more or less the picture is clear we may go somewhere now for the moment forget about the expansions playing a little bit geometry not only algebra that is very general but something not so easy to directly apply play a little bit with simple geometry and I am going to prove come back to the control system to the control system or not even to the control system yes q dot equals f u of q and let us consider just it's totally can be done in non-linear setting and I don't want to make any assumption about dependence on u and the nature of u will be just a set of indices if u is a set of indices I still can use piecewise constant control ok I can go sometime using u1 sometime then using u2 u2 u3 and so on when I fix foot guy here I have just autonomous time invariant ODE corresponding flow on parametric group then I switch to other everything is well defined with piecewise constant control and as you can imagine it's hard to write expansion but we know that we break it so crucial we break it so crucial and I am going now to prove the following very very general result that is equation by Kriner Kriner theorem now my idea not to write expansion but try to go exactly in the desired point from q0 and the general result says the following very general it almost nothing is needed just family but we allow to switch from one field time to time obtain some set theorem says that consider Li of our family FU that is just linear hull linear hull of all iterated brackets FU1 FUK Li at point q0 in particular so all linear brackets k is one two and so on all possible brackets and the theorem says that if if Li q0 FU this Li q0 of U is a subspace in the tangent space q0 okay we compute vector fields we break it the vector fields themselves and evaluate them if very general result but as you see the proof is simple nice and simple I have time to explain it and maybe I start even with I explain them give a color which says give a sufficient condition to get all the points by the way in such a setting we cannot do not have a chance to plane only we do not have a chance always to get all points from q0 because I could be okay so let me explain we consider this linear hull this linear hull and and I assume assumption main assumption is q0 FU is the same the whole vector field so by brackets we can get all direction all direction available by brackets at least at the level of linear hull what linear combination of then then then maybe I need let me call t of q0 the points when I can get all the points q of t it's called attainable set I need some terminology t u of dot where u of dot is just piecewise constant so that you can apply it always for any dependence on u piecewise constant so q of t of u of dot values where we can arrive and moment t switching in orbitary way from one dynamic to another that's attainable set we have started point q0 and we can go somewhere and the statement is so if it is like that we have more general statement we do it even for any q actually we can work locally it's somehow a kind of local result if we are if linear hull of all brackets is full-dimensional q0 of course it's full-dimensional in the neighborhood so it's not a big difference actually if we work and we can even think about many for this small neighborhood then I claim that that at has interior points and moreover and moreover it is closure of its interior points it has everywhere the same and full-dimension then a interior of at is not empty so if brackets are full-dimensional then also attainable set is for that for any t then for any positive t and it's and moreover at is contained in its it's closure of interior points so it may be the whole neighborhood it may be it may be kind of manifold with a border but it may be included or not this is at or it can have a casp it can have a casp but but you can have things like that of different dimension pieces of different dimension this is forbidden it's everywhere full so each point is reachable point is reached from interior as a limit of interior point so this is the result now let me explain for it such a generality looking only only we cannot we cannot I explain the proof of this and the next step perhaps in the beginning of the next lecture I will study the situation when we have a complete controllability starting from this result but in general we cannot do not have a chance to get everything to characterize complete controllability it's already a lot but to characterize complete controllability looking only on the algebra in particular let us consider the following situation for us it's very important for this course it's very important q dot is f0 of q plus sum of ui fi of q and ui say ui smaller than some constant we compute brackets we compute brackets but the bracket does not say anything in compute only linear hull about the force about the length of this vector with respect to this one so if f0 is much bigger than fi if f0 is like that and fi are not so big then all these velocities are contained in the cone this is a cone is f0 plus sum of ui fi where ui smaller than c so if we have a fine system and normally we have a fine system then the free term is dominating then of course we do not have a chance to actually this cone is quite low dimension but real linear hull real images of small dimension but there is a full dimensional cone where all velocities are contained all the time we just have a strong drift we have in one direction and we do not have a chance to come back if it is strong everything it may be even a constant vector field you take any linear system it may be a very big constant vector field and it moves it but also in this case if brackets have a full dimension then we guarantee that if it is not obtainable set that is a very important property universal variable why it is like that why it is like that some small lemo observation we have our manifold term and this sub-manifold 10 and m sub-manifold 10 and m the proof is based on very simple observation assume that we have a two vector fields vector field f and g they defined everywhere f and g some property, simple property of brackets f and g and assume that f and g restricted to n f of q and g of q belong to t qn for any q belongs to n so it's tangent assume that two vector fields are tangent to sub-manifold everywhere then I claim my lemo says that then also bracket is tangent so this lemo bracket is tangent let me explain so first of all if I solve my equation with some linear combination of f and g starting from n then trajectory at least for small time staying n if n is compact it's forever staying n if n is not compact then it can live n through the border staying n, why we have a vector field that here it is good to think with manifold because even if space was ran we have a manifold and we can think about f vector field you don't need any calculations to show that it's just the correctness of Cauchy problem you perhaps I will start next lecture with the proof of Kriner theorem based on this lemo because I don't have time at this I would like to explain well so trajectory for if you take any combination any control that is combination if you consider system q dot say u of t u of q plus v of t g of q combination like that then trajectory which stay in n at least for small time because you can think about this vector field if you are sitting in n you can think about this vector field as a vector field on n you can simply forget for the moment about ambient space it's a well defined vector field you have a tangent vector 2 so Cauchy problem is well defined there is a unique solution of this system in n there is some solution of this system in n but I say that this solution is unavoidable solution but it is also solution of the system in n because it satisfies the same equation and we know that there is just unique solution from the given point of a big system on mm it means that that solution because that solution it's almost autology but give great result now go to break it go to break it let us consider SID that the first lecture u of t v of t such that like a primitive it gives you with zero average assume that u of t dt on ok we can take upside u of t until the one u t dt equals v of t dt equals zero with zero average then and put epsilon here put epsilon here and consider this curve the curve which is q of one time one epsilon u v ok controls such a curves you have a curve here you have a curve here and computed at time one so you have a trajectory and for any epsilon now we take a curve which depend on other parameter time is always epsilon time is always epsilon and is always epsilon but time is always one but trajectory but control depends on epsilon we simply shrink it for any epsilon we need another manifold because trajectory belongs to the manifold when epsilon this is going to be call it the curve phi of epsilon this point this is phi of epsilon phi of epsilon so we have let me write we have a curve phi of epsilon on our sub manifold q zero and when epsilon is zero we arrive in the original point and we know how to compute expansion how to compute expansion this curve is this phi of epsilon is q zero plus epsilon square first order term is zero because average of control is zero and here we have area as you remember we can recover what I told in the first lectures from our expansion from second term of our expansion area of curves that the primitive of our this curve of gamma of t this is gamma, this is integral of u integral of v from zero to t gamma of t d tau and this is a gamma of gamma we have a lot of choice and here we have f1f of q zero and plus of epsilon cube so this is actually tangent vector as I already mentioned to our curve the whole curve stay in n and tangent vector is also belong to n and this is exactly what I told what I told if couple of vector fields are tangent to submanifold they are brackets and avoidably tangent to submanifold and iterated brackets tangent to submanifold I repeat I take control with zero average go time one and since trajectory must belong to my this is something which get obtained from control it looks like that this curve with parameter epsilon I get this control with average zero multiplied by epsilon so for each epsilon I have some point when epsilon go to zero I shrink to zero and since average is zero then the tangent to this curve is bracket and I have a lot of controls of course with nonzero area any choice is fine so I obtained that f1f2 belongs belongs to it ok, so I stop here because my time is over and from with this simple lemma I start next lecture with proof of Kriner series then I immediately give conditions, efficient conditions for complete controllability and some stable properties of this complete controllability and we go ahead we will have a lot of information of controllability and then also I will have time I am not sure that from the next lecture but on the last lecture to discuss seriously optimal control ok, thank you