 OK, well, thank you very much. Thank you very much for the invitation. So what I'm going to talk about is a joint work with Roy Kenyak, she's from McGill, Varga Nersisov, who is here from Versailles, and Claude Olympia, from Marseille. Well, the subject of this talk is closely related to the mini course, given by Sergei Kuksin. So I will not try to motivate the abstract form of the problem I'm going to discuss, but we'll do consider some examples later. So I will start directly from the setting of the problem in an abstract form, and I will try also to use the notation used in the mini course. So we consider two Hilbert spaces, H and E. So these are separable Hilbert spaces. And I consider continuous operator acting from a direct product into H. And finally, I will also need two subsets, one of them in H and the other K in E. And I will assume that these are closed subsets for the moment. So we'll consider the following random dynamical system with a phase space H. We have the relation UK equal to S of UK minus 1 E to K, or K greater than 1, or equal to 1, and the initial condition, U naught at time 0. So here, E to K is a sequence of IID random variables. And we assume that they take values in the space in the set K. I forgot to mention here the condition on the subsets. I will assume that the subsets are such that S of X times K is included in X. So this is an invariant subset, which means that if my random variable E to K is in K, for any initial condition which is in X, I will always remain in the set X. So E to K is IID with range in K. And due to this condition, equations 1, 2 generates random Markov process in X. Well, the goal is to study the large time behavior of trajectories of this Markov process. But very often, it is not the Markov process which we want to observe, but a functional of the Markov process. I'm not observable. So we'll fix F, which is a function from X to R, or can be also R B. And instead of the process UK itself, we consider the values of this F on the process UK. And we want to study the large time behavior of F of UK. Or very often, you deal with not this sequence itself, but you consider time average. So I will denote psi n of F, the time average of this sequence, because when you consider this sequence, you cannot say much. So you should take either the time average with respect to ensemble or the time average. So let's take the time average, which is 1 over n, the sum K from 0 to n minus 1 F of UK. So now the question is to study the large time behavior of psi n of F. Now there are at least three groups of result concerning psi n of F or concerning F of UK. So the first one is low of large numbers, or strong low of large numbers, even. It is about the existence of this limit when n goes to infinity. There is also what is called central limit theorem. It's essentially about the rate of convergence or after property normalization, it's about the limiting measure, and also the large deviation principle. And as the title suggests, I will discuss in this talk the large deviation principle. Just to mention briefly that this property, together with some additional hypotheses, always implies central limit theorem and central strong low of large numbers. So in a sense, this is a finer property than the first two of them, and a price to pay is that usually it holds under some additional conditions. So let me begin with the definition of large deviation principle. So there are, I will discuss in this talk, two levels of large deviations. So let's begin with level 1 LDP. So we say that satisfies level 1 LDP, or I will sometimes say just 1 LDP, if there is a function from R to 0 plus infinity, the plus infinity is not excluded, such that you have the following properties. First of all, if you consider the level sets or sub level sets, this should be compact subset of R. This is true for any C. And in addition, you have the following complicated inequalities. I take any Borel subset of R, and for any Borel subset of R, I should have the following. If I take the lower limit, as n goes to infinity, of 1 over n log Riff P u, this u means that I consider trajectory starting from the point u. It can be deterministic, or it can be random, but it should take values in x. And I compute the probability that this time average of the observable f belongs to gamma. So this is always no greater than the upper limit when n goes to infinity of the same expression. For the moment, I didn't write anything. The meaningful part is the following. So this should be greater than minus lower bound of i of x when x belongs to the interior of gamma. And this should be smaller than minus lower bound of i of x when x belongs to the closure of gamma. So let me write here where gamma dot is the interior of gamma, and gamma bar is the closure of gamma. Well, I didn't specify for the moment the class of functions I take. Well, you can think of continuous functions, bounded continuous functions, or just measurable functions. It depends. But when I talk about level 1 LDP, it means that I fix a function, and my rate function, this is called rate function, of course, depends on f. It will not be the same function for all of f. So there is a function if, which depends on f, which has compact level sets, and you have this rather complicated inequality. Well, if you are not familiar with this, this may indeed look complicated. But it is, in fact, the best you can hope for. So to see this, let us consider an example. Let us consider a trivial example when your phase space is the same as your noise space, and is equal to r. Then your map S is trivial. S of u eta is equal to eta. And your random variables, eta k, are normal random variables with mean 0 and variance sigma squared. And finally, your initial condition is a random variable, which is at the same low, and it's independent of eta k. Well, in other words, I just consider iidk's. So what I have written here is the iidk's, and you take a trivial function f of x is equal to x. This is just to show that iidk's is a particular case of this general random dynamical system. So in this case, everything can be computed. In this case, psi n of f, psi n of f, well, just a triviality, it will be just the sum of eta k, k from 0 to n minus 1 eta k. And for this random variable, everything can be computed. The low of psi n, the low of the sum from 0 to n minus 1 eta k, well, we have independent random variables with normal lows. When you take the sum of independent random variables with normal low, you again obtain a normal random variable with variance n sigma squared. And then, if you divide it by n, you get, again, normal random variable. But with variance divided by n squared, this should be divided by n squared. So you get sigma squared divided by n. So in other words, if you want to compute the probability of psi n being gamma, well, you have a density. You have explicit formula. You can write it explicitly. This will be square root of n divided by square root of 2 pi sigma. And then, you need to take the integral over gamma of exponential minus, so this is the variance. You will have x squared over 2 sigma squared multiplied by n dx. So everything can be computed explicitly. And you see that if you want to study log time behavior, the structure of gamma will play an important role. Because you have here exponentially decaying term. And of course, the maximal value of this one will determine the exponential decay. So you need to take the maximum of this one, because this will be the most important one. So if we introduce the function i of x, which is just this factor, x squared divided by 2 sigma squared, then everything will be determined, more or less, by the minimum of i on the set gamma. If gamma has a regular structure, however, let's consider the following two cases. So this is 0. Gamma is an interval here plus a point here. Then, of course, this point will not play any role. So if you want to have a lower bound, you need to consider only those points which are in the interior of gamma. On that end, if you want to consider the upper bound and you have a complicated counter-set of positive measure or counter-set minus something countable dense, then interior point will not be sufficient. You do need to consider closure of gamma to get the right asymptotics. In any case, it is a simple exercise to prove that for this particular case, you do have these type of asymptotics. If you take the logarithm and if you divide by n, then this will be described by i of gamma dot or gamma bar. So you have explicit formula. It is really easy to prove that you will have this and you cannot do better. So this complicated inequality is somehow justified. So this is what is called level 1 LDP. Now what about level 2 LDP? This is what is interesting for applications, but it turns out that in this form, it is more difficult to study. So the right way to proceed is to consider something more general, which is called level 2 LDP. So what is level 2 LDP? Let us consider the following sequence of random measures on x. So we have nu n. And I write 1 over n the sum k from 0 to n minus 1. And I take Dirac mass at the point u k. So u k is a random sequence. Dirac mass at the point u k is a random measure on x. So nu n is a random measure, random probability measure, on x. And then definition, so for each omega, for each realization, this is a probability measure on x. So definition, we say that 1 satisfies 2 LDP or level 2 LDP if there is a function i which acts on the space of probability measures on x. So this is the probability measures on x, which range in 0 plus infinity. Infinity is not excluded. Such that, again, you have this compactness property. If you take the level set, then this should be compact set in px. And since I'm talking about compactness, I need to specify the topology. So this will be always weak start topology. So weak start topology is defined by convergence on bounded continuous functions. So this should be compact for any c. And now for any Borel subset of the space of measures, because now we are working in the space of measures, you have this type of inequality. So let me write it again. So we have lim 1 over n pu. Let me remind you that u means that we start from point u. And here I'm writing nu n, this random measure. So this for each omega, this is a measure. So I can consider probability that this belongs to gamma. And I forgot the logarithm here. So this is, again, smaller than lim sup. As n goes to infinity, all the same expression, 1 over n, logarithm, probability for the trajectory starting from u that nu n belongs to gamma. And the left and right and leftmost and rightmost term are exactly the same, minus. So I will introduce an notation here, i of gamma bar, and greater than minus i of gamma dot, where i of something is the lower bound of i lambda when lambda belongs to a. So this is exactly the same inequality, except that now we are writing in for the space of measures. So we have level 1 LDP. It is for functions. And we have level 2 LDP. It is for measures. So what is the relation between these two? Relation is very simple. It is given by the following proposition. If you have level 2 LDP, then you have level 1 LDP for any bounded continuous function on x. And let me outline the proof. So this is a well-known result. And it's called contraction principle. So let us call, let us consider a function. Let us fix f, which is a bounded continuous function on x. So this is a case of bounded continuous function on x. And let us consider the following map from, it depends on f, and which goes to the space of measures into r. And it takes any measure lambda to the mean value of f with respect to lambda. So this is bracket mean, the integral of f with respect to lambda. It is straightforward to check that if you take psi n of f, then it is exactly the mean value of f with respect to nu n. Or in other words, psi n of f is the image of this sequence nu n under the map phi f. And this is a continuous map. Sorry, this is not what I want to say. Now if you have this, you have also similar relation for the lows. The low of psi nf, because large deviations involve the lows, right? We are considering the low of this random variable. So the low of psi nf can be expressed in terms of low of nu n. Now low of nu n is a probability measure on this space. So this is a probability measure on the space of measures. And I need to consider the image of this low under this mapping phi f. So you have this relation. And now this application, which sends the lows on px on the lows of r by this formula with respect to phi f, is continuous. And there is what is called contraction principle. It has nothing to do with the Banach contraction principle about fixed point. And it says that if you have LDP for nu n, LDP for nu n, this is exactly level 2 LDP. Then due to continuity of phi f, it should be continuous. Then you will have LDP for psi nf for any f, which is bounded and continuous. So this is really direct application of what is called contraction principle. And you can find contraction principle in any book on LDP. This is one of first result. This is really a soft result. In addition, you can write, so let me write here, the explicit form of the rate function. Rate function for 1 LDP. So i f of x is the infimum of i of lambda. So i f of x is the rate function for 1 LDP. i of lambda is the rate function for 2 LDP. Then you should take all the measures such that f integrated with respect to lambda gives you this value x. OK, so this is really a soft result which shows that level 2 LDP implies level 1 LDP. So from now on, I am going to consider only level 1 LDP. OK, so let me summarize what we have done because I'm going to erase something. So we are starting this random dynamical system. So let me write this system here. u k is s of u k minus 1 e to k. So this is system 1. 2, this is the initial condition. This is the initial condition. We have e to k is an iid in x. And this x, no, so it's iid in e. And I consider my system u k in x. And this hypothesis ensures that we do have a well-defined system in x. OK, so this was the first section. Now the second part is the main result. Well, from now on, until almost the end, I will assume that x and e, x and k, are compact subsets of h and e. So I consider the compact setting. This is not very restrictive if we are dealing with PDEs with bounded noise. And I will need some hypotheses which are very similar to those discussed by Sergey Kuxin. So the first hypothesis is just regularity condition for s. So I call it regularity. Let me emphasize the hypothesis so we can see them easily. Regularity, it says that there is a vector space compactly embedded in h such that s, which acts, s acts from h times e to v and is c2 bounded. But c2 bounded means in this context that it is bounded on bounded subsets. So we have a map which is twice continuous to differentiable. And I assume that the map itself and its derivative up to order 2 are bounded on bounded subsets. So c2 bounded means by definition that twice differentiable, OK, so then c2 bounded is just shortened for saying continuous to differentiable, twice continuously variable bounded on bounded subsets. OK, so this is the regularity condition. Now, the second condition is about global controllability. So let's say gc, OK, so there is no big difference between y and rho. So global controllability, it says that for any epsilon, there is an integer m such that for any initial point u0 and final point u hat, I can find eta 1, eta m in k such that the distance between the trajectory starting from u0 and driven by this control, eta 1, eta k m, is in epsilon neighborhood of my target u hat. So sm of u is the trajectory of this one at time k, right? So in other words, sm of u is um when you use controls eta 1, eta m. So this is the trajectory of 1, u equal to u0. So this is the usual concept of global approximate controllability in x. We start from any point. We want to go to the epsilon neighborhood of my target. And the property is that it should be uniform with respect to the size of the ball. Sorry, with respect to the initial and final point. But of course, it can increase when I decrease the size of the ball. So I fix the size. Then there is a universal integer such that this is true. Now, third condition, controllability of linearization, this is exactly the same condition as before. If you want, I will use b3 prime in survey cooks in stock. And it says that for any u in x, for any eta in k, the image of the derivative d eta s u eta. So I differentiated s with respect to eta. This means that this is a map linear application from e to h. This should be dense. OK, so three hypotheses, regularity hypotheses. It is c2 with a smaller space, compactly embedded in h. Global controllability, I can go from any point to any other point uniformly with respect to the initial and final points. And controllability of the linearized map, which says that the derivative of s has a dense image. So these are conditions on s. And we need also condition on eta on the low of eta. And I call it decomposability. And it says that there is an optimal basis in e such that two properties. First, low of eta, I'm taking the lows of these IID random variables, can be written as a Tesla product of its one dimensional projections onto the straight lines spanned by Ej. So I will write and make some comments. So Lj is the image of L on the one dimensional projection. So let me write here Pj. So we have an optimal basis, Ej. We have orthogonal projection Pj on the straight line spanned by Ej. So Pj is a one dimensional projection. So for each measure, for this measure L, I consider all its one dimensional projections. And the condition is that L should be Tesla product of these measures. Or if you want this in terms of the random variables eta k, it means that eta k can be written as the sum j from 1 to infinity of, well, it can always be written like this way, eta k j Ej. Well, I just decomposed eta k in the orthogonal basis. The condition of these random variables should be independent. So this is a rewriting of this decomposibility condition. So L is a Tesla product of its one dimensional projections. And in addition, I assume that this one dimensional projections possess c1 smooth densities. So this is true. And Lj of dr is equal to rho j of r dr. And rho j is a c1 function on r. And of course, it also has compact support because I assume that this is compact. So in particular, each projection has a compact support. So four conditions, regularity, global controllability, controllability of linearization in the sense that the image is dense, and decomposibility on these IID random variables. Then you have the following result. OK, main theorem. Well, it has two parts. The first part is very easy to formulate. If you have r, regularity, global controllability, controllability of linearization, and decomposibility, then you have 2LDP for 1. OK, this is really a theorem. You don't need anything else. Second property about the rate function. The rate function which enters this inequality, which determines the large time asymptotics of these probabilities. So the rate function has the form. The rate function is a function on the space of measures. So I of lambda is equal to, you need to take the supremum over all continuous functions, which are, let's say, greater than 1, of the following integral. Integral over x. And you need to take here logarith, g divided by p1 of g. I will write in a second what this p1 is. d lambda, this is, in fact, a Markov semi-group, a time 1, where lambda is an arbitrary measure, probability measure nx. And p1 of g, this is a function. For instance, I can write this way p1 of g at the point v is the main value of g calculated on s e to 1. Or in other words, this is just Markov semi-group, a time 1 associated with this process. OK, so this is the main result. So let us compare what with the results in the mini-course. Yeah, well, it is hidden here. So I believe, yeah, I mean, if you assume this b3 prime, which is a stronger condition than b3, then you don't need global fixed point. Globally stable fixed point. So I did assume b3 prime. So I don't need globally stable fixed point. What would be sufficient for Sergei? It would be just there is one point to which you can drive all your trajectories. Here, I need something more. I need really irreducibility from any point to any other point. OK, and this will be one of the open problems I'm planning to give at the end. OK, so this is the main theorem. So the next point is the applications. But before doing that, let me mention some references. In fact, the whole story about large deviations for stochastic equations started in the 70s by Don Sker and Varadan. They've written a series of four papers covering a lot of stuff. Then there was an entire series of papers about improving this result, proving it for more complicated equations, so on. So there's no way I can mention any of them. In the context of PDEs, there was a result in 2006 by Liming Wu and Gursi. Matthew Gursi, he was a student of Liming Wu. They studied Navier-Stokes equation or Burgers equation per term by rough noise. And if we translate into this language their condition is that here, the image of the derivative is not simply dense, but it is equal to h. So this is the difference between these results and what was done by Gursi and Liming Wu. And what I'm talking about is our two papers. One of them is published. The other is not. Well, in fact, there was another one for unbounded noise with my co-authors, Yakshic, Nershe Siam, Pia and myself. So this is the way it is written in the recent paper of 2019. OK, I think I have some 15, 60 minutes left. So let us discuss the application. All this is out and I will conclude by formulating three open questions in this subject. So application. Well, in fact, this was more or less done by Sergey Kugsin, so let me do it briefly. So we can apply this result to, roughly speaking, any reasonable parabolic PDE. So let me consider two examples. Navier-Stokes in 2D. Well, I consider Navier-Stokes in a bounded domain, in a compact domain in 2D. So let me remind the form of Navier-Stokes, DTU plus U nabla U. Well, of course, this is just a scalar product in R2. Minus nu delta U, nu is a, well, let me raise nu. It doesn't play any role. Plus nabla P is equal to eta of Tx. Divergence of U is equal to 0. And everything happens in a bounded domain D. Well, and we impose directly a boundary condition. U on the boundary is equal to 0. And some initial condition U of 0x is equal to U0 of x. Well, I will specify in a second what is the exact form of eta, but it's exactly the same as in the mini-course. Or you can deal with complex emotes-Landau equation. Again, in a bounded domain. But now we consider it in dimension 3. And it has the following form, DTU minus delta U plus I U to 2P U. Again, equal to function eta of Tx. Again, I impose directly a boundary condition. On the boundary is equal to 0. And some initial condition U0x is equal to U0 of x. And here P is a number. So P equals 0 is trivial. But we can do P equal to 1 and 2 in dimension 3. So these are two examples for which you can apply this theory. Now, let me specify the form of the random force. It is exactly as before. So you have a sum k from 1 to infinity. You have indicator function of the interval k minus 1k. And then you multiply it by independent random variables eta k. So let me first write T minus k plus 1x. So eta k is an IID in L2 on the interval 0, 1 times D, which range in R2. So I divide my interval, time interval, time axis into pieces of length 1. And here, I act with the force eta 1. Here, with the force eta 2, eta 3, and so on. Then, if I'm interested in the dynamics of my system at integer times, so if I denote by Uk the solution at time k, then obviously, to calculate solution at time k, I can apply the time 1 resolving operator for Navier-Stokes. No, for Navier-Stokes. And I need to apply to this value solution time k minus 1 and to the noise which acts on the interval k minus 1k. So we have exactly the system of that type. So to check that we have large deviation principle on level 2 for Navier-Stokes, we need to verify these four conditions. Well, a regularity condition is nothing because this is always true for PDs in sufficiently smooth functional spaces. Controllability is something. I will discuss it in a second. Controllability of linearization in this setting, it is trivial. Because I assume that my space E is, well, sorry, I didn't assume anything for the moment. But I will assume that e to k of tx as the following form. It is basically what was written there. j from 1 to infinity bj psi jk ej of tx, where ej is an orthonormal basis in L2 in exactly this space, which I will denote by E. This is my control space. And bj are non-zero numbers. I can assume that they are positive, such that the sum of bj squared is finite, so that the noise is regular. And psi jk are independent random variables with densities. The law of psi jk has a density, which is denoted by rho j of r dr. And this rho j are compactly supported functions, C1 functions on the interval minus 1, 1. So if we assume that your noise or your control in the right-hand side has a full range, and when you consider linearization, you will have a linear equation with a control which has a full range. So there is no problem for verifying controllability of linearization. So the only condition is this one, which is not clear how to do, because your noise is bounded. And even though you have a full range, you are not allowed to leave the support of the noise. Because here it is important to have something this eta j should be in the support of the noise. So you cannot leave the support of the noise. And this gives a restriction for our result to be applicable for the Navier's talks. We have to take the following setting. These conditions ensure that you have a unique stationary measure. So let, so this is about condition global controllability. So let, so this discussion about this one. So let mu be the unique stationary measure for one. Well, this we already know that it exists, and it is unique. Then it's really easy to prove, using the uniqueness, that the support of this measure is exactly the domain of attainability, let's say, from 0, or the closure of domain of attainability. So you can see that all the points in the phase space, which can be reached starting from 0 and using controls in the support of eta k, in the law of eta k. So you take the domain of attainability from 0 at time 1, time 2, time 3, and so on. You take the union, and you take the closure. Then this is really not difficult to prove that. This is domain of attainability from 0. And this will be exactly our space x. The advantage of taking x in this form is that you can reach from any point to any other point. Because you can reach from any point to 0, exactly due to the property Renova's mentioning. You just switch off the noise. Ah, sorry. In this setting, in the form of this one, I have to assume that rho j is positive at 0. Sorry, I forgot about that. Well, in the abstract theorem, we don't need that, because we just postulated it. But in this concrete application, we need to make 0 accessible from any other point. And this is done by assuming that the density at 0 is positive. So we can go from any point to the neighborhood of 0, and from neighborhood of 0 to any point here. So in this, if we take x to be the domain of attainability from 0, which is the support of mu, then you will have this global controllability for free. So this shows that the result can be applied to Navier's Tox system, provided that your initial condition is in the support of invariant measure. OK, and the same result is true for the Ginsburn-Land allocation. So this is all what I wanted to tell about applications. Now, three open problems. The first one is exactly here. So we know that there is a unique invariant measure. We know that any solution converges to this invariant measure exponentially fast. We know also that if you start a solution from the support of invariant or stationary measure, you have large deviation principle. And what about if you start from somewhere else? You can say that you will end up at random time, which is final exponential moment, in any neighborhood of mu. But you cannot say that it will hit the support of mu. In any neighborhood of a support of mu, but there is no guarantee that you will hit the support of mu. So it is an open problem to prove the LDP in the setting of man theorem for initial data, the complement of x. Let's take deterministic initial data, which does not belong to x. My system is defined everywhere. And let's assume that you can reach any neighborhood of mu in exponential time. Still, what all you can prove is the local LDP. Well, I don't want to give definition. But if you really want honest global LDP, this is an open question. Second open problem is the case when you don't have this assumption. So this is one extreme case. You have the other extreme case when x is h and k is e. So your support is everywhere. Your phase space is unbounded. Well, there are some partial results with the same authors in non-linearity in 2018. Well, it is not partial, it is about kick forces unbounded. Yeah, unbounded kick forces supported everywhere. But these are kick forces. These are not this multiplicative type force. So this is a LDP in the unbounded case. Well, the challenge is to really formulate a simple result. Well, I find it simple, at least as far as the hypotheses are concerned. So to formulate something like that in the case when you have unbounded noise, well, of course, you will need some extra condition, like Laponov function. But it would be interesting to have this type of result, which can be easily verified. As easy as this one. Roughly speaking, you add Laponov function and you have a result. Something like that. And finally, the third problem is related to Sergei's talk, Sergei's mini course. It is LDP, but when you don't have B3 prime, you have only B3. So prove the LDP when CL is replaced the following hypothesis, again in unbounded setting. For any U in X, there is KU, a subset in K, such that the measure of this KU is equal to 1. And the image of the derivative is dense in H. And this should be true for any U in X. Sorry, it's already there. Such that the image is dense for any eta in KU. I have to say something about the difficulties of this problem. So this is not, I don't even know that this is true or not. Well, this one, I believe, can be done because you already have something here. You just need to understand what exactly is used there in terms of controllability. And finally, this one, I believe, will be as hard as the problem of exponential mixing in this setting. OK, I think I'll stop here. Thank you for your attention.