 Okay, so I'd like to thank all the organizers for this beautiful meeting, and it's already quite late, this is the last talk, so I'll try to make it simple and hopefully brief. So this is going to be, I'm going to give you some history and what you are doing for the Boussinesq equation, and then I'll try to give you some proofs hopefully easily followable. All right, so I'm going to talk about the Boussinesq system. It's a system which is coupled Navier-Stokes equation with the Heade equation, and as usual there is a usual notation about the gradients and Laplacians. What's important as you as usual for the fluid flows is velocity, pi is the pressure and T is the temperature. So this is a thermal motion of a fluid, so imagine that you have a, well I'll explain it in the next slide, I won't make it brief. So first of all we have a usual Navier-Stokes equation. I'll explain that, I'll explain that right now. So first of all it consists of Navier-Stokes equation, so that's a usual equation for the evolution of the velocity. Then it has this additional term which is called buoyancy term and it's modeling that pretty much the warm fluid is rising and the hot fluid is dropping, so that's that's that's the responsibility of that term. Then there is a Heade equation for the heat transfer, okay, so there's a usual Heade equation T sub T is equal to Laplace of T plus we are assuming that the heat is transported by the particles that are distributed or going along the trajectories of velocity field U and now if you find them, if you write down this model in physics notation then it of course has millions of parameters but after some rescaling and randomization then you can actually, well you can reduce those parameters into two, one is called Prandtl number which is quotient of the viscosity and thermal diffusivity and the other is called Heade-Raley number which depends on many, many parameters, viscosity, thermal diffusivity, also on the size of the domain gravity and so on and so forth, so the Heade-Raley number is kind of complicated, Prandtl number depends only on the properties of the fluid, okay, so it depends on the viscosity and thermal diffusivity and know that the Prandtl number might be large or might be very, very large, for example, if thermal diffusivity is much smaller than the viscosity or can be very, very small if it's other way around, so there is no like the canonical good range for which, for which the Prandtl number is either small or large, okay, and then there is external forcing and this is stochastic conference, so of course the forcing is going to be eventually stochastic, okay, all right, so what about the boundary conditions, so the canonical model for the Boussinesq system that studied for, you know, 150 years now is the fluid confined between two plates, we are keeping the hot surface on the bottom and the cold on the top, so you can imagine it's like a heating coffee in the morning that they are heating water from below and cooling it from above that's cooled by the air, I'll get to that one, but I believe that F2 is more physically motivated, yes, so the reason for, well, I hope I didn't erase that slide, okay, so what we are going to assume that the boundary conditions is we are going to assume that there is a non-slip boundary condition on the top and on the bottom and the temperature is bigger on the bottom than on the top, okay, so by easy transformation we can assume that the temperature on the top is zero and temperature on the bottom is positive, then we can go to non-physical boundary conditions, it's a mathematically easier model that's going to be the torus when I'm going to somehow wrap up the top and bottom boundary, in the horizontal direction I'm always assuming periodic boundary conditions, okay, all right, so there's a very short slide about the motivation, so there are many, many models, there are many, many situations when we are modeling, we're modeled by the Boussinesq equations, for example, the atmosphere, the earth is warmer than the space, so there is a natural convection, there is any, the same is in the earth mantle where the core is much hotter than the lithosphere, so and this pretty much flow is responsible for the motion of continents and so on and so forth, okay, the same is in the sun and in other stars where the core is much warmer than the space and so we need to, we need to take thermal effects into account, okay, all right, so, all right, so the motivation, this erase that slide, so if you think about this situation and you try to decide where to put your forcing, then there's a natural stochastic forcing coming, for example, in the sun or in the earth mantle, it's the radiate active decay, this is inherently stochastic, okay, we know that and it's inducing some, it's adding some temperature to the fluid, so you can be modeled as a point kicking in the temperature equation or you can have other reasons for the stochastic forcing in the temperature equation, all right, so just very briefly, if we are trying to, this is the older result of ours, so if we put on the torus, then we can get a analogous result as Aaron Mattingly did for the Navier-Stokes equation, sorry, who is D? D, oh no, I erase that slide, I'm sorry, I'm sorry, this, you're completely right, so we are summing, so our stochastic forcing is going to be N is equal to 1 to D, actually that's the number of modes that we are forcing, something like an E, N, X, times W and T, okay, so these are stochastic forcing and this is the number of modes that we are forcing, okay, that's coming into, D is the parameter of the stochastic forcing, all right, so the usual method, well the usual culture in the field is that the smaller the D is, the harder the problem is to prove actually the uniqueness of invariant measure and we were able to establish the uniqueness of invariant measure when only four modes are, only four modes are forced and for example the modes X1 and X2 and this is actually less than is needed for the Navier-Stokes equation, okay, so for the Navier-Stokes equation if you force it only with this forcing it's not known if the, if the invariant measure is unique or not, okay, so the point is that although we need less forcing we have somehow more nonlinearity, so it's not only this Navier-Stokes term, we also have this nonlinear term in the, in the temperature equation that's, that's helping us a little, okay, so there is a long history of this problem with various, various versions of D, I want to go into details I'm probably going to skip at least half of the people that, that work in the field, so, so that's about the torus, okay, so what about if we take physical boundary conditions meaning that I really have where they are fixing non-slip boundary conditions on top and on the bottom, all right, so the problem is harder and there's like a subtle, subtle problem and we cannot, cannot do the hyperliptic case and we cannot do even the hyperliptic case in the, for, for the Navier-Stokes equation only, okay, so this is not really technical difficulty of the Boussinesq equation but it's inherent there, all right, nevertheless, we can prove that there exists, the question is can we get the existence and uniqueness of invariant measures, okay, and the other motivation is going to be to investigate what happens when the parallel number goes to infinity, okay, let me, let me talk about that, all right, so what we are going to work on is the dimension when the dimension is two or three, so both physical dimensions, okay, so this already looks very, should look very, very suspicious, how can we work for the 3D Boussinesq system when we, there's a Navier-Stokes equation, we don't even know it's well-posed, okay, so the problem what we are solving is a little bit easier, it's a little bit of cheating, so in 3D what we can prove that there exists a stationary state, that's probably not a big deal, the existence is known for these problems by the Krilov-Bogolivov procedure and these states are unique in 2D, okay, that's probably culture that, like many people would be able here to solve it maybe in 10 minutes, then what we can do is when parallel number is infinite, so that really means that this term is not there, then we also have unique invariant measure in two or three dimensions, okay, so we can prove that, which maybe it's again not too surprising because, okay, the bad term is gone, so we can proceed and the result that may be the most interesting of those is in 2D or 3D if we pass the limit, so we have a unique invariant measure that we call mu infinity when parallel number is infinite and then if we try to calculate the distance between parallel number for finite, sorry, the invariant measure for finite parallel number and the, sorry, the distance of the measure for the finite parallel number and the distance to the infinite parallel number, we can actually prove that they converge to each other, so technically we don't know that the measure is unique because it's a 3D problem and the problem is hard, nevertheless if the parallel number is getting larger and larger then all the all the invariant measures, no matter if it's unique or not, are going to converge to this unique invariant measure at the end, so this, okay, so the problem is, one of the problems is if you think about the system when parallel number is infinite, this is not evolution problem anymore, okay, we are losing time derivative, we are killing the Navier-Stokes term but we are also losing the derivative and at that moment your phase space is just one dimensional, it's only temperature equation, whereas where these invariant measures for the full system live is four-dimensional, it's one component in temperature and three components in the velocity, so it's not even clear what you mean by convergence, it's four-dimensional objects converging to the one-dimensional one, nevertheless it's possible to prove that such a statement holds, all right, so the state of art right now as I know for the Boussinesque is for the torus we have a complete answer for the hyperliptic case for the system which has boundaries, we need sufficiently many modes to prove uniqueness in 2D and in 3D we have these results about the convergence, all right, let me, all right, so now I would like to talk about what is going to be the main part of my talk, so now the question is what happens if we don't have that many modes even in the hyperliptic case, what do we are forcing just very very very few modes, so for the title of this conference there was a topic of this conference of course there's no way that we have any controllability that we can get anywhere in the space, so the question is is this problem anyhow interesting and what do we want to do about that, okay, so first of all what happens if we don't have, if we don't have enough lee brackets, if the noise is not spanning the whole space, so for example the motivation is if the noise is acting through the boundary then we can substitute it off and suddenly we have very degenerate noise that's coming into the equation, also, okay, so the answer that I'm going to talk about is what if our noise acts only in the vertical direction, so in that case we have a huge advantage because I'll show you in the next slide we can write downwards the invariant measure that's very explicit and another ingredient of this one is, okay, if we have unique invariant measure we saw this one in all those talks, we have this exponential mixing or attractivity to the unique invariant measure, so that's clear, but what we don't know and that was question of Sergei, our first questions of the conference is do we know anything about the distribution of invariant measures, do we know where they live, high modes, low modes, what's the distribution of those, so the point is that uniqueness is implying somehow stability, okay, so everything is attracted to this unique state, well if we are in the, in this very degenerate case then we have very explicit invariant measure, we know what it is exactly, we can write it down, it's just a bunch of OU processes, that's, we know exactly how, how that, how that, how that measure looks like, now if it's stable and we know that it must be unique and that's the limiting measure, okay, so that's the limit measure, now the question is for which parameters this measure is stable, meaning for which parameters we actually know how the invariant measure, the unique invariant measure looks like, and of course there is a, there is an open question if we are going out of these parameters, are we still close to the, to the invariant measure, so can we say something about the distribution of invariant measure, all right, so what I'm going to talk about, now take this again, the Boussineq system, now the slides, the slides are going to be pretty full but I'm going to write down quite a bit on the board, so what we are going to do, we are going to assume that the, that the forcing is acting only in the z direction, okay, so using the board, so this is not going to be that important yet, so okay, so stochastic forcing acts only in the z direction, so we don't expect the uniqueness of invariant measure but still uniqueness is related to the, to the stability, so, so first of all how does it look in the deterministic setting, so again we have this domain, just like between two plates, and in the deterministic setting meaning when sigma is equal to zero, no forcing, then if a Rayleigh number is small, then we have a globally attractive equilibrium, and that globally attractive equilibrium is equal that u is equal to zero, nothing is moving, and temperature is just a linear profile, so when you're heating your coffee in a, in a way for a while the fluid is not moving, all the, all the temperature is moving only by, only by conduction, there's no conduction, there's no motion of the fluid, okay, so, so this is when Rayleigh, so this is, this is, this is a solution, so this is a solution and what's important is stable is this Rayleigh parameter is small, okay, this Rayleigh parameter is small, then this is a, this is a, now this is a classical result, classical result of Rayleigh that what he proved is that there is a loss of stability for a large Rayleigh number, okay, I'm going to talk about this one a little bit later, so now in the stochastic setting, so now imagine that the stochastic setting is not zero, we still have something called, so this is deterministic, and now stochastic, so the stochastic one is we still have this conductive state, that's a huge advantage when u is equal to zero, and what's the equation for the temperature, it's just the usual OU process minus, this is the vertical direction of tau, is, is pretty much stochastic forcing, okay, so this is a nice, nice equation, everybody can solve this one, it's a heat equation with the stochastic forcing, it decomposes into the, into the Fourier modes, so the solution is completely explicit, it's just a sum of OU processes, here we can write down what the invariant measure is, we can, we know everything, so now the question is the uniqueness and stability of the full system, meaning when u is non-zero, well it will depend on parameters, if this, if this invariant state is stable, then we have uniqueness, everything is going to that one, if we don't have stability, then we are probably, we probably don't have uniqueness of invariant measure, okay, in this situation it's harder, yes, yes, but I'm interested in this, in this case of thermal, yes, all right, so what do people are doing, okay, so we have a, we have one state, so in order to assess stability, well we just calculate the fluctuations, okay, so what we are going to do is I'm going to write the, I'm going to decompose my solution temperature, in the temperature as my state that I know what it is, that's my OU process, plus some fluctuations, well if you, if you write down the equation for the fluctuations, I don't want to, it's, it's, it's rather trivial, and use, use the energy estimates, you just multiply the first equation by U, the second equation by theta, then you get this equality, okay, so now this, this is getting interesting, so you have one half of derivative of the L2 norm, so that's, theta square is the fluctuations, plus one over Prandtl Rayleigh, those are some parameters, the U2 square is actually equal, and that's going to be minus gradient of theta square plus gradient of U square with parameter Rayleigh, and then there is a nonlinear term plus integral UD is the D component, theta, the derivative with respect to xD of tau minus one, and there's an integral between zero and one, ah sorry, this is over the domain, over the domain D, all right, so the point is, so these are fluctuations and we'd like them to go to zero, so what I'm going to do, I'm going to just divide this guy by the, by the red hand side, theta square plus one over Prandtl Rayleigh U2 square, and multiply it back, a good habit, okay, so all's good, so now I'm going to denote this factor, I'm going to denote this factor lambda, which depends on tau, well it's going to be a little bit slightly different, so let's, let's call this one capital lambda, so these are stochastic, time dependent quantity, now there's a, there's a big step what was done in deterministic, in the deterministic case, I'm not going to calculate this lambda tau, it's a mass, okay, what I'm going to do, I'm rather going to denote lambda tau to be just in themum with respect to theta and U of this, of this quantity, gradient theta square plus one over Rayleigh H1 norm of U square plus integral over D UD theta, let me just erase this one, it's not going to be important, so over D UD theta and then, and then L2 norms, okay, so now this system is not, and for fixed, so given tau, I'm going to denote this quantity, and I'm sorry, this is very important, there's a minus sign here, okay, so, all right, so the problem is, so this is the infimum over all U and theta, so that really means that if I can prove somehow that lambda is positive, lambda is positive, then for my tau, say for every tau, then this is of course exponential decreasing, exponential decreasing differential equation, and it must be converging to zero, hoping that lambda goes to zero, lambda is always positive for any tau, is probably a little bit, a little bit too optimistic, but nevertheless, so so just, this is just a remark, when we are in deterministic setting, meaning that tau is equal to one minus z, then we also know that lambda negative, this variational problem implies instability, this is a non-trivial proof, okay, in our setup, we don't know what's the instability criteria, okay, so I'm going to prove only some criteria for stability, because you see for the stability, it's great if I take still infimum and have positivity, however, for the instability, I shouldn't take the infimum and I'm back in the full problem and that one is a mess, all right, so what you can do, and I'm just going to, I'm not going down this rabbit hole because it's hard, I think it's hard, so what you can do is, okay, you can find, you can decide, okay, it's a minimization problem, so I'll just write the Euler Lagrange equation, so if I write the Euler Lagrange equation for my variational problem, I'll get this coupled system and there's a coupled system of four equations, so if you're smart and you apply it curl twice to the velocity equation, just to get rid of non-divergence condition, then you get the system, like, eigenvalue problem that looks like this, okay, maybe there are some people that can do something with this problem, but it looks very, very bad to me, okay, so this problem I need to, the question is, can we estimate the distribution of lambda if I give you the distribution of tau? I literally don't know how to do this one, okay, so I don't want to go down this rabbit hole, all right, so here we come, so this is our minimization problem, so the good habit is we should have existence of minimizers that we know that we are solving something, okay, so in order to prove the existence of minimizers, it's rather easy because we know that this nonlinear term for if tau is l infinity in x, it's actually some or finitely many cosines or sines, so it's not a problem, then we can estimate the nonlinear term by the l2 norms and that gives us immediately that the functional is bounded from below, it's bounded from below functional, it's quadratic functional, it's like one line proof to prove there's weekly lower semi-continues and you open the book on the page one on variational methods and then you see the result there exists a minimizer, that's about that difficult, all right, so now the question is what are we after, what's the criterion, so notice again that our system seems like it took infimum, the ODE that we'd like to solve has this lambda on the right hand side, so now imagine that I fixed t tau, meaning that I fixed one trajectory in my noise and I get tau, and so I know that if this integral between zero and an infinity lambda is positive, then the solution is actually exponentially decreasing, strictly positive, all right, so that's good, but the problem is that tau depends on the noise and so on is four, so it's rather complicated, however, this is where argonicity kicks in, so what I can do is I'm going to take, I'm going to assume beaker criterion, so imagine that they take integral and divide it by t and I send t to infinity, if I know that this is positive, then I'm still good to go with exponential decay, that really means that this is, yeah, this lambda increases like some constant times t at least and this is going to give me the exponential decay, all right, but now you notice that here is only tau, which is the OU process, we know everything about OU process, lambda is continuous, depends continuously even as a Lipschitz function on tau, so I can invoke some results from argonicity and realize that actually I don't need to take these integrals between zero and t, I don't have to take time averages, I can take just expectation, so here is a general thing, I think, okay, I'm always afraid of this slide a little bit in front of any audience because somebody may tell me that that this result is known for 150 years, well maybe not but for 50 years, so I'm literally asking if you know any results in this direction, you have a variational problem where some coefficients depend on some stochastic parameter, say this is Gaussian, imagine that tau is Gaussian, so I have a Gaussian distribution and I have a quadratic function or any other functional, the question is, I'm asking what's the expectation of the minimum, are there any results about stochastic variational problems that anybody's aware of, I would be really happy if you can tell me, okay, so and there is maybe possibility for like doing like some, maybe it's hard but there is a possibility to start developing the theory of random variational problems with certain, with trying to find the distributions of the minimizers, here I'm looking only for the expectation and the point is that if I have that the expectation is positive then I have stability, if the expectation is negative I don't know but there is a good indication that there is an instability okay, all right, so what can we do about this one, so this this is actually a very very good problem because everything is quadratic there, so first of all the hope is that there exists something like a critical Rayleigh number as in deterministic case saying if I'm, if my Rayleigh number is small then the state is going to be stable, if the Rayleigh number is large then the system is going to be unstable, okay, that's kind of the hope all right, so this is actually something that we can prove well not with the instability but so let me tell you how to how to do that one, so I'll take just the minimizer so I'll take lambda with the fixed tau is actually equal to gradient of u square plus the other terms divided by theta bar square plus the other term so I'm just going to plug in the minimizer and since this is in p-moom with respect to all functions what I can plug in, oh let me just write down this this term the important term that I have here is integral of ud bar theta and some term here and this is the term that then I'm interested in so I choose a competitor and as a competitor I'm going to choose u bar minus theta bar so you plug this one in the value of the function must be bigger okay u bar theta bar is in p-moom so this must be bigger, all right, I am sorry I messed up a little bit just just give me bear with me for a for a second if you have this I need to do very very quick substitution what I do I rescale u by square root of Rayleigh so that really means that this Rayleigh from here will disappear Rayleigh from here will disappear and it will show up here as a square root of Rayleigh, so square root of Rayleigh there's no other no other Rayleigh there okay so you plug in the competitor then you see everything is in the norm so these two terms are the same denominator is exactly the same the only thing that will change is I'll get here minus Rayleigh integral u bar theta bar and so on so from here we immediately have that the integral of u d bar theta bar and then here is the derivative with respect to tau minus one must be non-positive and actually you can prove that it must be negative on some set when tau is large okay when tau is large then we can prove that this inequality must be actually strict okay and now we see that if I'm increasing Rayleigh the only thing that's changing here is this term well I see that this term is negative if I'm increasing Rayleigh then this minimum must be decreasing okay there's a little calculation like two lines long that that's going to give you that so it means that this is point-wise decreasing and since this inequality is strict with positive probability we know that this expectation must be strictly decreasing with the Rayleigh with the Rayleigh number it's not increasing it must be strictly decreasing all right so from that we know that of course there's only one zero of the strictly decreasing function so it might be finite or it might be infinite but we know that it cannot happen that we have stability instability stability we cannot have any switchings once we go from from positive to negative expectation then then we're done we cannot come back all right so so that was easy so so now the question is so so this is a very curious and somehow mind-bugging simulation so we did numerical simulation on this problem and we were comparing the stochastic forcing to the deterministic forcing okay so we took the stochastic forcing and then the deterministic one was when the when the forcing was assumed to be just one and this is what we found okay so deterministic forcing we have the the strength of the forcing there's the parameter in in front of the force and then we were looking at the critical Rayleigh number and not surprisingly maybe if the forcing was zero then it was exactly the same as it was calculated for the for the deterministic case and then as the forcing was increasing then our critical Rayleigh number was decreasing all the way to zero okay so if we are pumping more and more deterministic forcing this the system is somehow becoming more and more unstable fine probably believable all right so what happened with the stochastic forcing is it's kind of kind of strange because we start putting the stochastic forcing then it was kind of equally stable as if there was no forcing at all okay so in particular the stochastic forcing was somehow more stable than the deterministic one it seemed like the stochastic forcing was stabilizing the whole situation a little bit then on the on the when the forces was of order one there was very sharp transition and suddenly the stochastic forcing was destabilizing the whole situation so the stochastic forcing was requiring smaller and smaller Rayleigh number compared to them compared to the deterministic forcing so it seems that for small forces stochastic forcing is stabilizing for the large one is destabilizing compared to deterministic ones the question is do we understand can we say something about this one can we can we somehow fill this numerical picture all right so i go back to my functional and in order to get some something about uh so no so recall i'd like to say something about the expectation of lambda as a function of tau and tau is something like a it's something like a normal distribution is a stationary stationary measure for the OU process okay so i'd like to say something about this one and we all know this this is something like a lambda x times e to minus x square okay so i need to integrate the function against the Gaussians so in order to do some estimates on this one i would like to know something about tau sorry something about lambda that's that's that's that's kind of my goal so the problem is lambda is the minimum like a minimum of some variational problem it doesn't look that easy all right so but surprisingly there there are some things that we can say and actually conclude some some some some results so we can prove that the function is continuous hopefully believable and it has one-sided derivatives and it's not only that i'll show you it's actually the crucial one of the crucial calculations which is rather simple and but still reasonably powerful so what i'm going to do is i'm going to estimate what's the lambda tau plus delta times v i'm going to choose my direction and i'm trying to differentiate in the direction of v my lambda all right so what i can do i know that this is actually less or equal then if i take my functional let's call q this is the functional q at with the parameters tau plus delta v but i'm going to plug in the wrong function there i'm not going to plug in there the minimum for tau plus delta v but i'm going to plug it in u and theta for tau i'm going to plug in the minimum for the functional tau okay and of course there's a test function admissible so this is going to be just bigger hopefully not a big deal but now if i plug it in so you see there's a quadratic uh there's a quadratic everything is quadratic there so what i'm going to get is q at tau u tau and the only thing that's that's sticking out is is this term when when i here i don't have tau but i have tau plus delta v so the additional term that i get there is delta integral u tau theta tau partial derivative of x d of v dx but this term is immediately lambda tau so it's just a simple calculation just plug in the the other function multiply everything out and that's it but now this is good because you see this is almost if i move this one to the other side and divide by delta then i get something like lambda tau plus delta v minus lambda tau divided by delta is lesser equal than this integral that's there all right so i can pass to the limit and i get something about the limb soup so actually i can i can i can calculate that uh that the derivative from the right of lambda in the direction of v is actually equal to infimum yes infimum with respect to all minimizers of this integral okay so i can calculate that all right so i have one-sided derivatives i can also calculate that the derivative from the from the left is the same quantity only with the supremum same argument not not not not anything is different okay so i know the one-sided derivatives and on the top of that i prepared the calculation but i don't have time for that i can prove that the function is concave the lambda function is concave all right so notice again i need some estimates on the function lambda so i know that the function is concave and i have one-sided derivatives well what else do i need i also need to fix the value of lambda at least at one point well that's good because when i plug in tau is equal to zero then this trouble will disappear and the functional here i have the lambda is also concave now i can plug in tau is equal to zero and get lambda at zero is something like infimum of gradient q square plus gradient of theta square minus integral of ud theta divided by q square there is a square root of Rayleigh just to keep it here theta square all right there is nothing stochastic about this one anymore there's nothing there's a quadratic functional this this is it so you can do analysis and this time you can write Euler-Lagrange equation so if you write Euler-Lagrange equation let me just go back to to the Euler-Lagrange equations that were here so now this term is gone this tau is gone and you can write everything in the Fourier space and if you write down everything in the Fourier space that's actually very similar to to what Rayleigh did then write everything down in the Fourier space then you can i probably don't want to bother you with that so just believe me in Fourier space so imagine so these two times curl helped me in many ways it killed the divergence but it also reduced the problem from four equations to only two equations this equation for ud and theta only there's nothing else sticking out and tau is gone so in Fourier space this is going to give us a system of two algebraic equations all right now we are now it's getting ridiculous okay the analysis is not that easy still because it depends on the Fourier coefficients and we need to somehow minimize lambda with all the Fourier coefficients it's getting a little bit hairy nevertheless it can be done so what we can what we can do we can estimate the values at zero so this this is deterministic calculation done 100 years ago and we also have a global bound on the derivative look we also know that the derivative plus or minus derivative of lambda this is this calculation this is why i wanted to have a one-sided derivative is actually lesser equal than something like supremum of tau and i'll put l2 norm on theta and l2 norm of u also there's a derivative of tau not that important this is super smooth function and now i just estimate this one by the derivative of tau and then here i put theta square plus one over prandtl really u square square root of prandtl really okay and this one i can assume it's equal to one because it's in the denominator on my on my um on my variational problem and i have a global bound on the derivatives okay if you do the calculations carefully then this bound this constant c is comparable to the derivative at zero it's not much bigger it's not of different order of really or prandtl numbers all right so now we're in good shape okay so now you see this is kind of easy calculations more or less maybe and now what do we know here is my forcing h and here is going to be my lambda how lambda behaves so now i know the value at zero that's good i know the derivative at zero actually we can calculate the derivative at zero and i know that the function is concave so the derivative at zero is going to give me a line so this is the line with the derivative at zero this is another line this is when the this is the line with the derivative of lambda at zero this is a maximal derivative of lambda and i know the function is concave so it must be somewhere in between so i know that lambda is concave and it must lie somewhere in between now i'd like to have an estimate so i can estimate this point let's call this point h star i have an estimate how to where this h star is placed i have maximal derivative minimal derivative i'm sorry i know where it intersects zero roughly and now how do i get the estimate for lambda well lambda is estimated from below by the line connecting zero and h star the functions concave and on this line i take the maximal derivative i know that the functions cannot be increasing more so that's the lower bound and the upper bound do i have another color chalk i do the upper bound i'll take the tangent at the h star function is concave so functions from below and i know that the function is also decreasing it's not difficult to prove so i can just cut it off here so it's going to be my upper bound so that's and that's it so with some suffering and calculations again gaussians so i need to integrate it against gaussians we can actually prove that when Rayleigh number is very very small then is more stable than the deterministic one we can compare it to that one and by by by the estimating lambda actually from below estimating lambda from above we can prove that when the Rayleigh is large then the whole thing is more unstable than than the deterministic force so we can get all these upper bounds and get stability and instability the the result is of course suboptimal we need to do some estimates we need we have this only for like close to zero or close to infinity but nevertheless we can somehow justify the picture all right so there's the last slide like literally one minute the point is that this is a very flexible framework you see i want to point out this is an example but possibly there is a there must be many applications to other variational problems like stochastic variational problems they had to google it they didn't find almost anything okay so we were able to to use this to use this framework to other situations for example when when we are forcing on the boundary stability of quiet flows and so on and so forth okay so there are other things where you can you can you can you can proceed in this way and get some get some quantitative bounds on the on the stability of invariant measures all right so this is actually giving us that not only that we have the uniqueness of invariant measure we actually know what it is it's the OU process the question is what's happening after the bifurcation after you lose stability that's another maybe interesting maybe very hard direction it's like a dynamical system on invariant measures maybe i don't know how to name it hard to say instability is open we don't know how to do it it means that you are not working with a variational problem or somehow relate the variational problem to the full problem that's not clear how to do and also we'd like to prove continuity on invariant measures meaning that saying okay maybe we don't have the uniqueness of invariant measure anymore we don't know any example of that but maybe still when we are close to the line our our invariant measure should be close to the to the OU process that we constructed the one that that's that's another so meaning if i if i start adding forcing into into other not only vertical modes we know that the the measure is unique if the forcing is small if it's still close to the invariant measure that we constructed that's not clear all right so there are a couple papers and i thank you for