 Now I want to concentrate on the fair competition case. Yes, one comment about this computation with the second moment. Remember, this computation is only in the fair competition case, and by the way, there was some mistake in the constant. One thing was the numerator should be on the denominator. It doesn't matter. The final answer was correct. Good. So we were over here. I told you what is known for the classical color seal. Now let me tell you how to deal with this more general case. Let's start with k between minus dimension and zero. In this range, obviously, a good candidate to use here in order to control these terms is the famous hardy-leadable Sobolev inequality. Because for any potential modulus of x to the k, when k is between minus dimension and zero, these famous inequality for which we know even the optimizers. So in fact, this is precisely what tells you what is the critical parameter in this case. So let me first tell you what is the hardy-leadable Sobolev inequality. So if k is between minus dimension and zero, the m related to k by the relation of fair competition, m is between one and two. And then what you can write is that the double convolution of f x, f y with modulus of x to the k is bounded by the typical HLS. What it tells you is that what is there, that will not write here on the board, is bounded by a constant times the norm of f, the norm p of f squared, where p is what is 2d over 2d plus k. This is what you used, hardy-leadable Sobolev directly. The question is that here in the functional, what is involved is the lm norm of rho and not the lp norm. And then you can ask yourself what is the relation between p and m? So in fact, remember that we are in the fair competition case. So in the fair competition case, you can check that in fact this is less than 1 minus d over k. This is a direct inspection, just a check that this is true. And this is m. So in fact the p is between 1 and m and this happens again because we are in that range. In a sense. So the common is that this exponent is less than m, allows me to interpolate the lp norm between 1 and m. This is what I do. So I just interpolate the lp norm here between 1 and m, I compute the exponents and I write the exponents in that way. Don't forget that k and m are related always. Good. And now the fact is that for our densities, remember that for me rho is always of l1 norm equals 1. So this goes away. And then I can relate these to the lm norm of rho. So in fact I have a relation between the two terms exactly in the free energy. Okay? And then this determines the exact constant kai for which I can relate these two. Because also for that, let me make the comment for the HLS inequality, I know the optimizers. But this is not exactly the HLS inequality when I do the interpolation. Nevertheless, I can find the optimizers of that inequality. Okay? It's a different inequality and it makes a big difference because if you remember for the HLS inequality, the optimizers are always of the Talenti form, r1 over 1 plus x squared to a power. The optimizers of this one, which looks a naive generalization of the other one, they are comparably supported functions. So it's a big difference. Okay? And that's the interesting thing about this variation of the HLS inequality by interpolation. If you look at the optimal constant here, let's call C star, the optimal constant of that inequality. Then what you can write directly is that you identify somehow the critical parameter because the critical parameter will be related to the optimal constant here and then somehow the HLS inequality is true if and only if the corresponding free energy with that value of the critical parameter related to C star is larger or equal than zero. So this is the total, if you want, it's a total parallel with the Keller Siegel. For the Keller Siegel, the inequality involved was the logarithmic hard-delivals of 11 equality. And this was determining the constant 8 pi as the critical. Here, the function inequality that you need is this variation of HLS. Once you realize about that, then it's not too difficult to prove the assistance of the optimizers of inequality or if you want equivalently the system of global minimizers for that particular value of the parameter chi, for the critical chi. And yeah, okay. Let me then... So essentially what you can prove, let's make a summary here, somewhere. I'm gonna erase where probably here. Computation is not needed. So you have this variant of the HLS, gives you the short constant of that variant that reminds exactly the critical chi for which you have f of rho larger or equal than zero and the equality cases here are equivalent to the optimizers of the inequality or cases of equality of the variant HLS. Okay, once you realize about that, then you have the termine, the critical value. Here you have this infinitely many stationary states and I told you that they are comparably supported. I will give you an idea why is this the case, but this is the main difference with respect to the Keller-Segel case. They are comparably supported and they are like a helter function. So they are kind of C0-alpha at the boundary. So they look very much like bottom blood-like solutions of the porous median equation. Yes, yes, it's the same business. The same reason as for the HLS inequality, you can add all the dilations. Remember that the f of rho is homogeneous. So once you have f of rho equal zero, Rolanda will have also zero energy, they will be all of them minimizers. So in fact it's a one-parameter family given by the dilations of that. The same business as the Keller-Segel, sadly. Okay, and then what you can show also is that if chi is less than chi-c, the same control here, now just because chi is less than chi-c, you can see that you can control separately the two terms in the free energy and in fact you have then LM norms bounded for the minimizing sequences and you can check in fact that the effect of that is that you don't have stationary states but you have them in scale variables. So let me just explain you that for chi less than chi-c. For chi less than chi-c, what you can do is something very similar to what you do also for classical Keller-Segel. You rescale variables in order to make appear a confinement potential and then for this one you prove that there are minimizers. These in original variables, what it gives you are self-similar solutions. So the minimizer of these gives you the profile of the self-similar solution in original variables. So here what you can prove is in fact global assistance of solutions. You suspect, I mean you can prove. I said you can prove but we prove it only in a particular case. So in the whole generality I'm telling you here this hasn't been done. I hope that at some point somebody will do because all the ingredients are there. So global assistance of solutions and we have this for sure what we have is these self-similar solutions. Of course you suspect these self-similar solutions to be the global tinnocentotics of the solutions that happens also in the classical Keller-Segel. This is in particular one result that Mischler and Egania did recently. I mean in the sense that they did the best result known in that direction. So you suspect such a result also in that range here and what else I wanted to mention here? Yes. So for Kyla, I don't have anything there on the slides but for that I did this computation. So for Ky larger than Ky C, what happens? In fact, this tells you already what is happening. You see the critical case is where f of rho is larger than 0 and the minimizers are for f of rho is at least 0. For Ky less than Ky C, the same inequality proves you that in fact f of rho is extremely positive for every rho. What happens when you cross Ky equals Ky C and Ky is larger than Ky C is that you can prove that there are densities rho such that f of rho is initially negative. Once f of rho is initially negative, look at this computation. It's telling you the derivative of the second, so if you have initially something, this is rho at time t, okay? But the free energy is a Lyapunov functional. This is less or equal to d times m minus 1, the free energy at time 0. So if this is negative, all of this is negative and it's bounded by a negative constant, this tells you that the second moment should decay linearly. So it touches 0 in finite time and there is a contradiction in the case of the classical Kyler single. The second moment cannot be 0 unless you have a concentration before. So the whole HLS inequality gives you the full dichotomy again. It tells you that there will be the resist blow up in finite time and here, of course, what I'm telling you by this, I'm not telling you really that there is blow up. I'm just telling you that there are no global solutions with the right regularity and compute the second moment. Proving that there is blow up in finite time will be another business, okay? And we don't have a proof yet of this similar to the one of the Kyler single. But you expect, again, the same dichotomy. Okay. So at least with these computations I showed you what you have is this dichotomy at the level of stationary solutions and what you expect of the free energy to happen in each of the cases. So now, this I told you what happens when K is negative. Let me tell you a little bit what we know when K is positive. So what happens when K is positive? Okay, let's first look at the sign here in this functional. In the case K negative, this is negative term. M is between 1 and 2 in that case. This is positive term. When K is positive, we have the things change. This becomes positive. This becomes negative. Okay? And if you want to find a functional, I mean an inequality, again similar to HLS, it should tell you that you can control now this term by that term. It's a kind of reversed inequality of a hard-to-leave or sub-left time. So in fact, that range is much less understood. And what I can tell you is one thing that is interesting is that there is no for sure what we can prove is that there is no critical parameter chi. So why this? Because what we can prove is that just looking at the oil and orange conditions that I will do next on the board, you can show by fixed-point arguments that you have always a stationary state for every value of the parameter chi. So at least there is no this critical value chi at the level of assistance of a stationary state. Okay? In this particular for K positive, let me remind you that M is between 0 and 1. So since there are some experts in fast diffusion equation here, in that range, you spec the solutions to be supported on the whole line with some decay at infinity. In that case, you had to look for the solutions in the right space with the right decay at infinity. This is given by the K, by the K moment in fact. But you can prove by direct fixed-point arguments that you have a solution of the oil and orange equations. So at least you have a stationary state for every value of the chi. So let me just mention you in short, at least in 1D. This is how it looks like in the sense that K is minus 1 and 1. K equals 0 is the case of the log, so like the Keller single in 1D say, sort of. And then you have what is in red are values for which you may have finite time blow-up, red. The ones in U will have some similar solutions here. Here what you have are stationary states. I forgot to say you have stationary states and probably said they're wrong in scale variables. So these are similar solutions in the original variables. So here you have some similar solutions stationary states in scale variables. Sorry about that. So nevertheless we are still trying to understand the K positive case in other ranges which might be related to some HLS inequalities that are in the market. But this is something that we are just working at the moment. Okay, so this is everything I can say right now about the fair competition case. Let me repeat that for the evolution problem we have very little results. It's mainly about stationary states. But let me do a computation for you on the board now, which is about the oil and orange conditions. Well, I don't know if, yes. Okay, probably I should. Yeah. This is important for the next one. Are we here? No. So what are the oil and orange conditions of this function? Think about it. We have to do variations of the functional with densities in L1LM. So when the variations doesn't look that trivial here. So but let's do it. Okay. So what I want to do is take that functional and compute f of rho plus epsilon phi minus f of rho divided by epsilon and take the limit and as epsilon goes to 0 of this for sure if I want to find oil and orange conditions I will be assuming that rho is a local mean at least then this should be at least larger equivalence here. Okay. And I'm going to tell you which variations I want to use. So let's take phi related to rho. Let me take first phi of this form, rho psi minus the integral of rho psi where it's a given c infinity company supported function so as good as you want I'm going to assume that I don't need anything more. So let's check one thing here I need the integral of phi to be 0 because I want to preserve the mass the integral of rho plus epsilon phi should be 1. So it's clear from the form I chose it because the integral of rho is 1 so the integral of phi is 0 and also if you choose epsilon small enough then you can check that rho plus epsilon phi is larger than 0. Okay. So in fact you just need epsilon for instance in a brutal way epsilon smaller than twice the norm of psi in L infinity. So then you will get this if you do that and in fact if you want you can start by doing it with psi larger than 0 it doesn't really matter what you will get if you do this computation it's easy to see that what you get you do the formal computation of at least the the Taylor expansion of the power rho to the m at 0 and you do for the other term it's very easy because it's quadratic what you get is exactly the variation of f with respect to rho that I wrote before it's not here variation of f with respect to rho is somewhere here but it's m over m minus 1 rho to the m minus 1 plus modulus of x to the k divided by k with both rho ok so what you get is this multiplied by phi and this should be larger equal than 0 ok in principle good now what I'm going to use is this particular form I chose of phi and then you see that from here I can write this in the following way so I get variation of f with respect to rho then I get rho psi and then I will have minus the integral of rho psi times rho variation of f with respect to rho I can write this way if you want like this and this should be larger equal than 0 is it correct what I'm saying wait maybe I may yes correct what I'm saying what you get from here is in fact you can now compute if you want this is a constant in fact because this is something that depends on rho let me call it d of rho 1 ok whatever it is it can be written in terms of the free energy but I don't really need it now and then from here what you get is that the variation of rho minus d of rho times rho psi is larger equal than 0 you put it all together ok and in fact the sign of psi doesn't play any role in this computation so you can choose either psi positive or negative so at the end of the day you can put that you have also the equality ok the sign of psi doesn't enter at all in the computation so you can choose psi minus psi and then you will have the equality here so you see from this that on the support of rho this should be equals to a constant ok so from this computation what you get is that the variation of f with respect to rho should be equal to a constant which is determined by rho itself but a constant on the support of rho in fact you could have guessed this the question is that you determine the constant why you could have guessed this because the PD in fact is here divergence of rho gradient of the variation of f with respect to rho so you see that the steady state for sure is when this is constant on the support of rho either rho is zero or the gradient of this is zero ok but you get it from just this simple computation let me not do another variation you can do another variation but I'm going to just write the test function that I use and not do the computations you can do another phi now you play differently with the test function so you take a psi but instead of doing that you take psi minus rho integral of psi the difference being that now you need before since rho was multiplying so this only appears when you are inside the support of rho outside you don't have anything this is zero but here now outside the support of rho the only thing remaining is psi so psi has to have a sign and here is important so here psi has to be larger than zero in order to impose that rho plus epsilon phi is larger than zero so here it wasn't playing a role but there it plays a role so if you do the same kind of computation what you can see from here is that you get the following relation so you get the intel of psi variation of f with respect to rho minus this constant the same constant d of rho this has to be larger or equal than zero for every psi larger or equal than zero so what is the information that you get out of these variations is that this has to be true on the support of rho and what it has to be true is larger or equal all over say on rd okay so this reminds a lot what happens with the bottom blood profile and in fact from here you can write somehow what the all the critical points should look like from this computation here I can write now let me substitute what is the variation so I have m minus one to the m minus one plus modulus of x to the k divided by k combo of rho it has to be equal to a constant inside the support of rho and larger or equal than that constant outside so for minimal for critical points of that energy what you can write is that I can take rho outside from here so you can write rho so I put this on the other side d of rho minus modulus of x to the k divided by k combo of rho I multiply by the constant so let me put in m minus one over m here m minus one over m and then rho has to be what well it has to be in fact positive part of that to the one over m minus one so you can convince yourself that these two things tell you at the same time these two things are important to tell you at the same time that what you have is exactly that rho satisfies this so of course it's an equation an internal equation for rho but what do you learn from here now you learn how it can happen that this company support it because what you need to prove somehow that the growth of this well that this grows at infinity or I mean it has to be the right let me see case negative m here minus minus this is positive so this has to go to minus infinity the whole thing in such a way that this is company supported and in fact this is how you prove that this company supported okay in that range and also it tells you that in that case another way of proving stationary states would be to find solutions of this internal equation this is how we attack with in the scale variables the system of solutions in the k positive case so just to mention that these are the Euler-Larance equations now they look also quite a lot like the obstacle problems if somebody talk about the obstacle problems last week but I don't want to enter into that good so now that we learned a bit about why it could be that the solutions are company supported let me go now and discuss the other range the diffusion dominated case so in the rest of the talk now I want to concentrate in what I can say for this range and a first part will be more general than that range it's about stationary states for any m larger than 1 in fact so let's go now to the diffusion dominated case so we are in the case in which by dilations I have a minimizer in here and I expect the diffusion to dominate in the system but in which sense that's the diffusion win here so as I told you with VanSanCalves and years ago we subtly studied this particular case so in two dimensions you take k equals 0 m larger than 1 and we prove in that paper all solutions are globally and they have uniform bounds in time uniform in time but what happens with the long-time asymptotics that was the open problem there and of course we knew for a long time what was the conjecture but we were not able to prove it already by simple numerics at that time we realized that the expected asymptotics was somehow only given by a stationary state so what we were expecting is as t goes to infinity you converge towards a stationary state with the right center of mass because this is a translational invariant so let's assume the center of mass is 0 initially and with the right I mean if we are normalizing the mass with the right mass so this is somehow what it came out of the of the simulations always you converge towards something which is in fact comparably supported and here what I'm plotting is precisely this function I'm plotting the variation of f with respect to rho so you see that in the support is constant outside the support is larger equal than that constant okay so this is what we wanted to prove for our own time okay so let me tell you how we got into proving this after several years the first thing is what happens with global minimizers so again I'm going to do it in this particular case of the log today on the slides the log in two dimensions but very recently with my PhD student Franca Hoffman we just finished a paper that is on archive for a month right now which we do the same stuff for the whole range diffusion dominated for every k negative okay so let me explain to you how to get the global minimizers for this so what you want to do is to find the minimizers in the set of densities here in this part of the talk I'm not going to normalize the mass instead of having the chi since the chi doesn't play any role in the diffusion dominated case there is no chi in fact so I'm not going to normalize the mass I'm going to leave the mass equals m so I look for the minimizers for every m with the zero center of mass some of the results I introduced this particular subset which is the subset of the L1LM densities which are radially decreasing okay I used the notation of decreasing rearrangement with this sharp I will discuss something about the decreasing rearrangement later so I can explain to you a bit what it is it's a way of constructing a radially decreasing function out of a function rho and you have rho equals rho sharp radially decreasing and the thing well the statement that we can prove is that we have a unique global radial minimizer of the free energy for every mass m okay in this case so let me explain to you how this goes in fact in this case since I have the log the role of the HLS before it will be played by this variation of the HLS that was introduced by Carlin and Loss the log HLS so the first thing is that you can check that minimizers should be radial just because of the job that they did because they prove that the interaction energy with the log decreases with rearrangement and also by using the inequality of the log HLS that they prove in that competing symmetries paper which is exactly that that's the log HLS you control the interaction energy of the log with the integral of rho instead of the power with LP norm that's the kind of inequality you have and the mass enters into the constant here this constant is explicitly known depending on the mass and you know also the optimal functions for that inequality even if now this doesn't play a big role the interesting thing is that this inequality implies in this case that you can prove that this since M is larger than 1 you can prove that this free energy is bounded below directly essentially because since you know how to control this in terms of rho log rho what you add and subtract here rho log rho with the log HLS you kill the log interaction and then you just need to play with the behavior of rho to the M and rho log rho at infinity since M is larger than 1 then that is bounded below essentially that's the game so it's very easy to check with the log HLS that you have a bound from below the energy so you can talk about the infimum of the energy now if you want to pass to the limit in minimizing sequences we know our minimizing sequences are radial functions we know that this function is bounded from below so the next step that you can do can display some things here on the board about this remember that I'm discussing about the case with rho to the M 1 over M minus 1 double intra with the log probably there was a constant here but it doesn't really matter ok and what you know now for a minimizing sequence is that this is bounded from below and the thing is that from there you can convince yourself for the minimizing sequences that both terms because you have the bound from below both terms are separately bounded that's the information that you get because it's bounded from below and from there you have a bound on the LM which is good if you want to apply any compactness but also the problem that you want to that you have is that you want to keep the mass of the possible limit of the minimizing sequence so you need to avoid mass at infinity on the minimizing sequence and this is a kind of a neat here because it's not clear which of the two gives you that in fact it should be this but it's not so clear from here why there should be a confinement of the mass but the control of this gives you such a control on the mass outside a ball of radius R of the minimizing sequences you can show that this decays like 1 over the log and here is essential that the functions are radial here is essential that the functions are radial let me give you an idea about this because I think it's neat the R one first here you can show that for any R you have a control of these quantities you control also that why because for modulus of X minus Y less than R that part because you do a held there with the log and you have the LM norm you can bound it with the LM norm essentially the local part with the LM norm of rho you can control it so then you have a bound let me just assume if you want take this as an assumption that there is a constant that will be bounded on R if R is larger than say than 1 ok now the thing is that let me take now I fix an X with modulus of X larger than R ok and then I take any Y such that X dot Y is less or equal than 0 following is the following I take the ball of R I take an X here which is modulus of X less than R I'm taking this set of the Y's ok then what is easy to check is that X minus Y is always larger equal than modulus of X and then sorry and then larger equal than R and just to check that put the square then the scalar product has the right direction and then that's easy so then for any of this is included in that set so I also have a bound on this quantity I have a bound on the integral in R2 integral on let me see sorry, modulus of X larger than R integral X dot Y less or equal than 0 of that quantity so this is bounded now ok for any X well I mean this is bounded for any X modulus of X less than R I did that and then this integral is bounded and now here where I'm going to use that row is radial so modulus of X minus Y is larger than R R I'm taking it larger than 1 so then I can say that this is larger equal than the integral of modulus of X larger than R log of modulus of X integral row of X and here I can put the row Y well I mean put it there here the integral of X is 0 of row Y dy dx ok and then here what do I get? it's just the integral of row in a half space since row is radial the integral of row in the half space for any X I choose is 1 half is M over 2 is half of it so then this is M over 2 so then you control the log moment ok so this is a neat way of controlling the mass outside of all and then these together with the control of the LM norm as I said modulo very easy computations gives you the right compactness to achieve the minimum ok good so you have a minimizer now I'm sorry yes you have a minimizer minimizer sequence has a limit you can talk about the minimizer I mean a minimizer in principle and we'll talk about why it's unique later so a global minimizer will be radial and will be a solution of this Euler-Larance equation that I mentioned before here I'm writing for you what it is exactly the constant in terms of the free energy remember the free energy I call it G to make it different from the previous one because it's a particular case ok and as a consequence as I told you before your minimizer will satisfy this equation ok from there working on that equation that I will not do the details you can obtain that they are company supported continuous function in fact held there up to the boundary and smooth inside the support you have to work a bit on that it's not so easy but that's somehow with the main ingredient start from this relation and then with respect to uniqueness well uniqueness we didn't prove anything because it was already proven that there was uniqueness of radial solutions in fact this was already known for quite a long time so in short what this was this was done by Yao Yao and Inguang Kim in the thesis of Yao a variation of results of Li Ban Yao for some related functions and the idea is to use mass comparison in radial coordinates so I'm not going to enter into that so this gives you the uniqueness part of the theorem so what we got is the uniqueness of the radial global minimizer radially decreasing and nice and smooth so we have the good candidates for the long time asymptotics ok but what's the problem to really get the result of the long time asymptotics the problem is about the stationary states why? since I don't have that much time I don't think I will be able to get into the most recent results but let me mention to you what's the difference between the minimizers and the stationary states so now let's go back to the good candidate to be the long time asymptotics let's go back to the time dependent equation I write it in the short way so let's assume that we have all the a priori estimates that tells you that time divergent sequences of this have compactness let's assume that we have that then we can get any for any time divergent sequence we could get subsequences that converts something and that we can identify that as a stationary state of this equation ok so that first part in fact all the ingredients were in the literature and we just put them together so one can check that all possible candidates for long time asymptotics they should be stationary states and what do I mean by stationary states I mean that in fact that I have yes probably let me go here so that's what I mean by stationary states ok so I'm going to assume a bit more on the row just to simplify my life a bit so I'm going to assume L1 and infinity I'm going to assume that power m is in h1 log I'm going to assume that well in fact you can check that in that case gradient u combo with the row is in L1 log and then it satisfies this relation in the sense of distribution this relation in the sense of distribution is somehow rewrite this equals to 0 in a different way well the interesting thing is that in that case what's the difference the difference is that for a stationary state this guy the variation of f with respect to row the only information that you get is that it should be constant but the constant might be different in different connected components of row so nobody is telling you first that the stationary state is radial nobody is telling you that there is only one component in the in the support the information that you get from the stationary state is that you have the variation of the free energy which is written there is a constant with I being in some index set could be countable but it could be an infinite set of connected components and the constants may be different in the different connected components so a priori I don't know if it's a unique stationary state in the sense that the constant is the same everywhere like it happens for the global minimizer given by this formula or this formula here so the question is can there be other stationary states rather than the one that I find by global minimization there's a question about uniqueness of the stationary states if we want to get the results finally and my time is almost over in a few minutes like 5 10 so then let me tell you the main the main difficulty that we found the main difficulty is precisely to avoid this it's a question about when you can prove that stationary states are radial or not in a sense that's the main cornerstone of the proof how can we go around symmetry there so in my previous work that I was discussing before with Bruno Bolton and Daniela Castorina we were able to prove the radially symmetric the radial symmetry for the stationary states assuming that the stationary state is comparably supported if it's comparably supported we were able to do it and this was by a kind of non-standard moving plane technique that Daniela came with a very nice idea to overcome the standard difficulty for doing it with a non-local term there but the comparably supported was helping for doing that nevertheless, even if we know for comparably supported that the only one is radial if we want to do the long-term asymptotics because we don't know a priori we don't know how to prove that solutions even for comparably supported initial data they are comparably supported for all times for this PDE very good open problem if we knew that, that would be done at least for comparably supported initial data so we needed to let's say improve on the assumptions that we do to prove that it's radial so that's the main result in fact of the most recent work in this direction so let me at least state the theorem so in fact it's for more general than the one that I was discussing with the log and m larger than 1 so in fact this shouldn't be there for m larger than 1 for any m larger than 1 if I have a general aggregation diffusion equation of that form let's try not to read all the assumptions I'm going to just tell you in words what I assume m larger than 1 and u being less singular than Newtonian at 0 behaving at infinity like Newtonian in any of the corresponding dimensions so essentially these assumptions include the Newtonian in any dimension okay but the singularity can be less singular than Newtonian at 0 so for the whole family what we have proven with yaw, yaw, sorry probably in order, Savina Hitmayer Bruno Balsone and yaw, yaw what we have proven is that all the stationary states in this definition that I show you have to be symmetric so that's the main result that we have done and let me just mention in few minutes some of the ingredients so well this I told you already what the difficulty was the difficulty about the stationary states so let me tell you what are the main ingredients the main ingredient is in fact again the variational structure so for me it's the first time also that the variational structure enters in these kind of equations very crucially at the level of properties of symmetries so we use the liabono function that is behind the structure of the equation and it's an argument by contradiction so I'm sorry it's going to be a bit technical the next slide but I think it's worthy to mention how this is done the argument by contradiction is the following you assume that the steady state is not radially symmetric after any translation so let's fix that it's not radially symmetric about some hyperplane and let's put the hyperplane to be x1 equals 0 to simplify of course you can always do this but let's assume that it's not radially symmetric now I'm going to explain you how we did the contradiction I'm going to explain you how we construct a family of functions depending on one parameter so they are going to be perturbations of the steady state which I assume is not really symmetric I call rho epsilon and I'm going to prove well I'm going to tell you more or less that what we were able to prove is that if I compute the energy of this family of functions the difference of the energy with respect to the energy of the stationary state is less than a constant minus a constant so constant is positive times epsilon so it decreases linearly with respect to the parameter epsilon and the constant is quantitative we can compute a bound such a way that this is atletes ok so I'm going to tell you two things first I'm going to tell you how we did the construction of the rho epsilon and second why this gives you a contradiction so let me start by telling you how we constructed so we constructed using a different rearrangement technique that was already available in the literature this was used by people in minimization precisely of the in particular of the Newtonian potential and also was used by people like Kavol and Almut Burtcher in different province in elliptic equations and they were using this kind of continuous Steiner symmetrization let me explain you a few words what it is in one dimension is nice and easy and it's essentially in this picture so it's a way of constructing out of a non-radially symmetric function let's say in one dimension I start with this mu naught my non-radially symmetric function to construct something depending on a parameter sorry about the t should be epsilon something that depends on a parameter which is more radially symmetric than what you started so how do you do this so in one dimension it's easy to do it's not a business but thinking in terms of needles I will explain you why I call about needles so you do a kind of layer cake representation of the function and in each of the of the of the horizontal sorry of the vertical cuts here you compute for every of them its center of mass so the center of mass originally is the red ballad here each of these needles you're gonna move it towards the center of the mass a little bit so this needle here is a bit to the right so I'm gonna move it to the left so I put it on the blue square the same for this one here this one is center so I leave it this one is to the left so I move it a bit to the right and so on so it's like cut it into needles and make it a bit more radially symmetric putting the center of masses towards the towards the center of mass of course if I move all of them to this I get the standard kind of rearrangement and this gives you a way of doing small perturbation in fact making it a bit more radio symmetric so you can generalize this kind of idea for more dimensions and even it's not too difficult again always thinking of needles and playing with all the just in one direction moving the corresponding level sets and if you do this kind of symmetrization you keep the LM norm like in the usual decreasing rearrangement and it was known this was proven already by Brooke in the end of the 80s he proved that in fact the Newtonian potential he did it only for the Newtonian potential that then it decreases the interaction potential with the Newtonian decreases by doing this but he just proved it decreases what we had done is to check exactly how much it decreases with epsilon because for us it was important to know subtly quantitatively how it decreases and we proved that in fact it decreases linearly with epsilon so there is where we had to work quite a lot to prove that in fact this decreases linear ok so we found this direction in which we decreased the energy that goes linear now why this is a contradiction well the thing is that we can massage a little bit more this decreasing rearrangement in such a way that also we have this property for the curve that we construct so in order to have this property what I want is that the support of the mu epsilon doesn't change if I do exactly what I said in the previous slide this will change but what I'm going to do is near the support I don't touch and I increasingly touch for a very small height and then I do this continuous dynamism and their error that I make is not that small and I can still control that it goes like minus c epsilon and then on top I can tell you that the mu epsilon satisfies that property so it's not too far from the steady state and then we have this control exactly ok so this is an additional work that we need to do to massage a bit this mu epsilon so believe me that this we could do and now what's the contradiction the contradiction comes from the fact that being a steady state you somehow are a critical point of the functional so then the error in epsilon should be quadratic cannot appear at the first order so just because it's a stationary state you can check that the energy of mu epsilon minus the energy of rho s should be like of order epsilon squared and that's the contradiction now on one hand has to be epsilon squared for epsilon small but the difference of the energy in fact is linear and that's where we get the contradiction so it's the first time that I see the use of this continuous astyler symmetrization and use in variational arguments related to this gradient flows in order to get symmetries so with this we were able to fix the problem of the radial symmetry of the minimizer I mean the radial symmetry of the stationary states and then if you put together this with the previous result of minimization what in the particular case this is probably not what I wanted to say as yes for Newtonian potentials we know that for radial minimizers we knew that they were unique but let's not continue just to say that if you put together the two results in the particular case of this energy we know first because of the radiality of the stationary states we know all the stationary states are radio second we know the global minimizer of this free energy is radio and third we knew that it was unique and the uniqueness came from the particular case of being Newtonian okay so then we have a unique stationary state the same mass and center of mass that is radially decreasing in general for any potential you if you put together this result with the minimization that you can do in more general potentials what you get is that you have you reduce uniqueness of stationary states to uniqueness of radially stationary states and there is no general answer about that yet only in the Newtonian case okay so with this we get the right candidate to be the long-time asymptotic now without any doubt for the right definition of a stationary state and then just to finish let me tell you that yes then you can get the long-time asymptotic result and is what you expect and all of you have in mind the unique global minimizer of the free energy determines the global behavior as it goes to infinity for solutions in the with initial data at least in that space and of course with this kind of components arguments without any decay rate but at least we get the global asymptotics okay and with this I just want to conclude quickly the two talks sorry about the few minutes more but so I hope to have convinced you that the different regimes make sense for homogeneous pressure and kernels in the fair competition regime we know that we get a very similar picture as for the classical case at the level of a stationary states and sell similar solutions there should be much more work to understand if it's at the same business at the level of the evolution this hasn't been done yet for the efficient dominated regimes we understand now by the symmetries based on the free energy and we get the good candidates in some particular cases for the global asymptotics but and then we can get the long-time asymptotics in the 2D case if you want to go to the 3D case that will be very interesting what's the big open problem there we don't know how to for larger dimensions we don't know how to get the confinement the confinement is quite challenging the confinement of the mass for the evolution problem the confinement of the mass we know how to do it only in the 2D case I'm not going to show you why it's so but there you need the confinement of solutions to the evolution problem so you cannot assume any longer that they are radio that's the big issue and for dimension larger than 3 we don't know if there is confinement of the mass or not which means that as t goes to infinity we don't know if we lose mass at infinity or not so with that I stop and if there are any questions please write in the comments thank you