 Δεν είμαι ο οργανισμένος Φιλίξ. Πρέπει να προσπαθήσω όλα αυτά τα πράγματα εδώ και να προσπαθήσω ό,τι πιστεύω για να μιλήσω. Λοιπόν, σήμερα θα τελειώσουμε την συμφωνία για τις πρόσφυρες και θα ξεκινήσω μέχρι και ό,τι πιστεύω από σήμερα και σήμερα θα μιλήσω για την ομογενισμότητα και την ευλογία της πρόσφυρες. Είναι ένα τεκνικό δύσκολο, όταν θα το κάνω από τη σήμερα. Ξεκινήσω να πω την πρόσφυρή για το πρόσφυρες για να εξηγήσεις τη σύμφωση της ευλογίας της συμφωνίας της πρόσφυρες και τα μυσκότητας. Θέλω να εξηγήσω ότι αυτό είναι για ένα σχέδιο πρόβλημα με δύο πρόβλημα, Αλλαίτηση Я σημαίνω, ότι μεγάλωσε ένα πρόβλημα με δύο πρόβλημα. Θα δώσει και την ευλογία της πρόβλημας γιατί το αντιήθειο με κάπου το δύο πρόβλημα. Πρόβλημα με δύο πρόβλημα κι εσείς καθώςρος κοιμήθηκε, vraiment διότι σου διόδησε μεγάλωσετα. Πρώτας πρόβλημα, κοιμήθηκα αυτόήν σε αυτήν την συμφωνία, Πρώτα απ' όλα, διδάξω φορμανικά χρησιμοποιήσαμε αυτή η συγκλωτική χρησιμοποίηση της μεθότητας κελερουμπινστίνων και στεμβερκότητας. Είχω δείξει φορμανικά τη χρησιμοποιήση και δείξει why one expects mean curvature. Και μετά σας δείξω ένα σύμπροπο πρόβλημα για... Εντάξει πιο φαμμύριο να βλέπω F εδώ μέσα από W, για να βλέπω F, σας δείξω ένα σύμπροπο πρόβλημα για αυτό. Δείξω πως ξεκινάτε με καλύτερη αρχή δέτα. Ξεκίνησε να βλέπω F στους εύκολες. Μετά από τα άλλα. Δεν χρειαζόταν για τα δύο. Λοιπόν, πρέπει να είναι ένα ή άλλο. Το κυρίο είναι ότι αυτή η αρχή είναι η ίδια αυτή η αρχή. Είχω χρησιμοποιήσει όλα αυτά για να είναι 1, και μήνωσα 1, 1, μήνωσα 1. Και εγγράψω ότι δείχνει μόνο για ένα από τα δύο. Ξεκίνησε να βλέπω ένα από τα δύο. Ξεκίνησε να βλέπω με καλύτερη αρχή δέτα, δηλαδή ότι είναι like that. Δείχνει δείχνει δείχνει. Σήμερα θα είναι σημαντική. Ξεκίνησε να βλέπω ευκολογία, πρέπει να έχω ένα εντυριό, ένα εξ-τυριό, ένα πκείριο. Βλέπω την εξ-τυριό που αρχίσε να μην έχεις κλειδά με τον κειρο Fare Elefant. Ξεκίνησε να βλέπω ένα από τα δύο. Ξεκίνησε να βλέπω ένα από τα δύο. Υπήρχε με δυσκολογία, πλάλη με τον Υπ-τυριό καιτακρό. Φεκίνησε να βλέπω ένα από τα δύο. Φεκίνησε να βλέπω ένα από τα δύο. Επισηξη είχα καταγωνία... διαφορετικές, και καταφ plaster δείχνει ότι εγγερφωσία χρειάζει να δει η εγγκάριση και η εγγκάριση είναι η Τελφέρα από την εξαπλήτη εξαπλήτη μέσα στοιχερμό που μπροσφέυει σε μπροσφέρα, αυτά που έχουν να υπάθει βασικαίωση και τελφέρα που δει την δυοκλή, but while it was positive it satisfies minus dZ square plus one equal zero and I remarked although I did not prove that, that implied that distance dZ is the sin distance of the mean curvature flow starting u0gamma zero. Και φυσικά, όταν θεωρήσαμε στις θεωρές, θα always make the assumption that there is no interior and nobody asks me, what does the distance function approach recover when there is no interior. So the distance function gives you, depending on how you derive it, if you derive it the way I did by doing this crazy change of variables and making something to be minus infinity infinity, then it derives the maximal and the minimal of the front. So if you had something like that you find this and that. On the other hand, if you take anything in between and you can take the distance function from that, we'll also solve the mean curvature flow. That's the non-uniqueness. All right, so I show you the proof of that. I never show you that why this implies that, but the proof is complicated, so I don't want to get into that. An aside comment, when anything I proved I said I wanted to prove that something has a limit and I said okay, I assume there is the limit then blah blah. So I want to show here for you, as an aside comment, how one does that. Why am I allowed to say that if something has a limit, why am I allowed to say okay, I can prove there is a limit, so the limit satisfies that. Why can I always assume there is a limit? And this is now the power of the theory that is behind the viscosity theory, so typically if you have a family of function, let me call it U epsilon, and you want to prove that U epsilon converges to some limit locally uniformly because the theory is a theory of local uniform in a local uniform topology, how do you do it? There is only one theorem for that, that we learned earlier in school, and this thing is, unfortunately I learned it to say in the wrong order, I think it's in the Italian thing it will be as a last call, I learned from the beginning that it's as a last call, and then okay. Okay, and so what you need to prove, I'm sorry this is calculus, is that the U epsilon's are equi bounded and equi continuous, and if you look between the two, typically this is easy to get, just prove that something is bounded uniform in epsilon, and this is more difficult. As a matter of fact, the easy case of that is if you can show that the U epsilon's are uniformly Lipschitz continuous, so if somehow you have a way to prove that the U epsilon's, I'm sorry if you prove that they are Lipschitz, yeah what was I said, if you can prove somehow that this is bounded, then of course you have the okay continuity. The worst case is when you know the things are not Lipschitz, and then you have to control somehow uniformly the modulus of continuity, and that's difficult, it's easier to prove that something is Lipschitz continuous than proving it's Holder continuous, because you cannot differentiate. And I show you the other day that sometimes getting Lipschitz bounds are a problem, so how do you bypass that? So I will show you a method that works with viscose solutions that says that I can pass to the limit if I have, so if U epsilon's, if the U epsilon's, so this thing is something that goes back to Barl and Pertham and says the following. The idea is the following, and resembles a little bit this notion of gamma convergence, so it says write down the largest possible limit, so I have a family U epsilon, and this U epsilon's solve let's say an equation with comparison principle, namely if order data implies order solutions. And let's say that you have so, you know that the U epsilon's are uniformly bounded, so someone has to give you something. So what you do is you write down the largest possible limit, and here it's important thing, the largest possible local uniform limit, let's call it U upper star, you write down the smallest possible limit, let's call it U lower star. By definition, this is less than that, smallest, largest, and then the only thing you need to do is to prove that the U upper star is a sub solution of the equation, of the limiting equation, and the U lower star is a super solution. If you do that, because there is a comparison, sub has to be below, has to be below sup, but that's the opposite of this. So then you put them together, of course, and you get U upper star is equal to U lower star, and because you define equals U, and because you define that, and I'll show you how to do it, as local lim sup and lim inf, that implies that the U epsilon's converge to U locally uniformly. So at least when you are dealing with solutions of problems in L infinity that have a maximum principle, a weak maximum principle, you can bypass the need of the applied continuity by adjusting this. Now of course I'm cheating all these places, all the data and the comparison has to work for semi-continuous solutions. So what's the definition of the U upper star? So the U upper star at x, and I'm going to assume my functions U epsilon are only functions of x, is the lim sup, as you expected that, as epsilon U epsilon, and now I have to make it local uniform, and that's why I will put here y goes to x of U epsilon y. So this is what I mean by, and the lim inf is the same, and you need to take a little bit of care, anytime you write that you have to make a small argument to explain why this thing is independent of the order you take the epsilon, but I think it looks very much like a gamma limit that you're going to take. So anytime I say I converge, this is what I have in mind. So what happens is the lim sup of the z epsilons does this, the lim inf of the z epsilons does that, and then in the background, I didn't show you that, I have a theorem that says that if I have a function that does this when it's positive, then it's below the distance function. And when it does that, when it, I'm sorry, when it's negative, and when it does that, when it's positive, it's above the distance, but then they have to be the same. Alright, so that's how I do always the convergence, that's why I never argue, I never say, okay, let me show you how it converges. Alright, so now I want to write down a fake problem and make the problem a little bit complicated to show you that this thing doesn't work and show you how you fix it. So, and to make it a fake problem, I'm going to put here an x dependence, or I could put here a drift or whatever, but let's put the next dependence and maintain that for each x, f of x, u as this form, I mean may grow a little bit more, is all of them look alike. Say it again, it's limited, it's limited. Yeah, yeah, yeah, thank you, thank you. I said it, but I didn't write it. Maybe I didn't even say it, I don't know, we can have the replay to see whether I did it or not. The idea is very, very simple, okay, you write down the largest possible thing, the smallest possible thing, and then you hope you can order them. And that's what I'm saying, it's like the gamma has this, okay, so look alike in the sense that they are whatever they are, and then I keep the plus minus one the same to make like simpler. And I would like to get here a result that says that the u epsilon's go to one minus one, and there's going to be a front moving with certain velocity. So now I want to show you how one does this. And again, I will skip a lot of the technical details, and I try to show you how to adapt, how to make, if you like, rigorous, the formal calculation I show you, the calculation of Keller-Stenberg and Rubenstein-Stenberg, how to make it rigorous. So just for simplicity, let's assume again, so what happens here? We have a traveling wave, so as before, since I made that assumption, we have a solution to this problem, which is strictly increasing, q plus minus infinity of x is equal to plus minus one, but this solution, it's a function of, so for each x solves this problem, and for each x has this behavior, uniformly or whatever. So if I try to do the method I wrote there, I will have to do, if I put here is the epsilon of epsilon, I have to put here an x, okay, because it depends on x. And therefore, when I do the differentiation and whatever, I will create extra terms, and if you remember, this calculation was made exactly to, when I show you for that problem, was tailored made for the problem. I mean there was a limit, or if you like, this calculation told you that if you were to take the approach of Keller, Rubenstein and Stenberg, which was to write a U epsilon as some function plus some other function, what I show you here was a slick way not to worry about that. Yes, here, yes, thank you. All right, so the calculations, the proof I showed you last week, it was a slick way to avoid this, so it just worked. It's not going to work here. And the reason it's not going to work here, let me tell you the reason in advance. You remember, so that calculation, the idea of that thing was to get it to be a super-sub-solution of the equation that's satisfied by the distance function. But the problem here has an x dependence, right? Then you remember, if you remember, when I had the problem with the next dependence, the x, when I used the distance function, I had to evaluate things at such points. Remember that, I made the whole point that for the distance function, you can only see it on the front. So anytime you write down the distance function with the next dependence, you have to pull it on the front. And if I do that, it will mess up this nice equation I'm going to get from the z epsilon, because the z epsilon is not going to see this x minus distance to the front, so it's not going to work. So what do you do? Then you go back and you say, all right, I'll do this. And so the only thing I'm going to show you today is how can you rigorously justify the formal asymptotics. But for that rigor, I'm going to cheat a little bit and I'm not going to try to justify the integrated generality. I'm going to do it assuming that the distance function is smooth. And there's a story about that. So as I mentioned, some people, I mean, the Motonian Satsman had proved the convergence to mean curvature for the original problem up to the first time there are singularities. And then what Evans, Sonner, and I found a very elegant proof to prove it. We extended that proof, so here I'm showing you an extension of that proof with Sonner. And eventually, we hit a problem where that proof would not work, global in time. It didn't work and I will show you why. And so then we had to go, we went back with Barr and said, okay, what are we going to do now? We have this very slick proof, works perfect for everything, but it doesn't work in a very interesting problem. But it works if everything is smooth. And that's a scene in the theory of viscose solutions. If something works when it's smooth and you get something that makes sense in the end, there has to be a way to prove it. And so we had to go back and redo everything. And after redoing everything, we came up with a theorem that said basically, if you can do it when it's smooth, you are done. So the proof I'm going to show you today, although that's why I'm not showing you all the details, because in the end we do it when it's smooth. Okay. Let me emphasize one thing. There is this cube that comes in, the traveling wave. Now in the cubic case, why is the traveling wave in? The traveling wave, you can think of it like that. Far away, the limit is 1 and minus 1. The solution will look 1 minus 1. And you need a way to transition at the epsilon level from there to there in such a way that you don't screw up the problem. And that's where the traveling wave comes in. Because you need to find something that far away controls, it looks like it should be, and it goes down to a smooth way. Now you can say, look, I mean, if this is minus 1 and 1, I can find many functions that go from minus 1 to 1. In particular, let's use hyperbolic tangent for everything. But it has to adjust to the equation. It shouldn't create extra terms. So it has to be adjusted to F. And that's a real problem. If anybody can find a proof for that, that doesn't require the cube, I will be extremely happy because I will be able to prove other problems. Now, what's the difference between this and what, at least in the two-phase case of the things you saw in Felix's lectures, at least from the viscosity point of view? The algorithm that Felix has doesn't see that. The algorithm he has sees this and that. It's like seeing what the solution I wrote down, the minus 1, 1 solution, that looked crazy. Therefore, he never has to worry about that transition if you do it in the viscosity sense. And because you don't have to worry about that, at least the viscosity proof I will show you for this is very simple. But unfortunately, if you have epsilon positive, you need to find a way to deal with this boundary layer you create. Because there's a boundary layer created in the transition. Because remember, as epsilon goes to zero, the whole idea is this, we'll go to that. And so what you are trying to do is do that together while in the BMO scheme, somehow you have taken epsilon to zero and you have 1 minus 1 from the beginning. So I mentioned this because there's a big open problem that in some sense in this are not big, but there is one problem left in all this theory that nobody knows how to attack. It may not be true in full generality, but the place where it gets stuck, everybody gets stuck, is the lack of traveling waves. Okay, so you do that and you plug it in and then you start seeing X dependence. So if I plug that in, I'm going to get X dependence. So typically if you have another question, if you have a formal expansion, I mean Sigurd mentioned that, you can have a formal expression and somehow you put it together and prove at the end. How do you prove that the formal expansion you get actually does the job? How do you do mathematically? You have to find a way to control the errors you make. So you need to introduce in your problem something that will control this error. Otherwise it's not going to happen. It's going to be automatic. And the way to control this error here, and this is the essence of the theory, which allow us to do many other things, is the following. Let's assume that this is the correct nonlinearity, just to draw something. These are the other. So what I do is, and this is minus one, and this is one. What I do is I add a little bit to the F. How much bit I need, I add an epsilon alpha to the F. What does this do? This moves, if alpha is positive, let's say, the new equilibria will be this, will be minus one plus little of plus, and plus like that. So I moved a little bit, but this is going to give me some room to work. Of course if I move that, I lose the property that the two wells have the same depth. Therefore I lose the fact that I have a traveling wave like that. I will have a speed. So think of this as being a number, and then I will have a speed which will be c of epsilon alpha, a speed q dot minus q double dot plus F of q x equals zero and all the other properties. The theory says that because F is cubic, there exists a unique c plus epsilon alpha. There is a theorem that goes back to Weinberger and so that says that there is a unique c for that. Because this non-linearity is not any more an equal area of non-linearity. And it's also easy to prove in that case also that the c epsilon over alpha over epsilon converges uniformly to some other constant times alpha and this is a uniform constant which depends on the original F. This is one of the facts from traveling waves. This is the way I'm going to create a room. So instead of making an expansion like that, I'm making an expansion putting here the traveling wave that corresponds so this is just a parameter, the traveling wave that corresponds to this small perturbation. And now I put the p here and so on. I have to find a p and here let's make it from the beginning easy. Let's put the distance to the front and here I will put distance to the front over epsilon x minus epsilon alpha. And the hope is that this little alpha that I introduce will allow me to control all the errors as epsilon goes to zero. And then, since alpha is arbitrary, I will let alpha go to zero and I will get my mean curvature flow. If I don't let alpha go to zero, I get a front that moves by mean curvature in a little bit of velocity. So I create something that goes a little bit faster to take over what happens. And now you do again the calculation I did before, although we don't have to guess anything and I try to do it rigorously. So if I plug this thing in, you're going to say, okay, but how do you gain something? I will gain because to use the traveling wave in the original equation I have to add the epsilon alpha. So I plug it in. So I get one over epsilon instead of writing all these things I will write q epsilon. So I get one over epsilon q dot epsilon it will be this minus the Laplacian and it will be minus one over epsilon squared q double prime dd squared which of course is one. It will be plus one over epsilon f of q. Okay, so these are epsilon's and there will be some terms from the additional x dependence so when I compute I will get a term which will be like minus one over epsilon and it will come from when I take the mixed derivative here so this will be 2 dx q dot epsilon dd so this is a term that I pick up because I have the x dependence. Okay, so there's a mixed term there has to be one over epsilon. I ignore the term I get with two derivatives on x because it's of order epsilon I will never worry about order epsilon. If you don't like it like that just compute it and you're going to see that the dominant term is this and then that's it and then I get a bunch of terms from p and I only write down the ones that matter the ones that will matter will be it will be minus p double prime epsilon with one over epsilon in front and I guess that's the only thing that matters here. Maybe there is one more I think that's the only thing that matters. Okay, and of course when I write this thing down I make a mistake this is q epsilon plus p epsilon x and now I'm correct. There are more terms but they are higher order I don't care. So what did I do in the previous proof I wanted to kill this with that? The only way I can do it is if I add an epsilon alpha and epsilon alpha this together gives me a minus I may have messed up the signs on the alpha but we'll see that. This thing together give me a c epsilon over epsilon squared times q dot The traveling wave is this I added this Okay, so this together with that up to an error and the error is plus one over epsilon squared f prime q epsilon p epsilon times epsilon I do exactly what I did in the formal computation So this I get it because to put that in and out I have to eliminate the p epsilon So I add it, subtract it and I make an error There are more errors again I repeat there are more and more errors but you'll see what will happen in the end with the other errors So I do that These are the terms it comes Now having written like that so I replace this this now and now I'm correct This goes, has gone and gave us this I used the equation for that so let me write down what I have Is this clear or I made it Maybe I messed it up but It's a calculation that once you know the answer you cannot miss how to do it All right, so now let's put all this mess I did together, we get and this is one over epsilon here Okay, so what do we get We get minus one over epsilon p double dot epsilon So let's put one over epsilon minus p double dot epsilon I get plus f prime q epsilon p epsilon I leave here some room Equals and on the other hand I get all my errors q dot epsilon dt minus that minus 2 What else have I left I took that I have minus α and I have which I will keep outside and I have minus c epsilon α over epsilon squared and there is this term, minus one over epsilon α I keep it separately and this is epsilon here So all that together what did I do Yeah, it's equal to one over epsilon times that because there is no epsilon squared anymore So I have this plus garbage but the garbage will be taking into the α over epsilon α is fixed No matter what garbage I have here if it's of order one goes into that I will have all error terms but this is α over epsilon and nothing here will depend on either will not have an epsilon coefficient or will be multiplied by epsilon So anything goes there And now you see I can finish because this up to an error is what I call minus α minus α α Remember this assumption plus an error that I will put it here and now I have this expression and what do I do I say ok I need the p dot epsilon just to make it consistent with everything we have we can put here ok let me leave it like that and so what do we do we apply Fremont's alternative we say that for this thing to have a solution this part this thing has to be orthogonal to q dot so we need to have q dot ε square we need to have this thing orthogonal to q dot ε α So once we make that assumption we have we find the p and I also find the equation I want because this gives me the equation right the same way as in the formal argument I got the equation that's how I get the equation here I compute that for q dot ε α and I have fixed all the errors so what did I prove I prove this way that the u ε are in volts or this construction so I didn't prove that if I call this thing w ε and I put plus minus α I prove that this thing is a super or sub solution to the reaction diffusion equation therefore u ε is less equal w ε w let's say plus we are the respective signs and now I let ε go to zero and I find that the front I get is above or below a front that moves by my curvature because as ε goes to zero this equation goes back to the original one and I get a curvature and you see where the x dependence comes in there will be a term like that in the equation right which if you are trying to do it in the slick way you will lose it So this is Πρέπει να δημιουργήσουμε το πρόβλημα, που είπα, το πρόβλημα, να δημιουργήσουμε το πρόβλημα, όταν όλοι είναι σύμφωνοι. Παρακολουθώ να σας δείξω ένα πρόβλημα. Τι έχω αυτό το πρόβλημα. Παρακολουθώ να σας δημιουργήσω, έχω ξεκινότητα με ένα πρόβλημα ανσκέλο, και είμαι ενδιαφέροντας σε έναν χρόνο, έναν χρόνο μεγάλη εξαιρετική πρόβλημα. Λοιπόν, να πάμε back. Υπάρχει κάποιος να εσύ με αυτή την πρόβλημα. Αν η φ είναι κουβική με αρκετά θέματα, τότε κάνω το σκέλεγμα εξαιρετική, σε τέτοιο εξαιρετικό σκέλεγμα, και κάποιος μου πούρει, γιατί δεν κάνω εξαιρετική, σε τέτοιο εξαιρετικό. Αυτό που ξεκίναμε το πρόβλημα, το είχε ήρθει. Λοιπόν, εγώ, νομίζω αυτό, και αυτό είναι ένα αγγωμα, που υπάρχει σε κάποιες δημιουργήσεις της Πολ-5, που δεν είχα δει. Αλλά νομίζω την αρχή δημιουργή του πρόβλημα. Τι πρόβλημα είναι ότι, ουσιακώς, έχετε ένα φροντιό, που μείνει με κάποια δημιουργία. Λοιπόν, σε ένας λευκότος σημαντικότητας, μπορείτε να πιστεύετε ότι έχετε μια εξαιρετική στιγμή πριν την δημιουργία της εξαιρετικής, όπως θέμαστε να πει σε τέτοιο εξαιρετικό. Και αν θέλετε να σκέψετε αυτό, θα πρέπει να δείτε, ως α, ήχε, νομίζω, δεν είναι το ίδιο α. Ας δούμε θέμα,νά enfants τα παρναχłączά, εξαιρετικό κολοκα Wool, δώσ' νομίζω, σημαντικό κολοκαλμά, ας τραπατολούνρίileen ο άνθρωπον,�τι μεlli, επίψω, σε ένας «δίδος» με μία προσπάθεια, τότεού, να τα εξαγρίσουμε λίγο u, όπως η θ θερασία στις θερασία. Τώρα, δεν κάνουμε αυτή τη λαμπή. Αν κάτι, υπάρχει ένα θερασία που δεν είναι τελευταία, ήωρας κάτι. Και πού βρεις έναν θερασία που δεν είναι τελευταία, αν οι δυο γλώσεις δεν είναι the same. Αν δεν είναι η αγώηση, τότε θα γίνω θερασία. Αν οι δυο γλώσεις είναι the same, και αν κάνω τα άλλα σκήματα, θα γίνω τελευταία. Ξέρουμε εδώ, θα γίνω τελευταία, και θα πάρω στην επόμενη τέρνη. Και αυτό είναι λοιπόν why I do this. Τώρα, σε κάποιο στις επόμενη, Κεργεύα Ευαννήτρια σκέφτηκε για το επόμενο αποφασίδιο. Δεν για το κουμπί, για κάτι σύμφωνο, για την πραγματική κομμάτιση. Πήρα, πέρα, θα γίνει η πρώτη θερασία. Ποια είναι η θερασία που θα πάρουμε την πρώτη θερασία, να παρακολουθήσουμε την πρώτη θερασία, να παρακολουθήσουμε την πρώτη θερασία και να δοδηθεί. Αλλά αφού qu we could prove exactly the formal expansion. What if we could really not say, okay when I get zero I do the next scaling. What if I were to say, I have my velocity, I have my front, subtract from the front whatever that means. A front that moves and what you have left is the front moving by min curvature. I think the problem here is how you make sense out of this idea. Πώς μπορώ να κάνω εξαναγωγή στον κόσμο. Αυτό θα είναι έναν εξαναγωγή για να κάνω, γιατί θα λειτουργεί για πολλές πράγματα. Έχεις άλλες εξαναγωγές που ξέρεις, έχεις έναν εξαναγωγή, έχεις πρώτα εξαναγωγή, αλλά αν έχεις εξαναγωγή, έχεις εξαναγωγή. Μπορείτε να φτιάξετε αυτή η ιδέα της 5 εξαναγωγής. Όχι, αυτό δεν είναι το πρόβλημα, όπως είπα ότι είναι ανοίγη. Λοιπόν, πάμε back to that. Λοιπόν, αν δεν υπάρχει τίποτα εδώ, θα αυτόρώ το πρόβλημα, όταν είναι καλό στη δωση στην δωση. Αν θέλω να φτιάξω εξαναγωγή, θα σημαίνω ότι θα βρήκαμε εξαναγωγή, λοιπόν, το πρόβλημα που βρήκαμε εδώ, στήσετε, όταν έχω έναν πρώτα... το εξαναγωγή είναι όλος δεν είναι. Και όπως βλέπεις, έχει ένα εξαναγωγή. Άννον αν είναι εξαναγωγή, είναι ένας μόνος πληκτικός, αλλά είναι εξαναγωγή. Θέλω να φτιάξω ένα προβλήμα που είναι ένα ρεκτήριο προβλήμα. Αυτό λέει ότι αυτό είναι η εξαιρεία που θέλω να φτιάξω. Πολλογήω σε έναν λεκτήριο πρόβλημα. Αν κάνω το σκέληγμα, θα φτιάξω ένα προβλήμα. Μετά το σκέληγμα θα φτιάξω σε αυτό το προβλήμα. Εκεί έχω ένα αυτοίγιο σε X. Λοιπόν, θα πρέπει να κάνω κάποιες αυτοίγιες σε X. Θα πω ότι δεν είναι πολύ χρή. Είναι η X. Ας λέμε ότι είναι περιοδικό. Λοιπόν, θα κάνουμε έναν λίγο σκέληγμα. Θα πω ότι αυτό είναι ένα περιοδικό. Για τώρα. Αυτό είναι πρόβλημα σε εξαιρεία. Πολλογήω σε αυτό το σκέληγμα. Όταν βλέπεις αυτό το προβλήμα, θα υπάρχει ένα κομμήμα της προβλήματος, ό,τι θέλω να φτιάξω, η εξαιρεία μετά το κόσμο, ό,τι είναι... Δεν ξέρω πώς να πω. Μετά το κόσμο που οι άνθρωποι είναι χρησιμοποιήσεις, και πρέπει να το καταφέρει με αυτοίγιωση. Λοιπόν, τι γίνει, είναι... Λοιπόν, τι γίνει είναι, εξαιρεία μετά το κόσμο, αλλά στην τέτοια, θα πω. Λοιπόν, ας λέμε, περιοδικό... στην κομμή... στην κομμή. Λοιπόν, θα δεις το προβλήμα και δεις, okay, να δω πώς να μεταφέρει, δηλαδή είναι αυτοιγένι, δεν πω πώς να κάνω αυτές τις προβλήματα. Θα πω εδώ, εξαιρεία εξαιρεία, και θα πω εδώ, κομμή, μεταφέρει με αυτοίγιωση. Και θα πω εδώ, εξαιρεία εξαιρεία, και θα πω... Δεν ξέρω τι θα κάνω. Για κάθε εξαιρεία εξαιρεία, θα πω στο κομμή... είναι η εξαιρεία για να κάνεις σκέφτη, Αυτό που έγινε εκεί, και γυρίζω ότι έχεις δημιουργήσει πολλές ειδικές τέρνες που δεν μπορείς να δημιουργήσεις. Γιατί οι τέρνες που έγινε από εδώ και έγινε ότι η μόνη τέρνη που έγινε ήταν αυτή. Δεν θα κάνει, θα υπάρχει πιο τέρνες. Λοιπόν, αν έγινε αυτό, δημιουργήσεις ότι αυτό εδώ είναι ένα περιοδικό. Τι θα είναι το πρώτο τέρνι που θα γίνουμε. Τώρα θα είναι ένα τέρνι που θα έγινε πιο τέρνες. Πραγματικά πλοχή, πλασχεία λαπλάσια με τη σύγχροτηρία του ί. Θα οδηγήσω αυτό το αγγημένο y. Πραγματικά πλοχή, πλασχεία λαπλάσια με τη σύγχροτηρία του y. Let's say dd square minus that minus Laplacian to Laplacian with respect to y of q dot dot d plus f of q y. All this thing is going to be multiplied by one over epsilon square. It's going to be and you can see it easily that if you take two derivatives in next year you're going to pick up one over epsilon square. So this thing I said, this naive thing was going to kill only that, cancel only these two. But you have this extra term and remember the whole idea was you get rid of the one over epsilon square term. So that q is not good q. To do that you need a q that actually solves this equation. And these things have been studied going back to Jackson and reinvented many times by Bere Sticky without referring to Jackson, my politically incorrect statement, but goes back to the thesis of Jackson who figured out these things which they call them at that point pulsating waves. So what is that? For fixed y this looks like a traveling, think of it as a traveling wave in psi which oscillates in space. So you have two directions. It's a wave that way but also it's going up and down, not in that direction, in the way, okay? So think of a family of waves that go from minus one to one but they are repeating it after a while in space. And they satisfy this, they satisfy this equal zero. And as a matter of fact if you add, if f is non, let's make it like that, then there is also a term here, c, α, yeah. Okay now that's the definition of the pulsating wave. If I kill a little bit equal, that's then there is a c α here. Now these traveling waves notice they are not any more independent of the direction. They depend on that, that's why I kept it here. So the pulsating wave corresponding to a general f is something that looks like c, depending on the direction, c-qp2-2dyq.e-λπ q, plus f of qy equal zero. That's a pulsating wave. It is periodic in y increasing in psi and again I would put it plus minus one. I like to fix these things instead of making them also periodic, the equilibrium. So the question is, do these things exist? And yes, there is a dot here. Yeah, yeah, yeah, yeah, yeah, yeah. Thank you. All right, so for every e there is a c, the c of e is zero if the f is an equal area, and under some assumptions they don't always exist, but under some assumptions these things exist. It's a non-trivial way, a technical non-trivial way to prove their existence. And the reason is if you look at this operator here, because that looks like Laplacian in Rn plus one, Rn for the one, if you like to look at this problem, it's degenerate. It's elliptic but degenerate elliptic. So you need to do some work to fix that and this goes back to old work by Lyons, but you have to solve it in a big box because you have periodicity, you cut off in space, you do it there and then you regularize the problem, you can solve it there and then you want to show that as you let the size of the box go to infinity, you get a solution. It's not a very difficult way to do it, but it's one of these very tedious proofs, so you would like someone to come and tell you where this is true, because there isn't much you can do with that, but they are there. Okay, so what do you do? Going back to our problem now. So I did that, I know that there is a problem, but I don't know if you can solve it, but now I have to put a direction. Before I was stopping here, I have to put a normal direction. The normal direction is this. So I need to add the text that there, so I need to prove that somehow this gives me the answer. Formally at least, if I take a pulsating wave at the formal level, you're going to see that this is a normal direction, but if I take a pulsating wave at the formal level, you're going to see that this will be, I will get that to be zero, so it will go away, and then it will give me the linearized operator, the linearized operator that corresponds to this, the right side has to be orthogonal, and then the velocity of the traveling wave is going to be a mess when you write it down, and so on. But there is a problem. How do you prove it? If I try to repeat the proof I said I had earlier, but make it a weak proof, namely now, don't assume that the distance is smooth, but the distance is really a viscosity solution. So let's assume now that the distance is not smooth, it just solves the problem in the viscosity sense, so it's just a Lipsitz continuous semi-convex, semi-concave function, but has no second derivatives everywhere. How do I justify these calculations? Because when I start taking derivatives, at some point I will create terms, I will create one term when I take a Laplacian to that, I will create a term that will be, will look like QEE, let's say Laplacian, not QEE partial of Q with respect to the direction, Laplacian of the distance. And this thing is a sin, as I mentioned the other day in the viscosity theory. But it just works, nothing else. You don't know what to do. What saves you, you look at that more carefully and you say okay, but there's an epsilon in front. So at the limit it's not there. At the limit it's not there, do you handle it? You cannot make sense out of three derivatives. There's no maximal principle anymore. And this is a made-up problem, we came up with this problem when we studied hydrodynamic limits for anisotropic particle systems where this thing comes, this non-homogen, the dependence on the direction, the anisotropic comes up very easily. So how do you solve this problem? And I have to admit that we were stuck for some time. And then in the end, I think we discovered it at the same time as Giovanni and others and it was related to what the Georgie was suggesting at some point. It's like, I think even if it wasn't at the same time it was without knowing each other, doing it. Which we came up with another formulation of the notion of motion, a very geometric thing that depending on sets, moving sets and not functions, and then it works. So let me explain to you what I want. So definition number three, which I'm not going to write it down with all the details but I'm going to draw a picture. So I have a set, omega t, gamma t, the boundary and omega t, I mean t is like that. So there is a plus, let's say a minus in the boundary. And the plus, this family I will call it omega t. And I want to claim, I want to describe what it means for this thing to move with normal velocity v. So I'm going back to lecture one. I want to give you a definition why this thing is, I want to call that, believe me you don't want to see the definition, I want to write this as super flow. So I want to say that this is a super flow with velocity, it's a super flow but let's, super flow of that motion. So here's what I'm going to do. And I take this to be open. I come inside, really in the interior, and I take a nice set, a ball. Okay? Now I assume that no matter which motion you give me, I know how to find a short time existence. And if you don't you just write down the Euler approximation to the motion. But it's too late to write this single down so I assume I can propagate any ball with that velocity. You give me this velocity I can propagate it. Now I propagate the ball however with velocity, I'm drawing it like that too, with velocity v plus alpha. I'm inside, I'm inside, strictly inside, I give it a little bit more boost, whatever it's plus minus, whatever you need to, I think it's minus, whatever you need to make the velocity. So this thing, if I were to draw a picture, if this was schematic picture, if this was the omega t, what I do is I move it like that. Okay? Inside. And for short time, I have such a velocity and I demand the definition is for small h, I have this picture for every ball I can put inside. Then you work hard and you prove that this definition yet is equivalent to what I proved all this time. It has the advantage, I never gave you the definition of viscose solution. The viscose solution is what you do is you touch from above or below something by smooth functions. So what we do here is we touch from inside to outside, sets by smooth sets. Notice this alpha. This will play the same role of the alpha I saw you earlier. This extra speed will allow me to control all the errors I'm making. Okay, now why is that good enough? So you have a theory like that, that's the new definition. We address this problem. This is my U epsilon here. This is the problem I do. They pay me to speak, and appreciate the other lectures better. Which are very well written down and organized. After me, every lecture looks amazing. Okay? I know this slide because I remember both Sigurd and Felix's lectures which were on the blackboard, how well they were. Here I had to look into a mess to find my equation. Okay, so what do we do by star? Define two sets. Let's define ones. And this thing is going to be the set where the lim soup of the U epsilon is one. Okay, we know that U epsilon's will go to plus minus one because we multiply by epsilon squared that will go to that. So right here the set where they go to one. And we claim that that set is, again, it quotes mean it's either super or subflow moving I didn't tell you what the velocity is. It's not anymore mean curvature. It's going to be a velocity which will look like there will be some anisotropic. There's going to be an anisotropic coming up because of the exception of the dependent. So this is like the people call the Kubo type or if you want to impress people you call it the Einstein constant. I've seen that in many papers when someone wants to impress they put a coefficient and say we compute the Einstein constant. They say Einstein must be important. Anyway, so that's the as many names but it's the thing you get by averaging that coefficient. And you get that. So what do you do now? Now we go to the epsilon problem. So here we are at the level of the epsilon problem. This is the set omega t. You need to prove a little bit that this set has some interior. So you go there you put the ball like that and you solve the reaction diffusion equation starting with an initial data to be 1 here and minus 1 away. So what I do is I put my ball and now I solve this equation with an initial data to be 1. So if you like the picture I'm here at the slice t0 I put down a ball and now my solution is somewhere there and I put an initial data which is less than 1 it doesn't matter and like this but no matter what it is if it is minus 1 it will be below the U epsilon because the reactions will never be minus 1 and since I'm inside the ball I can say that if this is 1 inside the ball I can make everything to be like 1 minus delta so my solution being like that is below so that's the U epsilon. T0 I didn't put that 1 I put it a little bit less than 1 but that's not a problem so that's the picture I have but the reaction diffusion equation satisfies maximum principle so that implies that U epsilon T plus T0 is greater or equal let's call it phi epsilon of T where this thing is the phi epsilon everything is smooth I can do everything I claimed all these things this thing I can do it if everything is smooth I told you a little bit how it works and so what I have is I do this in equality and I know that in the smooth evolution of this with plus minus alpha this thing goes to 1 because that's the smooth proof I claim I can do so the phi epsilon they go to 1 as long ok but if this goes to 1 that goes to 1 which implies that this definition holds because I just proved that I have my set I think this should be lim inf where am I? because what I proved is the set that moves with velocity v plus or minus alpha is inside the set and I want to claim that is a super flow and that's it I verify the definition I have the result and in the end the only thing I had to do was do the proof for the smooth distance function nothing else provide I have this definition so I have to say this was a little bit anticlimactic because we worked very hard to find this very beautiful and slick definitions but we just had to not definition proofs and then in the end we said well that's it but I consider this that's something I did with Gibral I consider this a nice result because it took care of a lot of problems and ok here you may say this is a made up problem but these problems are not made up perhaps the homogenization is made up but the real case where one can find equations that are I will call them these equations no local in the sense that they see a lot of the y dependence but there are real equations let me give you an example of one that comes up time runs out so it's not possible I went to slow the first days and I'm paying for this let me write down for you an equation that you think that's made up now let me tell you what you are going to see ok so let's take j to be no negative and just to simplify things compact support and even let me write down this equation so if you don't see what I have here I have hyperbolic tangent beta j convolution u so you look at the equation and say ok it's early in the morning maybe I'm doing what Luis is doing with local equations I'm just creating jobs for people by writing down complicated equations and I want to do exactly the same problem I want to understand the long time behavior of that why do I want and I will explain to you what happens alright this equation is not made up ah what is phi phi is a function of course it's going to be very illuminating what I'm going to write down now phi is a function that has the property r is psi of minus 2 r 1 plus e to the minus 2 r now I wrote that people say ok now we understand what phi is I mean are we clear because I introduced psi ok so ah where is so where did this come from suppose you take the following particle system so now we forget all that stuff we are in Rn in Zn and we do discretely I don't want to write more points this is a lattice and we do if you like the discrete BMO scheme which by the way was discovered before also described the BMO this is due to to Griffith and Tarver and they did that as a simplification of the Ising model so suppose you have a discrete front this works very well in the united states they did it as a voting model but there is a version for multicolors right so suppose you have let's say we are in the US we are either republican or democrats and there is an island let's say west side of Austin near Northwell Hills Northwell Hills where people vote democrats that's a very small area actually the district goes all the way there is a district that starts from Austin and goes like that all the way to the Gulf of Mexico and just captures here the place that only votes democrat in Austin ok that's where you live and I live there so I understand that these are the elections and you are here and you decide how to vote so this has to be this are the so this is some set and you are here you decide how to vote let's say you are a little bit further there so what you do is you find a neighborhood that's too big I'm here I'm finding a neighborhood but I'm intersecting with what I have and if the points in the intersection is more than a threshold I go democrat and if I'm below I stay republican and I keep doing that this is exactly the BMO scheme the difference instead of having a neighborhood in the BMO scheme you have the gaussians instead of counting you integrate right so you do that but if you let this thing evolve and you scale it appropriately you get motion by mean curvature ok but nothing is probabilistic here now let me make this thing probabilistic the voters now will be spins so the voters are going to be either plus one or minus one or if you like the spins are either republican or democrats and I repeat this process so I have now a spin in the location X plus minus one so now I work on the space what is that I'm working functions from zd into minus one one I do exactly the same thing but I allow two randomness actually one randomness when I decide to check so I'm sitting there and then I have a clock that clock has the property to ring randomly it's called Poisson clock so at some point this thing goes then you wake up time to vote you count your neighbors so you have an energy now the energy decides what happens in your neighbors so there is an energy and also there is an external magnetization there is a magnetization which could be the advertisements you watch the advertisements magnetize you and decide to go in a certain way or something which depends on the number of neighbors that have the same spin and some external factor you decide whether you vote this way or the other now all these things I can write them down and this is where the psi comes in it's incredible to say how the heck do you get out of that hyperbolic tangent it's an amazing thing what you can do when you work with functions that have values only plus or minus one you can get a lot of things and in particular this comes up because in the rate you define you have to do it in such a way that there are Gibbs measures there are invariant measures so that problem was one of the big problems the people had in statistical mechanics and a lot there was work in Italy by Dimassi Presuti and others but I learned these things so if you take that model now how do you ok that looks nice but how do you make a pd out of that what you think when you write that probabilistically you can write all these things down considering a generator so I didn't tell you what the gamma this is j no there is no gamma here sorry so what matters here is now what happens in how you define your neighbors you can decide to look at only your immediate neighbors so if you are like that four points whatever nine points that is called short range interaction or you want to look at many points and that will be called long range interactions long range interactions has been understood much better than short range interactions because they are difficult ok I will tell you the result for that not the general result one case so here we are doing long range interactions so which means I am going to scale our potential by saying j of x-y of gamma gamma d I do something like a convolution and I am not defining times but I am knowing the support to go to infinity ok I just need maybe 5 minutes I know I am at the end but I entertain you a little bit I told you a lot of jokes I just need to finish this so I don't have to do it next time so you start something like that and what you evolve is a measure on the space of functions with value minus one one on the gd and so what you care about is the evolution of a measure so you have some initial configuration of spins and you just evolve it and you scale it to see what happens and the first scale is a mesoscopic limit in the modern language this will be mean field limit and you find how does the evolution so this is the measure configuration of the spins at time t you call that thing m of gamma xt and you try to see how this thing evolves in time and it turns out that this has a limit as gamma goes to zero and that limit solves that equation so probabilistically the way to think of it is that the measure this limit is more complicated but this quantity so if you like the evolving measures converge weekly to a Bernoulli measure where depending on where you are in space has a mean given by that equation but the probabilists really care to see again what is the shape you get so you have to do you say ok that's clear what you do you start with the particle system you let gamma go to zero you get the mean field equation the mesoscopic equation you scale it like before and you discover the Einstein formula you find a front like that but the statistical fields don't care about that statistical fields they don't know that they know it but they don't know that they want to go directly to here from the from the particle system directly to the evolving front the result there will be that properly scaled this thing converges to a Bernoulli measure which depending on whether you are inside or outside the front its mean is equal to the equilibria of the problem and this in data case if you have large inverse temperature so in the regime where this is the average there exists a raw star such that for every epsilon gamma going to zero slower than that if you scale the limit as gamma goes to zero or the evolution of new gamma of Pt epsilon gamma minus 2 goes to zero so this looks like the actual scaling I did this looks like 1 over epsilon 1 over epsilon square but it's not frozen it will depend on gamma and it has to go there is a whole continuum of scales you have which is due to the fact that you have the hyperbolic you have the particle system so there is more room and so that's what the actual result to prove it you need to do this thing because this thing is anisotropic right so I conclude with a picture take the discrete let's take the BMO scheme if you start initially zero the scheme will give you zero so there is no front probabilistically I will call that let's start with a great picture equal number of pluses and minuses so when you average you get zero there is nothing but in time probabilistically and this doesn't change because this is deterministic if you look at computations and actually this can be proved although I don't think anybody has made the connections with the two theorems if you do the same thing probabilistically so let me assume I have configurations that have mean zero at time t equals zero and you let it move in time that is like logarithm a very small time logarithmically small you are going to observe some random sets some blobs let's say of black or maybe white and black but you start seeing some crazy things this is a Monte Carlo computation you do and you see that and now if you go further in time these things will move with the anisotropic mean curvature so what happens is although we start with zero there is again some kind of random effect that forces you away from the equilibrium and then you start moving and then you get the mean curvature and that's the advantage of the probabilistic thing and that and finally real conclusion I told you long time interactions what happened with short term I think that's some work that Spawn has done I don't know the exact statement but one way to get reaction diffusion equation out of this model local in particular you can pick up this equation probabilistically with particle systems but with local interactions so you only care about what the neighbors do but you have to include an additional process and that's stirring so what you do is you have a way at every point not only you see what your neighbors do but you stir the thing and that's a way of putting in the Laplacian so the nearest neighborhood they give you this and you combine it with the Laplacian and that is called Kawasaki Dynamics was the first thing that and what was the name of the Italian who did it in the smooth case if you doesn't remember I refer to the person so it's Giovanni's fault that it's not name we don't have it ok I'll stop here and then next time I'll do a little bit of homogenization but next I will not resist and show you the viscosity proof of the two phase case of whatever ok so thank you very much go have coffee and whatever and if anybody has any question can ask me this is my most valuable person here you see he asked for the chalk I'm advertising this