 Zdaj. As I was telling you, there is now a change in the schedule. Professor Suganidi will teach now his third lecture in place of Felix Otto. And Felix Otto will teach tomorrow in place of Professor Suganidis. So thanks also to Takis for giving the possibility to make this change. Thank you for waiting. Can I start? Okay. I think I did. Where's my friend? He didn't come today. Where's my chalk? You move from the seat. There you drink for that. So I derived the formulation with a distance function. So in the case where we had motion by mean curvature, we said that a tripled was by mean curvature if and on if the sine distance function x, t satisfies inequalities. So it is a subsolution to the heat equation in the set where it's negative and a super solution to the heat equation in the set where it's positive. And in addition, when in this set also satisfies it's a distance function so the gradient will be 1 and in this set it's also a distance, I'm sorry. In this set it's a distance function so it will have a gradient 1 and here it's a distance function and it will have again a gradient 1 but since we are describing this in the viscous distance remember the actual condition is it solves this problem and in the viscous distance these are not equivalent. And I didn't mean that I meant minus, minus plus, yes. And I saw you waving my hands that that's equivalent to the level set formulation. That's equivalent level sets. And then I said that now we are ready to start studying the asymptotic behavior of reaction diffusion equations and I wrote down this the u prime of U epsilon equal to zero where w is a double well potential like that with wells of the same depth. Let's say to normalize them at minus 1,1 and that implies that w prime is something like that in the two areas that's the other way around. Anyway, it's this or the other, it's not this. All right. And I said that the result, the conjection actual result here is that U epsilon goes to 1 minus 1 inside outside in mean curvature flow. Make it precise at zero let's say is positive in omega zero and negative in omega zero complement and to simplify things let's assume that we have no interior at time t equal zero. So I promise to show you why this happens formally but right now let's accept it because it fits very well with what I did here and try to prove this in this very particular case using this result here is a nice cute proof that it's like magic so I'm going to make the assumption I have to make another assumption U zero is well prepared know that this is a very illuminating assumption which means that's a special about such a reaction diffusion equation. Special thing is that meets traveling wave solutions. So when you have a by stable nonlinearity because that's as I said it's either this or the other when you have an inequality like that there is a traveling wave there is a standing solution there is a solution to this problem of the form minus c minus q double prime plus w one prime q equal zero for q being plus minus one so any time you have a nonlinearity that comes from a potential like that there is a theorem that says there is a solution of that form so this is in one dimensional solution of the type that solution exists the speed c is unique and the solution is unique after translation by constants and the only condition is that it decays exponentially to the equilibria at plus minus infinity so that's a fact for any w like this there is a solution like that so I rewrite it and unique it's increasing it's unique after translations so let's make it unique if we assume that q of zero is zero and c is unique that's a fact of life and as a matter of fact you can even have an expression for c because if you multiply the equation by q prime and integrate so if I multiply by q prime integrate from minus infinity to infinity from here I get minus c q prime square that goes away and here I get plus w at one minus w at minus one equals zero so the speed is exactly like that and somehow it's unique although it's multiplied by this factor but it's unique so you see from here that if I'm in special situation where the two points are at the same depth these are the same so if w at minus one is the same as w plus one then the c is zero and it is in this case we are working here because I'm making that assumption and as I will indicate later on this is why I can put an epsilon square there ok, so we do that so my normalization here is that this will be where this is a traveling wave ok, so with this assumption let's see what we do this assumption suggests that perhaps we should make a change of variables you may think this is crazy but once I show you the formal proof you're going to see that that's exactly what the formal proof does but let's start with this and now let's compute what does the epsilon do so the epsilon solves this equation a little bit faster, just through the computation up to here without dividing by q prime you get what comes from the Laplacian in the time derivative then I use that f of q I use the fact that q double prime is equal to w w prime q I use this fact q double prime that q remember when c is zero the w solves that so that's the q I'm using and if you do here the calculation ok, let me do it slower so I compute the equation for z epsilon so I write down derivatives and I get 1 over epsilon q prime z epsilon t minus 1 over epsilon q prime Laplacian z epsilon minus 1 over epsilon q double prime epsilon square dz epsilon square plus 1 over epsilon w prime q and that should be zero on the other hand by the choice of q this term here is q double prime that's what q does so that eliminates w prime brings in the q double prime 1 over epsilon q dot z epsilon t minus 1 over epsilon q dot Laplacian z epsilon minus 1 over epsilon square so this is q double prime I bring it in and I get I simplify by the 1 over epsilon divide by q prime remember the travel wave is increasing so I can do that and I'm left with the equation z epsilon t minus Laplacian z epsilon minus q double prime over q prime evaluated at z epsilon over epsilon dz epsilon square minus 1 and I claim that this problem is much easier to understand that what was the original before so for the students this is a situation and it happens often in PDE where if you start with a linear problem or semi linear in this case in the form of w prime and you want to get some estimates and it turns out that occasionally it's better to make the problem nonlinear because once you make the problem nonlinear you break the symmetry of the linearity so you may be able to to get something more useful in another context when this thing comes up all the time in a very simple form but the problem is more general than that for example if someone gives you the Laplacian if you assume that u is positive often what you do is you make this change this score is known as the hope of cold transformation and that leads to that nonlinear PDE but because of the presence of this nonlinearity you can do more things not of course in terms of value but if you want to get estimates and so on you can do it this way and in some sense this is a transformation that we are doing here the only difference is the q double prime being a function that is like that you have different behavior for this factor at either end ok so this is perhaps one of the few proofs I'll give in full detail so clearly I want to send epsilon to zero and if I send there is an epsilon here because this was the only term with a 1 over epsilon squared so I want to let epsilon go to zero and the only thing I am going to get if I let epsilon go to zero is that if the z epsilon has a limit I will get that d z squared equals 1 multiply by epsilon let epsilon to zero this factor this coefficient goes to limits so let's say this goes to which are non zero you can compute it easily from here so if I let epsilon go to zero the best you hope to get is d z squared equals 1 let's assume that the z epsilon have a limit assume for now that the z epsilon has some limit and what do you expect at the limit to have you would expect that d z squared to be 1 that looks good because it looks like a distance and if you could solve that that goes to zero really fast you will get z t minus Laplacian z equals zero and somehow if you were on the front you had a front movie by me curvature of course this is a dream and we are not going to get it but let's see what we can get is the limit satisfies that so I will prove for you assuming there is a limit I don't want to get into these technicalities that the limit is a sub solution to the heat equation and it is a distance with a negative sign when it's negative satisfies that when it's positive and therefore it's motioned by main curvature so let's see how we do that don't want the quality at the limit I want inequality so somehow the only hope I have to get this inequality because I know equality will not hold is if I can create a sign here if this thing can we can find a sign for that quantity sign s i g n so the first claim I am making is that since d zero is less equal to one gradient of z is less equal to one if you say I know how to do it it's a parabolic PD there is a code for the kinds of things you can do for the equations to get Lipschitz bound and it's called Bernstein argument Bernstein type argument which suggests, which basically says write down this quantity you want to compute to estimate subtract something that depends on you write the equation satisfied by that and apply the maximum principle after you make a lot of changes here, there if you try to do this you are doomed and the reason is that once you hit this with derivatives you bring in more epsilon and you bring in cubic in some sense you create a cubic term in the gradient and that is a problem but there is a shortcut to that and this is I always do that when I teach course solutions it's a beautiful shortcut which says the following let's define the function and w is not good anymore psi of xt to be the soup of zepsilon xt minus x minus y I'm saying something stupid sorry for that zepsilon xt minus zepsilon yt minus x minus y look at that expression you double the variables we want to claim that we want to prove out of that this estimate it holds initially so the only thing I care is that look at the maximum of that for t positive because if the maximum happens zero then I have the answer so you look at that and then what happens at this maximum and then of course you have to assume there is a maximum so look at the maximum of that so two things will happen so let's call the maximum x bar, y bar either x bar is equal to y bar at the maximum that will imply that zepsilon xt is less equal zepsilon yt so that's one direction of that bound that's one direction of this bound or x bar is not equal to y bar there are two alternatives but if x bar is not equal to y bar remember the zepsilon has smooth functions gradient zepsilon would be equal to x bar y bar over x bar minus y bar because that's when x bar at the maximum minus y at the maximum that's a differentiable function if you're not at zero and since the zepsilon is a smooth function the gradient will be like that which of course is one and if you use the definition of the maximum that implies the answer because that implies that zepsilon xt is less equal zepsilon yt plus x minus y so that's the idea of the proof so you get a lepsich bound for free but it was important to have it initially that's why I said I have to prepare well my initial data once I have this bound once I observe that I have this bound now I'm home free why? because let's see what happens I let now epsilon to zero look at the set first z negative z remember is the limit of the zepsilons so look at the set where is negative if the limit is negative that means the zepsilon for small epsilon is negative somewhere which means that this thing is negative so when the limit is negative for small epsilon this quantity is negative because this quantity is positive I'm sorry what is this quantity? it's positive at minus infinity and it's multiplying by minus so negative negative positive again we know it's negative this is negative it's positive for small epsilon that means that actually I have this inequality because if this is positive this has to be negative and therefore letting epsilon to zero I get this inequality in the set where z is positive I still have this negative but now this is positive and so I get zepsilon t minus Laplacian zepsilon greater equal zero and after I let epsilon go to zero z t minus Laplacian z greater equal zero I have half of what I want but I did not use the big thing I had from the beginning that goes to zero so if you let epsilon also go to zero you get this goes to a positive so now we look what happens here you multiply by epsilon let epsilon go to zero this goes to negative constant back there and you get at the limit minus dz squared minus one less equal zero because dz squared is less equal one you get equality but with the minus which together with the fact that dz is less equal one will give us minus dz squared plus one equal zero and we get dz squared minus one greater equal zero and now we switch together with the dz less equal one gives us dz squared minus one equal zero this is in the z negative this is in the z positive and now I am done because you go back to the original chains and what we prove is that whenever the limit is positive this goes to one and whenever the limit is negative but when it's positive is inside the set that moves by mean curvature and it's negative outside the set that moves by mean curvature and that's in some sense the slickest proof you can have for that but in general when you have a very slick proof you don't hope this to work for everything in particular this proof will not work if I make the following simple chains so I allow next dependence on the potential or I put something anisotropic in front and the reason is that for this particular proof it's going to fail very quickly with when you write down the equation where it's going to fail it's going to fail because x by x the traveling wave now will depend on x the assumption is that for every x this is a cubic nonlinearity I mean it's a by step whatever is a double well potential with wells of the same depth so the q will depend on x this is a parameter so when I write when I make the change of variables q x z over epsilon I can do that nothing prevents me from making that change of variables once I start getting derivatives I'm going to start producing derivatives in q with x which will create additional terms here and if I create additional terms here so my trick is not going to work anymore are there any questions about this this proof I know it looks like magic and it's short but as I said it's life ok some history about this problem with proofs first proof for something like that was given by Kohn and Bronzer in the radial symmetry case then there were a result was a very nice result by Timotoni and Satsman up to the first time you had singularities and I mentioned Timotoni also because Timotoni had a house a little bit out of here and the last time I was interested he invited us and had a beautiful view and so on he died in an accident so years ago so Timotoni and Satsman and then Chen who is at pitch proof there are a lot of initials there also provide the same proof again up to the first time there was singularity the first proof past singularities by Evans and Sone and myself some time ago and then more work happened took place ok now let me go back I show you this proof, it looks like magic let me describe to you if you were an applied mathematician which in my department will be a sin but if you were an applied mathematician you would have figured out that motion by mean curvature plays a role in this problem so we forget this now nice rigorous proof and we go to the argument of this classification of mathematics into pure and applied is really awful because it suggests that people who do applied math are not clean or pure or whatever so together we decided that the better definition is classical versus modern mathematics you pick which one you prefer to be that and if you wanted to go even further you could say static versus dynamical mathematics anyway so so here is a formal which in the end turns out after a lot of a big cycle of work to be able to justify rigorously and as matter of fact there was a whole cycle which I will try to describe to you so the first proofs were very elegant and so on and then at the end with Bar we figured out a way to say that we don't need all these elegant proofs that we can do the problem in a different way faster ok so formal derivation we start with this problem we expect so W is let's say a potential double wells just to simplify things one minus one a general potential like that and you expect to have you want to see what happens as epsilon goes to zero so this is the same as saying I start with this reaction diffusion equation in study what happens in the solution as t goes to infinity of course this problem had been there even before Allen and Kahn Allen and Kahn made the connection with mean curvature but people like researchers like Aaronson and Weinberger had looked at the behavior as t goes to infinity of this problem and what you expect to see as t goes to infinity is the solution becoming either plus one or minus one and then of interest is to see what happens to the interface in between the plus one and the minus one which for t large whatever is a smooth transition so if you want to figure out if we want to figure out long time behavior it is often very convenient to introduce a scaling because the behavior of u as t goes to infinity is similar to the behavior of u epsilon as epsilon goes to zero and if you do that you have to figure out what do you space in principle if you do that x may go far away so what you are trying to do is you try to keep things in a compact set so you introduce the scale so what this does is it tells you what happens for long time and long time large space so this is long time large space behavior and basically what you expect is this to go to one and minus one long time it will go to the equilibria and for large space to the right and left you go there so it makes sense therefore to start with this problem and so what Keller Rubinstein I really recommend this paper it's a beautiful applied math paper with nothing rigorous but that's how you learn to do formal asymptotics if you don't know how to do and it takes a little bit of getting used to it so I don't mess up the notation what they make they make the conjecture that they can write a solution which will depend on epsilon they can have an asymptotic expansion for the solution and the asymptotic expansion will have a first term that will call it u0, x, tau bear with me I'll put all the variables in it and then there will be a next term in the expansion and of course other terms if you are brave so let me tell you what the things are the z here will be like some phi of x t over epsilon so this z there will be like the capital z epsilon I had before tau will be the time t over epsilon and eta will be the time epsilon t so we are looking something for a long scale we are putting in a scaling and this type of expansion allows for the possibility that we over scale the problem that actually what happens happens at slower t and that's the eta t there or it happens for longer t which is the t over epsilon writing it like that you capture both behaviors and the assumption is that in this phi in principle will depend on eta 2 and you make these ansatz the ansatz is that the solution should look like that and now we try to figure out what phi does so to do that we write down equations namely you do the calculation I did before with the z epsilon but now you do it for all this mess so if you do that you are going to get terms that have one over epsilon in front terms are going to have nothing in front in terms that are going to have epsilon in front and so on we care only about the terms that have one over epsilon in front and nothing so you get the following two equations u0 tau plus phi t u0z minus d phi square u0zz plus u0 equal 0 this is the term you are going to get the expansion for everything that has one over epsilon in front notice if I take a time derivative here I get a one over epsilon derivative with respect to tau I have the epsilon there and when I do this term I will get this same way I got a q double prime there so this is the thing that multiplies the one over epsilon and the thing that multiplies epsilon to the zero is u1 tau plus phi t u1z OK, where did this thing pop out? I do a formal expansion so if I plug that in I have w prime of u0 plus epsilon u1 and I will write this thing as w1 u0 plus epsilon w prime u0 u1 plus higher other terms there is a one over epsilon in front so when I find the terms that do not depend on epsilon I will get rid of that epsilon if you ask me why don't I go higher I know I don't need to go in more derivatives and everything we are assuming we can do these things so we got these two equations I didn't finish that this is going to be equal minus u0 t plus left plus and phi u0z plus 2 grad phi dot du u0z minus phi eta u0z OK, that's as general as it can go and here I didn't say at t equal zero is phi zero x over epsilon so that was our initialization does not depend on eta t and tau of course say it again, how do I get a second derivative this is dw alright so we do a formal analysis so basically what we need to do is we need to find u0 and u1 and we need to find phi and what is going to happen is phi will come up as what phi does is going to turn out to be a compatibility condition in order to be able to solve u0 and u1 so when you do something like that a formal expansion you are assuming that at some point you can take the variables to be independent of each other I know that's weird but that's in this is the applied aspect of what I'm doing is although everything here is mixed with each other I'm going to look at these things and I say once I have one variable the other variables are constant it's a completely formal argument and then you have an equation here this is a straight equation for u0 there is nothing else there so either you are able to solve it or not and then you have an equation for u1 which looks more or less the same up to here and there is a right hand side and anytime we learn in PDE to solve a problem with the right hand side we know that we can do it provided the right hand side is orthogonal to the kernel of the adjoint so we know therefore that in order to be able to find u prime this thing to be perpendicular to the kernel of the adjoint problem in that condition that compatibility condition will give us the equation for phi so you should always think of that because that's what's happening in these problems the equation satisfied by phi is a condition you need in order to be able to close this expansion that more or less says that I don't have to worry I can close the expansion, I'm done ok, now let's figure out how we do that and so now we make more assumptions we're going to make the assumption that, I mean somehow you know what you want to prove so we're going to assume u0 of zx tau t eta looks like a traveling wave t for tau large tau, which is this variable t over epsilon and of course we can think of tau large because it's fixed t, but it's like epsilon going to zero, so you make these answers now with these answers we try to simplify the equations now if you ask me how do you make these answers when you start something like that you have a little bit of an idea of what you're trying to get you know for these problems where you get the old results of Arons and Weinberger and others where you knew that there were traveling waves and the solutions are attracted the traveling waves attract the solutions and all this ok, so you make these answers and once you do that you can simplify the problem so the first equation if the u0 is like that the first equation becomes so now I replace the z derivative by derivatives q double prime becomes a prime so this is like that so now you see I start coming back to the thing I had, like the traveling wave so the q has to satisfy like that also the assumptions here will be that as z goes to plus minus infinity this should go to 1 or minus 1, the two equilibria or the w so I have this that's what the u0 has to be like the q which is like that now I do the same thing, I multiply by q prime integrate the xi let's say and I get that as before that phi t minus c should be equal to minus w plus 1 minus w minus 1 divide it this is the computation I did before so I multiply by q prime integrate and so on so I get this and now if you want to simplify things a little bit more if you make a change of variables because remember what is the q? what is the argument of the q? the argument of the q is phi minus ct over epsilon that's really the argument of q and so if you want to eliminate that, eliminate the epsilon from there so in principle the way it's written the q depends on epsilon of course it's not but there is an epsilon inside the q so to eliminate that you make an additional change of variables you write a z which is now z minus c tau over d phi that's just addressing now in this problem in that case you find that this problem becomes forget that so you make this change and if you plug it in and use the calculation you find that phi t minus over c over d phi is minus w plus minus w minus so if we write this as like that we keep it away ahead and so you see that if there is a non-trivial right hand side the phi solves this equation and the non-trivial right hand side has to do with whether this is zero or not if this is zero best have something here I think you get zero which is the case of the double well potential of the same depth then basically you get that phi is independent of t and the next dependence on phi is now on eta so that implies that phi is really a function of x and eta so to figure out what happens now we have to go into the next term so we go into the next term finally how the formal thing takes so much work but I decided to show you all the details so if this thing gives us zero information because if c is zero basically there is no c here this is gone so this is phi t minus c this is gone and what you find is the traveling wave I had before the one so now you make the same assumption in u1 bare with me the same assumption I made u0 now I'm making it for u1 and if I do that to simplify let me write directly what we get here we get minus d phi square p double prime plus w double prime q p is equal to a plus in phi plus 2 grad phi grad ok so now how does the Fragment alternative come in Fragment alternative is going to come down for this operator so the operator in which you are going to apply Fragment alternative is this one you have this problem let's linearize it it's easy to see by taking a derivative that q prime solves this problem and q prime goes here and the form of the w implies that the kernel of this linearized operator linearized operator is minus d phi square p double prime plus w prime of q prime the assumptions of w you imply that there is only one dimension in the kernel so in particular any solution and because the q prime does it the kernel of that operator the kernel the joint of L star is basically q prime which means which implies this mess here integrating from minus infinity to infinity Laplacian phi I'm messing up something here sorry for that just the correct right hand side it means that Laplacian phi minus phi eta dot square plus 2 q prime integrate from minus infinity to infinity grad phi grad q prime square d psi should be zero that's the orthogonality condition I mentioned that's Fragment alternative the q prime comes up to the factor here the big assumption since everything is formal is that the variables we can freeze any variable we want so we get that now this looks a little bit more weird but this because of the two this becomes minus q becomes d phi dot gradient with respect to x of q prime square now you say oh there is gradient we integrate in part remember the integration is in psi is not in x there is nothing to do with mean curvature the reason why this thing has to do with mean curvature is the following remember this change of variables we made that change of variables had the effect to tell us that the only x dependence on q was through that so q is really q of psi in x is really r psi that's what it is so now I can compute the derivative of this because I know exactly what is the x dependence and if I do it carefully using that in mind I'm going to get that q double prime square phi prime, phi eta minus delta phi minus grad phi grad phi over grad phi x i equals 0 so the only thing I'm doing is I'm using this form of q so when I compute q psi I bring in a gradient a derivative of 1 over d phi when I compute a derivative of 1 over d phi I'm going to create this term I mean this is d phi here but this is nothing else but the mean curvature so this is a constant this has to be 0 so that is 0 and that's exactly if you compute these terms this is exactly the motion by mean curvature equation so that's the connection in a very semi rigorous semi so just 5 minutes this approach that I presented I'm going to do next time next time I'm going to show you how to look at problems just to show you what else you need to do if there is additional dependence and the answer will be again now that we know how things work it will be that the u epsilon will look like some q let's make it simple x plus epsilon p distance over epsilon x the meaning of these things is the same and we're going to have to choose the p so that we can control the errors we make when we bring in x derivatives of q so it's the same argument there are however bunch of problems real interest where you also scale the x environment how do you get this equation let's go back for a minute I started with the idea that we obtain these problems by scaling a reaction diffusion equation so how do you get that you get this by scaling of course this problem but to get the next there you have to put the next there to get that you need to start with this problem which basically says that you are not far away from being homogeneous because as epsilon goes to zero you don't have it there what happens however if at that level I want to scale this problem now if I scale this problem this is going to bring me to that regime and now it's going to create a homogenization problem combined with front propagation but now I have to worry about two things I have to worry and let's assume that w is periodic in x so there are two mechanisms that are going to take place here one is again this front propagation an expansion in a direction orthogonal to the front and the other is the averaging that is going on from the x of ereptions so these two mechanisms will act together and combine to give you the answer the method I presented is going to collapse here because I cannot put anymore the chains q over distance epsilon plus there will be an x over epsilon here blah blah but there is an interaction with these two scales now so you cannot do it with an actual traveling wave so to solve this problem you have to use what is known as pulsating waves which I will explain next time but these pulsating waves will lead to expansions of the form and now there will be an additional gradient dependence because these pulsating waves are anisotropic so they depend on the direction you go and so if you try to make an expansion that will bring in a gradient of d which is the direction of the normal now forget any kind of previous argument if I try to write down an equation for this I get in trouble because I will have to compute the Laplacian of the gradient that will bring me a third derivative and that's a sin in this course the theory because there is nothing you can do with a third derivative if the distance is smooth of course that's not a problem it's multiplied by an epsilon so if I were in the smooth regime it's not a problem and on the other hand the whole argument, the whole theory is to go beyond the smooth regime and to do that there is an issue and so because of that I will introduce next time yet another way to describe the front where now instead of writing a PDE I'm going to think about the function of sets and they are instead of using smooth test functions to define viscosity solutions I'm going to use smooth surfaces evolving by the right velocity as test surfaces and once I do that everything will be reduced to dealing with smooth things and therefore I will not have a problem ok so I'll do that whatever tomorrow thank you and sorry for being late you can blame the people who keep the door alright thank you very much