 It's a pleasure for me to introduce Professor Carlo Mantegazza from the University of Naples. This is the first lecture you will talk about. He's an expert on the motion of partitions, so he will speak about the net motion of net rules by Min Corbucci. Thank you. Thanks. Thanks to the ICDP, to Giovanni Francesco Carlo and Claudio for their kind invitation. So as my title says, I'm going to discuss the motion of networks by curvature in the plane. And actually, it's a long project that started 15 years ago, more or less, with Matteo Novaga and Vincenzo Tortorelli when I was in Pisa. Then we had a period that simply we weren't able to find out new results. And we started with Hannibal Emani, my ex-PhD student, and Matteo Novaga again in 2010. And then, more recently, with Matteo Novaga and Alessandro Pluda, who is here, ex-PhD student of Matteo, now in Braggsburg. And Feli Schulze, which is one of the competitor groups, actually, well, composed by Feli Schulze, Andre Neves and Tommy Lvanen. And very recently, a couple of colleagues in Naples, Petro Baldi and Manuele Haus, they started to get interested in giving contribution to the subject, in particular, as I will discuss next lecture, to the classification of similar solutions to network flow. So what is all this project about? Well, you have something like this, a network, network curves in the plane. And, well, you want to let him evolve by mean curvature. So the motivation, well, not my mean curvature. Here, there is only one curvature, by curvature. And which means that you assume that every point of this curve, that every time, it moves in normal direction by the curvature of the curve passing by there. What are the motivations? Well, actually, there is one easy physical or actually engineering motivation, which is some kind of toy model, because it is one-dimensional and one-co-dimensional. For evolution, for the dynamics of an interface, in this case of a planar system, you can imagine that this region that contains immiscible fluids, for instance, water, oil, fuel, stuff like that, they cannot. And these curves are simply the interfaces between the phases of your system. And in case the energy of your system, which is maybe not too realistic in this situation, but it's the easiest case, then you can try to generalize to more complicated ones, it's simply given by the total length of the sum of all the length of the interfaces around. So you have only this energy, you have your system which is perturbed by the equilibrium, you let him evolve, and you want to understand what's going on. Actually, I'm not cheating, I'm not an engineer, I'm not a physicist, so I'm more interested in the mathematical side, the model part come from outside, so we only got the model of this problem that came from several situations, like for instance also evolution of grime boundaries in polycrystalline materials. So we concentrated more on the mathematical point of view. And actually, me and Matteo at the beginning, we were also very interested in the fact that after a lot of work of mean collature flow in the smooth situation for hyper surfaces, mostly done by Gerard Wiesken and Richard Hamilton, and the methods were actually differential geometry of curved surfaces and PDEs, in particular maximum principle, but at that point, there was a lot of work by analysts in order to generalize the motion by mean curvature to less smooth sets. For instance, at the end, you have generalized definition of motion by mean curvature, even for merely closed set in the Euclidean space. For instance, level sets formulation. But there are other generalizations, for instance with very faults or other kind of generalization of smooth surfaces, which are actually singular. So we're a little bit interested in seeing how much you can push the standard classic parametric approach developed mostly by Gerard Wiesken, in particular, to a situation which is actually singular. These are not curves. Around triple junction here, or multiple junction here, this network only has triple junction. And there is a reason. But in general, you can think of a network, a graph, with multiple junctions, with four or five points or whatever. And they are really singular sets. But actually, if you consider all the curves to be smooth, so the singularity is concentrated in a way in the multi junctions. So you are in co-dimension one, one dimension, and the energy of your system in a way, since the motion by curvature is in a way the gradient flow of the energy associated to the sum of a total length around, or if you want for a singular set to the H1 measure, the house of one dimensional measure of your set. So you want to, in a way, take the gradient flow by this energy. In a way, this is the simplest situation and the simplest geometric object, singular, but not so singular, the simplest singular object. So you have hope that if you, mostly of the techniques that you developed for the smooth case, could be even used also in this case. Actually, I have to deal with these guys, the triple junction. I think you can guess that if I use the techniques, standard techniques, in the interior of the curves, I can get results. Instead here, you have to develop something new. And actually, at some point, you have really to develop something new. And differential geometry and maximum principle in particular is not really sufficient to deal with the problems given by the triple junction. So we needed to use some extra variational method, which are, in a way, taken by the variational method we use in the fully singular situation. So trying not to use their full strength, but just to deal with these very special singularities. So in a way, there are not so many singularities in the network curves, less than in the general case. OK, so before going on, there are some simulations of the motion of this guy. One that I'm going to show you was done by Ken Bracke. And as I said, I'm not a physicist, not an engineer, but what I was told is that these simulations are very, very close to what you observe in real experiments. Several of these, you can find a lot of information about these, for instance, on the west side of Kinderlehrer at the Carnegie Mellon that has a full lab, which is interested in studying real experiments, simulation, and theoretical conclusion on this kind of motion of interfaces in general. So I'll show you this. This is a quite complicated network that evolves by this rule that I told you, so motion by curvature. You see that several things happen here. Regions are vanishing. New junctions are born after a region is vanishing. In particular, what you can observe is that apparently larger regions eat smaller regions. To be more precise, a region with more than six edges increase their area. Less than six edges decrease their area. One, maybe the easiest observation. Moreover, the exception of two peculiar phenomena. First, the easiest one. One region is disappearing. Vanishing and disappearing at all. And there is another phenomenon that without vanishing of a region, that one single curve is vanishing. Like I'm betting, could happen to this curve here now. So the region is not disappearing, but simple one curve is disappearing. So the two regions bordering that curve lose one edge. And actually, after this, there is an opening to other two triple junctions. And actually, if you look at the whole simulation, with the exception of these times where one region is collapsing or one curve is disappearing, in all the other moments, you have only triple junction around. So four or more junctions, they don't live. They can appear only at a single time when there is some kind of change of structure because of these two possible phenomena. And all the triple junctions, if you look, have always the same free angles of 120 degrees between the three curves that they are conquering at the triple junction. So you have these special times when something happens, when you can see four points, five points, whatever, and a region has collapsed or a curve has disappeared. And in all the other times, you only have triple junction with 120 degrees. Now, I hope that it's going to happen what I was saying, that this curve is disappearing. Here it is. There was no collapse of region. One curve is disappearing and an opening in the other direction with the other two triple junctions going more or less orthogonally in the other direction. And let's see the conclusion and then it stops. This simulation was, in a way, periodic. You can think you were in a torus. So it was the borders. Maybe it's not so easy to see it, but the borders are the same, so there are no boundary points. Actually, what I want to discuss in my lectures, a situation like this, I want to discuss the motion of a network inside always some domain which will be a convex set. This is only technical, but it's useful. Well, I assume, for instance, this guy, a very simple network that we call a triode, actually, where the points on the boundary of the convex set are fixed and the network moves inside. This kind of Dirichlet kind of problem. You can also change your problem. And instead of asking that these end points are fixed, that the networks arrive on the boundary, for instance, with the right angle, actually. Kind of Neumann boundary problem. Or that I'm going to see also for what we call open networks. Networks like they don't have the open set where we are discussing it. The wall are two. And we have some lines, asymptotic lines, where the network is asymptotic at infinity, becoming straight and straight asymptotically infinity close to these straight lines at infinity. And inside your network does whatever, things like this. So all I want to discuss will apply also to this situation, not only to this, but also to more complicated situation with fixed end points and open networks where you assume that the network is asymptotic to some fixed alphalines at infinity. And for simplicity, we're not going to discuss in general the behavior of the network at the boundary point. In this case, I will have to discuss the behavior of network at infinity. I won't discuss the details of this or of this in a way because reflecting the network to this point, taking your network to a reflection with respect to this point, this point is no more a boundary point. So you can deal with the point of the boundary in a way after this trick like they are the inside point of a double network or reflection of the union of these with the reflected network. So I will discuss what happens inside here. And I will forget several moments of the behavior and in some estimates, the contribution of the boundary point, I ask you to believe me that they are not relevant, and more or less, if you understand what's happening in the interior, you are able to deal with what is happening at the boundary points. Moreover, if you really don't want to deal too much with boundary points, you can also assume, like in the simulation, that you live on a torus so that your network, for instance, is living on a flat torus so there are no boundary points around, and you have kind of networks like that that they close. There are also networks without boundary points in the plane, for instance, this guy, without one of the free network, compact networks without boundary points, where all this theory can be applied. And moreover, all the theory can be extended to motion of networks on a compact surface, or on surfaces with even only complete that have a good behavior at infinity. Bounded geometry, for instance. OK, let me get back to the observation that I put more or less, the observation that I already told you, that larger regions, more than six edges, grows in the area, and less than six edges, the area decreases. With six edges, the area is constant in the flow. In a way, that means that the only region with less than six edges can collapse. Before collapsing, one region must lose, with the other phenomenon that I mentioned, one or more of these edges. In order, the edges becomes less than six. And as I said, with the exception of the times when this structural change, actually, apparently, all the flow is smooth. The curves are smooth. And you see only triple junction around. And the concoring curves at every triple junction is only 120 degrees. Without vanishing of a region, that instead can produce several situations when you get a multipoint. If you have only one curve collapsing, well, there is only one situation. You get a four point. And actually, if we look again at the movie, what we find is that there is always a situation like this. A four point with angles, not general angles, but this coming, this is actually one convergence here around. You have this angle of 120 because of this. And what you get here is a four point with angles of 120, 120, 60, and 60. Not the general. Always you observe this kind. And then immediately after, there is an opening in the other direction. And notice that this angle of 60 here becomes, in a way, an angle of 120. There is a discontinuity in the angles, which is not here. Here the angles are continuous, in a way. And here there is a discontinuity in the angles. The angles of 60 become similar in angles of 120. Here there is a situation where this guy here, which we call the theta guy, evolves with a collapsing of this in inner curve here. You get this guy, again, 120, 120, and 60. And then it opens like this kind of eyeglasses guy. OK, on this observation, you look at several simulations. You see that they are always there. Then apparently it's very, in my opinion, quite complicated. And you need a lot of technology from analysis and geometry to prove this observation actually And let me say, I'll tell you in advance, we still don't have a full proof. We have an open conjecture that I will mention tomorrow. In order to, well, one main open conjecture and another secondary conjecture that we believe, but at the moment we are not able to prove, in order to make rigorous this observation in the model of motion by courage. OK, since the networks, we say that, OK, if this observation here in particular is true, so this guy with only triple junction and 110 degrees are very important, are the basis of the structure of the problem. And so we started investigating how they behave during the flow. And we call them regular. So regular is a network such that every junction is a triple junction and the angles are always 120 degrees, the regular ones. That hopefully, at the end, it should turn out that with exception of this great set of times, all the networks of where there is a change of structure, all the next ones of the flow are done like this one, are regular. To be more precise, we are considering embedded network. So the curves cannot intersect. They are really, in a way, interfaces between the regions, the inner regions. You cannot consider crossing network. Moreover, for several technical, well, they can only intersect at the end of the curves. And on a point on the boundary, you can only have one curves right there, not another curve. This is more a technical assumption, but it's very good to have it. And this is the concurrency condition, which is something sometimes called herring condition, that you only have a triple junction and the angles between the three concurring curves, which are C2 curves till the boundary, till the end. They form angles 120 degrees. And this can be expressed analytically, saying that the sum of the inner unit normal to the free curve is equal to 0. Small or less, this is equivalent to say that the free angles are 120 degrees. There is only the possibility that they have 120 degrees. Example, easy example. The one is the trio that I depicted before. It's simply a regular trio, actually. When I say regular curves, I think at least C2, but very often it's an infinity curve, actually. And the three embedded curves, they don't intersect, they have three fixed points on the boundary, and they meet at a single triple junction with 120 degrees. And this other guy, another simple guy, which we call a spoon, which Braque called it a spoon, that it's given by two curves, one curve connecting a single fixed point on the boundary, and a curve which close to itself. And again, only one triple junction and 120 degrees there. So that's a regular spoon. Actually, if you want to see another example of these guys, more or less these are all the possible topologies. We have this guy, theta, with no points on the boundary and eyeglasses in two different versions. And this guy, they are the only two with two points on the boundaries. Here only two triple junctions around. These ones were the only guys with one triple junction around, the easiest. These guys are the only ones with the most two triple junctions around. So this island, and this tree, because it's the only one which doesn't have any loop inside. What does it mean moving by curvature? Trying to put things analytically. Well, as I said, well, you have a regular network, so this collection of curves, parametrizing time, because you want this to be your flow, in R2, with several conditions that you have kind of, by side you need to have your tables where you know where the curve intersect forming the structure of your network. But when they move, what you want is exactly that the time derivative of the curve, which means the velocity of the motion at every point, is in normal direction. So the normal velocity of your curve is equal to the curvature k is the curvature of the curve gamma passing by that point. If you write the curvature by elementary differential geometry curves in the plane, your curvature is actually given by taking the second derivative divided by the square. This curve are not parametrizing arc length. They're simply map giving your curve. You only want that they are regular, which means that the space derivative is different by zero. So you can divide here. So you take this guy here, the second derivative divided by the square of the first derivative, and project on the normal component, which is exactly what I'm doing here. This nu is the normal to the curve at the point x and t. So you take this guy, project it, and this is the projection operator. You take this, and this is the law telling you that all your curves are moving at every point, also the boundary points, by curvature. So now you want to find a solution. So you have an initial network, your initial set of curves with all their relation, and now you want to look if there is an extension in time of this set of curves satisfying this equation. More precise, you want something which is actually situ in space in order to speak easily, classically, of curvature, and see one in time in order to write down this time derivative. And moreover, you still want that every time your curve are regular, so the x derivative is different by zero. This is you want that you're like in the simulation, the herring condition, the fact that they are only triple junctions, and the angles between the three curves are 120 degrees, which is analytically expressed by this equation here. Still holds, still there in the evolution. And finally, in a way, you can rewrite any solution of this. This says that the normal component of time derivative of gamma must be equal to the curvature, which means that the full derivative of gamma must be equal to the curvature times the normal, plus some function times the normal. If you're able to find out a gamma such that time zero is equal to your initial network, and at every, for some short time, for some time, at time positive, it satisfies this free condition. Well, you say that you find, finally, you find out actually a solution of this, so a motion by curvature of your initial network for an initial regular network. Because if you want something which is continuous up to the time zero, this condition must hold also at time zero. So we are now concentrating in finding out the flow for short time, for a moment, at least, for an initial regular network. Only triple junction and herring condition of 120 degrees satisfied. For the other, non-regular network, which means multiple junction or triple junction with not the correct angles, we will deal later on. Actually, after you formulate this problem, the assumption that you have all this regularity till the boundary of your curves has immediately some consequences. Because actually, let me fix some notation. I will call tau the tangent to our curves, which means regama x, that they came right because this guy is different by 0, so the curves are regular. And we'll always consider the normal being a rotation of the tangent where r is the rotation of 90 degrees in the plane, just to set up the notation. So immediately, suppose that two curves gamma i and gamma j arrives at the same point, and they flow by. And I have a solution of my flow by curvature. Well, suppose that this point for the curves gamma i and gamma j are exactly gamma i of 0 t, and must be equal to gamma j of 0 t. Since you don't want the two curves to go in a different direction, this triple junction must stay there. So I have this. I can simply differentiate. So if I differentiate, this must hold for every time. So what you get here, if the last equation is satisfied, you must see that at the triple junction, k i, ni i plus lambda i tau i must be equal to k a j, ni j plus lambda j tau j. This for every couple of curves conquering at the triple junction. So there is also the third curve. So this also must be equal to k, sorry. I changed this. That's gamma m, k m, now m plus lambda m. And this must hold along your flow. Now, if you do some linear algebra, knowing that these are 120, always, so the sum of the tangent and the sum of the normal also, sum add to 0. You do some linear algebra using this. What you find out that you must have the sum of k 1, 2, 3. Let me write this in this. k i plus k j plus k m equals to 0. And also the tangential part of the velocities must add to 0. So you already have condition on your flow. And actually, you also find out, again, linear algebra, a good way to see these things is to call, OK, now forget for a moment about i, j, and m. Let's call it gamma 1, 2, 3. This must hold for every triple junction. At every triple junction, this must happen. And call bk is equal to the vector k i 1, k 2, k 3. And big lambda is equal to lambda 1, lambda 2, lambda 3. And, OK, can be seen. You can rewrite this to a equation like the bk vector product, the vector 1, 1, 1 equals to 0, which is also equal to big lambda 1, 1, 1 equals to 0. Oh, sorry, are perpendicular to these guys. And moreover, you can also do some other linear algebra that if you take the sum of lambda i, k i, this is equal to 0, which means, again, that big k is perpendicular to big lambda. So they form an orthonormal frame, an orthogonal frame. And to be more precise, you can also get out from this that k is equal to lambda, which 1, 1, 1 over square root of 3, which also means, by taking the modulus that the modulus of big k is equal to the modulus of big lambda. Which actually says that if you have a control on the free curvatures at the triple junction, which is the normal component of the velocity at the triple junction, you also have a control on the tangent velocity at the triple junction. So you can only, if you want to estimate it on the tangent velocity, the tangent velocity there that apparently doesn't get in the equation trying to solve. So apparently, you don't have too much control. This is partially true, because at least at the triple junction, the tangent velocity cannot be whatever she wants. It's related by this equation to the normal velocity, which is related to the curvature at the triple junction. OK, these are in a way. Sorry, I don't hear. Yeah, exactly, the last one. On the junction. No, this is everywhere. This one is everywhere, also on the junction. This is another way to write down the condition over there. I'm sorry for the bad notation. This is the set of the, well, the tau i, it's the sum of the three tangents getting to the triple junction. That o is the triple junction. That the free tangent must be 0, which means 120 degrees at every triple junction. And moreover, this is another way to rewrite the equation over there. I want that the time derivative of gamma, the normal part is equal to the curvature, then there is some other lambda tangential part that apparently I don't know at the moment. But at the triple junction cannot be whatever she wants because of this linear algebra computation. OK, then, so at the end we have, so I found out something taking the first condition, the structural condition that gamma i, I use this, gamma i of 0t is equal to gamma j of 0t or concurring curves. I also have this, that the sum of tau i in 0t is equal to 0. So I found out information taking the time derivative of this equation, structural equation here. Now we can find out other information taking the time derivative of this equation here. But to do that, I need to take the time derivative of the tangent vector. So I need some computation. So I have to find out the evolution equation for the relevant quantities on the curves moving by curvature. So we start with this, that time derivative of gamma is equal to k times the normal plus this lambda tangent. And let me put this, the s is the arc length along the curve. And the derivative in arc length is exactly the derivative in the parametrization divided by delta x of gamma. And there is, well, it's obvious by Schwartz theorem that if I can take the x and t derivative, I can switch them. But this instead, I consider the arc length derivative. I can no more switch them. Because the arc length derivative, the arc length is not independent of time. So I need the switching formula when I invert t and s derivative, which is useful in the following computation. And the formula is this. Let me write it like this, the time derivative. s derivative is equal to the s derivative. Time derivative plus k squared minus s derivative of lambda s derivative. You can get it. Well, I will let you the survey that we almost finished with all the details of the computation. But this is anyway a nice exercise for elementary differential geometry. And just let me show you an example how to use this in order to compute, for instance, the time derivative of a tangent vector doing the evolution. Then you can find out all the other evolution equations for the relevant quantities. So let's try to compute what is time derivative of the tangent. Well, what is the tangent? Well, the tangent is simply the s derivative of a curve since it is in our graph. So this is equal to time derivative of s derivative of gamma. I want to switch using this. So this is equal to s derivative time derivative of gamma plus k squared minus lambda s of s derivative of gamma, which is actually what is this? Well, I know this. It's given by the evolution equation. So this is equal to s derivative of the velocity of the motion, which is k times the normal plus lambda times the tangent plus this guy there, k squared lambda s. So now you have to distribute this derivative here. The first gives you ks times the normal. Then I have plus k times the s derivative of the normal. Well, elementary differential geometry tells you that the s derivative of the tangent is equal to the curvature times the normal. What is the s derivative of the normal? Well, the normal is the rotation of the tangent. Rotation is invariant, independent of s. So this is what gets out. So this is equal to the rotation of the s derivative of the tangent, which is k times the normal. And the normal is the rotation of the tangent of 90 degrees in the anti-cockwise. So if you take it two times, you simply put a minus in front of your vector. So this is equal to rotation of rotation. k is a constant. I can take it out of the tangent, which is simply minus curvature times the tangent. So here, the s derivative of the normal is minus k times the tangent. So minus k squared, the other k comes from here, tangent. Now I want to do the same here. I take s derivative of lambda plus lambda s tangent and s derivative of the tangent plus lambda s derivative of the tangent is written here. It's k times the normal. And then you have to add this. But you see, which is not unexpected, because if you derive a unit tangent vector, what you find out is another tangent vector, which is orthogonal. So there is a cancellation of the tangent pieces, one and two, that appear right here. And the conclusion is that what you get is ks plus lambda k times the normal. A little bit long, but straightforward. So if now I use this here, I take this now and I take the time derivative of this. It still must be 0. When I take the time derivative, I now can use this that I just computed. And what I've concluded is that 0 is equal to the sum of ks i plus lambda i k i. So you find out a new relation that must hold at every triple junction. So we have a 0 level relation, no derivative, 0 level. Tangent is one derivative, one level derivative, which actually we impose in our problem. And then what you follow is you get the second, what I wrote before, the sum of the k i plus b 0. k is s derivative of gamma. So we have a second level derivative, a second level condition, third level condition, derivative of the curvature. So three derivatives of your parameterization. And so on. Because if now I take this that must hold every time at every triple junction, I take time derivative again. OK, now I have to compute what is the time derivative evolution of the curvature. But I guess you can believe me that what you find out is ks s plus something. So you get another condition here of fourth order. And so on, taking time derivative repeatedly, you find out a lot of conditions. These conditions are called complementary conditions of your problem. And we'll get on that in a moment. But just because it's very important for the future, I'm not doing the computation. It's another very nice exercise along the same lines. If we, since the curvature is the most important object around here, the evolution of a curvature will satisfy this. Time derivative of the curvature is equal to ks s plus ks lambda plus k to the fourth power. This will be very important. And again, you do along the same lines, use the fact that the curvature is to space derivative in our clamp of the parameterization. OK. So if you have a smooth flow, sin infinity, which means that you can do all these derivatives. I didn't get into details if you can do or not. I simply took the derivative. Suppose you have a sin infinity flow, till the border to the triple junction. You can take whatever derivatives you want. You have a lot of conditions that must be satisfied at the triple junction for your flow. Infinity. For every order, you have one. OK. This is a priori computation. Now I'm going to give some hints on how to find a solution to this problem. So at least for regular initial, regular networks, how to find out the short time existence of your flow. If you look at this problem here, actually it's a parabolic problem, quasi-linear, since there is first derivative around. But actually if you take linearization and look at the symbols, because of the fact that you are killing one part of this operator here, you are taking this guy here and projecting only on the normal component, you will see some zeros in your symbols. So it's a degenerate quasi-linear parabolic system, actually. That actually doesn't fit directly in the standard theories. But the fact that you are really interested only in the normal component, but that you have some freedom in choosing the tangential component and solve the problem with the special tangential component that you set as you want, and that if you are able to solve that, well, you are able to solve this. Because the normal part is the same. So what is the right choice of tangential component? Well, I take all this, the wall operator, without the projection. Well, if you found the solution of this, simply project and you get the solution of the other ones, of the original, of the one that you are interested in. And this one is no more degenerate. You are not throwing anything away. It fits. It's a quasi-linear parabolic system, non-degenerate. But then you have to satisfy the condition at the triple junction. I'm forgetting to speak about the boundary point. Forget about that. You can also have to satisfy the condition there. There are similar conditions to this one at the boundary point. But forget about that for a moment. Let's concentrate on all these family of condition coming. It's gone. OK, all these conditions that I wrote there that you can imagine you can obtain going on differentiating in time. We call spatial curvature flows the ones that you can obtain solving this problem. OK, it's not going. Solving that problem over there that produce one flow by curvature, like in the original problem. Why, when you are able to find out a solution of this, OK, we did with Matteo Novak at the very beginning for the easy situation. Because, well, everything fits in a theory of Solonnikov, quite complicated ones, that if the operator is non-degenerate, like I said before. And now all the point is about the boundary condition. And the guy there is too compatible. What does it mean too compatible? It means that the condition of order 2 coming out by this computation is satisfied. Then you have a solution in the standard parabolic spaces of all the spaces C2 plus alpha, 1 plus alpha, C2 plus alpha space, 1 plus alpha in time. If the condition order 2, that means that if our initial guy there, that triad, this super easy situation, is made by C2 plus alpha curves joining 120 degrees. And the sum of the free curvature there, this is equivalent, sorry, the exact two compatibility is this one for the initial triad. Then you find out the solution if it is too compatible. What is actually true is that if you have this sum of the free curvature equal to 0, then you can choose a special reparametization of your initial triad such that this condition here is actually satisfied. So you have a very special class of triad that you are able to let evolve for a short time in this natural class over there. In fact, it's easy to see that this condition implies that the sum of the free curvature must be 0. Instead, the visa versa is that if you have this, there is always some reparametization, putting you in the situation that your sigma there, the reparametized sigma there, satisfy the condition there. So what you do, you have this condition, this geometric condition, reparametize in order that your triad become too compatible, let it evolve by means of this theorem, and then reparametize back. Because being a geometric problem, it's invariant by reparametization. If you reparametize, let the triad evolve, and reparametize back, you have the same curvature flow of the un-reparametized triad lecture. So you can always use it. Since because it's a geometric problem, the curvature is invariant by reparametization. Not the tangential component, only the curvature. But the tangential part of the velocity is not relevant here, because at the end we are looking for a solution which satisfies that equation there that involves only the normal component. OK, then it's not so difficult to generalize the result to a more general triad, a more general network. Again, too compatible. And actually, also, if you start with something which is smooth, and smooth means not only that all the curves are infinity, that all the condition which is infinity compatible in a way, then you find out a smooth solution of your flow up 2 times 0. This is all contained, fortunately, in something which was already there in the Solonicov theory, which is quite involved, actually. So actually, we more or less used the theorem as a black box, only checking that all the hypotheses were satisfied, in particular the compatibility condition in order to use the theorem. So this was more or less the line. The very special case, the idea, come back to Bronsa the right here, they did for a triad with some extra geometric assumption. We, in a way, take away the geometric assumption. And we did the general theorem for a regular network. And when you find out the special flow, as we said before, since the special flow is different by the curvature flow only by the tangent part, you actually has produced a general flow like that, which is a solution of the initial problem. Actually, if you find a regular to compatible initial network, you have a curvature flow, actually. Not only a special, but special is a technique in order to get real solution of the curvature flow. What about uniqueness? Well, actually, you see that there are a lot of choice here and now that I choose the way to produce a solution. I reparametrize, and then I get back. So you would like to have at least some kind of geometrically unique curvature flow, which means uniqueness up to reparametrization. So that if you look at your network as a wall, you don't see the difference between the network and this reparametrization. If you look as a max, you see that there are different objects. But if you look as a set, you don't see the difference. So you want a uniqueness looking at the set. Unfortunately, only some special situation. And the point is that you have geometrically uniqueness if you start with a C2 alpha in this class. The natural class where you look for uniqueness is C2 in spaces one in time, as it is natural. We are able, and it's an open problem, to show unique geometric uniqueness only if you restrict the flow in older spaces. In C2 plus alpha in space, C1 plus alpha over 2 in time, but not in the natural space. And the problem is related to the lack of maximum principle. Because since you have this boundary point, you cannot use an easily maximum principle. But you can use maximum principle only when your maximum is an interior of your domain. So only in this case, if the maximum of the quantity you are interested in is in the interior of the curves. If it is on the triple junction, you cannot use maximum principle. Maximum principle is the main tool in order to prove uniqueness for mean curvature flow in the smooth situation. Just to conclude, by approximation, after having this theorem for very special network, it's actually possible to let start an initial C2 regular network, only C2, without any compatibility condition, losing the continuity, actually, of the curvature function. Because this theorem tells you that immediately the flow is smooth at every positive time. Immediately the compatibility conditions are all satisfied. Immediately means for every positive time. So if, in particular, the one of order 2, if the curvature is continuous, means that also must be satisfied at this condition here, must be satisfied also at time 0. But if you don't start with something which is too compatible, like you can do it, because here you only ask your initial network to be C2, this cannot be true. So you lose continuity of the curvature, but still you have at least C1 convergence to the initial network. Find out the curvature flow which is regular and smooth for every positive time. Again, the geometric uniqueness, it's absolutely an open problem. But in general, as I said, you would like to have uniqueness in the natural space, and it's impossible at the moment. It's impossible because apparently there is only the line of maximum prints where there cannot be used. OK. Just a minute, sorry. Late. All this work for regular guys. Super smooth, super satisfying, satisfying all the compatibility conditions, or two compatibility conditions, or simply C2. But only triple junction, only 120 degrees. But instead, you want something to let evolve any network. Even if you start with something at the very beginning irregular, when there is a collapse, which is what we are going to investigate in the next lecture, well, you see, we had four points, or multi points. And you want to go on, to move on, after the collapse, like in the simulation. So even if you avoid it at the very beginning, you have really to deal with general networks, with general structure, not only irregular. For instance, you have this, you want to restart. So fortunately, there was a quite weak definition of curvature flow, which is was the, this theorem was developed by Ilman and Neville Schultz, that we're able to prove that there is a bracket flow, which is kind of a variational, geometric, major theory notion of flow by curvature, which is very special, which gives you an evolution of your network, which is a very special bracket flow. In particular, it is smooth, all the compatibility, all corvus infinity, irregular, 120 degrees, and only triple junction, for every positive time, which is exactly what we see in the simulation. So the beginning, the first, your network can be very messy, having a lot of multiple junction or bad angles. But this theorem tells you that it's quite weak definition, which is anyway immediately super regular for every positive time. So you can restart your flow starting from things like that, from with a four point, for instance, and having a curvature flow. The next lecture will be devoted to understanding this part, the collapse part that we saw in the simulation, and trying to describe more accurately as possible what happening when one curve collapse or one region collapse. With the idea that with this theorem, you can move on. We want to say that actually every collapse carries us in the hypothesis of this theorem, so it can be applied to continue the flow after the singularity.