 We're going to talk about, for me, curvature flow minimal surfaces, expanders, and shrinkers. So these are self-similar solutions. They are of the type that we looked at yesterday. And again, the point is to create examples that illustrate the theory or that show that certain things happen. And so the particular kinds that I'm going to look at today will be hyper surfaces in Rn, where Rn will be Rp cross Rq. And I'm going to look at surfaces that are rotationally symmetric, that are invariant with respect to rotations in Rp and rotations in Rq, which reduces the problem of finding minimal surfaces, expanders, or shrinkers, again to ordinary differential equations. So in this case, we have two dimensions to play with. And the class of examples is richer than what you have in the case that I looked at yesterday. So to begin with, all the surfaces I'll be looking at are invariant in the rotation. So these are the rotation groups in Rp and in Rq. So simply they look like this. You take a curve in the first quadrant. This will be x-axis, y-axis. So given such a curve C, I will consider a surface M that consists of all points x, y in Rp cross Rq, such that if you take the length of x and the length of y, that's a point in the first quadrant, two non-negative numbers. This has to belong to the curve C. And the idea is that you take this curve C and you swing it around this axis in Q dimensions. So the group SOQ rotates around this axis. And you also swing it around this axis. So this is where you swing it around this axis in P dimensions. And what you get is a hyper-surface in Rp cross Rq that is completely determined by this curve. So now if you consider mean curvature flow, then you consider the evolving surfaces. So if you have a family of these things, you can let the surface evolve by letting the curve evolve. And now you can ask, what condition do you have to impose on C for this to become mean curvature flow? For M to undergo mean curvature flow, and I'll write down the equation. You can do this in more generality. You don't have to look at. So I drew a graph of a function, which is what I'll be talking about most of the time. But these things, they can turn around like this. You're going to let them self-intersect, and they can do all kinds of things. So I'll stick to graphs most of the time. One condition is, so what are these curves? What does the curve C have to do? If you want this M to be a smooth surface, what can C do at the axes? So the condition is very simple. Your curve C has to hit the axis perpendicularly. So if this angle is 90 degrees, then the corresponding surface that you get by rotating it around the axis will be at least C1. And if this thing is, so if the curve C is C2 and hits the axis perpendicularly, then M will be a C2 surface. And in general, if you can extend it as a smooth even function, then the surface M will be a smooth hyper surface. So those are the boundary conditions that you have to deal with. And the same one appears here. So in particular, if you take a circle, the corresponding, so this is meant to be a circle, quarter circle, the corresponding surface will be a sphere. So mean curvature flow is equivalent to a PDE for the curve C. So if we assume that C is a graph, which this thing isn't, then mean curvature flow for the hyper surface M is equivalent to normal velocity equals the mean curvature. And that is equivalent to, and if you work out what these things are, sort of like what I did yesterday for curve shortening, you get a PDE. And the main terms of these PDEs that we get are always, the first term is pretty much always the same. It's this thing, which comes from the curvature of this curve, seen as a plane curve. And then you get two extra terms, which come from, which are the curvatures that are produced by rotating this thing around each of the two axes. So this is almost the same equation as yesterday. The only term that is new is this one in the middle, which so you could think of what we did yesterday as a special case of this, where P was equal to 1. If P is equal to 1, then you have the real line cross Rq, so that's Rq plus 1. You get an ordinary rotationally symmetric surface. If you set P equal to 1, this term disappears, and you get yesterday's equation. So we're going to look at this PDE. Boundary conditions would be the graph has to hit this axis perpendicularly, because if you do that, then the thing extends as a smooth hyper surface. And then we want you to stay away from zero, or if you, well, of course you could say, well, I can evolve a sphere by mean curvature flow, so I should be able to evolve this thing. And if you want to do that, then it is not good to write the equation in this way that you should choose different coordinates. You either regard part of the solution where y is a function of x, and then you regard the other part of the solution where x is a function of y. Very long calculus exercise. If you switch coordinates, just x and y coordinates, you make x the dependent variable, the dependent variable, and y the independent variable, then this is equivalent with a PDE for v, and it's easy to write down that PDE, because this is a geometric equation. So this setup should hold no matter which P and Q you take. And if you switch x and y around, what you're really doing is it's equivalent to switching P and Q around. So it should be this equation. If you ever have a calculus class where you think they should really practice the chain rule a lot, you ask them to derive that equation from this after you have inverted this function. If you're not worried about your teaching evaluations. OK, so in any case, that's a fun fact. These two things are equivalent. Geometrically, it should be clear. The calculation is sort of long and unpleasant. OK, so we'll deal with this thing. So let's start with minimal surfaces. What minimal surfaces are there? So minimal surfaces are surfaces whose mean curvature vanishes. In this context, they are solutions to mean curvature flow that don't depend on time. So it's a solution of this equation where u is independent of time. So I just set this side equal to 0. What you get is an ODE. So let's describe the solutions to this ODE. To begin, there is one solution that in hindsight is obvious. There is the stationary cone, which you get by assuming that A is a linear function. So if you substitute this in here, a very quick calculation, the second derivative is 0. So the first term is gone. ux is equal to A. u is equal to Ax. The x's cancel, you find. There's one particular angle A, aperture A, for which this thing is a stationary cone. So in particular, if p and q are the same, and at some point I'm going to switch to the case where p and q are the same, there's just one particular reason why I want to spend the first couple of minutes talking about allowing p and q to be different. But so once we've done that example, I'll just pick values for p and q. So there's a cone like this. It's a generalized cone because it sits in this. So the lowest dimensions that we're considering today is dimension four, because p and q each have to be at least two-dimensional. Otherwise, we're just dealing with rotationally symmetric surfaces like yesterday. So we're not looking at that case anymore. So p and q are both at least two. So that means the dimension of the space we're looking at, the total ambient space is at least four. So that makes it difficult to visualize these cones. But I'll call this thing A star because there will be other A's. OK, so that's a stationary solution to mean curvature flow. At the origin, it is singular. If you rotate this thing around this axis and that axis, you will find that at this point, it is singular. The mean curvature of this cone is zero, but the mean curvature is just the average or the sum of all the principal curvatures. The principal curvatures themselves are not zero. So this thing, actually, if you calculate the total curvature of this thing, it blows up like one over r squared. It blows up like one over r as you go up to one over the distance to the origin. So the actual curvature of this surface becomes infinite as you approach this point. But the mean curvature, which is this quantity, is zero. OK, so one famous example of this is Simon's cone, which is the one that you get by setting p and q equal to 4. So it's this thing in r8. And so in short, these things are stationary cones. They're sometimes called minimal surfaces. That does not mean that they actually minimize the area. It turns out that if the dimension is at least, so for example, Simon's cone and all the ones in dimension eight or higher, with one exception, are actually area minimizing, whereas the ones in dimension below eight are not area minimizing. So in the next half hour, I'll explain all that and show why that's true. So in any case, there's a difference between minimal cone and minimizing cone. A minimizing cone is one that actually has least area. They worry you can't find something that has smaller area. OK, so what other minimal surfaces are there? So let's analyze this minimal surface ODE. And one thing to notice about this is that it is invariant under scaling. If I multiply x and u by the same constant, then all terms scale in the same way. Yeah, I should not use thank you. Let me give this thing a name. Minimal surface ODE. OK, so this is a straightforward ODE analysis. So the equation is invariant under rescaling. It's a non-autonomous second order differential equation, so that's just a little bit beyond what the textbooks tell you how to solve. So what you do is you pick two scale-invariant quantities and you write down the ordinary differential equations for this. So you calculate wx and you calculate zx, and it turns out the equations are nicer if you calculate these things. This one was that, and this becomes, you get two. So in particular, if you let x is log, and so I shouldn't use t, so let me say theta, then x d dx is w theta. So this is a calculation that you can verify. It follows from that ODE, that second order ODE. So this gives you a system of two autonomous first order ordinary differential equations. So it leads to a phase plane. x is, sorry, theta is log x. Sorry. So dd theta is x dd x. And so when you're doing calculations with this kind of ODE, the actual theta is not an important parameter. And if you don't want to keep track of what theta is in relation to x, you could just add x as extra variable and say x theta is, well, what is it? x theta would be x dd x of x. Also, if you're programming this on a computer and you want to plot diagrams, you just run this system of three ordinary differential equations. And if you want to plot, say, w versus x, you plot the solution of this one versus the solution of that one, right? So this, OK. So in your standard undergraduate differential equation class where you study phase planes, what do you do? You try to find the fixed points. You linearize at the fixed points. You look for the lines where w theta is 0. You look for the curves or the curves where z theta is 0. And that gives you the following picture. Sorry. OK, so yeah, let me erase this. This equation is invariant under rescaling. If you multiply x and u by the same constant, then you get something that satisfies the exact same equation again. And so therefore, you look for, so the ratio between x and u is invariant under scaling. And the ratio between du and dx is also invariant under scaling. So we pick these. And so others that you could have taken, u divided by x, you could have taken. In a sense, this one would have been a better variable to choose because this one is singular when u is equal to 0. And so for today, that's not going to bother us. A different thing that you could have done is to say, here's the xy plane. Here's my solution to mean curvature flow. You could look at these two angles, write down differential equations for those. And that's equivalent to what we have over there. The derivation is a little longer. Trigonometric functions show up. But it is somewhat nicer. OK, so the phase planes that we get. So for w and zeta, z are the following. So let's see. z theta is 0, so I'll just draw this picture. So there are two fixed points. The origin is a fixed point, and here's a fixed point. So there's the fixed point 0, 0. And this fixed point, when you calculate what it is, it's the point a star, 1 over a star, and this is. And so what interpretation do these fixed points have? This one corresponds to the cone. So if you look back at the cone, on this cone, you have w is dy dx. That's a star, and z, that's x over y, that's 1 over a star. So if your solution was this cone, the w and z that you get are constant, and they give you this fixed point. If you have a different solution that starts out here and that hits this axis perpendicularly, then at this point, we have x over u is, well, u is not equal to 0, and x is 0. So this will be 0. The u dx is also 0. And the fact that here x is 0 and u is not 0 is the reason why we're looking at x over u instead of u over x. So this is a decision that was made in hindsight. You could have also chosen this one. If you choose this one, then the fixed point that corresponds to the right boundary conditions actually is at infinity, because u will be non-zero and x will be 0. So all these are trivial little facts that you discover by experimentation. So there are two fixed points. The differential equation goes like this. So here everything goes up. Here it goes like this. And now you could linearize at these two fixed points. The linearization tells you that the origin is a saddle point. And should I do that? Let me not. So what you find is a saddle point. Briefly, what you do is you assume you have w dot. So at the origin, the linearizing is easy, because it amounts to writing down the differential equations, throwing away all nonlinear terms. So w dot is 1 plus w squared times q minus 1 z minus q. Which is, so at the origin, w squared is roughly 0. So this whole thing is 1 plus something small. And these are linear terms. So you just get q minus 1. Get this. And z dot is z minus z squared w, which is roughly z. So at the origin, the differential equation is roughly given by this linear system of differential equations. And then you look at the eigenvalues and eigenvectors of this matrix. So you see the eigenvalues. There's one positive eigenvalue, one, and there's one negative eigenvalue. So that's why it's a saddle point. I'm going to reproduce this picture later. So at the origin, we have an unstable direction, which turns out to be in this direction. There's another one going in the other direction, but we don't care about that. We only care about w and z that are positive. And then the stable manifold turns out to be just the vertical axis. Now if you linearize at the other fixed point, this fixed point, it's a bit more complicated. So the results that you get are dimension dependent. So linearization. So if people's q is less than 8, it turns out only to depend on the sum of p and q. So I won't do the calculation. It's similar to this one. I'm finding the eigenvalues of a 2 by 2 matrix. So you get this situation, or there's this situation. And then for the algebra, it actually doesn't matter. The slower bound doesn't matter. But for us, p plus q will never be less than 4. Here you get complex eigenvalues, minus alpha plus or minus i beta. And there are formulas for those. And so for later on, the only important case is that at some point I'll switch just. And I'll assume that p and q are both equal to 2, which is the lowest dimensional example that we can consider. But it's still typical for the others in this range. In this case, alpha is 1 half. And beta is 1 half squared at 7. So these are very specific numbers. And if people's q is bigger than 8, then the eigenvalues that you find are real. And both of them are negative. And so when you have complex eigenvalues, they're complex, but the real part of the eigenvalues is also still negative. So what this means is that this is a sink. So it's attracting. If you're near this fixed point, the differential equation will draw you at an exponential rate into the fixed point. And exponential is exponential in the variable theta. So it'll be e to the constant times theta, e to the eigenvalue times theta, which in our case is x to the power lambda. OK, so let me draw the pictures. So for people's q, if the dimension is between 4 and 7, this thing has a complex eigenvalue. So the solution that has the right boundary conditions is a solution that starts here. Since this is a saddle point, there's only one of those. And what it does is it spirals into that fixed point. OK, so that's the phase plane. Now, what kind of minimal surface does that represent? I'll get to that in a second. But let me also draw the phase plane for the other dimensions. So let me do that here. So this is now has two real eigenvalues. So that means it's still exponentially attracting. And there are two eigenvectors. One of the eigenvectors has a larger eigenvalue than the other one. So most orbits will approach this thing according to the direction of the slow eigenvector. These eigenvectors have roughly this direction. So what happens if people's q is greater than or equal to 8 is most of the time the solution just looks like this. Bonnetonically, now just from the information of the linearization, you can't say that this is true. The only thing you can conclude is that the solution starts here. Once it approaches this fixed point, it will either go in from this direction, that direction, that direction, or that direction. And it could actually have looped around a couple of times before reaching there, getting there. So how do you know whether it loops around or not? And these are so one thing you can do is you could try to produce barriers. So if you could produce an explicit curve here, a sub-solution where the arrows are all pointing inwards, then you know that the orbit cannot cross this curve. And you would know that it doesn't loop around. On the other hand, if you could produce a barrier like this, then you know that this orbit cannot monotonically approach this point. And it has to loop around at least once. So since these are very specific differential equations, constructing these barriers is something that's trial and error. And so this is true for all. So the result is that this is true for all p plus q bigger than or equal to 8, except p is 2 and q is 6. So for p is 2 and q is 6, we have this situation. It does this. It loops around once. So in particular, it crosses this line exactly once. So let me mention some names. So there are a bunch of papers. So this phase plane is studied by Alenkar. There is, let's see, there's a book by Gary Lawler. It's an AMS memoir, which contains all this information and much more. So it's sort of a textbook after extending all earlier results. And there is, so that last one, the p equals 2 equals 6. This was done by a noise. And I'm a bit confused about the order in which this was done, because that's the oldest one that I know of. Then this came later. There is another paper by MQY that appeared a bit later. It contains more information. And then this book showed up. OK. What do the corresponding minimal surfaces look like? Oh, I'm sorry. Yeah. Yeah, no. So these are all pictures of the same phase plane with different values of p and q. Right, so now we'll draw the xz plane, then sort of the xy plane. And the easiest way to convert these phase diagrams into xy diagrams is just to keep track of the variable z, which is the ratio between, which is x divided by y. And I use y and u interchangeably. So let's start with this one. So, right. So initially, x over u is 0. That means that x has to be 0. And you can be any positive number. Now, which positive number should it be? We got these differential equations by getting rid of the scale invariant. So that means for each solution that we find here, we get a one-parameter family of solutions to the minimal surface equation. If you have a minimal surface and you magnify it or shrink it, it is, again, a minimal surface. So that means that the initial point here is something we can choose. And I'll just pick the number 1. OK, so now what does z do? z increases. I remember z is x divided by u. So in the x u plane, x over u is essentially this angle, or it's the tangent of that angle. So this angle increases, and then decreases, and then increases, and then decreases, and increases, and decreases, and so on, and converges to the aperture of the stationary cone. All the while along that curve, w, which is ux, is positive. So that means that the solution is strictly increasing. So we get an increasing function that oscillates around the minimal cone and converges to it. It's asymptotic to the minimal cone. And now any dilate of this thing is, again, a minimal cone. So if I take this thing and I try to, I shrink it by a factor of 2, drawing I shouldn't attempt. So any dilate of this thing is, again, a minimal surface. And they cross each other. So those are all the ones that you get from here. These are not area minimizing. Let's see, sorry, the cones are not area minimizing. So I'll show you in a second that this thing, so that, yeah. So I'll draw that exact picture that you're thinking of, yeah. OK, so here the situation is as follows. We're in this situation, what happens is that ux being w is still positive. So the solution, the minimal surface that we get is it's the graph of a strictly increasing function. And the angle decreases monotonically and converges to that of the stationary cone. So we get this. Note that this thing is star-shaped, meaning if I draw a straight line from the origin out, it will intersect this graph only once. That follows from the fact that it's convex. It's convex because why is it convex? It's convex because uxx is the derivative of ux is wx. And w is increasing along this curve, so therefore this is positive. Also, x divided by u is decreasing, so therefore every line intersects this thing exactly once. So if I look at this, and that's an important observation because it means that if I look at dilates of this thing, they're all this joint. OK, so the orange curve is the one that I showed that I got by starting here at height 1. All the other ones are just magnifications or reductions of they're all, you get them from the orange one by multiplying with a fixed constant. Particularly if I choose the constant to be a, then so this one is exactly a times the orange one. So in this case, what you see is that the complement of the cone is foliated by minimal surfaces, right? So if you swing these things around the axis, you get smooth minimal surfaces, you get a whole bunch of them. They fill up the whole space, and there's exactly one through each point. And then there is the exceptional case, the Seymour's case. This is the minimal cone, the exceptional one. What does it do? Z increases, hits this value, which is the aperture of the stationary cone, hits it once, and then goes back to converges to the cone. So this thing looks like that. If you take dilates of this thing, they intersect. So these things do not foliate the whole space. OK, so let's compare areas. So I'm going to erase. I think we're done with these pictures now, so I'll get rid of these. OK, so let's look at the case where people skews between four and seven. We have this cone, and then the thing about the foliation, well, yeah, we're in this situation, and these things don't foliate the whole complement of the cone because they all intersect each other. But if I only look at this particular piece, I take the segment of my minimal surface and just take the piece until the first intersection with the cone, then I could look at dilates of that, and those would foliate the complement, because why is that each line intersects this thing exactly once? And that's because along this, so we're looking at this part of the solution. This is the right height. Along this part of the solution, Z is strictly increasing, and Z was the angle from the vertical, with the tangent of the angle from the vertical. So this thing is star-shaped, and all dilates of this thing foliate the cone. So let me call this thing M, and then if this is A, then this one would be A times M. This part, I'll call that C. That's the cone, C. I want to compare the areas of M and C, and so now if you have a foliation like this, there's a trick that allows you to compare these areas that I want to explain. So I think in minimal surfaces, it's a well-known trick. It turned out to be useful in recent work on ancient solutions with Natasha Sesem and Tascallopoulos, where we applied something like this in a different context. So what you do is the following. If you have a foliation by surfaces like this, what you can do is you can look at the unit normals to these surfaces. So at each point, there is one of these minimal surfaces that goes through that point. It's unit normal that's new at this point. So it's a vector field that's defined on this whole domain, this whole region. And I'm making a two-dimensional drawing, but I'm now thinking of this region as you're sitting in our P plus Q. So that's new. An important thing about new is that it is divergence free. If you have a foliation and you take the unit normal vector field to the foliation, the divergence of that is minus the mean curvature. And so let me not explain why that's true, at least not in detail. So how do you get the mean curvature? How do you get the principal curvatures? You have a surface like this. And at each point, you have a unit normal. So this is epsilon times unit normal. And this is a unit normal. So you have a unit normal at each point on the surface. And as you move the point along on the surface, the unit normal rotates. So you can differentiate along the surface. You can differentiate the unit normal along the surface. And that gives you a linear map from tangent vectors to tangent vectors, because I should differentiate the normal. The tip also moves parallel to the surface. And the eigenvalues of that linear map are the principal curvatures. And the sum of those is the mean curvature. So the trace of that linear map is the mean curvature. So what you do, so those are the derivatives. So the divergence is the sum of all the derivatives of new in along all perpendicular set of directions. And that includes all directions tangential to the surface plus one perpendicular to the surface. And now that extra term doesn't matter, because if you move the vector field perpendicular to the surface, it may rotate. Because the surfaces are, they're not all parallel. They're not flat. So it will rotate a little bit. But its direction will be perpendicular to the vector in which you're moving. And when you take the trace, you take the dot product with that vector. So that extra term is 0. So there's an explanation like that for why this is true. OK, so in our case, these things are all minimal surfaces. So therefore, OK, so we have a divergence-free vector field. And that means that you must apply Geed-Green's theorem to something. So let this region be r. Then 0 is the integral over r of the divergence of new dx, or d volume. And that is the integral over the boundary of r of new. And now the unit normal, the output unit normal to r, can't call it new anymore, so we'll call it n with respect to area. And so the boundary of r consists of three components. In this picture, you might be tempted to include this part in the boundary, but it's not. We're rotating this around the axis. These are all interior points. So the boundary consists of this part. It consists of m and of the cone, c. And the unit normal n will be outward here and outward there. So if I, this is, so the difference between this unit normal n and that unit normal n is that they are all pointing upwards. So this one, this n is that one. This n is minus that one. So n, I'll draw the n's in orange. And now if you look at these two integrals, on m, we're taking the dot product of the unit normal vector field to the minimal surfaces and the output unit normal. But this part of the boundary is one of the minimal surfaces. So on this part of the boundary, in this term, new and n are the same. So it's just the length of a unit vector. It's just one. So the first thing is just the area of m of this piece. And the second part is minus the integral over c of n dot nu. So this is 0. We started by saying 0 is equal to all this. So the conclusion is that these two things are equal. The area of m is the integral of n dot nu over c. And so this shows you that the cone is not minimizing because it's a dot product of these two vectors. And that dot product is strictly less than 1 because they're nowhere parallel. And so this is the area of c. So the conclusion is that the area of m is less than the area of c. So the cone is not minimizing. This thing is minimizing. Or this thing has smaller area. And now a similar argument will actually show that if you prescribe this boundary value, this is the minimizer. And then as a curiosity for this, so this is a general argument. In this particular case, as a curiosity, there's this one observation about this integral, namely that the thing that we're integrating is the dot product of nu and n over this part of the boundary. So if I draw that again, I would have to take at each point here, this is n. And this will be nu. And I have to calculate that at each of these points. And so the point is that it's constant. It's the same two vectors at all points because these surfaces are dilates of each other. So the normals at all of these points are exactly the same. Instead of saying, so I could say, this is nu dot n. So that number is something that you can read off from solving the ODE. OK, so these things are not minimizing. These cones are not minimizing. This thing that I told you is true if the dimensions are between 4 and 7. And so the CMOS cone is curious in the sense that the same argument applies there. So remember, in the case PS2Q is 7, the minimal surfaces look like this. So we had that initial segment that can apply the exact same argument. So the curiosity of this example is the numerical value of that dot product, the angle between these two. So if you haven't seen this before, this is kind of weird. The picture that I drew for you for CMOS cone is that the thing goes around, it loops around, and then does this. So particularly the curve hits this thing. And somewhere over here, this is where it intersects the cone. This is UX. The cosine, UX is the slope of that thing. So you can read off that dot product. You can read off from the W coordinate at this point. OK, so how did CMOS prove that this happens? In 1975, it was a numerical calculation. He ran probably Fortran with punch cards. And the reason that this is very difficult to do by hand, I don't think anyone has done this by hand. And we're talking about this system of differential equations where q is six, so we have a five here, p is two, so there's a one. So it's this system of differential equations, extremely specific. This distance, so this distance is squared five, between two and three. This distance is about 10 to the minus five. So this drawing is really a gross exaggeration. What it really looks like is, it's indistinguishable. It does this, where if you magnify this 100,000 times, you see something like that. So there's a numerical, and then of course, there's a numerical calculation with interval arithmetic, so it's probably rigorous. And then other people have confirmed the same thing using different approaches. You can, with integer arithmetic, you can produce barriers, so this is, it's true. It's, so this distance is also something of order, 10 to the minus five. And to get this dot product, you have to take the cosine of that number, essentially. And so you get, the cosine is one minus that number squared. So for the CMOS cone, the ratio, so for the CMOS cone, this cone and this minimal surface, this minimal surface beats the area of the cone by an incredibly small number. So the cone is not minimizing, but it's almost there. It's really, and so, and I don't know if any reasonable explanation of why these really small numbers have to show up. So it's difficult to do that because it's just one example. Okay, so then the other cases, so if P plus Q is greater than or equal to eight and Q is not two and P is not two, then you get this foliation by minimal surfaces on the outside that never touch the cone, and if you flip them around, you also get them on this side. So using this foliation, you can show that the cone is actually minimizing, right? So, and the argument is very similar to the one that I erased, to the one that, yeah, to this argument. Again, you take the unit normal, so now all of space except the origin is foliated. You have this divergence for unit vector field everywhere. If you assume, if there were something like this that were minimizing, you apply Green's theorem and you'll find that the area of this thing is bigger than the area of the cone. And you don't need symmetry for that. So another application of these minimal surfaces is that if you look, if you remember the example of the cross that, you know, two crossing lines and you evolve them forward time by curve shortening, you get fattening. We're done with these. So under mean curvature flow, this is a picture in R2, two crossing lines. So this is the example that we saw yesterday. You get many different ways of evolving this in forward time and so you get a non-unique solution or you get the viscosity solution fatness develops an interior. If you replace this cone, so what is this cone? This cone is just a set of all x, y in R2 where x is equal to y in absolute value. If you replace this two, so this is R1 cross R1, right? So this is the case, p is one, q is one. If you replace p and q both by four or higher numbers and you could say, I have this cone here, how does it evolve in forward time? And the presence of all these minimal surfaces shows you that the cone, the forward evolution of the cone is the cone itself. It doesn't move. It's a fixed point for the flow. It does not fatten. Why is that? It follows from the maximum principle. And the reason is, so the quick explanation is, try this again. Suppose I start with this cone as initial condition for mean curvature flow. The viscosity solution, so the union which contains all possible other solutions has the property that it obeys the maximum principle and so it will avoid all smooth surfaces. If you have a smooth solution to mean curvature flow that is disjoint from the initial value of your viscosity solution, then they will always stay disjoint. And I'm lying a little bit because the smooth solution has to be compact, but so you can fix that in this example. Okay, so if you start with this thing, it's initially disjoint from this minimal surface and that minimal surface on the mean curvature flow does not move. Therefore, whatever solution comes out of this cone has to stay disjoint from that minimal surface. And I could, so we have many of these that argument applies to all the minimal surfaces that I have above here. So therefore the forward evolution of this cone, the stationary cone actually stays on this side of the cone. It could go to the other side, but in this condition we have on the other side you can flip the axes around and you have a foliation with similar surface on this side. The exceptional case where PS2 and Q is six has following feature. So this case, so this is the case PS2 and Q is six. And here if you switch the axes around, you have to switch the P and Q around. So the surfaces that come from this axis are the ones that correspond to Q is six and PS2. And those all stay on this side of the cone. So it's only the ones with PS2 and Q is six that cross the cone. So we do have a foliation on this side by minimal surfaces. So that means the forward evolution of this cone stays above the cone. On the other hand, I can approximate this cone by taking this segment and that segment. And now if you had a viscosity solution to this, any test function, so this thing would in forward time, you can make this thing arbitrarily small. So this thing in forward time will start evolving outward and go monotonically upward. So you can use this thing to prove that if you start with CMOS cone, it will fatten but only on one side. So the forward evolution of this thing stays, one boundary of the forward evolution is the cone itself and the other one is a solution that comes out in this direction. So yeah, that's everything for minimal surfaces in great detail, I think. So next, so I'll start a little bit now and then complete this tomorrow. We'll look at expanders and shrinkers. So what happens when you look at expanders and shrinkers? The ODE becomes a bit more complicated. So what are the extra two terms? And I'm going to stick to the case where people's Q is less than eight. So there should be no expanders, there are no, so the reason for that is that there are no expanders or shrinkers coming out of this cone in this particular case. So there are no expanders because an expander would be a forward evolution coming out of the cone, that was not the cone itself and this argument shows that those things don't exist. And there can be no shrinkers and that's because if there's a shrinker, the shrinkers are uniquely determined by their cone at infinity and the cone itself is a shrinker. So that's the only one there is. So I'm going to stick to the case people's Q less than eight and nothing is lost. So things simplify if you just assume that P is two and Q is two, so we'll do that case. So that simplifies this. And so if a solution is self-similarly expanding or shrinker as described yesterday, then the time derivative is equal to plus or minus one half U minus XUX. So this is if the solution is of the form square root plus minus T U. And so if you get rid of the time derivative you put these two ordinary derivatives on this side you find, so the ordinary, the expander shrinker equation is this thing where you have lambda is plus one half for expanders. So this equation is the same equation that we started with except for that we have something non-zero on the right-hand side. The trick that we used or the thing that inspired us to you to look at that two-dimensional phase plane with the Z and W variables is that the equation was invariant on the rescaling but now if you replace, so what happens if you replace X by AX and U by A times U, if you do that, then each of these terms, each of these terms, so as a physicist you would look at the dimensions, the units of each of these terms, right? So U has dimension of length, X has dimension of length. The second derivative is one over length. The first derivative has no units. Each of these terms has the units one over length. So if you do this scaling, each of these terms gets divided by A. These terms, this has dimension of length, this has dimension of, so X cancels length, cancels against length, this has dimension of length. These, this term gets multiplied by A. So if you do this scaling, what happens is that this whole equation gets an A squared here. So this equation is not invariant on the rescaling space. So that means that, so it is still useful to look at those, those scaling invariant variables, but you have to add a third variable because you won't get a closed system of ODE's for those. So, I think, yeah, I have one, yeah, so I should stop here. So let me say what will I do tomorrow? I'm going to analyze these equations and I'll show some particular, there are examples, two shrinkers that become a cone. So there are evolution in the mi curvature flows that they turn into a cone and then they have several different expanders. Converged to a cone and that particular cone will have many, many different ways forward evolutions. I'll discuss that and then as a continuation of that, I'll, there are also non-self-similar, non-self-similar solutions forward. So these examples show that they show that if you, so if you were going to look at very full solutions or solutions with not too bad singularities, all these solutions have only one singularity, namely at the origin. The, if you wanted to use these things to, say, construct minimal surfaces by looking, by using mean curvature flow as a gradient flow, there is a very, very severe non-uniqueness. The set of non-self-similar solutions, so here I'll construct expanders. There will be many of them for the stationary cone, there are infinitely many, but it's a discrete sequence. Here we get a whole continuum, so I can get the, so depending on the cone, you can get a, you can get an n dimensional family of solutions where n can be arbitrarily large. So I'll stop here and continue tomorrow.