Global smooth solutions for the inviscid SQG equations
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Cordoba, Diego
In this lecture I will discuss some recent work, joint with Angel Castro and Javier Gomez-Serrano, on the existence of a non trivial family of classical global solutions of the inviscid surface quasi-geostrophic equation (SQG).
 First of all, thank the organizers for this invitation. It's a real pleasure to be here. And yes, so I'm going to talk about the surface quasi-gestrific equation, and my collaborators are Angel Castro, who is in Madrid as well, and Javier Gomez Serrano, who is at Princeton. So what is the surface quasi-gestrific equation? Well, this is an equation in dimension two. We're in the plane. There are no boundaries. Theta is the unknown. Theta is a scalar function, and it's transported by an incompressible fluid. Since the velocity field is incompressible, we can define a stream function such that the velocity is equal to the gradient perpendicular of the stream function, and then the relation between the stream function and the scalar theta is given by the square root of the Laplacian. If we write down the velocity in terms of theta, then what we have is that the velocity are the risk transforms of theta. So the outline of my talk is, since I'm the only one speaking on SQG, I will give an introduction, small introduction of what is known as properties. Then I will present the theorem I want to introduce, and then I will give a brief sketch of the proof. And during the proof, I will then talk about patches, solutions of patch, patch solutions, because the motivation of the proof comes from the vortex patch problem. Okay, so the question is, is it a model that models the dynamics of big masses of air in the mid-latitudes where the Coriolis force plays an important role? And this is the quashity of the equation, it's a three-dimensional equation, and the dynamics that happens at the surface is what is known as the surface quashity of the equation, which is this one. So this model, and here's several references, and it's had been posted and studied in several of papers in the geophysical context, but it was not until 1994 that Peter Konstantin and Dimayda and Esteban Tabak realized that this was a very good model, it was a two-dimensional model of the 3D Euler equations. So let me describe why. Here we have 3D Euler and 2D QG. So I'm going to look at the 3D incompressible Euler equations in the vorticity form. So the vorticity is the curl of the velocity, and the equation for the omega, the vorticity, where omega is a vector and it's divergent free, and the velocity is divergent free as well. And the gradient of U is, in terms of the vorticity, a single integral operator acting on the omega. So we compare this with the surface quashity of the surface equation. What plays the role of the vorticity will be the gradient perpendicular of theta, which is the derivative with respect to x2 of theta, partial theta with the derivative with respect to x1. And the equation for this vector is equivalent to the equation that the vorticity satisfies, but this is a vector in three-dimension, this is a vector in two-dimensions. And of course, the gradient perpendicular of theta is divergent free, and the velocity is divergent free as well. And the gradient of U are single integral operators, risk transforms of the gradient perpendicular of theta. So but there is also a relation by, with the conservative quantities. In the case of Euler, the vortex lines move with the flow. Vortex lines are tangent to the vorticity in all points, and they actually move, they're transported by the flow. And the L-tune of the velocity, the energy is conserved. So in the case of SQG, the level sets of theta place the role of the vortex lines, they're tangent to the gradient perpendicular of theta, and since it's a transport equation, the level sets move with the flow. And the energy is conserved as well. There's, since there's a transport equation, and for SQG, the more quantities conserved is that you actually can bound the LP norm of the velocity for P bigger than 1 and strictly smaller than infinity, and this is bounded by the initial data. And this is because the velocity are the risk transforms of theta, and the LP norms of theta are conserved because of the transport equation. So what do we know? Here we have local existence in certain HK solvath norm, and in the case of SQG, similar to local existence, the K may vary because we are three or two dimensions. And the problem of the formation of singularities are open for both. Now the criteria for the formation of singularities for Euler, well I can mention too, the Bill Katomayda criteria, which says that if you control the integral in time of the L-infinity norm of the vorticity, then there's no blow up at time capital T. Well, similar thing is true for SQG. If you control the L-infinity norm of the gradient of theta, gradient perpendicular if you want to compare one with the other, yes. And there is the Konstantin-Meida, the Konstantin-Phefermann-Meida result that if you control the direction field of the vorticity, so if it remains lifted, then there's no singularities. So in the case of SQG, identical theorem is true for the direction field of the gradient perpendicular of theta. So if there's a singularity, the Lipschitz norm has to stop being bounded. Okay, so there goes the similarities with one equation and the other. And now, is there singularities for SQG? We have no idea. Numerics don't show any evidence for smooth initial data. There's no evidence at all. But in order to get a flavor of it, let me toy around and look at some 1D models. So we have a transport equation and the velocity is the risk transform. So in one dimension, let's take the velocity to be the Hilbert transform of, in this case, let me call it F, our unknown. So the usual Hilbert transform. And we look at, let's take this model where the F is transported by this flow. And let's take the initial data, F0 to be bigger or equal to 0. This is true initially, since the transport equation will maintain like this. And the energy actually will decay when F is positive. So for this, we have, we can prove local existence in some solar space. And there is finite time singularities for this model. For all initial data, this is strictly bigger or equal to 0. But of course, in this model, there's the incompressibility that doesn't play any role. So let's do the following. Let's change, take the divergence outside the nonlinear term. So the SQG, let's write, so this will be another model. Let's write the SQG as the divergence of U theta. Okay? So we just took the divergence outside. Then write the 1D model in this case. And what we have is this toy model. And for this, let's take for the Cauchy problem F of 0, bigger or equal to 0. So one, we have local existence in HK. But two, if the initial data is strictly positive, then there's global existence. Three, if there exists a X naught such that F of X naught is 0, then finite time singularities at the point X naught. Okay? For the same model, if the data stays positive for the model A, for A not for B. Yeah, and for B. What is B? B is this model. So this is SQG just to justify taking the, since I'm taking the divergence free outside. So the model will be the Hilbert transform inside and the derivative outside. So this model is the version of the equation is coupled. This one and this one? No, the derivative here is hitting. Oh, the derivative is outside. Yeah, yeah, this is a transport equation. And here the derivative is outside. Yeah, but just to see that the incompressibility may change, well, just to get the flavor. Okay, good. Can I ask a question again about, if you mention numerics of SQG, so there's no evidence of singularity? If the initial data is smooth, yes, no evidence of. So that I know of? Well, I just said for other various people playing. Yeah, yeah, yeah, yeah. So there's nothing like this? There used to be at a time for SQG at the time, but then it was ruled out. And numerically as well and then. But that's 20 years ago. Okay. So, but right now I don't know of any, anybody claiming numerical simulation where there are singularities for smooth initial data. Okay, so in this framework, I have to say that Resnik and his thesis in Chicago, he proved that as a weak solution is their global existence in L2. And the problem of uniqueness is a big open problem for this for SQG. Okay, so it may allow me to add a parameter to the system because I'm going to mention several references and in order to, to quote those references is good that I add a parameter. Okay. So the parameter is in the power of the Laplacian, I'm going to add an alpha. So I'm interpolating the system from to the Euler to SQG alpha equal to zero is to the Euler where theta is the vorticity and alpha equal to one is SQG. Okay. So, in for the global existence, to the Euler is critical. So one, this global existence for alpha equal to zero, but when alpha is strictly bigger than zero, it's an open problem. And, and radio for a book and so for the purpose of my talk, let me point out that the radio functions if theta is radio, then it's a stationary solution. And that has to do with the powers of the Laplacian preserves the symmetry and makes the nonlinear terms zero. Okay. So that's for the introduction and the theorem I want to present is that we prove that there exist non-trivial global solutions smooth for the SQG equation. How are these solutions? How do they look like? Well, okay, so let me take this for the modified one. This is for alpha equal to one, SQG. So the modified one, I'm going to call it generalized SQG. And anytime I don't mention SQG, I will say some alpha. So this is for SQG. And so the solutions are going to be a perturbation of a radial function, theta, where theta is going to be one inside this circle. Outside another circle, theta is going to be zero. And in between, it goes from zero to one in a smooth way. Okay. Actually, it's going to be a C2. Now it's going to be a perturbation of this. And the perturbation is going to have a three-fold symmetry. So we rotate by 120 degrees. We get the same structure. And these solutions rotate. And they rotate with constant angular velocity. Now the thickness in our solutions, this thickness is very relevant for the proof. And the thickness is 0.05. Later I will explain why. So these are the solutions. While we were working on this problem, we learned from Dritchell. He published a paper where he actually computed the specific solutions of the SQG of C1-1.5 regularity. And these are also, these are the same thing, but they're an ellipse. And they're explicit. So it's an ellipse. Outside the ellipse is zero. And inside, theta is smooth. And the way that he does it is that he goes to the 3D equation and he computed explicitly that the ellipsoids are rotating solutions for the 3D. So when you look at the surface, it gives you this ellipse rotating. But it's not a patch. Inside is smooth. Now where does the C1-1.5 come from when you cross the boundary? Okay, so let me, so I'm going to show you how, what are the main ideas and the tools in order to prove this theorem. So first of all, we're going to, in between theta equal to 1 and theta equal to 0, we're going to take the level sets of theta. And we're going to parameterize them by the alpha rho t. Rho tells me in which level set I am and alpha tells me in what place of the level set. And the function f is going to tell me how my function theta changes from one level set to another. So we plug this in, in the equations, in the transport equation and this is how it looks like. We are going to assume that these solutions are rotating solutions with constant angular velocity. So we impose that in our answer. So now z is our known, is going to be x, x of alpha rho. The function f is going to be given. I'm going to explicitly tell you what f looks like and it's independent of time. The unknowns here are going to be x of alpha rho and the angular velocity lambda. And since my picture is like this, I only have to show that the solutions, that the solution solves this equation in the support of f, the derivative of f. So how does f look like? It's going to be a function, c4 function, and has this structure where a, remember a is going to be 0.05. And it's more or less decreasing. And actually, it will be very important for the proof that in the middle, so there will be another beta, that when you add one minus a, you add a beta and you subtract a beta from one, in those regions in between, we are going to ask for a linear decay. And when a tends to 0, if we take a to 0, we recover the problem for the patch. So the solution will be a patch. So we plug this ansat in the equation. We use the fact that the velocity are the restransforms and we have the time derivative that disappears. And this is the equation for lambda and x alpha rho. And we have to solve for this. We have some freedom to choose the parameterization of x alpha rho. It's going to be a perturbation of the circle. So I'm going to define x alpha rho to be a function r that depends alpha rho times the vector cosine alpha sine of alpha. And now our unknown becomes r alpha rho and lambda. So we plug that in, that ansat in the equation above and what we get is this function that in order to be a solution, it needs to be 0. Okay, so one observation is that if alpha is independent of, sorry, r is independent of alpha, then what we are doing, we have this radial solutions. And f becomes 0. Okay? And what we're looking for is r's that depend on alpha. So if we take r alpha equal to rho or any radial solution, that is 0. Okay, so we are in the tool that we're going to use is the Kranta-Rabbi knowledge theorem. All right? So our parameter mu is going to be the angular velocity is lambda and the function r is r. Okay? So the goal is to find and so if we look at the plane of solutions where let's say this is lambda and this is r alpha rho and this line is r equal to rho, right? This line is a solution of the function. Then we have to find a lambda star and the space is x and y such that this condition satisfies so that the Cato derivatives around, we are linearizing around the solution rho r equal to rho. So around this solution and we have to find x and y such that they got to derivative exists and they're continuous and we have to find a lambda together with these spaces that the dimension of the kernel of the linearized operator is of dimension 1, the co-dimension of the image is also dimension 1 and then the fourth condition is the terms versality. So if those conditions are satisfied, Crandon-Rabinowitz says that you can bifurcate there's a solution parting starting from lambda star where it's a solution of the system f. Okay? So that's the goal. So in order to introduce the ideas that are behind this, this was used for the vortex patch problem. So I'm going to go and review or revise the vortex patch problem. Vortex patch problem is now we're dealing with a solution that the vorticity is let's say 1 here and 0 outside. It satisfies Euler equations. The curve, let's parameterize the curve of C alpha t, I'm sorry, like this, then it's possible to have, when you look at the evolution of z, you can close the equation in terms of z itself. So once you know how z looks like, then you can recover the velocity everywhere and it satisfies Euler equations, to the Euler equations. So this, since it's a transport equation, the values of omega will remain 1 inside the curve and 0 outside the curve. So this is very well understood and was for the Cauchy problem, Schemann was the first to prove that this problem is globally well posed for C1 plus alpha. And later there was another proof, a different proof by André Avertot-Temperature-Constantin of the same result. And then one can apply Eudovic's theorem to guarantee that this is actually, there's only one solution, it's unique. What about SQG? Well one can ask the same question for SQG. And one can close again a system in terms of the evolution of the curve and the curve itself. And this was studied first by José Luis Rodrigo in 2005, where he proved that it's locally well posed in the space C infinity. Why? Because this operator, it's actually, when you linearize, you lose one log of derivative. So his proof is using a Nash-Mosser scheme. Then Francisco Gancedo in 2008 showed that one can actually subtract anything in a tangential direction, any component that will not change the shape of the interface. And you can choose a component that gives you a special cancellation. And with this cancellation, you're able to close the a priori estimates in HK. So he proved local existence in HK. And recently this year, we showed that among all the patch solutions, this one is unique. And the global existence for this problem is open. Of course, all this works for alpha bigger than zero. And the question of global existence is open also for alpha bigger than zero. But in this case, we do have examples of finite time singularities. So let me briefly give you what is known. So in 2014, Francisco Gancedo and Bob Strain, they show that no splash singularities can be formed. So if you expect that a self-intersection between, of the boundary, then the curvature must blow up. Recently this year, Sascha Kicileth, Lenya Ritnik, Yal-Yal, and Flatos, some direct Flatos, they prove finite time singularities. But they add a boundary. So instead of considering the problem in the plane, they consider a problem in the half-plane. And they take two patches that self-intersect with the boundary at one point and they're symmetry. So already since it's touching the boundary, so the Cardart condition, all these local existence theorems is fundamental that you have a control of the Cardart condition. But they're proving local existence where the Cardart condition fails. And they managed to prove local existence but for a small range of alphas. And the range, I don't know, is 1 over 12 or 1 over 24, but it's more or less, that's the, it's a fraction. And they prove local existence. And moreover, they are able, since there's a lot of symmetry, that the patches attract each other. And they made the assumption, okay, let's suppose there's global existence. And they show a contradiction. If there's global existence, then both patches will self-intersect in finite time. If they self-intersect, Ganselos and Strang's theorem says the curvature has to blow up as well. Okay? But it's a contradiction, so it's really, it's telling you that there's a singularity but we don't know how the singularity looks like. Okay? This is with boundaries. And this is for alpha between 0 and 1 over 12. Excuse me, but this problem is bound to be a different problem. It is a different problem, but from the point of view of local existence. Oh, from those point of view, yes, from the point of view of singularity. Singularity, okay, yes, because, I mean, if there's no boundaries, you have a problem of controlling how the patches move and you don't have any control of it. So maybe the singularity cannot form because the patches are moving so much that it dissolves the singularity. Yes. But this theorem is saying, if you fix the boundary, you put this, then these patches do not move from there, but they detract each other and it has to intersect into, so I think it's very interesting. It's true that it doesn't tell us anything about the problem in without boundaries. Okay, so numerical simulation. So there is numerics showing evidence of formation singularities for the patch problem. And there is, so together with a team in Madrid, Fonteiros, Mancho, and José Luis Rodrigo, in 2005, we found this numerical simulations and we took a patch, the SSQG, and these patches were attracting each other. At the same time, they're moving, okay, but they attract each other and in finite time, they self-intersect and the curvature blows up at the same time they self-intersect in the same point. And what's really interesting is that it was a self-similar blow-up, locally self-similar blow-up. And this is the claim and this is the conjecture. So let's take an ellipse. Now one ellipse and see how it evolves. Well interesting, if you take an ellipse which is very close to the circle, then the ellipse, what happens is that it starts rotating, it changes the shape, changes the shape, but it remains convex and nothing happens. If you take a longer ellipse, then it starts rotating, you lose convexity, but you gain it again and you keep going and it seems like there's global existence. But when you take a longer one, then it loses convexity and it forms a finite time singularity at the same type as we got for the two patches. Okay so this, so Drichel and Scott predicted another type of singularities and this is another example of an ellipse and this is, so what Drichel and Scott took this ellipse at some point before the blow-up, before the singularity and this is how it looks like and they zoom, okay and the orange is the zoom. When they zoom, they have these instabilities. Then they zoom again, this is a fixed time, they zoom again and they get these instabilities. They zoom again, same thing, zoom, zoom, zoom, zoom. It's a cascade of instabilities. This is beautiful, okay. Okay so I'm going very slow. So let me hurry up. So some explicit examples for the patch problem, the circle, our stationary solutions for all alpha. Krzysztof proved that the ellipses for 2D Euler are rotating solutions, known also by these states. So it's an ellipse all the time. And we showed that for alphas bigger than zero, this is not true. Ellipses are not rotating solutions. And we were concerned about the change of convexity and we proved rigorously that the patch can actually change the convexity back and forth. Then in 1978, Dima and Thabuski predicted for 2D Euler the existence of rotating solutions with M-fold symmetry. 3, here 3, 4, but any M-fold symmetry. And this was a prediction from the numerics. And it was in 1982 that Burbea proved the existence of such solutions. And in 2013, 2D Smiddy, Joan Mateo and Joan Verdera gave another proof using the Krandar Ravinovich theorem. And they proved the existence, but they also proved that these solutions, they are actually C infinity. Together with Angel and Javi, we actually proved that they are real analytic. In 2000, Frankel showed that if you take the angular velocity to be zero, the only solution is the circle. And Tufik in the same direction showed that if the angular velocity is negative, the only solution is the circle. And then you can bifurcate from many other scenarios. For example, a patch inside another patch. And that's what they did. Or you can bifurcate from the lips. So Tufik and Joan, Mateo, they proved C alpha regularity bifurcating from the lips. And we actually proved that they are real analytic. And OK, so how did I start working on this subject? Why did I get interested? Well, when Tufik and the Joannes proved this theorem, the first theorem, so I invited Joan to give a talk in Madrid. And I got interested in this subject. And then I talked with Tufik. And I told him, why don't you try to prove this for SQG? I mean, for Euler, we know there is global existence. But for SQG, we have no any solution that exists globally that are non-trivial. So a year after, Tufik sent me a paper. And he worked with his former student, Hassainia. And they proved that for alpha bigger than 0 and alpha smaller than 1, these solutions exist and they are C alpha. But there was a problem. They couldn't reach alpha equal to 1, the SQG equation. So I invited Tufik to Madrid. He gave a talk with Chad. And he explained as what was the, why they couldn't do it. And they actually were not even to prove C infinity between 0 and 1. And then, Angel Javier and me, myself, we studied the problem with a different formulation. And then we realized that, I mean, the problem for going from alpha smaller than 1 to alpha equal to 1 is this loss of log of derivative. They were working on helter spaces. To implement this in helter spaces is quite hard. But to implement it in a sub-office space is trivial. So we changed our framework. And instead of working on the compass plane, we work on the real world and in sub-office spaces. So at first, we proved the existence of M-fold symmetry for SQG in the H3 plus log topology. And then we proved that the solutions are real analytic. Now one is true that Kran-Darabinovic tells you that it's unique, the branch. So where it overlaps, the solutions is real analytic. But Kran-Darabinovic doesn't tell you how far you can go in the branch. So maybe you have real analytic first in the beginning of the branch, and then you lose the analyticity and it becomes infinity. I don't know. So 10 to 3? Yeah. OK. A little bit more because you started a couple of minutes later. Thank you. Thank you. Thank you. OK. So let's look at this for the patch. So now we're going to bifurcate from the circle. And we assume it's a rotating solution, angular velocity lambda. So now our answer is r alpha cosine alpha sine of alpha. We plug that in the SQG equation and we get this function that has to be equal to zero. When r alpha is a constant, this is automatically zero. So we can apply the Kran-Darabinovic. We can go and try to apply the Kran-Darabinovic. So the spaces, and this is what I meant by adding the log of, lost of the log of derivative in the solve of the spaces. So this integral is just to assume that that integral is bounded in L2 and that's enough. And our spaces X is going to be, as it is for the V states for Euler, is going to be the sum of cosines. And we are going to ask this to be in the space hk plus log. And then the space Y is going to be the signs. We have the metronary for the Kran-Darabinovic. And now let me just point out one part of the Kran-Darabinovic is to see that the dimension of the kernel, the linearized part, is actually dimension one. So we take the linear part and we want to see an element in the kernel. We take the element h equal to cosine of m alpha. We substitute that in the linear part. And what comes up is that this equation has to be equal to zero, as simple as that. So it's giving me the eigenvalues and the, sorry, the eigenvalues, the velocity, angular velocity and it has for each n-fold you have this velocity. Of course now, so the problem is reduced to taking, to bifurcate from this angular velocity. And then of course one has to check the co-dimension of the image and the term versatility. But it's as hard as this. So the hard part here was to choose the right spaces in which you can work. The real analytic proof is that you choose spaces that are real analytic properly in order to framework to work. So let's come back and see what happens in the smooth case. Now in the smooth case, remember we are bifurcating from rho cosine alpha sine of alpha. So now this is a smooth profile. And f is going to be given. There's no loss of log derivative because it is smooth, our solutions are smooth. So we don't have to worry here. The mixed norms of the solid-volve mixed norms is okay. And but again we take the cosines to be our space x. This is the linear part and we have to look for the element, an element in the kernel. So again we do as before, we substitute our this ansatz in this function. And what we get is not a scalar equation, we get a function where I tilde looks like this, t tilde looks like this, and t tilde is an elliptical integral. Okay, so we need to solve this. The case of the patch is just a scalar equation. If we take b to be 1 and we take the limit at a goes to 0, we recover this, we recover the formula of the angular velocity for the patch. So quickly, what is the problem? t tilde m is not self-adjoint, but it's compact, it's a compact operator. We have a smoothness in i tilde, this is because of f. And using the properties, we are able to show that there exists a solution. How is this? Well, briefly, the reason is that the anti-symmetric part of the operator, it gets small as we take this guy small. So the goal is to find, to prove, you boil out the problem to end up showing that the eigenvalues, there's a gap between the eigenvalues. Here you don't know what any single eigenvalue, but for the symmetric part, you can have a guess, and this is a numerical guess, and you can rigorously prove the error you make. Now taking this sufficiently small, we can make the eigenvalue of the whole thing to be close to the symmetric one. How small this has to be? 0.05 is enough. And how do we do that? Well, we play with the f in order this to work out. So once we show that there's a solution, one has to show the regularity. So this is a bootstrap argument where we take advantage of the regularity of f. And we have to prove uniqueness, and we have to prove that the co-dimension of the image is one, and that the transversality is also applies. Also we managed to do that. And let me finish in this three minutes some remarks and some very recent results. So this is a, the solution is a perturbation, right? So if we perturb a little bit f, everything should work. And that f has that profile is not a requirement, okay? A perturbation of that works. And can we get, we prove that the solutions are C2? And that's just because we chose f to be C4. If we make our f smoother, but that will make everything, maybe we have to take us thinner this distance to be smaller, then we can make the solution to be of higher regularity. We, since this, in order to show that this is small, we use a computer-assisted proof, we have to do this for each m. So it's not a theorem for all m, we prove it for one single m, m equals to three. But we believe that if you put any other m, you have to correct the numbers. And maybe this 0.05 will be modified, but it will certainly work. And this can be extended for any alpha, particularly for 2D order, okay? And now, very recent results. So what I just talked about is in the archives. And what I'm going, this slide will be in the archives in a couple of weeks. And is, can we get the computer, get the proof of this, the existence of the solutions without the computer? And the answer is yes, by bifurcating from the patch. And the way we bifurcate from the patch is that now this A, A is the, is this distance, we are going to bifurcate from A equal to zero. A equal to zero is the patch, and at the same time if A moves to the branch, it opens up the smoothness of the function. And when, then, if, okay, so if we do this, the idea is that we rescale, we scale the profile f. And what is surprising is that this works for any m, and for any profile f. f can change sign, it does not have to be monotonously increasing, it can have any shape, it just have to go from zero to one. Okay? But, here we, since it's a random knowledge, we cannot tell you how, how much it opens up. So we cannot reach the solution. And that's all. Thank you. So, all of these solutions are periodic in time, you know, right? Yes, it is. This is, for this point, there are no solutions that are not periodic in time, they're global, but they're not. Oh. How important is it that it's compact support? Could you have a Gaussian, which is very narrow? Okay, so compact support makes, so I think, so the question is, for example, in the stationary solution, right? So the result of a Frankel that proves that for 2D, the vortex patch is, if it's stationary, it has to be a circle, right? But you can construct stationary solutions of 2D Euler that are not completely supported and they are not circles. Okay, so the fact that it's compactly supported, it's already telling you that there is some restriction, very strong restriction. Okay, so to the problem, but can you construct this from a Gaussian or something and the answer is yes. Okay, so yes. But it gives restriction, for example, in the stationary, so I think that Frankel's result is also true for SQG and it's also true for a smooth perturbation of the patch. Are these things rotating in the other direction? Okay, so it's open. So 2Fick's result says, so these solutions, the angular velocity is explicit and they're all positive. And you know, in order to prove, to use the Kandahar-Havinovich, right? I'm not saying maybe there's another perturbation that has this structure, the other symmetry and the angular velocity is different. I'm not saying that. In order to prove the Kandahar-Havinovich, but 2Fick's result for the 2D Euler says that if it's negative, there's, the only solution is the circle. I believe that can be extended to SQG and also to the case of the smooth case as well. But that's something to be done. Okay, then I go on with questions. Thank you.
 First of all, thank the organizers for this invitation.  It's a real pleasure to be here.  And yes, so I'm going to talk about the surface quasi-gestrific equation, and my collaborators  are Angel Castro, who is in Madrid as well, and Javier Gomez Serrano, who is at Princeton.  So what is the surface quasi-gestrific equation?  Well, this is an equation in dimension two.  We're in the plane.  There are no boundaries.  Theta is the unknown.  Theta is a scalar function, and it's transported by an incompressible fluid.  Since the velocity field is incompressible, we can define a stream function such that  the velocity is equal to the gradient perpendicular of the stream function, and then the relation  between the stream function and the scalar theta is given by the square root of the Laplacian.  If we write down the velocity in terms of theta, then what we have is that the velocity  are the risk transforms of theta.  So the outline of my talk is, since I'm the only one speaking on SQG, I will give an introduction,  small introduction of what is known as properties.  Then I will present the theorem I want to introduce, and then I will give a brief sketch  of the proof.  And during the proof, I will then talk about patches, solutions of patch, patch solutions,  because the motivation of the proof comes from the vortex patch problem.  Okay, so the question is, is it a model that models the dynamics of big masses of air in  the mid-latitudes where the Coriolis force plays an important role?  And this is the quashity of the equation, it's a three-dimensional equation, and the  dynamics that happens at the surface is what is known as the surface quashity of the equation,  which is this one.  So this model, and here's several references, and it's had been posted and studied in several  of papers in the geophysical context, but it was not until 1994 that Peter Konstantin  and Dimayda and Esteban Tabak realized that this was a very good model, it was a two-dimensional  model of the 3D Euler equations.  So let me describe why.  Here we have 3D Euler and 2D QG.  So I'm going to look at the 3D incompressible Euler equations in the vorticity form.  So the vorticity is the curl of the velocity, and the equation for the omega, the vorticity,  where omega is a vector and it's divergent free, and the velocity is divergent free  as well.  And the gradient of U is, in terms of the vorticity, a single integral operator acting  on the omega.  So we compare this with the surface quashity of the surface equation.  What plays the role of the vorticity will be the gradient perpendicular of theta, which  is the derivative with respect to x2 of theta, partial theta with the derivative with respect  to x1.  And the equation for this vector is equivalent to the equation that the vorticity satisfies,  but this is a vector in three-dimension, this is a vector in two-dimensions.  And of course, the gradient perpendicular of theta is divergent free, and the velocity  is divergent free as well.  And the gradient of U are single integral operators, risk transforms of the gradient  perpendicular of theta.  So but there is also a relation by, with the conservative quantities.  In the case of Euler, the vortex lines move with the flow.  Vortex lines are tangent to the vorticity in all points, and they actually move, they're  transported by the flow.  And the L-tune of the velocity, the energy is conserved.  So in the case of SQG, the level sets of theta place the role of the vortex lines, they're  tangent to the gradient perpendicular of theta, and since it's a transport equation, the level  sets move with the flow.  And the energy is conserved as well.  There's, since there's a transport equation, and for SQG, the more quantities conserved  is that you actually can bound the LP norm of the velocity for P bigger than 1 and strictly  smaller than infinity, and this is bounded by the initial data.  And this is because the velocity are the risk transforms of theta, and the LP norms of theta  are conserved because of the transport equation.  So what do we know?  Here we have local existence in certain HK solvath norm, and in the case of SQG, similar  to local existence, the K may vary because we are three or two dimensions.  And the problem of the formation of singularities are open for both.  Now the criteria for the formation of singularities for Euler, well I can mention too, the Bill  Katomayda criteria, which says that if you control the integral in time of the L-infinity  norm of the vorticity, then there's no blow up at time capital T. Well, similar thing is  true for SQG.  If you control the L-infinity norm of the gradient of theta, gradient perpendicular if you want  to compare one with the other, yes.  And there is the Konstantin-Meida, the Konstantin-Phefermann-Meida result that if you control the direction  field of the vorticity, so if it remains lifted, then there's no singularities.  So in the case of SQG, identical theorem is true for the direction field of the gradient  perpendicular of theta.  So if there's a singularity, the Lipschitz norm has to stop being bounded.  Okay, so there goes the similarities with one equation and the other.  And now, is there singularities for SQG?  We have no idea.  Numerics don't show any evidence for smooth initial data.  There's no evidence at all.  But in order to get a flavor of it, let me toy around and look at some 1D models.  So we have a transport equation and the velocity is the risk transform.  So in one dimension, let's take the velocity to be the Hilbert transform of, in this case,  let me call it F, our unknown.  So the usual Hilbert transform.  And we look at, let's take this model where the F is transported by this flow.  And let's take the initial data, F0 to be bigger or equal to 0.  This is true initially, since the transport equation will maintain like this.  And the energy actually will decay when F is positive.  So for this, we have, we can prove local existence in some solar space.  And there is finite time singularities for this model.  For all initial data, this is strictly bigger or equal to 0.  But of course, in this model, there's the incompressibility that doesn't play any role.  So let's do the following.  Let's change, take the divergence outside the nonlinear term.  So the SQG, let's write, so this will be another model.  Let's write the SQG as the divergence of U theta.  Okay?  So we just took the divergence outside.  Then write the 1D model in this case.  And what we have is this toy model.  And for this, let's take for the Cauchy problem F of 0, bigger or equal to 0.  So one, we have local existence in HK.  But two, if the initial data is strictly positive, then there's global existence.  Three, if there exists a X naught such that F of X naught is 0, then finite time singularities  at the point X naught.  Okay?  For the same model, if the data stays positive for the model A, for A not for B.  Yeah, and for B.  What is B?  B is this model.  So this is SQG just to justify taking the, since I'm taking the divergence free outside.  So the model will be the Hilbert transform inside and the derivative outside.  So this model is the version of the equation is coupled.  This one and this one?  No, the derivative here is hitting.  Oh, the derivative is outside.  Yeah, yeah, this is a transport equation.  And here the derivative is outside.  Yeah, but just to see that the incompressibility may change, well, just to get the flavor.  Okay, good.  Can I ask a question again about, if you mention numerics of SQG, so there's no evidence of  singularity?  If the initial data is smooth, yes, no evidence of.  So that I know of?  Well, I just said for other various people playing.  Yeah, yeah, yeah, yeah.  So there's nothing like this?  There used to be at a time for SQG at the time, but then it was ruled out.  And numerically as well and then.  But that's 20 years ago.  Okay.  So, but right now I don't know of any, anybody claiming numerical simulation where there  are singularities for smooth initial data.  Okay, so in this framework, I have to say that Resnik and his thesis in Chicago, he  proved that as a weak solution is their global existence in L2.  And the problem of uniqueness is a big open problem for this for SQG.  Okay, so it may allow me to add a parameter to the system because I'm going to mention  several references and in order to, to quote those references is good that I add a parameter.  Okay.  So the parameter is in the power of the Laplacian, I'm going to add an alpha.  So I'm interpolating the system from to the Euler to SQG alpha equal to zero is to the  Euler where theta is the vorticity and alpha equal to one is SQG.  Okay.  So, in for the global existence, to the Euler is critical.  So one, this global existence for alpha equal to zero, but when alpha is strictly bigger  than zero, it's an open problem.  And, and radio for a book and so for the purpose of my talk, let me point out that the radio  functions if theta is radio, then it's a stationary solution.  And that has to do with the powers of the Laplacian preserves the symmetry and makes  the nonlinear terms zero.  Okay.  So that's for the introduction and the theorem I want to present is that we prove that there  exist non-trivial global solutions smooth for the SQG equation.  How are these solutions?  How do they look like?  Well, okay, so let me take this for the modified one.  This is for alpha equal to one, SQG.  So the modified one, I'm going to call it generalized SQG.  And anytime I don't mention SQG, I will say some alpha.  So this is for SQG.  And so the solutions are going to be a perturbation of a radial function, theta, where theta is  going to be one inside this circle.  Outside another circle, theta is going to be zero.  And in between, it goes from zero to one in a smooth way.  Okay.  Actually, it's going to be a C2.  Now it's going to be a perturbation of this.  And the perturbation is going to have a three-fold symmetry.  So we rotate by 120 degrees.  We get the same structure.  And these solutions rotate.  And they rotate with constant angular velocity.  Now the thickness in our solutions, this thickness is very relevant for the proof.  And the thickness is 0.05.  Later I will explain why.  So these are the solutions.  While we were working on this problem, we learned from Dritchell.  He published a paper where he actually computed the specific solutions of the SQG of C1-1.5  regularity.  And these are also, these are the same thing, but they're an ellipse.  And they're explicit.  So it's an ellipse.  Outside the ellipse is zero.  And inside, theta is smooth.  And the way that he does it is that he goes to the 3D equation and he computed explicitly  that the ellipsoids are rotating solutions for the 3D.  So when you look at the surface, it gives you this ellipse rotating.  But it's not a patch.  Inside is smooth.  Now where does the C1-1.5 come from when you cross the boundary?  Okay, so let me, so I'm going to show you how, what are the main ideas and the tools  in order to prove this theorem.  So first of all, we're going to, in between theta equal to 1 and theta equal to 0, we're  going to take the level sets of theta.  And we're going to parameterize them by the alpha rho t.  Rho tells me in which level set I am and alpha tells me in what place of the level  set.  And the function f is going to tell me how my function theta changes from one level set  to another.  So we plug this in, in the equations, in the transport equation and this is how it looks  like.  We are going to assume that these solutions are rotating solutions with constant angular  velocity.  So we impose that in our answer.  So now z is our known, is going to be x, x of alpha rho.  The function f is going to be given.  I'm going to explicitly tell you what f looks like and it's independent of time.  The unknowns here are going to be x of alpha rho and the angular velocity lambda.  And since my picture is like this, I only have to show that the solutions, that the  solution solves this equation in the support of f, the derivative of f.  So how does f look like?  It's going to be a function, c4 function, and has this structure where a, remember a  is going to be 0.05.  And it's more or less decreasing.  And actually, it will be very important for the proof that in the middle, so there will  be another beta, that when you add one minus a, you add a beta and you subtract a beta  from one, in those regions in between, we are going to ask for a linear decay.  And when a tends to 0, if we take a to 0, we recover the problem for the patch.  So the solution will be a patch.  So we plug this ansat in the equation.  We use the fact that the velocity are the restransforms and we have the time derivative  that disappears.  And this is the equation for lambda and x alpha rho.  And we have to solve for this.  We have some freedom to choose the parameterization of x alpha rho.  It's going to be a perturbation of the circle.  So I'm going to define x alpha rho to be a function r that depends alpha rho times the  vector cosine alpha sine of alpha.  And now our unknown becomes r alpha rho and lambda.  So we plug that in, that ansat in the equation above and what we get is this function that  in order to be a solution, it needs to be 0.  Okay, so one observation is that if alpha is independent of, sorry, r is independent  of alpha, then what we are doing, we have this radial solutions.  And f becomes 0.  Okay?  And what we're looking for is r's that depend on alpha.  So if we take r alpha equal to rho or any radial solution, that is 0.  Okay, so we are in the tool that we're going to use is the Kranta-Rabbi knowledge theorem.  All right?  So our parameter mu is going to be the angular velocity is lambda and the function r is r.  Okay?  So the goal is to find and so if we look at the plane of solutions where let's say this  is lambda and this is r alpha rho and this line is r equal to rho, right?  This line is a solution of the function.  Then we have to find a lambda star and the space is x and y such that this condition  satisfies so that the Cato derivatives around, we are linearizing around the solution rho  r equal to rho.  So around this solution and we have to find x and y such that they got to derivative exists  and they're continuous and we have to find a lambda together with these spaces that the  dimension of the kernel of the linearized operator is of dimension 1, the co-dimension  of the image is also dimension 1 and then the fourth condition is the terms versality.  So if those conditions are satisfied, Crandon-Rabinowitz says that you can bifurcate there's a solution  parting starting from lambda star where it's a solution of the system f.  Okay?  So that's the goal.  So in order to introduce the ideas that are behind this, this was used for the vortex  patch problem.  So I'm going to go and review or revise the vortex patch problem.  Vortex patch problem is now we're dealing with a solution that the vorticity is let's  say 1 here and 0 outside.  It satisfies Euler equations.  The curve, let's parameterize the curve of C alpha t, I'm sorry, like this, then it's  possible to have, when you look at the evolution of z, you can close the equation in terms of  z itself.  So once you know how z looks like, then you can recover the velocity everywhere and it  satisfies Euler equations, to the Euler equations.  So this, since it's a transport equation, the values of omega will remain 1 inside the  curve and 0 outside the curve.  So this is very well understood and was for the Cauchy problem, Schemann was the first  to prove that this problem is globally well posed for C1 plus alpha.  And later there was another proof, a different proof by André Avertot-Temperature-Constantin  of the same result.  And then one can apply Eudovic's theorem to guarantee that this is actually, there's  only one solution, it's unique.  What about SQG?  Well one can ask the same question for SQG.  And one can close again a system in terms of the evolution of the curve and the curve  itself.  And this was studied first by José Luis Rodrigo in 2005, where he proved that it's  locally well posed in the space C infinity.  Why?  Because this operator, it's actually, when you linearize, you lose one log of derivative.  So his proof is using a Nash-Mosser scheme.  Then Francisco Gancedo in 2008 showed that one can actually subtract anything in a tangential  direction, any component that will not change the shape of the interface.  And you can choose a component that gives you a special cancellation.  And with this cancellation, you're able to close the a priori estimates in HK.  So he proved local existence in HK.  And recently this year, we showed that among all the patch solutions, this one is unique.  And the global existence for this problem is open.  Of course, all this works for alpha bigger than zero.  And the question of global existence is open also for alpha bigger than zero.  But in this case, we do have examples of finite time singularities.  So let me briefly give you what is known.  So in 2014, Francisco Gancedo and Bob Strain, they show that no splash singularities can  be formed.  So if you expect that a self-intersection between, of the boundary, then the curvature  must blow up.  Recently this year, Sascha Kicileth, Lenya Ritnik, Yal-Yal, and Flatos, some direct Flatos,  they prove finite time singularities.  But they add a boundary.  So instead of considering the problem in the plane, they consider a problem in the half-plane.  And they take two patches that self-intersect with the boundary at one point and they're  symmetry.  So already since it's touching the boundary, so the Cardart condition, all these local  existence theorems is fundamental that you have a control of the Cardart condition.  But they're proving local existence where the Cardart condition fails.  And they managed to prove local existence but for a small range of alphas.  And the range, I don't know, is 1 over 12 or 1 over 24, but it's more or less, that's  the, it's a fraction.  And they prove local existence.  And moreover, they are able, since there's a lot of symmetry, that the patches attract  each other.  And they made the assumption, okay, let's suppose there's global existence.  And they show a contradiction.  If there's global existence, then both patches will self-intersect in finite time.  If they self-intersect, Ganselos and Strang's theorem says the curvature has to blow up  as well.  Okay?  But it's a contradiction, so it's really, it's telling you that there's a singularity  but we don't know how the singularity looks like.  Okay?  This is with boundaries.  And this is for alpha between 0 and 1 over 12.  Excuse me, but this problem is bound to be a different problem.  It is a different problem, but from the point of view of local existence.  Oh, from those point of view, yes, from the point of view of singularity.  Singularity, okay, yes, because, I mean, if there's no boundaries, you have a problem  of controlling how the patches move and you don't have any control of it.  So maybe the singularity cannot form because the patches are moving so much that it dissolves  the singularity.  Yes.  But this theorem is saying, if you fix the boundary, you put this, then these patches  do not move from there, but they detract each other and it has to intersect into, so I think  it's very interesting.  It's true that it doesn't tell us anything about the problem in without boundaries.  Okay, so numerical simulation.  So there is numerics showing evidence of formation singularities for the patch problem.  And there is, so together with a team in Madrid, Fonteiros, Mancho, and José Luis Rodrigo,  in 2005, we found this numerical simulations and we took a patch, the SSQG, and these patches  were attracting each other.  At the same time, they're moving, okay, but they attract each other and in finite time,  they self-intersect and the curvature blows up at the same time they self-intersect in  the same point.  And what's really interesting is that it was a self-similar blow-up, locally self-similar  blow-up.  And this is the claim and this is the conjecture.  So let's take an ellipse.  Now one ellipse and see how it evolves.  Well interesting, if you take an ellipse which is very close to the circle, then the ellipse,  what happens is that it starts rotating, it changes the shape, changes the shape, but  it remains convex and nothing happens.  If you take a longer ellipse, then it starts rotating, you lose convexity, but you gain  it again and you keep going and it seems like there's global existence.  But when you take a longer one, then it loses convexity and it forms a finite time singularity  at the same type as we got for the two patches.  Okay so this, so Drichel and Scott predicted another type of singularities and this is  another example of an ellipse and this is, so what Drichel and Scott took this ellipse  at some point before the blow-up, before the singularity and this is how it looks like and  they zoom, okay and the orange is the zoom.  When they zoom, they have these instabilities.  Then they zoom again, this is a fixed time, they zoom again and they get these instabilities.  They zoom again, same thing, zoom, zoom, zoom, zoom.  It's a cascade of instabilities.  This is beautiful, okay.  Okay so I'm going very slow.  So let me hurry up.  So some explicit examples for the patch problem, the circle, our stationary solutions for  all alpha.  Krzysztof proved that the ellipses for 2D Euler are rotating solutions, known also by  these states.  So it's an ellipse all the time.  And we showed that for alphas bigger than zero, this is not true.  Ellipses are not rotating solutions.  And we were concerned about the change of convexity and we proved rigorously that the  patch can actually change the convexity back and forth.  Then in 1978, Dima and Thabuski predicted for 2D Euler the existence of rotating solutions  with M-fold symmetry.  3, here 3, 4, but any M-fold symmetry.  And this was a prediction from the numerics.  And it was in 1982 that Burbea proved the existence of such solutions.  And in 2013, 2D Smiddy, Joan Mateo and Joan Verdera gave another proof using the Krandar  Ravinovich theorem.  And they proved the existence, but they also proved that these solutions, they are actually  C infinity.  Together with Angel and Javi, we actually proved that they are real analytic.  In 2000, Frankel showed that if you take the angular velocity to be zero, the only solution  is the circle.  And Tufik in the same direction showed that if the angular velocity is negative, the only  solution is the circle.  And then you can bifurcate from many other scenarios.  For example, a patch inside another patch.  And that's what they did.  Or you can bifurcate from the lips.  So Tufik and Joan, Mateo, they proved C alpha regularity bifurcating from the lips.  And we actually proved that they are real analytic.  And OK, so how did I start working on this subject?  Why did I get interested?  Well, when Tufik and the Joannes proved this theorem, the first theorem, so I invited Joan  to give a talk in Madrid.  And I got interested in this subject.  And then I talked with Tufik.  And I told him, why don't you try to prove this for SQG?  I mean, for Euler, we know there is global existence.  But for SQG, we have no any solution that exists globally that are non-trivial.  So a year after, Tufik sent me a paper.  And he worked with his former student, Hassainia.  And they proved that for alpha bigger than 0 and alpha smaller than 1, these solutions  exist and they are C alpha.  But there was a problem.  They couldn't reach alpha equal to 1, the SQG equation.  So I invited Tufik to Madrid.  He gave a talk with Chad.  And he explained as what was the, why they couldn't do it.  And they actually were not even to prove C infinity between 0 and 1.  And then, Angel Javier and me, myself, we studied the problem with a different formulation.  And then we realized that, I mean, the problem for going from alpha smaller than 1 to alpha  equal to 1 is this loss of log of derivative.  They were working on helter spaces.  To implement this in helter spaces is quite hard.  But to implement it in a sub-office space is trivial.  So we changed our framework.  And instead of working on the compass plane, we work on the real world and in sub-office  spaces.  So at first, we proved the existence of M-fold symmetry for SQG in the H3 plus log topology.  And then we proved that the solutions are real analytic.  Now one is true that Kran-Darabinovic tells you that it's unique, the branch.  So where it overlaps, the solutions is real analytic.  But Kran-Darabinovic doesn't tell you how far you can go in the branch.  So maybe you have real analytic first in the beginning of the branch, and then you lose  the analyticity and it becomes infinity.  I don't know.  So 10 to 3?  Yeah.  OK.  A little bit more because you started a couple of minutes later.  Thank you.  Thank you.  Thank you.  OK.  So let's look at this for the patch.  So now we're going to bifurcate from the circle.  And we assume it's a rotating solution, angular velocity lambda.  So now our answer is r alpha cosine alpha sine of alpha.  We plug that in the SQG equation and we get this function that has to be equal to zero.  When r alpha is a constant, this is automatically zero.  So we can apply the Kran-Darabinovic.  We can go and try to apply the Kran-Darabinovic.  So the spaces, and this is what I meant by adding the log of, lost of the log of derivative  in the solve of the spaces.  So this integral is just to assume that that integral is bounded in L2 and that's enough.  And our spaces X is going to be, as it is for the V states for Euler, is going to be  the sum of cosines.  And we are going to ask this to be in the space hk plus log.  And then the space Y is going to be the signs.  We have the metronary for the Kran-Darabinovic.  And now let me just point out one part of the Kran-Darabinovic is to see that the dimension  of the kernel, the linearized part, is actually dimension one.  So we take the linear part and we want to see an element in the kernel.  We take the element h equal to cosine of m alpha.  We substitute that in the linear part.  And what comes up is that this equation has to be equal to zero, as simple as that.  So it's giving me the eigenvalues and the, sorry, the eigenvalues, the velocity, angular  velocity and it has for each n-fold you have this velocity.  Of course now, so the problem is reduced to taking, to bifurcate from this angular velocity.  And then of course one has to check the co-dimension of the image and the term versatility.  But it's as hard as this.  So the hard part here was to choose the right spaces in which you can work.  The real analytic proof is that you choose spaces that are real analytic properly in  order to framework to work.  So let's come back and see what happens in the smooth case.  Now in the smooth case, remember we are bifurcating from rho cosine alpha sine of alpha.  So now this is a smooth profile.  And f is going to be given.  There's no loss of log derivative because it is smooth, our solutions are smooth.  So we don't have to worry here.  The mixed norms of the solid-volve mixed norms is okay.  And but again we take the cosines to be our space x.  This is the linear part and we have to look for the element, an element in the kernel.  So again we do as before, we substitute our this ansatz in this function.  And what we get is not a scalar equation, we get a function where I tilde looks like  this, t tilde looks like this, and t tilde is an elliptical integral.  Okay, so we need to solve this.  The case of the patch is just a scalar equation.  If we take b to be 1 and we take the limit at a goes to 0, we recover this, we recover  the formula of the angular velocity for the patch.  So quickly, what is the problem?  t tilde m is not self-adjoint, but it's compact, it's a compact operator.  We have a smoothness in i tilde, this is because of f.  And using the properties, we are able to show that there exists a solution.  How is this?  Well, briefly, the reason is that the anti-symmetric  part of the operator, it gets small as we take this guy small.  So the goal is to find, to prove, you boil out the problem to end up showing that the  eigenvalues, there's a gap between the eigenvalues.  Here you don't know what any single eigenvalue, but for the symmetric part, you can have a  guess, and this is a numerical guess, and you can rigorously prove the error you make.  Now taking this sufficiently small, we can make the eigenvalue of the whole thing to  be close to the symmetric one.  How small this has to be?  0.05 is enough.  And how do we do that?  Well, we play with the f in order this to work out.  So once we show that there's a solution, one has to show the regularity.  So this is a bootstrap argument where we take advantage of the regularity of f.  And we have to prove uniqueness, and we have to prove that the co-dimension of the image  is one, and that the transversality is also applies.  Also we managed to do that.  And let me finish in this three minutes some remarks and some very recent results.  So this is a, the solution is a perturbation, right?  So if we perturb a little bit f, everything should work.  And that f has that profile is not a requirement, okay?  A perturbation of that works.  And can we get, we prove that the solutions are C2?  And that's just because we chose f to be C4.  If we make our f smoother, but that will make everything, maybe we have to take us thinner  this distance to be smaller, then we can make the solution to be of higher regularity.  We, since this, in order to show that this is small, we use a computer-assisted proof,  we have to do this for each m.  So it's not a theorem for all m, we prove it for one single m, m equals to three.  But we believe that if you put any other m, you have to correct the numbers.  And maybe this 0.05 will be modified, but it will certainly work.  And this can be extended for any alpha, particularly for 2D order, okay?  And now, very recent results.  So what I just talked about is in the archives.  And what I'm going, this slide will be in the archives in a couple of weeks.  And is, can we get the computer, get the proof of this, the existence of the solutions without  the computer?  And the answer is yes, by bifurcating from the patch.  And the way we bifurcate from the patch is that now this A, A is the, is this distance,  we are going to bifurcate from A equal to zero.  A equal to zero is the patch, and at the same time if A moves to the branch, it opens up  the smoothness of the function.  And when, then, if, okay, so if we do this, the idea is that we rescale, we scale the  profile f.  And what is surprising is that this works for any m, and for any profile f.  f can change sign, it does not have to be monotonously increasing, it can have any shape, it just  have to go from zero to one.  Okay?  But, here we, since it's a random knowledge, we cannot tell you how, how much it opens up.  So we cannot reach the solution.  And that's all.  Thank you.  So, all of these solutions are periodic in time, you know, right?  Yes, it is.  This is, for this point, there are no solutions that are not periodic in time, they're global,  but they're not.  Oh.  How important is it that it's compact support?  Could you have a Gaussian, which is very narrow?  Okay, so compact support makes, so I think, so the question is, for example, in the stationary  solution, right?  So the result of a Frankel that proves that for 2D, the vortex patch is, if it's stationary,  it has to be a circle, right?  But you can construct stationary solutions of 2D Euler that are not completely supported  and they are not circles.  Okay, so the fact that it's compactly supported, it's already telling you that there is some  restriction, very strong restriction.  Okay, so to the problem, but can you construct this from a Gaussian or something and the  answer is yes.  Okay, so yes.  But it gives restriction, for example, in the stationary, so I think that Frankel's result  is also true for SQG and it's also true for a smooth perturbation of the patch.  Are these things rotating in the other direction?  Okay, so it's open.  So 2Fick's result says, so these solutions, the angular velocity is explicit and they're  all positive.  And you know, in order to prove, to use the Kandahar-Havinovich, right?  I'm not saying maybe there's another perturbation that has this structure, the other symmetry  and the angular velocity is different.  I'm not saying that.  In order to prove the Kandahar-Havinovich, but 2Fick's result for the 2D Euler says  that if it's negative, there's, the only solution is the circle.  I believe that can be extended to SQG and also to the case of the smooth case as well.  But that's something to be done.  Okay, then I go on with questions.  Thank you.
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