So first of all, thank the organizers for this invitation. It's a real pleasure to be here.

And yes, so I'm gonna talk about the surface quasi-geostrophic 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-geostrophic 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, 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 solutions, because the motivation of the proof

come from the vortex patch problem. Okay, so the quasi-geostrophic is a model that models the dynamics of big masses of air in the mid-latitudes, where the Coriolis force plays an important role.

This is the quasi-geostrophic equation, it's a three-dimensional equation, and the dynamics that happens at the surface is what is known as the surface quasi-geostrophic equation, which is this one. So this model, and here's several references,

it had been posted and studied in several papers in the geophysical context, but it was not until 1994 that Peter Konstantin, Andy Maeda, 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, vorticity, is where omega is a vector and is 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 omega.

So we compare this with the surface quasi-geostrophic equation, what plays the role of the vorticity will be the gradient perpendicular of theta, which is the derivative with respect to x two of theta,

partial theta with the derivative with respect to x one. And the equation for this vector

is equivalent to the equation that the vorticity satisfies, but this is a vector in three dimensions, this is a vector in two dimensions. The, 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 to theta. So, but there is also a relation with the conservative quantities.

In the case of Euler, 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 L2 of the velocity, the energy, is conserved.

Well, in the case of SQG, the level sets of theta plays the role of the vortex lines, they're tangent to the gradient

perpendicular to 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 SQD, there are more quantities to conserve, is that you actually can bound the LP norm of the velocity for p bigger than one, and it's 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 solvents norm, and in the case of SQG, similar 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 criterions for the formation of singularities for Euler, well, I can mention too, and the Vilcatto-Maidah criteria,

which says that if you control the integral in time of the L-infinity norm of the vorticity, then there's no blowup 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, ds. And there is the Konstantin-Maidah, the Konstantin-Phetherman-Maidah 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, okay? So, if there's a singularity,

the Lipschitz norm has to stop being bounded. Okay, so there's goes the similarities

with one equation and the other, and now is there singularities for SQG? And the answer is we have no idea, okay? Numerics don't show any evidence for smooth initial data, okay? 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 wrist transform.

So, in one dimension, let's take the velocity to be the Hilbert transform of, in this case, let me call it f in 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 equal to zero. This is, if 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 solo space, and there is finite time singularities

for this model, for all initial data. This is strictly bigger equal to zero. But of course, in this model, there's the incompressibility 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 on the outside, and write the 1D model in this case,

and what we have is this toy model. And for this, and let's take for the Cauchy problem, f of zero bigger equal to zero.

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 x naught such that f of x naught is zero, then finite time singularities

at the point x naught, okay? For this model, if the data stays positive for the model A? Yes. For A, not for B? Yeah, and for B.

Oh, 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 a version of...

Are the equations coupled? This, this one and this one? No, the derivative here is hitting. All the derivatives outside? Yeah, yeah, this is a transport equation and here the derivative is outside.

Just to see that the incompressibility may change. Well, just to get the flavor. Okay, good. Can I ask one question again about, you mentioned 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 first for Euler and various people coming in. Yeah, yeah, yeah, yeah, yeah. So, there's nothing like this then. There used to be at the time, for SQG, at the time, but then it was ruled out.

Numerically as well and analytically. 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 in 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 problem, for SQG.

Okay, so, allow me to add a parameter to the system because I'm going to mention several references and in order to quote those references is good that I add the parameter, okay?

This parameter is in the power of the Laplacian. I'm gonna add an alpha. So, I'm interpolating the system from 2D Euler to SQG. Alpha equal to zero is 2D Euler, where theta is the vorticity and alpha equal to one is SQG, okay? So, for the global existence,

2D Euler is critical. So, one, there's global existence for alpha equal to zero but when alpha is strictly bigger than zero, it's an open problem.

And radial, well, and so, for the purpose of my talk, let me point out that the radial functions, if theta is radial, then it's a stationary solution and that has to do with the powers of the Laplacian preserves the symmetry, okay?

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 exists non-trivial global solutions, smooth for the SQG equation, okay?

How are these solutions? How do they look like? Well, okay, so, let me take. Excuse me, this is for the modified one or the? This is for alpha equal to one, SQG. So, the modified one, I'm gonna call it generalized SQG, okay? And anytime I don't mention SQG,

I will say it's some alpha, okay? So, this is for SQG. And so, the solutions are going to be a perturbation of a radial function, theta, where theta is gonna be one inside this circle.

Outside another circle, theta is gonna be zero. And in between, it goes from zero to one in a smooth way. Actually, it's going to be C2. Now, it's going to be a perturbation of this,

and the perturbation is going to have a threefold symmetry. So, we rotate by 120 degrees, we get the same structure. And these solutions rotate, and they rotate with a 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 Drichil, and he published a paper

where he actually computed the solutions of the SQG of C1 half-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 it's smooth. Now, where does the C1 half 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 one and theta equal to zero, we are going to take the level sets of theta,

and we're going to parametrize them by z alpha rho t. Rho tells me in which level set I am, and alpha tells me in what place of the level sets. 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 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 unknown 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. Then 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 solution solves this equation in the support of f,

f of the derivative of f. Okay, 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 monotonically 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 zero, if we take a to zero,

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 risk transforms, and we have the time derivative that disappears,

this is the equation for lambda and x alpha rho. And we have to solve for this. We have some freedom to choose the parametrization 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 functional

that in order to be a solution is to be zero. Okay, so one observation is that if alpha is independent of, sorry, r is independent of alpha,

then what we're doing, we have this radial solutions, then it becomes zero, 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 zero. Okay, so we are in the tool that we're going to use is the Crandall-Rabi knowledge theorem, right? So our parameter mu is going to be

the angular velocities 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 they got two 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 two derivatives exist, and they're continuous, and we have to find a lambda together with the spaces that the dimension of the kernel of the linearized operator

is of dimension one, the co-dimension of the image is also dimension one, and then the fourth condition is the transversality. So if those conditions are satisfied, Crandall-Rabi knowledge says that you can bifurcate,

there's a solution starting from lambda star, where it's the 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 gonna go and revise the vortex patch problem. Vortex patch problem is now, and we're dealing with a solution

that the vorticity is let's say one here and zeros outside, it satisfies Euler equations, the curve, let's parameterize the curve of t, c alpha t, I'm sorry, like this,

then it's possible to have, when you look at the evolution of c, 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 one inside the curve and zero outside the curve. So this is very well understood

and was for the Cauchy problem, Shemin was the first to prove that this problem is globally well posed for c, one plus alpha. And later there was another proof, a different proof by Andree-Avertochen, Peter Konstantin of the same result.

And the result, and then one can apply Udovitz 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 Jose Luis Rodrigo in 2005, where he proved that it's locally well posed. And 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-Moser scheme. Then Francisco Ganceto 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 Ganceto and Bob Strain, they show that no splash singularities can be formed.

So if you expect a self-intersection of the boundary, then the curvature must blow up. Recently, this year, Sascha Kisilev, Lenia Riznik, Yao Yao, and Flatos, Andrei Tartos,

they proved 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 their symmetry.

So already, since it's touching the boundary, so the chord arc condition, all this local existence theorems is fundamental that you have a control of the chord arc condition. But they're proving local existence where the chord arc condition fails.

And they managed to prove local existence but for a small range of alphas. And the range, I don't know, it's one over 12 or one over 24. But it's more or less, it's a fraction, okay? And they proved local existence. Moreover, they are able, since there's a lot of symmetry,

and 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, Gansetto's strange theorem says

the curvature has to blow up as well. But it's a contradiction, so it's really 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 zero and one over 12. Excuse me, but this one with boundaries is really a different problem.

It is a different problem. But from the point of view of local existence. Oh, from those point of views, 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've put this, then these patches do not move from there, but they attract each other and it has to intersect.

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's, so together with a team in Madrid, Fontello, Mancho and Jose Luis Rodrigo, when in 2005, we found this numerical simulations. And we took a patch, this is SQG,

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 blowup, locally self-similar blowup. 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.

So what Drichel and Scott took this ellipse at some point before the blowup, before the singularity, and this is how it looks like, and they zoom. 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, so for the patch problem, the circle are stationary solutions for all alpha. Kirchhoff proved that the ellipses for 2D Euler

are rotating solutions. So known also by these states, okay? 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 a patch can actually change convexity, back and forth, okay? So, then in 1978, Deem and Tabuski predicted

for 2D Euler, the existence of rotating solutions with M-fault symmetry. Three, here are three, four, but any M-fault 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, two things, Schmid, Joan Mateo, and Joan Berdera gave another proof using the Krandar-Rabinovich 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 Tuthick 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 Tuthick and Joan Mateo, they proved C alpha and a regularity bifurcating from the lips, and we actually proved that they're real analytic. And okay, so how did I start working on this subject?

Why did I get interested? Well, when Tuthick, Midi, and then the Joannes, proved this theorem, the first theorem. So I invited Joann to give a talk in Madrid, and I got interested in this subject.

And then I talked with Tuthick, 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, okay? That are non-trivial. So a year after, Tuthick sent me a paper,

and he worked with his former student, Hasainia, and they proved that for alpha bigger than zero, and alpha smaller than one, these solutions exist, and they are C alpha.

But there was a problem, they couldn't reach alpha equal to one, the SQG equation. So I invited Tuthick to Madrid, he gave a talk, we chat, and he explained to us what was the, why they couldn't do it, and they actually were not able,

not even to prove C infinity between zero and one. And then Angel, Javier, and myself, we studied the problem with a different formulation, and then we realized that, I mean, the problem for going from alpha smaller than one

to alpha equal to one, is this loss of log of derivative. They were working on Helder spaces. To implement this in Helder spaces is quite hard. But to implement it in Sobolev spaces is trivial. So we changed our framework,

and instead of working on the complex plane, we work on the real world and in Sobolev spaces. So at first we prove 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 Krander-Rabinowitz tells you that it's unique, the branch. So where it overlaps, the solution is real analytic. But Krander-Rabinowitz 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 analytic and it becomes infinite. I don't know. So, it's 10, two, three? Yeah, okay. A little bit more, because you started a couple of minutes later.

Thank you, Camilo, thank you, thank you. Okay, 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 ansat is R alpha cosine of alpha sine of alpha.

We plug that in the SQG equation, and we get this function has to be equal to zero. When R alpha is a constant, this is automatically zero. So we can apply the Krander, we can go and try to apply the Krander-Rabinowitz. So the spaces, and this is what I meant by adding the log of,

lost of the log of derivative in the sublattice spaces. Well, it's 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 these 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 sines. We have the imaginary for the Krander-Rabinowitz. And now let me just point out one part of the Krander-Rabinowitz, 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 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 m fold, you have this velocity. Of course now, so the problem is reduced

to taking, to bifurcate from these angular velocities. And then of course one has to check the co-dimension of the image and the temperature. 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 for the framework to work. Okay, 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 f is smooth, our solutions are smooth.

So we don't have to worry here, the mix norms of this, the solid volat mix norms is okay. 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 this ansat in this functional. 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 is a elliptic integral. Okay, so we need to solve this. For the case of the patch, it's just a scalar equation.

If we take b to be one and we take the limit as a goes to zero, we recover this, we recover the formula of the angular velocity for the patch. Okay, so quickly, what is the problem? t tilde m is not self adjoint,

but it's compact, it's a compact operator. We have the 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 to showing that the eigenvalues, there's a gap between the eigenvalues. Here you don't want 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 has this has to be? 0.05 is enough, okay?

And how do we do that? Well, we play with the f in order this to work out. So once we show that we can, there's a solution where it 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.

This is 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 solutions is a perturbation, right?

So 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. Can we get, we proved that the solutions are C2.

And that's just because we chose f to be C4. If we make our f smoother, but then we'll 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 a small, we use a computer assistant proof. We have to do this for each M. So it's not the theorem for all M, we prove it for one single M, M equal 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 in particularly for 2D order, okay? And now very recent results.

So what I just talked about is in the archives and this slide will be in the archives in a couple of weeks. And is can we get the computer, get the proof of 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 this distance, we are going to bifurcate from a equal to zero.

A equal to zero is the patch and at the same time 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 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 monotonically increasing,

it can have any shape, it just have to go from zero to one. But here we, since it's at Crandar-Rabinovich, we cannot tell you how much it opens up. So we cannot reach the solution.

And that's all, thank you. So all of the solutions are periodic in time eventually? Yes, yes. So for this problem, 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 Frenkel, 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 with Frenkel's result, it's also true for SQG, and it's also true for a perturbation of the, a smooth perturbation of the patch. Could these things rotate in another direction?

Okay, so it's open, so to fix results, says, and so these solutions, if the angular velocity is explicit, and they're all positive.

And you know, in order to prove, to use the Kanda-Rabinovich, 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. But in order to prove the Kanda-Rabinovich, but to fix the result for the 2D Euler, says that if it's negative,

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.