 Can you hear me? OK. At first, I'd like to thank the organizers for inviting me. What I have to talk about is fairly simple. But I thought it may be of interest to this audience, not just because of the physical problem, but because of the mathematical method that is being used, which is quite general. So we're talking about steadily. This is work with my PhD student, Tyoon Kim. So most of the credit problem goes to him. And so we're talking about a pair of translating vertices, oppositely rotating. As you can see, one occupies the space D1. The other one occupies the space D2. One with positive vorticity, the other with negative. And the vorticity is constant. Outside this domain D1 and D2, you have inviscid fluid. So when you have a pair of vertices, they move. That's the steady state we're talking about. So if you're in the frame of the moving vortex, you have a uniform stream at infinity. Oops. You have a uniform stream right there. Let's see. OK, that's fine. Here's the velocity u. And that has to be determined as part of the problem on top of finding the shape of the vortices. So these, in practical situations, most of you probably know that wingtip vortices behind aircraft, you get large vortices which roll up. And with the effect of viscosity eventually smearing out vorticity, you probably get pretty much like a vortex patch, a pair. And these are important technologically for safety reasons. So Pierre Humbert actually numerically calculated these shapes. And there is one parameter family of solutions. And I'm showing the shape of the upper half vortex. And the one parameter that's changing is actually the centroid distance between the two vortices. And until now, mathematical analysis has only been done, as far as we know, for small vortex size. But the question is, how do you actually prove the existence of this solution? And the method we use, as you'll see, is quite general, not just for this problem. And that's part of the reason I decided to choose this material for the talk. But going back to the vortex problem, it's fairly simple. What you have outside the vortex, you have a irrotational flow, irrotational potential flow. So you have a complex velocity potential in two dimension. So that's complex velocity potential is denoted by w. And so the problem is, for a given choice of length scale and vorticity, you can set each of them to 1. You want to determine a simple differentiable closed curve, delta d1. And because of the assumed symmetry between the shapes, you only have to determine the shape of the upper half vortex. So delta d1 is the boundary. You have to find that. And you have to find the mapping function z, which maps to this boundary, such that on the boundary, you have a streamline condition. That's essentially the streamline condition, what you see in equation one. This is the velocity is given by this expression that's quite easily derived. And this expression is valid on the boundary. This is the effect of the other vortex, located at the image location in the lower half vortex. And you have a uniform stream, which is right there. Now, let me talk about the quasi-solution method. The method is quite general, as you'll see. And the only way I can present it in a short way is to abstract it and give the, suppose you have any nonlinear problems, ODE, PDE, integral differential equation, whatever you call. We call a nonlinear problem, and think of a nonlinear operator n acting on v equals 0. Suppose somehow, some way through numerics or otherwise, you obtained v0 with this following property, that when you plug it into the equation, then the resi- I'm getting used to this. All right, where am I? OK, when you plug it into the equation, your residual or the remainder term is small. Further, suppose, of course, this operator comes with initial slash boundary condition. Suppose v0 comes close to satisfying those initial and boundary condition. Then the problem of finding solution to 1 is equivalent to the following problem for the quantity v minus v0, where all I did was I rewrote this in a slightly different way. I took the linearized part of this Frichet derivative, of this operator evaluated at v0, and I called this linear operator. Basically, in the ordinary language, if you're talking about Navier-Stokes, you're linearizing about a solution. That's all we are talking about. The linearized operator is out. So you put that on the left-hand side, and r is the remainder, and you put everything else on the right-hand side. Then if you examine these two, these are exactly the same equation. If L happens to be invertible, and in many cases, they are, and they can be proved to be, then what happens is you can invert this equation and write it in this form. So E0 happens. You need a little bit of E0, because remember, when you solve LE0 equals 0, it comes with perhaps small non-zero initial or boundary condition. So E0 takes into account the solution to this, which satisfies the missed boundary initial condition for v0. And L inverse is the inverse of this operator, and you can write in this way. One thing to be noted is that if you have small initial slash boundary condition, E0 will be small for any reasonable problem. So E0 will be small. If L inverse is bounded, and residual is something that you can control, depending on how accurate you make the solution, this is also small. So these two quantities, which I call E soup 0, is small. And then you have this non-linearity. So this becomes a weakly non-linear problem. So a strongly non-linear problem has been transferred into a weakly non-linear problem, which is very suitable for a Banach contraction theorem, proving existence. And in the process, you're able to come up with an analytical formulation of v0. And I'll talk about that, quasi solution, meaning approximated. So in fact, in more details, you could actually, if you are in some Banach space, if you write it down, as you can see over here, I'm not going to push this point too much, you can find estimates to prove that you have a contraction of this operator M in a small ball under certain conditions that guarantees existence of solution. Anyway, so now I want to make some remarks, because some of what I'm saying is related to past work. What I talked about is a very natural idea. So it's very natural that it has come, other people noticed it in different contexts. For instance, if someone is trying to prove that there is a solution to n of u epsilon equals 0 for small epsilon, where epsilon is some parameter, and they have an exact solution when epsilon equal to 0, what do they do? They do exactly what I mentioned, namely, they formulate an equation for E, E being u minus u naught. So it will involve the Frichet derivative of the operator, and you have to invert, and you exactly do what I said. So this idea also is the basis of rigorous error bound. There is a lot of literature and computer assisted proofs which basically rely on this idea. What appears to be relatively new, at least as far as we know, is that you can determine an accurate analytical, reasonably compact analytical expression for approximate solution with rigorous error estimate, and of course, proving that you have a solution. So there is a whole bunch of recent work. My collaborator, Ovidu Kostin, has been collaborating with me in many different things. He really got this ball started, and I've been, and my students and our students have been working on a number of these problems. So this is the latest example, and it has a number of twists which make it a bit harder to apply directly. First of all, let's go back to the problem we are trying to solve. We're trying to find a curve with this condition. With this condition. Now you see, in order to apply the quasi solution idea, I have to extricate the linearized part of the operator in a way that I can invert it properly. Now if I keep it in this form, on the boundary, then it is a difficult task to take this. For one thing, notice that there is a derivative of Z with respect to nu outside, and DWDZ, this also involves a derivative inside the integral, right? So in a sense, it looks like a fully non-linear equation. So how do I actually turn it into a form that I can do the inversion? The idea that will be pursued in this particular context is that we'll take this equation which is only true in the boundary and turn it into an equation which is true in the everywhere, not just in the boundary, okay? So these are the challenges I mentioned, and of course you have to choose the right space, right Banach space in order that you can deal with non-local operators. And also the other thing is, you not only have to determine the shape, you also have to find the velocity of translation. So in the equation I gave, it is not clear how does the velocity get determined. Of course numerically you can plug in and try and put in more points than unknowns and the extra point you put in for the velocity and try to solve that way, but analytically it is not clear how in the world the velocity gets determined. So what we do is we use a conformal mapping representation where let's consider for instance, because of the four and F symmetry and the top and bottom symmetry of the shapes, it's enough to consider what's going on in the domain D that you see over here. So D is this left, I mean D is actually infinity out here. So I'm only considering this quadrant, the shape outside the upper half vortex, okay? And what we look at the conformal map that sends this annular region into this exterior domain out here. So that these two points which are zeta equal to plus minus rho corresponds to D which is infinity and A which is at the origin. The smaller circle corresponds to straight line segments of this domain. Namely, CD corresponds to CD, D is up at infinity and the AB corresponds to this segment. And this segment A to D which is A to D right here, the real axis corresponds to the smaller circle, whereas the larger circle corresponds to the free boundary. Now from Schwartz's reflection principle, it immediately follows that if you took the full circle, annular region, that would correspond to the full upper half vortex. And if you actually, I mean, anyway. So let me, there are other things you can do to access the lower half vortex through reflection. Now, so the question is we need to remember, we need to find z as a function of nu. Nu happens to be the angle parameter for the circle, zeta equal to e to the i nu. So we choose that and because of well-known properties of conformal map, you know that you must have a singularity which is a simple pole at zeta equal to minus rho in order to put that point to infinity and the remaining part has to be analytic function in the unit disk, alpha of zeta is analytic function and you have to add an analytic function outside the circle of radius rho squared. So this, just from general theory, you know this must be the formulation for the conformal map. So what you, and this because it's analytic in the unit circle, a power series is a good representation and you have these coefficients which because of symmetry they happen to be real. With that, now remember we have the freedom of choosing one length scale to be one and we might as well choose a naught to be one as the choice of that length scale and vorticity omega is chosen to be one. So that's the choice we'll make. As far as the size parameter, we'll choose rho as the size parameter. When rho tends to zero, you get a nearly circular vortex and when rho becomes, tends to plus, tends to one, you get extremely deformed vortex. Now remember, our object is to take a condition which is only true on the boundary and then turn it into a differential, integral differential equation valid in the entire domain. In order to do that, what we do is we notice first a few things. This quantity minus u is actually behaves like z to the minus two as z tends to infinity. This is just an observation based on the symmetry of the two vortices. So the remaining part of the velocity after you take out the free stream part decays like one over z squared. Because of that, this and remember z equal to infinity is mapped to z equal to minus rho. Because of this property, this quantity in the zeta plane has a double zero at the location z equal to minus rho. That double zero completely cancels out the double pole that you have of this quantity because if you take this derivative with respect to zeta, you see a double pole. So this is a nice observation in the sense that this thing is singularity free even at minus rho. So it's singularity free in this domain. Now you need to do one more thing because we have a condition on the boundary. The other thing we notice is that just looking at this function where it is singular z equal to minus rho and using a method of successive reflection, it's easily possible to come up with this j function and j has no singularity in the domain of interest but it has the nice property that if you put that part in, this guy is singularity free. In other words, sorry, I didn't say it right. There is a singularity which is z equal to minus at minus rho which I've canceled out over here and the remaining part is singularity free. So this combination is singularity free and is real on the two little circles. Anyway, what's the point? The point is that at the end of the day, I don't want to bore you with all the details, at the end of the day, what used to be an equation on the boundary turns out to be something which is valid everywhere, which is this equation that you see. This is the equation which is valid everywhere. Now the good thing about writing it this way is you can then divide by this quantity and then isolate this higher derivative and then treat this non-locality in a way that you can deal with it. And so this is the equation that corresponds. This is the L that has been identified. It consists of the first derivative on the left-hand side and a very non-local operator which I'm going to write in a second. These are the residual and it involves the variation of the velocity from the quasi solution which is small u and you got all the non-linearity lumped on this side. Where they're fairly complicated, these are non-local operators. Anyway, I can talk about more details than, but I probably bore you. But what is the idea in this, what is the overall strategy to solve this problem? Or for given, so the way we start off is from given u, what is small u? Small u is the difference between the free stream velocity and u naught, which is, say, some numerically-computed velocity of the vortices. So the way we start off is you take a u in some open small interval and then you prove that this equation which is actually one, we are projecting out the k equal to zero mode. This, we're introducing this up. If you have h with the cosine series, p means that you take away the average quantity. So we have this equation which you take away the zero term, you solve this. And then, later on, come back to the zero-th mode and show that there exists a choice of the free stream velocity correction so that the remaining scalar equation is satisfied. And at the end of the day, you have to prove that the expression that you got, you have is univalent because you don't want vortex boundaries crossing each other. So with the space we choose, you have a cosine expression for the real part of the analytic function g, which we call e, and it has a cosine series and we choose a norm in the four-year space of this kind. So then, this norm is useful because if you have an analytic function and you have a real part, say in an annular region, a real part of the boundary is given by cosine series, then you have an almost Hilbert transform-like relation between real part and imaginary part. So if you can control this series coefficient, you can control both the real and imaginary part of that analytic function. And it also has the property, you have a Banach algebra, all the non-linear terms can actually be computed, estimated. So we need to find the constant c and we need to also find the inverse bounds for the inverse of the operator l. And that's the whole idea. What is the main result? The result is the following. For a given row, row equal to row naught, I'll say a little bit more on row naught. Row naught is somewhere between zero and one. There exists a unique solution of the vortex patch problem in the neighborhood of U naught, F naught, which I'm going to give an expression. U naught is, of course, a function of row and F naught is the conformal map. Because it has a, this is the similar part, which is only the analytic part in this way. Alpha has to be analytical. Alpha and this U naught. Where the diff, where the solution z minus the F naught is less than 10 to the negative five and U minus U naught equal to... Part of this solution F is univalent in the annular domain. So it does correspond to a physical work. In the case when row is small, then it's a fairly easy exercise. Formal perturbation expansion in powers of row. So any order we stopped at, we included, fairly simple only for the solution. All the way to point five. Point five happens to be reasonably deformed, but not hugely deformed. Now, in the case when row is larger, then it is possible to find a uniform representation, but then it becomes a bit ugly when you have a parameter as well as a variable because of how big the expression. So I'm expressing your, expressing particular cases. So row equal to row naught, when it's nine tenths, then the solution has this representation. You get the accuracy of 10 to the negative five for the derivative. Whereas if you want to have a similar expression just based on a truncation of the power series, you're 120th term. So this expression, you notice what is, what the origin of this expression, you have a little bit of this and this, and the way it came about is you take the, to a path approximation, and then, because you're doing path approximation for the derivative, which is the right thing to do, because if you do the other way, you remember that derivative is not a bounded operator. So you take a function, you make an approximation, its derivative will be bad. So you always start with the highest derivative in the problem. Come up with an expression for the solution, for the data, and then you integrate. And the reason I put this all in on rational numbers is because first of all, you need fewer digits, I mean fewer numbers to express something with high accuracy, and also the fact that you don't want any round of error or anything of that kind. So you have only, you have 12 numbers here and other 12 numbers here, and that's it. That describes this too high precision. The rule is that lambda j, of course, has to be outside the unit circle, otherwise it won't describe an analytic function. So the lambda j is that I have reported here are all in absolute value bigger than one. Right? Yeah, no, the sign I just chose because when you take the derivative of that, you get the pole, minus sign goes away. That's the reason I got the one. All right, so, and that's the shape of the corresponding vortex, all right? That's fairly deformed, I would say. So we can do that even for higher values of rho. There is no limitation, it's just that the solution expression becomes longer. So, so let me, do I have one minute? Yeah, yeah, it's a vortex pattern. Yeah, yeah, yeah, it's a constant on this and a minus constant on the other one, yeah, yeah, yeah, yeah. Okay, so, so now you see the non-local operators are not so nice, and they involve integrals like this. So how in the world do we do this in a nice way, all right? Which only involves sums, sums of maybe 50 numbers or 100 numbers. This is what is done in this, I'm just illustrating one term. You see, this guy is a known function, this is the quasi solution. So we use the fact that we know this as an analytic function of both eta and zeta, and we can prove its analytic properties on an annular ring for zeta and eta. And we can show, we can find uniform bounds for this expression. So I'm talking about this integral. Now, knowing it's bound on an annular region like this allows you to, tells you that if you actually replace this by a sum, say 50 terms, you will get exponential accuracy. In fact, in this case, we don't need too many terms, all right? You just add them, and you can bound it with precise exponential bounds. And so this is one of the things. So you basically need to look at the non-local operators which is t and try to bound them. So the way we do this is we want to be as close to the truth as possible, namely as close to what we observed numerically. If you just use bound brute force, you'll get extremely liberal bound, which will not help you in the proof. Okay, so what we do instead is we take this sum expression, we take this sum expression, express the four-year modes of this sum in the four-year space on the unit circle in zeta, after you integrate in eta, it becomes a function of zeta. And that's how you control the tails and are able to come up with very precise bounds. So let me, this is, I'm running out of time, but there is also a trick. You have to find the quasi-inverse. So you pose this problem in the sequence space, find an L prime analytically, so that L prime times L minus identity is small. You use the solvent identities to actually find the inverse of L inverse, so of L. Anyway, conclusion. We have our non-local, non-linear problem of steady transition can be analyzed using a quasi-solution method. One of the important steps was to use analytic properties of conformal map and complexity. What used to be, what used to be got transferred to an equation, and the method has no prior restriction on the border, on the central distance, or parameter, which is the row, and except that when row gets closer, the solution representation becomes far more complicated, as we have. So that the space where the power series coefficients that I, k squared and k is L2 method, by the way, is quite general. We have used it to know the boundary value problems, but it becomes, as the problem becomes more complicated, it becomes more computer-assisted, which is, we, thank you for it. Okay, thank you very much.