Spectral Theory for the Transmission Eigenvalue Problem
CC Attribution - NonCommercial - NoDerivatives 4.0 International:
You are free to use, copy, distribute and transmit the work or content in unchanged form for any legal and non-commercial purpose as long as the work is attributed to the author in the manner specified by the author or licensor.
Colton, David
The transmission eigenvalue problem plays a central role in inverse scattering theory. This is a non-selfadjoint problem for a coupled pair of partial differential equations in a bounded domain corresponding to the support of the scattering object. Unfortunately, relatively little is known about the spectrum of this problem. In this talk I will consider the simplest case of the transmission eigenvalue problem for which the domain and eigenfunctions are spherically symmetric. In this case the transmission eigenvalue problem reduces to an eigenvalue problem for ordinary differential equations. Through the use of the theory of entire functions of a complex variable, I will show that there is a remarkable diversity in the behavior of the spectrum of this problem depending on the behavior of the refractive index near the boundary. Included in my talk will be results on the existence of complex eigenvalues, the inverse spectral problem and a remarkable connection (due to Fioralba Cakoni and Sagun Chanillo) between the location of transmission eigenvalues for automorphic solutions of the wave equation in the hyperbolic plane and the Riemann hypothesis.
 I've been here several times. And each time I like it, this is much as the previous time. So as the title says, I'm going to talk about the transmission eigenvalue problem. And I anticipate that most of you don't know what it is or what use it is or anything about it. So part of my attempt here is to try to get you interested in it and hopefully get you new ideas, which the area very much needs. The transmission eigenvalue problem kind of came to the forefront of scattering theory in particular inverse scattering theory about 10 years ago, I'd say. So then before then, it became more significant recently. And basically what it is is that there's an eigenvalue problem, and you can detect the spectral data from scattering data. So you can send signals in, and you get the scattering data back. And from that, you expect spectral data. And from that spectral data, you'd like to find information about the scatterer. Now, the difference from what most are interested in probably is usually in this context, you know the shape. There are a lot of methods of determining the shape of a scatter without knowing anything else about it. So the real problem here is determining the material properties of the scatterer, which basically means you have a differential equation and you want to determine something about a coefficient. And the main problem with it is a non-self-adjoint problem. And it's very unusual and bizarre in a number of ways, which I think will become evident when I talk a little bit about the spectral theory. So I'm going to have my talk as going to be 53 sections. The first one, I want to spend some little while and just tell you where the problem comes from. You see why, you know, where it appears. Then I want to concentrate for the rest of the lecture on the simplest case, where you have a surrhythmic stratified media and a ball, basically. So basically that means you reduce it to a one-dimensional problem. And I want to talk about that problem, because that's the easiest one to talk about. And you'll get some idea what the problems are. And I hope you'll get some kind of flavor of unusual problems in some aspects. And then the third part of the talk, I want to talk about the in-respectal problem associated with that. This is similar in some ways to the in-respectal problem for the sternum-leval problem, which is, of course, a classic problem. But that's typically a self-adjoint problem. This is a non-self-adjoint. And strange things happen. And then the last two slides, I want to give a music connection between transmission eigenvalues and the Riemann hypothesis. In other words, if you know something about where the transmission eigenvalues are, then the Riemann hypothesis is true. And so I'll notice I made no claim that I'm going to prove the Riemann hypothesis. Just to make sure we're on the same page here. OK. So to start off, this is the basic scattering problem. I mean, we're a complicated one, but this is basically what I want to talk about. Here's a scattering medium. You sit in some incident field, which I call UI. Outside the medium, you satisfy the reducer wave equation. We factored out a term e to the minus i omega t. So we're assuming everything's time or money. Inside the medium, it's characterized by this coefficient n, which is called the index refaction, that has to do with the material properties of the medium. That's what you'd like to find information about. So again, we're assuming we know d. We'd like to find out, what is this medium medium? What's going on? Is it plastic? Or is it metal? And so in general, it's a function of the position. And these are just continuity conditions across the boundary. This is radiation condition. This makes the whole problem well posed. k here is the wave number, which is the frequency over the speed of sound. And again, as I said, this is an incident wave. So this is a basic simplified scattering problem in the acoustic case. There are analog results for all of this for elastic waves. And of course, as you notice, I'm sponsored by the Air Force. They're interested in Maxwell's equations. And so that's a concern to that also. And so this is the basic scattering problem I want to focus on. So to formulate the problem, and see where the transmission eigenvalues come from, we'll assume we have an incident wave e to the ikx.d, where d is the unit vector. And remember, we factored out e to the minus i omega t. So this is a plane wave moving in direction d. And it pushes upon the object. And then you have a scattered field. So for those of you not in scattered theory, the image to keep in your mind, although it's not fizzling too hard for it, is a beach ball and a hose. You turn the hose on the beach ball. That's the incident wave. And the water scatters out. And so this radiation condition means the water's going outwards, not into the ball. And a long ways away, the scattered field looks like a spherical wave. So that's e to the ikx, boidex, which in future are just called r, divided by r. And an amplitude, which depends on the observation direction, x-tap. That's x over the length. The direction d in the incident wave and the wave number. And the whole idea of inverse scattering theory is you measure this, you know what you're sent in, and you'd like to find something about n of x. That's the idea. And again, we're assuming that the shape's no. And so this thing is essentially inverse scattering theory. Of course, has a name. It's called the far-fleded plane. And for those of you who may be more in something like quantum mechanics, i minus the dose C per unit of closure related to the kernel of the scattering operator. So in particular, here's the far-fleded operator. Use this amplitude of the scattered wave as a kernel. This is called the far-fleded operator. And i minus 1 over 2 pi f is a scattering operator. So I'm just going to talk about the far-fleded operator. And the far-fleded operator corresponds to the scattered field corresponding to the incident which is weighted by g, just like this far-fleded pad is weighted by g. That's simply due to the fact that the scattering problems when you were inspected you. And this thing is known as the Hervdoss wave function, although it won't appear much in what is in the future. So this far-fleded operator is something that's intrinsic and basic to the whole scattering problem. That's the operator we're looking at. And from this, we're going to be able to determine a certain eigenvalue problem and certain eigenvalues, which is going to be the theme of this talk. So the basic theorem here, transmission eigenvalue problem in the general case, if you look at this far-fleded operator f, it's a mapping from the unisphere l2 into a cell. It's an objective. It has dense range. If and only if it does not exist a non-trivial solution to the transmission eigenvalue problem. So here's the transmission eigenvalue problem. Such that v happens to be a Hervdoss wave function. So for right now, don't worry about v being a Hervdoss wave function, because when I talk about spherical stratified problem, that's always going to be the case. Just focus on the nature of the problem. You have two elliptic equations, both defined in D, satisfied the same Cauchy data. Now in general, you can be shown that under appropriate conditions. In general, that simply means that v and w are 0. In fact, there's an old paper behind it many years ago, more generally. And when did the two elliptic equations, when can these here the same Cauchy data? And in general, you're not going to do it. So in a certain circumstance, there will be certain values of k for which there are non-trivial solutions, and those are called transmission eigenvalues. And because of this theorem, those transmission eigenvalues correspond to the place where the far-field operates on an objective. That means in principle, you can determine that. You have the operator in hand. You can do that because the kernel is a far-field pattern which you measure. And from the knowledge of the far-field operator, you find out where it's going to inject them, and there are ways to do that. That means you can determine the eigenvalues. And from that knowledge, the idea is I want to find out information about the indexed fraction. Now, there's been quite a lot of work done on this problem. There's a lot of information, for example. I took a mention of Professor Saccone, Professor Sunyear, both of whom had this problem. We know that the eigenvalues are discrete and they're appropriate assumptions, and that they exist, of course. And there's vials laws, accounting function, people have looked at. There's been a number of things they've looked at. But some of the most basic questions are unknown. For example, this problem in the appropriate spatial look at is not a self-atonite problem. And so it raises the question, are the complex eigenvalues or not? And if there are complex eigenvalues, what the hell do they mean? What are they? What have they to do with anything? And so I want to focus on a special case where we have n is a function only of the radial position and d is a ball. So that's the simplest case you could have with this problem and tell you about that. And then I think you get some flavor of what this problem is like. Again, here, the values of k, which may or may not be complex, we have to find out. Right now, for the general case, no one knows whether the complex eigenvalues exist or not, for the non-circulatory stratified case. It's an open problem. And I think that's one of the main problems that is now the story of, where are they? How do you compute them? And so forth. OK. Let's see. I'll just start. So I want to consider the simplest case, as I've been promised to you, in a case where n is a function only of r, where r is an absolute value of x, d is a ball radius a. In this case, you can show k is a transmission eigenvalue that in v is always a hertz wave function. So we don't have to worry about that. And you can further assume, since we have this nice circuit stratified medium, we can write w of r as y of r over r. The v of r is the y of 0 of r of r. Remember with the, let's go back here. Remember with the v of w r. Let us leave the link here this way. We can do this. And we can normalize it. We've got an eigenvalue problem. Normalize it so that y equals y not of 0 is 0. And y prime is 0 equals y not of 0 is 1. So we just normalize it. And then we're going to go away. And so in this case, we have that y just satisfies this problem. And basically separation of variables and elementary calculations. So you have a transmission eigenvalue of y of r and the solution of this, which of course is just a sine function. They're just constant c1 and c2. So this is true. This takes care of the Cauchy data at green. Remember we had u equals v on the boundary. And partial nu, partial nu is the unit outward normal. I forgot to mention that. It's 0 on the boundary. That's just those conditions. And so everything comes down to this. So in other words, case of transmission eigenvalue in a special case, if and only if this determinant is 0. So that looks like that should be a pretty simple thing to deal with. We have a problem in ordinary differential equations. And what I want to show you in a short minute, even in the case of n bar, just the constant, which you would think of it about the simplest case you could possibly imagine. You have two ordinary differential equations with constant coefficients, second order. You shouldn't be able to say everything in the world about them. But you always can, but it's not so trivial as you might think. And the main tool we're going to use here, since we're looking at this determinant, is the entire function theory, the entire function of a complex variable. Because you can easily show that y and y prime are entire functions of k. So we have an entire function here. And we should be able to use the results of the entire functions to set it up. And so one of the things I'm going to do, again, my assumptions here are a, the torture before. No one knows very much about transmission eigenvalues. But I'm also not too sure how much everyone knows about 30 of entire functions. So I'm going to, good. I'm glad you shook your heads. I mean, I won't be wasting my time. So I'm going to, the appropriate terms and entire functions I'm going to mention as we need them. So they'll get some flavor of entire functions. For me, I think entire functions are the most beautiful area of mathematics. And to me, in many ways, this is just an excuse to work an entire function. Although I didn't tell the Air Force that. Of course. That's another story. Do you have cameras on you? So here, so d of k first, that's an entire function of k. There's a real for real k. It's bound on the real axis. Then there's a theorem from the entire function 30 that says if dk is not identity zero, there exists a cannibalistic input set of transmission eigenvalues. So we know that, as long as dk is not zero, there are lots of eigenvalues. And there's a theorem by Akassin, Degeneres, and Papenikulow that if dk is zero, that means n of r is one. That means there's no scattering. So in this strictly stratified case, the fact that there exists an infinite number of eigenvalues pretty easy in these entire functions. In the general case, when you have n of x, that's not true. In fact, transmission eigenvalues were invented around 1988. And existence was proven in 2008, I think. Because about 20 years, a lot of people didn't even know whether these things existed. Although they talked about it a lot. But it always made, and I was among those people who talked about it, and always made me uncomfortable talking about something which I didn't know existed, as you can imagine. So it was very comforting. That was particularly resultant. We all have a second that existed in general. It was very comforting. But in the strictly stratified case, it's pretty easy. And this, again, is one of the reasons I want to talk about the simplest case so we can talk about some flavor without going through some detailed and rather technical proofs. So let's assume, not if you want. There's a scatter there. And if n of a is 1 and n prime of a is 0, that means the indexed fraction is varying smoothly across the boundary, which physically is not the most reasonable thing. If you have a medium, you expect there's a jump in the material properties. But does not worry. We'll come to that soon. Then you can do an elementary asymptotic analysis that shows that this determinant looks like this. I say elementary in the sense all you need is a little bit of asymptotic analysis of ordinary differential equations, which appear in the determinant. And you can be right away and get this. And so you can easily see if n of r is between 0 and 1, for example, n of r bigger than 1, you have infinite number of positive transmission eigenvalues. Just from what I thought was obvious there. Everything's real. But I made the assumption n of x is real. And n of r is real. The case when n of r is complex, which is a very important case, corresponds to the absorbent media, that's a whole other story, which we won't get into here. So it exists infinite number of positive transmission eigenvalues. And you can also show that's true if you don't make these assumptions. But to show this, you have to use a little bit of the theory of almost periodic functions, which I wanted to avoid here. So in one slide here, you can show there exists an infinite number of transmission eigenvalues. This takes quite a few pages in the general case. OK, so far? Now, first theorem, I want to talk about the case when n of r is a constant. If you're going to talk about transmission eigenvalue problem and a spherulite stratified case, the simplest thing you can imagine is when n of r equals n not, a constant. And I'm going to show you, that's all obviously you may think, the theorem I'm going to use is Lagrange theorem, which is one of my favorite theorems in the entire function theory, it follows the pattern of archfactorization theorem. It says that if you have an entire function of order less than 2, that means that e to the a r rho where rho is less than 2, in growth. And it's real for real z, and it's only real 0s. And the 0s of f prime are also real, and are separate each other from those of ethic. So it's a Rohl's theorem for n-line functions, basically. And just to give you an example of flavor, it's not true that you have order 2. Here's a function of order 2. And you look at this, you take its derivative, you find out that 0s of f prime of z are complex, even though the 0s of this are 0s. Obviously, it's not working. On the other hand, if you have this, the 0s here are at plus and minus 2. When you take a derivative, the 0s are real, but not done in delays. So the fact that it has to be order 2 is crucial. This does not much to do anything to the top. I couldn't resist talking about the garrison simply because I like it. I thought I'd show you that it's pretty sharp. So here's the result. It is as simple as possible to taste. And it's here to convince you that things are not so simple as you might think. If n of r, n not square, right, that's a positive constant, not equal to 1. If it was equal to 1, we're in trouble. Then if n not is an integer, or a simple integer, all the transmission eigenvalues are real. On the other hand, if it's not an integer, or a simple integer, then there are infinitely many real, infinitely many complex eigenvalues. Now that's pretty unusual behavior for a set of two ordinary differential equations, second order constant coefficients. I mean, I was very impressed by this theorem because although it says Colton-Leone, why do you lose a colleague of mine at Delaware? I can read, write, or join paper, and this is his theorem. So he did it. In fact, he did quite a lot of us here. And so I want to be sure he gets his appropriate praise as we go along. So just to give you an idea of what's happening, let's just take the case n not the integer, just to give you some idea of how the Gaertz theorem comes in here. Then it's easy enough, y of r is that. You can solve the OD right away. And as integer, the non-zero roots of d of k equals 0 are critical points in this entire function. Now, it's just the m of z. And it's in its entire because n is not an integer. That's the crucial part of n not being an integer. And it clearly has only real zeros. And so it's all set up for the Gaertz theorem. You take its derivative and you get the d of k. And so right away, you see that as 0 is a d, k must be real. So it takes a big hammer in a certain way to handle the simplest case when you have this constant coefficients. So let me move on. Now, just to give you some examples, if n not equals 1 half, so it's reciprocal of an integer, and then you have d of k exist, you can see there's a real zero complex. And n not equals 2 thirds. You have this, you have complex one. So just to convince you the theorem's actually true in this special case anyway, it's not some thing that you can do it simply. Now, what happens when you have n of r? I'm not sure how my hand got there anyway. That's supposed to be a delta there. So we have delta not equal to a. The square root of n of rho is between these bounds. There's an infinitely many real and infinitely many complex transmission eigenvalues. So here's the case where n on the boundary is not going to be 1. And if n were a constant, that would violate this situation here. And so this shows you that at least in these cases, we can say something about the case where n of r is not constant. What happens in between these things? I don't know. If I were given open problems, that would be open problem number 1. And you can also show that any n based on n of 1, and all the complex eigenvalues exist, the all line is just parallel to the real axis. Now this is what we will be relevant a little later on. Not this particular theorem, but the idea when I mentioned about the Riemann hypothesis, because this will have to do with where the transmission eigenvalues lie. So the best we can say about this point is they lie in a strip. And this is corresponding not to the case when n of r is going to have a jump. And n of a is not equal to 1 on the boundary. Now just recently, this year, in fact, is not spherically stratified. And Vodaf has shown that if complex eigenvalues exist, they all lie in a strip. So that's true now in the general case. And so what's been happening is a lot of the results we've had for spherically stratified media has given people motivation to try to say, is that true more generally, and try to prove it in the general case, which you'd like. And this is one consequence of that effort. So now let's go back to the spherically stratified case and assume that we actually have the case when things very smoothly across the boundary. So in the case when you have a jump across the boundary, we have those thirds here. Things are going to look different here. So what it's saying is the spectral theory of transmission eigenvalues depends on how the index of refraction behaves on the boundary. That's the point that you should take home here. So just to again give you a little bit about some entire functions, if you can talk about the order by this slip-soup condition, that means the function grows like e to the a r to the rho. So if a function or an entire function of order 2, rho would be 2. And this function of order rho equals 1 is called a function exponential type, and this is tau. So that means it grows like e to the tau mod z. And there's a whole faster entire function of exponential type, which is, again, in my opinion, the most beautiful area of mathematics, arguably, I guess, but in my view, that's the case. And you'll see how many one theorem we'll use to help with the ditch of that. It's coming up right now. So here's the number of 0s in the entire function in the right half plane. Then an entire function f of z, an exponential type, so this is true. So in particular, if it happened that f is bounded on the bound and on the real axis, then that certainly would be satisfied. And you suppose that this growth along the imaginary axis is tau. Then the number of 0s of rho r is tau over pi. And that works for any entire function that satisfies these two conditions, the number of 0s in the right half plane is the same as the type divided by pi. That's a hard theorem. I mean, that took a lot of work by Cartwright and Leibniz and to get. So this is not some simple consequence of atom r or anything else. And this is the number of power of pi. It's called the density of 0s in the right half plane. Now, I want to get a result here on showing that if you have atom r very smoothly across the boundary, that you have an infinite number of complex eigenvalues and real eigenvalues. There are no situations, and no values of n such like we had before where n's an integer. When you have an atom r, it had a jump across the boundary. There were some real ones. You always have real eigenvalues and complex eigenvalues. If you do that, I need a linear operator. And this is a classical Galphan-Leviton operator. So a lot of you know that. And so I'm just going to go over it rather quickly, because it's just an inner operator mapping the sine functions onto the solutions of y to a prouveless k squared n of y equals 0. So first you make the Louisville transformation, and you get this function z, where z is just this. And p is just the density function, which you don't have to worry about. And then z satisfies this condition. And therefore, z is represented in this form here, where this k satisfies this Gursau problem. The important thing to notice here is k does not depend on k. Big k does not depend on little k. There's a statement for you. k does not depend on k. I'm glad no student of mine said that. Big k does not depend on little k. And so once we have this representation to z, then of course we have a representation for y. And so now we have this determinant. d of k again. I'll give you four memories. Delta was just this n adult and so would a. And there's this infinite number of positive real zeros. And since the tides are delta minus a, the density delta minus a over pi. That does how many zeros there are in the right half plane? n of r over r, the limited stat. On the other hand, you take that integral representation and just integrate by part. So this is a cheap way of getting asymptotic expansions once you have that integral representation. So you integrate a couple times. And you look at this thing here and you see n double prime a is not 0. You have this sign delta plus a here. So what happens? This factor here generates an infinite of positive real zeros with this density, as we just said. And n double prime is not 0. This factor here gives a density in the right half plane of delta plus a over pi. It takes a little work to show that. This is just an asymptotic on the real axis. But you can do a little bit of a relation show that's true for the right half plane for complex k2. And therefore, that means that this exponential tie delta plus a, and then for by the Cartesian 11th theorem. So this part was just not to go into the details. This shows you how this representation gives you a quick asymptotic expansion and how crucial what the coefficient is. And so the theorem is that if n of a is 1, n prime a is 0, and delta is not equal to a, and the second derivative a just equals 0, there is infinitely many real and infinitely many complex zeros. So that's opposed to when it has a jump. Sometimes, for example, if n is a constant, you have only real zeros. That's not the case here. And you can show that they do not lie in the strip here of the axis. In fact, Godef again is shown they lie inside a parabola fact region. So here again, depending on how n of r behaves on the boundary, you have a very different distribution of the eigenvalues. In one case, it was a jump in a strip. In another case, in a parabola, it was not a strip. Now, I'm now ready for the last part of the talk. I'm just going to summarize. The first part said, here's a transmission eigenvalue. The problem why is important. You measure spectrodynamic from the far field operator, and that should hopefully give you information about n of r. So far, I've said nothing about information about n of r. I'm going to do that now. The second part of the question is, what is the value of that now? The second part of the talk was designed to tell you things that kind of funny the eigenvalue problem. Sometimes they're real zeros. Sometimes they're complex zeros. Where the distributor depends on how coefficient behaves on the boundary. This reflects it's not only not self-adjoint, it's a pretty bizarre problem. So if you got that message, that's the message you're supposed to get. OK. So let's talk about whether they determine n of r. Now, we know the general complex eigenvalues. So the question is, do all the eigenvalues, real and complex, determine n of r? So the first, there are many results. But the most recent ones are Ackerson, James Pappenichel, again, say that in this case, if n of a is 1, n prime a is 0. So no jump. And if n of r is less than 1, the transmission eigenvalues uniquely determine n of r. Remember, the transmission eigenvalues, again, are determined from the scattering data. Now, this condition, based on this, seems pretty important. There have been a number of papers written saying this results true for n of r greater than 1. All the papers being written have mistakes in them. No one's found any counter examples. So at the moment, the bin is necessary. However, these conditions are, presumably, you'd like to get rid of those. And so I'm going to tell you how to get rid of both these by a trickier idea that a student of mine has, who's working with me right now, and would still just only use the spectral data, the far field pattern. That's the data we have. But we're going to modify the far field of it. So the aim here is to avoid the restriction on n of r. So this head, here is a scattering problem. Don't worry about the actual details. It's a scattering problem. D, you remember, is the unit of law. So you can do this by separation of variables. This is a problem you can just sit down, do to undergraduates to, and then separate variables, and do it. Just make sure you got that message. Here it is again. And then this is a scattering problem. It also has a far field. It has some kind of behavior. It has a far field pattern. So that's it. All we've done, you don't need a computer anything. You just separate variables and do it. You write it out. And so instead of considering our old far field pattern, you can see there's a modified one, where u infinity is again. This is just, I just copied down the first slide for you. This is an original scattering problem. So this is the data you measure. This is the data you're physically interested in. Now you look at this new script F far field pattern. And you worry about where that's injected, instead of the old one. And it turns out, as you might expect, the far field script F is injected with a dense range. If and only if there's not a solution to this modified transmission problem. Where now you see this eta slipped in there. That other problem, the auxiliary problem, had an eta in there. Right, just eta there. And the result is that F is injected with dense range. If and only if this. So it's the same as the transmission eigenvalue problem. Now we have an eta there. So the whole purpose of introducing this modified transmission problem was to get this eta stuck in there. It was one before for the original one. And therefore, the transmission eigenvalue is like before. You ask this determinant. But you have a slide. You have an eta there. So the way you get that, again, is by looking at this modified operator instead of the original one. But you're still using the same data, U infinity. That's given to you. But you're modified by subtracting off this artificial far field pattern. And the first term is that if n of r is not necessarily one, but it's eta squared, there's just infinitely many real and infinitely many complex modified transmission eigenvalues. So instead of having the one here, now we have an eta. And then we now have, if you choose eta so it's bigger than n of r, so you need to know a priori of some upper bound for the coefficient, then the modified transmission eigenvalues through the multiplicity uniquely determine n of r. So note again, there's no restriction on what the data is on the boundary. And there's no condition n of rb1 because we've introduced this auxiliary operator. So that says that if you know these eigenvalues, you're in good shape. Now you're in bad shape because, now I have to tell you some bad news because you can't have only good news. The bad news is very difficult to measure complex eigenvalues from scattering experiments. The k is the wave number omega over c. You measure it. That means there's some noise on the data. So how do you get these complex eigenvalues? Well, there have been various ways to continue functions in exact data, but it's not anything anyone would want to do willingly. So that's a problem that has to be looked at. OK, another case. So in other words, the real complex eigenvalues, if you know them all, you can determine n of r. That's basically what the message is so far. There's nothing like this available for the case. It was not spurious to identify. There are various monotonous problems that your albos have developed and so forth that give you estimates for the industry fraction, but there's all kinds of stuff to be done for higher dimensions. OK, so now let me, in my last two slides, in my last two minutes, probably, I mean, the following advice I've always given my graduate students. No talk is too simple and no talk is too short. And so whatever you do when you go give a talk for a job, do not go over the top. No one's as interested in the subject as you are. So this is due to Fr. Albusaconi and her colleague, Danilo. I guess I have to say, if you have any serious questions about this, Fr. Albus right there, and she'll answer all of us. So you can also consider the idea of transmission eigenvias. You can have to scatter the theory for automorphous solutions of the wave equation and the hyperbolic plane with isometry corresponding to the modular group. So that's a well-known area of complex analysis. And this kind of scatter results in an interactive, an instant field with the boundary of the fundamental domain. So in fact, there's a whole book by Lax and Phillips that you may know about, scattered by automorphic forms, for example. And what comes out of that, they're mainly concerned with the poles of the scattered matrix. But if you want to see more details and so forth, this is a paper which has been submitted. It hasn't come out yet. But I'm sure you can get access to it. And the theorem is the following. This is slide last slide. So in this context, the remuner potholes is equivalent to the statement that all transmission eigenvias lie on this parabola, this red parabola here, except for trivial eigenvias, 0 and 1 quarter. So what's happening in this particular, now, here's, in general, the fact that it lie inside here, is the fact that the zeros of the remuner zeta function lie in a critical strip. So that's analogous to the zeros here lying inside a parabola. That's what. Now, if you knew somehow they were all this red strip, then the remuner potholes is equivalent. Now, so all you need to prove the remuner potholes is to find some, the way this is done, you explicitly exhibit the solution, you say, ah, that's what it has to be. So what you need is some other method to compute where the eigenvias lie, transmission eigenvias lie in a special case. And this other method shows you that the olive island's parabola, you've proven the remuner potholes. Now, I'd like to say for Alvin, I just finished up a paper verse, unfortunately not true. And now I'm not true. I have the foggiest idea of how to make it true. I do recall, do what I said earlier on in the talk. I did show, for example, that in the case I consider with scurvy stratified media, that all the eigenvias lie in a strip. So at least I said I had some result. And it's possible to show where eigenvias lie. It's not a hopeless type of situation. But in this context, it showed me that this particular parabola, that's more than I can say. So thank you very much for your attention. And that's what I have to say. Thank you. Thank you. So you mentioned that small changes in n near the boundary of the object can cause a very large changes in the far field behavior. The inverse problem, how stable is that? Like let's say eigenvalues are real, right? But you still have some noise in your measurement. Good question. Disaster is a very almost problem. The mapping from the index of refraction to the far field pattern, it takes something, let's say, C2 to C infinity. It's very smooth. So when you try to invert it, you have a very severely almost problem. So the main point is that probably you're not going to be able to measure all these eigenvalues at the beginning. You'll be lucky to measure three or four of them maybe. And you hope that that would say something. So the question is not so much maybe, do all the eigenvalues uniquely determine a var, but do the first three or four eigenvalues have something to say? So for example, for those of you who know some of us scattering the resonance, the first eigenvalues say something to volume, the second one says something. Well, in fact, you have vials, that type of idea. You'd like to have something that says the coefficients. There is a vials type theorem. You'd like to say something that coefficient has some physical meaning. That is not there. If you have real and non-real eigenvalues, sort of for an engineering situation, do you expect the components eigenvalues to dominate? Let's say you take the ball and the components play, and then you count. I don't know. Because for the general, let's say, for scattering of a splinter effect, you know roughly what a circle where you start, you have approximately constant times r-square. And you can count some real axis of them. And you can get something like that, how certain dimension of the limits that are like that. Can you identify it? But in general, I don't know what it is. I think a Voders' result in the paper, he has a counting function. And the first coefficient, he has a counting function with the first count in the asymptotes. And the coefficient involves n of x. So if you know asymptotically how it behaves, for example, counting function inside the ball, then you can, you know the first coefficient in the asymptotic that you can get, so in each row that involves n. It's in that paper. OK. In general, as I recall, the papers that are written as counting function include all the eigenvalues together. Inside the ball. Inside the ball, right. Inside the ball. Right. I guess it's a remark, a little bit more less slight, but I think the scattering pools for some look for the transmission eigenvalues, but they are known to be, let's say, 1 half zeros of zeta. So for more, you look for people. Yes. For more, for more. And for some other cases, which is kind of, that's what I did have time to show you. I took two notes also. Modular and I took two. Yeah, yes. So you have zeros of zeta appearing, and then, of course, you can formulate whatever question you ask about zero and zero zeta, you can formulate. Yeah. Yeah. It's not much more. So how does that sound similarly explicitly here? Yeah. So it's the same. But it's just for resonances. Did there a direct relation to resonances? Could you show an engineering case, perhaps, with the transmission eigenvalues, and you have resonances once? No, but what if you mentioned resonances and transmission eigenvalues? For example, if you look at a very simple case, for example, some say you imagine, say in R3, like I would have considered here, and you look at the scattering matrix correspond to that problem, the transmission eigenvalues correspond to the zeros of the scattering matrix, you use it properly by and on a project. The resonances correspond to the poles. So there's a connection between resonances and states in that sense. Then from the period of practical point, if you want just to do a computation for particular given time, you should be at the same level of difficulty as the bit, rather, basically need to find zeros of complex eigenvalues. And as usual, there are differences when zeros correspond to poles and so on. But the difference here is important. The scattering ranges, you know, they're all out of the lower half plane. You only know the scattering data for real take. So determining the scattering resonances, you have to, and they continue with inexact data to the lower half plane to find out where they are. Now for the transmission eigenvalues, remember, there are points where the partial output is no longer objective. So you have to find out where is that output going on objective. And in the case I've been considering, when any of our is real, these transmission eigenvalues all land on the, there are some transmission eigenvalues in the real axis, so you can hopefully be able to measure them. Now the problem in general is you don't know the complex eigenvalues exist and how to measure them. But the idea that I said before in the entire dimension, the hope would be that you could get specific geometric properties associated with each coefficient with each transmission eigenvalue in some way. For example, in scattering resonances and the asymptotic transmission of scattering resonances, the wild type law, each coefficient there corresponds to some geometric property. We don't have something like that for transmission eigenvalues. Now, it'd be nice to have it, but I don't know, but it doesn't exist at this moment. No, I'm just talking about computation. So given, and could you make a arrival over, it can be defined to a sufficient estimate. Is there a, I'm not sure. Yeah, in general, or for the, I mean, I have no doubt that, but there is a way, it's not stable, that there is a way to measure numerically, to compute numerically the transmission eigenvalues from the fact that the operator. And also there is also another connection with scattering metrics. So there are some strange behaviors of the eigenvalues of the scattering metrics if you are at the transmission eigenvalues. So you can catch that it's missionizing the numbers to measure that, yeah. But for general, for even a non-injective that work pretty reasonably well, that has to be stabilized. It's an ill-post problem, it has to be stabilized. The computation is really the only valid, I think for the first few or four probably, you know, trying to calculate more and that's probably not a good idea. So that's the situation I understand. All right, so, both groups, maybe some other people can example, so, Ioscali, the surfaces are kind of potentials which are different. So I wonder if you could back up such examples, you could get examples of the spectral transmission eigenvalue problem. So you would need to read the address next. Go, go. But it's, it's getting, so, examples like in scattering theory are also, you've got to go with the shape of a drop, so. Right, and it's a similar thing here. So, I think there's a very, one of the reasons I had a great deal of pleasure coming here is I think there's a lot of questions on transmission eigenvalues that I don't have a clue on how to think properly about, whereas people here who are more familiar with other types of spectral problems, particularly geometric spectral problems, are going to take, why don't you try filling some, or why don't you look at these situations. So, I agree. I know your answer. I'm sorry. I'm sorry. So, you can talk about either eigenvalues or some, who's that, actually. Okay, so it's a philosophical question. So, if we have problems with, say, real eigenvalues, like you said, Laplace-O'Cray, or if we talk about resonances, then there are often, is such thing as trace work. Right, we are on one hand, to some, say, Fourier transform of some function, of all eigenvalues, all resonances, and then on the right hand side, we get some of values of distribution, and something like, of those geodesics or whatever. So, are there some types of transformable, like this for transmission eigenvalues? No, I know. No, I don't. No, no, no. No, no, no. You have to remember, the serious study of transmission eigenvalues really only started, I think, roughly, with the Ralph's paper in 2010. So, this isn't something that people have been working a lot on. So, that's why I like to appeal to people here, that there's new ideas, though you can transfer some other aspect of structural theory to this, to the possibilities, that's what's needed, I think. I'm stuck in scattering theory, basically. It's a new idea, the outside. Well, sometimes, as we'll say, we'll go for some sort of question, like, what if the right hand is people? For example, Voda, for example, I mean, he made his living for a long time on scattering residence and this kind of thing. But what if it's just what she is? Or doesn't it need to be the right side,
 I've been here several times.  And each time I like it, this is much as the previous time.  So as the title says, I'm going to talk about the transmission  eigenvalue problem.  And I anticipate that most of you  don't know what it is or what use it is or anything about it.  So part of my attempt here is to try to get you interested in it  and hopefully get you new ideas, which the area very much needs.  The transmission eigenvalue problem  kind of came to the forefront of scattering theory  in particular inverse scattering theory  about 10 years ago, I'd say.  So then before then, it became more significant recently.  And basically what it is is that there's  an eigenvalue problem, and you can detect the spectral data  from scattering data.  So you can send signals in, and you get the scattering data back.  And from that, you expect spectral data.  And from that spectral data, you'd  like to find information about the scatterer.  Now, the difference from what most are interested in probably  is usually in this context, you know the shape.  There are a lot of methods of determining the shape of a scatter  without knowing anything else about it.  So the real problem here is determining  the material properties of the scatterer, which basically  means you have a differential equation  and you want to determine something about a coefficient.  And the main problem with it is a non-self-adjoint problem.  And it's very unusual and bizarre in a number of ways,  which I think will become evident when I talk a little bit  about the spectral theory.  So I'm going to have my talk as going to be 53 sections.  The first one, I want to spend some little while  and just tell you where the problem comes from.  You see why, you know, where it appears.  Then I want to concentrate for the rest of the lecture  on the simplest case, where you have a surrhythmic stratified  media and a ball, basically.  So basically that means you reduce it  to a one-dimensional problem.  And I want to talk about that problem, because that's  the easiest one to talk about.  And you'll get some idea what the problems are.  And I hope you'll get some kind of flavor of unusual problems  in some aspects.  And then the third part of the talk,  I want to talk about the in-respectal problem  associated with that.  This is similar in some ways to the in-respectal problem  for the sternum-leval problem, which is, of course,  a classic problem.  But that's typically a self-adjoint problem.  This is a non-self-adjoint.  And strange things happen.  And then the last two slides, I want  to give a music connection between transmission eigenvalues  and the Riemann hypothesis.  In other words, if you know something  about where the transmission eigenvalues are,  then the Riemann hypothesis is true.  And so I'll notice I made no claim  that I'm going to prove the Riemann hypothesis.  Just to make sure we're on the same page here.  OK.  So to start off, this is the basic scattering problem.  I mean, we're a complicated one, but this is basically  what I want to talk about.  Here's a scattering medium.  You sit in some incident field, which I call UI.  Outside the medium, you satisfy the reducer wave equation.  We factored out a term e to the minus i omega t.  So we're assuming everything's time or money.  Inside the medium, it's characterized by this coefficient  n, which is called the index refaction, that  has to do with the material properties of the medium.  That's what you'd like to find information about.  So again, we're assuming we know d.  We'd like to find out, what is this medium medium?  What's going on?  Is it plastic?  Or is it metal?  And so in general, it's a function of the position.  And these are just continuity conditions  across the boundary.  This is radiation condition.  This makes the whole problem well posed.  k here is the wave number, which is the frequency  over the speed of sound.  And again, as I said, this is an incident wave.  So this is a basic simplified scattering problem  in the acoustic case.  There are analog results for all of this for elastic waves.  And of course, as you notice, I'm  sponsored by the Air Force.  They're interested in Maxwell's equations.  And so that's a concern to that also.  And so this is the basic scattering problem  I want to focus on.  So to formulate the problem, and see where  the transmission eigenvalues come from,  we'll assume we have an incident wave e to the ikx.d,  where d is the unit vector.  And remember, we factored out e to the minus i omega t.  So this is a plane wave moving in direction d.  And it pushes upon the object.  And then you have a scattered field.  So for those of you not in scattered theory,  the image to keep in your mind, although it's not  fizzling too hard for it, is a beach ball and a hose.  You turn the hose on the beach ball.  That's the incident wave.  And the water scatters out.  And so this radiation condition means  the water's going outwards, not into the ball.  And a long ways away, the scattered field  looks like a spherical wave.  So that's e to the ikx, boidex, which in future  are just called r, divided by r.  And an amplitude, which depends on the observation  direction, x-tap.  That's x over the length.  The direction d in the incident wave and the wave number.  And the whole idea of inverse scattering theory  is you measure this, you know what you're sent in,  and you'd like to find something about n of x.  That's the idea.  And again, we're assuming that the shape's no.  And so this thing is essentially inverse scattering  theory.  Of course, has a name.  It's called the far-fleded plane.  And for those of you who may be more in something  like quantum mechanics, i minus the dose C  per unit of closure related to the kernel of the scattering  operator.  So in particular, here's the far-fleded operator.  Use this amplitude of the scattered wave as a kernel.  This is called the far-fleded operator.  And i minus 1 over 2 pi f is a scattering operator.  So I'm just going to talk about the far-fleded operator.  And the far-fleded operator corresponds  to the scattered field corresponding to the incident  which is weighted by g, just like this far-fleded  pad is weighted by g.  That's simply due to the fact that the scattering problems  when you were inspected you.  And this thing is known as the Hervdoss wave function,  although it won't appear much in what is in the future.  So this far-fleded operator is something  that's intrinsic and basic to the whole scattering problem.  That's the operator we're looking at.  And from this, we're going to be able to determine  a certain eigenvalue problem and certain eigenvalues, which  is going to be the theme of this talk.  So the basic theorem here, transmission eigenvalue problem  in the general case, if you look at this far-fleded operator  f, it's a mapping from the unisphere l2 into a cell.  It's an objective.  It has dense range.  If and only if it does not exist a non-trivial solution  to the transmission eigenvalue problem.  So here's the transmission eigenvalue problem.  Such that v happens to be a Hervdoss wave function.  So for right now, don't worry about v  being a Hervdoss wave function, because when  I talk about spherical stratified problem,  that's always going to be the case.  Just focus on the nature of the problem.  You have two elliptic equations, both defined in D,  satisfied the same Cauchy data.  Now in general, you can be shown that  under appropriate conditions.  In general, that simply means that v and w are 0.  In fact, there's an old paper behind it  many years ago, more generally.  And when did the two elliptic equations,  when can these here the same Cauchy data?  And in general, you're not going to do it.  So in a certain circumstance, there  will be certain values of k for which there are non-trivial  solutions, and those are called transmission eigenvalues.  And because of this theorem, those transmission eigenvalues  correspond to the place where the far-field  operates on an objective.  That means in principle, you can determine that.  You have the operator in hand.  You can do that because the kernel is a far-field pattern  which you measure.  And from the knowledge of the far-field operator,  you find out where it's going to inject them,  and there are ways to do that.  That means you can determine the eigenvalues.  And from that knowledge, the idea  is I want to find out information  about the indexed fraction.  Now, there's been quite a lot of work done on this problem.  There's a lot of information, for example.  I took a mention of Professor Saccone, Professor  Sunyear, both of whom had this problem.  We know that the eigenvalues are discrete  and they're appropriate assumptions,  and that they exist, of course.  And there's vials laws, accounting function,  people have looked at.  There's been a number of things they've looked at.  But some of the most basic questions are unknown.  For example, this problem in the appropriate spatial look  at is not a self-atonite problem.  And so it raises the question, are the complex eigenvalues  or not?  And if there are complex eigenvalues,  what the hell do they mean?  What are they?  What have they to do with anything?  And so I want to focus on a special case where we have n  is a function only of the radial position and d is a ball.  So that's the simplest case you could have with this problem  and tell you about that.  And then I think you get some flavor  of what this problem is like.  Again, here, the values of k, which may or may not  be complex, we have to find out.  Right now, for the general case, no one  knows whether the complex eigenvalues exist or not,  for the non-circulatory stratified case.  It's an open problem.  And I think that's one of the main problems that is now  the story of, where are they?  How do you compute them?  And so forth.  OK.  Let's see.  I'll just start.  So I want to consider the simplest case, as I've been  promised to you, in a case where n is a function only of r,  where r is an absolute value of x, d is a ball radius a.  In this case, you can show k is a transmission eigenvalue  that in v is always a hertz wave function.  So we don't have to worry about that.  And you can further assume, since we have this nice  circuit stratified medium, we can write w of r  as y of r over r.  The v of r is the y of 0 of r of r.  Remember with the, let's go back here.  Remember with the v of w r.  Let us leave the link here this way.  We can do this.  And we can normalize it.  We've got an eigenvalue problem.  Normalize it so that y equals y not of 0 is 0.  And y prime is 0 equals y not of 0 is 1.  So we just normalize it.  And then we're going to go away.  And so in this case, we have that y just satisfies this  problem.  And basically separation of variables and  elementary calculations.  So you have a transmission eigenvalue of y of r  and the solution of this, which of course  is just a sine function.  They're just constant c1 and c2.  So this is true.  This takes care of the Cauchy data at green.  Remember we had u equals v on the boundary.  And partial nu, partial nu is the unit outward normal.  I forgot to mention that.  It's 0 on the boundary.  That's just those conditions.  And so everything comes down to this.  So in other words, case of transmission eigenvalue  in a special case, if and only if this determinant is 0.  So that looks like that should be a pretty  simple thing to deal with.  We have a problem in ordinary differential equations.  And what I want to show you in a short minute, even in the  case of n bar, just the constant, which you would  think of it about the simplest case you could  possibly imagine.  You have two ordinary differential equations with  constant coefficients, second order.  You shouldn't be able to say everything in the world  about them.  But you always can, but it's not so trivial as you might  think.  And the main tool we're going to use here, since we're  looking at this determinant, is the entire function theory,  the entire function of a complex variable.  Because you can easily show that y and y prime are entire  functions of k.  So we have an entire function here.  And we should be able to use the results of the entire  functions to set it up.  And so one of the things I'm going to do, again, my  assumptions here are a, the torture before.  No one knows very much about transmission eigenvalues.  But I'm also not too sure how much everyone knows about  30 of entire functions.  So I'm going to, good.  I'm glad you shook your heads.  I mean, I won't be wasting my time.  So I'm going to, the appropriate terms and entire  functions I'm going to mention as we need them.  So they'll get some flavor of entire functions.  For me, I think entire functions are the most  beautiful area of mathematics.  And to me, in many ways, this is just an excuse to work  an entire function.  Although I didn't tell the Air Force that.  Of course.  That's another story.  Do you have cameras on you?  So here, so d of k first, that's an entire function of k.  There's a real for real k.  It's bound on the real axis.  Then there's a theorem from the entire function 30 that says  if dk is not identity zero, there exists a cannibalistic  input set of transmission eigenvalues.  So we know that, as long as dk is not zero, there are lots of  eigenvalues.  And there's a theorem by Akassin, Degeneres, and  Papenikulow that if dk is zero, that means n of r is one.  That means there's no scattering.  So in this strictly stratified case, the fact that  there exists an infinite number of eigenvalues pretty  easy in these entire functions.  In the general case, when you have n of x, that's not true.  In fact, transmission eigenvalues were invented  around 1988.  And existence was proven in 2008, I think.  Because about 20 years, a lot of people didn't even know  whether these things existed.  Although they talked about it a lot.  But it always made, and I was among those people who talked  about it, and always made me uncomfortable talking about  something which I didn't know existed, as you can imagine.  So it was very comforting.  That was particularly resultant.  We all have a second that existed in general.  It was very comforting.  But in the strictly stratified case, it's pretty easy.  And this, again, is one of the reasons I want to talk about  the simplest case so we can talk about some flavor without  going through some detailed and rather technical proofs.  So let's assume, not if you want.  There's a scatter there.  And if n of a is 1 and n prime of a is 0, that means the  indexed fraction is varying smoothly across the boundary,  which physically is not the most reasonable thing.  If you have a medium, you expect there's a jump in the  material properties.  But does not worry.  We'll come to that soon.  Then you can do an elementary asymptotic analysis that shows  that this determinant looks like this.  I say elementary in the sense all you need is a little bit of  asymptotic analysis of ordinary differential equations,  which appear in the determinant.  And you can be right away and get this.  And so you can easily see if n of r is between 0 and 1, for  example, n of r bigger than 1, you have infinite number of  positive transmission eigenvalues.  Just from what I thought was obvious there.  Everything's real.  But I made the assumption n of x is real.  And n of r is real.  The case when n of r is complex, which is a very important  case, corresponds to the absorbent media, that's a  whole other story, which we won't get into here.  So it exists infinite number of positive  transmission eigenvalues.  And you can also show that's true if you don't make these  assumptions.  But to show this, you have to use a little bit of the  theory of almost periodic functions, which I wanted to  avoid here.  So in one slide here, you can show there exists an infinite  number of transmission eigenvalues.  This takes quite a few pages in the general case.  OK, so far?  Now, first theorem, I want to talk about the case when n of  r is a constant.  If you're going to talk about transmission eigenvalue  problem and a spherulite stratified case, the simplest  thing you can imagine is when n of r equals n not, a constant.  And I'm going to show you, that's all obviously you may  think, the theorem I'm going to use is Lagrange theorem,  which is one of my favorite theorems in  the entire function theory, it follows the pattern of  archfactorization theorem.  It says that if you have an entire function of order less  than 2, that means that e to the a r rho where rho is less  than 2, in growth.  And it's real for real z, and it's only real 0s.  And the 0s of f prime are also real, and are separate  each other from those of ethic.  So it's a Rohl's theorem for n-line functions, basically.  And just to give you an example of flavor, it's not true  that you have order 2.  Here's a function of order 2.  And you look at this, you take its derivative, you find out  that 0s of f prime of z are complex, even though the 0s of  this are 0s.  Obviously, it's not working.  On the other hand, if you have this, the 0s here are at plus  and minus 2.  When you take a derivative, the 0s are real, but not  done in delays.  So the fact that it has to be order 2 is crucial.  This does not much to do anything to the top.  I couldn't resist talking about the garrison simply because  I like it.  I thought I'd show you that it's pretty sharp.  So here's the result.  It is as simple as possible to taste.  And it's here to convince you that things are not so  simple as you might think.  If n of r, n not square, right, that's a positive constant,  not equal to 1.  If it was equal to 1, we're in trouble.  Then if n not is an integer, or a simple integer, all the  transmission eigenvalues are real.  On the other hand, if it's not an integer, or a simple  integer, then there are infinitely many real,  infinitely many complex eigenvalues.  Now that's pretty unusual behavior for a set of two  ordinary differential equations, second order constant  coefficients.  I mean, I was very impressed by this theorem because although  it says Colton-Leone, why do you lose a colleague of mine  at Delaware?  I can read, write, or join paper, and this is his theorem.  So he did it.  In fact, he did quite a lot of us here.  And so I want to be sure he gets his appropriate praise as  we go along.  So just to give you an idea of what's happening, let's just  take the case n not the integer, just to give you some idea  of how the Gaertz theorem comes in here.  Then it's easy enough, y of r is that.  You can solve the OD right away.  And as integer, the non-zero roots of d of k equals 0 are  critical points in this entire function.  Now, it's just the m of z.  And it's in its entire because n is not an integer.  That's the crucial part of n not being an integer.  And it clearly has only real zeros.  And so it's all set up for the Gaertz theorem.  You take its derivative and you get the d of k.  And so right away, you see that as 0 is a d, k must be real.  So it takes a big hammer in a certain way to handle the  simplest case when you have this constant coefficients.  So let me move on.  Now, just to give you some examples, if n not equals 1  half, so it's reciprocal of an integer, and then you have d  of k exist, you can see there's a real zero complex.  And n not equals 2 thirds.  You have this, you have complex one.  So just to convince you the theorem's actually true in  this special case anyway, it's not some thing that you can  do it simply.  Now, what happens when you have n of r?  I'm not sure how my hand got there anyway.  That's supposed to be a delta there.  So we have delta not equal to a.  The square root of n of rho is between these bounds.  There's an infinitely many real and infinitely many complex  transmission eigenvalues.  So here's the case where n on the boundary is not going to  be 1.  And if n were a constant, that would violate this  situation here.  And so this shows you that at least in these cases, we can  say something about the case where n of r is not constant.  What happens in between these things?  I don't know.  If I were given open problems, that would be open  problem number 1.  And you can also show that any n based on n of 1, and all the  complex eigenvalues exist, the all line is just parallel to  the real axis.  Now this is what we will be relevant a little later on.  Not this particular theorem, but the idea when I mentioned  about the Riemann hypothesis, because this will have to do  with where the transmission eigenvalues lie.  So the best we can say about this point is  they lie in a strip.  And this is corresponding not to the case when n of r is  going to have a jump.  And n of a is not equal to 1 on the boundary.  Now just recently, this year, in fact, is not  spherically stratified.  And Vodaf has shown that if complex eigenvalues exist,  they all lie in a strip.  So that's true now in the general case.  And so what's been happening is a lot of the results we've  had for spherically stratified media has given people  motivation to try to say, is that true more generally, and  try to prove it in the general case, which you'd like.  And this is one consequence of that effort.  So now let's go back to the spherically stratified case  and assume that we actually have the case when things very  smoothly across the boundary.  So in the case when you have a jump across the boundary, we  have those thirds here.  Things are going to look different here.  So what it's saying is the spectral theory of transmission  eigenvalues depends on how the index of refraction behaves  on the boundary.  That's the point that you should take home here.  So just to again give you a little bit about some entire  functions, if you can talk about the order by this  slip-soup condition, that means the function grows like  e to the a r to the rho.  So if a function or an entire function of order 2, rho  would be 2.  And this function of order rho equals 1 is called a function  exponential type, and this is tau.  So that means it grows like e to the tau mod z.  And there's a whole faster entire function of exponential  type, which is, again, in my opinion, the most beautiful  area of mathematics, arguably, I guess, but in my view,  that's the case.  And you'll see how many one theorem we'll use to help  with the ditch of that.  It's coming up right now.  So here's the number of 0s in the entire function in the  right half plane.  Then an entire function f of z, an exponential type, so  this is true.  So in particular, if it happened that f is bounded on  the bound and on the real axis, then that  certainly would be satisfied.  And you suppose that this growth along the imaginary axis  is tau.  Then the number of 0s of rho r is tau over pi.  And that works for any entire function that satisfies these  two conditions, the number of 0s in the right half plane is  the same as the type divided by pi.  That's a hard theorem.  I mean, that took a lot of work by Cartwright and  Leibniz and to get.  So this is not some simple consequence of  atom r or anything else.  And this is the number of power of pi.  It's called the density of 0s in the right half plane.  Now, I want to get a result here on showing that if you  have atom r very smoothly across the boundary, that you  have an infinite number of complex eigenvalues and  real eigenvalues.  There are no situations, and no values of n such like we had  before where n's an integer.  When you have an atom r, it had a jump across the boundary.  There were some real ones.  You always have real eigenvalues and complex eigenvalues.  If you do that, I need a linear operator.  And this is a classical Galphan-Leviton operator.  So a lot of you know that.  And so I'm just going to go over it rather quickly, because  it's just an inner operator mapping the sine functions  onto the solutions of y to a prouveless k squared n of y  equals 0.  So first you make the Louisville transformation, and  you get this function z, where z is just this.  And p is just the density function, which you don't  have to worry about.  And then z satisfies this condition.  And therefore, z is represented in this form here, where this  k satisfies this Gursau problem.  The important thing to notice here is k does not depend on k.  Big k does not depend on little k.  There's a statement for you.  k does not depend on k.  I'm glad no student of mine said that.  Big k does not depend on little k.  And so once we have this representation to z, then of  course we have a representation for y.  And so now we have this determinant.  d of k again.  I'll give you four memories.  Delta was just this n adult and so would a.  And there's this infinite number of positive real zeros.  And since the tides are delta minus a, the density delta  minus a over pi.  That does how many zeros there are in the right half plane?  n of r over r, the limited stat.  On the other hand, you take that integral representation and  just integrate by part.  So this is a cheap way of getting asymptotic expansions  once you have that integral representation.  So you integrate a couple times.  And you look at this thing here and you see n double  prime a is not 0.  You have this sign delta plus a here.  So what happens?  This factor here generates an infinite of positive real  zeros with this density, as we just said.  And n double prime is not 0.  This factor here gives a density in the right half plane of  delta plus a over pi.  It takes a little work to show that.  This is just an asymptotic on the real axis.  But you can do a little bit of a relation show that's true  for the right half plane for complex k2.  And therefore, that means that this exponential tie delta  plus a, and then for by the Cartesian 11th theorem.  So this part was just not to go into the details.  This shows you how this representation gives you a  quick asymptotic expansion and how crucial what the  coefficient is.  And so the theorem is that if n of a is 1, n prime a is 0,  and delta is not equal to a, and the second derivative a  just equals 0, there is infinitely many real and  infinitely many complex zeros.  So that's opposed to when it has a jump.  Sometimes, for example, if n is a constant, you have only  real zeros.  That's not the case here.  And you can show that they do not lie in  the strip here of the axis.  In fact, Godef again is shown they lie inside a parabola  fact region.  So here again, depending on how n of r behaves on the  boundary, you have a very different distribution of the  eigenvalues.  In one case, it was a jump in a strip.  In another case, in a parabola, it was not a strip.  Now, I'm now ready for the last part of the talk.  I'm just going to summarize.  The first part said, here's a transmission eigenvalue.  The problem why is important.  You measure spectrodynamic from the far field operator, and  that should hopefully give you information about n of r.  So far, I've said nothing about information about n of r.  I'm going to do that now.  The second part of the question is, what is the  value of that now?  The second part of the talk was designed to tell you things  that kind of funny the eigenvalue problem.  Sometimes they're real zeros.  Sometimes they're complex zeros.  Where the distributor depends on how coefficient behaves on  the boundary.  This reflects it's not only not self-adjoint, it's a  pretty bizarre problem.  So if you got that message, that's the message you're  supposed to get.  OK.  So let's talk about whether they determine n of r.  Now, we know the general complex eigenvalues.  So the question is, do all the eigenvalues, real and complex,  determine n of r?  So the first, there are many results.  But the most recent ones are Ackerson, James Pappenichel,  again, say that in this case, if n of a is 1, n prime a is 0.  So no jump.  And if n of r is less than 1, the transmission eigenvalues  uniquely determine n of r.  Remember, the transmission eigenvalues, again, are  determined from the scattering data.  Now, this condition, based on this, seems pretty important.  There have been a number of papers written saying this  results true for n of r greater than 1.  All the papers being written have mistakes in them.  No one's found any counter examples.  So at the moment, the bin is necessary.  However, these conditions are, presumably, you'd like to get  rid of those.  And so I'm going to tell you how to get rid of both these  by a trickier idea that a student of mine has,  who's working with me right now, and would still just only  use the spectral data, the far field pattern.  That's the data we have.  But we're going to modify the far field of it.  So the aim here is to avoid the restriction on n of r.  So this head, here is a scattering problem.  Don't worry about the actual details.  It's a scattering problem.  D, you remember, is the unit of law.  So you can do this by separation of variables.  This is a problem you can just sit down,  do to undergraduates to, and then separate variables,  and do it.  Just make sure you got that message.  Here it is again.  And then this is a scattering problem.  It also has a far field.  It has some kind of behavior.  It has a far field pattern.  So that's it.  All we've done, you don't need a computer anything.  You just separate variables and do it.  You write it out.  And so instead of considering our old far field pattern,  you can see there's a modified one, where u infinity is again.  This is just, I just copied down the first slide for you.  This is an original scattering problem.  So this is the data you measure.  This is the data you're physically interested in.  Now you look at this new script F far field pattern.  And you worry about where that's injected,  instead of the old one.  And it turns out, as you might expect,  the far field script F is injected with a dense range.  If and only if there's not a solution to this modified  transmission problem.  Where now you see this eta slipped in there.  That other problem, the auxiliary problem,  had an eta in there.  Right, just eta there.  And the result is that F is injected with dense range.  If and only if this.  So it's the same as the transmission eigenvalue problem.  Now we have an eta there.  So the whole purpose of introducing  this modified transmission problem  was to get this eta stuck in there.  It was one before for the original one.  And therefore, the transmission eigenvalue is like before.  You ask this determinant.  But you have a slide.  You have an eta there.  So the way you get that, again, is  by looking at this modified operator instead of the original  one.  But you're still using the same data, U infinity.  That's given to you.  But you're modified by subtracting off  this artificial far field pattern.  And the first term is that if n of r is not necessarily  one, but it's eta squared, there's  just infinitely many real and infinitely many complex  modified transmission eigenvalues.  So instead of having the one here, now we have an eta.  And then we now have, if you choose eta so it's bigger  than n of r, so you need to know a priori  of some upper bound for the coefficient,  then the modified transmission eigenvalues  through the multiplicity uniquely determine n of r.  So note again, there's no restriction  on what the data is on the boundary.  And there's no condition n of rb1  because we've introduced this auxiliary operator.  So that says that if you know these eigenvalues,  you're in good shape.  Now you're in bad shape because, now  I have to tell you some bad news because you  can't have only good news.  The bad news is very difficult to measure complex eigenvalues  from scattering experiments.  The k is the wave number omega over c.  You measure it.  That means there's some noise on the data.  So how do you get these complex eigenvalues?  Well, there have been various ways  to continue functions in exact data,  but it's not anything anyone would want to do willingly.  So that's a problem that has to be looked at.  OK, another case.  So in other words, the real complex eigenvalues,  if you know them all, you can determine n of r.  That's basically what the message is so far.  There's nothing like this available for the case.  It was not spurious to identify.  There are various monotonous problems  that your albos have developed and so forth  that give you estimates for the industry fraction,  but there's all kinds of stuff to be done for higher dimensions.  OK, so now let me, in my last two slides,  in my last two minutes, probably,  I mean, the following advice I've always given my graduate  students.  No talk is too simple and no talk is too short.  And so whatever you do when you go give a talk for a job,  do not go over the top.  No one's as interested in the subject as you are.  So this is due to Fr. Albusaconi and her colleague,  Danilo.  I guess I have to say, if you have any serious questions  about this, Fr. Albus right there,  and she'll answer all of us.  So you can also consider the idea of transmission eigenvias.  You can have to scatter the theory for automorphous solutions  of the wave equation and the hyperbolic plane  with isometry corresponding to the modular group.  So that's a well-known area of complex analysis.  And this kind of scatter results in an interactive,  an instant field with the boundary of the fundamental domain.  So in fact, there's a whole book by Lax and Phillips  that you may know about, scattered by automorphic forms,  for example.  And what comes out of that, they're  mainly concerned with the poles of the scattered matrix.  But if you want to see more details and so forth,  this is a paper which has been submitted.  It hasn't come out yet.  But I'm sure you can get access to it.  And the theorem is the following.  This is slide last slide.  So in this context, the remuner potholes  is equivalent to the statement that all transmission eigenvias  lie on this parabola, this red parabola here,  except for trivial eigenvias, 0 and 1 quarter.  So what's happening in this particular, now, here's,  in general, the fact that it lie inside here,  is the fact that the zeros of the remuner zeta function  lie in a critical strip.  So that's analogous to the zeros here  lying inside a parabola.  That's what.  Now, if you knew somehow they were all this red strip,  then the remuner potholes is equivalent.  Now, so all you need to prove the remuner potholes is to find  some, the way this is done, you explicitly exhibit the solution,  you say, ah, that's what it has to be.  So what you need is some other method  to compute where the eigenvias lie,  transmission eigenvias lie in a special case.  And this other method shows you that the olive island's  parabola, you've proven the remuner potholes.  Now, I'd like to say for Alvin, I just finished  up a paper verse, unfortunately not true.  And now I'm not true.  I have the foggiest idea of how to make it true.  I do recall, do what I said earlier on in the talk.  I did show, for example, that in the case I consider  with scurvy stratified media, that all the eigenvias  lie in a strip.  So at least I said I had some result.  And it's possible to show where eigenvias lie.  It's not a hopeless type of situation.  But in this context, it showed me  that this particular parabola, that's more than I can say.  So thank you very much for your attention.  And that's what I have to say.  Thank you.  Thank you.  So you mentioned that small changes in n  near the boundary of the object can cause  a very large changes in the far field behavior.  The inverse problem, how stable is that?  Like let's say eigenvalues are real, right?  But you still have some noise in your measurement.  Good question.  Disaster is a very almost problem.  The mapping from the index of refraction  to the far field pattern, it takes something, let's say,  C2 to C infinity.  It's very smooth.  So when you try to invert it, you  have a very severely almost problem.  So the main point is that probably you're not  going to be able to measure all these eigenvalues at the beginning.  You'll be lucky to measure three or four of them maybe.  And you hope that that would say something.  So the question is not so much maybe,  do all the eigenvalues uniquely determine a var,  but do the first three or four eigenvalues  have something to say?  So for example, for those of you who know some of us  scattering the resonance, the first eigenvalues  say something to volume, the second one says something.  Well, in fact, you have vials, that type of idea.  You'd like to have something that says the coefficients.  There is a vials type theorem.  You'd like to say something that coefficient has some physical meaning.  That is not there.  If you have real and non-real eigenvalues,  sort of for an engineering situation,  do you expect the components eigenvalues to dominate?  Let's say you take the ball and the components play,  and then you count.  I don't know.  Because for the general, let's say,  for scattering of a splinter effect,  you know roughly what a circle where you start,  you have approximately constant times r-square.  And you can count some real axis of them.  And you can get something like that,  how certain dimension of the limits that are like that.  Can you identify it?  But in general, I don't know what it is.  I think a Voders' result in the paper,  he has a counting function.  And the first coefficient, he has a counting function  with the first count in the asymptotes.  And the coefficient involves n of x.  So if you know asymptotically how it behaves, for example,  counting function inside the ball, then you can,  you know the first coefficient in the asymptotic  that you can get, so in each row that involves n.  It's in that paper.  OK.  In general, as I recall, the papers  that are written as counting function  include all the eigenvalues together.  Inside the ball.  Inside the ball, right.  Inside the ball.  Right.  I guess it's a remark, a little bit more less slight,  but I think the scattering pools for some look  for the transmission eigenvalues,  but they are known to be, let's say, 1 half zeros of zeta.  So for more, you look for people.  Yes.  For more, for more.  And for some other cases, which is kind of,  that's what I did have time to show you.  I took two notes also.  Modular and I took two.  Yeah, yes.  So you have zeros of zeta appearing,  and then, of course, you can formulate whatever question  you ask about zero and zero zeta, you can formulate.  Yeah.  Yeah.  It's not much more.  So how does that sound similarly explicitly here?  Yeah.  So it's the same.  But it's just for resonances.  Did there a direct relation to resonances?  Could you show an engineering case,  perhaps, with the transmission eigenvalues,  and you have resonances once?  No, but what if you mentioned resonances and transmission  eigenvalues?  For example, if you look at a very simple case, for example,  some say you imagine, say in R3, like I  would have considered here, and you  look at the scattering matrix correspond to that problem,  the transmission eigenvalues correspond to the zeros  of the scattering matrix, you use it  properly by and on a project.  The resonances correspond to the poles.  So there's a connection between resonances and states  in that sense.  Then from the period of practical point,  if you want just to do a computation for particular  given time, you should be at the same level of difficulty  as the bit, rather, basically need  to find zeros of complex eigenvalues.  And as usual, there are differences  when zeros correspond to poles and so on.  But the difference here is important.  The scattering ranges, you know, they're all  out of the lower half plane.  You only know the scattering data for real take.  So determining the scattering resonances,  you have to, and they continue with inexact data  to the lower half plane to find out where they are.  Now for the transmission eigenvalues,  remember, there are points where the partial output  is no longer objective.  So you have to find out where is that output going  on objective.  And in the case I've been considering,  when any of our is real, these transmission eigenvalues  all land on the, there are some transmission eigenvalues  in the real axis, so you can hopefully be able to measure them.  Now the problem in general is you don't  know the complex eigenvalues exist and how to measure them.  But the idea that I said before in the entire dimension,  the hope would be that you could get specific geometric properties  associated with each coefficient with each transmission  eigenvalue in some way.  For example, in scattering resonances  and the asymptotic transmission of scattering  resonances, the wild type law, each coefficient  there corresponds to some geometric property.  We don't have something like that for transmission eigenvalues.  Now, it'd be nice to have it, but I don't know,  but it doesn't exist at this moment.  No, I'm just talking about computation.  So given, and could you make a arrival over,  it can be defined to a sufficient estimate.  Is there a, I'm not sure.  Yeah, in general, or for the, I mean, I have no doubt that,  but there is a way, it's not stable,  that there is a way to measure numerically,  to compute numerically the transmission eigenvalues  from the fact that the operator.  And also there is also another connection  with scattering metrics.  So there are some strange behaviors  of the eigenvalues of the scattering metrics  if you are at the transmission eigenvalues.  So you can catch that it's missionizing the numbers  to measure that, yeah.  But for general, for even a  non-injective that work pretty reasonably well,  that has to be stabilized.  It's an ill-post problem, it has to be stabilized.  The computation is really the only valid,  I think for the first few or four probably,  you know, trying to calculate more  and that's probably not a good idea.  So that's the situation I understand.  All right, so, both groups, maybe some other people  can example, so, Ioscali,  the surfaces are kind of potentials which are different.  So I wonder if you could back up such examples,  you could get examples of  the spectral transmission eigenvalue problem.  So you would need to read the address next.  Go, go.  But it's, it's getting, so, examples like  in scattering theory are also,  you've got to go with the shape of a drop, so.  Right, and it's a similar thing here.  So, I think there's a very,  one of the reasons I had a great deal of pleasure  coming here is I think there's a lot of questions  on transmission eigenvalues that I don't have a clue  on how to think properly about,  whereas people here who are more familiar  with other types of spectral problems,  particularly geometric spectral problems,  are going to take, why don't you try filling some,  or why don't you look at these situations.  So, I agree.  I know your answer.  I'm sorry.  I'm sorry.  So, you can talk about either eigenvalues  or some, who's that, actually.  Okay, so it's a philosophical question.  So, if we have problems with, say, real eigenvalues,  like you said, Laplace-O'Cray,  or if we talk about resonances,  then there are often, is such thing as trace work.  Right, we are on one hand, to some, say,  Fourier transform of some function,  of all eigenvalues, all resonances,  and then on the right hand side,  we get some of values of distribution,  and something like,  of those geodesics or whatever.  So, are there some types of transformable,  like this for transmission eigenvalues?  No, I know.  No, I don't.  No, no, no.  No, no, no.  You have to remember,  the serious study of transmission eigenvalues  really only started, I think, roughly,  with the Ralph's paper in 2010.  So, this isn't something that people have been  working a lot on.  So, that's why I like to appeal to people here,  that there's new ideas,  though you can transfer some other aspect  of structural theory to this,  to the possibilities, that's what's needed, I think.  I'm stuck in scattering theory, basically.  It's a new idea,  the outside.  Well, sometimes, as we'll say,  we'll go for some sort of question,  like, what if the right hand is people?  For example, Voda, for example,  I mean, he made his living for a long time  on scattering residence and this kind of thing.  But what if it's just what she is?  Or doesn't it need to be the right side,
201807021551-Colton_lrv_0_1427.png 201807021551-Colton_lrv_1428_1967.png 201807021551-Colton_lrv_1968_3647.png 201807021551-Colton_lrv_3648_3695.png 201807021551-Colton_lrv_3696_3971.png 201807021551-Colton_lrv_3972_9851.png 201807021551-Colton_lrv_9852_33395.png 201807021551-Colton_lrv_33396_36155.png 201807021551-Colton_lrv_36156_38351.png 201807021551-Colton_lrv_38352_45119.png 201807021551-Colton_lrv_45120_46727.png 201807021551-Colton_lrv_46728_46919.png 201807021551-Colton_lrv_46920_47411.png 201807021551-Colton_lrv_47412_53507.png 201807021551-Colton_lrv_53508_56579.png 201807021551-Colton_lrv_56580_58295.png 201807021551-Colton_lrv_58296_60179.png 201807021551-Colton_lrv_60180_60311.png 201807021551-Colton_lrv_60312_60875.png 201807021551-Colton_lrv_60876_61079.png 201807021551-Colton_lrv_61080_61871.png 201807021551-Colton_lrv_61872_61979.png 201807021551-Colton_lrv_61980_62027.png 201807021551-Colton_lrv_62028_62087.png 201807021551-Colton_lrv_62088_62183.png 201807021551-Colton_lrv_62184_65123.png 201807021551-Colton_lrv_65124_65255.png 201807021551-Colton_lrv_65256_65339.png 201807021551-Colton_lrv_65340_65579.png 201807021551-Colton_lrv_65580_65627.png 201807021551-Colton_lrv_65628_65699.png 201807021551-Colton_lrv_65700_66251.png 201807021551-Colton_lrv_66252_66563.png 201807021551-Colton_lrv_66564_66611.png 201807021551-Colton_lrv_66612_66731.png 201807021551-Colton_lrv_66732_66935.png 201807021551-Colton_lrv_66936_67415.png 201807021551-Colton_lrv_67416_67499.png 201807021551-Colton_lrv_67500_69095.png 201807021551-Colton_lrv_69096_69227.png 201807021551-Colton_lrv_69228_69299.png 201807021551-Colton_lrv_69300_70187.png 201807021551-Colton_lrv_70188_70235.png 201807021551-Colton_lrv_70236_70283.png 201807021551-Colton_lrv_70284_70331.png 201807021551-Colton_lrv_70332_70715.png 201807021551-Colton_lrv_70716_70883.png 201807021551-Colton_lrv_70884_71099.png 201807021551-Colton_lrv_71100_71315.png 201807021551-Colton_lrv_71316_71495.png 201807021551-Colton_lrv_71496_71543.png 201807021551-Colton_lrv_71544_71557.png