Lecture 08 Shocks and Patterns in Hyperbolic and Hyperbolic Parabolic Balance Laws

So last time we were talking about the Henry result that the the spectra of the end states + minus are the boundary of essential spectra and here okay here's my minus to plus graph to remind you we we're our operator is linearized around stationary wave if we just freeze and look at the constant limits and look at their spectra at the spectra of of those constant solutions In our case, we actually know they look like this like parabola that these will number one we proved part the curves lambda of of limiting spectra of spectra or the limit limiting states which by the way is the dispersion curves given by the 4A symbol. Those are in the essential spectra. That was part And now the more important part for us is to show that to the right of these There is either all the central spectrum or else only resolvent and point spectrum. That's what that's that's what we want to show. And to do that finally we need to know something more about the resolvement equation and spectra. Okay. So before do this, I'm going to take an excursion do some asmtotic OD theory because the resolment operator is the solution of an OD and okay so we're going to now develop two very important results for us. The first one which is something proved with seeoula and sanetta and these are essentially simultaneous in 1998 and here's it's called but we named it we gave it the catchy name call it the gap lema and that caught on. Okay, so I'll explain the name after. Okay. So let of XP be constant matrix plus theta of XP. And theta is order of to the minus little theta in derivatives on. So this is for on the half line and is parameter. So I'm thinking of this of course I'm thinking of the resolvant equation which is an example of an OD of the type we're going to look at. So okay. Okay. So but there are also other parameters right besides just the frequency lambda there are model parameters and it might be important to be able to change these. And let's see. And now let B+ of be an iggon vector of A+ with value mu of I'm going move over then the first order OD which is dx = of as solution of xp equals vp to of to the mu of times identity plus let's see guess said that in the in the wrong order. Well, okay, let's just write it like this. Okay. Minus theta tilda and this is for any let's see that's the guy. Thank you. Can you okay. think Okay, maybe. Let's see. No, we've got it off and on. Better. So, don't think it's frozen, but it just the camera doesn't show that last board. It only shows the first two. okay. That's need to go to number two. yes. We can see. Yes. Okay. good, good, good. All right. Thank you. You're okay. Yes. let me actually do want to write it like Okay, so this is the the constant coefficient solution that is if we just crossed out theta here, that would be an exact solution of the OD. So the idea is near any so if we have we've assumed that this is exponentially decaying and so long as we have fast have exponential decay this says we can find solutions that are near the constant coeffic the limiting constant coefficient solutions. And here this is for any Okay. Let's see. And to say B+ of to the mu plus of So it's just matching these guys, right? And where or any theta tilda less than theta and this is this is connected to the the issue had the other day. definitely need to give up something here. Theta tild is just matrix. It's it's matrix function. It's It's just I'm going to get let's see. Maybe should maybe should just say that the answer for this is the on this this is the answers for the solution. And does it does it pass some? let's see. Maybe need to just see what do want to say here. want to say it's little more yeah. Okay. let me just say it's easier. Let's just put it put it like this where when when yeah it's it's maybe better to write like let me write it like this. So it's plus of to the minus mu to the mu plus Okay. plus order of plus of to the plus of px times to the minus theta delta I'll just write it like that. This is this is really what want. or any and this this big depends on theta minus theta tilda. And it's going to blow up as let these come together. can't just take theta tilda equal to theta. Everything would explode. And where the solution is as smooth in as are a+, b+ and mu plus. So whatever smoothness have in the parameters, will inherit it in the solution. Okay. And this is closely related to result that can be found in Copel asmtoic OD book. In his setup, theta is just L1 not exponentially decaying but he gets similar result. But in order to get in order to but for his result he needs spectral gap between mu and the remaining values and this so this lama is basically saying exponential decay can substitute for spectral gap. So it's called the gap lema because of the absence of gap. It's the opposite. Okay. So, but the and and the proof it's really very similar to the proof in Capel, but it's actually somewhat simpler. It's harder with only L1. So, we just we're taking advantage of the fact we know we have exponential decay almost all the time and we want so we want little that gives us more information and we want to take advantage. Okay. So the idea is we we're going to do basically fix point mapping. So contraction mapping and the leoponov setup which is used in all of invariant manifold theory and all kinds of fixed point setups. So the idea here is we're going to set up okay let's see so we write dx minus plus of equals theta of xp and we'll consider this as forcing term and use the linear solution formula variation of constants but with different starting points for different parts of the solution. That's leopon. If you start at one point and treat the Koshy problem, that's just variation of constants, suppose. But if you do this then like the term leoponov. So I'm going to set I'm going to try to find equals TV. This is the mapping and and that is going to be defined as plus the very important start without loss of generality mu equals zero. can just do an exponential change of coordinates in to make mu equals 0. And haven't lost any of the smoothness, right? just a+ is converted to plus minus That's the only change. Okay. Now, so now what I'm looking for is something that's stationary at infinity. And so here's what how prescribe the stationary part. And now I'm going to integrate out to plus infinity the plus of - This will be less than zero. And so what want is something that's kind of unstable because I'm integrating back from infinity towards back. So here I'm going in the backwards direction. want things to decay. So want the unstable manifold. But actually want not just unstable. I'm going to say weekly unstable which I'll define below. This is the weekly this projection onto the weekly unstable space times theta of this is theta of of dy and then subtract the integral from to of to the + - this is now pos greater than or equal to zero pi strongly stable theta of yp of Okay, so what I've done is for the weekly unstable space, I'm going from out to plus infinity and initialize at plus I'm basically initializing at infinity integrating down toward This is what choose to happen at plus infinity. don't want anything blowing up. want only the neutral mode to remain. That's my prescription. And then the strongly stable. Well, okay, that's fine. don't have to the letter at the end of the at the bottom of the integral. Yes. Okay. So, okay. So, what I'm going to do now is I'm going to restrict. Yeah, forgot to say restrict to in infinity large. This is I'm going to try to construct it just out near infinity. If can do that, I'm fine because can integrate then then backwards to zero. can if find solution out near infinity, can integrate it back towards zero and any exponential bounds have will just accumulate some large constant in front of them because this OD does have bounded flow from to zero. It may be large, but it's bounded. So, really only have to control things out near plus infinity. And if do, I'm done. That's important though because I'm going to leverage this. So, okay. So, this is actually it was not zero. apolog. So, now I'm looking at from to infinity and will need to be large at some point to get contraction. That's that's the trick because why? Well, that makes theta small. mean, it's it's pretty clear why want to do that. Otherwise, it could be quite large. would have not really much control at all if don't restrict it somehow because have big constant in front of my my assumption. This is all that's driving the contraction and the big is constant. It could be like didn't say anything about that. Yes. And now Yes. Yeah. There are two projections. So okay and these are the are projectors or iggon projections onto the subspaces with okay the first one is so mu weakly unstable is greater than negative theta tilda strictly greater and mu strongly stable is strictly less than negative theta to okay well how can be sure this happens well it's okay for dense set of thetas happen to choose theta tilda that that landed on value can't do this because didn't allow for equality want strict difference. But okay, if if hit one don't like, I'll bump it. It'll only have another theta that's stronger. So it it's it's no harm, but it makes it so don't have to write as much. Okay. So yeah, so there's some strategy going on, right? So, if you had spectral gap, you'd you you'd yeah, you you actually you do much the same, but you you would just have zero here, and life would be lot harder. But you'll see this is going to work out nicely for us. Okay. So basically nice way to think about this is project the original equation onto onto these two subspaces and treat them separately. I've effectively you know I've diagonalized the system. It's coupled only through the forcing term which I'm treating as as as known. So diagonalize that as well. And now one of them start from infinity and integrate into the other one start from and integrate up to and okay and that's that's it and I'm going to yeah that's all that's all need and could take anything like could have initialized almost anything like believe at it would be okay but zero is good let's see is larger than yes I'm going out. I'm keeping only the the very asmtoic part. What's B+? B+ was I'm just the plus is just to indicate I'm on positive halfline. and V+ is just and A+ is just the limit at plus infinity. plus the of the matrix. plus is an igen vector of that matrix of the plus matrix. Mu plus is the igen value. So all the pluses are together at plus infinity is the that's the idea. Okay. Sorry. Yeah. In the in the goes to plus. Yeah. So for now we're just basically dividing the problem in two parts. We're just going to look on half rays because we can't, you know, we there's no reason to know what's happening near zero, right? We can only know what happens near infinity because all we've assumed is exponential convergence to infinity at infinity. In between anything could happen. So this is the best we could do. So that's what we're going to do. on the other side with the minus or yeah we would the same now we would do exactly the same but you you just yeah you you flip the whole argument there's no difference between plus and minus except that need to choose one so feel like going in the in the positive think positive we're we're we're not going to we're not going to glue we have matching condition yeah and the matching condition is just that they they do match, right? They have to exactly match. If we're looking for solutions of an resolant equation or an igen value equation, we insist that they be continuous across zero. So that's going to be the match. But but that that comes minute in minute. So let's see. So let me now show contraction. Hi Kevin. Yes. so is it okay that this sa tilda negative sedat tilda be spectrum so it's okay? No it's not okay I'll I'll yeah in fact I've I've excluded that by by saying these are my requirements. this your requirements but it's really also requirement on theta tilda. If happen to land, if happen to accidentally have theta tilda that hits an iggon value, I'll take slightly larger value of theta tilda that is still less than theta. I'll carry out the argument for that one and it implies the other one. So what if it's essential spectrum and everything are continuous? Would like there's no diff. So right now this is not okay. This is an OD. Okay. Now I'm I'm I'm really thinking about the resolvent equation. The resolvent equation is finite dimensional OD. Okay. Usual OD. So this is result that I'm going to use to to analyze the resolant equation. Okay. So yeah. So right here we've gone from infinite dimensions down into finite. Okay. So only point spectra. Yeah. Yeah. So there's no essential spectrum. The only spectra are the igen values of the matrix A+. Okay. Okay, plus is finite dimensional. Okay, thank you. byN matrix. Yeah, probably should have emphasized that guess, right? Okay, so now we want to prove contractions. So we look at TV1 minus TV2. Okay, and we take the just soup norm contraction is is what we're going to get. Okay, so this is less than or equal to. Okay. So, integral from to plus infinity to the + - pi weekly unstable time theta of dy. And then I'm just going to pull out v1 + minus v2 the supnorm of v1us v2. I'll just pull out of the whole thing because that that can come out of the integral. That's the supreum over the whole domain of integration from to plus infinity. Okay. So yeah, that's should write it more carefully. It's the supreum over to plus infinity. of of just the just the the usual soup norm that we use for for most OD things. Of course, for invariant manifolds, it's often weighted norm, but this is even easier because we removed the growth and decay that we want. We're just looking for stationary mode. So, we just work in the ordinary soup norm. Okay, so we've got this the the the plus is gone. It subtracts out the constant term is gone and all we have left is integral from to of to the + - pi strongly stable theta oops of dy and again times v1us v2 norm And this is this comes out really nicely. okay. So okay less than or equal to now I'm going to use the fact that okay because made these assumptions on theta have growth and decay bounds on to the know that that the weekly unstable will be bounded by big constant. That's why there's here. And can put to the theta delta - and then theta See, I'm going to actually pull out big constant big constant time v1 minus v2 and then the rest. Okay. And for theta, but I'm not even going to put theta because I'm going to put my bound assumed on theta, which is to the minus theta Let's see. And this is okay. But this is minus theta tilda and then have plus integral from to of to the minus theta tilda - eus theta And what I'm using is that this is this is less than or equal to zero here and this is greater than or equal to zero. And that's where get my two bounds. And the interesting thing is these are exactly the same expressions. It's very funny. So and that's what greatly simplifies this argument. So it's just integral from to plus infinity of to the minus theta tilda - Well, actually, let me pull out the I'm going to pull out what want, which is the part, the part not in the integral. And have to the theta tilda minus theta That's all that's left. And this is negative. is positive. This is bounded. So that means can actually I'm running out of board. So just cross this out and put it in with the But now notice well actually do know you know want want * to the minus theta to always be less than 1/2. So have contract less than strictly less than one but can achieve that by going back and re rechoosing because is bigger than or equal to by assumption. So this is large argument here. So no matter how small theta tilda is, can make this times can make * eus theta be uniformly less than 1/2 say and then it's contraction. So somewhere had to use that was out near infinity but yeah it's actually so that's done. said some other things said smoothness was you inherit all the smoothness and and things like that. That's actually that's true for any contraction mapping with parameter. Yes. of bus. Yeah. So this this one. Okay. So let's see. yes. Those are projections on to so so for instance the weekly the let's see weekly unstable. If wanted to be strongly unstable this would be greater than zero. But I'm going to allow my neutral mode that want to be in there so can prescribe it at at at infinity. And I'm going to also allow some more because it doesn't harm anything in in the estimates. mean what's the type of separation? They're on to the igen spaces whose igen values have these properties. Yeah. Of of plus. The igon the igen spaces of a+ that are associated to these igen values. The ones that grow grow or decay very weakly and the ones that decay well I'm calling it strongly even though this might be small but it's strict decay strictly. And then have some this is non-strict growth you have expensive where the Okay. So for the unstable ones it's because I'm going backwards. Yeah. Yeah. And the other one is going from the one from let's see. Yes, that's that's what's Yeah, that's what's saving me here. Okay, this is going on to plus infinity, right? And this is hypothesis on the yeah on the on on that that it's it's growing. Let's see that in the backward direction it would if it were strongly unstable it would be exponentially decaying in the backward direction. But it's not it's just not growing too fast. See, this is this is exponentially growing but not too fast. And it's being saved by this. But the funny thing is mean probably wrote four or five versions of this before noticed these are exactly the same. You can just put them together. And it's interesting to look at Capel's version where the two are are are different. They're inherently different. And then you have to treat this one pretty carefully. It's harder proof. But but this one it just becomes very simp. If you have exponential decay, life simplifies lot and you can prove more. And we we were in gardener and and Capu and Santo were also we were in scenarios where we needed more. we we could not accept we didn't have spectral gap and we still needed to define such thing and so yeah and you'll see one one of one of those cases is stability of viscous shock waves you're going to see where this comes in handy but right now it's just convenience we don't really need it for what we're doing Okay. Okay. So that gives us way to very you know kind of arduously construct all the solutions that we want. We can make we could go vector by igen vector and construct these things and then put them all together to get expression for resolving. And this is what did for many years. nearby with the same back with no because it's not an it's it's it's it's perturbed by function in right it's it's well you get something with the same growth rate let's say if you want to call that an but but it involves the the integration of the OD yeah it's solution it's not an it's solution to the OD not not not yeah so which is more it's not merely an igen vector of of XP it's the solution of the OD that has this property that it it asmtoically behaves like the constant coefficient limit which we would like intuitively to to think, but we need we need to prove it. And so this is one one proof that's pretty flexible. Okay. What about any questions online? Gia, hope you liked seeing Leoponov Peron. You gave us some lectures back home on using this method. Okay. all right. Okay. So, everyone okay with dilemma? It it's, you know, it's not so unusual, right? Just that it's kind of special situation. See where my Okay. So now, so used that lema for many years until Mativier and found great improvement to that. So here's I'm going to give you another one, corollary. It's actually quite an immediate coralary you'll see. And this one is Mativier. And this is in what 200 4 so bit later. and we called it the conjugation the conjugation lema. Okay. So now and claim this is huge improvement. So of xp equals + plus theta of xp. Same setup and this is finite dimensional od. Theta is order of to the minus theta with no tilda in derivatives. is however many we need. and okay and derivatives derivatives and say jp derivatives which could be and so it can extend up to if it's analytic in we will inherit that that's important because we used analyticity of the resolvent with respect to lambda we used that plenty We definitely need that property. Okay. Then there exists coordinate transformation. So we'll set I'm sorry the OD didn't write an OD. And the OD is dxv = of xp Again there's coordinate transformation. If take = of xp and is the identity plus theta tilda and theta tilda decays like to the constant time eus theta tilda any theta tilda less than theta. The same key assumption such that the equation in is constant coefficient plus of cable. It's the same setup but much better result. It includes the other result is the same same A+ but now didn't assume any don't have to know the structure of A+ just say can find change of coordinates that's so like to say it like this it's like an analog of Floques lema lema says if you have periodic coefficient OD there's periodic coefficient transformation that changes it to constant coefficient. This one says if you have an exponentially asmtotically constant OD there's an exponentially asmtoic coordinate transformation sorry asmtotically exponentially decaying coordinate transformation that converts it to constant coefficient. So it has the same right and these are two you know really big applications periodic things and things that are asmmptotically constant other than that things get you know lot harder so this allows you to treat to treat that case now if want to come up with the this function of xp that did before would just take the exact solution of the constant coefficient equation and would apply the coordinate transformation would get and have the same bounds. So it's immediate to get the other one. But it's better because what if don't have all vectors? What if have Jordan block or you know or or other strange things? What if I'm changing looking at what if 48 transform in some other direction so that I'm looking at family of resolve and odes that may change their structure dramatically through Jordan blocks and other things all those kind of topological changes they're subsumed in this okay so think you you can see that it's better and it implies the the old one but here's the really funny thing. The old one implies this one almost trivially. It's it's trick. It's one of my my favorites. And and the same is true for if you want to prove Lok's lema, you can do it in the same way. Of course, there are many ways to do that, but here's the proof. Okay. So, we're going to derive the homological equation. So we say dx equals That's our od. So that means dx of tz equals tz. That says dxt plus dx is equal to And now we put in but the xz is our goal is to make that plus Okay. So now let me write this out. Notice that now everything has This has to be true for all in fact, right? Because can give any data want. If this if this is if this coordinate transformation is to do what say, it has to apply for this equality has to hold for every And that means can just take out of the equation. And that gives me this homological equation which is dxt is equal to now it may look little bit familiar minus plus which shows up all the time in numeric and right so it leads to ricotti equations and interesting things. Yeah, also in normal forms. Yeah. Trying to find or or even in linear algebra just trying to find change of basis, right? All all the time. Okay. But now I'm going to view just look at this in in different lens. Okay. instead of matrix you know byn array I'll look at it as vector with squ entries and this says it's equal to some linear transformation applied to itself because I'm multiplying from the oops I'm multiplying from different sides both are linear operations on the matrix and it if you write it So now it's big this is big 2 by squ matrix effectively and it looks kind of weird and it's very sparse but it this is just take the the the columns of of roman and put them one after the other in large column vector. So now have squared entries in it's one this is one sorry an 2 by one vector and this is an 2 by 2 matrix operating on it. Well know there is one because it's linear transformation. So can always do this and and we've written these out plenty of times for various reasons. It's kind of amusing to do to see the structure, but anyway, all we need to know is that it's linear. So now I'm going to view this as as the previous problem like gap lumber type problem but on this enlarged matrix this lifted matrix big could write it down but it's actually that would just obscure what we're doing. It turns out it's better not to because here's what all want to point out. There's an iggon vector of of script is the identity which would be okay E1 E2 through the the standard uklidian basis written one after the other. How do know that? Well, because know the action of is minus plus. I'm sorry. It's an igen vector of plus. At plus, what is plus? plus applied to matrix going. So, going back to the the other form that we know is still there, it's going to be plus minus a+. So this is the behavior at plus infinity. It's just commutator of and And if take equals the identity, it's vanishes. So it's vector is identity and the igen value is mu plus of is identically zero. So it's actually given to us as zero. We don't have to modify it to make it zero. It's just zero from the beginning. So applying the gap lama we obtain solution of xp. So this is just did say yeah this this is plus of is this of xp is plus times plus order of G+ to the minus theta del delta let's put it like this but that now projecting down from the lifted description that's of is equal to the identity plus theta tilda and theta tilda which is just defined to be this piece is big of to the minus theta delta and that that's what all that's what the statement of the theorem was. You know, remember remember very clearly Ghee was visiting and sitting in little visitor's office and we were trying to do this for well maybe we can do it for the diagonal case or something you know some very special case of matrices and we wrote this down and looked at it and said but but doesn't this work for all the matrices that's very strange it's all it's something with almost no work you get lot just just viewing it from different perspective This is equally equivalent to the gap lema and yet much more convenient. Okay. And that it's almost tutology, but that's one of just the main tools in this business. Okay. So, see Perfect. Okay. So that's the exact end of the that part. So let's take five minute break and we'll come back. Come back. is identity. So it's almost that this transformation is the identity. Yeah, it's near near identity transformation. And also we get let's see we get that it is invertible as well. guess that's just by runskian computation. for example you have no you it gives you Right. Yeah. Yeah. So that's very familiar at high frequencies, right? People use this for KB and normal forms. But yeah, the simple Yeah. simple version of it. Yeah. That's the linear equation. It has writimar. You were just supposed to change again. you sort of get it was think did oversimplify it's pretty simple but oversimplified but you for instead of the exponential decay of the pubation you half. Yeah, that's that's tricky. yeah. So, if it's Yeah, that's hard. If if it's if it's an L1 perturbation, okay, you could apply the theory of coal, but you need gap. You need spectral gap. You're bit limited. Okay. And if it's not, so student of mine, Peter Howard, and he's the chair at Texas&M now. so he gave him this problem to treat the case of one over it comes up in in detonation. It's an important case. He treated it and you could look it up see what really hard In that case, mean like what would you expect if there's any positive out there? that's yeah. So from that one there's something like gap lever. Yeah. But think you're asking is there something like Yes, that's good question. That's great question because when Peter Howard was doing his course, we didn't have congregation. But he didn't try but maybe you could that would be great help. That's good So suppose you want to do this. Let's see. think in order so we've got this one to use is proof you need it to be simple item and that means that there that the values of possible mistakes. So now you have lot of strong all of them must be gaps but you have particular problem. No, not really. Yeah, think do. I'm always interested in just questions. Yeah. So, me too. You know, that's an excellent question. Don't know the answer, but I'll write it down. So, pure empowered WD around side fantastic pieces. gave him this problem which was supposed to be small variation of that. It was so hard that it took like three years to do it. It's really good work and but it was very, you know, they receive an email with the grace and stuff and also don't need to carry the booklets around. So it's have 180 papers in in room in my in my home and have to bring them back. That that's why I'm I'm crowd mark. We don't we don't want to to do that. think I'm going to take mean big luggage and bring there's cart. No, but have to bring them here first. to me from my home. have to have to take the car, not the subway for that. It's my my exam last term was from at night. So did that. my goodness. smart. never played the 188 still. Yeah, short changed you. That was good question about the inverse, Federico. Here's here's how answer it. If want to find the equation for the inverse, use the the derivative formula for matrices which says it's - inverse dxt inverse. know have this sort of commutator for the derivative of And now when multiply this out, get the reverse commutator for inverse. get an equation that's almost the same, but don't really care what it is other than it's linear. So there's sort of duality because for linear equation know have global existence of for OD. So can integrate both of these all the way to zero. And this this relation actually ensures that inverse is the identity all the way. How do what's the equivalent system? don't understand. We have system for capital then what's and okay so let's do okay yeah maybe maybe let's try small one and just see how it goes. So, so let's see if can can do this. Okay. So, I'm going to take script is going to be T11 T12 21 T2 2. Yes, know don't know. Yes, that's what want. want to just write the matrix in. no. said want to do columns. Maybe it's easier with columns. So, G21 G12 G22. Okay. And now want Okay. So should be minus a+ and then you know made calligraphic in other words want to write down the first column and the second column okay so let's see may have okay let's just try so this is going to be what does this Look like it looks like this one looks like T1 and that'll be the column first column minus one now I'm already getting let's see better write the dot okay so let's just write A11 T11 One 2 21, right? No, want the columns. so it's it's going to be T1 T2. Okay, this one's going to mix things up little bit because now have the row guess this and then have to take let's see. except just want the first Let's see. Is that what want? Trying to think of simple way to write this. Okay, think I'm don't have the block structure in mind, but you see you you multiply this out. You take the resulting columns and you put them one over the other and that's that's your answer. But yeah. So if right if is 2 by two then script is 4x4 it's not so bad. Yeah. square by square. Yeah. is two if Yeah. So for 2 by two is not not so bad. We could write it out but yes, but I'm botching it so won't. Yeah, they said the trouble is yeah it's kind of it tries to resist block structure because it's in reverse order. But yeah, you have to write it down and sit down and do it. think so. If you'll allow me, I'm just going to pass. Not that fun. okay. So now, okay. So now an application. This was for looking at the resolvement equation. So let's see what we can do with it. Okay. So, first of all, it's second order. So and the theory was you know as usual it's for the first order system it's easier to deal with. So, but we can always change to we can rewrite it as first order system first order equivalent is to take dx as the variables and then dx of dxv minus zero identity DXV. Okay, the first row is immediate. get zero on the right hand side. I'm going to get an over here if do it right. This should be lambda + dx bar. And this should be bar. And that's the usual trick. And so we'll call this even though it's not the same as the previous any kind of lifting I'm just going to call script and we'll call this script or minus script VA is script and script is zero little It has kind of special form but it doesn't matter at all for what what we're going to do. Okay. Now we're going to apply conjugation limit. So this is going to give us t+ on in 0 + infinity minus for for on minus infinity to zero. two different transformations that change change this equation star into constant coefficient. So this converts star to star. Okay. Star star is dx minus plus okay let's say plus or minus of and here what's our parameter is just lambda there's no other parameter in this setup and this will be Okay. So, plus or minus and that's okay. And that will be that's going to be plus or minus inverse time zero little if you want to know exactly what it is. Okay. And now we're going to we have star star but then we also have very important fact we have so we've sol we solve the equation on on two half lines but then we have to connect it. So it's important that we require the matching condition okay and this comes from = plus or minus and these are equal at the common point = 0. They have to agree. And that tells me that okay that tells me that at so minus at 0 minus with minus is equal to + plus 0 plus these have to be also have two different variables they live on the opposite half lines and they must have this matching condition which I'll could also write just as minus of 0 is equal to 1 minus inverse plus of 0 plus so 0 minus to 0 plus and could call this there's some mat there's transmission matrix Now that relates them. So in the original variables it was just matching the matches at the common point but in the new variables it's different. It's something else and something that we have we have no idea what it is. All the only thing we know is it exists and we know that because the inverse is is well defined. We just discussed it all the way down to zero and bounded. So this is just some matrix conditioned minus is just straight it's not back to this caligraphic no this is this is cal sorry okay this is I'm sorry this is okay well it now I'm running out of letters maybe should have made these something else this would be yeah some other it's calligraphic only because is calligraphic Yeah, really shouldn't be really shouldn't be. But yeah, let me let me leave it bold. Okay, it's going to act on the calligraphic guys augmented. Yeah, because wrote it as first first order. It's phase phase variable system instead of the original variables which needed in order to mean all the these things did used heavily that it was first order to solve by variation of constants etc. it doesn't you know work so nicely in higher order. So this is standard. Okay. And we so now we have two constant coefficient problems and boundary condition and the boundary condition is kind of unknown. This is matrix that depends only on lambda or let's say more generally of but it depends only on lambda and it's on 2n by 2n so it's in complex numbers 2 by 2n complex it's complex matrix because lambda is complex and that's what it is. Okay. So now we need to solve it. Well, you know, if you've worked with constant coefficient OD with boundary value problems, you know that the enemy of solution is if you have zero, if you have imaginary values, you won't have nice bounded solution. You need spectral gap. your matrix must have nice spectral gap between so we need no center subspace or the coefficient matrices in order to solve. And we'll see that quite explicitly when write down the solution formula. But up to now let's just take this as as for granted. We we'll see why it is in minute. But here's lema. So plus or minus has non-trivial center subspace if and only if this is for our plus equals plus of lambda if and only If lambda is lambda plus or minus of So one of this dispersion curves lies on dispersion curve or plus or minus. So I'm going to re So I'm relating back to these dispersion curves again. And it's kind of illuminating though. This this proof is pretty nice. And yes, lambda of So these dispersion curves that we drew these were lambda 1 plus of ik lambda 2 minus of ik they were dispersion curves lie I'm sorry that that was that is it lies on lambda lies on dispersion curve. So in other words, we've shown that these dispersion curves are bad. They have central spectra. but away from those, at least we have spectral gap for matrix. And here's how you show it. Okay. from the format. did write it down? see here it is star. This is the formula for So we have plus is zero identity lambda Yeah, there we go. just all that I'm lacking is the derivative term. And so if we have an if we have an imaginary iggon value that means that well the igen vector is equal to mu How do know that? Well, know that by the by the construction by the phase space construction, know that this corresponds to the derivative of this and know that if if integrate that let's see no know that this is the derivative of this by by in the integrating the OD but it's Yeah, could know there corresponds an exponential solution of the original equation with this derivative. Okay, so you can verify this algebraically. This this has to hold. You just put you put in and and some unknown. you'll find this is the unknown. Okay. But but this is now plug this in and find that * KR let's see is equal to + And this tells me magically that sorry lambda + minus a^ 2 equ= find that it solves the dispersion relation just because I'm just undoing wrote it as an the first order the second order system as first order. As soon as put an igen value in here, it goes back to essentially second order system. And then notice that because chose the igen value to be Well, it's exactly like the 4A transform. The derivative is So just get the symbol times and so this is an iggon value of the symbol 4A symbol and therefore it's in the dispersion relation. I'm sorry. this should be these should have all been times I'm sorry. There's no That the is the See what did leave out? Thanks. Yeah. Yeah. If if you write this out, you get the just the FA symbol and therefore and lambda. So now looking at things the usual way in terms of time coordinates and with as the parameter we see that lambda must be on one of the dispersion curves. So that is how they link and this comes up all the time. sometimes the lambda are are called temporal values or temporal modes and the muj of lambda are called spatial modes or spatial igen values. These are the igen values. When you consider the resolving equation as function of and if you consider it as function of then these are the familiar the dispersion value the values for the dispersion curve that are more familiar. This is like lelass transform. This is for transform. So it's it's natural. Okay. So this point right so know now where have spectral gap and where don't lose spectral gap exactly on these spectral curves now what happens in between okay but should yeah okay so any questions about this part is just manipulating some algebraic symbols Well, let's see. Let's just we could just write it out. 0 lambda + mu Okay, so we're going to get mu and we'll get lambda plus + mu and that is supposed to be mu which it is and mu ^2 So the first equation is automatically satisfied. That's because of the augmented system. And what is the second equation? If we write out it's write it out, it's ^2 sorry lambda plus + minus ^2 * = 0. So there's first order guy. And now I'm just there's mu here. That was mu. And now just substitute in for mu. what was the sorry. Star is the which where is star? The equation for the vector of the this star. Yeah, this is you mean this this star? The the original star. Okay. Yeah, but now I'm going now I'm only looking at the matrix at infinity, the limiting matrix. Yeah. Now, right now I'm only talking about the limiting matrices because that's all that's left. I've transformed away all the spatial dependence and now only have to talk about limiting matrices and that's all all can talk about for the zcoordinate because that's the only thing there. There's no more bar, just plus and minus. the first line has has non center this because that is hypothesis if and only if if and only if yes because yeah because when we when we write down what it means to have center subspace okay so let's see just just did the calculation showing that for any value this is the relation so it reduces to second order problem with mu corresponding to dx. But what if have what if if have center subspace that means mu= with real and when substitute in mu= this becomes the forier symbol. if have no if have no center subspace, automatically have space if you are relying on this, right? Yeah. No, no, you don't have it on this curve. Yeah. But right you having having non-trivial center subspace is equivalent to not having spectral gap. Yeah. Right. So I'm identifying so trouble spots are going to be where you have center subspace. And we already know but we already knew they were trouble spots by different argument. We already know this is essential spectra. It can't work. And yeah so it turns out though if if you do have spectral gap then it usually works. things work out. So that's the next step is to look at okay how do we then solve the equation center correspond existence of value on the imaginary axis any value so that's it could be any this is why take ik mu equals ik for hyperbolic system yes you you would be you would be really stuck, right? You would have all spectra on the imaginary axis and you would have to move away from the imaginary axis to be okay and you would get at most bounded stability. You won't get any decay. There's no hope. And that's all that that's what people people prove welloseness for hyperbolic equations. There are hyperbolic systems that have that are not conservation laws that have damping of some sort and those you can prove exponential not exponential stability but depending yeah you can do something but first you have to move off the something has to move you off of the imaginary reactions you're exactly right right yeah so have I've been mentioning hyperbolic systems sometimes those are the ones mean the the the nice ones Yeah. Okay. So, okay. So, now now comes well it's the up andoff peron again because we're just going to be we're solving boundary value problem. So, we'll just do it by Let's see. Okay. We'll do it much the same as we as we did in the gap proof, but now it's not You'll see it's not contraction mapping. It's just it's just linear system. Okay. So that this was by the way was to connect connect the spectral gap to this dispersion curves and see what's little bit what's going on. Now let's convert to an integral equation. And that's Leoponov Peron again. So, and here's how we can do it. So, we've got dx. Let's just do let's see, let's do plus infinity. and write it down. So we have of is equal to okay to the the stable part of so the projection onto the stable subspace and times the stable part of 0. So we're going to initialize the stable things with value at zero which we don't know the unstable. sorry. And here's the rest of the stable guys. Now, we have to put in the Okay, so there really Let me just put the pi like did before. And now this is exponentially decaying strictly because assumed there was spectral gap. So all values either have positive real part or negative real part. I'm dividing them into those two classes. The strictly negative real part are the ones in in in the stable projector and the strictly positive are in the unstable. And haven't written that down yet. And it wasn't even necessary for me to put an unstable or stable part there. could have. In fact, shouldn't. Let me not. So, can write the solution this way. And not notice that everything in sight is exponentially decaying. This is going forward and this is the stable part. So it's decaying like to the minus theta absolute value of this is - is is positive. Okay. Again this is to the minus theta - absolute value. This one's going in the reverse direction, but it's unstable. So, this is order of to the minus theta - again because the sign of - is is now opposite. So, everything's got good exponential decay and this is in fact okay. Yeah. So we can we can from this nicely estimate the say the L2 norm of by by holder just take the L2 norm of this whole quantity of each integral is bounded by the L2 norm of times the L1 norm of this exponential decay which is bounded So this this gives me nice bounded operator mostly and there's only one problem. don't know what this is. This is somehow going to be determined from plus because okay if know the stable value at zero and know F+ then know everything but don't know this. there's another formula on the negative side which gives me unstable this is likewise unstable of 0 plus is unknown and needs to be found out I'm so sorry yeah everything is yeah everything in sight should be Yeah, but luckily there weren't too many Z's where Yeah, only one. Yeah, yes. Yeah. Yeah. This is if and so it's Yeah, that's Yeah, and that's the point. Well, there's only one hope, right? have matching condition. mean, this this problem is not it's not well posed without the matching condition. You can see it. could have any value of ZS and would have nice solution afterward. It's quite non-unique. This is where use the bugs, right? Okay. So, notice Yeah. So the box says which is guess say for matching. We have okay of 0 minus is equal to of lambda of 0 plus but we don't need all of this. This is okay. of 0 minus is Okay, let me be careful here. So 0 minus want this was minus. want unstable and stable. And this is invertible. We know that because inverse is invertible. And this is stable and unstable of 0 plus and 0 minus. And the ones we really care about are zstable of 0 plus. and unstable of 0 minus. Those are the two that we need to find out from this relation. And the others are determined as functions of of and of zero of and of sorry unstable of 0 minus and stable of 0 plus. me what means. This is just the stable and unstable parts. So project onto the the the stable part the stable modes of the matrix minus and the unstable modes of minus and then likewise on on the other side. But notice that this is function of this at This is function of this at So can write this as as some okay some little equals sum of after rearranging things. This comes from know if have relationship like this that is invertible and that and these are functions the the the unboxed guys are functions of the ones in boxes and then can rewrite it in this form where okay don't know what this is but it's something it's not it's not that closely related to big it's just inherited by solving this big relation and taking into account the dependence of of the unboxed traces on and the boxed ones. So it's complicated though it's quite complicated because to find the dependence plug this in solve this integral equation and then evaluate at zero. So this one's gone but this one is here. So it's an integral equation, integral dependence on Yes, because have an equation such as this. And let's yeah, let's just let's look just look at this for moment. Okay, so what if what if want to find the unstable value at zero? Okay, there's this is in the stable subspace. It's not part of that. It could only be and so is this. It can only come from this and it's with equals So that's an integral from 0 to plus infinity of well it only depends on actually doesn't depend on misspoke. It does not depend on the other one. So that goes over onto the of side which is just some it's just two two vectors. Yeah. So substitute for this and this some functions some linear combination of sorry some linear functions of minus and plus rearrange will always get something of this form. could explicitly write it down, but it's not super explicit because it's an integral oper as an integral operator in there. but for what I'm going to say right now, it's not super important what it looks like. It's just that want want to have unique solution. And what do need for unique solution? need that well, need the right dimensions first of all. So need that these two these dimens this should be square. So this is this is by something right we we we don't know exactly what it is but we sorry 2 is that exactly right let's See? well, basically need that these that these two dimensions add up to This is what need. need that this that dimension of of the stable subspace plus dimension of the unstable subspace on the plus side is need that they add up to In other words, need that the stable subspace of has the same dimension on both at both infinities. Otherwise, it's just non-square matrix. I'm not going to be able to solve that uniquely. So, just by dimension count, I'm in bad shape. Now, it could be that there's some redundancy that allows me to solve it, but okay, that would be generically need it to be square. Okay, this is this condition is called consistent splitting. That's Alexander Gardner and Jones defined this term and it says the dimension of the stable subspace of plus is equal to the dimension of the stable subspace of minus. Once have this have nice square matrix here then expression. Yes. So that of But is it yes it's on the left it's expressed only in terms of stable then no this is both this is both this is this is both stable and unstable this both together the stable part yeah and let's look what is the stable part the stable part is well at equals can tell you the stable part it's this it's because this becomes identity and this is integral from 0 to 0 and this is all in the unstable part so know this really is the initial condition the I'm in initializing the stable component but don't know what is the unstable component at zero except that if pl if hit it with the unstable projector that kills the first line and leaves only the second 9 and I'm integrating from 0 to plus infinity something that's everything's known here's table of zero plus yes yeah that's how that's how know that every all of those things can be put over on the other side as functions of plus or minus and then have this left okay so there's there couple more things to say here. So, yeah. So, let's look at just at our picture of our dispersion curves. Okay. Off towards plus infinity, it's pretty easy to determine that the stable subspace has dimension one and the unstable subspace has dimension one on both sides. is because it's dominated by you could for you can forget about the off diagonal term is is actually dominated the spectrum of this matrix is dominated by the spectra of zero one lambda 0 which is nice hyperbolic matrix with values plus or minus square root of lambda that that for big lambda we can forget about this of one term so we can compute that consistent splitting holds out here and right and so it hold and it also can be seen to hold here curiously but in here yeah there's something it it just that should be all some kind of essential spectrum but what kind it's it's like very dispersed it doesn't have it doesn't have curve measure it has measure on all on space as spatial measure So that's very different from self adjoin operators we expect to when we talk about essential spectrum as just continuous spectrum thinking of curve at least do but this is has central spectrum that involves curves but also is supported on on regions of of nonzero measure. Okay. So we can define another definition. The Evans function is defined as of lambda defined as determinant little of lambda on the region of consistent splitting. automatically because all all on that region. Why why is it important that be on consistent splitting? Well, notice this formula depended on split specific splitting into stable and unstable modes. If cross one of these lines, the projectors change. It would not be analytic crossing over curve. Even even if it happened that even if it happened that these two curves coincided as they can for some cases so that it was all so I'm just hopping across from consistent splitting to consistent splitting the evidence function would change because the formula changes and this depended on separating out the plus and minus modes in special way. It doesn't even make sense if you jump across the curve. But now so long as is uniformly invertible actually have unique and bounded inverse have resolant point. So, so the proposition is on domain of consistent splitting. I'll just approximate it like this. row of is equal to set where determinant of does not equal zero. And spectrum of is equal to the point spectrum of or all points are in the essential spectrum of health. Okay. For our particular case, I've already excluded that. Have excluded that? No, haven't quite excluded that. mean know because why this is an analytic function it's zeros are isolated it's the inverse of this function is mamomorphic because it's rational function sorry it's not rational it's ratio of co-actor matrix and determinant so two analytic functions therefore it's mamomorphic therefore any zeros would be of finite multiplicity. So it's it actually is playing exactly the role that the characteristic determinant played in the matrix case in the finite dimensional case and it can be shown with with proper normalization. This is Fredo determinant for something if you're careful but it's finite dimensional. So we we found out lot. It it wasn't clear that you know that anything should be muromorphic before now, but now it is because we have such concise well maybe not concise. You might not consider this concise. It's in principle concise, right? Because it it's it's an algorithm for finding the solution. And with this algorithm, we could in principle estimate anything because it's all constant coefficient. And and we will have to estimate few things. But for right now, we just have we've just proved Henry's theorem because to the right of all the essential spectrum curves, we have either two cases. Either there's no point in the resolvant to begin with out here, in which case it's filled with essential spectrum by definition, or there is point in the resolant set. And and therefore we have right because the analytic function the determinant could vanish everywhere. We don't know that it doesn't but once we know it doesn't vanish everywhere then it's neuromorphic and that means there's no more essential spectra to the right which that was part that we were trying to prove. So that's the end of of Henry's theorem and good because that's the end of the of the time slot. but yeah, so so next time we're going to we're going to do another theorem of Henry and Sadinger which shows how to deal with the problem of essential spectrum coming right up to the origin. So now we know that's that's all there is and we're going to try to get rid of the rest of it in some way. So stop there and take any questions. Hi Kevin. but yeah so far we didn't really need any we didn't really do any computation right we just showed you could formulate it as linear system and from that you extract this Evans function that's crucial is it's and and some dimensional conditions which gives this consistent splitting this Evans function and now we'll go on Kevin yes so what about the points that not satisfy the consistent splitting for those lambda? They must be point spectrum. No, that that would be some kind of essential spectrum because if you think about what it it's actually reduces now this reduces to problem in matrix algebra. When is when is this problem unique uniquely solvable? Yeah. first of all need square matrix for that to happen. Otherwise, it's not even unique. So, so and if it doesn't have unique solution, that automatically puts it in the essential spectrum. Sorry, have to teach you. Yes. Yeah. So, so there's not only point spectrum but also maybe when the consistent splitting doesn't satisfy there may be essential spectrum that is on the right of the this value space. you say if if consistent splitting were not satisfied, yes, we would have problem, right? Okay. Yeah. But you know, then then your your equation would be the original PTE would be illposed. Okay. So, usually it will satisfy. Yeah, it it's right. Okay. always satisfied because if you have good solution theory, good linear solution theory, then Hilla theorem says you have resolvance points all the way, you know, everything to the right of some everything with real part bigger than equal to some value is in the resolant set. Yes. So if you don't if you don't have resolvance set out there, you don't have any linear bounded linear theory. You don't have linear existence theory. for the PD. So that that wouldn't come up in cases we look at depends on this condition doesn't depends on lambda right it's just minus so yes right right so as long as there's one point there is fine that's what you mean yeah as long as there's one point and there always will be one point if our PTE were well posed line linear level okay yeah thank Yeah. Did Did you use the So you you have plus or minus So with the little you were able to solve for two of them. The other two box. What happened to them? And what do we do? The ones in the box. These are the ones. They're supposed to be the ones. wrote the wrong ones. let's see. the ones in the box we need to solve for. did write it backwards? Let's see Zs at U+ should be Which one? This was the min. These are the We were trying to solve for the the ones in the Yeah, the ones that appear here. So, it's stable at plus. Did Did write it? Yeah, it's should be reversed them. It's stable at plus on the plus side, unstable on the minus side. Yeah, this little matrix allows you to to find them. Yes, but you know, nothing tells me this is invertible. And we know sometimes it isn't because there can be values for operators, but it does tell us that it's invertible except at isolated points just by the fact that it's analytic. So it gives lot of information with no computations. And once you know these two quantities, what do you do with them? Because you will have this expression explicitly. Give yourself. Yes. well right. So so then then have bound resolant bound and remember that is what needed to get my stability estimates. just needed resolvant bounds and this says can get them but it's little I'm going to discuss this more later, but it's little bit subtle in that yes, this gives bound, but it could it depends on lambda. It depends on everything. So on compact set of lambda, this gives uniform bound. But what about high frequencies? Let lambda go to infinity along the imaginary axis, then it's not clear, right? So you have to get those in different way. But there are very easy ways to do that energy estimates or well okay there's an easy way which is energy estimates and there's hard way which is WKB which we you know high frequency asmtoic OD analysis which is like this but as lambda goes to infinity this is the bounded lambda version and goes to infinity And the other one is lambda goes to infinity and stays frozen basically. Which one do you do you do the one? like to I've done both ways. Yeah. If want all the information do the KB because mean it gives you complete almost like solution formula. But if want to make the analysis simple use energy estimates. Yeah. I've been, you know, WKB is really fun once when you don't know what you're going to get, but then afterward when you sort of know what you'll get, but you still have to do it, maybe less fun. So, so then started looking for shortcuts. Giving you the shortcut version here. Yeah, won't won't need to do any high frequency stuff. Maybe it's very elegant. That's true. But there are even more symbols like it's doubling it's doubling the contraction mapping arguments doubling the symbols in this context. Yeah. Yeah. it's it's it's beautiful. Yeah. Yeah. And powerful. But it does involve at least this many computations, wouldn't you say? So whereas an energy estimate is just one line. Yeah. Also we we have to do an energy estimate anyway. So he's concerned. All right. Well nonjerator, right? Yeah. Yeah. That's think among PTE people you can safely Yeah. safely do that. Yeah. Yeah. Yeah. Well, thanks. Yeah, thank you for your your Yeah. See you. See you in while. See you. let's see. There is one coming up on Friday. colloquium in the department. didn't see maybe think there was pure like algebraic one or think saw that. Okay. Oops. Guys are still here. You're very patient. Thank you. How's it going back there? How's this the temperature? Oops. think you should look this way, right? Yeah. can't. It's cold. Yeah, super cold. But it's fine. Warming up. Yeah, it's fine. So, so, okay, have to do transformation from from the gallow side is, better and better means colder and colder and probably the other people are the reverse. Anyway, yeah. All right. okay. have call from Indiana actually. wait. See you later. All right. Bye. Bye. hi. Hi. just leaving the classroom and will talk to you in one minute because I'm grabbing my materials and no, no, no, no. Because remember I've been so can't can't very well come to my office but really one moment I'm quite available just putting things in my bag. All right. Yeah. So It's always this way for people