Sorry, the connection broke. So I had to restart this. So we have shown this relationship, what I call a star. >> This is true for all right? Okay? >> So, okay, so, you know, if you want, you can take the difference of these two things. So this, but this is enough to show F1 plus F2 is equal to F1, okay? >> This isn't enough. Okay? So you can, just, so you can go back and now this requires the positivity off. Okay? That's what is required here. >> So you can check that for all F1, F2, OK? >> And similarly, S alpha of cf is c times this alpha. You can check that, yes, alpha, it's linear, right? >> So we have, we have a linear map, a salvo from L2, y2, to which a solution map maps F2 u. >> So we have a bounded linear map. P is equal to integral F comma P for all that. >> This is the unique solution. >> Off >> Alright, so this hat satisfies the solution. >> Okay? Now of course, we also prove this in the solution exists for the star from the adjoint problem, right? >> So similarly you can construct bounded linear map, which I'll call T alpha to zeta one. You have to use g for this t alpha t. >> So it satisfies the adjoint problem, right? >> So this is t alpha t comma u equals integral g comma u, u for on revenue, that is t alpha. >> So that is v equals T. T is the unique weak solution of L star v minus alpha v equals to g on u and v equal to 0 delta u. >> That is the solution. All right? So we also have, so we have the alpha map, okay? Self-map, because the solution of this problem, okay? >> For this alpha, remember that alpha is fixed for which this, every half there's a unique solution. So S alpha as a solution for this problem, which is rigorously speaking, the solution of this problem. And then we have the T alpha map, which is the solution of the adjoint problem. Now, of course, this L star is just a formula that joint which has that Property. But we have a connection between b alpha star and d alpha, right? >> So we expect a relation between alpha and t alpha, right? >> So another question is, what's the connections? >> We know this relation k for all u and v. Is there a relation between S and T? >> So before we write the relation, notice that, you know, t alpha is going from L to one US alpha as being from okay? So this is not quite the way to look upon as alpha and t alpha, right? Really, what do you expect? Right? I'm expecting t alpha to be the adjoint of herself. But as alpha is not an operator in the sense it doesn't go, go back to L2, you. So really the way to look at as alpha and t alpha are that these things are actually sitting inside. >> These are operators on, okay? >> So since we can regard them as operators an LTI, since H is a subset of L2, y2, you can regard as t alpha as bounded linear operators from L2 U2 L2 u alpha d alpha g mu c times f u. >> And same thing, similar norm T alpha G. Okay? So we're really going to think of alpha and T alpha as operators from L2, U2, L2, you, except the range of these things are, is actually going to be contained in 801. You and actually even the continuity is there. But these homes, right? So now we can try to see what is the connection between us alpha t alpha, right? >> So >> So remember that. So from now on, for the rest of that, I'm going to rest up. This whole thing has bounded linear operator from L22, L2. >> Okay? >> So let's start with the two relations that we have which define this alpha. So first one as B, alpha of F b is, I'm going to write this, but the L2 in this notation, the L2 inner product for all V in it, right? >> And we have d alpha star d alpha g u is equal to g mu nu. >> Alright? So we have these two relations. Ok? >> Now, so v can be anything, and it can be anything here. And in particular to SUMIF and t alpha g are taking. >> So let me actually write this slightly differently. >> Actually, this one I leave as, I leave this as f comma p. And this I want to reverse because it's symmetric here. So just going to write this slightly differently. >> So we want to write this as u comma j, right? >> It's the same. >> So taking v to be d alpha g, and u to be S alpha F. >> Notice these two, both these things are needs grow anneal. >> So b comma v, right? So, so you put it in this relation here, right? And you know that you have to flip the two if you want to connect it to the alpha star, right? >> And then now if you use this relation, you get this is equal to ux herself. Okay? So I'm just using these two relations and the relationship between B alpha and alpha star. >> Ok? And this is true for all f and g and L2 you. >> So it tells us actually The TL phi as the Hilbert space had joined the operator at joint. >> Okay? So, so he prudent several things here. >> Ok, so let us try to summarize this. So remember alpha is that special lumber says that for lambda less than or equal to alpha, that eigenvalue with an F on the rents and has a unique solution. In particular, this has a unique solution for every F and LTE and L2 you. >> Okay? >> So SAL finds a solution operator for this problem, T alpha is a solution operator for the adjoint problem. But calling this an adjoint problem is just purely formal because you know, L and L star not operator, that's just a name we have given them, right? So it's not obvious that there will be a clear, easy connection between as alpha and t alpha, the solution operators for these two problems, because our calling this selfed the adjoint is just purely formula, right? But actually we show that the two are related. That this addition operator of the original problem and the solution of the operator for the adjoint problem, actually they are adjoins of each other, right? >> So I'm going to summarize all of this into one proposition, which we are going to be called upon later on for the boundary value problem, the ad joint first one. There is a bounded linear operator from L2, U2. You who with range, right? >> All these things >> With the unique solution off. Okay, number two, considering bounded linear operator from L2, U2, L2, you had choi then S star g, the unique solution of L star u minus alpha u equals g. Or knew the Soviets further to belongs to L2, y2 one y2 is equal to K times K. >> Okay? >> This is what we have shown. Shown there is a solution operator for the original boundary value problem with the alpha there. >> Okay? >> And it's from L2 to 0 to you. If you consider as phi as a bounded operator from L2 you dwell to you then let us tire alphabets. >> Ad joint. >> Then S star alpha actually gives us a unique solution for the ad joint problems. >> Okay? >> And of course the solution of the problems, and it's, and it's bounded even from that. It's not just as an L2 operator to be approved. >> All of this >> Just start a remark here. So the very fact that these are SIF has a weak solution of this problem. So that is of course given in terms of, so by definition of weak solutions, we have S alpha f comma p is F comma v, v and h 01, ooh, is mu comma G. >> Okay? >> So this is going to be useful for us. I'm going to actually, so notice I've only done this for alpha, right? And this problem, and we know for lambda less than alpha, we understand this, it's invertible. So the issue is what happens when lambda is greater than alpha, right? What happens in that case? That is what we need to understand. And we are going to understand it by writing everything in terms of this operator S alpha. So even when I have a lambda here, I will convert things too as alpha. And this alpha is a nice operator. Well, I mean it's defined operator. So by using properties of S alpha, I will be able to answer questions for this problem. >> Ok? >> So basically I'm interested in whether we can solve the boundary value problem. L u minus lambda I equals f on you who has a unique solution for lambda less than or equal to alpha for each unique weak solution. >> Okay? >> So we need to know what happens when lambda greater than alpha, meaning does this boundary value, okay? >> Does this boundary value problem have a unique solution when f is, when lambda is greater than alpha, for which lambda, okay? So we know we're going to answer it. So there are no eigenvalues of this when lambda which are less than alpha. So question is, what are the eigenvalues of this problem, right? And having eigenvalues and stuff related to this. >> So actually what basically we are looking at what is the spectrum of this operator L, even though L is not really an operator, but for which values of lambda as this have a unique solutions, right? >> That's what we're going to call this, the spectrum of a solution for every f. So we need to understand this. >> Okay? >> And we study this problem by reducing it to studying a spectrum of the operator alpha. >> Okay? N is not an operator, so you cannot really have a spectral theory for it. >> Okay? There is the spectral theory for it, but S alpha is proper operator, a bounded operator from L22, L12, and L22 81. >> So it has a spectral theorem and all that stuff right here. And it's actually even more specialized. I haven't yet mentioned very important property of Sappho, which I'll do in the next lecture. >> But so this is a nice operator. >> Okay? So if you can reduce this study of an unbounded operator, the spectra of study or the spectrum of a proper operator, regular operator. >> I mean, that's a big, big begin. Ok? >> So that's what we're going to do. We know the spectrum of this or at least the nature of it? And we'll use that to determine the spectrum of this unbounded operator L. >> All right. That's it.
The adjoint BVP II
From Rakesh Rakesh April 14, 2020
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