 Okay, record you are being recorded You can introduce yourself to I guess some of you might be here because you're hoping to hear cool things like Can't be here in the shape of a drum I'm not gonna do anything cool. I'm just gonna like what you're gonna get from me is some constraint of imitation and Of course the third matter of the cockles of variations. So yeah, I'm sorry about that Okay, I don't want to assume anything for this talk. So I'm gonna just go back to like step one and define everything If something is Okay, so That's just with like So The support of app is just the closure That means that the support is a complex that oh, okay, so you're far off from that. Okay We can define See, oh, yeah, this is very this is very easy You can you can approximate functions like absolute by your legs So It makes sense to Consider the completion of these days with respect to that The other object that we're going to We want to discuss But solving the plastic and value problem or like see to find things really hard so I need to Cook up some spaces in which you'll be able to like do things without caring This is Later As I remarked before the h1 norm it's just a stronger norm than the So what we're essentially what we're doing we're taking these days over here, and we're adding some And we want to get a better grasp on What kind of functions are in this space we have to Against I'm using the rest of my parts, and of course there is no So g is zero in the boundary and that's why the boundary So what is the clever idea that someone had If f is very good, this is always true but The right side of the equation doesn't require f to be any more regular than just So I wanted the idea to say that if f is We can define so this is essentially Yeah, thank you So what am I saying here if f is more than a two is maybe c1 then h is of course just the But actually this relation my own four functions that are not c1 less So if we want to rewrite H1 using this notation We actually end up getting the following Characterization, it's not a hundred percent precise, but it's good enough for the purpose of this So H1 of omega is the set of all f in F2 such that There exists the weak gradient of f according to that definition the weak gradient of f is a square interval function And here is where it's not entirely precise F is zero on the bottom of omega Why am I saying that this is not entirely precise? I'm saying something about the net two function on a set of the bedminder at zero. So It's kind of a baby like it's a little bit delicate, but In some sense f is a little bit more than just not to function because it has a weak gradient. So somehow F remembers that it comes from a sequence of c1c function. So in some sense remembers that it has to stay like I Help this make sense Do another thing we need is the Laplace operator. So if say u is smooth enough So C2 has the sum over i from 1 to n and is the dimension of the um, get space v squared u over so it's the It's the trace of the action maybe it's for you. Are you ready to consider the big embodying problem? Do I Roll in the abstract that I'm gonna end up proving this back up theorem for the pass operator? I'm not gonna like that's a little bit The time constraint doesn't allow me to I guess but we get close to it Be able to like through Things that maybe are interesting So The big embodying problem is As I was mentioning before Approaching this problem in this way is really harsh and I actually Not for general If I make it a square then yeah It's always So the way to go here is to Somehow find a weaker formulation for this problem and that's the following so So we look for a function in H1 0 such that in pieces I want this relation to all for every This problem Okay, um, but what do I mean exactly by weaker than the problem B? What I mean is the is that if we have a solution for problem B Then it's gonna be a solution for a problem And this is actually very easy to About you we know that it satisfies this relation everywhere in Omega, so it's any so for every function B in C1 C Omega, I will also have that Times B Just because this is identity But then if this is true Nothing changes if I Integrate these over Omega So then also the info over Omega So we have that for every one of these Now what do I want to do I want to show that you satisfy this into a relation for every B in H1 0 So it's this B in each one zero Omega and Remember what it's 20 is is the closure of C1 C under the 20 R. So and Bk B in C1 C Omega be such that converges to be Right, I can always find such a sequence, but Now for every one of these Bk, I have that this is satisfied So I have that then this is Always I can perform integration by bars on this one here Then you get that this is equal to minus So this is just Again, no No boundary condition no boundary integral because I'm assuming the big day Is So definition of my plush and use like that right because you is in each one zero not you it's Proving that it's a regular like I'm proving that piece But what did I get here There's a man somewhere which I can This shows that We And now I just remember that Bk converges to be in the H1 norm But that just implies that This goes to be over omega. That's exactly what I wanted Remember that we're doing this for fix B in each one zero, but like So this shows that If you solve speed then it's all stuff I should mention that Somewhat surprisingly The converse is also true If you use a solution of the weak problem Then you can actually prove that it's not just a 20, but you can prove that it's c infinity and it actually satisfies the problem p But this is very that it's regularity theory. I'm not going to go into that just just to mention it I'm solving the weaker problem is actually really solving the original problem and this is Frequently the way that People in PDE approach these kind of problems you go to a like weaker version of your equation that it's easier to solve and improve existence and then You if you want a classical solution, you end up trying to prove that the weak solution that you have is actually ready or enough to solve the Okay That's let's grab some So very easily If If you want I should mention that from now on every time I say Again functions and taking values. I mean of the problem value. I'm not gonna talk about the regular problem So if you want to do our a function with a To The great number one is our problem to the green Again functions relatively to different invites are Very easy all we have to do is We want to use so in the equation for UI We want to use uj as a test function Um Satisfy this for every being in each one zero So I want to use you to as a test function for you one and you want as a test function for you, too so the equation for you one Over omega green you want Is you want that one Omega of you And the celebration So What you get is that these terms are the same and then So I get Then this one has to be But that one what is it is exactly the other If I want to get this one Just look at the question one of these guys you have to do so But that's yet to in the front of the bridge So that one is for free time to do some constraint optimization Let's consider the following two functions And These are going to be found to know it's from h10 of omega into r And the first one f of u Is defined as the e2 over omega Green u squared So in other words, this is just the L2 norm squared of the gradient of u and g of u will be defined as u squared x squared of u Okay, and As we've seen aside on these two Functionals have a very close relationship with the eigen body of problem So we have the following result if u is A local extreme other dog amongst would mean Subject to equal to zero Is an eigen function eigen value okay, and This is pretty much an application of the Lagrange multi-values theorem in the Bonnack space calcified variation of the set piece and Like this so if we So before starting the group let me discuss briefly The setting so if f and g advanced space and in fact x with zero so These up in here are not necessarily the same as there, but of course we will apply it with f and g and with x H10 so the setting is exactly the same here. What we have is that for f What I want to do with this is constraint optimization, right? so I want to say that the In some sense, I want to say that the gradient of f has to be parallel to the gradient of g right, that's so I Don't really have gradients in this case because we're looking at functions not just function So the result is the same, but we have first variations instead of gradients And I'm gonna define first variations in assignment. It's gonna be very intuitive but Just bear with me with a second. So if the first variations of GOX Interaction five is different from zero for some y x in the direction y is equal to long though The gradient of g Okay, so So let's apply this with f and g That's the one that I have here. So let's compute the first variation of In the direction b This is the time as exactly what you would expect This is the limit as maximum cost of zero of f of U plus absolute b. It's just a direction of the reason we're going in the direction of a function We have f is the So Now if I expand the square here, I get This one Two times green of you two times times absolute times Green of you in the probability of B And I have plus Absence for simplify absent here and When I take the limit the first quantity stays here because there is no absolute then and This term of it here The same computation can be done. It's actually easier to show that the The gradient of g You the direction B. It's just twice Okay So what do we have now Notice that if We plug in you as a direction here We have the gradient of g at you in the direction you is twice the input of this bird But remember what we have a new you have the g of you is zero and but if you should you have you is The L2 normal view That's one. I don't know if you guys remember that's how I define g before But this means that the L2 normal view is one so So this is just So I can actually like I mean our position to apply the theorem and get that then there exists a lump That's just that this relation is satisfied So what this means is that Such that Let me just simplify the tool right away because there is a tool In the product is equal to what is this? This is exactly the eigenvalue equation. Just Just really being a product This is just in the world like that so What are we done? We have so that If you is a lower extremum that subject to this constraint then isn't a function Let's complete the eigenvalue But this is easy because Is equal to recognition right you But this by the equation is equal to lambda you And now again, this is the L2 norm square, which is an equal to one And this concludes the theorem One more I'll put the statement here Here there exists the constraint g of u. I hope you see where this is going So we have the lower extremum of f under this constraint are eigenfunctions with eigenvalue equal to f of u What we're doing here is by showing that f has a global minimizer The global minimizer is of course a lower extremum, right? So we're going to get an eigenfunction, but not just Not just an eigenfunction. We're going to get the eigenfunction that is relative to the smallest eigenvalue Because the eigenvalue is always equal to f of u. So we're minimizing So we're going to find the smallest eigenvalue Makes sense. That's where the fun part is So that proof, so that let's see functions such that g of w and Let the infimum over c feel thanks to notice F is Is the integral of something squared? So f is always positive So this number is going to be bounded below by zero And it's it's not just infinity because you can just plug in any c to function into f and get a number So I would be between zero and that number So So what we're going to do now is as I mentioned the direct method of the conquest operations, which is Get an infimizing sequence. So that Was one of our kids. We're going to use The properties of this page is when space it 20 to Have combatness So we'll get we'll get combatness from the invading This I guess this doesn't make much sense unless you have seen it just once before but I'm going to do it right now If you've seen this before, you know Okay, so Of this person we have this relation here. That means that the gradient of UK Square is bounded by this number I plus one over k But let me just go up and put one and have a uniform estimate for every case So I thought I would do it before I don't know I don't remember if I did it or not No, I'm ready here. I meant to do this after talking about each one I guess I'll do it now This should probably answer your question. What do we care about this place? We have the following two parts the first one which actually I don't know There exists a constant C that all depends on omega such that For every function at 20 The integral of x squared can be controlled by this constant C times the interval degree If you've seen this before you believe it You mean what like you're asking about any relics theorem. Yeah, you have to prove that The norm of use this one otherwise the limit might not be Okay, I was gonna say you should believe this one because what I'm saying is that I have a function at that is Like clamped on the boundary omega zero and you have an estimate on the gradients the function cannot go to opera So you should expect to control this And to we have graphics Which is what I stated there? a fun zero of Omega and that's Compactly into The way we use this is that if we have a bounded sequence in a zero, then it's gonna have a Subsea events that converges That's what okay Very is always equal to one Because we're taking you gay inside the set C which is the point where G is able to do and that means having And we have that the gradient of you gay is going to be only found This means that The sequence we gave Is bounded in each one this implies that then by relics theorem Okay, before I'm gonna just take a weekly convergence of sequence So this is a sequence that's found in a few first base then up to a sub sequence it has It convert like up to a sub sequence it converges to Some function in this space with respect to the week to budget this space right, so then That converges weekly These evens of course enjoys the same balance because it's just a sub sequence and there I can use these ready compactness theorem to say that there exists a further sub sequence that And That converges strongly It has to be the same you Okay, what do you want to do with this? That's all about compactness now that we have The one that is you is actually the gov only imagine if you remember if you remember by stress Like by stress theorem like existence of maximum meal all you need for your function to have a meal is lower semi-continuity right because you just need to like go down Let's do that for this case So playing first, this is the worst in utility of the norm. Yes, the norm of you Each one So let me Reliable everything like this last sub sequence that converts nice thing is just Which is equal to with converges to the limit of you And then you can just This thing with coaches words You Mean we just simply buy this one. Okay, so that proves that now for To finish up the group since you k converges to you Then the norm of you Has to stay to one and And This is all we need for the function you will to be in the set Remember the set C This means that then and the reason why I need this is that then I can say that I Is less than or equal than that? I is the mean of F over C But now what do I have? I have f of you is equal to reading you This one So the first term you bring of you plus your square is it's just the H1 So I have H1 norm Square minus And By using the claim over there, I get this thing is less than or equal than I'm using that I have strong convergence here to put together So what this gives me is that This quantity here looks exactly like this one Okay, so that is equal to the limit S k goes to infinity over omega of just reading a new day because the F2 norm part they just But now what is this? This is just f of you k And what do I know enough of you k? I know that it's less than or equal than I mean copy the mean So this shows that you can always achieve and use a So we prove the theorem Okay, as I as I remarked before what we have now is a Megan function of the Laplace eigenvalue problem with eigenvalue That is as small as possible because we have that use a global image so the Standard notation in this case So actually Minimizing f Minimizing f over the strength of the atom norm is one is the same as minimizing this question. So Which is the minimum I'm allowed to write an email now because that is actually that's the minimum of You Is equal to the minimum Zero, I'm removing that Radius heard over and this quantity here is called the problem without first pretty much is a Compact way of writing So what if you want to find the next smallest taken value? So what we do is we define X1 to be You one Such that These are close condition this basis close and it's a new space again So you can pretty much repeat the exact same analysis that we did with X1 in place of X X was it's one zero and If you do the same analysis you find In number number two that would be the infim or mean is this is the same over all of you in X1 You can repeat this process by just Adding you to here and then you find it's free and in general you get that lambda Say plus one Okay, so we have a Non-decreasing sequence of egging values and to each of these egging bodies we can associate In a game function that has had to mark one Note that I'm not saying that the egging space has dimension one I'm not using that anywhere Like I'm not gonna need these for Is that even true? That's fine that I'm using the Revenants of 30 or something like that But I don't know if you can get that one Anyway, I spent some time yesterday thinking about this but I don't know Anyway, so for each egging value we can find the egging function and But the way we have to find these all of these egging functions are gonna be our problem to each other Because we look for everyone in the Okay, so the last result for today is gonna be So as I said, this is an increasing sequence so So the only way this is not true is that if they are even from the top Suppose By contradiction that there exists a number M such that Lambda n is that circle then m for every n. Okay, then using the egging body question This is equal to lambda n of u n out to square but this is just equal to 1 so this is lambda n and this is uniformly bounded by n So again, I have But the gradients are uniformly bounded and Since also the add to norm is bounded by 1 if they are exactly equal to 1 then I have that u n Is bounded in Again But this allows me to do again is to use the rally compactness theorem to find a Convergent of students there exists that converges some functions. You know, this implies that then You and K In particular at least shows that You and K minus U and K plus 1 This one has to go to zero But remember that these functions are orthogonal to each other So if you use the dog, right here in this one, this is just equal to U and K minus twice U and K plus 1 plus U and K plus 1 Now they have a problem. So this is 0 and this is 1 So this one it has to stay constant to so you cannot go to zero. So What is not verified is that there exists And this proves that The students are making by this Yeah, so he had time to prove the spectrum theorem, but I think it's worth to just write it down It says that I think to justify all the work that we've done like that's see some useful application So for many we can write proof of this is you I'm just gonna like I have to say something The proof of this is just a lot So do it for F smooth smooth enough at first and then Prove it for every app in that tool by approximating it. So you first do it for F In each one zero because you want to you want to be able to say something about the derivatives of F And what you do is like you do an estimate on the truncation error So F minus the sum up to N And this goes to T. Do you guys have questions? Yeah, so just for this the spectral theorem thing Does it not work just to so we peeled off the smallest eigenvalue and We take a problem couple of people the next most high value eigenvalue We have these increasing eigenvalues and so if That didn't spam the whole space. Could we just continue to process? Ah So it's an even process, right? Like you got a sequence. So What you're seeing is that I Guess like my interest like my I mean, I'm probably thinking to my Intuitions that since you have this increasing eigenvalues condition that that says like one past through this process should Yeah, yeah, that's what we do exactly you Once like you go just keep going and of course, you're not going to end up It's like what we were wearing boys a shoulder So, yeah, what you're saying is that just pick this even so they can buy this and Look at like one We are just think about just interval that that is You can practice sign and And in that case you can play around