 Quindi, abbiamo cambiato il programma e invece di professore Cabrera, oh, invece di professore Baskets, ora, che si sta parlando di ora, abbiamo professore Valdinossi. Abbiamo avuto la seconda legge. Quindi, andiamo. Sì, sono felice che ci sono molte persone che vengono a vedere Ciavi e Juan Luis, per essere contenti un po'. Sì, sono felice di essere dopo Juan Luis, perché posso usare molte delle idee che ho introduzione, quindi potrei usare le metode di extensione e l'idea che, comunque, le modelli fraccionali corrispondono alle modelli di power, le case e cose così. E anche se useremo l'idea che queste modelli fraccionali devono diventare con cose importanti che vengono da realizzare. Quindi, una cosa che devo correggere da una legge di Juan Luis è che, ovviamente, Real Madrid è solo la seconda migliore squadra nel mondo e la prima è, ovviamente, Roma. Come vedete, la più importante cosa, ma le più importanti cose di quest'anno, è quello che è avuto in Stadio Olimpico yesterday e per raccontare un po' di questo, cosa ho fatto è l'ultimo. Penso che è una computazione instructiva se vuoi studiare problemi fraccionali. Questo è stato da Wikipedia. E la season-by-season hai il numero di gols scortati da Totty, che è un grande numero. Che ho fatto è che ho preso la differenza di gols da una season alla prossima. E ho aggiunto questo numero due anni in avvenzione e due anni dopo. E ho preso questo in un graffo, quindi non so se... Quindi, basicamente, cosa hai in la direzione verticale è la frequenza di quanto questo numero è accortato. Quindi, vi aspettate che i miei scorti, il stesso numero di gols, più o meno, in avvenzione come i previousi anni, quindi vi aspettate che i zero accortano con una grande frequenza, ma hai casi in cui, da un anno all'altro, hai una grande differenza di gols scortati sia con il sign positive o con il sign negativo, ok? E quindi i red dots sono la frequenza di queste ocurenze. Per esempio, in avvenzione, con il plus o il minus 2 anni, l'avvenzione scortata, il stesso numero di gols di i previous anni, 8 volte. Ok? E poi, cosa ho fatto, ho computato l'avvenzione e i start-up deviazioni di questi punti e ho portato il gaussian con questa avvenzione e il standard deviazione. E il risultato è terribile, perché la cura blu non sembra, ma questo è il gaussian e è il gaussian perché, quando hai agli oscillazioni e lungo taile di infinito, il gaussian tende a essere molto flattato e vicino a zero, perché che è il gaussian? Il gaussian è sempre un'exponazione negativa, ok? Quindi... Sì, minus sigma. Ok? Quindi quando hai bisogno di il gaussian per essere molto flattato e infinito per avere una lunga taile, il migliore che puoi fare è fare questa sigma molto piccola, ma questo significa che questa è quasi constantemente e poi, per mantenere il volume è uguale a 1, il grande c è veramente molto piccolo e quindi la piccola che hai è un gaussian con una piccola constante in fronte e una piccola sigma che è quasi una piccola constante e quindi non è buono per modellare qualcosa di questo. Ma poi puoi ricordare che non sei fracciano alla plascia e puoi fare questo, puoi mettere... So, let me say that I will call X the horizontal variable and Y the vertical variable and then I can put these plots in a log-log frame so I will call big Y to be the log of Y and big X to be the log of X. Now as a technical remark, the reason for which the number of goals with plus or minus two years is because I wanted to have positive X and Y and not to deal with the log of zero, something like this. Well, if I do that what I get is the following so these are the dots in the big X and big Y variables and the blue line is the gaussian in these variables, still pretty bad. But so instead of this this is the same picture of the red dots in the log-log frame but now I do something a little bit smarter so I try to put a straight line the blue line as close as I can to these plots. Again, this is not a very professional job I did it by hands but then let's see what happens if I plot back the original X and Y variables now the agreement looks much better so it's not perfect you see there is still some dots here and there but if you compare with what happens before it's a huge improvement and this is a power-tail distribution in the sense of Juan Luis where roughly speaking because in the log-log plot I was put in a straight line so a straight line is Y equal to a constant was a straight line which actually decreased in direction so there was a minus with some coefficient M times X if you scale in the original variable let me call big C like the log of small c you have log Y equal to log C minus M log X which is the log of C over X to the M so it's like Y equals C over X to the M ok now I promise to do some of the proofs of the claim of this morning only using pictures so I will only prove this in a very specific case because by picture you cannot do everything of course but I think they will kind of convincing so for this I will use a couple of information one thing is that one knows what the square root of a complex number is so I will use this notation if I take a complex number Z to be rho e to the i theta I will define the square root of Z to be the square root of rho e to the i theta over 2 now to avoid multiplicity problems I will restrict the argument theta to be between minus pi this is a plot of the real part of the square root under this notation and as you see you have some problems here when theta is equal to pi must be that and this is the plot of the imaginary part and you see here that when you do a circle you have 2 pi difference from one piece of the surface to the other as it should be from the expression there if you have if you put instead of theta theta plus 2 pi you get a pi here which gives you this continuity ok why we care about the square root we do care about the square root because this morning we made a very funny claim which was this that if you have us of x to be 1 minus x square to the s plus and then you take the fractional application of us this is equal to a constant probably I should have emphasized more this morning because some people asked me I have the inclination to forget about constants so all the constants sooner or later will become 1 and even in definition of the fractional application I didn't put any constant but if you want to be consistent for instance with the Fourier approach you have to put the right constants to do things in the correct ways and this was very strange for us because you have a picture like this equal to 1 because it looks like concave here so you have a minus sign should be 1 ok and this was very strange because it was producing a boundary behavior which was not the same as the classical case and another thing that it was producing somehow so this is in b1 another thing that was producing that was somehow strange is that if you fix this point here and you make a blow up picture then you have this situation that I was calling vs of x which is take for instance the last coordinate the positive part to the power s then vs is harmonic in the half space I mean somehow if you believe in the first you believe in the second because when you blow up your picture you scale us like in us of x over r and you do this when you make the scaling the one will go to a power of r and send in r to zero you kill the non linearity you kill the right hand side and you go to zero and this again was very strange because of several reasons the first is geometric because we have a strange object we have something which is zero here and then it grows like a casp to infinity and this doesn't really seem natural to think that this object is harmonic but it is in the fractional sense and the other at least for me the first time I saw this thing was a computational objection so somehow if I take the fractional application of order s it is an object which makes two s derivatives so if I make two s derivatives of an s power I should get a minus s power so the first time I saw a picture like this or a formula like that I said to myself that is for sure a mistake because there is no geometric reason for which this picture is true and there is an obvious analytic reason for which if I take the fractional application of this object well this should be not zero should be a constant times xn to the power minus s so let's discuss a bit whether or not the picture is correct and whether or not the analysis is correct so first of all if you think about the picture is correct well you can find these heuristic explanations and also some rigorous proofs in a small book I wrote with Claudia Bucur is a springer lecture note of the Union Mathematica Italiana so let me give a heuristic argument that convinces us that this picture should be correct which is somehow this let's consider for instance an interval here and let's try to understand what is the s-harmonic function with this data outside so if we are in the classical setting starting from this point and arriving to this point the natural thing to do in the harmonic case in the classical harmonic case would just to take a straight segment joining these two points and as Juan Luis taught us there is a probabilistic reason for this because we can see this data as the payoff that we obtain if we follow a random walk back and forward and what is the expected payoff that we expect starting from here is exactly the average value between this point and this point which is the point in the middle so for instance suppose that this is 0 and this is 1 well this point here is the average between 0 and 1 which is 1 half and then you can do the same starting from the point 1 4 and you say what is the expected payoff starting from here well I can either drift to this point and reach 0 or drift to this point and reach an expected value of 1 half so the average of the two is 0 plus 1 half divided by 2 which is 1 fourth ok and so you get 1 fourth and so on so this is a complicated proof if you like the straight segment is harmonic in the classical sense but it's an instructive proof because let's try to do the same argument in a local case then we see this that if I start from 1 half then I can follow the process and say exit to the right or to the left if I exit to the left I always get 0 if I exit to the right I can get 1 or I can get something bigger than 1 so this is very good because somehow it tells me that my expected payoff is not the average between 0 and 1 but it's a weighted average between the values on the left which are always 0 and the values on the right which are always bigger than 1 so the value that I have to get here is actually bigger than the straight line so I take something here and similarly if I do now the straight line joining this value with this value and I repeat the argument here well if I start from 1 fourth I can drift and get 0 to the left or drift towards the right and get a payoff which is in fact bigger than the payoff that I have here so I can get this value or I can get something more and so the value that I have here for my s-harmonic function must be above this segment and so on so now probably the picture gets a little bit spoiled if I keep going but what I wanted to say is that whenever I find the value and I make a segment joining with the origin the picture that I have is always to be above this segment so this is a kind of natural concavity that my s-harmonic function needs to have near the origin ok still the analysis plays against us because it tells us that if I take 2s derivatives of an s-power I should get a minus s-power so what's wrong with that and this is a little bit more tricky because the answer is that there is nothing wrong with that for instance if we apply this formula on the negative values of x that's exactly what we get we get a constant times x to the minus s the very strange thing is that if we apply this formula on the right hand side of the positive x then the constant that we get is equal to zero so for instance if you want proof of this in this line that is the idea of computing this constant by the integrals the singular integral defining the partial application can be computed explicitly it's not completely trivial one has to integrate by parts but if one does that can compute this constant and check that this constant is magically zero and for instance this proof is contained in this booklet if you want to look at it but today I promised no computations just picture so I will do the proof of this formula in dimension one and for s equal to one half I have to admit that this is the only proof no of this fact based only on pictures well based only on pictures and on the formula that one Louise pointed out that is the idea of the extension so somehow the idea of the extension is that you have a function small u that produces a function that I will call big u of x and y so since I'm doing picture for me x will be in R the x y will be in R times zero infinity and this big u is the harmonic extension of small u so somehow big u solves the class of u equal to zero big u of x zero equal to u of x and then I can reconstruct the fashion Laplaccia by taking the y derivative of big u and evaluating it at zero now there is a there is a little trick in this definition this definition is not perfect as it is because in a sense the harmonic extension is not unique because if I have u that solves this equation I can always sum a constant times y and it solves the same equation a constant times y vanishes when y is equal to zero and it's harmonic so in this I also have to follow one Lewis advice forget a bit the functional spaces and just rely on intuition I have to choose so whenever I choose a harmonic extension I am forced to choose the representation of this harmonic extension that has the best possible decay at infinity ok so we always keep in mind that whenever I draw a picture of a harmonic extension it better grows as much as possible at infinity so if I do that the approach for this so now I want to prove this formula just basically using pictures and the idea is that I have to use to find the harmonic extension for the square root of 1 minus x square so the natural thing to do is to say ok I know the complex square root and I know that if I take the real part of a holomorphic function I get the harmonic function so the natural thing to do would be well why not looking at u one half of x y to be the real part of the square root of 1 minus z square ok that could be a good idea the only drawback of this is that if I take c of the form I y y large the square root of 1 minus z square is equal to 1 minus i square which is minus 1 which becomes plus y square so it goes like y at infinity so this is very bad so the only way I can make this as a good candidate with decay at infinity is to subtract the y or if you prefer it's clear right I have to subtract the y I mean if you want to write it as a real part of a holomorphic function this is just the real part of so I have to put I z I put I z as promised this is the picture so as you see this is exactly the function big U here you can see this is the profile at y equal to 0 which is exactly my half-circumference and this is the decay at infinity well with this I'm done because if I want to compute the fashion laplation of small U I will just make the y derivative of big U and take y equal to 0 now this guy is quadratic and you can check that it is no contribution well this guy is actually minus 1 minus y when I take the derivative I get minus 1 but I have a minus sign here so I get 1 so if you are convinced that this is a proof do an even more fancy thing since we learned our method the other thing I claim today is that if you put minus one half so if instead of U one half you look at U minus one half in the sense that we put minus one half here and zero outside there is a minus one half I don't know if it's readable from far I change one half to minus one half so now the picture is much more dramatic because the function has now a shape like this but quite surprisingly the fractional application of this function is zero now I'm happy that also Francesco is surprised by this thing so when Nicola Batangelo told me this I said now come on it's not possible it looks quite surprisingly but it's true and actually if we learned our method now it's kind of easy to do it because I can just made a minor modification on the black part I want to look now for the harmonic extension of U minus one half which will be called B U minus one half well what can I do I knew from the previous example that it's a good idea to look at one over the square root of one minus z square now in this case I don't have to subtract anything at infinity because this thing is always decaying so it helps me this time I don't have to do anything the only thing is maybe to look at the picture and the picture is that one now the problem of pictures is that sometimes the computer is not intelligent and does not put the picture where I want so of course this part here that is the part in infinity is somehow the most nice thing of the picture but the computer cuts the picture at level 2 but it's kind of I mean you can see that this is going to infinity somehow yeah so it seems to me like the dome of a of a circus right you have two peaks here and the dome of the circus like this and so the trace function blowing up that I drew there it's going to zero at infinity and if I do this now so if I do the case with minus one half well as I was saying you believe me before that this was quadratic and it was producing a zero so this is just producing a zero ok so for next picture probably I have to discuss a bit the last difference that I didn't have time to deal with this morning which was somehow already highlighted by Juan Luis in his lecture so at some point Juan Luis complains say you said that you had eight differences but eight is not it's not a very nice number and so I thought a bit and they said well but the last difference is actually made of three differences because and so that would make ten if you count it like this so the difference eight for me was the decay at infinity we will see three types of differences because this morning I said ok I will do this three times one for the Schrodinger equation one for the Allenkahn equation and then I said and I don't remember the other but then I remember what was the third one when Juan Luis said speaking about the heat flow should include this as a new difference and in fact that was the other one so the the fundamental the heat kernel that was supposed to be the third one the solution of the fractional heat equation so let me start with this so let's discuss what is a fractional heat kernel well in the classical case you can decide that you study the heat equation like dTu minus the fractional application of u equal to zero say in rn times infinity or whatever and suppose that you prescribe u at time zero to be the delta of zero so all your heat is concentrated at the origin and you diffuse it and you see what happens then in the classical case as Juan Luis says the the real thing that you have to learn in mathematics is that the solution is a Gaussian you say something like this in the lecture so u over x ok and since I will mess up the scaling I compute what happens at time one everything is invariant on the space time scaling but since I am quite sure that I will put the t in the wrong place I scale the time to be one and I get up to constants the Gaussian and so I will call this of course I can do the same game in the fractional case so the fractional case would be just instead of the Laplaccia put the fractional Laplaccia and find what you can call gs of x again you can call the solution of this equation at time one and the question is how this gs looks like there is an easy answer the answer is take Fourier transform of it if you take Fourier transform you have that the derivative of the Fourier transform of u is equal to minus c to the 2s u hat forgetting constants and this says that well u hat is the anti Fourier transform of e to the minus psi to the 2s sorry u is the anti Fourier transform u hat is e to the minus and then u anti Fourier transform right well this is a good answer not very good answer in the sense that it's absolutely not evident from this expression that you have a nice object to deal with it can be proved as one always mentioned that u goes like one plus that g so again gs is going like a power at infinity so at infinity you have this heat kernel that still decays to zero but much much in a much much lower way than the classical case the classical case was a Gaussian so the exponential of a quadratic function here you only decay like k plus 2s well I don't have a simple proof of this for any s but there are simple proofs for s equal one half for s equal one half there are several proofs that one can do this would be just the anti Fourier transform of e to the minus x and the first proof is historical so the things have a name this guy is called Abel Kernel Fourier transform of the Abel Kernel is the Poisson Kernel but this is just because one knows the references is not really a proof it's just a name the fact is that if n is equal to one then one can just compute the integrals n equal to one is much easier in this case because the norm of x is just the absolute value so if I write down the definition of Fourier transform I can integrate and say the model of x is minus x for negative x x for positive x and compute the integrals of the exponentials and I get what I want so this I leave to you as an exercise again if you don't manage to do it let's chat about it but this exercise I would say is much simpler than the one I gave this morning about the master equation this is really an exercise that if you sit down five minutes probably you do it just computing the integrals of the exponentials when n is bigger or equal than two the situation is a little bit more tricky I have a proof using computing integrals but it's not nice so I have it somewhere and it takes some pages but above all it's not very instructive because it only uses clever tricks of integration so I will just show you it takes a little bit more than a page and if you see it from far it's not very instructive I mean it's just full of integrals and you integrate cleverly bypass three or four times you use one Lewis trick to remember that the Gaussian is the Fourier transform of itself and by magic it works but there is a nicer approach that is valid in any dimension and it's basically based on this picture is only two dimensional because of the limits of the white board but the argument is actually it works in any dimension and it's based on the idea somehow that when you look at the extension the variable y can play the role of a translation with respect to time so if you want if you are more advanced from a PD perspective the fact is that the Poisson kernel is the normal derivative of the fundamental solution and so you just have to put your fundamental solution in the extended space a little bit far from trace and to take derivatives we will be more clear now so the idea is that you write your fundamental solution let me call gamma of x to be so for me gamma is the fundamental solution in r n plus 1 so the fundamental solution is 1 over x to the n minus 2 but if n is n plus 1 the fundamental solution is 1 over x to the n minus 1 if n plus 1 is bigger or equal than 2 which means no if n is bigger or equal than 2 and the log so if I do this I also consider u of x, y and t to be the derivative of gamma in y plus t I look at the fundamental solution I translate it and I make a derivative so somehow I use y and t as interchangeable parameters but now I have a nice computation because you can compute the derivative of y of gamma this is just a simple computation in both cases you get y plus t divided by x squared plus y plus t squared plus 1 over 2 ok which means that your u big u of x0 t is actually the function che chiamavamo g1 half x0 1 g1 half exactly 1 over x squared plus t squared to the n plus 1 over 2 I think I wrote it here but I erased it at some point but that was the Gauss kernel the fractional heat kernel now the fact is that we have to prove that the fractional heat kernel in this way satisfies the fractional heat equation so forth actually here it is it's still written on the blackboard so just that I have to compute how much is dt u plus the fractional application of u well dt u is fine I I can compute it from here this would be just I take y equal to 0 would be the derivative in t of x squared plus where to the n plus 1 over 2 now this guy here will be minus the limit as y goes to 0 of the y derivative of bq but now the y derivative of bq bq is this bq is dy gamma so if instead of bq I put y plus t etc etc I have this expression but now you see this is 0 because here I'm differentiating in y here I'm differentiating in t but y and t plays exactly the same role in the computation so I don't even need to compute the derivative I know just looking at the symmetry that this is 0 ok and the picture you have above is exactly the plot of bq into d so the trace would be the Poisson kernel and you have the extension as drawn ok what else so this takes into account the striking difference between the decayed infinity of the heat kernel in the classical and fractional case and then what about the Allen-Kann equation so the Allen-Kann equation anche in 1d è una questione difficil in the fractional setting so the classical Allen-Kann equation is that minus la plus of u is equal to u minus u cubed and probably we will talk more about this equation later on right now just let me tell that it's a very important equation in many fields and it models phase coexistence situations and it's somehow one of the first equation in the realm of boundary of reaction diffusion that Juan Luis was talking about and you can consider it's fractional analog that would be this that can be considered somehow by extension method a sort of boundary reaction equation in the local setting well still you want to know what happens to the solution at infinity now in the classical case you can solve an ODE in 1d you can multiply by u and integrate and when you do this you have an explicit solution that is exactly of t say or of x if you prefer is the hyperbolic tangent up to the constant which is square root of 2 and now the hyperbolic tangent goes to the equilibria exponentially fast so what happens is that for large x well u converges to 1 but it converges exponentially fast to 1 ok as you imagine this is not true in the fractional case let me erase this we have to remember that we will discuss the Schrodinger equation later on if time allows so in the fractional case this is not true you can look at this equation in 1d now even in 1d I don't have a better notation for the fractional Laplacian I mean the Laplacian becomes two derivatives in 1d the fractional Laplacian there is no simple name even in 1d so I write it like this even if I'm doing the analysis in 1d and the analysis in 1d is unfortunately extremely complicated there are two papers the analog of this object in the fractional case one that I did with with Giampiero Palatucci and Dovid Yusavin and there are two papers by Ciavi Cabre and Yanix here and one of the results that was found in these papers was that you have a no to find the solution of these equations normalized in a way that u of 0 is 0 that u is increasing and that this limits plus and minus infinity equal to plus and minus 1 which would be the natural analog of this one so this is something that looks like this but you can still find such solution the only thing is that u does not decay exponentially to the equilibrium but it only decays polynomially fast so what you have in this case is that 1 minus u goes like a constant divided by x to the 2s come can we prove this at least for s equal to 1 half in a simple way well the idea is that we can look at a slightly modified equation which is instead of u minus u cube let's put the sine of u at the end sine of u is at the third order so let's look to a similar equation let's say s equal to 1 half minus u cube but sine of let me renormalize things so that I have 1 and minus 1 I put the pi just because I want 1 and minus 1 as equilibria and not the pi as equilibria ok e poi che c'è una soluzione esplicita di questa che scrivo come 2 over pi arcton ora l'arcton sembra simile al graffo per i tangenti hyperbolic ma è struttuale perché se trovo l'arcton ma se sei più attenzionato di capire che l'infinità di 1 minus u è solo 1 over x quindi non è sponenziale quindi le due funzioni sono struttuali molto differenti e se hai questo non puoi dire che questo deve essere un comportamento tipico anche per questa equazione o puoi costruire questo per costruire barriere per altre equazioni quindi puoi non pericolare ma cosa voglio fare è convincere che questa è la soluzione di questa equazione e questa non è completamente scopriere perché, per esempio, qui ho detto, multipliare con te e integrare e hai la soluzione ma in realtà potrei dire che questa è la soluzione che si tratta di derivativi quindi, se scrivo la soluzione nel caso classico che è una soluzione che può essere barri ma è un comportamento tipico perché ho bisogno di dare derivativi a questi errori qui, anche se qualcuno si tratta di dire che questa è la soluzione beh, ancora ho grande risposta per scoprire che è una soluzione quindi ho detto una possibilità una possibilità è che uno si trova per BU che è 2 over pi della tana e se pensate della tana è come l'argomento di un numero complex quindi, qui, basicamente si tratta di z plus i si tratta di la log e si tratta di la parte immaginata di questa quindi, se procede così puoi che questa funzione sia armonica e quando y è uguale a zero ovunque si trova questa e la foto per questa come vedete questa è l'arctangente che infatti sembra, infortunatamente, l'argomento hyperbolic che ho avuto qui ma sappiamo che non è uguale e poi ha di andare a zero a infinity quindi ha questa funzione che a questo punto è vicino all'argomento quindi si tratta di zero questo punto, qui, è molto vicino a minus 1, si tratta di a zero quindi questa linea è quasi orizzontale non molto, qui è un po' più più di qui perché non siamo arrivati a infinity ma ancora in principio stiamo tentando di mettere tutto in zero se facciate quindi completare l'argomento Y di Big U e osservare che è esattamente uguale all'argomento Pi U solo mettere tutto insieme e non lo faccio nel vero bordo perché diciamo che questo è un esercizio l'esercizio non è difficile ad essercizio che devi ricordare un po' di formula trigonometrica per scrivere l'argomento del arcomento perché ci sono molte fungioni trigonometiche quando compone una funzione trigonometica con un'argomento trigonometica ci sono semplici simplificazioni e questo è il motivo per cui quando mettete le cose qui il signo dell'argomento ci dirà esattamente cosa vuoi quanto tempo ho? finito? ah, ok, no no quindi discussing the decay of shreddingers in cypromised and we will try to prove that all functions are s-harmonic or s-caloric or whatever let's see if we manage to stay in time sorry for the delay