 So, let me just recall where we stopped, we were considering our transport equation, linear transport equation, where we have this new term, positive term, and h, with an inflow boundary condition, and because I was just stating that in the case the domain is strictly convex, then you can say that f is continuous outside the grazing set gamma zero. So, since f is continuous, now you can start thinking about derivatives of solution, and about the regularity of the derivative. So, you would like to take derivative in time, space, and velocity. So, actually let's start from the initial condition f is zero, and from the boundary condition g zero. If we want to compute the derivative of our function f, that if you recall we were really able to write it, you have to derivate all this term here, with respect to time, or with respect to x and v, like the coordinate x1, x2, v1, v2, v3. But we will need to find the value of the derivative of the initial function f0, and of the boundary condition g as well. So, everything is easy when you derivate the initial condition with respect to x, and with respect to v. But what about the derivative with respect to time of this function? This function is f0, it does not depend on time. So, in order to give some sort of compatibility, we will require that our f0 will solve in a way our linear transport equation. So, we will impose that derivative with respect to time of f0 will be given by the equation that I wrote here. So, just to make it clear, we have that dt f0, we suppose that f depends on time, f0 depends on time. If it has to satisfy the equation, it has to satisfy this, f0 equal to h. So, we do not know what is this derivative. Okay, we impose that dt f0 will be equal to h minus v scalar product f0 minus nu f0. So, we will impose this condition. Now, the same happens for g. g, we recall, was the value of the inflow boundary condition. So, actually g in this case is written here, it depends on time x and v, but is defined only on gamma minus. So, actually, it's not really depending on the whole x and v, it's just defined on the x that are on the boundary. So, it's defined on a surface. So, whenever I take the time derivative, I'm fine, it's easy. Whenever I take the v derivative, the gradient, I can do it. Whenever I try to derivative this function with respect to x, I can do it in a way using tangential vector. So, let me, I just wrote them here because I realized that they were written on the slide. So, we have t1, t2, the two tangential vector. I can derivative g with respect to them because g is defined on the surface, on the boundary. So, in that case, they exist. So, that derivative exists while the problem is the derivative of g with respect to the normal, because actually g is not defined on the whole space. So, it's not existing actually. So, I will do the same. I will take my transport equation and I will suppose that g satisfies the transport equation up to the boundary. So, that dtg plus v gradient of xg plus mu g equal h. So, from here, actually, I can recover the value of the normal, of the derivative with respect to the normal. So, it turns out that it's enough to take minus. So, I have that v gradient xg is equal to minus dtg minus mu g plus h. And then from here, recalling that I can take gradient x like d tau 1, d tau 2, d n, I have that the final result is actually this one. So, here you have the tangential multiplied by v. And here you have this term that actually gives a problem because it's 1 over n times v. And this thing can be 0. So, actually doing like this can create something that is a singular. So, we will have to see how, especially in our problem, how we have, how we cope with this term that can be singular. Okay. So, using this notation now, we can even write the derivative, the gradient with respect to x of g in this way. So, you just need to multiply by n or by tau 1, tau 2 in order to recover the derivative with respect to n and the derivative with respect to tau 1, tau 2. Okay. The tangential vector. So, what is possible to prove is that whenever you require that the derivative are LP in this way. So, the derivative of the initial datum and the derivative of the inflow boundary condition. And you had some integrability condition on the term h. Then you can prove that for sufficiently small t and for any p actually in between 1 and plus infinity, there exists a unique solution f that is in C0 LP such that all the derivatives belongs to this space. So, this is again for the transport equation, the linear transport equation with initial datum that are regular LP and LP even for the boundary condition. And actually you can also prove that the traces of the derivative of f with respect to time or with respect to velocity and with respect to x, all behave well on the boundary gamma minus. So, you have all these compatibility conditions. Everything is verified. So, you are really, I mean, everything is fine. So, that's what you have in the case of the transport equation. So, just to have an idea on how we prove this kind of theorem. So, I wrote before the definition of the function f like this. So, you can derive with respect to x with respect to time or with respect to x and with respect to v explicitly all these terms recalling that now nu that you have here. If you derivate with respect to time, for example, you have time here. You have time inside nu here. So, the derivative with respect to x of nu will give something that depends on time here. You have time here in the derivative with respect to f0 and then again the same for the other term. So, actually the time part is easier. Then whenever you derivate with respect to x, actually you have x even in the definition of tb, the backward exit time. And even in the definition of xb, the backward exit point. So, whenever you do the derivative of this term, you have to derivate here, for example, nu will depend on x as well. You have x even here. So, you have to produce even the gradient of tb, the gradient of xb with respect to x and as well with respect to v. So, it becomes much more complicated but what turns out is that actually you can do by direct computation. You can compute the derivative with respect to time here, with respect to x and with respect to v outside the set where t is equal to tb because in that case, you have the two conditions, the initial one and the boundary condition as well. So, it's much more complicated to compute the derivative since they are different according to which one you are considering. But it turns out that actually I don't want to explain you how they are recovered but you can derivate the tb and the xb with respect to x and with respect to v. So, these are the explicit formula that you have to put inside when you derivate all the terms. But it's very, very long. So, I don't want to enter more in the details. But up to now, this computation is fine but it's not enough because you have to know how to cope on the set of point such that t is equal to gb. But what it turns out is that actually the derivative that we can compute by direct computation here where this partial means the derivative with respect to time x and v according to which one we are considering. So, the one that we compute were precisely the weight derivative of our function because it is possible to prove that whenever you take a function phi that is the c infinity c and you try to compute the product, the integral between f and the derivative of d phi, you will obtain precisely this stuff by integration by part as usual. And this will give that actually this derivative is precisely the derivative, the weight derivative of the function. So, you don't need to add any term on the surface, let's say on the set where t is equal to gb. So, let me show you another things that we can recover using this transport equation. Actually, from the theorem before, we have some estimates that I haven't written here for the derivatives that can be recovered computing directly the derivative and then working a bit on the trajectory. But you can also do in another way that is you can see to the derivative at least if you consider for example the one dimensional derivative with respect to time or each derivative with respect to space like with respect to x1, x2, x3 separately and then v1, v2, v3 separately. The derivative, this derivative so in this way, take one of this value, you will still have a function and this function actually still satisfies something that can be seen as a linear transport equation like properly but satisfies this term and then what it turns out is that you have to put an h here that is no more the h that you had before but is given by this. Let me just explain why. So, we have that our f satisfies dtf plus v gradient of xf plus uf equal h and now you just have to derivate this by this delta being either dt, either dx1 or some fillet. So, you derivate all this by this and do the same here. So, what it turns out is that this is dtdf, no problem, then you have dv gradient f, gradient xf. So, you are derivating the first plus v times the gradient xdf plus d nu f plus nu df plus dv gradient xdf plus dv gradient xdf plus dv gradient equal dh. So, if I am looking at the equation satisfied by this, we still have the shape of a transport equation but I have to change a bit the term that I was supposed to be here with the term dh minus this one, minus this one, no, sorry, not this one, this one, d nu f and that's all. That is precisely what is written there, I forgot anything, okay. So, you see that this h actually depends still on f. So, we can say that this is a transport equation but in a way improperly because f is still inside here but if we are considering that now we know f, it's okay and then this term of course will change according to the derivative that I am considering because remember this d can be dt dv dx and actually if we are considering this term, this term will be different from 0 only if this d is one of the free derivative of the velocity because this v does not depend on t and does not depend on x. So, this is the only times where you have this term and it's precisely saying that the derivative the velocity derivative of f will depend also by the time the space velocity, the space derivative of f here while when you are considering this term our nu was depending on tx times so you have all the free derivative and even x h here was depending on space and time so you have everything. So, actually you can reapply all the reasoning that we were considering before for f to the derivative of f in this case and you can produce estimates even for the derivative like this one. So, let me just comment a bit around these estimates. Here you have that the p norm of df either time, space or velocity and then you have the trace norm of df on gamma plus the outcome in boundary is bounded by the initial condition the derivative of the initial condition, the p norm of the derivative, the condition on the boundary and then this term that actually depend on dh that is still contain something about f and df again. So, actually it's not really something that can give us a bound a proper bound but in a way we can use these estimates to cope with, to bound in the iteration scheme that I will show we will use for producing Boltzmann estimates. Okay, finally two Boltzmann equation. So, let me explain to you why I spent a lot of time talking about this transport equation in order to cope with the singularity and the regularity of the Boltzmann equation. So, the first theorem is an existing theorem that holds for this use condition, specular reflection condition and bounce back reflection condition and for a domain that is just smooth and bounded. So, we don't need to require strict convexity because we are just proving the existence of the solution. So, actually this theorem was already proved by Google in 2010 we just stated a version that is a bit that is an improvement because it gives us something that is polynomial and not exponential like the rate that you have here. Now, you suppose that your initial data f0 is greater than 0 satisfies the compatibility condition and such that there exists a theta in between 0 and 1 third that leads this norm the infinity norm less than infinity. So, we are waiting the initial condition by this exponential. I will let you understand why we need this condition in a minute. Now, the theorem says that there exists a unique solution of the Boltzmann equation on a interval of time 0 t star where the t star depend on this quantity this norm that we have here and this solution is such that we have this estimate. So, actually this estimate again as it was in the case of the inflow boundary condition for the transport equation you need to put here at theta prime this smaller. So, you have to reduce a bit here and actually it is just up to this star but let me understand under line that whenever the initial datum is very close to the equilibrium then this star can be taken equal to plus infinity. So, this estimate here in this case precisely says that the equilibrium the convergence the polynomial convergence to the equilibrium. This is more or less say it in a minute actually it is because of grad estimates like that you need to use maybe is in the next slide but okay. So, the idea of the proof. So, here you can see why I spent a lot of time on the transport equation because the idea finally is that we can use a positive preserving iteration. So, we start from f 0 that is actually our initial condition positive greater than 0 and then we solve the Boltzmann equation but we put here f 1 for example f 1 f 0 here and f 0 here. So, in the collision term the gamma gain here we put the condition that we are the function that we already know here as well we do the same and then we iterate. So, at every step actually our new is known because we are putting the function that we found at the step before and our H is given like this. So, actually in this case I am reducing to a linear transport equation as of the type that I just showed you before. Of course, we will need more estimates all the difficulty will be in giving finding estimates for this term and this term according to the different boundary condition and while I will take an inflow condition because actually I can state that the value of f m plus 1 the function that I am trying to solve to find here is equal to g m where this g m according to the boundary condition that I was considering is given like this. So, if for example I am considering diffuse reflection boundary condition I have to put f m here instead of f. So, the one that I found the step before and I call g m like this. If I have a specular reflection boundary condition I just have to call g m the value of f m evaluated at r x v here. If I am considering a bounce back reflection g m is like this. So, actually the free boundary condition can be reduced to the inflow boundary condition. If I have an inflow boundary condition actually it is easier because g m is g and that is all. So, exactly the point is that now I am reduced to this linear transport equation. So, I can use the Amel formula. So, this is precisely the formula that we computed before for a general f that was the solution of the linear transport equation. Now, it is the formula satisfied by the iteration f m plus 1 where our nu is precisely nu evaluating the square root of nu f m the previous iteration. The initial datum is always f 0 while the boundary condition now is g m. So, it depends on m here and the h that was supposed to be before is now gamma gain evaluated at f m f m. So, all the estimates that we were able to find before can be applied now in order to pass to the limit. So, in order to find the existence we need l 2, l infinity estimate. Actually, the estimate that I showed you before for the function hold for every p here. So, everything is fine. We can use them and then the difficulty relies actually in finding estimates for this term and this term again. So, I don't want to enter into the details because it is complicated. Actually, for the existence it is not really so complicated. But for the next step, yes, it is much more difficult. But okay, let me, trust me, everything comes from here. You take this Duamel formula and you take it and then you can produce estimates. Actually, we can comment upon this. What you are supposed to do at a certain point is a sort of change of variable. Go from t minus t b x b to t x or s y. Let's say it is not really the same t and x. So, this change of variable will have a Jacobian that involves the regularity of t b and x b. So, according to the different boundary condition that you have, you will have some difficulty that rise or not. That's all the difficulty are practically hidden here. This is the point in the change of variable that you do in order to produce estimates for this function. So, okay, this is the existing theorem. Okay, yes, the comment y theta. So, I told you that we need to find estimates for the h and u here. But now h is gamma gain f m f m. So, this is actually the collision operator, the gain collision operator. And we have estimates obtained by grad that says that whenever we have a theta, that we have a theta in between 0 and 1 first, this is precisely the one third that counts here. And for any p in 1 plus infinity, you can find this bound for the p norm of the gamma gain. And then again for the p norm of the new term that we have here with these estimates. So, the e theta that comes is this one and that's why we have to impose that 1 third, the theta is in between 0 and 1 third. I think we couldn't do better, but what you need to reproduce all these estimates. The gamma loss actually is here. This is gamma loss. I just call it in a different way because it's linear. So, it's much simpler. Yes, I did the computation before. Maybe I didn't stress it enough. But then, okay, using these two estimates, you can recover these estimates then from for f m plus 1. And then you can pass to the limit and obtain the estimate for the solution. This is how you recover the estimate for the solution when m goes to plus infinity. Okay, so, this was the existing result. And now we put pass to regularity, more regularity. So, for the derivative as well. So, we need convexity. So, again, this is the same slide I showed you before. Just to recall that our omega is a bounded open subset and that now we impose that the domain is strictly convex because we want to cope in this case with the regularity of the solution. So, we remove the propagation of the singularity along the grazing set. Okay. Now, we are looking now for the regularity of the derivative. So, the time derivative, in the case of diffuse specular reflection or bounce back reflection, is easy because actually, as I tried to make you understand before, you can compute from the Duhamel formula explicitly the derivative without interacting with the derivative of the time tb and xb, the backward exit time and the backward exit point. So, since the singularity is hidden down there, everything will be easier because you don't have to use them, actually. So, again, you take f0 positive that satisfies the compatibility condition. Now, you allow a different theta, but it can be the same. No, no way, no matter. And you require that these two will be less than plus infinity because actually, this condition is because we will cope with the derivative with respect to time of f0 and of the function as with f0, just with the remark that I gave you before. We define the derivative with respect to time of f0 in this way, imposing Boltzmann equation up to time 0 in this way. And then, using exactly the same strategy as before, you are able to prove that the derivative is regular and this is the estimate that holds for a theta prime that is less than the minimum of the two and less than one fourth. So, it's still a polynomial estimate. So, actually, you apply the same results to DTF instead of 0. Why? Because if you want to work a bit hard, you can compute precisely the derivative. So, just to show you that it's longer, it's very longer. So, this is the first term that comes when you derivate the first part. It's a pity that I erased the function, but this was the first part coming from the initial condition. So, it produced two terms, three terms, one, two, three, because you have the derivative. This one is plus two terms and then the derivative of f0 that is here. Then, this was the second part coming from the boundary condition and then the part coming from the integral to the minimum here. So, it's very long, but you see that you never need the derivative of the exit time or the exit point, actually. So, everything is fine. Okay. Now, we want to deal with the space and velocity derivative and now the problem arises. So, the point is that one of the crucial ingredients in order to cope with these two derivatives is this kinetic distance. So, we need to find something that will be a way that we will use to cope with the singularity of the derivative that is defined like this. This is alpha xv is defined as the modulus of the scalar product between v and the gradient of xc to square minus two times this quantity. So, first of all, you can immediately see that since the normal actually is at a boundary point is gradient of xc, gradient of xc like this, this distance precisely vanishes on the grazing boundary. So, actually using this, we can erase the grazing boundary. So, whenever we multiply our function f times alpha, we are making this term equal to zero on the grazing boundary. So, we can get rid of it. And then, we can have some property. Actually, the most important one is this velocity lemma that holds precisely when the domain is a strictly convex. That says actually that if you call alpha stxv, the value of alpha along a vector trajectory, this xcl and vcl were the generalized trajectory that I showed you one hour before, then you have that for any S1 S2 that can be one greater than the other in any order, you have these estimates. So, imagine that, for example, at S1 alpha is equal to zero because we are at S1 here on the grazing boundary. This will imply that even alpha S2 will be zero for all S that are in between zero and T. So, actually this is saying that whenever we are on the boundary, so that alpha is zero, for all trajectory going to and again is still zero. So, this implies that it's not possible when the domain is strictly convex that a trajectory touches the grazing boundary. That is actually what we were seeing very well from the definition of grazing boundary like this. If it's convex, it's not possible that any trajectory comes from outside because it's always zero and this distance is zero just on the point that are on the grazing set. So, this is the idea. So, gamma zero in the strictly convex case does not really participate or interfere with interior dynamics. But I mean, we knew from the geometric point of view without using the velocity lemma, but it turns out that actually this lemma is very helpful in order to find the regularity of the trajectory. But I mean, we won't really see this. Okay. But let me try to explain a bit why the space velocity and the space derivative and the velocity derivative are the problem when we would like to prove the regularity. The point is that as I, as we already noticed, if I want to define on the boundary the normal derivative like this on gamma minus, I need to divide by one over Nv. And this creates something that is a singularity. We did this computation for G before. Now we are considering this for F, but we are still on the boundary. So, actually it turns out that one of the estimate that we have to produce, you know, when we are doing this iteration scheme that we will be always used even in this case, we have to give an estimate of this term. So, we are going to look for W1P norms. So, we have to put the derivative to the power P. And since we are integrating on the boundary, we need to put this that is the integration on the surface, actually. So, it turns out that this term is raised to the power 1 minus P. So, it can be integrable only when P is less than 2. So, that is why we will be able to prove estimates on W1P norms of the derivative only with P in between 1 and 2, like directly, because this singularity is not integrable when P is greater than or equal to 2. While if you want estimate for P greater or equal to 2, we need to use this distant function that is able to cope erasing the grating set with this singularity. This is more or less the idea. So, I will try to explain a bit better, but this is what you have to remember. That actually you can have, as you see here, estimate in W1P direct estimates without any weight for the gradient of f with respect to x and the gradient of f with respect to v for P that is in between 1 and 2. So, why we lose the 1? Because we will pass through weak convergence and we have weak convergence only for P greater than 1. Because if you remember for the inflow condition and linear transport equation, the estimates for the gradient were for P from 1 to plus infinity. But since we are producing estimates for the iteration, then we have to pass to the limit and in order to recover the weak convergence, we need to ask that P is greater than 1. So, this is precisely the estimate you have whenever you suppose that the norm of f0 is well defined and it is well bounded in P and in P and here. So, this is as usual and then you have the estimates for the derivative and for the traces. These are the norm of the traces on the gamma boundary here and here. So, again, this is a polynomial convergence in this term and we can obtain this convergence without putting any weight. Let's see why. So, the idea of the proof is again used in the iteration scheme and actually is precisely the remark with which I started this hour. Because I am considering the equation satisfied by the derivative of f n plus 1 here where d is either dx1 dx2 dx3 or dv1 dv2 dv3 like this. So, I have now this new here and this H m where this H m, again as I underlined before, still depends on f n plus 1 for example here. So, these were precisely the estimates that I was stating before for the df, now will be for df n plus 1 and you see here that you have to cope with these estimates in order to produce nice estimate for this term and with this term here. Now, this term still depends on f n plus 1. So, you have to iterate in order to cope with this and this term here is precisely the value of f n plus 1 at the boundary. So, the value of f at the boundary is something even like this. So, it's singular with this singularity like here. So, that's why in order to bound this term we need to put that p is less than 2. So, this is what will give us these estimates for m for f m and then you can pass the weak convergence for p greater than 1. So, you will have the estimates for the derivative for p greater than 1. Okay, now just some words on what happens if you look for estimates when p is greater or equal to 2 and up to plus infinity. Now, as I told you, you cannot expect that you have a direct estimate because that part is not integrable because it was also over the p minus 1. So, you need to add a weight and it turns out that the precise weight is a power of alpha, the distant function that erases the singularity times this term here that is actually linked to the geometry of the boundary. For example, this omega bar can be taken equal to 0 if the boundary is a quadratic. So, for example, for a ball. And actually this term is linked to the derivative of f. So, just to give an idea, I don't really want to enter more in the details, but what it turns out is that you can prove the regularity, so the estimate for any p greater or equal to 2 up to plus infinity with a polynomial weight here and again using the iterative scheme. And actually still you can see using this weight, alpha, you can prove even infinity estimates like L infinity estimates that are necessary to prove the C1 regularity of the function away from the singular set, away from the grazing set gamma 0. So, you need to require more regularity, the compatibility condition for the initial condition as usual on gamma minus gamma 0. And you are able to recover these estimates again in W1 infinity and then they will give this one propagation. Okay. Just for this, actually, I forgot to say, all the estimates, yes, I forgot. The two estimates that I showed before, so this theorem, this one and this one were obtained for diffuse boundary condition because these are the easiest conditions that we are able to cope with. We have other results for specular and bouncing back condition, but if these were already difficult, that one are even more difficult than this. So, I don't want to enter into the details. Just let me, I mean, for that boundary condition we really don't cope with alpha every time, but we are able to prove the C1 regularity away from the grazing set, even in that case. Now, once we have proved this nice regularity for the derivative, we want to see what happens in the non-convex case. So, in the non-convex case we know that we have a singularity that propagates inside the domain. So, the first important fact is the fact that it's possible to prove that the singular set is a set of co-dimension 1. Actually, the idea was if you remember the sigma bs that was precisely where the singularity propagates from the gamma 0s, the singular, so from this point, so all the trajectory that propagates from point like this, non-convex point, form a sub-manifold of co-dimension 1. So, the idea is that if we want to look for regularity, we have to look for bv regularity, and no more than this. So, how can we work with this? So, first let me just recall you the norm of bv function, like this is the total variation of the function bv, usual definition, if you know it, like this one. So, bv function are functions for which the derivative is kind of a measure. So, they are not so regular, but still something. How can we cope with this? We will find using again the iterative scheme, we will find w11 estimates, so that can, when we pass to the limit, the limit function will be bv. And how can we find estimates for this iteration scheme? The point is that again, this is the theorem, so again we will do the iteration, but we need to cope really with this set, because we don't like it. So, the idea is that we will construct something like a tubular neighbor that has to be constructed properly in order to cover all the singularity. Then we will use a cutoff function to remove this from the iteration. Outside this set everything will be fine, so we can prove the regularity, and then we pass to the limit. So, the difficulty lies in constructing this boundary in a proper way, so that all the estimates are uniform, so that we can really cover all the singularity and things like this. And doing like this, we are able to prove this theorem. So, this is the bv theorem. Whenever we take a bv initial datum, and again we have this condition that is the condition that gives the existence of the solution, actually. Then we have that f is in L infinity bv in space and velocity, and that its derivative is actually a Radon measure on the boundary, and the estimate satisfies is of this type. So, usually it is always the same type of estimate, but for bv function. And the estimate for Radon measure, for the total variation of the Radon measure is given again by the f0 bv norm and this polynomial. So, the idea of the proof, we have to reduce again to a simpler linear problem, as usual. So, we consider this problem, but now the domain is no more convex, so it can have some non-convex point. So, if the domain is not convex, the theorem that I showed before just give the fact that the solution exists, but the solution is no more regular. So, actually, we cannot apply directly the iteration scheme to this equation if we don't remove the gamma 0 set. So, the set that is propagating from here. So, what we do is that we take a smooth cut-off function that is vanishing on an open neighborhood of this set, and we look at the equation satisfied by this f epsilon, that is f times this cut-off function. So, we take the initial datum, we remove the boundary set like this. We apply the cut-off function to the linear transport equation like here, to the boundary condition as well. And now here, we can use all the estimates that we showed before. The only fact is that we are looking for W11 estimates. So, the p for the derivative now is equal to 1, but we saw that all the estimates were from p equal to 1 to plus infinity in the linear transport case. So, the difficulty actually lies in the estimates for this term that we will have to cope with the gamma gain term of our transport equation. And we have to cope with the fact that since we have the presence of this cut-off function, we will have also the derivative of this cut-off function when we consider, actually, there is a mistake missing, but when we consider this derivative, it will involve even the derivative of the cut-off function. So, all the difficulties is actually in constructing this cut-off function in order that the estimates are uniform in epsilon and in order to be able to pass to the limit. I am a bit hiding all the difficulties, but it is precisely hidden here, actually, in the construction of this cut-off function. Okay, and then passing to the limit when you have this, you can pass to the limit and obtain that the function f will be like you apply the iterative scheme as usual and the function will be in BV and you will obtain the results. So, just to summarize what I showed to you, because now you will be completely lost, probably. So, in convex domain for diffuse boundary condition, let's just think about diffuse boundary condition in this case, we were able to prove that the solution is C0 away from gamma 0. I just showed you that we have W1P estimates from 1 to 2, actually, yes, it was not our results for P equal to 1, but you can obtain estimating P equal to 1 in another way. Then, the point is that, actually, you cannot think about H1 estimate and we found like a counter example for the transport equation. So, you cannot think about the simple W1P estimates without adding a weight function, but if you had a weight function that was the alpha times the E omega, then you can find estimate from 2 to plus infinity weighted estimates. This will allow us to prove C1 estimates and the C1 regularity of the solution away from the grazing set, while we produce a counter example for higher order regularity. So, we cannot expect much more regularity for the second derivative in space or the action actually in space or in velocity. While in non-convex domain, key is to prove that, actually, you can produce a singularity from this gamma 0, but even if singularity are produced, you can prove the BV regularity of the solution. So, the singularity are bad, but BV is okay. Now, just a few words on other boundary conditions. So, as I told you, we were also able to go with specular boundary condition and bounce back boundary condition, but this case, that case were more difficult. But again, as C1 regularity through the kinetic distance that we always have to put as a weight function is possible to be obtained for both cases. Actually, what it turns out with specular boundary condition is that, I mean, you can understand when a trajectory touches the boundary, then it's reflected, and then it's reflected again, and then again, and then again, and then again, and then again. So, it creates a lot of singularity. So, actually, if you want to remove them, you have to cover the ball, if it's a ball, for example, with charts that changes. So, the difficulty lies in producing charts that changes around the boundary of the singularity with regularity, actually, like uniform and in a continuous way. So, this was actually all the technical details that we have to use to prove the C1 regularity. And the bounce back boundary condition will be more or less the same. And then, there were results by Briant and Guo in 2015 on Maxwell boundary condition. So, maybe this is more related to what you were thinking about, Micaela. And even results on non-isothermal boundary. There are actually, probably they are not up to date, but I think there are other results by Esposito, Guo, Kim and Mara. There is maybe a new one that I forgot to write on non-isothermal boundary condition. And that's all. And I thank you for your attention.