 Good morning everyone Let me make an announcement first that the talk of Alexis Vassar on Friday Which was announced at two o'clock will be at 1 30. So it's going to be a talk between 1 30 and 2 30 So you have more time to get to Venice or something, right? So I'm We have now the next talk today the first talk today That's again Laurent de Villette and he will continue with his third lecture now on collisions in plasmas delanda equation, please Okay, so thanks a lot Let me recall a few things that we saw yesterday first if you remember we made a sort of general discussion of the contributions of various Colleagues to the Entropy method and then we Turned to the definition of the Boltzmann equation and After a certain number of computations and after having used a large number of assumptions We ended up with the abstract form of the Boltzmann equation which is written here on the blackboard so before Writing a parametrization of the of the operator and It's something that anyway we saw with Daniela yesterday. I wanted to Discuss a little the remark which is that there is an obvious Class of steady states for this operator which are the so-called Maxwellians So what I will call the Maxwellian systematically In this series of talks is a function which is the exponential of Second-order polynomial in the variable v but as v is a vector This is not just any kind of second polynomials which would give you a general Gaussian It is a second-order polynomial in which the matrix acting on the second degree Of the polynomial is a matrix which is the Constant times the identity so in that case we really speak of Maxwellians and This class of functions can be seen as a subclass of the Gaussian functions It's exactly the subclass of the of the Gaussian function which have a covariant matrix which is a Constant times the identity so it's a way you should you should see this so let me explain why this specific functions Play and a very important role in the theory The point is now suppose that you have this m of v defined like this and let's try to see What is the result of? Computing f of v prime f of v prime star and f of v f of v star So this is the remark here So if I take for f the quantity m and I compute m of v times m of v star you see that I Have to to compute the following quantity and using the property of the exponential I can write it like exponential of 2a plus b Scalar product with v plus v star minus c v square plus v star square Just like this Okay, and under this form Thanks to the direct masses which are written here. We know that For the v v star v prime v prime star which are written here V plus v star is equal to v prime plus v prime star inside the integral and also The kinetic energy before collision is equal to the kinetic energy after collisions. So I can write this Also like this So this is true of course only If you look at this integral and this is exactly m of v prime times m of v prime star And as you can see You get that this quantity here is equal to zero when f is n, okay so you get this class of Steady states for the equation df over dt is equal to is equal to this quantity here Okay, what's a very important remark for the for the rest of the of the lectures Let me now Present the Two typical parameterization For the Boltzmann equations. So this is another remark if you wish it is possible Recast the formula which is here which is quite abstract because of the direct masses inside The integrals let's see the term Which I called q both Be so the be here is referring to the Cross-section be here. Okay, so I parameterize the operator by This parameter here, which is something which is sort of a free parameter in the in the operator Okay, sorry. Anyway, I can recast this a term would be equal to the integral so now I Have an integral over three n In dimension three n and I remove a set which is of dimension n plus one So I will have an integral over a set of dimension To n minus one The one will be over our n and the other one will be over the sphere s and minus one So this gives you something which is of size to n minus one you still keep the same notations for the The part which represents the gain and the loss term and You remove the direct masses and you just write Be here and now be as we will see so let's call it be one if you wish It will be a function still of the modulus of the relative velocity But the second term will now the second variable will in fact be the Absolute value of the scalar product between the Direction of the relative velocity and this parameter omega which now lives in this sphere as n minus one and if you Remember the formulas presented by Daniela yesterday. This is exactly identical once I Write the definition of v prime and v prime star in terms of omega and this can be done in this way So this is usually the formula which is written down in textbooks on the Boltzmann equation and one can check that These formulas are really exactly equivalent to saying that V plus V star is equal to V prime plus V prime star and the same for the Kinetic energy So more precisely there is equivalence between both two things and the fact there exists a parameter omega on the sphere such that this It's not very difficult to see first. You write things in the reference frame of the center of math and then you say that two vectors have the same modulus if there is a symmetry Which goes from one to the other and this symmetry is exactly the symmetry related to the to the vector omega But actually I will never use this formula. So this is just to make the connection With with the talk of Daniela because I will really now look at the specific case of the dimension 2 Which is not the physical case but which is a case in which it's a little easier to do the computations So I would like to use a parameterization which holds only in dimension 2 for the same for the same kernel So when n equal to another Parameterization, which is actually much better behaved from the point of view of math and it's the following So it's a still the same formula But now since the dimension is 2 the integral is taken over R2 for the variable V star and the second variable will be chosen in S1 so in the in the circle and one good way to parameterize the circle is just to write an angle between Minus pi and pi so let's call it theta then the rest is identical and here Usually we take a parameterization of the cross section Which is such that it is the Absolute value of theta which appears in the second variable of of b which I call here b2 And the parameterization is given by the formula which I think It may be easier to understand if you are familiar with the with using The center of mass reference frame so the center of mass reference frame Means that you look at the average of the velocities before collisions. So it's v plus v star over 2 So once you are in this frame The two velocities are v minus v star over 2 and v star minus v over 2 So if you look in this frame the collision Happens with just two particles arriving like this And then they will get out like this And you will use the angle between those two lines to parameterize the collision So if you do that it just amounts to say that you use a rotation of angle theta Taken on v minus v star over 2 and actually this parameterization is Is probably the best unfortunately it works only in dimension 2 because it's not so easy to parameterize The sphere in dimension 3 With something which has the properties of a group When I say it's not so easy it's a euphemism So the the point here is that you have all the good properties with this parameterization because It's something which is linear in terms of dv star gives v prime v prime star And it is something in which the parameter appears in a very simple way Whereas here as you can see the omega Appears in a quadratic way okay, so this is Something which makes the computation usually quite easier And it's a reason why what I will present to you now Is presented only in dimension 2 because it's a little easier to do the computations But I promise to you that All what I will now say is still A true in dimension bigger than 2 provided that you accept to do the heavy computations Okay So let me do now By the way or those who wish to see another slide This is already written here. So let me now give a definition The definition of the so-called grazing collisions So if we consider that the Boltzmann equation in itself is a definition is actually the number 5 We say that grazing collisions are collisions such that the Velocity after the collision is close to the velocity before the collision which Also is equivalent to the fact that The same holds for the second partner So grazing collision is a collision in which the velocities are Changed by a very small amount And as you can see looking at the formulas that you have here This is equivalent to saying that v minus v star is almost orthogonal To omega so all of this is equivalent to V minus v star Is almost orthogonal to omega if you use the omega representation written here Or if you prefer it's also It also corresponds to theta which are close to zero But if you take theta is equal to zero You will just get v plus v star over 2 plus v minus v star over 2 which is equal to v Okay, so Another way of looking at grazing collisions consists In writing that theta is close to zero That's okay The remark Remark which is actually due to I think both Lando and Chapmanic and Colling So this is this is something which was understood between the 30s and the 50s is that For plasmas All collisions are in fact grazing collisions So in plasmas the main difference that you have Between for the particles with with respect to a neutral gas Is that the particles are charged so you have like electrons and ions And when they collide They collide because of the Coulomb potential And actually the Coulomb potential has a very long tail That is you can feel another particle from very far away and because of this You can You can guess that The in a in a typical collision with the Coulomb potential the Amount of velocity which is exchanged during the collision is very small and so This remark which was done let's say at the at the physical level Has deep consequences At the level of the operators so for plasmas all collisions When I say all Well, it's not really a mathematical statement. Let's say All collisions are grazing So this means that when you're looking at a plasma Basically, you should take The Boltzmann operator. Let's say this one in dimension two And you should perform a limit when theta goes to zero Inside the operator V prime and V prime star depends on theta through those formulas And you try to let theta go to zero So let's try to do this that is Theta go to zero so let's do the computation when N is equal to two so let's forget Forever the general case and let's concentrate on the case of dimension two If you let theta go to zero You see that you can write V prime as V plus V star over two plus V minus V star over two cos theta plus V minus V star over two orthogonal sin theta Where I use a sine orthogonal for The rotation of angle pi over two Of a vector, okay So by definition This is just this is just this and now if you expand The cosine and the sine Thanks to let's say Taylor expansion Um You see that you get V plus V star over two plus V minus V star over two Minus V minus V star over four Feta square Plus V minus V star over two orthogonal theta plus Something which is of order theta to the free When theta go to zero goes to zero, okay, and you can simplify The first Some here and get exactly V which is not surprising since we saw that This parameterization is chosen in such a way that when theta is equal to zero Then the collision Corresponds to no collision at all if you wish Okay, if we do the same for V prime star The result is given by the following formula Which I will not fail but it's Of course, very easy to check With a formula which is very close And now let's compute at the formal level F of V prime F of V prime star Sorry Using those formulas so this amounts to have F of V plus V minus V star over two orthogonal theta minus V minus V prime star theta square plus This times the quantity which is here and when you perform now they When you expand F in Taylor series around V here and V star here You can believe me That What you get is the following So I will try make no mistakes here So you have terms in theta You have terms In theta square Let me write it rather here It's F of V plus V minus V star gradient F of V theta minus V minus V star over four gradient F of V theta square Plus This is a sum over components of V minus V star orthogonal A and J this is divided by two And here I have The second derivative of F with respect to variables I and J theta square over two and you have Let's say A formula which looks like the same and which starts with F of V star But the rest has more or less the same shape So let's have a look to to what happens In the in the kernel We just computed this times this when theta is close to zero The first term is F of V times F of V star which will cancel With the F of V times F of V star which is here The second term Will be made of things which are proportional to theta But the point is that here the cross section depends upon theta in an even way and because of this The term with theta will give a contribution which will be zero in the integral Okay because of a parity argument And so what happens is that only the second order terms are not zero and You can check that Here you will have Things which look like the quantities that you see here so As you can see Inside you will have things Which look like this one or the same With gradient F of V star so you can expect to to have inside the result Terms in which you have first order derivatives of F And they will be multiplied typically by V minus V star This is cannot product And you will also have terms which look like this, okay now If you take The tensor product Of the orthogonal of two vector is exactly as taking identity Minus the tensor product of the corresponding vectors So you can also expect to have terms Which would look like this identity times V minus V star to the square minus The tensor product of V minus V star With itself We do the contracted product with the gradient of the gradient of F Let me say that this is just a shorthand notation for the matrix Which have the component V minus V star square delta ij minus V minus V star i V minus V star j And the contracted product consists in taking the sum over all possible components Of this quantity so if you don't like Let's say the abstract algebraic notation You can use the components and you will find this Okay, so I will not perform this computation because it would need like Alphanar more actually But I will write for you the result on the slide So maybe let's look at this slide When you do this Asymptotics, which is sometimes called the grazing collision asymptotics in the Boltzmann kernel you end up with the formula Which is written here and which is called the Lando operator for collisions in plasmath So the how does this relate to the computation that we have done here? Just have a look to this a here This a is made of One part Which is a number and one matrix And this matrix is exactly the matrix which is written here Up to the multiplication by z square Okay And this is applied to gradient f of v And then you have a divergence here. So this is a second derivative which Corresponds to the second derivative that you have here as for the first order terms You can See that they arise from the part of the operator which is here in which you have just one derivative Because this one can be absorbed by an integration by parts So you can believe me that when you do seriously this computation Yeah, I will It's just in the not in the right order But I will it's promised So what you get at the end is really this and what I will do now is just write it on the blackboard because we will Need this definition In the sequel so this is definition six Wish so The Lando operator defined by q Lando and of a parameter psi, which is a given function It is quadratic. So we typically denote the operator with The notation q of f f This is the divergence v minus W usually we do not use v star for the Lando operator. It's rather the common notation is rather w We have this pi of v minus w which is a matrix And this matrix is applied to the vector Which is obtained by Taking this kind of symmetric form in which both f and the gradient of f Are appearing so here pi of v minus w This is The projector on to The hyperplane which is orthogonal to v minus w And its components are given By The identity so delta ij minus V minus w i v minus w j divided by The modulus of v minus w to the square We can you can just Take it as a definition and in fact it's the Way It's really as a definition that it was proposed by Lando in 36 in his first paper on the Lando equation That is Lando made no link Between the operator which is defined here and the Boltzmann operator It just wrote it directly like that And the link which I sort of Let's say discussed a little Is something that appeared Basically in the book of chapan and calling but actually not the first edition is a very old book And it's in in an edition from the 50s that for the first time this link is established Unfortunately not in the vocabulary of math so you really have to To to read it carefully in order to understand what they really do but basically it's a computation that I have Presented to you so one way to make this computation precise is To actually Define A sort of rescaled version of the cross-section in the Boltzmann equation. So you define So you take the Boltzmann equation in dimension two and you define a new cross-section by concentrating the The collisions on the collisions which have very small Angled theta that are the grazing collisions. So you do it in this way. So here you divide theta by epsilon you extend This quantity by zero for theta different from Which is not between minus pi and pi And you put here one over epsilon to the free and the free here has nothing to do with dimension It is obtained by just putting one over epsilon in order to have something which is Of integral one let's say or a fixed integral and then the one over epsilon square which remains Correspond to the fact that the two first terms are disappearing In the computation. So if you since you go to order two You have to divide by one over epsilon to the square in order to get something which is finite at the end So this is the right way to do let's say the mathematically the computation which is done here and Let's say the proof That you have this property here that you can go from the Boltzmann equation Operator to the Boltzmann to the Landau operator for a given f It's something which was written down. I think first at the mathematical level in the early 90s There is a paper by Brigitte Lucca and Pierre de Gaulle and also a paper by myself on this topic But the real hard proof is actually Uh Can be found in in a series of paper by alexander and villainy So I think there are actually two papers which were written around 2000 in which They do not only prove that for a given f this is converging To this but that for the solution of the equations You have a convergence of let's say f epsilon towards f And they do it in the setting of renormalized solutions of Lyons and diperna So I will not say much more about this, but let's say the the the hard mathematics is done in this in this paper here Okay, so this is more to convince you that you can get Lando out of Boltzmann My my lectures are really not on this topic this time But at least looking at this at this Uh Link between the two equations you can guess something It is that since the maxwellians were steady solutions of the Boltzmann equation They are still steady solutions Of the Lando equation because the Lando equation is obtained as a limit of the Boltzmann equation of the Boltzmann operator. Yeah so it's uh I don't remember what is the exact hypothesis in in the results, but it seems to be Yeah, but uh in some sense when you do this you sort of forget About the fact that they are charged particles and you can look for any kind of b and any kind of c And at the mathematical level, of course, you can decide what are your hypotheses? So the important point is that at the end it should contain The Coulomb case which corresponds to c equal to 1 over z And if I remember it is a case in at least the second paper Of alexandre and villani so it contains the interesting case But of course for the mathematical point of view you can try for Any kind of cross sections well, so uh, it's not it's not a surprise that the maxwellian are still steady solution of the Lando equation, but I think it's important to Check this directly On the Lando kernel, so I will do that immediately Because this is really the first The first step to the entropy structure of the equation. So in order to do this, let's Let's look at the quantity gradient m over m when m is a maxwellian This is just the logarithmic Gradient of m and since this is an exponential in the logarithmic Derivative, of course, it disappears and then you just have to take The derivative of this or the gradient or the let's say the derivatives Of this and as you can see this is a second order polynomial So the derivative will be actually an affine function Okay, so gradient m over m It is something Which can be written like if I'm not mistaken b minus 2c Uh scarer product with v This is affine. Okay, and so when you compute now f of w So here m of w gradient m of v minus m of v gradient m of w You can put m of v m of w as a factor And you will get just gradient m of v Over m of v minus gradient m Of w over m of w. Okay, and so this is m of v m of w times This quantity at point v minus this quantity at point w What I get is just minus 2c v minus w here If f is a maxwellian This is just Uh something which is a multiple of v minus w But the operator pi Here is exactly the projector the projector onto The orthogonal of v minus w. So when you apply it on this vector here, you get zero Okay, so this is a way of seeing that maxwellians are steady solutions of the lando equation without knowing anything On the Boltzmann equation. Okay Let me at this level say one word about the This trend operator here So, uh, let me first say that this is not only, uh An object, uh with which mathematicians are playing It's really what is used in numerical codes when you want to Understand the effect of collisions in the plasma So it's something Which is really, uh of Interest for the for the physicists And, uh, let me add that it has a structure which is, uh, quite fascinating Because, uh, if you are familiar with parabolic equations or let's say with elliptic equations Then you can see that you have a structure Which is quite, uh, uh common that is, uh You have the divergence Of something times the gradient Plus something times the function. So it's like you have a diffusion coefficient and A drift coefficient in front of f. So it looks like The usual, uh, think you can find But they, uh, uh, the point is that Those coefficients are obtained through an integral of f in the variable w So they are themselves depending on f So it looks more like let's say a nonlinear diffusion But usually in nonlinear diffusion you look rather to things which are let's say local like laplacian of f square And here it's not laplacian of f square. It's like laplacian of f times f convoluted with something And so One, uh, way of looking at this is just to forget everything about kinetic theory And to try to use parabolic theory to to get, uh, some insight And there are a lot of works which are been done in this direction And then there is a second, uh, way of looking at things Which is to, uh, keep the symmetric form that we have here As you can see this is extremely symmetric And, uh, this means that somehow you forget about the parabolic nature, uh, of the problem And you really base your analysis on on on kinetic theory So there is some in some kind of, uh, two ways of looking at the same objects Which are completely completely different Uh, and let me say that what I will present now which is the Entropy structure of the equation is rather based on the kinetic way of looking at things And, uh, it's not so obvious to use the entropy structure if you look at it from the parabolic point of view Uh, anyway the main difficulty as we will see, uh, when we are looking Precisely at the most Important case from the point of view of physics That is when the the function psi here is z to the minus one and this corresponds To the case of plasmaz Is that somehow those two terms, uh, are, uh I mean the the drift term is not really dominated by the Uh, by the diffusion term So the symmetry when you take this Potential here the symmetry Uh, plays a very important role in the equation And trying to use a diffusion without looking at the drift for example leads nowhere Okay, so this is just a general A general remark, uh You can see this in some sense as some kind of interpolation between the Boltzmann equation And the Fokker-Planck equation that I presented during the first lecture From the Boltzmann equation it retains a quadratic nature It retains a non-locality And from the Fokker-Planck equation it retains of course the parabolicity Nature of the first term and the drift that you have in the in the second one Uh, and from both it retains the fact that you have a set of steady state which has to do with Gaussian functions Okay Well, all of this is a little, uh fuzzy, let's see So let's come back to, uh more practical, uh practical things Um, the, uh Next step consists in, uh looking at a weak form of the operator So here I will not write things down. I will just use slides The as usual in PDEs let's say or in the theory of integral equation actually the same It's quite important to understand what happens when you multiply the operator by a test function and you integrate So this is just, uh The definition of formulation if you wish So here if you take the Boltzmann kernel and you multiply by a test function phi Typically what you obtain is a formula in which you have the term which is appearing in the Boltzmann equation This f of v prime f of v prime star minus f of v times f of v star and this is multiplied by phi of v and integrated Okay, then what you do first is to make a change of variables which transform v in v star and v star in v When you do that the phi of v becomes the phi of v star Okay, and if you take the first formulation And the second one and you divide by two you will get just one half of phi of v plus phi of v star at this level Okay, once you have done that You use another symmetry of the equation which consists in changing v and v star in v prime and v prime star So this is a change of variables Which is quite easy to perform For example, when you use a parametrization, uh that was Uh presented by Daniela or the parametrization with theta in dimension two But when you do that, it's easy to say that you have a Jacobian which is one So it does not appear in the in the integrals And the phi of v plus phi of v star becomes phi of v prime plus phi of v prime star And of course here things are reversed That is v v star becomes v prime v prime star v prime time v prime star becomes v and v star and so the minus here Makes it that you have to change the signs at this level Okay, so by doing two changes of variables in the In the in the big integral here you end up with this Uh Formula here by the way, sorry about the this this is uh, this is not true. It's a different It's a different integral here. It's integrated over over the totality of the variables. It was lost in the So let's let's look at a direct consequence of of this formulation of the of this weak formulation Let's now take phi equal to one v i or v square over two So this corresponds to looking at mass momentum and energy On the world if you wish that is instead of looking at it for just one particle You just integrate over all the particles entering inside the kernel So when you do that if you take phi equal to one It's clear that you get one plus one minus minus minus one and you get zero If you take any component of v You will get here, uh, let's say v i plus v star i minus v prime i minus v star prime i And this will be equal to zero since you integrate On a set in which you have this direct mass here And of course the same will hold For the v square over two here because of the direct mass that you have here, okay So the first consequence of the weak formulation is that you have This global conservation of mass momentum and energy Which is no surprise because this is something that you imposed at the microscopic level at the level of each collision In the kernel. So you will recover it when you integrate over all possible velocities And now let's try to do the same for the Lando operator. So the Lando operator Is written here Okay Now you multiply by a test function phi of v and you integrate So as you can see here The first thing you have to do is an integration by part Which transforms the divergence here in a gradient on the test function So it's exactly what will happen at this level You will have this gradient phi of v which will appear And then you do As in the Boltzmann equation you do a change of variables which transform v in w and w in v So when you do that You you get gradient phi of w And if you look at this formula here, you see that this is anti-symmetric with respect to v and w But if you if you change v and w you get exactly the same Except it's a different sign. So because of this you will get a minus here Okay And finally to get the formula which is here what you do is just you just Put into factor The f of v and f of w. So you take them out Of the formula here you put them here And you get gradient f over f minus gradient f over f at points v and w The way to understand this formula is that you have a matrix here, which is the projection operator Which acts on two vectors as a quadratic form. The first one is this one And the second one is this one Okay Now let's try to see what happens if I take again phi equal to one vi or v square over two If you take phi equal to one, it's clear that you get zero because gradient phi is zero if you take Let's say any of the components of the momentum v Then the gradient will just give you Quantities which will be constants. So you will get a constant minus itself. So we will get zero And finally if you take the energy here You will get gradient phi of v equal to v and you get v minus w at this level And you remember that this is a projector onto the orthogonal of v minus w. So this gives you also zero Actually, you could get this formula directly from the Limit that you have From this link that you have between the Boltzmann equation and the Landau equation. Okay But it's better to to get it directly from the formula of the Landau equation And now it I still have five minutes to present to you The entropy structure itself That will be the main object for the two last lectures so This is the so-called h theorem of Boltzmann. So in the case of the Boltzmann equation It was proven by Boltzmann himself around 1860 and it consists of noticing that If the in the weak formulation of the Boltzmann equation There is something which happens if you take phi of v equal to log f of v So takes just take phi of v equal to log f of v So you replace here phi by log f What happens with the log is that you can transform log f of v plus log f of v star in log of f of v times f of v star So it's exactly what you will find in this slide Here I used log f of v as The test function and here you will get log f log of f of v prime f of v prime star minus log of f of v f of v star And you see that using the log is the only way to get here something which depends on these products Any other function would not give you that it will still give you Something which you cannot relate to the quantities which are here. Okay And now the the the observation of Boltzmann is that It's as if you had something like x minus y times log x minus log y And since the logarithm is an increasing function. This is non negative Okay and so if now you Define the Boltzmann equation as df over dt Is equal to q of f Or q of f f and you call h of f the integral Of f log f you see that d over dt of h of f Is the integral Of log f plus 1 df over dt of tv So this is always true and now if df over dt Is equal to the Boltzmann operator So if f is satisfying the Boltzmann equation then The quantity here is equal to two Boltzmann b of f f Of tv log f of tv Because remember that one Multiplied by the Boltzmann kernel and integrated give zero. This is a conservation of mass And you get exactly the quantity which is here and so you see that this is Negative okay, what you have obtained is the entropy structure for the Boltzmann equation that is uh This entropy This is an entropy So the proposition is that H is an entropy For the Boltzmann equation The entropy dissipation Which I will call d as I did in the first Lecture is given by this formula here, okay So this is the entropy structure for the Boltzmann equation I don't know if I still have one minute maybe At least to announce what I will do Tomorrow so the point is that Since the Lando equation the Lando kernel is a limit of the Boltzmann kernel You can only expect that by doing the same transformation that is you multiply by log f you integrate You will get something positive, okay This is a direct consequence of the link between the two but actually It's better to see directly on the on the kernel. So if you remember the weak formulation of the Lando equation Which is here What happens if now you put log f instead of phi? As you can see you will get gradient of log f of v which is gradient f over f And so you end up here with Gradient f over f of v minus gradient f over f of w And you have here a symmetric matrix Which is taken on The same vector as a quadratic form And so in order to see that This is positive It is enough to know that this matrix is definite positive or at least semi-definite positive And since it is a projection It is in fact semi-definite positive Okay, so you get that this is also bigger than zero and The proposition which is here is also true as a consequence for the Lando equation Finished by saying that Since now we have the entropy structure for both the Boltzmann and the Lando equation The point is to check that this entropy structure is strict And then to try to produce an entropy-entropy dissipation estimate Which if It exists Will lead to Convergence theorems for the Boltzmann or Lando equations when the time t goes to infinity And this will be The point of the two last lectures That I will present tomorrow