 So, good morning. My name is Giovanni Bellettini. I'm giving a course on partial differential equations and functional analysis. So, partial differential equations and functional analysis. So, this is a very huge topic. So, we have to choose various subjects and we cannot talk about everything, of course. So, the idea of the course is essentially to divide it into two parts. The first part we will study some special kind of partial differential equations. In particular, we will start with... So, the first part of the course we will start with first order PD and linear and nonlinear. Then, we will pass to study the second order, second order PDs. And this means that we will focus attention on Laplace equation where this symbol means the Laplacian. So, this is the basic so-called elliptic equation. Then, maybe not exactly in this order but this will be the other equation that we will study. This is the heat equation, which is the standard example of parabolic PD. And then we will pass to say something about the wave equation. So, this is the standard example of hyperbolic second order PD. So, after this kind of studies then we will pass to the second part of the course. So, the second part of the course is function analysis. This is more abstract. Therefore, there will be various results but essentially we will study some of the properties of Hilbert spaces where here the difficulty will be that this is a vector space of infinite dimension. So, this is exactly the difficult point. Then the Banach space. And then we will see how much time it remains to continue the part of functional analysis. So, I don't want to say anything more because I still don't know how much time I will have. For instance, to say something about Fourier transform maybe. Books that are suggested. So, concerning this part there is a book by Evans titled Partial Differential Equations. This is a standard reference at the moment for the first part of the course. This book contains a lot of things, a lot of informations. We will not at all cover the content of this book but it is a very good reference in my opinion for this. And then another very good reference concerning this more abstract part. Of course, in this book there are also several informations on functional analysis. So, you can use this maybe also to cover some of the topics in the second part of the course. But on the other hand, very standard and the important reference on this is the book of Brezis. Maybe the title is functional analysis and partial differential equations. Functional analysis and PD. And of course also in this book you can find a lot of informations about PDs. These are more or less standard references. There are a lot of other references. So when I will use some other reference I will let you know and I will write on the blackboard the book that I am using at that moment. So please feel free to ask everything you don't understand. That's important because also probably for all of us. If somebody does not understand something it is better to repeat it. So let us start with PDs. First order PDs. So first order PDs now I will start probably from the simplest one. So let me give a vector in our n. So this is a constant vector. To begin this partial differential equation. So what is this? U is a function of t and x. So t, say, is bigger or equal to 0. And x is in our n. U is a scalar we don't want in this course to talk about systems. So this is just one PD and not two or three coupled PDs. This is another problem. So it's just one PD. And this is the gradient. Here is the gradient with respect to space. So x is equal to x1, xn and this is equal to b over dxn. So this is the scalar product between the vector b and this vector here. This dot means the scalar product. And ut is the derivative with respect to t. So in the sequel I will also change the notation. But for the moment let us use this as a notation for the variables. So in this way in some sense I am giving the first variable some sort of special role. We will see in the examples what this does mean. So concretely this is bi du dx i equal to zero. B of course is b1, bn in components. Ok. So this is called linear transport equation. We will see why in a minute we will see why this is called transport equation. It is linear because you see the derivatives here appear linearly. This is clear. It is of course first order because we have just only first derivatives of the unknown which is a function u. It is homogeneous because there is not the right hand side is identically zero. Ok. So as it happens in ordinary differential equations. So we don't have anything on the right hand side so this is called homogeneous. And the coefficients are constant. Ok. Because the coefficients are one here in front of ut one and then the vector bi in front of du and dx i and this is not a function it is just a constant. Ok. So constant coefficients the coefficients are 1b if you want. So 1b in r 1 plus n and we are taking the scalar product between the time space gradient of u against u1b. Scalar product of this against u1 against 1b. Ok. And this dot means scalar product in r1 plus n. Ok. Is this clear? Ok. So this is maybe the simplest PD that we can imagine it is worthwhile to study it so to understand its structure. Ok. So what and we look for the point is to look for a C1 solution C1 in time space solution. Why C1? Well, it is clear because we write the equation in this form so we take partial derivatives and if the function is C1 these partial derivatives are continuous and so this is well defined at each point everywhere at each point. So the regularity of the solution we look for is exactly this. So what is to study so this is an equation without any kind of boundary condition. This is a free equation. But soon we will couple this with a condition of some sort of boundary conditions. This very soon we will do this. And this is also an issue because it is not clear in general which kind of boundary condition you have to impose on that specific equation. This is not completely obvious why one type of boundary condition and not another one for instance. So now we will discuss also that issue. So which are the main problems that you that one phase in a new PDE. So the main problems are the following. So first of all the standard way to proceed is one look for special solutions special solutions explicit. Look for special explicit maybe solutions means that you look for solutions with some kind of symmetry for instance. I mean you look for maybe radial solutions which are independent of time or something like that. Special, not general solution. But this is an important point but maybe sometimes you are not able or maybe you find one explicit solution but then you have a problem in domain open domain with some strange boundary condition and you are not able to find a special solution to that. So what do you do? Now of course we have the problem of a solution but existence of a solution means that you have to exactly define what do you mean by a solution. So this means that existence of a solution in some class. And of course the larger the class is the easier is to find a solution. The larger the class the better it is somehow. Of course when you enlarge the class if your solution is not very smooth for instance not c1 but say lip sheets or less then you have to declare what do you mean by being a solution because maybe you cannot differentiate point twice. So if you enlarge your class because you want to find existence you look for a very large class but then you have to define what do you mean by a solution. Because if your solution is not differentiable then what does it mean solving the PDs is not clear. But this is strictly related with another problem which is uniqueness of the solution. So sometimes it is very useful that the solution is unique but then if you want the solution is unique your class should be not very large because it is more difficult to have uniqueness in a large class. So if your class here is too large maybe you lose uniqueness too many solutions of the same PD. And so here this the smaller say the class the better it is. So you have to find a compromise between point two and point three to have at the same time existence and uniqueness. Is it clear? And then there is maybe the more difficult point usually so these two points are connected so you cannot really solve somehow separately. Very often you have to look at two and three almost at the same time. And point four is the regularity of solutions. This means the following assume that you are able to prove existence the solution and uniqueness for instance at the same time. Then maybe your solution is much more smooth than you expect only from point two. For some mysterious reasons it happens very often that the solution is much more smooth than you expect. And this is probably the more deep part of all the story. And again point four is somehow is strictly related to these two and you cannot think of point four separately from point two and three. Somehow you have to have in mind all these three points at the same time. So when you look for a class here for a solution then you have to imagine in advance which could be the regularity that you would like to have for your solution. So I mean these are all together. And this is the difficult point of the story. Ok, so for the moment we have a simple reasonable PD and we start looking for explicit solutions just to understand. Now we concentrate on point one and by the way it is in point two where you look for this large class it is a point where function analysis enter. Because in that large class you need no smooth functions in general maybe so ball of spaces for instance. So in this part in point two here there is function analysis entering the story in particular maybe so ball of spaces and distributions theory of distributions which are two typical arguments understanding the structure of the so ball of space is a typical problem in function analysis usually in Hilbert and Banach spaces. Now let us start from this PD here. So what do we see from here? This says in question one in question one says that you has sort of directional derivative do you see it? Can you see it? I mean here it is written that there is a sort of directional derivative of u which is constant in time space. Why? You see if I take the time space gradient of u along the direction one comma b I see that this derivative is zero. This is exactly the scalar product between the time space gradient and the vector one comma b and this is here it is written that this scalar product is zero. So this means that the derivative of u along the direction along the direction of the vector one b is zero. So it is natural to introduce a function let me call this z of s maybe with my notation which is this u of t plus s x plus sb. So this is u of tx plus s one b where s for the moment is an exterior parameter and this is this is looking at a possible solution of your problem along the line passing through the point tx and parallel to the vector one b. In my pictures now what I do is usually the following I put x here and t here so notice this kind of choice t here is considered as the first variable but when I draw it in a in a graph I put it vertically so it just a matter of convention just a convention so now so let me let me have the vector one b and then let me compute the derivative of z with respect to s which is of course ut at the point tx plus s one b plus the gradient of u at the point tx plus s one b scalar product with b and this is zero so this means that z is constant so this means that if I have say if this this is the direction of the vector one b then this means if I take a line here parallel to this vector and all lines say and I read the function along this line passing to the point tx what I see is that here z u is constant on these lines ok ok so now so what we can do is with the boundary condition so with my notation now let me call it yes so let me introduce some notation here so sigma is equal to t equal to zero into so this is sigma and then let me take a function u bar into c one of sigma so this is given maybe you can you read it here ok so now I am giving a hyper surface for instance this hyper plane ok and then I assign the following new condition here I look for a solution to the following problem so u t plus b dot grad u equal to zero in in zero plus infinity times rn u in c one say zero plus infinity times rn u equal u bar and therefore say c one maybe let me close it so now I assume that I want to find the solution of the free equation in the half space positive time solution of class c one maybe it would be enough to assume u continuous up to the boundary I have written c one up to the boundary but just for so continuous up to the boundary and then I want u equal u bar given u bar is given exactly on this boundary so now this is a problem with the boundary conditions more precisely in this case it is an initial value problem initial value problem meaning that I assume what happens at time zero u bar and then I start the flow so this in this form with this sigma this is called an initial value problem so now I want to find this solution so how can I argue so assume that I have a point here so let me write it as the point x and t so I have a generic point here in the half space and this is t index and then I want to find u a solution u at the point t index what do I do so what I know is that the solution here through this point is constant where along the line is vector 1b and passing through this point so now I fix this point here in the positive half space I take the unique line parallel to 1b passing through this point and then I know that u is constant on this line and therefore in particular its value is exactly the same as its value at time zero so t is equal to its value the value of u bar because I want u to be u equal to u bar at time zero but at a different point of course which point not the point x but the point in this point here where the point so that I indicated maybe by pi of tx so this is tx and this is a pi of tx this notation stands for say projection point in some sense even if it is well it's a sort of notation so at the point pi of tx so that the line passing through pi of tx in the direction 1b in the direction 1b contains tx so I have to find so given tx now I have to find pi of tx so given tx zero plus infinity say maybe this times rn find pi of tx into rn and s of tx into r such that tx is equal to pi of tx zero maybe plus s of tx 1b is it clear the point s tx is the parameter s somehow I have introduced an exterior parameter which say is the parameter parameterizing your line ok ah yes, s of tx is just a number is this written here parameterizing your line and so of course this point here to describe it in the new system of coordinates what do you need you need this point here and then the parameter s such that if I move from here of the quantity s along the direction then I reach exactly this point which is what is written here in this equation is it clear is it ok up to now so now I can solve maybe I can you read it maybe I then I can solve this you see what happens here well let me rewrite this so tx so 0 plus s tx time 1 so s tx comma and then pi of tx plus s of tx times b and remember that b is a vector s is a number pi is a vector and this is the first component components 1n 1n see is it ok well s must be equal to t s is equal to t and then x is equal to pi of tx plus t b because now s is equal to b s is equal to t so if I impose second component the second block of components x equal to this plus this I already know that s is equal to t from the first equation so let me substitute here t and so then I have this equal to this and then finally I can recover pi of tx which is my unknown essentially in this procedure I am inverting a map and this will be more clear in the next in the sequel of the lecture but one point is exactly to invert the change of variables ok in any case pi of tx you see given t and x pi of tx is x minus tb explicit so the inverse map say s of tx is equal to t and this is the expression and so we have found that necessarily solution of 2 interesting because we have found an explicit solution solution of 2 is necessarily u of tx at any point in the half space is equal to u bar of x minus tb so u bar is given from the problem b is given and this is an explicit solution the idea is simply the following what is the value of u here is the value of u bar here that's simply the idea because u is constant along exactly this line so it's very simple however it is interesting ok so this is some remarks in the order I think they are quite important so if everything is clear now I erase just keep the equation and the solution ok so at the end the exterior parameter s parameterizing the line is identified with time s is equal to t at the end ok let me write this in the order ok remark 1 maybe remark 1 is that u is c1 why u is c1 because u bar was assumed to be c1 on the sub space on the half plane on the hyper plane sorry on the hyper plane by assumption u bar was c1 and therefore u is a composition of c1 and c infinity in particular c1 ok so this condition is satisfied it is clear that at time 0 if I put here t equal to 0 so u is c1 up to the boundary up to sigma if I put t equal to 0 here it is clear that u is equal to u bar so u 0x equal u bar x and this means the third condition and then by construction u solves this so u is clearly a solution of our problem it is very smooth and that is exactly the same smoothness so the same smoothness smoothness u bar what does it mean this is obvious in this case but it is an indication of the following phenomenon that such a kind of equation does not regularize initial conditions if your initial condition is c1 your solution is c1 and not more than c1 not c2 or c3 it is just c1 so there is not a regularizing effect of the equation the equation has not a regularizing effect the reason for this is that this is first order actually linear pd the question is how can I prove that this equation has not a regularizing effect this sentence means initial condition as some kind of degree of smoothness is say c3 then the solution is c3 and not more than c3 this means the sentence ah yes you mean uniqueness well this is a proof because if you have a solution the question is what about uniqueness of the solution so well in u of this problem necessarily u is equal to that this is what I have proven so any other solution you have of this problem c1 et cetera must be equal to that because my constant I have proven this if you have a solution then your solution is explicit and necessarily is that one this is also uniqueness same smoothness as u bar next remark why I insist on this because this kind of these kind of pds are in some sense worse than other pds exactly because they do not regularize your initial condition on the other hand other equations such as for instance typically the heat equation second order pds has an immediate regularizing effect if you start with something which is not smooth at any positive time it immediately becomes extremely smooth and that is an irregularizing equation but this is not this is not so this is somehow a point that makes makes interesting the subsequent theory is not so trivial as it seems this remark is not so trivial because of its consequences ok then remark 2 we have used in the proof the following argument the following assumptions that the vector 1b is transverse to sigma what does it mean transverse means that non tangent 1b non tangent to sigma well this is obvious because sigma was this hyper plane this was time first coordinate put vertically and this is space ok and sigma this red yellow red was this red hyper surface and the vector 1b is of course non tangent exactly because there is one here so this one component independently of what b is this one component makes it non tangent is transverse is not like this and this makes the procedure work because we have used if this were tangent then our procedure clearly does not work so this is another remark transverse to sigma third remark maybe third remark is that this is a transport equation linear transport equation in the sense that we are transporting the initial condition so we have u bar and at subsequent times we are transporting the value of u bar in some sense because the value here is u bar here so in some sense this is then ah yes then there is a more delicate point this is much more delicate if u bar is non smooth for instance like this this continues with constant values 1 0 assume that this is u bar the graph of u bar 1 0 now if you put u bar here then u is not anymore c1 so the point is is it reasonable or not if u bar has this kind of profile discontinuous is it reasonable or not to accept this right hand side as a solution of our PD in some sense not c1 and this is not so easy to understand is it reasonable u bar of x minus tb solves 2 in some weak sense of course not in point twice c1 sense because if u is discontinuous I cannot differentiate so easily at each point I cannot so the answer is yes still it is reasonable but this requires a notion of solution of saying what does it mean that discontinuous object solves this what does it mean exactly what I have to put here in place of ut in order that this makes sense so this is long story and so I don't want to touch now this question just a hint that in this very special case you can ask non-trivial questions like this finally the last remark a notice remark 5 the lines lines the lines parallel to 1b are a solution solve the following point x dot equal 1b say x dot of 0 say equal to a given x if you want notation is x bar so if I solve so now capital X is a vector in 1 so x bar is a vector in rn maybe I could call it pi but let me be free of using different notation as far as this is understandable so now I have a point x bar here time and then I have to solve ordinary differential system of ordinary differential equations now it is a very easy system so capital X let me write it as x0 x1 xn so this is the first time component with the index 0 and these are the usual space components and so x0 of s from here is equal to s and x1 xn of s are equal to x bar and so it is clear that this is a parameterization capital X of s is a parameterization of the line passing through this point in the direction 1B so these lines in particular we have obtained them looking at the PD but in particular one can observe that these lines are obtained as solutions of a system of oddies which is this ok so this is called system of characteristics characteristics it is a system of oddies now it is not really an oddie because just an integration I mean there is not capital X on the right hand side so it is an immediate oddie just a remark to keep in mind for the future development now so let me pass so we have understood everything about I think the the simplest PD let me now modify it and make it a little bit a little bit more difficult so under the previous assumptions now let me consider the following problem so let be given now let F be as a right hand side so say a source term in C1 say 0 plus infinity times Rn and assume that I want to study the following problems A plus B grad U equal to F so now this is non-homogeneous case so linear first order PD with constant coefficients coefficients but non-homogeneous non-homogeneous and so assume that I have again the point so this is in the half space and this is well this is a little bit too much but anyway it doesn't matter the smoothness enough and then U equal U bar on sigma with the same assumptions on sigma and U bar as in the previous case sigma is the hyper plane and U bar is C1 ok on the horizontal hyper plane so now this is of course linear as before non-homogeneous because there is this source this is given is on the right hand side does not depend on U is given function of T and X so now the point is assume that ok now F is C1 maybe it's too much but you see U is C1 and so UT and grad U are continuous so maybe it would be more natural to assume that F is continuous doesn't matter let me to be more precise one could write here C0 ok anyway doesn't matter so now I want to solve this so I mean it's natural as before but how can I solve it now it is not so clear that we have still the transport interpretation as before maybe it's not so clear indeed and however I can proceed as before so I can introduce as before the function Z of S ok so which was U of T plus S for a given TX and then it was X plus SB for a given TX and let me compute the derivative with respect to S so Z dot of S is equal to UT evaluate that T plus S X plus SB plus the gradient of U plus B scalar product the gradient of U T plus S X plus SB and now this is not anymore if U is a solution this is not anymore equal to 0 it is not but it is equal to it is equal to F at those points F T plus S X plus SB ok so now we can look at this and we can integrate so now TX and X are given and we can integrate this between minus T and 0 pardon? in 0 plus infinity in the open half space usually this is a way to write down a PD with the boundary conditions so you want the PD to be to be satisfied in the interior of your domain now let me draw a rough picture you want the PD to be satisfied in the interior in this case the interior is the half space T positive and then you ask for boundary condition on the boundary and in this case the boundary is just the horizontal bottom of the half space and then of course you need U to be sufficiently smooth inside and you need that it makes sense that U has a trace here so say C0 up to the boundary for instance I have written for simplicity C1 up to the boundary you see here there is a square parenthesis but it would be enough to write C1 in the interior so round parenthesis and then C0 with the square parenthesis to be more precise this is because I want the PD to be in the classical meaning inside all derivatives on the boundary there is a continuous trace and so everything is well defined this is just for this now so we are here and so we can integrate now I integrate between minus T so that S T plus minus T becomes 0 and then 0 so that T plus S becomes T so I integrate this between minus T and 0 and I end up of course this is equal to Z of 0 minus Z of minus T and which is in turn equal to my Z of 0 T X Z of minus T is equal to U of 0 X minus T B is it ok? up to now degree ok so from this deduce that my solution U of T X so let me continue this here so U of T X is equal to this right hand side plus this so I deduce the following U of T X is equal to by the way the trace of U on the hyper plane is equal to U bar so I can substitute this with U bar at X minus T B plus ok so now let me change variable so let me call this as say sigma T is fixed so I consider this as a change of variable for S and so this becomes equal to U bar of X minus T B now when S is equal to minus so when S is equal to 0 sigma is T T plus S is just sigma and then S is equal S is equal to sigma minus T sigma minus T B in this sigma so let me just for maybe elegance let me call sigma equal to S so this is our solution again so we have found another explicit solution much more complicated than the previous one of course the previous one was this when F is equal to 0 so you see when F is equal to 0 this is the previous solution so we are clearly generalizing the case when F is non-zero and we have this strange expression integral so we don't have a clear interpretation as before using just only the initial condition U bar so now you have the point X here and the point T here and so your solution here in the future at this time T and at this point X is equal to your initial condition say X minus T B but then there is another contribution which uses all the values of F in between 0 and T so S is in between 0 and T so I need to know the values of F for all times T and then here X plus S minus T when S is equal to 0 this is just X minus T and when S is equal to T this is just X so I am using the values of F in an intermediate region here somehow on the left of X and for all times okay so this is somehow another interesting result so when we put right hand side here we have we lose somehow this transport interpretation and we use the values of F in a big region here so I have maybe another remark before leaving this non-homogeneous problem so let me consider again the system of what it is as written as before so X dot of S equal to 1 B X of 0 equal X bar equal 0 X bar maybe also the following mu 0 equal U bar of X bar so now I have the previous system of characteristics that I called characteristics so these systems give you the line the line parameterize with S equal to T so this is the previous system with S equal to T passing through the line passing through X bar and then I have another now this is 1 or D these are N plus 1 or D this is just 1 because this is a scalar and then I so and then I look at this together and what I see well they are not really coupled why? explicitly here then I can put the solution here and so I can find why just by integration so these actually they are not really coupled this one equation one problem, one Cauchy problem is not coupled with this system but anyway let me consider this ok so therefore is explicit because it is the integral and moreover maybe look at this part here so define so solve this solve this for X giving you the solution which is next depending on what on S and also on your initial condition X bar so this is again the line passing through X bar through 0 X bar in the direction 1 let me call this more implicitly as X depending on S the exterior parameter which at the end is time and X bar in the same object Y of S equal to U of X of S X bar so I claim that Y of S solve this so assume that you have given X assume that you have given your solution U define Y and then I want to go back assume that I have X I have Y and I want to find U of course I know what is U I am saying this because for more complicated equations I cannot explicitly find the lines they are not lines anymore they are curves and so on and in that case I have to in view of having more complicated problems where the solution will be more implicit now let me make this kind of comments that will help in more difficult cases so let us start from now from this system find the solution define this and look, assume that you is a solution of the PD and look at what kind of odd Y solves so Y of S is nothing that is nothing else U of S X plus X bar plus SB because we know that X of S X bar is exactly S X bar plus SB this is equal to capital X S X bar so Y dot is equal to U t evaluated at S X bar plus color product review at S X bar plus SB since U by assumption for the moment of our PD this is equal to F S X bar plus SB ok, which is F of X do you agree? so we have discovered that again given capital X given U solution of the PD this function Y satisfies this equation so, given U solution of PD of 2, maybe it was written 2 solve for X define Y as here then Y solves it is clear that Y of 0 is equal to U 0 X bar so that Y solves S Y solves 3 so now however we are more interested in the opposite direction so it is somehow so let me stress the following fact this is extremely important for the next lectures the solution of this very easy system of ODEs for the moment is of course a function of the parameter S but it is very important to consider it as a function also of the initial condition X bar is the initial condition and consider the solution X as a function of its parameter and the initial condition ok, this is very very important when you write when you look in the books chapters concerning the system of characteristics you will see that the solution of this system of ODEs are always indicated denoted also the independence also of the initial point you start the trajectory not only that very often more generally it happens that but this is not the case here but just to let you know it happens that you can write the characteristics capital X also as a function of the initial condition I don't write it here because I start from 0 but if I want to start the problem at some t0 t0 then the solution depend on S, X bar and t0 ok, so and very often you see in the books that capital X is capital X of S, t0 and X bar but now t0 is 0 so I don't use this in the notation it's important on the other hand to have this dependence on the initial condition so now given you solve X define Y and Y solves T now of course this is not our problem our problem is to find you I don't want you to be given I want to find you so what do I do so the idea is maybe also the other way around works namely and this is finally the beginning of the method of I'm slowly coming to the method of characteristics for first order PDEs very slowly maybe it works also the following find X solving this find Y solving this because once you know X then you can solve for Y and then try to find U try to define U at the point TX ok, so now U of TX what is the reasonable definition for U of TX in your opinion after this discussion so X because B is given X bar is given then you know Y because F is given so X and Y are given the trajectories, the lines are given the Y is given now you want to find U what could be a reasonable definition for U in order that U solves your PD Y, yes, exactly now the point is this what do I put inside the Y remember sorry, so let me call the solution of this again let me stress that Y is a function of S but is a function also of X bar so now once I have said this the definition now I have to put something here there should be T and X because T and X is here so if I want to write inequality I have to put the same variables on the left and on the right so what do I put here and so so you know why depend on S but S is a function of T and X and also X bar T and X so this is a possible definition again we have somehow to invert a change of variable ok so let me check that this is a solution of so check that this solves our problem ok, so what is Y of S, T, X so you see at the end it could be equal to T we know this so we have a change of variable essentially the idea is the following we have a change of variable for which any point here can be expressed as a projection point here and the parameter S and conversely if I give you S and the projection point T and X so there is this change of variable now you don't see very well this change of variable exactly because the situation is too simple and because this is flat but as soon as we will take an hyper surface sigma which is curved this change of variable will be more easy to understand now because here there are some identifications because of this that make the change of variable a little bit hidden but ok, so let me write this so this is X of bar T of X so this is equal to U bar of X bar plus the integral U bar of X bar of T X plus the integral in between 0 and S of T X of F of X of sigma d sigma is equal to U bar of X bar of T X plus the integral 0 S of T X now X of sigma is given sigma X bar of T X plus sigma B sigma and what do we know we know that S of T X is equal to T and X bar of T X is equal to X minus T B ok, this we know so, what do we have what I have done so I have defined U in this form ok now what is Y Y is the integral between 0 and sigma of F in the sigma plus its initial value so Y is equal to its initial value plus the integral in between 0 and S of F sigma and so on ok, now I substitute my expression of S and X so that finally I find that that is equal to U bar now in place of X I write X minus T B which is this ok plus now we know that S of T X is equal to T so in place of the extremum S I put T here then I have F sigma is sigma comma now X bar is X minus T B plus sigma B X minus T B plus sigma B and this is our previous solution that compatible with what we have already found ok so the message for today is for this ok, we have solved PD explicitly but maybe there is a more general way to solve the PD the general way is you want to solve the PD ok construct a system of OD X dot equal write down another OD and then define U as follows what is your solution here everything is explicit but this so the idea is to solve our PD what we do is to solve ODs and then with the solution of these we construct the solution of the PD and then this is the so-called method of characteristics which works more generally than this and we will study more in detail this method tomorrow in the next lectures ok