 OK. So yesterday, we saw the linear partial differential equations with constant coefficients of the form ut plus b value equal f, where f was a function of t and x. And b was a vector. And we have seen the explicit expression of the initial value problem u equal u bar is at time 0. This is in the half space. And then the regularity was u in c1. So yesterday, we solved this problem, provided that f was c1. OK, also continuous was enough. OK. And also, u bar was sufficiently smooth, c1 also. OK. Let me make just a comment, a geometric comment, on the situation that we have. So assume now that for simplicity, f is equal to 0. So we have here the line, say the time equal to 0. So this is t equal to 0. This is x. OK, so this horizontal now picture is r to the 1 plus n, say. And then let me draw here the graph of u bar. So your initial condition is defined at time 0. And sorry, this is time. This is time. So this is the x-axis. This is time. And so we have an initial condition u bar defined at time equal to 0. And so this is its graph. So we have a new variable here, so you call it y. OK. And we saw yesterday that at the point here, say, here you have t and x. So x here and t here. At this point here, the value of the solution u was the following. I take the value of u bar here, can you remember? So we have here the vector field b, say. So this is the vector field b. b was transverse to sigma. This was called this sigma time equal to 0. So for instance, in this case, this b is horizontal in the horizontal plane. And the value here, so I have these lines. And the value here at this point, tx here, is exactly the value at this x here when I go backward along the line constructed parallel to b. So you see here, in this picture, the value of the function utx is exactly the same height. Sorry for the picture, it's not perfect. The height, so the value of y, is exactly the value of u bar at this point. OK. And so I can imagine that I have a sort of piece of surface with constant height in the sense that along this segment, the vertical component is constant. So it's like this. I have a curve, and then I have sort of this constant. So what I am constructing here at the end, so given this graph above sigma, I am constructing a piece of surface meeting this graph and continuing a little bit with this property of being constant height. So this graph, of course, this piece of surface, is the graph of the solution u. So this is the graph of u. So I repeat it. So I have any point tx. I look at the value of u bar at this point here, where this point is along the line parallel to b. So I take this point here. Maybe yes, it was called x bar. So I have tx. Then I go to x bar. I look at the value of u at x bar, of u bar at x bar. And then the value of u at tx is exactly the value of u bar at x bar. And so if I draw this segment, I can do this everywhere. And so I can imagine a sort of piece of surface continuing the graph of u bar for some time. This was the geometric idea of yesterday. And everything was based on a change of variable, say, which was constructed solving the OD x dot equal, say, b. And x of 0, 0, 0 x bar. So everything was based on the solution of this, which gave the trajectories, in this case, lines parallel to b. OK. So the notation yesterday, this was called x bar. But it was also called, at some moment, it was also called maybe pi of tx. This was, I have two notations. Maybe x bar, maybe pi of tx. Doesn't matter. OK. So this is the geometric picture of yesterday. OK. Now, today, let us continue with this sort of PDEs with the linear transport equation with f equal to 0. So homogeneous for the moment, but with non-constant coefficients. With non-constant coefficients. So homogeneous means f equal to 0, but with non-constant coefficients. So and let me generalize a little bit. So now, you have to be careful, because now I change notation as follows. I don't want to distinguish any more time variable from space variable. OK. So I don't want to give a special role now at t. Here you see, it was a special role to t, because, essentially, we had one in front of ut here. OK. And so this system was x dot, more generally, x dot equal 1b, but this allows to identify s with t. But now, I don't want to give a special importance to the t variable. So now, my notation will be x in our n. And that's it. So x in our n corresponds. So x will be x1, xn. And yesterday, our notation was, yesterday, how to pass from one notation to the other was n equal 1 plus n. Huh? And x1 was equal to t. This is yesterday. OK. So now, today, the variable are called x. I don't distinguish between t and other variables. OK. In some sense, this is somehow, at some moment, a more easy notation. OK. Now, let us consider the following operator. So define L of u equal to the sum i from 1 to n bi of x du to the xi, where bi of x in omega. And consider the problem L of u equal to 0 in omega. Open set contained in our n, where bi, bi are, say, c1 functions. Bi are c1 in omega. For any i, now f is not 0. Now f is 0. And bi is for any i equal 1, capital N. OK? So now, my PD is this equal to 0. What is the relation between the notation of yesterday? So yesterday, yesterday was ut plus b dot grad u equal to 0. So this means that notation of yesterday becomes what? Become this when I choose b1 of x. First of all, x is actually t and x. And then b1 of x is equal to 1. And b2 bn of x is the b of yesterday, b of yesterday. OK, so simply, if you put b1 equal to 1, if you think about the first coordinate x1 equal to t, then you end up with ut here. And then plus b dot grad u provided that b2 bn, you identified it with a constant vector b. So it's just a change of notation. This is more general, clearly. Change of notation. This is more general. OK, so let us concentrate on this new notation. So forget the time now. So this is, of course, a first order PD because you have only just one derivative here. It is linear. This is clear. The coefficients now are the bi. They are non-constant because they depend on x. Non-constant coefficients. I assume bi to be c1. And this is a notation. It's just to remember linear operator L-like linear applied to a function u. So the problem is to find the solution of what? So now if I want to reason like yesterday, I have to give an hypersurface sigma and an initial condition. So now this hypersurface, yes, it was very special because we had time in mind. And this was time equal to 0. So it was like an initial condition. But now there is no time here. So how to assign now a boundary condition? We do this. So let sigma be a c1 hypersurface. Yesterday, omega was the half space. So yesterday, omega was the half space. This was yesterday. And yesterday, sigma was t equal to 0. Today, sigma is instead of a hyperplane. It's just a hypersurface. Maybe it is curved, for instance. It's not flat in general. And it's of class c1. And for simplicity, let me assume that without boundary. And then let u bar, maybe the notation was let u bar be a given function on sigma of class c1. So this is slightly more general than what we have said yesterday. This is the setting. The question is, what's the meaning of a surface without boundary? Do you know what is the boundary of a hypersurface? Maybe it's not the case now how to define it. But let me give you some example. For instance, if you have a piece of plane, just a piece of plane, just a sheet of paper in R3. So you have a sheet of paper inside R3. Then the sheet of paper is, of course, a surface. But the boundary of this sheet of paper is just the curve, which is, say, on the border. On the other hand, if the sheet of paper is infinite, it is a plane, then there is no boundary. In this sense, I'm thinking about the generalization of an infinite plane. So an infinite surface may be a little bit curved, but without border. I don't want to define now what is exactly the boundary of a surface, because I need some tools. But so you have to think about this. You have, say, this is sigma, and this is, say, omega, and this is sigma. And I want to work, say, here in omega intersection sigma. So the problem is, more precisely, problem, try to solve the following PDE, say, in omega, or maybe in omega minus sigma, for instance. So say here. And then u sufficiently smooth, say, c1 of omega. And then u equal u bar on omega intersection sigma. So you want the initial condition u bar here, u equal u bar here. Then you want to solve the PDE, say, here, out of the surface. Of course, here, if you think about time, this is space time, time space. Mm? Omega, now, if you want to remember time, if you like, it's not necessary. But if you like, then we are in time space here. Yesterday, indeed, the situation was this. This was time. This was space. And this was sigma. And omega, say, was the half space. So maybe this is asking too much, maybe. So we will see that we can maybe try to solve the problem not exactly on the wall, omega, but maybe in a strip, in a neighborhood of sigma. Maybe we can hope to find the solution in a neighborhood, not everywhere. Now we will see why. Anyway, this is the problem, generalization of yesterday. So, however, yesterday, there was a, yes, yeah, you're right. Exactly, thank you. So maybe we can do, thank you. Yeah, also, yesterday, you're right. Thank you, omega bar, say, intersection of sigma. And then, sorry, the other question was, yes, maybe yesterday we said, you see one up to the boundary, right? So you see, now, you is, now, because, well, because you are thinking about the half space, but on the other hand, you see yesterday, we construct a solution for positive time. But indeed, we could get the solution also for negative times. So maybe the better is to think about the whole space equal to omega, and now sigma is contained in omega. And I don't need this, OK? Yes, because physically, it's meaningful when t is time, maybe it is reasonable to look for a solution in the future. And so we think about omega half space. But on the other hand, as you see the proof, we can go also backward. So maybe omega is the whole space, if you wish. And then sigma is contained in omega, OK? So sigma, if you want, we can change this. So let me put it like this on sigma. And sigma is contained in omega. So we have a sigma. Now, the new sigma is this. So now, without boundary in omega. So I don't have to worry about these two points. This is my sigma, omega is open. And we are exact in this picture, OK? So now, what I was saying is that yesterday, we observed that one important condition was that the vector field 1b, b was constant. But the vector field 1b was transverse to sigma. Remember, we had the component that was 1. So this was 1 and this was b. And this vector field was transverse to sigma, exactly because of this component 1b. Transverse means that this is non-tangent to sigma. It's not like this. And this was very important for solving our problem. Can you remember? OK. Now, what could be the new condition now that it is natural to impose? So we cannot hope to solve this if we do not impose a sort of transversality. Because this problem generalize what we have said yesterday. And it was clear from yesterday that we need that 1b was transverse to sigma. But here, it is not written anything like that. So we need the transversality condition between the vector field by bi of x now and sigma, OK? So need c yesterday, some transversality condition between the vector field b of x and sigma. And we have to now give this notion now. Maybe I repeat. Yesterday, this was 1b and this was t equal to 0. And exactly because of this component equal to 1, these were transverse, OK? So which is the natural new assumption that we have to do? The natural assumption that we have to do now is we need some definition. So definition. So let x in omega define the set. Let me call it this way. So I take a point in omega. So b of x is well defined because b is defined in omega. And then I define the set of, say, characteristic vectors at x for the operator L, set of characteristic vectors for L at x. As follows, it is a non-zero vector, which is orthogonal to the vector field b of x. So you have a vector field at the point x. You have a vector. And then you take the orthogonal vectors to this, OK? Definition. So this is definition 1, definition 2. So let sigma be a surface of class c, hyper surface, hyper surface of class c 1 as before. We say that sigma is characteristic for the operator L at the point x on sigma if class c 1, c 1. c 1 because I need to talk about the tangent space or the normal vector, if you want. So there should be not coordinates on the surface or casps or something strange like that, OK? So we say, so this is sigma inside omega, this. And I have a point x on sigma. And then I say that sigma is characteristic for the operator L at x if the normal, one of the two, unit normal to sigma at x, where this is a unit normal to sigma at x. So it is simply this. So you have the point x, OK? Then you have the tangent space to sigma at x. Let me denote it by, this is the tangent space. Let me denote it by tan x sigma. This is a hyper plane. And then I have one of the two unit normals, new sigma, this or the opposite. Or this or this, one of the two. It's not important, it's one of the two. So this is new sigma, new sigma. So if this vector, this unit vector, new sigma, say, is a characteristic vector in this sense. So this means that new sigma, that is the scalar product between new sigma at x and bx must be equal to 0, OK? New sigma x dot bx is equal to 0. That is, a new sigma is orthogonal to bx. Or if you want, equivalently, bx is tangent to sigma. So if your vector field here, b, belongs to the tangent space. Say bx is like this, say, this is bx. Then you say that this x is characteristic, OK? Then, definition 3, we say that sigma is characteristic for l, that sigma is characteristic for the operator l, for l. If for any x into sigma, sigma is characteristic at x. So at each point of sigma, b of x is tangent to sigma, simply. If this happens, at all points of sigma, OK? Which is, by the way, the worst situation. Because remember, what we have said yesterday, this is sigma. Vector field was 1b. And the vector field was never tangent to sigma. Was never tangent to sigma. So this condition is bad. And we want to avoid it, OK? So we would say that sigma is non-characteristic, finally. So we say that sigma is non-characteristic for l at x. If this condition is not true, if new sigma of x does not belong to x at l, that is bx is not tangent to sigma at x. Is not tangent to sigma at x. And finally, we say that sigma is non-characteristic for l. If it is non-characteristic for l at any x. Yes. Ah, yes. The l, the question is, you don't see the l on the right hand side. Well, l is identified by the vector field. Because your linear operator l was this. Uniquely identified, you see. Because l was the sum from 1 to n of bi of x du over dxi. So once you know bi, you know the operator l. And this is l. So if you have no l, you have b. So this l is hidden here, OK? So what it is important to remember at the end, OK? So we say that sigma is non-characteristic at x, a point-wise definition. But then we say that sigma is non-characteristic for l. If it at any point x, it is non-characteristic for l. So, and this is what we want. So we continue now our analysis of the problem. So the new assumption that we do is the following. So assume to continue our analysis, assume sigma non-characteristic for l. That is, for any x in sigma, bx is non-tangent, does not belong to the tangent space. So this is now our assumption, OK? So this is an assumption relating sigma and the operator l. Or in other words, sigma and the vector of it b. So this is our assumption. Previous case yesterday, b of x was 1b, with the notation of yesterday, 1b. Always non-tangent to sigma. So it is clear now that this assumption generalizes the transversality condition that we observed yesterday. So let me make an example now. So let me make an example. Assume n equal to 2. And assume that l of u is equal to u over the x1. And assume that sigma is equal to, sigma is equal to, say, x2 is equal to 0. So an omega is equal to r2. So this is x1. This is x2. And we have this operator. So b of x now is simply 1, 0. Is it clear that b of x is equal to 1, 0? Because it is 1 times this plus 0 times the derivative with respect to x2. So b of x is this at any point. So this is sigma. And b of x is 1, 0, which is therefore tangent at each point. At each point of sigma, b of x is clearly tangent to sigma. So it is characteristic at any x. So sigma actually is characteristic for, now assume that I want to solve l of u equal to 0. And u in r2, say, and u equal u bar on sigma. So what I find here? Well, I find the following. If I find the solution, so I find that on sigma, if the solution is smooth, say on sigma, the u bar over the x1 must be equal to 0. Because this is the u over the x1 equal to 0, u bar here. And therefore, the derivative of u bar along x1 is equal to 0. But this is not reasonable. Unreasonable? Unreasonable. Why? Because u bar is your initial condition. And u bar is any initial condition of class c1. There is no reason for which u bar must have derivative equal to 0. No reason for this. u bar is any smooth function. u bar is given in c1, but nothing else. u bar has no condition on its derivative. u bar is given. u bar is given. So in other words, if you have understood this, from now on, it will be clear why we always assume that sigma is non-characteristic. So the question is, do you want sigma characteristic only that sum would be enough, say, or could you allow, say, sigma to be non-characteristic, unless, say, one point or two points? This is the question. No. To be as simple as possible, I will assume that sigma is non-characteristic at any point. This is to make things simpler in this one first-order PD situation. This is the simplest case. So let me comment once more, because this is important to understand. So in other words, we shall assume sigma non-characteristic for L, because otherwise, if sigma, say, is characteristic, at least characteristic in a neighborhood of x and sigma, we would assign a condition on the tangential derivative of your bar, a tangential derivative of your bar, which is unreasonable, which is not reasonable. In the previous example, the tangential derivative was the derivative along the direction x1, which was tangential to sigma. Actually, it was the unique direction of sigma. And we were asking for this to be true, which is not reasonable. So more generally, if sigma is any hyper-surface of c1, asking sigma to be characteristic in a neighborhood of a point x in sigma would mean assigning the tangential derivative of your bar along the vector field b to be equal to something for instance 0, which is not reasonable. So from now on, we will always assume this. Now, the experience of yesterday was going to consider, now consider the following system of oddies, ordinary differential equations. x dot equal b of x. x of 0 equals some x bar into sigma. Now, if you want to be really very, very precise, locally, you have the following picture. This is sigma. This is sigma. Then you have the vector field b of x, which is transverse. Say, for instance, is like this. Smooth. This is b of x. And then you have assigned the starting time, 0, and the starting point x bar. So x bar, what is exact x bar? So to be precise, you should. Since now sigma is a manifold, then you have to parametrize it locally. So you have to find sort of open set here and the parametrization, local parametrization. Let me call this, say, phi. Let me call the parameters here sigma. So phi is a local parametrization, local parametrization of sigma, of sigma. So taking some open set in r to the n minus 1 space of parameters, regularly into rn, and phi of sigma is, say, x bar. So this sigma is going exactly to this x bar. This is a way to describe your sigma. So locally, it is the image through a map, an embedding map, phi. And so any point x bar is coming from one parameter sigma through the embedding. OK, b of x is here. So this is rn. And so x bar, if you want, is now phi of sigma. So now consider the system of ODEs. Fine. Our problem, remember, let me write it here, is just the sum from 1 to n bi of x du over dx i equal to 0. This is a PDE. This is a system of ODEs. How many of these are these? And capital N, OK? Capital N of these, the equations. Because capital X is going into rn, OK? Yesterday, the system, remember, was yesterday was x dot equal 1b yesterday. And yesterday, this allowed to identify the s with time. And so at the end, what was interesting was to study just only the last n components equal to b. This was yesterday, OK? So now, which is the difference between yesterday and today? Yesterday was x dot equal to 1b, where b was constant vector. And now, we have x dot equal b of x. Where is the difference? Well, the difference is that now, well, this is really the E. I mean, it's not so easy to solve it. Yesterday, the solution was explicit. And the trajectories were lines, remember? Yesterday, the important fact was that we found a solution. The trajectory of the ODEs were flat, were lines. Today, there is no reason to say that now, the solutions to these are lines. Because the derivative is not constant anymore. Yesterday was a constant vector. Today, it is not constant. This is the first remark. So today, if we are able to find the solution, the trajectories are not lines in general. The second point that we have to be sure that we have a solution to this system of ODEs. So now, you have to remember the existence and uniqueness and the dependence on initial condition of ODEs. Maybe you already know this. So let me quickly say that. So let me call this point problem 1. So there is a theorem. And if you want to look at the proof for this, one possible reference, but very difficult, however. But anyway, I give you a reference for this. These lectures on nonlinear hyperbolic equations. Well, this is a theorem on ODEs. You can find it in several books. This is a possible book. Please keep in mind, it is a very, very difficult book. So you're not supposed to know it at all. Just if you are interested in, look at the first two or three or four pages. If you like. So the theorem says that theorem, so let me write here, theorem, so there exists a unique solution. Let me call it x, depending on s, which is the parameter to which I'm differentiating as usual. And then also, depending on the initial condition, x bar. Or if you want, if you prefer sigma. Since x bar is nothing else, any unique image of some parameter, sigma, I can make this depending on s or on x bar or sigma as you wish, as you prefer. Yesterday, we had this x bar, and this is a copy of our n minus 1. So the capital phi yesterday was simply, so what is the situation of yesterday? Who is capital phi of yesterday? Well, capital phi is simply x bar into 0x bar. So there is really an identification between x bar and sigma, which is still x bar. So really, so when this is flat, this is a plane, and it was a horizontal plane at level 0, height 0, then the phi was simply this phi, so trivial embedding. And so yesterday, we put the dependence on x bar because it was reasonable. Today, if you want, maybe we can consider dependence on sigma or x bar as you prefer. Let me write dependence on sigma. There exists a unique solution, x of s and sigma, for s less than or equal than some delta. So there exists delta. So there exists delta positive such that there exists a unique solution for short times s of your system of ODE's. So this is a local in time existence. Now, s is time, OK? And since you don't know, you know that B is C1, but you don't know, say, that B has some other magic properties like global lip sheets, for instance. So what you know at the moment is that this has a solution, unique, but just for some time, for some amount of time, but not for all times. Yesterday, it was for all times because the line was globally for any s, for any s. But today, we are not sure about this. We can only say that for some positive time, delta, we have a solution, unique. This is, of course, because the vector field B is of class C1. If B is not smooth like C1, then it is much more difficult. Not only this, so this is the first difference with respect to yesterday. Yesterday, the solution was global and was a line. Today, the solution is not necessarily a line, it's just a curve. Actually, which curve? Well, it's the integral curve of this vector field. You see, if I want to draw the solution for short time, what I have to do is exactly to take those curves such that B is tangent to that curve. So if I draw the vector field B in a neighborhood here, the curves solution to this are exactly the so-called integral curve, meaning that at each point of the curve, B is tangent to the curve, like this. These are the trajectories, is OK? Yesterday, of course, B was a constant vector, and the trajectory was the line, which is tangent to the constant vector. Is it OK? Are there other questions? Problems? Is it clear? So please feel free to ask if there is something which is not clear. I'm making, I'm happy to repeat. So now I'm drawing the integral curves, which are the solution of our system of ODEs. You see? We have a solution like this, OK? So these are called integral curves, integral curves of B. And I repeat, it means that at any point along the curve, B is tangent. This is B, et cetera. You see, what is a big difference now? Well, first of all, we have existence, but just for a fixed x-bar. What happens if I change x-bar to the existence time, delta? What happens? Well, because delta depends on x-bar. So if I am here, say I live for one second. But if I am here, maybe I live for 1 half second. And if I live for 10 seconds, I don't know. The dependence now, the time of life of the solution depends on the previously chosen x-bar. Well, it is possible to show, say at least in this local compact situation, say in a piece of sigma, that delta is uniform with respect to the choice to x-bar. So if I confine myself in a piece, compact piece of sigma, I can say that the solution lives for delta, but for all x-bar. So at any point now, I am sure that my solution exists for delta. And delta is not depending on the point that I am choosing. This means exactly uniform delta. So delta is not going to 0 when I move x-bar. Then, moreover, it is also possible to prove that the dependence of x from s and x-bar is smoothy c1. Actually, it is Lipschitz also with respect to this. So an x is c1, say, well, in the book that I put there, you find also the estimate which says that x actually depends in a Lipschitz way with respect to sigma. And the Lipschitz constant is related to the local Lipschitz constant b. This is not so important at the moment. So we know that this dependence is smooth, say, sufficiently smooth, at least. And next, there is another difference. You see, what happens here? OK, maybe the solution is like this. So it may happen that these trajectories, at some moment, meet together. Yesterday, our trajectories was this. And there were two interesting properties. First of all, that the trajectory was defined for all times. It's a line. Second of all, that this situation does not happen. Say like this. This was not the case of yesterday, because they were parallel, of course. But you see, if you think that what we are trying to do is to define a solution here as the value of your bar here, this is potentially very, very dangerous. Because if two of these objects meet at the point for positive time, then I don't know which kind of value I have to assign to the solution you, if the value of your bar here for the value of your bar here. So this is a very dangerous situation. And it is possible to prove that if delta is sufficiently small, this doesn't happen for delta sufficiently small. So for delta sufficiently small, not only we have existence of a solution, smooth dependence with respect to the initial condition, but we can also say that if delta is sufficiently small, the integral lines say delta sufficiently small implies that integral curves do not intersect. So we are exactly in a sort of situation like this. Here we have sigma. Here we have sigma. Then we are filling a neighborhood of sigma with integral curves, with curves which do not meet each other. And they fill a neighborhood everywhere of sigma. So we are filling a neighborhood of sigma with these curves. Sorry for the picture. It's not perfect like this. So what are these pictures? So this is simply the curve that at s, it associates x of s given that sigma. So if sigma corresponds to this point, so sigma is here. Sigma is mapped to this point here. Then this piece of curve is the image of this map. This is our situation. And finally, this gives us exactly what we like, because then by construction, the solution u at x is equal to u bar x. So now I have to say to make rigorous what I want to say is to say that now there is no f. The equation is homogeneous. So the value of the solution here, say, the value of the solution here, x, will be the value of u bar here. And what is here? Well, I have to take the integral curve passing to this point x, starting from a unique other point on sigma and solving the PDE. So to be more precise, before saying this, I have to observe that there is a change of variable. There is a map from. So let me call this u and neighborhood in Rn minus 1. So there is a map from minus delta delta times u into a neighborhood of sigma. So let me call this object here. This is n, neighborhood of sigma. And the map is the following. I associate to any s and any sigma here the solution x at time s starting from the point phi of sigma. So let me repeat it. I have the space of parameters, capital U, in Rn minus 1. Then I have some time delta. So I can think about this. Fix time s, fix a parameter sigma. Sigma is here. Then I have phi of sigma x bar is here. x bar equal to phi of sigma. And then I can consider the solution of my PDE starting from this point x bar for time s. And this gives me a diffeomorphism between this big rectangle and this neighborhood n by construction. This diffeomorphism between this and its image that I'm called n. I'm calling n the image of this. So given s and l and sigma, I can find a unique point. Conversely, this is invertible. So I can have the inverse object. Now I have a map from the neighborhood n into minus delta, delta times U. So I am the inverse map. The map is invertible with the inverse of class C1. And so at any x, I associate simply s of x and sigma of x. Of course, this diffeomorphism is not always explicit. But this inverse diffeomorphism is, it exists. And so my solution U of x will be exactly the solution U bar starting from phi of sigma of x. This is the solution that we have constructed. U bar of phi of sigma of x. So I remember, I recall once more, given any point x in this neighborhood n, what do I do? Well, this x corresponds to a point s and sigma in this rectangle, delta minus delta. So there is a point s and sigma here going to this point here. So the solution U out of sigma in n here is what? Is the initial condition U bar where U bar, sorry, here, it is even easier, sorry, is even easier, is the initial condition phi of sigma. So it is the initial condition U bar where here, what is this point here? Well, so the point x is going through this, so it's exactly this point here. So this is the parameter. The parameter is mapped to the embedding to this point, exactly the base point, the projection. And so U bar here, so let me recall once more. So I have a point x in the neighborhood. And what do I have to do? Well, I have to find exactly that curve which pass through x of our family. And then this curve which pass through x meets sigma in some point. And then I take U bar at this point. That's exactly what I want to do. So this is x. So this x is coming from one pair s sigma here. OK? In particular, sigma is mapped to the embedding to this point here. This is phi of sigma. And so if I take the solution of our system of these, it starts from here and pass through this x and fill a little bit. And so our U at x will be the value of the initial condition at this projection point x bar or phi of sigma. OK? So what is really important here is the invertibility of this map s sigma into the trajectory solution of the system of all these. OK? Starting from this physical point, phi of sigma leaving for some time. OK? So you see it's a little bit implicit. Anyway, now home exercise maybe would be generalize this to the problem, say, sum from 1 to n b i of x b over dx i of u x equal sum given function f given, say, c 1 of omega given. So the exercise consists in trying to find the solution with the method of characteristics. Yesterday, we studied this problem. But in the special situation, yesterday, remember, we have studied this problem, but with p i u of x equal 1 b. And we found that the solution was the u bar at the previous time, say, but then there was an integral between 0 and t of a function depending on f properly with proper variables. So and here is the same, but you have to try to find to reconstruct the solution. OK? Now, our linear operator that we have studied up to now, you see, is not, I mean, even if this PD is one of the simplest, the solution, a local solution, is not so easy to find, as you have seen. For me, it's not so easy, in my opinion. And of course, a problem is due to the regular, the smoothness of bi. Now, it would be very interesting to study the problem when bi is less smooth than c1, but this becomes quickly too difficult. So we will confine ourselves to this. So what can we do in order to make the problem more general? Well, it's to consider now the following PD instead of weakening the regularity, the smoothness of the coefficients. Let's still take c1, but consider a little bit more general problem equal to f, u, x, x. So now, I have given not only b as before, but I have given also a function c, c1 omega, and f. So this is still a linear operator. There is a new piece, the so-called zero-order piece, 0y, because this piece depends linearly on u, but there are no derivatives. So it's called zero-order term. And so if we define, again, l of u equal to sum, as before, bi of x bi u of u over dx i, we are interested in studying the following problem. L of u equal f minus c times u. So this is usual in PDs. You isolate the principal part of the operator, in some sense. Principal meaning the part where you have the highest derivatives. So here, the highest derivatives are first-order. Here, there are no derivatives. So I isolate this part, and then I consider this problem. And again, we want to couple this with an initial condition. So sigma, non-characteristic for l, for the principal part of the PD, as before, exactly the same definition. And then we want to study this problem, this coupled with u equal some c1 function, u bar on sigma. So this is what we want. So slowly, we are coming to more and more difficult first-order PD. We started from constant coefficients and homogeneous, and independent of u without this term. Then we passed constant coefficients, but with f, non-homogeneous. Then we passed to non-constant coefficient and with this equal to 0. Then we passed to non-constant. This is an exercise now for you. Exercise, non-constant coefficients with the non-homogeneous, non-homogeneity. And finally, more generally, any first-order linear, first-order PD with non-constant coefficients, so now there is a new term here. And this is not so, I mean, it's an important generalization. It's not so immediate. So now we want still to use the method of characteristics. So still we want to construct a local solution, local meaning in a neighborhood of sigma, just only in a neighborhood. But there is this more difficult, this new part. What do we do? So which is now the system of characteristics that we have to consider? Do you have suggestions for this? We have to find the correct system of oddies in order to solve this problem. So yesterday I think that we saw something. So you have to want to make an exponential change of variable, for instance, like, ah, this is maybe a good, if you are able to change, so his suggestion, he says, why don't we try to change you into a new variable, let's say v of x, which is sort of, is equal to, say, e to some coefficient times u of x or something like that, a transformation of u into some new v, so that v satisfies something without this 0 or the 10. Well, I think that, well, if you try to do this, if it is possible as a home exercise and try to find the transformation, OK? This would be, this, if this can be done, would be interesting so that you use the previous case to find v and then you invert your transformation and you go back to u, if it is possible. But I want now to, not to do such a sort of trick, because at the end I want to find out a system of oddies, which will be even more general, applyable to more general PDs, more general than this. What I would like to go, what I would like to study would be more general than this, sort going back, now I write it because this is classical, what is a very interesting equation is this. And so for this I still have to find a system of characteristics, which probably is suggested by this problem and which is the difference between this and this and this now is not linear anymore. It's not linear in you because you have the product u, ux. So I don't use the, I don't want to consider the trick because I want to find a system of characteristics that I hope to be applyable also in some nonlinear case, like this. And so, so suggestion on what could be, please if you have some suggestion on what could be now the system of oddies to consider. Remember that yesterday we wrote at some moment a system into in x and also in y. Can you remember? So I think that still we can try x dot equal b of x. And this x of 0 equals sum x bar equal phi of c. This is, I think it's somehow reasonable. But then this is not enough. What could be another equation that we have to couple to this, remembering the discussion of yesterday? At some moment we coupled this with another equation, y dot equal. It was yesterday was f of capital X, right? This was yesterday. And we observed the following. Well, this is uncoupled with respect to this. Namely, first I solve this. I find x, then I put here x, and I find y. This was yesterday. And yesterday this term was not present. Now, what is the reasonable modification of that oddie minus c of capital X times what? Times y. Initial condition y of 0 equal u bar of x bar. Do we agree? It is somehow reasonable to think about to consider this. Now, you see, which is the difference with respect to yesterday? Now these are really somehow coupled together, in some sense. Not so coupled, but I mean still more precisely. Yesterday, given capital X, I can simply integrate and find y. This was yesterday. Today is not so easy, not so difficult, however, because I can find capital X here. And then I have an oddie given capital X. Now this is a first-order oddie, linear first-order oddie in y. It's not just an integration problem. I have to solve another oddie. It is linear in y. So still, this seems doable. Still, I can first solve this and then this somehow, OK? So well, time is over. We will continue from the analysis of this linear first-order oddie. And maybe I can give you a preliminary exercise which is, by the way, maybe if you are not able to do this exercise, don't worry, because still we have to discuss a little bit this. But exercise, and we will do this exercise in the next lecture. So we will do this in class. So the exercise, so consider n equal to 3 and consider the following problem. First-order solve x1, sorry, du over dx1 plus 2, x2 du over dx2 plus du over dx3 equal 3 times u. So it is exactly in this form with f equal to 0 and c constant equal to minus 3 u equal u bar of x1 and x2. Now the notation is clear, because sigma on sigma, which is x3 equal to 0. So try to find the solution to this, OK? Check that the operator is, that the surface sigma is non-characteristic for the operator, of course. Check all assumptions, smoothness of coefficients, smoothness of sigma, smoothness of everything. And then I tell you which is the solution. The solution that you have to find is the solution that you have to find is u of x must be equal to u bar of x1 e to the minus x3 comma x2 e to the minus 2 x3. If I'm not wrong, the solution should be this one. So u bar is the initial condition given u bar c1. You see, then you have u bar at some points and then there is also this exponent exponential e to the 3 x3 if I'm not wrong, OK?