 We have seen last lecture the following theorem. So consider, so let now the initial conditions, now I don't remember the notation of the initial conditions. U0 bar and U1 bar. OK, thanks. So assume that this is C2 and U1 bar is C1. And consider the following problem. Utt minus C squared Uxx equal to 0 in r times r, say. This is time. This is space, time space. And then U in C2, U in C2 of r times r, U at time 0 equal to U0 bar, Utt at time 0 equal to U1 bar. Then there exists a unique solution. This problem has a unique solution given by the Dallenbert formula as follows. U of Tx equal, well, it is of the form capital F of plus capital G of x. If I'm not wrong, these were the notation. But more precisely, it is 1 half. And then we have U0 bar x plus Ct plus U0 bar x minus Ct plus 1 over 2c x plus Ct U1 bar, U1 bar of S. This was the Dallenbert formula. And this was the result. So the proof was that if you have an equation like this, then by the splitting property, then necessarily you can be written as the sum of two functions, capital F and capital G, depending on this variable and this variable. Where F is, for instance, U0 bar plus 1 over 2c, a primitive of U1 bar, any primitive. For instance, you start from 0 to this. And G is, of course, the other. So this was the result. So keep in mind that this result is one dimension. So it is essential here that we are in one space dimension. If you look for such a kind of formulas in any space dimension, then the situation is much more complicated. And you have to distinguish in between even dimensions and not dimensions. But so as far as we are concerned, we are interested for the moment just to space dimension 1. So and from this, we can infer some qualitative properties of the solution, qualitative property of U. In particular, the continuous dependence. In particular, I would like to say something about the continuous dependence of U on the initial conditions, on U0, U1. With respect to the L infinity norm, indeed, we have the following remark. So remark 1. So assume that there exists epsilon. Assume that U0 bar in L infinity is less than epsilon. And that U1 bar also is less than epsilon. For some epsilon, positive. So there exists epsilon positive such that the initial position and the initial velocity of the string is of the wire, say, of the wire, is small in the uniform norm. Do you know what is this L infinity norm? The supreme, yes, of the absolute value. OK, then what we can say? So what about the solution? So now U of tx is less than. So look at this. This is our explicit solution. So we have 1 half epsilon plus epsilon here. So we have 2 epsilon, and therefore epsilon plus. Then we have 1 over 2c. Then here, again, we can put epsilon times the length of the interval, which is, again, 2ct. So 2ct epsilon, epsilon. So we can write this as epsilon times 1 plus t. Absolute value, if we are interested in. We are going backward and forward here. Of course, positive times are the times of physical interest. But if we look at the solution, this makes sense also for negative times. So for this argument, there is no reason to exclude negative times. So I put an absolute value here. And so you see, this says that if U0 bar, U1 bar, are epsilon close to the origin in the L infinity norm, then on a strip of the form minus AA times R, so on a horizontal strip. So provided that time is bounded between two constants, then on a strip, also the L infinity norm of the solution is, say, epsilon close to 0 in L infinity. So this is continuity with respect to L infinity of initial conditions with respect to L infinity. Of course, here there is time. So if I want a uniform bound, I need time bounded between two constants. So this is small provided that time is bounded. Of course, this is not small for any time. But on a strip, on a horizontal strip, any fixed horizontal strip, then I have a uniform bound on the solution. So this says that we have continuity of initial conditions with respect to L infinity norm. Because there is here an assumption that I'm close just to the origin. But this is a linear equation. So actually, it is enough to say that this is close to 0. So to say continuity with respect to initial conditions, since the equation is linear, it's equivalent to say continuity at 0 with respect to the initial condition. So continuity at 0 because of linearity. Or if one wants, so smallness in L infinity implies a smallness of the solution at least on horizontal strips. Now, if you want to avoid this bound and you want a uniform bound on the whole time space, then you have to make an assumption. So the second remark is assume that there exists epsilon positive such that u0 bar in L infinity is less than epsilon, u1 bar in L infinity is less than epsilon, and the support of u1 bar, say, is compact. And so assume that the support of u1 bar is contained in a compact. And then you see what happens here. Then, well, actually, these integrations is only in K. And so we find then u is uniformly small, sorry, then u is less than epsilon 1 plus, how can I call, intervals of b, say, so assume that K is contained in minus bb. So this compact set, assume it is contained in some interval minus bb, then u is less than epsilon for any in R. OK, so we have this now. We have avoided the restriction to an horizontal strip, but we have assumed that the initial velocity is compactly supported. So now I left you some exercises. So I keep just this formula. And maybe another remark is useful. Remark 3, before discussing the exercise, is that we cannot apply, in the previous theorem, we cannot apply the Koshik-Kowaleski. Koshik-Kowaleski. Why we cannot apply the Koshik-Kowaleski theorem? Because the Koshik-Kowaleski theorem holds, of course, for non-characteristic surfaces, but also because it holds so when u0 bar and u1 bar are analytic. In this case, u0 and u1 were c2 and c1, not analytic. Therefore, we cannot say that this is the solution given by the Koshik-Kowaleski, because we miss the regularity assumption. u0 bar is only c2 and not analytic and not necessarily analytic. And u1 bar is only c1 and not necessarily analytic. Why this distinction, c2, c1? Well, you see, we look for a solution, u, which is c2, because we want to make two derivatives in time and in space. So if u0 is c2, then u1 bar is c1, then u1 bar is c2. So if u0 is c2, this part in parenthesis is clearly c2. If u1 is c1, u1 bar is c1, then its primitive is c2. Therefore, u is the sum of two c2 functions and the solution is classical. This is the reason why we distinguish c2 and c1. So we need slightly less regularity on the initial speed to have a classical solution. And this formula is not given. The Koshik-Kowaleski theorem gives us a local solution. First remark is local. And this is a global solution. Global means for any times, for all positive times, negative also. So global in time. So Koshik-Kowaleski gives you a local solution. This expression is explicit and gives you a global solution. Moreover, it holds also when these are not analytic, but in this regularity class. So this is the third remark. And then now we can pass to solve the exercises. I will make some pictures. So this is an extremely interesting explicit formula. So we have a traveling wave. We have a first traveling wave moving toward the right along the characteristic, this, x minus ct, say. And then we have a superposition with this traveling wave, with this other traveling wave, g, superposition. I'm moving, sorry, this was moving toward the right. It's moving to the left. And so we have, say, this is the direct characteristic. Let us call direct characteristic this and reverse characteristic this. And so we have the superposition of two fronts, of two profiles which remain always the same and move one toward the right with velocity c and one toward the left with velocity minus c. Or, by the way, toward the right with velocity c and toward the left with velocity c. Same velocity. So the exercises, so it was exercise one. I think it was assumed that u1 bar, the initial velocity is identically zero so that we have only to look at this part of the solution, just only at this part. And assume also that this was the initial u0 bar. So this is the graph of u0 bar. And our notation where I think u0 bar of x is equal to x plus alpha. So this is minus alpha. This is alpha. This is zero. x. OK. x plus alpha if minus alpha is less than or equal than x, less than or equal than zero. And then minus x plus alpha if zero less than or equal than x, less than or equal than alpha and then zero. OK. So this is u0 bar, which unfortunately is not c2 because it is just only lip sheets. You see it is lip sheets because we have, in the graph there are three angles. Anyway, let us suppose that still, so remember u0 is lip and not c2. But however, still we assume that the correct solution is this and this can be seen. So we keep still this as the correct solution. In any case, keep d'Alembert. So now let us try to depict. Initially, then I will try to do some movie, a sort of movie. So by movie, I mean that for different times, I draw the solution. So I have just pictures at different times. So let us call this movie. OK. So we have time is alpha over 4c. So alpha over 4c means that now we divide this into four parts. So let me put it here. OK. So this is alpha. This is alpha. And then I have. So I have. So let us consider the direct wave, direct meaning moving toward the right. This is the direct wave moving with velocity C toward the right. And then let us consider the inverse. So these are our two objects separately separately. And then we have two superpose. So I need another object. So let me try to do something like maybe I'm wrong. Maybe I'm doing something not so precise. Another pic another. So this is our solution in blue at time alpha over 4c. Now let us continue the picture at alpha time alpha equal t to alpha over 2c. OK. So my wave now is here. And then I have the other one. So this is alpha over 2 something like this. Sorry. OK. So that our solution is essentially say something like this. So in blue I have I have the solution you. OK. So let us continue. Well, then let me continue the movie. Say this is time 3 alpha over 4c. So that now our solution is something like something like something like this. OK. Something like this. So from this you already see an interesting fact from this picture you see that these are so that the maximum the point where the solution you is maximum like this part here are not preserved because you see now there are other maximum point. I mean this goes down but this goes up. And then finally there is an interesting time. It's an interesting time. So we have 1 4 1 half 3 4 half over c. And so we have since we have this one half in front. Now you see the supports are disjoint. The support of the direct wave is just this interval and the support of the inverse wave is just this interval. So they are maybe they just intersect at one point. But now we have to take the mean of the two because there is this one half in front. And therefore we have such a kind of solution. Then now so for all these times the two waves interact together. They interact because you see they partially overlap. So there is an interaction between the two after this time there is say that they have passed. So take t equal c. And so now we are in this regime. And so the blue one is the solution. And then they move one from the right to the right, the other to the left. And then this is time 3 half. Sorry, the picture is not very precise. So from this it is interesting because you have a flavor of what it's happening. It's already rather complicated as you can see even without the initial velocity. So remember that all this picture has an assumption for us just to understand the interaction of two waves, one moving toward the right, the other moving toward the left keeping the same form. So the same profile. And another obvious fact which is of course written here is that as I have already said to you is that there is no regularizing effect. If u0 is Lipschitz, u0 bar is Lipschitz, then u is no more than Lipschitz. As you can see from the pictures you see this is Lipschitz. It's not c1. But it is immediate from the expression. So there is no regularizing effect. So this is the movie. I hope it is more or less, I'm sorry for the pictures, it is more or less clear. Everything follows from the explicit expression. And so we can try to understand the similar properties of looking at another picture, which is the picture now in time space. So as I told you there are six, so let us assume exercise two. Assume now u0 bar is any initial condition but the support of u0 bar is contained say in alpha 1, alpha 2. So if you want you can think about the previous u0 bar. In the previous case u0 bar was this and the support was contained in minus alpha alpha. More generally now assume that u0 bar is an initial condition with the support contained in some interval alpha 1, alpha 2. So for instance the previous u0 bar is OK. So now we have alpha 1 and alpha 2 and so we can write the inverse and direct characteristics in time space. And so we have six regions. The number was, now this is number six, this is number one, this is number two, this is number three, this is number five, this is number four. So first remark is that after some time which can be computed maybe is alpha over c at least in the previous example. In this region here time space region u is equal to zero. And this is, you see starting from this time u is zero here and then afterwards for subsequent times, so after this time u is zero in a region which is increasing in time. So you see this region increases. As time go on the space increases here because this is x and this is time and in this time space region the solution is vanishing and it's exactly this phenomenon. So starting from this time then there is a larger interval where the solution is zero, you see. Moreover we know that the solution must be zero also here, so zero, zero, zero. And this can be understood from the previous picture, you see the solution where zero here and here, here and here. So in this part of the time space zero, interaction of the two waves, direct and inverse happens in this region one. Here there is a direct wave, here there is an inverse wave and they interact in this small triangle, region one, which is here. And then here there is only the direct wave in this part and here there is only the inverse wave. Sorry for the pictures. This is an attempt to make clear the situation which is not easy because even if this is a one space dimension PD still there is time. So times and space are at least two coordinates. It's not so immediate even to draw a picture. So now, you see how important is this formula because everything is deduced from the explicit expression. Now let me try to suppose now that exercise three assume now that the initial position of the one of the assume now that you zero bar is equal to zero, but you one bar is non-zero. So that u of tx is just only one over two c integral x minus ct x plus ct u one bar s ds. So now maybe this we can write this as alpha. Our notation maybe is alpha of x minus ct minus alpha of x minus ct where alpha is a primitive of one over two c u one bar. For instance, alpha of tau alpha of r is equal one over two c integral say from zero to r of u one. Now therefore, we are almost in the previous case almost which is the difference between initial position equal to zero and initial velocity equal to zero. You see still we have a superposition of the same profile like this, but now there is the first difference. There is this minus here. Alpha one. Yeah. One over two. One over two. Ah, yeah, yeah, yeah. Sorry because thank you for doing this. Because I put, thank you. OK, so now as I, what I'm saying is that now there is the first difference here. There is a minus. Therefore, it's not exactly like the previous case because one of the two change signs. It's not exactly as before. Because here there is a plus and here there is, it is true that here we have the same function alpha alpha and here u zero bar and u zero bar. OK, but there is a difference here. There is a plus and a minus. So it is still, you can imagine again a picture as before, but now you have to change. If you have it, if your inverse wave is alpha going through the left to the left, then the direct wave is minus alpha going to the right. So there is this difference here. Not only this, but also maybe I can give you this. So think about the previous case. So what about here? So now the question may be homework. Is it true that now that, so take, take, take, is it true that, that u is equal to zero in region six? Is it true? Is it true? Take for instance, take for instance, for example, say u one bar equal to, assume for instance that u one bar plays the role of the u zero bar in the previous case. This was u zero bar in the previous case. Now, now we are in this exercise, we are assuming u zero bar equal to zero. Therefore, take u one bar like this, for instance. Well, what happens? Maybe I can, this is not true. It's not true. Why is not true? Well, the idea is the following. You are taking a primitive of this. For instance, a primitive of this is a function say, which is maybe zero here. Then the derivative is positive. So it is something like this, say. It is the primitive, maybe zero here. Then this is the derivative of this. So the positive derivative means that u is increasing, this is increasing, and then it reaches some constants. So you see, if this is your profile moving, say, toward the left, and assume that this is zero. Well, the point is now that here is not zero anymore. And even then, if you change sign, well, try to convince by yourself that now in general we cannot expect in this region that the solution is zero. Think about this example. Now, some modifications of this problem. So, OK, we have studied an ideal problem because our space was the wall line. But more realistic is that the space is zero. Our space was the wall line. But a more realistic problem is the half line. Half line means that we are interested in, say, r, maybe, times zero plus infinity. Now we want to work. So our space domain is just a half line. Instead of all the wall line. However, here there is a problem because this object has a boundary. And so it is not enough anymore to assign an initial position and an initial velocity. But you have to specify also a boundary, so-called boundary condition. So this makes the problem more realistic. We have to impose a boundary condition. And so we can do this. So let us consider the following problem, say in r times zero plus infinity. Then u of zero equal u of zero dot u of zero bar dot u in c2, OK, as usual, ut zero dot u1 dot. And then u of t zero. So this is the initial position because this is time. This is the initial velocity given as before, given functions. OK, this is, say, in c2. This is in c1. And then now time is going, but space is fixed. Therefore, we now fix this, say, for instance, equal to zero. So our object, our one-dimensional wire is fixed at the point, exactly, for any times. So we could, in principle, assign this. But for simplicity, take h1 equal to zero. Now this formula is not true anymore. And moreover, for simplicity, we could also work for positive times, maybe. Positive times just for, OK, this is for any positive time. We are looking for a smooth enough solution. This means that if you want a smooth, do you see that there is something missing here? If we look for a smooth enough solution, there is something that we have to add. Can you see? So we have space, time. Here, our u must be fixed. Directly condition equal to zero. And there is something which we need to ensure u to be smooth of Class C2. Can you see it? I mean, we want u to be equal to zero for all times at the origin. So here, u must be equal to zero. h1, u must be equal to h1, which is equal to zero. So in the simplest case, u is equal to zero. So our chord, say, is fixed at the origin. Not only this, this is this condition. Next, we want that at time zero, on this half line, u is equal to u0 bar. And the vertical component of the gradient must be equal to u1 bar. So, necessarily, there should be something that we have to require here. These are just the boundary and the initial conditions. I hope it is clear. I mean, these are the two boundary conditions, so-called boundary conditions. Meaning that at time zero in space, u initially is u0 bar and its velocity, which is the gradient in the vertical direction, is equal to u1 bar. Since now our space domain is the half line, it has a boundary here. And here, also, we assign a condition. In particular, we have decided to assign u equal to zero. h1, t. So there should be some compatibility, at least, in order to have a solution u, which is c2. In this domain, then there should be reasonably some compatibility between u0 bar, u1 bar, and h1. Here, ut is equal to ut0 is equal to h1 of t. I mean, u0 bar, which is a function of x only. u0 bar is a function of x only. So now it is reasonable, at least, to impose that u0 bar, which is a function of x only, of class c2 up to this point. So it is, in particular, defined on this point. It is reasonable to assume that u0 bar, at this point, is equal to h1 and zero, which, for our choice, for our simple choice, h1 is zero. Yeah, it is not, indeed, but there is another one. And then, what about u1 bar? Yes, exactly. So h1 is also, h1 is zero, by the way, but assume that now is of class c2, 0 plus infinity. Now, this 0 plus infinity refers to time, because the domain of h1 is time, is not space, is this one, h1 prime. This is equal to zero. So with our choices, this is equal to zero. And maybe this is still not enough. And so we have to impose h1 second of zero. So this is ut, t is equal to uxx c squared bar second. This is the second. These are the compatibility conditions. Let me check just. OK, so these are called compatibility conditions in order to have a c2 solution, u. OK, so maybe I can give you the following not easy exercise. Not easy exercise. So which is the explicit expression of the solution in this case, the solution u. So this is home as the following expression. The usual one, this is the usual one, u1 bar of s, if x minus ct, if x minus ct is bigger or equal to 0, t is positive, x is positive. And if, so if instead x minus ct is less than or equal to 0 and yet x is positive, t is positive, then there is a difference, which now I tell you. So the difference is this, this, and this. We are working in this, in this x is positive, t is positive. And then if this left extremum of the integral is positive, then this is the usual Dahlenbert formula. But if in this region this left, this x minus ct is negative, then we have to change it and this is the situation. So how to prove this? Well, hint the solution to the more realistic semi-infinite problem one. Still this can be written as, still a solution u to 1 can be written as f plus g. Some f and g c2. Still this. Then once you know this, then you can try to impose the boundary conditions. For instance, then you have some, so you have u. Now you want to find capital F and capital G. Let me summarize. Assume that you have a c2 solution of this problem. Since this operator splits into two operators as we know, two first order operators, still our solution, if it exists, must be of this form. For some f and g of class c2. Now we have to find f and g because at the end the solution of the exercise is this. Well, the only thing that we can do is to impose the boundary conditions and the initial conditions. For instance, at time 0, we know this in particular, I put 0 here, so u of 0x must be equal to u0 bar of x and therefore this must be equal to f of x plus g of x. This is given by this. This is the first relation between f and g. Not only this, but we have also ut. Therefore ut of tx, which is the derivative with respect to t of this, which is cf prime of x plus ct minus cg prime of x minus ct. This gives us another relation because now I have this for any times. My solution is c2 up to the boundary and therefore I can put here time equal to 0. And therefore u1 bar at x, ut at time 0x, which is equal also to u1 bar at x, must be also equal to cf prime of x minus cg prime of x. And then we have another relation. So if we integrate here, for instance, in between 0 and x, say u1 bar of s ds, we get f prime of x minus and so on. I mean, you can integrate here because this is homework. You can integrate here. And so you have f of x minus f of 0 minus g of x minus g of 0. And therefore then you can couple this. I mean, you can try to put this and this together. And then go on using also that this is equal to 0, et cetera. So this is the exercise. It's not easy. And then there is final, maybe I want to say two things more. Maybe let us go to this one, to the last one. The solution. I mean, you start from a solution. Then necessarily this solution will be this. So it becomes the solution. I mean, if you have any solution of this regularity class c2, then that solution is always this. No, thank you for the remark. Maybe it's not c2. The question is maybe it's not c2 along the characteristic line. No, because of our compatibility conditions, you can check that this solution is actually c2 because of our compatibility conditions. It's actually c2 everywhere in the quadrant. Positive times and x positive and t positive. x positive and t positive. I'm working on x positive and t positive in this moment. OK. Yes, the uniqueness issue is always delicate. But if we assume sufficiently regularity of the data and in this one dimensional linear case, the solution is explicit and that is the solution. Classical solution. The classical solution. Meaning that at any point tx, ut at tx is equal to c2 uxx at tx. This means to be a solution at any point. OK. Now a remark concerning uniqueness. This is theorem on bounded domains. Uniqueness on bounded domains. So this theorem in principle does not apply to what we have set up to now because now we will work on a bounded domain. So omega is say a bounded smooth domain in a random. The previous case was n equal to 1 was the previous case. But in the previous case, omega was either the line or the half line. This on the other hand is any dimension but bounded smooth domain. Say bounded smooth open set. It means that its boundary locally can be written as a graph of a c infinity function, the usual stuff. OK. So if you write it in local charts, the charts are c infinity. So you can write it locally as a graph of a function of n minus 1 variable. So consider the solution. Let us consider the problem utt minus c2 Laplace of u equal to f. So let me give an also f. Smooth enough. Be given also. This is the Laplace of u. So in one space dimension this is uxx. But this argument now works in any dimension. So then u of 0, let me call it u0 bar, ut of 0 u1 bar at t equal to 0. So this is omega. And assume also that we want now we need also u equal h1. So this is in 0 capital T times omega. This is at time 0. Initial conditions here. Next I want to give a boundary condition h1. On, say, boundary of omega. On, say, 0 t times boundary of omega. So this is the boundary of omega consists of two points. So these two points in this picture is the boundary of omega. So on this, see this is capital T. We have a solution u inside this cylinder, half cylinder, a portion of cylinder, say. And then assume that we have compatibility conditions between h1 and u0 and u1 plus compatibility condition between h0, between u0, u1, and h1. So then 2 has at most one solution, one classical solution. My classical I mean c2. So this is uniqueness is not existence. It does not say that the solution exists. It says if it exists, it's unique. It means nothing about existence. But there is a statement on uniqueness. Yeah, the case n equal 1 would be the case on an interval. Say 0l, for instance, would be. Yes? Ah, yeah, yeah. In any, so the question is about the smoothness of the boundary of an interval. Any interval, say 0l, the boundary of this interval is of course the point 0 and the point l. And this is perfectly smooth. It is perfectly smooth. In one dimension, it is trivial that it's smooth if your omega is just an interval. If, for instance, is a countable number of intervals, it is less trivial. But this is just, now assume in one dimension take omega as a one interval. So this is automatically satisfied. It is automatically smooth. Now you will see why I need smoothness and boundedness because the proof runs. So let us prove this result. So assume that you have two solutions. u1 and u2. Assume that u1 and u2 are two solutions, two classical solutions. Then take the difference. Let me call it maybe w, w1 u1 minus u2. Of course I want to show that w is identically 0. The t's is, so define w. t's is, w is identically 0. Because once w is identically 0, it means that u1 is equal to u2. It means that those solutions are the same. So now I introduce the following for any t, say in 0 capital T. I introduce the following quantity. Let me call it ut square t dot square plus u square. So let me introduce this sort of energy. What is this? It is ut evaluated at that t. So t is fixed. I mean, I am fixing t. I have this time slice. And I integrate on this time slice. What are, sorry, what are they? In this? Two. They are always the same condition as before. Say c2 of omega bar c1 omega bar. I think they are enough. I think that they are enough. Also c2. But how regular must be the initial condition then we understand from the proof. So what I was saying is this. So fix any positive time and then consider this integral quantity ut square plus grad u square. So everything is well defined because we are assuming that the solution is c2 up to the boundary. Therefore these are finite. Ut is c1, grad u is c1 up to the boundary. So the integrals are finite. Continuous function up to the boundary. So these integrals are finite for the moment. So let us consider this sort of energy and let us try to see what happens. Now this is differentiable in time. And so we can try to differentiate e in time. Is this symbol clear? This is the gradient of u in space. So e dot t. Maybe I can put one half here just for elegance. One half just for elegance. So now I can differentiate this with respect to time. Everything is smooth enough so that the derivative I can put derivative inside the integral. And so this is equal to ut utt plus scalar product. So let me use a symbol for the scalar product. Scalar product grad u grad utt dx. Remark, before continuing let me do this remark. W is in c2, et cetera, et cetera. And solves the homogeneous problem. For this argument let me normalize things so that c is equal to 1 just for simplicity. I normalize with c equal to 1. So there is no c here, no c in the equation. This is equal to 0. W0 is equal to 0. Wt of 0 is equal to 0. And also time 0. And also W is equal to 0 on 0t times 1. Well, because the problem is linear. W is the difference of two solutions. The problem is linear. So I take the difference of the two equations. F minus F here gives 0. U0 bar minus U0 bar gives 0. U1 bar minus U1 bar gives 0. H1 minus H1 gives 0. So W is a solution of our PD with 0 boundary and initial conditions. The point is to prove that W is 0. Necessarily 0, not any other function. So let me summarize. I take any two solutions, U1 U2. The difference I call it W. I observe that W solves an homogeneous problem by difference. And I want to show that W is identically 0. I want to introduce this integral quantity for any time, this object. And I first, maybe Wt. No, OK. This is Ut, yes. This integral quantity, Wt. I want to show that W is identically 0. First of all, I will try to show that E is equal to 0. So if I show that E is equal to 0, then Wt is equal to 0 and grad of V will be equal to 0. OK. Can you follow? Is it clear? So let me say once more. I introduce this integral quantity. My aim is to prove that W is equal to 0 so that these are equal. To prove that this is equal to 0, I will try to prove first of all that E is equal to 0. This. So this is W everywhere, of course. So if I prove that E is equal to 0, this does not imply that W is equal to 0. But at least it implies that W is constant. Then I will use the boundary conditions which says that W is 0 at the boundary. So necessarily W will be 0. This is the strategy of the proof. OK. Is it clear? So now E dot of t is this. And now I can make an integration by parts. So this remains as it is. I cannot integrate by parts here because the integral is just only in space. So I cannot integrate by parts this. But this is the gradient in space. I have, of course, used the following, by the way. I have, of course, used by smoothness. I have, of course, used this. That I can interchange time and space derivatives like this. OK. This I have already used. I have this theorem because W is C2 by assumption. So now I am here. And now I can integrate by parts. I hope you that you remember now the integration by parts formula. I think that they have already written down. Of course, I don't remember the symbols. Under smoothness assumption, the integration by parts formula was something like scalar product. Do you remember the symbols that they have used? Maybe we can try to use the same symbols. OK. So let us go back to the, I don't remember, the versions of eta, omega bound and smooth and so on. And then this was also the exterior unit normal. New, same symbol, new. OK. Fine. This is an integration by part. Ah, yes. Sorry. Yes. And this is the house of measure. OK. So now from this, we can deduce an integration by far formula slightly more general, but actually equivalent. So eta was a vector field. Now replace to eta the product u times eta, where u is scalar, smooth enough function. So u, scalar, smooth enough function eta, smooth enough vector field. So this product is again a vector field. Scalar time vector. So I replace, in place of eta here, I put u times eta here. So let me rewrite it, u eta. So our formula says this, u scalar product, just the previous formula with u eta in place of eta. And then, however, I can write with this. So this is actually grad u times eta plus u divergence of eta. Therefore, our formula substituting this vector calculus inside here the following u divergence of eta in the x is equal to minus grad u scalar eta in the x plus u eta dot mu here, which is minus 1. OK. I have simply substituted this. Therefore, I have u divergence of eta here. And then this integral term is now on the right-hand side. Is it OK? So maybe we can keep this as our integration by part formula. Is it OK? OK. I mean, you know this formula, in my opinion, very well. For instance, in the one-dimensional case, n equal to 1, this is u eta prime is equal minus u prime eta plus, as you call, u eta. I don't know. This is simply the usual stuff. They simply, how do you call this integration by parts? Changer? No, no, integration by parts. You see, because now if your object, your domain omega is just an interval, the unit normal is just outwardly normal as this. So you surely know at least this formula, but this is just the generalization of that. OK. OK. Now, it is also useful to keep this formula when eta is not any vector field, but the gradient of something. So when eta is the gradient of something, this is u Laplace of eta equal minus grad u grad v, the x plus u dv over. I have simply substituted this vector field in place of eta here. The divergence of the gradient is the Laplacian. This is the scalar product between two gradients. And this is the normal derivative of v. This is simply, now it's very late. Sorry. So I can, this is the directional derivatives of v in the direction u omega. So now, sorry, it's very late now. We will start tomorrow. We will start from exactly where we are now. So remember, I have written this formula because now I want to apply it. This integration by path formula will be applied into, you see, what happens here? I have a gradient here, you see? Gradient against gradient. And here we have gradient against gradient. So it means that this object can be written as this plus this. And therefore, one gradient, one gradient, sorry, this is Laplace of v. Sorry. So one gradient here on u disappears and it goes twice on v. And then there is a boundary term. So the point here, I want to work on this integral expression e of t. And this, and now I want to put this gradient here, here, changing sign. So you see, we can apply this formula. And then there is a boundary term. Boundary of omega, this one. So I have d plus of v, a plus of v dv over dn wt vh a minus 1. So this is e dot of t. Now, is it OK up to now? Now you see, what is this? Well, is this? It is 0. Why is it 0? Because w satisfies the homogeneous equation inside our cylinder. So w satisfies our pd inside this portion of cylinder. In particular, on this time slice, t here, everything is c2. So on this time slice, the pd is identically satisfied. And therefore, this is identically 0. So this is 0. This is also 0. And then you have to try to see why this is 0. Using the fact that w is 0 on the boundary. Also on the, hm? So since w is 0, also wt will be 0. So e dot is 0. This means that t is constant. Next, we have to conclude that not only w is constant, because w is constant. So e t, we will continue tomorrow.