 So today, we have concluded the part for the moment, the part of first order equations. So we start now with a new kind of equation, which is, by the way, very, very useful in physics in particular. So what I would like to start today is to study the following PDE. So utt equal some constant, positive constant, plus of u. So let me introduce the symbols. So we are, our variables are t and x now, t and x into r1 plus n, which is split as follows. So t is a scalar variable, x is a vector. And we will, so if you want to compare with the previous notations, maybe we use this as, we use this capital N in the sense of 1 plus n. So t is real, x is a vector. We will be mainly interested in the case n equal 1, OK? Yes, we will study mainly. u is a function, and also for the region of interest is maybe not the wall space, but we will often restrict to this half space. So t maybe will be non-negative, positive times, say, and x will be in our n, usually. Even if we could also go back in time, but physically maybe is more meaningful to consider non-negative times. Our pictures will be usually putting x in the horizontal axis and t in the vertical axis, despite the fact that t is the first variable, as usual. Remember this strange convention that we took. So what does it mean, ut? Ut, for us, is du over dt. And therefore, so this is the partial derivative. And therefore, utt is the second derivative. So c is given, is a constant given. And this is called Laplacian of u and is defined as the sum from 1 to n du over dx i. Maybe these two we can put it here. So it's the trace of the action. It's the sum of the pure second derivatives. In one space dimension, our equation is utt equal c square uxx. This is our interest. Sometimes this can be written as follows, 1 over c square utt equal Laplace of u. And or if you wish, d over dt, d over. So sometimes the time variable is substituted by this constant c times the time variable. And very often, it happens that people take the normalization c equal 1. Very often, one normalize things so that c is equal to 1 normalization. So for many qualitative results, we can take c equal 1. Physically, maybe sometimes it is better to keep the c because c will be very large. So let us start. So this is a linear, first of all, which is the difference with respect to the previous equation. Well, this is a second order PD to the derivatives. So it is a completely different object with respect to first our quasi-linear PD that we have studied before. So this is, of course, linear, but first order, one derivative here, one time derivative, one first order spatial derivative, non-homogeneous. For the moment, this is, on the other hand, second order in time, second order in space, and homogeneous. So if the operator is written as follows, then this is the PD operator, and it is clearly homogeneous because there is no f on the right hand side. Don't make the confusion of this equation with another completely different equation, which I can write, for instance, like this. So don't confuse the sign. It is very important here that the sign is minus. Otherwise, if the sign would be plus, and if you take, for instance, c equal 1, then this is simply the Laplace operator in r1 plus n. So it's another equation. It's a so-called elliptic equation. So please keep in mind the sign. So the first arguments that we took for the study of first order linear PDs was to define the characteristic surface because that was the point on the question was, what does it mean given initial condition? And we discovered that we can give an initial condition for this equation provided that there is a transversality, say, between the vector field 1b and the sigma that we are considering where we want to give the initial condition, the hyperplane t equal to 0, for instance. Do you remember this? So if our initial condition was given at time 0, then our vector field b, say, at least in the simplest case, say, in the simplest case, when b is constant, for instance, then should not be tangential to sigma. Do you remember this? So now, and that was important for giving, because it's never immediate to say which are the natural boundary conditions for this PD or that PD. So any PD must be understood with this particular initial condition. So now we want to consider this free equation in the space, in time space. But we want to give also a notion of initial condition. So what does it mean initial condition in this case? That is the first question. And again, we need, as before, for these linear problems, we need the notion of characteristic surface. So it is interesting that also here, there is again the notion of characteristic surface. So let me give you some definitions. So definition. Let me call the operator, maybe the operator L. So let me call this L of u. So L is an operator taking, in this viewpoint, L take a function in some functional space that now, for the moment, I don't specify. So takes the function u into L of u, which is this combination of second derivatives. For instance, if u is c2, this is c0. So it is a linear operator from, say, for instance, c2 into c0. Define. So let me use this notation. So a vector, say, actually a covector xi into r1 plus n, let me split it into, so this actually would be a covector. But I don't want to insist in the geometrical meaning on this, but xi hat, maybe xi prime, xi prime. And so xi is in r, xi prime. So I split xi into the first component and the last components. xi prime are all the remaining components. And so xi 0 square equal c square xi square. So let me, for simplicity, in this argument, take c equal to 1, which is not, for the moment, important. So let me normalize constants so that c is equal to 1. And so let us consider this hypersurface in this space. And this is called, so a vector here is called the set of characteristic vectors for the operator L for L. So this is the set of characteristic vectors. So and I say that let sigma be a hypersurface into rn plus 1, hypersurface, c1 hypersurface, so that I can speak about its normal. And I say that x into sigma is characteristic if the, so I have maybe already used this notation, if the normal to the surface, one of the two unit normals, one of the two unit normals to the surface at x is a characteristic vector. I say that it is non-characteristic, non-characteristic. This point here, x is non-characteristic if sigma of x does not belong to car. So for instance, let us, so example, so first of all, maybe it is important to draw this set here. Can you tell me which kind of hypersurface is this, which kind of hypersurface is this hyperbolic? Do you think so? Hyperboloids, let us, for instance, consider we are interested mainly, as I said, n equal to 1. So let us try to understand what does it mean. Oh, sorry, x1 squared. What is this? Exactly, it's not a hyperbola. What is this? It's a cross. This is a cross. But more generally, if we are in three dimensions, what is this now? In three dimensions, this is the norm, by the way, the Euclidean norm. So this is simply, and what is this? Is that? And so what is this? Now, he says cylinder, no, it's not a cylinder, right? Only half cone, double cone, it's just the cone. So this is a cone in general. Now if you restore the C here, then you modify the aperture of the cone. Depending on C, then the cone change. But if C is equal to 1, this is exactly 45 degrees in two dimensions. If you change C, then maybe you can end up with some different cross like this or like this, depending on C. For C equal to 1, it's just at this 45 degree angles. This is the origin, by the way. This is the origin of this theory. The surface sigma is non-characteristic if all x in sigma are non-characteristic. So examples, exercises, say. So exercise one, take sigma equal this sigma. So you have to imagine this sigma here. Now is this sigma characteristic or not? Thank you. Take, well, then take only half of this, say without the origin. Or just a part of it. Just say this half line or just one line instead. Yeah, C equal 1 just for simplicity. Let us take C equal 1 for the moment. Or if you want to restore the current, well, take this without the 0, however. Remove the origin from the cone because it's a singular point. And the observation is that it's not in hyper-surface in the classical sense. OK, remove the origin from the cone. Take the cone without the origin. You see also here. So now what do we have to do? We have to compute. So this is sigma for the moment. Let us compute the normal to sigma. One of the two normals, say either this or this. One of the two. OK? How can I compute it? Well, let me introduce, for instance, one idea is to look at this as the level set of a function in 1 plus n variables. And then take the gradient. So if I look at this as the 0 level set of a function with non-vanishing gradient, at least out of the origin, then the gradient will be normal to the surface, right? Because the gradient is orthogonal to the level sets. So let me obviously introduce. Sorry, this is xi prime. So my notation was to consider the spatial part of xi, xi prime. So this is xi prime here. This is xi prime. And so two xi primes. OK, so now this is 1 half xi square minus xi prime square. It is clear that now sigma is the 0 level set of this. So the level set of sigma are, how are the level set of sigma? By the way, can you imagine? So this is the 0 level set. Yes, now yes. OK, but it doesn't matter. We are interested just only in the 0 level set. And so we know that now let us take the gradient of psi in the whole space. This is differentiable, of course. Let us take the gradient of psi everywhere. So psi is defined everywhere. This is the domain of psi. But psi vanishes here on the cone. But it is defined everywhere, also outside the cone, everywhere. Now this is xi 0 minus xi prime, right? Now this, of course, is 0 in the origin, at the origin. But out of the origin is not 0. So out of the origin, this vector is orthogonal to the cone. Is it clear? Is this clear? The cone is a level set. The gradient is not 0. The gradient is orthogonal to the level set, OK? Fine. Now, but the gradient, now take a point where, so for instance, in two dimensions, take this point here. This is sigma. And now what happens to the gradient? So this is psi 1, psi prime, this is psi 0. And now take the gradient of psi at this point, which means simply taking the gradient and restrict it at this point, evaluate it at this point. And where is it now? Now I change the sign of psi prime. So I take this psi prime and the same psi 0. And this is again, so this is the normal, which is here with our choice. This is the first example. Now take another example, is what we will always take, will be t equal to 0. Sigma is t equal to 0. And t equal to 0 is, again, this as usual. The normal is this. And in this case, the normal does not belong to the cone. This is more lucky. This surface now is such that the normal does not belong to the cone. In the previous case, the normal was belonging to the cone. As you see here, the normal was belonging to the cone. But now the normal does not belong to the cone. So this is OK. It's non-characteristic. The horizontal line is non-characteristic. The cone, on the other hand, is a problem because you see, take the normal here, then the normal belongs to the cone. So it's characteristic. This is not, is non-characteristic, and the normal is, of course, nu sigma at x is 1, 0, 0, which does not belong to the cone. So this suggests that maybe it is reasonable to take sigma as a surface where studying the Cauchy problems, where assigning initial conditions. So we will never assign initial conditions on surface sigma like this, because the normal to this belongs to the cone. But we will assign initial condition on sigma like this. So let me summarize. This set here, sigma equal t squared equal to x squared is characteristic. So let us forget the origin for the moment, which is not so important, but sigma t equal to 0 is non-characteristic. So we will consider having the experience of first order equation. So we will consider u t t equal, say, c squared laplace of u in, as I said, one could also work for backward in time. But let us consider for the moment just for simplicity. And then u 0 equal, say, 0 bar, u bar, u 0 bar, maybe, at t equal 0 on sigma. Now, the point is that one in this kind of PD, one has to assign also on sigma not only the initial position, the initial condition, but also the initial velocity. This is because the PD has the following structure. u t t equal something like this. Because here there are two times derivatives. The time derivatives, as we will see, requires a natural way to impose initial conditions, like for ODEs, in some sense. If you have a second order ODEs, you specify initial position and initial velocity, usually. Now you have not, this is not an OD, this is a PD, but it has two derivatives in time, and therefore it seems reasonable to specify not only the initial function at time 0, but also it's time derivative at time 0. So sigma now is non-characteristic. And I will assign u bar 0 and u bar 1. These are given, smooth enough. So now the point is to try to solve this, OK? But before doing this, I want to mention just one big theorem. Theorem. And for instance, you can look at the proof. For instance, in the book of Evans, and it is called Koshy Kovaleskaya. The theorem is called Koshy Kovaleskaya. And it says the following. Assume sigma non-characteristic. Actually, one could give this in a local version, but for simplicity, I don't want to state it locally around the point. I state it globally just for simplicity. So assume that sigma is non-characteristic and analytic. Assume u 0 bar, u 1 bar, analytic, real analytic. This means that around each point these functions coincide with their Taylor series around each point. And this means that you can look at this, for instance, in a parametric way or as a graph locally of an analytic function, as usual. So everything is analytic. And assume that you want to consider utt equal f of t x ut grad u, say, h of u, say space h, for instance. All second derivatives with respect to space. So the important is that here on the right hand side, you don't have a second time derivative here. And you don't have derivatives of u of the order higher than 2. For instance, in one space dimension, f, t, x, ut, ux, uxx, for instance. So on the right hand side, no third derivatives, not second derivative with respect to time. Just in place. Yes. And u equal u 0 bar and ut equal, say, for instance, say take this t equal to 0, for instance, ut equal u u 1 bar on sigma. So if everything is analytic, and f also is analytic, assume that you consider f analytic in its arguments, its arguments, utt equal this. Then, say, take this for simplicity. Then this problem, one, has a local in time solution, has a unique one, has a unique analytic solution u, solution u for short times. What does it mean? It means that sigma is clearly analytic in this case, because it is a hyperplane. If I give you this problem with initial condition, which is analytic, and initial velocity, which is analytic, and this is clear, this f is analytic in its argument, so then this has a local in time solution, unique which is analytic. I don't prove this theorem here. It's not easy, it's long, but just to let you know that, at least, when everything is analytic and sigma is non-characteristic, that we have a local solution. One can give this statement around the point of sigma without assuming this, but taking just sigma and non-characteristic, and so locally, you have a unique solution. But I don't want to insist on this, because I don't prove it. Just to let you know once that there exists this theorem, which is actually much more general than this. It is actually, it holds for a very large class of PDs, not only for kind of wave, sorry, this is called the wave equation, by the way, wave equation. This is called the wave equation. And this is a typical example of hyperbolic second order PDs. Hyperbolic second order linear PD. So this theorem, I was saying, this theorem is valid not only for hyperbolic type PDs, but it is much more general than this. But you need the analyticity of everything, initial conditions, sigma, f, and so on. If you don't have analyticity, then you cannot use this theorem. So this is just to let you know that, OK, even for PDs, in a large class of PDs, if everything is locally analytic, then you have a local solution analytic. Then you can expand in the Taylor series. It's not an easy theorem. And if you want, if you are interested just for curiosity, then, for instance, you can look at the book of evidence. t equals 0. So yes, thank you. What does it mean, existence for short times? Oh, it means that the PD is satisfied, for instance, in 0 t times rn, say, for some t, but not in principle for all t's, just in a strip. Ah, yes, but times is here. All i's, OK. So to study this PD, it is better to start the first one-dimensional space case. So we start with n equal 1. n equal 1 means this, OK? n equal 1 means this problem here. So now, for the moment, consider just the PD without conditions, and there is the very interesting structure of this PD is the following. In one space dimension, we have that this can be split into the composition of two operators, OK? And so composition of, so let me introduce v maybe. v, let me follow this notation v. v of tx equal du over dt minus c du over dx. So let me introduce this, and let us observe that dv over dt plus c dv over dx is equal to 0. So remark, assume that you have a solution, OK? Assume that you have a solution. You define v as follows. Then v satisfies this. Let us check this, d over dt. Indeed, d over dt u t minus c ux t plus c u t minus c ux plus c. So this is this. And so what happens? You see u dt. This is u dt minus c ux t plus c u tx. Well, we are talking about C2 solutions, OK? So the problem is also, let me write also the regularity class. So we look for a solution, which is u. Say in C2, 0 plus infinity times r, so that everything is smooth enough. Assume that u0 is maybe C1 or C2 also. Assume that this is C1. Let us take this. So let me go back here. So u dt minus c ux t plus c u tx minus c squared uxx. Since by assumption, u is a solution. And this C2, u tx and uxt are equal. And so this is equal to this. So this means that this operator, L of u, actually splits into the composition of two operators. So L of u splits into, first you do, say, this, and then you do this. So it is the composition of two operators. So L of u is, say, d over dt plus c d over dx of d over dt minus c d over dx of u. This is a little bit formal, but it is clear what it means. So first I do u t minus c ux here. First thing to do, for instance. And then to what I obtain that I have called v. Yes, this is v. And then to v, I apply this other first order operator. You see? So splitting property, this is an extremely important fact. So splitting property. So the operator L, remark, it's interesting to observe that L splits into the composition of two linear first order PDEs, operators, first order operators, first order operators. But which kind of operator? You see? This is linear transport operator with constant vector field. And this is, again, another linear transport operator, first order with the minus c, is now replaced by plus c. So first you apply a step in which you do a linear first order with minus c. And then to the result, you apply, again, linear first order with plus c. And this is very, very, very nice property of this second order operator. And this, in one space dimension, this is the trick that allows to find the solution. Now we will see how. So let me keep the blackboard. Let me keep this and this also. OK, this is not necessary anymore. We can start from studying this. And what do we know about this? We know the solution. So start, we start with this. And so now we know that v of tx, it is equal to what? Maybe we could write, say, v at time 0, just to be precise. And then we chart the, now, what is what? Now, exactly, x minus ct, do you agree? So let us call maybe the notation, maybe a, a of x, let us define it as v of 0x. Define this just to simplify notation. So this is actually simply a of x minus ct. Of course, this a is neither u0 bar nor u1 bar. You see? This is not. Because at the end, what is this? So I have to put, say, time 0 here. And so here you see there is 1 u1 bar minus c something. OK? So for the moment, let us just simply define this. So let me write it just not to forget it, the notation. So a of x is defined as v of 0x. And so v of tx is this. OK. Now, so we have solved this for v. And therefore, now, our problem is however this, which becomes ut minus cux equal a of x minus ct. Right? And what is this? Now, this is linear transport equation with constant coefficient, but, however, is non-homogeneous. So we need the solution. We have studied this with constant coefficients, but maybe it is better to, if we are able to remember it, maybe we should write it here. If I'm not wrong, so if I take ut plus bux, say v dot at u equal f of tx and u of 0, call it u and phi, then the solution was phi of x minus bt plus the integral from 0 to t. And then we have now f of tau. Let's please check this, the tau. Check, please, this formula. Is it correct? OK. Fine. Please check it, because otherwise, if it is not correct for some problems in the sign, then we don't arrive at the end of the reasoning. So this is linear transport with constant coefficients, but non-homogeneous. OK. Is it correct? OK. Fine. Now, therefore, now we can apply this to our problem. Our problem is this, exactly. And I simply have to change letters, symbols, and to keep a little bit, pay attention not to make mistakes. So this means that we know that the solution u of tx has the following expression. Is the initial condition, now b is, of course, minus c. OK, b here is. So when there is b here, I have to put it minus c. Look, b minus c. So change b into minus c in this formula. Take one dimension, space. And then, so phi is the initial condition. So now, finally, phi is actually u0 bar. So u0 bar. Now I have to take this. And then I have plus the integral from 0 to t. Then in place of f, I have, so let me write from here. I don't want to, I try not to make mistakes. So f of tau x plus b tau minus t is, in our case, is equal to what? Is equal to a of x plus b tau minus t. But b actually is minus c. So this is the new x. Is the new x is, and then, minus ct. This is our, if I'm not wrong. So let me check it. So our f is this, depending only on x, on the space variable. So f of tau, this is the object. So f of tx is equal to a of x minus ct. And therefore, f of tx is this plus this minus this. Ah, thank you. Thank you. Thank you. Thank you. Thank you. Let me check it if it is correct. Minus c tau. Yes, this is thank you. So now this is equal to a. And then I have x minus 2c tau plus ct. Therefore, our solution is a of x minus 2c tau plus ct in the tau. Now, let me change variable here. So let me now make a change of variable. And so x minus 2c tau plus ct, I call it for the moment, say, y. OK, I call it y. And therefore, u tx, now this integral, the integral from 0 to t of a x minus 2c tau plus ct d tau becomes a of y. Then d tau is minus 1 over 2c. So and then 1 dy. And then when tau is equal to 0, this is x. And then when tau is equal to t, this is x minus ct. And therefore, actually my solution u, read as follows. Now I raise this part. So I end up with u of tx equal to u0 bar of x plus ct plus. There is a minus here, so I change the integral. So 1 over 2c x minus ct x plus ct a of y dy. So what remains now? Now it remains exactly. It remains to express a in terms of u0 and u1. And therefore, what we know about a? So we know that a actually is what? Is u bar 1 minus c u0 x? Because you see we have u is equal to u0 bar on sigma. So on sigma, they agree. By smoothness, u actually is smooth up to sigma. So the derivatives of u in this direction is equal to derivative of u0 bar in the same direction. So our solution actually is smooth. And so a is this. Do you agree? By smoothness, u is c1 up to t equal to 0. And we can therefore almost, we are almost at the end. So utx actually is equal to what? To u0 bar of x plus ct plus 1 over 2c, the integral plus ct. And then I have u1 prime bar y dy. And this I keep alone. And then 1 over 2c. And then I have x minus ct, x plus ct. By the way, now we have, there is this minus c actually, OK? So I erase this. And then apparently there is also a minus. And then I have u0 x bar y dy. And then I end up with what? With 1 half u0 bar x plus ct. Because this 1 half cancels 1 half of this. Plus 1 half u0 bar x minus ct, plus 1 over 2c integral from x minus ct. This is a very, very interesting formula, dy. Let me check that I have, it's correct, surely correct. But yes, it is correct. So we have proven the following theorem. So we have proven the following theorem that it is called, no, well, this formula is called the Dallenbert formula. Dallenbert formula. Maybe Dallenbert, Dallenbert, Dallenbert. OK, the theorem that we have proven is the following theorem. So let say u0 bar in c2, u1 bar in c1 of sigma, define u of tx as 1 half u0 bar of x plus ct plus u0 bar of x minus ct, plus 1 over 2c, x minus ct, x plus ct, u bar 1, y dy. Then u is c2 times r. You see c2, because u0 bar is c2 by assumption. u1 is, u bar 1 is c1. So it's integral, actually, c2. And that's important, because we want to solve a second or the other PDE. So utt, uxx, we need c2, is a solution to our PDE. And say the limit as tx goes to 0x of utx is equal to u bar 0x limit as tx goes to 0x utt, tx equals u bar 1 of x. This is called the Dallenbert formula. It's remarkable to have such an explicit expression of the solution, rather remarkable. Remember, it is essential. We have used, in essential way, the one space dimension. The problem of having such a kind of formulas in two, three, four dimensions is another story. But in one space dimension, at least, we have this. Yes, it is unique. It is possible to prove that it is unique. That's right. You write here that it is a unique solution to our PDE. Yes. Indeed, you can prove the following. So the question was, is this solution unique for remark? So you see, what we have done up to now in this course is to look for more or less explicit formula, more or less explicit, maybe sometimes implicit formula for representing solutions. This is not possible, in general, for any PDE. But these are linear PDEs. And therefore, it's very interesting to start with a special solution, with an explicit solution, so that you can understand the qualitative properties. It's extremely important. Then, if you have a more difficult problem, for instance, assume that you want to solve this PDE, but with strange boundary conditions, or instead of the wall half space, just only in a piece, then maybe you don't have an explicit solution like this. And so you cannot hope to have such a kind of formula. But at least you know that in these special simple cases, you can understand a lot of things from this. And then you can hope that in more general cases, maybe qualitatively, there are the same properties, maybe. It's always important to have the largest possible number of explicit solutions at hand. This is very important. Remark concerning your question, then maybe this is some work is not so difficult. I think that assume u in C2 is a solution to utt minus C square uxx equal to 0, then necessarily in, say, 0 plus infinity times r, then necessarily there exists two real functions, fg of class C2 from r to r of class C2 such that u of tx is equal to f. Once you know this, this is not difficult, by the way. Why it is so? Because the operator splits into the composition of two operators of the form x minus Ct and x plus Ct. So this is homework. It's not difficult at all. This is very easy. But once you know this, then it turns out that necessarily f must be this. Now if you add initial conditions, now add your initial. This is in general without conditions on the boundary. This is in general, in the half space. So in the half space, a C2 solution must be the superposition of something traveling on the left. C is positive. And something traveling on the right. I will be more precise on this. Superposition of these two objects. Once you know this, if you want to solve your PD with u0 equal u0 bar at time 0, derivative at time 0 equal u1 bar, then necessarily you end up with this. This is uniqueness. This is uniqueness. Imposing boundary conditions, initial conditions, initial conditions implies that u is s. So now let us try to make some exercises on this expression. So I keep, maybe I keep, OK. So you see, this is of course of the form f of x plus ct plus g of x minus ct. No, not necessary. Not necessary. Not necessary. u0 x plus ct. So you claim that f of g are equal? No, I don't think so. Because you see? So who is f now? f equal 2. f is equal to, y is equal to? Yes. u0 bar, y plus 1 over 2c, integrals from x to y. So x now, maybe you have to just fix some point, 0, y. I mean, you have to just to fix. Because this depends on y, you know? Ah, yes. OK. 0, y, say. u bar 1 of s in ds. OK. So this is f and g. It's the same you claim. Maybe you're right. Let's see. So if I take now f of x plus ct plus g of x minus ct, what do I find? I find. This is correct, of course. And then 1 over 2c equal to 0, y, sorry, 1x plus ct, plus integral 0x minus ct. But then there is a minus here. Minus x minus ct 0, u bar 1, u bar 1. And so I have just the problem of the sign here, right? There should be plus. Let me check one. So this is the part concerning this is this, from 0 to this. And then if I take plus 1 to c, 0x minus ct, u bar 1, y. Is this right? If I just sum them. So it seems to me that, well, the idea is OK. In order to construct f and g, you take this. This is OK. And then I take a primitive, or u 1 bar exactly. So having only f would, it is already interesting to take initial velocity equal to 0. It is already very interesting to start the problem with initial velocity equal to 0, which is only f, as you said, essentially. OK. So now, let us just argue. Let us just, so for the moment, let us just try to understand the proper qualitative properties of our wave knowing only that it has this particular structure. So the exercise, so let me write down the following exercise. So this is z equal g of x, z equal g of x, z equal f of x, for instance. And then, OK, ct. So c is positive. So this is the graph of capital G, say. And therefore, this is the graph of g of x minus ct. This is the graph of g of x minus ct, moving in this direction, on the right, with velocity ct. So this is a sort of wave. You can think of this. So you have the form of the graph of g. This is given. And then your solution, the part of your solution is something which travels on the right with constant velocity on the right, given by c, keeping the same shape. So it translates on the right, like this. This is, on the other hand, something which translates on the left with keeping the same form of graph of f. This is the graph of f. So it is the superposition. Our solution is the superposition of two objects traveling. For instance, if the initial speed is 0, which is more easy to understand, if the initial speed is 0, then this is not present anymore. And we can think about f to be the initial condition, u0 bar, essentially. Essentially. And so you have u0 bar here and u0 bar here. Just only the superposition of u0 bar of x plus ct and u0 bar of x minus ct. It is, to understand the Dallenbert formula for me, it's interesting to analyze the case u1 bar equal to 0. So that these are, as he said, they are equal to u0 bar, 1 half, 1 or 1 half. So I leave you some homework for now. The next week I will not be here. So we will continue. So you have a lot of time. So maybe it is better to give you some exercise. So homework 1, u of t0 x0 depends on the values of u0 bar, u1 bar, only in x0 minus ct0 x0 plus ct0. It does not depend on the global properties. You have to think about this. This says, this is very easy to see from your solution. It says that your solution does not really depend on the global shape of the initial conditions, but the solution here does not depend on how u bar 0 and u1 bar are very, very far, but just only on an interval, which is this. So the value of u is insensitive of modification of the initial conditions out of this. If you change the initial conditions out of this, you don't see it at the level of the value of u at that point. Yes, I know. It is very easy, but it's immediate. Think about this. Write down the interval if you have here, say, t0 x0, write down here where is the interval. And you have to meditate on this, because this is an essential property of wave equations. It's immediate, I agree. Second immediate exercise. It's very easy. So fix an interval of the form x minus a less than or equal than r x-axis. Fix this interval. Draw the set of tx such that u of tx depends only on the values of u0 bar, u1 bar on the interval. So here, you have, I'm sorry, this is a homework. So third homework, draw and prove that this is equal to x minus a less than or equal to r minus ct. Maybe we can go also backward in time. More interesting exercise is the following. Take u, so assume u1 bar equal to 0. Take u0 bar of the following form x plus alpha if minus alpha is less than or equal. Minus x plus alpha if 0 less than or equal to x less than or equal to alpha and 0 else. Depict, draw, say depict, the graph of u tx at times u t dot at times t equal to alpha over 4c, alpha over 2c, 3 alpha over 4c, alpha over c, 5 alpha over c, and larger. And by the way, maybe a remark, this initial condition is not c2, as you see. It's just leapsheets. So this creates a little bit of a problem because our formulas for the moment work only for u bar 0 in c2. This is the corners in the graph, in these corners. By the way, the Dallenbert formula is OK also in this leapsheets case. So the exercise number three says, take the Dallenbert formula as it is also in this leapsheets case and draw the solution for these times. And then, suppose this is alpha 1, this is alpha 2, characteristic lines. These are called the characteristic lines of this x minus ct equal to constant. Draw the characteristic lines. Suppose u bar 0, in this case, again, is equal to 0 in alpha 1, alpha 2. Then divide this into six regions, time space into six regions, 5, 1, 3, 4. This is time space. Here, the initial condition is 0 in the complement of this. Otherwise, it is 3 in the complement of this. So here you have superposition of two ways. Try to look at that number three exercise. To interpret the number three exercise into the time space graph here, try to realize that here you have superposition of two waves. One going to the right, the other going to the left. Here in region two, here it is 0. Try to prove that the solution in region six is 0. Try to prove that the solution in region five is 0. Try to prove that the solution in region four is 0. Here, there is superposition of two waves. And here, there is just one wave. And here, there is just one other wave. So here, there is just direct wave. So called direct wave, just that one traveling to the right, and the other is called the inverse wave. So if you look in time space, there are these six regions, where here you have superposition of a direct and inverse wave. They sum up together. Here you see just in region two, you see just only one wave. Here you see the other wave only. And in the other regions, the solution is compass is 0. So this is interesting, this exercise, because the first part, you need to understand the shape of the solution. Given a time, then you depict the graph. And for that time, the graph in space, a fixed time. But this is a more, more, is another viewpoint. So the previous one is just the time slice of this. So if you cut this time, say alpha over 2c, you draw a horizontal line at that time. Then the graph you're looking for there is just what you see from here at this time. But this is in time space, so it's more complicated. It's more global. Yes, u1 bar is 0. This is still exercise 3. This is a continuation of exercise 3. And u1 bar 1 is 0. Yes.