 It's time to start. We have the uniqueness theorem to conclude the proof. So the theorem was, so let me rewrite the statement quickly. Assume omega is open, bounded, smooth, and assume that u is a C2 solution. And I will consider and consider the problem u tt minus C squared Laplace of u equal to 0 in 0t times omega. And then we have the boundary condition. So u equal h on 0t times the boundary of omega. And then u equal u 0 bar, say, in C2. And u t 0 u 1 bar in C2 at t equal 0. We're assuming also compatibility condition on h u 0 bar and u 1 bar plus compatibility conditions. So that we are discussing about C2 classical solutions. So assume and consider that problem. Then one has at most one solution. So let me recall that now this is a uniqueness statement. So nothing is said about existence. It is a uniqueness statement. It does not pretend to construct a solution. And what we have done up to now was in one dimension. Here we are in n dimensions. But we studied the problem on an unbound domain. Because in one dimension we studied the case omega equal to r. And therefore, this statement does not apply to the one dimensional case. So the proof was the following. First of all, we remember that we have the classical integration by parts formula. So if everything is smooth enough, omega is bounded, smooth, and so on, then we know that u f divergent, sorry, maybe you can also consider the non-homogeneous case. Say C2 also, but C0 would be enough, but smooth enough. OK, so let me call phi divergence of eta dx. We know that this is equal to minus grad phi eta dx plus the integral over the boundary of eta dot nu omega phi dh n minus 1. In particular, we know that therefore, if eta is a vector field which is a gradient, so if eta is the gradient of something, say eta is the gradient of v, say, then this formula becomes phi Laplace of v dx equal minus integral omega scalar product between grad phi and v dx plus phi the directional derivative of v in the direction of the exterior unit normal. So we know this. This is preliminary, so let me keep just this form of the integration by parts formula. And so let now u1, u2 be two classical solutions to problem one and define w as yesterday the difference. This is a definition. So w satisfies by difference. This is a linear problem. So wt t minus c squared w v equal to 0 and v equal to 0, v equal to 0, v equal to 0, v equal to 0 on the lateral boundary, say here at time maybe w0 at t equal to 0 and wt 0 equal to 0. So it satisfies the homogeneous problem. Everything is 0 initially, initial position, initial velocity, and also the condition on the lateral boundary of the cylinder. So let us introduce for any t0 capital T. Capital T is given also positive. Let us introduce this quantity. Maybe we can put enough. So we have wt squared plus the gradient of v tx squared dx. Therefore, we are looking essentially at the energy at time slice. So this is omega. And then we have capital T. And so at some time, we are integrating on this slice these quantities. So we are integrating at this level. So for any t, we can compute the derivative of this quantity with the aim of proving that this is 0. So we can differentiate under the integral sign so that we find wt wt t plus scalar product between the gradient of v and the gradient of v t. So wt, all objects here are evaluated at tx. And I have already used the fact that the derivative with respect to time of the derivative with respect to x commutes. So I have already made this passage so that this is the gradient of v t. In this form, we can apply. So notice that this is just an integral in space so that we can apply this theorem where now t is considered as fixed. So in this form, there is no time. But this now can be considered as an object like this provided that t is just a fixed parameter. And therefore, I can apply this with the following choices. So this remains the same. And then wt t. And then I apply this with the following choices. Phi equal to w at the time t. So phi of x is w at time t fixed x. And v is now wt at fixed time t considered as a function of x. And therefore, this quantity is exactly the first addendum up to the sign on the right-hand side. So this is equal to minus this plus this. So let me put the sign. So let me put the sign. OK, so before we have that this is equal to, let me write it here specifically. So grad v, grad vt is equal to minus vt Laplace of v plus the integral of vt dv d nu. So let me check that this is correct. Maybe it's this. So let me check once more. So v is Laplace of v, grad v, dv dv d nu. OK, this is OK. Is it OK? OK, so now by replacing this, the x plus the integral over the boundary of omega, wt, scalar product between grad v and normal. Now c squared, c squared. Now this is equal to 0 inside. Therefore, for any t, this is equal to 0. We can take c equal to 1 if you want. Doesn't matter. Because, yes, because I have to put the c square here. Yes, yes, yes, yes. Thank you. Because yesterday we decided to take c equal to 1. OK, thank you. So let me put c squared here just to, OK. So c squared, c squared, c squared, c squared, c squared. And maybe there is a c squared. Where is it? A c squared everywhere here. Fine. So this is equal to 0. Not only this, but also we know that w is equal to 0 here. The lateral boundary is equal to 0. And therefore, wt is also equal to 0. w is equal to 0 in the lateral boundary. And therefore, wt is equal to 0. Therefore, this is also equal to 0 on the lateral boundary here on the, how can I say, on the boundary of the portion of the cylinder. Hence, for any t, 0 capital T, we have that e dot t is equal to 0. OK. This implies that et, by the way, et, this just for simplicity, we have made this computation in the open interval, just to avoid the right and left derivatives. But this implies that et is equal to e0 for any t. Now, e0, now e0, OK. So we have this. This is constant with this. Now, e0, what happens to e0? Everything is smooth enough. So what happens to e0? Let me simply observe that e0 is 1 half wt0 square plus e0 square. But again, w is 0 on the bottom. And wt is also 0 on the bottom. And grad v is 0 on the bottom. And therefore, we obtain that for any t, 0 capital T, et is equal to 0, which says, therefore, that wt is equal to 0 in the interior of the portion of the cylinder. And the gradient of v is also 0 in the interior. Because this is, I mean, if this is 0, necessarily this is 0 everywhere and everywhere on the slice, on the time slice. But this is true for any time slice. I mean, we have shown that here wt and grad v are 0 just on this time slice. But then this is true for any time slice. And therefore, it is true inside everywhere in the open space. Hence v is constant. Hence v is constant. But v is 0 on the boundary. And therefore, necessarily v must be 0. This concludes the proof. Because if v is 0, this implies it's equivalent to say that u1 is equal to 2, OK? So we have the proof of the uniqueness theorem. So you see, for the moment, what we have done in this course is the following. We have considered a class of PDs, maybe first order. Now it is second order. We have found some more or less explicit expression of solutions under assumptions on the domain, on the initial conditions, on the non-homogeneity, and so on. And then we have only considered smooth classical solutions. And then we have made some theorem on uniqueness, but nothing on existence for general domains. This is for the moment the spirit of the course. So what we have done is we have a PD. Maybe we have studied first order, quasi-linear. And we have studied second order model of the wave type, linear, wave equation, linear. Then we have found explicit as most as possible solutions so that we can deduce qualitative properties under, however, explicit classical solutions. Classical meaning C1 in first order and C2 in the second order case. Classical solution, we have studied qualitative properties. And under, of course, assumptions on F, which is the right hand side, initial conditions, initial conditions, and importantly also on omega. Maybe in the whole line, for instance. And then we have said something general, however, now more general uniqueness result. So it is clear that in this picture there are several things missing. In particular, what is missing is construct existence of solutions in some class for bounded domains, for instance. It is immediately clear that we have never touched this problem. So this has been our approach because the field is very huge and so one have to choose exactly some topics. Now, yes, yesterday maybe I leave you an exercise, exercise homework, which was assumed that you have to consider in one space dimension. In the half line, you have utt minus c square uxx equal to 0 in the first quadrant. Then you have your initial conditions, say u0 equal u0 bar, smooth enough, c2. Ut of 0 equal u1 bar, c2, c1, maybe, enough. And now we have a boundary. So this is x. So we are working in this first quadrant. And we take, for simplicity, u equal to 0 here plus compatibility conditions. Assume also, assume compatibility conditions on u0 bar, u1 bar, so that what we are looking for will be c2 up to the boundary of the quadrant. And then we want to find the solution to this, the solution, actually. So one way can be the following. So extend u0 bar, as follows, on r. Because you know, this is c0 just on the half line. And this is c2, sorry. And this is c1 just on the half line. Now we can extend u0 bar, call this uxt u0 bar x. Define it as follows. So u0 bar x, if x is bigger or equal than 0, one compatibility condition on u0 bar says that u0 bar 0 is equal to, u0 bar 0 is equal to, by a sum. I mean, u0 bar is defined here. It continues up to the boundary. And its value at 0 must coincide with our condition that we assign coming from this direction. And coming from this direction, we are assuming 0. So one compatibility condition, one of the three compatibility conditions is this. Minus u0 bar of minus x, if x is negative. Then there is also the other compatibility condition on the derivative, which is 0 and 0. And then also the second derivative. Anyway, we have written this compatibility yesterday. So this is an even extension, sorry, an odd extension on the wall. We also extend u bar 1 as the same. So u bar 1 x is u bar 1 x, if x is bigger or equal than 0. And minus u bar 1 minus x, if x is less than 0. Also, this is an odd extension. So that now we can consider the problem. So let me denote it maybe by u tilde. We can now consider the problem on the wall line. By compatibility, everything is smooth enough. So we can consider now a new problem, u tilde tt minus c squared u tilde xx equal to 0 in t. But now we have this. And then u tilde 0 equal u0 bar x. And u tilde t of 0 equal u bar 1 x. Because now our initial data are given on the wall line. Therefore, we know, and they are smooth enough by compatibility, c2 and c1. And therefore, we can write down who is u tilde tt. So u tilde tt of tx is equal to 1 half. Then we have u bar 0 x of x plus ct plus u 0 bar x x minus ct plus 1 over 2c integral. u bar 1 x s ds. So this is, sorry, u tilde tx. So this is the expression of our solution. So this is an old fashioned problem. It is the problem that we have studied on the wall line two lectures ago. We are allowed to do this because we know that the initial conditions are defined as smooth enough in the wall line. And so we know that the solution of problem, so this is problem say 2. And this is problem say 2 extended. So the solution to 2 extended to x is this by the previous lectures. Now we have just to see how this solution looks like on the first quadrant. Because this on the first quadrant will solve this. So what do we have to do? Is to restrict this just only on the first quadrant. And we will have the solution. OK? Oh, by the way, notice also that u tilde of tx is also minus u of t minus x. u tilde tilde tt minus. So u tilde tt at tx will be minus u tt at t minus x. And u tilde xx at tx will be minus u xx at t minus x. I mean, if I define u tilde as this, if I define this, then I have that u tilde tt minus c square u tilde xx is equal to 0 on the wall half plane. Because here you see the minus remains. And also here the minus remains. And therefore, when I add these two with the proper sign, the same is true. OK. So we have to restrict our u tilde just to. And maybe this could be home. Or let me see if it is so difficult now. Homework. So for instance, when x minus ct. So x bigger than 0, t bigger than 0. And also this. Then we have that u of tx under these assumptions is just 1 half. Then, OK, this is positive. So the extension coincide with I can remove the x here. This is positive. And therefore I can remove the x. And these are both positive. And therefore I can remove the x. So at least we have shown that if x is positive, t is positive. And also x minus ct is positive. Then our solution is u0 bar of x plus ct plus u0 bar of x minus ct plus 1 over 2c integral x minus ct x plus ct u1 bar sds. Now what happens in the remaining case if x is positive? t is positive, but we are in this region. Now we have this solution. This is true for any without restriction. So now x is positive and t is positive. So surely this is positive. And so here I can remove the x. However now this is negative. So this is negative. And therefore u0 bar x is minus this with the opposite sign. Minus u0 bar of ct minus x. Now maybe I can leave you this home. It's not difficult. I think u1 bar of sds. It's not difficult because you can write this. I show you the hint. So let us consider just this part, just the integral part. So let me rewrite it as the integral x minus ct0 u1 bar x sds plus integral from 0 to x plus ct d u1 bar x sds. This is just this integral part. Now we know that x plus ct is positive. Hence we can remove this. Now what is this? So this remains the same. Now what is this? By definition is 0 minus. Now I recall the definitions. This is negative. This is 0. So I'm integrating in negative region. And so I have to change. I have to use the definition of this. So minus u1 bar minus sds. OK, now I make a change of variable. I call sigma equal minus s. And therefore, I get the integral from ct minus x and 0 u1 bar sigma s. Let me call now sigma, let me recall it sds. And so you see now when you sum this with this, you end up exactly with this. By the way, so the exercise is complete. Well, a remark is that this exercise is more realistic than, so this situation of a half line is maybe slightly more realistic than the whole line. But what it is interesting about this is that this result, this expression, allows to find useful. Let me just remark this. I don't want to enter the details because the course is very huge. And we don't have time to do too many things about just one equation. We have to say several things about various different equations. But the remark is that this formula, say formula star, is useful for finding explicit solution. The explicit solution of utt minus c square Laplace of u equal to 0 with initial conditions in, say, 0 plus infinity times r3 plus initial conditions. What I'm saying is that this formula, if you are interested in, you can look in the book of Evans. If you want to find the solution now on the free problem, the whole space in three space dimensions, then you can use passing to a tree on polar coordinates and so on. Passing to polar coordinates, you can use this result to attack the problem of finding explicit Dallenbert type, sort of Dallenbert type expression, but much more complicated in this more physical situation, three space dimensions. So finally, we can study the problem. So maybe at home work also, let me see, home work maybe. So let L be positive number. And let us consider the problem of, so this is the last exercise on the wave equations, just the last one. So utt minus c square uxx equal to 0, t, say, bigger than 0, and x in between 0 and L. So we have really a string. Now, this is realistic. We have a string of finite length. And I want to see the propagation of waves inside this string. But the problem is, of course, the string is finite. So when a wave goes, for instance, in this direction, then there will be a boundary. It bounces back in this direction. So there is this phenomenon, very interesting physical phenomenon. So this is the wave equation. And then I have the initial condition, as usual, u0 bar with the smoothness, smooth enough, ut0 u1 bar. And then we keep the string fixed at the two extreme. So the string is 0 Dirichlet boundary conditions. So for any time at 0, we have 0. And for any time at L, we have 0. So these are 0 Dirichlet, so-called Dirichlet boundary conditions. So we have a sort of function like this, say, 0 and L. And now the idea is to extend it as follows, for instance, to L in an even way, and then periodically, continue periodically. So this is just and so on. OK, so we start in an interval 0 L. We know that u is 0 here and here and here. So we can continue in an odd way here and here. So we find now a 12 periodic periodic. So this we can do for u0 bar and u1 bar. So u0 bar of x minus u0 bar minus x for any. And then u0 bar x minus u0 bar 12 minus x. OK, there are the compatibility conditions as usual. So we extend u0 bar in this way and also the same for u1 bar. So now the idea is to look for a solution to this problem. Now you see we have periodic functions, u0 and u1 smooth enough. So that's smooth enough so that what I will say will converge. Is there something that comes in mind when you have such a periodic situation? Is there something? Can you repeat the question? So now you have a 12 periodic function, u0 bar and u1 bar. The same for u1 bar. So what it is natural to do when you have a linear equation and periodic conditions. Yes, yes. One possibility is to write u0 bar and u1 bar in Fourier series. OK, so u0 bar of x. Now this is odd. And so let me denote it by an and pi x over l for any x to 0l, for instance. And the same with u1 bar. And let me use the notation bn pi over lx, u1 bar and u0 bar. OK. Where n over 2, 1 over 2l minus 0 and 2l of u0 bar s as sinus 1. OK. And the same for bn. The same for bn with u1 bar here. Now what says the exercise? OK. We know that this operator splits into the composition of two operators. So we know that we have the formula, the usual stuff. So u of tx is equal to 1 half u0 bar x plus ct plus u0 bar x minus ct plus 1 half 1 half x minus ct x plus ct u1 bar sds. Because we have extended everywhere our functions. So we know that now on the whole line we have such a kind of solution. And then the point of the exercise is to rewrite this expression of the solution in terms of the coefficients an and bn. And so the exercise says prove the following. So prove that, so the expression of the solution u. So prove that u of tx is the following superposition. Is the sum an cosine pi n ct over l plus bn sine of pi n ct over pi n ct over l, OK? Sine, OK? Where an is equal to 2 over l integral sine of pi n s over l, yes? n bn equal to pi n c. I hope that the numbers are correct. Sine of pi n s over l, yes? OK, prove that u of tx has the following expression. What you have to do is, so u0 bar has this expression. However, here you have x plus ct and x minus ct. So you have to use the formula sine of alpha plus beta equals sine of alpha cos beta and so on. Here also you have u1 bar, which has this expression. And you can integrate the series term by term. So you can integrate. So you can, here you have a series. But the series goes outside the integral because we have smooth enough assumptions on u1 bar so that the series is sufficiently strongly uniformly converging so that you can interchange the series and the integral. The smoothness assumption is useful also for this. And so you can put the series outside the integral. So you can integrate term by term. And the integration is easy because it's very easy to find a primitive of this. Once you have done this, then I think that it is just a matter of cupping together the terms. And then you should have this expression. So the exercise is just maybe not so easy, but just a computation using the fact that you can interchange the integral with the series. Fine. So this is, I think that we have concluded now the wave equation. And now we will change type of equation. So this concludes this part. Now we want to go through the study of another equation, so the so-called heat equation. So let me now make some quick introduction. So the heat equation is now completely different from the previous one, even if apparently it is very similar. So k here is a constant. k is a constant. k is a constant. And for us, it is very important that k will be always positive. So keep in mind, in particular k equal 1 will be the usual k's. So k equal 1. If k is equal to minus 1, this is another style. If k is equal to minus 1, so avoid to consider the case k equal to minus 1. So remember that the coefficient in front of the diffusion of the Laplacian here is positive. So it is very important not to change sign in this coefficient. Now, apparently this is a very similar. If you put here c squared in place of k, you could think that this is similar to the previous one. Actually, it's a completely different equation because here, the difference is that you have just one derivative in time instead of two derivatives in time. So this is a very big difference. So completely, we will see that qualitatively very different from utt minus c squared Laplacian. So the equation is called a hyperbolic equation. While this is called a parabolic equation. It's called a parabolic equation. Now, remark, maybe first remark. Remark one is that if u is independent of time and solves your PDE, so if you find for some reason a solution of this, which actually does not depend on t, then by solution, I always mean a classical solution, minus Laplace of u is equal to 0. And this is another gain very different PDE, which is called elliptic equation. So stationary solutions, where stationary means that it is independent of time, stationary solution of the heat equation solves an elliptic equation like this. This is just remark. What we will do in the course, we will study a little bit solutions to this and then a little bit solutions to this. This is the program of the next, say, four or five lectures. So there is a relation, an immediate relation between this and this, at least. Stational solutions to this solve an elliptic problem. Notice that once more that this is not taking c equal 1. This is not Laplacian, minus Laplacian of u in one higher space, one higher dimensional space. This is not an elliptic equation. If I want this to be an elliptic equation, this should be minus utt minus c squared Laplace of u equal to 0. This would be sort of Laplacian in a higher space dimension. But this is not. Here there is a plus and here there is a minus. So this is definitely not an elliptic equation in a higher space dimension, because you have a plus and a minus. This is elliptic and this is parabolic. If u is independent of time, in this case, that's true that it satisfies this. However, this is very rare. So stationary solutions are usually considered for the parabolic case. Yes, where stationary means independent of time. It is true, however, that if you are lucky enough and u is independent on time, then still you have this. But in general, you don't have stationary solutions. And so this is a PDE, which is completely different from this, and this is just a remark. Then, remark 2. What about the initial conditions? How to assign initial conditions? So what is our experience up to now? For first order, we have assigned, for first order PDEs, even quasi-linear, we have assigned an initial function, u bar. For the wave equation, we have assigned initial position and initial velocity, like in physics, when you have an ODE of second order. Acceleration equals something, you assign initial position and initial velocity. And also, it was important here that on our hypersurface sigma, we had some sort of transversality condition between the PDE, the operator, and the sigma itself to be, say, non-characteristic. Here, and also here, we have seen that the time t equal to 0 is non-characteristic for the wave equation. And this was at least related to the Koshy-Kowalewski theorem. Now, what about the initial conditions in this different PDE? What do you think? Can we assign u0 bar as before and u1 bar as before? No. This cannot be done. And there is a clear reason for this. Why? Why I cannot say, why in general I cannot study this problem, say, ut minus Laplace of u equal to 0, say, and then at time 0, I cannot, in general, study this. Because if I give this, then the equation now, if the solution is c2 up to the boundary, then I have a relation between u1 bar and the Laplacian of u1 bar, which is not reasonable. Because u1 bar is. So this says that we cannot study this. And indeed, this is usually a way that we can put the initial condition. So there is an initial condition just only on the position. You cannot assign the initial velocity. This is in strike contrast with the problem of the wave equation in this case. Now, next is the following is that psi in n minus 0 is characteristic for the PDE, for the heat equation. This is for l. If when you have a second order PDE, you have to look at the principal part of the operator. And it turns out that, ah, yes, k. Thank you, thank you, sorry. Thank you. OK, the diffusion coefficients. So k. If, by definition, now I'm considering just only this part of the operator. And the definition is that, so it is x1 plus n. x1 squared plus xn squared is equal to 0. So you have n plus 1 components. The first x is x0. And then this must be equal to 0. And the point x on sigma is non-characteristic for the operator l if nu sigma x is non-characteristic. So this is the usual definition. I am going quickly here because the definitions are exactly the same as in the wave equation. Just only the, in the wave equation, remember this was psi squared. Take c equal 1, psi 0 squared minus psi 1 squared plus x. This was the wave equation. Noxie 1. Here, you say? No, no, no. There is noxie 0 here because when you look at the operator, you just look at the principal part, namely the part of the linear operator where the higher derivatives are. If you look at the higher derivatives, so you have here two derivatives. So you define the characteristic taking into account only this part of the operator. No, now you see. Now I show you. So first of all, just only the principal part of the operator here you have two derivatives and two derivatives. So c0 and c1, cn. Now here, you have noxie 0 because you have just only c1, cn. So there is noxie 0 in the k. So you don't have this. And so you have minus this equal to 0, which is this. This is the definition. Now let's see how they look like. For instance, take a plane t equal constant. Take a plane and hyperplane t equal constant. t corresponds to c0. So t equal constant. For instance, t equal to 0. For instance, t equal to 0. Now one of the two unit normals, so this is sigma. This is one of the two unit normals. This or the opposite. And you see that this is, which are the coordinates. In this case, nu sigma at x is clearly equal to 1, 0, 0, 0. And these are n components. And so what we see is that, unfortunately, actually, this is characteristic. This is another difference with respect to the previous case. We will usually study our PDE on a characteristic surface now, because we will assign our initial condition at time 0. So all these planes, t equal constant, are characteristic. Nevertheless, we will study this problem on t equal to 0, parentheses, characteristic. This does not exclude that we can study our problem. We can study, but what we cannot invoke is a Koshy-Kowaleski type theorem. This we cannot invoke. We cannot invoke to have a local solution a Koshy-Kowaleski result. So all this discussion, already from this very initial, starting discussion, you can already realize how different is this PDE from the previous PDEs that we have studied. So OK. So let me come back to our new PDE. So let me try to observe something about the scaling properties of the PDE. Of course, one remark is the following. If you change t into minus t, then your PDE changes radically, because you have a change of sign here, but nothing happens here. So it is something completely different, as I said. It is like to change k positive into k negative here. So you cannot think about this PDE as modeling something which is reversible in time, in some sense. Indeed, this is called the heat equation, because it is a model for expressing the propagation of heat in a medium, which propagates as time goes on only. On the other hand, when you had utt here, if you differentiate twice, you change t into minus t, then you don't have any difference. So now here, there is a direction of time, which is, well, so time is going on, is going in this case. Next remark is that there is an invariance here, scaling invariance. Namely, if you have a solution, u of t and x solution, then you can multiply it by some constant, let me call it a of ax, bt ax, is still a solution for some b and a constant. Well, there is an important scaling here, underlying this PDE. And so you see that b must be equal. So you see b must be equal to a squared, essentially. This suggests that it is interesting to look at the following quotient, because you see it is like you take half derivative. So this equation has the following property. Two derivatives here, but one derivative here. And so this is the reason for these two here and this one here. And so this is the so-called parabolic scaling. And indeed, one can look for special solutions, depending on this variable. So maybe this suggests that maybe a function v of this, we could look for a solution special. Maybe we don't know what to do, because this is a new equation. Maybe we can look for a solution, not depending on t and x, but maybe depending on x squared over t. So maybe v of psi, psi. This could be an attempt just looking at the scaling and symmetry properties of the operator. Actually, it is more convenient to look for a solution where there is another t in front here. Let me call this constant t to the alpha, maybe. So t to the alpha v of psi. It is more convenient to do this. And so let us start to look for special solutions to utt minus k uxx in one space dimension, for instance, for simplicity of the form, for special solution of the form, one of the square root of t, t is positive, remember, v of psi, t is positive. So maybe this could be really a very good exercise, homework that we will do tomorrow, of course. But I believe it's instructive. So home, look for an even v so that v prime of 0 will be equal to 0, even function, v prime equal to 0, and show that maybe Maya has slightly changed the notation. I would be more consistent with the usual notation, maybe more usual. Instead of calling this v, let me call it capital U. Sorry, a small change of notation. So look for an even real function u, which is even u prime of 0. Show that u prime plus 1 over 2k psi u is equal to 0. And hence, u of psi is some constant. Let me call c. We cannot use c because c was used. So let me call some constant b, e to the minus c square over 4k. Hence, u of tx is some constant e to the minus x square over 4k t. So there is a sort of miracle. There is an explicit solution. In the whole half space, half time space, b is a constant. And then try to, so this is home. And again, home. At which points of say 0 plus infinity times r is u singular, 1 is u in L1 log. Well, if you are not able to do, don't worry, because we will do this tomorrow at the beginning of the lecture. This is a very, very important function, which is an explicit solution out of the origin of your PD, and has a name. It's called the Fundamental Solution of the Heat Equation. And it is very useful for constructing other solutions and also to understand the qualitative properties in general. So again, you see, we will prove tomorrow that we are able to find some explicit solution in some domain. Of course, there will be some singularity here. There is some denominators or something. But try to see if you are able to do something about. Try also to understand the shape of this function. Maybe try to depict a graph in time space when you are in one space dimension. Try to depict the graph of this. So one space dimension, one is time, one is space, so you have a graph in R3. Maybe it's not impossible. Just to understand what is the shape of this. This is a Gaussian, of course.