 So let us start. So let me remember that we are considering now this equation here. And we are considering it in, say, in t positive and x in rn. So k will be for us essentially always positive. So this is the heat equation. And we are trying to look for some special solution, for instance, starting with dimension 1, space dimension 1. So we look for special solutions. We have already observed that x squared over t is a nice scaling. Because here we have just one derivative in t and two derivatives in space. So this is natural parabolic scaling, so-called. And therefore, yesterday we have defined the function, maybe capital U, if I remember correctly. And then of the variable xi. So where xi is exactly this, by definition. And maybe divided by some t to the alpha. And we have taken alpha equal minus. And we take alpha equal minus 1 over 2. So this is essentially 1 over square root of t, capital U of xi. I hope that this is where the notation of yesterday. So we look for solutions capital U so that, so we look for U, capital U so that 1 over square root of t, t is positive. And capital U of x squared over t solves our PDE. And so what do we have to do? So let me call this maybe U. So U t. We insert now this ansatz inside the PDE. And we look for an equation on capital U, which is the advantage. The capital U is a function of just one variable only. And so this is a way to reduce our PDE to an ODE, actually. So this is minus t over minus t to the minus 3 half 1 half. So minus 1 half t to the minus 3 half U of xi plus U prime of xi xi t. And therefore it is equal to minus 1 half t to the minus 3 half U xi plus. So xi t is equal to, this is xi. So we have t to the minus t to the minus 2 x square. And so this is equal to 1 over square root of t times t to the minus 2. Then we have x square. And then we have U prime of xi. Now t to the minus 3 half. So let me now isolate t to the minus 3 half here. So I have minus 1 over 2 t to the minus 3 half U xi minus t to the minus 3 half. So if I make some mistakes, please correct me. x square over t U prime of xi. And so this is actually xi again. So this is xi again. So let me write it minus 3 half U xi minus t to the minus 3 half U prime of xi. Next we have U xx. So U xx, if this is U, U xx is 1 over square root of t. And then I have, I think, so U x, what is U x, is 1 over square root of t, then U prime of xi. And then let me recall what was xi x. This is 2x over t. So this is 2x over t. And then U xx is equal to 2 over square root of t. And then we have U second of xi multiplied by x square over t square plus 2 over square root of t U prime of xi over t. So let me again isolate 2 over t 3 half. And then we have, by the way, we are in one space dimension. So sorry, this x square is exactly the modulus of x square. So one space dimension. So this is 2 t to the 3 half. Then we have U second and then again xi, I think. And then we have 2 t to the 3 half U prime of xi. So please check the computations if there are mistakes or not. So now we have this is U t and this is U xx. And therefore, this is equal, this left hand side is exactly equal to. So minus 1 over 2 t to the 3 half U minus 1 over t to the 3 half U prime. And this is U t. This is U t. So let me put maybe 1 over t to the 3 half everywhere here. So this is this, this 1 over t. And then let me put maybe also a minus outside. So we have a plus here, plus here. And then we have plus 2k U second plus, thanks, thanks, thanks, 2k xi U second. There is xi also here. Thank you. 2k xi U second plus 2k U prime, 2k U prime, OK? 2k, let me check. So we have 2. So let me check. So the derivative, the time derivative? There is? This? I have to check this. So this 2k xi is 4. It is 4. Your exacts, it is 4. This is actually a 4. So this is a 4. Thank you. So that is a 4. This means that here there is a 4, OK? So let me check if the computations are wrong or correct. So U t is equal minus 1 half. This is minus t. I have a minus 1 half here, apparently, also in my notes. So U t is this xi t. There is not minus 1 half here. So there is no minus 1 half here, t to the minus 1 half. Apparently, in my computations, I have minus 1 half there. And U second and U xx U prime, U xx, 1 over square root of t, 4. So at the end, let me put outside, there is a 4 here, 1 half U. This is OK. This is apparently wrong. U prime side. So I have, are these computations correct? Let me check. Ah, no, no, no, no. No, no, this is, the point is that I make this change of variable. Sorry, let me, so I take this U. I take this of x over square root of t instead of x square. So let me take this. Sorry, this is xi. And this is U of tx. Let me do this. Otherwise, we have too many terms. So this is still the parabolic scaling, but we don't need the square. So let me compute again once more. Now it should be correct. So U t is therefore equal minus 1 over t to the 3 half, 1 half. And then U plus 1 over square root of t U prime of xi. And this is xi t. Now, so xi t, so this is correct. But now xi t is slightly different. So xi t, if this is the new xi, xi t is equal to minus 1 over t to the 3 half x. So this is, this remains the same. And this is U prime. So let me correct this. So let me correct this. It is simpler, actually. U prime, 1 over square root of t, t to the 3 half, 1 over square root of t, x. Is it correct now? And so we can continue. This we write as before, U of xi. And then we isolate again as before, 1 over t to the 3 half. And what remains is x over t over square root of t. So what remains here is x over square root of t. And x over square root of t is the new xi. So we have this. And this is now correct. So now Ux should be actually much simpler because we don't have x square. So that was the, so what is Ux? Ux is, Ux is equal to 1 over t. Now U prime of x over square root of t. And Uxx actually is simpler. That was point. So now this is simple. It is 1 minus, it is 1 over t to the 3 half, U second of xi. Do you agree? Fine. Now let us see if now finally we end up with the correct conclusions. So let me rewrite the left hand side. So the left hand side is what? So let me put 1 over t to the 3 half in front of everything. Utt is, and let me put also a minus as before. So Utt is equal to 1 half U, which is this. Then we have plus 1 half xi U prime. And then we have here a k. And then we have U second. And this is the, if I'm not wrong now, is the, so k U second with the minus. And then we have this. And this. Do you agree? Is it OK? OK. So OK. So this must be equal to 0. And therefore we are looking for capital U solving t is positive k U second plus 1 half. And then you recognize that here there is the derivative of xi U with respect to xi, OK? Because this is clearly OK. It's just the derivative of phi. This says that k U prime plus 1 half xi U must be a constant. Can you follow the computation? Sorry? This? No. Minus on a 1, 2, 3, 4, 3, 2. This one? Next. This one? p squared half. Why not? No, no, it says 3 over 2 over 2. 1, 2, 3, 4. OK. So please copy. And so we have this. Now, let us look at this expression. Let us look at this expression. This is invariant under the change xi into minus xi, OK? So this is invariant under the change xi into minus xi. And therefore, well, we look for actually an even solution, U. In particular, we look for some solution, U, which satisfies this, because of this argument of invariance under this change. This says to us that if you look for solutions to this with this further condition, then this constant is 0. Because if we take xi equal to 0 here, constant equal to the value at xi equal to 0. But then xi equal to 0, we look for solutions so that this is 0. And then U prime is 0 by our new assumption. So the constant must be equal to 0. And therefore, we have U prime plus 1 over 2k xi U equals 0. OK? OK. And then we find the capital U, which must be equal to some constant. Let me call this C. And then we have an exponential of minus, say, xi square over 4k, maybe. So you can check that, of course, U prime is equal to U. Is this OK? OK. So and this is U is smooth, is even, smooth, and bounded, as you can see. Fine, because now, at least in one space dimension, now let me recall which is the relations between U. U of tx was defined as 1 over square root of t capital U of xi. And this was the new xi, say. And therefore, we have found the solution U of tx equals some constant C divided by square root of t. And then e to the minus x square over 4k t. OK, and t was positive. And this was the result for some constant C, for any constant C. Now, notice that, by the way, we have not really used the positivity of k in this discussion. We have not used. But we can use it here. Let's follow this equation. No, I mean k is non-zero, of course. k is non-zero. Why do you think so? I mean, no, we have just non-zero. But of course, if k is zero, this is a trivial PDE, OK? So we can assume that k is non-zero. However, k positive says that this is actually a decrease. I mean, it's going to zero rapidly as xi goes to infinity. On the other hand, if k, say, is minus 1, then this is actually blowing up at infinity. So if we are looking for solutions in some special classes, for instance, it is natural to take now k positive. But by the way, for the moment, this is true also for k negative. OK, this is. So now let me give you an exercise homework. So let us first give you the following definition. So now let n be greater or equal to 1. t be greater or equal to 0. x in Rn. And define, I need some symbol. Let me call this phi, maybe t of x is equal to some constant that I will specify. Some constant that we will specify in a minute. OK, let me specify it. OK, some constant t over n over 2 e to the minus x. So let me also normalize things. Otherwise, so let me, for now on, take k equal 1. So k now is 1, and so here there is 4t. So if t is positive, x in Rn. And then if t is equal to 0, let me define it 0. If t is equal to 0, x in Rn and say tx different from the origin. Just to, OK, just maybe we can also, if this maybe is not useful, but we can also extend this also for negative times. So c is a specific constant. I think that c is 1 over 4 pi to the n over 2. And this is called the fundamental solution of the heat equation. Now, the homework consists in the following. So we have checked that this, for positive times, is a solution of our PD in dimension 1, when n is equal 1. So remember that here we have square root of t. So this corresponds to n equal 1 here. This is just a definition. No, so let me, so what we have done in the previous computation is that when n is equal to 1, means here you have square root of t, this solves our PD. This is a solution, OK, by the previous computation. This is a explicit solution in this half space. Now, the homework consists in checking that still in the positive half space, check that phi solves ut minus Laplace of u equal to 0 in 0 plus infinity times rn. So check that this is still a solution in any dimension, in any dimension. The difference here is this factor, n over 2. OK, so now the, OK, actually any constant is OK. Now we try to see why we need exactly this constant. Any constant would be OK. But this is just for a normalization reason. We are normalizing to area equal to 1. Decarn this object here so that its integral in space is always equal to 1. So let me do this computation so that I try to justify. So let me compute the integral over rn of e to the minus x square over 40 dx. So we make the following change of variable, y equal to x over 2 square root of t, OK? So that this, with this change of variable, becomes exactly what? Becomes e to the minus y square. Then we have maybe 2 to the n t to the n over 2. So please check again the computations, so dy. Now this is 2 to the n t to the n over 2 product from 1 to n. So OK. So and we know what is this, right? We know how to compute this, right? So let me compute it. Now this quantity is here e to the minus s square ds. I compute it to the square, which is equal to e to the minus s square plus sigma square ds d sigma. So we do this in polar coordinates, OK? So this is equal to integral between 0 and plus infinity. Then we have 2 pi. And then we have rho e to the minus rho square d rho. And this is actually we can do this. So this is 2 pi. And then we have, I think, 1 half, 1 half. Because the primitive is 1 half e to the minus rho square between 0 and plus infinity, OK? So this is pi. Hence, this object is square root of pi. And therefore, let us check the computations. 2 to the n, t to the n over 2, pi to the n over 2. Pi to the n over 2, this means that, let us see, OK. So this is exactly 1 over 2 to the n, pi over n over 2. And so you see that now, if I take, so this implies that 1 over 4 pi, t to the n over 2, integral over n over 4t in the x, this is exactly equal to 1 to 1. This is the reason for the choice of the constant, just a normalization reason. So with this choice, this is a kernel in the sense that the area, for any time, if you compute the integral in space of this quantity with this choice of g, this is always equal to 1. This is the reason for the constant. Now, is it OK? So now, let us try to understand some properties of this function, phi, which is actually the first example of a non-trivial solution, at least for positive times. So let us do this. So try to depict the graph. So here we have, say, time. Here we have space. So maybe we can put time here, this here. So first of all, what happens to our function? So this is x equal to 0. This half line here is x equal to 0. So what happens here? What do we see here? Well, we have a sort of profile in this plane, in this plane containing this line, x equal to 0. And this vertical line, there is 1 over square root of t, because you will put x equal to 0 here. And so apart from this constant, c, which in one dimension is 1 over 2 pi square root, and so on. So let me, for the moment, forget about this constant, take it equal to 1, n is equal to 1 in this picture. So here there is a profile, which is 1 over square root of t. So what we see here is something like a sort of function like this. So if we take a movie, namely, we cut this graph at different times, what we see is something like this. At some time then, as time goes towards 0, we see something like this. And we have normalized things so that the area below this graph is always equal to 1. Always equal to 1. So this is essentially the graph of the function phi. Only in one space dimension, of course. We cannot draw it in two space dimensions. So now, what about smoothness of this function? So remark, it is clear that phi is always positive if t is positive. Then for convenience, we have defined it equal to 0 here by definition. So this is for positive times. At the origin, we don't have defined anything, but for time equal to 0, apart from the origin, apart from this point here, but here we have set the function phi equal to 0 by definition. And if one is interested to extend it also for negative, times where one possible definition is to extend it equal to 0. Now, so what about the regularity? So is it true that the function, for instance here, far from the origin, but on this line, is it true that the function is smooth or not? So we have to understand a little bit this. Because this is a definition. We have to check that at least this definition is continuous here and maybe even more. So 0, 0, 0, so here, 0, 0. And here is sort of singularity. And then this is actually our function, OK? Now, what about the regularity? Of course, for positive times, it is always strictly positive. Not only this, but the support of this function is non-compact. It doesn't have compact support. Next, phi is surely c infinity, at least here. For positive times, it's surely c infinity. Actually, it is even more. It is analytic, real analytic, in 0 plus infinity times rn. Now, what happens here? So let us take this is t, this is x. Let us take a point x bar and compute the limit. So x bar is different from 0. So tx converges to 0x bar of phi of tx. So we have a neighborhood here. We want to take the limit of the function phi go on converging to this point, which is not the origin. And phi is 0 here, 0 here by definition. So look at this expression. If t goes to 0 and x is non-zero, what is this limit? It is 0, right? It is clear? I think it's quite clear, so this is equal to 0. Not only this, but also all derivatives, for the same reason, this kills any negative power of times here. So phi is also, we can say that phi is also, sorry, c infinity minus the origin. And also, well, remember that phi of tx dx is always equal to 1 for any positive t. So these are the first properties that we can infer from the explicit expression of the kernel. Sometimes it is called kernel of this function phi. On the other hand, at 0, there is a singularity. At this point, time equal to 0 and x equal to 0, there is a singularity. So the function is singular at that point. So now, exercise, well, I mean, the function is there, obviously from the picture of the graph, the function is there. How do you say it? There does not exist a limit as tx goes to 0, 0 of phi, for instance. This is clear. OK. So for any delta positive, the limit as t goes to 0 of phi of tx is equal to 0 uniformly in x bigger or equal than delta. Exercise 1, exercise 2, for any delta bigger or equal than 0, the limit as t goes to 0 of the integral phi of t y, y is equal to 0. OK. Maybe I can leave you this as homeworks. Do you understand what does it mean this uniformly with respect to x? Is it clear? So the supremum over x, bigger or equal than delta of this limit, is going to 0. So this is a way these two exercises say to you that, of course, the main contribution of phi is close to the singularity point. Because you see, when you go as t goes to 0, phi is always 0, but far from the origin. So in some sense, close to the origin, I mean what you see for very small times is something like this. And also, this says that if you look at the integral of this function phi, essentially here, there is nothing. And remember that this must be equal to 1. For all positive times, this integral must be equal to 1. But this says that, so essentially, it says that all contribution to the integral, as time goes to 0, is concentrated in an extremely small neighborhood of the origin. No, it cannot be 1. Because I'm integrating out exactly this. I'm integrating outside the origin. So this says that it should be 1. But where is it concentrated? I mean, this says that since it should be 1, it is concentrated exactly. If you compute the integral here, then you'll find anything, OK? So for any delta positive? Yeah, delta is small enough, but any delta is OK. Of course, if it is small, the exercise is more interesting. If delta is large, that is not very interesting. But in any case, it is true for any delta. So we have a strange function which has these properties. This leads to, we will see, I hope, to the notion of distributions and Dirac deltas. It's very close to that notion. So maybe I can leave you this as exercises that we will do the next time, OK? So don't worry if you are not able to do this. Well, this is easy. This is not so easy. But try. And then we will do this together in the next lecture. Again, in the spirit of the course, we don't construct solutions on any domain, but we look for uniqueness properties of solutions. So let me now discuss this very important issue, which is the weak maximum principles on bounded domains, on bounded domains. So now we are in the following situation, essentially. So we have, say, some domain. So this is time. This is space. And let me call this q. Then we have some t. My notation are the following. t, tx in q, such that t is less than t. So we have this bounded domain q. Assume qt is bounded and non-empty. So we have sort of this is qt, q is open and bounded. q is open, say, into r1 plus n. So qt is also open. But it is what is below this time capital T here. Then let me introduce also some other name, like gamma, which is this, gamma t is tx into q, such that t is equal to t. This is gamma t. q is open. And then let me introduce gamma t is this red part. Gamma t is this red part. And then let me also introduce st, maybe. The notation was st, maybe, as tx into the boundary of q, such that t is less than t. So then for we have, so this is the boundary of q. And so we have this part here, this is gamma. In general, it's quite often the problem is studied in cylindrical domains. So in general, specific example, for instance, q is the product. q is a portion of a cylinder. So this is q. This is t, any t. And this is omega, bounded domain over n. So actually q is the product of, say, 0 sum capital T, say, some big t times omega. This is the usual situation where we study parabolic problems like this, namely, on a portion of cylinders over omega. This is omega. And we have this situation. And therefore, let us look at our, so this is the upper part of the boundary. In this case, it is just this. So this is, say, gamma t. This is gamma t. So here's q a little bit large. So this is t. This is s. So gamma t is, if we look at times before t, gamma t is this part here. It's the upper part inside the open set q. And then s of t is just, s of t is the lateral boundary, so-called lateral boundary. So it is the lateral boundary of this cylinder and like this, with this bottom. Is it OK? Look at the definition. This is the definition. So it is clear that this situation is less general than this, because this is a special situation in case of products, when q is a product. However, this is the usual setting of our, in any case. Now, theorem. So let u be a continuous function in qt, closure, intersection. Now, with this symbol, I mean that u has one time derivative, which is continuous, and two space derivatives, which are continuous. So one, so you see, the equation is this. This is the operator. So it is natural to require to have smooth objects, classical objects. It is natural to require inside the open set qt. It's natural to require one time derivative, well-defined, and two space derivatives, well-defined. So let you, in this class, satisfy qt minus Laplace of u less than or equal than 0 in qt. So this is so-called smooth sub-solution, because there is less than or equal, so it is not necessarily a solution. It is a sub-solution, and this is smooth enough. So u, smooth, sub-solution. Key, normal? Yes, for simplicity, normalize it so we know what is k. We know that k is positive. We know that in the fundamental solution, it goes to the denominator. But for these arguments for simplicity, let us take k equal to 1. Of course, everything works the same. The important is that now k is positive. So let us take k equal to 1. So this is a smooth sub-solution. Now, the word is clear, because it is less than or equal than 0. And smoothness, we have by assumption. Then, so under these assumptions, then the max of, so now u is continuous, and q bar t is bounded. So u assumes on the closure of qt a maximum. This is not plus infinity. It's well-defined as a maximum. And then this is equal to the max of u over the closure of the lateral boundary. So the lateral boundary was defined like this. So smooth sub-solutions have the following properties. Now, let me continue the statement. So under all these assumptions, under the boundedness of qt. So in the statement, there is assumed qt is bounded. This is moreover, furthermore. So this is called weak maximum principle, weak maximum. Weak maximum principle for sub-solutions. And then, furthermore, let v continuous up to the closure. So that these are well-defined, the two derivatives, one derivative, b such that satisfy is a smooth super solution, qt, then this v is a smooth super solution. In particular, so in particular, if u is a solution, solves u t minus laplace u equal to 0, then we have both together. So then max over q bar t of u is max over s bar t of u, mean q bar t of u. OK? So let me summarize the statement. The statement is a statement which says something about smooth sub-solutions, weak maximum principle, and the smooth super solutions, weak minimum principle, weak minimum principle. In particular, if u is at the same time a smooth sub-solution and a smooth super solution, namely, it satisfies the quality. So it is both. Then we have both the b-sessions. OK. So there is no role of gamma t. No. In principle, yeah, there is no role of gamma t. There is actually the role of gamma t in the sense that this theorem does not exclude, OK, remark. So remark does not exclude that take a function that take a solution u, that the solution u takes its maximum on gamma t or on in qt. Does not exclude this. It just says that if you look at the value of u, take a solution. If you look at the values of u on this yellow part, then there will be the maximum value. It is written here, for instance, for a solution. But in principle, it could happen that also at some point here, you could take its maximum value or also in qt. It could take a maximum value. This is not excluded by the theorem. This is why the reason of the word weak, because there is also the strong maximum minimum principle. So this is called weak maximum principle. And it does not actually, it is possible to exclude. So actually, it is possible to exclude the content of this remark. But this is much more difficult to do. Gamma t bar. Well, qt will, because qt bar, essentially, I mean, no, because qt bar contains also the lateral boundary. We cannot put the bar there. So now we will prove this if there is time. But so the important fact is to understand that this situation up for the moment is not excluded. So let me continue the remark. Maybe this can be excluded with other arguments. This can be excluded with other arguments that for the moment we do not do. OK. Say for subsolutions. For, yes, for subsolutions. So take t bar, x bar, inside qt. So we have, say, a point here. Take an intermediate time between t bar and capital T. Take epsilon positive. And we construct, starting from u, from u, we construct a strict subsolution, as follows. Now we will understand what does it mean strict. So we define v of tx equal to u of tx minus epsilon t. So epsilon is positive. Say that everything here is positive. For definitiveness, t is positive. Everything is positive. t bar is positive. Tau is positive, just to fix ideas, like in the picture here. And define this v so that, of course, in, say, in q tau, then vt satisfies as the following property. vt is equal ut minus epsilon. Epsilon is positive. Laplace of v is equal to Laplace of u. Therefore, vt minus Laplace of v is equal to ut minus Laplace of u minus epsilon, which is less than or equal than minus epsilon, because this is a subsolution. Remember, in the wall of qt, this is a subsolution assumption. We are doing the proof for subsolutions. Then what is the inequality satisfied by v? Well, this is strictly negative, because epsilon is positive. Hence, v is a strict subsolution. In this sense, v in c, q bar, tau, intersection, c1, 2, q tau, is a strict subsolution. Actually, it's also a strict subsolution. This is a strict subsolution. Now, claim the maximum point of v cannot be on q tau union gamma tau. The maximum point of v cannot lie on the upper part of the boundary here and in the interior. Let us check this. This is a claim. Assume by contradiction that t0, x0 is a maximum point. Assume by contradiction that this is a maximum point. We want to find a contradiction with using our strict subsolution property. First of all, what happens to the Laplace of v? Maybe this definition can be given everywhere. And this is OK, and this is also OK. So what happens to the Laplace of v at the point t0, x0? And what happens to the derivative of v at t0, x0? So I'm looking, this derivative is not necessary 0. It is 0 if t0 is less than tau. And in principle, I'm looking at the problem inside here. Now, tau, this is tau. I'm looking at the problem here. And if I am inside, it is clear that this derivative is 0. So if t0 is less than tau, here it is clear that this time derivative is 0. But if t0 is equal to tau, then this derivative is what? Not necessarily equal to 0. It is a maximum point. So it is increasing there, bigger or equal than 0. And what about the Laplacian in general? What about the Laplacian of v at the maximum point? Do you know? It's a non-positive. By the way, the defect? What is it? Laplacian is less than tau. At the maximum point? At the maximum point? Well, this is all the action. The action at the maximum point is the trace of the action in particular. So the action has a sign at the maximum point. At the local maximum, the action has a sign. In particular, the trace has a sign. So this is surely non-positive. And this is also non-negative. So in general, we conclude the following two properties, this. And this is always true also, t0, x0, bigger or equal than 0. At the maximum point. At the maximum. Now look at this. Look at this. At the maximum point, so evaluate this at the maximum point, this must be negative. But it is the sum of two non-negative quantities. Because this is bigger or equal than 0, and minus this is bigger or equal than 0. So at t0, x0, we have a contradiction with the strict sub-solution property with this. So we have proven our claim. The four-hour claim is proven. So maybe time is over. So maybe it is better to continue the next time, because time is over. And so the next time we will start. First, by doing the two exercises that I left to you on the decreasing properties of the kernel of the heat equation, phi. And then we will conclude the proof of the weak maximum and minimum principles, unbounded domains.