 So let us start. Today, I would like to show you this example of taking off concerning the non-uniqueness of solutions to ut minus the plus of u equal to 0. So we are now in n equal 1 space dimension. So we work here in r. And then u of 0 equals some function equal 0 at time 0. And you also, so it is clear that the 0 solution is a solution of this problem, because at time 0. So u equal 0 is a solution, obviously. So the point is to construct another solution. The point of the example is to construct another non-zero solution to 1 so that, at the end, there cannot be uniqueness of solutions to this problem. Remember that, on the other hand, on bounded domains, yesterday we saw that in, say, 0 capital T times omega, omega bounded, the solution is unique. Remember, in the solution, if it exists, it's unique. So the point here is that the domain is unbounded. OK, so the idea of T-conoff is to first reason formally, look for a solution for u of Tx of the form some coefficients depending on T x to the j for any Tx in 0 plus infinity times r. So we try to find the solution of this form. Of course, we will have to see that the series is converging suitably strongly. But for the moment, let us forget for the moment the problem of convergence. So let us reason formally. And assume that we also require, so this is r, x, this is T. We also require, look for such that on this yellow half line, u is equal to some given g given to be chosen on x equals 0. It is on this line, x equals 0 on, say, 0 plus infinity product 0. That is on this half line. And ux equal to 0 on same line. So this seems to be too much. But if one is able to construct a solution with the following properties, this will be a solution. So now you will see why we are asking that the derivative in this direction along this line is 0 on this line, and u is g. So I mean, if one is not able to do, then one cannot construct anything. Otherwise, one can try to look for solution with these properties. And of course, one has also to impose this at the end. So let us see what happens if we formally compute ut. So assuming convergence sufficiently strong, I differentiate under the design of series. So formally, we have g prime j of t xj formally. ux formally is equal to sum from 1 to plus infinity jj of t x to the j minus 1. And uxx is equal to sum from 2 to plus infinity jj minus 1 gj of t x to the j minus 2. OK? This is formal. In principle, one cannot do this. But at the end, assume that I am able to construct coefficients. Suitable coefficients. Assume that I am able to construct coefficients gj, depending on g, so that this converges uniformly, for instance. And all derivatives converge uniformly, for instance, on, say, for bounded sets, at least in space. Then it will turn out that this can be done. I can differentiate under the derivative of the series. It is actually the series of the derivative. OK? So for the moment, I have this and I have this. So what happens if I impose such a new to be a solution of this? What happens? Well, let me rewrite this as the sum from 0 to plus infinity of j plus 2, j plus 1, gj plus 2 of t x to the j. This is a simple change of variables, OK? I change variable. I call j minus 2. I call it k. OK? This is a change of variable. OK, so if I impose u to be a solution, I am imposing that this is equal to this. This must be equal to this. OK, now let's see what happens if I require u equal to g on this line. So u of t, 0, is actually equal to what? u of any time, 0 is 0 of t. Hence, I want my first requirement is on the coefficients. Of course, I'm looking for the coefficients. So I require the first coefficient to be equal to g that I still have to choose. So this condition is due to this requirement. Now, I am assuming also a derivative in this direction along this line. I'm looking for a solution having 0 derivative here, so some sort of even solution. So ux, ux, so this says that g1, you see, so ux at t, 0, ux at t, 0, I see it from here. It is just g1 of t. And therefore, on the second unknown coefficient, I require g1 equal to 0. So let me write now what happens. So for the moment, I have imposed this condition and this condition. From this condition, I deduce this. From this condition, ux on this half line, I deduce g1 equal to 0. Now I have to impose the validity of the equation. The validity of the equation says, you see, this is a series. This is another series. I want this to be equal. And therefore, I require that all coefficients of the same xj coincide. So I require that gj prime of t must be equal to j plus 2, j plus 1, gj plus 2, gj plus 2. That is, I require the following condition. gj plus 2 of t to be equal to gj prime of t divided by j plus 2 times j plus 1 for any j bigger or equal than 0. So I have a condition on the first coefficient. Condition that the 0th coefficient must be g. The first coefficient must be 0. And therefore, I see what, and then this. So what I deduce from this? So g0 equal to g, let me erase this. I deduce that, first of all, on all odd indices, this must be equal to 0. Is it OK? Because you see, we have that the first odd coefficient is 0. And then, for instance, g3 for j equal to 1 is equal to g1 divided by 0. Therefore, g3 is 0. And therefore, g5 is 0, g7 is 0, g9 is 0. Yeah, but g1 is identically 0. And therefore, g1 prime is identically 0, and so on. This is always the case when you have a sort of problem and you are looking for a solution written as a series. And then you have that one coefficient is 0. And then you have a relation between not 1 and the previous 1. But this differs by 2, j plus 2 and j. Then you have this. On the other hand, so what happens to g2, for instance, which is the first even number index, j equal to 0? This says that this is equal to g0 prime divided by 2, OK? But g0 prime is equal to g. And what happens to g4? What happens to g4? g4 must be equal to g2 prime divided by 4 times 3, which is equal to g. So g2 prime is actually this. And therefore, I have g2 divided by 4 factorial, OK? And in this way, it's not, it is. 4 times 3 times 2, OK? So by induction, one shows, therefore, that g for any. So g2k of t is equal to gk divided by 2k factorial, OK? So now, we keep only, therefore, this information on the blackboard. So we have this, and then we have this. So what happens to our series? Our series, our function u, we are looking for. You see now, what is this? So this is j from 0 to plus infinity gj of t x to the j, which is equal now to what? It is equal to k equal 0 plus infinity gk 2k factorial x to the 2k, gk of t. Please check this, because this is actually, why is it true? Because this sum is actually only on the even indices, because the odd indices, the coefficients are 0. So g2k, so this is simply sum over k. j is equal to 2k, and so it is this, divided by this. This seems to be correct, OK? Now, you see, what happens, what does it mean? So our u, now, our candidate, because for the moment, everything is formal. There is no proof. Everything is formal. But our candidate at the moment has coefficients, which depend only on one function g, because this is the k derivative of only one function g. So now, we have to choose g. And we hope, we choose g, and we hope that with this choice, this object satisfies all required properties. Is it clear the line of reasoning? So now, choose g. Now, choose g, so that this sum satisfies hopefully all required properties. What does it mean? Satisfies all required properties. Well, choose g, so that this sum is converging. All derivatives in time and space are converging. This is smooth enough. And by construction, it will satisfy, so this object automatically, by construction, satisfies u equal g on the yellow half line, and ux equals 0 on the yellow half line. So what we have, so this is all required, namely, the sum is converging sufficiently well, strongly, sufficiently strongly. And u is smooth enough. And u at time 0 is equal to 0, or required properties so that I can differentiate under the sum and in these holes. OK, so let now, so up to now, it is clear, otherwise, please ask, is it OK? So I keep the expression of the function u on the blackboard. So let right now, the expression just only here, u of tx, sum of, now we have gk of t, then we have 2k factorial x equal to k. Which equality? This, yes, yes, OK, yes, it's OK. Sufficiently strongly. Uniformly, we'll see this at the end, OK. So this is what we have proven up to now. If such a solution exists, it must have the following form. Now, choice of g define g of t equal to e to the minus 1 over t squared for t bigger than 0 and 0 for t equal to 0, OK. Which is a c-infinity function, OK. Now the point, you see what happens also. If everything is smooth with this choice, if everything is smooth formally with this choice, I have that this is gk of 0. And gk of 0 is 0. So if you want extend it, if you like, but this is not important, but you can extend it like this. And then this is equal to 0 formally. By the way, notice also that this function g is now c-infinity, say, but of course, is not analytic. It is not real. It is a standard example of a c-infinity function which is not analytic. All derivatives in 0 are 0. So if there is a hope that this is the correct object, so there is a hope because this function g satisfies, for instance, this property. Now the point is to study the convergence of this series. So now we are trying to estimate gk of t to see the behavior of gk of t. Study gk of t. Estimate, say. Now consider, so we have for positivity. So now we have positivity. Consider now an extension of g as follows. g of z is equal to, say, e to the minus 1 over z square for any complex z different from the origin and 0. So now g is g is holomorphic out of the origin. So we can now compute, in particular, if t is positive. If t is positive, I am out of the origin. And I compute this quantity here. Let me call the notation that I'm using. gk, let me use omega. Omega minus t to the k plus 1. So this is omega, complex number. k plus 1 in the omega. What is this? OK, now take gamma. Gamma is a parametrization of the circle centered at t. This is the Cauchy formula for holomorphic functions. I mean, the Cauchy formula usually is for k equal to 0. But then we know that it extends to all derivatives. So gamma is a parametrization of the circle centered at t0, the complex number t0 of radius, small enough such that we don't touch the origin, because in the origin, there is a problem. Actually, it's not analytic. It's not homomorphic. Say, for instance, radius t over 2. So I have t0. And then I have this circle. And then I have by Cauchy formula for holomorphic functions. So in this way, I will try. So remember that I want to estimate this. So why I'm doing this? Because I want to estimate this. This is the goal. I want to see that the series converges. So I have this. Hence, I can say that gk of t is less than or equal to k factorial over 2 pi, the integral over gamma of g of omega divided by omega. These are complex numbers. So this is k plus 1. And then I have d omega. So this is a circle centered at t. So let me define this circle. So the circle is the image of the map gamma. That is the circle is the set of all z such that z is equal to t plus 1 of t. Namely, z minus this real number is a circle of radius t over 2. So this is the circle centered. So z minus t. So any point of the circle centered, so this is t over 2. This is t0. Any point of this circle can be written as t plus t half e exponential. So this is the circle. So therefore, this is what? k factorial over 2 pi, the integral over gamma. Now, omega minus. So omega is on this circle. So omega minus t is equal to 1 half, omega minus t is equal to t over 2 times this unit vector. So this is just t over 2 to the k plus 1. Then I have 1 half t theta. And then I have e to the minus 1 over omega square. Is it OK? So what do I have now? This t over 2 cancels with this, with one of these. Therefore, what remains here is equal to. So let me check. We have to be careful about the numbers, of course. So this is 2 over t to the k, the integral of gamma. So now, e to the minus 1 over omega square. Now, e to the minus 1 over omega square is equal to e minus the real part, minus the imaginary part, 1 over omega square. And therefore, this is equal to this, which is equal to the product of the two exponential. And 1 is 1. So it is just e to the minus real part for 1 omega square. Is it OK for the moment? Now, remark. So if z, so this is t, then I have the circle of radius t over 2. Now, let me make a larger picture. This is t. This is t over 2. And this is 3 half t. And then I have this circle here, which is the image of gamma. And I am integrated on this set. But I am integrating something which depends on 1 over z. So this is z. And then I have to see where is 1 over z. Because now I am integrating 1 over omega z or omega square, whatever you want. OK, the claim is now that if z belongs to the circle, the circle centered at t of radius t over 2, then 1 over z. So assume that t is very small. So 2 over t is very large. So I have this 2 over t here, which is 1 over this. Then I have this is 3 half t. So this point through the map 1 over z goes to 2 over t. So the claim is that all points then lie on a circle. Lie belongs to the circle centered at 1 over t. Not sorry, not 1 over t. Centered at 2 over t minus this divided by 2. So centered at 2 plus. Yes. So what is this difference? So this difference is 2 over t minus. So it is 4. This difference is this. And therefore, I think this difference is this. So 1 half of this plus 2. So centered at 4 over 3 t and radius 2 over 3 t. Radius, sorry. Radius, yes, 2 over 3 t. This is the remark. Well, the remark should be proven. So 1 is to prove that if z belongs to this, then 1 over z belongs to the w complex such that w is equal to the center 4 3 t plus the radius. So assume for the moment that we have proven this remark. Maybe we can leave this remark as an exercise. So for the moment, assume this remark. Assume. OK, so now 1 over omega here. Now omega travels along this circle. So 1 over omega travels by the remark along this new circle. And therefore, 1 over omega is of this form. So it is of this form. And therefore, it is of the form 4 over 3 t. Multiply it by 1 plus 2 e to the phi if omega belongs to gamma. Then 1 over omega by, sorry for the notation. Let me call this z again, z maybe. Then if omega travels on this circle, then by this remark that we leave as an exercise, 1 over omega lives on this circle. So now I put this in front of the parenthesis, so this is twice. Now 1 over omega square is equal to 16 over 90 square 1 plus 4 e to the 2 phi plus 4 e to the phi plus 4 e to the phi 1 over omega square. 16 over 90 square. OK, there is some number that now I don't find, unfortunately. Is it OK? This z square? It should be 1 up. There is 1 up, where? 1 over omega is 4 over 3 t plus this. OK? Sorry, I have made a mistake here. Thank you. Thank you. I have 1 over 2. This is the claim, sorry. And therefore, now we have this is, thank you, is 1 over 4, and this is 1, maybe. Hopefully that now the constants are correct. OK? So now the real part, this is the problem of complex integrals. By the way, so the real part of this is equal on this circle. The real part of this is, therefore, t is real. So 90 square times, now we have 1, 1 4 cosine of 2 phi plus cosine of phi for any phi in 0 to phi. So now I want something like this, right? Therefore, I need that this bigger or equal than something. OK? So let me try to find the minimum of this function. So we minimize now the own parenthesis, minimize now min. So this is, so let me consider now the function say f of phi equal 1 plus 1 4 cos 2 phi plus cos phi on its domain. So let me minimize it. So f prime of phi is equal to minus 1 half sin of 2 phi minus sin of phi, which is equal to minus sin of phi cos phi minus sin of phi. OK? So, hm? So f of 0 is equal to 1 plus 1 4 plus 1, which is equal to f of 2 phi. Now let me see what happens at the interior critical points. So we have sin of phi must be equal to minus sin of phi sin of phi. So solutions to, so f prime equal to 0 in the interior. Then solutions are sin of phi equal to 0, which means phi equal to, so we have already checked what happens at the boundary. Now, phi equal to phi is a solution, phi. And on this, f of phi, f of phi is equal to 1. Then the cosine is equal to minus 1, I think. It's 1 4? F of phi is 1 over 4, like this? OK. So if sin of phi is equal to 0, and then otherwise we have cosine of phi, so sin of phi difference from 0. The other solution is cosine of phi equal to minus 1. Sin of phi equal to minus 1. And so we are, again, phi equal to pi. So the minimum point is phi equal to pi, where the value is 1 over 4. OK, so what I'm saying is simply that phi, f of phi, is always larger or equal than 1 over 4 for any phi. Then it is also probably less than or equal than 9 over 4. But this is, for the moment, I mean, we also have this. So this is larger. So hence, this object here, can I erase this part? OK. So this is larger than or equal to this I keep. It is positive. And then 1 over 4. 4 over 90 squared. Now, we can continue this inequality, getting that gk of t is less than or equal to what? k factorial divided by 2 pi. So then I have 2 over t to the k e to the minus 490 squared times 2 pi. Is it OK for the moment? OK, this is, I mean, there is an exponential, which does not depend on theta. So it comes out of the integral. Then I add the length of this, which is 2 pi k factorial. So let me check if I'm correct or not. I don't know. Apparently, it seems to be correct. OK. Therefore, this is the estimate that we were looking for on the coefficients. So now u of tx, therefore, is less than or equal to what? k from 0 to plus infinity. Now we have this 2k factorial x to the 2k. And then I have this e to the minus 4 over 90 squared outside. Then I have 2 to the k, k factorial here. And then I have t to the k. Do you agree? Is it OK? Of course, if I made some mistake, please let me know. Otherwise, we do not converge to the solution. Now, claim for any k to k factorial over 2k factorial is small. Actually, smaller is small. By the way, you see 2k factorial is very large in comparison to this and this. But let us try to see whether this is true for any k. Do you immediately see that this is true? How can you see it? I mean, you have to show that this 2k factorial k factorial is less than 2k factorial. Is it so immediate? Why? We open this 2k factorial. Yes. And 2 for until 2k, it goes with 2k. So you open this, and so you cancel the last k factorial with this? No. We cancel even numbers with 2 degree k multiplied by k. Yeah. Yes. If you don't see it, maybe we can prove it by induction. Just check it by induction. k equal to 1, k equal to 0 is OK. k equal to 1 is OK. So assume this, then you want to show k plus 1 factorial. k plus 1 factorial, you want to show this 2k plus 2 factorial. And maybe you can use induction so that this is less than or equal to what do you have? You have a 2 here, then you have 2k factorial. Because 2k, 2 to the k, k factorial, k factorial is less than this 2k factorial. What remains is this 2 and k plus 1 squared. And so you have to show that this is less than this. So you have to show that this is less than this. And to show this, you simply divide it by 2k factorial. So it is enough to show that 2k plus 1 squared is less than or equal to 2k plus 2 times 2k plus 1, which is obviously true. Because this is equivalent to say that k plus 1 is less than or equal to 2k plus 1. So it's true. I mean maybe there is a quicker way. Just induction works very well. OK, so this is true. So we can continue our inequality by putting here e to the. So now I rewrite this. And then I have here k from 0 to plus infinity 1 over k factorial x squared over 2 to the k. Now you see why we like 1 over k factorial, because you can recognize that this is what is the exponential. So this is equal to the e x squared over t minus 4 over 90 squared. That is, if you want, e to the 1 over t x squared minus 4. So actually, our series is converging even uniformly. We have a bound e 1 over t x squared minus 4. We have this bound, if I'm not wrong. So you see, so the series, the sum, the infinite sum, is converging uniformly for bounded x for any positive t. Hence, u is continuous. For any non-negative t, u is continuous. And u, therefore, u of 0 is equal to 0. Fine, so what we have proven up to now is that u defines a continuous function, which is 0 and 0. Next, what remains is to show that similar properties are true for the ut, ux, and uxx. So there are similar kind of computations for the series. The infinite sums given by g prime of t and uxx. So at the end, and now I don't repeat the computations, maybe. So at the end, u turns out to be even more c infinity r. By construction, once you have this regularity, you can differentiate under the infinite sum. And by construction, u satisfies the PDE, up posterior by construction, because your coefficients have been constructed like this. So u, sums, OK. So this way, Iconov found another solution in an infinite half space, non-zero solution of the heat equation. So this is an example that shows that uniqueness can be very tricky. So what we have not proven, but maybe I can leave it as a homework, we have not proven the claim that if z is in that circle, then one over z is in the other circle. I leave it. OK, so now I would like to start a discussion. I don't know if I'm able to conclude today. Last discussion on parabolic equations. So on y, the notation for the heat kernel was phi, this fundamental solution of the heat equation. It was phi. Y phi of tx is so useful. So remember that we have some constant here, what is important, n over 2. And then, minus x squared over 4t. And this is defined, say, for any tx in 0 plus infinity times rn. And if you want, we can extend it at time 0 with the value 0, but not at the origin. Remember that the origin in time space 0, 0 is a singularity. So why this is so important? So this will conclude our discussion on parabolic equations. Then we will say something on elliptic equations the next lectures. And then we will start functional analysis. So there will be a big parenthesis on functional analysis. And maybe at the end of the course, we will be able to come back to PDs as in applications of functional analysis. This is the scheme of the course. For the moment, we have not used any functional analysis. We have worked in smooth space continuous c1, c2, whatever. No theorem seems to abstract the linear function. OK, so now, yes, we have the following theorem. So let u 0 bar be a function in this class, a bounded function and contiguous. The function u of tx defined as follows. u of tx equal to the integral t y minus x u 0 bar of y dy for any tx 0 plus infinity times rn. Satisfy the following properties. 1 u is c infinity. ut minus Laplace of u is equal to 0 times rn and 3 limit. As tx goes to 0x of u of tx is equal to u by x. For any x, you see? This is very interesting result, which says that if you want to solve for initial condition, proper initial u 0 bar, you want to solve the heat equation with this initial condition. One solution is given by this expression here. This is called, do you know what is the convolution? OK, so this is the convolution in space of the heat kernel at time t together with the initial condition. So we can produce a smooth solution taking the convolution. Remark. And so, I mean, this is quite important. Maybe it is also the reason why phi is called fundamental solution. You can construct using phi, you can construct solutions to this kind of problem by convolution. Remark, before proving this, assume, in addition, u 0 bar is not identically 0, but is, say, bigger or equal than 0. Then, u of tx is bigger than 0. What does it mean? So assume that your initial condition, Professor, what if u is less than 0? You use 0 bar? Yes, u 0 bar is 0 in measures. u 0 bar is this. So the striking fact. So this is x, and this is time, and this is u 0. So u 0 is identically 0 up to a small bump, very small, somewhere. Of course, u 0 is the bump is such that u 0 is non-negative. Then the striking fact of this solution, which makes me make it. Well, the striking fact is that now take any positive time. Any positive time. Then you look to this object here, and it happens that the solution. So let me do it a little bit larger. It's impossible to define a small bump, maybe very far. Maybe very, very far. And then you take any positive time. Well, at any positive time, your solution will be something like, but positive here. Positive, positive, not 0, strictly positive. So this means that if somewhere very far, the solution is positive, then instantly, and the initial condition, sorry, is positive, then instantly your solution will be positive here, almost at minus infinity. So this says that this comes from this expression. This says that there is this fact that the so-called infinite speed of propagation of the signal. Something that happens very far initially at time 0, then produces a disturbance here very far for any time, even for t equal 10 to the minus 1 million. So this is a property of parabolic equations, like the heat equation. And of course, it's not anymore true for the wave equation. We have seen, with example, remember, we have studied this initial condition. And we have seen that there were large regions where the solution remains 0. We have two waves interacting, but at the end, the solution was 0 in a large region. So the compactness of the support here was preserved. On the other hand, here, the compactness of the support is not anymore preserved. So this is maybe one of the most important differences between parabolic and hyperbolic second-order equations. In some sense, but maybe this is questionable, say, parabolic equations are not so physical, maybe, because it is difficult to imagine a disturbance that propagates with infinite speed instantly in real physics. This is maybe questionable. Of course, for the wave equation, we know that we have a finite speed of propagation. This is not anymore true in the case of parabolic. So this is really very, very qualitative important fact concerning the difference between parabolic and hyperbolic. And this is immediate from the fact that this kernel here is strictly positive everywhere. Of course, it goes to 0 very quickly and exponentially, but it is always strictly positive. And this fact has this implication. So we have this. So we can now prove the theorem. No, we cannot, because there is no time. So maybe tomorrow, so please, for tomorrow, try to do the part of the claim on the circles concerning the tickle of example. And then tomorrow, we'll prove this, which will be the last theorem on parabolic equations.