 OK. So the theorem of today, so let me recall the definition of the heat kernel, 1 over 4 pi t to the n over 2 into the minus x squared over 4 t. So the theorem is the following. So let u 0 bar. Maybe we can, since now we have just only one initial condition, let me call it u bar, maybe. In n infinity, intersection phi u of tx phi t x minus y u bar y dy for any t index in 0 plus infinity times rn. Then 1, u is infinity of 0 plus infinity times rn. 2, ut minus laplace of u equals 0 in 3. The limit for any x0 into rn, the limit as tx goes to 0 x0 of u tx is equal to u bar. Where now t is positive. So this is a way, hence, we can produce a solution to the problem, to the Cauchy problem, to the Cauchy problem, ut minus laplace of u equal to 0 and u equal u bar in the whole space. Produce one solution. We know that if the domain is unbounded, then the solution is not necessarily unique. This produces one solution. Let us prove this theorem. So let us consider uts plus hx minus utx divided by h. This incremental quotient is the integral over rn phi t plus h yx minus y, u bar of y dy, 1 over h, minus u of tx, minus u of tx, minus the integral phi t x minus y u bar y dy, which is equal to 1 over h integral phi of t plus h x minus y minus phi of tx minus y, u bar of y dy, which is equal to 1 over h phi prime phi t into some intermediate point by the Lagrange theorem bar y dy. By some for some theta 0, 1. So now let me compute the difference of ut plus hx minus utx divided by h minus the integral over rn phi t t x minus y u bar y dy. So I have to take the difference between this and this, so which is equal to the difference between phi t t plus theta h x minus y minus phi t t x minus y phi t t x minus y u bar. So this is less than or equal, say less than or equal than this times u bar y dy. Now we know that u bar is in L infinity, so this is bounded by a constant, the L infinity norm of u bar. So a constant times the integral phi t t plus theta h x minus y minus phi t t x minus y dy, which is equal to h, the constant u bar, the integral over rn phi t t, say, t plus theta prime h for some other, again the Lagrange theorem for some other theta prime, which is finite, by the way, this is finite. Therefore, by the property of phi, phi is rapidly decreasing. For any positive time, this is integrable on the whole space. Therefore, you see the limit. Therefore, we deduce that the limit as h goes to 0 of this is 0, because I have this h in front. And therefore, u is differentiable in time with such a derivative. So u is differentiable in time, and the derivative is this, and then you can see it is continuous with respect to x. In this way, you do this kind of argument for all derivatives in time space, and you deduce that u is infinity. Now, so we know that u is smooth enough, and therefore, now we can now start to show that actually, u is also a solution to the problem. Well, this is almost immediate, because let me now erase this. We have seen that ut, so u is infinity in the interior, and ut, with this kind of argument, is this phi t x minus y. And also, with the same argument, so you do this kind of incremental quotient for all derivatives in space, and therefore, you deduce that also this. Do you agree? It is enough to repeat the argument instead of taking the incremental quotient in time, you take the various incremental quotient in space. So that, for instance, the partial derivative of u with respect to xi will be the integral of the partial derivative of phi with respect to xi. The second derivative with respect to xi xj of u will be the integral of the second derivative of phi with respect to xi xj integrated, and so on. OK? So in particular, the Laplacian is the integral of this Laplacian. OK? Is it OK? So may I go on? OK. OK, so now, but then it is immediate to see that 2 is true. Why? Because ut at the point tx minus the Laplacian of u at the point tx is equal to the integral over rn of phi t at the point tx minus the Laplacian of phi at the point tx multiplied by u bar y dy by this simply subtracting these two formulas. But then we know that phi is an exact solution, is a solution of the PDE for positive times. You remember? Phi is a solution in the positive half space. But there is no x minus 1. Ah, because this is, thank you. Because it is a mistake, so thank you. There is always x minus 1. Sorry. Thank you. OK, x minus y, x minus y everywhere. Hence for any t, for any x, for any y, this parenthesis is identically 0 by the properties of the heat kernel phi. And therefore, u is a solution of the PDE. For any positive time, phi t of tz minus Laplacian of phi of tz is equal to 0. OK? Now you call z equal to x minus y, and you have the search. This is, remember, well, we found the exponential, et cetera, et cetera, as a solution of this, but just only for positive times. And then we have a singularity at time space equal to the origin. So therefore, we have proven also number 2. The most difficult part, actually, in the proof is to show that the initial condition is attained in this sense, in the pointwise sense, in the sense that this limit is equal to this. So for the moment, we have just only used, on your bar, for the moment, what we have used is just this. This is the only assumption. We have two assumptions on your bar, remember, because for us, it was bounded and continuous. OK, for the moment, we have just used to have that integral converges and so on and so on. In the proof of point 1, we have just used that u is bounded. So up to now, up to now, we have used only this. Now, so fix, so now the point is to show that actually the solution takes the initial condition in this pointwise sense, so pointwise. So fix x0, let epsilon positive. So we know and let epsilon positive and take delta such that u bar of x0, such that x0 minus y less than delta. Take such that if y is in Rn and x0 minus y is less than delta, then u bar of x0 minus u bar of y is less than epsilon. So we just, so we know, so u now is continuous. U bar is continuous. So since u bar is continuous, we can take delta. So now we want to estimate this difference minus u bar of x0. So we want to show that if x is delta closed, so it's closed in terms of delta to x0, and so epsilon is fixed. If for the same delta, if x and x0 are closed, say of delta over 2, for instance. And t is sufficiently small, because t is going to 0. And if this is less than epsilon or 2 epsilon or whatever, then we have the search. So let us estimate this. So this is the integral over Rn of phi t x minus y u bar of x0. But we know that the integral, if you remember, over Rn of phi of t z, say t z, d z is equal to 1. So what we can do now is to use this and to put an integral here, because we integrate with respect to another variable. So let me use the correct symbols. So x0 is OK, because the symbols are the same. Correct? So let me write this simply as the integral over Rn phi of t x minus y times u bar y minus u bar of x0. And this is the symbol that I have. Do you agree? Because also by translation, I mean not only this, but also for any point, there is this phi t z minus y z is equal to 1. Not only this, you can also translate there. OK. So now we are in this situation. And therefore, this is clearly less than or equal. So this is clearly less than or equal than the integral over Rn phi t x minus y u bar y. OK, so now we are focusing around the point x0. And therefore, we split this integral into the ball of radius delta, centered at x0 of the same quantity, plus the complement with delta of x0 phi t x minus y u bar of y minus u bar of dy. OK. Just split the integral into the ball and outside the ball. Why I do this? Because I can estimate immediately the first integral this way. Because this is less than or equal. Yes, epsilon, because now you see I have the difference between y and x0. And y in this integral is in this ball. The ball is of radius delta. Therefore, there this is less than epsilon. Therefore, this is less than epsilon, the integral over B delta of x0 phi t x minus y dy plus the same object here. So now I raise everything up to this. So this difference is less than or equal. Therefore, it's less than or equal than this plus this. OK. So this is less than or equal than this plus this. But now this, so I continue here. This is epsilon positive, and this is less than the integral over the whole space, which is 1. So plus, and I rewrite it, the integral out of the ball phi. So now I have the ball. I have x0 here, and I have the ball of radius delta. And I have the point y, which is outside. Now, if I take, the idea is the following now. The idea is rather simple. So now I have to see that this is small. When? I hope that this is small. Well, you see. Here I have the difference x minus y. And what I know about phi is that everything is concentrated close to the origin. So when t is very close to 0, and x and y are almost the same. OK. So now, assume that take now x in the ball of radius delta half of x0. So x is here now. So I have three points. The center, y, the integration variable, ranges out of the ball. This is delta over 2. And then I have the third point, x. And now I am assuming that x minus x0 are delta closed, say delta over 2. So now what happens to y minus x? So y minus x, yes, y minus x, which is this that I want to estimate. Exactly, this. I want to say that it is not small. If this is not small, then this is not small. And therefore, this will make the integral converging to 0. So I want to say that this is not small. This is the instance. I claim that this distance is larger than 1 half. For instance, this distance, this is not difficult. You see from the picture, this is out of the ball of radius delta. This is inside the ball of radius delta over 2. So this distance is larger than 1 half y minus x0. This is maybe homework or maybe we can try to prove it. So y minus x0 is less than or equal than y minus. So this is less than or equal than 1 y minus x plus x minus x0 by the triangular property, which is less than or equal than y minus x. Now x minus x0 is delta over 2. And then delta is also larger than y minus x0. So y minus x. So from here, we get y minus x larger than y minus x0 minus delta over 2 minus delta over 2. Ah, yes, yes. It's OK because now y minus x0 is larger than delta because I am out of the ball. So y minus x0 is larger than delta. And therefore, this is less than or equal than y minus x plus y minus x0 divided by 2. And therefore, this is larger than this minus 1 half of it. OK, sorry. So let me repeat it, the argument. I am estimating the distance from this to this. I use the triangular property inserting this point here. Now x minus x0 is by assumption less than delta over 2. So I continue here. But y is outside the ball. So y minus x0 is larger than delta. And therefore, this delta is less than y minus x0. Then, therefore, this is larger than this minus 1 half of itself. So why I'm doing this? Because now we'll take x. So this is, you should remember, what is this? Epsilon plus the integral out of the ball delta x0. 1 over 4 pi t to the n over 2 e to the minus x minus y squared divided by 4t u bar y minus u bar x0 dy. And in that set, we know that this is larger than this. Therefore, minus is less than. And therefore, this is, say, less than or equal than epsilon plus 1 over 4 pi t to the n over 2 integral out of the ball. And then I have e to the minus, finally, 1 half y minus x0 squared over 4t times u bar y minus u bar x0 dy. Now, of course, we cannot estimate this difference. But this is less than or equal to twice the infinity norm of your bar. 1, there is why? Sorry, 1 over 2. Because there is this 1 over 2 here. 1 over 4. Sorry. Thank you. Sorry. You're right. This is 4. Thank you. 1 over 4. Do you follow? Is it OK? So now this is one of our exercises that we made two lectures ago, maybe. For any, we wrote that for any positive delta, if I integrate out outside the ball of radius delta centered at the origin, then the integral of the square kernel of the heat kernel goes to 0. Now, of course, it's simply a translational argument. We are not integrated out of the ball of centered at the origin, but out of the ball centered at x0. But here we have y minus x0. So it's exactly the argument of the second exercise of last time. This goes to 0 as this product goes to 0 as t goes to 0 by one of our exercises. Hence, we conclude. Because what we have shown is that take a point x0, fix epsilon, take delta as the continuity of u bar at x0, corresponding to u bar at x0. So epsilon is given, delta is given. Now take x minus x0 less than delta over 2. Now take t so small such that this is less than epsilon. So take t bar positive such that t less than t bar t into 0 t bar implies that that object there is less than epsilon for any less than. Then the proof is concluded because if x is less than x0 of delta over 2 and t bar is sufficiently small, then we have that the difference u tx minus u bar of x0 is less than 2 epsilon. And therefore, the proof is concluded. Hence, the spirit of this proof is that close to x0, this is small because u is continuous. Far from x0, we cannot control this difference. But far from x0, the kernel is not concentrated. I mean, the kernel is going to 0 very rapidly. Because of that trick of the points, if I am outside here, y is here, x0 is here, then this must be large. This must be large. Once this is large, once there is some room here, then the kernel goes to 0. So this concludes the proof. Just one remark. The proof, I mean, it seems to me that the proof is correct under the following less stringent assumptions on u bar. So remark, proof is OK. So assume u bar in an infinity. Take x0, a point where u bar is continuous, then u, then the same conclusion of the theorem, same conclusion. Assume u bar, then u is infinity, ut minus laplace of u equal to 0, just only under these two assumptions. Next, take a point where u bar is continuous, then the limit as tx goes to t of u. So at all continuity points of the initial condition, we have that the initial condition is taken by the solution. Do you agree? Yeah, we use the continuity only in the third part if our proof is not wrong. OK? In particular, for instance, with this remark, one can consider an initial condition, which is, for instance, the characteristic function of a bounded set. For instance, u bar equal 1 if x is inside bounded set b, 0 else. So it is continuous initial condition. Just bounded, however, with two values. Then, where the function is continuous, the initial condition is taken. OK? So this is allowed. This is an initial condition, which is allowed. Then we can, so this concludes the part on the heat equation, on parabolic equation for the moment. So I hope that we will have time to discuss, again, the problem of existence once we have started some functional analysis. OK, so now we pass to some other argument. OK, so the next argument are, say, something about elliptic equations. Say something about the Laplace equation. So now we want to discuss the following PD minus plus of u equal to 0 in omega open set contained in Rn. So this is, you see, now in this problem, there is no time. So this is just a problem where time variable is not present anymore. So the solution, so definition, if u is c2 in omega and minus Laplacian of u is equal to 0 in omega, then u is called harmonic in omega. u is called harmonic. If u is c2 and minus Laplace of u is less than or equal to 0 in omega, this is called sub-harmonic. And of course, if u is minus Laplace of u is larger than or equal to 0, it is called superharmonic. These are just names. So in one space dimension, it is immediate to integrate u second equal to 0. So and already from this, I mean, in one space dimension, linear functions are the harmonic functions, one space dimension. And u second larger than 0 are sub-harmonic functions. So u second in one space dimension, u second larger than or equal than 0 sub-harmonic, what does it mean? Yes, it means convex. So already from here, we see that we can expect the maximum principle for sub-harmonic functions. You see, the maximum cannot be attained in the interior for a sub-harmonic function in one space dimension. But the minimum, yes, but the maximum, no, as we can see from this profile. And indeed, maybe I left you as an exercise to show a sort of weak maximum minimum principle for sub-harmonic and superharmonic function. We will discuss it. So in one space dimension, it is clear what is a smooth harmonic sub-harmonic, concave and convex and concave functions. Now, in two dimensions, in two space dimension, do you know how to produce solutions non-trivial? Non-trivial, I mean. Of course, linear functions are always solutions to this in any space dimension. But do you know in two space dimensions, for instance, how to produce non-trivial, non-constant, non-linear solution harmonic functions? Yes, if you know the theory of holomorphic functions, you know that the real part and the imaginary part of a holomorphic function is harmonic. Do you know this? So it is plenty of harmonic functions. I mean, I don't know, say, so what is this? Say, e to the x plus e y is equal to the x cos y of x and y. So take the real part, for instance, into variable u of x and y, and then you realize that if you take the Laplacian, it is equal to e to the x times cos y minus e to the x times cos y. So for instance, this is a non-trivial harmonic function. So if you want to think of examples of harmonic functions in two space dimensions, maybe it is very good to go to complex analysis. Indeed, the theory of harmonic function is strictly related to the theory of holomorphic functions. One space dimension. OK, now, again, we have a new PDE. And of course, any harmonic function is a solution of the heat equation once you add a new time variable. So I mean, if u is harmonic, then v of tx equal u of x, if you add a fictitious new parameter, this solves the heat equation in omega times 0 plus infinity. If u is harmonic, because v does not depend on t, so vt is equal to z minus Laplace of u is equal minus Laplace of v, which is equal to z. And therefore, this is a solution of the heat equation in this cylindrical domain, for instance. So this is a way to produce also stationary solutions to the heat equation, stationary in the sense that they do not depend on time. So now the point is that let us try to find, so look for new solutions. Look for radial, for instance, solution to minus Laplace of u equals 0. Meaning that I'm looking for a solution, ux of the form, say, v of x. Again, this is a way like when we discussed the heat equation, we looked for special solution in one space dimension of the form. It was like v of x over square root of t, maybe. This was our claim. And this claim reduced the PD to an OD. Well, now, again, we expect that if there are radial solutions, then we have to solve an OD. And so let us compute the Laplacian in this radial coordinates. So the derivative of u with respect to xi, it is v prime of x times the derivative. Everybody knows, I think, what is the derivative of this. Do you know what is this? Over the norm of x. I hope that this is clear for everybody. If it is not clear, it is immediate because this is by definition. Now I assume, of course, that I'm outside the origin. So look for radial solution, at least for x different from the origin, so that everything is smooth enough. And so you see, from this, you immediately compute this. So now the second derivative of u with respect to xi and xj is, therefore, the derivative with respect to xj of this. And this is equal to what? v second norm of x times this plus v prime of x. Then I have, I think, there is plus eta ij times this. And then there is another term here, plus v prime of x, xi. And then I have to differentiate 1 over x, which gives you a minus, I think, xi over x cubed xj. So x minus 2? Yes. Is it correct? Yes, this is the same symbol. This is the identity. It is equal 1 only if i is equal to j, 0. Just a symbol, delta ij, OK. Symbols. So now we take the Laplacian. Now, OK, we take the Laplacian. So please check the computations. So this is x over modulus of x. This is OK. So let us take now the Laplacian. It means that we have to take the sum over the trace of dash. So the sum for i equal to j. So summing the i equal to j, what is this? It's x squared, which divides with this. Therefore, I surely have this. Then I have v prime, v prime. And then I have, OK, 1 over x. Maybe I keep it. And then, which is the trace of this matrix? And because we are in n dimension. And then, what happens to this part? Again, this is an x squared, which divides this cube. And so it remains 1, OK? Is it OK? So we have to solve now an OD. So call for simplicity this equal to rho. So we have to solve v second of rho plus n minus 1 over rho, v prime of rho, equal to 0. Maybe this is known to you, which is the Laplacian in polar coordinates, in the rho coordinates. OK, so now, well, we can, if you want, v prime, call this f. So we have f prime of rho plus n minus 1 over rho f, equal to 0, f of rho. So assume that f never vanishes. So if it never vanishes, we can divide. This is equal to minus n 1 minus n over rho. So this is log, OK, assume that f is positive, OK? Log f prime is equal to 1 minus n log rho prime. And then we have that log of f is equal to 1 minus n log of rho plus a constant, which gives f of rho equal rho, another constant, rho to the 1 minus n, f of rho, equal to. So this is this. I think it was f of rho equal to 1 minus n. Some constant. And f was v prime, OK? f was v prime. Therefore, v of rho. Now, let us distinguish the case n equal to 2, with the case n bigger than equal to 3. This is a constant. Let me call alpha divided by rho to the n minus 2, plus a constant, gamma log rho plus delta. In two dimensions, if you differentiate, this comes to be this is 1 over rho. So n equal to is 1 over rho. Otherwise, it's this. So now we fix the constants. So then we give a definition. So the definition, the fundamental solution of the Laplacian, I think it is called, fundamentals, yes, for Laplacian, is defined as symbol. Let me use the same symbol, so we don't confuse it with the heat kernel now. We are in a different setting, OK? By using the same symbol, but then I have some constant, 1 over 2 pi, the log of 1 over x. And then I have some other constant over x to the n minus 2. I will explain what is this. We will understand the role of the constants. So we have fixed special constants. So this is, well, this is this in two dimension. So this is harmonic. This is an example of harmonic function out of the origin in the plane. And this is an example of a harmonic function out of the origin in space. What are the constants now? Omega n is the volume, the back measure of the bull of radius 1 centered at the origin. This is the back measure by definition. Small omega n, omega n. This is volume. Volume, the back measure, definition. So we have n omega n. So by scaling, omega n rho n is the volume of b rho, OK? And then I think also that if I am not wrong, n omega n is the area of the boundary. So maybe we should check this. Well, but for instance, take n equal to 2. So take n equal to 2 just to check something. So omega 2 is the area of the unit disk, which is pi. And the length of the boundary of the unit disk is 2 pi, 2 pi. In three dimensions, let's check the volume of the bull of radius 1 is what? It's 4, 3 pi, right? And the area of the surface, the surface area of the bull, of the solid bull, it is 4 pi, which is 3 times 4, 3 pi. So this is probably correct. Anyway, it is an exercise, an exercise. Maybe home, OK? So we have the, so notice the big difference between 2 and 3 and 4 dimensions, OK? In particular, the physical situation is n equal to 3, where we have 1 over x. So then maybe the Newtonian case. So remarks, let us see whether or not phi is smooth out of the origin, where at the origin there is a singularity. This is clear. So this is a singular, I mean, it's a smooth solution out of the origin. At the origin there is a singularity, like this, for instance, in this case. Now, how strong is this singularity? So is it true or not that phi is in L1 lock, for instance? I mean, can this object be integrated around the origin or not? Exercise. Is it possible or not to integrate it around the origin? Well, we compute. So let's see, in n equal to 2, the integral over a ball centered at 0 of the log of rho, then this is, say, dx. This is equal to the integral from 0 to rho of rho log rho to pi, whatever, the behavior up to a constant. I mean, 2 pi and so on, in two dimensions. So there is this factor coming from the Jacobian of the change of coordinates in polar coordinates, which makes this integrable, OK? So this is integral in two-space dimension. Let us see in n equal to, in any dimension, 1 over rho to the n minus 2, then I have to take the Jacobian, which goes like rho to the n minus 1. And therefore, this is surely, actually, is rho. And hence, it's integrable, OK? So maybe I leave you as homework to show, so home shows that the gradient of phi, there exists a constant such that we have this. And out of the origin, there exists a constant such that this is the action, c over x n minus 2 n. Sorry, it is worse and worse, worse and worse. And so the question is then, so this is the first exercise, just an immediate exercise. And then is this, no, OK, this is enough. OK, now, before going on, so as you can see, the estimates are worse when we increase the degree of derivative, close to the origin. Now, before going on, I think that I need something about Green's identity, so Green's identity. So the first Green's identity, so assuming that everything is smooth, integration by parts can be done, assuming smoothness enough on omega, bounded, and so on, and so on. So we have already seen that we have this fundamental theorem, which I don't remember the notation now. I have used about something like this. Are these the notation? So this is the outward exterior unit normal to the boundary. If you, of course, want to use the interior, normal, you change sign, OK? Now, this gives us the following formula, in particular, the Laplacian of u, if everything can be done. It's just d u over d v integrated over the boundary. This follows from this, just simply taking eta equal the gradient of u. Not only this, but also we can do this. If we have also this, assume that now you apply this with the choice v grad u, v grad u. So you find this is equal to v Laplacian of u plus scalar product grad v grad u. I don't remember if I have used the dot or the two indicators, scalar product. So maybe let us keep the dot. Sorry, dot. OK, then dot grad u. So expanding this divergence, this is grad of this, dot grad of this, plus v Laplacian of u, which is that one. And by this theorem, this must be also equal to the boundary of omega. And then we have, this is the new eta. Therefore, scalar product v grad u, dot, new omega. So we have this equality here. Maybe this is called the first green identity. First green identity. OK? So let's write the first green identity here v Laplacian of u. Sorry, I don't put the x just for making the formula shorter. Minus grad v grad u, plus v grad u, dot, new. OK? This is the first green identity. Very important to observe, it is obvious, that this is an integral on the boundary and these are solid integrals. Surface integral and solid integrals. OK? Now if I exchange v with u, I also now exchange in the symbols. I have v, sorry, u, u grad v times new. And therefore, we now deduce the second green identity. So this is the first one. Now this is the first green identity and then the second green identity is obtained usually by taking the difference. So second green identity says the following. This minus this, say, u Laplace of v minus this minus this, v Laplace of u, must be equal. So now you see these cancels because they are symmetric with respect to u and v. When I take the difference, this cancels with this. And then I have what remains is this minus this, u, dv over the new minus minus v du over the new. OK? So this is called the second green identity. OK, so this is the end of the lecture. And the next week, I will be here, I think, only on Friday. And we will continue, on Friday, we will continue the theory of elliptic equations, Laplace equation and Poisson equation.