 OK, so let me recall the notation for the fundamental solution. So I will use this notation here today. This is the fundamental solution for the heat equation. So it is 1 over 2 pi log 1 over norm of x minus y, et cetera. So this is the fundamental solution defined for x different fundamental solution. The dimension here is larger equal than 2. And we want to show the usefulness of this function by proving the following theorem. So assume that f is a continuous function, actually c2 function, with compact support in the whole space. Then so define u for any x. Then for any x, the function u of x, which is equal to phi of x minus y, f of y, y, solves, is a solution, Laplace of u equal f u n. Maybe it is better to write first the regularity. So this is the usual convolution between, sorry, minus. So remember that this is in L1 log. This is in L1 log, phi is in L1 log, f has compact support. Therefore, we can compute, we can define the convolution between f and phi. There is no problem. And then this is one solution to this elliptic PDE in the whole space. We want to show this. So the proof. Now maybe I can leave you as homework showing that u is c2. Well, this follows as we have already seen something similar, I think, in the study of the heat equation. You fix a direction, a EI, a unit vector in top. This is a unit vector. And then you fix a point, and then you consider the quotient. And then remember that you can always write this as. And then you can therefore write this quotient simply in terms of a quotient here. OK, so the homework consists in the following. You write this quotient writing here f of x plus h EI minus y. So this is actually is equal to the integral over a rem f x plus h EI minus y minus f x minus y. Now the point, now we know that f is in c2. Therefore, we know that as h goes to 0, this converges to the partial derivative with respect to xi of f at the point x minus y. And therefore, we can pass to the limit because f has compact support of class c2. This converges to the partial derivative. We can pass to the limit under the integral psi. So this converges as h goes to 0. So this is actually homework, but just a sketch. df over dx i x minus y. OK, hence, there exists the limit also of this left-hand side as h goes to 0, and this limit is this. But this function on the right-hand side is continuous with respect to x, et cetera, et cetera. So this is the sketch of the proof. And so you do this also for second derivatives and so on. Now let us prove, so the regularity is OK. Now the point is to show that the PDE is satisfied. OK, so now the idea is the following. So remember that at x equal to y, we have that this is a singular point of phi. So what do we do? We start by computing this object here at any point x. And this actually, the best is to look at the convolution in this form and to differentiate twice f. And then, so let me write it like this, laplacian of f x minus y dy. Now the idea is, of course, now maybe it is better to use a symbol. So this laplacian is considered with respect to x. I am differentiating with respect to x. So this is clear. And this formula follows arguing as here. So for instance, the partial derivative from this kind of argument for homework, you see that du over dx i at x is equal to usual property of the convolution. And therefore, also the second derivative you will prove that also the second derivative xi xj at x is equal to second derivative xi xj dy dy. And therefore, in particular, the laplacian. So from this kind of argument, it follows this and this. In particular, u is c2, not only, but if and how I take the sum over e equal to j of this expression, I obtain this. Is it OK? This implies this. Take the sum. This equality implies this. OK, now the point is that we would like to use integration by parts here. But we have to be very careful, because phi is not differentiable. It's singular at the point. Particular phi is singular at the origin here. Therefore, it's not clear how to compute the integration by parts. So the first thing to do is to split this integral into two integrals. First, an integral out of a small ball centered at the origin. So remember that the origin now is a singularity. Now I isolate a small ball b epsilon centered at the origin of radius epsilon. So and I write this integral as the sum of two contributes, namely these two objects here. So that here I can surely integrate by parts. So now let me follow the notation. I call this equal to i epsilon. And this equal to j epsilon, maybe. No, sorry. This is i epsilon in my notation. And this is j epsilon. This is i epsilon. And this is j epsilon. Now, what about i epsilon? We can start to estimate i epsilon as follows. So in two-dimension, I have that i epsilon is less than or equal than what? I have a constant here, a constant here. And then I have a constant, which is the infinity norm of this function. This is c2 of compact support. Therefore, this is bounded. i epsilon, yes. So this is the absolute value of this is less than or equal than a constant. And I call this constant c. Then I have the integral over b epsilon of the log of epsilon. Right, essentially. Log of y, maybe. Log of y. Is it OK? For n equal to? Yes, I'm taking the situation, the fundamental solution. This constant, maybe there is some pi 2 pi or so. I have this infinity norm times 2 pi and so on. And then this is less than, say, this is, so this goes as epsilon squared, because it is two-dimensional. This goes as a log. Therefore, this goes to 0, as epsilon goes to 0. So when epsilon goes to 0, this is negligible. And when n is larger than or equal than 3, again, I have a constant. Then I have epsilon to the n, which is the volume. And then I have something which is 1 over epsilon to the n minus 2, which again goes to 0. So that this is negligible as epsilon goes to 0. So let us concentrate attention on this object here. Now we can integrate by parts here, because now everything is smooth. Because phi is smooth, because I am outside. I'm far from the singularity. Epsilon far from the singularity. So now I can integrate by parts. And so I can write it that J epsilon is equal to what? So it is minus the gradient. No, maybe there is another trick here. I want to integrate by parts. But if I want to integrate by parts, I need the derivative with respect to y and not with respect to x, right? But they are equal. Because two derivatives, minus times minus, equal plus. So I can use this small trick so that now I can now really integrate by parts. And so I have the gradient of f against the gradient of phi with the minus in front. And then I have also the boundary term. dy. So I have phi of y, scalar product, where nu is outward. So I am integrating. I'm working on Rn minus a bowl. Actually, not quite, because f has compass support. Therefore, actually this integral is not on the wall Rn minus a bowl, but is somehow here, say, if this contains the support of f, I am integrating in this annular region, essentially. So this is an open, smooth bounded set. Therefore, I can apply the integration by parts. And therefore, this volume integral becomes a volume integral with the minus. And then there is a boundary contribution, but there is no contribution on this part of the boundary, because f is 0. And the only contribution is here. And nu is exterior. Is this clear? If I put the plus here, nu is exterior to the domain. So exterior to this circular annulus on this boundary means invert with respect to the bowl. So this is exterior. Unit and exterior. So let me give you another notation. So let me call this, I call this k epsilon. And then let me call this L epsilon as to follow the notation of Evans. Now, again, for the previous, arguing as in the previous case, look at this. So this is bounded, because this has length 1, and this is bounded. Therefore, this is bounded. Then I have to integrate this. And again, I have, now it is worse than before, because this is length in one dimension. This is just length epsilon. But this is a log. Therefore, this again goes to 0. L epsilon, which is the length of this in two dimensions, epsilon times a log. Let's see in any dimension, epsilon to the n minus 1 divided by epsilon to the n minus 2, which still goes to 0. Therefore, also L epsilon goes to 0. So this goes to 0, and this goes to 0. So what really matters here is, in the contribution of the integral at the end, when I will let the limit as epsilon goes to 0, what really matters is just this integral here. So let me concentrate just on k epsilon, k epsilon. k epsilon, again, I can integrate by parts once more. So I'm using again, just remember, the usual formula, divergence of u eta. The integral is equal to integral of u eta scalar nu on the boundary of omega. And it is also equal to the eta plus u divergence of eta. So I am applying this with the choice of u equal to f, so that I have grad u times grad eta equal grad phi. So grad u times grad phi with the minus is equal to u divergence of eta, which is the Laplacian. Thank you. Sorry. Thank you. Thank you. R minus the ball. And then I have the scalar product between f of x minus y, scalar product of grad phi y mu. Do you agree? Now there is a big advantage. I claim that this first integral is exactly equal to 0. Why is this so? This is the first volume integral is identically equal to 0. This is 0, because this is harmonic out of the origin. In particular, it's harmonic out of the ball centered at the origin. So you see, essentially, what we are doing here is trying to put this Laplacian here. But if we do simply this, then it happens that this is 0. And indeed, this is false. It's not true, because we have to be careful. The point is that phi is singular. So we cannot really put this Laplacian here, and then this is 0 identically. Because of the singularity. And therefore, so this is 0, but this is not 0. And this is the origin of our computation. So OK. So we end up with this object here. So now I keep only this. Let me work a little bit on that integral. So let me rewrite it so that I see it more clearly. So our Laplacian is equal to minus the integral over the boundary of the ball of f of x minus y, scalar product. So we have to compute explicitly the gradient of phi. So now we compute explicitly this quantity. So we do this with the constant, with the proper constant. So which now I will write here. So the gradient of phi with the proper constant r, so phi was equal, say for instance, when n is larger than equal to 3, was I think n minus 2 omega n 1 over y to the n minus 2. This for instance, when in dimension larger than 2. And then this, if I take the gradient of this, then this is equal to minus 1 over omega n. Then I have here y n minus 1. And then I have the gradient of y, right? We agree? And therefore, this is equal to minus 1 over omega n, y divided by y to the n. No, there is a mistake. There is an n missing here. Here multiplication. I don't remember the constants. Maybe you're right. So check the constant, please, on the normalization constants. There is also an n, n times n minus 2. OK, maybe yes. So apparently there is another n here. So just a matter of constants here. This is in dimension larger than or equal than 3. So that is grad phi and the new n. And new is actually, we have said that new is exterior to the complement. Therefore, it is equal to minus. So new at y is equal to minus y divided by y with the minus. Hence, if I'm not wrong, this is equal to minus. Boundary over B epsilon, f of x minus y. Then I have the constant. There is a minus n omega n. And there is another minus here. Minus and minus is plus. So I have 1 n omega n. Then there is also this minus, which I keep. Then I have the scalar product between this and this. OK. So this is equal to y divided by y times y divided by y to the n. So this is equal to 1 over y to the n minus 1. 1, OK. But y is epsilon. It's epsilon because I am on the boundary of the ball. For it seems to me that this is equal at the n minus 1 over n minus 1 over n omega n epsilon to the n minus 1. d h n minus 1 y. This is very nice because this is the mean value up to the sign. This is exactly the length, the area of this surface. Because this is omega n minus 1. Remember? This is omega n minus 1. Therefore, this times this is exactly the area. Therefore, this is the mean value of the function, which now f is smooth. When I let epsilon goes to 0, this converges to minus the value of f at the center. Because this is simply, if you wish, this is simply minus the mean value by a change, by a translation. You center it at x. So you translate the coordinates. And then you have to take the mean value, the surface mean value around x of f. When epsilon goes to 0, this converges to, since f is continuous, this converges to the value at the center, which is x. Therefore, I keep this minus. But this was plus laplacian of u. And therefore, so we have shown that minus laplacian of u is equal to x. Because this is a minus. And we start from plus laplacian. So you see now it is very useful, this notion of fundamental solution. Because, for instance, it gives us one solution of the Poisson. This is called Poisson equation in the whole space. So this shows the theorem. And now we can continue with other similar computations. So now I want to give you a representation formula. So theorem, maybe this is called the third green formula. So let me remember maybe the second green formula. Please check u laplace of v minus v laplace of u dy if omega is smooth and bounded. And if u and v are c2 of intersection c1 omega bar, we have this kind of formula here, u dv over d nu minus v du over d nu. So please check in your notes if the signs are correct. Lipschitz or omega is smooth and bounded. Or also, Lipschitz is enough. But let me take a smooth c1, c2, whatever for the moment. So this is maybe the second green's identity. It's just a simple application of integration by parts, usual stuff. Now we want to write. And notice that here there are no assumption on harmonicity, just valid for any u and v sufficiently smooth. No PDs. And again, also here there are no PDs. So let me take a function u, c again c2, under the same assumption on omega, c1 omega bar. Then for any x in omega, there is a representation formula of u without assuming any PD. And the representation formula is the following. So I have a first volume integral that maybe I write here. Now phi of y, phi of x and y, Laplace of u. Y, plus the integral of omega of phi x minus y, u over u minus u. So maybe in some books, the notation is slightly different. Maybe in some books, this is indicated by phi of x and y. Actually, it's a function of the difference. But we can more generally write like this. So this is also sometimes written as phi of x and y. And this is also phi of x and y. And here, what does it mean? This means taking the gradient of phi with respect to y, scalar product with nu. Because now the variable of integration is y. Therefore, this is the gradient of phi with respect to the variable y, scalar product, nu. Let me check the signs if they are correct. The plus, the plus, and then minus. So how to prove this? So notice it is interesting this result because there is no assumption on nu. Of course, in particular, if u is harmonic, then this is 0 in particular. And say if u is harmonic, and this is interesting because if u is smooth enough, and say in particular, if u is smooth enough, u is harmonic in omega. Say, and u is equal to g on the boundary, for instance. In particular, if I am able to prove that this has a solution, if it has a solution, then u must be equal to 0 plus this is g, and this is what it is, u over d. So this, of course, does not prove that this problem has a solution, but gives a representation of a solution. If you are able to prove that there is a solution, then this is a representation theorem. It's not an existence theorem, just a representation. But before going to this, let us prove this kind of formula. So fix x and take a usual small ball b epsilon centered at x. Small enough so that this ball is contained in omega. And therefore, we can now compute, again, similar trick as before, starting from this equality section. So we integrate over omega minus b epsilon x. The Laplace of u times phi of x and y. We compute this quantity. So let us compute this quantity here. Well, to compute this quantity, we can use this formula. So we replace. Now, we use this formula with the following notation. In place of omega, we take omega minus the ball. Yes, I use this notation. So in place of omega, I take this new omega. OK? u is u, and v is phi. v is phi. x is fixed at this moment. x is fixed, and I'm integrating with respect to y. So now I apply this formula with the following. x is fixed. u is equal to u. And v of y goes to phi xy, x fixed. Everything is smooth enough, because on this new omega, I am just throwing away a small ball around the singularity, which is x. And therefore, this u and this new phi are smooth enough. And omega also is smooth enough. OK? So from that formula, I end up with the following integral. Omega minus b epsilon x of u of y, laplace with respect to y, phi xy minus dy minus integral. The epsilon x is equal to what? Is equal to u of y over the new phi xy minus phi xy, u over the new y, plus the integral over the ball, the same quantity, u of y. Now, OK, nu is exterior, again as before. Here, this nu is like this. Minus phi of x and y, u over the nu, y, dh, n minus y. OK? Now, laplace of u, laplace of phi. I hope that the signs are correct. So let me check first what happens to this. Again, using exactly the previous arguments, arguing as before, u is smooth enough, so it is bounded. And so this goes to 0, as epsilon goes to 0. And this goes to 0, as epsilon goes to 0. OK? Therefore, these are negligible, exactly with the previous argument. Now, I have a minus here. I hope that the signs are correct. You see, now let me look at this. So this with the minus is this, because it's easy to see that if you compute this integrand on the small ball, again, this goes to 0. So this, as epsilon goes to 0, converges to this. Is this clear? It is clear, because this is a long lock. This is C2. So this converges to this, with the minus. Then, this with the plus is this. If I put this on the left-hand side, I find it with the plus. And then, if I put this on the left-hand side, I find it with the minus. And therefore, I just only have to show that this with the minus, with the minus. So it remains to show that minus integral over omega minus epsilon x u of y, plus phi xy dy, converges to u of x, as epsilon goes to 0. Do you agree? Do you agree? So let me repeat it. I have the second Green's identity, which I apply with omega in place, with this in place of omega, and this in place of u. Now, I first recognize that these contributions goes to 0, exactly for the previous reasons. Because u is bounded, and the grad u is bounded. I integrate over epsilon, an object which goes as epsilon to the n minus 1. This goes, one shows that this times this, and this times this, again, goes to 0. Therefore, what remains is this, plus this, plus this. And this, again, also, this goes to 0. It is immediate. Therefore, this red quantity goes to this red object. Now, what remains to show is therefore that if I put this on the right-hand side, so I put this with the minus, this converges to u of x. And this is the previous computation, computation of the previous reason with the minus. So let me keep this statement here on the blackboard, third Green formula. So now, from this, we want to continue the theory. So now, let us slightly modify this. This will be useful for the introduction. So also, let me keep the standard notation phi of x and y. So let I have it phi of x and y, phi of x and y. So remember that this is the gradient of phi with respect to y, scalar product mu. So there is no, I mean, everything is well-defined. OK, now let me work a little bit on this and assume that now h, let me call it h, c2 omega intersection, c1 omega bar, is harmonic in omega. Let me call this h so that we have the following 0. I have raised, unfortunately, the second Green's identity. So from the second Green's identity, I have h du. We have this from the second Green's identity, right? OK. Do you recognize it? Because, let me write it very quickly, h Laplace of u minus u Laplace of h equal h du mu minus u dh du. OK? So if Laplace of h is identically equal to 0, this is 0. And therefore, h du over du minus u dh over du minus h Laplace of u is identically 0. OK? I have this equality here. Now I can add this to this. So let me call this formula 1. Let me call this formula 2. OK, and let me introduce the following notation. Notation g of x and y phi of x and y plus h of y. So let me call capital G is this formula. And so now let us add 1 and 2 together. So let me put right first the end. So phi plus h, which is g, du over du minus u phi plus h dh minus 1. And then I have minus g. OK? Minus g. OK, so let me keep this information on the blackboard. And this. So this is a representation formula. And now let me make this comment. Suppose, remark. So this formula, when this formula is valid, is valid when u is sufficiently smooth, when I am able to find such an harmonic function with this regularity. And so under the omega is bounded, this book. So under this assumption, if I define capital G like this, I have always this, OK? Now, suppose we can show that g of x and y is equal to 0 for any y on the boundary of omega. Assume that we are able to do this. Namely, assume that we can find h. So assume that we can determine h so that g, defined as s, follows as the following property. Assume that we can do it. So assume that we can determine g so that, in addition, the following property solved. Notice that x is inside and y is on the boundary. Therefore, x is inside. y here, in this equality, is on the boundary. So therefore, it never happens that x is equal to y, where phi is not defined. This never happens in this. Now, if I am able to determine capital G so that this is true, then I look to my representation formula. And so I have, then, for any x in omega, u of x is equal to the integral. Now, here you see y is on the boundary, and this is 0. Therefore, what remains is very interesting, because what remains is minus u of y, gradient of g with respect to y, scalar product with nu, minus g of x and y, Laplace of u, y. Now, you see what this is more. This now makes more precise in my previous comment. Now, assume that you are able to show, hence, if we are able to solve the following problem. Minus Laplace of u equal to, say, f, no, sorry, u into c2 omega intersection c1 omega bar, comma. Minus Laplace of u equal to f in omega. And u equal to g on the omega. If we are able to solve this, then we have a representation formula, which is quite interesting, because, necessarily, if this is true, then, necessarily, u is not explicit, I mean. But it is representable using only the data of the problem. Namely, it is equal to minus g plus. Minus Laplace is equal to f. And so we have proven a sort of representation theorem under several, several assumptions. So let me now list the assumptions. So the main assumptions are, so let me now, OK? So do you agree with this? You see now that on the right hand side, there is not u anymore. There is not u, because this is f and is given. This is g, which is given. And then, of course, there is this function, capital G, which is very difficult to find usually. So which are the assumptions? The assumptions are existence of this problem, essentially, with this regularity. We always want this regularity, because we always use the second Green's identity. And the second Green's identity, you need the derivative of u on the boundary. And therefore, it is better to have uc1 up to the boundary. Next, we also are supposing that we can determine capital G so that this following property holds. So under all of these assumptions, we have somehow implicit representation of the function of the solution of a Poisson equation of this sort, OK? We can find capital G, yes. Yes. Yes, at the end, yes. This is actually a good comment. I mean, this is not an existence theorem. It's just a representation. But it suggests you how to prove existence, because then you try this, assume that you have G, then you try to prove that this actually is a solution. And indeed, this is what we do usually. It is also, this suggests also how to prove existence. OK, maybe now I keep this. This is very, very difficult, usually. Yes, it has, now let me, yes. How to find G usually depends on omega. It depends on omega. And so for particularly simple omega, it is possible to find explicit G. For omega generic, it's usually not. So now let me give you a definition. Definition. So let fix x in omega. A Green's function G of x and y, Green's function for the operator Laplacian on omega is a function of the form G of x and y equal phi of x and y plus x is fixed now. So if I consider x is fixed, now I look at h as a function of y, where phi is the fundamental solution, Laplacian. h, maybe this is enough, h is such that it is better to give a formal definition to what we have deduced up to now. So this is the famous Green's function for omega. So now as he has observed just five minutes ago, essentially, the problem of solving minus Laplacian of u equal to f on some domain omega, u equal to g on the boundary, is essentially reduced to find the Green's function. Maybe homework, if g exists, then it is. Try to show this. Use the maximum principle, because you have to show, essentially, that this h is unique. But this h satisfies a PDE, because it is harmonic. And on the boundary, it has a suitable phi. So this is almost immediate, right? I mean, h is unique, because the plush of h is equal to 0. And on the boundary, this plus this is equal to 0. And therefore, this on the boundary is minus this. This is the proof. OK. So I would like also to rewrite once more what we have shown up to now, theorem, that omega be open, smooth, bounded. Suppose that maybe sometimes in the book, the Green's function is denoted also by g of omega, maybe, sometimes. OK. Just to remember that it depends on omega. Suppose that there exists the Green's function for that omega. Then if u is a solution, if u solves u into c2 minus laplace of u equal to f, u equal to g, then OK. So now we want to make an example for this. And we want to find the Green's function for the sphere. So do you agree for this statement? So there are a lot of assumptions. It is clear. Regularity assumptions, the existence of the Green's function, which is usually difficult problem, and the existence of this, regular enough up to the boundary, also is non-trivial, and so on. Of course, say that g and f here are continuous, j and n, so that everything is well-defined. f and g are continuous. So this is the main theorem of today. And now it is important to make at least one example of the computation of the Green's function. So there will be no time to conclude this example today. So we just start, and then we see continue tomorrow. So here everything is strictly related to potential theory in physics. So the Green's, the fundamental solution is the potential generated by a point-wise charged particle at the point x. That is, the electrostatic potential generated by this, which is singular attacks, of course. And so if you have studied electromagnetism, electric field, the magnetic field, you surely recognize a lot of analogies between this and the potential theory. u is the sort of potential generated by distribution of charges. So now let us take, as omega, the ball of radius r. And let us do the computation just only in dimension and larger or equal than 3. So now take a point x in omega and define the following new point, x star. x star actually is a new point depending on x, which is along the direction. Now, home, check that x star is outside. So if x is inside, this point is outside. It's sort of a specular point. Now we will see in which sense it is specular. Home, 1, 2, x. Therefore, you see, if this is closed, sorry, x is not the origin. So take a point which is not the origin. From this relation, you see also that, so take this point, and now we'll make a picture. If x is very close to the origin, so this is very small, then this must be very large, because the product must be constant. It's sort of inverse. Now let me try to do a sort of picture. So this is Br. Then I have a point x here. This is the origin. And this point, say, is x star. You see, x star is in this along x, same direction, but with a different length, depending on x. Sort of fine. Now we look for g omega of the following form. There is a singular part here, which is the phi of x and y. And then we look for some, now we look claim. We look for g, a special form, a fundamental solution, minus some constant to be determined. Fundamental solution, however, here there is x star. Alpha to be determined, and alpha must be determined so that alpha to be determined so that the boundary of the ball. So now, home, home so that tomorrow we can be more quick. So take a point y here. So take a point y on the boundary of s, of the ball, and consider the following triangles. So the triangle I'm considering are o, y, x star. And the other triangle is o, y, and x. y is a point on the boundary of omega. Once you have x, x star defined as here, sort of specular point of x with respect to the boundary. Yes, also you can do, but generically take this situation. So, OK, assume that y is not this. Take all the other y's. Now, check that these two triangles, these triangles have same angles. Check this, that these two triangles have the same angles. Of course, you have to use this definition of x star. So they are similar. These triangles have the same angles. Of course, they have an angle which is common to both. But then I claim also that the two adjacent edges to this angle are proportional. So try to check this so that they have one angle. And the two adjacent sides of each triangle then they have the same angles, OK? Namely, this is equal to this. So check this, and tomorrow we will compute. Now, maybe I can anticipate for you what is the green function at the end for the sphere. Maybe you can try to do by yourself. It's not easy. It's not easy. But the green function for the sphere has the following expression. 1 n minus 2 omega n 1 divided x minus y n minus 2 minus, OK? So this is much more difficult. Try to show also these two, three, and four.