 Tako. So let me recall the results of yesterday so. So given an open smooth bounded subset of Rn. We found that maybe our notation was this. The signs were, yes, correct, minus u of yd. This was the result of yesterday for a function u in C2 intersection C1 of the closure. And G was called the Green's function. G was the Green function associated to the open set omega and to the operator minus laplacian. Okay, this was the result of yesterday. For today, we would like to compute the case of the sphere. So omega, so assume just for simplicity that n is larger or equal than 3. Assume that omega is the ball of radius. Let me check R centered at the origin. And we want to find essentially the Green function. And the idea, as I told you yesterday, was to claim that so this is the fundamental solution. So for any x and y in Rn, for any x different than y. This is the fundamental solution of the laplacian. And then there may be some constant alpha, yes, alpha. Phi of x star y was claimed. We look for g of the following four. Okay, this is just a claim. And alpha to be chosen. Alpha to be chosen so that g of x, y, if x in omega, g of x, y equal to zero. Okay, so alpha must be chosen. Okay, so now our domain is the sphere. So let me make the picture of yesterday, so zero. Then there is some x here. So x is different now from the origin. X star, let me remember that x star is a long x. This maybe was x star. Was a long x. And then was the definition of x star. So given any x different from the origin inside omega, this is the bowl of radius r, vr. I define x star as a function of x. Therefore this function, it actually is a function of x and y because x star depends on x. X is different from the origin. And now we see that now if I take any point y, yesterday y was here more or less, any point y on the boundary, different maybe take y not lying on this line and consider the following triangles. Okay, so one check. So this angle, so let us consider the triangle of x, y angle o x star. Then thanks to this relation, so they have a common angle. Okay, and then they have also the same angles. And this angle is, so this is r. From here you see that x star divided by r is equal to r divided by x. Namely, x star, x star is equal to r square. From this relation, just keep the norm of this vector, the norm of this. There is a cancellation between one power and one power. And therefore the norm of this divided by r is equal to r divided by x. So this says that this divided by this, so this divided by this is equal to this divided by r itself by x star. So you see if you consider this angle, then you take the quotient of this divided by this, sorry, divided by this. So consider this angle, take the quotient of this divided by this. So r divided by x star, consider r divided by x star for instance, so this divided by this, then consider this angle and this is equal to, by this relation is equal to this divided by this, x divided by r. Okay? So you, and therefore these two, so these two, the adjacent edges in mon angle are proportional. This means that these three angles have the same angle. Everything is consequence of this definition. Okay? Okay, therefore now consider the following quotient, r divided by x star minus y, r divided x star minus y. So now this divided by this must be equal, this divided by this, now must be equal to this divided by this, right? So that r divided by x must be equal to x star minus y divided by x minus y. Okay? So, now let me go back to this. What is this? This is equal to 1 divided by n, n minus 2 omega n, 1 divided by x minus y to the n minus 2, because now the Newton, the fundamental solution is we are in three dimensions or more. Minus alpha divided by x star minus y divided by n minus, to the power n minus 2. Okay? Of course, yesterday we already observed, sorry, remember that this is trivial. I mean x in omega and x star is outside omega, say. Okay, now for which choice of alpha now we realize from this relation here we realize that if we put now alpha equal. So if I choose alpha equal r divided by x to the power n minus 2 what happens? Well, it happens that let me, it happens that, okay, n minus 2 omega n over x minus y minus 2. So this is the expression of g minus 2 and we see that if now I just substitute this quantity with that there is a cancellation of this denominator with that denominator there. Therefore, this is exactly equal to zero when y is on the boundary, y is on the boundary. You see? By construction now y is on the boundary and see, I simply substitute this with this then there is a cancellation of this against this and what remains is exactly this denominator to the correct power n minus 2 which cancels with this. So okay, when y is on the boundary we have that this is satisfied. So let me so at the end this is the green function for the sphere so maybe I should I keep this, I have to keep something because otherwise. Okay, so now assume that we want to solve assume want to solve the following problem the plus of u equal to zero in Br and u equal, which is the notation that I use g on the boundary g on the boundary of Br. So for simplicity I take just zero the right hand side to be to keep the discussion more simple. Okay. So now assume that I want to solve this and so we have a candidate for solving this. Candidate and this is in the direction of answering equation of yesterday. So now let me remember that we have this representation formula for any function u with the proper smoothness. Okay. So the candidate now is the following g. We know who is g and we know that this g is zero when y is on the boundary. Before this is not present. We also want to solve the harmonic function inside. So this Laplacian is zero and therefore the candidate is the following for any x in omega our candidate solution is the following minus u of y g over d nu x, y Yes. Well, it doesn't matter I think that it doesn't matter because at the end this function is smooth also at x. Because you see y is on the boundary if x is the center Yes. Yes. Because otherwise this goes at infinity. But this function here I believe is smooth also at the origin. Without of course. So you can you don't see any singularity there and therefore now the point which is the point so this is a candidate we don't know if it is a solution I mean if the solution exists it is necessary this. We have to show that this is a solution. Right? So if it solves if it is the solution if it is a solution it is the solution because we already know uniqueness. So what do we have to do? Well we have to show that the Laplace of this is zero we have to show that 2 is smooth enough we have to show that it takes the boundary condition so now this I don't need anymore I have my candidate and the what does it mean? Inside the integral ah yes excuse me thank you yes yes sure yes of course we must have something on the right hand side there is not anymore you of course only the data of the problem so the boundary datum and something which depends on omega ok of course maybe one comment before continuing you should at home try to make the same picture that we have made but changing the problem as follows assume now today we will work on this but assume that we want the other hand so this is called the Dirichlet ok this is called Dirichlet boundary condition boundary condition ok so the question is a little bit vague but you should try to think a little bit about it so the question is what happens to this picture that we have considered up to now when instead of considering the Dirichlet problem on the sphere you consider the so-called Neumann problem so the Neumann problem is different it is this this is so-called Neumann boundary condition so you should try to redo to do once more the same reasoning but with this new kind of boundary condition ok so but for today we confine to the Dirichlet so u equal to g to g on the boundary ok so now the point is first of all to compute this derivative here this derivative here so the idea is to write this expression in a more transparent way computing directly this derivative so let us try to do the computation without hoping make mistakes so now I have to erase this cancel everything in order to find the famous Poisson formula so now the point is to compute dg over dne on the boundary of omega ok so this is g so let us go slowly first of all we have that x minus yy square is equal to x square plus y square minus 2 xy so let me indicate the not by gamma the angles the angle between so maybe in the picture it is better to have it once more so ok this is x this is y and this is x star so this is gamma therefore what is this that is equal to the gradient with respect to y of g scalar product with nu ok I don't remember if a scalar product is a dot or what else is the dot scalar product with nu ok so I have to compute now the gradient of x minus y with respect to y and so what is this it is this times what times y y minus 2 x y over y ok so it is equal to y minus x y over y gamma x minus y please if I make a mistake so now so so this is this and I hope it is correct then I have to take by the way the gradient of this ok so the gradient of 1 divided by x minus y to the n minus 2 what is this this is minus n minus 2 x minus y to the n minus 1 and then there is the gradient sorry I go slowly because I I don't ok so it is this and therefore this is equal to minus n minus 2 divided by x minus y to the power n and then I have multiplied by y minus x y divided by y cos gamma ok please check so now I have to take the scalar product of this with y scalar product of this with this 3 unit normal so the scalar product so is equal and then I want to compute this of course on the boundary of the sphere but anyway this is equal to minus n minus 2 divided by x minus y to the power n and then I find multiplied by y minus x y ok let me check if I am correct now there is a mistake here mistake is that yes there is a mistake when I take the scalar product of this with this this is one sorry this is not present ok so this is not this concerns sorry there is also the constant here this is minus and then multiplied by so there is a minus this cancels cancel with that there is one and there is and there is n omega n this for the first part ok this concerns just the first part ok and also I have to remember there is also this constant so I have just computed the gradient the normal derivative of this part for the moment ok now I have to compute the other part now the angle between x star and y so now I need to do the same computation the derivative is with respect to y so the same computation and the angle is also the same between x star and y as always gamma because I want to differentiate this so with the same computation finally I have the following result the following result that is dg over d nu remember the constant is nothing else is nothing else minus 1 over n omega n 1 over n omega n and then I have here what do I have I have y minus x divided by x minus y to the n plus minus rx n minus 2 y minus x star divided by x star minus y to the n y equal r so let me check if this is correct because otherwise we cannot continue this is r because let me just substitute r is the norm of y which is r is just the norm of y it seems to me that this should be okay so now I want to rewrite this so this seems to be correct so for the moment it seems to be correct so now I want to I want to use something which depends only on x and y so now I have to remember that I have this condition and also the other condition which unfortunately I have erased which lies this norm with the norm of x minus y could you please so r over okay so we have this now let me try to substitute this and here this so the second one gives me x star minus y equal r x minus y divided by x okay so what is this? minus 1 over n omega n and what do we have? this I keep it as it is x okay x minus y dn minus r x dn minus 2 then I have r minus r squared divided by x divided by x minus y to dn and then I have also r to dn here and then finally also x to dn okay so let us try to simplify a little bit omega n x minus y to dn then I have r minus cos gamma I hope that this cos gamma at the end will cancel then I have this what do we have? I have minus I have r squared due to this against this then I have x squared due to this against this and then the rest remains r minus r squared over x okay so from this what I see I see so let me write just this part so this is equal to what? it is equal to x squared divided by r minus x okay do you agree? please check this so this minus cancels with this plus and therefore I end up with this expression which I rewrite as follows minus 1 over n omega n r x minus y to dn okay it seems to be okay so this is just a computation concerning this quantity here so our candidate actually is a little bit more transparent therefore u of x is equal to minus is equal to minus r squared minus x squared divided by n omega n r let me check this r yes divided by r and then I have the integral over the boundary of the r g of y divided by x minus y to dn there is a mistake minus and minus a plus so generally I have to check yes okay this is the candidate this is a famous formula is called Poisson Poisson okay again we don't have proven yet that this is a solution this is just the candidate of our solution but it is a little bit more transparent than before okay so now I don't need finally now this is x is inside omega is open so x is inside the ball and y is on the boundary therefore this actually is never zero and this shows also that u actually is very smooth inside because this is smooth so u actually see infinity inside in the interior yes indeed it is very important to realize that y is on the boundary coincident okay so u is infinity is infinity on br so now what do we have to show well there are essentially two things to show one thing is to show that this is harmonic and the other thing so we have to show two things one and the other is that for any x bar now on the boundary of omega the limit as x converges so once we have shown these two things we have solved our problem namely we have solved minus laplace of u equal to zero inside and u equal g on the boundary g is continuous therefore we still have to check these two once we have this then we know that this is the solution okay so what do we have to do let's try to compute the Laplacian with respect to x now I have to take the Laplacian of this with respect to x so I have to show that the Laplacian of this with respect to x if I show this the function will be harmonic in the ball so you see you can produce a lot of harmonic functions in the ball and also you can so this is called also the harmonic extension of g so g is given just on the boundary and you extend it inside which is not trivial at all far from being trivial and this is the extension so now what remains to show is this and then this okay so let us try to compute this d over dx i of this again I have to go very slowly because there is always the risk of make mistakes so now the derivative with respect to xi of this is equal to minus 2 xi divided by x minus 1, x minus y to the power n and then what do we have I have plus and then the derivative of this with respect to x n then I have x minus y to the n plus 1 and then I have the derivative of x minus y okay which is equal to minus 2 xi divided minus and what is this so this is equal n r squared minus x squared by x minus y n plus 2 times xi I think we agree so I say that the derivative of this is just equal to xi divided okay so let me check if I am making mistakes xi okay okay maybe maybe this is incorrect because let me write explicitly this is the derivative of the sum x1 minus y1 square plus xn minus y1 square and so sorry there is okay now take the derivative with respect to xj the derivative with respect to xi of now r squared minus x squared divided by x minus y to the power n so what do I find so I have to differentiate with respect to xj this quantity this so let us try not to make too many mistakes so this is equal to minus 2 identity ij divided by x minus y to the n and then I have minus n x minus y to the n plus 1 times the derivative with respect to xj of the norm so I have minus and then I have minus plus actually because this minus against this minus so I have plus n to n xj divided by x minus y to the n plus 2 multiply by xy minus y so this is the derivative of this so ok then minus n r minus x squared divided by x minus y to the n plus 2 then I have to differentiate that with respect to xj so I have ij and then I have plus n times n plus 2 r minus x squared times x i minus y y divided by x minus y I think this should be n plus 4 and then here there is xj minus yj ok so first mistake is that this is a plus because I have a minus here ok then other mistakes n n plus 2 nxj so in my notes here there is n plus 2 correct this is n plus 2 why n plus 1 n plus 2 this is a 2 so this is an expression of the action of the function and so what we have to do is now to take the trace and let us see what happens so take the trace ok so the mistakes were here a plus and here a 2 instead of 1 so now therefore the Laplacian of u at the point x is equal to where is it r squared 1 2 3 4 there is ah thank you yes of course is r squared ok take the trace of this matrix so now we have minus 2n divided by x minus y to the n now here there should be also where is the i here there cannot be only j there should be also i so there should be also i there so it was the derivative of this ah I have forgotten to rewrite maybe this 2 xi because there is a 2 xi check ok now I have to sum over i equal to j here i equal to j everywhere so this is n then I have plus i to the n plus 2 then there is 2nx square which is this against this and then minus 2n scalar product between x and y and this for what concerns just only this part then I have plus 2n again x square minus 2n xy x minus y to the n plus 2 everything has a plus plus minus n square minus x square divided by x minus y to the n plus 2 n plus 2 n plus 2 n plus 4 plus so this is the norm square which cancels with 2 of this so this is n plus 2 times n r square minus 2 square divided by n plus 2 correct? n plus 2 this is ok this seems to be ok so this apparently has a different power so I have to isolate by itself so this is equal to minus 2n divided by x minus y to the n remains alone then I can put plus 1 over x minus y to the n plus 2 rentesis so these are equal but unfortunately with the plus so 4n x square minus 4n xy then I have minus n square n square plus n square n square r square minus n square x square plus 2n r square minus 2n x square so there is a cancellation here there is a cancellation here so this is equal to minus 2n divided by x minus y to the n plus 1 x minus y to the n plus 2 multiply then what do we have 4n minus 2n is equal to 2n x square 2n x square so maybe also the n goes here so x square minus 2xy plus s square so and we want this equal to 0 meaning that this is equal to 0 if minus x minus y square plus if this is equal to 0 then also this is equal to 0 ok and this is fortunately seems to be 0 because this is equal to minus x square minus y square plus 2xy therefore this cancels with this this cancels with this and we are on the boundary of the board therefore this cancels with this so this is harmonic ok so this is harmonic so we have checked the first point and now we have to check the second point now this argument is similar to an argument that we made for the heat equation so if can I ok so as you as you can see for the moment concerning the PD part we have never considered the problem of existence essentially just representation but not existence why? because for the existence theory we need weak solutions we need souble spaces we need tools that for the moment so the idea of the course is that today will conclude the first part on PD so no existence just only representation or clever computations like this and then we start with the function analysis part which is more abstract gives us the notion of souble space and if there will be time enough then we will go back to the existence of solutions to be these using souble spaces ok this is the idea so today we conclude with this so now maybe it is better that I leave you as an exercise but I have to give you a hint for proving that now as I told you I mean we have to show is to show that the solution you keeps the boundary condition in the pointwise sense this means this now this sort of argument because of the fact that there is a kernel here which is singular is not trivial to show this remember that we have done similar computation for the heat equation when we show that if we take the convolution with the heat kernel of the initial condition then the solution was keeping the initial condition and that was that was an argument that is similar to this one so how to show this the hint is what we did for the heat equation we were able to write here there was the convolution with the heat kernel and not with this of course and was an integral also not a surface integral but in any case what we did in that case was to write g of y minus g of x bar and why we could do so in that case because the heat kernel is normalized to have integral equal to 1 in space and so in that case we found the g of y minus g of x bar and therefore since g was continuous then we could take epsilon delta and so on but here there is no g of x bar so how can we do how we can face this problem and write instead of this g y minus g of x bar well, the trick is the following you consider this formula with the function identically equal to 1 so you have the following interesting formula the following curious formula one is harmonic so we can write n omega n r integral over b r 1 divided by the trick is to use this strange this strange equality now instead of one you put the constant here and the constant is g of x bar and the constant now goes inside and outside the integral therefore if you want to consider the difference u of x minus g of x bar we have to consider the difference we have to consider the difference u of x minus g of x bar we have to consider the following now fix epsilon positive there is delta such that for any y v r y minus x bar less than delta g of y minus g of x bar is less than epsilon so this suggests to split so give an epsilon this suggests to split to split this integral into two contributions one contribution so let me call i delta this is i delta this is x bar on the boundary of the ball then I have all points this is the ball of radius delta centered at x bar and then I have all these points here so all points on the ball at distance less than delta so let me call this i delta and therefore let me split the integral in i delta and outside i delta so this is equal to r square minus x square divided by n omega n r the integral delta g of y minus g of x bar divided by x minus y to the power n plus r square minus x square divided by n omega n r integral outside so the boundary minus i delta g of y minus g of x bar divided by x minus y now given epsilon if y is in i delta then we have this therefore at least the first sum so at least the first sum can be bounded estimated as follows this i1 so this is this is a and this is b so estimate of a is less than epsilon because this is less than epsilon when y is in i delta and then what remains is r square minus x square sorry minus x square divided by n omega n r actually i delta 1 divided by x minus k to the n ok, this is less than or equal than the integral over the whole sphere and this is equal to 1 by the previous observation 1 equal to if I put 1 here so this is equal to epsilon because this is equal to 1 hence we only have to estimate b so we have x bar delta this was i delta then let me take a smaller ball say delta over 2 omega x in dr and assume that x minus x bar is less than delta over 2 ok, so x is somewhere here x and y now however concerning b y is outside of i delta so y is here you agree? y is here this should be small given epsilon because ok, this is now I cannot use anything about g, g is continuous I don't know if this is smaller or larger just a constant twice the infinity norm of g ok, so for the moment I forget g so I have just to estimate r squared minus x squared this is a constant so I have just to estimate this product 1 over ok I mean, if I estimate this product then I am ok because I throw it away this difference because g is bounded constant and infinity norm so I have to show that this is small if this is small then everything is small so ok when x is sufficiently close to x bar say less than delta over 2 now x minus y is larger than delta over 2 because I have this intermediate region which gives me delta over 2 so x is here y is outside therefore the difference is at least x minus y is this so 1 over x minus y to the m is less than or equal than 2 over delta to the m to the n, sorry n well and hence this is controlled by this and then why this is small well because this is does not blow up but this is small because x is convergent to x bar and x bar is not equal to r so give an epsilon there exists delta such that if x minus x bar is less than delta over 2 then this product becomes smaller than epsilon over 2 or whatever this goes to 0 this goes to 0 as x converges to x ok so this I think concludes the so you have seen this part is complicated harmonic functions it's very huge there are books of this size only for harmonic functions and elliptic equations so this is just the beginning of the story and I think that you can find all these arguments in the book of Evans more or less in different order there you can find everything so this concludes the part on PDs so we are about half of the course more or less and so now there is no time to begin but I just try to know what do you know about function analysis do you know what is a Hilbert space so you already know what is a Hilbert space and the Banach space also fine and you have studied LP space surely ok and so maybe of course you don't know what is the hand Banach theorem extension theorem ok so next time we will start with a quick list of definitions maybe on Hilbert and Banach spaces but very quickly because you already know have you already studied the spaces of sequences the most important one maybe is this have you ever seen this space so do you know it is complete it is separable fine no, you don't know but you know what is a Hilbert space and Banach space this you know ok so I would say maybe two words about this actually this is very very important because it is the typical Hilbert space typical separable Hilbert space so it is very it is very very complicated this object here space of sequences in L2 so I would say maybe something about this and then we will start with hand Banach theorems and linear functional analysis ok