 OK. So, today we will define what is a distributional derivative, and then we will do some exercises on distributional derivatives. So, let omega as usual be an open subset over n, let t be a distribution, and let i be an index between 1 and n. OK. So, we give the following definition. We define dt over dx i as follows for any definition. So, we give the following definition. For any test function phi, this object here, tested against phi, is by definition t, the minus in front, tested against this derivative, partial derivative. OK. So, it is clear that the phi over dx i is again an element of the omega. The four, the right hand side is well defined. And also it is clear that this is a distribution. How can we check that this is, so, the remark is a distribution. How can we check this? So, do you remember the criterion for any k compact set contained in omega? There exist two constants mk, such that there is no space enough, sorry, less than or equal. This was the criterion, right? Therefore, and this, so, this is if and only if t is a distribution. OK. Remember this. Now, if we do, if we now compute this, then we use the definition. But t was a distribution. So, there were some k, some mk and some small mk, such that this holds. So, this is less than or equal than mk. OK. OK. So, this is, so it is enough to take mk plus one, so that this shows that the criterion holds with mk plus one in place of mk. Therefore, this is still a distribution. In addition, by this simple computation, we have that if the order of t is finite, then the order of t over dx i is less than t over t plus one. OK. So, this is a distribution. So, in this sense, you can differentiate a lot of things. And so, today we want essentially to make exercise on this. OK. So, first of all, one has to see whether this is a good definition. So, we have the following theorem. So, if you see one, then, so, let me recall maybe the notation. So, do you remember who is tu? Everybody remembers? OK. OK. So, then, du over dx i is equal to du over dx i. So, with some abusing a little bit in notation, where one can usually write something like this with some abuse of notation. But this is simply to say that if you have a function u, of course, a function u in this class is in l1 log. OK. So, this is, this tu is well-defined. And therefore, whenever the function is smooth, distributional derivatives are nothing else, the standard derivative in this sense, in the prime. So, coincidence means that as soon as your object is smooth, your weak derivatives actually is the strong derivative. You want to differentiate a lot of things, but for smooth objects, your operation coincides with the old one. So, this is the first property that you have to check in order to give a good definition. If you want to generalize derivatives to something which is not smooth, you have to ensure that if it is smooth, you have the classical derivative. OK. In this sense, I write here coincidence. OK. So, let us see the proof. What does it mean that these are equals in the sense of the prime? It means that for any point phi, they are equal. So, take a test function phi and consider the over the xi of phi. So, this is by definition t u of the phi over the xi. This is our definition. Yes? Which is the question? Sorry? This? 5 in d. 5 in d. Test. So, the omega was, remember, the omega was test functions with the topology that I wrote last time. Not with topology, with the convergence that I wrote last time. OK. This is just a symbol. On the other hand, when I write inequality in this sense, I exactly mean that for any phi, this against phi is equal to this against phi. This is the meaning of this. Which was the question, sorry? Is it OK? So, now, by definition, by the previous definition, this is equal to this. Now, let me recall who is t u. So, t u against psi, if psi is a test, is the integral of psi u, in the sense of Lebesgue, for any test psi. OK. The fourth, if I now apply this remat, this definition, with psi replaced by this, then the right-hand side is exactly equal to minus the integral over omega of u, sorry, the phi over the psi times u dx. OK. Do we agree? So, now, I integrate by parts. Now, I can integrate by parts. Because for the Lebesgue integral, I have integration by parts. So, this is actually equal to, the derivative on u. And I have no boundary contributions, because u is compactly supported. So, I have phi du over dx psi dx. OK. And the minus cancels with this minus. And what is this, if you remember, is exact. Now, it is clear that du over dx psi is in l1 log, because this is continuous. Therefore, this object is well-defined. And therefore, this is nothing else, phi. So, as you see, for any test phi, this distribution is equal to this distribution, which is exactly the meaning of the quality that we want to show. And of course, once you have learned to differentiate once, you can differentiate twice, three times, and so on. You can make, you can inductively define all derivatives in the sense of distributions. OK. So, we know, from this theorem, we know that there is nothing new if u is c1. Now, it is interesting, in compute derivatives, when u is not c1. But still, it admits a derivative in the sense of distributions. So, let h be equal to the heavy side function, which is by definition, the following h of x is equal to zero, if less than zero and one, if x is larger than zero. OK. With this symbol one, I denote the characteristic function of the half line, of the right half line. So, one on the right half line and zero else. OK, just the characteristic function. Maybe in measure theory, somebody writes this, well, it is just a notation, characteristic. So, this is the value one, this is the value zero. OK. Then, we can, actually, it is interesting to know that this, OK, this is in L1 log, zero. No, no, it doesn't matter. I mean, the value, so, this is a big theory. It doesn't matter the value that you assign at one point. You could put any value you want in zero. So, this is just almost everywhere defined as a function of L1 log. OK. So, you have to remember that now, in functional analysis, we have equivalence classes of functions with respect to the Lebesgue measure. OK. So, h is defined at all points except one. In particular, it is almost everywhere defined and it is clearly in L1 log. It is interesting to remark, then, sorry, my notation. What does it mean, this? Well, this means, OK, this is again my small abuse of notation. So, if you want to be really, really rigorous, don't take into account this and just write that the distribution associated to h derivative, if you want, OK. Then, this actually can be differentiated, but its derivative is not represented is a distribution, delta Dirac delta at the origin is a distribution and is not represented by an L1 log function. So, this is an extremely interesting result. So, we learn how to differentiate these continuous functions. Of course, we cannot do this point-wise. It is clear that this cannot be done, but now we will prove this. OK. So, this equality holds in the prime. OK. So, what do we do? Well, take any test and then write this. We have to compute this, but this by definition is equal to minus t h applied to phi prime. This is our initial definition. t h is represented in L1 log. Sorry, h is in L1 log, so we have an integral representation of this distribution, which is just equal to the integral of h phi prime dx. But now you remember who is h. So, actually, this integral is not on the wall of r, but just only on the right half line, because h is 0 on the left and 1 on the right, so this is just the integral minus the integral between 0 and plus infinity phi prime dx. OK. Well, now phi is in compatibility supported, therefore we can integrate by parts the contribution of phi at plus infinity is 0 because this is compatibility support. On the other hand, there is a contribution on the left extremum with the minus, and so this is just minus of minus of phi of 0. But now we recall who is phi of 0. Phi of 0 is the Dirac delta at 0 evaluated at phi. So, you see, it's not so difficult once one has given the correct definition of derivative. It is clear that, so we have proven that for any test phi, th prime is equal to delta naught, and therefore this is the exercise. So the idea behind the definition of distribution of derivative is somehow integration by parts. Essentially. Is a way to enforce integration by parts on something which is not smooth. So really the basis of this is the Gauss Green theorem, if you want. The divergence theorem. Now, other exercises, maybe you can do, or maybe I can ask you now. Well, just a small modification, let us take a more symmetric function, which is this. So, minus 1 1 instead of 0 1. This is, for instance, it is called the u equal this function here. Ok. So, can you tell me who is now u prime with again, sorry, this is equal to u prime. Prime. Can you tell me now who is u prime? I mean u is minus 1 1. Minus 1 1. This is the new function. So, it is the sign, so called the sign function. If x is positive, the sign of x is 1. If x is negative, the sign of x is minus 1. Plus and minus. Just the name sign is this. Either plus or minus. So, what do you think to be? What do you think to be this distribution on the river? So, the height of the jump is now twice the previous jump. Because the previous jump was just 1. Now the height of the jump is 2. I am jumping by 2 instead of 1. What happens at the level of the distribution on the river? It is 2. If everybody agrees, it is twice. It is 2. Delta 0. If you are not convinced, you redo this proof. Now there is a new term here. Because now when you write this, you have a contribution on the negative offline. Because there u is minus 1 and not 0 anymore. So, there is another term here which is the integral on the left half line of 5 prime. Then you integrate by parts also that term. And then signs are so that there is twice here. They do not cancel. So, please do this at home. OK? So, home. So, this is interesting. There is a coefficient in front of the derivative depending on the height of the jump. Of course, we can multiply a distribution with a number. Remember, you can multiply by a number, a distribution. You can multiply also by other things, a distribution. But surely you cannot usually multiply a distribution with another distribution. If you do this, you don't have a distribution in general. So, remember you can multiply distributions by numbers, also by other things, tests. But you cannot in general multiply Dirac delta with the Dirac delta. You don't get any distribution. So, anyway, this operation is permitted and the computation do by yourself. It's very similar. Almost the same than this. Maybe another exercise. Take u of x equal x. Then what do you think? That we have u prime of x. Again, again I am using abuse of notation. The claim, I mean the conjecture is this in the prime. Therefore, if this is true, the second derivative of u is twice the Dirac in the sense of distribution is twice the Dirac delta. You see? Because the first derivative of this, the second derivative is the derivative of the sign. We have already seen that the derivative of the sign is twice the Dirac delta. So, if this is true, we can actually differentiate twice the function, which is Liebschitz, but not C1, of course, and we end up with these kind of results. Now, it is almost clear that if there is a candidate then this should be true, because you see, out of the singularity, if I localize out of the singularity, the derivative here, the function is actually smooth. So, if you look inside with a classical derivative, so minus 1 on the left and plus 1 on the right. So, if there is a distribution here, it is something that essentially pointwise is plus 1 on the right and minus 1 on the left. However, there is a singularity here. So, here there is a new contribution. For the same reason, one expects somehow something, differentiate here something, the test is supported here, I should have zero. When the test is supported here, I should have zero. But when the test is supported here, I cannot have zero. On the other hand, and the result is the Dirac data here. Well, we have to compute to see whether we have zero. So, what do we have to do? So, take t u prime against the test. OK? By definition, we know that this is this phi prime. OK. Now, t u is in one lock, so minus integral over r u of x, sorry, yes, u of x, dx, which is therefore, by definition, equal to minus absolute value of x phi prime of x dx. OK? And then it is better now, we want to use integration by parts, so it is better to, so this is a Lebesgue integral, so we can split, so the origin is actually not important in this integral. I can neglect the origin, integrate on the left open half line, so the modulus of x for x negative is minus x, so I have a plus here, and I have x phi prime of x, dx, and then I have minus, sorry, and then I have the integral on the right of open half line, again x phi prime. So, check the designs, please, because if I make a mistake of the designs, then I don't reach the conclusion. Now it is clear, the strategy is always the same. So, in this case, I still do a integration by parts, since phi is compatible support, I don't have any contribution at minus infinity, but I surely have a contribution so this means that I have the integral minus the integral minus infinity and zero phi of x, because I put the derivative on x, therefore which is one here, and I change sign, and then I have plus, I think I have zero, because I have x phi of x zero minus infinity, which is zero. Ok? And then I have this, so now I again integrate by parts, so I have this and I have minus x phi of x which is still equal to zero. So, what do I see? Well, I simply see that this is nothing else that sign of x, phi of x. Ok? So, let me make a small comment here related to the so let some a alpha d alpha be a differential operator a little bit formal and at this level is not so important differential operator with constant coefficients a alpha. So, we call and let t be a distribution so this is a constant differential operator with constant constant maybe complex coefficients but we have worked over the reals so for us maybe it is better to take real coefficients and let d be a distribution then we say that t is a fundamental solution for this operator so let me call this operator p fundamental solution for p if p applied to t is equal to delta naught Ok? What we have seen up to now is the following d sorry h prime is equal to delta naught So, this is this h is a first example of fundamental solution of a differential operator with constant coefficients the differential operator is just the first derivative in one dimension in this case just one dimension just one derivative the operator is p u equal u prime Is it clear? So, p applied to a function u is just u prime one dimension just one derivative p t equal delta naught so you see so h is that distribution which is the fundamental a fundamental solution the fundamental solution for u prime So, this is the first example of fundamental solution they are called fundamental solution because they are really fundamental so we would see in a moment other examples Ok? Ok, so exercise still on distributional derivative so take n equal to 3 and take u of x equal 1 over norm of x for any x in the open set omega which is this compute minus laplacian of u in the prime minus laplacian of u in the prime Actually, we will see that we will see that there is a constant claim there is a constant c such that minus this constant times the laplac of u is equal to delta naught so I am abusing notation everywhere here you see this would be tu and this would be tu so let me just this is quite common I mean we know that l1 log embeds injectively in the prime so in place of u ok so please allow me to do this if you want to be precise write tu everywhere ok, if you want to be precise right plus of tu ok Is this clear? So if this is true it means that the Newtonian potential in three dimension is another example of fundamental solution for which kind of partial differential operator minus laplacian and I hope that at this moment you will recognize that we have already studied such a kind of function actually we have studied remember something like log of x times some constant in two dimensions to the n minus 2 times some other constant in n dimensions larger than 2 so in three dimensions this is this and what we saw several weeks ago was that these are examples of harmonic functions out of the origin so it is not bad I mean this equality is not so bad because this says that essentially if I could roughly out of the origin that a plus of u is 0 at the origin I cannot really do this because this is not a function so I cannot say that at the point outside the origin this is 0 I cannot write this is not possible but morally morally there is a harmonic out of the origin at the origin there is a singularity so the first thing that you have to check is that this is in L1 log so that t u is well defined so to check we have already done these computations but it is better to repeat so if I have one over rho then I have rho square which is the Jacobian of the change of variables and I see that that is integrable in L1 log locally integrable this is like rho so on a compact set it is integrable so u is in L1 log and also actually there is no we could also write here omega equal to the wall of R3 again and we want to define this almost everywhere so we can define this almost everywhere on R3 and everything can be written here R3 in place of omega so actually maybe it is I can put here R3 it is better to put here R3 on the other hand this is defined up to one point so almost everywhere in R3 so actually here omega is not really necessary so compute so the exercise consists in doing this well we have understood how to do such a kind of exercise now we have to see what we have to do so let me compute minus Laplace of u against any test so take a test in V over 3 R3 and compute this so if I have a distribution the Laplacian of a distribution apply to phi do you agree with this do you agree with this it is clear because now there is not the minus anymore because minus times minus is plus so actually we simply have u applied to Laplace of phi so please correct here R3 everywhere R3 just only here if you want so everywhere here is R3 so now sorry for this abuse of notation this is a little bit strange u applied to Laplace of phi if we want to go back to the old notation minus t u of Laplace of phi which is maybe more clear and so we can write here minus the integral over R3 of u Laplace of phi dx so now u is in a log of R3 Laplace of phi so this is minus the limit as epsilon goes to 0 so this is actually an integral of a compact set ok at the end not only that but it is also the limit of say minus the integral of out of the bowl of the same quantity why is this so well this is maybe the dominated convergence theorem of whatever you want ok no no the minus is here so minus is here ok I was starting with this minus so let me check if I'm correct minus minus I think yes it's ok I'm starting from the minus from the beginning so ok so apparently the minus is correct and so this is just the inter this is the compact set containing the support of Laplace of phi this is the origin so what I'm doing exactly is the integral in this region actually the integral is there ok so this is k this is well out of the small bowl of radius epsilon centered at the origin this ok so we can do integration by parts and we have to remember the second green side identity I think so we are slowly coming back to partial differential equations as you can see unfortunately time is the penultimate lecture so we cannot do really anything but this would be just the beginning of the real theory of PD so remember u Laplace of v minus v Laplace of u can you remember this formula now I'm not writing the conditions on omega, u, v and so on so do you remember which is the formula u what do we have here u, some normal derivative I don't remember the notation on the this was the notation nu du, du nu ok, so this is the normal derivative and nu is exterior exterior to omega and unit length so now we have an exterior unit normal so we have now something like this exterior to our domain 3 dimensions so so now we apply that green's identity there to this with phi in place of u so we have minus the limit epsilon goes to 0 plus sorry, plus here and so what do we have is equal to our domain so our domain omega is just so this is omega smooth with bounded and so on so x large not equal to epsilon with Laplace of phi u times 2 dx and then I have the integral over now again we have already seen this there are no contributions on the exterior boundary because phi is compactly supported so all that we have are the is our contribution on the boundary of the ball ok so we have u over the nu minus phi u over the nu vh2 and the limit is outside everything so there is a big parenthesis here sorry here Laplace of u and phi Laplace of u phi do we agree with this? ok so we have this equality now what do we do give me some hint to continue the proof ok now u is smooth out of the origin so in particular is smooth in omega and is harmonic in omega so u is harmonic in this annular region with Laplacian pointwise defined identically equal to zero so we surely have this therefore now let me continue here maybe so this is equal minus limit so the minus let me put the minus inside so I have integral x equal epsilon phi du over the nu vh2 minus integral x equal epsilon u phi du over the nu vh2 now and the limit is outside now there is a term which is easily seen to be infinitesimal among these two there is one term which is easily seen to be infinitesimal maybe is the second one because you see one over rho, one over epsilon right this is a constant then I have the area of the sphere which is epsilon square right so comment the second so let me let me call this 2 to epsilon is less than or equal than what well the infinity norm of the gradient of phi which is a number independent of epsilon then I have u which is one over modules of x but there is one x is epsilon, the modules of x is epsilon therefore u is integrated there is just one over epsilon so what remains here is 4 pi epsilon square and you see that this is of epsilon which goes to zero so actually our computation says that the only term that we have to take into account is the first one so this is just the limit as epsilon goes to zero plus of this surface integral here phi u over the limit and if computations are correct this limit should give us phi of zero up to a constant sorry phi of zero up to a constant so what do we do, now we have to compute so u of x is one over x so the gradient of u of x is what is minus x to the minus 2 x over x of x so this is minus x over x to the 3 this is minus 1 homogeneous and therefore the gradient should be minus 2 homogeneous ok so this should be correct now the normal the exterior normal is because now exterior to our angular region means pointing toward the origin ok, so there is a minus in front it is minus x over the normal of x this is our unit normal so what do we have at the end so this is the limit sorry, I need more space so you see this is the limit as epsilon goes to zero then I have this phi of x now the scalar product between there is a minus there but there is also another minus here so minus against minus is a plus are you following so and so I have x over epsilon this is new actually this is minus u and there is another minus so x over epsilon cube ok, scalar product dh2 and then things arranged perfectly well because this is actually x square which is again epsilon square so what remains here is 1 over epsilon square you see at the numerator we have epsilon square at the denominator we have epsilon to the 4 so what remains is epsilon square at the denominator so this is equal to the limit as epsilon goes to zero plus so 1 over epsilon square the integral on the surface of the ball phi x dx now the argument is sort of mean value theorem phi is smooth so you multiply you take the mean value so you just multiply and divide by 4 pi which is the area of the surface 4 pi so I have the mean value of phi on the surface of the sphere of radius epsilon multiplies by constant 4 pi in continuity as epsilon goes to zero this converges to phi at the center of the sphere so this limit is exactly phi of zero multiplied by 4 pi which is of course 4 pi delta zero applied to phi so what we have shown we have also found the constant and so maybe now for you is much more clear the meaning of this normalization constants it was also clear at the beginning of the course but so we have shown that minus 1 over 4 pi Laplace of u is equal to delta naught in three dimensions so the Newtonian potential 1 over norm of x in three dimensions if you multiply by 1 over 4 pi which was I think the constant you have to check that this is c3 the constant that we have considered at the beginning of the lecture but probably it is so and so we have found another fundamental solution this is called the fundamental solution from the Laplace equation in three dimensions so for the moment and we have seen also the importance of such a function for the moment we have just two examples of fundamental solution one is the heavy side and the other one is this in three dimensions so home work you should do so home but this kind of computation actually we have already done at the beginning of the course but anyway you can imagine which is the exercise so home that I don't remember the symbol that we used for the function 1 over modulus to the x then minus 2 it was capital phi I don't remember at all I don't remember solely but some constant c2 in two dimensions and cn 1 to the x I don't remember now the symbols that we have used if you can find in your notes let me know check that this with the correct constant so if you can go back to our notation our fundamental solution a small phi like a test so this phi is better than small phi but also the constants so this is I'm sorry if I write small phi then it seems a test so in this context we cannot use small phi and this was something 1 over n omega n time ah yes yes n minus 2 sorry n omega n times n minus 2 check these are fundamental solutions of the operator with constant coefficient second order operator minus laplacian finally the last exercise of today ok fine last exercise hoping that I'm using the the usual notation I don't know I don't remember so consider the following function U of t capital phi of tx equal to 1 over 4 pi t n over 2 e to the minus if x is in rn t is bigger than 0 and then 0 say if t is less than 0 so this is the heat kernel the standard heat kernel ok so now show that d over dt minus laplac phi is equal to delta 0 in d prime still we have another example of fundamental solution now for a different operator this constant coefficient operator which is the heat equation heat operator so this we have already seen that capital phi is in one lock of time space so this I think we have already seen that on a time slice phi of tx dx is equal to 1 I think that we have seen this normalization for any positive t so from this for any t so if you integrate over rn so from this it is clear that this is l1 lock even more than 1 lock ok so in this sense we can consider again abuse of notation this would be you have to understand this as t capital phi this symbol here again you have to understand as since this is l1 lock this is a distribution as usual so what do we have to do what do we have to do is the following against any so given a test in dr times rn we have to compute this sorry for the notation where there is capital phi small phi everywhere I'm sorry ok now what is this by definition this is minus t capital phi of what so let me do it separately so d over dt t phi capital phi is by definition minus minus t phi d over this this is the first object then I have minus laplacian of t capital phi applied to phi is equal minus t capital phi to laplac of phi so I think that I am not wrong if now I write the following I think that I am not wrong please check this sign in particular check this plus so well this is equal to what where this is equal to the integral over this there is a minus then I have capital phi of tx capital phi of tx d phi over dt plus laplac of phi dt dx so actually now we remember the expression of capital phi which is zero for negative times so actually this is the integral over positive times phi of capital capital phi tx d phi over dt plus laplac of phi dt dx phi is a function of t and x by t bigger than zero I mean the set of all tx it is true this is an n plus one dimensional integral ok this is just an abbreviation to denote the in time space this is t, this is x this is now well you can already imagine what would you do here what would you do that is this plus this you have to remember that at some moment you have to use that capital phi satisfies the heat equation out of the for positive times so maybe what I would do is the following let us try this small phi is compactly supported so this integral actually is an integral somewhere in this region and the function is in L1 lock therefore if now I take say sort of small epsilon here if this is epsilon I can take the limit as epsilon goes to zero of t larger than equal to epsilon so minus limit as epsilon goes to zero plus t larger than epsilon say of capital phi tx d phi d t plus n plus phi d t dx do we agree so I'm some again the principle is always the same you have a singularity somewhere and then you enlarge a little bit the domain and you integrate a little bit far from the singularity and then you let epsilon goes to zero let go to zero of course we need some theorem passage the limit under the integral sign here so what do we do now I think that we have to minus limit now we remember that there is a double integral so actually it is an integral on the support of but doesn't matter so t bigger than epsilon so now sorry from epsilon phi d phi over d t plus d t dx plus integral epsilon to plus infinity integral over rn capital phi Laplace of phi dx dt and we have this maybe this was your comment to split into two or less was your commentating now we can integrate by parts the first with respect to time so our first integral so this is equal to minus the limit as epsilon to goes to zero plus now then I put derivative here but I have a contribution at zero at epsilon sorry so this must be minus integral over rn integral from epsilon to plus infinity d phi capital phi over dt small phi dt dx plus double integral there is a minus here so please be careful there is a plus here and then we have integral over rn integral over from epsilon to plus infinity then what do I have I have capital phi capital phi from plus infinity epsilon dx is it okay up to now just integration by parts in one dimension in the first coordinate so let me check once more I have a minus here and then the plus and then the boundary contributions and then I have to integrate by parts also this one okay so but let us first so then there is then there is this and here we have the identity that we know but there are no contributions so integral from x plus infinity integral over n Laplace of phi phi plus nothing I think plus nothing if I am not wrong let me check that I am doing well maybe I am forgetting something there is nothing no other contribution I think here because the Laplace of phi capital phi integral over rn there is no other contribution I think also because in this case there is no contribution the gradient of phi with respect to x is always orthogonal to the time direction so I think that this is okay so there are no other contributions hopefully now there is a minus in front here and the plus here so we know that this sum because capital phi solves the heat equation on this region that solves the heat equation so we have that here d over dt if you want minus this plus Laplace of phi is equal to zero in this region so minus this the minus in front plus this they add you can add together and this is equal to zero okay the full what remains now let us be careful with the sign there is a minus here noting there and so it seems that there is a minus limit as epsilon goes to zero plus integral over rn capital phi of epsilon now take the contribution at infinity here there is nothing because small phi is compactly supported also in time clearly and so this is just contribution at time epsilon so with a minus in front so there is this minus cancels and there is a plus so phi epsilon x phi epsilon x dx okay phi epsilon x so the claim is this this must be equal to phi of zero now to do this I think that we should change variable time space zero is real and this zero is a vector space okay the change of variable now let us look who is capital phi at time epsilon time epsilon is equal to 1 over 4 pi epsilon to the n over 2 so I think that for epsilon I think that we have to change variable as usual x over square root of epsilon equal y x over square root of epsilon this should be the correct the correct change of variable and when you do this so that this becomes e to the minus y square over should divide by 2 you are right because yes you are right before we end up with e to the minus y square over 4 doesn't matter doesn't matter I have done this so that at the end you have so that phi at epsilon x what does it become it becomes capital phi it becomes it becomes 1 over 4 pi epsilon to the n over 2 e to the minus y square over 4 with this change of variable phi epsilon epsilon square root of epsilon y with this change of variable so if I now do this change of variable I have the limit as epsilon goes to 0 plus the integral over rn then I have I have put a square root of epsilon times y there so I have 1 over 4 pi epsilon to the n over 2 and then I have the integral over rn then I have e to the minus y square over 4 then I have phi of epsilon and then again square root of epsilon y and then I have all the epsilon should cancel because dx should cancel this so I have just 1 over 4 pi dy do you agree? and so now it is again another exercise to see that this should really go so that you try it by yourself you have an epsilon here square root of epsilon here and this is equal to phi of course we are using the fact that this normalization that this this is a convolution and you just normalize to 1 the integral so this is what remains to do please do it by yourself so this shows that there is another example of fundamental solution where now the Dirac delta maybe your question was so the Dirac delta in 0 is Dirac delta in time space ok so that's all