 Tako, zelo smo prišličiti dve odliče. Zelo, ki sem prišličila, je to, da je prišličen, za vseh deltapozitiv. Liničnje st. je 0 plus. Zelo je 5. Zelo je uniformil. Uniformil. Faj je hitkernel. Faj je 1 over 4 pi t, n over 2, into the minus x square over 4 t for any t equal to 0. Taj je hitkernel, zelo u t minus n plus u in 0 plus infinity times. Taj je soluzija do hitkernelj za pozitivne tajme v zelo. Zato, kaj x je nekaj delta, taj je i minus x square over 4 taj je nekaj delta, taj je i minus delta square over 4 taj je pozitivne taj. Mno sa 4, superimum over x, taj min, ga je uvanje delt, taj pi taj x. Faj je tez produča pozitivne nיתaj taj taj poliži. Zato danes mi je 1 over 4 pi t, n over 2. In tudi je to izgleda z E do minus delta over 4 t. In, tudi, je limit s t goz do 0. S t goz do 0 od obovoj. Vsrednji suprimum over x. vzakratimo z delt из deltimi deltoj tx. To je 0, kaj je kajkosk casualitet. Svok, usingo poato na sej deltih poseljoni, od todas za vse. To je zelo urat. Zelo... Iseko je to wi? Želi. Češtje izgleda je, češtje je zelo, da je Faj koncentriračnja v tem spasih. Češtje izgleda je, da je nekaj delta pozitiv, da je nekaj stjera, da je zelo izgleda. Češtje izgleda je, da je zelo izgleda. Češtje izgleda je, da je izgleda. Češtje izgleda je, da je se v taktom formu, zelo izgleda je, da je zelo izgleda. Češtje izgleda je, da je izgleda. The s t goes to zero from above, we have a sort of function like this, where here the area below the graph and above the x axis, so the area here is always equal to one. So this says that if I fix now delta, so minus delta, delta, and then I integrate the function phi, when t goes to zero, this area, only this part, this area here, when t goes to zero, goes to zero. This is the geometric picture, okay? So let us first do the computation in one dimension. To solve exercise number two, let us first do a computation in one dimension. So assume for the moment n is equal to one, take a positive number a. So let us integrate over x bigger or equal than a of e to the minus y squared dy for the moment. Let us do first this one dimensional computation. So what is this? So this is, of course we cannot, I mean this is equal to two integral on this, oh this is obvious, but unfortunately we cannot do this integral unless the domain of integration is say the whole r. But surely we are not able to do it in this interval here a plus infinity. However, if y is in a plus infinity, so we know that y divided by a is bigger than one. So if then, and therefore here I can put the number larger than one, obtaining the following. So I have y divided by a, y divided by a e to the minus y squared dy. Okay? This is clear because this is larger than one on this interval. So this is equal to two over a the integral e to the minus y e to the minus y squared dy. And you see now the reason why I made this trick because now I am able to do this integral. So this trick is a way to make appear the term y here that allows to make the integral now explicitly. Of course I have less than or equal. So I am losing something, but at least I am able to do the integral. And this I do. And this is equal to what? So the primitive is one half e to the minus y squared with a minus. And so this is equal to, I think it is equal to two over a. And then I have one half e to the minus if I am not wrong. Okay? So this is equal to e minus a squared divided by. So please correct me if I make some mistake in the computations. Okay? So let me write it here so that I will remember. So for any positive a, the integral over these two half lines of this object here is less than or equal than e to the minus a squared over a. Now I want to pass to any dimension. And now I want to integrate this. So the set of all x less than, bigger or equal than a is maybe contained in the set of all, the set of all product. In the set of x such that xi is less than or equal, bigger or equal. So let me check if this is true. So assume that I have a point in the complement of this. So indeed if xi is less than or equal square root over n for any i, then for any i from one to n, then the norm of x square, which is sum of x square is less than or equal. So any xi is less than, so the square, let me take the square for simplicity. So this is less than or equal of what? Sum a square over n. So if xi is less, for any i, if xi is equal, is less than this, then the Euclidean norm, which is the sum of all xi square, is less than or equal to the sum of a squared divided by n, which is n a squared divided by n, which is equal to a squared. So if before we have proven that the set of all x such that xi is less than or equal, then a over square root of n for any i one to n is contained in the set of all x so that x is less than a, less than a. We have proven this more precisely, so we have proven this. And therefore if I pass to the complement, the set of all x having norm larger than or equal to a, the complement of this is contained in the complement of this. The set of all x for which there exists an index xi for which this is larger than this. So this is equal, less than or equal, than the integral. Now I can take, I can use this one-dimensional result into the integral of, x one, bigger or equal, than a over square root of n of e to the minus y squared dy. So this is now one-dimensional integral. It is multiplied by n because I have n factors. Now the only difference is that I have, instead of a, I have this a over square root of n. And therefore now I can, so now I can use that result, one-dimensional, where now in place of a I have this new a, this new, in place of a, so I have e to the minus a square over n now. And then here I have square root of n and a here, if I am not wrong. So this at the end is equal to n to the n over 2, which is a number. So it's not really very important e to the minus a square over n. And then here I have another number, which is now a to the n. Hoping that the constants are correct. Now I have, my initial question was, now take a equal to delta. And so I have to compute 1 over 4 pi t to the n over 2, the integral over this set here of e to the minus y square over 4t in dy. So now I apply this result taking in place of a simply delta. And so this is less than or equal to n over n over 2. Then I have 1 over delta to the n. Then I have 4 pi, yes, is it correct? So I have e to the minus a square here, right? I have, thank you, thank you, because I am, yes, thank you. Yes, I have this to the power n, thank you. So this is a square. Then I have, so let me recall, I am taking now a equal to delta. So I have 1 over 4 pi t to the n over 2, which is this term. Then I have delta to the n, n to the n over 2. And then I just have e to the minus delta square, OK? And then I have, I have also this 4t, right? OK, this 4t, which maybe, so I can, so let me. First you have to change the variable. Yes, I could change variable, but maybe here is not needed, because, so x is bigger or equal than delta. So, so I could put delta here. First you have to change this y to x, because you are integrating yx, e-grid should be x. No, this, I want to use this result, I want to use this result. Yes, yes, thus here you have to change first integral. So e to the minus a square, y square. I have this 4t here. So let me first try to see what happens if I put delta here. Because if x is less than or equal than delta, then this is less than or equal than minus delta. And therefore minus this is bigger or equal than minus t. Minus, so y is less than or equal than delta. I have to change variables. Maybe I think that I have to change variables. I think that I have to change variables. Let me see. I think that maybe here it is the case to change variable. Otherwise I cannot manage this factor. So let me change variable and define zeta equal over 2 square root of t. So before doing this passage, I need to make a further computation. So let me change variable. And this is what happens to this. This is 1 over 4i t to the n over 2. So now this becomes what? It becomes e to the minus now z square. OK, here the domain of integration is, if y is larger than this, then z is larger than delta, then z is equal to z to square root of t. And therefore z is larger than or equal than delta, z larger than or equal than delta over square root of t. Then I have the z, the y, the z. So this is equal to 2 to the n square root of t to the n zeta. Fine. Now I will try to apply the previous computation. So I have this 2 to the n, maybe this 4pi to the n over 2. It doesn't matter, these are constants. Now I apply this result, where now in place of a I have delta over 2 over square root of t. So n over 2 e to the minus delta square over 4t e to the minus delta square over 4t. And then, is it OK? Because, I mean, I repeat, I take a equal now delta over 2 square root of t. So this is a factor. This is a the minus delta over 4t. And now I have this factor here, which is equal to delta to the n. And then I have here 2 to the n. And then I have here t to the n over 2. Hoping it is correct. Is it correct? This t. Why? Because I have e to the n here at the denominator. No, it's OK. So it seems to be OK. Pi. And this goes to 0, as you can see. Now you take the limit as t goes to 0. And because of this quantity, this limit goes to 0. Yeah, yeah. OK, seems too easy. Seems to be too easy. But this apparently is at the numerator, and not at the denominator. So let me check once more. I have this at the denominator. And now at the denominator I have a new quantity, which is this delta over 2 square root of t. And so this goes to 0, as t goes to 0. Square root of n. This is not important. Let me see e to the n. Square root of t delta e square root of t. This converges to 0, as t goes to 0 plus. OK, this is the second exercise. So now let me come back to the problem of the weak, very important problem of the weak, maximum, minimum principle in bounded domain. So let me state the following theorem. Let, yes, we have already the following notation, qt. If you remember from the previous lecture, qt. So this is time space, this is qt. So this is qt, open and bounded, non-empty. I gave also probably a name, which now, of course, I don't remember to this upper part. This was gamma t. OK, and this lateral, so this is gamma t. And then I gave a name also to this lateral part of the boundary, as t. In the usual case, we are in the following situation. This is part of a cylinder, so I have over omega. So I have omega here into our n. And then I have, say, capital T. And so q is just the product of 0 capital T. So you have to think to this situation. qt is just the product of 0 capital T times omega. And then there is an open bounded set over n. And therefore gamma t is just the interior of this facet. So this is just gamma t. While s of t is just the union of this lateral boundary and also the bottom. So now let u be continuous up to the closure so that it admits maximum and minimum finite. And assume that it is also one differentiable, one time, once in time and twice in space with, say, continuous derivatives in the interior, open set here. And assume that ut minus laplace of u is less than or equal than 0 in qt. So this is, in words, namely u is a smooth subsolution of the heat operator. The max of u over the closure of st is equal to the max of u over the closure of qt. So please, check that the symbols are the same. OK, so let us prove this result. So first of all, we want a strict inequality in one. So to have a strict inequality in one, we slightly modify u. And so we define v of tx as equal to u of tx minus, for instance, epsilon t. Where epsilon is positive. This is for any tx into the open set qt. Now let me compute vt. So vt is equal to ut minus epsilon. So this is a way to produce a strict subsolution because this is equal to this. And on the other hand, v is obviously equal to the laplacian of u, plus epsilon t. Now we adjust the sign. Therefore ut minus laplace of u is equal to vt. So take a plus epsilon. vt minus laplace of u is equal to ut plus epsilon minus laplacian vt minus laplacian v is equal to ut plus epsilon minus laplacian of u. The sign were correct. So vt is equal ut minus epsilon minus laplacian v is equal to minus laplacian v. And therefore this minus this is equal to this, which is less than or equal to minus epsilon, which is less than or equal to zero. OK. Do you agree? Before v is a strict subsolution. Now the claim. Now fix now any time tau, fix a point, fix a point, any point arbitrary t bar x bar into the open set ut, fix a point here t bar x bar. And then fix tau in between, take tau in between t bar and capital T. The claim have maximum v considered in q bar t considered in q bar tau. Now fix tau. Now I consider v strict subsolution into this compact set, q bar tau. I cannot take remark maybe because you can wonder why we need this tau. Why we don't take tau equal to capital T. We cannot take tau equal to capital T because we are assuming smoothness up to here and here there is not the closure. If I would have the closure here, then tau would not be required and I could take tau equal capital T. Just a small detail. So I cannot take for the moment tau equal capital T because the function is not differentiable on this yellow part in principle. It is not. It is just differentiable inside but not on the top. Just continuous but not differentiable. The maximum in gamma tau union q tau cannot have a maximum. Assume by contradiction that T naught x naught belonging to gamma tau union q tau is a maximum. Laplace of v at T naught x naught at a maximum inside the Laplacian has a sign. And here the Laplacian is no positive. Namely minus Laplacian of v. T naught x naught is bigger or equal than zero. Now what about, so this is, I want to find the contradiction so you see. Now this is bigger or equal than zero at T naught bigger or equal than zero at T naught x naught. We know however that everywhere in the open set this must be negative. And this term is bigger or equal than zero at that point. You see, this is a point inside somewhere. Now what about dv over dt at T naught x naught. So what about this? Well if T naught is less than tau certainly again this is a maximum point in time space so this is equal to zero. If T naught, we are working on this compact set, blue one, in principle T naught x naught could be on this upper part. But there we only have like this, I mean it's a maximum, reached at the boundary therefore the derivative is bigger or equal than zero. Now in any case, either this, if this is true or this is true in any case this is bigger or equal than zero. Therefore this is also bigger or equal than zero at T naught x naught. This gives a contradiction. Therefore the claim is proven. So any maximum point of v in q bar tau lies on the lateral part in this remaining part in which I denoted by c. Now we have this u of T bar x bar so let me come back to this arbitrary point u of T bar x bar. What is this? It is equal to v, v T bar x bar plus epsilon T bar by definition. Here is T bar, tau is larger so this is certainly less than or equal than the maximum over q tau bar of v plus epsilon T bar because v at this point is of course less than or equal than x maximum obviously. But it maximum is also equal to the maximum of v just only on s bar tau plus epsilon T bar. Now let me now remember who is v and now I want to replace u here so the maximum now s bar tau is surely s bar tau is just this part of the boundary that is surely contained in s bar T. So this is less than or equal than the max of u in s bar T u is what? This plus the max over T in s bar T plus epsilon T bar. Is it clear this passage? The maximum of v is the maximum of this sum but the maximum of the sum is less than or equal than the sum of the maximum. So I can take the maximum here plus the maximum here. And then I even take the largest part, the largest part of the boundary. I was taking the maximum and now I say it is less than or equal than the maximum in the red. And this is clearly capital T. So we have proven hence u of T bar x bar is less than or equal than the maximum of u over the lateral boundary plus epsilon T bar plus say capital T. Well, in all the pictures capital T was positive. Everything was positive. Time was positive. So if you don't like to take the absolute value you can take capital T without the absolute. And so now what do we have? Now v is not necessarily necessary anymore. There is no v here, there is only u. V was just a trick to have a strict sub solution. But now v is not present anymore. And hence we can take the limit as epsilon goes to zero to obtain that u at T bar x bar is less than or equal than the max of u. Since this is true for any positive epsilon let epsilon goes to zero. And so I have this. Now what do I have? Now you see, this is true for any T bar x bar which does not appear on the right hand side. Before I can take the soup over T bar x bar into qT of u T bar x bar which is less than or equal than the max of u. But then u is continuous so this supremum is surely the maximum over the closure. Hence we conclude that the maximum over the closure of u is less than or equal than the maximum over the closure of gamma T. S t, sorry. Is it okay? Now the opposite inequality is immediate. So you remark this is obvious because s bar T is clearly contained in this. So we conclude that actually we have equality which is the statement of the theorem. So this concludes the proof of the maximum weak maximum principle and now several comments are in order because this is the major tool in the study of parabolic equations. So I think it's important to make a lot of comments here. So first of all remark one why weak? Why the word weak? Because a theorem does not exclude theorem does not exclude maximum point of u, qt union that the upper part was called gamma this does not exclude. In principle there could be other maximum points here in principle. Actually this is not possible it can be excluded however this can be excluded but well this can be excluded but by other arguments which have a name and are a consequence of the strong maximum principle. So for the moment we just have the weak maximum principle and not the strong maximum principle but you have to know that there cannot be maximum points on this part. Second remark weak minimum principle principle for super solutions is a smooth super solution it says the following so let v v continuous intersection c12 over qt be a smooth super solution in qt of v is equal to mean over the proof is home work home work but in this case is not difficult because the idea is simply to define u equal minus v ok this transformation transforms super solutions into smooth super solutions into smooth sub solutions so u is a smooth sub solution and therefore the max of u over q bar t is equal to the max of u over s bar of t but this is also equal to the max of minus v over q bar t which is equal to minus mean over v over q bar t and this is also equal to max over s bar t of minus v which is equal to minus mean and this concludes the home work actually so I was rather quick but trying to check this computations so these are in particular for u for a smooth solution solution u since a smooth solution is in particular a smooth super solution and a smooth sub solution we have at the same time the weak maximum and minimum principle satisfy we have the weak principle together they are true so this is the remark number 2 now what are the assumptions that allow to prove the theorem excuse me home work maybe now home work try to prove statement let omega be bounded open and bounded and let u be a continuous function such that now there is time there is no time anymore then try to prove sort of this max now remark number 3 the previous result the weak max of mean principle is false namely boundedness boundedness qt was essential boundedness of qt was essential to have the analog of weak maximum minimum principle one needs to require some growth condition at infinity so in unbounded domains if you don't make any assumption on the growth of the solution u for instance the heat equation then there can be problem and so maybe before remark 3 let me go home remark 2.5 maybe maybe this the corollary before doing remark 3 so let me just do remark 2.5 corollary let u respectively be smooth sub respectively super assume that u is less than or equal than v so the lateral part was called s then u less than or equal than v in qt so let me prove this corollary the corollary is very interesting it says assume that you have two solutions or more generally a sub solution and a super solution take two solutions and assume that at time zero and laterally you know that this is less than this then necessarily this remains true everywhere in the domain this is called weak comparison principle weak comparison principle this is called weak comparison principle it is very interesting it says that if initially you are a solution is below another solution and not only initially but also on the lateral part of the boundary this is true then it must be true also inside let us prove this corollary weak comparison principle what did you do to prove this w equal u minus v define w equal to u minus v ok so this is in the good class is continuous in the closure and c12 because it is the difference of two continuous functions in the closure and c12 in the domain in the interior and what about u wt so wt wt is ut minus vt Laplace of v is equal to Laplace of u minus Laplace of v and therefore wt minus Laplace of v is equal to ut minus Laplace of u and then we have minus vt minus Laplace of v ok this is by assumption less than or equal to zero because ut u was a smooth sum solution v was a smooth super solution therefore this is inside the parenthesis is larger than or equal to zero and with the minus in front is less than or equal to zero therefore this is less than or equal to zero ok and then by the weak maximum principle so this is a sub solution is a sub solution so we can apply the maximum principle so by the weak maximum principle applied to to w the weak maximum principle applied to w implies implies that the max over the q closure of w is equal to the max of w but on st we have that the maximum of w is not positive by assumption therefore this is less than or equal to zero and therefore the maximum of w everywhere is less than or equal to zero and therefore u is less than or equal to d everywhere so remark 2.5 which is actually very important is not remark is a theorem is a corollaries comparison principle has the following consequence so consequence uniqueness so let u and v under the previous assumptions two solutions be two solutions of say u t minus laplace of u equal to zero in side this then u equal to say u equal to u zero bar time zero on the lateral part so assume that you have this the in the corollary super solution yes u is a smooth sub solution and v is a smooth super solution so that this is less than or equal to zero and this and this is larger than or equal to zero so with the minus is still less than or equal so this plus this is less than or equal to zero so you can compare sub solutions with with super solutions ok is it ok? so this is maybe so important this we compare in principle that I prefer to leave it on the blackboard by the way remember that in some lecture ago we have proven sort of theorem like this for one dimensional conservation law remember the proof that we made of the comparison principle for c1 solutions of the Burbers equation so this is the analog for the heat equation so at least ok so now assume that you have continuity and the compatibility condition between phi at time zero and u at the boundary so plus compatibility conditions between phi and u at zero bar compatibility conditions and assume that phi is continuous u zero bar is continuous and so let u and v v two solutions of this under the previous stated hypothesis then u is equal to b so if one solution exists in this class under this assumption on omega, phi, u, zero and so on, this solution is unique again no claim on existence we have not constructed the solution that the solution exists in this class then it is now uniqueness is immediate why is immediate because we have comparison therefore u is a sub solution and v is a super solution they are equal on the boundary therefore u is less than or equal than v but u is also a super solution sub solution I can interchange because they are both solutions they are equal on the boundary and so u is also larger than equal than v so the proof is homework but proof u is less than or equal than v and u is larger or equal than v by the weak comparison principle fine now now I can come back finally to remark 3 let me be a little bit more precise remark 3 more precisely let me put it like this more precisely there exists a non-zero solution a non-zero solution it's infinity of u t minus laplace of u equal to zero in zero plus infinity times r times r u of zero equal zero on so it is possible to construct of course a solution to this problem is the function identically zero it is immediate because the function identically zero satisfies this problem but however it is possible to construct another solution non-zero non-zero solution to this problem this shows that we don't have uniqueness in this unbounded domain case so uniqueness so uniqueness is so we have constructed two solutions of the same pd of the same koshi problem therefore there is no uniqueness if we don't put any assumption on the growth condition of our solution when you have an unbounded domain you would like to put boundary conditions what about boundary conditions in this domain boundary conditions means that your solution must have at most growth such and such of course the boundary is empty at infinity in time for instance so if you want a solution u equal to zero here in this unbounded half plane you would like to put some boundary condition so the way to put quotation boundary conditions is to require that you must growth not more than this and this such and such if you don't put any growth condition you don't have uniqueness so this is an example due to Tikonov maybe tomorrow we can say towards more details about this uniqueness if you want this is remark 3 now let me go now remark for assume that now I want to consider such a kind of operator now assume that I want to generalize with the following assumption iij is symmetric matrix symmetric matrix is uniformly positive definite positive definite uniformly for tx into q bar t and aij are continuous in q bar t assume that bi is continuous in q bar t so then so this is you see if this is the identity then we have the Laplacian and if bi is equal to 0 we have the previous equation so previous case previous case is equal to identity and bi equal to 0 now if more generally we have this kind of linear notice also just one remark that if aij is 0 this cannot be considered because we are assuming strict positive definitiveness so we make this equal to 0 if this would be 0 then this would be the transport equation that we have studied in the first lectures but now we don't allow to be in particular the transport equation because this is positive definite then the remark is not difficult now not very difficult to prove that the same results hold namely weak maximum and minimum principles are true for some solution and super solution weak maximum principle for sub super solution on bounded domains well, why this is true you see if you remember the proof of the if this is true then also the weak comparison principle true and so on and all consequences since this is true then we have weak comparison weak comparison is true et cetera now, why this should be true if you remember the proof of the weak minimum and maximum principle we have constructed a strict sub solution this we do and then essentially everything follows from the fact that a strict sub solution cannot have a maximum point inside or on the top why it cannot have because the Laplacian there has a sign and ut also has a sign that is the principle essentially but now look at this take a maximum point this is zero at a maximum point this creates no problems when you have a gradient part there is no problem in the weak maximum principle interior maximum point are stationary the gradient is zero so this is zero, no problems this is as before has a sign and this has a sign because this is positive definite is like the Laplacian more or less so at a maximum point still the same essentially you have to change variables but essentially the idea is that this creates first order term no problems strict positive definite no, it works as in the Laplacian case if you want to see the details of the proof here you can look at the book in chapter 7 maybe of the book of Evans I think that the book of Evans you should probably make the proof in this case but the spirit is exactly the same as before now there is however a question mark now assume that I want now to consider slight remodification so for instance this but for simplicity take bi equal to zero and aj equal to identities so the Laplacian but then we add apparently in no equals zero order term no derivatives where c say for instance could be a constant but also continuous is the proof of the weak so take a solution or take a sub solution take equal to zero the weak maximum principle for you still working there is a problem here so please think a little bit about this question you will see so try to re prove the maximum principle for sub solutions and then you will see in moment there is a problem in the proof and so what kind of analog of the maximum principle could hold when there is this zero order term this is the homework what kind of generalization say your analog of the weak maximum principle when you have this zero order term can you expect so this is the the answer so the answer well maybe it is better that you don't look at the answer just try to think of it anyway the answer is again in the Evans book for instance in this book for instance maybe chapter 7 so the spirit of this is the following when you have a well behaved second order operator like this in the first order then you can expect when there is the zero order term then you have to be more careful much more careful concerning the weak maximum and minimum principle and tomorrow we will continue about this