 Zdaj smo pričali prišličiti prvičjev v komparizmu prinsipu. Zdaj mi pričeli, kaj je taj situacij, kaj je to način, kaj je taj bar, a je to b, a je to e. In v tem pokazati. Zato smo ... Zato da se počujem, na glasne teorem, in na teorem da se vseče. Gatsne teorem ne ne ne ne ne, tako omega je sveča, da se zeprešel. C1, da se toreče, na vseče omega, na teorem da je eta. In početno, da bomo početne traje v zelo, da smo vzeli, da je 1 omega bar. To je početno, da imamo početno vzelo. To je zelo, da je zelo, da je zelo, da je zelo, da je zelo, da je zelo. Tako to je omega, to je nu, omega. Tako to je nekaj za nas, ker smo božemo zapraviti to, z našem vektorom. Vse, da vse, bi se načno bil taj nekoliko, da je skazajis glasbeno, da je je to zelo v blaži, pa je se paje na vse, da je to dobe vse, da je to za taj nekoliko, da je to za taj nekoliko, staj, en da smo vznikaj vsega terapezoidalj vznik. Taj je vznik. Vznikaj, da smo taj. Zelo, ki je, da je... E je ljpeščit. Boundari v E je ljpeščit. In je zelo, da so da malo zeloته vzupi konditione na eta, ker, išle tudi, da tega vse je pozitivna vzupi v V. Zelo, da točno, da je neč nekaj nekaj vzupi v klah C1. Kako V je C1, to je tudi prejmo, da, da veče ste vzupi vzupi v C1, tudi točno je lipšič. In tega, to je zelo vsočno, kako smo zelo vsega vsega vsega. Vsega, lepšiče, kontynuoče in kontynuoče vsega. In tega, različno, že to je lepšič, to je vsega vsega vsega vsega in ne vsega, ali taj nezavrši, bo izdaril je, da je vse vse zelo občak, in nezavrši vse zelo občak. So, tudi, da Ita je vse zelo občak in tudi, da je zelo občak vse zelo občak, in zelo sem oddišla, da receivingi, da je jo zaradi vse ob insecure Sunuje. To je s nekaj vse ulebego vse, da je več vse vse ulebego. En omega tukaj kaj je nekaj vse ulebego vse. Nekaj bil vse občast. Vseh to srenje unice, do radoče. Preplačaj, da počutek je završutno zobčatila nowo, e-gredne deljezme počutek. Tako proč, to je objeva omega, in to je wheel 1, počutek je z vrne, ne redoj ta vzela, tegao grin tjorem, z otvoru, z njo je zelo vzela, prizela, zelo početno, da zelo je vzela, početno je izgleda, da je zelo, da početno je izgleda, taj mislimo, da je zelo zelo, ta njo je zelo, Tako, to je plus f of v, w plus, okej. Tako, in tko in tega spliti v dve pravje. Tako, je to tega, kaj je ovo tega, kaj je ovo tega, kaj je ovo tega, kaj je ovo tega, Over the two yellow segments, so it's... So, let me quickly denote by this, okay? And then, so the H1 plus the integral over the two lateral segments. Is it clear? Daj, zato tazza se tako zelo površana. Sete, so nekaj zelo površana. Zato je vesel, tako da se je zelo površana iz tej viznej segmenti, pri toj stri, a je to boto so. Zato, da sem sem način, da je to je nekočnico. To je nekočnico, da je tvoj nekočnico. Vznikaj, vznikaj, je vznikaj na t-bar, ker je t-bar, in vznikaj, vznikaj, je vznikaj, taj vznikaj, taj vznikaj, je vznikaj. I zelo trebowamo, delošno, načo je l consequence vznikaj, pa je taj nino, vznikaj vznikaj nevertheless je vznikaj. To je doskri, kar nervnimo vznikaj vznikaj. Zelo vznikaj, da vznikaj, da vznikaj je vznikaj, v, and this was the definition of W, and at time zero, this was the definition, and at time zero, this is obviously equal to this, which by assumption was non-positive, because this function u of zero, remember that u of zero by assumption is less than is less than or equal than V of zero. This is assumption. Therefore, this is non-positive, and therefore, this is non-negative, the minus in front. Yeah, in particular. Therefore, this is equal to the integral over this shorter segment. I could write integral between a and b, actually. W plus of t bar, the other variable is, say, x dx. Do we agree? So we have for the moment that so zero is larger than or equal to the integral on the interval a, b of w plus of t bar x dx plus 2. Because 2 is exactly this integral. So now, so this is 2. Assume that we are able to prove that 2 is assume that we can prove that 2 is non-negative. If we are able to prove that 2 is non-negative, then we have the inequality this w plus of t bar x dx, non-negative. So we have this. Assume that we are able to prove this, then we are here. And therefore, we conclude, since w plus. So we conclude that, necessarily, for almost every x in a, b, w plus of tx is equal to zero. Because w plus is non-negative. But its integral is non-positive. And therefore, necessarily, this must be zero, almost everywhere. If, OK, there is a problem of regularity here, because this should be defined exactly at this slice. And maybe, well, no, this is lep sheet. So no problems. It's defined also at this slice. So OK, this is OK. We agree? But then a and b were a, b, and t bar were arbitrary. T bar was arbitrarily chosen between zero and capital T. T bar is arbitrary. A, b were arbitrary. Therefore, for any t bar, for any a, b, we have this. For almost every x in a, b. And this implies that w plus is zero. But t bar is arbitrary. A and b are arbitrary. And this is actually exactly what we want to prove. Because saying this is equivalent to say that u is less than or equal than v. So this is our thesis, actually. This is equivalent to say that u is less than or equal than v. So what remains to show is that 2. So if you accept this, we need to show that 2 is non-negative. Once we know that 2 is non-negative, then the proof is concluded. So the point is to show that the integral. Now we have to compute the unit normal. So these lines are so here. This is equal to so nu t. And nu t here. So we have this is integral omega plus c. Because in one case, the normal is parallel to 1c. And in the other case, the normal is parallel to 1 minus c. So in this case, dividing by the length of these vectors, we find that this is equal up to the denominator of the length of these vectors v plus. And the other part is equal to, so there is the whole 1 divided square root 1 plus c square here. 1 divided square root 1 plus c square here. So and then there is the other integral w plus minus c f of u v v plus. So, which is equal up to this factor, which is, of course, as a sign is positive. So then I can write it as follows w plus c plus f of u v plus integral w plus. Or maybe I made a mistake on the sign probably here. Let me check. F of u v. So probably I made a mistake on the sign. So that probably there is a minus here in the plus here. Yes, indeed. So sorry, there is a mistake here. There is minus. It is minus plus minus. Now, this is, of course, larger or equal than 0. This is, of course, larger or equal than 0. And now you have to remember the definition of c, because c was the maximum of f prime of y, such that y is l l. And also the definition of f of f of y, u and v was an integral of f prime, sort of s u plus 1 minus s v ds. And so you see that c is so large that c plus f of v is necessarily larger or equal than 0. This is maybe homework, but it's very easy from the definition. So this is so large so that this, even if this is negative, this is so large that this is non-negative, positive. And similarly, this is so large that, even if this is positive, then this is negative, but still this is so. Because of the choice of c made in the beginning, we have this. And therefore, this is larger than or equal than 0. This is also larger than or equal than 0. And therefore, 2 is larger than or equal than 0. And this concludes the proof of this delicate sketch, just sketch of the proof of this comparison result. Maybe I left you an exercise yesterday. So the exercise was the following. Assume that b is equal of b of tx, depending on time space. And let x of s tx solve the following characteristic system. Our notation, who I think was this, and x of t is equal to x. And then define u of tx as the solution, as the function u bar at x, starting at time t at the position x, and up to time t. This was the claim. This was the definition of u. And the claim was that then ut plus b grad u is equal to 0. And u equal to bar. This was the exercise. So the idea was to start at some point here tx. And then let run the trajectory of the solution of this system, actually of this equation. Now we are in one space dimension, actually. Because n is equal to 1 of this equation, starting at time t at the position x. And let us flow it up to time capital t. So we reach a point here. And then the claim is that this function here solves our linear transport equation with final condition. This was the claim, the homework. OK, first of all, let us see what happens at the final time. So if this is small u, then this is equal to u bar at x t tx. And therefore, I am here. I start at time capital t at the point x. And I run for 0 time, because I start at time t. And I look at the same time t. So I am at x. I'm starting at x. And so by definition, this is u bar of x. Because my ode, my equation there, capital x at capital t, what is it? It is the solution of the ode, which at time capital t start at this point x. Time capital t is started. At this point x. And then nothing happens. I mean, there is nothing to solve here, because it's just an initial condition. So by definition, x at capital t, capital t, x is x. Fine. Now, so what does it mean? It means that our claim, so I define u as follows, then at least u satisfies this condition. So the boundary condition is OK. OK, so u satisfies this. Now we have to check that u, defined as follows, satisfies the PDE also. Also. Well, one way, let us try to do this. So maybe in the exercise, sorry, b maybe was just b of x. Sorry, so that we write this. Maybe check the notes of yesterday, but maybe it was b of x. OK, sorry, b equal b of x. Now, let us make the following remark. So sorry for the notation, because t is smaller than s, exactly opposite as in the alphabetic symbols. Sorry. But I used this notation because from the beginning, here this dot was d over ds. So this was a function of s. So now I take it again as a function of s and I fix t, but t is smaller than s. Usually in books maybe you find exactly the opposite. This is d over dt. And then you fix s, which is smaller than t, like in the alphabetic order. But since we started from s, OK. So let us remark the following semi-group property of capital X. So now I want to use specific symbols. So let us take 0 less than or equal than t, less than or equal than s, less than or equal than capital T. And then I put something inside, OK. So as I said, t is less than s. So this is in agreement with our notation, OK. And then I insert a tau in between. And so the idea is the following. Now I have t, tau, and s. Now if I consider the solution of the OD, which starts at time t and arrives at time s starting at the point x, so x evaluated at the final time s, starting at the initial time t from the point x, by uniqueness of the solution of the ODs, is nothing else. Now I let run the trajectory for some time up tau. So and then I take as initial condition the time tau at the position, this position, and I let it go up to time s, by uniqueness. So I take an intermediate point on the trajectory, and I take that intermediate point as the new initial condition. That is, if I make a jump, say, from here to here, then it's the same that if I go from here to here along the solution, and then from here to here. This is because we have uniqueness of solutions of the OD. And now I want to express this. Let us try not to make confusion with the symbol. So I start. Now, what is this point here? What is this intermediate point? Well, this intermediate point is what? I'm starting from t at the point x, and then I let it go for time tau. Then I take this as the initial point starting at time tau, and then I let it, the solution, go for time, at final time s. So we have this equality. Let me check that I am correct, hopefully. Apparently, yes. So do we agree? So this is just the semi-group property. So now we can differentiate this. This is valid for any tau in between t and s. So we can differentiate it with respect to tau. So do we agree? Have you understood this equality? Are there questions on this? Of course, this is a sort of group property. OK, anyway, it is called like this. So I have this for any tau in between, which is the smaller t and s. Therefore, since this is an equality valid on an interval, everything is smooth. I can differentiate it with respect to tau, and still the quality of derivatives remains. So differentiating this with respect to tau gives us 0 equal, and now we have x, t evaluated at s tau x tau tx plus xx s tau x tau tx. So this is multiplied by the derivative of x with respect to x s tau tx x dot. And this x s, which is x dot, actually, x s is x dot, because dot is equal to d over ds, remember? And remember our notation. Now, so this is b of what? Of x tau tx. So now take, so we have to remember this equality, and take tau equal to t. Let us rewrite it in particular for tau equal to t. And so I rewrite it here, so x tau t s t. Now, if I take tau equal to t, this here, what is this object now? x. Thank you. xx s tx times b of tx. And this is equal to 0. That is useful. Let me check it's OK. x t s tx b tx apparently is OK. Now we can forget this. Just remember that equality differentiating the semi-group property, we can go back to our definition and compute separately ut and ux. So from this, we have that ut at tx is equal to u bar prime evaluated at x t tx times the derivative x t t tx, b of t. Ah, sorry, thank you, sorry, sorry, b of x, sorry. Just for simplicity, thank you. And then ux of tx is equal to u bar prime of x capital t tx x of t capital tx, OK? So now you see. Now let us put small s equal to capital t here. This is true for any s. So in particular, for s equal to capital t. You see, if you substitute, now that you can multiply this by b of x, if you want, OK, multiplying by b of x is equal to b of x. Now if you substitute this with minus x t capital t tx, then you immediately see that this plus this is equal to 0, because they are opposite, exactly, one or the opposite of the other. There is a minus. So maybe I have, yes, maybe we can just, did you try the exercise of yesterday for the Berger's equation? So maybe we can go through quickly through the exercise. I left yesterday for the Berger's equation. So the exercise, remember well, was ut plus f prime of u ux equal to 0, where so let us take the flux, the flux function f of y equal to y square over 2, so that this is nothing else, the Berger's equation u x equal to 0. And u of 0 was u bar, u bar and u bar at time 0, t plus r. And u bar of x was now 1 divided by 1 plus x square. So actually, we are identifying sigma. Sigma is going to 0 sigma. And sigma is identified with x. So sigma is identified with x. With our notation. So this was the embedding, remember, phi from r into r2. This was the parametrization of sigma, capital sigma. And now, so the system of characteristics, so we have n equal to 1, so we have x1 dot b, the vector field b of x of say t, x and y is what? With our notation. The vector field b, 1, y. So x dot is equal to x1 dot is equal to 1, x2 dot is equal to y, y dot is equal to 0, x1 of 0 is equal to what? x1 of 0, 0, x2 of 0 is equal to sigma, which actually is equal to x. And y of 0 is equal to u bar of phi of sigma. And so is equal to 1 divided by 1 plus sigma square. This is our situation. We can solve these solutions. We can solve this, for instance, giving y of s and sigma is equal to what? So is 1 plus sigma, because this must be square, simply. Doesn't depend on s. And then we have, say, x1 of s and sigma is just s, I think, because I'm starting from 0. So at the end, this will be identified with t, as usual. And x2 dot must be equal to 1 divided 1 plus sigma square, which gives us x2 of s and sigma, actually equal to s divided by 1 plus sigma square. And then we have the initial condition, which is sigma. So we have this. Now, our change of variable is the map taking s sigma. So maybe can I now s call it t. So s can be identified with t now, as usual. So, well, if you don't like s, s divided 1 plus sigma square plus sigma. And this is our diffemorphism around the strip containing sigma, capital sigma. And this we want to invert it. But maybe the inverse is not so easy, because we want to solve for, say, t equal s. But then we want to solve for, say, x equal to t, divided plus sigma. So, given t, maybe I write it as equal t. So given t and x, now I know it is invertible. I know it is invertible locally. And I want to find s of tx and sigma of tx. So, given t, s is simply equal to t. It's OK. But given x sigma, so given x, I want to solve for sigma. So, I want to find 1 plus sigma square x minus t minus sigma times 1 plus sigma square. So, given t and x, solve for sigma, which is not so easy, actually. No? So, actually, it's not easy. And therefore, we leave it as it is. We don't solve it explicitly. Just leave it implicit. We know that locally, for small time, we can solve. Because we know that locally d is a difemorphizm. So, this should give for t less than some capital T. And x in r, say, this gives us sigma solution of tx. One solution only for small times. So, at the end, u of x is equal to y, u of tx, sorry, u of tx, is equal to y, y of s of x, s of tx, sigma of tx. And therefore, which is equal to 1 divided 1 plus sigma square of tx, and u of tx, and x is t divided. So, we have sort of implicit representation of our solution, u, locally. Is it OK? OK, this is not u of tx equals something, depending on tx, unfortunately, it remains a little bit implicit. With other symbols, since, OK, t was identified with s, t was equal to s. And we have already said that we can identify the parameter sigma with x. So, we can change names, and place x in place of sigma, maybe, is more, more. So, u of t at the end, u of t, t divided by 1 plus x square plus x is equal to 1 plus x square. Yes, we have to be a little bit elastic with these changes of names of the symbols. And this is, so, the expression of x of t, the expression of x of x1, x2 was, say, t, and then t plus x. This was x2 of t, x2 of s. This was x2 of s, in the defied s with t. So, let us try to see, to write down, so these are curves in the plane, actually lines, if you want. So, the parameterization is just t into this, t into this. And x can be considered now as a parameter. So, fix x, say, x here. And then I have a parameterization of a line, which is, I don't know, x positive. So, I pass through here, say, when t is 0, I am here positive. And the coefficient here is changes when I move x, which is now a parameter, but, say, it is sort of positive, so sort of this, like this. Now I move x. And remember that our solution is the value of u bar. Our solution here is the value of u bar. I mean, u is sort of constant along this line. Constant depending on x, because we know that, well, remember our expression. So, u along the line is a constant, it depends on this choice of x. It is written here. I mean, the right hand side is independent of t. So, once I fix x, and I draw this line, then I take the value of u bar at the value x, at the number x, this is u bar of x. Now, I change line. So, I take another x here, and I want to understand what happens to the line. I think that it happens to the line. I mean, x is here now. So, I am passing through here at time zero. But the coefficient now, x is larger. The coefficient is larger. One of it is smaller. What happens to the line? Yeah, it becomes, the point is here, unfortunately, is that t is vertical. But when the coefficient here is smaller, the line becomes, you try to become, say, more parallel to the vertical axis. If you put t here, then you understand, you may be more familiar. So, when the coefficient is smaller, then you try to become horizontal, if t is here. And now t, for some reason, strange reason, is vertical. So, this means that now my line is this. Problem. There is a problem, right? So, we can say some sort of this, maybe, okay, maybe sort of this, this. No, this is sort of this, and so on. So, there is, in this picture, a first time, there is a first initial, first time, such that two of these lines meet, there is a first time, t-sing, where two of these lines meet, and before it, they never meet. No, in this case, t-sing is not equal to zero. In this case, t-sing is positive and should be computed. I don't know exactly what it is, but I think it can be computed. Yes, t-sing is non-zero. Yes. The question is, well, t-sing equal to zero would mean instant singularity. You start from an initial condition, which is u-bar, and the question, if t-sing is equal to zero, this means that instantly the solution makes something very strange. I don't know, it breaks the graph, makes a single jump, something like that. This is not the case for this equation. T-sing appears, is a singularity time. We don't know what happens after that, but it is positive. Well, this should be proven. You are right. It's not obvious, you should prove. Is it important to create some line or some line? I think that this intersection point, yes, some line here, but I'm taking the first point, the first time where this line starts. Line of singularity. I'm saying that before t-sing here, in this region our method of characteristics perfectly works. And the solution is this one. A little bit implicit, but this can be uniquely solved. And so in principle if I am able to solve cubic polynomials which in principle can be done I can solve this and write u of tx equals something depending on t and x. But then there is t-sing and it's very interesting try to understand what happens after t-sing. But the point is that we don't have solution of solution. We do not have it. If we continue to I think that if we continue to our our lines so actually t-sing is sort of of this picture here. Nothing happens for x negative. So now, this is not the right place to continue the analysis of singular solutions to Berger's equation which is by the way very interesting. Unfortunately we cannot do this here. But let us give you just another viewpoint of the same problem. So this is our u bar. And let us look to our equation which says that u t if everything is smooth so if this is c1 u is c1 this is continuous and so on so we can take the product because you see assume that you have a singularity and assume that this singularity means that u jumps becomes discontinuous because singularity should be defined what it is. But assume that for us now a singularity means that u creates a jump. Then this is a big problem because ux is a measure so sort of I don't know if you know it but it is called the delta Dirac delta and the measure cannot be multiplied by this continuous function. This cannot be done in general. So it is not easy at all obvious how to interpret this product measure times this continuous function. When you will study measure theory probably you have studied some measure theory then you have learned how to integrate a continuous function maybe it comes apart, I don't know but continuous function with respect to the measure this you can do but the test must be continuous. Now what does it mean to integrate this continuous function with respect to the derivative because the derivative is a measure is not clear and indeed maybe one more natural way to write it is this so that at least this is in conservation low form and we can integrate by parts as I have already tried to explain maybe one time ago or two times ago but anyway so let us assume that everything is smooth so look at this this says that the horizontal velocity u t of the points of the graph of u solution assume that u is positive like in this case u bar is positive and assume that you can prove that if u bar is positive u also is positive actually it is immediate but you know that u is positive ok now the velocity in the horizontal direction depends on the height because there is u of x so this is the height is y, is the variable y if I am here I have a velocity if I am here I have another velocity because u is higher here positive and higher then it turns out that that points here higher points it turns out that our solution is something like this so the profile after some time is sort of this so this higher points move toward the right with faster than this and so like a wave exactly like a wave you have a wave and then as the c proceed in front of you the waves try to become vertical and so at the end what you see after the t-sing is a time where you are in this situation this is what happens at time t-sing in this case now you see what is the problem and what do you now is discontinuous after so I mean once you are in this situation then this is not part of the graph in the usual sense because this is vertical so this is actually is sort of discontinuous function now that you have now you want to let discontinuos so you don't know exactly what to do because our initial condition you bar up to now was c1 so we don't know exactly what to do what is the right notion of solution starting from this initial condition is not clear at all and we do not accept I think I don't know maybe you have this wave then maybe this is not acceptable this is not a function it's a multifunction it's not a function so maybe we do not accept this we accept something which are graphs so this is for us for some reason we do not accept this so it's not clear how to continue the solution of course from here there is a story starting from here it's a long story just to have some flavor on what happens at this thing now I would like to leave you at homework for the next so homework let so take b in Rn just to this is let you be the solution to ut that we know is just the first example interesting example of PDE this is now now t is larger than s u of s now I start at time s so this is just the linear transport equation with constant coefficient so we know everything just we start not at time equal to 0 but at time equal to s f is smooth given ok and define so homogeneous but with an initial condition this means the following what is dot u of sx is equal to fsx for any x is just to indicate a variable we know that this is a function of x maybe in some books if f is a function of two variable maybe in some books again this is a function of x so we could write f of s the important fact is that the notation is clear it is clear what we mean dependently of the notation maybe in some books on semi group theory parabolic equations one uses also this this kind of anyway now define u oh sorry v maybe I change name v of tx equal integral from 0 to t of u so let me denote this solution I need a symbol this solution to this I denote it as u t sx starting at time s at a generic larger time t t sx ds so let us superpose sum somehow the solution of this homogeneous problem starting however not from 0 but from f so I make a sort of superposition of solutions sort of some generalized sum the homework is then then v solves vt plus v grad v equal f and v at 0 equal 0 in 0 t times r t ok this is related to the so called duhamel principle or duhamel I don't know duhamel principle which is a principle which is very general actually not only valid for linear transport equation but also for oddies ordinary differential equation also for other kind of linear parabolic PDEs other equation in particular for the linear transport equation it says the following assume that I want to solve non-homogeneous problem with zero initial condition so non-homogeneous with zero initial condition the idea is that I can solve this as a superposition of what solutions you of homogeneous PD with f as initial condition at time s sort of general principle ok this is not a difficult exercise I don't think it's so difficult I mean you can simply differentiate with respect to t you have t here and t here so you have two contributions then you differentiate with respect to x of course you can differentiate inside the integral and then immediately you see that this at time 0 if you put t equal to 0 then this is 0 clearly this is not so difficult maybe we can do this remark at this point maybe we can do this remark so this is the homework ok maybe we can do this remark now let us consider ut plus b equal f let us consider this and u equal u bar then we have the expression of the solution we remember it so we have u of tx is u bar of x minus bt plus the integral between 0 and t f of s and then we have x plus s minus t ts this was the solution ok so this solution u which has right hand side and an initial condition is the sum of this plus this and this the first one just solves the homogeneous PD with the initial condition u bar so this is say the solution of the homogeneous problem and this what is this well this should be the solution of the non-homogeneous problem with 0 initial condition time is over now sorry