AI Assistant
Transcript
00:07:580Paolo Guiotto: Last time we,
00:26:530Paolo Guiotto: discuss existence uniqueness. We did determine the domain where the right-hand side of the equation is defined, which is in this case, the planar 2.
00:37:940Paolo Guiotto: We check that the derivative with respect to Y Together with death.
00:43:00Paolo Guiotto: is continuous on the domain, and this ensures that we're getting solution to each puship rather than in each domain.
00:51:360Paolo Guiotto: What the initial condition is.
00:54:580Paolo Guiotto: Any point of the play.
00:57:240Paolo Guiotto: Second, determine the stationary. That means constant solutions and regions where the
01:03:460Paolo Guiotto: the solutions are increasing, decreasing. So constant solution is easy, because…
01:08:360Paolo Guiotto: Yes, it's solving an equation, but in a very special…
01:12:210Paolo Guiotto: Type of solutions that transpose the equation into a
01:16:460Paolo Guiotto: of the Bright equation, we found that there is a unique solution, constant equal to zero.
01:22:220Paolo Guiotto: Then we studied when solutions are increasing or decreasing, Checking…
01:29:160Paolo Guiotto: Well, the derivative of Y is pi.
01:33:290Paolo Guiotto: This turns out that this gives a condition on Y, general, on TY,
01:39:100Paolo Guiotto: So that there is a region of the plane TY where solutions, when they are in this region, are increasing, and complementary will be decreasing. For this case.
01:50:680Paolo Guiotto: we realized that wherever is the solution, it is always… So, in particular, we have now to study the solution of the Cauchy problem with the initial condition Y0 equals Y0, Y0 is positive.
02:06:00Paolo Guiotto: This initial condition, is on the y-axis.
02:11:840Paolo Guiotto: In any case, the solution will be increasing, so we can, start now solving, answering the third question. Determining monotility, concavity, discuss, if the initial time is,
02:27:310Paolo Guiotto: Minus infinity or finite. And then, what about the final time, beta?
02:33:430Paolo Guiotto: And the limits of the solution.
02:36:390Paolo Guiotto: when time goes to alpha and beta. So, let's see what we can say here. So, since, well…
02:46:460Paolo Guiotto: scenes.
02:48:730Paolo Guiotto: solutions.
02:50:940Paolo Guiotto: Always.
02:55:880Paolo Guiotto: white crime.
02:57:520Paolo Guiotto: So,
03:01:850Paolo Guiotto: Always Y prime T greater or equal than zero. Whatever is our solution, Y will be increasing, okay, as we already…
03:11:680Paolo Guiotto: Now, let's discuss the concavity. So, this is about the monotonicity, there is nothing else to be said. Then, let's see what about the concavity.
03:22:280Paolo Guiotto: So concavity.
03:27:240Paolo Guiotto: we needed to compute the second derivative, because Y has the concavity pointing upward, so it's convex, technically.
03:39:610Paolo Guiotto: If and only if Y second is paused.
03:44:570Paolo Guiotto: from the equation, from… equation…
03:51:370Paolo Guiotto: Y prime is equal to YE to Y minus 1. Remind followers that here Y is Y up here.
04:00:580Paolo Guiotto: So that Y means Y of P is a function of P.
04:04:690Paolo Guiotto: And when I say Y prime, we mean Y prime of t, the derivative of this Y with respect to t, and of course, Y second will be the derivative, the second derivative of Y with respect to T.
04:19:290Paolo Guiotto: So we take this equation, we differentiate, So, differentiating…
04:25:510Paolo Guiotto: we get y second equal. So if you want, it is the derivative of this thing, YE to Y minus 1, but not the derivative with respect to Y, okay? It is the derivative with respect to t, so you have to imagine that here is written Y of T here, Y of T, etc.
04:43:720Paolo Guiotto: So, first of all, this is a product, y times the parentheses, so it will be Y prime times e to y minus 1.
04:53:20Paolo Guiotto: plus Y times the derivative of E2Y minus 1.
04:58:450Paolo Guiotto: What is this derivative?
05:00:310Paolo Guiotto: Well, the derivative of 1 is 0, so it remains the derivative of e to y.
05:06:660Paolo Guiotto: Which is, the derivative of an exponential is the exponential, so e to y, times the derivative of the argument of the exponential, which is y prime.
05:16:370Paolo Guiotto: So at the end, we have Y prime, E toy minus 1 plus Y, and this comes E2Y, Y prime.
05:27:70Paolo Guiotto: Let's, factorize.
05:30:420Paolo Guiotto: DY prime.
05:32:640Paolo Guiotto: So we get E2Y minus 1 plus y e2y.
05:40:710Paolo Guiotto: Okay, now, so this is way second, and we have to determine, if possible, the sign of this tank. So what do we know?
05:48:910Paolo Guiotto: We know that this is… Positive, because it is always positive.
05:56:50Paolo Guiotto: So now everything depends on the sign of this parenthesis.
06:00:840Paolo Guiotto: So what can be said about the sign of that parenthesis?
06:04:860Paolo Guiotto: You cannot solve the inequality because, you don't know what is why, okay? So we have just to think, to argue with what we know.
06:14:60Paolo Guiotto: Wait, definitely this is a positive turn, and that's for sure.
06:18:940Paolo Guiotto: What about Y?
06:23:640Paolo Guiotto: Well, we don't know yet, too much. We know that Y is increasing. As the picture suggests, Y should be positive.
06:34:470Paolo Guiotto: But this demands a little argument. Why this solution Y cannot be negative?
06:42:10Paolo Guiotto: Because to be negative, it should cross the red line, which is another solution. That's forbidden because of the uniqueness. That's the argument. So, we have a first claim.
06:53:240Paolo Guiotto: claim, it's not trivial, it's not evident, it is not written anywhere, we have to prove. Y is always
07:03:780Paolo Guiotto: Positive.
07:06:930Paolo Guiotto: So, suppose that we have done this, we know that that Y will be positive.
07:13:140Paolo Guiotto: And since, again, Y is positive, what about this term?
07:20:900Paolo Guiotto: How is E2Y with respect to 1? If Y is positive, E toy is…
07:29:760Paolo Guiotto: Exponential of a positive number is…
07:33:720Paolo Guiotto: Yes, thank you, but that happens also for negative numbers.
07:38:530Paolo Guiotto: But here, I have to subtract 1, so I have to compare to 1.
07:43:920Paolo Guiotto: yeah, this is greater than 1 because of this, so this is again positive. So you see that once we know that Y is positive, we can say that that parenthesis is now positive, Y prime outside is positive, everything is positive, this is positive.
08:01:810Paolo Guiotto: or let's stay larger, greater or equal than zero, nothing changed. So it means that Y second will be greater or equal than zero, and therefore, our solution
08:11:220Paolo Guiotto: is convex.
08:12:850Paolo Guiotto: So we know that in this figure, the two possibilities cannot be both true. This is not possible.
08:22:80Paolo Guiotto: the plot will be something like that, okay? So from this, from this claim, it will follow that Y is convex.
08:33:950Paolo Guiotto: So all our goal becomes to show that Y must be always positive, that we said this comes from the fact that if Y were negative somewhere, it should cross the red line. So let's see precisely this argument.
08:53:900Paolo Guiotto: So… In fact.
08:56:910Paolo Guiotto: This is the situation. So you see that the constant solution is not important by its own for the solution Y, it's just another solution, but it's important because it's a simple solution we have, and we can use somehow to get some information about our solution.
09:15:180Paolo Guiotto: Which is this one. Now, we take our solution Y of T. So…
09:21:380Paolo Guiotto: If… so this is classical, contradiction argument. If Y is negative somewhere.
09:31:870Paolo Guiotto: So, formally, if there is a time T, say, T hat.
09:36:930Paolo Guiotto: Such that Y at T hat is negative.
09:42:990Paolo Guiotto: So let's say that there is a time. Of course, in this video, since the function is increasing, this time cannot be after 0, because after zero, the function is increasing, so if it is positive at time zero, it will be positive.
09:55:940Paolo Guiotto: In the future, but it will be in the past, however, it's irrelevant. Assume that there is a time t where the solution is negative here, so this is the value YT negative.
10:08:470Paolo Guiotto: Then, by continuity.
10:12:580Paolo Guiotto: And our solution is continuous, because it's differentiable. So there will be another time, T double hat, such that Y at that time, T double hat, will be equal to 0. So there is another time, T double hat.
10:30:470Paolo Guiotto: Well, maybe we're right here, D double hat.
10:35:240Paolo Guiotto: Well, the solution is here. So we have a solution that crosses the red line at time t hat.
10:43:990Paolo Guiotto: But then… our Y, should coincide with the zero solution wherever both are defined by uniqueness.
10:56:200Paolo Guiotto: And since the red solution is defined for every T, the solution of our Cauchy problem, Y, is not necessarily, for the moment, defined for every T, but on the interval alpha beta. So the common times are these ones, for every T in interval alpha-beta.
11:14:580Paolo Guiotto: And this follows by uniqueness.
11:21:680Paolo Guiotto: But this is a contradiction. Then I should have that also at time t equals 0, Y should be 0. But…
11:29:500Paolo Guiotto: Because of the Cauchy problem, this is why not, which is positive.
11:33:870Paolo Guiotto: So this is impossible.
11:40:300Paolo Guiotto: So, as you can see, we get a contradiction.
11:43:820Paolo Guiotto: If it is negative somewhere, it must be zero.
11:47:160Paolo Guiotto: And then it must be 0 also at the initial time, which is not the case. So this closes this, argument, and then we can… we can conclude about the concavity. So, what we know so far is this figure.
12:05:270Paolo Guiotto: We have, the plane TY.
12:09:30Paolo Guiotto: the constant solution.
12:12:90Paolo Guiotto: And our solution Which is decreasing convex, like that.
12:20:210Paolo Guiotto: Now, the next point concerns what?
12:25:710Paolo Guiotto: What is it?
12:27:210Paolo Guiotto: We have to discuss about alpha.
12:31:770Paolo Guiotto: The initial time of life for this solution.
12:36:00Paolo Guiotto: Is it minus infinity, or…
12:39:700Paolo Guiotto: or not. It cannot be plus infinity, because it is the left side, no? It's the past.
12:45:420Paolo Guiotto: So we have to determine, whether this alpha is, like, here.
12:52:670Paolo Guiotto: Or, about here, alpha equal minus infinity.
12:57:300Paolo Guiotto: Now, the point is that they cannot be there because of the argument of the exit from compact sets. We can easily find a compact box that contains the solution, if alpha is finite, in the past, and the solution never leaves this compact K.
13:17:340Paolo Guiotto: So, this K… so, if… alpha, well, find it, then this solution points TYT
13:29:230Paolo Guiotto: Which are the points on this.
13:31:780Paolo Guiotto: on this line, these are points with abscity and ordinate YT.
13:37:920Paolo Guiotto: So they are points of the graph, belong to the box from alpha to 0,
13:45:820Paolo Guiotto: And, the, the… The quote is 0, and the upper court is Y0.
13:53:750Paolo Guiotto: This, for every time T, Between alpha and initial time, 0.
14:01:450Paolo Guiotto: So, in particular, this says that the solution won't escape this compact K,
14:08:910Paolo Guiotto: in the past. It will do in the future, but it must happen that it escaped both in past and future. And what, what happens here is that it doesn't leave this K in past, in the past, okay? So, solution…
14:28:680Paolo Guiotto: does… Not… Leave.
14:34:370Paolo Guiotto: K… Ian.
14:37:420Paolo Guiotto: beats boss.
14:40:230Paolo Guiotto: And this is, impossible. It's a contradiction.
14:45:850Paolo Guiotto: But if this is impossible, it means that
14:49:150Paolo Guiotto: Alpha must be equal to minus infinity.
14:53:20Paolo Guiotto: Okay?
14:54:110Paolo Guiotto: So, that's how we, replied to the question. What about alpha? Is it finite? Is it minus infinity?
15:02:990Paolo Guiotto: and we are here now, let's compute also the limit. So we know now that we can say that this diagram descend
15:12:860Paolo Guiotto: positively.
15:15:950Paolo Guiotto: when we go to minus infinity. So now, what about the limit?
15:21:790Paolo Guiotto: So let's call L the limiter of the solution when T goes to minus infinity.
15:30:550Paolo Guiotto: Now, this limit exists because the function is increasing, so as any monotone function, there are leads.
15:37:730Paolo Guiotto: It… exist.
15:42:50Paolo Guiotto: Because… Y is.
15:48:50Paolo Guiotto: Increasing, or decreasing, is the same.
15:52:940Paolo Guiotto: Now, what else can be said? The function is positive, so the limit won't be negative because of the permanence of sine.
16:02:60Paolo Guiotto: So, L is greater than or equal than zero, because
16:07:690Paolo Guiotto: Y, it doesn't matter if Y is strictly positive. Y can never be zero, because if Y is 0, again, we cross the red solution.
16:16:360Paolo Guiotto: But, the limit can be zero, no?
16:20:480Paolo Guiotto: 1 over X is positive when you go to infinity.
16:24:280Paolo Guiotto: plus infinity, 1 of X goes to 0, you know, being positive, subtly positive.
16:29:700Paolo Guiotto: Okay, this is not yet an exact value. We can also say that, if we look at the figure, definitely the limit will be less than why not, right?
16:40:780Paolo Guiotto: Because the function is decreasing going to the left. So we can say that there is also this other information. So we have a range for the limit, but not yet the exact value. And now it's the equation to give the answer, because if we look at the equation, Y prime.
16:57:910Paolo Guiotto: Now… Y prime of T, well, let's say Y prime equal.
17:04:460Paolo Guiotto: YE2Y minus 1. If we send the T, to minus infinity.
17:11:740Paolo Guiotto: We can say that the right-hand side has a limit, because Y goes to that value L, so also here, in the argument of the exponential, will go to L, the exponential is a continuous function, this quantity will go to… sorry, L,
17:30:200Paolo Guiotto: E to L minus 1.
17:33:550Paolo Guiotto: And now the point, as I said last time, this value cannot be anything less than zero. If you want, we can state as a general factor, general.
17:44:820Paolo Guiotto: factor.
17:47:600Paolo Guiotto: If we have a function.
17:50:190Paolo Guiotto: If we have that Y of T goes to L, when T goes, this works for minus infinity or plus infinity is the same.
18:00:30Paolo Guiotto: plus or minus infinity. And Y prime of t
18:05:670Paolo Guiotto: Be careful, because I want to stress you something, it's not trivial fact. And Y prime of t for the same values of T, so when t goes to plus or minus infinity, goes to some value M,
18:20:660Paolo Guiotto: And you know that this cell is finite, you need to know this.
18:25:150Paolo Guiotto: But necessarily, that M must be equal 0.
18:31:480Paolo Guiotto: There cannot be any other possibility.
18:34:220Paolo Guiotto: Notice that, warning.
18:39:470Paolo Guiotto: Normal, this is a very common, error.
18:45:00Paolo Guiotto: a common…
18:49:300Paolo Guiotto: Error.
18:51:700Paolo Guiotto: is to think… that, you just need to say that if Y prime
19:01:110Paolo Guiotto: Sorry, if Y of T goes to L,
19:04:630Paolo Guiotto: and this cell is finite. This alone implies that Y prime of D is
19:11:310Paolo Guiotto: Because, intuitively, Y prime goes to a finite value means the function has an assigned thought at plus, minus, infinity, no? So if you imagine a function doing like that, you imagine a function like this, no? Where this is the limit value
19:30:180Paolo Guiotto: L.
19:31:300Paolo Guiotto: No?
19:33:30Paolo Guiotto: So you would say the tangent must become horizontal at infinity.
19:38:410Paolo Guiotto: Well, this is wrong.
19:45:600Paolo Guiotto: Because you can easily build examples where you have a finite limit at infinity, but the derivative is unbounded, for example, and
19:54:310Paolo Guiotto: does not have any limit. Take this. Y of T equals… we take something like 1 over T times sine of, t cubed.
20:06:650Paolo Guiotto: For example, Now, what happens when t goes to plus infinity?
20:12:430Paolo Guiotto: If you send t to plus infinity, 1 over t goes to 0. Sine has not a limit, but it is bounded.
20:21:290Paolo Guiotto: It is between minus 1 and 1, so you multiply by something that goes to zero. It is clear that the product will go to 0, okay?
20:29:730Paolo Guiotto: But if you look at the derivative, y prime of t, this is… well, we have derivative of 1 over T, which is minus 1 over t squared, sine t cubed.
20:40:120Paolo Guiotto: And this is fine, because when t goes to infinity, we have the same situation as above. But now, when you do the other factor, 1 over T, we have cosine T cubed times… the derivative of T cubed is 3T squared.
20:56:500Paolo Guiotto: So this term here, which is 3P cosine The Cuba.
21:04:500Paolo Guiotto: So when t goes to infinity, this goes to 0. So this part of the derivative goes to zero. But this part here, when t goes to plus infinity, or minus infinity, you see that there is this T, which becomes bigger, and this cosine that obstinates between minus 1 and 1.
21:24:260Paolo Guiotto: So this problem can be bigger, arbitrarily bigger.
21:28:60Paolo Guiotto: When this cosine is, the value of P,
21:31:390Paolo Guiotto: Such that cosine is 1, this product will be huge, no? Positive minus 1, the product Negative. Big.
21:41:630Paolo Guiotto: So, in fact, this is unbounded.
21:48:410Paolo Guiotto: And, of course, there is no limit
21:53:580Paolo Guiotto: when T goes to plus or minus infinity.
21:56:940Paolo Guiotto: So the simple existence of a finite limit is not enough to ensure that the derivative goes to zero.
22:05:730Paolo Guiotto: But this is saying something more. If you know two things here, that there is a limit of the function.
22:13:300Paolo Guiotto: plus… Here, it's an assumption. You know also that there is the limiter for the derivative.
22:20:390Paolo Guiotto: It's not that you can say necessarily that unit is zero.
22:24:330Paolo Guiotto: You see, this second condition is not verified in this example, so I'm not in that case, no? Because the second requirement here is not true.
22:35:80Paolo Guiotto: But in this case, if we have this.
22:38:660Paolo Guiotto: And we can easily see that this happens, so let's close this remark. Let's see the one-line proof of the fact of…
22:51:800Paolo Guiotto: Generally.
22:53:80Paolo Guiotto: factor. It's actually a consequence of
22:57:240Paolo Guiotto: your beloved rule, orbital rule, so we know we know.
23:05:70Paolo Guiotto: that Y of T goes to L when T goes to infinity, and at the same time, Y prime of t goes to M.
23:15:30Paolo Guiotto: This is finite.
23:17:20Paolo Guiotto: when T goes to plus infinity, or minus infinity.
23:21:410Paolo Guiotto: So I wanted to say that this M is 0.
23:25:260Paolo Guiotto: Okay, let's do this.
23:27:440Paolo Guiotto: Then, If we compute the limit.
23:33:470Paolo Guiotto: for T going to plus or minus infinity, depends on the case, of Y of T divided by T,
23:41:890Paolo Guiotto: this thing. Well, we know what is the value of the limit, because the numerator goes to a finite value L, this is finite.
23:52:200Paolo Guiotto: by assumption, and you divide by t, that goes to plus-minus infinity, so the fraction goes to Zero.
24:01:90Paolo Guiotto: So we know this.
24:02:860Paolo Guiotto: But if we apply the orbital rule, this would be the limit also when t goes to plus-minus infinity.
24:09:350Paolo Guiotto: Of the ratio of the derivatives.
24:12:170Paolo Guiotto: And here we have Y prime T divided the derivative of T, which is 1.
24:18:250Paolo Guiotto: So it means that that limit must be equal to this one, which is our value, M.
24:24:770Paolo Guiotto: So the conclusion is n equals 0.
24:32:230Paolo Guiotto: So this is the general fact that we can always use in this kind of situations. So, returning back to the exercise, we knew that the solution has a finite limit. It's very important. The limit must be finite. If the limit is infinite, the argument doesn't work.
24:49:340Paolo Guiotto: There is a limit for the solution, and… and here we get this information from the equation.
24:55:180Paolo Guiotto: If Y prime is the right-hand side, if you are able to show that the right-hand side moves somewhere.
25:01:510Paolo Guiotto: We have also the limit for the derivative. We are exactly in this context. We have the limit for the solution, the limit for the derivative, and the limit for the solution is 5.
25:12:120Paolo Guiotto: Then, necessarily, this is the M.
25:16:800Paolo Guiotto: So… going… Beck.
25:26:910Paolo Guiotto: Exercise… So we know.
25:34:950Paolo Guiotto: that Y prime T goes to… sorry, YT.
25:39:960Paolo Guiotto: goes to some limit L that belongs, Between 0 and Y0.
25:46:730Paolo Guiotto: Y prime T is YE to Y…
25:51:360Paolo Guiotto: minus 1. This goes to L, E to L minus 1, and since L is a finite value, this is a finite value. This is the number M,
26:05:740Paolo Guiotto: by the test we have just proved that M must be equal to 0, and from this we gain an equation for L. L times e to L minus 1 must be equal to 0.
26:19:430Paolo Guiotto: This is not yet L equal, but solving this equation, we get the value of L. In this case, it's easy, because that product can be 0 if and only if either L is 0, or E2L minus 1 is 0.
26:34:750Paolo Guiotto: which is, in fact, E to L, equal 1, So again, L equals 0.
26:41:540Paolo Guiotto: So this is a unique solution, L equals 0.
26:46:620Paolo Guiotto: So it means that our limit cannot be anything else than 0. The limit is 0. So L…
26:53:990Paolo Guiotto: Equals zero.
26:56:440Paolo Guiotto: So now we gained a new information, which is…
27:04:860Paolo Guiotto: This is the plane TY.
27:07:190Paolo Guiotto: This is the constant solution. The initial condition, Y0.
27:12:560Paolo Guiotto: the solution is decreasing in the past, going down to zero at minus infinity. It is convex. And this also happens for the future.
27:22:760Paolo Guiotto: Now, the next part concerns pita.
27:27:610Paolo Guiotto: Okay?
27:29:00Paolo Guiotto: Now, this is a little bit more difficult, because, for example, we have the solution is increasing.
27:36:700Paolo Guiotto: Now, what can be said is that, well, what cannot happen, what could happen?
27:43:200Paolo Guiotto: So we could have these alternatives. Alternative number one.
27:47:60Paolo Guiotto: The solution just ends at some time beta here.
27:51:620Paolo Guiotto: Is it possible?
27:55:960Paolo Guiotto: Is it possible that this is the solution?
28:01:880Paolo Guiotto: No, but why?
28:05:200Paolo Guiotto: You remind the solution can never die inside the domain, and this is the argument of compact set. So, I will find a box that contains the solution in the future. So, this is case one.
28:19:920Paolo Guiotto: Well, let's say that this is, let's do for beta a discussion. What about beta?
28:27:340Paolo Guiotto: So we can, we can distinguish, we.
28:31:620Paolo Guiotto: have,
28:35:840Paolo Guiotto: Following… Okay, exists.
28:41:620Paolo Guiotto: We have to understand which one is the possibility? So, let's do the… quickly the figure…
28:50:200Paolo Guiotto: In a… in small, let's say. That…
28:53:460Paolo Guiotto: Well, beta can be finite or infinite, that's the first alternative, but…
28:58:610Paolo Guiotto: Bitifying it could be with this solution, so a solution that ends just into the domain. This won't be possible.
29:06:930Paolo Guiotto: or beta, File it.
29:10:520Paolo Guiotto: And what can happen if the solution does not enter into the domain?
29:16:590Paolo Guiotto: What can do the solution? It is convex, so pointing upward, it is increasing. What can happen if it dies at beta?
29:27:610Paolo Guiotto: Yeah, it has a vertical asymptote. It blows up.
29:31:640Paolo Guiotto: Okay?
29:33:90Paolo Guiotto: So this is in Asanto. This is the case beta finite. The alternatives for beta-finite. It cannot be any other alternative.
29:41:370Paolo Guiotto: Then we have a beta plus infinity.
29:46:60Paolo Guiotto: So the case beta equal plus infinity, let's see what are the possibilities.
29:51:310Paolo Guiotto: So now we imagine we have a convex function, increasing convex.
29:55:820Paolo Guiotto: Is this possible?
29:58:330Paolo Guiotto: No.
29:59:780Paolo Guiotto: So, this cannot be. Well, so you may imagine that the solution, we grow, grow, grow forever, so we like an exponential, like, something like this.
30:10:860Paolo Guiotto: Okay, so these are the three possibilities.
30:13:820Paolo Guiotto: The point is, which of this is the true? Is this one? Is this one? Is this one?
30:21:540Paolo Guiotto: Well, we can exclude at least the first one immediately.
30:25:950Paolo Guiotto: So let's call 1… 2… And three, these three scenarios.
30:33:950Paolo Guiotto: So, the Scenario 1… is impossible.
30:42:340Paolo Guiotto: because… Off.
30:48:390Paolo Guiotto: exit.
30:51:320Paolo Guiotto: from… compact sets.
31:02:200Paolo Guiotto: you can easily determine a box. So, imagine that this is bit, imagine that this is the solution, So…
31:11:720Paolo Guiotto: you see that this value is a finite value, okay? So if this is the value L, let's call it again L, we can easily write a box like this one.
31:26:340Paolo Guiotto: So this K is literally the internal zero beta.
31:31:70Paolo Guiotto: times 0 to that value L.
31:34:760Paolo Guiotto: is a compactor.
31:37:280Paolo Guiotto: Closed and bounded, contains in the domain B, which is the plane R2, so that's easy.
31:46:370Paolo Guiotto: And this solution… That's not… leave.
31:56:330Paolo Guiotto: this K, now this time is in the future, not in the past. In the past, of course, it lives, but in the future, it doesn't live, because this is the box that takes all possible times and all the range of the solution, so there cannot be any exit from that.
32:12:940Paolo Guiotto: Kate.
32:14:340Paolo Guiotto: in… Future.
32:18:220Paolo Guiotto: And this is impossible.
32:25:470Paolo Guiotto: Now, the point is, which one, so the first alternative has been excluded.
32:33:430Paolo Guiotto: So now we have these two. Beta, fine, little beta infinite.
32:37:940Paolo Guiotto: Which one? This is difficult, because it's tricky. We have to do something here.
32:44:820Paolo Guiotto: Okay, so if you look at the two figures, 2 and 3, they have a common point.
32:51:290Paolo Guiotto: Which is, the solution goes up to plus infinity.
32:54:930Paolo Guiotto: So… 2 and 3.
32:57:980Paolo Guiotto: In.
32:59:610Paolo Guiotto: Scenarios, too.
33:01:570Paolo Guiotto: And… 3… The solution Y of T goes to plus infinity when t goes to beta.
33:12:750Paolo Guiotto: So now, how can we use this factor?
33:15:840Paolo Guiotto: Now, this is a very particular case, it's not a general argument. Sometimes can be used, but it's not granted. I noticed that the solution, Y prime, the equation y prime equals y to y minus 1,
33:31:750Paolo Guiotto: In principle, it's a separable variable equation we could solve, but in fact, this is impossible to be solved explicitly.
33:39:740Paolo Guiotto: However, since our solution is, Y is positive, so this quantity is, we say that it's positive, it's never zero, we can write in this form, Y prime over Y
33:53:340Paolo Guiotto: times C2Y minus 1 equal to 1. And we do this now. We integrate this identity between the initial time and the final time beta, both sides. What do we get? We have done this last time.
34:08:250Paolo Guiotto: But here is more tricky, because when I integrate from 0 to beta this identity, I get on the right easy, integral of 0 beta of 1, this gives beta, which is the value
34:20:730Paolo Guiotto: of the final time. It is equal to the integral from 0 to beta of this Y prime. Let's write the letter, because here we have to understand
34:29:760Paolo Guiotto: Which is the integration variable?
34:33:920Paolo Guiotto: So the integration variable is T, because we are integrating in the variable here. This is a function of T, so we are integrating in this variable.
34:43:790Paolo Guiotto: Now, of course, I don't know the solution. I cannot compute that integral. I don't… I don't even know what is beta. But here there is the trigger.
34:52:400Paolo Guiotto: Which is a trick that you would do by solving… you would try to solve this equation with the method of separation of variables. Call U equal Y of T. Do a change of variable. What happens here?
35:05:880Paolo Guiotto: Now, let's first see inside. So, first of all, wherever you see Y, it becomes U. So I have 1 over U times E to U minus 1. And remind that if you do this, change your variable, u equal y of t.
35:21:490Paolo Guiotto: DU becomes Y prime TDT.
35:25:570Paolo Guiotto: So, Y prime infinity is what is U.
35:30:140Paolo Guiotto: This is the technique of change… the technique of separation of variables. Now, what happens to the range of the new variable, u?
35:40:720Paolo Guiotto: So, if this is 0, you vary from Y of 0, which is the value Y naught.
35:46:940Paolo Guiotto: and the value at beta. What is the value at beta? Well, we say that when t goes to beta, y goes to plus infinity, so this is…
35:55:850Paolo Guiotto: plus infinity.
35:58:230Paolo Guiotto: So we get this formula. Beta.
36:00:990Paolo Guiotto: is equal to the integral form Y0 to plus infinity of 1 over, what is U times E2U minus 1.
36:10:710Paolo Guiotto: Now, I won't compute this integral, because it's not possible to be done. If I could compute this integral, I would have the value of beta explicitly, but unfortunately, this is not elementary.
36:24:130Paolo Guiotto: to be computed, even if it looks not particularly complicated. But the only point is that I want to establish if beta is finite or not, because these two cases is… one is beta finite, the other is beta is finite.
36:36:820Paolo Guiotto: And, so beta is finite or not, if and if this integral is finite or not.
36:43:600Paolo Guiotto: And what about this integral? Is that integral finite or not?
36:53:520Paolo Guiotto: No, the point is not Y0, this is what is a generalized integral from Y0 to plus infinity.
37:00:110Paolo Guiotto: At Y0, that function that you see here, so Y0 is, here, it is positive, right? Because the initial condition is positive.
37:11:40Paolo Guiotto: the Y0 is a positive value, okay? So I am integrating that function, from Y0 to plus infinity. The function is 1 over UE to U minus 1.
37:26:640Paolo Guiotto: Now, due to this… now, I hope that you have seen generalized integral last year, right?
37:33:340Paolo Guiotto: So now the point is, do you know how to discuss convergence, so the fact that the integral is finite or not?
37:42:890Paolo Guiotto: Well, if you can compute, you can see directly it is finite or not, okay? But we cannot compute.
37:49:110Paolo Guiotto: So we have a, let's say we could have a… two possibilities.
37:54:660Paolo Guiotto: One is the asymptotic discussion. We can see that when U goes to plus infinity, that function 1 over UE to U minus 1 is asymptotic.
38:07:680Paolo Guiotto: Well, U is going to infinity. E to u is going to infinity. 1 is 1, so this is like E to U.
38:14:820Paolo Guiotto: When U is big. So it is like one of a U, E to U, or if you want, it is E2 minus U divided by U.
38:24:410Paolo Guiotto: Now, this function here, as a function of U, is integral at plus infinity.
38:30:670Paolo Guiotto: Because it goes past to zero, there is the exponential. If I have just 1 over U, it's not integrable at plus infinity, because it is the log of U, and when you send U to plus infinity, you go to plus infinity. But this one is integral.
38:48:40Paolo Guiotto: At plus infinity. And this would say that that integral is finite.
38:54:130Paolo Guiotto: Not infinity. So this would say that the integral from Y naught to plus infinity of 1 over ue to mu minus 1, this quantity is finite, and this means beta finite.
39:08:600Paolo Guiotto: In Alternativo.
39:11:340Paolo Guiotto: in this particular case, I can do a little trick to not to compute, but to estimate that integral.
39:19:890Paolo Guiotto: Alternative?
39:22:70Paolo Guiotto: I have beta equal integral from y naught to plus infinity, 1 over u, e to u minus 1,
39:31:230Paolo Guiotto: This is a trick that works for this example, so these are not general
39:35:700Paolo Guiotto: ideas, so the question here, the last point is difficult. I noticed that the U is between Y0, which is positive, and plus infinity, right? So this denominator here is larger than Y0. What happens if I replace that U
39:54:830Paolo Guiotto: with, the value Y naught?
39:59:830Paolo Guiotto: I take a smaller value, the fraction, the denominator.
40:04:810Paolo Guiotto: diminishes, so the fraction increases. So this is less than the integral from Y0 to plus infinity of 1 over Y0, which is now a constant, so I can carry outside of the integral, so I can put outside.
40:21:690Paolo Guiotto: So this is 1 over Y0 integral from y naught to plus infinity, and the point is that I can now compute this integral.
40:30:870Paolo Guiotto: Because,
40:36:00Paolo Guiotto: Because this is a rational function of E to U, so if you want, you can compute, this comes,
40:45:410Paolo Guiotto: So we should, do another change of variable. Let's do without the variable. E to U minus 1, EU,
40:53:950Paolo Guiotto: If you set D equal E to U, this means DB equal E to U, DU.
41:02:690Paolo Guiotto: Or, sorry, U equal… log of V.
41:09:50Paolo Guiotto: VU is equal 1 over VDB. This becomes the integral of 1 over
41:16:890Paolo Guiotto: E to U is V minus 1, this is the function, du becomes 1 over V, DVD. So you have to integrate this thing.
41:25:900Paolo Guiotto: Which is a rational function, so you split this,
41:31:140Paolo Guiotto: say, minus, minus, you put the plus B minus V,
41:35:180Paolo Guiotto: So this becomes minus primitive of 1.
41:41:400Paolo Guiotto: of a V when I divide the CR.
41:45:440Paolo Guiotto: Yes, minus plus…
41:49:450Paolo Guiotto: primitive of 1 over V minus 1 dB. So it becomes, what is log of V minus 1 over V,
41:59:530Paolo Guiotto: Because this is minus log of V, V is positive, so log of 1 minus 1 of V.
42:07:790Paolo Guiotto: Going back to U, D was E to U log of
42:13:400Paolo Guiotto: 1 minus, 1 over E to U.
42:17:220Paolo Guiotto: Okay, so this is the value of that integral. So if I want now to evaluate from Y0 to plus infinity, I have to do Y0 to plus infinity of 1 over e to nu minus 1DU,
42:30:870Paolo Guiotto: It is the final value, log of
42:34:290Paolo Guiotto: 1 minus 1 over E2 plus infinity.
42:38:980Paolo Guiotto: Minus the initial value log of 1 minus 1 of e to y naught.
42:45:430Paolo Guiotto: E2 plus infinity is plus infinity, 1 over e to plus infinity, that's 0. So log of 1 is 0, and so what remains is minus log of 1 minus 1 over E.
42:59:510Paolo Guiotto: Y0, which is finite.
43:02:680Paolo Guiotto: So at the end, I… I… I know that this value is 5.
43:10:20Paolo Guiotto: which is multiplied by this value, which is finite, I get, in any case, beta finite.
43:20:320Paolo Guiotto: So, the conclusion is that. So, conclusion…
43:26:290Paolo Guiotto: Beta is finite, and the solution goes to plus infinity when t goes to beta.
43:36:40Paolo Guiotto: And therefore, the graph of the solution Is finally of this type.
43:44:10Paolo Guiotto: So let's say that without this final part, you don't know what to do here.
43:50:260Paolo Guiotto: Okay, you could say it goes up to plus infinity, but…
43:54:640Paolo Guiotto: If it goes… does it go in finite time, in infinite time? Now we know that it goes in a finite time, so the function is like that.
44:03:540Paolo Guiotto: So that's the solution of the equation.
44:08:690Paolo Guiotto: Okay.
44:10:560Paolo Guiotto: Do you want to take a little breaka?
44:14:810Paolo Guiotto: Okay.
44:15:940Paolo Guiotto: Before we take a new problem.
44:19:660Paolo Guiotto: Okay, so let's take, 5 minutes.
44:30:570Paolo Guiotto: So, let's do another problem.
44:34:930Paolo Guiotto: I do the exercise 4-10-4.
44:40:220Paolo Guiotto: The equation we have here is Y prime equals 1 over logarithm of y.
44:47:650Paolo Guiotto: plus t. Be careful, because if you don't see any parentheses, it means that the argument of log is…
44:58:820Paolo Guiotto: Is it Y or Y plus D?
45:01:980Paolo Guiotto: Just Y. If I want Y plus T, I have to write this, okay?
45:08:80Paolo Guiotto: So, always be careful when you read an assignment. If you have doubts about the writing, please ask. I'm talking about the exam in particular.
45:19:80Paolo Guiotto: So, question one is, domain… of local…
45:28:720Paolo Guiotto: existence and uniqueness. Well, let's write questions one by one.
45:33:150Paolo Guiotto: So, here… The right-hand side is the function
45:40:10Paolo Guiotto: 1 over, if you want, let's write it this way, T plus log of Y, so there is no doubt about the interpretation of the argument of log, okay?
45:49:290Paolo Guiotto: This function is defined defined.
45:54:880Paolo Guiotto: on D.
45:56:910Paolo Guiotto: which is the set of points, TY in R2, such that… what are the conditions? We have log.
46:06:250Paolo Guiotto: A log asks her for Y positive.
46:09:820Paolo Guiotto: Then, we have also a fraction.
46:13:150Paolo Guiotto: That denominator must be different from zero, so t plus log of y must be different from 0.
46:21:230Paolo Guiotto: So let's see what this says in the Cartesian plane.
46:24:950Paolo Guiotto: So, if we go in the Cartesian plane, Y must be positive, so it means that the plane is the plane TY,
46:32:100Paolo Guiotto: There cannot be this part of the plane.
46:36:110Paolo Guiotto: Okay.
46:37:300Paolo Guiotto: and even the axis.
46:39:310Paolo Guiotto: So the… the t-axis is the Y equals 0.
46:43:460Paolo Guiotto: Then, we have that t plus log y must be different from 0. What does it mean? Well, it is, if possible, better to write this as a condition where y is some function of T. Because if you see, t plus log of y is equal to 0,
47:01:830Paolo Guiotto: Well, I can say t is minus log of Y, but that T is the absence and Y the ordinate, so perhaps it is better to say log of Y is minus T,
47:13:640Paolo Guiotto: So it means, Y.
47:16:600Paolo Guiotto: equal E2 minus T, right?
47:19:960Paolo Guiotto: So, that quantity is 0 when Y is equal to E minus T. Now, this is, be careful, because the domain is that quantity different from 0. So, the TYs that verify this condition are not in the domain.
47:34:670Paolo Guiotto: So they are, let's say, bad points. So, let's mark with the dashed the line, because the line Y equal E to minus T would be this one, no? Well, let's… I do first the plot in this way. This is the line Y equal E minus T.
47:51:20Paolo Guiotto: But these are not points of the domain, because,
47:55:110Paolo Guiotto: What is in the domain? Points where T plus log Y is different from 0, not equal to 0. So it means that these are points not in the domain, so we have to max some… in some way to recognize that they are not in the domain, so I will dash this line in this way.
48:13:910Paolo Guiotto: Okay? So what is at the end of the domain? What remains?
48:18:600Paolo Guiotto: So the up half plane without that dashed line. So this is the domain V, so it's somehow divided into parts.
48:28:990Paolo Guiotto: Now, on this domain, you know.
48:32:180Paolo Guiotto: The function is clearly made of continuous functions, so there is no problem.
48:37:990Paolo Guiotto: F is continuous on D.
48:44:520Paolo Guiotto: And of course, also DYF, if we want it. We can compute, it's 1 over, so it means minus the square of denominator, T plus log of Y,
48:57:10Paolo Guiotto: square, and then we have the derivative with respect to Y of the denominator. The denominator is T plus log y, derivative with respect to Y is minus… sorry, it's not minus, it's just 1 over Y.
49:09:560Paolo Guiotto: So whatever it is, as you can see, this quantity is well-defined and continues on the domain, because the domain is Y positive, so you have no danger for the numerator.
49:19:810Paolo Guiotto: and for the denominator, and the parentheses, the argument of the parenthesis is different from zero. So this is a well-defined and continuous function on domain B.
49:30:200Paolo Guiotto: So… Local.
49:37:320Paolo Guiotto: Existence and uniqueness also.
49:42:20Paolo Guiotto: on… That's for question number one.
49:47:440Paolo Guiotto: Now, question 2.
49:51:560Paolo Guiotto: It asks, determine, determine… stationary solutions.
49:58:290Paolo Guiotto: If any.
50:02:270Paolo Guiotto: And, regions.
50:07:990Paolo Guiotto: of B, Where?
50:12:470Paolo Guiotto: solutions.
50:14:860Paolo Guiotto: R.
50:15:980Paolo Guiotto: increasing.
50:18:130Paolo Guiotto: or decreasing.
50:21:750Paolo Guiotto: Now, for stationary solutions, these are two questions in one.
50:25:460Paolo Guiotto: So, Y constantly equal to C is a solution?
50:30:970Paolo Guiotto: If and all if, here it is easy, you just plug into the equation.
50:35:730Paolo Guiotto: The equation reminds that it is Y prime equal 1 over T plus log of Y.
50:41:940Paolo Guiotto: So you plug the constant there. Of course, Y prime will always be 0, so 0 equals 1 over T plus log C.
50:52:610Paolo Guiotto: What do you think about this equation?
51:03:450Paolo Guiotto: Huh?
51:05:290Paolo Guiotto: It's not possible, because 1 over anything cannot be zero, you see?
51:10:210Paolo Guiotto: So, impossible.
51:13:600Paolo Guiotto: So, the conclusion is, simply, there are no constant solutions.
51:22:80Paolo Guiotto: Okay, it can be.
51:24:00Paolo Guiotto: Then, let's see where solutions are increasing, decreasing. So, we say that Y is increasing if and only if Y prime is greater or equal than 0. This is 1 over P plus log of Y
51:39:850Paolo Guiotto: Now, if you want that fraction be positive, you need…
51:56:480Paolo Guiotto: Why the derivative?
52:01:230Paolo Guiotto: the denominator must be positive, because the numerator is 1. So T plus log of Y must be positive.
52:09:260Paolo Guiotto: Okay, let's write this as something in Y function of t, that is.
52:14:350Paolo Guiotto: better to be seen. So, I have log of Y greater than minus T. It's similar to the equation we saw one minute ago. So, since we now apply the exponential, the exponential E is increasing. So, this is greater than this if and only if e to log of Y
52:33:670Paolo Guiotto: is greater than E2 minus T.
52:36:600Paolo Guiotto: E to log of y is y, so Y greater than E to minus 2. That's the condition. So, above that exponential solutions are increasing, and
52:49:30Paolo Guiotto: Of course, below, they will be decreasing, because the other sign will apply. So if I go back here, above means here, so when I am here, a solution will move up, okay? It means your diagram is here, you will see the solution increasing.
53:07:330Paolo Guiotto: Okay? Because whenever you are above that line, the derivative is positive. So here, here, wherever, above that line, solution is increasingly.
53:18:430Paolo Guiotto: Why? When I am below, solution is decreasing. Now, be careful, because the… I don't know why there should be some… some particular way of functioning of the brain, I suppose. Someone writes this.
53:36:650Paolo Guiotto: It's nice.
53:38:430Paolo Guiotto: But it may be confusing.
53:42:250Paolo Guiotto: Some of you will do that.
53:44:260Paolo Guiotto: Because if you take this adult in this way.
53:47:850Paolo Guiotto: Now, that could communicate well, that when you move to the left, you go down, so when you move to the right, you go up, so they are still increasing.
53:57:790Paolo Guiotto: So please be careful if you do this kind of thing.
54:01:580Paolo Guiotto: I know that, statistically, about 20-30% of people who put the arrows, yeah, they put the arrows in that way.
54:10:120Paolo Guiotto: I suppose that they have a specular weight.
54:12:860Paolo Guiotto: to work with the brain. I know these things are quite common, so it's better if you use,
54:20:610Paolo Guiotto: the arrows in this way.
54:22:710Paolo Guiotto: Because here, you understand that they are
54:26:200Paolo Guiotto: Increasing, decreasing are referred always. When the variable moves to the right, the function goes up, increasing goes down, decreasing, okay? So that's why I'm saying this.
54:38:410Paolo Guiotto: Okay, so now we have question 3. Question 3 now takes a specific Cauchy problem.
54:46:750Paolo Guiotto: It says, Y… From alpha-beta. Well, alpha is missing, there's a typo.
54:55:150Paolo Guiotto: Yeah.
54:56:360Paolo Guiotto: is the, maximal solution, name maximal solution is the solution.
55:05:900Paolo Guiotto: solution of the Cauchy problem.
55:09:190Paolo Guiotto: Well, it's written CP11.
55:12:310Paolo Guiotto: Probably there is a notation for this.
55:15:290Paolo Guiotto: Actually, it's written. That is, Y of 1 is equal to 1.
55:20:140Paolo Guiotto: But before we continue, let's see where is it this initial condition, because here we have some different behavior of the solution, depending on when the condition is, okay? So, let's see where is it.
55:34:450Paolo Guiotto: So let's do the figure, let's leave some space to finish right there.
55:39:500Paolo Guiotto: the question. So, we have nothing down here. We have the bedline of E minus P. Now, what is it? Y of 1 equals 1?
55:51:220Paolo Guiotto: So T is 1 is about here.
55:54:380Paolo Guiotto: And the Y is 1. Of course, the point is we are above or below that line.
56:01:250Paolo Guiotto: Now, this is the line… E2 minus T, so this is the value e to minus 1.
56:08:30Paolo Guiotto: So, certainly, this is 1 over E is less than 1.
56:12:300Paolo Guiotto: So the one… actually, this is one, by the way. The one is here. So our passage condition is here. So it is above… it means that we are in the region where solutions are increasing.
56:27:950Paolo Guiotto: So, even if you don't see the remaining part of the question, I already know that my solution is there. So, at least for a while, the solution will be increasing. Then, let's see what happens for the entire solution.
56:43:500Paolo Guiotto: Okay, let's continue with the question.
56:47:70Paolo Guiotto: The first question is determine monotonicity and concavity… off.
56:55:940Paolo Guiotto: the solution, why?
56:58:180Paolo Guiotto: Well, about the monotonicity, what do you think? Well, we know that we are, initially, at least, in the region where solutions are increasing.
57:10:150Paolo Guiotto: So?
57:13:150Paolo Guiotto: Do we stay all time in that region? Because if we stay all time in that region, we will be increasing.
57:21:420Paolo Guiotto: The unique possibility to change the behavior, we must leave that region and cross the dashed line.
57:29:60Paolo Guiotto: Can we do that?
57:35:40Paolo Guiotto: So, I don't know if you are following me. I'm saying.
57:39:410Paolo Guiotto: As soon as I stay in this region, above the dashed line, I am increasing.
57:46:150Paolo Guiotto: So, if I stay all time in that region, the solution will be increasing all time.
57:52:950Paolo Guiotto: So the question is, do we stay all time in that region?
57:57:880Paolo Guiotto: or not.
58:01:740Paolo Guiotto: Why not? Why the solution? You understand that this looks to be impossible, no? Something like this. Why?
58:09:30Paolo Guiotto: Because, you see, to go down, I need to descend in a region where I am increasing. But why I cannot do this? I can cross and maybe… well, that sounds difficult, because you see that there should be…
58:21:930Paolo Guiotto: Something like that, that's…
58:29:890Paolo Guiotto: Yeah, that's exactly the point. This dashed line is not solution, so I cannot say that I cannot clock this line, because otherwise I would cross the solution, so I would lose uniqueness. This is for me.
58:43:200Paolo Guiotto: No, that's not the case. But, there is another reason why I cannot cross that line. That line is not in the domain, so if I would cross the line.
58:52:860Paolo Guiotto: would be that at the moment when I cross, I am outside of the domain, and this can never happen.
58:59:810Paolo Guiotto: Your solutions are always in the domain.
59:03:160Paolo Guiotto: Never outside. Even if the outside is made of one single line, who cares?
59:08:640Paolo Guiotto: Now, let's put down this argument formally. So, we say now that we stay all time above that line, okay? So, the claim is
59:19:20Paolo Guiotto: Claim.
59:21:190Paolo Guiotto: My solution, Y,
59:23:540Paolo Guiotto: Y of T is above, so greater, than E minus T for all times t in the time interval of the solution, alpha betta.
59:35:10Paolo Guiotto: Once this is achieved, from this it will follow that solution is automatically increasing, because I stay… so from this it follows Y is increasing.
59:47:20Paolo Guiotto: Do you see the point?
59:48:790Paolo Guiotto: Now, let's see the why.
59:50:960Paolo Guiotto: And the argument is similar to the argument of the crossing we have seen in the previous exercises with the constant solution. It's similar, but different.
00:01:630Paolo Guiotto: So, if false… Of course, sir.
00:08:650Paolo Guiotto: when you write a solution, yes, you could be,
00:14:990Paolo Guiotto: Strong enough to be able to write such detailed arguments, okay?
00:20:610Paolo Guiotto: But it's difficult to write precisely, because you must be able to transform an intuition into a formal argument, which is a very important step in your process, in your learning process, but
00:35:00Paolo Guiotto: Let's say that even if you are not able to write exactly, for example, if you write by words the argument, as we said, we cannot cross the dashed line, because otherwise we would get out of the domain.
00:48:590Paolo Guiotto: it's already… let's say that the idea is there, you are not able to formalize, but the idea is correct. Okay, so it will be appreciated somehow. So, if false, it means that there exists a time, let's say T hat, where this thing is non-true.
01:07:50Paolo Guiotto: So, well, Y of T h hat
01:10:150Paolo Guiotto: It is less or equal to E minus T hat, because if that is not true always, there is a just one time where this is false, so it means that the inequality is reversed.
01:22:100Paolo Guiotto: Now, we have two cases. Case one, if Y at T hat is equal to E minus T hat.
01:33:220Paolo Guiotto: It means that our solution is here.
01:36:900Paolo Guiotto: is crossing the, the boundary, no? So, the point T hat, YT hat.
01:46:410Paolo Guiotto: is not in the domain indeed.
01:50:10Paolo Guiotto: And that's impossible, because we say that solutions are… you remind the definition of solution, we have given a
01:58:190Paolo Guiotto: Don't remember, couple of times ago? One time? Couple of times…
02:07:310Paolo Guiotto: No, probably yeah.
02:14:520Paolo Guiotto: The basic, basic definition of solution. Yeah. You see?
02:19:290Paolo Guiotto: that path in yellow. The solution must always be in the domain of F, otherwise F is not defined, no?
02:28:400Paolo Guiotto: So… And this is impossible.
02:36:520Paolo Guiotto: But, what if,
02:39:260Paolo Guiotto: we have the strict inequality. If Y at T hat were strictly less than e to minus T hat, then what is the case? The case is when I am now here.
02:51:680Paolo Guiotto: I'm saying there is a time, T hat, where my solution is now below the line.
02:59:520Paolo Guiotto: Well, by continuity, if you were at time t equals 1R above, at some other time R, below, by continuity, somewhere in the middle, you must be equal. That's the intermediate values theorem, the same, always the same story. So there should be, by continuity.
03:19:190Paolo Guiotto: There should be another time, T double hat, such that they equal false.
03:25:680Paolo Guiotto: Y, at time TXX is exactly on the line.
03:31:380Paolo Guiotto: And therefore, we get… we go back to this case, to another contradiction. So it means that
03:37:710Paolo Guiotto: If the conclusion is false, whatever is the case, we get the contradiction, so it means that the conclusion is true.
03:44:10Paolo Guiotto: So the claim is true.
03:46:650Paolo Guiotto: So, Y of T is always greater than E minus T for every T, and this means that Y is increased.
03:56:160Paolo Guiotto: So we determined the monotonicity. Concavity.
04:03:460Paolo Guiotto: For this, we need to go to the second derivative, no? So Y is pointing upward if and only if Y's second is positive.
04:13:310Paolo Guiotto: And pointing downward, if and only if Y7 is next.
04:18:340Paolo Guiotto: So let's see about Y second.
04:21:180Paolo Guiotto: We have the equation y prime equals 1 over t plus log of Y.
04:27:940Paolo Guiotto: So now we differentiate this equation, and we get away second.
04:32:450Paolo Guiotto: Why second is, be careful, because this is still… when I say prime, I always means I do not… that there is a reason I do not write
04:43:80Paolo Guiotto: Because I think this is the derivative with respect to T, which is the variable of 1, okay? So, it is the derivative with respect to t of this.
04:52:140Paolo Guiotto: So…
04:53:50Paolo Guiotto: to avoid the confusion, and to forget something, and to do the wrong derivative, write slowly these things. So we now have to derive this thing with respect to t, reminding that here there is Y of T.
05:05:730Paolo Guiotto: Okay? So this is a fraction. I use the formula that 1 over G, the derivative, is minus G prime over G squared.
05:16:210Paolo Guiotto: So it is minus the square of denominator, T plus log of Y,
05:21:770Paolo Guiotto: square. Then I have the derivative of this thing with respect to T. What is it? Which is not the derivative I computed above,
05:31:810Paolo Guiotto: You see, it is not this one. You don't have to confuse with this one, because this is the… it starts to Y.
05:38:900Paolo Guiotto: Okay, it's another place.
05:41:420Paolo Guiotto: This is the derivative with respect to t, so we have to do the derivative with respect to t of that argument. What is it?
05:48:920Paolo Guiotto: 1… Now it comes to difficulty.
05:54:650Paolo Guiotto: Why Brahm over white, that's right.
05:57:50Paolo Guiotto: So, because the derivative of log gives 1 over Y, then you have the derivative of the argument of log, which is not 1, it's Y prime of t. Okay, we have this.
06:08:290Paolo Guiotto: Now, since we don't know who is Y, and we don't know who is Y prime, well, actually, we could use the equation for Y prime, and…
06:15:890Paolo Guiotto: do something, but let's see if we can… the scope here is to determine the sine of this, okay? We don't have to determine values of this, but sine.
06:26:220Paolo Guiotto: So, let's see if we can say anything about the sign from this expression. What do we see here? First of all, we have a square, so this one is
06:36:960Paolo Guiotto: is positive.
06:39:290Paolo Guiotto: Then we have a minus in front, and then we have a numerator, 1 plus this thing. 1 is positive. What about this fraction?
06:50:100Paolo Guiotto: What can be said about, we discussed what is Y prime, and now, one minute ago, we discovered that Y is increasing, so Y' is positive. So this number here is a plus. What about Y?
07:04:490Paolo Guiotto: But why?
07:11:980Paolo Guiotto: Because we are in the domain, and in the domain, we are always this golden rule. In the… we are in the domain. What do we know about the domain? Domain is made of Y, which are positive.
07:24:110Paolo Guiotto: Cannot be negative.
07:25:880Paolo Guiotto: Okay, so we know that this is positive, so positive and positive is positive.
07:31:680Paolo Guiotto: Numerator is 1 plus positive, it is positive, so we can put together all these things. Minus positive over positive, the fraction is positive, with the minus in front is negative.
07:43:690Paolo Guiotto: So we can say that this is less than zero.
07:46:750Paolo Guiotto: You see, we don't know what is Y explicitly, we don't need to compute anything, we just, we just deduce this information from the equation and from what we have seen about the solution.
08:00:90Paolo Guiotto: So this means that the solution has second derivative negative, and this means that the concavity is pointing downward, so it's concave.
08:11:990Paolo Guiotto: Okay, so now we can do… A new figure, yeah?
08:18:229Paolo Guiotto: So, again, we discard this part. This is the E minus T exponential. This is plane TY.
08:26:310Paolo Guiotto: This is the initial condition, about here. The solution is increasing and concave, so something like this.
08:37:470Paolo Guiotto: So, let's say, roughly, like a logar, no?
08:42:29Paolo Guiotto: the shape of the log, more or less. It's not the log, and it does not go down, but to have an idea, it's a functional thing.
08:48:800Paolo Guiotto: Okay, this, is a question…
08:53:840Paolo Guiotto: 3, now there is a question 4 that concerns alpha and beta. So, question 4… What about Alpha…
09:03:760Paolo Guiotto: What about pita?
09:05:689Paolo Guiotto: So, this question is not determining the exact values, but are they defined or not, okay?
09:14:350Paolo Guiotto: And, It asks also, what about the limit when t goes to alpha?
09:22:340Paolo Guiotto: Not of the solution, but of the derivative, Y prime T.
09:26:800Paolo Guiotto: And what about the limit when T goes to beta?
09:32:359Paolo Guiotto: of white tea.
09:36:310Paolo Guiotto: And then we will have the last question to the plot, but the plot, actually, the final question comes after this.
09:44:649Paolo Guiotto: So, Namda, let's discuss first about the alpha.
09:48:800Paolo Guiotto: What do you think about Alpha?
09:51:350Paolo Guiotto: Is it finite? Well, the left time can be finite, or… minus infinity.
10:00:820Paolo Guiotto: What do you think?
10:03:400Paolo Guiotto: It is funny, because this solution is going to crash Yeah.
10:09:00Paolo Guiotto: So, very soon, if this is time 1, you can even say that if this is time 0, the solution will crash for a time alpha between 0 and 1.
10:21:870Paolo Guiotto: If you want to see formally, I can say that the claim is a claim…
10:30:850Paolo Guiotto: Alpha is between 0 and 1.
10:36:00Paolo Guiotto: Well, this is trivial, okay?
10:40:860Paolo Guiotto: Because, The solution is defined at time 1,
10:45:720Paolo Guiotto: it's the initial condition, the alpha must be less than 1, so that's not the point.
10:50:820Paolo Guiotto: Well…
10:53:210Paolo Guiotto: Y alpha must be finite, and in particular, positive. Because, as you can see in the figure, the solution here is going down when we move to the left.
11:05:680Paolo Guiotto: And why I say that exactly is a positive, because if you look, this is not, let's say, is not,
11:12:820Paolo Guiotto: incidental, because that's exactly the value where this liner, the dashed line, crosses the y-axis. So what I can say is that, since we know that the solution is always above the exponential.
11:32:220Paolo Guiotto: Okay, so when we send,
11:36:290Paolo Guiotto: When we send T2 alpha, this inequality remains, so we get Y alpha must be… well, when you do the limit, normally the sign becomes weaker. You cannot be sure that it is greater, no?
11:50:930Paolo Guiotto: the function, which is positive, but the limit is zero, okay? You can say that this is greater or equal than e to minus alpha.
11:58:30Paolo Guiotto: But, we know that the function Y is increasing, so the y of alpha, which is this value here.
12:07:740Paolo Guiotto: must be less than Y of 1, because the function, when we move to the left, is descending.
12:14:360Paolo Guiotto: And since this is increasing, we get that Y of alpha is less than Y at 1, which is the initial condition, is the unique exact value we know about the solution, is 1.
12:27:970Paolo Guiotto: So we get that e to minus alpha is less than 1. And when this happens.
12:35:450Paolo Guiotto: Because this must be less than this, which is less than this. 1E2 alpha is less than… e to minus alpha is less than 1.
12:46:60Paolo Guiotto: When minus alpha is negative, that means alpha is positive.
12:51:810Paolo Guiotto: Okay.
12:53:80Paolo Guiotto: Okay, let's work still on alpha, and let's take this limit. What about this limit?
13:02:390Paolo Guiotto: So, limit when T goes to alpha.
13:07:190Paolo Guiotto: of the su- not the solution, but the derivative, Y prime.
13:12:430Paolo Guiotto: Well, of course, we will have to do the limit using the equation. Limit when t goes to alpha over… it is 1 over t plus log of Y.
13:24:680Paolo Guiotto: Now, what happens when t goes to alpha?
13:27:890Paolo Guiotto: Well, as the figure suggests, this solution should crash on the dashed line. Well, here there is a little important point. Why the solution crashes on the dashed line and, for example.
13:42:990Paolo Guiotto: why the solution does not do this. So, for example, this is the dashed line.
13:47:730Paolo Guiotto: This is the initial condition why the solution does not end there.
13:52:970Paolo Guiotto: Away from the dashed line.
14:02:860Paolo Guiotto: Yeah, yeah, but the argument we are trying to learn to use is the argument of the compact set, because if this happens, if the solution ends above the black line.
14:17:220Paolo Guiotto: So I would have a box, like this one, that contains the solution for all times in the past.
14:24:720Paolo Guiotto: So let's see precisely the argument. So, the claim is that, claim,
14:31:880Paolo Guiotto: is that I want to say that the solution crashes on this line.
14:37:620Paolo Guiotto: So, how do I write this? That the time when the solution dies, time alpha.
14:43:880Paolo Guiotto: This point, the final point, alpha Y of alpha, must be here.
14:51:560Paolo Guiotto: So, it must be that Y of alpha is equal to E2 minus alpha.
14:59:400Paolo Guiotto: If not.
15:04:570Paolo Guiotto: Since we know that the point is always above, hmm, So… if Y of alpha
15:16:90Paolo Guiotto: where greater than e to minus alpha, so I'm saying the solution dies exactly as in the figure, so let me cancel out all these things.
15:27:780Paolo Guiotto: And let's emphasize this. You see, the solution dies a little bit away from the, the, the line, while, Y equal E2 minus T.
15:41:570Paolo Guiotto: So this is the situation. This is alpha, this is Y to alpha, and as you can see from this figure, Y to alph… Y of alpha here is greater than the value here, which is E2 minus alpha.
15:56:430Paolo Guiotto: Okay?
15:58:00Paolo Guiotto: then I… I can take this box, K, What is this?
16:03:830Paolo Guiotto: Well, it is from alpha to 1.
16:08:900Paolo Guiotto: process this, the minimal quote is Y alpha.
16:14:830Paolo Guiotto: And the maximum is the quartered one, which is, by the way, Why?
16:20:160Paolo Guiotto: Now, this is compact, closed and bounded.
16:24:770Paolo Guiotto: Compacta.
16:26:700Paolo Guiotto: In this case, it's less trivial than the previous one, because in the previous cases, we had domain is R2.
16:33:930Paolo Guiotto: So the box is always contained in the domain. Yeah, you have to be careful, because the domain is not a tool, but because of this condition, we are not touching the dashed line, so we are definitely in the domain D.
16:47:860Paolo Guiotto: And the solution Y… does not… leave… K… in the past.
17:02:270Paolo Guiotto: You see, because we include all times until the…
17:06:560Paolo Guiotto: the initial time of life of the solution, which is alpha.
17:11:750Paolo Guiotto: So, this cannot be… this is impossible.
17:17:720Paolo Guiotto: So, it means that the solution must die.
17:22:240Paolo Guiotto: On the dashed line, huh?
17:24:980Paolo Guiotto: And this is important because it means that if you take the log, you have that log of Y alpha is equal to minus alpha, or if you want, alpha plus log of Y alpha must be equal to 0.
17:42:720Paolo Guiotto: So, going back to the problem, when we do the limiter for C going to alpha, that denominator goes to zero.
17:49:980Paolo Guiotto: So we have 1 over 0.
17:52:770Paolo Guiotto: It should be infinite.
17:55:670Paolo Guiotto: A little… a little point, because you cannot say that 1 over 0 goes to infinity.
18:03:360Paolo Guiotto: You need to… you need the one more information about that zero.
18:08:920Paolo Guiotto: You need to know if it is a 0 plus or a 0 minus, because if you are going to 0 plus, denominator goes to 0, positive, the fraction is positive, and the limit is plus infinity.
18:20:220Paolo Guiotto: If this is 0 minus, you have minus infinity. If the sign is not determined, you don't have the limit, okay? So you have to be careful. But here we know that it is 0+, because that quantity is positive. We come from the region
18:33:620Paolo Guiotto: Where the solution is above, so this is a positive quantity going to zero, so it's a zero plus, and this means that the limit is equal to plus infinity.
18:45:500Paolo Guiotto: So, now we know that the solution Crashes on the exponential.
18:53:90Paolo Guiotto: Lioner?
18:55:280Paolo Guiotto: And it crashes within finite, positive infinite delivered, so it means that it comes like that.
19:02:860Paolo Guiotto: Okay?
19:04:500Paolo Guiotto: This is a function with vertical tangent at that point, so it does not come like that, to be clear, okay? It goes at that point with the positive infinite slope.
19:16:160Paolo Guiotto: like the root of… you have… do you have in mind the plot of root of X? Root of X goes to zero with the… with the steep positive and infinite.
19:25:380Paolo Guiotto: So it is like that.
19:27:00Paolo Guiotto: Whoa.
19:29:400Paolo Guiotto: And so this solves, basically, the first half of the question.
19:33:870Paolo Guiotto: Then what about beta?
19:38:350Paolo Guiotto: Let's now discuss about beta.
19:41:870Paolo Guiotto: Beta is finite or infinite.
19:50:130Paolo Guiotto: Well, let's think about… How can we continue? Imagine that beta is fine, like here.
19:57:560Paolo Guiotto: So, what should happen?
19:59:400Paolo Guiotto: to the solution.
20:04:100Paolo Guiotto: The solution does not survive after time beta.
20:08:120Paolo Guiotto: Remind that we know something about the solution. The solution is increasing and
20:15:340Paolo Guiotto: Concave, that's very important, because if it is concave.
20:19:600Paolo Guiotto: It does something like this. It cannot do something like that.
20:24:520Paolo Guiotto: So, you see…
20:27:30Paolo Guiotto: And so this would say that if beta is finite and the solution is concave, we should die into the domain, and that's not possible, okay? So this would say beta equal plus infinity.
20:38:740Paolo Guiotto: Okay, so the claim is, claim… Is that… Beta equals plus infinity.
20:48:360Paolo Guiotto: If beta were less than plus infinity, then…
20:54:520Paolo Guiotto: Well, here we could use an interesting argument that comes from the concavity. That is the following.
21:02:910Paolo Guiotto: So now the function is concave. I want to say that this is the figure, so the solution dies inside the domain, which is forbidden.
21:12:750Paolo Guiotto: So, what can I say? I compute the limit of the solution when I go to beta, so let's call L be the limit when T goes to beta of the solution.
21:24:120Paolo Guiotto: And of course, this argument will work, as the figure suggests, if L is finite, because if L is infinite.
21:31:490Paolo Guiotto: it won't work, but this can be the unique case. So, I claim… that L is fine.
21:41:690Paolo Guiotto: Why? Well, for example, here we can use the concavity to tell this, because a particular property of concavity, by concavity.
21:52:970Paolo Guiotto: we have this general fact that the solution Y is geometically below all of its tangents, is below all…
22:06:860Paolo Guiotto: of… It's tangent.
22:10:610Paolo Guiotto: That's a dramatic consequence of the concavity. So, in particular, I can write Y of T is less that I write the equation of the tangent at the unique point I know for this solution, t equal 1.
22:26:630Paolo Guiotto: In general, the tangent is Y of 1 plus Y prime of 1 times T minus 1, right?
22:35:380Paolo Guiotto: So if you want, we can plug numbers here, but it is not important, because this says that when you send the right-hand side to this beta unknown.
22:47:60Paolo Guiotto: This will go to the value Y1 plus Y prime 1 times beta minus 1.
22:53:770Paolo Guiotto: Which is supposed, since beta here is final, this is final.
22:59:160Paolo Guiotto: And what about Y of T? But Y is below.
23:02:350Paolo Guiotto: So you cannot go to plus infinity. So from this, it follows that the limit of Y of T is finite as well. So, you see, I claim that L is plus infinity, and this is the justification. Then, we can use the argument of the compact set. Here, I can take this compact.
23:22:620Paolo Guiotto: That, traps the solution all times in the future.
23:27:10Paolo Guiotto: So what is this, K? K is… this is time 1.
23:31:850Paolo Guiotto: So, from time 1 to time beta, we are assuming that beta is finite, so that's a closed bounded interval, and then for the Ys, this is Y equal 1 to Y equal L. We proved that L is finite, so this is now a closed and bounded, so compact
23:51:150Paolo Guiotto: Compact content into the domain.
23:54:500Paolo Guiotto: and the solution.
23:56:610Paolo Guiotto: solution.
23:58:70Paolo Guiotto: Why? Never.
24:01:220Paolo Guiotto: leaves.
24:04:360Paolo Guiotto: this K in the… Future. And we got a contradiction.
24:10:980Paolo Guiotto: Impossible.
24:13:30Paolo Guiotto: So… Where is the error? The error is the initial assumption.
24:18:150Paolo Guiotto: If beta is less than plus infinity, it follows this. So, the conclusion is that beta must be
24:26:590Paolo Guiotto: Equal to plus infinity.
24:30:580Paolo Guiotto: Okay?
24:34:500Paolo Guiotto: So beta is plus infinity, well, let's leave the limit of the solution, which is complicated, and also time is over.
24:43:20Paolo Guiotto: So, let's finish with the… the plot of this solution. So, what are the informations,
24:51:20Paolo Guiotto: In any case, whatever the information you get from this analysis, you can always plot a graph with the information you know. At this stage, we know that the solution at time 1 equals 1,
25:07:610Paolo Guiotto: Passes to this point, it is increasing, it is convex, it survives until time plus infinity, and we can say that the solution will be likely something like that.
25:19:500Paolo Guiotto: Okay, so this is the plot of Y.
25:23:650Paolo Guiotto: Okay, guys, let's stop here.
25:27:630Paolo Guiotto: Let… let me just give you a couple of,
25:32:680Paolo Guiotto: Let me say a couple of things about, before we stop this recording, Has it stopped? Not yet.