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00:00:00Paolo Guiotto: Okay, good morning, nice to see you again.
00:12:330Paolo Guiotto: So, we have to give an end to this course, and,
00:18:760Paolo Guiotto: We have to do, just,
00:22:720Paolo Guiotto: at least a bit of the chapter on differential equation. I think we do just the first three sections.
00:32:240Paolo Guiotto: Which are mostly made of example, apart for, a type of problem we…
00:38:400Paolo Guiotto: I hope we are able… we will be able to study.
00:42:00Paolo Guiotto: Soap.
00:43:80Paolo Guiotto: The topic is differential equation, a topic which is not new to you. I suppose that you have seen some differential equation in first-year calculus, isn't it? So, I suppose that everyone knows what is a differential equation.
00:59:140Paolo Guiotto: However, a differential equation
01:06:110Paolo Guiotto: Is, first of all, an equation.
01:12:180Paolo Guiotto: So, this means, identity.
01:18:900Paolo Guiotto: Weed that… Let's say N.
01:23:330Paolo Guiotto: Unknown.
01:28:970Paolo Guiotto: And, This equation, and that's, let's explain this part of the name, huh?
01:37:660Paolo Guiotto: It's called differential equation because the unknown
01:46:630Paolo Guiotto: is a function.
01:49:990Paolo Guiotto: for this type of equations, we consider the function. We normally use the letter Y for the unknown. It's a function, and we use the letter T for the variable, because most, but not necessarily
02:04:410Paolo Guiotto: the unique applications of these equations are to problems where the functions are a function depending on time, so that's why we use letter T, but it's not necessarily
02:16:710Paolo Guiotto: They… the… it's not a restrictive.
02:21:390Paolo Guiotto: condition, this one, can be whatever the variable of the function.
02:26:580Paolo Guiotto: that… appears…
02:35:620Paolo Guiotto: in the equation.
02:44:90Paolo Guiotto: We don't… some… off.
02:50:830Paolo Guiotto: It's derivatives.
02:58:730Paolo Guiotto: So the general, let's say, the general form is something like this. It's an identity that we can always write.
03:06:660Paolo Guiotto: something equals zero, we can write everything in one side. What is in the left side is an expression that contains the function and some of the derivatives. So, in general, it's something like a relation between the quantity Y of T, which is the function, the derivative of Y, the second derivative, etc, up to a certain
03:30:120Paolo Guiotto: maximum order derivative, let's say YN of t, and possibly even of T explicitly. So this is the general shape of an equation of this type. So, for example, we can write, I don't know, Y of T
03:46:400Paolo Guiotto: plus Y prime of t equals 0. This is a linear… First up… or the…
03:55:490Paolo Guiotto: equation that you started last year, I suppose.
03:59:790Paolo Guiotto: For example, Y second T minus 3TYT, this is a linear second order equation.
04:08:790Paolo Guiotto: Linea, second… Order.
04:12:680Paolo Guiotto: equation.
04:15:30Paolo Guiotto: Now, this order means the maximum order of the derivative that appears in the equation. This contains the Y second, which is the second derivative.
04:26:380Paolo Guiotto: And there is not the third derivative, there is no fourth derivative, and so on. So this is called the second order. If you want a fourth order, I could write Y differentiated four times with respect to T plus 3Y prime T,
04:43:530Paolo Guiotto: plus, for minus… I'm just writing at random minus Y second t…
04:51:540Paolo Guiotto: equals 0. This is an equation of fourth order, okay?
04:57:650Paolo Guiotto: These equations are linear because the dependence on Y, Y prime, etc. is true first-degree polynomials, but we could have also impressions like this, Y prime T plus Y squared, YT square.
05:16:290Paolo Guiotto: Plus, sine T… YT inside the argument of sine.
05:23:520Paolo Guiotto: This is a nonlinear equation.
05:26:110Paolo Guiotto: Okay? Because it depends on Y, Y prime, etc. You see the quantity Y squared is not linear in Y, and also this thing is not linear in Y. It's not a linear function, it's not a degree…
05:40:160Paolo Guiotto: a polynomial of degree 1 in the quantity Y. Normally, the real world where we live is nonlinear, so it's driven by this kind of equations, and normally also nonlinear means harder.
05:53:930Paolo Guiotto: Difficult to treat.
05:55:570Paolo Guiotto: For example, you have seen that, for some of these equations, probably
06:03:40Paolo Guiotto: For this type of equations, you have a formula to write the solution. So, you can write this, like, second-degree equation, you have the formula for the roots, no?
06:12:920Paolo Guiotto: X equal,
06:14:940Paolo Guiotto: what is minus B plus minus root of delta divided 2A, now, for the equation AX squared plus BX plus C equals zero, no? You have a formula.
06:25:270Paolo Guiotto: That's the best thing you can have, no? There is basically nothing better than explicit formula, because you can compute, you can compute by a computer exactly, you can do by hand, you can have the maximum information from a formula.
06:43:420Paolo Guiotto: But not always it's possible to solve the equations explicitly. For example, when the equations are nonlinear, this is difficult, or impossible.
06:55:40Paolo Guiotto: And if these equations represent some physical, or economical, or whatever biological phenomenon, what we want to know is to have an idea about the solution. So…
07:08:280Paolo Guiotto: If we know the solution explicitly, we have the formula, we can look at the formula, we can draw the graph, we can give to a computer the function, and we have a plot of the solution, we can see the solution.
07:21:950Paolo Guiotto: Now, but in general, for an equation of this type, there is not an explicit formula, and the problem how… what kind of things can be said about solutions of that equation becomes a non-trivial problem.
07:36:940Paolo Guiotto: Okay, so let's say that for our purposes, we will consider a very particular sub-case of this.
07:45:130Paolo Guiotto: First of all, we consider an equation of order 1, so where at most there is the first derivative.
07:53:860Paolo Guiotto: And second, so we have to see that in this formula, you cover everything that contains Y second, Y…
08:01:740Paolo Guiotto: these derivatives are not present. And second, we also consider an equation where the highest order derivative, which is Y prime, is expressed explicitly in terms of all the other quantities. So, here we consider
08:19:990Paolo Guiotto: Yeah.
08:22:690Paolo Guiotto: We… We'll… consider… equations… of type, So this is the shape.
08:38:260Paolo Guiotto: Y prime T?
08:40:440Paolo Guiotto: equal, we will write preferable in this way, equal a function f that depends on T and on Y of T.
08:49:570Paolo Guiotto: These are older… Fast.
08:55:810Paolo Guiotto: Order.
08:57:800Paolo Guiotto: equations.
09:01:70Paolo Guiotto: in… normal, form.
09:09:890Paolo Guiotto: So where normal form is just the form where you have the highest order derivative explicit in terms of all the other quantities. So this is the type of equation. For example, so we consider these equations.
09:25:470Paolo Guiotto: example, you already considered these particular cases of this. For example, if you take this equation, Y prime of t equals y of t, this is, again, a first-order linear equation. Easy, you can solve.
09:39:290Paolo Guiotto: Y prime of t equal Y of T
09:43:610Paolo Guiotto: square, where the square is… the square of Y,
09:47:650Paolo Guiotto: This is nonlinear, but it is solvable. It's a separable variable equation. You have… do you know that?
09:56:550Paolo Guiotto: Okay?
09:57:820Paolo Guiotto: First… or the… Sit about a boat.
10:04:280Paolo Guiotto: invaluable.
10:06:110Paolo Guiotto: equation.
10:08:20Paolo Guiotto: Here we have a method to… I will refresh, because sometimes we will use… we have a method to…
10:16:880Paolo Guiotto: Compute exactly the solution.
10:20:10Paolo Guiotto: That's not always possible, because, for example, if I write this, which seems to be just a slight modification of this one, Y of T,
10:32:670Paolo Guiotto: plus T squared.
10:34:930Paolo Guiotto: For this one, this is not a separable variable, and we don't know exactly how to solve this equation, even if it seems
10:42:950Paolo Guiotto: Little variation of the previous case.
10:47:230Paolo Guiotto: But it's still in this form. You see, Y prime t equals something that depends on y of t and possibly on T also, okay?
10:57:20Paolo Guiotto: So if you want to see the functions here, the function f, which is
11:01:790Paolo Guiotto: Going to be the key ingredient in this equation.
11:05:340Paolo Guiotto: all the story, basically, will depend on how that F is made. So if you want to see this function F here, F of T, Y of T, now this F will be a function of two variables, T and Y, two numbers, number T and number Y.
11:24:230Paolo Guiotto: And in this case, the function is just FTY equal Y, because when you plug Y equal Y of T, you get that thing, no? For example, here, if I want to see this as F of P, Y of T,
11:39:830Paolo Guiotto: This means that my F depends on TY as Y squared.
11:46:190Paolo Guiotto: And, for example, here, this is F of T, Y of T,
11:51:840Paolo Guiotto: DF is… you know what is it? You see?
12:07:330Paolo Guiotto: Oh, it's January… what is 7, not January 1st, not December 21st. So, are you awake, or no?
12:17:310Paolo Guiotto: So what is the function FPY? Such that way that when you put T and Y of T in the argument, you get Y of T plus T squared. It will be Y plus T squared, because when you plug here y of t, you get that.
12:33:460Paolo Guiotto: So you have to be careful, because here we use the letter Y with a little bit of ambiguity, but there is no ambiguity, in fact.
12:41:600Paolo Guiotto: For two different things.
12:43:490Paolo Guiotto: Y of T is a function of T, is the solution of the differential equation, so it is a function. The letter Y, simple, denotes just the numerical value, Y, no? So F of TY, F is a function of two variables, T and Y.
13:01:370Paolo Guiotto: So it's a classical R22R function, the kind of function we introduced at the beginning of this course, and in this case, it is Y plus T squared.
13:12:920Paolo Guiotto: Okay? So, a solution of the equation is a function Y of T, that when you plug into that identity, you get, the identity is verified.
13:21:960Paolo Guiotto: Okay?
13:24:150Paolo Guiotto: So, let's give, first of all, this definition of solution. We have to be a little bit precise, because the function F, so, the function f…
13:40:970Paolo Guiotto: F as function of TY will be defined then on a domain D in general, which is contained into R2,
13:53:920Paolo Guiotto: And it is real valued.
13:57:70Paolo Guiotto: So, we can… it is useful to have,
14:00:790Paolo Guiotto: an idea of this, this is the Cartesian plane. Since the points are not with these two different letters, here we use T and Y to keep in mind that the horizontal axis is the axis of T, which is also
14:16:120Paolo Guiotto: the domain for the variable of the solution. The solution is a function of P, and this is the y-axis. So the domain of F will be a subset of this Cartesian plane, so here it is where F is defined.
14:33:760Paolo Guiotto: Now, this set here is also called the domain of the equation, because… let's introduce the concept of solution definition.
14:43:730Paolo Guiotto: So, a function… Y equal function of T.
14:52:150Paolo Guiotto: that will be defined. This is a function of real variable, there is one variable, T,
14:58:120Paolo Guiotto: Normally, we consider this as defined on an interval I.
15:03:920Paolo Guiotto: So, I contained in our interval.
15:11:340Paolo Guiotto: with values in R, so is… a solution
15:20:110Paolo Guiotto: of the equation, Y prime t equal FTYT.
15:27:860Paolo Guiotto: on the interval I, If… we have to specify something here. So, basically, if whatever is written makes sense.
15:41:670Paolo Guiotto: And what… what this means, first.
15:44:540Paolo Guiotto: Well, you see at left, there is a derivative. So, this function cannot be any function, but must have a derivative. If Y is differentiable.
15:56:790Paolo Guiotto: on the interval I, and this is one variable function, so the derivative is the standard derivative, you know, since the first-year calculus.
16:06:870Paolo Guiotto: Number 2.
16:10:250Paolo Guiotto: You need also that the right-hand side makes sense.
16:13:940Paolo Guiotto: And so this means that the point TY of T is a point in the Cartesian plane, belongs to the domain where F is defined.
16:23:990Paolo Guiotto: So, the point TYT…
16:27:780Paolo Guiotto: belongs to the domain of definition of the function f, which is this set D, for every time t in the interval i.
16:36:790Paolo Guiotto: So this is a geometrical interpretation, because it means that if B is the domain of that function.
16:43:220Paolo Guiotto: And you look and say that the solution is defined on this interval, i.
16:49:480Paolo Guiotto: and you take a time t in that interval, the point T, weight, is a point in the Cartesian plane. Where is it, this point? It must be inside the domain, so this is the point TYT.
17:04:710Paolo Guiotto: And this must be for all times when this solution is defined.
17:09:819Paolo Guiotto: Now, since when you variety this point, TYT, describes a line, which is the graph of the solution, so you will have something like this.
17:21:319Paolo Guiotto: So this is the graph of the solution.
17:28:10Paolo Guiotto: So this condition number 2 means exactly that the graph of the solution belongs to the domain of the function f, cannot be outside. So if that is the domain.
17:40:250Paolo Guiotto: So this means, the graph… of why… is… contained.
17:55:90Paolo Guiotto: in the… So you cannot see this, for example. This is the plane TY.
18:02:960Paolo Guiotto: This is the domain D.
18:05:770Paolo Guiotto: You cannot see, this is the solution.
18:09:690Paolo Guiotto: this… And not… B… a solution.
18:17:990Paolo Guiotto: Okay? There cannot be solutions outside of the domain, because what does it mean, F evaluated on this point? Nothing, because if that is the domain, there is no F there, okay? So this is, not a load, this figure.
18:36:90Paolo Guiotto: So…
18:37:330Paolo Guiotto: Keep in mind this, because you see, generally, this F will have a domain that is not necessarily R2. We will see examples easily that this domain is not the full Cartesian plane R2.
18:52:570Paolo Guiotto: So it's a subset. Solutions can be contained only in that subset, and they cannot be outside of the subset.
19:00:710Paolo Guiotto: So, we said, Y is differentiable? The graph of the solution is contained in the domain, and of course, the identity holds. So, number 3…
19:10:990Paolo Guiotto: Y prime of T.
19:14:650Paolo Guiotto: must be always equal to what you get by computing F at point TYT for every T in capital I.
19:23:90Paolo Guiotto: So this is what formally is a solution of a differential equation.
19:28:890Paolo Guiotto: So, in particular, example, so let's say that you have the equation y prime t equal, I don't know, log of YT
19:40:850Paolo Guiotto: Lastly.
19:42:310Paolo Guiotto: What is the domain of this?
19:44:660Paolo Guiotto: determine… domain… Off… East… equation.
19:57:520Paolo Guiotto: So, here…
20:01:830Paolo Guiotto: the function F, we have first of all to identify what is the function F. Now remember that this that you see at the right is F evaluated at TYT.
20:15:290Paolo Guiotto: Okay? Now, then, to make easier, this we will do very soon,
20:22:520Paolo Guiotto: an agreement on how we write the equation. However, look at that formula, and you see that FTY is log of Y plus T.
20:35:10Paolo Guiotto: Right?
20:36:590Paolo Guiotto: So what is the domain of this? Domain D is the set of points, TY in R2,
20:44:410Paolo Guiotto: such that the log of Y plus T makes sense. Log is log, so it asks that the argument Y plus T be greater than zero. So we get this condition.
20:55:970Paolo Guiotto: Now, we have to represent this set in the Cartesian plane, so we have to draw this, if possible, in the plane TY.
21:05:130Paolo Guiotto: How can we see? Now, T is the abscessa and Y is the ordinate here, so I could say that this is equivalent to Y greater than minus T.
21:15:00Paolo Guiotto: Y equal to minus T is this straight line that I just dashed because it is not in the domain, because y must be greater, strictly greater than minus T.
21:26:830Paolo Guiotto: So it means that the Y must be above that red dashed line, so it means all this half plane is the domain.
21:36:130Paolo Guiotto: For this equation.
21:38:320Paolo Guiotto: So it means that solutions for this equation, if any, will be in that red region. You cannot see a solution down here. This is not possible.
21:49:500Paolo Guiotto: Okay.
21:52:520Paolo Guiotto: Okay, so that's it for this, on.
21:56:420Paolo Guiotto: So, let's do this, agreement.
22:00:180Paolo Guiotto: So, normally, if you take any bulk, you won't see differential equations written in this way.
22:12:280Paolo Guiotto: But you will just see the equations.
22:18:370Paolo Guiotto: Y' prime… T equal F.
22:22:690Paolo Guiotto: YT.
22:25:330Paolo Guiotto: is… Usually, written.
22:35:30Paolo Guiotto: S… Instead of making explicit the letter T as the argument of the unknown.
22:42:970Paolo Guiotto: Everybody, we know that,
22:46:970Paolo Guiotto: Y is a function in this game. So when we talk about the differential equation, we know that Y is a solution of the equation, it's the unknown of the equation called as you like, but it is a function. So, we can write that equation in this way, Y prime equal F of TY, suppressing the dependence, the variable t from Y.
23:08:920Paolo Guiotto: Of course, we must be aware that Y is Y of T, it's not Y, the number Y.
23:15:580Paolo Guiotto: But this makes easier, for example, to do this kind of jobs, because if you write the equation y prime equals log of y plus t, you immediately see what is F, no? You have an F in phase.
23:28:240Paolo Guiotto: Okay? So…
23:30:590Paolo Guiotto: where we remind that Y stands for Y of T. So Y is a function, it's not a variable. So Y prime means Y prime of T.
23:44:680Paolo Guiotto: Okay, is the derivative of that Y, despite the Bible.
23:50:400Paolo Guiotto: the viability. So we will, of course, use also here this agreement, and therefore we continue with that, but
23:58:200Paolo Guiotto: For the definition, we must keep clear that
24:01:960Paolo Guiotto: A solution is a function that verifies the equation so everything makes sense. It is differentiable, it is contained in the domain of definition of F, otherwise you cannot compute F of PYT, and of course, it verifies the identity that gives the equation.
24:20:740Paolo Guiotto: Okay,
24:23:170Paolo Guiotto: So, we know also this is not a new phenomenon. Let's just quickly refresh from this equation. I have the equation y prime equals Y. It's a very simple equation.
24:36:10Paolo Guiotto: that you can solve. You don't need to remind formulas, because you can… this is a separable variable equation also. We can see that Y constantly equal to 0 is a solution.
24:54:130Paolo Guiotto: This is a particular solution of the equation. In fact, if you plug this into the equation, the derivative of 0 is 0, so you get 0. It is verified.
25:04:660Paolo Guiotto: And if Y is different from 0, And it is a solution.
25:11:440Paolo Guiotto: Well, you can rewrite the equation, this is the method of separation of variables, in this way, you can divide by the Y at the right-hand side, you get Y prime divided y equal 1.
25:23:400Paolo Guiotto: And the idea is that you can recognize that the left of this identity a derivative. That's the derivative of log of modulus of Y. If you do the derivative of log of modulus of Y,
25:38:100Paolo Guiotto: Be careful, because this is not the derivative with respect to Y, but it is the derivative with respect to T, okay? So this is log of modulus yt that you are differentiating with respect to t.
25:53:180Paolo Guiotto: So what is the derivative of this? Formally, it is the derivative of the log, which is 1 over the argument, so one over the modulus of YT,
26:02:830Paolo Guiotto: times the derivative of modulus of YT, which is not immediate, because it's a composition. There is Y of T followed by the modulus. So you have to do the derivative of the modulus. What is the derivative of the modulus?
26:19:840Paolo Guiotto: It is the sign… of the argument, which is Y of T here.
26:25:720Paolo Guiotto: And then I have the derivative of the argument, which is Y prime t. So this is the derivative of log of modulus yt.
26:34:270Paolo Guiotto: Now, you see that this seems complicated, but when Y of T is positive, sine is positive and the modulus is Y.
26:41:410Paolo Guiotto: So that… fraction is just 1 over Y.
26:46:290Paolo Guiotto: And when Y is negative, sine is minus 1, and the modulus is minus Y. So the minus disappear again, I get 1 over Y. So this quantity
26:57:20Paolo Guiotto: is always equal to 1 over Y of T, whatever is Y of T, positive or negative. And therefore, we get that this is Y prime over Y of T.
27:08:430Paolo Guiotto: And so, since y prime over Y of T is equal to 1, this derivative is 1.
27:15:300Paolo Guiotto: Now, derivative equals 1 makes easy this to be solved, because we know that log of modulus of Y is a function whose derivative is 1. This means that log of modulus of Y
27:29:30Paolo Guiotto: of T, let's keep T, is…
27:33:660Paolo Guiotto: a function whose derivative with respect to t is 1.
27:38:150Paolo Guiotto: What is the function whose derivative with respect to t is 1?
27:42:960Paolo Guiotto: who has derivative 1 when I do the derivative with respect to t?
27:47:570Paolo Guiotto: T, for example.
27:49:170Paolo Guiotto: Yes.
27:50:560Paolo Guiotto: Is there any other function for which we have this derivative 1?
27:56:610Paolo Guiotto: Yeah, that's very important, because if we are looking to all possible solutions, not just a solution, when you want to solve an equation normally, you want to solve for all the possible solutions to that equation, not just find a solution.
28:11:310Paolo Guiotto: Maybe it's the solution you don't, you are not interested. So this, there is a constant.
28:18:760Paolo Guiotto: Okay, and this means that we can now extract Y with a couple of steps, so this means modulus yt equal exponential T plus C.
28:29:270Paolo Guiotto: or if we want e to c times e to t, we relabel this constant as K, for example, so K e to t.
28:40:50Paolo Guiotto: And so, if models of Y is K to T, Y of T will be either plus minus k to t.
28:49:600Paolo Guiotto: Now…
28:50:870Paolo Guiotto: that K is E to C, which is a positive number, so this number is positive. With the plus-minus in front, since K is arbitrary, we have any number different from 0. So we can also say that this is the same of K, E to T, where k is any real
29:10:210Paolo Guiotto: except 0. We cannot take a k equals 0, because k equals 0 makes Y equal 0, and this is not allowed by this argument, because we are saying, if y is different from 0, and it is a solution, then, after all these steps, Y of T turns out to be
29:30:330Paolo Guiotto: K… constant times C to T, where k is any arbitrary number.
29:36:570Paolo Guiotto: So if Y is different from 0, it is of this form. If Y is 0 is also a solution, because we verified directly, so this can be obtained… sorry, this should be with K different from 0, so I should write also this.
29:53:60Paolo Guiotto: But when I put k equals 0, I get back the other solution. So I can say that with the unit formula, I could represent this way all the possible solutions of the equation, no?
30:05:520Paolo Guiotto: Okay, so this process, we have, quickly reviewed here.
30:10:10Paolo Guiotto: Starts from an equation, and at the end, yields all the possible solutions to this equation.
30:19:960Paolo Guiotto: There is the plural, because there is not a unique solution. There are infinitely many solutions, and that's not pathologic, it's normal, because if you take the trivial equation, Y prime equals zero.
30:32:520Paolo Guiotto: What are the solutions of this equation?
30:37:210Paolo Guiotto: The constants. So, why?
30:40:320Paolo Guiotto: T constantly equal to K, K real, is a solution.
30:45:430Paolo Guiotto: How many solutions? How many constants? So, infinitely many constants, infinitely many solutions, so it's not a strange fact that an equation can have infinitely many solutions.
30:55:890Paolo Guiotto: Sine X equals 0.
30:58:160Paolo Guiotto: has infinitely many solutions. So that's not a pathology, necessarily.
31:05:490Paolo Guiotto: in general, In general.
31:10:820Paolo Guiotto: a differential equation Oz.
31:16:580Paolo Guiotto: infinitely.
31:22:130Paolo Guiotto: Man… Solutions.
31:25:970Paolo Guiotto: There are lots of solutions.
31:28:350Paolo Guiotto: Normally, In real-world applications, any differential equation can arise from a physical problem, or…
31:40:530Paolo Guiotto: problem coming from engineering, economics. So you describe a system which is driven by certain physical, economical, biological, whatever, laws.
31:52:570Paolo Guiotto: And by imposing these laws, you get an equation like that. You get that the configuration of your system must obey to an equation of this type.
32:05:500Paolo Guiotto: Then, the next step is you want to solve the equation to see what is the configuration of the system that verifies these physical, economical, biological, etc. conditions, no? And that's the solving.
32:20:950Paolo Guiotto: To do what? By, for example, if, as in these cases, we interpret, if our Y, the unknown, is a function of time, for example, this could be used to do predictions on the future, no?
32:33:750Paolo Guiotto: It's plenty of examples. I use the equations of the physics of atmosphere to predict the forecast for tomorrow, no? So, in principle, I have the equations coming from physics, they are differential equations.
32:50:930Paolo Guiotto: I solve, I determine the forecast for tomorrow, okay?
32:55:860Paolo Guiotto: This is just in principle, because in practice is impossible, but the example works now.
33:03:240Paolo Guiotto: So, it is clear that if this message says, well, there are infinitely many solutions, it means there are infinitely many possibilities for the future, it's basically useless to do this, okay?
33:19:60Paolo Guiotto: Now, the point is that, normally, there are ways to make, to extract from all the solutions of a differential equation a very particular solution, which turns out, in many cases, to be also unique. And this,
33:36:100Paolo Guiotto: leads to the so-called Koshi problem.
33:42:670Paolo Guiotto: What is the Cauchy problem?
33:44:880Paolo Guiotto: Well, the Cauchy problem consists in searching for a solution of a certain equation that verifies an extra condition, okay? So, we consider the
33:58:590Paolo Guiotto: Cauchy problem.
34:01:470Paolo Guiotto: consists… Bien.
34:07:250Paolo Guiotto: searching.
34:13:679Paolo Guiotto: 4.
34:15:719Paolo Guiotto: Well, let's say, for the moment, air solution.
34:19:580Paolo Guiotto: Of a differential equation.
34:24:260Paolo Guiotto: verifying…
34:29:590Paolo Guiotto: some… Extra.
34:34:300Paolo Guiotto: Conditions, and in fact, just one condition.
34:39:429Paolo Guiotto: And this condition is the following. You take your equation, Y prime equals FTY,
34:45:750Paolo Guiotto: This is the differential equation. And you impose that the solution passes through a certain point. So you say that you want that Y at time t0 be equal to Y0.
34:59:380Paolo Guiotto: Okay? So where T0 and Y 0?
35:04:250Paolo Guiotto: known.
35:08:890Paolo Guiotto: So, you know, this is also called, if you think in terms of time.
35:13:490Paolo Guiotto: t is time, this means that you give the value of the solution at time t0, which is normally considered… could be considered an initial time. So I know that today, the state of the atmosphere is this one, and I want to predict what is the atmosphere tomorrow.
35:31:990Paolo Guiotto: So this makes sense, because if I take all possible solutions of the equation of the atmosphere, this could contain also solutions which are of no reality for us, okay? Because they contain a state of today, which is entirely different from what we see now. There are also solutions, let's say, today is knowing.
35:51:950Paolo Guiotto: But that's not the case, you see. So, this makes sense, or if you think in terms of,
35:58:870Paolo Guiotto: classical physics, you want to determine the trajectory of motion of a particle, and the idea is that… well, that's not the… specifically for this type of equation, but
36:10:10Paolo Guiotto: The idea is that if you give the initial state of motion, position, and velocity, you should be able to determine the interior future motion of the trajectory.
36:19:350Paolo Guiotto: Okay? Cool.
36:21:130Paolo Guiotto: This is the problem. Geometically, this problem means the following. We can also give a some goodness.
36:29:900Paolo Guiotto: We can have some theoretical insight into this problem. This is, as usual, the domain where F, the right-hand side of the equation, is defined, which is also the domain where you can see the graphs of the solutions. You cannot see outside of that domain.
36:48:80Paolo Guiotto: And, so you, in particular, that condition, YT0 equals Y0,
36:54:290Paolo Guiotto: Well, that point, the point, let's say that if this is T0, well, T0 cannot be, let's say, here, because,
37:03:270Paolo Guiotto: the point Y is 0, the point is 0. Y0 should be somewhere on this vertical line, so it should be out of the domain D.
37:13:930Paolo Guiotto: So, if the solution at the time T0 must be Y0, and we remind that for solutions, we just say that the point, condition 2, the point TYT,
37:27:430Paolo Guiotto: is always all time in the domain D, in particular at time P0, P0Y P0, Which is why not?
37:37:150Paolo Guiotto: is also in the domain D, must be in the domain D. So when you give an initial condition.
37:43:120Paolo Guiotto: Since TYT should be, indeed, for all time T,
37:51:290Paolo Guiotto: That belongs to the domain of definition of the solution. In particular, T0, why T0, Which is,
38:00:670Paolo Guiotto: Why not? This is known, this point, these two numbers, T0, Y0 are known, must be also in D. So the initial condition must be somewhere in the domain B, not outside of D, because it is one of the points of the solution.
38:18:260Paolo Guiotto: One of the… and that's why we call passage point, because the solution must pass through that point.
38:25:280Paolo Guiotto: Whatever it is, it must pass through that point at time T0, so maybe it is like that.
38:31:90Paolo Guiotto: Okay?
38:32:320Paolo Guiotto: So the problem consists in searching for a solution that at some specific time takes a certain specific value, or equivalently, it passes through a specific given point, no? If you apply this to the Cauchy problem.
38:49:250Paolo Guiotto: Example.
38:50:570Paolo Guiotto: You want to solve, The Cauchy problem
38:55:180Paolo Guiotto: Since we have already computed the solutions, we take this one. Y prime equals Y? Y at 1 equals 3. What is the solution
39:05:120Paolo Guiotto: Or these solutions, what are the solutions of this problem?
39:08:680Paolo Guiotto: Now, you can see that,
39:13:10Paolo Guiotto: We already know what are all the possible solutions of this equation, okay? So we must look for, among all possible solutions, if there are solutions such that the second condition is verified. So, since
39:32:540Paolo Guiotto: all…
39:35:440Paolo Guiotto: possible.
39:38:910Paolo Guiotto: solutions.
39:40:820Paolo Guiotto: of the equation, Y prime equals y r.
39:45:200Paolo Guiotto: YT equal KE to T, okay? So these are all the possible solutions. We look, for
39:55:510Paolo Guiotto: among them, if there is any one of them that verify also that condition. So… imposing.
40:05:960Paolo Guiotto: That…
40:07:690Paolo Guiotto: Y of 1 be equal to 3, we get… so Y , if we look at that formula, is KE to 1 must be equal to 3.
40:20:110Paolo Guiotto: So this becomes an algebraic equation for K, simple algebraic equation that can be disordered. It yields K equals 3 over E.
40:29:730Paolo Guiotto: So it means that among all infinitely many solutions, there is just one solution that verifies the passage condition.
40:39:610Paolo Guiotto: So, the solution… D.
40:43:290Paolo Guiotto: solution of the Cauchy problem.
40:47:650Paolo Guiotto: is unique, Y of T equals that specific constant, 3 over E, E to T.
40:56:810Paolo Guiotto: So that's very nice, and then it can be proved in general for linear equations, that the Cauchy problem has always a unique solution.
41:07:500Paolo Guiotto: So, I don't know if you have seen this, but let's hope that you have seen it. In general.
41:17:310Paolo Guiotto: So, it can be proved, huh?
41:25:480Paolo Guiotto: That.
41:27:830Paolo Guiotto: The… Because she problem.
41:31:940Paolo Guiotto: 4.
41:34:550Paolo Guiotto: first… Order.
41:39:980Paolo Guiotto: linear.
41:41:960Paolo Guiotto: equation.
41:43:420Paolo Guiotto: So let me write in the general form. So this is Y prime equals A
41:49:590Paolo Guiotto: t. A can be a function, depending on T, times Y plus B of T.
41:57:140Paolo Guiotto: with the passage condition Y is 0 equals Y0, Has… Always.
42:09:130Paolo Guiotto: unique.
42:13:80Paolo Guiotto: solution.
42:16:20Paolo Guiotto: Now, when in this… in which circumstances? You see that there are these two coefficients, so this… these two coefficients make… this is the right-hand side is the function of TY, where this function is defined.
42:31:380Paolo Guiotto: As you can see, as function of Y, since it is a first-order polynomial in Y, it is defined whatever is Y. Y, here I have two functions, generic functions of T, so they will be defined somewhere.
42:45:990Paolo Guiotto: So let's assume that these two functions be both defined for P that belongs to some interval to make easy decision.
42:54:570Paolo Guiotto: So… We assume that A and B are continuous function on some interval capital I.
43:04:240Paolo Guiotto: So this means that the natural domain for this equation is, in the plane TY,
43:12:190Paolo Guiotto: So the set of pairs, TY, for which the equation is defined, you need that TB in the interval I, otherwise the two coefficients A and B are not defined, and for Y, you can take whatever. So this means that if i is an interval on the real line, let's say that this is I,
43:33:880Paolo Guiotto: The domain is always a subset of the plane TY, is the subset made of
43:40:470Paolo Guiotto: points TY, where T is in I, and Y is in R. And this is a vertical trip like that.
43:50:10Paolo Guiotto: So this is the set of points where
43:53:930Paolo Guiotto: D is the set of pairs, PY,
43:58:30Paolo Guiotto: Such that P is in the interval I, and Y is 3 in R. Because for Y, for the linear equation case, there are no restrictions.
44:08:870Paolo Guiotto: So the domain in this case is this infinite vertical strip.
44:14:430Paolo Guiotto: So this, this result says that whenever you take a point T0Y0, In this stripper.
44:26:710Paolo Guiotto: So, for average, T0, Y0 in the strip, that means T0 is in I, and Y0 is arbitrarily chosen in I.
44:36:610Paolo Guiotto: So no matter how you take a point in that vertical strip.
44:40:400Paolo Guiotto: That theorem says that there is always a solution which passes through that point, and that solution is unique, and also it is defined on then-tier interval I.
44:51:790Paolo Guiotto: So there exists, as always, a unique solution, Y equals Y of T.
44:59:410Paolo Guiotto: defined… for all times t, in the interval, capital I.
45:07:360Paolo Guiotto: Okay, so this is nice because that's the domain, you take an initial condition, wherever you want, there is a solution, and actually just one, that passed through that point.
45:19:860Paolo Guiotto: Okay? That solution is unique, and moreover, that solution is defined for all possible times where it makes sense that are in this interval I, so we will have another solution like that.
45:33:180Paolo Guiotto: In particular, this means that the following situation is impossible.
45:38:820Paolo Guiotto: So you cannot have that you pick a point, like the blue point there.
45:43:280Paolo Guiotto: And there is a solution that does this.
45:47:300Paolo Guiotto: Why, this is impossible.
45:49:960Paolo Guiotto: The impossibility is the crossing point, because in this point here…
45:55:470Paolo Guiotto: If I consider this as an initial condition.
45:59:420Paolo Guiotto: There would be two solutions of the differential equation, the blue and the red, that pass through the same point.
46:07:800Paolo Guiotto: And this is forbidden because of the uniqueness, okay? So this, fact that, because she probably has always, a unique solution, tells that, yes, there is a solution.
46:21:60Paolo Guiotto: And there cannot be any other solution that passed through that point.
46:25:350Paolo Guiotto: But this means that, in other words, when you have two solutions, they can never cross.
46:31:780Paolo Guiotto: At some point.
46:35:10Paolo Guiotto: Because at that point, I repeat, if you consider that point as the passage point of the Cauchy problem, the Cauchy problem tells there is only one solution that… of the equation that passes to that point.
46:50:400Paolo Guiotto: Okay, since the blue is the solution of the equation, and the red is the solution of the equation.
46:57:370Paolo Guiotto: This is impossible. So this means that any two solutions can never cross.
47:04:70Paolo Guiotto: At some point.
47:05:650Paolo Guiotto: Okay, so this is, say, impossible.
47:14:890Paolo Guiotto: It's a very important thing, okay?
47:18:170Paolo Guiotto: This is also important for application purposes. The fact that you have a unique solution and two solutions cannot cross means that there is a unique possible future, a unique possible path. Imagine that this is today, okay?
47:34:30Paolo Guiotto: And this is the value of money between the investment in the financial market, and it says that this is the behavior in the future of this.
47:45:130Paolo Guiotto: But what means that there are two lines like that?
47:48:650Paolo Guiotto: Well, it means that if I am here, and this is sometime tomorrow, maybe, or in one year, I don't know, in the future, at this point, if this is the case, that will be the situation where I have two possible futures.
48:04:880Paolo Guiotto: In one future, I'm losing my money. In the other future, I'm… I'm losing money. So, which one is the right future? So, if this cannot happen, it means that I'm always safe, that whatever is the prediction, it is the unique possibility.
48:20:680Paolo Guiotto: Okay, so you see, that's why it's so important that you have uniqueness. Otherwise, you cannot use differential equation to do any prediction.
48:31:900Paolo Guiotto: And it doesn't mean that, well, let's use something else, because if this is the language you have to use to model some real-world problem.
48:43:460Paolo Guiotto: It means that you have always to, to…
48:50:850Paolo Guiotto: know what is the prediction you can do, if it is a unit possible prediction or not. Well, for linear equation, it happens that everything go nicely.
49:02:20Paolo Guiotto: For a nonlinear equation, unfortunately, the story is very different.
49:06:460Paolo Guiotto: Let's take, bracknow, 10 minutes, and then let's see what…
49:12:810Paolo Guiotto: Happens for nonlinear equations, through some examples.
52:15:850Paolo Guiotto: Is it okay now?
52:17:620Paolo Guiotto: Yeah, should be.
52:19:420Paolo Guiotto: Okay, we say that, we start solving the equation, finding all the possible solutions, then step number two, we see which one verifies the, passage condition.
52:34:940Paolo Guiotto: We start… solving… the equation.
52:46:620Paolo Guiotto: So Y prime equal Y squared. This is a parable variable equation.
52:51:450Paolo Guiotto: So we could divide by Y squared, but we have to be careful, because Y can be 0, so we can say that if, Y is… well, we notice that… better, we notice…
53:09:840Paolo Guiotto: That.
53:11:750Paolo Guiotto: Y identically equal to 0 is a solution.
53:16:770Paolo Guiotto: You agree?
53:18:690Paolo Guiotto: You put Y equals 0, the derivative is 0, Y squared is 0, so the equation is verified.
53:24:930Paolo Guiotto: So, by the way, this is the solution.
53:28:670Paolo Guiotto: We are talking about. This is Y constantly equal to zero. It's one of the solutions.
53:34:760Paolo Guiotto: So, for example, this could be the solution for any Cauchy problem with a condition like this one, no? Y of T0 equals 0, the solution could be Y equals 0 constantly.
53:48:520Paolo Guiotto: If Y, different from 0 is a solution.
53:56:690Paolo Guiotto: We can separate the variables, so we can rewrite the equation y equals Y squared as
54:03:660Paolo Guiotto: since Y is different from 0, Y squared is different from 0, we can divide by Y squared, so Y prime divided by Y squared equals 1. And the idea of this method is that we recognize that at left we have a derivative.
54:19:500Paolo Guiotto: So, this could be written as, if you want, as a power, Y of T to exponent minus 2,
54:27:820Paolo Guiotto: times Y prime of t. I write the T here to emphasize that I want to see this as a derivative with respect to t of something. The prime means derivative with respect to t.
54:39:890Paolo Guiotto: When do you get this by doing the derivative?
54:44:720Paolo Guiotto: So that's a power, this should come from a power. The derivative works that when you do the derivative of a power, we… you, you reduce the exponent by 1, okay? So to obtain minus 2, you have to start from minus 1.
55:02:270Paolo Guiotto: So the idea should be, let's look at YT at minus 1. If we do the derivative of this, we get minus 1 is the exponent, Y of T at exponent minus 1, minus 1, minus 2 times the derivative of Y, Y prime.
55:18:780Paolo Guiotto: So, it's basically the same, it becomes exactly, if I put the minus, minus Y of T to minus 1.
55:26:100Paolo Guiotto: Okay? So that equation can be also written as minus Y of T to minus 1 prime equal to 1. So the derivative of this box is 1.
55:39:260Paolo Guiotto: So when the derivative is 1, when the box, which is minus Y of T to minus 1, is t plus a constant, as above.
55:50:130Paolo Guiotto: So this means, literally, minus 1 over Y of T equal T plus constant, so we can put the minus this side. I do not put the minus in front to the C, because C is a constant, so I can always say plus C minus C is the same thing.
56:09:150Paolo Guiotto: And then I can extract Y of T by taking the reciprocal, this is 1 over minus T plus constant. Constant real.
56:20:60Paolo Guiotto: Okay, so you see that this function is never equal 0, so it is separated from the previous one.
56:27:630Paolo Guiotto: And, I cannot have a formula that,
56:31:650Paolo Guiotto: contains the tuna, so I have to say that the general solution
56:37:90Paolo Guiotto: general solution of the equation y prime equals Y squared is either Y constantly 0, or Y equal
56:49:830Paolo Guiotto: 1 over minus T plus constant, okay? So these are the solutions.
56:57:530Paolo Guiotto: Okay, now… If we want, we can do a figure.
57:04:730Paolo Guiotto: this one, where we draw all the other possible solutions. These are a little bit more complicated, because
57:12:330Paolo Guiotto: They are 1 over minus T plus C. What kind of function are these? Well, let's write in this way. Maybe it is better if I write a C minus T.
57:23:850Paolo Guiotto: Now, these functions are, something like 1 over Ateno. They are bad at point C. So, for example, this function, like, if C is this one.
57:36:350Paolo Guiotto: 1 over C minus T should be positive when T is less than C, so it is like that.
57:44:140Paolo Guiotto: So this is… 1 over C minus T, and negative… yeah.
57:52:70Paolo Guiotto: 1 over C minus D.
57:54:860Paolo Guiotto: Well, here there is a first point that we should have clear. Normally, when we talk about the solution of a differential equation, we mean a function defined on an interval, because
58:06:800Paolo Guiotto: think to real-world application. Imagine that the black line represents anything, any physical phenomena. What happens, if you look at here.
58:16:390Paolo Guiotto: This is time.
58:18:460Paolo Guiotto: It means that when you move to the future in this way, the solution basically blows up, goes to infinity, so it's like an explosion.
58:29:850Paolo Guiotto: So, at this moment, the solution is not defined. Simply, there is no solution at time t equals C, and then the solution restart anew, but coming from another blow up at minus infinity.
58:44:870Paolo Guiotto: So, physical, it's something that doesn't make any sense. What makes sense is maybe this is a phenomena that, at times, see, there is an explosion, no? And then there is no more solution. We just do not consider. Or maybe we can look at this part, for example, imagine that we are here.
59:03:590Paolo Guiotto: And we look at the past, we go back in the past, and we see an explosion in the past, big bang, no? How can we know that there was a big bang? Because we look at the solutions of differential equation, and we discover that in the past, there should be something, some critical phenomenon, like this.
59:20:770Paolo Guiotto: But there is no sense to talk about the solution before the Big Bang. We don't know what is it, that does not make sense. So, we can say that if we consider solutions only defined on intervals.
59:34:200Paolo Guiotto: We can see that here, the black line is not a unique solution, it's not a solution. Actually, this could be two solutions.
59:42:180Paolo Guiotto: Is this one, and the other is this one.
59:45:330Paolo Guiotto: Now, if we consider the Cauchy problem, So…
59:51:820Paolo Guiotto: Let's look at the cushy problem, and fix an initial condition.
59:56:690Paolo Guiotto: Well, now, I have two possibilities. My initial condition is here.
00:02:140Paolo Guiotto: So this is the Cauchy problem. Y prime equals Y squared, Y of P0 equals 0.
00:11:520Paolo Guiotto: So what is the solution? Is there a solution of the equation that at time t0 has value 0?
00:18:920Paolo Guiotto: well, let's look at this figure. The black lines are never equal zero, because they are
00:26:240Paolo Guiotto: 1 over something. So this quantity cannot be zero, yeah?
00:31:570Paolo Guiotto: No? So… The black line cannot be a solution for that equation, because one passed by that point.
00:40:180Paolo Guiotto: There won't be a T such that this quantity is 0. You see this or not?
00:44:950Paolo Guiotto: So, the black line cannot be a solution for this problem. Instead, the red line can be, because it passes to this point. So, I can say that if I take that point, there is just one solution that passes to that point, and it is the red line. So, the unique
01:07:270Paolo Guiotto: Solution… is, YT… identical equals zero. Which is, by the way, defined for every T. Defined
01:21:430Paolo Guiotto: for all tea.
01:23:470Paolo Guiotto: Let's say, from minus infinity plus infinity. If t is time, we may say that this solution is defined forever, in the future, forever in the past.
01:34:90Paolo Guiotto: Okay?
01:35:730Paolo Guiotto: Now, let's take a sec on the problem. Let's now take a condition which is not on the axis.
01:42:810Paolo Guiotto: So let's imagine that, for example, to fix ideas, it is here. So we have a T0,
01:49:870Paolo Guiotto: Why 0 positive?
01:52:820Paolo Guiotto: So in this case, it is clear that it cannot be the red solution, because the red solution is constantly equal to zero, so it cannot pass through that point. So in this case, if there is a solution, it must be one of these black lines.
02:08:10Paolo Guiotto: So let's see which one is it, okay?
02:11:680Paolo Guiotto: So, in this case, I want to look, in this case.
02:20:520Paolo Guiotto: We look… At a solution.
02:27:680Paolo Guiotto: of cypher.
02:31:910Paolo Guiotto: Y of T equals 1 over constant minus T, where, of course, the constant has to be determined in such a way that we, we pass through that point, where
02:48:330Paolo Guiotto: C is determined.
02:52:380Paolo Guiotto: Bye.
02:54:190Paolo Guiotto: the passage condition
02:58:340Paolo Guiotto: passage condition, and the passage condition is Y at time t0 must be Y0. So this means that 1 over C minus T0 must be Y0. This is a simple equation for C. You can solve it
03:16:530Paolo Guiotto: Well, first, since I have to isolate C, I have to multiply by C minus stay 0 and divide by Y0. Why can't divide by Y0? Because Y0 is not 0. It's not 0 because
03:30:400Paolo Guiotto: the case Y0 equals 0 has been already considered here. So in this case, Y0 is different from 0.
03:38:940Paolo Guiotto: the figure is positive, it could be negative. For the moment, there is no distinction. So this is different from zero, so I can say that this equation is equivalent to C minus T0 equals 1 over Y0, and from this, I get C equals T0 plus
03:57:550Paolo Guiotto: 1 over Y0.
04:00:200Paolo Guiotto: So this is the unique value.
04:02:720Paolo Guiotto: Not 2. Not 3 is a unique value, so there exists… there is… a unique…
04:14:910Paolo Guiotto: solution, which is now this, Y of T equals 1 over, the constant C is that quantity.
04:23:190Paolo Guiotto: T0 plus 1 over Y0, minus, T. C minus T, okay?
04:37:320Paolo Guiotto: The solution is Y of T equals 1 over C minus T. We solved for C, imposing the passage condition. We found C equals P0 plus 1 over Y0,
04:49:860Paolo Guiotto: And finally, we got… we plugged the C inside the function, we get Y of C is this form, okay?
04:57:680Paolo Guiotto: Now, let's see what is in this. This is still the same type of function we have here.
05:03:170Paolo Guiotto: No? This hyperbola that exploses at point C. What is this C?
05:09:310Paolo Guiotto: Well, this is important now to understand the figure if Y0 is positive-negative.
05:14:720Paolo Guiotto: Because we have two cases, as we will see.
05:17:960Paolo Guiotto: So let's start with the case we considered above. So this was T0, and Y0.
05:29:800Paolo Guiotto: Now, remind that that solution explodes at this point here.
05:34:790Paolo Guiotto: when T is equal to that value, the denominator is 0, so the fraction is 1 over 0. Where is it, this one? This is T0 plus 1 over Y0, right?
05:48:150Paolo Guiotto: So, what can be said in this case? Y0 for this figure is positive.
05:54:600Paolo Guiotto: So this number is positive. So it means that this is T0 plus something positive, so you are at right of T0. So the critical moment where the solution blows up is this one, T0 plus 1 over T0.
06:09:880Paolo Guiotto: So it means that since the black line is made like that.
06:14:400Paolo Guiotto: So before the blows up, the solution increases, and after the blows up, the solution increases from minus infinity. It means that here.
06:24:370Paolo Guiotto: What we see, this is the solution, is something like this, that passes through this point, and it blows up here.
06:33:650Paolo Guiotto: At this moment, we have an explosion, and then we say that there is no meaning to have a solution after that time. So the solution is what you see. So this is the solution.
06:45:60Paolo Guiotto: It is Y of T. The formula is the same, 1 over T0 plus 1 over Y0 minus T, and it is defined for time t from minus infinity to the explosion time, which is T0 plus 1 over Y0.
07:03:190Paolo Guiotto: So these are the alpha and the omega of the solution, no? The solution was alive forever in the past, but we'll explore some time in the future.
07:14:200Paolo Guiotto: Okay?
07:15:870Paolo Guiotto: And similarly, you can see what happens in the other case.
07:21:260Paolo Guiotto: What if I now take T0 here, but Y0 negative?
07:27:580Paolo Guiotto: So this is the point, the passage point.
07:30:230Paolo Guiotto: Well, we can do the same kind of argument, because the solution is still Y of T equals 1 over T0
07:40:620Paolo Guiotto: plus 1 over Y0 minus T.
07:43:930Paolo Guiotto: That… that time, because this is a time in this interpretation, is the explosion time.
07:52:320Paolo Guiotto: Which is now, since Y0 is negative, you see here, Y0 is negative.
07:58:950Paolo Guiotto: So the explosion time, T0 plus 1 over Y0, since 1 over Y0 is also negative.
08:07:910Paolo Guiotto: you are now subtracting something to dime T0.
08:11:990Paolo Guiotto: So it means that this time is at left now, with respect to T0. I don't know where, maybe it's very close, maybe it's very far, I don't know. Let's put, however, here, for example, T0 plus 1 over, Y0.
08:28:69Paolo Guiotto: So that's this explosion time, and we know that these kind of functions are increasing from 0 to plus infinity at left, and increasing from minus infinity to zero at right. So it means that our solution is, in this case, this one.
08:48:229Paolo Guiotto: Okay?
08:49:520Paolo Guiotto: For this solution, the time, the lifetime, is this one.
08:54:240Paolo Guiotto: There is an initial time, this one.
08:58:580Paolo Guiotto: And the future is forever.
09:03:130Paolo Guiotto: Now, what this figure tells is that, in any case, we can say there is a unique solution. So, still, we have existence of the solution and uniqueness. We also found the solution.
09:16:109Paolo Guiotto: But, something interesting to be noticed, which is different with respect to the linear case.
09:23:859Paolo Guiotto: In the case of linear equations, all the solutions are defined in the same interval I, so they have the same life, time life. They were born at certain time, and they will die at certain time, and this is the same for each of the solutions.
09:42:899Paolo Guiotto: Here it is different, because there is a solution, the constant solution, which is defined forever.
09:50:899Paolo Guiotto: Okay? So, it cannot be killed, this solution. It is a solution with an infinite time life in the future, in the past.
10:03:200Paolo Guiotto: There are solutions, like this one, that are alive forever in the past, but they will explode somewhere in the future. At finite time, I can also know when they explode. Time T0, which is the time of today, plus 1 over Y0, which is the value of today, Y0.
10:22:280Paolo Guiotto: There are solutions that will survive forever in the future, but they will explore somewhere in the past.
10:30:470Paolo Guiotto: Okay?
10:31:750Paolo Guiotto: Which are these kind of solutions when Y0 is negative.
10:35:790Paolo Guiotto: Okay, so if you want a complete representation of solutions, you can do the following figure. You can say that
10:44:390Paolo Guiotto: We have this solution, the constant solution.
10:47:990Paolo Guiotto: Then we have solutions of Cauchy problems for the positive value of the passage conditions. They are of this type.
10:58:720Paolo Guiotto: Well, maybe let's use… a caller.
11:03:400Paolo Guiotto: And they will finish their life somewhere at this time. So this is the solution when Y0 is positive. This is the solution when Y0 is 0.
11:18:190Paolo Guiotto: And then there are solutions, it's not necessarily positive, the explosion time, that, can be like these ones.
11:28:260Paolo Guiotto: These are the solutions with Y0 negative.
11:31:960Paolo Guiotto: So the figure is this one. And in particular, you can see that it is not true that all the solutions are defined in the same, in the same time.
11:44:60Paolo Guiotto: Okay.
11:45:660Paolo Guiotto: Now, let's look at a second example, which is similar to this one.
11:50:770Paolo Guiotto: where we have a sort of nightmare example, because we take this equation, Y prime, well, I take Y…
12:03:490Paolo Guiotto: to one half.
12:06:500Paolo Guiotto: or to avoid the problem with sine wave, let's take the cubic root, Y to 1 third, okay? This is the cubic root of Y. Whatever is Y, it is defined.
12:19:560Paolo Guiotto: So Y, of, Y0, as usual, T0 equals Y0.
12:29:310Paolo Guiotto: Also, here you have the function, FTY is this one, is independent of T, so T can be whatever.
12:38:830Paolo Guiotto: Y to 1 third is the cubic root is always defined. Positive, negative, no problem. So in this case, we have, as domain of definition of the equation, the anterior plane R2.
12:57:290Paolo Guiotto: Let's repeat the procedure we have seen before. So let's determine, first of all, all the possible solutions for this equation.
13:05:680Paolo Guiotto: Let's… determine… all… Solutions.
13:14:100Paolo Guiotto: of the equation, so not the Cauchy problem, the equation. The equations
13:19:430Paolo Guiotto: the equation will have infinitely many solutions, so Y to 1 third?
13:24:630Paolo Guiotto: We start realizing that there is one spatial solution. This is a separable variable equations. What I want to do is to divide by Y to 1 third, to carry everything in Y at left, and leave the constant at right. But to do that, I must be sure that Y is non-zero.
13:41:40Paolo Guiotto: I noticed that Y equals 0 is a solution, because if you plug 0 inside the equation, you get 0 equals 0 at left and right. So, Y constantly equals 0 is
13:53:900Paolo Guiotto: a solution.
13:55:600Paolo Guiotto: And it is, again, this one.
13:58:510Paolo Guiotto: So, if Y… Different from zero is a solution.
14:05:290Paolo Guiotto: Then, we can separate variables y prime equal y to 1 third.
14:11:80Paolo Guiotto: is equivalent. Since Y is different from 0, I can divide by y cubic root of y, which will be different from 0. I get y to minus 1 third Y prime equal 1.
14:23:890Paolo Guiotto: Again, that's very similar to the previous case. This comes by doing the derivative of what?
14:39:540Paolo Guiotto: It's a power. If it is minus 1, it's a logarithm, but for the other exponents, it's still a power.
14:46:280Paolo Guiotto: It's… you have to add the minus 1 third plus one, it's 2 thirds, so you will get that by doing Y to 2 thirds, more or less, because we have to check with the coefficients. When you do the derivative, you get 2 thirds times y to 2 thirds minus 1 is minus 1 third times Y is a function, it's not the variable. 2 times Y prime.
15:10:140Paolo Guiotto: So you see that we need just to add the 2 thirds here, and so we can multiply both sides by 2 thirds, we don't change anything. And then we have that derivative of Y to 2 thirds.
15:23:430Paolo Guiotto: prime equal to 2 thirds, which is much better, because now we know that that function has derivative 2 thirds. So this means that that function, Y to 2 thirds, is
15:35:490Paolo Guiotto: a primitive to 2 thirds, which will be 2 thirds t plus a constant.
15:42:180Paolo Guiotto: Okay?
15:43:250Paolo Guiotto: Now we can extract Y,
15:47:120Paolo Guiotto: So, to do that, we have to do the cube, so Y squared equal this, 2 thirds T plus C.
15:55:290Paolo Guiotto: 2 power 3, and then we have to take the root, and here we can have some problem, because when we take the root, we have to distinguish positive, negative. So let's say for the moment that we take Y of T equal plus minus the root of this quantity.
16:14:760Paolo Guiotto: 2 over 3T plus C cube.
16:18:870Paolo Guiotto: will be something like this. It means that there is a Y positive, there is a Y negative.
16:24:870Paolo Guiotto: how these functions are made. Let's try to have an idea. But now, to simplify things, I will start with
16:34:120Paolo Guiotto: Well, I will do a little trick, which is, you see the constant C? The constant C is any constant. I can always put a 2 thirds here, because 2 thirds C is C, it's the same thing. But this helps, because I can factorize the 2 thirds, so I have plus root of 2 thirds, which is
16:52:620Paolo Guiotto: Scalar, and then, well, to power 3, so formally, it should be 2 to 3, 8 over 27, whatever it is. Then we have the root of t plus constant cube, which is maybe a little bit better.
17:10:540Paolo Guiotto: I don't know if we gained anything from this, but… Let's write this word.
17:16:450Paolo Guiotto: Because in this way, we have an idea on how these functions are made.
17:23:220Paolo Guiotto: Another, maybe, before we do that, let's rewrite this as we have this in innocent constant. I prefer to see this as a power, so this is 3 half.
17:37:710Paolo Guiotto: You will see in a moment why.
17:40:170Paolo Guiotto: Now, let's do a plot of these type of functions. How are they made?
17:45:760Paolo Guiotto: Well, these are… well, let's start with the T23F. What is T23F?
17:54:50Paolo Guiotto: If I have to plot T to 3 half.
17:57:610Paolo Guiotto: What is it? Well, it's a power.
18:00:860Paolo Guiotto: The exponent is 3 half, it's greater than 1, so T to 1 is a straight line. When the exponent is greater than 1, it looks like a parabola, okay? It is not different from T-square. Of course, it's not T squared, but… so it's made about like that.
18:19:340Paolo Guiotto: What happens if I put the C now?
18:22:150Paolo Guiotto: Well, this is T plus C to exponent 3a.
18:27:200Paolo Guiotto: Well, that's just a translation, no?
18:30:150Paolo Guiotto: Well, it means that you can start when the argument is zero. Well, there is no function for T negative, you see, because the root is not defined.
18:40:710Paolo Guiotto: So, if, say that C is this one.
18:44:840Paolo Guiotto: Well, actually, the important value is minus C, it's not C. So let's do another, trick. Instead of writing plus C, I write minus C. It's the same, because plus C minus C is the same thing. So I will write minus C.
19:00:870Paolo Guiotto: PR?
19:02:140Paolo Guiotto: PR.
19:03:410Paolo Guiotto: This to make easier to draw these functions, not… Any particular…
19:12:210Paolo Guiotto: Because in this way, I see that this function starts when T is C. It cannot be before C, otherwise T minus C would be negative. So it will be at… if C is this value, this function is like this one, but starting at point C.
19:29:340Paolo Guiotto: Now you have the root of 8 over 27, which is a factor. You just scale this, does not change the qualitative figure.
19:38:560Paolo Guiotto: And you add the plus-minus. This a little bit changed, because it means that you have also this one.
19:43:770Paolo Guiotto: Okay, so we can say that these are plus minus the root of 8 over 27p minus C to exponent 3F.
19:53:830Paolo Guiotto: These are the functions, okay?
19:57:40Paolo Guiotto: Now, let's, see… well, this C can be wherever, so you… you can have also C there, but why not? C here? So these functions are made like that.
20:10:210Paolo Guiotto: Okay, now you can return on the Cauchy problem.
20:15:860Paolo Guiotto: Was she proper?
20:18:820Paolo Guiotto: Well, this time.
20:20:870Paolo Guiotto: Instead of starting from the apparently simple problem, which is the Cauchy problem where the initial condition is on the axis, as before.
20:31:00Paolo Guiotto: we will see that that's a bad problem. We start from the Cauchy problem, where the division condition is here, is not on the axis. For example, it's not important where is T0, it's important if the condition is
20:46:210Paolo Guiotto: positive or negative. So let's discuss if there is any solution that passed through that point.
20:52:800Paolo Guiotto: Well, I can say.
20:54:750Paolo Guiotto: first of all, cannot be the solution constantly equal to zero, no? Because that solution won't pass to that point. So you have to look at these functions here.
21:05:540Paolo Guiotto: Blue-red are the same, it depends on the value of C, no?
21:09:220Paolo Guiotto: So, I look for a value of C for which that solution passed through that point. That's what I'm going to do.
21:17:630Paolo Guiotto: Well, since the point is with ordinate way 0 positive, I will take the plus solution, okay? So I look for a solution like plus root of 28 over 27,
21:33:510Paolo Guiotto: T minus C to X plane 3 halves, because that one with the minus will never pass there, it's negative, okay?
21:42:490Paolo Guiotto: So, only one of these types. And now, let's find the C for which I have that, again, by imposing the condition. So, imposing that, this is my Y of P. The unique possibility is this one. Y of T0 is equal Y0 if and only if root of 8 over 27
22:02:470Paolo Guiotto: then T0, whatever it is, minus C to exponent 3 half must be equal to Y0.
22:09:600Paolo Guiotto: Now I have to solve for the constant C, so you plug this in the other side, root of 27 over 8,
22:16:740Paolo Guiotto: Then, to extract the argument there, I have to invert that power, so I have to square to eliminate the half, so this squared.
22:27:730Paolo Guiotto: And then I have to take the cubic root. I can do that, whatever is the value, because the cubic root is well-defined. So to explain two-thirds, I am almost done, because I get…
22:39:770Paolo Guiotto: So, C equals T0 minus that value. Y0, root of 27. I don't want to simplify, it's not the point, we have to just see that there is a unique value of C. If you take this C,
22:56:710Paolo Guiotto: you have the solution that passed through that point. The unique interest here is that the C also represents the time where this solution starts. You can see that it is T0,
23:09:80Paolo Guiotto: minus something, and since my Y0 is positive, this number is positive, that 2 thirds
23:18:680Paolo Guiotto: So square of the cubic root is certainly positive, so I am subtracting to T0, so this value, C, is about here, is before T0. So the solution will be this one.
23:33:690Paolo Guiotto: Okay?
23:35:800Paolo Guiotto: That passed through that pond.
23:37:960Paolo Guiotto: And in a similar way, you can solve for the negative axis, okay? So you will have the other solution.
23:46:440Paolo Guiotto: So, similarly.
23:52:670Paolo Guiotto: Sweet.
23:53:610Paolo Guiotto: than… So, vote.
23:57:610Paolo Guiotto: for… Y0 negative.
24:00:980Paolo Guiotto: And this says that, basically, to each passage point, there is a unique solution.
24:10:900Paolo Guiotto: But that's… well, that's an algebraic equation. You have root of 8 over 27, that's 0 minus C to 3 halves.
24:18:990Paolo Guiotto: equal Y0. I move the root on the right-hand side, becomes root of 27 of A. I have to invert the power, so I have the cubic root, and to raise to power 2. So I get T0 minus C equal Y0 root, etc, to exponent 2 thirds.
24:39:430Paolo Guiotto: So now that's C equal, you carry C this side, the stuff on the other side, you get that formula.
24:46:250Paolo Guiotto: Okay? In a similar way, you can solve the case when Y0 is negative. So let's say that this is T0. Let's say that this is Y0. You will have that here, there will be a solution like that. Okay? Same calculations.
25:04:200Paolo Guiotto: But let's see what happens when we take now the case Y0 equals 0.
25:11:220Paolo Guiotto: Finally, Okay, he's up.
25:15:840Paolo Guiotto: Y0 equals 0.
25:18:720Paolo Guiotto: So, we have… The passage condition now on the axis.
25:25:370Paolo Guiotto: This is T0 and Y0 is 0. The solution must pass through that. We already know that there is a solution, right? That solution is the solution Y constantly equal to 0.
25:38:350Paolo Guiotto: But here, I can show you that there are many other solutions, and actually infinitely many solutions.
25:46:440Paolo Guiotto: Let's see why.
25:47:780Paolo Guiotto: Because if you take any C at right of T0,
25:53:290Paolo Guiotto: And you consider this line. You take the line, which is 0,
25:59:180Paolo Guiotto: Until that moment C. And then you follow that, that function, what is, root of 27 over a t minus C to exponent 3 up, if I'm not wronger. That's the solution.
26:14:290Paolo Guiotto: So the total green line is made of gluing two solutions, no? This is a piece of solution that is zero exactly here, because when T is zero to 3F is zero.
26:28:720Paolo Guiotto: It's just here on the axis, and then you continue going back in the past with the zero solution.
26:35:500Paolo Guiotto: This is a function which is a solution of the equation, because at every point
26:41:100Paolo Guiotto: The constant verifies the equation, this one verifies the equation. KR solution, you are in the domain, the domain is everything, so there is no problem.
26:49:820Paolo Guiotto: And they pass through the same point. So the green is a solution of the cushy point, exactly as the red.
26:57:390Paolo Guiotto: With the difference that now you have two solutions, and since this C can be whatever, you can choose a different C and have a different solution, maybe like this one.
27:07:570Paolo Guiotto: I don't know, let's change color. You can have also this one.
27:13:620Paolo Guiotto: that take off some future time, another value of C.
27:19:130Paolo Guiotto: So for any value of C larger than T0, the unit point is that must be greater than T0, you can glue the constant solution with another, with one of these, let's say, parabolas, which are not parabolas, and you get, again, a solution.
27:38:610Paolo Guiotto: So it means that if you want a solution that passed through this point, you have infinitely many options.
27:46:640Paolo Guiotto: And this makes the prediction completely useless. So, for this problem, no uniqueness.
27:57:790Paolo Guiotto: So, this is an important example of an equation.
28:03:640Paolo Guiotto: for which the Cauchy problem has not a unique solution.
28:09:140Paolo Guiotto: So this does not happen with linear equations, but it happens, it may happen, with nonlinear equations. So it becomes important to understand when it happens, or at least to have
28:22:850Paolo Guiotto: Some conditions to ensure that this does not happen.
28:27:10Paolo Guiotto: So that there is a unique solution.
28:30:140Paolo Guiotto: But this will be the argument of next time.
28:33:780Paolo Guiotto: Let's stop here for today.
28:35:980Paolo Guiotto: we'll see on Friday.