AI Assistant
Transcript
00:25:330Paolo Guiotto: Good morning.
00:28:510Paolo Guiotto: Today we start a new topic.
00:33:230Paolo Guiotto: Which is actually an application of differential calculation.
00:38:320Paolo Guiotto: We talk about vector field, so what is a vector field?
00:43:370Paolo Guiotto: Well, a vector field…
00:54:580Paolo Guiotto: It's just, the function, A function of vector, valuable, vector, value, so it's a type of functions.
01:05:110Paolo Guiotto: We considered, so far.
01:07:760Paolo Guiotto: But with a particular feature, that…
01:10:650Paolo Guiotto: both domain and co-domain are in the same dimensional, in the same dimension vector space RB. So we go from RD into RD itself.
01:25:150Paolo Guiotto: So, this is the same, no? So, for example, this one, F… XY.
01:36:320Paolo Guiotto: This is a function of a vector of two components, so it is defined, let's say, on R2, defined as XE to Y log of X plus Y.
01:52:630Paolo Guiotto: He is, Two-dimensional.
01:58:920Paolo Guiotto: Backed up… Hilda?
02:03:660Paolo Guiotto: For example, this one, F… XYZ.
02:09:729Paolo Guiotto: equal, X, Z, Y, I don't know, just…
02:13:880Paolo Guiotto: Like, this… this is a function that takes a vector with three components and gives a vector with 3 components.
02:21:450Paolo Guiotto: Is it three-dimensional?
02:26:870Paolo Guiotto: Backdoor.
02:29:120Paolo Guiotto: Hilda.
02:31:220Paolo Guiotto: But, for example, this function.
02:36:660Paolo Guiotto: FXY.
02:38:670Paolo Guiotto: equal X plus YX times Y…
02:43:560Paolo Guiotto: sine XY. This is not a vector field, okay?
02:57:70Paolo Guiotto: Because you see that here.
03:01:640Paolo Guiotto: We take a vector with two components, and we produce a vector with three components, so they are different.
03:08:270Paolo Guiotto: Now, why vector fields are interesting, functions?
03:13:50Paolo Guiotto: Vector fields, huh?
03:19:340Paolo Guiotto: R… Important.
03:27:520Paolo Guiotto: In, physics, huh?
03:34:530Paolo Guiotto: Well, in general, in mechanics, also in engineering.
03:42:100Paolo Guiotto: And many others. So, let's see a few examples of vector fields, or very important vector fields.
03:50:220Paolo Guiotto: Well, as you know, for example, we are, we are subject to the action of a very important factorial field, which is the gravitational vector field. So, for example, you know that,
04:05:330Paolo Guiotto: the Newton…
04:11:220Paolo Guiotto: Lope.
04:12:740Paolo Guiotto: of gravitation.
04:15:600Paolo Guiotto: Says that, if you have, in general, two masses, so let's say a mass
04:21:190Paolo Guiotto: M, a point mass M, and a point… a little… a point mass little m.
04:28:480Paolo Guiotto: Each of them, acts onto the other with a force.
04:33:10Paolo Guiotto: Now, the force is an attractive force.
04:37:940Paolo Guiotto: That can be represented by a vector, F.
04:42:510Paolo Guiotto: And this force is a positional force. It depends on the position.
04:46:730Paolo Guiotto: And for the gravitational force, we have this… if we assume that, conventionally, the mass M, the capital M mass, is
04:56:380Paolo Guiotto: Positioned at the origin of the reference.
05:00:580Paolo Guiotto: system.
05:02:40Paolo Guiotto: Now, at this point of coordinates XYZ, The reason I wrote?
05:09:260Paolo Guiotto: force.
05:11:00Paolo Guiotto: applied to that mass at that point, which is basically proportional to the product of the two masses. Well, let's say that it is
05:22:20Paolo Guiotto: Proportional, so proportional.
05:26:320Paolo Guiotto: So… The product of the two masses, And,
05:31:980Paolo Guiotto: Inversely proportional to the square of the distance.
05:36:730Paolo Guiotto: Now, the square of the distance, if this mass m is in the origin.
05:42:870Paolo Guiotto: 0.000, and the other mass is at point XYZ, the distance is just the norm of XYZ, so this squared is the square of the distance.
05:55:770Paolo Guiotto: And the direction direction direction is the direction pointing to the origin.
06:07:720Paolo Guiotto: Pointing.
06:10:800Paolo Guiotto: 2…
06:12:670Paolo Guiotto: Direction zero, no? Now, what is the direction pointing to the origin from the point XYZ? Well, if you have the point XYZ, the vector XYZ is this one.
06:24:480Paolo Guiotto: it points to the point XYZ. So, if you want to point to the origin, you have to invert this, so this is just, say, if I imagine that the arrow is centered at point XYZ minus XYZ is the vector pointing
06:41:530Paolo Guiotto: to the origin in this way, so this is the vector minus XYZ.
06:49:340Paolo Guiotto: Now, if I want to combine all these, and I want to produce a force at point XYZ,
06:57:920Paolo Guiotto: which is proportional to that quantity. So it means that there is a factor, which is denoted with the letter G, which is called the gravitational constant.
07:11:50Paolo Guiotto: and I want that this force be proportional to that quantity and directed there, pointing to the origin. What I have to do is to take this.
07:21:410Paolo Guiotto: I have, minus G, M.
07:25:840Paolo Guiotto: times little m divided, if you want, norm of X, Y,
07:31:740Paolo Guiotto: Z squared, so this is saying I am proportional. So what I add here is now a vector that should point just to the origin, shouldn't have any length, so it should have length 1 pointing to the origin. Otherwise.
07:47:200Paolo Guiotto: I… I… I wouldn't have that it is proportional to that one. So this is a unit vector at point XYZ that points to the order. So, how do I do this? I take… well, let's say that we put minus here, XYZ,
08:04:590Paolo Guiotto: And we just divide by its length to make it a norm 1, a length 1 vector. So at the end, I get this formula, F of X
08:16:690Paolo Guiotto: YZ is minus G times capital M, little m divided by norm of X
08:27:450Paolo Guiotto: YZ. Now, this becomes cube times XYZ.
08:35:730Paolo Guiotto: So this is the formula for the gravitational field generated by…
08:44:190Paolo Guiotto: Well, this is the gravitational force generated by a point mass, capital M, at the origin over a mass M at point XYZ, no?
08:56:650Paolo Guiotto: So, as you can see from the point of view of a function, this is a function that takes a three-dimensional vector of XYZ and gives you a three-dimensional vector. That part here is just a numerical coefficient. Guys, it's color, I call this one.
09:15:20Paolo Guiotto: But of course, the coefficient depends on this poly.
09:19:810Paolo Guiotto: So force fields are, normally…
09:24:530Paolo Guiotto: vector fields, because if I have any force acting in space, I will have, at every point of space a vector that describes the force at that point. This formula is also used for electric force and other… many others.
09:44:280Paolo Guiotto: A second important example is, for example, the velocity
09:55:140Paolo Guiotto: field, huh?
09:58:450Paolo Guiotto: So imagine that I want to describe the movement of
10:03:750Paolo Guiotto: a flow of water in a river, so I may imagine that at each point, I have a little particle of water which is moving somewhere.
10:14:30Paolo Guiotto: So, for example, it could be interesting to know what is the speed, the velocity better, of this particle passing through this point. So, if we imagine that this is a plane.
10:28:360Paolo Guiotto: description, so there is, not three dimensions, just two, so this point is the point XY.
10:36:110Paolo Guiotto: At that point, XY, I may imagine that when a particle of water passes through that point, we'll have a certain velocity
10:45:220Paolo Guiotto: V, which is a vector, and this vector will be different depending on which point I am. Now, for example, this would be the velocity here, or maybe there are, I don't know, something like vortexes here.
10:58:710Paolo Guiotto: So I will have things like that.
11:02:200Paolo Guiotto: Or something spinning around, no? So I may imagine that the velocity field is a function V at point XY that is a vector with the two components.
11:15:430Paolo Guiotto: let's say V1XY,
11:18:830Paolo Guiotto: V2, XY, and these two components represent the velocity, the two components of the velocity, at that point, XY, okay?
11:31:390Paolo Guiotto: So this kind of function is, these are the example of vector fields. From the mathematical point of view, vector field is a function of this type.
11:41:70Paolo Guiotto: Now, the most important problem, one of the most important problems with vector fields is the following, and this problem arises from physics.
11:51:900Paolo Guiotto: So, it consists in determining what is called a potential for the vector field. So, given
12:02:960Paolo Guiotto: a vector field.
12:10:30Paolo Guiotto: Okay, boom.
12:12:330Paolo Guiotto: function of that, sir?
12:16:600Paolo Guiotto: determine… potential.
12:27:450Paolo Guiotto: Aw.
12:28:300Paolo Guiotto: F.
12:29:710Paolo Guiotto: What is the potential? That is a function.
12:34:250Paolo Guiotto: So let's introduce a definition. Definition.
12:40:750Paolo Guiotto: Potential.
12:46:590Paolo Guiotto: off.
12:47:650Paolo Guiotto: F.
12:49:870Paolo Guiotto: Back to a field defined on some domain.
12:54:750Paolo Guiotto: of RV into.
12:57:640Paolo Guiotto: I agree.
13:00:960Paolo Guiotto: is a function.
13:09:200Paolo Guiotto: Little f, so it is a numerical function of vector variable.
13:17:20Paolo Guiotto: defined on the main D.
13:19:800Paolo Guiotto: So, then in RD, with values in R, so it is a numerical function, Such that,
13:28:500Paolo Guiotto: the gradient of F At point X, huh?
13:33:70Paolo Guiotto: Now, if the gradient of F exists for this type of function, is what object?
13:40:750Paolo Guiotto: The gradient of F for a function of this type is… It belongs to…
13:47:140Paolo Guiotto: RD, no? Because it is the array made of the D partial derivatives. Now, what we want is that this be equal to the function, even function M.
14:00:260Paolo Guiotto: So, a function whose gradient is your assigned function at the left, is called the potential of the vector field F.
14:11:310Paolo Guiotto: We will see later why potentials are so important.
14:16:780Paolo Guiotto: But let's stay focused on this, let's say, which is a mathematical problem.
14:21:700Paolo Guiotto: I just want to notice that we already basically know something about this, because if the dimension D is 1,
14:31:490Paolo Guiotto: So, this means that the function f
14:34:830Paolo Guiotto: is a function from R1 to R1, so R to R is the usual function, numerical function of real variable.
14:43:820Paolo Guiotto: So, D contains in R into R.
14:48:720Paolo Guiotto: a potential… off.
14:52:400Paolo Guiotto: F.
14:53:690Paolo Guiotto: is a function little f, which is, again, a function of variable x, which is, in this case numerical, so it's an ordinary function.
15:04:220Paolo Guiotto: defined on domain, the real value there.
15:08:20Paolo Guiotto: Such that, now, if there is just one variable, the gradient reduces the derivative.
15:13:360Paolo Guiotto: F prime of x is equal to capital F of X.
15:18:270Paolo Guiotto: Now, we already know what is this problem, because given capital F, searching for little f, such that the derivative of F is capital F, means to look for a primitive of the function capital F.
15:31:800Paolo Guiotto: the solution… of, 2.
15:41:520Paolo Guiotto: this.
15:42:940Paolo Guiotto: problem.
15:44:750Paolo Guiotto: is, F of X.
15:48:30Paolo Guiotto: equal to a primitive of little f of x.
15:52:570Paolo Guiotto: And we know that, in general, this problem is not a unique solution, because if you add the constant and you…
15:59:590Paolo Guiotto: differentiate, that the constant disappear. So we may say that this plus a constant is again a solution.
16:07:970Paolo Guiotto: And these are all possible solutions when the domain D is elinkable.
16:13:90Paolo Guiotto: is this? If… the… is an interval.
16:24:820Paolo Guiotto: So we may say that if dimension is 1, the version in dimension 1 of this problem is a sort of well-known. You have seen last year this problem, basically. Now, let's see what happens with the functions of several variables. Let's see if it is the same or not, what can we say?
16:45:300Paolo Guiotto: Now… In general, I may notice that gradient F is equal to F,
16:56:320Paolo Guiotto: If and only if, so what should happen?
17:00:340Paolo Guiotto: Gradient of F is a vector made of partial derivatives, so there will be the derivative with respect to the first variable of F, derivative with respect to second variable of F, etc, derivative with respect to the last variable of F,
17:16:60Paolo Guiotto: And this must be equal to the vector F.
17:19:319Paolo Guiotto: So we may assume that since F is a vector field.
17:23:720Paolo Guiotto: It is a function FB to FB, as you can see, F of X will have certain components, say, F1, F2 in this case, no? So, if F, in general.
17:36:400Paolo Guiotto: If F is the vector made of components, F1… these are functions, okay? So F1 is a function of a variable X, which is an array here, F2 of X, etc.
17:52:50Paolo Guiotto: F, D, there are D components.
17:56:630Paolo Guiotto: Now, this must be equal to the array F1, F2, etc.
18:02:170Paolo Guiotto: FB.
18:03:930Paolo Guiotto: Now, two vectors are the same, if and only if the components are the same, and this is a system.
18:11:120Paolo Guiotto: Which is the following.
18:13:490Paolo Guiotto: The partial derivative with respect to the first variable of F, potential, must be equal to the first component of the field.
18:23:660Paolo Guiotto: The partial derivative with respect to the second variable of the potential must be the second component of the field, and so on. The partial derivative with respect to the last variable of F must be the last component of the field.
18:39:210Paolo Guiotto: Okay?
18:40:290Paolo Guiotto: So, if we want to solve the equation, gradient F equals capital F, I call it the equation because
18:49:360Paolo Guiotto: we have to determine this little f given F, capital F, so let's say that this is Known.
19:03:390Paolo Guiotto: And this is… the little f is unknown.
19:12:710Paolo Guiotto: I have to solve this system.
19:16:300Paolo Guiotto: Now, the first thing to do is to check on some example.
19:21:60Paolo Guiotto: What does it mean to solve this system? So let's take a very simple example.
19:28:440Paolo Guiotto: So, example, 1, which is the number 315,
19:34:520Paolo Guiotto: It's really one-star example, so very basic calculations.
19:40:190Paolo Guiotto: So… Determine D.
19:45:240Paolo Guiotto: potentials…
19:51:300Paolo Guiotto: of this vector field, which is a vector field in R2, so F of XY is by definition, YX.
20:04:180Paolo Guiotto: 4. XY in S2, which is the domain.
20:09:840Paolo Guiotto: Of this problem.
20:12:910Paolo Guiotto: So let's see what… what this means, huh?
20:18:150Paolo Guiotto: So we may say that F… is a potential.
20:23:490Paolo Guiotto: for Captain F.
20:26:600Paolo Guiotto: If and only if… well, of course, F is differentiable, because when we write gradient, we mean F is differentiable.
20:37:990Paolo Guiotto: And the… gradient of F is capital F.
20:44:730Paolo Guiotto: So this, in this case,
20:47:260Paolo Guiotto: As you can see, there is a number of equations equal to the number of variables. If I have D variables, the equations.
20:55:470Paolo Guiotto: If I have two variables, two equations. These two equations are the two partial derivatives. Here we use the letters X and Y for the two partial derivatives of our unknown function F, so this is function of XY, X, Y.
21:12:20Paolo Guiotto: So, the derivative with respect to X, this is this line here. The derivative with respect to the first variable must be equal to the first component of the field. The first component of the field is this one. So, this must be equal to Y.
21:28:190Paolo Guiotto: And this must be equal to the second component of the field. Of course, for every point XY of
21:35:600Paolo Guiotto: R2.
21:38:710Paolo Guiotto: Okay, so now, you see, it seems, not…
21:43:390Paolo Guiotto: say, immediate trivial problem, because we have two equations for the same unknown F.
21:50:10Paolo Guiotto: to this…
21:53:170Paolo Guiotto: would look as something like an impossible problem. Something that verifies two equations. But remember, this is a function, and in fact, it's not an impossible problem. Let's take the first equation. So let's look at this.
22:10:500Paolo Guiotto: Now, this says that the partial derivative with respect to X of this function f we are looking for is Y. Now, we know that the partial derivative is a derivative with respect to X as if the function is a function of X.
22:24:240Paolo Guiotto: So that's very similar to this problem.
22:27:210Paolo Guiotto: Looking for a function whose derivative with respect to x is something given.
22:32:650Paolo Guiotto: So, you would say that, okay, so the function f of xy
22:38:630Paolo Guiotto: should be a primitive in the variable X of this guy, Y, Plus a constant.
22:48:330Paolo Guiotto: Now, be careful, because in this problem, there is Y, which is variable, but respect to X is the parameter.
22:55:900Paolo Guiotto: So I may imagine that.
22:57:860Paolo Guiotto: That constant that I see there is a constant in X, but in general could be variable in Y.
23:05:880Paolo Guiotto: You see? Because if this is C of Y, when you do the derivative with respect to X, this is 0.
23:13:110Paolo Guiotto: Because it does not depend on X.
23:15:910Paolo Guiotto: Okay, now what is the primitive of this? Y is a constant for X, so I could carry outside, because it's like to have 3 in the primitive. So I could say this is Y, the primitive of 1 in variable X plus a constant of Y,
23:31:30Paolo Guiotto: Now, a primitive of 1 is X, so I get YX plus C of Y.
23:39:160Paolo Guiotto: Okay, so now I know that my F of XY should be this type of function. Should be YX plus
23:47:710Paolo Guiotto: And not yet determine the function C of Y. How can I determine now this guy, this C of Y?
23:55:200Paolo Guiotto: Well, I have a second equation here, so let's impose also the second equation. Now, we have a
24:02:50Paolo Guiotto: The first one says, no? The first equation says, if F is such that DXF is Y, then, necessarily.
24:10:400Paolo Guiotto: F must be YX plus a constant of Y. So we know that from the first equation, this must be F.
24:17:670Paolo Guiotto: Since we want that also the second equation be verified, we impose… now we have an F. Instead of solving also the second equation and then trying to match the two, we could say, well, since from the first we know that F must be like that, let's put into the second, let's see what happens. Now.
24:37:00Paolo Guiotto: If we put into the second, DYF equal X, this means that
24:44:970Paolo Guiotto: doing the derivative with respect to Y of this F, I get the derivative with respect of… well, let's write DY of YX plus C of Y. This thing must be equal to X. Now, this means that…
25:01:800Paolo Guiotto: The denial in respect of Y to Y of YX is X, plus the derivative in respect of Y of C of Y would be C prime.
25:11:210Paolo Guiotto: of wine.
25:13:340Paolo Guiotto: the ordinary derivative of C with respect to Y. This must be equal to X. So we can now cancel this, and we get that C prime of Y must be equal to 0.
25:26:720Paolo Guiotto: So what does it mean there? Since C prime of Y is 0, C, as function of Y is
25:35:180Paolo Guiotto: is a constant, but be careful that here, it does not mean that it is a constant in X.
25:40:400Paolo Guiotto: Because C of Y is already a function that does not depend on X, because it has been determined here, no? When we say plus a constant, which is independent of X, it's constant in X, but not in the remaining variable, perhaps, no? So it's just a constant.
25:56:850Paolo Guiotto: And therefore, we get that now F of XY takes this shape. X times Y plus constant, and the C is an arbitrary value for the constant.
26:09:180Paolo Guiotto: Okay, now, from the logical point of view, what we proved is…
26:13:960Paolo Guiotto: If F is a solution of that sequence system. So, in other words, if F is the potential of capital F, then F must be of this type.
26:27:470Paolo Guiotto: This does not necessarily mean that this is a potential. We have to do the vice versa to check that if F of this type, the gradient is that one.
26:38:160Paolo Guiotto: Okay? Now, we can easily check this.
26:42:10Paolo Guiotto: Now.
26:44:660Paolo Guiotto: But since…
26:47:50Paolo Guiotto: The derivative with respect to X of F is… when you do the derivative with respect to X, you get Y,
26:52:730Paolo Guiotto: the derivative with respect to Y of F is X, we have that this F is a potential All.
27:03:40Paolo Guiotto: Topical X.
27:06:80Paolo Guiotto: So, apparently, we solved the problem.
27:09:500Paolo Guiotto: There are a few details here.
27:12:140Paolo Guiotto: Now, I will, clarify a few points.
27:17:550Paolo Guiotto: That must be, understood.
27:20:740Paolo Guiotto: Now… hmm…
27:26:220Paolo Guiotto: Okay, before we see the details, let's try to see a second example. Now, so apparently, it seems easy, no?
27:34:410Paolo Guiotto: So, you see what we have done, no? We have the field, we've wrote the two equations, no, that are equivalent to the conditioned gradient ethical.
27:43:960Paolo Guiotto: capital F, so we have the two partial derivative of the potential, F, equal to the two components, and then we use the one of the two, for example, at first, integrating in X this identity, this equation.
27:58:850Paolo Guiotto: deriving a candidate and solving with that a known part, the C of Y, by using the second condition.
28:08:20Paolo Guiotto: At the end, we have,
28:11:880Paolo Guiotto: A class of functions, which are all potentials of the function capital F.
28:18:500Paolo Guiotto: Now, let's see a second example to see if we have understood the problem.
28:24:480Paolo Guiotto: Example 2, we take the same thing as before, but we do just an innocent modification, which is we change one of the sign of the two components.
28:37:830Paolo Guiotto: So, same problem.
28:43:570Paolo Guiotto: for this feed.
28:45:770Paolo Guiotto: F of XY.
28:48:230Paolo Guiotto: equal, for example, minus YX. Let's see what happens.
28:54:210Paolo Guiotto: So… F is a potential for… capital F,
29:04:270Paolo Guiotto: If and only if we have what? Gradient F equals capital F, That is…
29:12:880Paolo Guiotto: The derivative with respect to X of little f is the first component of the field, so here, minus Y.
29:20:50Paolo Guiotto: And derivative with respect to Y of F is the second component of the field, which is X.
29:25:510Paolo Guiotto: looks pretty much as the previous case. The previous case, we got derivative with respect to X of F equals Y, derivative with respect to Y of F equals X. It just…
29:35:420Paolo Guiotto: an apparently innocent change. So let's again take this equation.
29:41:440Paolo Guiotto: So we have that DXF of XY.
29:46:520Paolo Guiotto: equal minus Y. I repeat this point here once, again.
29:52:130Paolo Guiotto: Now, reminder, the partial derivative of a function is actually an ordinary derivative of that function, as if the function depends on that variable, and all the other variables are parameters.
30:07:890Paolo Guiotto: So, it's like I'm saying F' equals something, no? So, what is the F? So, F of XY will be F, primitive, which is actually a symbol, primitive of minus Y.
30:23:930Paolo Guiotto: in the variable X,
30:26:110Paolo Guiotto: So now forget that Y for X is a constant. It is another variable, plus a constant.
30:35:230Paolo Guiotto: Which is a constant in X, but since there are other variables, it might bend on the other variables, so let's write C of Y.
30:43:300Paolo Guiotto: So this means that the minus y can be written out of the primitive, because it's a constant.
30:49:870Paolo Guiotto: And then we have primitive of 1, which is X in the variable X. So we have this.
30:55:990Paolo Guiotto: Exactly as before.
30:57:880Paolo Guiotto: Now, imposing…
31:03:900Paolo Guiotto: DYF equals X, we get what?
31:08:850Paolo Guiotto: We do dy of minus YX plus C of Y.
31:14:120Paolo Guiotto: And we want that this be equal to X. What we got is…
31:18:840Paolo Guiotto: DY of minus YX is minus X,
31:23:410Paolo Guiotto: plus C prime Y equal to X.
31:27:840Paolo Guiotto: So now, this time, we got…
31:30:640Paolo Guiotto: C prime y equals to X.
31:35:220Paolo Guiotto: I may say, okay, so this means,
31:39:350Paolo Guiotto: C of Y is 2XY, no, because
31:44:700Paolo Guiotto: C prime as a function of Y as derivative 2X.
31:49:20Paolo Guiotto: So it should be 2XY plus a constant. What kind of constant? You may say, constant in X, but no. Well, actually, this is a nonsense, because C of Y is independent of Y, and you see the X here, how it can be?
32:07:750Paolo Guiotto: C of Y… does not… It depends.
32:15:530Paolo Guiotto: on X.
32:17:370Paolo Guiotto: Because it comes from this, right?
32:20:450Paolo Guiotto: So, it was built as a function, as a constant in X, maybe a function in the adults. But, so this means what, then?
32:30:890Paolo Guiotto: So, it means that this thing is simply impossible, because if the function c of y is independent of x, is a constant.
32:40:690Paolo Guiotto: Also, C prime of Y is a constant in X, and the right-hand side is not constant in X.
32:47:330Paolo Guiotto: So, it means that the second condition is impossible, and what does it mean this? It means that there cannot be any potential, because if you have a potential, then from the first equation, you get that this should be X, right?
33:03:570Paolo Guiotto: But by imposing the second condition, you get this. But this is impossible with a function C that is independent of X, because this says C prime is dependent of X, therefore also C must be dependent of X.
33:19:930Paolo Guiotto: So, it's a contradiction. So, this means that… There is not.
33:27:460Paolo Guiotto: F, such that gradient of F is this capital F.
33:34:200Paolo Guiotto: Now, why this example can be disturbing? I'm sure that probably it won't bother you too much, but you see what is going here.
33:45:950Paolo Guiotto: So here we have a field f equals YX.
33:49:870Paolo Guiotto: Dog.
33:51:520Paolo Guiotto: Apart for constants, we have a field where the two components are very easy functions, the coordinates themselves.
34:02:880Paolo Guiotto: Okay, this one has a potential, but just flipping one side to one of the two components of the field, I have a field with the same features, I have two very simple components for the field, but I do not have any potential.
34:23:810Paolo Guiotto: So this means that, for example, it's not a question of complexity, because this simple example shows you that you can have very simple fields without a potential. It's not a question of
34:38:489Paolo Guiotto: complexity, but it's a question of something else that we want to understand. Because the basic question we want to respond here is,
34:48:190Paolo Guiotto: To understand when
34:50:659Paolo Guiotto: Under which conditions can we say that a field has a potential? So, can we say something by looking the field
34:59:140Paolo Guiotto: And without necessarily determining the potential, because this argument here was easy to decide that there is not a potential, because we could solve the system.
35:10:470Paolo Guiotto: But we could show that the system cannot be solved by hands, no? We took the system, we solved the first equation, and then we derived this contradiction.
35:20:780Paolo Guiotto: But in general, this won't be the case, so…
35:24:30Paolo Guiotto: we need to understand is if… is there a test that can tell us this field has a potential, this field has not a potential, okay? So what is the reason behind this… this fact?
35:42:430Paolo Guiotto: Okay, now, there is an interesting remark.
35:47:280Paolo Guiotto: Which is the following.
35:49:510Paolo Guiotto: So, let's assume that… so… Letter F…
35:56:820Paolo Guiotto: Let's, again, return to the general discussion. So, we use F1…
36:03:200Paolo Guiotto: capital F1, capital FB for the components of F.
36:09:280Paolo Guiotto: B… Feel that.
36:15:120Paolo Guiotto: Weird, that… Well, then shout.
36:21:260Paolo Guiotto: By the way, we use, also, we borrow a terminology that comes from physics. We will return on later on this to understand where it comes from.
36:30:810Paolo Guiotto: a field, Well, let's say such a field.
36:42:560Paolo Guiotto: is to also… Hold up.
36:48:150Paolo Guiotto: concern of you.
36:57:140Paolo Guiotto: There is a reason, because potential… well, the opposite, in the case of a force field.
37:04:910Paolo Guiotto: If there is a potential, the opposite of the potential, so minus the potential, is what is called the potential energy.
37:12:930Paolo Guiotto: And it can be proved, we will see later this, that if you have a particle moving in a false field.
37:23:30Paolo Guiotto: with a potential.
37:25:380Paolo Guiotto: So there is a quantity which is called the mechanical energy, which is the sum of kinetic energy plus potential energy. This quantity here, along the trajectory, is a constant.
37:36:740Paolo Guiotto: So this expresses fact that something is preserved along the trajectors. And so that's why we say that these particular things are conservative, because… but, however, we will return later on this. For the moment, these are just labels we put. Okay, so conservative means that there exists a potential.
37:55:560Paolo Guiotto: So this means that, so…
37:59:280Paolo Guiotto: the J component of the field is the partial derivative with respect to the j variable of X.
38:09:990Paolo Guiotto: Now, suppose that the components of the field can be differentiated again. So, if…
38:18:550Paolo Guiotto: the components of the field, FJ, R,
38:24:260Paolo Guiotto: differentiable, all them, so for J, then 1 to D, all the companies are differentiable.
38:30:880Paolo Guiotto: Then we can talk about the derivatives of these components FJ.
38:36:500Paolo Guiotto: So we can write,
38:38:400Paolo Guiotto: the derivative with respect to the height variable of the jade component of the field. Now, this guy is what? Is the derivative of the it component
38:49:710Paolo Guiotto: respect… is the derivative with respect to the it-th variable of the jade component, which is the derivative, respect to the j variable of the potential.
39:00:80Paolo Guiotto: So it's a second derivative.
39:02:800Paolo Guiotto: It's a derivative, First, with respect to the jth variable, second, with respect to the i variable of F.
39:12:300Paolo Guiotto: Okay.
39:13:430Paolo Guiotto: And if we flip the indexes, so we do the derivative with respect to the j variable of the i component.
39:22:70Paolo Guiotto: This is DJ of… since the field is conservative, the ith component is the derivative with respect to the ith variable of the potential F. So this is, again, the second derivative of F in the order with respect to i and j variable.
39:41:230Paolo Guiotto: So these two quantities, you see.
39:44:270Paolo Guiotto: we have already mapped this when we computed the Asian metrics of a function f, no?
39:51:530Paolo Guiotto: And we noticed that In common cases, flipping the order of the two derivatives, you get the same result.
40:02:340Paolo Guiotto: I told you that this is not always true. This is true when these derivatives, these second derivatives, are continuous, and that is called… it's called the Schwartz Theorem. So, but if you apply this fact here, so recall that.
40:26:230Paolo Guiotto: If the second derivatives with respect to all possible
40:32:490Paolo Guiotto: pairs of indexes are continuous, then…
40:37:810Paolo Guiotto: The second derivative with respect to JIF is equal to the second derivative with respect to iJ of F, identically for every i and j. This is the Schwarz
40:57:710Paolo Guiotto: That's Sprunk.
41:00:170Paolo Guiotto: Watts.
41:02:160Paolo Guiotto: Here.
41:06:190Paolo Guiotto: So, if this happens, it means that this yields an identity on the components of the field. So…
41:15:710Paolo Guiotto: Since the second derivative with respect to JI, in this order, of F, is the derivative of FJ with respect to the i variable, first line, and the second one is the derivative of Fi with respect to the J variable, we get this
41:33:890Paolo Guiotto: Did this condition.
41:37:940Paolo Guiotto: D, FI.
41:41:860Paolo Guiotto: XJ, E42D, FJ, Respect to XI.
41:49:360Paolo Guiotto: for every indexes IJ from 1 to… Peace.
41:57:330Paolo Guiotto: So, what is… we are saying is the following factor. We proved, basically, this.
42:03:900Paolo Guiotto: we proved that.
42:07:590Paolo Guiotto: this.
42:09:940Paolo Guiotto: Fact.
42:12:180Paolo Guiotto: So, proposition.
42:17:990Paolo Guiotto: If… F.
42:20:770Paolo Guiotto: Equal components F1.
42:24:120Paolo Guiotto: to FD.
42:26:10Paolo Guiotto: Ease.
42:27:100Paolo Guiotto: conservative.
42:32:720Paolo Guiotto: So there exists a potential with… potential.
42:39:990Paolo Guiotto: F in C2, Then, necessarily, this condition must be verified.
42:49:240Paolo Guiotto: DFI with respect to XJ must be equal to DI with respect to FJ for every indexes IJ from 1 to D.
43:01:300Paolo Guiotto: And this identity will be an identity at every point where I have a potential, so let's say on B.
43:10:330Paolo Guiotto: say, conservative on D, with potential F, which is C2, then this these derivatives must coincide.
43:22:490Paolo Guiotto: You see?
43:23:780Paolo Guiotto: This is a sort of necessary condition that must be verified in order you are conservative, because look at the direction of the arrow is this one. It says, if the field is conservative, then this happens.
43:42:480Paolo Guiotto: It does not say that if this happens, the field is conservative, but it says that to be conservative, this must happen.
43:51:570Paolo Guiotto: Let's take the example we have seen before, where we do not have the potential. Take this one.
43:59:390Paolo Guiotto: Let's see what happens here. Example… If, F,
44:06:160Paolo Guiotto: is the field. We have minus YX, right? This was the field. So this is F1, this is F2, okay?
44:16:940Paolo Guiotto: So we need, to verify If these conditions are verified.
44:23:110Paolo Guiotto: Now, notice that these conditions here that I'm boxing now.
44:29:80Paolo Guiotto: Are interesting only when the two indexes are different.
44:34:160Paolo Guiotto: No? Because if i is equal to J, what this says, that the derivative of Fi with respect to the ith variable is equal to the derivative of FI with respect to the ith variable, so it's a triviality, okay?
44:49:590Paolo Guiotto: So this is interesting when the variables are different, because in that case, you have, for example, the derivative of the first component with respect to the second variable must be equal to the derivative of the second component with respect to the first variable.
45:04:30Paolo Guiotto: So that's interesting.
45:05:760Paolo Guiotto: Now, the condition…
45:10:270Paolo Guiotto: DY of F1 equal to DX of F2 is equivalent of… what is DY of F1?
45:23:360Paolo Guiotto: minus 1, this must be equal to DX of F2. What is DX of F2? 1.
45:29:760Paolo Guiotto: Is that true? No.
45:32:390Paolo Guiotto: So, since this is false, you see that that condition in red is not verified.
45:39:930Paolo Guiotto: And so, in particular, that field cannot be conservative. Otherwise, this condition would be verified, okay?
45:48:280Paolo Guiotto: So… F… Cannot.
45:56:40Paolo Guiotto: the… own server.
46:01:500Paolo Guiotto: Although wise.
46:05:750Paolo Guiotto: condition star.
46:09:880Paolo Guiotto: Whoa.
46:13:800Paolo Guiotto: verified.
46:17:960Paolo Guiotto: So now, this is interesting, because it provides a test
46:23:400Paolo Guiotto: Which is, for the moment, only a necessary condition that we can easily check, instead of all this argument we have done here.
46:33:210Paolo Guiotto: Because this here was easy, because the components of the field were easy. But normally, checking when these derivatives, these crossed derivative, are equal, it's easier because you have just to compute derivatives, so you don't have to compute integrals.
46:50:160Paolo Guiotto: This method is based on computing integrals, and you know that we cannot compute any integral in general. Our possibilities are extremely limited.
47:01:630Paolo Guiotto: While we can differentiate whatever, so it's much easier to check this condition. So this is a good condition, seems to be. So let's give a name to this definition.
47:17:40Paolo Guiotto: a field… Now, you may wonder where this name comes from.
47:23:950Paolo Guiotto: Be patient, because we will give a meaning to these words in a… Phew.
47:32:00Paolo Guiotto: minutes, let's say. F in F, with components F1, FD.
47:42:20Paolo Guiotto: is… cold.
47:45:970Paolo Guiotto: irrotational.
47:56:80Paolo Guiotto: If… the… Tests.
48:04:160Paolo Guiotto: Awful.
48:05:430Paolo Guiotto: the… It is… this is the name, how it's called, the cross, the… derivatives.
48:20:40Paolo Guiotto: codes. And what is the test of the cross derivatives? Is D, doesn't matter the order of letters, DIFJ is identically equal to DJFI
48:34:880Paolo Guiotto: on D for every indexes I, J, from one
48:41:120Paolo Guiotto: to D, and as I said, since when I is equal to J, that's a trivial identity, no? If you have the same index, i is equal to J, that DIFI equals DIFI. That's evident, no? It's not… it's something which is always verified.
48:59:390Paolo Guiotto: The condition is non-trivial when the indexes are different, so… this.
49:07:430Paolo Guiotto: condition… is non-trivial.
49:15:960Paolo Guiotto: when… I is different from J. So, that's the case that we should check in examples.
49:27:320Paolo Guiotto: Now, yup?
49:39:200Paolo Guiotto: No, we cannot say.
49:42:170Paolo Guiotto: Unfortunately.
49:46:690Paolo Guiotto: Yes, exactly. This is the spirit of this condition. This condition tells you that, for sure, the field is not conservative if the condition is not verified.
50:01:70Paolo Guiotto: So it's a… it's a test to exclude.
50:04:920Paolo Guiotto: Okay? The existence of a potential. If it is not verified, definitely that cannot be a potential.
50:13:420Paolo Guiotto: If it is verified, unfortunately, we will see that it is not sufficient, always, at least, to ensure the existence of a potential, but adding something, it can become also sufficient. Okay, we will see later.
50:28:610Paolo Guiotto: Okay, now it's a good moment to take a break, so let's take 10 minutes, and then we will start. No, 10 minutes to… we have just 5 minutes today. So, let's take until 9.35.
50:50:420Paolo Guiotto: insist according… Okay, so what you see here… is… This is not recording.
51:01:890Paolo Guiotto: Yes, please report.
51:04:890Paolo Guiotto: it says that it is recorded. Okay, so what you see here is the plot.
51:11:440Paolo Guiotto: as I said, you… we… we…
51:14:510Paolo Guiotto: hardly can have plots of these things. This is… this is not the plot of the function of the vector field. What we see here is, like the picture I gave you for the flukes of water. So imagine that at every point, XY, you have an arrow.
51:32:740Paolo Guiotto: which is the value of the field. I cannot plot a function f from F2 to F2, because I would need four axes to plot this function. So this is not the plot of the function.
51:44:940Paolo Guiotto: is rather another figure, so here you see a teach point of the plane, an arrow. That arrow is the value of the field at that point. This is the field YX, the first one for which we determine the potential, okay?
52:00:180Paolo Guiotto: So look at this figure. Look at what happens when I put the minus here. So this is now the plot of minus YX. So this is the case of the field for which we do not have a potential.
52:13:170Paolo Guiotto: Now, you see that this is rotating somehow, in some sense.
52:17:690Paolo Guiotto: So, the fact that it is rotating around the point is the problem, is the why there is not the potential, and why we call that condition irritation. Irrotational when the rotation is zero means, and when rotation is zero, more or less means there is a potential. In this case, there is a rotation, as you can see.
52:39:660Paolo Guiotto: It's like a fluid which is rotating around.
52:42:640Paolo Guiotto: Avoid it, on that point.
52:45:570Paolo Guiotto: So, that's, why there are these names. Now, of course.
52:51:210Paolo Guiotto: This does not explain why, in this case, there was not a potential. We had just tested the condition that the field is not irritational.
53:00:910Paolo Guiotto: So, since this is an essential condition to be conservative, you can say it is not conservative, okay? But it does not fully explain this figure.
53:11:40Paolo Guiotto: However, I wanted to show you these two figures to show you the difference between this field, which is irrotational, and actually it has a potential.
53:22:270Paolo Guiotto: So, since we proved that there is a potential, this one is irrotational.
53:27:830Paolo Guiotto: And this one, which we proved that there is not a potential, and we discovered it is also irritation, okay? This was just to…
53:37:190Paolo Guiotto: show you this. Now, I want to finish with this, so… Let's see… share screen…
53:49:730Paolo Guiotto: Okay, now,
53:56:430Paolo Guiotto: I cannot…
54:03:430Paolo Guiotto: Okay.
54:04:710Paolo Guiotto: Now, we say that this condition seems to be a good condition to test, at least to exclude the case when the field cannot have a potential.
54:17:90Paolo Guiotto: Now, we may wonder if it is a sort of if and only if, no? So we say that if there is a potential, if the field is conservative, it must be rotational. This is what this implication says in a few words.
54:33:20Paolo Guiotto: Is the vice versa true?
54:35:980Paolo Guiotto: Unfortunately, this is false, so, we may say that. Conservative?
54:44:680Paolo Guiotto: F.
54:46:220Paolo Guiotto: conservative.
54:50:280Paolo Guiotto: So this means that F is the gradient of some
54:54:350Paolo Guiotto: function F, who is a potential, implies F irrotational.
55:05:680Paolo Guiotto: So this is test of the cross derivatives, words, so DIFJ identically equal to the JFI.
55:17:860Paolo Guiotto: on the… Also, this song.
55:24:540Paolo Guiotto: But this one is unfortunately false in general.
55:30:650Paolo Guiotto: And let's see an example.
55:36:410Paolo Guiotto: So take this field,
55:38:650Paolo Guiotto: Still a playing field, FXY.
55:41:790Paolo Guiotto: equal.
55:44:600Paolo Guiotto: again.
55:45:710Paolo Guiotto: No excuse.
55:52:310Paolo Guiotto: Okay, take this… It's similar to the minus YX. In fact, you see that there is minus YX.
56:01:440Paolo Guiotto: But, you, you can see this, either putting this operation into the coordinates, I prefer to write outside, like this, 1 over X squared plus Y squared. This is a scalar factor. If you want to see into the components, these are minus Y divided X squared plus Y squared.
56:21:320Paolo Guiotto: and the X divided the X squared plus Y squared.
56:26:950Paolo Guiotto: Now, this cannot be defined at 00, so XY will be on the full plane R2, except .00.
56:39:770Paolo Guiotto: Now, if we take this field, we can prove that. We show that.
56:49:110Paolo Guiotto: this field, F,
56:55:150Paolo Guiotto: F.
56:56:120Paolo Guiotto: is irrotational.
57:06:160Paolo Guiotto: But, to… F… is not.
57:15:120Paolo Guiotto: conservative.
57:21:420Paolo Guiotto: So, unfortunately, you cannot say that it is conservative.
57:26:330Paolo Guiotto: even though if it is irrational. If this were true, we would have…
57:31:270Paolo Guiotto: an easy check to do to say if a field is conservative, so it has a potential or not. We should just checking the cross derivatives are the same, okay?
57:44:180Paolo Guiotto: And this would be easy, because it's just a matter of completing a few derivatives. Unfortunately, this example shows that this is not true. Let's see why.
57:55:510Paolo Guiotto: So, first, let's check 1. And as you will see, even if the calculation, if the field is not trivial, let's say, this calculation is really trivial.
58:07:620Paolo Guiotto: So, in this case, F has just two components, F1 and F2.
58:15:850Paolo Guiotto: Which are these ones. This is F1.
58:19:780Paolo Guiotto: This is F2.
58:22:270Paolo Guiotto: So the cross, the test of the cross derivative reduces just to checking that the derivative of F1, respect to the second variable.
58:33:90Paolo Guiotto: is the same of the derivative of F2 with respect to the first value.
58:37:860Paolo Guiotto: Because the other ones are derivative of F1 with respect to X1 equals derivative of F1 with respect to X1, which is trivial.
58:45:260Paolo Guiotto: As well as the derivative of F2 with respect to X2 equal derivative of F2 with respect to X2. So these are trivial, no? Because I'm saying? Something is equal to itself. What is non-trivial is when the two indexes are different.
58:59:90Paolo Guiotto: In the case of a plane field, there is only one possibility. One index is 1, and the other is 2, so one variable is X, and the other is Y. So, D…
59:12:950Paolo Guiotto: Best.
59:15:930Paolo Guiotto: Fall.
59:18:00Paolo Guiotto: Cross.
59:21:110Paolo Guiotto: derivatives?
59:27:490Paolo Guiotto: reduces 2…
59:32:690Paolo Guiotto: So, I take F1, which is the first component, I derive with respect to the second variable, and I have to check if this is equal to the derivative of F2 with respect to the first variable.
59:44:430Paolo Guiotto: So let's do this calculation. At the left, we have DY of the component is minus Y over X squared plus Y squared, so minus Y over X squared plus Y squared.
59:58:350Paolo Guiotto: So this is minus. We do this derivative, as if this is a function of Y, we do not look at X. X is a constant, so we are
00:07:850Paolo Guiotto: The, square of the denominator
00:11:700Paolo Guiotto: Then we have derivative of numerator, which is 1 times the denominator, X squared plus Y squared, minus the denominator, times the derivative of the denominator with respect to Y. So derivative is 2Y.
00:27:710Paolo Guiotto: So, at the end, we get X squared plus Y squared minus 2Y squared, so minus X squared minus Y squared over X squared plus Y squared squared.
00:44:560Paolo Guiotto: Okay? This is the derivative with respect to Y of the first component.
00:49:20Paolo Guiotto: What about the derivative with respect to X?
00:52:360Paolo Guiotto: of the second component. So, it is the DX of second component is X over X squared plus Y square.
01:01:130Paolo Guiotto: X over X squared plus Y squared. Well, if I'm smart, I don't need to do the calculation, because it's exactly the same, as you can see.
01:12:280Paolo Guiotto: Exchanging letter X with the letter Y, and I do not have the sine minus. So if you look at the end, I will have not the minus.
01:20:960Paolo Guiotto: So I want to have this and this to be Y squared minus X squared.
01:25:760Paolo Guiotto: It is exactly this one. However, if you don't see, let's do the calculation.
01:31:650Paolo Guiotto: So, we have the square of denominator.
01:37:00Paolo Guiotto: Then I have derivative of numerator, 1 times denominator, X squared plus Y squared, I'm now differentiating in X minus numerator times derivative of the denominator, which is 2X.
01:50:70Paolo Guiotto: So at the end, I have X squared plus Y squared minus 2X squared, and so this gives Y squared minus X squared divided X squared plus Y squared squared.
02:04:310Paolo Guiotto: As you can see, the two are the same.
02:07:300Paolo Guiotto: whatever is XY, apart for XYZ. So, for every point, X, Y,
02:15:750Paolo Guiotto: Apart for the unique point, I cannot take 00, so this means for every point, XY, that belongs to the domain for this field.
02:25:790Paolo Guiotto: So, the conclusion is that, this field, F… is irritational.
02:34:750Paolo Guiotto: Okay?
02:36:620Paolo Guiotto: Now, we checked that this is not conservative.
02:40:40Paolo Guiotto: And for the moment, the unique thing we have is… go back to the definition. So we have to say, assume that it is conservative. Let's see what happens, okay?
02:50:450Paolo Guiotto: So, F is… conservative.
02:55:400Paolo Guiotto: If and only if…
02:57:450Paolo Guiotto: F is the gradient of some function f, that is, there exists a function f such that it exists an F.
03:07:20Paolo Guiotto: function of XY.
03:09:800Paolo Guiotto: Such that the derivative with respect to X of F, XY,
03:14:630Paolo Guiotto: Well, this function must be defined for XY in the domain B,
03:18:800Paolo Guiotto: which is the plane, are to accept the origin.
03:25:600Paolo Guiotto: Such that the derivative with respect to X of this is the first component, so it is minus Y divided X squared plus Y squared.
03:35:440Paolo Guiotto: And the derivative with respect to Y of this F is the second component, which is X divided by X squared plus Y squared. This for every XY in the domain D, so for every XY apart for the .00.
03:52:340Paolo Guiotto: And now, how do we proceed? We look at the first equation, the two are the same, as you can see, and we say this. So, if this is true, we can say that
04:03:440Paolo Guiotto: with a little K, as you will see, but we can say that F of XY must be a primitive in the letter X,
04:13:420Paolo Guiotto: of the component minus Y divided the X squared plus Y squared, plus a constant that is a constant in X, but it can be a non-constant in Y.
04:27:550Paolo Guiotto: Now, We have to compute that primitive in variable X, it's not too complicated.
04:35:00Paolo Guiotto: We can take out the minus Y, which is a constant, then we have this, 1 over, let's write this way, Y squared plus X squared dx.
04:44:500Paolo Guiotto: This reminds of what?
04:48:380Paolo Guiotto: More or less.
04:51:400Paolo Guiotto: Huh?
04:52:700Paolo Guiotto: Yes, the arctangent. If you put Y equal 1, this is 1 over 1 plus X squared, which is arctangent of X. Now, in general, what could I do? I could factorize the Y, okay? So, here I should be careful, because the Y could be 0.
05:10:250Paolo Guiotto: If Y is 0, the story changes. So let's say for Y different from 0,
05:14:930Paolo Guiotto: This is minus Y. I take out Y squared, so it becomes here. I get primitive of 1 over 1 plus, when I factorize this Y, this goes into X over Y squared dx.
05:31:350Paolo Guiotto: Now, this is one of the Y.
05:34:340Paolo Guiotto: I give back this 1 over Y to the X here, I write this like this, and I change variable. I put then plus C of Y,
05:45:420Paolo Guiotto: Or if you want, you can, you can, you don't need to do all this,
05:50:390Paolo Guiotto: Because if you look at this, this is the arctangent
05:56:100Paolo Guiotto: a tangent of X divided Y.
05:59:760Paolo Guiotto: There's at least for Y different from 0.
06:04:190Paolo Guiotto: Okay? For y equals 0, however, this integral would be 0, so I would have only the function C of Y.
06:11:430Paolo Guiotto: So a constant C of 0.
06:13:830Paolo Guiotto: So, I have this arc tangent of X divided Y plus C of Y, this for Y different from 0.
06:27:450Paolo Guiotto: for Y equals 0, FX0.
06:33:270Paolo Guiotto: Well, if I put Y equals 0, this stuff becomes 0, because you see, when you put Y equals 0 here, you get 0 divided X squared, so you get 0. You are integrating 0, you get 0. So this will become a constant.
06:51:870Paolo Guiotto: in Y, but Y is 0, so it's C of 0.
06:56:750Paolo Guiotto: So we got this F of XY, Equal arctangent
07:04:660Paolo Guiotto: of X over Y plus C of Y for Y different from 0.
07:11:940Paolo Guiotto: And, C of 0 for Y equals 0.
07:16:730Paolo Guiotto: No.
07:21:430Paolo Guiotto: what do we do with this? This is the candidate. It's like, in the simple example.
07:27:350Paolo Guiotto: after the first, the first, let's say… I don't like to say integration, because integrals is another story. These are primitives, so the first calculation, the solution of the first equation. Now, what we do, impose… we impose the second equation, okay?
07:47:230Paolo Guiotto: So here, now, imposing…
07:54:140Paolo Guiotto: That.
07:55:970Paolo Guiotto: The partial derivative with respect to Y of F be equal to
08:00:530Paolo Guiotto: Now, we use the second equation, this one.
08:04:180Paolo Guiotto: be equal to X over X squared plus Y squared, so…
08:08:580Paolo Guiotto: X over X squared plus Y squared, let's see what we get. We get…
08:16:200Paolo Guiotto: So, of course, we will differentiate this one. So, dy of f is the derivative of this arctangent of X over Y.
08:29:399Paolo Guiotto: plus C of Y.
08:33:410Paolo Guiotto: We get, what? One other…
08:37:380Paolo Guiotto: We have the derivative of the tangent, so 1 plus the square, X over Y to power 2. Then there is the derivative of X over Y with respect to Y.
08:48:490Paolo Guiotto: And this is what? So you see derivative with respect to Y of X of the argument.
08:54:960Paolo Guiotto: Of course, down here, we will have the C prime of Y, right?
09:00:160Paolo Guiotto: Now, what is the derivative with respect to Y of X over Y? That X can be written outside, it's just a constant, so we put here X. Derivative with respect of Y, of 1 over Y is…
09:22:220Paolo Guiotto: Minus…
09:27:10Paolo Guiotto: One… Like this.
09:34:979Paolo Guiotto: Now, that's correct. Sorry, if you have any other… that's not correct. Okay, so we have, at the end, let's put the minus here. So, we have minus…
09:47:180Paolo Guiotto: Let's do the common denominator. Put Y squared here, we have Y squared plus X squared down here, then we have another 1 over Y squared plus C prime y, okay? We simplify this with this, and we get…
10:01:620Paolo Guiotto: minus X divided Y squared plus X squared plus C prime of Y. That's the derivative with respect to Y of F, okay? We got this.
10:15:590Paolo Guiotto: Right?
10:16:800Paolo Guiotto: And what should be equal?
10:19:320Paolo Guiotto: It should be equal to this.
10:21:650Paolo Guiotto: So let's… this should be equal to X over X squared plus Y squared.
10:28:530Paolo Guiotto: But now, as you can see, it happens something similar to the other example, because this, if you carry on a unique side, says that C prime of Y must be equal the two sums, and we have 2X over X squared plus Y squared.
10:45:330Paolo Guiotto: Which is impossible, because that function at left is constant in X. It's just a function of Y. Whatever it is, is constant, is independent of X, so it's a constant, and the right-hand side is not constant in X.
10:59:280Paolo Guiotto: So this is, constant.
11:03:480Paolo Guiotto: In fact, this is… not.
11:07:530Paolo Guiotto: constant.
11:09:460Paolo Guiotto: in X.
11:11:140Paolo Guiotto: apart exceptions. So, like, if I take only X equals 0, that's become a constant. But this means only on the y-axis, but in the domain.
11:21:890Paolo Guiotto: R2 minus 00, so when I allow X and Y be everywhere on the plane except the .00, the right-hand side is not constant, I hope that you see that.
11:32:530Paolo Guiotto: Okay?
11:33:620Paolo Guiotto: And so, a constant cannot be equal to something which is not constant. This means that that identity is impossible.
11:44:420Paolo Guiotto: But then we have a contradiction, because we said, suppose that there is a potential.
11:51:760Paolo Guiotto: So there is a function f that… such that each gradient is our field, capital F. This means that derivative with respect to X of F is that component, the green equation, and the Y is the other component, the blue equation.
12:05:550Paolo Guiotto: We took the first one, we said, then, necessarily, that F must be, after calculations, this thing, arctangent of X of Y plus C of Y, at least for Y different from 0. Y equals 0 is the x-axis.
12:19:360Paolo Guiotto: So, for every point in the plane, apart for those of the x-axis, that is the function. On the x-axis, we get just a constant.
12:28:390Paolo Guiotto: But, if we take this function, and we impose now this second condition, that this function must verify to be a potential.
12:37:760Paolo Guiotto: So, if we compute the DY alpha.
12:40:990Paolo Guiotto: taking Y different from 0, which is basically everything except the x-axis.
12:47:110Paolo Guiotto: we get this, this derivative, and this should be equal to X over X squared plus Y squared. This means that C prime of Y should be equal that right-hand side.
13:00:150Paolo Guiotto: But, on the plane, except the y equals 0, so the x-axis, the right-hand side is not constant.
13:09:60Paolo Guiotto: is constant only when X is 0, so when you are on the y-axis, but that's just a little part of the domain. Domain is everything, except the origin, and the right-hand side is non-constant, the left-hand side is necessarily constant, because
13:24:340Paolo Guiotto: arose as a function only of Y, independent of X, so it's constant. Whatever is the way it depends on Y, it's constant in X. And if you read this, which is constant in X, must be equal to that, which is not constant in X, it means that you have a logical contradiction. And this means what that initial
13:44:70Paolo Guiotto: the initial, assumption that is, suppose that this feed is conservative, leads to a contradiction, means that this cannot be true, okay? So.
13:57:50Paolo Guiotto: So… since… F.
14:03:360Paolo Guiotto: conservative.
14:05:830Paolo Guiotto: implies… a contradiction.
14:13:740Paolo Guiotto: We did use that.
14:15:290Paolo Guiotto: F.
14:16:880Paolo Guiotto: pan out.
14:19:690Paolo Guiotto: B.
14:21:10Paolo Guiotto: conservative.
14:23:860Paolo Guiotto: Okay, so this finishes our example. So this example has shown what? Has shown that we can have… the field is rotational, but not conservative.
14:36:110Paolo Guiotto: That's a sort of disappointment, because it would be nice if we have a test to say directly on a field, this is conservative, this is not conservative, an easy test.
14:49:970Paolo Guiotto: A test like this one, that, as you can see, it's a direct test, that means computing derivative should be a test like that.
14:59:410Paolo Guiotto: Okay, so let's, for the moment, we have still 15 minutes. Let's see some of the exercises.
15:07:730Paolo Guiotto: The exercises are more or less on the… yeah.
15:22:230Paolo Guiotto: 3 is the same, because the test… the test of the cross derivative
15:27:610Paolo Guiotto: is general for any number of variables. Of course, if you have 3 variables, for example, so X, Y, Z to be clear.
15:36:370Paolo Guiotto: remind always that these conditions are meaningful when the two indexes are different. So, let's write here. Maybe we will see later. For the moment, we will focus mostly on examples with two functions.
15:52:200Paolo Guiotto: evident simplicity, but remark… If,
15:59:100Paolo Guiotto: I have a field, F, with a field of three variables, XYZ,
16:06:800Paolo Guiotto: with three components, so these three components will be F1 over XYZ. Then there will be a second component, function of XYZ, and the third component, again, function of XYZ, right? This is the shape of a generic vector field in R3.
16:27:190Paolo Guiotto: The… Fast.
16:31:670Paolo Guiotto: off.
16:32:740Paolo Guiotto: crossed.
16:36:80Paolo Guiotto: derivatives.
16:40:590Paolo Guiotto: So that one that says DIFJ must be identically equal to DJFI,
16:47:860Paolo Guiotto: for all I and J, in this case, 1 to 3.
16:52:470Paolo Guiotto: This is meaningful when the two indexes are different, and let's see how many conditions we get.
17:00:440Paolo Guiotto: So, boils… Down… tools.
17:07:70Paolo Guiotto: Now, how many paths we can do with J123 with these indexes? With the two indexes different, okay?
17:17:140Paolo Guiotto: So, for example, I can do D1, F… I have… I need to have different indexes, so F2, and this must be equal to D2 of F1.
17:28:120Paolo Guiotto: A second pair is D1F3 equal to D3F1.
17:34:20Paolo Guiotto: And then there is another possibility, which is D2F3.
17:38:270Paolo Guiotto: equal D3F2.
17:40:720Paolo Guiotto: You see?
17:41:940Paolo Guiotto: So we get these three conditions. That must be verified together.
17:47:690Paolo Guiotto: Okay?
17:49:330Paolo Guiotto: Because this means for every, for every, not just the first condition, and not the second and the third, the three conditions. So if one fails, for sure, the test fails.
18:04:260Paolo Guiotto: Okay, there are a number of exercises, so…
18:15:380Paolo Guiotto: Okay, let's start with some, easy exercises. So, exercise 3, 4, 1. This is a very simple exercise. Simple because there are not
18:27:610Paolo Guiotto: particular difficulties with the components. So let's do the number 4, which is a field in R3.
18:35:930Paolo Guiotto: F is the field XYZ equal
18:39:850Paolo Guiotto: Y plus Z, X plus Z, X plus Y.
18:46:230Paolo Guiotto: for XYZ in arten.
18:53:510Paolo Guiotto: So… determine… if F… is irrotational.
19:03:900Paolo Guiotto: And conservative.
19:07:520Paolo Guiotto: And, in this case.
19:14:620Paolo Guiotto: Computer.
19:20:520Paolo Guiotto: 14 shells.
19:26:100Paolo Guiotto: Okay, so let's see here…
19:29:130Paolo Guiotto: just this test we wrote here, no? So these are the three components, capital F1, capital F2, capital F3.
19:38:630Paolo Guiotto: Okay, so, F… Ease.
19:42:470Paolo Guiotto: irrotational.
19:45:220Paolo Guiotto: If and only if…
19:47:300Paolo Guiotto: So, we rewrite, we take the first component, D, F1, respect not to the first variable. Two indexes must be different. So, this will be, for example, second index, this must be the same of D1, F2.
20:03:50Paolo Guiotto: So D, F1, or respect to the third variable, D, F3 with respect to the first variable.
20:10:970Paolo Guiotto: DF2 with respect to the third variable, equal to DF3 with respect to the second variable. These are the three conditions.
20:19:60Paolo Guiotto: Let's see what… what do they ask. So, D…
20:23:110Paolo Guiotto: 2. 2, of course, here means Y, okay? So the three variables are X, Y, Z. So what is D2 of F1?
20:35:520Paolo Guiotto: the derivative with respect to Y of, so this is DY of the second comp… the first component is Y plus Z.
20:44:780Paolo Guiotto: This must be equal to DX of F2X plus Z.
20:50:360Paolo Guiotto: Then, the second is D3, so this DZ of F1, again, Y plus Z. This must be equal to D1, so DX, of the third component, which is X plus Y.
21:04:640Paolo Guiotto: Last, we have the derivative of the second component with respect to the third variable, so X dz of X plus Z equal to the derivative with respect to the second variable of the third component, X plus Y.
21:20:30Paolo Guiotto: So this is what we have to check. Now, what is the Y of Y plus Z? 1.
21:27:130Paolo Guiotto: What is the X of X plus Z? 1. So is it 1 equal to 1? Yes, so this is checked. What is DZ of Y plus Z? It's again 1.
21:38:60Paolo Guiotto: What is BX of X plus Y? It's 1, so this one is checked.
21:42:710Paolo Guiotto: And therefore, the same here, 1 equals 1, also this one is verified.
21:47:910Paolo Guiotto: So you see that three conditions are verified. We conclude that F is irrotational.
21:56:140Paolo Guiotto: These are always verified, whatever is the point XYZ, on the domain V that, for this example, is at 3.
22:05:630Paolo Guiotto: However, we know irritational is not sufficient to say, for the moment, conservative. So, if we want to determine if it is conservative, we have to solve the equation for the moment.
22:17:170Paolo Guiotto: So, F is… conservative.
22:22:270Paolo Guiotto: If and only if F is the gradient of a function f, where this function is a function of three variables, X, Y, Z, but numerical.
22:34:50Paolo Guiotto: And this means that we have a…
22:36:940Paolo Guiotto: a system. So, the derivative with respect to X of F, derivative with respect to Y of F, derivative with respect to Z of F. These will be the three components of the field.
22:47:760Paolo Guiotto: So we have to go back to the field. The first component is Y plus Z, so we put Y plus Z here.
22:55:590Paolo Guiotto: then DY of F will be X plus Z,
23:02:790Paolo Guiotto: And the Z of F will be X plus 1.
23:07:580Paolo Guiotto: Now…
23:08:690Paolo Guiotto: At this stage, we have to integrate this. It is indifferent which one we take to start, because they are similar. If you see. There is the derivative of F with respect to a certain variable, and the right-hand side, you don't have the variable.
23:25:70Paolo Guiotto: Okay, so they are the same. Now, let's take the first one. Maybe the novelty here is that we have three variables. Let's see what happens. It's about the same. So, from this, I get that my function
23:38:200Paolo Guiotto: F, X, YZ. This says that if you look as a function of X, the derivative of F is that quantity. So.
23:46:730Paolo Guiotto: F will be a primitive of that quantity in the variable X,
23:52:980Paolo Guiotto: plus a constant, which is a constant in X.
23:56:180Paolo Guiotto: But…
23:57:260Paolo Guiotto: it might not be a constant in the remaining variables. Here, there are two remaining variables, okay? So this means that this constant is a constant.
24:06:340Paolo Guiotto: but function of Y and Z, constant in X, but function, or possibly a function of Y and Z. Now, this YZ is a constant for the integration here, so we got Y plus Z times X plus constant YZ. Okay.
24:23:880Paolo Guiotto: So now what do we do with this? We take this and we plug into the second equation, okay? Since the second must be verified.
24:32:840Paolo Guiotto: Now… imposing.
24:40:580Paolo Guiotto: DYF equal X plus Z, we get…
24:47:240Paolo Guiotto: So what is the DY of F?
24:50:560Paolo Guiotto: What is the DY of this?
24:56:770Paolo Guiotto: So, at left, we have X, X plus…
25:08:510Paolo Guiotto: What should I write here?
25:12:350Paolo Guiotto: I have to do this, right?
25:14:880Paolo Guiotto: DY of. Well, if you want YX plus ZX plus C of YZ.
25:24:220Paolo Guiotto: All this must be equal to X plus Z.
25:28:820Paolo Guiotto: So what do you say? Yeah? This is equal to… X…
25:39:560Paolo Guiotto: What is C prime? C is a function of two variables.
25:43:830Paolo Guiotto: It's not the… the derivative with respect to Y.
25:48:590Paolo Guiotto: of CYZ is equal to X plus Z. This is…
25:55:330Paolo Guiotto: So this means that, canceling this, we get the DY of CYZ, is equal to Z.
26:05:150Paolo Guiotto: So now, this means that it's usually the same problem. You know, the derivative with respect to Y of that function is Z. It means that that function, CYZ, is a primitive of this in the variable Y, because we are doing the derivative with respect to Y,
26:22:440Paolo Guiotto: plus a new constant, let's call it K, Constant in Y, that function, Potentially.
26:32:900Paolo Guiotto: In Zed, also in X,
26:40:440Paolo Guiotto: We don't have X because C is independent of X, so it cannot be depending on X here. So just a function of Z. So here, this is a constant, so we get ZY plus K of Z.
26:52:700Paolo Guiotto: So this means that now we know a bit better what is this C. So this means that we now know that the function F, XY, that was…
27:02:600Paolo Guiotto: So far, XY plus XZ,
27:09:300Paolo Guiotto: plus that function C of YZ. Now, C of YZ turns out to be ZY plus a function now only of Z.
27:20:440Paolo Guiotto: You'll see that… Step by step, we are eliminating the uncertainty. Finally.
27:28:820Paolo Guiotto: We impose the third equation, because we have a third equation here, this one.
27:34:850Paolo Guiotto: imposing.
27:39:540Paolo Guiotto: that DZF be equal to the third component, which is X plus Y, we get.
27:50:590Paolo Guiotto: So when we do the DZ of that thing, we get the BZ of XY is 0. The DZ of XZ is X,
27:58:710Paolo Guiotto: plus the dzn of ZY is Y.
28:01:850Paolo Guiotto: plus the DZ of KZ is now K prime, huh?
28:05:960Paolo Guiotto: Zeda?
28:07:210Paolo Guiotto: This must be equal to X plus Y.
28:10:520Paolo Guiotto: We can cancel this.
28:13:190Paolo Guiotto: And we get that K prime z must be equal to 0.
28:19:250Paolo Guiotto: So this means that K is,
28:22:570Paolo Guiotto: Case of Z is just a constant.
28:26:390Paolo Guiotto: So, the conclusion is that sorry, here I'm missing a letter.
28:33:180Paolo Guiotto: The conclusion? The potential exists.
28:37:440Paolo Guiotto: And it is equal to F of XYZ. It is equal to X times Y plus XZ plus ZY plus a constant where k is just
28:50:710Paolo Guiotto: A constant, but… Okay.
28:55:370Paolo Guiotto: Okay, so, let me just write the exercises to do. Do… The Exercise 341.
29:05:950Paolo Guiotto: Do also the number two.
29:09:50Paolo Guiotto: The number 3, these are one-star exercises, so… Number 4…
29:16:950Paolo Guiotto: Well, let's say that the first non-trivial exercise is 3, 4, 5. Not difficult, but you have to work a little bit more.
29:26:760Paolo Guiotto: I don't know tomorrow what I will do, but I will do definitely this one, plus maybe a couple of the others, but please do these exercises this afternoon.
29:40:750Paolo Guiotto: Okay, thank you, have a nice day.
29:45:770Paolo Guiotto: Let's stop the recording.