▶️ Watch on 3Speak - Telegram - YouTube - Summary - Hive Notes - Sections playlist - MES Links --- In this video I derive the Laplace Operator or Laplacian in polar coordinates, which will come in ...
Transcript
Let's look at the Laplace operator in polar coordinates. All right, just some notes on this Laplace operator in polar coordinates, and in the general spherical harmonics. So I will generally follow the derivations shown in videos by Michael Dabowski, from the Mike, the mathematician YouTube channel for this right here, so one converting the Laplacian in polar coordinates. converting the Laplacian in spherical coordinates, and solving the Laplacian to obtain a spherical coordinates. Again, Laplace operator or Laplacian is the same thing. So I've also used some help from GROC AI for one of the above videos are lacking, and yeah. And I also may be using the Leo AI as well, on the high of blockchain. So to help in obtaining the Laplacian and spherical coordinates, let's first obtain it for the simpler but related polar coordinates. So that's basically, the way he uses is he does it for polar coordinates, and then he just throws it in there for spherical coordinates. So let's solve the Laplacian in, or the Laplace operator. Solve the Laplacian, I'll say the Laplace operator. plus operator or Laplacian and spherical coordinates. Let's first obtain it for the simpler but related polar, polar coordinates. All right, so in polar coordinates, what this means is, well, we're going to have, actually, the Laplacian, delta F, this equals to, again, you can do is the gradient setup or divergent setup, this like this. This equals two, and it's only going to be in, so we basically have to convert rectangular two dimension into the polar, polar coordinates. This is going to be the partial derivative of F, the second partial derivative of F in terms of X, plus the second partial derivative of this is going to be of F in terms of Y. This is going to be just like that. Don't put a 2, make it like that. This is going to be, in the way it's written, it's like this, equals to 0. So that is the... That's, again, this is in two dimensions. This is in three dimensions. This is three dimensions, and then this is two dimensions, and now we need to convert this into polar coordinates, and recall polar coordinates, and then also find the relation between X and the polar coordinates, X, Y. So if we have, I just derive this. So this is the X and Y, like that, this is X, this is Y. Let's say we have a point P right here. and the distance is R. So we have point P, and this is the angle, this is a polar angle, this is going to be theta. So this is R and theta, like that. And so now, this distance, let's say this distance is X, this right here, this full thing is X, that's Y, so in other words, this would be X equals 2, and then recall that the sign, yeah, and this would be the Y. is y is equal to, so recall that the trigonomic, basic trigonometry, and this is the origin, yeah, basically, this is the origin, yeah, and then basic trigonometry, you have the sine of theta is equal to opposite over hypotenuse, so in other words, we get y over R, and then we just move this y over to, I mean, R to the other side, so we get y is equal to R sine. Theta. And likewise, X is going to be, well, cosine Theta equals to adjacent of our abundance, X over R by definition, move this over to the other side. So we get x is equal to R cosine Theta. So that's a fastening. So what that means is, and also we have Pythagoras as well. So Pythagoras, or Pythagorean theorem, also a Pythagoras. but it's put it like that so we also have now the side squared equals r squared so in other words r squared equals to x squared plus y squared so we have that so put it all together we have x is equal to oh yeah put this like this x is equal to r cosine because we're again we're converting to polar coordinates so we got to put these x and y in terms of polar accordance and y is equal to r sign and then we have r squared is equal to x squared plus y squared box this in all right now the next up is well let's write out the first partial derivative of f in terms of x and y by using the chain rule for partial derivatives and its associated tree structure and that's by first of all if you look at this right here because we want to get this in terms of r and theta and this is right here, so we'll take the first partial derivative, just to show you what this means. So this is the second partial derivative in terms of x, and then we're going to use this associated tree structure, so I haven't covered tree structure before, but the general idea is pretty much the same, and it's using chain rule for partial derivatives, or just chain rule in general. So if we have, for example, a function F, And then you make a tree, so it is a function of R and theta in a spherical coordinates. So we want spherical coordinates. So it's a function of R and theta, but then R and theta, so R is a function of X and Y. Theta is a function of X and Y. So there's, there's Theta here, you could write this as a function of X and Y. So then this is going to be X and Y. And this is going to be X. and y. So you have to use a chain rule. So basically, this is the tree, sorry, I'll put this right here. I'll just put this, yeah, I clearly just moved it over there, so I'm going to go tree structure. Just so we have the things in point there. So now the partial derivative of F now, in terms of X, so we have to go, we have, well, there's two parts right here. So we're going to have to have this equal to a partial root of, so we have to go over to this part, so we have to go, it's going to be partial root of f, and then part of r, and then holding this x, this part constant. We'll put this right here, this is what the guy, that math video was, how would you do, in his derivation. So we put this right here, this means hold constant, or keep constant. Keep constant. All right, so we keep that constant, so we have, that's thing, it's partial derivative, the reason it's partial, because we're going from initially this part to here, and then we're going to go the reverse way and add those up. And then we're going to have to multiply using, again, chain rule, this is going to be just multiplied by, this is going to be partial of, no, now it's partial of r, so it goes to function of x, so now we go partial of r, partial derivative of r in terms of x, and where we hold this y content. So keep constant. And then we add the next term. So then we go partial f of theta. And where the r is constant, and then dot partial r, and partial theta, and then this is going to be partial x. and this is at Y is constant. So there it is. Alright, so that's the partial F, partial X, so now the next one's going to be, I'll just say likewise. Put it here, and then this is going to, but now we have taken the derivative of F in terms of Y. But again, we want the R and Theta converted to it, so now these are both functions of Y. Yeah, so it's going to be actually pretty much the exact same thing, but we just replace X and Y. So what I'll do is I'm going to copy this. I'm going to paste this here, and I'm going to move this here. So we have this, and then this just becomes, yeah, we just flip everything. This can be partial derivative of F in terms of Y. This is going to be, then this is going to be right here, Y, and then this X is constant. And then this one here is going to be Y, and that's X. This is exactly the same thing. So because both directions go from here, and now you have to split up to go left or right so we go here now this one goes here and then it goes straight down but in both scenarios we use the R is basically the first one's always the same that's so fascinating stuff All right, now we have this, let's go for this, and I just want to make a note, I will cover partial derivatives more closely in future videos, but the chain rule is similar to that for normal derivatives from my earlier videos, so recall, just for just general functions, the chain rule, if G is differentiable at X and F is differentiable at G of X, then the composite function, F, G right here, yeah, F of G, basically it's like, F is a function of G, is defined by F of X, equals to f of g of x is differentiable at x and f prime is given by the product so basically you take the product f prime of x equals to lowercase f of yeah f prime of g of x times it by now the inside derivative g prime of x So in Lebanon's notation, if y is equal F of U and U is equal to G of X are both differentialable functions, then DY over DX, so we do the inside one, is equal to DY over DU, so we take the outer derivative times it by the inner, then DU over DX. Again, exactly the same thing we did here, so we had partial F in terms of partial R. and then multiplied by partial R in terms of partial X, and likewise for the Y side and the theta side. So, fascinating stuff. Anyways, so we can solve the first partial derivative of F in terms of X as follows from our polar coordinate formulas. Yeah, so if we go, so we have this setup, all right, and now we have, and we basically throw them all in here, so we have, this is the first partial derivative of F in terms of X, so we'll go partial derivative, of X. Actually, we could just copy, paste the whole thing, so it saves time. Otherwise, I've taken many hours for this video. So it's just a lot of terms. So it's more complicated than it looks, is what I'm saying, but spherical homics. I think just the terms are just so big. So there it is. Look at that. All right, yeah, so we have this, and then we, from our polar corner formulas, well, we take a look at what we have. So what I'll look at is this here. I'm going to look at this R squared right there. So that r squared, if we squared both sides, and also basically I want to get this term right here first. Let's get this term. There's R, R and X, because we have you, R and X there. So this equals two, because we want to convert these ones, basically, to polar coordinates, this one to polar coordinates. Because this one is pretty much there in polar coordinates. So now what we're going to get is this equals to, or, I mean, R, square root, is equal to x squared plus y squared. That's just basically this one right here, using Pythagoras. So that is that one, we square root both sides. So now this also, just write a bit neater, this is going to equal to x. This is x squared, y squared. This is power of, power of 1 over 2. I mean, you have 1 divided by 2 or 1 half. Just put it like that, it's easier to take the derivative. All right, so now we have this, basically, now this partial derivative right here, the partial derivative are in terms of x while holding y constant. We can solve that by going, well, partial derivative of r in terms of x while y is constant so in other words we just take the derivative of this but keeping y constant so in terms of x so this is recall the power or a law this goes down so it's going to be one half times it by i'm just going to do the full just to understand us and we subtract one here so we're going to go x squared plus y squared and then one half minus one so that in other words minus two over two uh this is one and basically one minus two is negative negative um one so we're gonna get a negative one half like that so when you have one half subtract by one just one negative one half and then we take we keep going by using a chain rule inside derivative of the inside where y is constant that goes to zero so we're gonna get a two x so in other words this two cancels and this one is just to put that in the denominator so we're gonna get a x over square root X squared plus Y squared. And, well, we know what this is, this is just R. That's just this part here. So that's just equals to that. So in other words, this equals to X over R. And we already know what X, we all, yeah, we already know what X over R is. That's just basic trigonometry. So that's just X over R is just equal to the cosine theta. So adjacent over our podness is just cosine theta. So this equals to, so we've replaced the X now with a cosine theta. So we have literally got rid of it and included a polar coordinate. So we got rid of the x, now we have a polar coordinate. Alright, so we have that term. But now the next term that we can try to solve is this one here. This was an R. So the next one we want to get rid of is this x right there. So this partial root of theta there. So now, before we get to that, I'm going to go as note. that tan of theta, just basic trigonometry, well, 10 of theta, a recall definition is just over, is just going to be sine theta over cosine data, or just by definition, opposite over adjacent. So in other words, going to be Y divided by X. Like that, so that's about Y divided by X, because we want to take the derivative of theta in terms of X. So we need to solve this one. So basically, this is just Y over X. So in other words, theta, this equals to recall inverse functions, theta is equal to the inverse of 10, so 10, negative 1, like that, of y of x. Yeah, so that's what we have there. And so now, basically, we could take the partial derivative of theta in terms of x while keeping y constant. So we're going to go, so thus, yeah, basically I'll just put it in narrow like this, so that we got a partial derivative. of theta in terms of x while keeping y constant. This equals to, well, recall that this equals to, yeah, I'll show that in a bit. Basically, the derivative of this one here, this just equals to 1 divided by 1 plus the inside function, y over x squared. So that is the derivative of this inverse function there. and as I proved that in my earlier video, and I'll show you a screenshot soon, like that. And then now we take, using chain rule, the derivative of this y over x in terms of x while keeping y constant. So in other words, this is just going to be a, this part right here. Yeah, this is the same thing as writing again, you get right it as y times cosine negative 1. So the derivative, is going to be put the negative 1 down. So we're going to get a negative. I'll put a negative 1. I'll just put a negative, and then we subtract 1. So we're going to go y, and then minus that is going to be minus 2, or just to put a square like that. Yeah, just put it like that. That's it. Or yeah, I put this power negative 1, like that, just easier for me to solve the derivative inside. So now what we'll do is multiply this x squared inside everywhere. So multiply the dot on side and see what we get. So now we're going to get equals 2. We're going to get a negative y in the top. Let's move the y in the top there. The bottom's going to be, well, 1 times x squared is just x squared. And now this is y squared about x squared. In other words, we're going to have y square there. The x squared cancels. So this cancels with this one. So this just equals 2 plus y squared. And we know again, x squared plus y squared. That is just r squared. So again, that's just part right here. R squared equals X squared plus Y squared. So, in other words, we got rid of, well, one of the stuff we got rid of, we got a negative Y over R squared. This is, again, this is R squared. Equals R squared, let's put it like that. Yeah, that's R squared. Yeah, but, so we have this, but going further, we could write this as is, well, this just equals to negative y, and then divide it by R, So put a negative y divided by r, let's put a y, divided by r, divided by 1 over r. Yeah, so in other words, this is just R squared separate that, so we have this, and then we know that negative, we know what y over r is, scroll back up. Y of R is just sine theta. So that's just equal to, yeah, this whole thing is equal to negative sine theta. So this equals to, negative, sine theta, theta, divide by R. All right, so we solve this. So this is negative. Let's put it above, yeah. Yeah, so anyway, so that is what is, and we got rid of the X and Y terms. So, so put a line in here, so thus, the partial derivative of F in terms of X, and in polar coordinates, is going to equal to, so we have the first term. This is cosine, theta. So we have this part here, is this be partial F in terms of partial R, and then where this is constantly, we just ignore that, that's a given. And this one's going to be cosine data. And then this one is going to be over here, negative sine theta divided by R. Yeah, and then we have this, this Dell F over del Theta. So in other words, partial F in terms of theta. So this is going to be cosine theta. and then we have partial F, or partial R. Yeah, it's partial F in terms of R. And then we go subtract, and now this one's replaced their sign theta, R. Now we have a partial F, partial theta. And again, these ones, you could put this here and put a R, a constant, I'll just ignore it, you can do the same thing here, but it's this, but it's pretty, yeah, it's given, just assumed to be the case. So that is that. And just so quick an out, recall from my earlier video, the derivative of inverse tangents, this is from my earlier video, screenshots from the calculus book. So the formula for the derivative, the arc tangent function is derived in a similar way. If y equals to inverse 10 or arc tan is 10 negative 1 x, then tan y equals x. Basically, that's just what, that's just what it means, where you solve this in. In our case, it was data. and then differentiating this latter equation implicitly with respect to y'a x we have again we use implicit differentiation of this setup we get this the whole setup and then basically we get this over here derivative is equal to one by x squared and that's the one i used right over here this is a derivative of this one of this one over the x squared and in this case x or y over x like that and also here's the figure of this is for uh completeness. So figure 11 shows the graph of fx equals inverse tan and its derivative right here. Notice that F is increasing and F prime is always positive. The fact that this one here, 10 of the inverse tan is between plus or minus pi or two as, yeah, the fact as X goes to plus or minus infinity is reflected in the fact that that f prime of x equals zero as X goes to plus or minus affinity. So yes. The derivative. So basically this function right here so here is the this is the function is tangent here's inverse goes from negative 1 to yet to 1 as x goes to plus or minus infinity and this one goes to 0 as plus minus infinity this is the derivative notice that's peaked at the center and it goes there so it's always positive even though this one goes negative positive fascinating so anyways let's go further Similarly, for the partial derivative of F in terms of Y, we have, so we basically do the same thing and we get something similar to that. We'll just take this for completeness and just do this. I'll take this thing here, copy it, and this is the Y component, so put it there. So we have the same exact setup there. We know that r is equal to, so r is equal to square root x squared plus y squared, and then basically the partial derivative of r in terms of y, while keeping x constant. Yeah, like that, this is equals two, this is going to be The same thing, it's going to be negative. I mean, it's not negative, one half, and put this at the bottom, is going to be square root x squared, plus y squared, and derivative of this is 2y there. I got, again, exactly the same setup we did over here. So, derivative, this goes down below, put the y on top, there's a 2Y, and then 2s cancel, and we're going to go, this cancels, this cancels, we're going to, and this is R. So this equals to y over r, and this equals to move some space here. And this equals to, well, sign theta. All right, just move that over this box descent. So that's sine theta. And again, yeah, that is, yeah, that's our sign theta. And that is because Y of R, as we showed way earlier. That is just signed there. So it's signed there. Alright, now the next setup is, yeah, the next setup is going from the portrait of theta in terms of y. We'll just take this again. I'll just take this, just for a couple and just take both. So…