So, the last exercise dealt with writing the Laplacian in three-dimension in the cylindrical coordinates.
Whatever cylindrical coordinates, as the one says, is another way of writing the three-dimensional space
in terms of vertical components z, which is basically fixed, and in the two-dimensional plane x and y,
instead of the variables x and y, we consider the variables r, theta, where r, given a point here, is the distance from the origin
and theta is the positive angle with, sorry, x and y, with respect to the x-axis, the positive x and y-axis.
In other words, we represent x, y, and theta as r cos theta, r c theta, and z remains the same, r positive theta in zero to pi,
and, of course, then r.
Now, of course, one way to write the Laplacian with respect to these coordinates is, which, by the way, is useful for many problems in which you know that you have, I don't know, a radial symmetry only in the x-y variables, for instance.
So, I was saying, it is useful, I mean, the way to write it is to find a way to represent r and theta as functions of x and y.
Clearly, r is the square root of x squared plus y squared.
T, instead, is a little more tricky because it is the inverse tangent function, r tan of y over x, if x is positive and y is larger than zero.
But in the other cases, it doesn't have a definition, but then it is pi over 2, if x is zero and y is positive.
Of course, this function is, if defined in the origin, that's normal, but the origin we are not taking for this reason, so it's fine.
But then it's pi plus r tan y over x, if instead x is negative and y is r, then we have 3 over 2 pi,
if x is equal to zero, and y is negative, and finally 2 pi plus r tan y over x, if x is positive and y is negative.
Why so? Because of the choice, because we say that theta should be from zero to 2 pi, that's a problem.
So that this definition works only in the first quadrant of the plane x, y.
In this one, we have to translate by pi and then here by 2 pi because otherwise we will have negative values.
Which is exactly, I mean, if you keep going like this, this is minus theta, and this is the theta you want to evaluate.
And clearly the sum is 2 pi, no?
Now, so, telling that theta is only y over x is not correct, as a matter of fact.
However, calculations in the end can still be performed only in the simpler case, x and y positive, because in other cases,
I mean, first of all, this function is clearly continuous, and secondly, it's also differentiable,
as long as you take away x equal to zero, away from the y axis, as again, as one would expect.
And the difference between these three definitions that we consider these three main pieces is only constant.
So whenever I take a derivative, it does not matter.
And this is why many of you wrote these, which is not correct, and it will take into account, but nevertheless, we're able to perform the same calculations.
One could tell me also, I don't know, up to adding some pi constant according to the quadrant, theta is theta, y over x,
and therefore its derivative are the same. It was not necessary to write all this.
Some explanations why you can actually consider only this case and everything we define, however, goes required.
So the point is, we will write our function, u, is it true, as u of x, y, z, now is u of r,
cos theta, r, sin theta, z, bearing in mind that r, cos theta, and r, sin theta, tell me r and theta depends on x and y.
And so, the derivative with respect to the x, the derivative with respect to the x variable will be equal to du in the r times dr in dx plus du in the theta times d theta in dx.
I mean, this is simply the chain rule. And now, obviously, we have for me, let me write all of this just to make you see.
We'll do the second derivative. So, some of you wanted to do a total compress, I think it's better to take your time, because sometimes there are mistakes.
If one tries to skip some steps, it's better always not to do it.
So, here we have the second derivative with respect to r times the derivative of r in x squared plus du in the r times the second derivative of r in x squared plus, again,
we already phrased it in this term. And so, we would also need to differentiate it in theta.
So, we have the mixed derivative with u in the theta dr times dr in dx d theta in dx.
And then we pass to a second term. So, also here we have the mixed derivative. That is, we go to differentiate also in r.
So, basically, let me not write it. We get twice this term over here.
And, differentiating is that d theta gives us du squared d d theta d theta in dx squared plus du d theta, now we have to differentiate d theta in x.
So, the second derivative of theta in x. And, of course, for y, we get exactly the same result.
Very well. Now, the question is, what is the first derivative of r in x and in theta in y and so on?
Here should have enough space. So, well, dr in dx is simply x over square root x squared plus y squared, and this is if you want x over r or if you even want cos theta.
And now, cos theta in dy is y over r, and therefore it is sin theta.
The second derivative of r in x is, well, 1 over r minus, so it should be x times the derivative of r in x over r to the power 2.
And the derivative of r in x is exactly x over r. So, we get 1 n of 1, r squared minus x squared over r to the power 3, but this is y squared over r to the power 3.
And then, obviously, the second derivative of r in dy is x squared over r to the power 3.
Again, all this you could be written also in terms of sine and cosine, but this is more useful for later.
Theta in dx. So now, as I said, we can work with the first definition, so without loss of generality we can assume that both x and y are positive.
So, the differentiation of this gives us, of course, x squared over x squared plus y squared times minus y over x squared, simplified, minus y over r squared.
And the analogy was d theta in dy was instead plus x over r squared, because the roles are in that range.
And then, let me go over here. Let's just see it, yes? The second derivative of theta in x will become, well, minus y doesn't change.
That's nice. The derivative of r squared is simply 2x, and there is a plus coming from there, so 2xy over r to the power 3.
Sorry, for the force.
And then, obviously, the second derivative in y will be minus 2xy over r to the power 4.
Yeah. Okay. And now, we have everything to do to conclude the exercise, because, let me free some space over here, it is enough to sum up the second derivative in x.
Yeah, we need also the big true stuff over here, and the second derivative in y. I will keep only the other calculations so you can find it.
So, what do we have is that the Laplacian of u is the second derivative in x plus the second derivative in y. Very well.
But then, let's bring together the terms. So, there is the second derivative of u in r times the derivative of r in x squared plus the derivative of r in y squared plus the, oh, sorry.
This was, of course, r.
So, the first derivative of u in r times the second derivative of r in x plus the second derivative of r in y plus 4 times the mix of, no, let me see.
Twice the mix of the derivative of u in r in theta times the r in dx in d theta in dx plus the r in dy plus d theta in dy.
Plus the second derivative of u in theta times the first derivative of theta in x squared plus the second derivative of theta in y squared.
Plus the first derivative of u in theta times the second derivative of theta in x plus the second derivative of theta in y.
And now, we are done. It's enough to check what are these sums. Let's do it over here.
First one, the second derivative of r, the first derivative of r squared plus the first derivative of r in y squared.
This is x squared plus y squared over r squared. It is y. And so, we get the factor of 1 there as we want.
Then, here, the second derivative of r in x plus, by the way, notice, this is the norm of the gradient of r squared and this is the Laplacian of r.
This is the scalar product between the gradient of r and the gradient of theta. This is the norm of the gradient of theta squared and this is the Laplacian of theta.
It did the Laplacian of theta. It was not necessary. Anyway, the second derivative of r in y, what do we get here?
Well, we get x squared plus y squared divided by r to the power 3, which is 1 over r, as we mentioned.
The mixed term. The mixed term has a zero coefficient. Why? Because, basically, dt times dx is not even needed to do, but dt times dx is a negative sign and dt times y is a positive sign.
And that is in both cases, x, y over r to the power 3.
So, this term here, let's call it star. It's too long. Star is immediately zero. And let's finish with the other two calculations.
The first derivative of theta squared plus the derivative of theta in y squared, the norm of the gradient of theta, is equal to, well, y squared plus x squared over r to the power 4, and this is 1 over r squared.
And yes. And finally, to conclude, this term here, and that's zero, because also this term here is zero, because here they are plus 2xy minus 2xy over r.
So, all in all, we conclude that the Laplacian of u is equal to the second derivative of u in r plus 1 over r, u r plus 1 over r squared.
Oh, sorry. This is the Laplacian in x and y only, of course. The Laplacian of 1 over r squared, u dt times theta plus, of course, the second derivative in z, which had nothing to do with all these calculations, because that's the only variable that remained over there.
And so, well done.