1 Mixed boundary conditions for 2d Laplace's...

1 Mixed boundary conditions for 2d Laplace's equation L 0 Solve Laplace's equation $\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0$ In the region $0 \le y \le L$, $0 \le x < \infty$ with boundary conditions $T(0, y) = y^2 - 2yL$, $\frac{\partial T(x, y)}{\partial y}|_{y=L} = 0$, $T(x, 0) = 0$, $\lim_{x \to \infty} T(x, y) \to 0$. These are Dirichlet boundary conditions at y = 0 and Neumann boundary conditions at y = L. You can follow these steps: (a) Using the separation of variables technique, find all solutions of the form $T(x, y) = X(x)Y(y)$ that satisfy the boundary conditions: $\frac{\partial T(x, y)}{\partial y}|_{y=L} = 0$, $T(x, 0) = 0$, $\lim_{x \to \infty} T(x, y) \to 0$. Note that we do not require $T(0, y) = y^2 - 2yL$ yet. (b) The solutions in part (a) are labeled by an odd positive integer n, which we can write as $n = 2m - 1$ for $m = 1, 2, \dots$. Denote the $m^{th}$ solution of part (a) as $X_m(x)Y_m(y)$. We now look for a solution of the form $T(x, y) = \sum_{m=1}^{\infty} C_m X_m(x)Y_m(y)$, where $C_m$'s are unknown constants. Write down explicitly the equation that we get by setting x = 0. This part doesn't require much computation but is necessary for part (c). Hint: you should have found in part (a) that $Y_m$ is a simple trigonometric function, and $(2m - 1)$ enters as part of its argument. The boundary conditions require that m be an integer. (c) For any positive integers m, m', calculate the integral $\int_0^L Y_m(y)Y_{m'}(y)dy$ and show that it is zero for $m \ne m'$. What is the value for $m = m'$? (This depends on your normalization of $Y_m$, of course.)

Transcript

00:01 Okay, so we're interested in a two -dimensional wave equation that we're going to express in polar coordinates or cylindrical coordinates.

00:11 So our first separation of variables, we're going to take psi, which is a function of rho, phi, and t, and we're going to write it as a product of a function of rho and phi and another function of t along.

00:28 We'll substitute that in.

00:29 So for instance, on the left -hand side, the t factor just goes right through because none of these derivatives involve time.

00:40 On the right -hand side, we're only taking the time derivative, so rho and phi don't matter.

00:49 So we can divide by the product of x and t, should be a c squared down there.

01:03 And so when we write this, the thing all the way to the left is a function only of rho and phi.

01:13 The term in the middle is a function only of t.

01:19 And so the whole idea of separation of variables is that those two functions have to be equal, but they're not, they don't have any variables in common.

01:30 So the only possible thing that they could be equal to is a constant, which we're going to call minus k squared, where k is going to be some number.

01:39 Then we can take the t equation by just taking the two terms to the right and multiply it through by capital t times c squared.

01:49 You get an equation that looks like that, which is easy enough to solve.

01:53 Could write it, for instance, as cosine of kct, although later on, and the phi here is a phase angle.

02:06 It's not the variable phi that we're using up above.

02:10 And later on, we'll change this into being a complex exponential in time.

02:18 But either way works, as cosine and sine are part of the complex exponential.

02:23 So either way will work.

02:32 And then we can write out the x equation.

02:37 So let's go to cylindrical coordinates.

02:40 We'll suppress the z direction, so we're only thinking in terms of rho and phi.

02:46 And then i'll write this out, this laplacian, in terms of rho and phi, looks like this.

03:08 Then i'm going to separate variables again.

03:12 I'm going to write x as some r of rho times some capital phi of phi.

03:43 And then i'm going to divide through by the product, r phi, which is kind of our original x function.

03:53 And so i get something that looks like this.

03:55 So one of the things i could do is i could basically put all the phi's on one side and all the rho's on the other side.

04:27 But i can just notice that this term here in the middle depends only on phi.

04:34 Everything else depends on rho.

04:37 And so again, this thing, this middle, the phi dependent part has to be a constant.

04:46 I'm going to call it minus m squared, where m is an integer.

04:59 And phi x is going to be the exponential of i m phi.

05:03 So and again, we can plug that in and see that it works.

05:11 But here's what's really important here.

05:15 M has to be an integer.

05:17 And the reason that m has to be an integer is that you want this thing, the capital phi, to be continuous.

05:27 And the problem with phi is because we're basically going around in a circle.

05:37 When you get to 2 pi, if you've gone all the way around, your function had better be the same as it was at zero.

05:47 And the only way that can happen is if m is an integer.

05:51 If m is not an integer, then the function itself won't be continuous when you go from 2 pi back to zero as you're going around the circle.

06:02 So that's why m has to be an integer.

06:10 There are weird situations where maybe m isn't an integer.

06:15 But for the problem we're working on that we're told has circular symmetry, m has to be an integer in order for this thing to be continuous as we go around.

06:37 So we'll plug that back in.

06:39 And this is something called bessel's equation.

06:51 So it turns out that we can write down infinite series solutions for r as a function of rho.

07:03 So we can write down a series solution for this.

07:12 We can actually get a pair of them.

07:14 And they're infinite series.

07:18 They're not sines and cosines or any kind of functions like that.

07:21 They're called bessel functions.

07:23 So bessel functions are kind of like, they're kind of periodic.

07:31 They're not exactly periodic like sines and cosines are, but they're kind of periodic.

07:37 They have some other features as well.

07:41 But let's write out our solution in terms of kind of what we call the standard bessel functions, jm and ym.

07:53 So the solution we want is the jm solutions and not the ym's.

08:00 And the reason we don't want the ym's is that the ym's are what we call irregular.

08:13 They're what we call the irregular solution.

08:17 And the irregular solution goes to infinity as rho goes to zero...

Question

Please give Ace some feedback