TheFourierTransformAndItsApplications-Lecture05 Instructor (Brad Osgood):Okay, ready to rock and roll? All right. We’ve been getting a lot of – as I understand it, we’ve been getting a lot of email questions about when and where to turn in the homework. You can certainly bring it to class, and that’s fine. But if you find yourself wanting to fill in those last minute comments, I would say that – the policy is the homework should be due by 5:00 on Wednesday, and you can turn it in to the magic filing cabinet. Across from my office – my office is 271 Packard, and there’s a little hallway sort of across from there, and there are several gray filing cabinets, one of which has my name and the course number on it. And if you open that drawer, you will see a wire basket in there with a little sign on it that says something like, “Turn in homework here.” So you can just drop your homework in there, okay. So by 5:00 today the first problem assignment is due. All right. Any questions? Anything on anybody’s mind? Okay, all right. Anything else that came up that I should address? No, okay. All right. Today, I wanna finish off our discussion Fourier series. And sometimes we don’t really finish the discussion of Fourier series because it will always be a touchstone for reference for some of the other things that we do. But I wanna finish up the discussion we started last time about using Fourier series to solve the heat equation. And then I wanna talk about the transition from Fourier series to the Fourier transform, and how one gets from the study of periodic phenomenon to the study of non-periodic phenomenon, which is exactly what the Fourier transform is concerned with, by means of limiting process, all right. So that’s where I wanna finish up. But first, let me go back to the discussion we started last time about the heat equation. So this is the use of Fourier. This a classic example, the classic example one might say, of Fourier series, and also shows in a particular case, a very general principle that we will be seeing constantly throughout the course. That’s really one of the reasons why I wanna talk about it. So this is your Fourier series to solve the heat equation, one particular case of a heat equation. And I’ll remind you what the setup is. We have a ring, we have a heated ring like that with an initial temperature distribution of – we’re calling F of X. So the initial temperature – so I’m thinking of X here as a spacial variable. I wanted to mention a special variable, although the ring is sitting in two dimensions. So I can think of the ring, if I want, as an interval from zero to one with the end points identified. At any rate, the fundamental – the important fact here is that since there is periodicity in space, since the ring goes round and round and round, the function F is periodic as afunction of the spacial variable, the position on the ring. And we can normalize things to assume the period is one, all right. So that’s how Fourier series comes into the picture. So we take F to be periodic of period one. Then we let the U of XT – now, the temperature is varying, both in position and in time. The temperature change is a function in time, and the temperature is different at different points along the ring. So I let U of XT be the temperature at a position X at time T. Then U is also a function – it’s a periodic function in this spacial variable. U is a periodic function of X. So U XT is periodic in X. That is U of X plus one T at any time T is U of X at the same instant of time. When T is fixed, this periodic is a function of X. And the physical situation is described by the heat equation, which is also referred to as the defusion equation, and governs many similar processes that involve the diffusion of something through something else [inaudible]. So heat through a region charged through a wire is governed by this sort of equation. I mentioned these last time, holes through a semiconductor. It really has quite a variety of applications. And some of the techniques that we’re talking about here, although they’re specialized to this case, can be applied in various forms to many different situations. So you have the heat equation [inaudible] equation, so it relates to the derivatives in time to the derivatives in X in space. So it says UT is equal to one-half UXX. Here I’m just choosing the constant – there’s a constant on the right-hand side of the heat equation. I’m just choosing things so the constant is one-half just to simplify the calculations. So this is – the first derivative with respect to time, and this is the second derivative with respect to X. All right, now because the function is periodic in X, we can expand it as a Fourier series. So the rigger police are off duty. I’m assuming that everything converges here and there’s no question about writing down the sums. This is sort of a formal operation whose particular – the particular manipulations that I’m gonna do can be justified under reasonable assumptions, but that’s not the point. The point is just to see how the techniques can be used to solve the equation. So I can write the function as a Fourier series. Now remember, if periodic is a function of X, so the [inaudible] on T is on the coefficient. K one from minus infinity to infinity, that’s – I’ll call it C’s of K of T times the periodic term, either the two pi IKX, ano I’m assuming period one. That’s the basic assumption. Or rather, that falls from periodicity. You can expand it as a Fourier series. That’s how Fourier series get into the picture. Then you plug into the equation – we did this last time. I’m just reminding you of the setup. Plug into the heat equation and equate coefficients and you get an ordinary differential equation for the C’s. The C’s are the unknowns there.So the – you get this equation, you get CK prime of T is equal to minus two pi squared, K squared, times CK of T. That’s a simple ordinary differential equation, and we know how to solve it. That’s a simple [inaudible]. And solution is CK of T is E to the minus – is the initial condition TK at zero times equal the minus two pi squared, K squared T. That’s easy. That is easy. And this is where I think we got to last time. Now, what is the initial condition? I haven’t brought in the initial temperature here, but here’s where it comes in. So what is CK of zero?