What I am going to derive from now would be very unique, not easy. But I'm sure it will provide you solid basis of theoretical acoustics. So, anybody who has interest on theoretical acoustics, please look at carefully at what I am, what I will do, okay? One dimensional case, okay. Let's start to look at the sound field that would be generated by the piston located at some point. And I am measuring the position from here, x. This is zero. [SOUND] To be general, let's suppose I have impedance over here. That is Z0 and also I have impedance over there, that is ZL. Okay, if ZL is zero, then, for example, if ZL is zero that means what kind of boundary position? The ZL is zero that means acoustic pressure is zero. Therefore, pressure release boundary condition. If ZL is infinity, that means the velocity on the surface is zero. Therefore, this is rigid boundary condition. In between, as it, as described in the text, it represent any arbitrary boundary condition, of course. We go back to the issue of real part and imaginary part that describes the physical boundary condition over there and over there. And we want to know the pressure at any point. Okay, we note that [SOUND] the pressure field, where P, is I assume that, it has the pressure depends on x, space, and exponential minus j omega t. So if I, if you plug this equation to wave equation, we will obtain Helmholtz equation [SOUND] which look like this. And if there is no excitation, then it should be zero, homogeneous equation. If there is some excitation in this boundary, in this space, this term is no longer 0. And it should have some function f(x) which represents spatial excitation. f(x) can be anything. Okay? So I, I would like to start with this equation, because this equation and that boundary condition express every possible boundary condition as well as excitation in one dimensional case. And then I'd like to expand what I learned from this one-dimensional case to three-dimensional case as we did before. And the idea is f(x) can be written as integration of delta x minus x0 and something p(x0), dx0. Or I reduce, [BLANK_AUDIO] Okay, or f(x), whatever. Right? Introducing this very interesting function, delta function that has shfting property. In other words, it shift out the property at x equals x0. 'Kay. This is always true. And this special function, data function, is a very interesting function. Has a characteristic that at x equal x0, it goes to infinity. But, if integrated delta x -x0, along either x or x, it will give us 1. Okay. And this interesting function provides many interesting theoretical foundation. Now, suppose, okay, this suppose has a very long mathematical history. This delta function excites the acoustic field and the solution. In other words, the pressure that is induced by this delta function, let's call it Green's function. [SOUND]. I mean, if you go to the past, I used a minus sign over there, and a minus sign over there, but some students say it is not very convenient, so we can use a plus sign. Or minus, it doesn't matter, right? We can put the 4 pi over there, or 4 pi over there. It doesn't matter, I mean, so. If you look at some textbook, you could see there's a 1 over 4 pi and a 1 over 4 pi over there, but didn't explain why. Well, instead of going to the, trivial argument why we should put the minus 1 over 4 pi or two pi, something related with pi. You just leave that arbitrary constant. Doesn't really matter. Later on, we will see. Okay, then. Then, this is what we have. [SOUND] I emphasize again, double emphasize, G is a Green's function that satisfied this equation, okay? This Green's function does not have to satisfy any other thing, boundary condition. That means theoretically or computationally, you can select any Green's function that satisfy this equation. That means you generate the some excitation delta function like this, at x equals 0, G is the pressure induced by this excitation. So, in fact, G, is often used this symbol, because it expresses the cause and effect relation. G is the pressure that is due to the delta function excitation at x equals zero and observe the pressure at x equal x. Okay now, I have two equations. So, let's denote that this is equation one and equation two. Then let's multiply G to the equation 1. Then I get G. Sorry. d square P, dx square, plus k square GP is equal to G f(x). And I do have another equation. So, let's, let's multiply P over there. [SOUND] I got two equations. The reason why I did it? Mathematically, I can, I will, I can eliminate this common term, okay? Okay,the reason why we call this is Green function or Green's functions [SOUND]. It's to honor famous mathematician long time ago Green, English mathematician. He was not full time mathematician. During the daytime, he worked at the, bakery, you know shop, all the days. And then, he spend his time sometimes in the afternoon or the weekend in the library and he invented the famous Green's function. And then he went to Europe, France, and he tried to publish his paper based on Green's function. And then he got some fame from his function. And that is a very interesting approach. And I hope you will be excited about this Green's function too. Oh okay. The, the, the mechanical engineer, mechanical engineering trained the students are not very familiar with this Green's function. Therefore I would like to emphasize that, today's lecture is very important. Okay then, what I can do is maybe I subtract this. [SOUND] Then, what I will get is G d square P, dx square minus p d square P d square G, dx square is equal to G f(x) minus P delta x minus x0. That's interesting. Okay. Suppose G is, I mean P is monopole, I mean the solution of a monopole. Now, what is this? Strange, right? Right? And always second derivatives are not easy to understand compared with first derivatives. First derivative just expresses rate of change of something. d/dx we can see. What is d/dx over here? Smooth. What is d/dx over there. It's very rapid. We can see, we can feel it. So it's always nice to reduce the order of derivatives. Because it is simple. And it is a physically sensible. All right? So let's reduce this order of, you know, derivative. How, then how to reduce the second derivative to first derivative. What is it, hm? Integration by part. I mean, to reduce the first, the second derivative to the first one, the only way we can do is integration. [SOUND] With respect to what? Because the domain we have is from 0 to L. We have to integrate this from 0 to L. With respect to what? dx, dx, okay? And this is okay because we are handling again linear acoustics. The summation is, the integration is simply summation. Therefore, no further assumption is required. We just are simply summing up all the, all the, all this expression with respect to x. So this is okay. And then let's look at integration 0 to L, G dP dx square, d square P dx square. Okay, this the, this one. We can write this one as, using integration by part. And I can write this is G dP dx, and I have to evaluate at 0 and L. And then minus dG dx and dP dx. Integration 0 to L dx, and I will get the same one over here because what I will have from this using integration by parts. I will have P dG dx and I will have dP dx and dG dx and we have a common one. And then we have a minus over there. So, therefore those common factor will be cancelled out. So this is what I have, minus, and I have to integrate. So let's, I'm, I'm going to rewrite the result I obtained by using the integration by part. And the results will look like this. So, okay, integration 0 to L, G. No, not, I'm sorry about that. I have a G dP dx, evaluating on 0 to L. This is the boundary. Minus P dG dx evaluating on 0 to L. That has to be equal to integration G f(x) minus P delta x-x0 with respect to dx. And this one. Integration minus P delta x minus x0 would be P x0. Okay. Minus. So this minus this is this is equal to the integration of that term and minus P x0. Therefore I can write, moving this to the left-hand side, I can write P x0 is equal to okay P, dG dx evaluating on x and L and minus G dP dx evaluating on 0 to L. And I have plus integration G f(x)dx. What does it mean? Because this is linear, I can change, I can change [SOUND]. x and x0. Okay, let's exchange x and x0. Then, this equation says the pressure at x which we do want to have pressure at x. Let me draw the picture again. I have, sorry. I have one dimensional, and I have excitation f(x). I have boundary condition. [SOUND] This is the coordinate I'm using. And the pressure at any point is equal to P dG dx0, 0 to L minus G dP dx0, 0 to L plus G f(x). Sorry. Integration 0 to L, dx0. [BLANK_AUDIO] Okay. So, that is quite interesting. Physical interpretation of this equation could be following, okay. The pressure at x is contributed by three factors. One is this. And the other one is this. What is P dG dx? What is G dP dx? If G is monopole, that is dipole contribution. And the P is pressure. Therefore, the pressure is propagating into space like dipole. Physical meaning of this. Okay. Suppose G is monopole because monopole satisfies the equation that is governed by, I mean, the equations d square G, dx square plus G, k square G has to be delta x minus x0. This the equa-, this is the equation that governed the Green's function. So, this term is, dP dx. What is dP dx? Recall Euler equation says the, the pressure difference in space induce the motion, induce the acceleration of, of, of, of, of, of a particle. So this is related. Because this is the pressure difference between the x0 small element. The pressure difference induced a velocity change. So in harmonic excitation which we assume, this we would say, pretty much related with a velocity. And the velocity is propagating as the monopole, right? This is dipole propagation, this is monopole propagation, this is pressure, and this is the velocity. So physically means that anything related with a velocity component on each boundary or pressure component, each boundary will be propagating through the observation point at x in this manner. That's great. That's great. So, if you don't have any excitation inside, everything will be resulted by the sound coming from this boundary. So, if there is no excitation in, in, in, in, in, in, in the middle, say if f(x) is zero then I can write very interesting and famous equation. Like, Px, pressure at any point, is a two contribution, one is P dG dx 0 to L and then minus G dP dx 0 to L. Wow. That is great. Okay, let me expand this to 3D. Suppose I have certain boundary. Maybe, this is r0 and I want to predict a sound r then I can write, expanding what we obtain from one dimensional case to P at r any point as to two component. P dG dr and this r, has to sorry, this r has to be x0. [SOUND] So r0 and this contribution, that is G dP dr0, okay? Because this is 3D case, this is no longer the value at discrete point and this has to be the value along the surface. So this has to be integration and let's denote this is surface S0. And this has to be gone to surface S0. [SOUND] And I am integrating surface integral. And this is a very well-known Kirchhoff Helmholtz equation. Kirchhoff Helmholtz integral equations, strictly speaking. Wow. This is nice. And this is [SOUND] headache [LAUGH] in other word. It is nice, but this is really difficult to use if you do not know very well [BLANK_AUDIO]
