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Green’s Functions
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Lots of systems are governed by differential equations, which is why we want to solve them systematically. Green’s function is a very powerful and clever technique to solve lots of them. To illustrate intuitively how it works, we will consider electrostatics, one of the problems Green was interested in, and Newtonian mechanics, specifically an oscillator under some variable force. What is the common thing in these two equations? Well, they can
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be written in a very general form, Ly(x) = f(x). L is what’s called a linear differential operator, which I swear is not as scary as it sounds, y is a function of x to be solved, and the right hand side is some other function which we call the forcing term, which is given. Green’s functions can deal with any equation like this. But first we have to talk about what “linear differential operators” are. [CHAPTER 1: Linear differential operators]
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We are going to explain each word separately. Operators in this context means a function that has a function A as input, and the output being another function B. For instance, differentiation is an operator, because the squaring function is mapped to the doubling function under differentiation. So when we see an equation like this, it means that the function y(x), when fed into the operator L, becomes f(x). By solving a differential equation, we mean recovering y(x) from f(x).
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So what Green’s functions try to do is to sort of construct another operator that when the forcing term is fed, it outputs the desired solution y(x). So this explains “operator”. “Differential” doesn’t have too much meaning: it simply refers to relating to differentiation. But here comes “linear”. *Algebraically*, this is the definition of linearity in this context. Essentially the left hand side says that we first consider a linear combination
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of two functions, and then apply the operator L. The right hand side does things in the reverse order, by applying the operator to both functions first, then consider the linear combination. The operator being linear simply means that the order of these two steps doesn’t matter. But *intuitively*, what does this mean? Let’s take a closer look at this specific equation for electrostatics. For this video, you don’t need to know what this inverted triangle thing is, just that it is a linear
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differential operator. Phi is actually the electric potential, which is a function of position. Rho is the charge density distribution, which again depends on position. Epsilon_0 is simply a constant. Essentially, if you have a charged object with some charge density distribution rho, it generates an electric potential. An object with different charge distribution would yield a different potential as dictated by the equation. If the two objects
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are brought together, you simply add up the potential generated by each object. The fact that we simply have to add up the potential when putting two charge densities together defines linearity. And as we will see, Green’s functions exploit linearity to the fullest. [CHAPTER 2: Dirac delta “function”] Let’s return to the electrostatics example. More specifically, we want to take a look at the charge density distribution. From a practical point of view, how do we determine
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the charge density at position x? Well, the charge density is charge over volume in an infinitesimal region. Let’s consider this relatively small region first, which we call D. For the numerator, what is the charge within the region D? Well, it could be found by integrating the charge density over the entire region. And I just denote V as the volume of the region D. However, integrating over D is not ideal, because the region is going to change in the
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limiting process. Instead, we use a trick. We integrate throughout the entire space, but we make a small change to the integrand. We multiply by a simple function denoted by 1 with subscript D. This is called an indicator function, in this case, indicating whether you are in region D or not. The function returns 1 if you are in region D, and 0 if not. This trick works because the contribution of the integrand anywhere outside region D is 0, so you end up only considering inside region
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D anyway. Finally because V does not depend on position, we can push it inside the integrand. So now, we have this expression of the charge density function, well, technically, an approximation, but equal if the volume shrinks to 0. The good thing is that the only thing changing is this white boxed expression. This function tends to the infamous Dirac delta function when the volume shrinks to 0. This depends on the position xi, because the original functions
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before the limiting process depend on xi, but also depends on x, because the region shrinks towards the point x. Anyway, now we finally get this expression involving the Dirac delta function. (IMPORTANT CAVEAT) This might not look weird at first, but as we explore a little bit more about the delta function, it will be weirder. But then we are going to see why this is not *that* weird. Let’s go back to the measurement of charge density. As said just now, the delta function
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is the limit of this family of functions when the volume V tends to 0. These functions are functions of position xi. If xi is not x itself, then the value of the function might initially look something like this. The numerator is 1 because xi might still be inside the region D, and the denominator is the volume of the region. However, as we shrink the region, D eventually would not enclose xi anymore, so the numerator is 0 in the subsequent limiting
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process. So the limit is 0 and the delta function returns 0. But what if xi is x? Then the indicator function would always return 1 throughout the limiting process while the volume shrinks to 0, because by definition, the region has to include the point x. So in the limit, the delta function shoots off to infinity. Simply looking at these two cases, isn’t the delta function weird? Actually, this isn’t even a function, because you can’t define the
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value of a function to be infinity, at least not in this context. The weirdness continues. What if we try to integrate the delta function throughout space? We can use the formula at the top, substitute in the charge density rho being 1 everywhere, then in particular, the charge density at x is also 1. In other words, integrating the delta function throughout space gives you 1. These 3 things define the delta function,
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and again, I know this can be weird, but I am going to convince you that it might be something you have seen before. Let’s return to how we introduced Dirac delta in the first place. These functions of xi before the limit are legitimate. What if we now consider this function to be our charge density distribution? Well this means a uniform charge density within the region D, while the total charge is simply charge density, which is 1/V, multiplied by the volume
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V, which turns out to be constantly 1, regardless of what the region is. To obtain the delta function, we shrink the region down to nothing. So viewed as a function of xi, the delta function represents the charge density of a *point charge* at x. This gives us some intuition on this expression. In a very general sense, (slow) a charged object can be thought of as a continuous sum of point charges. And the charge of each point charge is governed
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by the charge density rho. This is a more mathematical expression of a pretty obvious concept, because nowadays we have the image that all objects are made of particles, in this case, point charges. However, this is only a useful mental image. By no means is this an accurate depiction of a charge density distribution. Otherwise, the charge density at every point is infinite because there is a point charge there. Instead, think of the reality as some blurred version of this.
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[CHAPTER 3: Principle of Green’s functions] Let’s return to the problem at hand. Given the charge density distribution function, we want to find the electric potential generated. And the method of Green’s functions exploits the linearity of the operator. Remember that the operator being linear means that if we want to find the potential generated by two objects, we can find the electric potential generated by each object first, then simply add them all up. Now given the previous interpretation
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of any charge density being a weighted continuous sum of point charges, we can first consider *a* point charge, and find the potential generated. Do the same for any other point in the charged object. Then from being a continuous sum of the point charges, the electric potential can also be continuously summed. That is the overall idea of solving with Green’s functions. Let’s rephrase this a bit more precisely.
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The potential generated by a point charge in this context is a rescaled Green’s function. The Green’s function depends on both the location of that point charge, xi, and the position where you find that potential, x, so we denote the function as G with two independent variables x and xi, well technically a rescaled one by epsilon_0 as I will explain shortly. The Green’s function can be thought of as a series of functions of x (wait), and which
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function to consider depends on xi, and we want a continuous sum of these different functions. A continuous sum is more precisely an integral with a weight, the charge density, so the potential function should also undergo the same summing procedure, which means that we replace the delta function with Green’s function. Then now, the potential is a weighted continuous sum of the Green’s functions. But why do we have to rescale the Green’s functions? It all boils down to how we normally
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define Green’s functions. Recall the related equation, when we consider the potential for a point charge, we essentially substitute rho to be a delta function, so the right hand side has an epsilon_0. However, we normally define the Green’s function to be only having the delta function on the right, so we have a factor of epsilon_0 to deal with. That brings us to a sort of summary and a more algebraic way of looking at all these.
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The equation we are interested in looks like this, and most importantly, L is *linear*. First off, we find the Green’s function, which is the equivalent of finding the electric potential generated by a point charge. The Green’s function and the delta function should be viewed as just different functions of x, which means something like this: different potentials generated by point charges in different positions. After finding the Green’s function,
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we multiply both sides by f(xi). Now this f(xi) does not depend on x, and so it is a constant in L’s perspective. In other words, we can swap the L and f(xi). This is true for all xi, and so we do the so-called continuous sum of the previous relation. The right side is precisely the continuous sum of point charges thing I’ve said in the previous chapter, so the box on the right hand side is precisely f(x). However, the
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left side is just slightly trickier. Remember that L being linear means that the steps of applying the operator L, and linear combination can be swapped without affecting the final result, so here on the left hand side, since the integral can be thought of as an infinite linear combination, it can also swap with L, and so we just swap them! Now compare with the equation we want to solve. If we call the thing inside the bracket to be y(x), then
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this precisely solves the equation! Isn’t that genius? If we know the Green’s function G, we can just plug f and G into this to find the solution. There are a few quite important questions that we will *briefly* go over in the remainder of this video, but before that, I think it is a good exercise to think through another example from physics with the same framework in this video. In this example, x refers to the extension from an equilibrium position; and F refers
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to the applied force at time t. As an exercise and, honestly, a kind of summary, first explain why the applied force can be written this way, directly from the fact that force is the rate of change of momentum. Then come up with an intuition of the delta function, which should be some analog of point charge. Then reinterpret the formula as some sort of continuous sum, and what the Green’s function in this case means. Finally, what
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does linearity of the operator mean in this context, and thus interpret the final solution involving Green’s function. Remember the context is an oscillator with an applied force. Pretty much the same procedure in the video, just with a different context and equation. [CHAPTER 4: Sadly, DE is not as easy] To solve the differential equation by applying the formula in this video, you first have to find that Green’s function. But how? Here is a very brief overview. Remember that
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by definition, the Green’s function satisfies this equation, and the method hinges on the properties of the delta function. The delta function has these defining properties, in particular these two. It is 0 almost everywhere, and integrating throughout space gives 1. Well no, precisely because it is 0 almost everywhere, the integrand gives 0 contribution almost everywhere as well, so we don’t need to integrate throughout space, just a small
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region D, as long as it contains the point x, and then we simply apply these two properties to the equation for Green’s function, i.e. solving the equation where the right hand side is 0 almost everywhere, and integrating it in any region D that contains x would give 1. The region D is cleverly chosen, in this case, a sphere, to make life easier, and sometimes, you might even want to shrink the region D as small as possible. Another trick to simplify
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would be to consider symmetry. Like in this case, a point charge. However you try to rotate, you see the same physical setup, i.e. the setup has a rotational symmetry, and so we expect the potential to also inherit the same kind of rotational symmetry. But actually in practice, you don’t even need to find the Green’s function yourself - this Wikipedia page already lists the Green’s functions for most common linear differential operators.
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Looks like everything’s good now, right? In order to solve a differential equation, as long as we have the Green’s function, we are ready to go; and the Green’s functions can be found pretty easily online, but there is usually something much, much messier to deal with. Let’s reuse the point charge example to illustrate further problems with finding the Green’s function. From a purely solving equation perspective, this is not the only
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solution. For instance, adding a constant would actually still solve the equation. Well you know what? There are even crazier changes that can make the equation still satisfied, but we only want one, so we impose some extra restrictions, like boundary conditions. In this case, the implicit boundary condition is that the potential dies when it is very far away from the point charge, and it turns out there is only one solution after we impose
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this extra restriction, but starting with different boundary conditions, we would obtain different Green’s functions. For physical systems where we are interested in time evolution, we also need to impose some extra restrictions so that there is only 1 solution. In this case initial conditions, like the initial position, and the initial velocity, and the usual convention is that they are both 0. Again, after imposing these initial conditions, we finally get a unique
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solution. This solution looks a little bit intimidating, but really it’s just a delayed and rescaled sine wave starting at time tau. Again, different initial conditions would result in different Green’s functions. So when you look up the table for Green’s functions online, it is important to see what kind of boundary or initial conditions they are assuming, explicitly or implicitly. In these two cases, the extra restrictions are quite nice, being 0 on the infinite boundary,
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or 0 initially. These conditions are called “homogeneous”, where the zero function satisfies. As we know, the ultimate solution is a weighted continuous sum of these Green’s functions. If we insist that all the individual Green’s functions obey this boundary condition, as long as the weight is sufficiently confined, adding up these Green’s functions would also obey the same boundary condition, simply because adding 0 a bunch of times still gives
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0. This is something good *if* we wanted this boundary condition in the first place. If, however, we want a different, more complicated, *in*homogeneous boundary condition, like the potential being 4 on the top face, but all other faces having potential 0, then there isn’t an easy way to directly apply the Green’s function technique. If you have found a Green’s function, i.e. the scaled potential for a point charge, for
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which the potential is 0 on the entire surface on the cube, then adding these potentials together would end up making the potential on the entire boundary 0, and the upper face is not 4. These cases will be much trickier to deal with. As a small summary of this chapter, basically we need to first ask whether the problem has homogeneous boundary or initial conditions. If not, then the method of Green’s functions can’t directly help, but it might
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help in some other ways. If, however, we have a homogenous condition, then we want to find the Green’s function with the exact same boundary or initial condition, but because there are so many possible conditions, the condition you have might not match the ones online. However, sometimes there are certain techniques to convert from Green’s functions on some tables to the Green’s function satisfying your boundary conditions, but we are not getting there in this video.
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I know this is only a very general overview of linear differential equations, not even requiring too much calculus knowledge, but these equations do appear in lots of places. And I do hope that this is somewhat useful or inspiring, and I encourage you to look more into solving differential equations, because that is definitely not something a 20 minute video could cover. If you have made it here, congratulations. Thanks to these patrons, and as always, like
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and subscribe with notifications on, comment and share the video as well! See you next time!
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