My 2019 Mathematics A To Z: Green’s function


Today’s A To Z term is Green’s function. Vayuputrii nominated the topic, and once again I went for one close to my own interests.

These are named for George Green, an English mathematician of the early 19th century. He’s one of those people who gave us our idea of mathematical physics. He’s credited with coining the term “potential”, as in potential energy, and in making people realize how studying this simplified problems. Mostly problems in electricity and magnetism, which were so very interesting back then. On the side also came work in multivariable calculus. His work most famous to mathematics and physics majors connects integrals over the surface of a shape with (different) integrals over the entire interior volume. In more specific problems, he did work on the behavior of water in canals.

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Green’s function.

There’s a patch of (high school) algebra where you solve systems of equations in a couple variables. Like, you have to do one system where you’re solving, say,

6x + 1y - 2z = 1 \\  7x + 3y + z = 4 \\  -2x - y + 2z = -2

And then maybe later on you get a different problem, one that looks like:

6x + 1y - 2z = 14 \\  7x + 3y + z = -4 \\  -2x - y + 2z = -6

If you solve both of them you notice you’re doing a lot of the same work. All the same hard work. It’s only the part on the right-hand side of the equals signs that are different. Even then, the series of steps you follow on the right-hand-side are the same. They have different numbers is all. What makes the problem distinct is the stuff on the left-hand-side. It’s the set of what coefficients times what variables add together. If you get enough about matrices and vectors you get in the habit of writing this set of equations as one matrix equation, as

A\vec{x} = \vec{b}

Here \vec{x} holds all the unknown variables, your x and y and z and anything else that turns up. Your \vec{b} holds the right-hand side. Do enough of these problems and you notice something. You can describe how to find the solution for these equations before you even know what the right-hand-side is. You can do all the hard work of solving this set of equations for a generic set of right-hand-side constants. Fill them in when you need a particular answer.


I mentioned, while writing about Fourier series, how it turns out most of what you do to numbers you can also do to functions. This really proves itself in differential equations. Both partial and ordinary differential equations. A differential equation works with some not-yet-known function u(x). For what I’m discussing here it doesn’t matter whether ‘x’ is a single variable or a whole set of independent variables, like, x and y and z. I’ll use ‘x’ as shorthand for all that. The differential equation takes u(x) and maybe multiplies it by something, and adds to that some derivatives of u(x) multiplied by something. Those somethings can be constants. They can be other, known, functions with independent variable x. They can be functions that depend on u(x) also. But if they are, then this is a nonlinear differential equation and there’s no solving that.

So suppose we have a linear differential equation. Partial or ordinary, whatever you like. There’s terms that have u(x) or its derivatives in them. Move them all to the left-hand-side. Move everything else to the right-hand-side. This right-hand-side might be constant. It might depend on x. Doesn’t matter. This right-hand-side is some function which I’ll call f(x). This f(x) might be constant; that’s fine. That’s still a legitimate function.

Put this way, every differential equation looks like:

(\mbox{stuff with } u(x) \mbox{ and its derivatives}) = f(x)

That stuff with u(x) and its derivatives we can call an operator. An operator’s a function which has a domain of functions and a range of functions. So we can give give that a name. ‘L’ is a good name here, because if it’s not the operator for a linear differential equation — a linear operator — then we’re done anyway. So whatever our differential equation was we can write it:

Lu(x) = f(x)

Writing it Lu(x) makes it look like we’re multiplying L by u(x). We’re not. We’re really not. This is more like if ‘L’ is the predicate of a sentence and ‘u(x)’ is the object. Read it like, to make up an example, ‘L’ means ‘three times the second derivative plus two x times’ and ‘u(x)’ as ‘u(x)’.

Still, looking at Lu(x) = f(x) and then back up at A\vec{x} = \vec{b} tells you what I’m thinking. We can find some set of instructions to, for any \vec{b} , find the \vec{x} that makes A\vec{x} = \vec{b} true. So why can’t we find some set of instructions to, for any f(x) , find the u(x) that makes Lu(x) = f(x) true?

This is where a Green’s function comes in. Or, like everybody says, “the” Green’s function. “The” here we use like we might talk about “the” roots of a polynomial. Every polynomial has different roots. So, too, does every differential equation have a different Green’s function. What the Green’s function is depends on the equation. It can also depend on what domain the differential equation applies to. It can also depend on some extra information called initial values or boundary values.

The Green’s function for a differential equation has twice as many independent variables as the differential equation has. This seems like we’re making a mess of things. It’s all right. These new variables are the falsework, the scaffolding. Once they’ve helped us get our work done they disappear. This kind of thing we call a “dummy variable”. If x is the actual independent variable, then pick something else — s is a good choice — for the dummy variable. It’s from the same domain as the original x, though. So the Green’s function is some G(f, s) . All right, but how do you find it?

To get this, you have to solve a particular special case of the differential equation. You have to solve:

L G(f, s) = \delta(x - s)

This may look like we’re not getting anywhere. It may even look like we’re getting in more trouble. What is this \delta(x - s) , for example? Well, this is a particular and famous thing called the Dirac delta function. It’s called a function as a courtesy to our mathematical physics friends, who don’t care about whether it truly is a function. Dirac is Paul Dirac, from over in physics. The one whose biography is called The Strangest Man. His delta function is a strange function. Let me say that its independent variable is t. Then \delta(t) is zero, unless t is itself zero. If t is zero then \delta(t) is … something. What is that something? … Oh … something big. It’s … just … don’t look directly at it. What’s important is the integral of this function:

\int_D\delta(t) dt =  0, \mbox{ if 0 is not in D} \\  \int_D\delta(t) dt = 1, \mbox{ if 0 is in D}

I write it this way because there’s delta functions for two-dimensional spaces, three-dimensional spaces, everything. If you integrate over a region that includes the origin, the integral of the delta function is 1. If you integrate over a region that doesn’t, the integral of the delta function is 0.

The delta function has a neat property sometimes called filtering. This is what happens if you integrate some function times the Dirac delta function. Then …

\int_D f(t)\delta(t) dt =  0, \mbox{ if 0 is not in D} \\  \int_D f(t)\delta(t) dt = f(0), \mbox{ if 0 is in D}

This may look dumb. That’s fine. This scheme is so good at getting rid of integrals where you don’t want them. Or at getting integrals in where it’d be convenient to have.

So, I have a mental model of what the Dirac delta function does. It might help you. Think of beating a drum. It can sound like many different things. It depends on how hard you hit it, how fast you hit it, what kind of stick you use, where exactly you hit it. I think of each differential equation as a different drumhead. The Green’s function is then the sound of a specific, uniform, reference hit at a reference position. This produces a sound. I can use that sound to extrapolate how every different sort of drumming would sound on this particular drumhead.

So solving this one differential equation, to find the Green’s function for a particular case, may be hard. Maybe not. Often it’s easier than some particular f(x) because the Dirac delta function is so weird that it becomes kinda easy-ish. But you do have to find one solution to this differential equation, somehow.

Once you do, though? Once you have this G(x, s) ? That is glorious. Because then, whatever your f is? The solution to Lu(x) = f(x) is:

u(x) = \int G(x, s) f(s) ds

Here the integral is over whatever the domain of the differential equation is, and whatever the domain of f is. This last integral is where the dummy variable finally evaporates. All that remains is x, as we want.

A little bit of … arithmetic isn’t the right word. But symbol manipulation will convince you this is right, if you need convincing. (The trick is remembering that ‘x’ and ‘s’ are different variables. When you differentiate with respect to ‘x’, ‘s’ acts like a constant. When you integrate with respect to ‘s’, ‘x’ acts like a constant.)

What can make a Green’s function worth finding is that we do a lot of the same kinds of differential equations. We do a lot of diffusion problems. A lot of wave transmission problems. A lot of wave-transmission-with-losses problems. So there are many problems that can all use the same tools to solve.

Consider remote detection problems. This can include things like finding things underground. It also includes, like, medical sensors. We would like to know “what kind of thing produces a signal like this?” We can detect the signal easily enough. We can model how whatever it is between the thing and our sensors changes what we could detect. (This kind of thing we call an “inverse problem”, finding the thing that could produce what we know.) Green’s functions are one of the ways we can get at the source of what we can see.

Now, Green’s functions are a powerful and useful idea. They sprawl over a lot of mathematical applications. As they do, they pick up regional dialects. Things like deciding that LG(x, s) = - \delta(x - s) , for example. None of these are significant differences. But before you go poking into someone else’s field and solving their problems, take a moment. Double-check that their symbols do mean precisely what you think they mean. It’ll save you some petty quarrels.


I should have the ‘H’ essay in the Fall 2019 series on Thursday. That and all other Fall 2019 A To Z posts should be at this link.

Also, I really don’t like how those systems of equations turned out up at the top of this essay. But I couldn’t work out how to do arrays of equations all lined up along the equals sign, or other mildly advanced LaTeX stuff like doing a function-definition-by-cases. If someone knows of the Real Official Proper List of what you can and can’t do with the LaTeX that comes from a standard free WordPress.com blog I’d appreciate a heads-up. Thank you.

A Summer 2015 Mathematics A To Z: well-posed problem


Well-Posed Problem.

This is another mathematical term almost explained by what the words mean in English. Probably you’d guess a well-posed problem to be a question whose answer you can successfully find. This also implies that there is an answer, and that it can be found by some method other than guessing luckily.

Mathematicians demand three things of a problem to call it “well-posed”. The first is that a solution exists. The second is that a solution has to be unique. It’s imaginable there might be several answers that answer a problem. In that case we weren’t specific enough about what we’re looking for. Or we should have been looking for a set of answers instead of a single answer.

The third requirement takes some time to understand. It’s that the solution has to vary continuously with the initial conditions. That is, suppose we started with a slightly different problem. If the answer would look about the same, then the problem was well-posed to begin with. Suppose we’re looking at the problem of how a block of ice gets melted by a heater set in its center. The way that melts won’t change much if the heater is a little bit hotter, or if it’s moved a little bit off center. This heating problem is well-posed.

There are problems that don’t have this continuous variation, though. Typically these are “inverse problems”. That is, they’re problems in which you look at the outcome of something and try to say what caused it. That would be looking at the puddle of melted water and the heater and trying to say what the original block of ice looked like. There are a lot of blocks of ice that all look about the same once melted, and there’s no way of telling which was the one you started with.

You might think of these conditions as “there’s an answer, there’s only one answer, and you can find it”. That’s good enough as a memory aid, but it isn’t quite so. A problem’s solution might have this continuous variation, but still be “numerically unstable”. This is a difficulty you can run across when you try doing calculations on a computer.

You know the thing where on a calculator you type in 1 / 3 and get back 0.333333? And you multiply that by three and get 0.999999 instead of exactly 1? That’s the thing that underlies numerical instability. We want to work with numbers, but the calculator or computer will let us work with only an approximation to them. 0.333333 is close to 1/3, but isn’t exactly that.

For many calculations the difference doesn’t matter. 0.999999 is really quite close to 1. If you lost 0.000001 parts of every dollar you earned there’s a fine chance you’d never even notice. But in some calculations, numerically unstable ones, that difference matters. It gets magnified until the error created by the difference between the number you want and the number you can calculate with is too big to ignore. In that case we call the calculation we’re doing “ill-conditioned”.

And it’s possible for a problem to be well-posed but ill-conditioned. This is annoying and is why numerical mathematicians earn the big money, or will tell you they should. Trying to calculate the answer will be so likely to give something meaningless that we can’t trust the work that’s done. But often it’s possible to rework a calculation into something equivalent but well-conditioned. And a well-posed, well-conditioned problem is great. Not only can we find its solution, but we can usually have a computer do the calculations, and that’s a great breakthrough.