In this post, we will introduce Green's functions as a tool for solving ordinary differential equations. Green's functions allow us to convert differential equations that we don't know how to solve into integral equations that we do know how to solve. Let's see how this works by diving into an example. We consider a physics problem that one might encounter in a kinematics course.
Alice is driving from work back home on a long, straight road on a very dark night, so she can't see how far she has driven just by looking out the window. Nonetheless, the lights illuminating her clock and speedometer still work, so she can see how fast she is going at any given time. If Alice's work is located at position along the road, and if she leaves work at time , then can Alice use the information from her clock and speedometer to determine her position along the road at a given time ?
The answer to this question is yes and amounts to Alice needing to solve a certain differential equation. To see this, let denote Alice's position as a function of time along the road, and let denote her velocity as a function of time. Alice knows because she can determine her velocity at any time by looking at her speedometer and clock, but she doesn't know . However, Alice knows that the derivative of her position with respect to time is equal to her velocity at that time. This fact can be mathematically written as
and the fact that she left work, which is located at , at time can be mathematically written as
Therefore, to determine , Alice needs to solve the first order differential equation above subject to this initial condition. In math, solving such a differential equation subject to an initial conditions is called an "initial value problem" for this reason. You may already know how to solve this equation without using Green's functions, but we want to use Green's functions here to explain how they arise in a simple example. Here's how we proceed.
First, for each real number , we find a function that has the following property
where is the Dirac delta function. I've used a "G" to label this function because turns out to be an example of a Green's function. I won't get into how we find such a function just yet because I want to show you how it would be useful first. Next, let be given, and notice that given any satisfying we have
The integral on the left can be integrated by parts to give
Putting these two facts together, and evaluating the boundary term coming from integration by parts at the endpoints and , we get
Since we were originally given the initial value of but don't know its final value for a given , it would be helpful also satisfied the following condition:
We will see soon that it is always possible to pick in this way, so we assume that this has been done and that therefore, the equation for reduces to
In the beginning, I said that Green's functions can be used to convert differential equations into integral equations that we know how to solve. If we were to have an explicit form for , then we would have achieved this because we have written an integral equation in which is written as a expression involving integrals of known functions and . This integral may still be hard to perform in practice, but at least we have solved for in principle.
We now turn to the question of how we determine an explicit expression for . In particular, we are looking for that satisfies
Recall that if , then , so the differential equation satisfied by gives
This implies that is constant in both the region and in the region . These constants must be different because otherwise the derivative of at would not be infinite as is necessary for it to behave like a delta function. Thus, there are real numbers and with such that
In other words, we see that for a given , plotting as a function of gives a step function that jumps from the value for to the value for . A nice, compact way of writing this is
where is the Heaviside step function defined as
The condition for applied to our latest expression for gives
so . Using this and the fact that the derivative of the Heaviside step function is the Dirac delta;
which means that by our original equation . Putting this together gives
If we plug this into the integral equation we got for , then we find
which for any gives
This is precisely the solution you would have obtained if you were to have integrated both sides of our original differential equation from to . Although that method of direct integration would be much simpler for this differential equation, there are instances of other differential equations where such direct integration is not possible, and Green's functions become very useful.
At this point, I'd like to try and help the reader develop some physical intuition for the Green's function we just derived. Let us think of as the position of an object as a function of . Then for all times , the object is just sitting still at , then at time , the position of the object jumps discontinuously from to , and for all times , the object is standing still again. This would happen if somehow that object could instantaneously accelerate from zero velocity to infinite velocity precisely when has value , and then instantaneously decelerate back to zero velocity. You might think of this as a spaceship that fires it's rear thrusters on full blast very abruptly and for a very short period of time, moves a certain amount essentially instantaneously, and then very abruptly fires its front thrusters to stop essentially instantaneously.
Last updated: February 9, 2013