Recently, we have seen that Laplace transforms can be used to reduce ordinary differential equations to algebraic equations.

Today, we will have a look at another technique that allows reducing a differential equation to the evaluation of a single integral. This method is called “Green’s functions”.

Consider again the differential equation

with initial conditions 𝑦(0)=𝑦′(0)=0. Last time, we had 𝑓(𝑡)=cos4𝑡. This time, we leave the forcing function completely general. So we are trying to find the solution for a complete class of differential equations. And our solution will be based on the solution of the specific simpler equation

where 𝛿(𝑡−𝑡′) is the Dirac delta function that we have met recently in this article:

The forcing function here is a single force impulse at time 𝑡′. For all other points in time, the force is zero. The idea is: once we have a solution for this equation, we can “add up” (integrate) similar solutions for impulses at various times. We call the solution of a differential equation, where the inhomogeneous term is a Dirac pulse, Green’s function , 𝐺(𝑡,𝑡′). And the idea of adding up amounts to:

It’s easy to see that this is true. We just insert this solution in the differential equation:

So the integral in equation 3 really satisfies the differential equation. All that remains is to evaluate the integral for given 𝑓(𝑡). But wait, we still need to determine 𝐺(𝑡,𝑡′). That’s what we turn to now.

Start a Jupyter notebook and define our symbols and differential equation:

To solve this equation, we use the method of Laplace transforms. So we transform the equation, first the left-hand side

and insert the initial conditions

Then we transform the right-hand side:

So the transformed equation reads

Now we solve for the Laplace transform of G:

and transform back to normal space:

So we have that Green’s function for differential equation 1 with initial conditions 𝑦(0)=𝑦′(0)=0 is

So we immediately get the solution for any forcing function 𝑓(𝑡) by solving the integral

Let’s try this for the special case 𝑓(𝑡)=cos2𝑡:

Let’s check, if this really satisfies the differential equation:

which is the right-hand side f(t) . So that’s ok. Does it also satisfy the initial conditions?

Yes, it does, and we are done. For other differential equations or other initial conditions, we get different Green functions. But once we have a Green function, we basically have the solution, no matter what forcing function we apply.

This is really just a teaser and a tiny step into the huge field of Green’s functions. In fact, a large part of quantum field theory is an applied theory of Green’s function. So it is definitely worthwhile to look deeper into the topic. If you want to dig deeper, I can highly recommend A. Zee’s wonderful book “ Quantum Field Theory in a Nutshell ” (no, I don’t get any money from that link).