Solving of Inhomogeneous Differential Equations

Let L be a differential operator.

Say we want to solve the equation L u(x) = f(x) for u

Then what we'd like to say is L^{-1} L u(x) = L^{-1} f(x), that is u(x) = L^{-1} f(x).

However, if L U(x) = 0 has nontrivial solutions (i.e. more than one solution: the trivial solution and another one), then L apparently cannot be injective. Hence this doesn't work.

So if L is surjective, then at least there is a way to do:

L (L^{-1} u) = u (called right inverse)
L G = 1 (the right-inverse, note that 1 is an operator)
(L G) u = 1 u
L (G u) = 1 u

So we make the problem worse, so to speak, in order to solve for u.

One of the possible ways to construct G (which you need in order to make u worse) and (G u) is the Convolution. (G is called the Green Function)


Convolution

(f*g)(x):=∫\limits_{-∞}^{∞} dx' f(x - x') g(x')

Properties of Convolution

f*g = g*f

δ*f = f*δ = f


Let's return to the Green Function.

Let the solution to the following be known:

L G(x) = δ(x)

Then this G(x) is a possible Green Function.

L (G u) = 1 u
L y = f
L y = δ*f
L y = (L G)*f
L y = L (G*f)
y = G*f

Example

Let's say we want to solve a linear differential equation with constant coefficients (i.e. a translation invariant equation)

Let's say we have a inhomogenity f(x) acting on it:

L u(x)=f(x)

If we know a functional G so that the following holds, we can go much further:

L G(x,s) = δ(x-s)
∫(L G(x,s))⋅f(s)⋅ds = ∫δ(x-s)⋅f(s)⋅ds
L ∫G(x,s)⋅f(s)⋅ds = ∫δ(x-s)⋅f(s)⋅ds
L ∫G(x,s)⋅f(s)⋅ds = f(x)

L ∫G(x,s)⋅f(s)⋅ds = f(x)
L u(x) = f(x)

L ∫G(x,s)⋅f(s)⋅ds = L u(x)

So maybe

∫G(x,s)⋅f(s)⋅ds = u(x)