# Green's Function / Impulse Response Confusion

"In mathematics, a Green's function is the impulse response of an inhomogeneous linear differential equation defined on a domain, with specified initial conditions or boundary conditions.

A Green's function, $G(x,s)$, of a linear differential operator $L = L(x)$ acting on distributions over a subset of the Euclidean space $ℝn$, at a point $s$, is any solution of $$LG(x,s)=\delta(x-s)$$ where $δ$ is the Dirac delta function."

But it seems to me that the impulse response, $I$, would be

$$L(\delta(x-s))=I(x,s)$$

Then $I$ and $G$ are not the same. That is, $I$ is the output from $L$ due to an impulsive input and $G$ is the input to $L$ that produces an impulsive output.

My Question: How can the impulse response be the same as the Green's function given these definitions?

You are simply reading it backwards. When they say "impulse response of an inhomogeneous linear differential equation defined on a domain, with specified initial conditions or boundary conditions" they mean the function $G(x,s)$ such that $L(G(x,s)) = \delta(x-s)$.

I know it is a little counterintuitive at first since one thinks on the differential operator $L$ as something that you feed with a function and gives back a function. But in this case the operator we are dealing with is in fact the solution operator of the differential equation. That is $S:X\rightarrow Y$, where $X$, $Y$ are function spaces such that $S(f) = u$ when \begin{equation} \begin{cases} Lu=f &\text{ in } \Omega\\ u = 0 &\text{ in } \partial\Omega \end{cases} \text{ (It can also be a different boundary condition depending on the problem)} \end{equation}

Think of the Green functions and the $\delta$ in the following way to notice why this is useful, the $\delta$ is "kind of a base of the functions spaces" since you can "write" any function as \begin{gather} f(x)"=" \sum_s f(s)\delta(x-s)\\ \text{ (It really is an integral not a sum, in fact is a convolution integral)} \end{gather} And, since the solution operator is linear, then to solve the problem for a general $f$ it is enough to solve it for the deltas and then you just sum using superposition. The Green functions are just the solutions of the deltas, that is \begin{equation} G(x,s) = S(\delta(x-s))\\ \end{equation} so \begin{equation} u(x) = S(f)(x) "=" \sum_s f(s)S(\delta(x-s)) = \sum_s f(s)G(x,s) \end{equation}

Notice that $s$ is a parameter not the variable of the delta function so $f(s)$ is a constant for the solution operator. Notice also that this is not a really formal answer, but I hope it is useful to star understanding what the Green functions are about.

• Sorry for the short delay in accepting, although i am still confused--so is my equation for the impulse response wrong? Aug 21 '19 at 18:22
• It seems to me that in $L(G(x,s)) = \delta(x-s)$ , $\delta(x-s)$ is occupying the place of a forcing function in a non homogeneous equation. Aug 21 '19 at 18:29
• You may think that $\delta$ is the input of the solution operator. Not the equation itself. By solution operator I mean that which takes the forcing term and gives you a solution for that forcing term. Think on the linear matrix equation $Ax=y$ then the solution operator is the one that gives you $x$ when you provide $y$. In this case you alsao have a linear equation $L(u) = f$ and the solution operator gives you $u$ if you give $f$.
– I.C.
Aug 22 '19 at 19:09