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From: https://en.wikipedia.org/wiki/Green%27s_function

"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?

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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.

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  • $\begingroup$ Sorry for the short delay in accepting, although i am still confused--so is my equation for the impulse response wrong? $\endgroup$
    – user45664
    Commented Aug 21, 2019 at 18:22
  • $\begingroup$ 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. $\endgroup$
    – user45664
    Commented Aug 21, 2019 at 18:29
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    $\begingroup$ 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 $f$. $\endgroup$
    – I.C.
    Commented Aug 22, 2019 at 19:09

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