6
$\begingroup$

This is a curious question about the way George Green could have defined his Green's function. All the definitions I see have only Dirac-delta $\delta(x−x′)$ function as their source on the RHS. But the Dirac-delta was defined about a century after Green's work.

I just want to know, how George Green have defined his function (in case he defined it)? In case he didn't define his Green's function, how did he give a hint about this method (without Dirac-delta function) ?

$\endgroup$
8
$\begingroup$

It is by no means necessary to introduce the Dirac delta. For instance, $G(x,x')$, singular for $x=x'$ is a Green's function of $\Delta$ if, defining, $$\phi(x):= \int_A G(x,x')f(x') dx'$$ we have $$\Delta \phi(x) = f(x)$$ where further details on the behaviour of $\phi$ and the regularity of $f$ and the nature of the integration domain $A$ are assumed and I omit them here...

The general idea is that $G$ is the integral kernel of the inverse of a given linear operator, nothing further. In some cases the inverse exists and is described by an integral kernel. Under some regularity hypotheses this integral kernel is a distribution in a proper sense and the equation defining it can be recast into the modern language of distributions.

$\endgroup$
  • 1
    $\begingroup$ Thanks. The definition in the first equation makes perfect sense. However, I hope Green would have just defined this by first definition you have said, since the definition of linear operators and their inverses was quite far ahead of his time. $\endgroup$ – user35952 Jun 28 '14 at 15:40
  • $\begingroup$ @V: I am curious as to why the name kernel was chosen for the inverse. Does it have any relation with how the "kernel" (set of points in domain which map to 0) of a map behaves? $\endgroup$ – user7757 Jun 29 '14 at 10:47
  • $\begingroup$ Actually I so not know. The full name is actually "integral kernel". I do not think there is a relation with the algebraic kernel... $\endgroup$ – Valter Moretti Jun 29 '14 at 12:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.