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I noticed that most Green's functions I have used take on the following functional form $G(x_1,x_2)=G(|x_1-x_2|)$. I assume these subsets of Green's functions are translationally invariant? Correct me if I am wrong.

Does there exist a green's function that does not have translation symmetry?

For example $G(x_1,x_2)=G(x_1x_2)$?

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    $\begingroup$ Can you explain why you want $G(x_1,x_2)=G(x_1x_2)$? Because Green functions are only a tool to solve PDE, and it will be highly depends on which PDE you have and then you decide what green functions you want. $\endgroup$
    – spatially
    Jan 18, 2015 at 1:18
  • $\begingroup$ @wisher I do not have a specific pde problem, yet. I just noticed that in the general problem translational symmetry may not be obtainable. I was unaware of any such green's function. Knowing one would give a me a tangible solution. $\endgroup$ Jan 18, 2015 at 1:29
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    $\begingroup$ You call $G(x_1,x_2)=G(x_1x_2)$ as "translation symmetry"? Sorry but should it be $G(x_1,x_2)=G(x_2,x_1)$? $\endgroup$
    – spatially
    Jan 18, 2015 at 1:33
  • $\begingroup$ @wisher No. $G(x_1x_2)$ is my example of a greens function without translation symmetry. I assume translation symmetry means ${x_1,x_2}\rightarrow{x_1-a,x_2-a}$ leaves the greens function unchanged. $\endgroup$ Jan 18, 2015 at 1:46

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Yes, take any fundamental solution of a PDE which doesn't have translational symmetry, ex: the fundamental solution of $Ly=xy', y'(1)=0\,.$ If $Ly=f\,, f(1)=0$ then $\displaystyle y(x)=\int_{-\infty}^\infty \chi_{[1,x]}(y)\frac{f(y)}{y}\,dy\,.$ So the fundamental solution is $E(x,y)=\frac{\chi_{[1,x]}(y)}{y}$ which isn't translationally invariant.

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Ok, this is not an exact answer but it will give you an idea.

Usually we build up green's function based on the Fundamental solutions of certain PDEs. For example, the laplace operator $-\Delta u=0$ is the most basic example. Notice that Laplace operator is translation invariant and hence the Fundamental solution of laplace operator is translation invariant as well.

If you look at how we build up the green functions, you will find that Green functions actually just a modification of Fundamental solution around the boundary, in order to match the boundary condition of your PDEs. Hence, your problem could be reduced to ask: is there a Fundamental solution that does not preserve translation invariant? Or evan just ask that is there a PDE operator which does not preserve translation invariant?

The answer I would say is no. Just think in this way: your PDE is actually does not depend on where is your coordinate system is, i.e., I could move the origin of my coordinate system wherever I want but without change the behave of my PDE, and hence the behave of Fundamental solution will not be changed.

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  • $\begingroup$ Why doesn't the PDE doesn't depend on where the coordinate system is? $\endgroup$
    – JLA
    Jan 18, 2015 at 3:10

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