Does there exist a green's function that does not have translation symmetry? 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 $G(x_1,x_2)=G(x_1x_2)$?
 A: 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.
A: 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.
