What does "translation invariant" mean for linear differential functionals? I am not super experienced in solving PDEs. Today I was reading about Green's functions, from Wolfram Mathworld and Wikipedia, since the notes in my course focus a bit to much on a special case and I would like to have something a bit more general. I understand most of the theory, but there is a particular part that I can't seem to find any good resources for. So first, I'll present the problem and then discuss my issues with what Wikipedia has.
Given a linear differential functional $\mathscr{L}:F\mapsto \mathscr{L}[F]$ and a corresponding PDE
$$\mathscr{L}[u]=f\tag{1}$$
What is the most general solution for $u:\mathbb{R}^n\to\mathbb{R}$ ?
I'm sure many of you are familiar with Green's functions and whatnot so I won't go through the theory. It's the following part I am having trouble understanding -

If the operator is translation invariant, that is when $\mathscr{L}$ has constant coefficients with respect to $x$, then the Green's function can be written as a convolution kernel,
$$G(x,s)=G(x-s)$$

First question - what is the exact meaning of "translation invariant" with regard to functionals? To me based on intuition of linearity from linear algebra it seems pretty obvious that $\text{constant coefficients}\implies \text{translation invariant}$ but I am doubtful of the converse, namely
$$\text{translation invariant}\overset{?}{\implies}\text{constant coefficients}$$
Second question - What is the meaning of $G(x,s)=G(x-s)$ ? $G$ has gone from being a function $G:\mathbb{R}^n\times \mathbb{R}^n\to\mathbb{R}$ to instead being a function $G:\mathbb{R}^n\to\mathbb{R}$. What's going on here?
Final question - Once we have found the Green's function, what is the most general solution $u_{\text{G}}$ to the above? Letting $u_{\text{H}}(x)$ be the solution of $\mathscr{L}[u]=0$ and $u_{\text{P}}$ be the solution to (1), is it simply
$$u_{\text{G}}=u_{\text{H}}+u_{\text{P}}$$
As in the case of linear ordinary differential equations? Or is it not that simple?
Thanks everyone.
 A: 
First question - what is the exact meaning of "translation invariant" with regard to functionals?

Many, many people write "invariant" when they mean "equivariant," and these are very different conditions. Write $L_t$ for the operator given by translation by $t$, which means
$$(L_t u)(x) = u(x - t).$$
Invariance would mean that the differential operator $\mathscr{L}$ satisfies
$$\mathscr{L}(L_t u) = \mathscr{L}(u).$$
This is not the intended condition; for example, it is already not satisfied by $\mathcal{L} = \frac{d}{dx}$ when $n = 1$. Actually this condition is not satisfied by any nontrivial differential operator. The intended condition is equivariance, which means the differential operator satisfies
$$\mathscr{L}(L_t u) = L_t \mathscr{L}(u).$$
It's a nice exercise to check that (under mild hypotheses, e.g. $\mathscr{L}$ has smooth coefficients or something like that) this condition is equivalent to the condition that $\mathscr{L}$ commutes with every partial derivative operator $\frac{\partial}{\partial x_i}$. This in turn is equivalent to the condition that every coefficient of $\mathscr{L}$ (writing $\mathscr{L}$ as a linear combination of monomials in the partial derivatives, with coefficients given by smooth functions) is annihilated by every partial derivative operator $\frac{\partial}{\partial x_i}$, which gives that they must be constant.

Second question - What is the meaning of $G(x,s)=G(x-s)$ ? $G$ has gone from being a function $G:\mathbb{R}^n\times \mathbb{R}^n\to\mathbb{R}$ to instead being a function $G:\mathbb{R}^n\to\mathbb{R}$. What's going on here?

This is imprecise. What is meant here is that there is some function (really a distribution in general) $F : \mathbb{R}^n \to \mathbb{R}$ such that $G(x, s) = F(x - s)$.

Final question - Once we have found the Green's function, what is the most general solution $u_{\text{G}}$ to the above? Letting $u_{\text{H}}(x)$ be the solution of $\mathscr{L}[u]=0$ and $u_{\text{P}}$ be the solution to (1), is it simply
$$u_{\text{G}}=u_{\text{H}}+u_{\text{P}}$$
As in the case of linear ordinary differential equations? Or is it not that simple?

There is something to be careful about with boundary conditions here. I know very little about PDEs so I'll wait for someone else to comment, though.
A: I have been working with Green's function for the heat equation lately, so maybe I can help with some points of your question. I'm not sure about the translation part, thus, I will jump straight to your second question.
The heat equation $$u_t -\Delta u = 0$$ is in my opinion translation invariant. The fundamental solution to the heat equation is sometimes stated as $$\Gamma(x,y,t) = \frac{1}{(4\pi t)^{n/2}} e^{-\frac{|x-y|^2}{4t}}$$
for $x,y\in \mathbb{R}^n$, $t>0$. We can now build a solution for the heat equation with initial data $u(x,0)=u_0$ as described in the Wikipedia article $$u(x,t) = \int_{\mathbb{R}^n} \Gamma(x,y,t) u_0(y) \, \mathrm{d}y.$$ But often the fundamental solution of is stated as $$\Phi(x,t) = \frac{1}{(4\pi t)^{n/2}} e^{-\frac{|x|^2}{4t}}.$$ In this case we have $\Gamma(x,y,t)=\Phi(x-y,t)$ and the solution becomes $$u(x,t) = \int_{\mathbb{R}^n} \Phi(x-y,t) u_0(y) \, \mathrm{d}y,$$ thus, the solution has the form of a convolution which wasn't the case for $\Gamma$. But a general parabolic differential operator might not have a fundamental solution $\Gamma$ that can be written that way.
For your last question, I'm not quite sure what you mean by most general solution, but solutions using Green's function can often be represented as you describe. For exmaple for evolution equations this is basically Duhamel's principle. Duhamel's principle states that a solution of the problem \begin{align} u_t(x,t)-\Delta u(x,t) &= f(x,t) \\ u(x,0)&=u_0(x)\end{align} is given by $$u(x,t) = \int_{\mathbb{R}^n} \Phi(x-y,t) u_0(y) \, \mathrm{d}y + \int_0^t\int_{\mathbb{R}^n} \Phi(x-y,t-s) f(y,s) \, \mathrm{d}y\mathrm{d}s,$$ where the first term solves the homogeneous heat equation with initial data $u_0$ and the second term solves the non-homogeneous problem with zero initial data.

Edit: In general if $L$ is linear and you have a solution $L u_H=0$ and a solution $L u_P = f$, then $u=u_H+u_P$ is another solution for $Lu=f$.
In the setting of the Wikipedia article, where $L G(x,s)=\delta(x-s)$ we can build a solution for $L u(x)= f(x)$ by setting $$u(x) = \int G(x,s) f(s) \, \mathrm{d}s.$$ This means if you want to build a solution for $L u=0$ then $$u(x) = \int G(x,s) \cdot 0 \, \mathrm{d}s =0,$$ thus, using a green's function this results in $u_H=0$ in your setting.
