I'm interested in a criterion that determines whether a linear scalar PDE (arbitrary order) has a unique solution given vanishing boundary conditions at spatial infinity. I'll try to formulate the question more precisely below.

Consider a PDE of the form $L[u]=0$ where $u(t,x,y,z)$ is a scalar function of one time $(t)$ and three spatial variables $(x,y,z)$, though this choice of dimensionality is not central to the question. The function $u$ is required to vanish "sufficiently" fast if the $(x,y,z)$ variables are taken to infinity, keeping $t$ fixed. [If that's not enough, it can also be required to vanish at infinity along any hyperplane that is space-like with respect to the Lorentzian metric $\mathrm{diag}(-1,1,1,1)$.] However, no requirements are put on the behavior of $u$ as $t\to\pm\infty$ for fixed $(x,y,z)$. The linear differential operator $L$ can be assumed to have constant coefficients, but could by of any order. Though, I'd also like to know how the answer generalizes to the case when the coefficients and the background Lorentzian metric are no longer constant.

> So, my question is this: for which operators $L$ does the equation $L[u]=0$ have a unique solution?

Let me give some examples.

* Equation $\partial_z u=0$ has a unique solution. An arbitrary solution comes from integrating the rhs wrt to $z$ and adding any function that's constant wrt to $z$. From the boundary conditions, it is easy to see that both pieces must be zero. Hence, $u=0$ is the unique solution.

* The same argument does not work for $\partial_t u=0$. For any given solution, I can get another solution by adding a function of $(x,y,z)$ only that vanishes at infinity, and there are plenty of those.

* The equation $(\partial_x^2+\partial_y^2+\partial_z^2)u=0$ is uniquely solvable: ignore $t$ dependence and invert the Laplacian, with uniqueness given by the same argument as in the first example.

* The equation $(-\partial_t^2+\partial_x^2+\partial_y^2+\partial_z^2)u=0$ is not uniqely solvable: solutions are parametrized by Cauchy data on, say, the $t=0$ hyperplane, and so are definitely not unique.

These examples make me think that the answer is some version of an ellipticity condition. Unfortunately, I'm only aware of how to formulate this condition for second order systems. Any help appreciated!

<b>Status Update:</b> Willie Wong provided some good, relevant information below. Let me summarize my understanding of it here and then sharpen my question in light of it.

If the partial differential operator (PDO) $L$ contains no time derivatives, then the Fourier transform $\hat{u}$ of a nontrivial solution $u$ must be supported on the zero set of $P(\xi)$, the symbol of $L$. If this set is of measure zero, then $\hat{u}$ can only be a distribution. On the other hand, local regularity of $\hat{u}$ is controlled by the decay of $u$ at infinity. In particular, if $u$ is in the Schwarz space, then $\hat{u}$ cannot be sufficiently singular to be a distribution. Hence, $L[u]=0$ would admit only the trivial solution, and hence be uniquely solvable.

In principle, I can use the same argument for any $L$ that is expressible only in terms of derivatives parallel to a given space-like hyperplane, by appealing to the fact that I've imposed the same boundary conditions at infinity (say Schwarz) for each such hyperplane.

In principle, the above reasoning gives a nice large space of PDOs that satisfy my criteria. But are there any more?

Now, suppose that I cannot ignore time derivatives. Willie's suggestion is to write the equation $L[u]=0$ in evolution form $\partial_t v + Av=0$ and exclude $L$'s for which the evolution equation is well posed as an initial value problem. But not all such $L$ can be excluded, since not all initial data generates solutions that satisfy the boundary conditions (decay at infinity along any space-like direction). I'm thinking that there should be a geometric condition involving the background null cone and the characteristics of $L$. <i>TO BE CONTINUED...</i>