PDEs, boundary conditions, and unique solvability

Question

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!

Status Update: 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? I think there are (see below).

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 $L$ or the zero set of $P(\xi)$. For instance, if $L=-c^{-2}\partial_t^2+\partial_x^2$ with $c>1$, then the corresponding equation has infinitely many solutions parametrized by Cauchy data on $t=0$. However, these solutions would correspond to waves pulses propagating along the $x$-axis at a speed faster than the background speed of light. On the other hand, the solution $u(t,ct)$ must vanish for large $t$ as $(1,c)$ is a space-like vector. I believe this is enough to show that this $L$ also admits only trivial solutions. My reasoning here is based on the exact representation of solutions via the D'Alambert formula, but that doesn't generalize easily. Any idea what kind of geometric condition could be used for more general operators?

Ultimately, I'd like to know something about the geometry of the space of operators that satisfy my criteria. Say I fix the maximal order to make things easier. The space is clearly not linear, but could it be convex or the complement of a convex set? (These last guesses are probably not right. I'm just throwing out ideas.) I'd be happy if I could understand this space for just first and second order operators, preferably with hints of how this understanding could generalize to higher orders.

Answer 1

Hi, I am adding another answer because this suggests a rather different approach then what I have outlined before, and this is targeted at the fact you are willing to grant smooth with compact support on any space-like hyperplane.

If you are willing to let your solutions vanish in such a large set, then the proper tool for the analysis is the theory of unique continuation for solutions to linear differential equations. For example, solutions to elliptic equations tend to have the property of strong unique continuation: if the solution vanish on any open set it must vanish everywhere. Using the geometry you can weaken ellipticity to have cases where no derivatives are taken in certain directions ($\partial_x^2 + \partial_y^2$, for example).

Similarly you can address parabolic type operators. (They have infinite speeds of propagation as well.)

For hyperbolic type, if the coefficients of the operator are real analytic, then you can use Holmgren's Uniqueness theorem if the inner-most characteristic cone of the operator remains outside of the Minkowskian null cone: the compact-support criterion on space-like (to back ground metric) slices will guarantee that there will be a time-like surface (relative to the differential operator, not to the background metric) to one side of which the solution vanishes, then Holmgren kicks in and tells you that the solution must vanish everywhere.

(A version of Holmgren's also applies to ultrahyperbolic operators, with suitable changes in the geometry.)

Outside of these trivial cases, you need to actually consider the geometry of the characteristic cones and compare them against the geometry of the support set of your function. The keywords you will be looking for are Carleman estimates, unique continuation, and pseudo-convexity. There's a nice, comprehensive treatment in Lars Hormander's Analysis of Linear Partial Differential Operators (I think volume III, may be IV?); there are also some nice notes by Daniel Tataru on his website.

Answer 2

I am somewhat doubtful that the question as posed as any sort of reasonable answer. (Also, I don't really see how the Lorentzian metric even enter into the problem.)

(a) ANY hyperbolic PDE in (1+3) dimensions will, by definition, not have unique solution for your problem.

(b) Even elliptic PDEs are not guaranteed to have unique solutions, even for nice ones that come from minimization of a strictly convex energy functional: Euclidean space simply has too large a symmetry group, and if the equation itself is invariant under translations, any finite translate of a solution that decays at spatial infinity is another solution.

(c) If you restrict yourself to linear scalar PDEs $L[u] = 0$, by linearity, $u \equiv 0$ is necessarily a solution that satisfy your conditions. So your question as posed reduces to "which linear operator admits only the trivial solution". For constant coefficient $L$'s you can simply take the Fourier transform of both sides and study the zero-set of the PDO.

If you clarify your motivation for considering the question your are asking, maybe more reasonable answers can be given.

Edit: To say a bit more about the Fourier transform method. For example, if you assume that your solution is in Schwarz space on spatial slices, and assume the the PDO $L$ does not have $\partial_t$ terms, then you can ignore the time component. Restricting to the spatial slice and taking Fourier transform, you see that $P(\xi)\hat{u}(\xi) = 0$ where $P$ is the symbol for $L$, and is a polynomial in $\xi$. Using that the Fourier transform is injective on $\mathcal{S}$, you immediately have that $\hat{u}$ can only be supported on the zero set of $P$, but as $P$ is a polynomial, its zero set has vanishing measure unless $P\equiv 0$. Thus you see that in Schwarz space any non-trivial constant coefficient $L[u] = 0$ only has trivial solutions. Now, it is well known that the smoothness of $\hat{u}$ is related to the decay properties of $u$. By considering restriction theorems/Strichartz type estimates, there can be solutions for certain PDOs in $L^p$. (As an example, let the space-dimension be 4 [only because I remember the Lebesgue exponent explicitly in this case], Then the PDO on $(t,w,x,y,z)$ given by $-\partial_w^2 + \partial_x^2 + \partial_y^2 + \partial_z^2$ admits infinitely many solutions in $L^4(\mathbb{R}^4)$, with Fourier transform supported on the set $\hat{w}^2 = \hat{x}^2 + \hat{y}^2 + \hat{z}^2$. You can even ask that the solutions be smooth: they just cannot decay too fast in the $w$ direction.)

Edit 2 [After the updated question, one should probably read the comments below first before reading this]: A few things

(i) Intuition from 1+1 dimension can be misleading. Waves don't decay there. In the 1+3 case, solutions to the wave equation disperses. So for the equation give by $L = -c^{-2}\partial_t^2 + \triangle$, even when the wave speed is bigger than the speed of gravity, the solution is not incompatible with mere decay at spatial infinity. (Though of course, since the equation is no-longer Lorentz invariant, speaking of Spatial Infinity is somewhat of a red-herring: the conformal compactification of the space-time does not give a compatible compactification of the solution to the equation.) If to rule out such cases you need to also impose a rate of decay.

(ii) When I said local well-posedness of $L$, implicitly I mean on a suitable function space on space-like slices. For (strictly/symmetric/regularly) hyperbolic equations, you can of course study the characteristic cones and compare against the back ground null cone to have the decay automatically also satisfied for boosted slices.

(iii) As a trivial example, also note that in the $\partial_t v + Av$ formulation, you can take the Fourier transform to get the solution to be $\hat{v}(t,\xi) = e^{-A(\xi)t}\hat{v}_0(\xi)$. If the matrix $A(\xi)$ has only eigenvalues with negative real parts, then Schwarz data will lead to Schwarz solutions, and lead to non-uniqueness of the solutions. This is sort of the explicit version of the Hille-Yoshida type theorems.

Answer 3

You appear to be considering only constant coefficient linear differential operators where every term is of the same order. Is that right?

If so, constant coefficient operators were studied pretty extensively in the 60's and 70's, notably by Leon Ehrenpreis. I recommend that you dig through this literature.

Also, an instructive non-elliptic example to study is the operator $\partial_x^2 + \partial_y^2 - \partial_z^2 - \partial_t^2$. I forget the name of this operator (maybe "hyperlaplacian"?), but for some reason I associate the name Peter Lax with it and maybe there is a discussion in Garabedian's PDE book?

As always, I am curious about why you are interested in this question.