Let us consider the Klein-Gordon equation $$(\Box +m^2)u=0,$$ where $u$ is a scalar valued function, $m\geq 0$, $\Box=\frac{\partial^2}{\partial x_0^2}-\sum_{i=1}^d\frac{\partial^2}{\partial x_i^2}$.

In the physics literature, in particular in QFT, one sometimes says that if the unknown function $u$ satisfies boundary conditions then $u=0$.

Question. Under what decay assumptions at infinity are the solutions of the above equation necessarily trivial?

I am mainly interested in such conditions of physical interest, say used in QFT.

UPDATE. Below is a concrete example of a situation of interest. In comments I briefly describe yet another situation which is typical for QFT and more interesting, but mathematically unrigorous. As I was told, the second situation is not appropriate for this site.

Consider now the classical (i.e. not quantized) free electromagnetic field $A_\nu$. It satisfies the equation $$\Box A_\nu-\partial_\nu (\partial\cdot A)=0.$$ It is gauge invariant, i.e. $A_\nu+\partial_\nu f$ satisfies the same equation for any function $f$. Under a change under such a gauge transformation we may assume that $\partial\cdot A=0$ (Lorentz gauge). Then the equations become $$\Box A_\nu=0,\,\,\, \partial\cdot A=0.$$ It is still invariant under adding to $A_\nu$ the expression $\partial_\nu f$ where $\Box f=0$. What are the natural boundary conditions on all the electromagnetic fields which would guarantee that $f=0$?

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    $\begingroup$ In QFT it is usually either compact support funtions or Schwartz functions. $\endgroup$
    – Slereah
    Jun 24, 2016 at 9:01
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    $\begingroup$ That's rather strange statement and it's hard to clearly understand what it means out of context. Can you provide a source where such a statement actually appears? One possible interpretation is that $u(x_0,x_i) = 0$ when $u(0,x_i) = 0$ and $\frac{\partial}{\partial x_0} u(0,x_i) = 0$, by the well-posedness of the initial value problem for the Klein-Gordon equation. $\endgroup$ Jun 24, 2016 at 9:32
  • $\begingroup$ @IgorKhavkine: Let us consider the case $m=0$. Let $A_\nu$ be an EM field (it does not have to be free). When one computes various path integrals over $A_\nu$ it is necessary to fix a gauge. Let's fix the Lorentz gauge $\partial^\nu A_\nu=0$. One integrates over such fields satisfying some boundary conditions I would like to understand. One can still make a gauge transformation $A_\nu\mapsto A_\nu+\partial_\nu f$ where $\Box f=0$. One would like to have that the boundary conditions implied that $f=0$, i.e. there is no non-trivial gauge transformations. $\endgroup$
    – asv
    Jun 24, 2016 at 9:42
  • $\begingroup$ @IgorKhavkine: As far as I understand, when one computes amplitudes of various scattering processes the boundary conditions may depend on the choice of In and Out states. $\endgroup$
    – asv
    Jun 24, 2016 at 9:46
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    $\begingroup$ Unfortunately, since you have not actually formulated a concrete mathematical question, your post is not really appropriate for MO. From the context, I can tell that a better forum for your question would be physics.SE. But, just to guide your thinking: the heuristic that you mentioned comes from thinking of a mechanical system with generalized position $q(t)$ on an interval $[t_0,t_1]$, where fixing the values of $q(t_i)$ usually uniquely fixes $q(t)$ (which could be $q(t)=0$). $\endgroup$ Jun 24, 2016 at 9:57

1 Answer 1


There are a few things that can be said about this, and it depends on how one "approaches" infinity.

  1. Wave equations satisfy conservation of energy. Let $$E(t) = \int_{\{t\} \times \mathbb{R}^d} |\partial_t\phi|^2 + |\nabla \phi|^2 + m^2 \phi^2 ~\mathrm{d}x$$ then it can be shown that $E(t) = E(0)$ for every $t$. And hence if your "decay assumptions" is sufficiently strong to guarantee that $$ \lim_{t \to \infty} E(t) = 0 $$ then you can conclude that $E(t) = 0$ for all $t$ and hence the solution vanishes.

  2. Wave and Klein-Gordon equations satisfy "finite speed of propagation", which means that if you require decay conditions only at spatial infinity (for example, of the form $$ \forall t \in \mathbb{R}, \forall \omega \in \mathbb{S}^{d-1}, \lim_{r \to\infty} f(t,r\omega) = 0 $$ which states that for every time, in any direction, the solution decays to zero as you move away toward infinity along the appropriate ray) you cannot conclude that $f\equiv 0$: in fact if the solution has compact spatial support at any time-slice, the solution will have compact spatial solution for every time slice.

  3. On the other hand, wave and Klein-Gordon equations are dispersive, this means that the spatial support of their solutions tend to grow over time, so decay conditions giving strict rates of decay can be used. For example, it you require that for all times, the solution vanish outside a fixed ball of radius $R$, then you can conclude that the solution must identically vanish. (There are many ways of doing this, Holmgren's Uniqueness Theorem is one way; John's Plane waves and spherical means book has some other discussions.)

    This result can be somewhat strengthened: for example, let $\epsilon > 0$ and $R > 0$ be fixed, and suppose that your solution is "concentrated" in the following sense: $$ \forall t \in \mathbb{R}, \int_{\{t\} \times B_R(0)} |\partial_t\phi|^2 + |\nabla\phi|^2 + m^2 \phi^2 ~\mathrm{d}x \geq \epsilon E(t) $$ in other words, you assume that at any given time, a fixed positive fraction of the total energy remains within the ball of radius $R$. Then you can conclude that the solution must be identically zero. (This is a consequence of the "integrated local energy estimates".)

  4. So far we have discussed decays toward spatial and time-like infinities. For wave equations (the statements in the following do not apply as well to the Klein-Gordon equations for technical and not-completely understood reasons) there is a third natural notion of infinity, which is that of null or light-like infinity. In the relativists' language this is the space-time boundary portion usually denoted $\mathscr{I}^\pm$.

    Free waves are expected to radiate to null infinity. This much was well known since the 60s. And so one can study the "rescaled limit" of the solution at null infinity, which at times goes by the name of "Friedlander's Radiation Field" (for reference, one can consult F.G. Friedlander's The Wave Equation on a Curved Spacetime or Lax and Phillips Scattering Theory). The short summary is that when this rescaled limit is identically zero (which implies a particular rate, depending on the dimension (roughly speaking faster than $r^{(1-d)/2}$), of decay of the solution to null infinity), then the solution must vanish identically.

    More recently the analyses have been improved by Alexakis, Schlue, and Shao who prove, using unique continuation methods, a large class of "sufficient decay conditions" toward null infinity that can guarantee the vanishing of solutions in the interior which hold for not just the homogeneous wave equation on Minkowski space but also for suitable perturbations.

  • $\begingroup$ Nice comment, but I'm puzzled by point 3). In what sense is the support of the solutions growing over time? In fact, isn't the solution the overlapping of two travelling waves, each with support that is moving but not increasing in measure? $\endgroup$ Jul 7, 2016 at 21:37
  • $\begingroup$ @DelioMugnolo what happens in one dimension doesn't always hold in higher dimensions. Imagine the ripples after you through a stone into a lake. $\endgroup$ Jul 7, 2016 at 21:46
  • $\begingroup$ Sure, but I understood your answer as a statement about general properties of the wave equation. Now I see that you mean that dispersivity may hold. $\endgroup$ Jul 8, 2016 at 6:28
  • $\begingroup$ @Dello: well, yes and no. You are correct concerning (3), in that the wave equation in one spatial dimension is not dispersive. However, the first sentence is only a rough description that motivates the discussion. The actual examples given (starting the second sentence) does not depend on dimension. $\endgroup$ Jul 8, 2016 at 13:18

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