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Igor Khavkine
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Methods for determining domains of influence

Given a hyperbolic PDE, the domain of influence of a spacetime point $x$, say $I_x$ though $x$ could be replaced by any set, can be defined in two ways. Lets call one of them geometric ($I_x^G$) and the other analytical ($I_x^A$). In Lorentzian geometry, the geometric domain of influence consists of the interior of the cone of null geodesics emanating from $x$ (let me not bother about whether to include the boundary in the definition of $I_x^G$ or not). In general, a similar definition can be given using characteristic cones instead of null cones. The analytical domain of influence can be defined as the set of all spacetime points $y$ for which there exists a neighborhood $O$ of $x$ ad two solutions $u_1$ and $u_2$ satisfying the condition $u_1(x')=u_2(x')$, for all $x'$ on a Cauchy surface passing through $x$ except for $x'\in O$, and also the condition $u_1(y)\ne u_2(y)$. The latter one is the definition used in Lax's book on Hyperbolic PDEs.

Similar defintions can be given for the geometric and analytical domain of dependence, say $D_S^G$ and $D_S^A$. Such a definition should capture the desired equality $D_K = I_{S\setminus K}$, for a Cauchy surface $S$ and $K\subset S$ (once again, being sloppy with boundaries). I know that energy methods can be used to establish that $D^G_K \subseteq D^A_K$ (the analytical domain of dependence is at least as large as the geometric one). Hence, by duality, the same methods establish $I_K^A \subseteq I_K^G$ (that the geometric domain of influence is at least as large as the analytical one).

My question is about the reverse inclusion, $I_K^G \subseteq I_K^A$ or by duality $D_K^A \subseteq D_K^G$. Maybe it's too much to ask for the analytical and geometric definitions to coincide. But when they do, what methods are used to establish that? When they don't what methods can identify the obstruction? Except briefly in Lax's book, I don't know what references discuss this problem explicitly, so those would also be appreciated!

Igor Khavkine
  • 21.6k
  • 2
  • 61
  • 114