1
$\begingroup$

Let $(M,g)$ be a globally hyperbolic Lorentzian manifold with Lorentzian volume density $dV$ and $\Sigma$ a Cauchy hypersurface, i.e. each inextendible causal curve hits $\Sigma$ exactly once. Furthermore let $P: C^\infty(M) \rightarrow C^\infty(M)$ be a formally self-adjoint wave operator, that is a $2^{nd}$-order differential operator with principal symbol given by the metric $g$ and $$(P\varphi,\psi)_{L^2(dV)} = (\varphi,P\psi)_{L^2(dV)}, \qquad \varphi,\psi \in C^\infty_c(M).$$ For each fixed $q\in M$ consider the Cauchy problem $$\left\{ \begin{array}{cl} Pu(\cdot,q) = 0 \qquad & \text{on} \enspace M \\[2mm] u(\cdot,q) = u_0(\cdot,q) & \text{on} \enspace \Sigma \\[2mm] \nabla_\nu u(\cdot,q) = u_1(\cdot,q) & \text{on} \enspace \Sigma \end{array} \right., \qquad (*)$$ which by well-posedness determines the function $u$ on $M\times M$ uniquely. Here $\nabla_\nu$ denotes the covariant normal derivative along $\Sigma$. My question now is:

When I have symmetry of the initial data on $\Sigma$ in the sense $$u(\sigma,q) = u(q,\sigma) \quad \text{and} \quad \nabla_\nu^{(1)} u(\sigma,q) = \nabla_\nu^{(2)} u(q,\sigma), \qquad \sigma \in \Sigma,~q \in M, \quad (**)$$ does this imply symmetry of $u$, i.e. $u(p,q)=u(q,p)$ for all $p,q\in M$? Btw $\nabla_\nu^{(1)}, \nabla_\nu^{(2)}$ stand for differentiating w.r.t the first and the second entry, respectively.

An equivalent question would be whether solutions of $(*)$ are solutions of the corresponding Cauchy problem for $Pu(q,\cdot)=0$ with initial data given by $(**)$.

All cases I checked, where one can give a certain solution operator propagating the solution from the initial data, symmetry directly holds true, which makes me quite optimistic. Thanks!

$\endgroup$

1 Answer 1

0
$\begingroup$

I think this just follows from linearity.

The condition that $u(\sigma, q) = u(q,\sigma)$ implies that $P u(\sigma,\cdot) = 0$. In fact, you have that $q\mapsto u(\sigma,q)$ is the unique solution to the initial value problem for $Pu(\sigma,\cdot) = 0$ with data prescribed on $\{\sigma\} \times \Sigma$ under your hypotheses.

Denote by $P^{(i)}$ the operator $P$ acting on the $i$th factor, then by linearity you have $P^{(2)}u(p,q)$ solves the initial value problem (*) with initial data given by $P^{(2)}u(\sigma,q)$ and $\nabla_\nu^{(1)} P^{(2)}u(\sigma, q)$ both of which vanishes. ($\nabla_\nu^{(1)} u(\sigma,q)$ can be defined as the solution to $P^{(2)}u = 0$ with data given by $\nabla_\nu^{(1)} u$ and the value of "$(\nabla_\nu^{(1)})^2 u$" induced from the initial data and the formal compatibility condition $P^{(1)}u|_{\Sigma\times \{\sigma\}} = 0$.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.