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Let $u(s,t,x)$ solve the equation $$ i \partial_s u +\partial^2_t u - \partial^2_x u =0$$ on the set $[0,1]^3$ and suppose that $u(0,t,x)=0$ on $[0,1]^2$ and that $$ u(s,0,x)=\partial_t u(s,0,x)=0$$ for all $x,s \in [0,1]$

My question is whether it follows that the function $u$ must also vanish in the typical domain of dependence $$\{(s,t,x)| s \in (0,1)\quad t \in (0,\frac{1}{2}) \quad x \in (t,1-t) \} $$

My intuition tells me that since the principal symbol is hyperbolic, this property must hold but I’m not sure and since I could not prove it.

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  • $\begingroup$ The principal part of your operator is not hyperbolic. Using $\sigma, \tau, \xi$ for the dual variables, the principal symbol is $P_2(\sigma,\tau,\xi) = - \tau^2 + \xi^2$. This is not hyperbolic in the $\tau$ direction as for any $\sigma \neq 0$, the mapping $\tau \mapsto P_2(\sigma,\tau,0)$ has a double root. $\endgroup$
    – Willie Wong
    Commented Oct 2, 2019 at 19:23
  • $\begingroup$ Yes you are correct, and that is why heuristically I only expect the finite speed of propagation to hold in the region that I defined. It could still easily be wrong $\endgroup$
    – Ali
    Commented Oct 2, 2019 at 19:28

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