The <telegraph process> has backward survival equations
$$
\boxed{\begin{aligned}
\partial_tQ^+&=s\partial_yQ^++\lambda(Q^--Q^+),\\
\partial_tQ^-&=-s\partial_yQ^-+\lambda(Q^+-Q^-).
\end{aligned}}
$$
Initially $Q^\pm(y,0)=1$. Only a left-moving trajectory reaches the target, while reflection reverses a right-moving velocity at the outer wall, so the hyperbolic boundary conditions are
$$
\boxed{Q^-(0,t)=0,
\qquad Q^+(L,t)=Q^-(L,t).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 356 / 1 / c / Solution

{wiki}

The telegraph process moves at one of two opposite constant velocities and reverses direction at Poisson event times. Its position has finite propagation speed and persistent increments.


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