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Assume $\omega >0$. The characteristic roots of the damped harmonic oscillator are $-k\pm \sqrt{k^2-{\omega }^2}$. Its general homogeneous solution isThese are respectively the underdamped, critically damped and overdamped regimes.

For the velocity-switched driven oscillator, the initial acceleration is $-{\omega }^2{x}_1<0$. Its velocity is therefore negative on the first half-cycle, and the forcing is zero:Indeed $\dot{x}=-({\omega }^2{x}_1/p)e^{-kt}\sin pt<0$ in that open interval. Set $r=e^{-k\pi /p}$; at the first turning point, $x_2=-r{x}_1$ and $\dot{x}=0$.

Put $\tau =t-\pi /p$. On the second half-cycle the velocity is positive, so the particular solution is $a/{\omega }^2$. Matching position and velocity givesIts derivative is positive for $0<\tau <\pi /p$. At the next turning point,The turning-point return map therefore repeats the initial state exactly whenThis value exceeds $a/{\omega }^2$, so the subsequent acceleration initiates the required negative-velocity half-cycle again. The switched solution is continuous with continuous velocity, and is piecewise twice differentiable; acceleration has one-sided jumps where the forcing switches.

For $k>\omega$, the specified positive initial value givesThere is no next turning point: the forcing stays zero and $x$ approaches zero monotonically. The motion from the specified initial state cannot be periodic. More generally, a solution started at any positive maximum has this same obstruction; a nonconstant periodic orbit would require such a maximum. The zero equilibrium is a separate constant solution.