Laplace Transforms — Question 4

Question 4 – Solution

Strategy. The shifted impulse produces a shifted homogeneous response; stopping requires cancellation of the complete pre-impulse state.

Step 1: Solve in the transform domain. The equations are $sF-G=1+\alpha e^{-as}$ and $F+sG=\beta e^{-as}$. Thus, for $\operatorname{Re}s>0$, $$F=\frac{s}{s^2+1}+e^{-as}\frac{\alpha s+\beta}{s^2+1},\qquad G=\frac{-1}{s^2+1}+e^{-as}\frac{-\alpha+\beta s}{s^2+1}.$$ This half-plane suffices for all parameter choices, even when cancellation makes the actual response have compact support.

Step 2: Invert and check jumps. For $t<a$, $(x,y)=(\cos t,-\sin t)$. For $t\ge a$, with $\tau=t-a$, $$x=\cos t+\alpha\cos\tau+\beta\sin\tau,\qquad y=-\sin t-\alpha\sin\tau+\beta\cos\tau.$$ Hence $\boxed{[x]_a=\alpha,\ [y]_a=\beta}$, where $[f]_a=f(a^+)-f(a^-)$. Integrating the original equations across $a$ gives these same jumps: the ordinary bounded terms contribute zero in the shrinking interval, while the impulses contribute their masses. Away from $a$ the ordinary equations hold.

Step 3: Cancel the full state. The state just before $a$ is $(\cos a,-\sin a)$. The only kick making the right-hand state zero is $\boxed{(\alpha,\beta)=(-\cos a,\sin a)}$. The ensuing homogeneous IVP then stays zero by uniqueness. Conversely, an identically zero future requires that same right-hand state, so the kick is unique.

Step 4: Restrict the available impulse. With $\alpha=0$, cancellation requires $\cos a=0$. Thus $\boxed{a=\pi/2+k\pi,\ \beta=(-1)^k,\ k=0,1,2,\ldots}$. The earliest is $a=\pi/2$, $\beta=1$. There $x$ is continuous at zero, while $y$ jumps from $-1$ to $0$. Both vanish afterward. The vertical dotted segment marks a jump, not values traversed continuously by the solution.