Dirac Delta Function — Question 4

Question 4 – Solution

Strategy. One impulse can adjust velocity only. Inspect the displacement at the chosen impact time before attempting to cancel the entire state.

Step 1: Find the impulse response. The zero-data transform is $$Y=\frac{e^{-s}+Ke^{-bs}}{(s+1)^2}.$$ Since $\mathcal L\{te^{-t}\}=1/(s+1)^2$, set $h(t)=te^{-t}$ for $t\ge 0$. Then $$\boxed{y=H_1(t)h(t-1)+K H_b(t)h(t-b).}$$ Here $h(0)=0$, $h'(0)=1$, and $h''+2h'+h=0$ for positive age. Thus displacement is continuous, and velocity jumps by $1$ and $K$.

Step 2: Test the terminal displacement. Put $d=b-1>0$. Immediately before the second kick, $$y_b=de^{-d}>0,\qquad v_-=(1-d)e^{-d}.$$ The second impulse leaves $y_b$ unchanged. A solution identically zero after $b$ must have zero displacement there, so $\boxed{\text{no finite }b>1\text{ and }K\text{ can produce complete rest}.}$ The strict positivity of $h(d)$ is the obstruction, not a shortage of algebraic manipulation.

Step 3: Minimize the post-impact energy. For fixed $b$, $$E_+(K)=\tfrac 12[(v_-+K)^2+y_b^2].$$ This strictly convex quadratic has the unique minimizer and value $$\boxed{K_*=(d-1)e^{-d},\qquad E_{\min}=\tfrac 12d^2e^{-2d}>0.}$$ The minimizing kick makes $v_+=0$ but leaves positive stored spring energy.

Step 4: Follow the resulting motion. For $\tau=t-b\ge 0$, the critical homogeneous solution with state $(y_b,0)$ is $$\boxed{y(b+\tau)=de^{-d}(1+\tau)e^{-\tau}.}$$ Its derivative is $-de^{-d}\tau e^{-\tau}$: it starts horizontal and then moves strictly downward. Its acceleration at $\tau=0+$ is $-y_b\ne 0$. Substitution verifies the equation and the terminal state. Thus the system eventually approaches rest, but is not at rest throughout the post-impact interval. The positive polynomial-exponential tail has transform domain $s>-1$.