Dirac Delta Function — Question 8

Question 8 – Solution

Strategy. An impulse at the initial time requires a declared convention. Either transform pre-impact data with the impulse or restart after the jump.

Step 1: Compute the post-impact data. There is no displacement jump under an ordinary force impulse. Integrating the equation across the origin gives $2[y']_0=J$; the damping contribution is $4[y]_0=0$. Thus $$\boxed{y(0^+)=1,\qquad y'(0^+)=2+\frac J2.}$$ The factor $2$ is the mass and must not be omitted.

Step 2: Transform the pre-impact formulation. The stipulated derivative rules give $$2(s^2Y-s-2)+4(sY-1)+2Y=J,$$ so $$Y=\frac{2s+8+J}{2(s+1)^2} =\frac 1{s+1}+\frac{3+J/2}{(s+1)^2}.$$ Writing $c=3+J/2$, the inverse is $\boxed{y(t)=e^{-t}(1+ct),\quad t\ge 0}$. It has $y(0^+)=1$ and $y'(0^+)=c-1=2+J/2$, exactly as the jump calculation.

Step 3: Check the equivalent restart. For $t>0$ the equation is $y''+2y'+y=0$. Using the post-impact data $1,2+J/2$ gives the critical solution $e^{-t}[1+(3+J/2)t]$ again. This restart has no remaining initial impulse: its effect is already encoded in the velocity. Inserting $J\delta(t)$ again with those data would add a second velocity change $J/2$.

Step 4: Classify positivity and monotonicity. Since the exponential is positive, nonnegativity for every $t\ge 0$ requires and is ensured by $c\ge 0$. Also $$y'=e^{-t}(c-1-ct).$$ For $c\ge 0$, this is nonpositive for all $t\ge 0$ exactly when $c\le 1$. Therefore $$\boxed{0\le c\le 1\quad\Longleftrightarrow\quad -6\le J\le-4.}$$ At $J=-6$, $y=e^{-t}$; at $J=-4$, $y=(1+t)e^{-t}$ starts horizontal and then decreases strictly. For $J<-6$ the response eventually becomes negative, and for $J>-4$ it initially increases. All these nonzero polynomial-exponential responses have exact real transform domain $s>-1$.