Laplace Transforms — Question 4

Question 4 – Solution

Strategy. Separate the impulse kernel from the input derivative, then match distributions rather than imposing false continuity conditions.

Step 1: Transform the distributions. For positive delay, $\mathcal L\delta(t-1)=e^{-s}$ and $\mathcal L\delta'(t-1)=se^{-s}$. Zero prehistory therefore gives $$Y=e^{-s}(s+3)H(s),\qquad H(s)=\frac 1{s^2(s+1)^2},\qquad \operatorname{Re}s>0.$$

Step 2: Invert the kernel. Since $H=-2/s+1/s^2+2/(s+1)+1/(s+1)^2$, $h(r)=r-2+(r+2)e^{-r}$ and $h(0)=0$. Thus $sH$ transforms $h'$, without an additional origin impulse. For $r=t-1$, $$\boxed{y(t)=u(t-1)k(r),\qquad k(r)=3r-5+(2r+5)e^{-r}.}$$

Step 3: Match every singular coefficient. Put $J_j=[y^{(j)}]_1$. The coefficients of $\delta''',\delta'',\delta',\delta$ in the left side are, respectively, $$J_0,\quad J_1+2J_0,\quad J_2+2J_1+J_0,\quad J_3+2J_2+J_1.$$ Matching $0,0,1,3$ gives $\boxed{(J_0,J_1,J_2,J_3)=(0,0,1,1)}$. Directly, $(k,k',k'',k''')(0)=(0,0,1,1)$; away from $r=0$, $D^2(D+1)^2k=0$, completing the distributional verification.

Step 4: Interpret the motion. The response vanishes before $t=1$ and approaches $3(t-1)-5$ afterward. It is $C^1$, but its one-sided second derivatives differ. Changing a function at one point cannot change those one-sided limits or its distribution.