Dirac Delta Function — Question 7

Question 7 – Solution

Strategy. Cancel the growing mode’s coefficient, then inspect the remaining mode and the error produced by imperfect cancellation.

Step 1: Transform and invert. For sufficiently large $s$, $$(s^2-1)Y-s=Je^{-as},\qquad Y=\frac{s+Je^{-as}}{s^2-1},\qquad \boxed{y=\cosh t+JH_a(t)\sinh(t-a).}$$ The added term starts at zero with derivative $J$, so it preserves displacement and produces the correct velocity jump. Away from $a$ both terms solve the homogeneous equation.

Step 2: Select and interpret the decaying mode. After $a$, the $e^t$ coefficient is $(1+Je^{-a})/2$. Decay requires $$\boxed{J_*=-e^a,\qquad y_*(t)=\begin{cases}\cosh t,&0\le t<a,\\ \cosh a\,e^{-(t-a)},&t\ge a. \end{cases}}$$ The pre-impact velocity is $\sinh a$ and the post-impact velocity is $-\cosh a$; their difference is $-e^a$, as required. The response increases strictly before $a$ and decreases strictly after it. Its unique global maximum is $\boxed{y_*(a)=\cosh a}$.

Step 3: Quantify the sensitivity. Linearity or direct subtraction gives $$\boxed{y(t)-y_*(t)=\Delta H_a(t)\sinh(t-a).}$$ Because $\sinh\tau$ increases strictly on $[0,L]$, the desired error bound is equivalent to $$\boxed{|\Delta|\le\frac{\epsilon}{\sinh L}.}$$ This is necessary as well as sufficient, and equality attains the bound at $a+L$. The selected solution still approaches zero only asymptotically; the impulse does not put it at the zero state.

Step 4: Check convergence and stability. The selected tail is a positive multiple of $e^{-t}$, so its exact real transform domain is $\boxed{s>-1}$. The apparent singularity at $s=1$ in its transform is removable. For every $\Delta\ne 0$, the error contains a nonzero growing mode and the actual domain becomes $\boxed{s>1}$. The differential equation’s unstable homogeneous mode is unchanged. The plot uses $a=1$, with a dashed curve for $\Delta=0.05$.