IVP’s With Step Functions — Question 7

Question 7 – Solution

Strategy. A finite input can cancel one unstable trajectory exactly. Keep the cancellation coefficient explicit to measure its sensitivity.

Step 1: Transform and invert. Here $f_B=-1+(B+1)H_1(t)-BH_2(t)$ and $$Y=\frac{1+[-1+(B+1)e^{-s}-Be^{-2s}]/s}{s-1},\qquad s>1.$$ The zero-start response to a unit step for $y'-y=1$ is $r(t)=e^t-1$. Adding the initial response gives $$y=e^t-r(t)+(B+1)H_1(t)r(t-1)-BH_2(t)r(t-2).$$ This formula satisfies $y(0)=1$ and is continuous at both switches.

Step 2: Select the exact cancellation. Before $1$, $y=1$. On $[1,2]$, $y=(B+1)e^{t-1}-B$, so $y(2)=e+B(e-1)$. Consequently $$\boxed{B_*=-\frac e{e-1}},\qquad \boxed{y_*(t)=\begin{cases} 1,&0\le t<1,\\ (e-e^{t-1})/(e-1),&1\le t<2,\\ 0,&t\ge 2. \end{cases}}$$ Direct substitution gives $y_*'-y_*=-1,B_*,0$ on the three intervals, and the values match at $1$ and $2$. The zero terminal state remains zero under the homogeneous equation after $2$.

Step 3: Quantify the error sharply. The error is zero before $1$, then equals $\delta(e^{t-1}-1)$ until $2$. After $2$ the homogeneous equation amplifies its terminal value: $$\boxed{y(t)-y_*(t)=\delta(e-1)e^{t-2},\qquad t\ge 2.}$$ Its magnitude increases on $[2,5]$ when $\delta\ne 0$. Thus the required necessary and sufficient tolerance is $$\boxed{|\delta|\le\frac{0.01\,e^{-3}}{e-1}.}$$ Equality attains the limit at $t=5$.

Step 4: Distinguish trajectory cancellation from stability. The perfect output has compact support and a transform for every real $s$. For any $\delta\ne 0$, a nonzero $e^t$ tail remains and the exact real domain is $s>1$. At the perfect setting the apparent singularity at $s=1$ is removable, as is the $s=0$ expression after cancellations. Small level errors eventually grow without bound; selecting one compact response does not change the unstable homogeneous equation.