IVP’s With Step Functions — Question 10

Question 10 – Solution

Strategy. The step series gives the transient solution, while the cycle map identifies a periodic state and its exact attraction rate.

Step 1: Transform the repeated switches. At each fixed time only finitely many terms are active: $$f=\sum_{k=0}^\infty(H_{2k}(t)-H_{2k+1}(t)),\qquad F=\frac{1-e^{-s}}s\sum_{k=0}^\infty e^{-2ks} =\frac 1{s(1+e^{-s})},\quad s>0.$$ Integration of the nonnegative disjoint pulses justifies the sum. With $r(t)=1-e^{-t}$ for $t\ge 0$, $$Y=\frac 1{s(s+1)(1+e^{-s})},\qquad y=\sum_{k=0}^\infty[H_{2k}(t)r(t-2k)-H_{2k+1}(t)r(t-2k-1)].$$ This time series is locally finite; each activated response starts at zero and satisfies the required first-order equation on its active intervals.

Step 2: Solve the cycle recurrence. During the on-half, $y(2k+1)=1+(x_k-1)e^{-1}$. During the off-half this value is multiplied by $e^{-1}$, so $$x_{k+1}=e^{-2}x_k+e^{-1}(1-e^{-1}),\qquad \boxed{x_k=m(1-e^{-2k}),\quad m=\frac 1{e+1}},$$ using $x_0=0$. The cycle map has the unique fixed point $m$.

Step 3: Build the periodic state and exact error. Write $t=2k+\tau$, $0\le\tau<2$, and put $M=e/(e+1)$. Then $$y_{\mathrm{per}}(t)=\begin{cases} 1-(1-m)e^{-\tau},&0\le\tau<1,\\ M e^{-(\tau-1)},&1\le\tau<2. \end{cases}$$ The pieces match at $M$, and the value at the end of the cycle is $m$. Both solutions have the same forcing, so their difference satisfies $d'+d=0$ with $d(0)=m$. Continuity propagates this identity across every switch: $$\boxed{y_{\mathrm{per}}(t)-y(t)=m e^{-t},\qquad t\ge 0.}$$

Step 4: Interpret the limiting cycle and tolerance. The periodic minimum is $m$ and maximum is $M$. Integrating its equation over a cycle gives $\int_0^2y_{\mathrm{per}}\,dt=1$, so its mean is $1/2$. The even and odd integer samples of $y$ tend to the different values $m,M$; there is no single limit. The error decreases strictly, giving the sharp time $$\boxed{T=\ln\!\left(\frac{100}{e+1}\right).}$$ Both responses have exact real transform domain $s>0$, since their positive persistent cycles prevent convergence at or below zero.