IVP’s With Step Functions — Question 5

Question 5 – Solution

Strategy. The two endpoints have opposite sine phases. Rewriting both phases first prevents a sign error in the delayed inverse.

Step 1: Shift the phases. For delayed time $\tau$, $\sin(\tau+a)=\cos\tau$ and $\sin(\tau+b)=-\cos\tau$. Therefore $$f=H_a(t)\cos(t-a)+H_b(t)\cos(t-b),\qquad F=\frac{s(e^{-as}+e^{-bs})}{s^2+1}.$$ The plus sign at $b$ cancels the continuing negative-phase input; it does not add a second physical forcing window.

Step 2: Derive and invert the response pair. The zero-data equation gives $$Y=\frac{s(e^{-as}+e^{-bs})}{(s^2+1)^2},\qquad s>0.$$ Differentiating $\mathcal L\{\sin t\}=1/(s^2+1)$ with respect to $s$ gives $\mathcal L\{t\sin t/2\}=s/(s^2+1)^2$. With $q(\tau)=\tau\sin\tau/2$, $$\boxed{y=H_a(t) q(t-a)+H_b(t) q(t-b).}$$ Here $q(0)=q'(0)=0$ and $q''+q=\cos\tau$, directly verifying the equation, the zero initial data and both matching conditions.

Step 3: Evaluate the residual motion. For $t=b+v$, $v\ge 0$, the two ages are $v+\pi$ and $v$. Hence $$y(b+v)=q(v+\pi)+q(v)=\boxed{-\tfrac\pi 2\sin v}.$$ In particular $y(b)=0$, $y'(b)=-\pi/2$, and the post-shutoff energy is $$\boxed{E=\tfrac 12(y'^2+y^2)=\frac{\pi^2}{8}.}$$ The pulse ends precisely at a zero crossing with nonzero velocity, so the system continues oscillating with amplitude $\pi/2$.

Step 4: Inspect the switches and convergence. The input jumps from $0$ to $1$ at $a$, and from $-1$ to $0$ at $b$. Continuity of $y$ gives $\boxed{[y'']_a=[y'']_b=1}$. There are no jumps in $y$ or $y'$. The nonzero sinusoidal tail has exact real transform domain $\boxed{s>0}$: its undamped primitive fails to converge at zero, and exponential growth of fixed-sign tail integrals rules out negative $s$.