IVP’s With Step Functions — Question 9

Question 9 – Solution

Strategy. A second-order system smooths a step more than the input itself, but only by a finite number of derivatives. Check one-sided limits directly.

Step 1: Transform and invert. With zero data, $$Y=\frac{e^{-s}}{s(s+1)(s+2)} +\frac{e^{-2s}}{s^2(s+1)(s+2)}.$$ Partial fractions give the two zero-start responses $$r(\tau)=\tfrac 12-e^{-\tau}+\tfrac 12e^{-2\tau},\qquad p(\tau)=\tfrac{\tau}2-\tfrac 34+e^{-\tau}-\tfrac 14e^{-2\tau},$$ so $\boxed{y=H_1(t)r(t-1)+H_2(t)p(t-2)}$. Their transforms are the two unshifted rational factors displayed above.

Step 2: Verify the equation and all jumps. Direct differentiation gives $r''+3r'+2r=1$ and $p''+3p'+2p=\tau$. Moreover $r(0)=r'(0)=p(0)=p'(0)=0$, verifying the initial data and $C^1$ matching. For the next derivatives, $$r''(0)=1,\quad r'''(0)=-3,\qquad p''(0)=0,\quad p'''(0)=1.$$ The previously active response is smooth at the next switch. Thus $$\begin{array}{c|rrrr} \text{switch}&[y]&[y']&[y'']&[y''']\\ \hline 1&0&0&1&-3\\ 2&0&0&0&1 \end{array}$$ These are jumps of one-sided ordinary derivatives, not assigned values at the switch.

Step 3: Relate onset orders to regularity. Taylor expansion of the newly added responses gives $$r(\tau)=\tfrac 12\tau^2-\tfrac 12\tau^3+O(\tau^4),\qquad p(\tau)=\tfrac 16\tau^3+O(\tau^4),\qquad \tau\downarrow 0.$$ Joining the first to zero preserves one derivative but creates a second- derivative jump. Joining the second to zero preserves two derivatives but creates a third-derivative jump. These leading coefficients reproduce the jump sizes above.

Step 4: Resolve both claims and the convention. The first claim is false: $y$ and $y'$ stay continuous at $1$, while $y''$ jumps. The second is also false: $y$ is $C^2$ near $2$ but not $C^3$ there. Changing an input at one isolated time does not change its Laplace integral or the solution obtained by continuous state matching. At switches the equation is interpreted through its one-sided limits, not by demanding an unavailable classical derivative. Away from them the solution is smooth. The growing linear tail gives exact real transform domain $s>0$.