Basic Concepts — Question 10

Question 10 – Solution

Strategy. Carry both position and velocity across the switch, and distinguish a piecewise classical solution from one that is $C^2$ everywhere.

Step 1: Propagate the first phase. With zero acceleration and initial velocity 1, $y'=1$ and $y=t$ for $0\le t\le 1$. The arriving data are $$\boxed{y(1)=1,\qquad y'(1)=1}.$$

Step 2: Use both matching conditions. Writing the second-phase antiderivative about the switch gives $$y=(t-1)^2+A(t-1)+B\quad(t>1).$$ Continuity of position sets $B=1$; continuity of velocity sets $A=1$. Thus $$\boxed{y(t)=\begin{cases}t,&0\le t\le 1,\\t+(t-1)^2,&t>1,\end{cases}} \qquad y'(t)=\begin{cases}1,&0\le t\le 1,\\1+2(t-1),&t>1.\end{cases}$$ These formulas satisfy all the stated requirements and uniquely determine both phases.

Step 3: Check the actual regularity. The derivative of velocity at 1 has left-hand value 0 and right-hand value 2. Therefore $y''(1)$ does not exist. Assigning a forcing value at that single point cannot change the unequal one-sided limits. The trajectory is $C^1$ and piecewise $C^2$, but is not a classical $C^2$ solution across the switch.

Step 4: Reject an unmatched velocity. Adding $k(t-1)$ on $t>1$ keeps the position continuous and does not change the second derivative on that open phase. However, the right-hand velocity at 1 becomes $1+k$, while the left-hand velocity stays 1. Hence the $C^1$ requirement forces $\boxed{k=0}$.

A second-order problem carries two pieces of state information. Position matching alone would admit a jump in velocity that is excluded by the problem’s stated regularity; checking only the differential equation away from the switch would miss it.