Solutions to Systems — Question 4

Question 4 – Solution

Strategy. Keep both endpoint times: a time-dependent equation need not evolve according only to elapsed time.

Step 1: Integrate from the actual starting time. The second component is $y(t)=y(s)$, and $x(t)=x(s)+y(s)\int_s^t2u\,du$. Hence $$\boxed{T(t,s)=\begin{pmatrix}1&t^2-s^2\\0&1\end{pmatrix}.}$$ Its derivative with respect to $t$ is $\begin{pmatrix}0&2t\\0&0\end{pmatrix}$, which equals $A(t)T(t,s)$, and $T(s,s)=I$.

Step 2: Check composition and inversion. Multiplying two such shears adds their upper-right entries: $(t^2-r^2)+(r^2-s^2)=t^2-s^2$. Thus $$\boxed{T(t,r)T(r,s)=T(t,s),\qquad T(t,s)^{-1}=T(s,t).}$$ The intermediate time cancels only because it is the end of the first journey and the start of the second, in the correct order.

Step 3: Test the one-parameter shortcut. The upper-right entry of $Y(t)Y(s)$ is $t^2+s^2$, whereas that of $Y(t+s)$ is $(t+s)^2=t^2+s^2+2ts$. They agree only when $ts=0$, not for all times. For example, $Y(1)Y(1)$ has upper-right entry $2$, while $Y(2)$ has entry $4$. This does not contradict the two-time composition identity.

Step 4: Apply the transition to the IVP. The solution is $$\boxed{X(t)=T(t,-1)(2,3)^T=(3t^2-1,3)^T.}$$ It returns to $(2,3)^T$ at $t=1$, and $X'(0)=(0,0)^T$. Nevertheless $X'=(6t,0)^T$ is nonzero away from $0$, so the solution is not constant. In a nonautonomous system, a repeated state at different times or a zero derivative at one instant does not force stationary evolution.