Nonhomogeneous Systems — Question 6

Question 6 – Solution

Strategy. A rotating coordinate frame removes the oscillation and leaves scalar damping toward a constant vector.

Step 1: Pass to the rotating frame. Let $J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$, so $R'=JR$ and $A=-I+J$. Substitution cancels the rotational terms, giving $Z'=-Z+(1,0)^T$. Hence $$\boxed{X(t)=R(t)\left[\binom 10+e^{-t}\left(c-\binom 10\right)\right],}$$ where $c=X(0)$ is an arbitrary real vector. This gives every solution.

Step 2: Identify and prove the periodic response. Taking $c=(1,0)^T$ gives $\boxed{P(t)=(\cos t,\sin t)^T}$. If another solution were $2\pi$-periodic, its difference from $P$ would be $e^{-t}R(t)d$ and would satisfy $d=e^{-2\pi}d$ at times zero and $2\pi$. Thus $d=0$, proving uniqueness. Damping prevents a secular term despite the matched rotation frequency.

Step 3: Quantify zero-state convergence. For $c=0$, $$\boxed{X(t)=(1-e^{-t})(\cos t,\sin t)^T\quad(t\ge 0).}$$ Since rotations preserve length, $\|X(t)-P(t)\|=e^{-t}$. Its radius is $1-e^{-t}$, so its distance to the unit circle is also $e^{-t}$. Its initial velocity is $(1,0)^T$, agreeing with the input at zero.

Step 4: Distinguish asymptotic response from exact return. The radius is strictly increasing at every finite forward time, so the solution neither repeats a state nor becomes exactly periodic later. Along $t=2\pi n$ and $t=2\pi n+\pi$, its states tend to $(1,0)^T$ and $(-1,0)^T$, respectively; there is no single point limit. The drawn curve is a nonautonomous state path approaching a periodic response, not an autonomous phase portrait or a claim of a finite-time merger with the circle. Only a finite initial segment is drawn.