Systems of Differential Equations — Question 5

Question 5 – Solution

Strategy. A moving basis contributes its own derivative; keep that contribution before multiplying by the inverse.

Step 1: Differentiate the whole change of coordinates. Since $X'=P'Z+PZ'=APZ$, invertibility of $P$ gives $$\boxed{Z'=P^{-1}(AP-P')Z.}$$ Dropping $P'$ would treat a moving basis as fixed and generally produce incorrect coordinate derivatives. Here $\det P=1$ for all real $t$.

Step 2: Compute the transformed system. The matrices are $$P^{-1}=\begin{pmatrix}1&-t\\0&1\end{pmatrix},\qquad AP-P'=\begin{pmatrix}0&0\\-1&-t\end{pmatrix}.$$ Consequently $$\boxed{Z'=\begin{pmatrix}t&t^2\\-1&-t\end{pmatrix}Z, \qquad Z(0)=(0,1)^T.}$$ This system is linear and homogeneous, but nonautonomous. Explicit time dependence can arise entirely from a time-dependent choice of coordinates.

Step 3: Check the supplied trajectory in both descriptions. For $X=(\sin t,\cos t)^T$, $X'=(\cos t,-\sin t)^T=AX$, with the stated data. Multiplication by $P^{-1}$ gives $$\boxed{Z=(\sin t-t\cos t,\,\cos t)^T.}$$ Its derivative is $(t\sin t,-\sin t)^T$. The transformed right-hand side is also $(t(\sin t-t\cos t)+t^2\cos t,-(\sin t-t\cos t)-t\cos t)^T$, which simplifies to the same vector.

Step 4: Interpret the instantaneous difference. At $0$, $X'=(1,0)^T$ but $Z'=(0,0)^T$. The identity $X'=P'Z+PZ'$ explains this: $P'(0)Z(0)=(1,0)^T$. The basis motion accounts for the entire instantaneous change. The figure compares the first components; the coordinate curve starts with a horizontal tangent while the physical component has slope $1$.