Nonhomogeneous Systems — Question 4

Question 4 – Solution

Strategy. For time-dependent coefficients, propagation depends on both the departure time and the arrival time.

Step 1: Normalize the homogeneous matrix. Solving $y'=0$, $x'=2ty$ gives $$\Phi(t)=\begin{pmatrix}1&t^2\\0&1\end{pmatrix},\qquad \Phi(t)^{-1}=\begin{pmatrix}1&-t^2\\0&1\end{pmatrix}.$$ Its determinant is one, and direct differentiation verifies $\Phi'=A(t)\Phi$ and $\Phi(0)=I$.

Step 2: Derive and integrate the varying coefficients. Substituting $X=\Phi C$ cancels $\Phi'C=A\Phi C$, leaving $C'=\Phi^{-1}f=(t-t^2,1)^T$. Zero initial data gives $C=(t^2/2-t^3/3,t)^T$. Therefore $$\boxed{X=(t^2/2+2t^3/3,t)^T.}$$ Indeed $x'=t+2t^2=2ty+t$ and $y'=1$.

Step 3: Use the genuine two-time transition. Matrix multiplication yields $$T(t,s)=\begin{pmatrix}1&t^2-s^2\\0&1\end{pmatrix}.$$ Thus $X(t)=\int_0^tT(t,s)f(s)\,ds =\int_0^t(s+t^2-s^2,1)^T\,ds$, reproducing the same formula. This expression also follows directly from $\Phi(t)\int_0^t\Phi(s)^{-1}f(s)\,ds$.

Step 4: Expose the incorrect shortcut. Using $\Phi(t-s)$ instead gives $\widetilde X=(t^2/2+t^3/3,t)^T$. Its residual is $$\boxed{\widetilde X'-A(t)\widetilde X-f(t)=(-t^2,0)^T,}$$ which is not identically zero. The two matrices differ because $t^2-s^2\ne(t-s)^2$ in general. A fundamental matrix normalized at zero cannot automatically be shifted in time when the coefficient matrix itself changes with time.