Systems of Differential Equations — Question 3

Question 3 – Solution

Strategy. Separate the measured Jordan block from the invisible unstable coordinate.

Step 1: Find the true measurement equation. From $y'=-y+X_2$ and $X_2'=-X_2$, $$\boxed{y''+2y'+y=0,\qquad y=(a+bt)e^{-t}.}$$ Both $e^{-t}$ and $te^{-t}$ occur as measurements and are independent. Any constant-coefficient annihilator must have a double root at $-1$, so order two is minimal.

Step 2: Recover the visible and hidden states. Every compatible state is $$\boxed{X(t)=((a+bt)e^{-t},\,be^{-t},\,ce^{2t})^T,\qquad c\in\mathbb R.}$$ The first two coordinates always decay. Therefore the full state is bounded if and only if $c=X_3(0)=0$. The measurement does not determine $c$.

Step 3: Remove the spurious scalar mode. The characteristic polynomial gives $$(D-2)(D+1)^2y=0,\quad\text{equivalently}\quad y'''-3y'-2y=0.$$ Its general solution adds $de^{2t}$ to the actual measurement family. Put $r=(D+1)^2y$. Then $r'=2r$ and $r=9de^{2t}$, so the precise exclusion condition is $$\boxed{y''(0)+2y'(0)+y(0)=0.}$$ The characteristic-polynomial equation is necessary, but without this constraint it is not sufficient for a scalar function to be a measurement.

Step 4: Test what observation can certify. The initial states $(1,0,0)^T$ and $(1,0,1)^T$ both produce $y=e^{-t}$ forever. Their full trajectories are $(e^{-t},0,0)^T$ and $(e^{-t},0,e^{2t})^T$; one is bounded and the other is not. Thus even an exact infinite measurement record cannot certify boundedness without additional information about $X_3(0)$.