Systems of Differential Equations — Question 2

Question 2 – Solution

Strategy. Invert the map from state coordinates to the measurement’s derivative coordinates.

Step 1: Test the measurement map. With $c=(1,2,0)$, the rows of $O$ are $c,cA,cA^2$. Hence $$O=\begin{pmatrix}1&2&0\\0&1&2\\-12&-22&-11\end{pmatrix},\qquad \det O=-15\ne 0.$$ The derivative triple determines the state uniquely.

Step 2: Convert and reconstruct. The identity $A^3+6A^2+11A+6I=0$ gives $$\boxed{y'''+6y''+11y'+6y=0.}$$ Solving $OX=(y,y',y'')^T$ gives $$\boxed{\begin{aligned} X_1&=(-33y-22y'-4y'')/15,\\ X_2&=(24y+11y'+2y'')/15,\\ X_3&=(-12y+2y'-y'')/15. \end{aligned}}$$

Step 3: Solve the recovered IVP. The measured data give $X(0)=(1,0,0)^T$. The scalar roots are $-1,-2,-3$; imposing the measured derivative triple yields $$\boxed{y=-3e^{-t}+9e^{-2t}-5e^{-3t}.}$$ The state is $X=(f,f',f'')^T$, where $f=3e^{-t}-3e^{-2t}+e^{-3t}$. In particular, $f+2f'=y$ and $(f,f',f'')(0)=(1,0,0)$.

Step 4: Prove full equivalence. If $Z=(y,y',y'')^T$, then $Z'=BZ$ with last row of $B$ equal to $(-6,-11,-6)$. Direct multiplication gives $OA=BO$, so for any scalar solution, $X=O^{-1}Z$ satisfies $X'=O^{-1}BZ=AX$. Its measurement is $y$ by the first row of $O$. Conversely every state solution produces this scalar equation. The recovered $f$ satisfies $f'''+6f''+11f'+6f=0$, checking the original last row directly. Invertibility of $O$ proves that no hidden state was discarded.