Real Eigenvalues — Question 8

Question 8 – Solution

Strategy. Two real exponential modes can cancel in an endpoint observation even when the full state remains uniquely determined by its initial data.

Step 1: Find the two modes. The characteristic polynomial is $\lambda^2-1$. Eigenvectors are $(1,1)^T$ for $1$ and $(1,2)^T$ for $-1$. The data $(1,q)^T$ give coefficients $2-q$ and $q-1$, so $$\boxed{X=(2-q)e^t\binom 11+(q-1)e^{-t}\binom 12.}$$

Step 2: Locate the singular observation time. The endpoint condition becomes $$b=2e^T-2e^{-T}+q(2e^{-T}-e^T).$$ The coefficient of $q$ vanishes exactly when $e^{2T}=2$, hence $\boxed{T_*=\tfrac 12\ln 2}$, the unique positive exceptional time.

Step 3: Classify compatibility at that time. At $T_*$, the equation reduces to $b=\sqrt 2$. If $b=\sqrt 2$, every real $q$ gives a solution; otherwise none does. There is never exactly one at this time. Distinct $q$ specify distinct initial states. In fact $x(T_*)=(3-q)/\sqrt 2$, so the full states remain different even though their second components agree. Each individual IVP still has its unique solution. The figure shows three such second components.

Step 4: Quantify sensitivity away from the exception. For $T\ne T_*$ there is exactly one solution, with $$\boxed{q=\frac{b-2e^T+2e^{-T}}{2e^{-T}-e^T},\qquad \Delta q=\frac{\delta}{2e^{-T}-e^T}.}$$ The denominator tends to zero as $T\to T_*$, so the absolute amplification factor diverges. This is a poorly conditioned recovery of missing initial data, not loss of uniqueness for a fully specified IVP. For a fixed nonzero $\delta$, the recovered initial error can become arbitrarily large.