Repeated Eigenvalues — Question 4

Question 4 – Solution

Strategy. Keep the difference of nearby exponentials together; an integral representation exposes the limit without subtracting divergent pieces.

Step 1: Solve before the roots merge. Eigenvectors are $(1,0)^T$ for $-1$ and $(1,-\varepsilon)^T$ for $-1-\varepsilon$. Decomposition of $(0,1)^T$ gives coefficients $1/\varepsilon,-1/\varepsilon$, hence $$\boxed{x_\varepsilon=e^{-t}\frac{1-e^{-\varepsilon t}}{\varepsilon}, \qquad y_\varepsilon=e^{-(1+\varepsilon)t}.}$$

Step 2: Pass to the repeated-root solution. Since $(1-e^{-\varepsilon t})/\varepsilon=\int_0^t e^{-\varepsilon s}\,ds$, its limit at each fixed real $t$ is $t$. Thus $\boxed{X_0=e^{-t}(t,1)^T}$. Direct differentiation gives $x_0'=-x_0+y_0$, $y_0'=-y_0$ and the stated initial data. The polynomial factor is the limit of the combined modes.

Step 3: Prove uniform finite-interval control. For $0\le t\le T$, the supplied inequality yields $$0\le x_0-x_\varepsilon =e^{-t}\int_0^t(1-e^{-\varepsilon s})\,ds \le\tfrac 12\varepsilon t^2e^{-t}\le\tfrac 12\varepsilon T^2.$$ Similarly $0\le y_0-y_\varepsilon=e^{-t}(1-e^{-\varepsilon t}) \le\varepsilon te^{-t}\le\varepsilon T$. Both bounds tend uniformly to zero on every fixed finite interval.

Step 4: Separate relative and absolute effects. At $t=1/\varepsilon$, $\boxed{x_\varepsilon/x_0=1-e^{-1}}$, so the relative discrepancy is $e^{-1}$, independent of $\varepsilon$. This time moves to infinity, and $x_0=\varepsilon^{-1}e^{-1/\varepsilon}\to 0$, so the absolute discrepancy still vanishes there. The separate modal coefficients diverge, but their exponential contributions cancel to a finite limit. Neither divergent coefficients nor this moving-time relative error contradicts the proved finite-interval convergence.