Repeated Eigenvalues — Question 5

Question 5 – Solution

Strategy. The exponential controls overall growth, while the polynomial factor can reverse one coordinate and dominate the limiting direction.

Step 1: Recover the defective solution. The double eigenvalue is $1$, with eigenspace $y=0$. Solving $y'=y$ and then $(e^{-t}x)'=3$ gives $$\boxed{x=3te^t,\qquad y=e^t.}$$ The factor $3$ is part of the coupling and cannot be omitted from the generalized mode. Both components satisfy the stated initial data.

Step 2: Find the curve and minimum. Since $y>0$, the orbit is $\boxed{x=3y\ln y,\quad y>0}$, traversed toward increasing $y$. Its derivative $x'=3e^t(t+1)$ vanishes only at $t=-1$, changing from negative to positive. The global minimum is $x=-3/e$ at $y=1/e$. The only finite zero of $x$ is $t=0$; its backward limit zero is not another attained zero.

Step 3: Find the two limiting directions. The normalized state is $$\frac{X}{\|X\|}=\frac{(3t,1)^T}{\sqrt{9t^2+1}}.$$ It tends to $(-1,0)^T$ backward and $(1,0)^T$ forward. Although the limiting line is the eigenline $y=0$, every finite-time state has $y>0$ and lies off that line. The origin is an unstable defective node; this IVP tends to it only as $t\to-\infty$.

Step 4: Distinguish reversal from repetition. The origin is never reached at finite time because $e^t>0$. Also $y=e^t$ is strictly increasing, so two different times cannot give the same state. The single minimum of $x$ comes from the factor $t e^t$, not recurring oscillations. There is no periodic orbit or repeated cycle: a coordinate can reverse even with a positive repeated real eigenvalue. The equal-scale figure marks the minimum and the initial state.