Real Eigenvalues — Question 5

Question 5

For a real parameter $-1<a<1$, consider $$X'=A_aX,\qquad A_a=\frac 12\begin{pmatrix}a-2&a+2\\a+2&a-2\end{pmatrix}.$$ Treat the zero-eigenvalue case explicitly; do not assume that every equilibrium is isolated.

Tasks

1. Find the eigenvalues, an eigenbasis independent of $a$, and the solution with arbitrary initial state $(p,q)^T$.

2. Classify the origin for $a<0$ and $a>0$, and identify the initial states that converge to it in each case.

3. At $a=0$, find every equilibrium and the limiting state of every solution. Is the origin attracting all nearby states?

4. For initial state $(2,0)^T$, compare $\lim_{a\uparrow 0}\lim_{t\to\infty}X_a(t)$ with $\lim_{t\to\infty}\lim_{a\uparrow 0}X_a(t)$. Explain the difference.

Question 5 – Solution

Strategy. Keep the parameter-dependent mode visible; its decay disappears at the zero eigenvalue.

Step 1: Use fixed eigendirections. The vectors $(1,1)^T$ and $(1,-1)^T$ have eigenvalues $a$ and $-2$. They remain independent, and their eigenvalues are distinct throughout the stated parameter interval. Setting $m=(p+q)/2$, $d=(p-q)/2$ gives $$\boxed{X_a(t)=m e^{at}\binom 11+d e^{-2t}\binom 1{-1}.}$$ This formula satisfies the system and the initial data for every allowed $a$.

Step 2: Classify the nonzero-parameter cases. For $a<0$, both modes decay, so the origin is a stable node and all initial states converge to it. For $a>0$, the origin is a saddle; convergence occurs exactly when $m=0$, that is, on $y=-x$. The $y=x$ mode then grows. There are no other equilibria in these two cases because $\det A_a=-2a\ne 0$.

Step 3: Resolve the equilibrium line. At $a=0$, every point on $y=x$ is an equilibrium and $X_0(t)\to(m,m)^T$. This is the orthogonal projection of the initial state onto that line. The origin does not attract all nearby states: nearby nonzero points of the equilibrium line remain fixed. Nevertheless it is stable in the distance sense, since the orthogonal constant and decaying modes never increase the Euclidean norm.

Step 4: Compare the orders of limiting. For the specified data, $m=d=1$. At each fixed $a<0$ the time limit is zero; hence the first iterated limit is $\boxed{(0,0)^T}$. At each fixed finite $t$, taking $a\uparrow 0$ instead gives $(1,1)^T+e^{-2t}(1,-1)^T$, whose time limit is $\boxed{(1,1)^T}$. The decay $e^{at}$ becomes arbitrarily slow as $a$ approaches zero from below. It is not uniform over such parameters on an unbounded time interval, so these two operations need not commute.