For each \(t\in\mathbb R\), let \(P(t)\) be a real symmetric \(n\times n\) matrix whose entries depend continuously on \(t\). Suppose every eigenvalue of \(P(t)\) is at most \(-1\) for every \(t\). If \(x(t)\) solves \[ x'(t)=P(t)x(t), \] prove that \[ \lim_{t\to\infty}x(t)=0. \]
Solution
Let \[ E(t)\coloneqq\norm{x(t)}^2, \] where \(\norm{\cdot}\) is the Euclidean norm on \(\RR^n\).
1. For every \(t\in\RR\) and every \(v\in\RR^n\),\[ v^{\mathsf T}P(t)v\le-\norm{v}^2. \]
Because \(P(t)\) is real symmetric, it has an orthonormal eigenbasis. In that basis, \[ v^{\mathsf T}P(t)v = \sum_{j=1}^n\lambda_j(t)v_j^2. \] Every eigenvalue satisfies \(\lambda_j(t)\le-1\), so \[ \sum_{j=1}^n\lambda_j(t)v_j^2 \le -\sum_{j=1}^n v_j^2 = -\norm{v}^2. \]
2. The energy satisfies\[ E'(t)\le-2E(t). \]
Using \(x'(t)=P(t)x(t)\) and step 1, \[ \begin{aligned} E'(t) &=2x(t)^{\mathsf T}x'(t)\\ &=2x(t)^{\mathsf T}P(t)x(t)\\ &\le-2\norm{x(t)}^2\\ &=-2E(t). \end{aligned} \]
3. For every fixed \(t_0\) and every \(t\ge t_0\),\[ \norm{x(t)}^2 \le e^{-2(t-t_0)}\norm{x(t_0)}^2. \]
Step 2 gives \[ \frac{d}{dt}\left(e^{2t}E(t)\right) = e^{2t}\bigl(E'(t)+2E(t)\bigr) \le0. \] Hence \(e^{2t}E(t)\le e^{2t_0}E(t_0)\) for \(t\ge t_0\), which is exactly the displayed estimate.
4. Therefore\[ \boxed{\lim_{t\to\infty}x(t)=0}. \]
The right side of the estimate in step 3 tends to zero as \(t\to\infty\). Thus \(\norm{x(t)}^2\to0\), and hence \(x(t)\to0\) in \(\RR^n\).
5. Q.E.D.
Step 4 is the required limit.