Conditions for map $F: \mathbb{R} \to \mathbb{R}^2$ to tend to $0$ at infinity

Question

Suppose we have a map $F: \mathbb{R} \to \mathbb{R}^2$ such that $D[F] = XF$ where $D$ denotes the derivative and

$X = \begin{bmatrix} -1 & 1 \\ c & -1 \end{bmatrix}$ where $c \in \mathbb{R}$.

I want to find all $c$ such that $\lim_{t \to \infty} F(t) = 0$.

Proof.

If we write $F(t) = (u(t), v(t))$, then we know $u'(t) = 0$ and $v'(t) = cu(t) - v(t)$.

Since $u'(t)$ vanishes, it follows that $u(t)$ is constant. But then in order to have $\lim_{t \to \infty} F(t) = 0$, it follows that $u(t) = 0$.

This makes $v'(t) = v(t)$, so that $v(t) = e^{t}$. But then $\lim_{t \to \infty} F(t) \ne 0$.

Therefore, there is no such $c$.

I'd appreciate if someone could tell me if I've made a mistake somewhere.

Answer 1

Your mistake: you wrote $u'(t)=0$, but this is false.

We have $u'(t)=-u(t)+v(t)$.

Answer 2

For this kind of linear equation, $$x'=Ax,$$ where $A$ is a matrix of $n\times n$, the general solutions are $x=\exp(At)$.

The condition that $x\rightarrow 0$ is equivalent to that the eigenvalues of $A$ are negative.