The asymptotic behavior of a solution to the matrix differential equation $\frac{du}{dt} = A u(t)$, where $A$ satisfies certain criteria

Question

$\mathbf {The \ Problem \ is}:$ Let, $\operatorname {u(t) = (u_1(t),u_2(t))}$ where $t\gt 0$ be the unique solution of the differential equation
$\operatorname {du/dt} = Au(t)$ where $\operatorname {u(0)}= (1,1)$ and $A= $$\begin{bmatrix} a & c \\ c & b \\ \end{bmatrix}$ is a $2×2$ symmetric matrix with $tr A\lt 0$ and $\operatorname {det}A\gt 0$ , then evaluate $\lim_{t \to \infty} \operatorname {u_1(t)} .$

$\mathbf {My \ approach}$ : I only could think that by the given information both the eigenvalues of $A$ are real and negative, and hence $A^{-1}$ exists, and $$ \begin{bmatrix} \operatorname {du_1(t)/dt}\\ \operatorname {du_2(t)/dt}\\ \end{bmatrix} = \begin{bmatrix} au_1(t)+cu_2(t) \\ cu_1(t)+bu_2(t) \\ \end{bmatrix} $$ , but I can't approach further.

Answer 1

Hint In analogy with the scalar setting, we can verify by differentiating that the general solution of $$\frac{d{\bf u}}{dt} = A {\bf u}$$ is $${\bf u}(t) = \exp (tA) {\bf u}_0 ,$$ where ${\bf u}_0 = {\bf u}(0)$. Now, since $A$ is symmetric, it is diagonalizable, say, $$A = P^{-1} \Delta P , \qquad \Delta := \operatorname{diag}(\lambda_1, \lambda_2) ,$$ where $\lambda_1, \lambda_2$ are the eigenvalues of $A$

So, $${\bf u}(t) = \exp (t A) {\bf u}_0 = P^{-1} \exp (t \Delta) P {\bf u}_0 ,$$ and $$\exp (t \Delta) = \operatorname{diag}(\exp \lambda_1 t, \exp \lambda_2 t) .$$ But $\lambda_1 + \lambda_2 = \operatorname{tr} A < 0$ and $\lambda_1 \lambda_2 = \det A > 0$ implies that $\lambda_1, \lambda_2 < 0$.

Answer 2

Find the eigenvalues of $A$ if they are $m_1,m_2$ then $$u_1(t)= C_1 e^{m_1t}+ C_2 e^{m_2 t}$$ Put this soln in the equation $$\frac{du_1(t)}{dt}=au_1(t)+bu_2(t)$$ to get $u_2(t). $ Next from $u_1(0)=1, u_2(0)=1$ get $C_1, C_2$ and hence you are done.$