Consider the following system of coupled differential equations
$$
\dot{x}_1(t) = -x_1(t) - \cos(\omega t)x_1(t) + \cos(\omega t)x_2(t), \ x_1(0)\in\mathbb{R},\\
\dot{x}_2(t) = -\gamma x_2(t) - \cos(\omega t)x_2(t) + \cos(\omega t)x_1(t), \ x_2(0)\in\mathbb{R},
$$
where $\omega$ and $\gamma$ are positive real constants. Observe that $\bar{x}=(\bar{x}_1,\bar{x}_2)=(0,0)$ is an equilibrium of the above system.

It is almost trivial to see that if $\gamma=1$ then $\bar x$ is attractive. Indeed, in this case, we have that $x(t)=[x_1(t), x_2(t)]^\top$ can be explicitly computed as
$$
x(t) = \exp\left(\begin{bmatrix}-t &0\\ 0 & -t\end{bmatrix} + \frac{1}{\omega}\sin(\omega t)\begin{bmatrix}-1 &1\\ 1 & -1\end{bmatrix}\right)x(0),
$$
so that $x(t)\to 0$ as $t\to \infty$.

However in case $\gamma\ne 1$ proving the attractiveness of the origin is not obvious (and perhaps not even true!).

>In particular, numerical simulations seem to suggest that for $\gamma$ and $\omega$ sufficiently small (e.g. $\gamma=0.001$ and $\omega=10$) the equilibrium $\bar{x}$ is **not** attractive.

I've struggled a lot to find a way of formally proving this, with no luck. So I decided to post the problem here hoping that some of you will provide some useful suggestions or tips. Thank you!


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I post here the Mathematica code that I've used in my simulations:

    (* nominal values for simulation *)
    values = {gamma -> 0.001, w -> 10};
    
    equations = {
       {x1'[t], x2'[t]} == {-x1[t] - Cos[w*t]*x1[t] + Cos[w*t]*x2[t], -gamma*x2[t] - Cos[w*t]*x2[t] + Cos[w*t]*x1[t]},
       {x1[0], x2[0]} == {0.1, 0.1}};
    
    {x1t, x2t} = NDSolveValue[equations /. values, {x1[t], x2[t]}, {t, 0, 1000}];

    Plot[x1t, {t, 0, 1000}, PlotRange -> {-0.2, 0.2}]
    Plot[x2t, {t, 0, 1000}, PlotRange -> {-0.2, 0.2}]