Question: where are my mistakes? (I was told that this does not prove stability)
In my thesis I am trying to show that this system is stable only when $w> \widetilde{w}$:
$\DeclareMathOperator{\sech}{sech}$ $\begin{align} \dot{y}(t) & =\big(n(t)f(w, \widetilde{w}) + \dot{n}(t)m( \widetilde{w} )\big) \tanh\Big(\lambda_1 \big( y_B-y(t) \big) \Big) \\ \dot{n}(t) & =a (w-\widetilde{w})\tanh \Big(\lambda_2 \big(n_B-n(t)\big)\Big) \end{align}$
Since my Jacobian matrix is "too wide" lets look at each element seperately. So my Jacobian matrix:
$\begin{align*} J=\begin{bmatrix} J_{11} & J_{12}\\ J_{21} & J_{22}\\[0.5cm] \end{bmatrix} \end{align*}$
Thus,
$J_{11} = -\big(n(t)f(w, \widetilde{w})+\dot{n}(t)m(\widetilde{w}))\big) \lambda_1 \sech^2(\lambda_1 (y_B-y(t)) $
$J_{12} = -m(\widetilde{w})a(w-\widetilde{w})\lambda_2\sech^2(\lambda_2 (n_B-n(t)) \tanh\big(\lambda_1 (y_B-y(t))\big) $
$J_{21} = 0$
$J_{22} = -a(w-\widetilde{w})\lambda_2\sech^2(\lambda_2 (n_B-n(t)) $
Here, $a>0$, $\lambda_1>0$, $\lambda_2>0$, functions $f$ and $m$ can only be positive values. We have an equiliubrium values $(y_B,n_B)$:
$\DeclareMathOperator{\sech}{sech} \begin{align*} J_{(y_B,n_B)}= \small{\begin{bmatrix} -n(t)f(w, \widetilde{w})\lambda_1 & 0 \\ 0 & -a(w-\widetilde{w})\lambda_2 \\ \end{bmatrix}} \end{align*}$
We know that $-n(t)f(w, \widetilde{w})\lambda_1$ is always negative. Lets say that $w>\widetilde{w}$ then $ -a(w-\widetilde{w})\lambda_2$ is also negative. Then we have an asymptotic stability.