I am trying to figure out how to find the jacobian of the following:
$$du/dt = w$$ $$dw/dt=B(1-u^2)w-u$$
I think the jacobian would be: (Let d be a partial for ease) \begin{bmatrix} du/dt \\ dw/dt \\ \end{bmatrix}
But I feel like I am missing a column but I can't figure out what it would be :/ Would it be wrt u maybe?
We have $$du/dt = w=f(u,w)$$ $$dw/dt=B(1-u^2)w-u=g(u,w)$$ The Jacobian matrix is the matrix of the linearized system
$$ \begin {bmatrix} f_u&f_w\\g_u&g_w\end {bmatrix}= \begin {bmatrix} 0&1\\-2Buw-1&B(1-u^2)\end {bmatrix}$$