a) To find equilibria, use Solve
:
eq = Solve[{A - B*x - x*y^2, A*(x*y^2 - y)} == {0, 0}, {x, y}]
b) Linearizing around the point $(\hat x,\hat y)$ means making a new, linear system $$\dot{\vec{z}}=J\vec{z}$$ where $\vec{z}=(x-\hat x,y-\hat y)$ and $J$ is the Jacobian matrix $$J=\begin{bmatrix}d\dot x\over dx & d\dot x\over dy \\ d\dot y\over dx & d\dot y\over dy\end{bmatrix}$$ evaluated at $(\hat x,\hat y)$.
It's easy to calculate the Jacobian:
j = D[{A - B*x - x*y^2, A*(x*y^2 - y)}, {{x, y}}]
(* {{-B - y^2, -2 x y}, {A y^2, A (-1 + 2 x y)}} *)
The only thing left is to evaluate $J$ at the different equilibria:
j /. eq[[1]]
(* {{-B, 0}, {0, -A}} *)
j /. eq[[2]]
j /. eq[[3]]
To check the stability of the equilibria, calculate the eigenvalues of the Jacobian. If all eigenvalues have negative real part, then the equilibrium is stable. j/.eq[[1]]
is easy to understand: the eigenvalues are on the diagonal. j/.eq[[2]]
and j/.eq[[3]]
are uglier. For those you might want to try the Routh-Hurwitz stability criteria: stable if $Tr(J)<0$ and $Det(J)>0$.
Sorry, I don't know about c).
References
Strogatz SH. 2014. Nonlinear Dynamics and Chaos: With Applications to Physics, Biology, Chemistry, and Engineering.
Ellner SP, Guckenheimer J. 2006. Dynamic Models in Biology.
"Linear Stability Analysis" at Wolfram MathWorld
NonlinearStateSpaceModel
inStateSpaceModel
(which is always linear) and it will convert automatically. Or you just define the non-linear equations inside ofStateSpaceModel
. Look in the documentation ofStateSpaceModel
under "Scope" > "Basic Uses" for an example. $\endgroup$(x,y)
point that you want to linearize around? Is the goal some kind of stability analysis? Of an equilibrium or of a limit cycle? $\endgroup$StateSpaceModel
. There isnt a specific example which has something to do with my nonlinear system... @ChrisK: The exercise has three parts: a) find the stationary points b) linearize the system c) find a lyapunov-function I think you have to linearize this system with the stationary points, or at least with one of the two points. $\endgroup$