I have two equations:
$$\frac{dx}{dt} = v(k-ux) – \delta x, \hspace{3mm} \frac{dv}{dt} = v(r-px)$$
I want to calculate the Jacobean so I can analyse the stability of the fixed points.
Does it matter if I have my Jacobean as $A = \begin{pmatrix} \dot{v}_x & \dot{v}_v \\ \dot{x}_x & \dot{x}_v \end{pmatrix}$ or $A = \begin{pmatrix} \dot{x}_x & \dot{x}_v \\ \dot{v}_x & \dot{v}_v \end{pmatrix}$ or $A = \begin{pmatrix} \dot{v}_v & \dot{v}_x \\ \dot{x}_v & \dot{x}_x \end{pmatrix}$ and so on?
When I say does it matter, I mean will it affect the stability analysis? I think it shouldn't, but I'm not sure why because I think switching rows in a matrix does affect its eigenvalues and a change in eigenvalues would change my stability.
Any help is appreciated.
Best Answer
You have the two dimensional dynamical system as $$\frac{dx}{dt} = v k - u vx = f(v,x)$$ $$\frac{dy}{dt} = vr - vpx = g(v,x)$$ The actual definition of Jacobian matrix is given by $J = \begin{bmatrix} \frac{\partial{f}}{\partial{v}} & \frac{\partial{f}}{\partial{x}} \\ \frac{\partial{g}}{\partial{v}} & \frac{\partial{g}}{\partial{x}} \end{bmatrix}$.
whose entries you can calculate.
Observe that the $det(A) = det(A^T)$, where $A^T$ is the transpose of the matrix $A$.
So you will get the same result regarding the eigenvalues and eigenvectors even you consider $J^T$.
But as you have mentioned switching rows. Recall that if you interchange two rows or (columns) the determinant will change sign and since the determinant is the product of the eigenvalues, so the eigenvalues changes. And then you check for the negative real part of all the eigenvalues, implying asymptotically stability of the fixed points.
If you swap the two rows two times then you get the original matrix back consistent with multiplying $(-1)^2$ to your original determinant.
Explaining your last comment:
For the first case $f = 2x+3v; g=4x+5v$, the Jacobian matrix is $\begin{bmatrix} 2 & 3\\ 4 & 5\end{bmatrix}$, we are talking of the stability of fixed point(see here that the fixed point of the system is $(u^*,v^*)=(0,0)$), the Jacobian evaluated at the fixed points remains unchanged,so we look into the eigenvalues of the Jacobian which are $\lambda_{1} = \frac{7 + \sqrt{57}}{2} >0$ and $\lambda_{2} = \frac{7 - \sqrt{57}}{2} <0$, so in this case the fixed point is a saddle(Mixed stability). Observe that the determinant of the Jacobian is $-2$.
Now considering the second case(when you interchange the expressions for $f$ and $g$) $f = 4x+5v;g = 2x+3v$. The Jacobian matrix is given by $J = \begin{bmatrix} 4 & 5\\ 2 & 3\end{bmatrix}$, which also has the fixed point $(u^*,v^*) = (0,0)$. Observe here the determinant is $2$ (negative of the first case!) The eigenvalues of the Jacobian matrix is given by $\lambda_{1} = \frac{7 + \sqrt{41}}{2}>0$ and $\lambda_{2} = \frac{7-\sqrt{41}}{2} >0$, so the fixed point is unstable.
So you conclude he effect of interchanging right! always be sure to check the nature of eigen values in order to check for stability of fixed points instead of the statement - the change in jacobian results in change in eigenvalues and hence change in stability(as it may be the case that the eigenvalues change when you interchange the rows but still in both cases the real part of eigenvalues may be negative implying in both the cases the stable fixed points )