# Deriving a Hopf Bifurcation – Perturbation Method vs. Jacobian Matrix

When deriving a Hopf bifurcation of a dynamical system, the usual process is:

1. Find a fixed point $$(x_0, y_0)$$
2. Perturb the system about the fixed point $$(x_0+\tilde{x}, y_0+\tilde{y})$$
3. Linearize, neglecting terms quadratic or higher in the perturbation
4. Find the solution of the linearized system

For example, starting with $$\frac{dx}{dt} = −y+(a−x^2−y^2)x$$ $$\frac{dy}{dt} = x+(a−x^2−y^2)y$$ we deduce a fixed point at $$(x,y) = (0,0)$$, perturb the solution by substituting $$(x,y)$$ for $$(0+\tilde{x}, 0+\tilde{y})$$, and linearize the result to obtain:

$$\frac{d\tilde{x}}{dt} = a\tilde{x}− \tilde{y}$$ $$\frac{d\tilde{y}}{dt} = \tilde{x}+a\tilde{y}$$

The matrix associated with this linear system is: $$\begin{bmatrix} a & -1 \\ 1 & a \\ \end{bmatrix}$$

The resulting eigenvalue problem yields eigenvalues of $$s = a \pm i$$. However, another method to deduce the Hopf bifurcation here is a little more straightforward:

1. Find the Jacobian matrix $$J(x,y)$$
2. Evaluate $$J(x,y)$$ at the fixed point $$(x_0, y_0)$$
3. Find the eigenvalues of $$J(x_0, y_0)$$

Indeed, the Jacobian matrix of our example, evaluated at $$(0,0)$$, is $$\begin{bmatrix} a & -1 \\ 1 & a \\ \end{bmatrix}$$

This is the same matrix derived from perturbing the system, and of course its eigenvalues will be the same. My question is: Why go through the machinery of perturbing the solution, when directly finding and evaluating the Jacobian produces the same result? In what cases will/won't it produce the same result?

• Oh, and what I was trying to say about the differences between the Jacobian and the perturbation method is that the latter can be useful for analyzing the behavior of the differential equation under unusual dependencies. In your notation, it would involve changing $(x,y)$ to $(0+f(\tilde x,\tilde y),0+g(\tilde x, \tilde y))$, where both $f$ and $g$ are non-trivial functions of $\tilde x$ and $\tilde y$. However, in the end, you still have to linearize the equation again, and in these cases, it's generally better to use the Jacobian approach. 2/2 Commented Jul 6, 2023 at 18:55