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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?

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1 Answer 1

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Both methods, the perturbation approach and the Jacobian approach, share a common underlying concept, which is to linearize the differential equation in the vicinity of the equilibrium point and analyze the trajectory behavior in that region. The perturbation approach is typically employed when the linear component of the differential equation is easily identifiable. Also, one may consider different kinds of perturbations, at different rates, which could be useful to study certain families of solutions of the differential equation.

On the other hand, the Jacobian approach is more commonly used as it provides a means to linearize a system of differential equations through differentiation alone, making it valuable even in cases where the linear components are not immediately apparent. This is particularly useful when dealing with systems containing non-polynomial terms that complicate the identification of the linear parts upon initial inspection.

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  • $\begingroup$ Thanks for your response! I've been thinking on it, and I realize that one result of the perturbation analysis is a differential equation/system for the perturbation itself. I suppose that having the linear portion easily identifiable, as you say, allows us to better understand to first order the behavior of the perturbation. Since the Jacobian method doesn't give us this, it seems more a method of last resort – appropriate as you say in more complicated cases. 1/2 $\endgroup$ Commented Jun 21, 2023 at 21:53
  • $\begingroup$ Do you find this accurate? Do you care to comment any further on use cases of perturbation method vs. the Jacobian? For example, are there any cases where they produce different results, or using one is much preferred over the other? 2/2 $\endgroup$ Commented Jun 21, 2023 at 21:54
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    $\begingroup$ Sorry for the late response @StevenBasmith. In my opinion, the Jacobian method is usually more useful because it can be applied to many cases easily since differentiation is typically straightforward. On the other hand, traditional perturbation is only helpful for simple cases, so its application is more limited. 1/2 $\endgroup$
    – Julian
    Commented Jul 6, 2023 at 18:48
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    $\begingroup$ 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 $\endgroup$
    – Julian
    Commented Jul 6, 2023 at 18:55
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    $\begingroup$ I appreciate the discussion and insight with you. Cheers! $\endgroup$ Commented Jul 10, 2023 at 6:28

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