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April 4th, 2019, 12:43 PM  #1 
Newbie Joined: Mar 2019 From: USA Posts: 4 Thanks: 0  Steady States & Eigenvalues
For an example problem to solve in an engineering class I was given the following systems and told to find the steady states of each and the eigenvalues for each steady state. 1) $\displaystyle dx/dt = x(12yx)$ $\displaystyle dy/dt = yx$ 2) $\displaystyle x' = x(10xy)$ $\displaystyle y' = y(302xy)$ I know that for steady states I need to solve for when the derivatives are both equal to zero. For both systems I know that x=y=0 is a trivial solution. I found another solution to be x=y=1/3 for System #1 and x=20, y=10 for System #2. My problem is, I remember how to find eigenvalues of matrices from Lin. Alg. but I'm not sure how to get these equations into a form where I can solve for the eigenvalues. How would I go about finding the eigenvalues for these systems? 
April 4th, 2019, 04:37 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics 
You are correct that the equations here don't technically have eigenvalues. This is a fairly common abuse of notation/expression which I suppose can be confusing the first time you encounter it. Let me try to explain it carefully. If $f: \mathbb{R}^n \to \mathbb{R}^n$ is a (autonomous) vector field, then it gives rise to a corresponding differential equation $\dot x = f(x)$. As you have already worked out, the steady states (also called equilibrium solutions) are just constant solutions of this differential equation which means the derivative is zero i.e. $p \in \mathbb{R}$ is a steady state if $f(p) = 0$. Now, if $f(x)$ is a linear function, then it makes sense to talk about its eigenvalues. If $f$ is nonlinear but at least differentiable, then the next best thing is to talk about the eigenvalues of its derivative. Keep in mind that $f$ is a vector field on $\mathbb{R}^n$ so its derivative at $p$, denotes by $Df(p)$ is linear transformation on $\mathbb{R}^n$. Commonly this function is represented as a matrix and the eigenvalues we are referring to are just the eigenvalues of this linear transformation. The important property is that for "most" differentiable vector fields, the stability of the origin in the linearized differential equation is identical to the stability of $p$ in the nonlinear ODE. What this means is, we consider the new differential equation $\dot x = Df(p) x$ which is a linear ODE. This is what is called the linearized ODE for $f$ at the equilibrium $p$. It is well known from your first ODEs course which tends to cover linear ODEs in depth, that the origin is an equilibrium solution for this equation and its stability can be determined from the eigenvalues of the matrix $Df(p)$. Well, the term "most" here means that if $Df(p)$ is hyperbolic, meaning every eigenvalue has a nonzero real part, then the stability of $p$ in the nonlinear system is the same as the stability of the origin in the linearized ODE. 
April 4th, 2019, 04:48 PM  #3  
Newbie Joined: Mar 2019 From: USA Posts: 4 Thanks: 0  Quote:
 

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