Steady States & Eigenvalues for dx/dt = x (1 - 2y - x) and dy/dt = y - x

asaiz

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Apr 4, 2019
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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)
\(\displaystyle dx/dt = x(1-2y-x)\)
\(\displaystyle dy/dt = y-x\)

2)
\(\displaystyle x' = x(10-x-y)\)
\(\displaystyle y' = y(30-2x-y)\)

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?
 

HallsofIvy

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Jan 27, 2012
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Only linear transformations have eigenvalues (which is why you learn about them in Linear Algebra). What this problem is asking you to do is to find eigenvalues for the linearized systems, linearized close to the steady state.

Yes, (x, y)= (0, 0) is a steady state and you "linearize" a system by arguing that higher powers of small numbers can be ignored. Around (0, 0) we can linearize by ignoring higher powers of x and y (which includes products of x and y). The original system,
\(\displaystyle \frac{dx}{dt}= x(1- 2y- x)= x- 2xy- x^2\)
\(\displaystyle \frac{dy}{dt}= y- x\)
linearized around (0, 0) is
\(\displaystyle \frac{dx}{dt}= x\)
\(\displaystyle \frac{dy}{dt}= y - x\).
The vector form of that could be written
\(\displaystyle \frac{d}{dt}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix} 1 & 0 \\ -1 & 1\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}\).
The "eigenvalues" referred to are the eigenvalues of the matrix \(\displaystyle \begin{bmatrix} 1 & 0 \\ -1 & 1\end{bmatrix}\)

To linearize around the other steady state, (x, y)= (1/3, 1/3), change variables: let u= x- 1/3 and v= y- 1/3 so that (x, y)= (1/3, 1/3) corresponds to (u, v)= (0, 0). Then x= u+ 1/3 and y= v+ 1/3 so the equations become
\(\displaystyle \frac{dx}{dt}= \frac{du}{dt}= (u+ 1/3)(1- 2(v+1/3)- u- 1/3)= (u+1/3)(1- 2v- u)\)\(\displaystyle = u- 2uv- u^2+ 1/3- (2/3)v- (1/3)u= (2/3)u- (2/3)v- 2uv- u^2)\) and
\(\displaystyle \frac{dy}{dt}= \frac{dv}{dt}= v+ 1/3- u- 1/3= v- u\)
Now, (x, y) close to (1/3, 1/3) means that (u,v) is close to (0, 0) so we can "linearize" by ignoring higher powers of u and v:
\(\displaystyle \frac{du}{dt}= (2/3)u- (2/3)v\) and
\(\displaystyle \frac{dv}{dt}= v- u\)
That can be written as
\(\displaystyle \frac{d}{dt}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}\frac{2}{3} & -\frac{2}{3} \\ -1 & 1 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}\)
The "eigenvalues" are the eigenvalues of the matrix \(\displaystyle \begin{bmatrix}\frac{2}{3} & -\frac{2}{3} \\ -1 & 1 \end{bmatrix}\)
 
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