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I'm trying to express a fixed point when x1 and x2 are both not equal to zero for this system of equations. I know that one fixed point is $(0,0)$ when $x_1 = 0$.

$$ \begin{split} x_1 &= \lambda x_1(1 - x_1) - 0.2x_1x_2 \\ x_2 &= 0.9x_2 + 0.2x_1x_2 \end{split} $$

Where $\lambda \in (0,3)$ is a scalar. Initially, $x_1^{(0)} = 0$ and $x_2^{(0)} = 1$.

I built the Jacobian matrix A with both equations for $x_1$ and $x_2$. This gives me a $2 \times 2$ matrix $$A = \begin{bmatrix}-λ-0.2x2&-0.2x1\\0.2x2&0.9+0.2x1\end{bmatrix}$$

When I try finding the eigenvalues of $A$ using $\det(A - \lambda I)$, it doesn't seem to work out.

$$(A - \lambda I) = \begin{bmatrix} -\lambda-0.2x_2-\lambda & -0.2 x_1\\ 0.2 x_2 & 0.9 + 0.2x_1-\lambda \end{bmatrix} $$ Any ideas what I'm doing wrong?

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You are confusing the constant in your problem with the matrix eigenvalues. Rename the matrix eigenvalues to $\mu$ to get $$A - \mu I = \begin{bmatrix} -\lambda-0.2x_2-\mu & -0.2 x_1\\ 0.2 x_2 & 0.9 + 0.2x_1-\mu \end{bmatrix} $$ so $$ \begin{split} \det (A-\mu I) &= -(\lambda+0.2 x_2+\mu)(0.9 + 0.2x_1-\mu) + 0.04 x_1 x_2 \\ &= (\mu+ \lambda+0.2 x_2)(\mu - 0.9 - 0.2x_1) + 0.04 x_1 x_2 \\ &= \mu^2 + \mu \left[ \ldots \right] - \ldots \end{split} $$ and use the quadratic formula

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  • $\begingroup$ Correct, and this results in: $$-0.9λ - 0.2λx1 + λu + 0.18x2 + 0.04x1x2 - 0.2x2u + 0.9u + 0.2x1u - u^2 + 0.04x1x2$$ How can this be factored to solve for u? $\endgroup$ – stevetronix Oct 20 '15 at 15:38
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    $\begingroup$ @stevetronix please see the last couple of lines in the update $\endgroup$ – gt6989b Oct 20 '15 at 15:59
  • $\begingroup$ Thanks. So that would be $$u^2 + u[-0.9 - 0.2x1 + λ + 0.2x2] - 0.9λ - 0.2x1λ - 0.18x2$$ with $$a = 1, b = (-0.9 - 0.2x1 + λ + 0.2x2), c = - 0.9λ - 0.2x1λ - 0.18x2$$ $\endgroup$ – stevetronix Oct 20 '15 at 16:20
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    $\begingroup$ @stevetronix yes $\endgroup$ – gt6989b Oct 20 '15 at 16:35

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