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[SOLVED] stable spiral but math says saddle

dwsmith

Well-known member
Feb 1, 2012
1,673
At my fixed point of (1,1), I should have a stable spiral but I keep getting analytically that it is a saddle.

$$
\begin{cases} \dot{x} = x(3 - 2x - y)\\
\dot{y} = y(-5 + 5x)
\end{cases}
$$
The fixed points are $(0,0)$, $\left(\frac{3}{2},0\right)$, and $\left(1,1\right)$.
Also, the Jacobian is
$$
\mathcal{J} = \begin{pmatrix}
3 - 4x - y & -x\\
-5y & -5 + 5x
\end{pmatrix}.
$$
At $(0,0)$, we have $\mathcal{J} = \begin{pmatrix}3 & 0\\0 & -5\end{pmatrix}$; that is, $\text{tr} = -2$, $\det = -15$, and the discriminant is greater than 0.
Therefore, we have an saddle at $(0,0)$.
For $\lambda = 3$, the eigenvector is $e_1$, so when $\lambda = 5$, the eigenvector is $e_2$.


At $\left(\frac{3}{2},0\right)$, we have $\mathcal{J} = \begin{pmatrix}-3 & -\frac{3}{2}\\0 & \frac{5}{2}\end{pmatrix}$; that is, $\text{tr} = -\frac{1}{2}$, $\det = -\frac{15}{2}$, and the discriminant is greater than 0.
Therefore, we have a saddle at $\left(\frac{3}{2},0\right)$.
For $\lambda = -3$, the eigenvector is $e_1$, so when $\lambda = \frac{5}{2}$, the eigenvector is $\begin{pmatrix}-\frac{3}{11}\\1\end{pmatrix}$.


At $\left(1,1\right)$, we have $\mathcal{J} = \begin{pmatrix}-2 & -1\\-5 & 0\end{pmatrix}$; that is, $\text{tr} = -2$, $\det = -5$, and the discriminant is greater than 0.
Therefore, we have a saddle at $\left(1,1\right)$.
For $\lambda = i\sqrt{5}$, the eigenvector is $\begin{pmatrix}\frac{i}{\sqrt{5}}\\1\end{pmatrix}$, so when $\lambda = -i\sqrt{5}$, the eigenvector is $\begin{pmatrix}-\frac{i}{\sqrt{5}}\\1\end{pmatrix}$.

phasebsecond.jpg
It is obviously spiral. I have re-checked and done the problem three times and nothing is working out. What is going wrong?
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
Check the [2,1] entry of your Jacobian matrix. I see a sign problem.