Node: Type, Stability, Slope at origin, Trajectories. Linear system.

113 Views Asked by At

I have a system of equations:

$$\begin{pmatrix}\dot{y}_1\\\dot{y}_2\end{pmatrix}=\begin{pmatrix}2&0\\4&-1\end{pmatrix}\begin{pmatrix}y_1\\y_2\end{pmatrix}$$

Looking at matrix $A$ I can see a value not on the diagonal equal to zero. So the eigenvalues here are $2$ and $-1$, meaning I have a saddle node. Next I look at the eigenvectors:

$\lambda=2, \left[\begin{array}{cc|c}0&0&0\\4&-3&0\end{array}\right]$, $y^{(1)}=\begin{pmatrix}3\\4\end{pmatrix}$

$\lambda = -1,\left[\begin{array}{cc|c} 3&0&0\\4&0&0\end{array}\right]$,$y^{(2)}=\begin{pmatrix}0\\1\end{pmatrix}$

$y=Ae^{2x}\begin{pmatrix}3\\4\end{pmatrix}+Be^{-x}\begin{pmatrix}0\\1\end{pmatrix}$

So I have an attractive trajectory on the $y$-axis, and a repulsive trajectory on the line $y=\frac{3}{4}x$

With the slope at origin being $\frac{\dot{y}_2}{\dot{y}_1}=\frac{4y_1-y_2}{2y_1}=\frac00=\pm\infty$?

Does all of this seem correct?

1

There are 1 best solutions below

0
On

It is indeed correct, but the slope you computed is the slope of the orbit on the phase space, because each coordinate $y_1$ and $y_2$ have an exponencial growth/decay rate. also the origin is not a node, but a saddle equilibria