Question

classify stationary points

Answer 1

\[x'=-(2+y)y, y'=x(1-y)\]

Answer 2

I used Jacobian method

Answer 3

stationary points = (0,0) and (0,-2)\[Jf(x,y)=\left[\begin{matrix}0 & -2-2y\\ 1-y & -x\end{matrix}\right]\]

Answer 4

how to determine if it is saddle or center?

Answer 5

@wio

Answer 6

I need to find the roots, I guess?

Answer 7

\[z^2-tra(A)+ \det(A)\]

Answer 8

\[Jf(0,0)= \left[\begin{matrix}0 & -2 \\ 1 & 0\end{matrix}\right]\]

Answer 9

I believe saddle has a positive and negative eigenvalue while a center has complex eigenvalues?

Answer 10

\[z^2- 0z+2\]
\[z=\sqrt{2}i\] complex roots!

Answer 11

so this is center with anticlockwise

Answer 12

where as the Jf(0,-2)= \[\left[\begin{matrix}0 &2 \\ 3 & 0\end{matrix}\right]\]

Answer 13

\[z^2-6=0\]
\[z=\pm \sqrt{6}\]

Answer 14

therefore a saddle point

Answer 15

and unstable and repelling. OFFFHH, am I right

Answer 16

first of all what are e-values? I know how to calculate but don't know what it represents

Answer 17

eigen values and eigen vectors are used to diagonalize a matrix

Answer 18

Well eigenvalues are just scalars that you could multiply an eigenvector by rather than the matrix to get the same result. That's what makes them special, it doesn't rotate them, only stretches or compresses vectors.