Question

linearize the system about the equilibria or critical points. Identify the critical points, e.g. saddle, node, center,…, but do not sketch.

Answer 1

\[\frac{ dx }{dt }=x-3y+2xy,\frac{ dy }{ dt }=4x-6y-xy\]

Answer 2

I think I can use the Jacobian matrix

Answer 3

hint:
we can write the Taylor series of the functions at the right side, up to the first order in x and y

Answer 4

\[f(x)= f(x_{0})+f'(x_{0})(x-x_{0})+...\]

Answer 5

We have not had a full lecture on taylor series yet, I am not familiar with it

Answer 6

@Michele_Laino

Answer 7

we have functions in 2 variables, so you have to write the Taylor series for functions of 2 variables

Answer 8

I do not know how to do that.

Answer 9

I understand! Nevertheless the function:
\[x - 3y + 2xy\]
can be viewed as a function of 2 variables, namely:
\[f\left( {x,y} \right) = x - 3y + 2xy\]
and x=x(t), and y=y(t), similarly for the other function at the right side of the second differential equation

Answer 10

Okay.

Answer 11

\[g(x,y)=4x-6y-xy\]

Answer 12

?

Answer 13

ok!

Answer 14

now, you have to apply the formula which gives the Taylor series for a function in 2 variables

Answer 15

Consider the nonlinear system:
\( \Large{ x ' = f(x,y)
\\ y ' = g(x,y)
}

\)

Near the critical points \( \large {(x_0,y_0) }\) satisfying

\( \large{ f(x_0,y_0)=0
\\ g(x_0,y_0) =0 }
\)


we can approximate the non linear system by the linear system :

$$\Large {
\begin{pmatrix}
x' \\
y' \\
\end{pmatrix} = \begin{pmatrix}
\dfrac{\partial f}{\partial x}(x_o,y_o) & \dfrac{\partial f}{\partial y}(x_o,y_o) \\[1em]
\dfrac{\partial g}{\partial x}(x_o,y_o) & \dfrac{\partial g}{\partial y}(x_o,y_o) \end{pmatrix}
\begin{pmatrix}
x \\
y \\
\end{pmatrix}

}
$$

Answer 16

We can then analyze these linear systems to conclude what we can about the behavior of solutions near the critical points.

Answer 17

we can use Michel's approach as well.

Answer 18

its up to you

Answer 19

I don't really know Michele's approach. I was trying to follow along with the Taylor series.

Answer 20

Can we try your approach?

Answer 21

ok i will continue

Answer 22

we will use the post above as a template

Answer 23

when I solved, I used the Jacobian matrix

Answer 24

and okay!

Answer 25

We are given
\( \Large{

\frac{ dx }{dt }=x-3y+2xy,\\
\frac{ dy }{ dt }=4x-6y-xy

}

\)

First task is to solve x' = 0 , y ' = 0 simultaneously

Answer 26

Solve the system:
x - 3y +2xy = 0
4x - 6y -xy = 0

clearly x=0, y=0 is a solution

Answer 27

the other solution we can find by substitution

Answer 28

Using the second equation
4x - 6y -xy = 0
4x - xy = 6y
x ( 4 -y ) = 6y
x = 6y / (4-y)

substitute this into the first equation for x

Answer 29

First equation: x - 3y +2xy = 0
substitute:
6y / (4-y) - 3y + 2 (6y / (4-y)) * y = 0

Answer 30

$$
\large {
x - 3y +2xy = 0
\\ \therefore \\
\frac{6y}{4-y} - 3y + 2 \cdot \frac{6y}{4-y} \cdot y =0
\\ \therefore \\
\color{red}{(4-y )} \cdot \left( \frac{6y}{4-y} - 3y + 2 \cdot \frac{6y}{4-y} \cdot y \right)=0 \cdot \color{red}{(4-y )}
\\ \therefore \\
6y - 3y(4-y) + 2 \cdot 6y \cdot y =0
\\ \therefore \\
15y^2-6y = 0
\\ \therefore \\
y(15y-6) = 0
\\ \therefore \\y = 0, ~~y = \frac{6}{15}= \frac 25
}
$$

Answer 31

$$
\Large{
x' =f(x,y) = x-3y+2xy\\
y' =g(x,y) = 4x-6y-xy \\~\\

\begin{pmatrix}
f_x & f_y\\
g_x & g_y \\
\end{pmatrix} =

\begin{pmatrix}
1+2y & -3 + 2x\\
4-y & -6 - x \\
\end{pmatrix}


}

$$

Answer 32

now plug in the critical points

Answer 33

\[\left[\begin{matrix}1 & -3 \\ 4 & -6\end{matrix}\right]\]

Answer 34

thats for 0,0

Answer 35

\[\left[\begin{matrix}1.8 & -3+2x \\ 3.6 & -6-x\end{matrix}\right]\]

Answer 36

thats for (x, 2/5)

Answer 37

actually the two critical points are (0,0) and (2/3, 2/5)

Answer 38

\[\left[\begin{matrix}1.8 & -1 \\ 3.6& -6.667\end{matrix}\right]\]

Answer 39

\[\left[\begin{matrix}1 & -3 \\ 3.6 & -6.667\end{matrix}\right]\]

Answer 40

disregard the 1st one

Answer 41

\[\left[\begin{matrix}1 & -3 \\ 4 & -6\end{matrix}\right]\]

Answer 42

$$
\Large{
x' =f(x,y) = x-3y+2xy\\
y' =g(x,y) = 4x-6y-xy \\~\\

\begin{pmatrix}
f_x & f_y\\
g_x & g_y \\
\end{pmatrix} =

\begin{pmatrix}
1+2y & -3 + 2x\\
4-y & -6 - x \\
\end{pmatrix}
\\~\\
\text{The jacobian matrix associated with (0,0) is :} \\~\\
\begin{pmatrix}
1+2(0) & -3 + 2(0)\\
4-0 & -6 - 0 \end{pmatrix}
=
\begin{pmatrix}
1 & -3\\
4 & -6 \end{pmatrix}




\\~\\ \text{The jacobian matrix associated with (2/3, 2/5) is :} \\~\\
\begin{pmatrix}
1+2(\frac25) & -3 + 2(\frac23)\\
4-\frac25 & -6 - \frac23 \end{pmatrix}

=\begin{pmatrix}
\frac9 5 & -\frac 53\\
\frac{18}{5} & -\frac{20}{3} \end{pmatrix}


}

$$

Answer 43

\[\left[\begin{matrix}1.8 & -1.667\\ 3.6& -6.667\end{matrix}\right]\]

Answer 44

now we can use these matrices to make linear equations

Answer 45

ex?

Answer 46

x'=(9/5)x-(5/3)y
y'=(18/5)x-(20/3)y

Answer 47

$$
\Large{ \text{ near (0,0)}\\
\begin{pmatrix} x' \\ y' \end{pmatrix} =\begin{pmatrix}
1 & -3\\
4 & -6 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}
\iff
\begin{cases} x' = 1x -3y \\
y' = 4x - 6y
\end{cases}
\\ ~~\\~~\\
\\
\text{near (2/3, 2/5) } \\
\begin{pmatrix} x' \\ y' \end{pmatrix}=\begin{pmatrix}
\frac9 5 & -\frac 53\\
\frac{18}{5} & -\frac{20}{3} \end{pmatrix}
\begin{pmatrix} x \\ y \end{pmatrix}
\iff
\large \begin{cases} x' = \frac95x -\frac53y \\
y' = \frac{18}{5}x - \frac{20}{3}y
\end{cases}

}

$$

Answer 48

Ok, so now I have to identify the critical points?

Answer 49

well now we can look at the eigenvalues

Answer 50

Cases:
Sources or Repellers: both eigenvalues have positive real part.
Sinks or Attractors: both eigenvalues have negative real part.
Saddles: one eigenvalues is positive and the other negative.

Answer 51

for (0,0) --> sink ?

Answer 52

we know the critical points are (0,0) and (2/3, 2/5)

Answer 53

one moment, calculating eigen values

Answer 54

for the first matrix associated with (0,0) i get eigen values -2, -3

Answer 55

thats what I got

Answer 56

so thats a sink :)

Answer 57

yay!

Answer 58

for the next matrix, i got eigenvalues

Lambda = 1.019362877,-5.886029543

Answer 59

Exactly:
-73/30+1/30 sqrt(10729)],-73/30-1/30 sqrt(10729)

This implies the critical point is a saddle

Answer 60

thank you