Find equilibrium points and stability system of ODE

justme.h

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May 24, 2019
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Hello,
I need a help with finding the equilibrium points and state of stability of the following system of ODEs:
dx/dz=x(E - Ax - y);
dy/dz=y(-1 + x);

Thank you in advance. I will be thankful for any advice.
 
Hint: If [math]y' = y^{2} - 9[/math], can you find the equilibrium solutions?
 
Hint: If [math]y' = y^{2} - 9[/math], can you find the equilibrium solutions?
Yes i can. If y' is 0 then y=−3 or y=3 which are the equilibrium
But in my case i have:
1) (x,y)=(0,0)
2)x=0 then y=0 or x=1
3)y=0 then x=0 or x=E/A

Is that right or im doing it wrong? and is there any cases left? And how can i find the state of stability? Thank you!
 
Yes i can. If y' is 0 then y=−3 or y=3 which are the equilibrium
But in my case i have:
1) (x,y)=(0,0)
2)x=0 then y=0 or x=1
3)y=0 then x=0 or x=E/A

Is that right or im doing it wrong? and is there any cases left? And how can i find the state of stability? Thank you!
Forget the "but in my case" business. This thought is only confusing. Do EXACTLY the same thing.

You need ALL the derivatives to be zero simultaneously. "ALL" just means more when there is more than one.

Solve the system:
x(E - Ax - y) = 0
y(x - 1) = 0

You did not just write, "2) x=0 then y=0 or x=1"
 
Forget the "but in my case" business. This thought is only confusing. Do EXACTLY the same thing.

You need ALL the derivatives to be zero simultaneously. "ALL" just means more when there is more than one.

Solve the system:
x(E - Ax - y) = 0
y(x - 1) = 0

You did not just write, "2) x=0 then y=0 or x=1"
Thank you! You mean that i have also a case with x=1
4)x=1 then y=E-A
and
5)x=1 then y is 0 or y is everything but not 0
 
I did this:

Solve the system:
x(E - Ax - y) = 0
y(x - 1) = 0

Ex - Ax^2 - xy = 0
xy - y = 0

Ex - Ax^2 = xy
y = xy

y = Ex - Ax^2 <== To me, this suggests a more interesting insight than the case listing. Others may disagree.

I'm not sure you understood my chastisement. If x = 0, then "or x = 1" is not a good thing to write. :) Be more careful.
 
Thank you so much i understand you. Can you help me with the state of stability?
 
Hello,
I need a help with finding the equilibrium points and state of stability of the following system of ODEs:
dx/dz=x(E - Ax - y);
dy/dz=y(-1 + x);

Thank you in advance. I will be thankful for any advice.
Equilibrium occurs when dx/dz= 0 and dy/dz= 0.
y(-1+ x)= 0. Either y= 0 or x= 1.
x(E- Ax- y)= 0. If y= 0 x(E- Ax)= 0. Either x= 0 or x= E/A.
If x= 1 E- A- y= 0 so y= A- E.
Equilibrium points are (0, 0), (E/A, 0), and (1, A- E).
The "Jacobian", the matrix with the partial derivatives, is \(\displaystyle \begin{bmatrix}E- 2Ax- y & -x \\ y & -1\end{bmatrix}\).

At (0, 0) that is \(\displaystyle \begin{bmatrix}E & 0 & 0 & -1\end{bmatrix}\). That has eigen values 0 and -1. Since there are no positive eigen values, this is a stable equilibrium point.

At (E/A, 0) that is \(\displaystyle \begin{bmatrix}-E & -E/A \\ 0 & -1 \end{bmatrix}\). That has eigenvalues -E and -1. Assuming E is positive that is a stable equilibrium point.

At (1, A- E) that is \(\displaystyle \begin{bmatrix} 3A & -1 \\ A- E & -1 \end{bmatrix}\). The eigenvalue equation is more complicated so I will leave that to you!
 
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