Understanding Equlibrium Points in differential equations

[Martyn Arthur]

Homework Statement

Find equilibrium points of differential equations

Relevant Equations

dx/dt =1-y (1)
dy/dt =x^2-y^2 (2)

For dx/dt = 0 (1)
1-y =0 , y= 1
substituting
For dy/dt = 0 (1)
for dy/dt x= sqrt 1 =1
so I have points (0,1) and (0,1)
But this isn't correct!
'cant figure out where to go from here?
Thanks
Martyn

 

[pasmith]

How do you solve x^2 - 1 = 0?

 

[Martyn Arthur]

Thank you....
x^2 =1
so x =1
not +/-1 just 1 as per my calculator.
I hate blaming ADHD, I rate quite high on chess.com its ridiculous but on occasions I get into this situation.
We have an x value of 1
I need to fit it in here, but can't see how
dx/dt =1-y (1)

 

[pasmith]

x^2 - 1 = (x + 1)(x - 1) = 0 has both x = \pm 1 as solutions.

 

[Martyn Arthur]

AH substituting 1 for y^2 and factorising gives me , for x +/-1

So thus, I believe we have (-1,1) and (1,1)
Thank you, onwards to the Jacobian maitrix.
Martyn

 

[Martyn Arthur]

Following on am I correct in saying that for the Jacobian maitrix the partial derivatives of dee u / dee/x and dee u / dee y are -y and 1 respectively
and the partial derivatives of dee v / dee/x and dee v / dee y are -2y and 2x respectively?

 

[Martyn Arthur]

Apologies, the question should have been accompanied by my workings I will try again shortle.

 

[SammyS]

Following on am I correct in saying that for the Jacobian maitrix the partial derivatives of dee u / dee/x and dee u / dee y are -y and 1 respectively
and the partial derivatives of dee v / dee/x and dee v / dee y are -2y and 2x respectively?

While we wait ...

Notice that you have not defined $u$ or $v$.

I suspect the definitions are : $\displaystyle u=x+y$ and $\displaystyle v=x-y$ .

 

[Mark44]

Relevant Equations: dx/dt =1-y (1)
dy/dt =x^2-y^2 (2)

For dx/dt = 0 (1)
1-y =0 , y= 1
substituting
For dy/dt = 0 (1)
for dy/dt x= sqrt 1 =1
so I have points (0,1) and (0,1)
But this isn't correct!

I'm late to the party as I didn't notice this thread, which was posted in the Adv. Physics HW section. I've moved it to Calculus & Beyond, which is a better fit.

@Martyn Arthur, your notation is confusing with the numbers that indicate the equations possibly being mistaken for something else.

From the first equation,
$\frac{dx}{dt} = 0 \Rightarrow 1 - y = 0 \Rightarrow y = 1$

From the second equation,
$\frac{dy}{dt} = 0 \Rightarrow x^2 - y^2 = 0$

For both equations to be satisfied, the second equation becomes $x^2 = 1 \Rightarrow x = \pm 1$
I should add that when you're solving a quadratic equation, you should be thinking about the possibility of two solutions.
Therefore he two equilibrium points are (-1, 1) and (1, 1).

There is no need to invoke Jacobians and partial derivatives for the solution to this problem. In fact, calculus is not even necessary -- just straightforward algebra.

 

[Martyn Arthur]

Thank you; particularily reference to possibility of 2 solutions