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- Thread starter thecoop
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- Feb 5, 2012

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Hi thecoop,Hello guys . I obtained a 3 dimensional dynamical system , how can I find its critical points with using software ? I tried it handy but its too involved to compute handy .

Welcome to MHB! What are the equations governing your dynamical system?

Kind Regards,

Sudharaka.

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- Jan 26, 2012

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I can't help you with your question but do have a question for you. What software did you write the equations with in your attachment? Was it Latex by any chance? Or was it with Microsoft Word? If either of those then I can show you how to use or convert to Latex that can be used on MHB.

If you already know Latex then this thread will show you how to use it on MHB. Someone will help you soon enough I am sure but in the meantime welcome to MHB!

Jameson

Generally, I believe you want to uncouple one equation from the other two but I don't know if that is possible with your equations.thank you

- Feb 13, 2012

- 1,704

$\displaystyle x^{\ '} (t)= f_{x} \{ x(t), y(t), z(t) \}$

$\displaystyle y^{\ '} (t)= f_{y} \{ x(t), y(t), z(t) \}$

$\displaystyle z^{\ '} (t)= f_{z} \{ x(t), y(t), z(t) \}$

... and its 'critcal points' are the solution of the system of equations...

$\displaystyle f_{x} ( x, y, z ) =0$

$\displaystyle f_{y} ( x, y, z )=0$

$\displaystyle f_{z} ( x, y, z )=0$

Kind regards

$\chi$ $\sigma$

- Jan 26, 2012

- 183

$ \left(-1,\dfrac{2}{5},0\right)$ and $\left (\dfrac{25}{8}, \dfrac{5}{18},-\dfrac{5}{8}\right)$.

This is how I got these. First set each of the right hand sides of your three equation to zero as Chisigma said noting that $y \ne 0$ due to the division by $y$. From the second we have

$-\dfrac{1}{4}\left(x+4z-4 \right) y+\dfrac{1}{4}\,x-\dfrac{1}{4}\,{x}^{2}-\dfrac{5}{4}\,xz+z-{z}^{2} = 0$

which we can solve for $y$ provided that $x+4z-4 \ne 0 $ so we consider this case first.

If $x = 4 - 4z$ then the second equation becomes $3z-3 = 0$ giving $z = 1$ and in turn $x = 0$. The first and third equations becomes $-3$ and $5$ respectively which is inadmissible since these are both to be zero. Thus, we can conclude $x+4z-4 \ne 0 $.

Solve the second equation for $y$ gives

$y = \dfrac{(x+4 z) (x+z-1)}{x+4 z-4}$ noting that $x + 4z \ne 0$ as this would give $y = 0$.

Simplifying equations (1) and (3) gives

$-{\dfrac {{x}^{2}+9\,xz+x+12\,{z}^{2}}{x+4\,z}} = 0$ and $\,{\dfrac { \left( x+5\,z \right) z}{x+4\,z}}=0$.

These second of these gives rise to two cases and the critical point fall out.