Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

i have this exercise :

we consider the following model : $$ \begin{cases} x'& = x(4-x-y)\\ y'&=y(2+2\alpha-y-\alpha x) \end{cases} $$

a) Find the critical point $P$ does not depend on $\alpha$ and having coordinates strictly positive.

-> The critical points are : $(0,0),(0,2 +2 \alpha),(2,2),(4,0)$ then P = $(2,2)$

b) Assuming $\displaystyle\frac{dy}{dx}$,solve the system for the value $ \alpha_0$ of $\alpha$ and such that $P$ is not hyperbolic,

Draw the phase portrait for $\alpha = \alpha_0$

c) what is the nature of $P$ for $ 0 <\alpha <\alpha_0 $ and $ \alpha_0 <\alpha $, Draw the shape of the phase portraits for these values ​​of $\alpha $ "

Can someone tell me how to solve b), why they use $\displaystyle\frac{dy}{dx}$?

Please help me

Thank you

share|improve this question
1  
I have no idea what "assuming $dy/dx$" might possibly mean, but $P$ is hyperbolic for $\alpha>1$. –  user8268 Sep 18 '13 at 8:30
    
It's not a book it is an exercise derived from an examination subject –  Vrouvrou Sep 18 '13 at 15:44
    
@user8268 , how you have found this ? –  Vrouvrou Sep 19 '13 at 14:29
add comment

1 Answer 1

For system of the shape $$\begin{cases} x' = f(x,y)\\ y' = g(x,y) \end{cases}$$ If one had a solution such one could write $y(t) = \tilde y(x(t))$ then one would have that $$\frac{d y}{d t} = \frac{d \tilde y}{d x}\frac{d x}{d t}$$
So, with an abuse of notation one writes $$ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{f(x,y)}{g(x,y)}$$ So assumming $\frac{d y}{d x}$ could simply mean that $g(x,y) \ne 0$ for a solution, and thus the previous is well defined. With this trick you turn a PDE into a ODE (provided the above works). Now you have a fraction of polynomial, which maybe you can integrate and get something nice, hence they say''solve''.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.