Create ODE Phase Portrait with MATLAB

[Dollydaggerxo]

Homework Statement

i need to create the phase diagram for the equation \ddot{x}=x^{3}-x
using MATLAB

Homework Equations

Well i have worked out that there is a centre point at 0,0 and a saddle point at -1,0.

The Attempt at a Solution

i found a site on the internet which shows me what the diagram should look like:

(http://fds.oup.com/www.oup.com/pdf/13/9780199208241.pdf ) page 19

but i have no idea how to do this in Matlab.

I tried to do it using a quiver but its not looking anything like this.

Then i tried to do it using ode23 package but i keep getting an error saying "Unable to meet integration tolerances without reducing the step size below the smallest value allowed "

would really really appreciate any help, thanks

 

[hunt_mat]

To compute the phase portrait, compute:

<br /> \frac{dy}{dx}=\frac{\dot{y}}{\dot{x}}=\frac{x^{3}-x}{y}<br />

This is an integral which is do-able, so do it and see what you get.

 

[Dollydaggerxo]

thanks for the reply!

do you mean just straight forward integrate it w,r,t, x
for example it would be

\frac{x^4 - 2x^2}{4y} ??

then what would i do with that though??
thanks so much for your help

 

[hunt_mat]

No, the integral wold be:

<br /> y\frac{dy}{dx}=x^{3}-x\Rightarrow\int ydy=\int x^{3}-xdx<br />

 

[Dollydaggerxo]

i see -- ok sorry.

so
\frac{y^2}{2} = \frac{x^4-2x^2}{4} + C ??

i still don't really know what to do with it though forgive me!

thanks again

 

[hunt_mat]

So now you plot y against x for certain values of C, that is your phase diagram.

 

[Dollydaggerxo]

for any values of C??
all i am getting is what looks like a square graph translated a lot I am not sure what its mean to look like

thanks

 

[hunt_mat]

Look at where there are turning points and if they're max/min.

For different values of C.