# Phase plane problem

• Jun 9th 2010, 02:43 AM
TheFirstOrder
Phase plane problem
Hi there, for the following equation that is in the phase plane (x,y),

$\displaystyle d^{2}x/dt^{2}-4x+x^{3}=0$ <--- sorry, there was a typo originally, now correct

where $\displaystyle dx/dt=y$

I need to prove that the solutions of the coupled, nonlinear system are given by

$\displaystyle y^{2}=1/2(C+4-x^{2})(C-4+x^{2})$

where C is a constant. So what I have done so far is:

$\displaystyle d^{2}x/dt^{2}-4x+x^{3}=0$

So the equivalent system is:

$\displaystyle F(x,y) = dx/dt=y$ (as stated in the question)
$\displaystyle G(x,y) = dy/dt-4x+x^{3}=0 => dy/dt=4x-x^{3}$

So from this, I tried letting

$\displaystyle dy/dx=(dy/dt)/(dx/dt)=G(x,y)/F(x,y)$

This upon solving this, I got $\displaystyle y^{2}=4x^{2}-(x^{4})/4+2C$

which seems close, but I am not sure this is the right method or not (Angry)
• Jun 9th 2010, 02:57 AM
Ackbeet
I don't think you're reducing the problem to two first-order systems correctly. If you look carefully, when you take the derivative of $\displaystyle y$, that's really the second derivative of $\displaystyle x$, which does not show up in the original DE.
• Jun 9th 2010, 03:08 AM
TheFirstOrder
Quote:

Originally Posted by Ackbeet
I don't think you're reducing the problem to two first-order systems correctly. If you look carefully, when you take the derivative of $\displaystyle y$, that's really the second derivative of $\displaystyle x$, which does not show up in the original DE.

Oh yes, sorry, that was a typo as the second derivative of x is meant to show up in the equation. So, I am right after that?
• Jun 9th 2010, 05:32 AM
Ackbeet
Well, this step:

$\displaystyle (dy/dt)/(dx/dt)=G(x,y)/F(x,y)$

is an incorrect middle step, but you fix it in the next step. What you really mean is this:

$\displaystyle \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{4x-x^{3}}{y}$.

Let me see. I'm not good at skipping lots of steps, so forgive me for reproducing the separation of variables:

$\displaystyle y\,dy=(4x-x^{3})\,dx$, so
$\displaystyle \frac{y^{2}}{2}=\frac{4x^{2}}{2}-\frac{x^{4}}{4}+C$, and therefore
$\displaystyle y^{2}=4x^{2}-\frac{x^{4}}{2}+2C$.
If you like, you can absorb the 2 into the $\displaystyle C$.

So I got a different constant multiplying the $\displaystyle x^{4}$ term.

Now, I dislike factoring. I would take your target expression, and simplify:

$\displaystyle y^{2}=1/2(C+4-x^{2})(C-4+x^{2})=1/2(C^{2}-4C+Cx^{2}+4C-16+4x^{2}-Cx^{2}+4x^{2}-x^{4})$
$\displaystyle =1/2(C^{2}-16+8x^{2}-x^{4})$.

If you compare this to the result we got above, you can see that the result follows by simply re-defining the constant of integration.
• Jun 9th 2010, 05:40 AM
TheFirstOrder
Thank! I didn't think of redefining the constant! thank you so very much!
• Jun 9th 2010, 05:44 AM
Ackbeet
You're very welcome. Have a good one.
• Jun 9th 2010, 07:34 PM
TheFirstOrder
Sorry, another question in regard to this one:

Let's say I let C=4 and then C=5. How do I go about drawing the phase portraits for these values?

What I have done so far is for the formula

$\displaystyle y^{2}=1/2(C+4-x^{2})(C-4+x^{2})$

I simply let C=4

$\displaystyle y^{2}=1/2(8-x^{2})(x^{2})$

and then moving the right hand side to the left, I solved and got the equilibrium points as (0,0), (8^1/2,0), (-(8^(1/2)),0) and then how do I figure out what sort of points are they (i.e. saddle points, centres)? And then, how do I draw it??? I've tried drawing in WolframAlpha expect that only give a basic outline. how do I go about figuring out when I would get periodic solutions? Your help is much appreciated.
• Jun 10th 2010, 02:55 AM
Ackbeet
Well, you can always draw the phase portrait by hand as a last resort. I would think Mathematica would be able to draw a phase portrait. Try the StreamPlot command. What you need to know is the x and y components of the field at each point.