Question

Which of these has a periodic solution? Support your claim:
a) dx/dt=x^2-xy and dy/dt=y^2+xy+1
b) dx/dt=3x^3+sin(y) and dy/dt= (x^4)y+y+tan-inverse(x)

Answer 1

d

Answer 2

Find the nullclines first.
\[\begin{cases}
\dfrac{dx}{dt}=x^2-xy\\\\
\dfrac{dy}{dt}=y^2+xy-1
\end{cases}\]
\[\frac{dx}{dt}=x^2-xy=0~~\iff~~x(x-y)=0~~\iff~~x=0,~x=y\]
\[\frac{dy}{dt}=y^2+xy-1~~\implies~~\begin{cases}y^2-1=0\\2y^2-1=0\end{cases}~~\iff~~\begin{cases}y=\pm1&\text{for }x=0\\\\y=\pm\dfrac{1}{\sqrt2}&\text{for }x=y\end{cases}\]
You'll also need to know the equilibrium points, which you can find by determining the intersections of all the nullclines.

Answer 3

you lost me a little bit.

Answer 4

For a nonlinear system such as this, you need to be able to determine some sort of basis for the behavior of the system.

The \(x\) and \(y\) nullclines are the paths over which either dependent variable's derivative is zero. That's why I have that set of equations with \(\dfrac{dx}{dt}=0\) and \(\dfrac{dy}{dt}=0\).

From here, you find the equilibrium points (points for which both derivatives are simultaneously zero). Here you have four: \(\left(0,1\right)\), \(\left(0,-1\right)\), \(\left(\dfrac{1}{\sqrt2},\dfrac{1}{\sqrt2}\right)\), and \(\left(-\dfrac{1}{\sqrt2},-\dfrac{1}{\sqrt2}\right)\).

Answer 5

so from here do I .......turn them into polar coordinates....? Sorry I am trying to solve this strictly from reading the notes and his notes are complicated. We didn't go over this in class in great detail nor did he give examples. My notes say to find the linearization and then solve the non-linear system by turning it into polar coordinates. I'm not sure what it means.

Answer 6

I don't know anything about converting a system like this into polar coordinates... As I understand it (it being the way I learned to solve such a system), you determine the Jacobian matrix, which has the form
\[J(x,y)=\begin{pmatrix}\frac{dx}{dt}&\frac{dy}{dt}\\\frac{\partial}{\partial x}\left[\frac{dx}{dt}\right]&\frac{\partial}{\partial y}\left[\frac{dy}{dt}\right]\end{pmatrix}\]
which in this case, would be
\[J(x,y)=\begin{pmatrix}x^2-xy&y^2+xy-1\\2x-y&2y+x\end{pmatrix}\]
You would then evaluate the Jacobian at each equilibrium point and use the resulting matrix as a usual coefficient matrix.

For example, take the equilibrium point \((0,1)\), then you have
\[J(0,1)=\begin{pmatrix}0&0\\-1&2\end{pmatrix}\]
and you would then be tasked with solving the linear system of ODEs
\[\begin{cases}\dfrac{dx}{dt}=0\\\\
\dfrac{dy}{dt}=-x+2y\end{cases}~~\iff~~\begin{pmatrix}x\\y\end{pmatrix}'=\begin{pmatrix}0&0\\-1&2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\]
The reason for this change from nonlinear to linear is that the nonlinear system behaves similarly to its linear counterpart near its equilibrium points.

Unfortunately, I think this method only gives you a means of finding solutions that are locally valid (in particular, near the equilibrium points), so it might not be the intended way to solve this...

Answer 7

Then again, it might be the case that you're not looking for an explicit solution, but rather a phase portrait.

Answer 8

are the phase portrait and the periodic solutions the same thing?

Answer 9

No, but the phase portrait can give you an idea of whether or not the solution is periodic. http://en.wikipedia.org/wiki/Phase_portrait

Answer 10

Notice that something exhibiting periodic behavior, like the pendulum in the first figure, has a cyclical phase portrait. If your phase portrait has these kinds of cyclical patterns, then you can say a periodic solution exists.

Answer 11

ok thank you! So i just need to do this same thing for the second system for b?

Answer 12

Yep, same procedure. Find your nullclines and equilibrium points, then approximate the nonlinear system with a linear one.