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 September 28th, 2010, 03:43 PM #1 Newbie   Joined: Sep 2009 Posts: 10 Thanks: 0 Dynamical Systems Question. Hello, I have a tough dynamical systems question and I was hoping someone could point me in the right direction. Let $\phi(t;x_0)$ be the flow of the differential equation $dx/dt=1+x^{2n}$ for $n \geq 2$ and let $\psi(t;x_0)$ be the flow of the differential equation $dx/dt=1+x^{2}$. Verify that $\phi(t;x_0) \geq \psi(t;x_0)$ for $x_0 \geq 1$ and$t \geq 0$. What can you say about the comparisons of solutions for $x_0 < 1$. Also, show that the trajectories of $\phi(t;x_0)$ go to infinty in finite time. i really do not know where to begin. Can someone help me out at all? I tried representing the trajectories with the fundamental theorem of calculus, but I dont know where to go, or even if thats the right idea. Thanks Richard.
 September 28th, 2010, 07:34 PM #2 Senior Member   Joined: Dec 2009 Posts: 150 Thanks: 0 Re: Dynamical Systems Question. Note that $(1+ x^{2n}) > (1 + x^{2}) > 0$ for any $n \geq 2$, which means that 1) both $\phi(t;x_{0})$ and $\psi(t;x_{0})$ are strictly monotonically increasing for all t, and that 2) for all t > 0 we have $\phi(t;x_{0}) \, > \, \psi(t;x_{0})$ because the derivative of $\phi(t;x_0)$ is strictly larger than that of $\psi(t;x_{0})$. We can actually solve the second differential equation explicitly using the standard technique of separation of variables: $\int^{x(t)}_{x(0)} \frac{ dx}{1 + x^{2}} \,= \, \int^{t}_{0} \,dt \, = \, t$ But the integral on the left hand side gives $arctan(x) - arctan(x_{0}) \,= \, t$, adding $arctan(x_{0})$ to both sides, and then taking the tangent of both sides gives us that: $x(t)= tan(t + arctan(x_{0}))$. The tangent function becomes infinite at $\pm \frac{\pi}{ 2}$ so $x(t)$ becomes infinite at $t \,= \, \frac{\pi}{2} - arctan(x_{0})$ which is finite. Since the derivative for the other flow is strictly larger than the one we explicitly solved for (that is for $n \geq 2$) we know that it must also go to infinity in finite time as it grows faster than a function we already showed exhibits that behavior. I'm not sure about what they're looking for as far as the different initial conditions go, but here is some food for thought: The value of $arctan( 1) \,= \, \frac{\pi}{4}$, and $arctan(x)$ is an odd function, so that $arctan(-x) \,= \, -arctan(x)$ and $arctan(0)= 0$ Hope this helps a bit
 September 28th, 2010, 07:56 PM #3 Newbie   Joined: Sep 2009 Posts: 10 Thanks: 0 Re: Dynamical Systems Question. It does help a bit, but I am still confused as to how you are certain that what you listed as 1) and 2) hold true. What makes them true, exactly? Also, how do I show its true for $x_0>1$? This is getting more confusing by the minute

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