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From my textbook (Nonlinear Oscillations, Dynamical Systems, and Bifurcations of Vector Fields). Consider the following system \begin{equation} \left( \begin{array}{c} \dot{u} \\ \dot{v} \\ \dot{w} \end{array}\right)= \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & -(1+\sigma) & 0 \\ 0 & 0 & -\beta \end{array} \right)\left( \begin{array}{c} u \\ v \\ w \end{array}\right)+ \left( \begin{array}{ccc} -\frac{\sigma}{1+\sigma} (u+\sigma v)w \\ \\ \frac{1}{1+\sigma}(u+\sigma v)w \\ \\ (u+\sigma v)(u-v) \end{array}\right) \end{equation}

My textbook says: "u axis is not invariant because the equation for $\dot w$ includes the term $u^2$". It is not clear to me, maybe because I'm not English speaking. But I have made the following reasoning. If $v=w=0$ we have:

\begin{equation} \left( \begin{array}{c} \dot{u} \\ \dot{v} \\ \dot{w} \end{array}\right)= \left( \begin{array}{c} 0 \\ 0 \\ u^2 \end{array}\right) \end{equation} from which

\begin{equation} \left( \begin{array}{c} u \\ v \\ w \end{array}\right)= \left( \begin{array}{c} u_0 \\ v_0 \\ u_0^2 t \end{array}\right) \end{equation}

If $u_0=1$ and $v_0=0$ then: $(u,v,w)^T=(1,0,t)^T$ that does not belong to the u axis for $t\neq 0$.

What do you think of my proceedings? Is my procedure right? I made ​​a mistake?

Thank you very much

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1 Answer 1

up vote 3 down vote accepted

The step "from which" is not quite right, since once you leave the $u$ axis, the derivative will no longer be as simple as $(0,0,u^2)$.

But on the other hand, you don't need a conclusion that strong; it's enough to note that since the derivative (evaluated on the $u$ axis) is not of the form $(\text{something},0,0)$, you are going to leave the $u$ axis immediately.

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