Saddle-node behaviour of planar systems

Question

I'm teaching a course on ODEs using Lawrence Perko's book. Having assigned Q3(c) of section 2.11, and both working out the solution using theorems in the sections and consulting the solution manual, I get an answer that seems to conflict with other calculations. The details are as follows:

The system is: $$ \dot{x} = y, \qquad \dot{y} = x^4 + xy. $$

According to a Thm 2.11.3 in Perko (see screenshot of Thms 2.11.2 and 2.11.3 below), taken from Andronov--Leontovich--Gordon--Maier, the critical point $(0,0)$ is a saddle-node, "because" the lowest order monomial of the form $x^k$ in the $\dot{y}$ equation is of even degree, the lowest order monomial of the form $x^ny$ has $n = 1$, with coefficient $b_1 = 1 \not= 0$, and $4/2 > 1$. The solution manual concurs.

Now from the equation, $\dot{x} > 0$ on $\{y > 0\}$, $\dot{x} < 0$ on $\{y < 0\}$. Moreover $\dot{y} = xy + O(|x,y|^4)$, which means $\dot{y} > 0$ in the 1st and 3rd quadrants, and $\dot{y} < 0$ in the 2nd and 4th quadrants close to $(0,0)$. And any way I draw this, it looks like a cusp near the critical point.

Finally, winplot confirms this: .

So my question is, is it known that Thm 2.11.3 is in fact not a theorem, or have I missed something?

Answer 1

If we plot the phase portrait of our system of ODEs $$ \dot x = y ,\qquad \dot y = x (x^3+y) , $$ we see two quite obvious separatrices, and it looks like they divide the plane into two hyperbolic sectors, so that there's a cusp at the origin:

However, a closer look reveals that there is actually also a third separatrix, intersecting the $x$-axis at $x_0 \approx -0.297$, such that the trajectories intersecting the $x$-axis for $x_0 < x < 0$ all converge to the origin, forming a parabolic sector:

If this is correct, there is in fact a saddle-node at the origin (in agreement with the theorem).

Now the question is of course how we can be sure of this, and the trick is to change to new variables $(u,v)$ to “blow up” the phase portrait at the origin. Exactly what new variables to take depends on the exponents appearing in the normal form of the system, but with the exponents in this particular example, the suitable choice turns out to be $$ x = u ,\qquad y = u^2 v , $$ or equivalently $$ u =x ,\qquad v = y/x^2 ,\qquad \text{for $x \neq 0$} . $$ This change of variables of course behaves “badly” along the $y$-axis (the line $x=0$), but that's the whole point. It is already obvious what the trajectories of our original system $(\dot x, \dot y) = (y, x^4+xy)$ do along the $y$-axis (away from the origin): they just cross it horisontally, since $\dot x \neq 0$ and $\dot y = 0$ there. Any such trajectory, with $y$ approaching some $C \neq 0$ as $x \to 0$, will be transformed into a curve (with two branches) in the $(u,v)$-plane that looks approximately like $v = C/u^2$ when $u$ is close to zero. The interesting thing to investigate now is what becomes of the trajectories approaching the origin in the $(x,y)$-plane.

From the change of variables, we have $\dot x = \dot u$ and $\dot y = 2uv \dot u + u^2 \dot v$, and hence $$ \dot u = \dot x = y = u^2 v $$ and (for $u \neq 0$) $$ \dot v = \frac{\dot y - 2 uv \dot u}{u^2} = \frac{u(u^3 + u^2 v) - 2 uv \cdot u^2 v}{u^2} = u(u+v-2v^2) . $$ So in the new variables, the system of ODEs becomes $$ \begin{pmatrix} \dot u \\ \dot v \end{pmatrix} = \begin{pmatrix} u^2 v \\ u (u + v - 2 v^2) \end{pmatrix} = u \begin{pmatrix} u v \\ u + v - 2 v^2 \end{pmatrix} . $$ Since we're only really interested in the behaviour of the trajectories for $u \neq 0$, we may as well remove the factor $u$ and consider the rescaled system $$ \begin{pmatrix} \dot u \\ \dot v \end{pmatrix} = \begin{pmatrix} u v \\ u + v - 2 v^2 \end{pmatrix} $$ instead; this will not change the shape of the trajectories, since it only amounts to a time reparametrization, plus a change of orientation for the trajectories in the left half-plane $u <0$. The phase portrait for this system looks like this, with equilibria at $(0,0)$ and $(0,1/2)$:

Here the third separatrix (purple) is quite clearly visible. It, and the orange one too, emanates from the equilibrium at $(u,v)=(0,1/2)$, while the red one approaches $(u,v)=(0,0)$, but when you map things back to the $(x,y)$-plane, the whole $v$-axis gets squished into the point $(x,y)=(0,0)$ so that all three meet there.

A typical trajectory in the parabolic sector arises as follows (and I hope you'll excuse my ugly drawings here):

There's a pair of trajectories, one with $u>0$ and one with $u<0$, that are both asymptotic to $v = C/u^2$ with the same negative value of $C$ (more precisely, with $C_0 < C < 0$, where $C_0$ is the value such that the lower part of the purple separatrix is asymptotic to $v = C_0/u^2$). When you map back to the old variables, and change the orientation of the left curve, they will join nicely at $(x,y)=(0,C)$, and form a single origin-bound trajectory of the original system:

Similarly, the curves in the left hyperbolic sector in the $(x,y)$-plane are formed by joining the images of trajectories in the $(u,v)$-plane asymptotic to $v = C/u^2$ with the same value of $C$, both for $C>0$ and for $C < C_0$.

Answer 2

Let's take the scalar equation $$ \ddot x=x\dot x+x^4 $$ and divide it up into a slightly different first order system with $v=\dot x-\frac12x^2$ so that \begin{align} \dot x&=v+\frac12x^2\\ \dot v&=x^4 \end{align} A minimal perturbation of the second equation to $\dot v=x^4-\varepsilon$ results in two close-by critical points $(\pm\sqrt[4]ε,-\frac12\sqrtε)$ with Jacobians $$ J=\pmatrix{x&1\\4x^3&0}=\pmatrix{\pm\sqrt[4]ε&1\\\pm4\sqrt[4]ε^3&0} $$ with eigenvalues $\frac{x}2(1\pm\sqrt{1+16x})$, so that for $ε<2^{-16}$ one eigenvalue is around $x=\pm\sqrt[4]ε$ and the other $-4x^2=-4\sqrtε$. For $ε>2^{-16}$ the eigenvalues for the left point become a complex conjugate pair with real part $-\frac12\sqrt[4]ε$.

This gives one saddle point at $(+\sqrt[4]ε,-\frac12\sqrtε)$ and one attracting node at $(-\sqrt[4]ε,-\frac12\sqrtε)$. For $ε\to 0$ they melt together, giving a saddle-node.