2x2 Nonlinear system spiral critical point(by linearization) remains a spiral(of the nonlinear system)

Question

I took this exercise from a book, it has to do with the behaviour of a nonlinear phase portrait around a critical point based on the behaviour of the linearized system. I've seen the proofs of many of the usual behaviours that are conserved but couldn't find the proof for the case of a spiral critical point. To be more precise, the exercise is the following.

Let's suppose we have an autonomous system of the form \begin{cases} \dot{x}=ax+by+f(x,y)\\ \dot{y}=bx+by+g(x,y)\\ \end{cases} with $f(x,y)=o(r)$, $g(x,y)=o(r)$ as $r=\sqrt{x^2+y^2}\rightarrow 0$.

Suppose that $(0,0)$ is a focus in the linearized system. (i.e. the matrix of the linearized system has two conjugate eigenvalues with non zero real part, we can assume negative real part for simplicity). Show that in the full system, the orbits are also spiralling near $(0,0)$, i.e. for each orbit near $(0,0)$ the polar angle takes all values in $[0,2\pi)$ (I assume it means congruence modulo $2\pi$) for $t>a$ and each $a\in\mathbb{R}$.

I've seen this stated as a theorem in several books but I'm not able to find a proof of it. So far I could do the following, lets write the system into polar coordinates to obtain \begin{cases} \dot{r}=r(a\ cos^2(\theta)+b\ sin(\theta)cos(\theta) + c\ cos(\theta)sin(\theta)+d\ sin^2(\theta))+o(r)\\ \dot{\theta} = (-a\ cos(\theta)sin(\theta) - b\ sin^2(\theta)+ c\ cos^2(\theta) + d\ sin(\theta)cos(\theta)) +\frac{o(r)}{r} \end{cases}

Notice that in both cases the part between brackets is the corresponding equation of the linearization in polar coordinates. I know that since $r(t)\rightarrow 0$ in the linear system then It has to do the same for the nonlinear, that leaves the second equation above in the form

\begin{equation} \dot{\theta} = \beta(\theta) +\varepsilon(t) \end{equation}

where $\varepsilon \rightarrow 0$ when $t\rightarrow 0$ but I cannot give any estimate of the rate of convergence (I think). So my guess is that I should somehow compare $\theta$ to the solution of an equation of the form

\begin{equation} \dot{h}=\beta(h) \end{equation}

Since the last would be the angle of the linear system, and this by hipothesis should grow(in module) up to infinity. I've tried substracting both equations, integrating, assuming Lipschitz on $\beta$ and using Gronwall, but I cannot have a good estimate on how near both functions are in order to prove what I want. I could only have an exponential estimate of the difference (and not the good exponential, something like $|\theta(t) - h(t)| \leq e^{Lx}$, with $L>0$).

Answer 1

Your analysis is missing just one final observation. If we want to get that $\theta$ increases or decreases monotonically in some domain, we have to estimate that $\dot{\theta}$ has definite sign in that domain. Let's be optimistic and ignore the $o(r)$ term: if the first terms are separated from zero, $o(r)$ term can't spoil that. The expression $$ c \cos^2\theta + (d-a) \cos \theta \sin \theta - b \sin^2 \theta $$ is a quadratic form with respect to $(\cos \theta,\, \sin \theta)$. It is separated from zero iff the quadratic form is definite (doesn't matter whether it is positive or negative definite). Let's check this: if this quadratic form is indefinite or degenerate, the determinant of matrix

$$H = \left ( \begin{array}{cc} c & \dfrac{d-a}{2} \\ \dfrac{d-a}{2} & -b \end{array} \right ) $$ would be zero or negative. This determinant equals $$-bc - \dfrac{(d-a)^2}{4}= -\dfrac{1}{4}\left ( a^2 - 2ad + d^2 + 4bc \right ).$$ Okay, at this moment we can't say anything about its sign, but we haven't used anything that we know about matrix $$A = \left ( \begin{array}{cc} a & b \\ c & d \end{array} \right ) $$ yet. We know that this matrix has complex eigenvalues, i.e. the discriminant of characteristic equation $\lambda^2 - {\rm tr}\, A \cdot \lambda + {\rm det}\, A = 0$ is negative. This discriminant equals $$(a+d)^2 - 4(ad-bc) = a^2 + 2ad +d^2 - 4ad + 4bc = a^2 - 2ad + d^2 + 4bc,$$ which is negative and which is the same expression as in ${\rm det}\, H$. Thus ${\rm det}\, H > 0$ and quadratic form attains non-zero minimum and maximum value on the compact set of vectors of form $(\cos \theta,\, \sin \theta)$. These values are of the same sign. One final step would be to prove that there is some small neighbourhood of the equilibrium such that all trajectories that start in it would remain there. This would prove the estimate for $\dot{\theta}$ and the spiraling behaviour.