Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2008 / 9

A critical point is an equilibrium: $\boxed{f(a,b)=g(a,b)=0}$. Put $X=x-a$, $Y=y-b$. The Jacobian matrix $J$ gives the first-order linearized equation $(\dot{X},\dot{Y}{)}^{\mathsf{T}}=J(X,Y{)}^{\mathsf{T}}$. The following eigenvalue tests classify that linearized system. For a continuously differentiable nonlinear system, the cases with nonzero real parts persist as the usual local node, saddle or spiral classifications. The purely imaginary case needs extra nonlinear information, as explained in (iv).

Both eigenvalues are real and negative: $\boxed{{\lambda }_1<0,{\lambda }_2<0}$, with distinctness as assumed. The two independent real eigendirections carry exponentially decaying factors $e^{{\lambda }_{j}t}$, giving a stable node.

Both eigenvalues are real and positive: $\boxed{{\lambda }_1>0,{\lambda }_2>0}$. The real eigendirections carry exponentially growing factors, giving an unstable node.

The eigenvalues are real and have opposite signs: $\boxed{{\lambda }_1{\lambda }_2<0}$. One eigendirection contracts and the other expands, giving a saddle.

For the linearized system a centre occurs when $\boxed{{\lambda }_{1,2}=\pm i\beta ,\beta \ne 0}$. Its real solutions are periodic ellipses after an invertible linear coordinate change.

As explained by a centre in the linearization need not be a nonlinear centre, for a nonlinear system these eigenvalues alone do not guarantee a centre. For example $\dot{X}=-Y-X(X^2+Y^2)$, $\dot{Y}=X-Y(X^2+Y^2)$ has eigenvalues $\pm i$ at zero but polar equations $\dot{r}=-r^3$, $\dot{\theta }=1$, so trajectories spiral inward. Removing the cubic terms gives a genuine centre with the same Jacobian. Thus there is no eigenvalue-only sufficient condition for a nonlinear centre in the stated generality.

The eigenvalues are a nonreal conjugate pair $\boxed{\alpha \pm i\beta ,\alpha <0,\beta \ne 0}$. The factor $e^{\alpha t}$ contracts while the imaginary part produces rotation, giving a stable spiral.

The eigenvalues are a nonreal conjugate pair $\boxed{\alpha \pm i\beta ,\alpha >0,\beta \ne 0}$. The exponential factor expands while the imaginary part produces rotation, giving an unstable spiral.

The equilibrium equations are $x(1+x-2y)=0$ and $(x-1)y=0$. If $x=0$, the second equation gives $y=0$. Otherwise either $y=0$, giving $x=-1$, or $x=1$, giving $y=1$. Thus there are exactly three equilibria. The Jacobian isAt $(0,0)$ it is $\mathrm{diag}(1,-1)$: a saddle. At $(-1,0)$ it is $\left(\begin{array}{cc}-1& 2\\ 0& -2\end{array}\right)$, with eigenvalues $-1,-2$: a stable node. At $(1,1)$ it is $\left(\begin{array}{cc}1& -2\\ 1& 0\end{array}\right)$, with characteristic polynomial ${\lambda }^2-\lambda +2$ and eigenvalues $(1\pm i\sqrt{7})/2$: an unstable spiral. All three are hyperbolic, so the nonlinear local classifications follow from linearization.

Suppose a nonconstant periodic orbit exists for the continuously differentiable planar vector field. Uniqueness implies that its image over a minimal period is a simple closed curve, enclosing a region $\Omega$ of positive area. Along the orbit, the differential form $fdy-gdx$ vanishes because $dx=fdt$ and $dy=gdt$. Green's theorem, with positive orientation, therefore givesa contradiction. This is the Bendixson-Dulac criterion in the whole plane; the fact that the enclosed region also lies in the domain is essential.

For the second-order equation set $y=\dot{x}$. The equivalent system is $\dot{x}=y$, $\dot{y}=-(1+x^2)y-\cos x$. Its divergence is $\boxed{-(1+x^2)<0}$ throughout the plane. Hence there are no nonconstant periodic solutions.

The nonconstant qualification is necessary: the equilibrium solutions $x(t)\equiv \pi /2+k\pi$, $k\in \mathbb{Z}$, do exist and are periodic in the literal sense of satisfying every positive period. The source's usual dynamical interpretation of “periodic solution” here is a nonconstant periodic orbit.