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One suitable autonomous planar system isIts equilibrium points are exactly the four combinations $(x,y)=(\pm 1,\pm 1)$. Its Jacobian matrix isThus $(-1,-1)$ and $(1,1)$ have eigenvalues $\pm 2$ and are saddle equilibria. At the other two points the eigenvalues are $\pm 2i$. To establish the nonlinear centre equilibrium classifications, use the first integralIts Hessian matrix is $\mathrm{diag}(2x,-2y)$, a negative-definite matrix at $(-1,1)$ and a positive-definite matrix at $(1,-1)$. Thus each is a strict local extremum, and nearby regular level sets are closed curves supporting periodic orbits, by the planar first integral criterion for a centre.

Immediately to the right of $(-1,1)$, the vector field points down because $x^2-1<0$, so that centre is clockwise. Immediately to the right of $(1,-1)$ it points up because $x^2-1>0$, so that centre is counterclockwise. This verifies both the nonlinear classifications and the prescribed orientations, with exactly four equilibrium points.

For a twice continuously differentiable function, an interior local extremum requires $f_x=f_y=0$. At such a critical point, put $D=f_{xx}{f}_{yy}-f_{xy}^2$. The second-derivative test gives a strict local minimum if $D>0$ and $f_{xx}>0$, a strict local maximum if $D>0$ and $f_{xx}<0$, and a saddle point if $D<0$. When $D=0$, the test is inconclusive. More generally the Hessian matrix is necessarily positive semidefinite at a local minimum and negative semidefinite at a local maximum, but these non-strict signs alone are not sufficient.

For this polynomial,Subtracting the two stationarity equations gives $y(2x-y)=0$. If $y=0$, then $x=0$; if $y=2x$, the remaining equation gives $x=0$ or $x=3/2$. Thus the only critical points are $(0,0)$ and $(3/2,3)$.

At $(3/2,3)$, $\det H=-18<0$, so it is a saddle point. At $(0,0)$ the determinant is zero, but along $y=-x$ the value is $-x^3$, which takes both signs arbitrarily near zero. ThereforeIn particular a vanishing Hessian determinant does not exclude a saddle point.

On the constraint $x^2-z^2=1$, substitute $x^2=1+z^2$ to obtainTherefore the two global minima areEvery pair $(y,z)$ is allowed on either smooth branch $x=\pm \sqrt{1+z^2}$, so these are also the only local extrema. There is no local maximum or global maximum: the displayed positive quadratic in $(y,z)$ has no other stationary point and is unbounded above. At the minima $F_z=-2z=1$, consistent with the hypothesis in part i.

At an interior local extremum of a differentiable function, $f^{\prime }(x)=0$. The second-derivative test classifies a stationary point: a positive second derivative gives a strict local minimum, and a negative second derivative gives a strict local maximum. A zero second derivative is inconclusive: $x^4$, $-x^4$, and $x^3$ at zero illustrate a minimum, a maximum, and neither.

HereThus $f^{\prime \prime }(1/3)=-8$ and $f^{\prime \prime }(3)=8$. Substitution into $f$ gives

Set both first partial derivatives equal to zero:Thus $x=0,\pm 1/\sqrt{2}$ and $y=0,\pm \sqrt{2}$, independently, giving exactly nine stationary points. The Hessian matrix is diagonal everywhere:Its diagonal entries are its eigenvalues, so the second-derivative test immediately classifies the points:

In particular,All nine stationary points are nondegenerate, so no higher-order test is needed. Neither type of local extremum is global: $f(x,0)=x^4-x^2\to +\infty$ as $|x|\to \infty$, whereas $f(0,y)=-y^4+4y^2\to -\infty$ as $|y|\to \infty$.

To organize the contour plot, complete the squares:The level set through all four outer saddle points of a scalar function factors exactly:It is therefore the union of the circle $x^2+y^2=5/2$ and the hyperbola $y^2-x^2=3/2$, which cross at those four saddle points of a scalar function. Near the origin, $f=-x^2+4y^2+O((x^2+y^2{)}^2)$, so the zero contours have tangents $y=\pm x/2$. These two exact structural features, together with the nested contours around the four local extrema, determine the requested sketch. They are an example of a separable quartic landscape with nine stationary points.

Let $F$ be a continuously differentiable planar vector field with an isolated equilibrium point $p$, and let the twice continuously differentiable first integral $H$ satisfy $F\cdot \nabla H=0$ and $\nabla H(p)=0$. If the Hessian matrix $D^{2}H(p)$ is positive or negative definite, then $p$ is a centre equilibrium.

For a positive-definite matrix, sufficiently short rays from $p$ have strictly increasing $H$ with radial derivative bounded below by a positive constant times distance. Every sufficiently small positive level difference therefore meets each ray exactly once, and the implicit function theorem makes these intersections a simple closed curve. These regular level sets shrink to $p$; use $-H$ in the negative-definite case. The vector field is tangent and nonzero on each such compact curve. Its speed is bounded away from zero, so a traversal takes finite time and each curve is a periodic orbit. An arbitrary strict local extremum without these regularity and nondegeneracy conditions does not by itself supply this argument.