Past exam of the mathematical methods II course of the University of Oxford / 2017 / 11 / d / Solution

The gradient equations are $y=x^3$ and $x=y^3$. Their real solutions are $(0,0)$, $(1,1)$ and $(-1,-1)$. The Hessian matrix is $\left(\begin{array}{cc}-12x^2& 4\\ 4& -12y^2\end{array}\right)$, so the origin has eigenvalues $\pm 4$ and is a saddle point, with value zero. At either other point the eigenvalues are $-8,-16$, giving strict maxima with value two. They are global maxima becauseand equality occurs precisely at those two points.

In polar coordinates, the zero contour isBesides the saddle $r=0$, its nonzero branches satisfyThe denominator is positive; real radii require $\sin 2\theta \ge 0$. The branches form two closed lobes in the first and third quadrants, meeting at the origin. Positive contours lie inside these lobes around the maxima, and negative contours extend outside. Along the coordinate axes the function is negative away from the saddle, whereas along $y=x$ it is initially positive; these signs show the crossing geometry in the sketch.