Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 2 / 12

Near a nondegenerate stationary point, the quadratic part of the Taylor expansion determines the local contour plot. For a strict local maximum, the model

f = f_{0} - X^{2} - Y^{2}

gives nested closed level sets, with values decreasing as one moves outward. For a strict local minimum, the model

f = f_{0} + X^{2} + Y^{2}

gives the same nested shape with values increasing outward. Unequal nonzero quadratic coefficients stretch the circles into ellipses after a rotation to the Hessian matrix's eigenvectors.

f = f_{0} + X^{2} - Y^{2}

gives hyperbolas. Levels above

f_{0}

open along the positive-curvature direction; levels below

f_{0}

open along the negative-curvature direction. The critical level set is the two crossing lines

Y = \pm X

. Degenerate stationary points can have different local geometry, so these are the standard nondegenerate sketches.

b

f_{x} = 4 x^{3} - 2 x = 2 x (2 x^{2} - 1) = 0, f_{y} = - 4 y^{3} + 8 y = - 4 y (y^{2} - 2) = 0.

Thus

x = 0, \pm 1/ 2

and

y = 0, \pm 2

, independently, giving exactly nine stationary points. The Hessian matrix is diagonal everywhere:

H_{f} = (12 x^{2} - 2 0 0 8 - 12 y^{2}) .

Its diagonal entries are its eigenvalues, so the second-derivative test immediately classifies the points:

At $(0, \pm 2)$ , the Hessian matrix has eigenvalues $- 2, - 16$ . These two points are strict local maxima, each with $f = 4$ .
At $(\pm 1/ 2, 0)$ , the Hessian matrix has eigenvalues $4, 8$ . These two points are strict local minima, each with $f = - 1/4$ .
At $(0, 0)$ , the Hessian matrix has eigenvalues $- 2, 8$ , so it is a saddle point of a scalar function with $f = 0$ .
At $(\pm 1/ 2, \pm 2)$ , with all four sign combinations, the Hessian matrix has eigenvalues $4, - 16$ . These four points are saddle points of a scalar function, each with $f = 15/4$ .

In particular,

2 local maxima, 2 local minima, and 5 saddle points.

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

∣ x ∣ \to \infty

, whereas

f (0, y) = - y^{4} + 4 y^{2} \to - \infty

∣ y ∣ \to \infty

To organize the contour plot, complete the squares:

f (x, y) = \frac{15}{4} + (x^{2} - \frac{1}{2})^{2} - (y^{2} - 2)^{2} .

The level set through all four outer saddle points of a scalar function factors exactly:

f - \frac{15}{4} = (x^{2} - y^{2} + \frac{3}{2}) (x^{2} + y^{2} - \frac{5}{2}) .

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} + 4 y^{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.