Past exam of the mathematical methods II course of the University of Oxford / 2013 resit / 11

For a twice continuously differentiable function, a finite second-order Taylor expansion about an interior point iswhere $A=f_{xx}$, $B=f_{xy}$, $C=f_{yy}$ at that point. A differentiable interior extremum requires $f_x=f_y=0$. For a strict minimum guaranteed by the quadratic term, the Hessian matrix must be positive definite. Completing the square gives $A{a}^2+2Bab+C{b}^2=A(a+Bb/A{)}^2+(AC-B^2)b^2/A$. Thus sufficient nondegenerate minimum conditions areIf $D=0$, the second-order test is inconclusive; for example $x^4+y^4$ has a strict minimum at zero despite a zero Hessian. The finite Taylor argument only needs $C^2$ regularity. The source's infinite Taylor representation additionally needs suitable analyticity: a smooth flat function, such as $e^{-1/(x^2+y^2)}$ extended by zero at the origin, has every Taylor coefficient zero there but is positive nearby.

At an interior stationary point, let $A=f_{xx}$, $B=f_{xy}$, $C=f_{yy}$ and $D=AC-B^2$. The quadratic change in the Taylor expansion is negative in every nonzero displacement when the Hessian matrix is negative definite. Applying the minimum test to $-f$ gives the sufficient nondegenerate maximum conditionsThe quadratic term then dominates the smaller Taylor remainder in a sufficiently small neighborhood, proving a strict local maximum. A vanishing determinant requires higher-order analysis instead.

At an interior stationary point, a negative determinant of the symmetric Hessian matrix means its two eigenvalues have opposite signs. Along their eigenvectors the leading quadratic change has opposite signs, and the higher-order remainder cannot reverse these signs sufficiently close to the point. Hence the nondegenerate saddle condition isWhen the determinant vanishes, a saddle is still possible but is not determined by second derivatives alone: $x^4-y^4$ is a simple example. Thus these tests are decisive for nondegenerate stationary points, not an exhaustive classification from first and second derivatives of every smooth function.

The stationarity equations are $4x^3-16y=0$ and $4y^3-16x=0$. Thus $y=x^3/4$ and $x=y^3/4$, so $x(x^8-256)=0$. The real solutions are $(0,0)$, $(2,2)$ and $(-2,-2)$. The Hessian matrix isAt the origin its determinant is $-256$, giving a saddle. At either nonzero stationary point it has eigenvalues $32,64$, so both are strict minima, with value $-32$:They are also global minima, because $f=(x^2-y^2{)}^2+2(xy-4{)}^2-32\ge -32$, with equality exactly at those two points. This is the special coefficient choice in a quartic double well with bilinear coupling.

The contours have exchange symmetry $f(x,y)=f(y,x)$ and inversion symmetry $f(-x,-y)=f(x,y)$. Negative levels between $-32$ and zero form separate loops around the two minima. The zero contour meets itself at the origin, reflecting the saddle there; positive levels form a single outer closed contour. At large radius the positive quartic terms dominate, so the contours remain bounded. There are no contours below $-32$.

The diagram marks the two minima and central saddle. The restrictions are classified separately below: a minimum or maximum along a single line need not be a two-dimensional extremum.

On $y=0$, the restricted function is $g(x)=x^4$, so $g^{\prime }(x)=4x^3$ changes from negative to positive at zero. HenceThe restricted second derivative vanishes there, but the quartic term determines the minimum. This does not contradict the two-dimensional saddle: values decrease along other directions, including the diagonal near the origin.

On $y=x$, the restricted function is $g(x)=2x^4-16x^2$. Its derivative is $g^{\prime }(x)=8x(x^2-4)$, so the turning points are $x=-2,0,2$. The restricted second derivative is $g^{\prime \prime }=24x^2-32$, negative at zero and positive at $\pm 2$. ThusThe corresponding points in the plane are $(0,0)$ and $(\pm 2,\pm 2)$. Along the diagonal the derivative is $f_x+f_y$, not a single arbitrary partial derivative; here $f_x=f_y$, so their zeros and sign changes agree with the restricted-curve calculation.