Past exam of the mathematical methods II course of the University of Oxford / 2015 / 9

The first partial derivatives areIf $\cos x=0$, then $z_x=-\sin x{e}^{-x^2-y^2}\ne 0$, so no stationary point is lost by excluding those values. The remaining condition $z_y=0$ forces $y=0$, and $z_x=0$ then givesThere are infinitely many such points. On every tangent branch $(n\pi -\pi /2,n\pi +\pi /2)$, the function $\tan x+2x$ has derivative ${\sec }^{2}x+2>0$ and goes from $-\infty$ to $+\infty$. The intermediate value theorem and strict monotonicity therefore give exactly one root on every branch. The central root is $x_s=0$, and oddness pairs the other roots at opposite signs.

For increments $h=x-x_s$ and $k=y-y_s$, the second-order multivariable Taylor expansion isAt a stationary point $(x_s,0)$, $\sin x_s=-2x_s\cos x_s$ and $z_s=\cos x_s{e}^{-x_s^2}$. Direct differentiation givesThus, since this particular function is analytic, the expansion can be writtenThere are no linear terms because the gradient vanishes, and no mixed quadratic term because the mixed derivative vanishes on the axis.

At a stationary point of a twice continuously differentiable function, write the Hessian matrix as $H=\left(\begin{array}{cc}A& B\\ B& C\end{array}\right)$ and its determinant as $D=AC-B^2$. The quadratic term in the Taylor expansion determines the local nature whenever $H$ is nonsingular.

If $D>0$, its two eigenvalues have the same sign. With $A>0$ the quadratic form is positive definite, giving a strict local minimum; with $A<0$ it is negative definite, giving a strict local maximum. If $D<0$, the eigenvalues have opposite signs and the point is a saddle, with nearby increasing and decreasing directions. ThusIf $D=0$, the test is inconclusive. For example, $x^4+y^4$, $-x^4-y^4$ and $x^4-y^4$ all have zero gradient and zero Hessian matrix at the origin, yet give a minimum, a maximum and a saddle respectively. One must inspect higher-order terms or use another argument in such a case.

For every stationary point found above, $\cos x_s\ne 0$ and hence $z_s\ne 0$. The Hessian matrix is diagonal, with entries $-(4x_s^2+3)z_s$ and $-2z_s$, soThe entries always have the same sign. ThereforeThe root on tangent branch $(n\pi -\pi /2,n\pi +\pi /2)$ has cosine sign $(-1{)}^n$. Thus even-numbered branches supply maxima and odd-numbered branches minima, symmetrically paired about the origin. There are no saddle points among these nondegenerate stationary points. The origin is a maximum with $z_s=1$, and is in fact the global maximum since $\cos x\le 1$ and $e^{-x^2-y^2}\le 1$.

For $w=x^3{e}^{-x^2-y^2}$, the first derivatives are $w_x=x^2(3-2x^2)e^{-x^2-y^2}$ and $w_y=-2y{x}^3{e}^{-x^2-y^2}$. Both vanish at the origin. The local expansion isThere are no constant, linear or quadratic terms, so every second derivative vanishes at the origin and the Hessian determinant is zero. The second-order test therefore cannot classify the point.

The cubic term does determine the behaviour: along $y=0$, $w(x,0)$ has the sign of $x$. Every neighbourhood contains values above and below $w(0,0)=0$, so it is neither a maximum nor a minimum. It is a nonisolated stationary point with nearby values of both signs. Indeed the whole line $x=0$ is stationary, so this is not an isolated nondegenerate quadratic saddle.