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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.

Put $S={\theta }_1+{\theta }_2$ and $D={\theta }_1-{\theta }_2$. The gradient has components $3\sin S+\sin D$ and $3\sin S-\sin D$, so a stationary point requires $\sin S=\sin D=0$. In the closed square, $S\in [0,2\pi ]$ and $D\in [-\pi ,\pi ]$, and the compatible possibilities are precisely the four corners and $(\pi /2,\pi /2)$.

The Hessian matrix isIts eigenvectors $(1,1)$ and $(1,-1)$ have eigenvalues $6\cos S$ and $2\cos D$. The second-derivative test gives the following classification:

The corner classifications also hold for the smooth extension beyond the square, and hence for the restricted region. The global bounds $-4\le U\le 4$ show that the stated minima and maxima are global within it. The central point has opposite curvature along the two diagonal directions.

The gradient vanishes at the origin. Its Hessian matrix there iswith eigenvalues $1$ and $-1$. The quadratic term is therefore indefinite, so the origin is a saddle point of a scalar function. Equivalently, the leading term $xy$ is positive along $y=x$ and negative along $y=-x$.

At a differentiable interior stationary point, all first-order changes of the scalar field vanish. In particular, the directional derivative along each coordinate axis vanishes, so $f_x=f_y=0$ and hence $\nabla f=0$. Equivalently, the first-order differential $df=\nabla f\cdot d\mathbf{r}$ vanishes for every displacement. A differentiable interior maximum or minimum must satisfy this condition; a stationary point need not be an extremum.

The standard nondegenerate local contour plot models are $-x^2-y^2$ for a maximum, $x^2+y^2$ for a minimum, and $x^2-y^2$ for a saddle point of a scalar function. Near a maximum, nested contours have lower values farther from the centre; near a minimum they have higher values. Near a saddle, hyperbolas occur on opposite sides of the critical level, whose contour has two intersecting tangent directions. More general nondegenerate quadratic forms deform these circles into ellipses; degenerate stationary points can have more complicated contours.

The level set $f=c$ can be rearranged, without changing its domain, asThe requested special contours are thereforeThe first is a parabola, the second a single point, and the last two intersecting straight lines. Moreoverwith equality only at $(0,1)$. Hence $(0,1)$ is a strict global maximum. At $(0,-1)$ the crossing critical contour separates sectors with values above and below $-1$, so it is a saddle point of a scalar function.

As a Hessian matrix check, $H_f(0,1)=\mathrm{diag}(-3,-1)$ is negative definite, while $H_f(0,-1)=\mathrm{diag}(-1,1)$ is indefinite. There is no stationary minimum.

The determinant of the Hessian matrix is $4\cdot 2-3^2=-1<0$. Hence it has one positive and one negative eigenvalue (explicitly $3\pm \sqrt{10}$), so its quadratic form is indefinite. The cubic terms are of higher order and cannot remove these opposite local directions of change. ThereforeFor example the quadratic term is positive along $y=0$ but negative along $y=-3x/2$, confirming the saddle point of a scalar function classification.

The Hessian matrix is diagonal:Its eigenvalues are $c$ and $d$. Hence the second-derivative test gives a strict local minimum if $c,d>0$, a strict local maximum if $c,d<0$, and a saddle point of a scalar function if $cd<0$. No degenerate case occurs because both are nonzero.

The local minimum is the unique global minimum, since both separated summands are strictly convex functions and tend to $+\infty$ at either end of the real line. Similarly, when both signs are negative there is a unique global maximum.

Now $h=e^x-e^y-x+y$ and the origin has stationary value zero. Near it,The level curves therefore resemble hyperbolas near the saddle point of a scalar function. Along the $x$-axis $h=e^x-x-1>0$ for $x\ne 0$, while along the $y$-axis $h=1-e^y+y<0$ for $y\ne 0$. The zero level set includes the exact diagonal $x=y$ and another branch crossing it at the origin; this second branch is tangent to $y=-x$ but is curved away from the origin. The right panel of the contour figure displays these features.

Solving $f_y=0$ gives $y=-1-\alpha x$. Substitution into $f_x=0$ gives $x\{3x+2(1-{\alpha }^2)\}=0$. Therefore $(0,-1)$ is always a critical point. When $\alpha \ne -1$ as well as the printed restriction $\alpha \ne 1$, there is a second critical pointAt $(0,-1)$, the Hessian matrix is $\left(\begin{array}{cc}2& 2\alpha \\ 2\alpha & 2\end{array}\right)$ with determinant $4(1-{\alpha }^2)$. The second-derivative test gives a strict local minimum when $|\alpha |<1$ and a saddle point of a scalar function when $|\alpha |>1$.

The allowed value $\alpha =-1$ requires higher-order terms. With $v=y+1+\alpha x$, completing the square gives the exact identityThe coordinate change is invertible, so the Hessian at a critical point under a change of coordinates preserves the classification. At $\alpha =-1$, this is a cubic degenerate saddle: along $v=0$, the difference from the critical value is $x^3$ and takes both signs. Thus, throughout the printed parameter range,There is no local maximum. When the second point is distinct, it is a local minimum for $|\alpha |>1$ and a saddle point of a scalar function for $|\alpha |<1$: in the $(x,v)$ coordinates its Hessian matrix is $\mathrm{diag}(-2(1-{\alpha }^2),2)$. The printed instruction to split only at $\alpha =1$ misses the negative-parameter cases.

At $x=0$, $g^{\prime \prime }(0)=36>0$, so the second-derivative test gives a strict local minimum. In fact $g^{\prime }(x)=4x(x-3{)}^2$ is negative for $x<0$ and positive for $x>0$ except at $x=3$, so $g(0)=5$ is the global minimum as well.

At $x=3$, the second-derivative test is inconclusive because $g^{\prime \prime }(3)=0$. But $g^{\prime \prime }(x)=12(x-1)(x-3)$ changes from negative to positive across $3$, while $g^{\prime }(3)=0$. Hence this is a stationary inflection point. Equivalently, the leading nonconstant term $4(x-3{)}^3$ in the Taylor polynomial changes sign across $3$. The paper explicitly uses “saddle point” to mean an inflection point, so in its terminology the classification isHere the “saddle point” is a one-dimensional stationary inflection point, rather than the usual multivariable use of saddle point of a scalar function.^[ref]

A saddle point of a scalar function is a stationary point with nearby values both greater and less than its value. An indefinite Hessian matrix proves this by choosing directions with opposite signs of the quadratic term. A degenerate Hessian matrix may require higher-order terms: $x^2-y^4$ has a saddle point of a scalar function at the origin despite a zero quadratic coefficient in the $y$ direction. This is distinct from a saddle equilibrium of a dynamical system.

Let $f$ have continuous second partial derivatives near an interior critical point $x_{\ast }$ of an open subset of ${\mathbb{R}}^n$, with $\nabla f(x_{\ast })=0$. The Hessian matrix gives the following sufficient conditions:

A positive semidefinite or negative semidefinite Hessian matrix with a zero eigenvalue does not settle the classification. The higher-order test for separated extrema gives two quartic examples with exactly the same zero Hessian. These conditions apply to an interior critical point; a boundary optimum or a constrained optimization problem requires attention to the allowed directions. ^[ref]

The proof outline is the Taylor expansionFor a positive-definite matrix, the smallest eigenvalue supplies a positive bound proportional to $\Vert h{\Vert }^2$, which dominates the remainder for small enough $h$. Negate the function for the local maximum case. For opposite-sign eigenvalues, choose the two corresponding eigenvectors: the leading terms have opposite signs along their lines.

There are also necessary conditions. At an interior local minimum, the Hessian matrix is positive semidefinite; at an interior local maximum, it is negative semidefinite. Restriction to each straight line reduces this to the one-variable second derivative condition. Since the Laplacian is the trace of the Hessian, it is nonnegative at such a local minimum and nonpositive at such a local maximum. The sign of the Laplacian alone is insufficient: $x^2-2y^2$ has negative Laplacian but a saddle point of a scalar function at the origin.

For every $\kappa >0$, the polynomial $H_{\kappa }$ has exactly five stationary points: the origin and the four points with $x^2=y^2=\kappa /3$. Its gradient factors asIf a coordinate vanishes at a stationary point, both do. Otherwise put $u=x^2$, $v=y^2$ to obtain $u^2=(\kappa /3)v$ and $v^2=(\kappa /3)u$. Dividing gives $u=v$ and then $u=v=\kappa /3$.

At the origin, the Hessian matrix vanishes. Axis restrictions give $H_{\kappa }(t,0)=t^6>0$, while diagonal restrictions give $H_{\kappa }(t,\pm t)=2t^6-\kappa t^4<0$ for small nonzero $t$. Thus the origin is a degenerate saddle point of a scalar function. Its leading quartic term is nonpositive, showing why the sign of the first nonzero homogeneous term alone need not settle the problem when that term vanishes along some directions.

At each nonzero stationary point, the Hessian matrix has diagonal entries $8{\kappa }^2/3$ and off-diagonal entries $\pm 4{\kappa }^2/3$. Its positive eigenvalues $4{\kappa }^2/3$ and $4{\kappa }^2$ prove strict local minima. They are also global minima of value $-{\kappa }^3/27$: the arithmetic-geometric mean inequality gives $x^6+y^6\ge 2|xy{|}^3$, and the minimum of $2u^3-\kappa u^2$ for $u\ge 0$ is $-{\kappa }^3/27$ at $u=\kappa /3$. Equality requires $|x|=|y|$.

An inflection point of the graph of a differentiable real function at which its derivative is zero. It has a horizontal tangent and a change of concavity, but need not be a local maximum or minimum. For a three-times continuously differentiable function, $f^{\prime }(a)=f^{\prime \prime }(a)=0$ and $f^{\prime \prime \prime }(a)\ne 0$ suffice: the expansion $f^{\prime \prime }(a+h)=f^{\prime \prime \prime }(a)h+o(h)$ gives opposite signs on the two sides. The elementary example is $f(x)=x^3$ at zero. The word “saddle” is sometimes used for this one-dimensional situation in elementary examination papers, but the usual saddle point of a scalar function has a multivariable meaning; do not merge the two concepts.

For a twice continuously differentiable real function, define the Newton map in optimization $N(p)=p-H(p{)}^{-1}\nabla f(p)$ wherever its Hessian matrix $H$ is invertible. Every stationary point in this domain is a fixed point, because its gradient is zero. This includes a local maximum and a saddle point of a scalar function, as well as a local minimum. Thus Newton's method in optimization needs a separate minimum classification or a method for ensuring descent when it is used for minimization problem.

For example, $f(x,y)=(x+y^2{)}^2+(x^2+y{)}^2$ has a stationary point at $(-1/2,-1/2)$, with Hessian matrixIts eigenvalues are $-1$ and $7$, so it is a saddle point of a scalar function, but starting Newton's method in optimization there gives a constant sequence. Its value is $1/8$, whereas the global minimum zero is attained at $(0,0)$ and $(-1,-1)$.