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The gradient and constant Hessian matrix areAt ${\mathbf{x}}_0=(1,0{)}^T$, $\nabla F=(10,-8{)}^T$ and $H_F^{-1}\nabla F=(1,-1{)}^T$. Hence one multivariable Newton iteration givesAt this point the gradient vanishes, so every subsequent iterate is $(0,1)$. The one-step result follows because the gradient of a quadratic polynomial is affine and its linearization is exact. Since the leading principal minors of $H_F$ are $6>0$ and $8>0$, it is a positive-definite matrix. Alternatively,directly proves that $(0,1)$ is the unique global minimum, with value $-1$.

For a twice continuously differentiable function $F$, its gradient and Hessian matrix areA critical point solves $\nabla F=0$. The Jacobian matrix of this vector equation is $H_F$, so the multivariable Newton iteration isEquivalently ${\mathbf{x}}_{k+1}={\mathbf{x}}_k-H_F({\mathbf{x}}_k{)}^{-1}\nabla F({\mathbf{x}}_k)$ when the Hessian matrix is invertible. Solving the linear system is the direct computational form. A critical point found in this way need not be a local minimum; its type requires a separate check.

The displayed vector field is defined on the punctured plane $U={\mathbb{R}}^2\setminus \{(0,0)\}$; it is undefined at the origin. Take the counterclockwise unit circle $\mathbf{r}(t)=(\cos t,\sin t)$, $0\le t\le 2\pi$. Along it,By the fundamental theorem for line integrals, it has no single-valued potential of a conservative vector field on $U$. In particular it cannot be the restriction of the gradient of a smooth function on all of ${\mathbb{R}}^2$.

For $(x,y)\ne (0,0)$, direct differentiation givesThere is no contradiction with Green's theorem: $\mathbf{F}$ is not defined, let alone continuously differentiable, throughout the disk enclosed by $C$. On an annulus the theorem does apply; the counterclockwise outer integral is $2\pi$, and the clockwise inner integral is $-2\pi$, so their sum is zero. The hole prevents this curl-free vector field from being a conservative vector field on its entire domain.

The middle component suggests a scalar potential with $-{\varphi }_y=2y{\cosh }^2(xz)$, so $\varphi =-y^2{\cosh }^2(xz)+h(x,z)$. The chain rule and $2\sinh t\cosh t=\sinh (2t)$ giveMatching the other two components requires $h_x=h_z=0$. ThereforeIts negative gradient is the given vector field in all three components; the additive constant has no effect on that gradient.

For two independent smooth constraints $g=h=0$, the tangent direction is orthogonal to both gradients. At a constrained stationary point, the gradient of the objective is also orthogonal to that direction, hence lies in their span: $\nabla f=\lambda \nabla g+\mu \nabla h$. This is the two-constraint Lagrange multiplier condition. Alternatively, distance extrema on a central plane section of an ellipsoid follow from a two-dimensional eigenvalue problem.

Minimizing distance is equivalent to minimizing $f=x^2+y^2+z^2$. Put $g=x-y+z$ and $h=x^2+(y^2+z^2)/4-1$. The Lagrange multiplier equations areAdding the latter two gives $(2-\mu /2)(y+z)=0$. If $\mu =4$, those equations give $\lambda =x=0$; the constraints then give $y=z=\pm \sqrt{2}$ and $f=4$. Otherwise $z=-y$ and $x=2y$. The ellipsoidal constraint gives $y=\pm \sqrt{2}/3$, and henceThe intersection is compact. The two constraint gradients are independent there: if $\nabla h=t\nabla g$, then $(x,y,z)=(t/2,-2t,2t)$, and $g=0$ forces $t=0$, contradicting $h=0$. Thus every global extremum is among these candidates. Taking square roots gives

In Cartesian components, $\mathbf{F}=(z-y,x,-x)$, soIt cannot be a conservative vector field. For the second field, put $\Phi (\mathbf{r})=-(1+|\mathbf{r}{|}^2{)}^{-1/2}$. Its gradient is $\nabla \Phi =\mathbf{r}/(1+|\mathbf{r}{|}^2{)}^{3/2}=\mathbf{G}$. This smooth global scalar function proves $\mathbf{G}$ is conservative, with $\boxed{\nabla \times \mathbf{G}=0}$. The following path integrals exhibit the different behavior directly.

The field is smooth on the simply connected domain ${\mathbb{R}}^3$. Its curl is $\nabla \times \mathbf{F}=(2\lambda -2,2\mu ,0)$, so a necessary and sufficient condition for a conservative vector field isAt those values it is explicitly the gradient of $\Phi =xy-x^2/2-yz+z^2/2$. Equality of the two previous path integrals would impose only $\lambda -\mu =1$, which is weaker: coinciding path integrals need not imply a conservative field. For example $\lambda =2,\mu =1$ makes those two integrals equal but leaves nonzero curl.

Expand $u=3x^2-2xy+3y^2$. The Lagrange multiplier equations for $u-\lambda v$ are $(6-2\lambda )x-2y=0$ and $-2x+6y+\lambda =0$, together with the constraint $y=x^2-1/4$. The first two imply $y=(3-\lambda )x$ and $(6\lambda -16)x=\lambda$. The value $\lambda =8/3$ would make the latter equation inconsistent, so division is allowed:Insert these into $x^2-y-1/4=0$ and multiply by $(6\lambda -16{)}^2$. Simplification gives $2(3{\lambda }^3-21{\lambda }^2+48\lambda -32)=0$, henceThe gradient of the constraint $(2x,-1)$ is never zero, so the Lagrange multiplier condition is necessary at every constrained extremum. To establish a global minimum, parametrize the whole parabola by $y=x^2-1/4$. The restricted function becomesIt is coercive: $q(x)\to \infty$ as $|x|\to \infty$, so it attains a global minimum. Moreover $q^{\prime \prime }(x)=36(x-1/6{)}^2+2>0$, making it strictly convex and ensuring that this minimum is unique. This proves more than the permitted assumption of one real multiplier root.

The level sets of $u$ are nested ellipses centered at the origin. In the rotated coordinates $p=(x+y)/\sqrt{2}$, $q=(x-y)/\sqrt{2}$, they have equation $2p^2+4q^2=\text{constant}$, with the longer axis along $x=y$. The smallest one meeting the parabola is tangent to it at the unique minimizing point, approximately $(x,y)=(-0.12654,-0.23399)$, where $u\simeq 0.15307$ and $\lambda \simeq 1.15085$.

For $h=x^2/a^2+y^2/b^2+z^2-1$, the gradient $\nabla h=2(x/a^2,y/b^2,z)$ is a normal vector to the ellipsoid. The outward unit normal vector is $(x/a^2,y/b^2,z)/\sqrt{x^2/a^4+y^2/b^4+z^2}$. The conditions $a,b>\sqrt{2}$ ensure that $z=\sqrt{1-x^2/a^2-y^2/b^2}>0$ everywhere on the rectangular projection.

The dot product $F\cdot \nabla h=2xy(1/b^2-1/a^2)$ vanishes pointwise when $a=b$. In that case $F$ has no normal component, so the surface flux is zero. For unequal axes, the given vector surface element givesPut $u=x^2/a^2$, $v=y^2/b^2$. Then $xydxdy=a^2{b}^{2}dudv/4$, andTwo elementary integrations now giveThis is the rotational flux through an ellipsoidal graph patch with unit vertical semiaxis and unit projection lengths. It also tends continuously to zero when $a=b$, agreeing with the pointwise argument.

Set $H(x,y,z)=x^3+xy+yz-1$. The gradient $\nabla H=(3x^2+y,x+z,y)$ is a normal vector to its level set. Since $y\ge 1/2$ on the given patch, division by its last component is valid and gives the unique normalized fieldIts dot product with $\mathbf{k}$ is one, as required; it is not generally a unit vector.

Write the surfaces as level sets $f_1=x^2{y}^{2}z-3x-z^2=0$ and $f_2=3x^2-y^2+2z-1=0$. Their gradients are normal vectors to their tangent planes. At the specified point they are ${\mathbf{n}}_1=(5,-4,2)$ and ${\mathbf{n}}_2=(6,4,2)$, with dot product $18$ and squared norms $45,56$. The angle between intersecting regular surfaces is the angle between their tangent planes, soBoth gradients are nonzero, ensuring that the tangent planes are defined.

Let $D=1+x^2+y^2$. The gradient has components $f_x=[D-2x(x+y)]/D^2$ and $f_y=[D-2y(x+y)]/D^2$. At a critical point, subtraction gives $(x-y)(x+y)=0$. The case $x+y=0$ cannot make either numerator zero, so $x=y=t$. The remaining equation is $1+2t^2=4t^2$, givingThese are the only stationary points; the denominator is positive everywhere.

At $(t,t)$ with $t^2=1/2$, differentiating the gradient yields the Hessian matrix $H=-tI$. At the positive point it is negative definite, and at the negative point it is positive-definite; in both cases $f_{xy}=0$ and $\det H=1/2$. Therefore the positive point is a maximum, the negative point is a minimum, and there are no saddle points.

They are global extrema: with $r=\sqrt{x^2+y^2}$, the Cauchy-Schwarz inequality gives $|f|\le \sqrt{2}r/(1+r^2)\le 1/\sqrt{2}$, with equality at exactly those two points. This is the extremum bound underlying the level sets of a linear numerator over a radial quadratic.

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 is $(f_x,f_y)=(2-2xy,1-x^2)$. Both partial derivatives vanish at $(1,1)$, so it is a stationary point.

The second component of the gradient forces $x=\pm 1$. The first then requires $xy=1$. Apart from $(1,1)$, the only stationary point is

Taking the negative gradient of the potential energy gives the force

Set $g(x,y)=y-(x-2{)}^2$. The gradient of $g$ never vanishes, so the Lagrange multiplier equations $\nabla f=\lambda \nabla g$ apply at a constrained stationary point. They are $4x=-2\lambda (x-2)$, $2y=\lambda$, and $y=(x-2{)}^2$. Eliminating $\lambda ,y$ givesWith $t=x-2$, this becomes $t^3+t+2=(t+1)(t^2-t+2)=0$. The quadratic factor is $(t-1/2{)}^2+7/4>0$, so the only real possibility is $t=-1$. ThereforeThe level curve through this point is the ellipse $2x^2+y^2=3$, meeting the coordinate axes at $(\pm \sqrt{3/2},0)$ and $(0,\pm \sqrt{3})$. Its tangent line at $(1,1)$ has slope $-2x/y=-2$, equal to the derivative $2(x-2)$ of the constraint parabola there. The two curves are tangent.

The parabola touches this ellipse from outside; every other point on the constraint lies on a larger contour. An exact check of this geometric argument iswith equality only at $x=1$. Thus the point is the unique constrained global minimum.

The gradient is $\nabla f=(y(1-2x+y),x(1-x+2y))$. At the three specified points it is respectivelyEach vector is perpendicular to its straight boundary edge and points out of the triangle, where $f$ increases from its negative interior values. The arrows in the preceding figure mark these directions.

The gradient is $(-y^2-2x-3x^2,2y(1-x))$. Its second component vanishes only if $y=0$ or $x=1$. The latter gives the impossible first-component equation $-y^2-5=0$. With $y=0$, the first component gives $x=0$ or $x=-2/3$. The Hessian matrix isAt $(0,0)$ it has eigenvalues $-2,2$, giving a saddle point. At $(-2/3,0)$ its eigenvalues are $2,10/3$, giving a strict local minimum, of value $-4/27$. These are all real stationary points. The minimum is local: for example $g(x,0)\to -\infty$ as $x\to +\infty$.