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The rows of the Jacobian matrix $Dg(a)$ are $\nabla g_1(a{)}^T,\dots ,\nabla g_k(a{)}^T$. Rank $k$ means precisely that these $k$ gradients are linearly independent. The Lagrange multiplier condition is that real numbers ${\lambda }_1,\dots ,{\lambda }_k$ exist withGeometrically, $\nabla f(a)$ is perpendicular to the tangent space of the constraint level set, whose tangent vectors form $\ker Dg(a)$. The full-rank hypothesis ensures this level set is a smooth local constraint manifold.

Take the curl. A gradient has zero curl, and the supplied identity, with $\nabla \cdot u=0$ and $\nabla \cdot \omega =\nabla \cdot (\nabla \times u)=0$, givesTherefore $D\omega /Dt=(\omega \cdot \nabla )u$. In planar flow $\omega =(0,0,{\partial }_{x}v-{\partial }_{y}u)$ and ${\partial }_{z}u=0$, so the stretching term vanishes. ConsequentlyVorticity is carried unchanged with each fluid particle. It need not be spatially uniform or independent of time at a fixed spatial point.

Take a smooth variation $u+\epsilon \eta$ with $\eta$ vanishing on the boundary; compactly supported test functions suffice for the interior equation. At an extremum, the first variation is zero:Integration by parts removes the derivatives from $\eta$. The boundary terms vanish, givingThe fundamental lemma of the calculus of variations gives the Euler-Lagrange equationThis is the classical equation for sufficiently smooth admissible extrema; for weaker regularity the first-variation identity supplies its weak form.

For the squared-gradient integrand, the equation is $2(u_{xx}+u_{yy}+u_{zz})=0$, namely the Laplace equation. Put $u_0=ax+by+cz$ and write any admissible function as $u=u_0+v$, with $v=0$ on the boundary. The energy identity isBy the divergence theorem, the middle integral is ${\int }_{\partial V}v(a,b,c)\cdot ndS=0$; equivalently it vanishes by weak integration by parts for zero-trace functions. The final term is nonnegative and vanishes at $v=0$. Thus this is a global minimum, not merely a stationary value:

Let $g(x,y,z)=x^2+y^2-z^2$. The upper cone away from the apex is the part of $g^{-1}(0)$ with $z>0$. ThereThe regular level set theorem says that a level set of a smooth scalar function with nonzero gradient is an embedded submanifold of codimension one. Intersecting it with the open half-space $z>0$ preserves this local property. Thus $M$ is a two-dimensional submanifold of ${\mathbb{R}}^3$.

The denominator is positive on $B$ because $x_3>-1$. The quotient rule gives the gradientIn particular $g$ is continuously differentiable throughout $B$.

The total derivative of $f$ at $x$ is a linear map $Df(x):{\mathbb{R}}^n\to \mathbb{R}$ satisfyingThe gradient is the vector of partial derivatives, $\nabla f(x)=({\partial }_{1}f(x),\dots ,{\partial }_{n}f(x))$. When $f$ is totally differentiable, these partial derivatives exist andThus the derivative is a linear functional represented by the gradient under the Euclidean inner product. Existence of partial derivatives alone does not imply total differentiability.

For every $z\in B$, $z_3+3>2$ and $|z_1+z_2|<2$. Consequently each of the three partial derivatives in the preceding gradient has absolute value below $1/2$, soThe segment joining $x,y\in B$ lies inside $B$. The mean value theorem and Cauchy-Schwarz inequality therefore giveFor $x=y$ both sides vanish. This proves the required Lipschitz continuity uniformly on the ball.

The denominator is positive on $B$ because $x_3>-1$. The quotient rule gives the gradientIn particular $g$ is continuously differentiable throughout $B$.

The total derivative of $f$ at $x$ is a linear map $Df(x):{\mathbb{R}}^n\to \mathbb{R}$ satisfyingThe gradient is the vector of partial derivatives, $\nabla f(x)=({\partial }_{1}f(x),\dots ,{\partial }_{n}f(x))$. When $f$ is totally differentiable, these partial derivatives exist andThus the derivative is a linear functional represented by the gradient under the Euclidean inner product. Existence of partial derivatives alone does not imply total differentiability.

For every $z\in B$, $z_3+3>2$ and $|z_1+z_2|<2$. Consequently each of the three partial derivatives in the preceding gradient has absolute value below $1/2$, soThe segment joining $x,y\in B$ lies inside $B$. The mean value theorem and Cauchy-Schwarz inequality therefore giveFor $x=y$ both sides vanish. This proves the required Lipschitz continuity uniformly on the ball.

The defining function $h(x)=x_1^2/a^2+x_2^2/b^2+x_3^2/c^2-1$ is smooth and has gradient $2(x_1/a^2,x_2/b^2,x_3/c^2)$. This cannot vanish on its zero level, since the origin does not lie there. The regular level-set theorem therefore makes $M$ a two-dimensional submanifold, with tangent space $\ker Dh(x)$ and normal space

The Lagrange multiplier condition says that at a constrained local extremum of a continuously differentiable objective on a regular surface $h=0$, its gradient is normal to the tangent plane: $\nabla f=\lambda \nabla h$ for some scalar multiplier.

First the box must be centred at zero. If the two face coordinates in one direction are $u_i,v_i$, subtracting the ellipsoid equations for two vertices differing only in that coordinate gives $u_i^2=v_i^2$. A nondegenerate box has $u_i\ne v_i$, hence $u_i=-v_i$. Its positive half-sides $x,y,z$ therefore satisfy $x^2/a^2+y^2/b^2+z^2/c^2=1$ and its volume is $8xyz$.

The feasible closed octant is compact, the boundary gives zero volume, and some interior box has positive volume, so the maximum is interior. Apply multipliers to $\log (xyz)$ there:Thus the three squared scaled coordinates are equal and sum to one. The unique positive solution gives

The defining function $h(x)=x_1^2/a^2+x_2^2/b^2+x_3^2/c^2-1$ is smooth and has gradient $2(x_1/a^2,x_2/b^2,x_3/c^2)$. This cannot vanish on its zero level, since the origin does not lie there. The regular level-set theorem therefore makes $M$ a two-dimensional submanifold, with tangent space $\ker Dh(x)$ and normal space

The Lagrange multiplier condition says that at a constrained local extremum of a continuously differentiable objective on a regular surface $h=0$, its gradient is normal to the tangent plane: $\nabla f=\lambda \nabla h$ for some scalar multiplier.

First the box must be centred at zero. If the two face coordinates in one direction are $u_i,v_i$, subtracting the ellipsoid equations for two vertices differing only in that coordinate gives $u_i^2=v_i^2$. A nondegenerate box has $u_i\ne v_i$, hence $u_i=-v_i$. Its positive half-sides $x,y,z$ therefore satisfy $x^2/a^2+y^2/b^2+z^2/c^2=1$ and its volume is $8xyz$.

The feasible closed octant is compact, the boundary gives zero volume, and some interior box has positive volume, so the maximum is interior. Apply multipliers to $\log (xyz)$ there:Thus the three squared scaled coordinates are equal and sum to one. The unique positive solution gives

For $x\ne y$, apply the scalar mean value theorem to $h(t)=⟨f(x+t(y-x)),z⟩$ on $[0,1]$. It is continuous on the closed interval and differentiable inside by the stated segment and regularity assumptions. The chain rule givesThus $h(1)-h(0)=h^{\prime }(\theta )$ for some $\theta \in (0,1)$, precisely the asserted identity. If $x=y$, both sides are zero when the zero-direction derivative is naturally taken as $Df(x)0$.

For the final claim, all partial derivatives tending to zero imply $\Vert \nabla g(x)\Vert \to 0$ away from the origin. For a nonzero small vector $h$, the one-variable function $t\mapsto g(th)$ is continuous on $[0,1]$ by continuity at zero and differentiable on $(0,1)$. Thereforefor some $\theta \in (0,1)$. For any $\epsilon >0$, if $\Vert h\Vert$ is sufficiently small, the entire punctured segment has gradient norm below $\epsilon$. Hence $|g(h)-g(0)|\le \epsilon \Vert h\Vert$, provingThis argument does not presuppose differentiability at the origin.

For a scalar-valued differentiable function, its gradient is the column vectorIt represents the total derivative by the Euclidean inner product:Thus $J_f(x)=\nabla f(x{)}^T$ is a row matrix; the gradient and the total derivative represent the same first-order information but have different types.

At $(-1,0)$ the gradient is $(1/2,0{)}^T$. Therefore the total derivative is the linear maprepresented by the Jacobian matrix $\left(\begin{array}{cc}1/2& 0\end{array}\right)$.

Differentiating the reciprocal of $1+x^2+y^2$ gives the gradient

The free surface is material, so applying the material derivative to $y-\eta (x,t)$ gives the exact kinematic boundary conditionFor an incompressible fluid with constant density, the Euler equations areSince the flow is irrotational, $\mathbf{u}=\nabla \varphi$ and $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\nabla \varphi {|}^2/2)$. Integrating the resulting gradient equation gives a function of time on the right of Bernoulli equation. Adding a time-dependent constant to $\varphi$ removes that function, without changing velocity, and choosing the atmospheric-pressure reference givesThe pressure immediately above the surface is atmospheric; the liquid-side pressure includes the stated surface tension jump.

At first order in wave amplitude, the products ${\varphi }_x{\eta }_x$ and $|\nabla \varphi {|}^2$, the slope-squared correction to curvature, and displacement corrections in evaluating $\varphi$ at $y=\eta$ are all higher order. Hence, on $y=0$,For $\eta =A\cos (kx-\omega t)$, a harmonic velocity potential decaying as $y\to -\infty$ and satisfying the first condition isIndeed $\Delta \varphi =0$ and ${\varphi }_y{|}_0=A\omega \sin (kx-\omega t)={\eta }_t$. Substituting in the second condition yields the dispersion relation for a gravity-capillary wave:Taking the positive-frequency branch, the phase velocity and group velocity areSubtracting givesSince $\omega >0$,The threshold is also the wavenumber at the minimum phase speed of a gravity-capillary wave.

For an increment $h$, direct expansion and the chain rule giveFor example, $\Vert x+h{\Vert }^2-\Vert x{\Vert }^2=2x^{T}h+\Vert h{\Vert }^2$, so the quadratic remainder divided by $\Vert h\Vert$ tends to zero. The corresponding gradients are $\nabla g(x)=2x$ and $\nabla k(x)=2A^{T}Ax$; the derivative of $F$ is the constant matrix $A$.