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A Runge-Kutta stage consists of a trial state $Y_i$ at the time node $t_n+c_{i}h$ and the corresponding vector field evaluation $k_i$. In the slope convention, $Y_i=y_n+h{\sum }_j{a}_{ij}{k}_j$. Stages belong to the construction of one Runge-Kutta method step; they need not have the same accuracy as its final output. Stage states can leave the physically allowed region even when the endpoint lies inside it, so $f$ must be defined at every stage used.

Saturating positive feedback limits the reinforcing rate as activity increases. Here is an explicitly synthetic ordinary differential equation for normalized nonnegative input $u$, response $x$, and return strength $\beta$:The return is an increasing, nonproportional function of $x$, while removal is proportional to $x$. Time and activities are dimensionless variables. At $x=0$ the vector field points into the nonnegative domain, and $dx/dt\le u+\beta -x$ prevents unbounded growth there. This makes the toy useful for asking about a bounded switch; it supplies no molecular mechanism for the chosen rate law.

At an equilibrium point, $f=0$. For a simple equilibrium point, the sign of the derivativedetermines local asymptotic stability. If $\beta =1$, the maximum reinforcing slope is $3\sqrt{3}/8<1$, so $f$ decreases strictly: every $u\ge 0$ has one asymptotically stable equilibrium point despite positive feedback.

Three equilibrium points become possible only if the reinforcing slope exceeds removal's slope somewhere. Its maximum is $3\sqrt{3}\beta /8$, at $x=1/\sqrt{3}$. Consequently this toy requiresfor two distinct folds. This is a necessary threshold for a bistability interval in this toy, not a universal biochemical number. At a fold, both $f=0$ and $f_x=0$, givingFor $\beta =1.8$, the folds are at $(x,u)\simeq (0.350003,0.153563)$ and $(0.887240,0.094405)$. Between these input values there are two asymptotically stable equilibrium points separated by an unstable one. At $u=0.12$, the three activities are approximately $0.171331$, $0.621230$ and $1.127439$; the low and high ones are stable.

A sufficiently slow increasing sweep follows the low branch to $u_{\mathrm{on}}\simeq 0.153563$ and then jumps upward. A decreasing sweep follows the high branch to $u_{\mathrm{off}}\simeq 0.094405$ and then jumps downward: hysteresis. These are limiting quasistatic thresholds, with relaxation between input changes and negligible perturbations that could cause earlier switching. Finite sweep rates and fluctuations can alter observed thresholds. A switchlike response with only one stable equilibrium point does not have this equilibrium hysteresis mechanism.

A scalar surface integral integrates a scalar field against the unsigned surface area element. For a regular parametrization $\mathbf{r}(u,v)$, it is $\int f(\mathbf{r}(u,v))\Vert {\mathbf{r}}_u\times {\mathbf{r}}_v\Vert dudv$. Reversing orientation does not change it. For example, the scalar surface integral of the constant one is the surface area; a signed flux instead integrates the normal component of a vector field.

For a smooth function $f$ and vector field $X$,on every tensor field, with $X\otimes df$ interpreted as the endomorphism $Y\mapsto Y(f)X$. Both sides vanish on functions; on vector fields this follows from $[fX,Y]=f[X,Y]-Y(f)X$. Both are contraction-compatible tensor derivations, so agreement on functions and vector fields proves equality on all tensor types. In particular ${\mathcal{L}}_{fX}\omega =f{\mathcal{L}}_X\omega +\omega (X)df$.

The vector field $S=t{\partial }_t+x\cdot \nabla$ generates simultaneous spacetime dilation. For the flat d'Alembert operator, $[\square ,S]=2\square$. Consequently it commutes with the homogeneous wave equation at the level of its solution set, although its operator commutator is nonzero.

A smooth positive Hermitian bundle metric admits local orthonormal frames by the smooth Gram-Schmidt process. In such a frame, metric compatibility gives $A_{ji}(V)+\overline{A_{ij}(V)}=0$ for real vector fields $V$, so the connection matrix is skew-Hermitian. The orthonormal frame is smooth; a holomorphic orthonormal frame is not generally available.

Suppose two oriented surfaces have the same oriented boundary and $S_1\cup (-S_2)$ bounds a volume inside a domain where the vector field $\mathbf{F}$ is smooth and solenoidal. The divergence theorem gives zero flux through their closed union, proving the displayed equality. If a global vector potential $\mathbf{F}=\nabla \times \mathbf{G}$ exists on the relevant surfaces, the Stokes theorem proves the equality directly from their common boundary. The domain condition matters: a singularity inside the joining volume can obstruct the divergence argument.

An automorphism preserving the physically specified background structure of spacetime. With only a Minkowski metric and fixed orientations, the appropriate external symmetries are the corresponding Poincaré group elements. An added preferred vector field reduces this group. This is distinct from merely changing coordinates.

The vector fields ${\partial }_t$ and ${\partial }_i$ generate time and space translations and commute with the flat d'Alembert operator. They form the unweighted part of the commutation vector fields for the wave equation.

To deduce $D{u}_t=I+tDF+o(t)$ from $u_t(x)=x+tF(x)+o(t)$, the remainder must be small in a topology controlling its spatial derivatives, such as $C^1$ locally. Small displacement alone is insufficient: $u_t(x,y,z)=(x+t^2\sin (x/t^2),y,z)$ differs uniformly from the identity by $o(t)$, but has Jacobian determinant $2$ at $x=0$ for nonzero $t$. A smooth flow map generated by a smooth vector field has the required derivative control.

The vector field ${\Omega }_{ij}=x_i{\partial }_j-x_j{\partial }_i$ is an infinitesimal spatial rotation. It commutes with the flat d'Alembert operator and supplies angular derivatives in the Klainerman-Sobolev inequality.

A spin texture describes how a spin expectation value or spin polarization varies with position or wavevector. It specifies a vector field, including its magnitude when that matters; a direction field alone is undefined wherever that magnitude vanishes. A Rashba spin texture is an example in which spin expectation values wind around circles in the plane of wavevectors.

For an upward-oriented band of the circular elliptic paraboloid between radii $r_0$ and $R$, the vector field $B=(-y^3,x^3,z^3)$ has vertical curl $3(x^2+y^2)$. Its surface integral is $3\pi (R^4-r_0^4)/2$. Stokes theorem gives the same value from the counterclockwise outer circle minus the counterclockwise inner circle. Equivalently the induced inner boundary direction is clockwise. Forgetting the inner boundary or its reversed orientation changes the answer.

Let $S$ be an oriented smooth surface with piecewise smooth boundary, and let $F$ be a continuously differentiable vector field on a neighborhood of $S$. With the compatible boundary direction determined by the right-hand rule,For $G=(-y,x,0)$ on the horizontal unit disc with upward normal, the left side is $2$ times its area, giving $2\pi$. The counterclockwise boundary has tangential field of magnitude one and circumference $2\pi$, giving the same line integral. Reversing the surface normal reverses the boundary direction and both signs. The theorem replaces a circulation calculation with a surface integral of the curl, whichever is easier.

Stokes theorem applies to an oriented surface with several boundary components. Cutting a cylindrical side along one generator produces a parameter rectangle: the two copies of the cut are traversed in opposite directions and their line integrals cancel for a single-valued smooth vector field. The upper and lower rims retain opposite directions when viewed from the same side. This proves why cylindrical topology does not require a modification of Stokes theorem; it requires inclusion of the entire oriented boundary.

Near a point where a smooth vector field is nonzero, there are coordinates in which it is $\partial /\partial x_1$. Flow a transverse coordinate slice by the local flow; the inverse function theorem makes the resulting map a coordinate chart. This is the first step of the coordinate proof of the Frobenius theorem.

The surface divergence of an ambient vector field is the tangential trace ${\nabla }_s\cdot \mathbf{v}=({\delta }_{ij}-n_i{n}_j){\partial }_j{v}_i$. For a tangent vector field it is the intrinsic divergence. For $\mathbf{v}=\mathbf{n}$ it gives the sum of the two principal curvatures; a sphere with outward normal has ${\nabla }_s\cdot \mathbf{n}=2/a$.

For a smooth tangent vector field $v$ on a compact orientable surface with smooth boundary and outward conormal of a surface boundary, the integral of its surface divergence is its boundary flux. Extend the unit normal vector to a unit field $m$. The curl of a cross product gives $m\cdot \nabla \times (m\times v)=\nabla \cdot v-m_i{m}_j{\partial }_j{v}_i$ on the surface: $m\cdot v=0$ there and $m\cdot {\partial }_{j}m=0$ by unit length. Apply Stokes theorem and use $(m\times v)\cdot d\mathbf{r}=v\cdot (d\mathbf{r}\times m)$ to obtain the formula.

For a differentiable three-dimensional vector field $F$, split its gradient tensor into symmetric and antisymmetric parts. The symmetric part is $S_{ij}=({\partial }_i{F}_j+{\partial }_j{F}_i)/2$. Contracting the antisymmetric part with the Levi-Civita symbol gives the axial vector $u_k={ϵ}_{kij}{\partial }_i{F}_j/2=(\nabla \times F{)}_k/2$, proving the decomposition. If $\nabla \cdot F=0$, the divergence of the symmetric part is ${\partial }_j{S}_{ij}=\Delta F_i/2$. This lets a vector Laplacian be converted into a boundary integral of a symmetric tensor by the divergence theorem.

A vector field on a symplectic manifold is symplectic when its flow preserves the symplectic form. Infinitesimally this is ${\mathcal{L}}_X\omega =0$. By Cartan's magic formula it is equivalent to $d({\iota }_X\omega )=0$. A Hamiltonian vector field has an exact contraction with the form, so is symplectic. On the standard two-dimensional torus, ${\partial }_x$ contracts to the closed but nonexact form $dy$, providing a symplectic non-Hamiltonian example.