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For $V=\xi (x,u){\partial }_x+\eta (x,u){\partial }_u$, its prolongation to derivatives through order $n$ isIt generates a symmetry of $\Delta =0$ exactly when ${\mathrm{pr}}^{(n)}V(\Delta )$ vanishes on the equation manifold.

Prolongation lifts a vector field to the jet bundle of maps so that its flow acts on derivatives of the dependent variables as well as on their values.

For a vector field $X$ on $M$, its pushforward by a smooth map between manifolds $f:M\to N$ is the vector field along a map defined by $(f_{\ast }X{)}_p=d{f}_p(X_p)$. Acting on a function $h$ on $N$, it satisfies $(f_{\ast }X{)}_p(h)=X_p(h\circ f)$. For a diffeomorphism this defines a vector field on $N$ by evaluation at $p=f^{-1}(q)$. For a general map there need not be a single vector at each image point: $f(x)=x^2$, $X={\partial }_x$ gives $2x{\partial }_y$, with opposite values over $y>0$.

A reflection acting as $(A,B)\mapsto (A,-B)$ forces an equivariant vector field to have $\dot{A}$ even and $\dot{B}$ odd in $B$. With the zero solution preserved, its quadratic normal form is $\dot{A}={\lambda }_{1}A+a_1{A}^2+a_2{B}^2$, $\dot{B}={\lambda }_{2}B+a_{3}AB$. For positive coefficients, every mixed equilibrium with $B\ne 0$ has Jacobian determinant $-2a_2{a}_3{B}^2<0$, hence is a saddle equilibrium. The two invariant-axis equilibria meet in a transcritical bifurcation, and the mixed branches meet them in pitchfork bifurcations.

Let ${\nabla }_{\tan }F=\nabla F-{\mathbf{e}}_r{\partial }_{r}F$ be the tangential surface gradient on a sphere of radius $r>0$. For smooth scalar fields, set $\mathbf{f}=R{\mathbf{e}}_r+{\nabla }_{\tan }G+{\mathbf{e}}_r\times {\nabla }_{\tan }F$. Its radial curl detects only $F$:To prove this, use the radial curl formula $(r\sin \theta {)}^{-1}[{\partial }_{\theta }(\sin \theta f_{\varphi })-{\partial }_{\varphi }{f}_{\theta }]$. The radial component $R$ never enters it. The $G$ terms cancel as $G_{\varphi \theta }-G_{\theta \varphi }=0$ by symmetry of second derivatives, while the $F$ terms give the unit-sphere Laplace-Beltrami operator. Taking the dot product with the position vector $\mathbf{r}=r{\mathbf{e}}_r$ gives the displayed $r^{-1}$ version. This tangential gradient decomposition does not assert that the whole vector field is solenoidal.

Differentiate the displayed homogeneity relation in $\lambda$ at $1$. On every regular point of $H=0$, the radial vector field $x$ is therefore perpendicular to the normal gradient and tangent to the level surface. Its flow $x(t)=e^{t}x(0)$ preserves that zero set. At a singular vertex this flow formulation remains meaningful even if there is no tangent plane.

In a normal neighbourhood $U={\exp }_p(V)$ with $V$ star-shaped, define $r(q)=|{\exp }_p^{-1}(q)|$. Away from $p$, the Gauss lemma gives $\nabla r=(d{\exp }_p{)}_v(v/|v|)$ for $v={\exp }_p^{-1}(q)$. Thus $|\nabla r|=1$ and this unit vector field is tangent to unit-speed radial geodesics. The function $r$ is not smooth at $p$, but $r^2/2$ is smooth and $\nabla (r^2/2)=(d{\exp }_p{)}_{v}v$ extends by zero. On a sufficiently small normal ball, $r$ equals the Riemannian distance from $p$.

// Target: riemannian-geometry.bigb

Let $-1<s<1$ and let $S_s$ be the outward-oriented boundary of $x^2+y^2+z^2\le 1$, $z\le s$. For a continuous function $h:[-1,s]\to \mathbb{R}$, the vector field $\mathbf{F}=h(z)(x,y,z{)}^T$ has outward fluxOn the curved unit sphere, the outward unit normal is $(x,y,z{)}^T$, so the normal component is $h(z)$. A horizontal spherical band has area $2\pi dz$, producing the first term. On the top disk, the outward unit normal is $(0,0,1{)}^T$, the normal component is $sh(s)$, and the area is $\pi (1-s^2)$, producing the second term.

For $h=f^{\prime }$ with a continuous first derivative, the fundamental theorem of calculus givesThis direct surface integral proof needs no second derivative of $f$. For $h=1$, it reduces to $\pi (2+3s-s^3)$; the divergence theorem identifies this with three times the enclosed volume.

On a fiberwise star-shaped neighborhood, let $H_s$ scale fibers by $s$ and let $E$ be the radial vector field. For a closed differential form $\delta$ with zero pullback to the zero section, the homotopy primitive $\eta ={\int }_0^1{s}^{-1}{H}_s^{\ast }({\iota }_E\delta )ds$ satisfies $d\eta =\delta$. For a two-form, it vanishes as an ambient covector along that section, providing the fixed-submanifold condition in the relative Moser theorem.

For $R>0$ in a normed vector space, define $P_R(0)=0$ and use the displayed formula elsewhere. This continuous map fixes the closed ball of radius $R$ and maps every point outside it to the boundary along its radial line. If a continuous map $T$ is defined only on that closed ball, $T\circ P_R$ is a continuous extension to the whole space. This extension is useful when applying a theorem whose hypotheses concern a globally defined vector field.

A vector field directed along the radius from a fixed origin, with magnitude depending only on radius. In three-dimensional Euclidean space and away from the origin, $\mathbf{F}=f(r)\mathbf{r}=rf(r)\hat{\mathbf{r}}$. The product rule gives $\nabla \cdot \mathbf{F}=3f+r{f}^{\prime }$, and the curl vanishes because the derivative matrix is symmetric. These identities alone do not establish regularity at the origin.

The unit radial vector field is $E=\nabla r$ on the punctured normal neighbourhood, where $r$ is the radial distance in a normal neighbourhood. Along ${\gamma }_u(t)={\exp }_p(tu)$ with $|u|=1$, it satisfies $E({\gamma }_u(t))={\dot{\gamma }}_u(t)$. The scaled radial field $W=rE=\nabla (r^2/2)$ is smooth at the centre and vanishes there. These conventions have different associated radial functions.

// Target: riemannian-geometry.bigb

Two closed symplectic forms agreeing as ambient bilinear forms along a compact embedded submanifold are equivalent near it by a diffeomorphism fixing it pointwise. Shrink until their affine interpolation is nondegenerate. A relative homotopy primitive $\eta$ for their difference vanishes as an ambient covector on the submanifold. Solving ${\iota }_{Z_t}{\omega }_t=-\eta$ gives a vector field vanishing there; its local flow fixes the submanifold and makes ${\omega }_t$ constant under pullback.

A vector field singular on an axis need not satisfy the classical divergence theorem on the original closed region. Remove a thin tube $\rho <\epsilon$, apply the theorem to the remaining smooth region, then examine the added cylindrical boundary. If its flux tends to zero and the remaining volume integral converges, the improper boundary flux equals the improper integral of the divergence. For $\mathbf{F}=(\log {\rho }^2,\log {\rho }^2,z/\rho )$ with $\rho =\sqrt{x^2+y^2}$, the cylindrical unit normal has zero $z$ component, so the added flux is $O(\epsilon |\log \epsilon |)$ on a bounded axial interval. The divergence $(2\cos \varphi +2\sin \varphi +1)/\rho$ is locally integrable under the cylindrical coordinates measure $\rho d\rho d\varphi dz$. Hence no finite line-source correction survives, despite the field itself being unbounded.

A Riemannian metric satisfying the displayed equation for a vector field $X$ and constant $\lambda$. A gradient Ricci soliton has $X=\nabla f$; it is steady when $\lambda =0$.

On an oriented Riemannian manifold with boundary, the divergence of a Riemannian vector field obeys the displayed integral identity for smooth compactly supported vector fields $X$, or for arbitrary smooth $X$ if the manifold is compact. Here $N$ is the outward unit normal vector and the boundary has the outward-normal-first boundary orientation. Its induced Riemannian volume form is $j^{\ast }{\iota }_N{\omega }_g$. Splitting $X$ into its normal and tangential components gives $j^{\ast }{\iota }_X{\omega }_g=g(X,N){\omega }_{\tilde{g}}$, since a top form vanishes on $n$ boundary-tangent arguments. Apply the Stokes theorem to $d{\iota }_X{\omega }_g=(\mathrm{div}X){\omega }_g$.

Without a support or decay condition, flux at infinity can invalidate the result even if the integrals converge. On $[0,\infty )$ with Riemannian metric $d{x}^2$, $X=(\arctan x){\partial }_x$ has boundary flux zero but integral of divergence $\pi /2$.

For a smooth function $f$ on a Riemannian manifold, its Riemannian gradient is the unique vector field satisfying $g({\nabla }_{g}f,V)=df(V)$ for every vector field $V$. It depends on the Riemannian metric, and its negative generates the downward gradient flow.

A smooth local basis of vector fields whose inner products under a Riemannian metric are the identity matrix. Tracing the Riemannian Hessian in such a frame gives the Laplace-Beltrami operator, with the chosen sign.

The right-invariant vector field associated with a Lie algebra element $\xi$ is obtained by right translation of its value at the identity. Its flow is $g\mapsto \exp (t\xi )g$, hence consists of left translations on a Lie group. Right invariance follows because left and right multiplication commute.

Take the upward-oriented graph $z=c\sqrt{1-x^2/a^2-y^2/b^2}$ over $0\le x\le L_x$, $0\le y\le L_y$, where $a,b,c,L_x,L_y>0$, $A=L_x^2/a^2$, $B=L_y^2/b^2$ and $A+B<1$. The surface flux of the vector field $F=(-y,x,0)$ is the displayed formula. Indeed the vector surface element is $(-z_x,-z_y,1)dxdy$, soThe change of variables $u=x^2/a^2$, $v=y^2/b^2$ changes $xydxdy$ to $a^2{b}^{2}dudv/4$. Integrating $(1-u-v{)}^{-1/2}$ over $[0,A]\times [0,B]$ gives $\frac{4}{3}[1-(1-A{)}^{3/2}-(1-B{)}^{3/2}+(1-A-B{)}^{3/2}]$, proving the result. When $a=b$, the value of the vector field is a tangent vector to the ellipsoid at every point, and the surface flux vanishes pointwise.

A Runge-Kutta method advances an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, by sampling its vector field at several trial states within one time step. These samples are Runge-Kutta stages; a weighted combination of their slopes produces the next state. The construction improves on the single tangent line used by the Euler method while requiring evaluations of $f$, rather than explicit higher derivatives of the solution. It applies to scalar equations and systems of ordinary differential equations alike. The coefficient choices distinguish methods such as Heun's method, the classical fourth-order Runge-Kutta method and implicit Runge-Kutta methods.

For $s$ stages, write $A=(a_{ij})$ for the stage matrix, $c_i$ for relative time nodes and $b_i$ for output weights. A time step of length $h$ computesHere $k_i$ is a slope, with units of state per time; $h{k}_i$ is a state increment. Another common convention writes $K_i=h{k}_i$, so its stage states are $Y_i=y_n+{\sum }_j{a}_{ij}{K}_j$ and its output is $y_n+{\sum }_i{b}_i{K}_i$. These are equivalent conventions: including $h$ both inside $K_i$ and again in the output would count the time step twice. The textbook formulation uses the increment convention.^[ref]

Usually the Butcher tableau obeys $c_i={\sum }_j{a}_{ij}$, called internal consistency. Stage $i$ then combines slopes over an effective time $c_{i}h$. The second-order conditions with independent Runge-Kutta abscissae explain what changes when time nodes are specified independently. A Runge-Kutta stage is generally an auxiliary approximation, not a separate accepted solution value at its node. Several stages can even have the same time node but different state arguments, as in the classical fourth-order Runge-Kutta method.

The useful idea is to approximate the integral $y(t_n+h)-y(t_n)={\int }_{t_n}^{t_n+h}f(t,y(t))dt$ without knowing the intermediate exact trajectory. Trial states provide places to sample $f$, and Butcher order conditions make the resulting combination agree with the exact Taylor expansion. More stages alone do not guarantee a higher order of a numerical method: their coefficients must satisfy the appropriate conditions. Embedded Runge-Kutta pairs reuse these same samples for a second output and a practical local error estimator.