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An activator in an activator-inhibitor reaction–diffusion system is a species whose local feedback enhances activation. Near a homogeneous equilibrium, positive self-feedback is reflected in the corresponding diagonal entry of the reaction Jacobian matrix. Cross-coupling to an inhibitor and differing diffusion rates can generate a Turing instability.

Along the antipodal two-cycle of $F(x,y)=(y,a(x-x^3/3)-y)$, the Jacobian matrix is $B=\left(\begin{array}{cc}0& 1\\ -2a& -1\end{array}\right)$ at both points. Its two-step periodic-orbit multipliers are squares of the roots of ${\lambda }^2+\lambda +2a=0$. The Jury criterion makes the cycle attracting precisely in the displayed interval, and repelling for $a>1/2$. This stable family is distinct from the always unstable nonzero-sum cycles.

The annual pulse-breeding predator-prey model is an area-preserving map in logarithmic population coordinates: $u^{\prime }=u+\log r-\kappa e^v$, then $v^{\prime }=v+\log s+\log (1+b{e}^{u^{\prime }})$. Each step is a shear of determinant one. At a positive fixed point, the ordinary-population Jacobian matrix has determinant one and trace $2-(1-s)\log r$. Eigenvalues lie on the unit circle for $0<(1-s)\log r<4$, but this linear neutrality does not prove nonlinear stability at every resonance. Area preservation excludes an isolated attracting positive periodic cycle.

A quadratic activator can have both basal production and activation inhibited by a second species. A normalized example has reaction rates $f(u,v)=1+r{u}^2/v-u$ and $g(u,v)=q(u^2-v)$, with $r,q>0$ and positive concentrations. At its spatially homogeneous equilibrium, $v=u^2$, so $u_{\ast }=1+r$ and $v_{\ast }=(1+r{)}^2$. Writing $a=(r-1)/(r+1)$, its reaction Jacobian matrix is $J=\left(\begin{array}{cc}a& -r/(1+r{)}^2\\ 2q(1+r)& -q\end{array}\right)$. Thus $\mathrm{tr}J=a-q$ and $\det J=q>0$, giving linear asymptotic stability when $q>a$ and instability when $q<a$. Basal production makes $a$ change sign at $r=1$; a positive activation feedback needed for a conventional Turing instability requires $r>1$.

A secant-updating method for a vector equation $F(x)=0$, named for Charles George Broyden. Solve $B_k{d}_k=-F(x_k)$, choose $x_{k+1}=x_k+{\alpha }_k{d}_k$, and form $s_k=x_{k+1}-x_k$, $y_k=F(x_{k+1})-F(x_k)$. For a nonzero step the displayed “good Broyden” update satisfies $B_{k+1}{s}_k=y_k$, learning the Jacobian matrix action along the observed step rather than evaluating a new full Jacobian matrix.

The update also has a least-change characterization in the Frobenius norm. Write $u=y_k-B_k{s}_k$. Any correction $E$ satisfying $E{s}_k=u$ has squared row norms bounded below by $u_i^2/\Vert s_k{\Vert }^2$, by the Cauchy-Schwarz inequality. Summing these bounds gives $\Vert E{\Vert }_F^2\ge \Vert u{\Vert }^2/\Vert s_k{\Vert }^2$, with equality for $E=u{s}_k^{\mathsf{T}}/\Vert s_k{\Vert }^2$. This proves the rank-one correction is the least-change update satisfying the secant constraint. Superlinear convergence needs local regularity and suitable initialization; it is not a global consequence of this algebraic identity.^[ref]

A uniform Newtonian gravitational field inside one spherical void can be cancelled at a point by the repulsive field representing another removed sphere. The latter field has a Jacobian matrix proportional to $I-3\hat{r}{\hat{r}}^T$: its axial eigenvalue is negative and both transverse eigenvalues positive. The equilibrium is restoring along the axis but unstable transversely. A zero force is thus not enough for stable rest.

The change of variables formula expresses an integral in coordinates that make its integrand or its region simpler. Three ingredients change together: the region, the function being integrated, and the element of area or volume. The last change is measured by the absolute Jacobian determinant; forgetting it amounts to counting transformed cells as though their sizes had stayed the same.

In one dimension, substitution is the integrated chain rule. If $g$ is continuously differentiable on $[a,b]$ and $f$ is continuous on its image, thenThese are directed definite integrals: if $g$ decreases, the transformed endpoints reverse. This identity does not require $g$ to be injective, because an antiderivative composed with $g$ already accounts for any backtracking.^[ref]

For a concrete computation, set $u=1+x^2$, so $du=2xdx$. The endpoints $x=0,1$ become $u=1,2$:The numerator is precisely the derivative of the expression in the denominator. Substitution reveals the familiar natural logarithm hidden inside this rational function. Keeping the original endpoints after changing the integrand would describe a different integral.

For the elementary two-dimensional theorem, let $T:U\to V$ be a $C^1$ diffeomorphism between open sets in the plane, let $S\subset U$ be a bounded region with piecewise smooth boundary, and let $f$ be continuous on $T(S)$, with the displayed integrals finite. ThenHere $DT$ is the Jacobian matrix. The diffeomorphism assumption means a bijection with a continuously differentiable inverse, and ensures a nonzero Jacobian determinant. This excludes overlapping sheets; the more general area formula handles multiplicity.^[ref]

Take $S=[0,1{]}^2$ and the linear mapIts inverse is $(u,v)=((x-y)/2,y)$. The square becomes a parallelogram with vertices $(0,0),(2,0),(3,1),(1,1)$: base two, height one, area two. Consequently ${\iint }_{T(S)}1dxdy={\iint }_{S}2dudv=2$. A nonconstant integrand changes too: ${\iint }_{T(S)}xdxdy={\int }_0^1{\int }_0^1(2u+v)2dudv=3$.

The direction of the map matters when selecting the factor. Here $T$ goes from the new coordinates to the original coordinates, so the factor is two. Its inverse has Jacobian determinant $1/2$, appropriate for converting area in the opposite direction. Writing down the map before computing its Jacobian matrix prevents a reciprocal-factor error.

Now reflect the image vertically: $\tilde{T}(u,v)=(2u+v,-v)$ has Jacobian determinant $-2$, but still produces area two. The sign records reversal of the orientation of a vector space; ordinary area uses its absolute value. For oriented differential forms, the pullback of a differential form instead retains the signed determinant.^[ref]

The geometric beauty is that a nonlinear differentiable map has a local linear approximation: tiny rectangles behave approximately like these parallelograms. Summing their scaled contributions gives the change of variables formula. In polar coordinates this factor is $r$, explaining $dxdy=rdrd\theta$; equal angular steps cover more area farther from the origin. Choosing suitable coordinates can simplify both the boundary and the integrand.^[ref]

A chemostat steady state has constant nutrient concentration and biomass concentration even while nutrient and cells flow through the vessel. It is an equilibrium of the chemostat ordinary differential equations, rather than a culture in which all growth has stopped. Assume $D>0$ and the Monod equation. Setting both derivatives to zero gives the always-existing chemostat washout equilibrium $(s_{\ast },x_{\ast })=(s_{\mathrm{in}},0)$. A positive biomass concentration instead requires growth to balance removal, $\mu (s_{\ast })=D$, whenceThe positive chemostat steady state exists exactly whenThe weaker condition $D<{\mu }_{\max }$ only makes the denominator positive; it does not ensure $s_{\ast }<s_{\mathrm{in}}$ or $x_{\ast }>0$. Because the Monod equation is increasing, the feed must permit a specific growth rate greater than $D$. At a positive chemostat steady state, increasing feed nutrient concentration at fixed $D$ increases the biomass concentration without changing $s_{\ast }$, within the one-nutrient assumptions.

For a concrete chemostat, let ${\mu }_{\max }=1{\mathrm{h}}^{-1}$, $K_s=1\mathrm{g}{\mathrm{L}}^{-1}$, $s_{\mathrm{in}}=9\mathrm{g}{\mathrm{L}}^{-1}$ and $Y=0.5\mathrm{g}\text{ biomass}/(\mathrm{g}\text{ nutrient})$. At $D=0.5{\mathrm{h}}^{-1}$,The nutrient supply minus outflow is $D(s_{\mathrm{in}}-s_{\ast })=4\mathrm{g}{\mathrm{L}}^{-1}{\mathrm{h}}^{-1}$, matching nutrient consumption $\mu (s_{\ast })x_{\ast }/Y=4\mathrm{g}{\mathrm{L}}^{-1}{\mathrm{h}}^{-1}$. Biomass production and removal both equal $2\mathrm{g}{\mathrm{L}}^{-1}{\mathrm{h}}^{-1}$. The chemostat washout threshold is $D_c=\mu (9)=0.9{\mathrm{h}}^{-1}$. For example, $D=0.95{\mathrm{h}}^{-1}<{\mu }_{\max }$ formally gives $s_{\ast }=19\mathrm{g}{\mathrm{L}}^{-1}$ and $x_{\ast }=-5\mathrm{g}{\mathrm{L}}^{-1}$, which cannot describe a physical culture.

There is a useful exact simplification: the nutrient-equivalent concentration $z=s+x/Y$ satisfies $\dot{z}=D(s_{\mathrm{in}}-z)$, so $z(t)=s_{\mathrm{in}}+[z(0)-s_{\mathrm{in}}]e^{-Dt}$. This combines dissolved nutrient and the nutrient equivalent of biomass; it does not assert that these two variables account for every chemical constituent. In coordinates $(z,x)$, the Jacobian matrix at the positive chemostat steady state has eigenvalues $-D$ and $-{\mu }^{\prime }(s_{\ast })x_{\ast }/Y$, both negative. Hence the positive chemostat steady state is locally asymptotically stable. In the worked example the eigenvalues are $-0.5{\mathrm{h}}^{-1}$ and $-2{\mathrm{h}}^{-1}$.

Chemostat washout means the culture's biomass concentration tends to zero because growth cannot sustain replacement of the cells leaving the vessel. In the ideal Monod equation chemostat with $D>0$, its equilibrium is $(s,x)=(s_{\mathrm{in}},0)$. Its Jacobian matrix has eigenvalues $-D$ and $\mu (s_{\mathrm{in}})-D$: a small inoculum grows when $D<\mu (s_{\mathrm{in}})$, whereas chemostat washout is locally asymptotically stable when $D>\mu (s_{\mathrm{in}})$. For nonnegative initial concentrations, chemostat washout occurs for $D\ge \mu (s_{\mathrm{in}})$; the boundary case also loses biomass asymptotically.

At $D=D_c$ the positive chemostat steady state meets the chemostat washout equilibrium, and one eigenvalue is zero, so linearization alone does not decide the boundary case. On the invariant nutrient-equivalent balance $s+x/Y=s_{\mathrm{in}}$, small positive $x$ obeys $\dot{x}=-[{\mu }^{\prime }(s_{\mathrm{in}})/Y]x^2+O(x^3)$: nutrient consumption lowers the specific growth rate below $D_c$, leading to slow decay rather than a positive biomass concentration. The continued mathematical branches exchange stability in a transcritical bifurcation; the positive-biomass branch becomes physically inadmissible beyond $D_c$.

The differential operators $Z_j$ commute with $D={\partial }_t+a(v)\cdot {\nabla }_x$. Indeed $[D,{\partial }_{v_j}]=-{\sum }_i({\partial }_{v_j}{a}_i){\partial }_{x_i}$, canceled by differentiating the explicit time factor in $Z_j$. Hence $Z_{j}f$ satisfies the same transport equation as $f$, and obeys the same L2 norm conservation when boundary fluxes vanish. For the standard Jacobian matrix $(Da{)}_{ij}={\partial }_{v_j}{a}_i$, the vector of these derivatives is ${\nabla }_{v}f+t(Da{)}^T{\nabla }_{x}f$.

For a conformal bijection $f:D\to \tilde{D}$ and smooth $u,v$ of compact support on $\tilde{D}$, $\boxed{(u\circ f,v\circ f{)}_{\nabla ,D}=(u,v{)}_{\nabla ,\tilde{D}}}$. The Jacobian matrix of $f$ is a rotation times $|f^{\prime }|$. Consequently the chain rule introduces $|f^{\prime }{|}^2$ into the gradient pairing, while the change of variables formula introduces precisely the same factor into area. They cancel. The identity extends by completion to the Dirichlet energy spaces; it does not preserve the $L^2$ term of an inhomogeneous Sobolev norm.

The constant rank theorem separates the directions a smooth map preserves from the directions it forgets. If a smooth map from an open subset of ${\mathbb{R}}^m$ to ${\mathbb{R}}^n$ has derivative of constant rank $r$ on a neighbourhood of a point, there are local coordinate transformations in source and target in which it becomesIt keeps $r$ coordinates, discards the other $m-r$, and inserts $n-r$ zeros. This is stronger than simplifying the Jacobian matrix at one point: it straightens the actual nonlinear map on a neighbourhood. ^[ref]

For a worked example, takeThe two columns agree, and the first row is nonzero, so the rank is exactly one everywhere. The image of a function is the parabola $b=a^2$. Every line $x+y=a$ is a fiber mapping to the same point $(a,a^2)$.

Choose input coordinates $u=x+y$, $v=x-y$, and output coordinates $\alpha =a$, $\beta =b-a^2$. Both coordinate transformations are diffeomorphisms: their inverses are $x=(u+v)/2$, $y=(u-v)/2$ and $a=\alpha$, $b=\beta +{\alpha }^2$. In these coordinates,The curved image has become a straight line, and the unused coordinate $v$ is explicit.

Constant rank on a neighbourhood is essential. The map $(x,y)\mapsto (x,y^2)$ has rank one when $y=0$ and two when $y\ne 0$, so the constant rank theorem does not supply one rank-one form across that line. By contrast, the full-rank square case of the theorem forgets no coordinates and recovers the inverse function theorem. The usefulness is that curved fibers and images can often be studied through this simpler local form.

A continuation method follows a solution branch of a parameter-dependent equation from a known solution, using nearby solutions to guide the next solve. This is useful when one wants to understand how solutions change, rather than obtain just one isolated answer. A continuation predictor-corrector step follows a tangent vector and then uses a Newton method correction on a transverse section. The same geometry supports both pseudo-arclength continuation through a continuation fold and transversal surface intersection tracing.

Consider the original scalar polynomial exampleIts solution branch bends back twice in $\lambda$. Increasing $\lambda$ alone eventually stops following the same nearby solutions; following the curve in the $(x,\lambda )$ plane can continue around both bends. The continuation folds areFor $|\lambda |<2/(3\sqrt{3})$ there are three distinct real zeros of a function at that fixed $\lambda$, all lying on this one connected curve. A continuation method keeps their geometric relationship visible.

Every continuation method needs a seed, supplied by an independent solve, an explicitly known solution, or another search. Following one solution branch does not guarantee finding every connected component of the solution set. A large step can miss detail or make the correction converge to a distant solution. At branch intersections or a loss of rank of the full Jacobian matrix, the regular-curve argument needs separate analysis. The seed and component limitations also matter in transversal surface intersection tracing.

For the cubic polynomial, an oriented, normalized tangent vector isIts Euclidean norm is one and $DF(z_k)t_k=0$. The sign here follows increasing $x$; for a general solution branch, choose the sign of each new tangent vector so its dot product with the previous one is positive. First predict $z_{\mathrm{p}}=z_k+h{t}_k$, then apply Newton method starting at $z_{\mathrm{p}}$ toThe second equation keeps the correction on the transverse line in this two-dimensional example; in higher dimensions it is a hyperplane. Solving the linear system $DH(z^{(j)})\delta z=-H(z^{(j)})$ and updating $z^{(j+1)}=z^{(j)}+\delta z$ corrects both coordinates together.

This step gives $z_{\mathrm{p}}\approx (0.617556,0.424641)$ and $z_{k+1}\approx (0.635532,0.378840)$. Check both equations: the solution must satisfy $x^3-x+\lambda =0$ and $t_k^{\mathsf{T}}(z-z_{\mathrm{p}})=0$. Checking only the first could accept any point on the solution branch. The new $x$ exceeds $1/\sqrt{3}$, so the step crosses the continuation fold, even though the predictor's $\lambda$ exceeds $2/(3\sqrt{3})$.

At the seed, set $a=3x_k^2-1$. The determinant of the correction Jacobian matrix with the exact unit vector above isAt either continuation fold, $a=0$, $t_k=(1,0)$ and $DH(z_k)=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$: the correction remains invertible. More generally, if $DF$ has full rank and a one-dimensional kernel spanned by $t_k$, adding $t_k^{\mathsf{T}}$ gives an invertible matrix, since a vector annihilated by both rows must be zero. This is a local argument; the old transverse hyperplane need not work arbitrarily far along the curve. Reduce $h$ and retry when correction fails, and check the equations and orientation after successful steps. Scaling state and parameter coordinates also matters, because it changes the Euclidean norm used to normalize the tangent vector.

// Target: analysis.bigb

A vector field on an open subset of Euclidean space whose component functions have continuous first partial derivatives. Its Jacobian matrix is continuous; this is a standard regularity assumption for the divergence theorem and Stokes theorem.

A manifold chart with coordinates $x^{\mu }$ induces a smooth local frame of the tangent bundle, with tangent vectors ${\partial }_{\mu }=\partial /\partial x^{\mu }$ forming a basis of each tangent space. Their Lie brackets of vector fields vanish because mixed partial derivatives of smooth functions commute. Under a change of coordinates they transform by the chain rule: $\partial /\partial y^{\nu }=(\partial x^{\mu }/\partial y^{\nu })\partial /\partial x^{\mu }$, with coefficients given by the Jacobian matrix.

For a system $\epsilon \dot{x}=F(x,y)$, $\dot{y}=G(x,y)$ with fast variable $x$ and slow variable $y$, the critical manifold is the set of equilibria of the fast problem $dx/ds=F(x,y)$ with $y$ frozen. It is attracting where the fast Jacobian matrix has eigenvalues with negative real parts and repelling where those real parts are positive. A smooth graph can cease to be attracting at a fold point of a critical manifold.

Imagine folding a sheet along its horizontal axis by the smooth map $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$, $F(x,y)=(x,y^2)$. The inputs $(a,r)$ and $(a,-r)$ have the same output. Along $y=0$, horizontal motion still changes the output, but vertical motion produces no first-order change: $F(a,\epsilon )-F(a,0)=(0,{\epsilon }^2)$. These inputs are critical points of a smooth map even though their differential is not the zero linear map. The missing direction in the target tangent space is what matters.

Precisely, let $f:X^m\to Y^n$ be a smooth map between smooth manifolds without boundary. An input $p\in X$ is a critical point of a smooth map when its differentialis not surjective, equivalently when its rank of a linear map is less than $n$. An input where it is surjective is a regular point; a map with no critical points of a smooth map is a submersion. The threshold is the target dimension, not $\min (m,n)$. In manifold charts, compute the matrix rank of the $n\times m$ Jacobian matrix; changes of charts preserve that matrix rank. ^[ref]

For the fold,Thus its set of critical points of a smooth map is the entire horizontal axis. Away from that axis, its differential can produce every infinitesimal target displacement. At the axis, it can produce only horizontal ones. This diagnoses where the two inverse branches in the picture merge.

For a scalar smooth function $f:X\to \mathbb{R}$, the target dimension is one. A linear map to $\mathbb{R}$ fails to be surjective exactly when it is zero, so here a critical point of a smooth map means $d{f}_p=0$. In Euclidean space this is precisely the familiar critical point condition $\nabla f(p)=0$: every first partial derivative vanishes. On a Riemannian manifold, it is equivalently ${\nabla }_{g}f(p)=0$ for the chosen Riemannian metric; the Riemannian gradient depends on that metric, but whether it vanishes does not. See Loring W. Tu, An Introduction to Manifolds, Proposition 8.23 for the scalar coordinate criterion. The scalar smooth function $f(t)=t^3$ has a critical point at zero without a local maximum or local minimum.

A source of lower dimension gives a different warning. For the smooth map $G:\mathbb{R}\to {\mathbb{R}}^2$, $G(t)=(t,t^2)$, the Jacobian matrix is the column $(1,2t{)}^T$, of matrix rank one at every input. Every input is a critical point of a smooth map, since the target dimension is two, even though this parametric curve has nonzero velocity everywhere and is an immersion. More generally, if $m<n$, no differential $T_{p}X\to T_{f(p)}Y$ can be surjective, so every input is critical. The terminology measures the ability to fill target directions, rather than whether the parametrized curve looks smooth.

For this cubic vector field, every non-axis trajectory has the first integral $I(x,y)=1/y-y/x^2$. Direct differentiation gives $\nabla I\cdot (\dot{x},\dot{y})=0$ where $xy\ne 0$. Equivalently the trajectories satisfy $x^2-y^2=C{x}^{2}y$. The axes are separate invariant trajectories, and the diagonals are the cases $C=0$. For each fixed finite $C$, approach to the origin forces $y^2/x^2=1-Cy\to 1$, so non-axis trajectories have diagonal limiting directions.

The origin is repelling in forward time. For $V=x^2+y^2$, $\dot{V}=2x^4+2x^2{y}^2+4y^4\ge V^2$, so every nonzero trajectory eventually exits any small disk. The zero Jacobian matrix at the origin leaves the linearized system constant, demonstrating why nonlinear terms are needed for this stability classification.

The Jacobian matrix of a deformation map carries an infinitesimal reference line element to its current image. Its determinant $J=\det F$ is the local volume ratio. An orientation-preserving deformation has $J>0$, and incompressibility imposes $J=1$.