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Stokes theorem converts circulation around an oriented boundary into curl flux through a spanning surface. For a sufficiently smooth vector field on an oriented piecewise smooth surface $S$,The boundary orientation follows the right-hand rule relative to $\mathbf{n}$. The field must be smooth on the surface, not merely on its boundary; a singularity crossing the surface cannot be ignored. For a fixed loop, applying the theorem to Faraday's differential law gives the integral flux law of induction.

The planar form is Green theorem. For a region $D$ with positively oriented boundary and sufficiently smooth $P,Q$,The outer boundary is counterclockwise and any holes have clockwise boundaries. For instance $P=-y/2,Q=x/2$ makes the area of $D$ equal to half the line integral of $xdy-ydx$; the unit circle gives area $\pi$.

A closely related use of “Green's theorems” in field theory is Green's identities, obtained by applying the divergence theorem to $\varphi \nabla \psi$:and, after subtracting the expression with $\varphi ,\psi$ exchanged,These relate interior sources to boundary data in electrostatics and other boundary-value problems. For example, a harmonic potential difference vanishing on a Dirichlet boundary condition has zero integral of its squared gradient by the first identity, establishing uniqueness. Both the planar circulation theorem and the field identities are included here to make the convention clear.

Expand the vector field about the reference trajectory:Over a sufficiently short interval, the tangent-linear propagator isConsequentlyIf $S=J+J^T$ has $S\mathbf{e}=\lambda \mathbf{e}$ and the perturbation is initially parallel to $\mathbf{e}$, its squared Euclidean norm obeysPositive $\lambda$ means instantaneous norm-squared growth, negative $\lambda$ decay, and zero $\lambda$ no first-order change. The corresponding norm growth rate is $\lambda /2$, not $\lambda$. These are eigenvalues of the symmetric part governing the chosen error metric, not generally the eigenvalues of the nonsymmetric Jacobian matrix or the long-time Lyapunov exponents. The direction and the reference Jacobian matrix evolve during a finite forecast.

Take two time derivatives at fixed reference position $\mathbf{X}$:The constraint $\det \mathbf{F}=1$ ensures that the deformation gradient is invertible, so $\mathbf{X}={\mathbf{F}}^{-1}(\mathbf{x}-\mathbf{c})$. Thus the spatial material acceleration is the affine vector fieldIts spatial gradient is $\mathbf{A}$, and its curl isPart (b)(iii) therefore requiresEquivalently, multiplying on the left by ${\mathbf{F}}^T$ and on the right by $\mathbf{F}$ givesThe translation $\mathbf{c}(t)$ contributes no spatial curl and is unrestricted by this condition. A useful equivalent first integral isthe antisymmetric matrix in parentheses must be constant, and need not vanish. The separate constraint $\det \mathbf{F}=1$ is still required; differentiating it also gives $\mathrm{tr}(\dot{\mathbf{F}}{\mathbf{F}}^{-1})=0$.

The condition is understood in the setting of part (b), where a static family already exists under the prescribed body force. For a homogeneous elastic deformation gradient, the elastic stress $\mathbf{S}(\mathbf{F})$ is spatially constant. Static balance of linear momentum consequently gives $\nabla p_0=\rho \mathbf{b}$, ensuring that $\mathbf{b}$ has a scalar potential. If the static-family assumption were dropped and an arbitrary time-independent body force allowed, the general necessary condition would instead be $\nabla \times \ddot{\mathbf{x}}=\nabla \times \mathbf{b}$.

The identity for the killed deterministic flow follows directly from the chain rule and the fundamental theorem of calculus, independently of the construction issue in part (b). For a continuously differentiable $f$, put $z(t)=\phi (s,x,t)$ and $H(t)={\int }_s^t\alpha (\phi (s,x,u))du$. ThenHence, on such test functions,The time subscript on $L_t^{\alpha }$ is necessary because the vector field may depend on time; it is suppressed on the right-hand side of the source's displayed operator identity. The killed deterministic flow transports an observable along the flow of a vector field and multiplies it by the no-jump survival probability.

Varying the vector field gives the Proca equation,Its divergence uses antisymmetry of $F^{\mu \nu }$ to cancel the first term. Since $m_A\ne 0$,For an on-shell momentum this imposes $p_{\nu }{ϵ}^{\nu }=0$ on the four polarization components, leaving three independent solutions. In the rest frame, $p=(m_A,0)$ makes ${ϵ}^0=0$, leaving precisely the three spatial polarizations of a massive spin-one particle.

There is no fourth independent oscillator hidden in $A^0$. Its canonical momentum is zero because the Lagrangian has no ${\dot{A}}^0$, and the temporal equation constrains $A^0$ in terms of the spatial data. Equivalently the primary and secondary second-class constraints remove two variables from the eight-dimensional phase space, leaving six phase-space variables, or three physical degrees of freedom.

For $r>0$, differentiate $r^2=x^2+y^2$ and use the two differential equations:Substituting $x=r\cos \theta$, $y=r\sin \theta$ gives the required equations in polar coordinates. The origin is handled in Cartesian coordinates, where it is an equilibrium point; an angular coordinate is not defined there.

A Lyapunov function near an equilibrium point is a continuously differentiable positive-definite scalar function $V$ whose derivative along solutions is nonpositive. This implies local Lyapunov stability under the usual comparison bounds; if its derivative is negative definite, the Second Lyapunov theorem gives local asymptotic stability. Here takeIf $a<f(0,0)$, continuity gives a small ball and $\delta >0$ such that $f(x,y)-a\ge \delta$ inside it. Consequently $\dot{V}\le -2\delta V$. A sufficiently small initial sublevel set stays inside the ball, and $V(t)\le V(0)e^{-2\delta t}$. The origin is asymptotically stable, in fact locally exponentially stable. If $a>f(0,0)$, continuity instead gives $\dot{V}\ge 2\delta V$ in a small ball. Every nonzero initial point there exits that ball in forward time, proving instability of the origin. These inequalities hold along all local solutions; continuity of $f$ alone need not supply uniqueness for a general vector field.

If $f=g(r)$, a nonzero constant radius $r_0$ satisfying $a=g(r_0)$ gives a periodic orbit of period $2\pi$, since $\dot{\theta }=1$. Linearizing the scalar radial equation there givesThus the periodic orbit is orbitally asymptotically stable for $g^{\prime }(r_0)>0$ and unstable for $g^{\prime }(r_0)<0$. The direction along the periodic orbit is neutral, as expected for an autonomous periodic orbit; orbital asymptotic stability means attraction in a dynamical system to the orbit with a possible phase shift.

For the specified radial response,The origin is therefore stable for $0<a<3/2$ and unstable for $a>3/2$. The nonzero-radius equation has roots of an equationretaining only positive radii. For $1<a<3/2$, the square root exceeds one, so only $r_{+}>1$ exists and $g^{\prime }(r_{+})<0$: the periodic orbit is unstable. For $3/2<a<2$, both radii are positive; $0<r_{-}<1$ gives $g^{\prime }(r_{-})>0$ and a stable periodic orbit, while $r_{+}>1$ remains unstable.

As $a$ increases through $3/2$, the stable small-radius periodic orbit emerges from the origin while the origin loses Lyapunov stability. This is the supercritical Hopf-type transition with a nonsmooth radial response requested. It follows directly from the explicit radial dynamics. There is a regularity qualification: $g(r)=3/2+r/2+O(r^2)$ near zero, so the vector field in Cartesian coordinates contains $x\sqrt{x^2+y^2}$ and $y\sqrt{x^2+y^2}$ terms. It is continuously differentiable but not generally twice differentiable at the origin, and the periodic orbit radius grows as $r_{-}\sim 2(a-3/2)$ rather than the square-root law of a generic smooth Hopf normal form. One should therefore use the explicit computation here rather than invoke a smooth Hopf bifurcation theorem with unmet hypotheses. At $a=3/2$ itself the origin is still weakly attracting, because $a-g(r)<0$ for small positive $r$.

Let $S(x,y,z)=(-x,-y,z)$. In the first two components of the vector field, replacing $x,y$ by their negatives changes the sign of the whole component; in the third, the product $xy$ is unchanged. Thus $F(Sv)=SF(v)$, establishing the symmetry, with solutions mapped to solutions by $S$.

At an equilibrium point, positivity of $a$ makes $y=x$. The remaining equations give $x(2c-a-z)=0$ and $z=x^2/b$. Hencewhere the two nonzero equilibrium points exist when $0<a<2c$. They merge with the origin at $a_p=2c$, in a pitchfork bifurcation.

At the origin the Jacobian matrix and its eigenvalues areAt $a=2c$ these are $0,-c,-b$. To see the local pitchfork bifurcation dynamics more explicitly put $\delta =2c-a$, $y=x+v$, and use $z=x^2/b+\cdots$ on the center manifold. The equations becomeAt $\delta =0$, invariance at cubic order gives $v=-x^3/(bc)+\cdots$, hence the reduced equation isIt has the supercritical pitchfork bifurcation form as $\delta$ increases, with the two branches $x\simeq \pm \sqrt{b\delta }$ attracting near onset while the origin becomes unstable.

For $0<a<2c$, the determinant of the planar block is $a(a-2c)<0$, so the origin has one positive and two negative eigenvalues and is a saddle equilibrium. At $a=c$, the planar eigenvalues are explicitly $\pm c$. For $a>2c$, the planar trace $c-a$ is negative and its determinant positive, so the origin is asymptotically stable. To distinguish its type putFor $2c<a<a_{\ast }$ both planar eigenvalues are real negative, giving a stable node. At $a=a_{\ast }$ they coincide at $-c/\sqrt{3}$; the planar block is not a scalar matrix, so it is a defective stable node. For $a>a_{\ast }$ they are nonreal with negative real part, giving a stable spiral in the planar block together with the stable third direction. This refines the requested node-versus-saddle classification.

The origin cannot have a Hopf bifurcation: a planar imaginary pair would require trace zero, namely $a=c$, but the determinant there is $-c^2<0$, so the pair is real and of opposite signs.

For either nonzero equilibrium point put $s=\pm \sqrt{b(2c-a)}$. The Jacobian matrix isIts characteristic polynomial isA nonzero imaginary root of an equation $i\omega$ requires, by separating real and imaginary parts, ${\omega }^2=B$ and $AB=C$. Equivalentlyso the candidate Hopf bifurcation values areAt such an admissible value the factorization is $p(\lambda )=(\lambda +A)({\lambda }^2+bc)$, giving the pair $\pm i\sqrt{bc}$ and a negative third eigenvalue $-A$, since $A=C/B>0$ on the nonzero branches.

The mathematical model parameter restrictions matter. A real candidate requires $b/c\le 17/8$, and the nonzero equilibrium points also require $0<a_h<2c$. The upper-sign value is admissible for $0<b/c<17/8$; the lower-sign value additionally requires $b/c>1$. For fixed $b,c$, differentiation of the characteristic equation at the imaginary root of an equation gives the eigenvalue crossing speedThus the transverse eigenvalue crossing condition holds except at the coincident-root case $b/c=17/8$, $a_h=3c/4$. At a transverse eigenvalue crossing, a generic nonzero first Lyapunov coefficient gives a Hopf bifurcation. The displayed radical is the spectral threshold; it does not justify an unconditional generic-Hopf claim for all positive mathematical model parameters, nor does it by itself exclude a degenerate Hopf bifurcation point. For example $b>17c/8$ yields no real candidate at all. The cubic Routh-Hurwitz stability criterion gives local Lyapunov stability on the side with $A>0$ and $AB-C>0$, where

For the modified equation, the planar block at the origin is $\left(\begin{array}{cc}-a& a\\ c& c\end{array}\right)$, with determinant $-2ac<0$ for the stipulated positive mathematical model parameters. Thus the origin is always a saddle equilibrium for nonzero $a,c$ in that mathematical model parameter range. The nonzero equilibrium points now satisfy $y=x$, $z=2c$ and $x^2=2bc$, so they areFor $b,c>0$ these branches stay a positive distance from the origin and do not merge with it as $a$ varies. The origin also has no zero eigenvalue for $a,c>0$. Therefore the pitchfork bifurcation is absent in the modified system.

Use $x=r\cos \theta$, $y=r\sin \theta$ with $r>0$. Dotting the vector field with the radial direction and taking its angular cross-product givesThe angular factor ranges from one to three. Hence every radial component is outward for $0<r<1/\sqrt{3}$, and every radial component is inward for $r>1$. The limiting radii areThere is a minor strictness point in the wording. At $r_1$ the outward inequality is non-strict, with tangencies where ${\sin }^2\theta =1$; at $r_2$ the inward inequality is non-strict, with tangencies where $\sin \theta =0$. Thus these are the supremum and infimum for strict crossing at all angles. The weak inequalities at the endpoints are exactly what is needed for the closed trapping annulus in the next part.

Put $P(N)=(a_1-N)(a_2-N)(a_3-N)$. The harvested vector field is $N[P(N)-E]$. Define the extinction threshold $E_T=P(0)=a_1{a}_2{a}_3$, and let $N_M\in (a_2,a_3)$ be the positive local maximum of $P$, with $E_M=P(N_M)>0$. Writing $s_1=a_1+a_2+a_3$, $s_2=a_1{a}_2+a_1{a}_3+a_2{a}_3$ givesPositive equilibria are intersections of $P(N)$ with the horizontal line $E$. A lower stable branch in $(0,a_1)$ exists exactly for $0<E<E_T$. A pair in $(a_2,a_3)$ exists exactly for $0<E<E_M$: its smaller member has $P^{\prime }>0$ and is unstable; its larger member has $P^{\prime }<0$ and is stable. At a positive root the linear eigenvalue is $N{P}^{\prime }(N)$, which proves these claims. Extinction has eigenvalue $E_T-E$, becoming stable for $E>E_T$.

The harvested bistable population thresholds determine the behaviour. If $E_M<E_T$, the upper stable population and its unstable separator first disappear in a saddle-node bifurcation. Populations then tend to the remaining lower stable state until that branch reaches extinction at $E_T$; beyond it extinction attracts every positive initial state. If $E_T<E_M$, the lower stable branch reaches extinction first. For $E_T<E<E_M$, extinction and the upper positive state are both stable, separated by the middle unstable root. Increasing effort through $E_M$ then destroys the remaining positive attractor, causing a sudden collapse to extinction. If $E_T=E_M$, both events occur at the same effort at different populations.

At $E=E_T$, near zero the leading term is $P^{\prime }(0)N^2<0$, so extinction attracts from the physical right side algebraically, despite its zero linear eigenvalue. At $E=E_M$, the double positive root is semistable: locally $\dot{N}\simeq -\frac{1}{2}{N}_M|P^{\prime \prime }(N_M)|(N-N_M{)}^2$, attracting from above and repelling from below. These critical cases matter for exact initial data. The figure shows examples of both strict threshold orderings.

A steady population $N_{\ast }$ is a root $h(N_{\ast })=0$. To linearize one needs $h$ differentiable near that root, an extra local regularity assumption beyond the source's continuity. Setting $N=N_{\ast }+\eta$ gives $\dot{\eta }=h^{\prime }(N_{\ast })\eta +o(\eta )$, hence $\eta (t)=\eta (0)e^{h^{\prime }(N_{\ast })t}$ at linear order.

Thus $h^{\prime }(N_{\ast })<0$ gives linear asymptotic stability and $h^{\prime }(N_{\ast })>0$ gives instability. If $h^{\prime }(N_{\ast })=0$, linearization is inconclusive. In a one-dimensional phase line, the more general nonlinear test is whether the vector field points towards the root from both sides. At the physical boundary $N=0$, only the right-hand side matters. Continuity alone does not guarantee uniqueness of trajectories, so a standard well-defined flow also presumes a local uniqueness condition, such as local Lipschitz continuity.

The planar divergence is $\nabla \cdot F=-(\Omega +x^2)<0$ for $\Omega >0$. Since the plane is simply connected, the Bendixson-Dulac theorem excludes any nonconstant periodic orbit: the flux across a closed trajectory would be zero because the vector field is tangent, whereas Green's theorem makes it the strictly negative integral of the divergence over the enclosed region.

Alternatively define $E=y^2/2-x^2/2+x^4/4$. Direct differentiation givesA nonconstant periodic solution would require the integral of this strictly dissipative expression over a period to vanish. That forces $y\equiv 0$ and hence an equilibrium, a contradiction. This energy argument identifies the damping mechanism behind the geometric criterion.

The Lorenz vector field has constant divergence $\nabla \cdot F=-\sigma -1-b$. Liouville formula for a fundamental matrix for the derivative of the flow therefore givesThis concerns transported three-dimensional phase volume, not the Euclidean distance between each pair of trajectories.

A quasiperiodic orbit with the usual smooth two-frequency torus closure would lie on an invariant embedded two-torus. That torus bounds a solid torus of positive finite volume in ${\mathbb{R}}^3$. Since the vector field is tangent to its boundary, its outward flux is zero, contradicting the divergence theorem, whose volume integral is $-(\sigma +1+b)V<0$. This negative-divergence obstruction to invariant tori excludes such quasiperiodic tori. The argument uses global negative divergence and the enclosed volume; saying only that an attracting torus has zero volume would not exclude it.

At an equilibrium the sum of eigenvalues is $-(\sigma +1+b)<0$, so they cannot all have positive real parts: no equilibrium is a fully repelling source. For a periodic orbit of period $T$, one Floquet multiplier equals one in the flow direction. The product of the other two is $e^{-(\sigma +1+b)T}<1$, so they cannot both have modulus greater than one: no periodic orbit is fully repelling either. Saddle equilibria and saddle periodic orbits remain possible; contraction of volume does not imply contraction in every direction.

Convexity of the components of a vector field is not the decisive criterion for a root-finding problem. Neither method becomes globally reliable merely because each component is convex.

Prefer the Newton root-finding iteration when the Jacobian matrix is available accurately and at reasonable cost. Solve $J({\mathbf{x}}_k){\mathbf{s}}_k=-\mathbf{f}({\mathbf{x}}_k)$ rather than explicitly forming an inverse matrix. Near a root with a nonsingular Jacobian matrix and a locally Lipschitz Jacobian matrix, the usual exact iteration has quadratic convergence.

Prefer Broyden's method when evaluating or differentiating the vector field makes a fresh Jacobian matrix expensive. It learns an approximate Jacobian matrix through secant updates using function differences. Under appropriate local regularity and initialization assumptions it has superlinear convergence, generally slower asymptotically than the quadratic convergence of exact Newton steps, but often with cheaper iterations.^[ref]

Dimension matters through storage and solves. A full dense Newton factorization costs order $d^3$ arithmetic operations; a dense Broyden approximation uses order $d^2$ storage, and maintained inverse or factorization updates can cost order $d^2$. A naive fresh dense solve can still cost order $d^3$. Sparsity can favour Newton, while very large $d$ can make either dense formulation unsuitable. Neither local rate guarantees convergence from an arbitrary starting point.

A line integral is path dependent when two admissible paths with the same starting and ending points give different values. This disproves path independence and, for a vector field, shows that it is not a conservative vector field on the relevant domain.

If a periodic vector field has a periodic potential of a conservative vector field, the line integral between points separated by a period vector must vanish, since it is the potential difference. A periodic gradient field need not satisfy this: the constant field $(1,2)$ on the plane has potential $x+2y$, but its integrals over the two unit translations are $1$ and $2$. On the periodic quotient these translation integrals are nonzero periods, obstructing a single-valued potential there.

Let $F$ be a continuously differentiable planar vector field with an isolated equilibrium point $p$, and let the twice continuously differentiable first integral $H$ satisfy $F\cdot \nabla H=0$ and $\nabla H(p)=0$. If the Hessian matrix $D^{2}H(p)$ is positive or negative definite, then $p$ is a centre equilibrium.

For a positive-definite matrix, sufficiently short rays from $p$ have strictly increasing $H$ with radial derivative bounded below by a positive constant times distance. Every sufficiently small positive level difference therefore meets each ray exactly once, and the implicit function theorem makes these intersections a simple closed curve. These regular level sets shrink to $p$; use $-H$ in the negative-definite case. The vector field is tangent and nonzero on each such compact curve. Its speed is bounded away from zero, so a traversal takes finite time and each curve is a periodic orbit. An arbitrary strict local extremum without these regularity and nondegeneracy conditions does not by itself supply this argument.

Every smooth closed differential form of positive degree is locally an exact differential form. On a coordinate ball centered at zero, put $F_t(x)=tx$ and let $R=x^i{\partial }_i$ be the radial vector field. Define the degree-lowering operatorFor a $p$-form with $p>0$, the integrand has a factor $t^{p-1}$ in its coordinate coefficients and is smooth at zero. Cartan's magic formula and differentiation of the pullback of a differential form give ${\partial }_t(F_t^{\ast }\alpha )=t^{-1}{F}_t^{\ast }(d{\iota }_R\alpha +{\iota }_{R}d\alpha )$. Integrating from zero to one gives $dK\alpha +Kd\alpha =\alpha$, since $F_0^{\ast }\alpha =0$ in positive degree. If $d\alpha =0$, take $\beta =K\alpha$. The local potential of a closed differential one-form gives the explicit degree-one construction. Global exactness can fail through nonzero periods, measured by de Rham cohomology.

Let $W$ consist of real-linear maps $\alpha$ from smooth vector fields to smooth functions such that $X_p=0$ implies $\alpha (X)(p)=0$. These are exactly evaluations by smooth differential one-forms. Define ${\omega }_p(v)=\alpha (X)(p)$ using any global field with $X_p=v$; the vanishing condition makes it well-defined. Bump-extended local frame vectors make its coefficients smooth. Also $\alpha (fX)(p)=f(p)\alpha (X)(p)$, so real linearity and pointwise vanishing automatically give $C^{\infty }$-linearity. The evaluation correspondence is a natural module isomorphism.

For the upper curved half of $x^2/a^2+y^2/b^2+z^2/c^2=1$, take the vector field $(x{y}^2,y^2,C)$ and outward unit normal. Its divergence is $y^2+2y$. The odd term integrates to zero over the half-ellipsoid. Scaling a unit half-ball gives $\int y^{2}dV=2\pi a{b}^{3}c/15$. The equatorial closing disk has downward flux $-C\pi ab$. The divergence theorem therefore gives the displayed curved-surface flux. The constant vertical component is not detected by the divergence, but is recovered by accounting for the closing disk.

A vector field $X$ is projectable through a smooth map between manifolds $f:M\to N$ if there is a smooth vector field $Y$ on $N$ with $d{f}_p{X}_p=Y_{f(p)}$ for every $p$. Agreement on fibres is necessary. For a surjective submersion it is also sufficient: smooth local sections of the submersion express $Y$ locally as $df(X)$ and prove its smoothness. An arbitrary non-surjective map may instead give a field only along its image, with extension to $N$ requiring additional choices.