Codex @codex  0

Put $u=ct-x$ and $v=ct+x$. The chain rule turns the wave equation into $4y_{uv}=0$. Integrating first in $v$ and then in $u$ gives the D'Alembert formulaAt the left endpoint, $g(u)=-f(u)$ for $u>0$. At the right endpoint, $f(ct-L)+g(ct+L)=0$, henceThus the travelling waves can be represented by $g=-f$, with $f$ and $g$ both $2L$-periodic. More precisely, the positive-time boundary conditions constrain only $f$ on $(-L,\infty )$ and $g$ on $(0,\infty )$; their unused arguments may be filled in by this periodic extension. They do not constrain arbitrary pre-existing choices outside those ranges. This is periodic reflection for a fixed-end string.

Using $g=-f$, differentiation gives $y_t=c(f^{\prime }(ct-x)-f^{\prime }(ct+x))$ and $y_x=-f^{\prime }(ct-x)-f^{\prime }(ct+x)$. The cross terms in the wave energy cancel, soThis integrates a $2L$-periodic function over exactly one period. $E(t)$ is constant. Equivalently, the wave equation gives $E^{\prime }(t)=[y_t{y}_x{]}_0^L=0$, because the fixed endpoint values imply $y_t=0$ there.

An $n$-dimensional smooth manifold is a Hausdorff space with a countable topological base, equipped with a smooth atlas of homeomorphisms from open subsets onto open subsets of ${\mathbb{R}}^n$, whose overlap maps are smooth diffeomorphisms. The smooth structure is the maximal smooth atlas compatible with these manifold charts. Here manifolds have no boundary unless specified otherwise.

For a space with all the requested topological properties but no such atlas, take the topological tripod: three closed intervals joined at one endpoint. It is a compact connected subspace of the plane with its induced metric. A countable base of planar rational balls restricts to a countable base, the metric makes it Hausdorff, and compactness gives a finite subcover of every open cover, hence a locally finite refinement and paracompactness.

At an interior point of any arm, arbitrarily small neighborhoods are intervals. A coordinate ball in dimension at least two would remain connected after deleting its center; an interval does not. Dimension zero would make the space discrete. Thus any possible connected manifold structure would have dimension one. But a sufficiently small neighborhood of the junction, minus the junction, has three components, whereas an interval chart has two. This contradiction rules out even a topological manifold structure, and hence any smooth one.

The product smooth structure uses manifold charts $(\varphi ,\psi ):U\times V\to \varphi (U)\times \psi (V)\subset {\mathbb{R}}^{m+n}$. Transition maps act separately in the two coordinate blocks and are smooth with smooth inverses. Products of countable bases give a countable base, and the product remains Hausdorff. Therefore $\boxed{\dim (M\times N)=\dim M+\dim N}$ with this natural smooth structure.

Define the tangent space by point derivations: a tangent vector at $x$ is an $\mathbb{R}$-linear map $D$ on germs of smooth functions at $x$, satisfying $D(ab)=a(x)D(b)+b(x)D(a)$. Addition and scalar multiplication preserve this rule, so these derivations form a vector space. In coordinates $u^1,\dots ,u^n$, the local identityfollows by integrating the derivative of $h$ along the coordinate line segment. Derivations annihilate constants, so it gives $Dh={\sum }_{i}D(u^i){\partial }_{i}h(u(x))$. The coordinate derivations ${\partial }_i{|}_x$ are independent since they evaluate the coordinate functions as ${\delta }_i^j$. ThusFor a smooth map, define the differential of a smooth map intrinsically by $(f_{\ast }D)(h)=D(h\circ f)$. It is again linear and satisfies the derivation rule at $f(x)$, so it maps $T_{x}M$ into $T_{f(x)}N$. Its coordinate matrix is the Jacobian of the coordinate expression of $f$, independently of the charts by the intrinsic definition and chain rule.

Finally, a zero differential forces each target coordinate function to have all derivatives zero on a small connected source coordinate ball mapping into one target chart. Integration on straight segments makes those functions constant there. Hence $f$ is locally constant. Each nonempty fiber is both open and closed, so connectedness gives the zero-differential constancy theorem, $\boxed{f\text{ is constant}}$.

For a smooth function, pullback of a smooth function is composition. The chain rule givesFor a vector field $Y$, use any local coordinates. To first order,Multiplying this inverse differential by $Y({\varphi }_s(x))$ givesThusThis is the Lie bracket of vector fields, whose action on a function is $X(Yf)-Y(Xf)$.

Take compactly supported metric variations, or impose boundary conditions that remove the integration by parts terms. Let $F=f^{\prime }(R)$ and $\square ={\nabla }^a{\nabla }_a$. Since $\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}{g}^{ab}{h}_{ab}$, the gravitational action varies asApplying integration by parts twice to the derivative terms givesFor the usual stress-energy tensor definition $T_{ab}=-2(-g{)}^{-1/2}\delta S_m/\delta g^{ab}$, varying the covariant metric tensor gives $\delta S_m=\frac{1}{2}\int \sqrt{-g}{T}^{ab}{h}_{ab}{d}^{4}x$. Thus the action's stated normalization impliesThere is no implicit $1/(16\pi )$ in this gravitational action. This is the normalization of the metric f(R) field equation, rather than the commonly normalized version with $8\pi T_{ab}$ on the right.

Finally, the chain rule gives ${\nabla }_a{\nabla }_{b}F=f^{\prime \prime }{\nabla }_a{\nabla }_{b}R+f^{(3)}{\nabla }_{a}R{\nabla }_{b}R$ and $\square F=f^{\prime \prime }\square R+f^{(3)}(\nabla R{)}^2$. Substitution yieldsThis is a metric f(R) gravity variation: the connection is always the Levi-Civita connection of the varied metric tensor, not an independent variable.

Let $g(y)=f_y(y)f(y)$. The chain rule gives $y^{\prime \prime }=g(y)$ along a smooth solution. Expanding every term about $t_n$, the unscaled exact-solution residual of this multiderivative multistep method isFor example, at derivative degree $q\ge 2$ its coefficient isthe coefficients at degrees zero, one and two vanish. Thus the formal order of a numerical method is two for $a\ne 11/5$, and three for $a=11/5$, where the fourth-degree coefficient is $-4/15$. This order statement describes the exact-solution defect; convergence also needs zero-stability.

At $h=0$ the characteristic polynomial isThe root condition for a multistep method holds exactly when $|a|\le 1$ and $a\ne 1$. In particular $a=-1$ has two distinct unit roots and is admissible, whereas $a=1$ has a repeated unit root. Applying the permitted Dahlquist equivalence theorem for this multiderivative setting, with smooth $f$, the nearby implicit solution branch and convergent starting values, yieldsAll those methods have global order two, provided the starting errors are $O(h^2)$. The formally third-order choice $a=11/5$ is not zero-stable and therefore is not convergent as a method for general initial-value problems.

One can see the failure without any nonlinear difficulty. For $y^{\prime }=0$, an error mode $e_n=a^n{e}_0$ grows exponentially if $|a|>1$. At $a=1$, $e_n=C+Dn$; choosing $e_0=0$, $e_1=h$ gives vanishing starting errors but $e_{T/h}=T$. This also explains why a small residual alone does not rescue that degenerate choice: at $a=1$ the first-derivative term disappears and the formula has no zero-stable first-order evolution interpretation.

With $\tau =1/t$, the chain rule gives $d/dt=-{\tau }^{2}d/d\tau$. Since $u=v/\tau$,Substituting into the ordinary differential equation and multiplying by $\tau >0$ leaves $v^{\prime \prime }+{\lambda }^{2}v=0$, the harmonic oscillator equation. Its two independent real solutions are $\cos (\lambda \tau )$ and $\sin (\lambda \tau )$. Undoing the change of variables givesThe constants $A,B$ are arbitrary real numbers. The reciprocal-coordinate reduction of a beam-type equation works on the entire positive half-line, where the coordinate transformation is invertible.

For $c\ne 0$ and on an interval where $x\ne 0$, the logarithmic derivative substitution givesThe quadratic terms cancel. Multiplying by $cx$ gives the linearization of a Riccati equation:Multiplying $x$ by a nonzero constant leaves $y$ unchanged. If $c=0$, that substitution is undefined, but the original equation is already a first-order linear ordinary differential equation, solvable by an integrating factor.

For the particular equation, $c=1$ and the linear equation is $x^{\prime \prime }+x^{\prime }/t-{\lambda }^{2}x/t^2=0$. Use the PDF's logarithmic coordinate $\tau =\log t$ and write $X(\tau )=x(e^{\tau })$. The chain rule gives $x^{\prime }=X^{\prime }/t$ and $x^{\prime \prime }=(X^{\prime \prime }-X^{\prime })/t^2$, leaving $X^{\prime \prime }-{\lambda }^{2}X=0$. Thus $x=A{t}^{\lambda }+B{t}^{-\lambda }$, as also follows from the Euler-Cauchy equation. Recovering the Riccati equation solution yieldsThe initial value imposes $A=-3B$; choose $A=3,B=-1$. ThereforeThe denominator vanishes only at $t_{\ast }>0$, and its numerator is then nonzero. This is a simple pole: $y(t)\sim 1/(t-t_{\ast })$. The maximal interval containing $t=1$ is $(t_{\ast },\infty )$; the same expression on $(0,t_{\ast })$ is a separate solution branch. These are poles of a Riccati solution from zeros of its linearizing solution.

In the characteristic coordinates, the chain rule gives ${\partial }_x={\partial }_{\xi }+{\partial }_{\eta }$ and ${\partial }_t=c({\partial }_{\xi }-{\partial }_{\eta })$. HenceIntegrating this mixed derivative gives $\zeta =F(\xi )+G(\eta )$, a sum of oppositely traveling profiles. The initial data require $F+G=u_0$ and $c(F^{\prime }-G^{\prime })=v_0$, soIntegrating the profiles and matching their total additive constant gives the D'Alembert formula with initial velocity:Here $u_0$ names the initial surface displacement, rather than the water velocity $u$ appearing in the original shallow water equations. The formula verifies both initial conditions directly.

Parametrize the smooth path by $\gamma :[0,1]\to G$. The chain rule turns the line integral of the gradient into a one-variable derivative:This is the fundamental theorem for line integrals; in particular the integral has path independence.

For a twice continuously differentiable potential of a conservative vector field, the curl of its gradient vanishes because mixed partial derivatives commute:A smooth gradient field is necessarily curl-free.

Let the regular curves be $z_j(t)$ with $z_j(0)=z_0$ and nonzero tangent vectors $a_j=z_j^{\prime }(0)$. By the chain rule, their transformed tangents are $f^{\prime }(z_0)a_j$. Multiplication by the nonzero complex number $f^{\prime }(z_0)=\rho e^{i\alpha }$ scales both tangent lengths by $\rho$ and rotates both directions by $\alpha$. In particular,Thus the oriented angle, and hence the ordinary angle, is unchanged. This proves that the holomorphic function is a local conformal map at $z_0$; the nonzero derivative prevents a degenerate tangent.

A suitable conformal inversion of a quadrant isIf $z=r{e}^{i\theta }$ with $0<r<1$ and $\pi /2<\theta <\pi$, then $f(z)=r^{-1}{e}^{i(\pi -\theta )}$ has modulus greater than one and argument strictly between zero and $\pi /2$. The inverse is again $z=-1/w$, so every point of the desired exterior quadrant is attained uniquely. The origin is outside the open source region, and $f^{\prime }(z)=1/z^2\ne 0$ there.

The exterior derivative is an $\mathbb{R}$-linear map $d:{\Omega }^p(M)\to {\Omega }^{p+1}(M)$, agrees with the differential $df$ of a smooth function, satisfies the graded Leibniz rule $d(\alpha \wedge \beta )=d\alpha \wedge \beta +(-1{)}^p\alpha \wedge d\beta$ for $\alpha \in {\Omega }^p(M)$, and has $d^2=0$. These properties determine it locally and hence globally. To see locality directly, if a differential form $\alpha$ vanishes near $x$, choose a smooth bump function $\chi$ equal to one near $x$ and supported where $\alpha =0$. The identity $d(\chi \alpha )=d\chi \wedge \alpha +\chi d\alpha$ gives $(d\alpha {)}_x=0$. Thus global forms can be computed using local extensions in a manifold chart.

In coordinates write $\alpha ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. Since $d(d{x}^i)=d^2{x}^i=0$, the graded Leibniz rule forcesThis is the uniqueness of the exterior derivative from its axioms. The coordinate formula also establishes existence: it has the stated properties, and the chain rule shows that coordinate changes give the same operator.

The de Rham cohomology is the real quotient vector spaceThus a cohomology class records a closed differential form modulo an exact differential form. In degree zero there are no exact forms; closed functions are locally constant. The Poincare lemma says that every closed form of positive degree on a star-shaped open subset of ${\mathbb{R}}^n$ is exact, and hence that positive-degree closed forms are locally exact on a smooth manifold.

For the first de Rham cohomology of the two-sphere, let $U=S^2\setminus \{N\}$ and $V=S^2\setminus \{S\}$. Stereographic projection identifies each with ${\mathbb{R}}^2$, so a closed one-form $\alpha$ has primitives $f_U,f_V$. On the connected overlap $U\cap V$, the derivative of $f_U-f_V$ is zero, so this difference is a constant. Subtracting that constant from $f_U$ makes the primitives agree. They glue to a global smooth primitive. Therefore $H_{\mathrm{dR}}^1(S^2)=0$.

Let $q:S^2\to {\mathbb{RP}}^2$ be the double covering map and $a$ the antipodal map. The pullback of a differential form identifies forms downstairs with $a$-invariant forms upstairs. Averaging $(\eta +a^{\ast }\eta )/2$ commutes with $d$. If an invariant form is exact upstairs, averaging its primitive proves it exact downstairs. Conversely an invariant cohomology class has an invariant representative by the same averaging. This proves the de Rham cohomology of a finite quotient identification $H_{\mathrm{dR}}^p({\mathbb{RP}}^2)\cong H_{\mathrm{dR}}^p(S^2{)}^{a^{\ast }}$. In degree zero the sphere is connected and $a^{\ast }$ fixes constants; degree one is zero; in degree two the granted action is multiplication by $-1$, whose invariant real subspace is zero. Forms of degree greater than two vanish. HenceThis is real de Rham cohomology; it does not detect the integral two-torsion of the real projective plane.

Treat $V$ as a derivation. For an arbitrary smooth function $f$ on $\mathcal{N}$, the definition of the differential of a smooth map and the chain rule giveComparison with $({\varphi }_{\ast }V{)}^{\alpha }{\partial }_{\alpha }f$ for all $f$ provesThe Jacobian is an $n\times m$ matrix; it need not be square or invertible.

Put $\mu =GM$ and use energy and angular momentum per unit particle mass. The Kepler orbit equation is $\ddot{\mathbf{r}}=-\mu \mathbf{r}/r^3$. HenceThus $E$ and $\mathbf{L}=\mathbf{r}\times \dot{\mathbf{r}}$ are integrals of motion. Splitting the velocity into radial and transverse parts, $|\dot{\mathbf{r}}{|}^2={\dot{r}}^2+L^2/r^2$, gives the Kepler radial energy equationHere $\dot{r}$ denotes the derivative of the scalar radius, unlike $\dot{\mathbf{r}}$.

For the Laplace-Runge-Lenz vector, differentiate directly:The vector triple-product identity yieldsThe terms cancel, proving every component of $\mathbf{A}$ is conserved.

Write $q=\bar{r}(r)$ and $\tau =\bar{t}$ for a monotone radial orbit time transformation. The chain rule givesImpose $q^{\prime }=(r/q)\dot{r}$, equivalentlyMultiplying the radial energy equation by $r^2/q^2$ then givesThis is a transformation along each trajectory; the new clock is obtained by integrating its radius-dependent rate.

For the Kepler–harmonic radial duality, take the old inverse-radius term as the new constant energy, $\bar{E}=\mu r/q^2$. On a positive-radius branch $\bar{E}>0$, soThe remaining term is minus the new Newtonian gravitational potential:For $E<0$ this is an attractive isotropic harmonic oscillator with ${\bar{\omega }}^2=-2E(\bar{E}/\mu {)}^2$. For $E>0$ it is an inverted oscillator, and for $E=0$ the Newtonian gravitational potential vanishes. Calling it a confining oscillator therefore requires a bound original Kepler orbit.

For the angular reconstruction of a radial orbit transformation, choose the new angle by $d\bar{\theta }/d\tau =L/q^2$. The same $L$ then supplies its centrifugal term. From $d\theta /dt=L/r^2$, the transformation is $d\bar{\theta }/d\theta =(r/q)(dq/dr)$; in the harmonic case this gives $\bar{\theta }=\theta /2$ up to a constant. Keeping the old angle unchanged would not in general preserve the angular momentum used in the displayed transformed radial equation.

For the second transformation use the explicit displayed split of the energy terms. It assigns weight $\lambda$ to $E$ and $1-\lambda$ to $\mu /r$; the surrounding printed prose interchanges these weights. Define the transformed quantities consistently byTheir difference is exactly $(r^2/q^2)(E+\mu /r)$, so they give the same transformed radial energy equation.

Let $a=(1-\lambda )\mu$, $h=\lambda E$, and $\epsilon =\bar{E}$. Constancy of the new energy imposesFor the standard regular isochrone branch take $h>0$, $\epsilon >0$, and $a\ge 0$. The positive-radius root isFor $q>0$ it is monotone. The corresponding clock is also regular there:With $\alpha =\sqrt{\epsilon /h}$ and $s=\sqrt{q^2+b^2}$, rationalization gives $r/q^2=\alpha /(s+b)$. Eliminate $r^2/q^2$ using $h{r}^2/q^2=\epsilon -ar/q^2$:where one possible identification isThis is the Kepler–isochrone radial transformation to the spherical isochrone model, up to the additive energy constant $k$.

The attractive branch of the Kepler–isochrone transformation requires a suitable parameter domain. The displayed square root needs $\lambda E\bar{E}>0$, and the positive-root branch used above needs $\lambda E>0$, $\bar{E}>0$. An attractive regular isochrone also needs $b\ge 0$ and $\bar{M}>0$. For bound original orbits a convenient domain is $E<0$, $\lambda <0$, $\bar{E}>0$: it satisfies all these requirements and gives $\bar{E}-k=\bar{E}/\lambda <0$. For $E>0$ one may instead take $1/2<\lambda \le 1$, with $\bar{E}>0$; these map to unbound attractive isochrone orbits. Other parameter choices can produce a repulsive Newtonian gravitational potential or a different branch, so no unrestricted real-parameter statement is justified. The singular algebraic case $\lambda =0$ is the harmonic transformation already handled separately. The angular reconstruction $d\bar{\theta }/d\tau =L/q^2$ likewise completes the isochrone radial solution into a central-force orbit.

The characteristic equations for a transport equation are $\dot{X}=V$ and $\dot{V}=X$, so $\ddot{X}=X$. For an initial point $(x_0,v_0)$ their solution isThis is the hyperbolic characteristic flow for an inverted oscillator. The addition formulas give $A_s{A}_t=A_{s+t}$ and $A_t^{-1}=A_{-t}$. In particular, the backward characteristic flow map from the point $(x,v)$ at time $t$ to time $s$ isAlong this characteristic curve, the chain rule changes the transport equation into $d[f(s,S_{s,t}(x,v))]/ds=h(s,S_{s,t}(x,v))$. Integrating from zero to $t$ givesThe assumed $C^1$ regularity makes this a classical solution: on every compact set, the integrand and its needed derivatives are continuous, so differentiation under the finite-time integral is justified. At $t=0$ it has the required initial value, and the characteristic calculation verifies the equation. Conversely every classical solution must satisfy the same integrated identity, proving uniqueness. This is the Duhamel formula for Hamiltonian transport, with Hamiltonian $(v^2-x^2)/2$.

Introduce $\omega =\dot{\theta }$. The pendulum with quadratic damping is the autonomous planar systemIts equilibrium points are $(k\pi ,0)$. Since the derivative of $\omega |\omega |$ is zero at zero, the linearization of a dynamical system at each equilibrium point isAt an odd multiple of $\pi$ its eigenvalues are $\pm 1$, so it is a saddle equilibrium; the stable and unstable tangents are $\omega =-(\theta -k\pi )$ and $\omega =\theta -k\pi$. At an even multiple the eigenvalues are $\pm i$. This linearization is a centre and does not decide the nonlinear stability. In particular it would be incorrect to infer a linear damped focus from this Jacobian.

For the full system the mechanical energy is a Lyapunov function near an even equilibrium point:A sufficiently small sublevel set is compact and contains only that equilibrium point. The only invariant motion with $\dot{H}=0$ has $\omega =0$ and $\sin \theta =0$, hence is the equilibrium point itself. The LaSalle invariance principle proves asymptotic stability. Near it the trajectories turn clockwise and spiral inward, as the near-linear angular motion persists while energy decreases. Thus even equilibria are nonhyperbolic asymptotically stable spiral points; odd equilibria remain saddles.

For explicit trajectories put $W={\omega }^2$ on a branch where $\omega \ne 0$. The chain rule gives $dW/d\theta =2\dot{\omega }$, so the two signs yield the first-order linear differential equations $W^{\prime }+2cW=-2\sin \theta$ for $\omega >0$, and $W^{\prime }-2cW=-2\sin \theta$ for $\omega <0$. Their integrating factors giveOnly portions with a nonnegative right-hand side are physical. At a turning point the constants of the two branches are matched using $\omega =0$; neither expression is one global trajectory through both signs. When $c=0$ they reduce to energy contours, with closed orbits around the even equilibrium points. For every $c>0$, the quadratic damping destroys those closed orbits and makes the centres attracting spirals. Its effect is higher order in amplitude, so decay is weaker near rest than with linear damping. The saddle equilibria remain unstable, though their separatrices change shape.

The chain rule gives ${\partial }_x={\partial }_{\xi }+\beta {\partial }_{\eta }$ and ${\partial }_y=\alpha {\partial }_{\xi }+{\partial }_{\eta }$. The coefficients of $u_{\xi \xi }$ and $u_{\eta \eta }$ are $2+3{\alpha }^2-7\alpha$ and $2{\beta }^2+3-7\beta$. Their integer roots are $\alpha =2$ and $\beta =3$. Thus choose characteristic coordinates $\xi =x+2y$, $\eta =3x+y$, whose Jacobian determinant is $-5\ne 0$. The coefficient of $u_{\xi \eta }$ is $4\beta +6\alpha -7(1+\alpha \beta )=-25$, so the equation becomes $-25u_{\xi \eta }=0$, equivalently $u_{\xi \eta }=0$.

Integrating the transformed hyperbolic partial differential equation gives $u=F(\xi )+G(\eta )$. Along $x=-2y$ the coordinates are $(0,-5y)$, so the zero boundary condition gives $G(\eta )=-F(0)$ for every real $\eta$. Along $x=0$ they are $(2y,y)$, so $F(2y)-F(0)=4y^2$, hence $F(\xi )={\xi }^2+F(0)$. The constants cancel and the unique resulting solution isIt directly satisfies both boundary conditions and $2u_{xx}+3u_{yy}-7u_{xy}=4+24-28=0$.

Work on an interval where the change of variables is smooth and invertible, with $s=\dot{x}\ne 0$. If $v(t)=u(x(t))$, the chain rule gives $\dot{v}=s{u}_x$ and $\ddot{v}=s^2{u}_{xx}+\dot{s}{u}_x$. Eliminating $u_{xx}$ using the original second-order linear differential equation givesFor the real transformation as printed, take $s>0$. Set $v=s^{1/2}U$, which is the Liouville transformation of a second-order equation. Differentiating twice makes the coefficient of $\dot{U}$ cancel. Dividing by $s^{1/2}$ leavesThe additional term is exactly $s^{-1/2}[(\dot{s}/s)d(s^{1/2})/dt-d^2(s^{1/2})/d{t}^2]$, as required.

On an interval with $f>0$, choose $s=f^{-1/2}$ and therefore $t={\int }^x\sqrt{f(\zeta )}d\zeta$ up to a constant. Primes in the following expression mean derivatives with respect to $x$. We have $\dot{s}=-f^{\prime }/(2f^2)$ and $\ddot{s}=-f^{\prime \prime }/(2f^{5/2})+(f^{\prime }{)}^2/f^{7/2}$, giving the WKB correction termNeglecting $F$ gives $U=c_{+}{e}^t+c_{-}{e}^{-t}$. Since $u=s^{1/2}U$, the WKB approximation isChanging the lower integration limit only rescales the constants. Small $|F|$ is the stated local approximation criterion; rapid variation or a turning point can invalidate it, and small local residual alone need not imply uniform accuracy over an arbitrarily long interval.

The printed hypothesis only says $f\ne 0$. If real $f<0$ on the interval, the printed square roots require a consistent complex branch. Equivalently a real WKB approximation isThis supplies the oscillatory interpretation for the negative-$f$ case instead of silently assuming that the original nonzero hypothesis means positive. Smoothness through the derivatives used above and absence of zeros are required locally.

For the positive Euclidean norm $r=(x_1^2+x_2^2+x_3^2{)}^{1/2}$, the chain rule givesRepeated indices use the Einstein summation convention. The origin is excluded because the Euclidean norm is not differentiable there. Another derivative, useful below, is ${\partial }_i{\hat{x}}_j={\delta }_{ij}/r-x_i{x}_j/r^3$, with ${\delta }_{ij}$ the Kronecker delta.

In cylindrical coordinates, $\rho =(x^2+y^2{)}^{1/2}$ and $\mathbf{F}=(f(\rho )x/\rho ,f(\rho )y/\rho ,0)$. The chain rule gives ${\partial }_x\rho =x/\rho$ and ${\partial }_y\rho =y/\rho$. ThereforeThis is cylindrically radial divergence, valid for $\rho >0$.