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Purely imaginary nonzero eigenvalues make a real planar linearization a centre equilibrium. They do not control the nonlinear radial drift. For example,has the same linearization as a uniform rotation. In polar coordinates, however, $\dot{r}=-r^3$ and $\dot{\theta }=1$. Every sufficiently small nonzero orbit spirals towards the origin, so the origin is an asymptotically stable weak focus, and no nonzero orbit is periodic. Reversing the signs of the cubic terms gives a repelling weak focus. Removing them gives a centre. Thus a nonlinear centre equilibrium requires a further argument, such as a suitable first integral; the imaginary pair alone is insufficient.

Write $a=D/2$ and wavenumber $k=2\pi /\lambda$. The pupil Fourier transform, normalized by pupil area, gives the Fraunhofer diffraction wave amplitude. In pupil polar coordinates,The angular average is $J_0$ because $J_0(z)=(2\pi {)}^{-1}{\int }_0^{2\pi }{e}^{-iz\cos \varphi }d\varphi$. The Bessel function of the first kind identity $d\{z{J}_1(z)\}/dz=z{J}_0(z)$ then reduces the radial integral to $A(u)=2J_1(u)/u$. Taking the squared magnitude yields the Airy disk light intensity profile.^[ref]

The apparent division by zero is removable. The power seriesgives $A(u)=1-u^2/8+u^4/192+O(u^6)$ and $P(u)=1-u^2/4+5u^4/192+O(u^6)$. At the centre, every pupil contribution has the same wave phase, so their normalized sum is one. The first dark ring instead corresponds to $J_1(u)=0$ with $u>0$, rather than to the removable zero of $J_1$ at the origin.

For an annulus $a<r<1$ with $0<a<1$, these factors interpolate the two boundary coefficients of a Fourier harmonic of positive integer order $n$ in a harmonic function. They obey $H_n(1)=1$, $H_n(a)=0$, $K_n(1)=0$, $K_n(a)=1$. Thus a mode with outer coefficient $v$ and inner coefficient $w$ has radial factor $v{H}_n+w{K}_n$.

To derive them, separation of variables in Laplace's equation in polar coordinates gives $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$, whose solutions are $r^n,r^{-n}$. Imposing the two Dirichlet boundary conditions on $R=\alpha r^n+\beta r^{-n}$ gives $\alpha +\beta =v$ and $a^n\alpha +a^{-n}\beta =w$. Solving yields $\alpha =(v-a^{n}w)/(1-a^{2n})$ and $\beta =a^n(w-a^{n}v)/(1-a^{2n})$, proving the formula. For the zero angular mode, the analogous factor is $v+(w-v)\log r/\log a$, since its radial solutions are $1$ and the logarithm $\log r$.

For square-integrable periodic boundary data and the Fourier coefficients used in the Poisson kernel for the unit disk, the displayed identity gives the squared area norm of its Poisson extension $u$. At fixed radius $r$, Parseval's identity givesMultiply by the polar coordinates area element factor $rdr$ and integrate from zero to one. The constant term contributes $\pi |a_0{|}^2/4$, and ${\int }_0^1{r}^{2n+1}dr=1/[2(n+1)]$. Interchanging the nonnegative sum and radial integral is justified by the monotone convergence theorem. The resulting factor $1/(n+1)$ shows how higher angular frequencies contribute less to the interior area norm.

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]

For the displayed metric tensor, its inverse has components $g^{rr}=e^{-2\chi }$ and $g^{\theta \theta }=e^{-2\chi }/r^2$. The nonzero second-kind Christoffel symbols are ${\Gamma }^r{}_{rr}={\chi }^{\prime }$, ${\Gamma }^r{}_{\theta \theta }=-r-r^2{\chi }^{\prime }$ and ${\Gamma }^{\theta }{}_{r\theta }={\Gamma }^{\theta }{}_{\theta r}=1/r+{\chi }^{\prime }$. They follow by differentiating the two diagonal metric entries and raising the first index of the first-kind connection coefficients. Polar coordinates require treating $r=0$ separately.

A circular sector is the portion of a disc between two radii and the connecting circle arc. If its radius is $R$ and its central angle is $\theta$ radians, its area isIndeed, polar coordinates give $A={\int }_0^{\theta }{\int }_0^{R}rdrd\varphi$. The full disc corresponds to $\theta =2\pi$, a semicircle to $\theta =\pi$, and a quarter-disc to $\theta =\pi /2$. In a solid with horizontal sector cross-sections, radius $R(z)$ and angle $\theta (z)$, its volume is $\frac{1}{2}\int \theta (z)R(z{)}^{2}dz$. Holding the angles and heights fixed while scaling every radius by a factor $s$ scales that volume by $s^2$.

For independent $N(\mu ,{\mu }^2)$ observations with $\mu >0$, the sufficient statistics are their sum and squared norm. Condition on $a=\sqrt{n}\Vert y\Vert /{\sum }_i{y}_i>0$ and write $v=\hat{\mu }=\Vert y\Vert /(q\sqrt{n})$, where $q=(\sqrt{1+4a^2}+1)/(2a)$. Polar coordinates supply radial measure $\Vert y{\Vert }^{n-1}d\Vert y\Vert$, while the conditional angular factor does not depend on $\mu$. This proves the displayed shape. Its normalization depends on $a,n$ but not $\mu$ after the substitution $v=\mu z$, and agrees with the normalized P-star approximation.

For an isolated fluorescent molecule at displacement $\mathbf{r}$ from the scanning optical focus, write $E(\mathbf{r})$ for relative excitation light intensity and $D_a(\mathbf{r})$ for the fraction of its collected emission passed by a centred detection pinhole of radius $a$. The expected response, apart from a common brightness and collection factor, isHere $\mathbf{u}$ is position in the detection-pinhole plane and $p_{\mathrm{em}}$ is the emitted light intensity distribution there conditional on collection, with units of inverse area. This point spread function normalization retains throughput: $D_a$ is a fraction, not a curve rescaled to have peak one. The product expresses excitation followed by detection; it is an intensity/probability product, not multiplication of coherent wave amplitudes from different molecules. ^[ref]

Assume linear, unsaturated fluorescence, independent emitters whose light intensities add, a stationary specimen, fixed collection efficiency and quantum efficiency, and negligible changes from scattering, absorption or bleaching during a scan. Different excitation and emission wavelengths generally give different excitation and detection responses. For a sufficiently small pinhole, $D_a$ is approximately its area times $p_{\mathrm{em}}$ at the centre; this is the familiar excitation-times-detection point spread function. It is not universally the square of one point spread function. A spatially invariant response further allows the image to be expressed as a convolution with the emitter distribution; depth-dependent optical aberrations can invalidate that approximation.

For an explicitly synthetic axial example, put $t=z/z_0$ and choose an on-axis excitation $E(t)=1/(1+t^2)$, inspired by a Gaussian beam. Stipulate a circular Gaussian function for the detection-plane light intensity, with standard deviation $s(t)=s_0\sqrt{1+t^2}$. In dimensionless coordinates $\mathbf{v}=\mathbf{u}/s_0$, put $\alpha =a/s_0$. ThenThe expression for $D_{\alpha }$ follows by integrating $p$ over a disk using polar coordinates. The assumptions specify a toy model with matched axial scales; its parameters are not a calibration for a real objective optical lens. It omits Airy disk rings, vector diffraction, and unequal excitation and emission wavelengths.

For a vanishingly small pinhole, the peak-normalized synthetic response tends to $(1+t^2{)}^{-2}$, whose full width at half maximum is $2\sqrt{\sqrt{2}-1}{z}_0\simeq 1.2872z_0$. With all collected emission accepted it tends to $(1+t^2{)}^{-1}$, of width $2z_0$. Meanwhile $H_{\alpha }(0)=1-e^{-{\alpha }^2/2}\to 0$ as the pinhole closes. The narrower normalized curve therefore accompanies a vanishing detected signal; peak normalization must not conceal that loss.

For positive twice differentiable $f$, the only nonzero Christoffel symbols are ${\Gamma }_{yy}^x=-f{f}^{\prime }$ and ${\Gamma }_{xy}^y={\Gamma }_{yx}^y=f^{\prime }/f$. In the convention $R^d{}_{abc}={\partial }_b{\Gamma }_{ac}^d-{\partial }_c{\Gamma }_{ab}^d+{\Gamma }_{ac}^e{\Gamma }_{eb}^d-{\Gamma }_{ab}^e{\Gamma }_{ec}^d$, substitution gives $R^x{}_{yxy}=-f{f}^{\prime \prime }$. Dividing $R_{xyxy}$ by $g_{xx}{g}_{yy}-g_{xy}^2=f^2$ gives the Gaussian curvature $K=-f^{\prime \prime }/f$. Thus $f=x$ gives the flat metric in polar coordinates away from its axis, $f=\sin x$ gives unit positive curvature, and $f=\sinh x$ gives unit negative curvature.

The parameter $e\ge 0$ in a focus-centred polar coordinates equation $r=\ell /(1-e\cos \theta )$ for a conic section. It distinguishes an ellipse ($e<1$), a parabola ($e=1$), and a hyperbola ($e>1$).

For a noncircular conic section, the eccentricity of a conic is the constant ratiowhere $F$ is a focus of a conic, $L$ the corresponding directrix of a conic, and $X$ a point on the curve. For an ellipse with semi-major axis $a$ and distance $c$ from the focal midpoint to either focus of a conic, $e=c/a=\sqrt{1-b^2/a^2}$. Its circle case has $e=0$; a finite directrix does not give that circle through the displayed ratio. The circle is the limit as $c\to 0$ and its directrices of a conic move arbitrarily far away.

The eccentricity of a conic is intrinsic to the cutting plane, and is not generally its slope in a chosen three-dimensional drawing. For the family $z=1+mx$ in conic section, with $0<m<1$, completing the square givesA displacement $dx$ along this plane has Euclidean norm $\sqrt{1+m^2}|dx|$. Thus the intrinsic semiaxis ratio satisfies $b^2/a^2=(1-m^2)/(1+m^2)$, givingFor $m=3/5$, this is $3/\sqrt{17}$, rather than $3/5$. This normalization separates the curve's geometry from a distorted projection of it onto the $xy$ plane.

A fundamental Gaussian beam is an ideal beam of light whose transverse wave amplitude has a Gaussian function profile. Here consider a circularly symmetric beam in vacuum, with propagation along $z$, two transverse coordinates $x,y$, and radius $r=\sqrt{x^2+y^2}$ from the optical axis. With the beam waist at $z=0$, total power $P>0$, and vacuum wavelength $\lambda$, its time-averaged light intensity, meaning irradiance in $\mathrm{W}{\mathrm{m}}^{-2}$, isHere $w_0$ is the minimum beam radius, and $z_R$ is the Rayleigh range. Far from the beam waist, $w(z)\sim \theta |z|$ with beam divergence half-angle $\theta =\lambda /(\pi w_0)$. These are the fundamental circular Gaussian beam laws within the scalar paraxial approximation, appropriate when $w_0\gg \lambda$ and the beam divergence is small. The underlying envelope solves the parabolic wave equation exactly; this does not make it an exact vector solution of Maxwell equations. ^[ref].

The radius convention matters. Writing $U$ for the complex wave amplitude envelope, the same $w(z)$ is the wave amplitude $1/e$ radius and the light intensity $1/e^2$ radius:At $r=w(z)$ these ratios are $e^{-1}\approx 0.367879$ and $e^{-2}\approx 0.135335$, respectively. The light intensity $1/e$ radius is instead $w(z)/\sqrt{2}$. None of these radii is a sharp edge: the ideal Gaussian function has nonzero tails at every finite radius. Kogelnik and Li explicitly use the wave amplitude $1/e$ radius in their fundamental-mode treatment. ^[ref].

For a numerical example, take vacuum wavelength $\lambda =632.8\mathrm{nm}$ and beam waist radius $w_0=0.500\mathrm{mm}$. Substitution gives Rayleigh range $z_R\approx 1.24114754\mathrm{m}$ and far-field beam divergence half-angle $\theta \approx 0.402852992\mathrm{mrad}$. At $z=z_R$, the beam radius is $w=\sqrt{2}{w}_0\approx 0.707107\mathrm{mm}$ and the on-axis irradiance is half its waist value. At the same wavelength, halving the beam waist radius to $0.250\mathrm{mm}$ quarters the Rayleigh range to $0.310286885\mathrm{m}$ and doubles the beam divergence half-angle to $0.805705984\mathrm{mrad}$. At fixed power, that smaller beam waist also quadruples the peak irradiance.

The contours in the left panel are $r=\pm w(z)$ in a plane through the optical axis; they mark an light intensity level, not an outer boundary. The right panel compares the exact model radii with the far-field asymptotes $w=\theta z$. The horizontal and vertical axes use different length units, so the drawn slopes are not physical angles.

The normalization expresses conserved power during free propagation. In transverse polar coordinates, an annulus has area $2\pi rdr$, soA proof outline is the substitution $s=2r^2/w(z{)}^2$ in this integral. Spreading reduces peak irradiance as $w^{-2}$ while increasing transverse area as $w^2$, leaving power unchanged. In particular, the disk of radius $w(z)$ contains $1-e^{-2}\approx 0.8646647168$, or $86.4665\%$, of the total power; the local light intensity at its rim is only $13.5335\%$ of the on-axis light intensity. These are different quantities.

The left panel compares normalized wave amplitude magnitude and irradiance; the right panel integrates the transverse area to show enclosed power. For an ideal absorbing circular aperture of radius $a$ centered on the optical axis, the same fraction gives the power passing the aperture at that plane. An aperture of radius $w$ clips about $13.5335\%$ of the incident power; radius $2w$ clips only $e^{-8}\approx 0.000335463$, or $0.0335463\%$. After clipping, extra diffraction means the transmitted field no longer obeys the unmodified ideal Gaussian beam law. Every finite aperture cuts some ideal tail. Kogelnik and Li distinguish their large-aperture beam analysis from their finite-aperture diffraction analysis. ^[ref].

The useful tradeoff is that a small bright spot cannot remain narrow indefinitely. An optical lens can create a smaller beam waist, but diffraction then makes the beam spread faster. The particularly simple relation $w_0\theta =\lambda /\pi$ ties localization directly to beam divergence. A proof outline for the propagation law starts with $U(x,y,0)=U_0{e}^{-(x^2+y^2)/w_0^2}$. The parabolic wave equation separates into two transverse factors, and a Gaussian integral in each coordinate givesHere $q$ is dimensionless and $k$ is the wavenumber; the carrier convention is $e^{i(kz-\omega t)}$, with angular frequency $\omega$. Taking the squared complex modulus gives the stated $w(z)$ and its $w^{-2}$ light intensity scaling. Each transverse factor contributes $q^{-1/2}$, explaining the $q^{-1}$ prefactor here and the different prefactor in the preserved Gaussian beam with one transverse coordinate.

Historically, Gaussian beam theory developed alongside early laser resonator theory and lens-guided transmission. In their 1966 review, Kogelnik and Li credit Boyd and Gordon, and Goubau and Schwering, with early beam treatments; they summarize earlier work rather than claim the whole construction as a new discovery. Their paper brings beam propagation, focusing and finite-aperture diffraction into a common practical account. ^[ref].

For diffusion with a constant diffusion coefficient $D>0$ and no sources, the model flux is $J=-D\nabla u$. A continuity equation gives $u_t=-\nabla \cdot J=D\Delta u$. This proportionality is a constitutive assumption about transport; the continuity equation expresses conservation. If $D$ varies with position, it must remain inside the divergence, giving $u_t=\nabla \cdot (D\nabla u)$ instead.^[ref]

An exact constructed heat equation example on the whole plane uses dimensionless coordinates with $D=1$ and the smooth initial Gaussian function $u(x,y,0)=e^{-r^2}$. Set $s=1+4t$. For $t\ge 0$,Direct differentiation verifies both the initial condition andThe center decreases, while the region $r^2>s$ increases at that instant. The sign-change circle expands with radius $\sqrt{s}$. A fixed outer location can initially increase and later decrease as this circle passes it.

At time $t$, the peak is $1/s$, and the radius where the profile is $1/e$ times its peak is $\sqrt{s}$. The width grows as the height falls. Integration in polar coordinates givesThus a decreasing peak is compatible with a conserved spatial integral. Dividing $u$ by $\pi$ makes a probability density function with coordinate variances $s/2=1/2+2t$, connecting the same spreading to Brownian motion with diffusion coefficient one. Equivalently the profile is evolved by the Gaussian heat kernel with time parameter $2t$, since that article uses generator $\frac{1}{2}\Delta$.

On the punctured plane ${\mathbb{R}}^2\setminus \{(0,0)\}$, consider the vector fieldWith $r^2=x^2+y^2$, its partial derivatives satisfyThus the field has zero scalar curl everywhere it is defined. Nevertheless, on any counterclockwise circle ${\gamma }_R(t)=(R\cos t,R\sin t)$ with $R>0$,The radius cancels exactly.

Applying Green theorem to the enclosed disk would be invalid: $G$ is undefined at the origin, inside that disk. Zero curl on the path alone is insufficient.^[ref]

Now take the annulus $A=\{(x,y):1\le r\le 2\}$. Its positive boundary orientation is counterclockwise on the outer circle and clockwise on the inner circle. The complete boundary line integral is thereforeHere Green theorem applies: an open set containing the closed annulus can avoid the origin. The inner contribution remains $-2\pi$ even for an arbitrarily small positive inner radius. Decreasing the hole's area does not decrease this line integral.

The cancellation has a useful proof picture. Cut the annulus along two radial segments into two half-annuli. Each piece has a single simple closed curve as its boundary; along each cut, the two induced boundary orientations oppose, so those line integrals cancel when the two applications of Green theorem are added. Only the outer and inner circles remain.^[ref]

Locally, a choice of angle $\theta$ in polar coordinates has gradient $\nabla \theta =G$. Globally, following a circle once increases that angle by $2\pi$, so it cannot be a single-valued potential of a conservative vector field on the punctured plane. The nonzero closed-path line integral also rules out such a potential by the fundamental theorem for line integrals.

On a simply connected open region, a continuously differentiable planar vector field with zero scalar curl does have a global potential. The hole is what separates the local derivative condition from the global conclusion.^[ref]

Invert the two-dimensional Fourier transform in the long-path projection of a magnetic correlation, then perform the angular integration in polar coordinates. The angular integral is $2\pi J_0(kr)$, a Bessel function of the first kind, giving the displayed Hankel transform with the stated Fourier-Hankel normalization for an isotropic spectrum. The result recovers the scalar tensor coefficient $H$, whose shell-integrated magnetic energy spectrum has an additional factor $k^2$.

The potential energy $V(x,y)=\frac{1}{2}(x^2+y^2)+x^{2}y-y^3/3$ in the Hénon-Heiles system has the following expression in polar coordinates:so it is invariant under rotation through $2\pi /3$. Solving $\nabla V=0$ gives a strict local minimum at the origin and three saddle points, at $(0,1)$ and $(\pm \sqrt{3}/2,-1/2)$. All three saddle points have value $1/6$. Their common level set factors exactly:Thus this level set is the union of the three lines $y=-1/2$ and $y=1\pm \sqrt{3}x$. They bound an equilateral triangle containing the local minimum. The Hessian matrix is the identity at the origin; at each of the other three stationary points it has eigenvalues $3$ and $-1$. For energies $0\le E<1/6$, the central accessible region is bounded; above that value the three saddle passages allow paths from the centre to arbitrarily large radius. This follows directly from the radial potential: in each direction with $\sin (3\theta )<0$, its radial maximum occurs at $r=-1/\sin (3\theta )$, with value $1/[6{\sin }^2(3\theta )]\ge 1/6$.

A Hopf bifurcation creates or destroys a small periodic orbit near an equilibrium point as a pair of complex conjugate eigenvalues crosses the imaginary axis. It explains how a resting state can give way to sustained oscillations: the linearization of a dynamical system supplies rotation and growth, while a nonlinear term can saturate that growth at a nonzero amplitude.

The simplest supercritical Hopf bifurcation is the exact Hopf normal formHere $z$ is a complex number, $\mu$ is the bifurcation parameter, and $\omega$ is the angular frequency. In polar coordinates $z=r{e}^{i\theta }$, the equations become $\dot{r}=\mu r-r^3$ and $\dot{\theta }=\omega$ for $r>0$. At $\mu <0$, each turn shrinks towards the equilibrium point. At $\mu >0$, small radii grow and large radii shrink towards $r=\sqrt{\mu }$, producing a limit cycle. For example, $\mu =0.04$ and $\omega =2$ give radius $0.2$, period $\pi$, and frequency $1/\pi$ cycles per unit time. The limiting angular frequency stays nonzero while the amplitude tends to zero: a tiny newborn oscillation can still repeat at a finite rate.

For a precise generic statement, consider a smooth function $f$ defining an autonomous differential equation $\dot{x}=f(x,\alpha )$ on ${\mathbb{R}}^n$, with a smooth branch of equilibrium points $x_{\ast }(\alpha )$ and critical parameter $\alpha =0$. Require:Then a locally unique branch of small periodic orbits, up to phase, emerges on the side $\beta a<0$. On the two-dimensional centre manifold, the branch attracts for $a<0$ and repels for $a>0$. Attraction in the full phase space additionally requires all remaining eigenvalues to have negative real part; any transverse unstable direction makes the periodic orbit unstable there.^[ref]

An eigenvalue-crossing bifurcation test alone cannot establish these nonlinear conclusions. For example, the purely linear equation $\dot{z}=(\mu +i\omega )z$ has the same crossing but no isolated periodic orbit: at $\mu =0$ every nonzero radius is constant, and at $\mu \ne 0$ none is. If the cubic coefficient vanishes, higher-order terms must be examined; a generalized Hopf bifurcation is one possible degeneracy. The usefulness of the generic theorem is that a large system can have its small oscillatory onset understood through one Hopf normal form, subject to these hypotheses and local limits.

A hyperbolic spiral is a plane curve whose radial distance in polar coordinates is inversely proportional to the angular displacement from a fixed angle: $r=a/(\theta -{\theta }_0)$, on a branch where $r>0$. For $a>0$ and $\theta >{\theta }_0$, it winds infinitely often towards the origin as $\theta \to \infty$. Its geometric winding alone does not determine whether a moving particle traverses it in finite time: that also requires a law for the angular velocity.

For a two-dimensional potential doublet of fixed complex strength $M$ outside a stationary circle of radius $a$, the Milne-Thomson circle theorem gives the image contribution $\bar{M}z/(a^2-\bar{b}z)$. If the singularity is stipulated to follow the regular fluid velocity at its position, differentiating this term and using the complex velocity convention gives the displayed motion law. Thus $\mathrm{Im}(b\bar{M})$ is constant: the trajectory is a straight line parallel to $M$. With $M=m{e}^{i\alpha }$ and initial location $b_0>a$ real, its polar coordinates satisfy $r\sin (\theta -\alpha )=-b_0\sin \alpha$. This statement assumes fixed orientation; the translational rule alone supplies no equation for orientation evolution.

The inverse function theorem answers a practical question: can nearby outputs be used to recover nearby inputs uniquely and smoothly? Its useful surprise is that a nonlinear function can be tested using its linear derivative. If the first-order changes preserve all input directions, then, on a sufficiently small region, the nonlinear map can also be reversed. This is the basis for treating a nonlinear map as a coordinate transformation.

Precisely, let $f:U\to {\mathbb{R}}^n$ be continuously differentiable on an open subset $U$ of Euclidean space, let $a\in U$, and put $b=f(a)$. If $Df(a)$ is an invertible linear map, there are open neighbourhoods $U_0$ of $a$ and $V_0$ of $b$ such that $f:U_0\to V_0$ is a bijection with a continuously differentiable inverse function $g$. Thus this restriction is a diffeomorphism. In coordinates the test is $\det Df(a)\ne 0$, using the Jacobian determinant. The derivative of a local inverse map is obtained by inverting the Jacobian matrix, at the corresponding input. ^[ref]

For a one-dimensional example, $f(x)=x^2$ has $f^{\prime }(1)=2$. Near $x=1$, the inverse function is $g(y)=\sqrt{y}$, and the inverse function rule gives $g^{\prime }(1)=1/2$. But $x^2$ has no inverse function on the whole real line, since $x$ and $-x$ give the same output. The theorem supplies a local restriction, not global injectivity. At zero, the derivative test fails; the nonnegative branch still has a square root inverse, but that inverse has no finite derivative at zero. ^[ref]

Here is a two-dimensional example with an invertible derivative everywhere and infinitely repeated outputs:The exponential function turns $s$ into a positive radius, while $\theta$ is the angle in polar coordinates. Restricting, for example, to $|s|<0.35$ and $|\theta |<0.6$ produces the reversible patch pictured below. On its image, with Cartesian coordinates $(u,v)$ and $u>0$, the inverse function isThe arctangent selects the angle near zero. Without an angle restriction, $f(s,\theta +2\pi k)=f(s,\theta )$ for every integer $k$, so there is no single global inverse function. This map is a local diffeomorphism but is not injective.

The short proof idea is to correct a guessed input using the inverse linear response. With $A=Df(a)$, solving $f(x)=y$ is equivalent to finding a fixed point ofAt $a$, the derivative $D{T}_y(a)=I-A^{-1}Df(a)$ vanishes. Continuity makes $T_y$ a contraction mapping on a sufficiently small closed ball, which it maps into itself when $y$ is sufficiently close to $b$. The contraction mapping theorem then gives existence and uniqueness there. Establishing the smooth dependence on $y$ completes the inverse theorem. This explains why the nonlinearity can be tolerated: its correction has a small local slope.^[ref]