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Adopt the streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$. Integrating the first component of the velocity field in $y$ gives $\psi =-xy+\frac{1}{2}\Omega e^{-2t}{y}^2+f(x,t)$. The second component then requires $f_x=\Omega e^{2t}x$. Thus, omitting an arbitrary function of time,At a fixed instant, the streamlines are the connected curves $\psi =K$. To classify this quadratic form, introduce the invertible scaled and rotated coordinatesThenFor $\Omega >1$, both coefficients are positive. Every $K>0$ therefore gives an ellipse, and $K=0$ is the stagnation point at the origin. For $0<\Omega <1$, the quadratic form is an indefinite quadratic form: each $K\ne 0$ gives a hyperbola. The level $K=0$ consists of the straight separatriceswhose rays approach or leave the origin; the origin itself is a stagnation point. At the transition $\Omega =1$, $\psi =d^2$, so the nonstationary streamlines are parallel straight lines and $d=0$ is a line of stagnation points. These are instantaneous streamlines; the time dependence means that they need not coincide with pathlines.

Taking the imaginary part of the complex potential gives the streamfunctionThe corresponding velocity field isPut $\zeta =z{e}^{i\alpha /2}=\xi +i\eta$. Then $w=\beta {\zeta }^2$ and $\psi =2\beta \xi \eta$. Thus the streamlines are the previous rectangular hyperbolas rotated clockwise by $\alpha /2$. Their separatrices have angles $\theta =-\alpha /2+n\pi /2$; the extension axis is $\theta =-\alpha /2$ and the compression axis is perpendicular to it. The middle panel of the figure shows this rotation.

For pure advection, a marked fluid element obeysConsequently the dye circle becomes an ellipse with semiaxes $R{e}^{2\beta t}$ and $R{e}^{-2\beta t}$. Its area isStretching in one principal direction exactly compensates compression in the other. Equivalently the flow-map Jacobian determinant is one, as required by incompressibility. Negligible diffusion is needed to regard the sharp dye boundary as a material curve.

Use the Joukowski map with $|c|<a$, in particular its intended range $0\le c<a/2$. On the original circle,Define the exterior inversewhere the holomorphic square root is chosen so that $\sqrt{Z^2-4c^2}\sim Z$ at infinity. Then $f(Z)\sim Z$ is the branch mapping the exterior of the ellipse to $|z|>a$. Composition of the zero-circulation complex potential from part 4(c) with this inverse givesThe ellipse is a streamline because its inverse image is the circle, and $W(Z)\sim \beta e^{i\alpha }{Z}^2$ at infinity. An additive constant in the complex potential has no effect on the velocity.

The transformed complex velocity obeys the chain rule:Since $|c|<a$, the denominator is nonzero on and outside $|z|=a$. The four stagnation points therefore come fromTheir physical locations areThey all lie on the ellipse, and their positions are independent of $\beta >0$, which only sets the overall speed scale.

The printed restriction that $c$ is real and $c<a/2$ is insufficient literally: $c=-2a$ satisfies it but makes $dZ/dz=0$ at $z=2a$ in the exterior, so there is no exterior conformal map of the claimed type; the stated minor semiaxis would also be negative. The intended positive-parameter range, or more generally $|c|<a$, is needed for the above solution. The reference to a sphere in the source is a wording slip: the geometry is two-dimensional flow around the mapped cylinder's elliptical cross-section.

Isaac Newton's achievement was not to introduce mathematics into a previously nonmathematical study of nature. Astronomy, optics and mechanics already had substantial mathematical traditions; Two New Sciences by Galileo Galilei, for example, connects geometrical reasoning with accelerated motion. The distinctive achievement of the Philosophiae Naturalis Principia Mathematica by Isaac Newton was to connect mathematical results about motion and forces with an ambitious, quantitatively constrained account of terrestrial and celestial phenomena.

The method separates a conditional mathematical proposition from its application. Given a specified force law and initial conditions, determine a trajectory. Conversely, given a suitably idealized trajectory and time law, infer what force would produce it. This makes the explanation more demanding than fitting a geometrical curve to positions: the timing and dynamical assumptions matter as well.

A modern calculation illustrates the connection. For a planar orbit under a central force, let $h=r^2\dot{\theta }$ be the conserved specific angular momentum and put $u=1/r$. An ellipse with the attracting point at a focus has $u=(1+e\cos \theta )/p$. Thus $u^{\prime \prime }+u=1/p$, and Binet's equation gives the radial accelerationThe orbit and equal-area timing therefore imply an inverse-square acceleration under the stated assumptions. The modern notation is explanatory; it is not an assertion that Newton wrote this particular differential-equation derivation. The geometrical treatment of central forces in Book I and its astronomical application in Book III supply the historical connection.^[ref]

The programme also joined apparently different domains. Newton's law of universal gravitation connects orbital motion with terrestrial falling motion through the same inverse-square dependence, with distance and time scales accounted for. That unification produced further tests rather than merely attaching one name to two phenomena. Departures from a perfect two-body ellipse, including disturbances by other gravitating bodies, became quantitative problems for the theory instead of automatic refutations of it.

Measurement did not disappear into deduction. Astronomical observations provide imperfectly known inputs; assumptions about mass distribution and neglected interactions restrict an ideal model. A discrepancy can expose a mistaken approximation, an observational error or a deficiency in the dynamical account. The mathematical proof alone does not decide which. The laws and initial data therefore create a disciplined relationship between measurement and explanation, not a proof of all nature from definitions.

Newton also did not claim that successfully describing a force mathematically supplied a complete mechanical account of its ultimate cause. Distinguishing a quantitative law from a speculative mechanism allowed a powerful programme of prediction while leaving explanatory questions open. The achievement was a new integration of mathematical demonstration, empirical inference and common dynamical principles. Its strength lies in the constraints this integration places on physical explanations, rather than in a simple opposition between mathematics and experiment.

The second moment of area about an axis weights each element of cross-sectional area by its squared perpendicular distance from that axis. The ellipse extends farther along ${\mathbf{d}}_2$ than along ${\mathbf{d}}_1$, so its material lies, on balance, farther from the ${\mathbf{d}}_1$ axis. ThereforeWith zero intrinsic curvature of an elastic filament, the respective bending moments required by the constitutive equation are $E{I}_1\kappa {\mathbf{d}}_1$ and $E{I}_2\kappa {\mathbf{d}}_2$. Thus their magnitudes satisfyfor $\kappa \ne 0$: bending about the minor axis is harder. Here $E$ is Young's modulus, and $E{I}_i$ is the corresponding bending stiffness.

The characteristic polynomial is $(5-\lambda )(3-\lambda )-3=(\lambda -6)(\lambda -2)$. Corresponding normalized eigenvectors areThese form an orthonormal basis. Write $x=u{e}_6+v{e}_2$; the quadratic form becomesThe locus is an ellipse centred at the origin. Its minor semiaxis has length $\sqrt{2/3}$ along the direction $30^{\circ }$ from the positive horizontal axis; its major semiaxis has length $\sqrt{2}$ along $120^{\circ }$. The right panel of the figure in part 1 shows these principal directions.

Split the mixed quadratic terms evenly across a real symmetric matrix:Its characteristic polynomial isso $c=2$. Choose the orthonormal eigenvectorsThe orthogonal transformation $\mathbf{r}=P\mathbf{R}$ givesIndeed $P^{T}P=I$ and $P^{T}AP=\mathrm{diag}(-1,-5,4)$; preserving $Q_2$ is precisely the length-preserving property of an orthogonal matrix.

The requested plane has normal $(2,1,-2)$. An orthonormal basis within it isWrite a point of that plane as $\mathbf{r}=u{e}_1+w{e}_2$. Restricting the quadratic form gives $Q_1=-u^2-(13/9)w^2$, so its section with $Q_1=-5$ isIt is an ellipse centred at the origin, with semi-axes $\sqrt{5}$ along $e_1$ and $\sqrt{45/13}$ along $e_2$. Normalizing the in-plane basis is essential: coefficients in nonorthonormal coordinates would not directly give physical lengths.

The complete surface is the one-sheet hyperboloidIt is connected, extends indefinitely in both directions along its axis $u_3$, and has an elliptical waist at $Z=0$ with semi-axes $\sqrt{5}$ and $1$. At larger $|Z|$ its transverse sections widen; its asymptotic cone is $4Z^2=X^2+5Y^2$. The specified plane is oblique to that axis, rather than the waist plane.

Substituting $\omega =U{\omega }^{\prime }$ and using the diagonalization givesFor $I_0>0$ and energy level one in the chosen units, its ellipse equation isThe major semiaxis has length $a=\sqrt{2/I_0}$ along $u_1=(1,1)/\sqrt{2}$; the minor semiaxis has length $b=\sqrt{2/(3I_0)}$ along $u_2=(-1,1)/\sqrt{2}$. The full axis lengths are $2a$ and $2b$. The original-coordinate curve is the same ellipse tilted by $45^{\circ }$, elongated along ${\omega }_y={\omega }_x$; an orthogonal coordinate change preserves lengths and does not change the ellipse itself.

Let $r_A=|\mathbf{P}-\mathbf{A}|$, $r_B=|\mathbf{P}-\mathbf{B}|$, and define the outward-from-focus unit vectors ${\mathbf{u}}_A=(\mathbf{P}-\mathbf{A})/r_A$ and ${\mathbf{u}}_B=(\mathbf{P}-\mathbf{B})/r_B$. The ellipse is the nondegenerate level curve $r_A+r_B=s$, with $s>|\mathbf{A}-\mathbf{B}|$. The previous gradient result givesIf $\mathbf{t}$ is a unit tangent at $P$, orthogonality to the level-curve gradient impliesTherefore $|\mathbf{t}\cdot {\mathbf{u}}_A|=|\mathbf{t}\cdot {\mathbf{u}}_B|$. These absolute dot products are the cosines of the acute angles made by the two focal lines with the unoriented tangent line. Thus those angles are equal.

Equivalently, the outward unit normal is $\mathbf{n}=({\mathbf{u}}_A+{\mathbf{u}}_B)/|{\mathbf{u}}_A+{\mathbf{u}}_B|$, the bisector of the two unit focal directions. Its denominator is nonzero for a nondegenerate ellipse. Using $|{\mathbf{u}}_A|=|{\mathbf{u}}_B|=1$, the specular reflection formula becomesThis algebra proves the equal-angle statement with the incident and reflected ray directions correctly oriented, rather than confusing the line $BP$ with the outgoing direction $PB$.

A ray leaving focus $A$ and reaching a boundary point $P$ has incoming unit direction ${\mathbf{u}}_A$. By the law of reflection, its normal component reverses and its tangential component stays fixed. The result just derived shows that the reflected direction is $-{\mathbf{u}}_B$, pointing exactly from $P$ to focus $B$. Thus all such in-plane rays from one focus converge to the other after one reflection, in the ideal geometrical optics model.

Moreover, every reflected path has the same total length $AP+PB=s$. In a uniform medium their propagation times are equal as well, so different reflection points do not introduce a geometrical path-length spread. Finite wavelength, surface errors and incomplete reflecting coverage limit a real focus; the ideal argument concerns rays in the plane of the ellipse. It does not assert that an arbitrary three-dimensional surface with only one elliptical section focuses every out-of-plane ray.

Take $C$ to be a simple closed, piecewise smooth curve oriented counterclockwise around its interior $D$. Green's theorem applied to $P=-y$ and $Q=x$ givesThe orientation is essential: a clockwise unit circle gives $-2\pi$, whereas twice its ordinary unsigned area is $+2\pi$. For a nonsimple curve, the corresponding formula concerns signed area with winding multiplicity rather than one unsigned enclosed area.

For positive semiaxes $a,b$, parametrize the ellipse counterclockwise by $x=a\cos \theta$, $y=b\sin \theta$, $0\le \theta \le 2\pi$. ThenHence $2\mathcal{A}={\int }_0^{2\pi }abd\theta =2\pi ab$, and

The gradient vanishes when $3x^2-2=3y^2-2=0$. Put $s=\sqrt{2/3}$. The Hessian matrix is $\mathrm{diag}(6x,6y)$, so the second-derivative test gives:

The positive-definite and negative definite cases give small closed level curves around the minimum and maximum. At each saddle, the zero contour crosses itself. In fact,so the entire zero level curve is the union of the line $y=-x$ and the ellipse $x^2-xy+y^2=2$. This factorization fixes the topology of a rough contour sketch. Neither local extremum is global, since the cubic is unbounded above and below.

Assume a simple, piecewise smooth closed curve oriented counterclockwise around its enclosed region $D$. Apply Green's theorem with $P=-y$ and $Q=x$:The orientation assumption is essential. A clockwise unit circle, $x=\cos \theta$, $y=-\sin \theta$ for $0\le \theta \le 2\pi$, gives $-2\pi$ although its enclosed area is $\pi$. For a self-intersecting curve the same line integral measures twice the signed area, with multiplicity, rather than necessarily twice an ordinary enclosed area.

For positive semiaxes $a,b$, the ellipse has the counterclockwise parametrization $x=a\cos \theta$, $y=b\sin \theta$. ThenThereforeIf nonzero signed constants are allowed in place of positive semiaxes, the geometric area is $\pi |ab|$.

The two planes bound $z-1\le y\le 1-z$, and the elliptical cross-section has $-1\le z\le 1$. Under $x=2r\cos \theta$, $z=r\sin \theta$, the absolute Jacobian determinant for $(x,z)$ is $2r$. ThereforeAs a check, the cross-sectional ellipse has area $2\pi$, while the part of the thickness proportional to $z$ integrates to zero by reflection. The mean thickness is therefore two.

The equation of motion in a rotating frame with constant angular velocity $\mathbf{\omega }$ isWith $\mathbf{F}=-4m{\omega }^2\mathbf{x}$ and $\mathbf{\omega }=(0,0,\omega )$, this becomesThe given initial data make $z=0$. For the complex coordinate $\zeta =x+iy$, the planar equations combine intoIts characteristic roots are $i\omega$ and $-3i\omega$. Applying $\zeta (0)=1$, $\dot{\zeta }(0)=0$ gives $\zeta =\frac{3}{4}{e}^{i\omega t}+\frac{1}{4}{e}^{-3i\omega t}$. The triple-angle identities reduce this astroid motion of a harmonic oscillator in a rotating frame toIt traces an astroid. Differentiating gives the relative speedFor $\omega >0$, its maxima occur at $t=(2n+1)\pi /(4\omega )$, withThis is speed measured in the specified rotating frame. The same physical motion has inertial complex coordinate $e^{i\omega t}\zeta =\cos (2\omega t)+(i/2)\sin (2\omega t)$, an ellipse; its initial velocity is the frame-rotation contribution even though the initial relative velocity is zero.

An ellipse here is a nondegenerate ellipse, hence a smooth embedded circle. Write the first ellipse aswhere $A$ is invertible. Write the second as $Q(x)=0$, wherefor a positive-definite symmetric matrix $B$. Its gradient is nonzero on the ellipse. Let $g(t)=Q(\gamma (t))$. At an intersection, transversality says the tangent vector ${\gamma }^{\prime }(t)$ is not tangent to $Q=0$, soEvery intersection is thus a simple zero of $g$, and $g$ changes sign there. The intersection set is finite: it is closed and compact, and transversality makes each of its points isolated. Start at a parameter with $g\ne 0$ and traverse one full period. Returning to the same sign requires an even number of sign changes. Therefore the number of intersection points is even, including zero when the ellipses are disjoint.

For the upper bound, substituting the parametrization into the quadratic gives the trigonometric polynomialIt is not identically zero: otherwise the first ellipse would lie on the second, and its tangent would be tangent to the second everywhere, violating transversality. Put $z=e^{it}$. Multiplying by $z^2$ turns this function into the polynomialIt is nonzero and has degree at most four, so it has at most four distinct roots. Distinct points on the first ellipse correspond to distinct parameters modulo $2\pi$ and hence distinct points $z$ on the unit circle. ThusBoth the parity and the upper bound follow without using an algebraic intersection theorem. The four-point bound is attained by a unit circle and a concentric ellipse whose semiaxes lie on opposite sides of one.

Write $k=GM$ and $r=|\mathbf{r}|$. The gravitational central force has zero torque, so the specific angular momentumis constant. For a nonradial orbit $h=|\mathbf{h}|>0$, motion lies in the plane normal to $\mathbf{h}$, and plane polar coordinates give $r^2\dot{\theta }=h$. With $u=1/r$ and primes denoting $\theta$ derivatives,The radial equation $\ddot{r}-r{\dot{\theta }}^2=-k/r^2$ is therefore the Binet equationIts general solution, after choosing the polar origin of angle, isThe origin is a focus of this conic section. To identify which conic section occurs, use the conserved specific energyA bound, nonradial orbit has $\epsilon <0$, hence $0\le e<1$ and is an ellipse. For example, putting $x=r\cos \theta$, $y=r\sin \theta$ in $r+ex=p$ givesan explicit ellipse equation. This is the intended low, bound Kepler orbit. The inverse-square equation alone does not imply an ellipse: an initially tangential speed larger than $\sqrt{2GM/r}$ gives $\epsilon >0$, $e>1$ and a hyperbolic Kepler orbit. Zero angular momentum instead gives radial motion. Those qualifications are necessary for the elliptic conclusion.

In a circular orbit, centripetal acceleration balances the gravitational acceleration:Here $\Omega$ is the positive angular speed; either sense of circulation has this period.

With the stated drag force, use the full angular momentum $\mathbf{L}=m\mathbf{r}\times \dot{\mathbf{r}}$. Gravitational torque is still zero, while the drag torque isTherefore $\boxed{\mathbf{L}(t)=\mathbf{L}(0)e^{-At/m}}$ for constant $A>0$. Its direction remains fixed, and its magnitude has exponential decay with time constant $m/A$; the orbit remains planar although its shape is no longer a fixed ellipse.

Outside a spherical planet, Newtonian mechanics gives the inverse-square central force equationThe satellite's mass cancels. The angular momentum is conserved, so the orbit lies in a plane. In that plane the general equations in polar coordinates areA synchronous circular orbit fixed above a surface point must be equatorial, prograde and have angular velocity $\omega =2\pi /T$. Substitution of constant $r$ into the radial equation givesThe radius is measured from the planet's centre; an external orbit requires it to exceed the planet's radius.

Now let the release speed be $v=(1+\eta )v_s$, with $0<\eta \ll 1$, in the same tangential direction. At release $\dot{r}=0$, butThus the satellite initially moves outward. Its specific total energy is $v^2/2-GM/r_s<0$ for this small increase, so the Kepler orbit is an ellipse, with the release point its pericentre. To make this explicit, put $b=v^2/v_s^2\in (1,2)$. The radial turning-point equation from energy and angular momentum conservation has rootsIts semi-major axis is $a=r_s/(2-b)>r_s$. The speed follows from the same conserved energy:It is greatest at pericentre and least at apocentre. In particular, the tangential speed at the two turning points satisfies $r_{\min }{v}_{\max }=r_{\max }{v}_{\min }$. The orbital period becomes $2\pi \sqrt{a^3/(GM)}>T$, so the orbit is no longer fixed over the same surface point.

Put $S=u^2+v^2$. At a nonzero equilibrium point the two equations form the homogeneous linear systemIts determinant must vanish, giving $S^2/2-(\mu +\sigma )S+{\mu }^2+{\sigma }^2-1=0$. ThusEach positive real polynomial root gives an antipodal equilibrium point pair; a negative polynomial root does not describe a real equilibrium point. A direction is specified by $\sin 2\theta =S/2-\mu$, $\cos 2\theta =S/2-\sigma$, where $u=\sqrt{S}\cos \theta$, $v=\sqrt{S}\sin \theta$. This is the polar form of the cubic confinement model. The origin is always an equilibrium point.

The origin's eigenvalues are ${\lambda }_{\pm }=\mu \pm \sqrt{1-{\sigma }^2}$. A simple zero eigenvalue occurs on ${\mu }^2+{\sigma }^2=1$, except at $\mu =0$. The symmetry of an odd function $(u,v)\mapsto (-u,-v)$ makes the generic steady bifurcation a pitchfork bifurcation. At a point $(\sigma ,{\mu }_0)$ on this circle with ${\mu }_0\ne 0$, vary $\mu ={\mu }_0+\delta$. Expanding the radius equation gives $S\sim 2{\mu }_0\delta /({\mu }_0+\sigma )$. In a signed, unit-eigenvector centre manifold coordinate, the cubic coefficient therefore has sign $-({\mu }_0+\sigma )/(2{\mu }_0)$.

On the lower semicircle ${\mu }_0=-\sqrt{1-{\sigma }^2}$ the transverse eigenvalue is negative: the pitchfork bifurcation is supercritical for $\sigma <1/\sqrt{2}$ and subcritical for $\sigma >1/\sqrt{2}$. On the upper semicircle it is supercritical for $\sigma >-1/\sqrt{2}$ and subcritical for $\sigma <-1/\sqrt{2}$ in the centre manifold direction, but its transverse eigenvalue is positive, so the bifurcating equilibrium points are not fully attracting. At $(\sigma ,{\mu }_0)=(\pm 1/\sqrt{2},\mp 1/\sqrt{2})$ the cubic vanishes. The radius equation instead gives $\delta =-S^2/(4{\mu }_0)+o(S^2)$, a nonzero quintic degeneracy with amplitude proportional to $|\delta {|}^{1/4}$: these are degenerate pitchforks in the cubic confinement model.

A Hopf bifurcation occurs at $\mu =0$, $|\sigma |>1$, with frequency $\sqrt{{\sigma }^2-1}$. To determine its direction, first note that the divergence is $2\mu -2S$. The Bendixson-Dulac criterion excludes every nonconstant periodic orbit for $\mu \le 0$, since this divergence is negative except possibly at the origin. We can also calculate a nonzero saturation coefficient. Set $a=|\sigma +1|$, $b=|\sigma -1|$ and $V=a{u}^2+b{v}^2$. At $\mu =0$ the linear orbits are ellipses of constant $V$. Direct differentiation givesOn a linear ellipse, write $u=\sqrt{V/a}\cos \varphi$, $v=\sqrt{V/b}\sin \varphi$. The angular speed is constant; the averages of $u^{3}v$ and $u{v}^3$ vanish, while $⟨S⟩=V|\sigma |/({\sigma }^2-1)$. Averaging the cubic terms, or removing their oscillatory parts by a periodic change of coordinates in its normal form, gives for $\varrho =\sqrt{V}$The cubic coefficient is strictly negative. Hence a small attracting limit cycle exists for $\mu >0$, andFinally $(\sigma ,\mu )=(\pm 1,0)$ have a nonzero nilpotent matrix as their linearization with two zero eigenvalues. They are reflection-symmetric double-zero bifurcations where the Hopf bifurcation and steady thresholds meet; an ordinary nonzero-frequency Hopf bifurcation calculation does not apply there. These, the two degenerate pitchfork bifurcation points, and the generic pitchfork bifurcation/Hopf bifurcation loci above exhaust the origin's local loss-of-hyperbolicity possibilities.

Assume $P,Q$ are continuously differentiable on a neighbourhood of a bounded region with a positively oriented piecewise smooth simple boundary. To prove Green theorem, first take a region between the graphs $y=f(x)$ and $y=g(x)$, $a\le x\le b$, with $f\le g$. Counterclockwise orientation traverses the lower graph to the right and the upper graph to the left; vertical end pieces contribute nothing to $Pdx$. Thus the fundamental theorem of calculus givesLikewise, for a region between graphs $x=u(y)$ and $x=v(y)$, the right boundary is traversed upwards and the left one downwards. Horizontal end pieces contribute nothing to $Qdy$, soFor a polygonal region, triangulate its interior. Each triangle is of both graph types with piecewise linear bounds, so adding the formulae gives the result: the double integrals add and the integrals on internal edges cancel because the adjacent triangles traverse them in opposite directions. For a regular piecewise smooth simple boundary, choose inscribed simple polygonal approximations to its piecewise smooth parametrization. Their line integrals converge to the original line integral, since the parametrizations converge uniformly and their piecewise constant tangent vectors converge in the integral norm on each smooth arc. Their regions converge in area: the symmetric difference lies in a shrinking neighbourhood of the original boundary, which has planar measure zero. Boundedness of the continuous integrand then gives convergence of the area integrals. Passing to the limit proves the formula for the stated boundary class. Combining the two identities givesPositive orientation is essential for the sign.

Choose $P=-y/2$ and $Q=x/2$, for which $Q_x-P_y=1$. The boundary formula for planar area isFor the ellipse, take $a,b>0$ so the given increasing-angle parametrization is counterclockwise. Then $dx=-a\sin \theta d\theta$ and $dy=b\cos \theta d\theta$, soIf signed nonzero parameters are allowed, the unsigned area is $\pi |ab|$, with orientation corrected accordingly.