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Classical mechanics describes the motion of bodies by assigning positions and velocities and specifying how their interactions change those quantities. A point mass is the simplest model: its size and internal structure are ignored, leaving a mass and a moving position. A rigid body retains orientation and rotational kinetic energy. Choosing a model means deciding which features affect the question being asked; a sliding block may be treated as a point mass, whereas predicting its rolling without slipping requires its moment of inertia.

Three complementary questions organize the subject. Newton's second law asks how the net force determines the instantaneous acceleration. The work-energy theorem asks how that force, acting through a displacement, changes kinetic energy. Conservation of mechanical energy asks which combinations of position and velocity are accessible when the interactions admit a time-independent potential energy and no other force supplies net power. These descriptions are compatible, but answer different questions: an energy calculation may find a maximum speed without first solving for position at every time.

The harmonic oscillator below makes this unity visible. A restoring force produces a repeating position trace; the corresponding phase plane has a closed ellipse; kinetic energy and potential energy exchange while their sum stays fixed. Lagrangian mechanics and Hamiltonian mechanics provide further ways to organize the same dynamics, especially when coordinates or constraints make individual forces awkward to track.

The scope of classical mechanics also matters. Newtonian mechanics approximates motion at speeds small compared with the speed of light; special relativity changes that description when this approximation fails. Quantum mechanics is needed for phenomena for which a classical assignment of simultaneous position and momentum is inadequate. These limitations do not prevent classical mechanics from being a useful predictive model within its regime. ^[ref]

At an equilibrium $(v_{\ast },k{v}_{\ast })$, set $p=f^{\prime }(v_{\ast })$. The Jacobian matrix, its trace and its determinant areThe eigenvalues and their discriminant areThus the complete local linear stability test isIn the last two cases, $\mathcal{D}>0$ gives a node, while $\mathcal{D}<0$ gives a focus; the sign of $T$ distinguishes inward and outward trajectories. At $\mathcal{D}=0$, a nonzero repeated eigenvalue gives a degenerate node. At $p=\gamma <k$ the eigenvalues are purely imaginary, so linear stability analysis alone does not decide asymptotic stability; with the required crossing and nondegeneracy this is a Hopf bifurcation. At $p=k$ there is a zero eigenvalue, requiring a nonlinear test and generically marking the saddle-node bifurcation in the preceding answer.

For a unique simple equilibrium, $p<k$, so it can be a sink equilibrium or a source equilibrium. The number of equilibria alone does not decide stability. If $k\le \gamma$, the unique simple equilibrium is always a sink equilibrium. If $\gamma <k$, it can become a source equilibrium where $\gamma <p<k$. The same conclusion follows particularly simply if $\gamma >M$: the trace is negative at every nonsaddle equilibrium.

For three simple equilibria ordered by their $v$ coordinates, the middle one has $p>k$ and is a saddle equilibrium; the two outer ones have $p<k$. When both outer equilibria are sink equilibria, each has an attracting neighborhood, giving bistability. In the usual two-basin portrait, the stable manifold of the saddle equilibrium separates their basins. Each branch of its unstable manifold initially leaves towards a different side; its eventual destination requires a global argument. If $k\le \gamma$, both outer equilibria must be sink equilibria. If $\gamma <k$, either or both outer equilibria can instead be source equilibria; the same test $p\lessgtr \gamma$ applies separately to each. The three lower panels illustrate all three possibilities up to exchanging left and right.

The arrows in the phase plane follow directly from the nullclines: below $w=f(v)+I$ the horizontal component points right, and above it left; below $w=kv$ the vertical component points up, and above it down. Near a stable focus trajectories spiral inward; near an unstable focus they spiral outward. Near a saddle equilibrium they approach along its stable manifold and leave along its unstable manifold. These directions, together with the local tests, determine the sketches.

There is also a useful global conclusion when the unique equilibrium is a source equilibrium. Withthe leading term in $v$ is $-\beta v^4$, so sufficiently large ellipses are trapping regions. A small repelling neighborhood of the hyperbolic source equilibrium can be removed to form an invariant annulus. The Poincaré-Bendixson theorem then guarantees at least one periodic orbit in that annulus. The numerical sketch illustrates the resulting sustained oscillation; this argument does not assert uniqueness of a limit cycle. A stable equilibrium can instead represent a resting membrane potential, with perturbations producing damped relaxation; near a threshold, a sufficiently large perturbation can produce a large excursion before recovery. Global basin boundaries and coexistence of periodic orbits need more information than the number and local stability of equilibria.

Eliminate $y$ using $y=1-x^{\prime }$ to get $x^{\prime \prime }+x=-1$. The initial data give $x(0)=1$ and $x^{\prime }(0)=0$, so the simple harmonic motion about the equilibrium $x=-1$ isIndeed, $x^{\prime }+y=1$ and $y^{\prime }-x=1$ identically, and both initial coordinates equal one. This also describes a circle of radius two around $(-1,1)$ in the phase plane.

Take $\gamma >{\gamma }_c$ and let $\epsilon \to 0^{+}$ with the other parameters fixed. The nullclines are the cubic polynomial graph $w=f(v)$ and the line $w=\gamma v$. In the phase plane, $v$ increases below the cubic polynomial graph and decreases above it; $w$ increases below the line and decreases above it. The fast ordinary differential equation, in time $s=t/\epsilon$, freezes $w$ and gives $dv/ds=f(v)-w$.

At $w=0$, the fast roots are $0,a,1$. Their linear stability follows from $f^{\prime }(0)=-a<0$, $f^{\prime }(a)=a(1-a)>0$ and $f^{\prime }(1)=a-1<0$. Thus a fixed $0<v^{\ast }<a$ falls towards the resting branch, whereas a fixed $a<v^{\ast }<1$ rises towards $v=1$. If $v^{\ast }>1$, the first fast segment instead settles down towards $1$, followed by the same recovery excursion. The threshold $a$ is the leading singular perturbation threshold; finite-$\epsilon$ corrections matter for initial values approaching $a$ with $\epsilon$.

LetThese are the roots of $f^{\prime }(v)=0$. The outer branches $v<v_L$ and $v>v_R$ of the critical manifold are attracting under fast motion; the middle branch is repelling. The full excursion has four stages.

For this cubic polynomial, the landing point is explicit. Since $f(v)-f(v_R)=-(v-v_R{)}^2(v-v_J)$, comparison of the quadratic coefficients givesThus a graph of $v(t)$ has a rapid rise, a slowly falling elevated plateau, a rapid fall through zero to a negative undershoot, and a slow approach back to zero. The figure shows both the phase plane and a computed time trace. The smaller initial perturbation stays near the resting branch and produces no elevated plateau.

The four-stage description concerns the singular limit. Close to a simple fold point of a critical manifold, balancing the local quadratic activation term with the slow drift gives $\Delta v=O({\epsilon }^{1/3})$ and $\Delta t=O({\epsilon }^{2/3})$ for fixed parameters. These layers shrink compared with the slow phases and do not alter the leading time integrals below.

Put $A=g\sin \alpha >0$ and $v=\dot{x}$. Multiplying the equation by $xv$ givesso its first integral isThese level curves give the positive phase plane for the avalanche front with linear mass entrainment. For $c<0$, the curve starts at $v=0$, $x=(-3c/(2A){)}^{1/3}$ and rises. For $c>0$, it diverges as $x\downarrow 0$, has a minimum at $x=(3c/A{)}^{1/3}$, and then rises. The $c=0$ curve is $v=\sqrt{2Ax/3}$. Forward-time arrows have increasing $x$; at a zero-velocity endpoint the equation gives $\ddot{x}=A>0$.

Every forward-moving trajectory has $x\to \infty$: after it has entered $v>0$, a bounded limiting $x$ would force a zero of the displayed positive-branch speed ahead of it, but its only possible zero is the lower endpoint. For large $x$,Expanding the equation as $x\ddot{x}+v^2=Ax$ and substituting the first integral givesThus both the asymptotic phase plane trajectory and the limiting acceleration are independent of the initial data.

Introduce the velocity variable $v=\dot{x}$. The second-order differential equation becomes the autonomous planar systemA critical point of a dynamical system is a point where both components of the vector field vanish: here $v=0$ and $f(x,0)=0$. It is a stationary solution in the phase plane. The Jacobian matrix at each such point gives the local linear stability of a planar equilibrium. For the equations below, a conserved energy also determines the nonlinear phase portrait; this is important because purely imaginary eigenvalues alone do not generally prove that an equilibrium is a center equilibrium.

The potential factors as $V=x^2(3-2x)$, so its zeros are $0$ and $3/2$, with a double zero at the origin. Its derivatives are $V^{\prime }=6x(1-x)$ and $V^{\prime \prime }=6-12x$. Thus there is a local minimum $(0,0)$, a local maximum $(1,1)$ and an inflection at $(1/2,1/2)$. The potential tends to $+\infty$ on the far left and $-\infty$ on the far right. These features determine the left-hand sketch.

Put $v=\dot{x}$. Newton's second law gives the phase plane system $\dot{x}=v$, $\dot{v}=-V^{\prime }=6x(x-1)$. Its conserved mechanical energy and trajectories areThe linearization at $(0,0)$ has eigenvalues $\pm i\sqrt{6}$, and the surrounding energy curves are closed, so it is a center equilibrium. At $(1,0)$ the eigenvalues are $\pm \sqrt{6}$, giving a saddle equilibrium. The figure shows the cubic potential barrier phase portrait; arrows point right when $v>0$ and left when $v<0$.

For $0<E<1$, there are three turning-point roots of $V(x)=E$. The two leftmost enclose a closed orbit around the origin: the particle oscillates between them. There is also a separate open component to the right of the third root, where an incoming particle turns and escapes to $+\infty$. The closed orbits run clockwise in the $(x,v)$ plane.

For $E=1$, the factorization $1-V(x)=(x-1{)}^2(2x+1)$ gives the separatrixThe branch with $-1/2\le x<1$ is a homoclinic orbit: it leaves the saddle asymptotically, turns at $x=-1/2$ and returns asymptotically. On $x>1$, one branch approaches the saddle and the other departs towards infinity. Approaching the saddle takes infinite time. The saddle itself is also an equilibrium trajectory.

For $E>1$, the particle can cross the barrier. There is one left turning point below $-1/2$; an incoming particle passes through the well, turns on the left and escapes to the right. For $E<0$, only a right-hand open component exists, with its turning point beyond $3/2$. At $E=0$, the stationary centre coexists with a right-hand open trajectory turning at $3/2$. Thus negative energy does not imply trapping here: the cubic potential is unbounded below on the right.

Assume the decreasing functions are continuously differentiable so their local gains are defined. In the positive quadrant, the nullclines are $x=f(y)$ and $y=g(x)$, both decreasing. The phase plane arrows point right when $x<f(y)$ and left when $x>f(y)$; they point up when $y<g(x)$ and down when $y>g(x)$. Both axes point into the quadrant. Since $f,g$ are bounded, sufficiently large $x$ or $y$ decreases, giving a bounded forward-invariant rectangle containing all equilibria and eventually containing each trajectory. Thus mutual repression can have one attracting intersection or several intersections separating different basins of attraction.

An equilibrium obeys $x_{\ast }=f(y_{\ast })$, $y_{\ast }=g(x_{\ast })$. Its Jacobian matrix and eigenvalues areThe product is nonnegative. Hence an intersection is a stable equilibrium with two negative real eigenvalues when $f^{\prime }{g}^{\prime }<1$, a saddle equilibrium when $f^{\prime }{g}^{\prime }>1$, and nonhyperbolic when $f^{\prime }{g}^{\prime }=1$. This is the mutual repression stability criterion. To see why an intervening saddle normally accompanies multistability, eliminate $y$ and set $F(x)=f(g(x))-x$. Bounded positivity gives $F(0)>0$ and $F(x)<0$ for large $x$, whileTwo downward, stable crossings require an upward crossing between them. For simple roots that middle crossing has $f^{\prime }{g}^{\prime }>1$ and is a saddle equilibrium. Its stable manifold separates the attracting states. Thus robust multistability requires the loop gain to exceed one at an intervening equilibrium, with the stable equilibria themselves having gain below one. A tangency at gain one is the threshold where equilibria can merge; degenerate crossings require higher-order analysis. A gain exceeding one somewhere is necessary for multiple intersections, but by itself does not prove that a particular pair of nullclines has multiple stable intersections.

Define the logarithmic repression sensitivitiesBoth are negative. At equilibrium $f(y_{\ast })=x_{\ast }$ and $g(x_{\ast })=y_{\ast }$, so their product is exactly $f^{\prime }(y_{\ast })g^{\prime }(x_{\ast })$. The condition can therefore be expressed asEquivalently, the product of the magnitudes of the two negative logarithmic gains must exceed one at that saddle equilibrium. The divergence of the vector field is identically $-2$, so the Bendixson-Dulac criterion excludes periodic orbits in the positive quadrant. The relevant stable long-time states in this bounded phase plane are thus equilibria, rather than attracting limit cycles. The following sketches illustrate the single-attractor and two-attractor possibilities; the saddle in the second panel lies on the diagonal, with its stable diagonal separating the two basins.

Let ${\mu }_i=⟨X_i⟩$. Applying the Markov jump-process generator to either coordinate gives the exact moment equationsIn each equation the joint reaction removes one molecule, at reaction propensity function $C{X}_1{X}_2$. Because the two reactants are different chemical species, the propensity uses their product; no same-species factorial correction is needed. The exact identity $⟨X_1{X}_2⟩={\mu }_1{\mu }_2+\mathrm{Cov}(X_1,X_2)$ shows explicitly where the equations fail to close. Neglecting this covariance relative to the product of means is the small-fluctuation moment closure. It gives the mean-field joint removal of two chemical species:The difference equation $d({\mu }_1-{\mu }_2)/dt=-\beta ({\mu }_1-{\mu }_2)$ actually holds exactly as well as after closure, because both the constant production and the joint removal terms cancel. The nonlinear moment closure will be used for the subsequent phase plane and stationary linear noise approximation calculations, as requested.

By symmetry the normalized covariance matrix has the form $\eta =\left(\begin{array}{cc}V& W\\ W& V\end{array}\right)$. Substituting the matrices from part (c) into the normalized stationary fluctuation-dissipation relation for a reaction network, $M\eta +\eta M^{\mathsf{T}}=D$, givesHence $V+EW=1/\mu$ and $W+EV=E/(2\mu )$. Solving these two equations gives the linear noise of two species with joint removal:For $E=0$ the normalized variance is $1/\mu$ and the covariance is zero, as expected for independent immigration–death processes. For $E>0$, $V=1/\mu +E^2/(2\mu (1-E^2))$ exceeds this independent value and $W<0$. The correlation coefficient isThis negative stationary correlation is compatible with positive instantaneous joint-removal noise: the slow difference mode amplifies fluctuations of opposite signs in the two populations.

The symmetric and antisymmetric eigenvectors diagonalize all three matrices. Their normalized covariance eigenvalues areThe corresponding restoring rates are $(1+E)/\tau =\beta +2C\mu$ and $(1-E)/\tau =\beta$. Thus the sum mode retains finite fluctuations, with ${\eta }_{+}\to 3/(4\mu )$, whereas the difference mode becomes arbitrarily noisy. In particular, at a fixed nonzero mean, or along $\beta \to 0$ at fixed $\lambda ,C>0$ so $\mu \to \sqrt{\lambda /C}$,The Fano factor of either population is $\mu V=(2-E^2)/(2(1-E^2))$, which diverges as $E\to 1$ regardless of how the mean is varied. Unnormalized variances are ${\mu }^{2}V$ and covariance ${\mu }^{2}W$, so their absolute magnitudes also depend on the parameter path through $\mu$.

The phase plane in part (b) explains the singularity. At $\beta =0$, the mean-field system has a whole hyperbola of equilibria, rather than an isolated restoring state. A perturbation in the species difference selects another point on that curve. In the actual stochastic reaction network, joint complex formation still leaves $X_1-X_2$ unchanged, but the two constant-rate births change it by $+1$ and $-1$, each at rate $\lambda$. Therefore exclusive joint removal has a diffusing difference mode:Its continuous-time symmetric simple random walk on the integers has no normalizable stationary distribution, so there is no finite stationary covariance at $E=1$. The divergence as $E\uparrow 1$ signals both the loss of difference-mode restoration and eventual failure of the small-fluctuation moment closure; near the limit, the computed normalized variances need not remain small even when the individual mean populations are large.

Introduce $v=\dot{x}$. The phase plane system for the damped pendulum isAt an equilibrium point, $v=0$ and $\sin x=0$, so all equilibrium points are $\boxed{(x,v)=(n\pi ,0),n\in \mathbb{Z}}$. Angles identified modulo $2\pi$ give one downward and one upward equilibrium point.

The linearization of a dynamical system at $(n\pi ,0)$ has Jacobian matrix and eigenvaluesWhen $n$ is even and $k>\omega$, both eigenvalues are real and strictly negative, since $0<\sqrt{k^2-{\omega }^2}<k$. These downward equilibrium points are stable nodes. When $n$ is even and $k<\omega$, the eigenvalues are $-k\pm i\sqrt{{\omega }^2-k^2}$, so the downward equilibrium points are stable foci, approached with decaying oscillations.

When $n$ is odd, $\sqrt{k^2+{\omega }^2}>k$, so one eigenvalue is positive and one is negative in either parameter range. Every upward equilibrium point is consequently an unstable saddle equilibrium. All these equilibrium points are hyperbolic equilibria in the two requested cases, so their linearization of a dynamical system gives their local nonlinear classification.

For a nonconstant invasion travelling wave, the relevant end states are $w(-\infty )=k$ and $w(+\infty )=0$. These fronts do not exist at every nonnegative speed: the phase plane calculation gives the threshold $2\sqrt{k}$. The constant solutions $w\equiv 0$ and $w\equiv k$ are, of course, nonnegative travelling waves for every speed, including zero.

Put $z=x-ct$ and $p=w^{\prime }$. The travelling-wave reduction of a reaction-diffusion system givesThe equilibrium points in the phase plane are $(k,0)$ and $(0,0)$. At $(k,0)$ the characteristic equation is $r^2+cr-k=0$, so there is one positive and one negative eigenvalue. Thus it is a saddle equilibrium. Its descending unstable manifold has $w<k$, $p<0$ and locallyAt $(0,0)$ the eigenvalues areFor $c>2\sqrt{k}$ it is a stable node; for $0<c<2\sqrt{k}$ it is a stable spiral. The repeated-root case $c=2\sqrt{k}$ must be included in the existence argument rather than inferred from distinct roots.

Suppose $c\ge 2\sqrt{k}$. Choose $a>0$ with $a(c-a)\ge k$; for example $a=(c-\sqrt{c^2-4k})/2$. Consider the closed triangular regionOn the horizontal side $p=0$, the vector field has $p^{\prime }=-w(k-w)\le 0$, pointing into $\mathcal{T}$. On the vertical side $w=k$, it has $w^{\prime }=p\le 0$, again pointing inward. For the sloping side use $h=p+aw$, so that on $h=0$,Thus trajectories cannot cross into $h<0$. The remaining boundary point $(0,0)$ is an equilibrium point and cannot be crossed in finite time by a nonconstant trajectory, by uniqueness theorem for ordinary differential equations. This proves forward invariance of $\mathcal{T}$. The argument works unchanged at $c=2\sqrt{k}$ with $a=\sqrt{k}$.

The descending unstable manifold of $(k,0)$ initially enters $\mathcal{T}$. Its forward trajectory remains in this compact region, so it exists for all positive $z$. Within the interior, $p<0$: a first contact with $p=0$ at $0<w<k$ would have $p^{\prime }<0$, contrary to arrival from below. Hence $w$ decreases strictly and has a limit $\ell \in [0,k)$. To identify the limit, its omega-limit set lies on the line $w=\ell$ and is invariant. Any trajectory remaining on this line has $p=w^{\prime }=0$; its second equation then forces $\ell (k-\ell )=0$. Since $\ell <k$, it follows that $\ell =0$, and the only possible omega-limit set is $(0,0)$. Backward in $z$, the unstable manifold tends to $(k,0)$ by its construction. We have therefore obtained a positive decreasing heteroclinic orbit and hence a nonnegative travelling wave for every $c\ge 2\sqrt{k}$.

For completeness, the threshold is necessary for such a front. When $0<c<2\sqrt{k}$, put $\nu =\sqrt{k-c^2/4}$, $X=w$, and $Y=(p+cw/2)/\nu$. Near the zero equilibrium point the system becomesFor a nonzero trajectory its polar angle satisfiesA trajectory approaching zero therefore makes infinitely many rotations and must cross $X=w=0$. It cannot be a nonnegative front. At $c=0$, the first integralis conserved. It has values $k^3/6$ and $0$ at the two desired end states, so no heteroclinic orbit connects them. ConsequentlyThis is the phase-plane construction of a Fisher travelling front, including the critical speed. The statement for all $c\ge 0$ must distinguish these nonconstant fronts from the two constant travelling waves.

Combine the real variables into $z=x+iy$. The linear system of ordinary differential equations becomes $z^{\prime }=(1-i\mu )z$, hence $z=(A+iB)e^{(1-i\mu )t}$ for arbitrary real $A,B$. Taking real and imaginary parts givesThe two independent constants cover every real initial condition. In the phase plane, the distance from the origin grows as $e^t$ and the polar angle changes at rate $-\mu$. For $\mu \ne 0$ the trajectories are outward logarithmic spirals; for $\mu =0$ they are outward straight rays.

The phase plane system is $\dot{x}=y$, $\dot{y}=-cy-\sin x$. Its equilibrium points are $(k\pi ,0)$ for every integer $k$. At such a point the Jacobian matrix and characteristic equation areFor even $k$, the equilibrium point is a stable focus for $0<c<2$, with eigenvalues $-c/2\pm i\sqrt{1-c^2/4}$; it is a stable node for $c>2$, with two distinct negative real eigenvalues. The excluded value $c=2$ gives the repeated critical case.

For odd $k$, the eigenvalues are $(-c\pm \sqrt{c^2+4})/2$. One is positive and one negative, so each is a saddle equilibrium for every $c>0$. These classifications follow from the linearization of a dynamical system, since all the relevant eigenvalues have nonzero real part for $c>0$. The even equilibria remain hyperbolic at $c=2$ too, although their linearization then has a repeated eigenvalue; that critical case is excluded from the requested classification.

Use $y$ for height above the horizontal wall. In the lubrication approximation, vertical momentum is hydrostatic and the interfacial stress balance with variable surface tension givesThe horizontal equation is $\mu u_{yy}=B$, with $u(0)=0$ and $\mu u_y(h)={\gamma }_x=-A{C}_x$. ThusIntegrating across the film and adding surface diffusion to surfactant advection gives the dimensional thin-film mass flux and insoluble surfactant flux:These include Marangoni stress, hydrostatic leveling, capillarity and surface diffusion with their signs fixed by the surface traction. In particular the derivative acts on the product of surface tension and curvature, not just on curvature.

Choose $h_{\ast }=(A{C}_0/(\rho g){)}^{1/2}$ and $u_{\ast }=A{C}_0{h}_{\ast }/(\mu L)$. The dimensionless definitions areTogether with the stated concentration and horizontal scales, these giveIn steady flow $Q,J$ are constants by liquid and surfactant conservation.

With capillarity and diffusion neglected, solve these two linear equations for the gradients, in the region $H>0,\Gamma >0$:For positive $Q,J$, the phase plane nullclines are $H=3Q\Gamma /(2J)$ for ${\Gamma }_X=0$ and $H=2Q\Gamma /J$ for $H_X=0$. Above both lines, trajectories go left and upward; between them they go left and downward; below both they go right and downward. There is no positive-quadrant equilibrium. The axes are singular boundaries of this positive-flux reduction, not regular equilibria.

The phase portrait shows these trajectories for one choice of positive fluxes; the two nullcline slopes rescale with $Q/J$.

Write $z=x+y$. When ${\mu }_1={\mu }_2=0$ the system gives $\dot{z}=y-z^3$ and $\dot{y}=-z^3$. Choosingmakes $V$ positive definite and radially unbounded, while its orbital derivative isThe set $\{\dot{V}=0\}$ is $z=0$. A trajectory remaining there must also have $\dot{z}=y=0$, so its largest invariant subset is the origin. The LaSalle invariance principle therefore proves that the origin is globally asymptotically stable. In the phase plane, $\dot{x}=0$ on $y=0$ and $\dot{y}=0$ on $x+y=0$; every nonstationary trajectory crosses the nested level curves of $V$ inward and tends to the origin.

At an equilibrium point, $y=0$ andHence the origin always exists, while $x=\pm \sqrt{{\mu }_1}$ exist for ${\mu }_1>0$. At $(x_{\ast },0)$ the Jacobian matrix isA stationary bifurcation occurs when a real eigenvalue passes through zero, hence onA Hopf bifurcation requires zero trace and positive determinant. At the origin this givesand on either nonzero branch it givesThus the Hopf locus consists of the negative ${\mu }_1$-axis and the ray ${\mu }_2=3{\mu }_1$ in the first quadrant; the stationary locus is the ${\mu }_2$-axis.

Now set ${\mu }_2=-1$, write $\mu ={\mu }_1$, and append $\dot{\mu }=0$. At $(x,y,\mu )=(0,0,0)$ the extended centre manifold for a parameter is tangent to the $(x,\mu )$-plane, so put $y=h(x,\mu )$. Its centre-manifold invariance equation isUnder the ordering $\mu =O(x^2)$, solving through cubic order givesThis is the pitchfork bifurcation normal form: $x=0$ is stable for $\mu <0$ and unstable for $\mu >0$, while the two stable branches $x=\pm \sqrt{\mu }$ emerge for $\mu >0$. The bifurcation is therefore a supercritical pitchfork bifurcation.

Simple harmonic motion is motion in which the acceleration is proportional to the displacement from equilibrium and points back towards it:Here $x$ is measured from equilibrium, and $\omega$ is the constant angular frequency. For a point mass attached to an ideal spring, Hooke's law gives the restoring force $F=-kx$, with stiffness $k>0$, and Newton's second law yields $\omega =\sqrt{k/m}$. The minus sign matters: for $x>0$ the force is negative, and for $x<0$ it is positive. A restoring force need not oppose velocity; on the return towards equilibrium it increases speed.

The general solution isThe amplitude $A\ge 0$ and oscillation phase $\phi$ encode the initial conditions. We use $P$ for the oscillation period to distinguish it from kinetic energy $T$. For $A>0$, the oscillation period is independent of amplitude, the property of isochronous harmonic motion. At $A=0$ the body remains at equilibrium. ^[ref]

Choose $m=0.5\mathrm{kg}$, $k=2N/m$, and release the mass from rest at displacement $A=0.3\mathrm{m}$. Then $\omega =2{\mathrm{s}}^{-1}$, $P=\pi \mathrm{s}$, and $\phi =0$. Writing time $t$ in seconds,At $t=0$ the velocity vanishes, but the acceleration is $-1.2m/{\mathrm{s}}^2$: a turning point is not a state of rest that persists. At $t=\pi /4\mathrm{s}$, the body crosses equilibrium with velocity $-0.6m/s$ and zero acceleration. At $t=\pi /2\mathrm{s}$ it reaches the opposite turning point, $x=-0.3\mathrm{m}$.

Taking potential energy zero at equilibrium, the same solution givesAt each turning point all mechanical energy is potential energy; at each equilibrium crossing it is kinetic energy. Both repeat twice during one complete position cycle because squaring removes the sign. At $t=\pi /8\mathrm{s}$, for example, each contributes $0.045\mathrm{J}$.

The physical position trace covers a line segment, $-0.3\le x\le 0.3$ in metres. In the phase plane, the paired position and velocity instead trace the ellipseThis is precisely $m{v}^2/2+k{x}^2/2=E$, rewritten: the single conserved mechanical energy fixes the amplitude and the whole ellipse. It does not fix the starting oscillation phase. At $x=0$ both $v=-0.6m/s$ and $v=+0.6m/s$ lie on it; position alone cannot distinguish them. Replacing $v$ by momentum $p=mv$ gives the equivalent phase-space trajectory. The arrows show increasing time, clockwise in $(x,v)$ coordinates.

This closed ellipse assumes a time-independent spring and no damping or applied drive. Adding a resisting force $-bv$, $b>0$, gives $m\ddot{x}+b\dot{x}+kx=0$ and $dE/dt=-b{v}^2\le 0$. An underdamped damped harmonic oscillator spirals inward in the phase plane, crossing successively smaller energy contours. The ideal harmonic oscillator is also useful beyond springs: near a smooth potential energy minimum with $V^{\prime \prime }(0)>0$, its Taylor expansion begins with $V(0)+V^{\prime \prime }(0)x^2/2$. For sufficiently small displacements, this gives the same restoring force to leading order, explaining why small oscillations share one solvable pattern.