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For oscillators of frequencies $1$ and $2$, the perturbationpreserves $N_A+2N_B$. On the energy-$9/2$ basis $|3,0⟩,|1,1⟩$, it is the matrixso the symmetric and antisymmetric combinations are exact eigenstates with shifts $\pm \sqrt{6}\lambda$.

A disturbance decaying over depth $1/k$ accelerates fluid mass of order ${\rho }_0/k$ per unit area, giving a frequency-dependent added inertia.

For a massless spring-supported membrane adjoining a fluid half-space, kinematic and pressure balance give the dispersion relation $A{s}^4+s^2-1=0$ with $s=c/c_0$.

Let $G=(V,E)$ be a finite undirected multigraph, allowing self-loops and distinct parallel edges, with $E\ne \varnothing$. Thenwhere $G_{+}$ is the edge support of a graph, and vertex degrees count each self-loop twice. There is no lower bound of three on the number of vertices. If $E=\varnothing$, the zero-length convention gives an Euler circuit exactly when $V\ne \varnothing$.^[ref]

The two requirements are independent: two separate vertices, each carrying a self-loop, have even vertex degrees but disconnected edge support of a graph, so no Euler circuit can cover both edges.

Necessity follows because a single trail cannot cover edges in different connected components of a graph, while a closed trail pairs all arriving and departing incidences, including those at its starting vertex. For sufficiency, extend a trail using unused edges until no extension is possible. Even vertex degree parity prevents it from stopping anywhere except its starting vertex: a different stopping vertex would have an odd number of used incidences and hence at least one unused incidence. The result is a closed trail.

If unused edges remain, connectivity of the edge support of a graph supplies a vertex on the current closed trail with an unused incident edge. Removing a closed trail leaves even unused vertex degrees, so the same argument constructs another closed trail from that vertex. Splice it into the existing record. The two trails use disjoint edge identities, so the splice repeats no edge. Each splice uses more edges; finiteness ensures termination with an Euler circuit. This is the constructive idea of Hierholzer's algorithm.

In the drawing, arrows indicate the chosen traversal direction along undirected edges. The first closed trail uses $AB,BC,CA$ and the second uses $AD,DE,EA$. Insert the second at the final occurrence of $A$ in $A,B,C,A$, giving $A,B,C,A,D,E,A$. A reader can check consecutive endpoints, closure, and the six distinct edges directly. With multigraphs, perform the same check on edge identities, since endpoint pairs alone cannot distinguish parallel edges.

The Euler theorem for homogeneous functions applied to an extensive quantity, together with $dE=TdS-pdV+\mu dN$ give

The Gibbs-Duhem equation constrains changes of the intensive quantities describing one uniform thermodynamic phase. For a mixture with $k$ independently specified component amounts, its molar form isHere $S$ is entropy, $V$ is volume, $T$ is temperature, $p$ is pressure, $n_i$ is the amount of substance of component $i$, and ${\mu }_i$ is its molar chemical potential. This assumes a simple compressible system whose bulk internal energy is an extensive quantity; additional work coordinates require additional terms.^[ref]

The proof exposes the source of the constraint. For internal energy $E(S,V,n_1,\dots ,n_k)$, replication of the whole thermodynamic phase by a factor $\lambda$ gives $E(\lambda S,\lambda V,\lambda n_1,\dots ,\lambda n_k)=\lambda E(S,V,n_1,\dots ,n_k)$. The Euler theorem for homogeneous functions and the fundamental differential of internal energy therefore giveTaking the differential of the first equation produces all the terms in the second, plus $SdT-Vdp+{\sum }_i{n}_{i}d{\mu }_i$. Subtraction proves the Gibbs-Duhem equation. Thus the constraint comes from two descriptions of the same state function, rather than an assumed microscopic interaction model.

At fixed temperature and pressure, divide by $n={\sum }_i{n}_i>0$ and introduce the mole fractions $x_i=n_i/n$:The chemical potentials cannot be varied independently. In a two-component mixture, put $x_A=x$ and $x_B=1-x$. Within a smooth single-thermodynamic phase region with $0<x<1$,One chemical potential fixes the other's change, but determining its absolute value still requires a reference value. This does not say ${\mu }_A={\mu }_B$: equality at phase coexistence concerns the same chemical species in different thermodynamic phases. Nor does the constraint say $dG=0$ whenever temperature and pressure are fixed; changing composition still contributes ${\sum }_i{\mu }_{i}d{n}_i$ to the Gibbs free energy.

For a rigid body with three distinct positive principal moments of inertia $0<I_1<I_2<I_3$ and zero external torque about its centre of mass, steady rotation about the first or third principal axis is Lyapunov stable in angular velocity, while rotation about the intermediate principal axis is unstable. Here steady rotation means $\mathbf{\omega }=\Omega {\hat{e}}_i$ in the principal body frame, with constant $\Omega \ne 0$. A small sideways perturbation remains small about the extreme principal moments of inertia, but generically grows about the intermediate one. Stable does not mean that the wobble decays: without damping, there is no asymptotic stability. Coincident principal moments of inertia and a stationary body are excluded.^[ref]

No external torque is needed for a body to tumble. Conservation of angular momentum fixes $\mathbf{L}$ in the fixed space frame, but the body can turn relative to $\mathbf{L}$, so its angular velocity components in the principal body frame change. The torque-free rigid-body invariants constrain this motion. A supported gyroscope with gravitational torque has different governing assumptions.

For principal moments $I_1=I_2=I$ and symmetry-axis unit vector $e_3$, angular momentum and angular velocity obeyThus $L$, $\omega$, and the symmetry axis are coplanar at every instant.

The Euler equations for a torque-free rigid body conserve rotational kinetic energy and squared angular momentum,Their common level sets are the intersection of two ellipsoids in angular-velocity space.

Thermal fluctuations smooth the zero-temperature pitchfork in filament buckling: the mean compression remains analytic and nonzero below the mechanical threshold, with rounding controlled by $L/L_p$.

For a fixed point a distance $a$ from the centre of mass and inclination measured from the upward vertical,The spin and precession angles are cyclic.

This is the Lagrangian of a Lagrange top, one of the Lagrange, Euler, and Kovalevskaya tops.

A finite point set is cyclic transitive when a cyclic group of isometries acts transitively on it.

A finite Euclidean configuration is edge Ramsey when every finite edge coloring of a suitable finite Euclidean set contains a monochromatic isometric copy. The edge Ramsey configurations are exactly the equidistant sets, equivalently the vertex sets of regular simplices.

Every regular polygon is a Euclidean Ramsey set. Its cyclic rotation acts transitively on the vertices, so this is a special case of the Kriz theorem for cyclic transitive point sets.

The two endpoints of a nonzero line segment form a Euclidean Ramsey set. The finite configuration consists of those two points, rather than every point of the continuous line segment.

For a finite colouring with $q$ colours and a desired distance $d>0$, take the $q+1$ vertices of a regular simplex of side length $d$ in ${\mathbb{R}}^q$. The pigeonhole principle gives two vertices of the same colour. Every pair of distinct vertices has distance $d$, so this pair is an isometric copy of the prescribed endpoint pair. This also gives an explicit sufficient ambient dimension for every number of colours.

Every nondegenerate triangle is a Euclidean Ramsey set.

If finite point sets $X$ and $Y$ are Euclidean Ramsey, then their orthogonal Cartesian product $X\times Y$ is Euclidean Ramsey. The proof first chooses a finite Ramsey witness for $X$, colors a witness for $Y$ by the complete color pattern it induces on the first witness, and then applies the two Ramsey properties in succession.

A finite set $X=\{x_0,\dots ,x_m\}\subseteq {\mathbb{R}}^d$ is spherical if all its points lie on one sphere: there are a center $c$ and radius $r$ with $\Vert x_i-c\Vert =r$ for every $i$. Subtracting the squared-distance equation for $x_0$ gives the equivalent linear equationsThus consistency of a finite linear system tests whether a common center exists. A nondegenerate triangle is spherical, with center its circumcenter. Three distinct collinear points are not spherical: the equal-distance equations along their line are inconsistent. This is the geometric obstruction behind three-term unit arithmetic progression is not Euclidean Ramsey.

Every Euclidean Ramsey set is spherical. The converse was historically conjectured. A September2026 preprint by Dömötör Pálvölgyi reports the seven-point configurationas a counterexample. These points do lie on the circle $(x-r{)}^2+y^2=r^2$; the non-Ramsey assertion is Theorem1 of that dated preprint.^[ref]

For $p\in C$,

An equivalent local martingale measure makes every discounted asset price a local martingale. In continuous-time markets this is the usual measure appearing in the no-free-lunch-with-vanishing-risk form of the fundamental theorem of asset pricing.

For a hyperbolic equilibrium of a smooth planar system, the eigenvalues of the Jacobian matrix determine local stability. If its determinant and trace are positive, both eigenvalues have positive real part and the equilibrium is a repeller.