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The Poisson bracket is $\{f,g\}=\mathbf{r}\cdot (\nabla f\times \nabla g)$. Since $\nabla {\rho }^2=2\mathbf{r}$, the scalar triple product vanishes, andEquivalently, check the identity on $x,y,z$ and extend to every polynomial using the Leibniz rule. Thus ${\rho }^2$ is a Casimir function of a Poisson manifold of this rotational Lie-Poisson structure on R3.

The Hamilton's equations areLinearization about $(1,0,0)$ gives $\dot{\alpha }=0$, $\dot{\beta }=(C-A)\gamma$, $\dot{\gamma }=(A-B)\beta$. The two nonzero exponents satisfy ${\lambda }^2=(C-A)(A-B)$. ConsequentlyWhen $A$ is outside this interval the transverse motion is oscillatory. At equality the linearization is nonhyperbolic; if just one transverse coupling vanishes, the other permits secular linear growth, rather than an exponential instability. This is the familiar intermediate-axis instability described by the Euler equations for a torque-free rigid body.

Substitute the Laplace contour-integral method for a linear differential equation into the differential equation and use $z{e}^{zt}={\partial }_t{e}^{zt}$. Integration by parts around a closed contour avoiding singularities givesIt suffices to solve $(1+t^2)f^{\prime }+4tf=0$, givingNonzero contour integrals arise by enclosing one or both double poles $t=\pm i$. At $t=i$, the residue isThe conjugate pole gives the conjugate expression for real $z$. Taking real and imaginary linear combinations yields the real solutionsTheir derivatives are $y_1^{\prime }=z\sin z$, $y_2^{\prime }=z\cos z$, and their Wronskian is $-z^2$, so they are linearly independent on every interval avoiding zero. Both extend analytically across zero and remain independent as functions; the vanishing Wronskian at zero reflects the singular leading coefficient of the printed differential equation there.

At an endemic equilibrium, $I_{\ast }>0$, so $\beta S_{\ast }=\mu +\nu$. Substitution into the susceptible equation givesThese values are physically positive exactly when $R_0>1$. The Jacobian matrix at this equilibrium isBoth eigenvalues have negative real parts, so the endemic equilibrium is locally asymptotically stable. Its decay exponents areThe regime $\nu \gg \mu$ means recovery is rapid compared with demographic replacement. For fixed $R_0>1$ away from the threshold, the exponents form a complex conjugate pair with frequency approximately $\sqrt{\mu \nu (R_0-1)}$. HenceStrictly, oscillations require $\mu (\mu +\nu )(R_0-1)>{\mu }^2{R}_0^2/4$; $\nu \gg \mu$ alone does not ensure this uniformly for every $R_0$ arbitrarily close to one or arbitrarily large.

Vaccinating a fraction $v$ of newborns replaces the susceptible input by $\mu N(1-v)$. The effective basic reproduction number is $R_v=(1-v)R_0$, and the same calculation replaces $R_0$ by $R_v$. Provided $R_v>1$ and the oscillatory approximation remains applicable, vaccination lengthens the oscillation period, because $R_v-1$ is smaller. Extremely near eradication the exponents become real and there is no oscillation period to extrapolate indefinitely.

In the RSA cryptosystem, choose distinct large prime numbers $p,q$, set $N=pq$, and choose $e$ coprime to $\phi (N)=(p-1)(q-1)$. Choose $d$ with $ed\equiv 1(\mathrm{mod}\phi (N))$. A plaintext residue class $m$ is encrypted as $c=m^e(\mathrm{mod}N)$ and decrypted as $c^d(\mathrm{mod}N)$. Fermat's little theorem modulo $p$ and $q$, followed by the Chinese remainder theorem, proves correctness even when $m$ is not coprime to $N$.

Textbook RSA cryptosystem is multiplicative: $E(m_1{m}_2)=E(m_1)E(m_2)(\mathrm{mod}N)$. For a chosen-ciphertext attack, multiply an intercepted $c$ by $r^e$ for a known invertible $r$. A decryption oracle returns $mr$, from which multiplication by $r^{-1}$ recovers $m$. Likewise, multiplying two textbook RSA signatures produces a signature of their product. These are homomorphism attacks, exploiting algebra rather than factoring $N$.

For an ElGamal signature, choose a prime $p$, a primitive root $g$ modulo $p$, a private exponent $a$, and public $y=g^a$. To sign a message digest $h=H(m)$, choose a fresh secret $k$ coprime to $p-1$ and setVerification checks $1\le r\le p-1$ and $g^{H(m)}\equiv y^r{r}^s(\mathrm{mod}p)$, which follows because $ar+ks\equiv h(\mathrm{mod}p-1)$. Authenticating the ciphertext and its context with such a digital signature, before decryption, prevents the simple multiplicative modification from being accepted without a fresh valid signature. One needs an appropriate cryptographic hash function and fresh secret cryptographic nonces: this is not a claim that every algebraic attack is impossible. In particular, the unhashed historical ElGamal signature scheme admits existential forgeries of specially chosen message exponents; mere multiplication of encrypted messages is not an authentication mechanism.

Assume incompressible inviscid layers, gravity $-g{\mathbf{e}}_y$, no surface tension, and irrotational perturbations of the uniform base flows. The full bulk equations are $\nabla \cdot {\mathbf{u}}_j=0$ and ${\partial }_t{\mathbf{u}}_j+{\mathbf{u}}_j\cdot \nabla {\mathbf{u}}_j=-{\rho }_j^{-1}\nabla p_j-g{\mathbf{e}}_y$. At the rigid walls $u_{1y}(h)=u_{2y}(-h)=0$. At the material interface $y=\eta (x,t)$ each layer obeys $u_{jy}={\eta }_t+u_{jx}{\eta }_x$, and $p_1=p_2$. There is no tangential-velocity matching condition across this vortex sheet.

Write ${\mathbf{u}}_j=(U_j,0)+\nabla {\varphi }_j$, with $U_1=U$, $U_2=-U$. Then ${\nabla }^2{\varphi }_j=0$, and the wall conditions are ${\partial }_y{\varphi }_1(h)={\partial }_y{\varphi }_2(-h)=0$. Linearizing the interface conditions at zero givesThe second follows from linear Bernoulli pressure $p_j^{\prime }=-{\rho }_j({\partial }_t+U_j{\partial }_x){\varphi }_j$ and the hydrostatic pressure shift $-{\rho }_{j}g\eta$.

For $k>0$, use amplitudes ${\varphi }_1=B_1\cosh k(y-h)e^{ikx+\sigma t}$ and ${\varphi }_2=B_2\cosh k(y+h)e^{ikx+\sigma t}$. Kinematics gives $B_1=-(\sigma +ikU)A/[k\sinh kh]$ and $B_2=(\sigma -ikU)A/[k\sinh kh]$. The dynamic condition becomesCompleting the square yieldsA positive real part occurs precisely whenThe function $\tanh (kh)/k$ decreases strictly from $h$ at $k=0$ to zero at infinity. Indeed its derivative has numerator $kh{\mathrm{sech}}^2(kh)-\tanh (kh)<0$, since $\sinh (2kh)>2kh$. Therefore the minimum speed producing instability at every finite nonzero wavenumber isAt equality every $k>0$ still satisfies the strict growth inequality, since $\tanh (kh)<kh$; the limiting infinite-wavelength mode is marginal. A strict uniform long-wave margin requires $U>U_{\mathrm{crit}}$.

Ignoring magnetization, electric polarization charge transport gives bound current ${\mathbf{J}}_{\mathrm{b}}={\partial }_t\mathbf{P}$. The assumed weak dielectric contrast therefore gives ${\mathbf{J}}_{\mathrm{b}}\simeq -i\omega {ϵ}_0({ϵ}_r-1){\mathbf{E}}_0{e}^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ inside the sphere. This is the first Born approximation: the incident field supplies the electric polarization to leading order in ${ϵ}_r-1$.

For $\mathbf{x}=r\mathbf{n}$, the far-distance expansions are $|\mathbf{x}-{\mathbf{x}}^{\prime }|=r-\mathbf{n}\cdot {\mathbf{x}}^{\prime }+O(a^2/r)$ and $|\mathbf{x}-{\mathbf{x}}^{\prime }{|}^{-1}\simeq r^{-1}$. Substitute the retarded time into the phase in the Lorenz gauge vector potential. The phase becomes $kr-\omega t+(\mathbf{k}-k\mathbf{n})\cdot {\mathbf{x}}^{\prime }$, so, using ${\mu }_0{ϵ}_0=c^{-2}$,The assumptions $r\gg a$ and $k{a}^2/r\ll 2\pi$ control amplitude and phase errors respectively. In the radiation zone the supplied electric and magnetic fields satisfy ${\mathbf{B}}_{\mathrm{scatt}}=c^{-1}\mathbf{n}\times {\mathbf{E}}_{\mathrm{scatt}}$. Complex phasors therefore give the time-averaged radial Poynting flux $|{\mathbf{E}}_{\mathrm{scatt}}{|}^2/(2{\mu }_{0}c)$. The incident flux is $|{\mathbf{E}}_0{|}^2/(2{\mu }_{0}c)$, soOrient spherical coordinates along $\mathbf{q}$ to evaluate the form factor:ConsequentlyAt $q=0$ the bracket is interpreted by its limit $a^3/3$, not a singularity. The transverse factor accounts for polarization of an electromagnetic wave; for complex polarization of an electromagnetic wave its squared norm means the Hermitian norm.