Linearized mass conservation, momentum balance and the isentropic pressure-density relation are ${\rho }_t^{\prime }+{\rho }_0\nabla \cdot \mathbf{u}=0$, ${\rho }_0{\mathbf{u}}_t=-\nabla \tilde{p}$, and $\tilde{p}=c_0^2{\rho }^{\prime }$. Differentiate mass conservation in time and take the divergence of momentum balance to eliminate velocity. This gives the acoustic pressure wave equationFor the convention $\tilde{p}=\mathrm{Re}{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, momentum balance gives the complex velocity amplitude $\mathbf{k}/({\rho }_0\omega )$, with $\omega =c_0|\mathbf{k}|$. Thus the incident velocity field is $\mathrm{Re}[{\mathbf{k}}_I{e}^{i({\mathbf{k}}_I\cdot \mathbf{r}-\omega t)}/({\rho }_1\omega )]$ in the lower gas.

Let $a=k_I\sin \theta$, $l_1=k_I\cos \theta >0$, $\omega =c_1{k}_I$, and $l_2=\sqrt{(\omega /c_2{)}^2-a^2}$ with nonnegative real or imaginary part. This chooses an outgoing propagating wave or an evanescent wave decaying as $y\to +\infty$. Conservation of tangential wavevector is the acoustic Snell law for elastic and acoustic waves. Write the complex pressures asInviscid kinematic matching requires equal normal velocities at the plate, not equal tangential velocities:Define the normal acoustic impedance ratio $\delta ={\rho }_2{l}_1/({\rho }_1{l}_2)$, so $T=\delta (1-R)$. For a propagating transmitted wave this is ${\rho }_2{c}_2\cos \theta /({\rho }_1{c}_1\cos {\theta }_2)$. The dynamic condition giveswhich is exactly the printed dimensionless parameter. Solving yields the pressure reflection coefficient and transmission coefficientAt $l_2=0$ use the preceding velocity equations, or take their limit, rather than substituting an infinite $\delta$. For real $l_2>0$, the reflected energy fraction is $|R{|}^2$, the transmitted fraction is $|T{|}^2/\delta$, and these sum to one. An evanescent transmitted wave carries no mean normal energy flux.

When the gases match, $\delta =1$, so $R=-i\beta /(2-i\beta )$ and $T=2/(2-i\beta )$. For a nongrazing incident wave of nonzero frequency, perfect transmission means $R=0$, equivalent toThe plate's inertial and bending terms cancel: the incident frequency and tangential wavevector coincide with its free flexural wave dispersion relation. Then $T=1$, including its phase. With positive $m,B$, this condition is possible only at angles and frequencies satisfying that relation; for normal incidence at nonzero frequency it cannot occur unless the mass is zero.

For $k\ne 0$, put $t=k^2>0$ and use the positive-definite Lyapunov function $V=x^2+t{y}^2$. Its derivative isFor $V>0$ set $u=x^2/V\in [0,1]$. The brace divided by $V^2$ is $A{u}^2+2Bu(1-u)+C(1-u{)}^2$, where $A=1$, $B=\beta (1+t)/(2t)$ and $C=1/t$. Since $\beta >2$, $B>A,C$. Completing the square gives its maximumHence $\dot{V}\le -2V+2M{V}^2$. If $V(0)<1/M$, this sublevel region is positively invariant: $V$ decreases, and $\dot{V}\le -2(1-MV(0))V$. The exponential bound makes $V\to 0$. The trajectory is bounded in a compact ellipse, so smoothness of the vector field also guarantees existence for every positive time. ThereforeFor $k=0$ the displayed strict inequality has no solutions; negative $k$ gives the same ellipse as $-k$. Taking the union of the nonempty ellipses is valid because each point needs only one Lyapunov certificate.

At $k=1$ the certified disk has squared radius $2/(\beta +1)$. Along $y=0$, the limiting threshold as $k^2\to \infty$ is $4(\beta -1)/{\beta }^2$, which is strictly larger, since their difference is $2({\beta }^2-2)/({\beta }^2(\beta +1))>0$ for $\beta >2$. Choose $x^2$ strictly between these thresholds and then a sufficiently large finite $k$. This point is certified by that ellipse but lies outside the $k=1$ disk. Thus the union gives a strictly larger certified region. It need not equal the whole basin of attraction.

The equilibrium equations are $x(-1+x^2+\beta y^2)=y(-1+\beta x^2+y^2)=0$. For $\beta \ne 1$ the fixed points are the origin, four axis points $(\pm 1,0),(0,\pm 1)$, and four diagonal points $(\pm s,\pm s)$ with independent signs and $s=(1+\beta {)}^{-1/2}$. The hypothesis $\beta >-1$ ensures that $s$ is real. The Jacobian matrix has diagonal entries $-1+3x^2+\beta y^2$, $-1+\beta x^2+3y^2$ and off-diagonal entries $2\beta xy$.

At the origin both eigenvalues are $-1$, so it is a stable node. At each axis point the eigenvalues are $2,\beta -1$. At each diagonal point they are $2,2(1-\beta )/(1+\beta )$, along the radial and transverse diagonal directions respectively. ThusThe axes and diagonals are invariant lines. For $\beta =1/2$ the axis saddle points have stable directions transverse to their axes and unstable directions along them; the diagonal points are unstable nodes. For $\beta =2$ these roles switch: the diagonal saddle points have stable transverse directions and unstable radial directions, and the axis points are unstable nodes. Their stable manifolds separate trajectories attracted to the origin from outward-escaping trajectories. Symmetry gives the other quadrants.

For $\beta =1$, the entire unit circle consists of equilibria. In polar coordinates $\dot{r}=r(r^2-1)$ and $\dot{\theta }=0$: every ray inside the circle moves toward the origin, and every ray outside moves outward. The circle has neutral tangential and unstable radial directions; its points are not isolated saddle points or unstable nodes.

All three phase portraits are shown below. The potential $\Phi =-(x^2+y^2)/2+(x^4+y^4)/4+\beta x^2{y}^2/2$ satisfies $(\dot{x},\dot{y})=\nabla \Phi$ and $\dot{\Phi }=|\nabla \Phi {|}^2$, ruling out nonconstant periodic orbits. Blue curves show the saddle stable separatrices, integrated backwards from their local stable directions. The flow arrows and invariant lines, together with the equilibrium classification, specify the topology of the sketches.

The bounded-time optional stopping theorem states that if $X$ is an integrable discrete-time martingale and $\tau$ is a stopping time bounded by a deterministic integer $N$, then $X_{\tau }$ is integrable and $\mathbb{E}{X}_{\tau }=\mathbb{E}{X}_0$. Indeed $|X_{\tau }|\le {\sum }_{j=0}^N|X_j|$ proves integrability, andThe event $\{\tau \ge j\}=\{\tau >j-1\}$ belongs to ${\mathcal{F}}_{j-1}$. Conditional expectation of each summand given ${\mathcal{F}}_{j-1}$ is zero, soMore generally, if $\sigma \le \tau \le N$ are stopping times, $\mathbb{E}[X_{\tau }\mid {\mathcal{F}}_{\sigma }]=X_{\sigma }$. To prove this, test against $A\in {\mathcal{F}}_{\sigma }$ and apply the same argument to $1_A{1}_{\{\sigma <j\le \tau \}}(X_j-X_{j-1})$; its coefficient is ${\mathcal{F}}_{j-1}$-measurable. Boundedness of the stopping times makes the sums finite, so no limiting optional-stopping hypothesis is being omitted.

Inductively, $X_0,\dots ,X_n$ are functions of ${\Delta }_0,\dots ,{\Delta }_n$, so ${\mathcal{F}}_n\subseteq \sigma ({\Delta }_0,\dots ,{\Delta }_n)$. The average $f_n(X_0,\dots ,X_n)$ is therefore ${\mathcal{F}}_n$-measurable and independent of ${\Delta }_{n+1}$. If $X_0,\dots ,X_n$ are integrable, thenStarting with $X_0={\Delta }_0$, induction proves integrability of every $X_n$. Adaptation holds by definition of the natural filtration. Independence and zero mean now legitimately giveHence the process is a martingale. Only first moments are used; no second-moment or boundedness assumption on the increments is needed.

A martingale with respect to a filtration $({\mathcal{F}}_n)$ is an adapted process: $X_n$ is ${\mathcal{F}}_n$-measurable for every $n$. It must satisfy $\mathbb{E}|X_n|<\infty$ andBy the tower property this is equivalent to $\mathbb{E}[X_m\mid {\mathcal{F}}_n]=X_n$ for all $m\ge n$. Integrability and adaptation are part of the definition, not consequences of writing a formal conditional-mean identity.

For observation $X=k$, the likelihood is proportional to $p^k(1-p{)}^{n-k}$. For $0<k<n$, its logarithmic derivative is $k/p-(n-k)/(1-p)$, vanishing uniquely at $k/n$; the logarithm is strictly concave. At $k=0$ or $k=n$, monotonicity gives the boundary maximum. Thus the maximum-likelihood estimator isA uniform prior multiplies this likelihood by a constant. The beta-integral normalization therefore gives the posterior distributionIn particular the posterior mean $(k+1)/(n+2)$ is not automatically the optimal estimator for the weighted loss in the next part.

Write $a=\mu /\lambda <1$. The embedded birth-death chain moves upward with probability $\lambda /(\lambda +\mu )$ and downward with probability $\mu /(\lambda +\mu )$ away from zero. The gambler's ruin formula, obtained by solving its two-point difference equation on $\{0,\dots ,N\}$ and then letting $N\to \infty$, gives ${\mathbb{P}}_k(\text{hit }0)=a^k$. Translation gives return probability $a$ from $r+1$ to $r$. From $r-1$ the chain hits $r$ almost surely: before that hit it remains in a finite reflecting interval, from every state of which a uniformly positive finite-step chance of hitting $r$ is available. This also proves that starting at zero it hits each $r\ge 1$ almost surely.

After departure from $r\ge 1$ the return probability is thereforeThe holding rate is $q_r=\lambda +\mu$, so the exponential occupation-time parameter is $q_r(1-r_r)=\lambda -\mu$. Applying the Strong Markov property at the almost-sure first hit givesAt zero the holding rate is $\lambda$ and return probability after departure is $a$, again giving parameter $\lambda -\mu$. From initial state $k$, the expected occupation time at zero is $a^k/(\lambda -\mu )$, since the time is zero unless zero is reached. Averaging over the initial law with probability generating function $G$ yieldsFor $\mu =0$, the interpretation is $a^0=1$, $a^k=0$ for $k>0$, and $G(0)=\mathbb{P}(X(0)=0)$; the same formulas hold.

Let $r_i$ be the probability of returning to $i$ after its first departure. Transience means $r_i<1$; in particular $q_i>0$. By the Strong Markov property, the number $N$ of visits, including the initial visit, has the geometric distribution $\mathbb{P}(N=n)=(1-r_i)r_i^{n-1}$, $n\ge 1$. The holding times of these visits are independent $\mathrm{Exp}(q_i)$ variables and are independent of the jump-chain return decisions.

For the total occupation time of a continuous-time Markov chain $T_i$ and $s\ge 0$, summing over $N$ givesUniqueness of the Laplace transform identifiesStarting elsewhere adds an atom at zero if the state might never be hit; the purely exponential conclusion uses the specified initial state $i$.

A time-homogeneous continuous-time Markov chain on a countable state space is a stochastic process with the Markov property: conditional on the current state, the law of its future depends on neither its past nor the current time. In the usual conservative, nonexplosive setting its sample paths are right-continuous step functions. Its Q-matrix $Q=(q_{ij})$ has $q_{ij}\ge 0$ for $j\ne i$, $q_i=-q_{ii}={\sum }_{j\ne i}{q}_{ij}<\infty$, and infinitesimal transition probabilities ${\mathbb{P}}_i(X(h)=j)=q_{ij}h+o(h)$ for $j\ne i$.

On entering $i$, the holding time has distribution $\mathrm{Exp}(q_i)$; independently, the next state is $j$ with probability $q_{ij}/q_i$. The successive states form the jump chain, whose transition probabilities areIf $q_i=0$, the state is absorbing, its holding time is infinite, and the jump-chain convention is $p_{ii}=1$. Nonexplosion ensures this construction defines the process for all finite times rather than accumulating infinitely many jumps in finite time.

The set of finitely disagreeing pairs is ${\bigcup }_{N\ge 0}{F}_N$, where $F_N$ consists of pairs whose digits agree at every position beyond $N$. For $M>N$, requiring agreement just at positions $N+1,\dots ,M$ defines a union of dyadic rectangles of total area $2^{-(M-N)}$: among the $2^{2M}$ pairs of length-$M$ prefixes, exactly $2^{M+N}$ meet these $M-N$ matching constraints, each rectangle having area $2^{-2M}$.

Since $F_N$ is contained in each of these finite-prefix agreement sets, its measure is at most $2^{-(M-N)}$ for every $M$, hence zero. Countable subadditivity now givesThus almost every pair disagrees in infinitely many digits. Dyadic endpoints, treated with the stipulated expansion convention, do not change any rectangle measure.

Each function $(x,y)\mapsto |R_n(x)-R_n(y)|$ is Borel measurable on the product square, by composition with its coordinate projections and arithmetic of measurable real functions. Its value is one exactly when the $n$th digits disagree. Therefore the partial disagreement countsare nonnegative Borel functions and increase to the extended-real function $f$. For every finite $a$, $\{f>a\}={\bigcup }_N\{f_N>a\}$ is Borel. HenceInfinite disagreement is allowed, so restricting the target to finite real values would lose part of the stated function.

The chosen non-eventually-one binary expansion fixes the ambiguity at dyadic rationals. Its $n$th digit isIn particularThis is a finite union of Borel sets; the zero-digit preimage is its complement in $[0,1)$. Since $R_n$ only takes the values zero and one, every Borel preimage is a union of these two sets. Thus it is Borel measurable. The half-open intervals implement the terminating-zero convention at dyadic endpoints.

Let $f_n$ be Borel measurable and converge pointwise to the finite real value $f$. For every real $a$,If $f(x)>a$, choose a rational $r$ strictly between them; convergence puts $f_n(x)>r$ eventually. Conversely eventual strict inequality $f_n(x)>r>a$ implies $f(x)\ge r>a$. The right side belongs to $\mathcal{E}$ by its countable union/intersection operations. Since these half-lines generate the Borel sigma-algebra, $f$ is a Borel measurable function.

The Borel sigma-algebra $\mathcal{B}(\mathbb{R})$ is the smallest sigma-algebra containing all open subsets of $\mathbb{R}$. A real-valued Borel measurable function on $(E,\mathcal{E})$ is a map $f$ with $f^{-1}(B)\in \mathcal{E}$ for every $B\in \mathcal{B}(\mathbb{R})$. It suffices to test the sets $(a,\infty )$ for rational $a$, because these generate the Borel sigma-algebra.

Here “Borel function” describes measurability into the Borel structure of the target; the domain need not itself be a topological space.

The semicubical parabola has coordinate ringunder $x\mapsto t^3$, $y\mapsto t^2$. The isomorphism is valid in every characteristic: after division by the monic polynomial in $x$, a remainder $P(y)+xQ(y)$ maps to a sum of distinct even powers and odd powers at least three, and can vanish only when both polynomials are zero.

Its fraction field contains $t=t^3/t^2$, which is an integral element over $A$ because it satisfies $z^2-y=0$. However $t\notin A$, since a polynomial in $t^2,t^3$ has no term of degree one. Thus $A$ is not an integrally closed domain. The ring $k[t]$ of the affine line is a principal ideal domain and therefore an integrally closed domain. This property is preserved by ring isomorphism. By coordinate-ring functoriality,The parametrization is a bijection on points over an algebraically closed field, but that does not make its inverse a morphism at the cusp.

An isomorphism $\varphi :X\to Y$ of affine varieties and its inverse $\psi$ pull regular functions back to regular functions. By the preceding identification this gives k-algebra homomorphismsThey preserve sums, products, and every constant in $k$. Contravariance of pullback and $\psi \circ \varphi ={\mathrm{id}}_X$, $\varphi \circ \psi ={\mathrm{id}}_Y$ give ${\varphi }^{\ast }{\psi }^{\ast }={\mathrm{id}}_{A(X)}$ and ${\psi }^{\ast }{\varphi }^{\ast }={\mathrm{id}}_{A(Y)}$. Hence

Let $A=A(X)=k[x_1,\dots ,x_m]/I(X)$. Polynomial restrictions are regular functions, and a restriction is zero precisely when its class in the coordinate ring is zero. We prove surjectivity.

Using the usual irreducible-variety convention, view a regular function $f$ as an element of the fraction field of $A$. On a neighbourhood of every point it has the form $a/b$, where $a,b\in A$ and $b$ does not vanish at that point. Equality on a nonempty open subset of the irreducible variety is equality of rational functions. DefineThis is an ideal and contains a denominator nonzero at each point. Thus its zero locus is empty. The Hilbert Nullstellensatz implies $J=A$, so $1\in J$ and $f\in A$. ThereforeThe result also holds under a convention allowing reduced reducible affine varieties. One may cover by finitely many principal opens $D(b_i)$, represent $f$ in each localization, and increase powers until the local representatives glue after clearing denominators. Since these opens cover, the ideal generated by those powers is the unit ideal by the Nullstellensatz; a unit linear combination glues their numerators to one element of $A$. Irreducibility is used only for the shorter fraction-field proof, not for the assertion itself.