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Forcing constructs new first-order models while controlling which ground-model properties survive. Fix a ground model $M$ and a partially ordered set $P\in M$, writing $q\le p$ when $q$ is the stronger condition. A generic filter is upward closed, directed towards stronger conditions, and meets every ground-model dense subset of a forcing order. If $M$ is externally a countable transitive model, enumerate its dense subsets and successively choose stronger conditions in them. The resulting filter is $M$-generic. External countability is crucial: this is not a claim that the ground model regards its own universe as countable.

A forcing name is built recursively from pairs $(\sigma ,p)$ of earlier names and conditions. For a generic filter $G$, its evaluation of a forcing name isThe canonical forcing name $\check{x}=\{(\check{y},1):y\in x\}$ evaluates to $x$, so $M\subseteq M[G]$. Recursive evaluation also proves transitivity. Name ranks bound the ranks of their evaluations, so an ordinal in the extension is below some ground ordinal and is already a ground ordinal. Thus forcing preserves ordinals.

The central link between syntax and generic interpretation is the forcing theorem. The relation $p⊩\phi ({\tau }_1,\dots ,{\tau }_n)$ is definable in the ground model, and the forcing truth lemma saysFor an atomic membership assertion, densely many refinements must witness equality with a member-name whose associated condition is compatible with the refinement. Equality is obtained by recursive mutual inclusion of names. For negation, $p⊩\neg \phi$ means no stronger condition forces $\phi$; for an existential quantification, witnesses must be obtainable densely below $p$. Induction on name rank establishes the atomic clauses, and induction on formulas establishes definability and the truth lemma. The generic filter meets the relevant dense sets at the existential step. An arbitrary semantic assertion about one chosen generic filter is insufficient: the definable relation must work uniformly for all generic interpretations.

This explains preservation of ZFC. Extensionality and foundation follow from transitivity; canonical names supply infinity. Pair and union names give pairing and union. For separation, restrict a name's members to conditions forcing the defining formula. For replacement, the definable forcing relation lets ground axiom schema of collection bound a set of names for all possible witnesses to the unique outputs; evaluating their collected name gives the range. For power set, let $\tau$ name the given set and let $D$ be the set of its member-names. Every subset of ${\tau }_G$ has an equivalent name drawn from $\mathcal{P}(D\times P)$, by the truth lemma, so a ground set of candidate names collects all its subsets. Finally choose a ground well-order of the relevant names; its evaluated image surjects onto the desired set, and choosing the first name for each value well-orders that set. This is why ordinary forcing over a choice model preserves choice.

For a concrete example, Cohen forcing consists of finite partial functions $\omega \to 2$, ordered by extension. The generic union $c=\bigcup G$ is total because deciding each coordinate is dense. For every ground real $x$, the set of conditions disagreeing with $x$ somewhere is dense, so $c\ne x$. This construction changes the universe by adding a genuinely new real without changing its ordinals.

The size and compatibility of conditions determine what else changes. With the countable chain condition for forcing, an ordinal-valued name has only countably many possible values at each coordinate: take a maximal antichain deciding that value. Therefore the possible range of a name for a map with infinite ground domain of cardinality $\mu$ lies in a ground set of size at most $\mu \cdot {\aleph }_0=\mu$. This prevents cardinal collapse and changes of uncountable regular cofinalities. Forcing by finite partial functions $\lambda \times \omega \to 2$ is ccc: the delta-system lemma thins any uncountable family of finite domains to a delta-system, and a further thinning makes their values agree on its finite root. Two remaining conditions are compatible. Distinct coordinates add distinct reals.

For example, start with GCH and take $\lambda ={\aleph }_2$. This forcing has size $\lambda$ and adds $\lambda$ distinct reals. A nice name for a real uses countably many antichains, and each antichain is countable. The ground arithmetic ${\lambda }^{{\aleph }_0}=\lambda$ bounds the number of such names by $\lambda$. Hencegiving a relative-consistency example of the failure of CH. The ground choice of GCH is available through the constructible universe; no claim that every starting model has this arithmetic is needed.

By contrast, countably closed forcing adds no new countable sequences of ground ordinals. Given a name, recursively strengthen a condition to decide each coordinate, then take a lower bound of the descending sequence. That condition decides the entire sequence. Closure and chain conditions thus control different features; neither should be silently substituted for the other.

Further applications include changing cofinalities, adding branches through trees, and constructing prescribed continuum patterns. If the aim is to violate choice, an ordinary generic extension of a choice model will not do: one restricts to a suitable symmetric collection of names, checking the axioms again. These examples also mark a limitation: forcing proves relative consistency, not absolute consistency of the background theory. The countable-transitive-model picture gives a concrete semantic construction; the definable forcing relation and Boolean-valued formulation supply the corresponding formal consistency arguments without asserting that a transitive ground model can be proved to exist.

An ultraproduct converts coordinatewise finite information into a single first-order structure. Let $(M_i{)}_{i\in I}$ be nonempty structures in a common first-order language, and let $U$ be an ultrafilter on $I$. In ${\prod }_i{M}_i$, setThe ultraproduct ${\prod }_i{M}_i/U$ has these equivalence classes as its elements. Functions are interpreted coordinatewise; a relation holds of classes exactly when it holds on a $U$-large set of indices. Intersecting finitely many large agreement sets shows that these interpretations are independent of representatives.

The Łoś theorem, which explains the construction's usefulness, saysHere is the proof. Induction on terms establishes the coordinatewise interpretation of atomic formulas. Finite intersections handle conjunction, and the ultrafilter property handles negation, since precisely one of a set and its complement is large. For an existential quantification, a witness class gives coordinate witnesses on a large set. Conversely, when the coordinate existential truth set is large, select a witness in each of those structures and a default element elsewhere. Its class is a witness in the ultraproduct. This uses choice. Structural induction on formulas now proves the theorem, with universal quantification obtained by negation.

Compactness. If every finite subset of a first-order theory $T$ has a model, take $I$ to be the finite subsets $\Gamma \subseteq T$ and choose $M_{\Gamma }\models \Gamma$. The cones $\{\Gamma :\Delta \subseteq \Gamma \}$ for finite $\Delta$ have the finite intersection property. Extend their filter to an ultrafilter by the ultrafilter lemma. For each $\phi \in T$, its cone is large, so the Łoś theorem makes the ultraproduct satisfy $\phi$. This proves the compactness theorem, including theories of arbitrary set size.

Elementary extensions and nonstandard elements. An ultrapower uses a fixed structure at every coordinate. The diagonal map sending $a$ to the class of the constant function $a$ is an elementary embedding by the Łoś theorem. For a nonprincipal ultrafilter on $\omega$, the class of $i\mapsto i$ in an ultrapower of the natural numbers exceeds every standard integer, because each corresponding tail is large. In an ultrapower of the real ordered field, the reciprocal of this element is positive and smaller than every positive standard reciprocal integer: an infinitesimal. These examples implement ordinary finite statements in a larger structure without claiming that every external subset or second-order assertion transfers.

Saturation. Such an ultraproduct over $\omega$ is countably saturated. To see the actual diagonal argument, list a finitely satisfiable type ${\phi }_0(x),{\phi }_1(x),\dots$ with parameters in the ultraproduct and fix representatives for those parameters. For each $n$, the set $E_n$ of coordinates admitting a simultaneous witness to its first $n$ formulas belongs to $U$ by the Łoś theorem. Arrange that the $E_n$ decrease and put $D_n=E_n\cap \{i:i\ge n\}$. The $D_n$ decrease, belong to $U$, and have empty intersection. At coordinate $i$, choose a witness to the first $m(i)$ formulas, where $m(i)$ is the largest $n$ with $i\in D_n$, or choose a default element if there is no such $n$. For each fixed $n$, $m(i)\ge n$ on the large set $D_n$. Therefore the resulting class realizes every formula. This proves countable saturation of an ultraproduct over omega; it does not assert saturation at all larger cardinalities.

Algebraic limits. Take finite fields ${\mathbb{F}}_p$ indexed by the primes and a nonprincipal ultrafilter. Their ultraproduct is a field of characteristic zero: for every fixed positive integer $n$, only finitely many primes divide $n$, so $n\cdot 1\ne 0$ on a large set. Every first-order sentence true in all finite fields holds in this infinite field, and more generally so does any sentence true for all but finitely many of the factors. This is a way to create an infinite field retaining uniform first-order information from finite fields. It does not make the field algebraically closed: the first-order sentence that every element is a square already fails in infinitely many of the factors.

Large cardinals. A countably complete ultrafilter gives a well-founded membership ultrapower of the universe. If it had an infinite descending chain represented by $f_n$, each setwould be large. Countable completeness makes ${\bigcap }_n{A}_n$ nonempty, giving an actual infinite membership descent at any index in the intersection, contrary to foundation. The Mostowski collapse therefore gives a transitive target and an ultrapower embedding $j:V\to M$.

For a measure on a measurable cardinal $\kappa$, every function $\kappa \to \alpha$ with $\alpha <\kappa$ is constant on a large set: otherwise intersect the fewer than $\kappa$ complements of all its fibres. Consequently $j$ fixes the ordinals below $\kappa$. The class of the identity function is below $j(\kappa )$ and above every constant class below $\kappa$, since all tails are large. Thus $j(\kappa )>\kappa$ and its critical point is $\kappa$. Fine ultrafilters on $P_{\kappa }(\lambda )$ that are normal ultrafilters on small subsets and kappa-complete yield the stronger embeddings witnessing supercompact cardinals.

These applications share a single principle: an ultraproduct preserves precisely the first-order information seen by its ultrafilter. Principal ultrafilters merely select a factor; nonprincipal ones can produce new sizes, elements and saturation. Existence of the needed ultrafilters and selection of coordinate witnesses are set-theoretic assumptions, and changing the ultrafilter may change the resulting model.

Let $P_{s,t}>0$ denote the time-$s$ price of a zero-coupon bond paying one at maturity $t$, with $P_{t,t}=1$. Information available at $s$ is the sigma-algebra ${\mathcal{F}}_s$ in the market filtration. For a curve differentiable in maturity, the instantaneous forward rate and its diagonal short rate areThe last identity follows by integrating the maturity derivative and using $P_{s,s}=1$. The continuous-time bank account reinvesting at the short rate is $B_s=\exp ({\int }_0^s{R}_{u}du)$, and its reciprocal is the discount factor. Modelling forward rates means specifying a random maturity-indexed curve with an evolution consistent with the same pricing measure for every bond. The expectation below is under that money-market risk-neutral measure, not an unspecified physical measure.

For fixed maturity $t$, if $P_{s,t}/B_s$ is a true martingale on $0\le s\le t$, its terminal value is $1/B_t$. ConsequentlyConversely, this representation gives $P_{s,t}/B_s={\mathbb{E}}_Q[B_t^{-1}\mid {\mathcal{F}}_s]$, a conditional-expectation martingale by the tower property. For example, for $r<s$, conditioning the right side on ${\mathcal{F}}_r$ returns ${\mathbb{E}}_Q[B_t^{-1}\mid {\mathcal{F}}_r]$. These claims assume ${\mathbb{E}}_Q{B}_t^{-1}<\infty$. A positive discounted local martingale alone need not equal the conditional expectation of its terminal value, so the word “martingale” cannot be weakened without an extra integrability argument.

In the Vasicek model, the Brownian motion $W$ and coefficients in $d{R}_s=\alpha (\beta -R_s)ds+\sigma d{W}_s$ must be understood under this pricing measure. Multiplying by $e^{\alpha s}$ and integrating gives, conditionally on ${\mathcal{F}}_s$,Assume first $\alpha \ne 0$, and put $\tau =t-s$ and $B(\tau )=(1-e^{-\alpha \tau })/\alpha$. Integrating the explicit solution over $s\le u\le t$ yieldsThe interchange of the stochastic integrals is justified by their deterministic square-integrable kernels on the finite triangular domain; equivalently approximate the kernels by step functions, interchange finite sums, and pass to the limit using the Itô isometry. The last integral conditionally has a centered normal distribution independent of ${\mathcal{F}}_s$, with varianceA deterministic Brownian integral is Gaussian because its step approximations are sums of independent normal increments; their variances converge by the Itô isometry. Completing the square in the normal density proves $\mathbb{E}{e}^{-Y}=e^{-m+v/2}$ for a normal variable of mean $m$ and variance $v$. Hence the zero-coupon bond price is exponential-affine:An equivalent familiar form isAt $t=s$, both coefficients are zero and the bond value is one. These formulas are valid for negative as well as positive nonzero $\alpha$ on finite horizons, though a genuinely mean-reverting Vasicek model has $\alpha >0$. For $\alpha =0$, direct integration gives $B(\tau )=\tau$, $J(\tau )={\tau }^3/3$ andAll the exponential moments used above are finite, so the prices defined by the conditional expectation do give true discounted martingales.

To fit an initial bond curve, let its observed prices $p_0(t)$ be positive, satisfy $p_0(0)=1$, and have a sufficiently smooth instantaneous forward rate $f_0(t)=-d\log p_0(t)/dt$. Set $R_0=f_0(0)$; this compatibility is forced by the infinitesimal-maturity bond price. Replace $\beta$ by a deterministic function $\theta (t)$, with $\alpha \ne 0$, so thatIts mean $m(t)={\mathbb{E}}_Q{R}_t$ satisfies $m^{\prime }=\alpha (\theta -m)$, $m(0)=R_0$. The variance of the integrated rate remains ${\sigma }^{2}J(t)$, since changing deterministic drift does not change the centered Gaussian process. ThereforeThe required mean is consequently $m(t)=f_0(t)+{\sigma }^{2}B(t{)}^2/2$. Solve its mean equation for $\theta$:Combining the last two terms gives the explicit Hull-White model calibration in the question's parameter convention:With $R_0=f_0(0)$, the specified mean solves the same linear differential equation and initial condition as the model mean, so it is the actual mean. The resulting model forward curve equals $f_0$; integrating and using $P_{0,0}=p_0(0)=1$ proves $P_{0,t}=p_0(t)$ for every maturity. If the drift is instead written $\tilde{\theta }(t)-\alpha R_t$, its calibration parameter is $\tilde{\theta }=\alpha \theta$, not $\theta$ itself.

The printed allowance of arbitrary constants needs a qualification for this final claim. If $\alpha =0$, replacing $\beta$ by $\theta (t)$ changes no coefficient at all: $R_t=R_0+\sigma W_t$ regardless of $\theta$. Its initial curve is fixed at $P_{0,t}=\exp (-R_{0}t+{\sigma }^2{t}^3/6)$. With $\sigma \ne 0$, the compatible smooth observed curve $p_0(t)=e^{-R_{0}t}$ is already a counterexample to the unconditional fitting claim. Thus arbitrary compatible smooth positive initial curves can be fitted by the stated replacement when $\alpha \ne 0$; it cannot fit arbitrary curves when $\alpha =0$. In that degenerate case one can instead introduce an independent deterministic drift $k(t)$ in $d{R}_t=k(t)dt+\sigma d{W}_t$, choosing $k(t)=f_0^{\prime }(t)+{\sigma }^{2}t$ to recover the desired initial curve. Likewise an externally fixed $R_0\ne f_0(0)$ is incompatible with exact fitting, and an arbitrary nonsmooth set of prices need not define the differentiable forward curve assumed in the model.

A field $u(x,t)$ has infinitely many degrees of freedom: its initial state is a function rather than a finite list of coordinates. For a Hamiltonian evolution, the analogue of the finite-dimensional construction seeks an infinite sequence of independent conserved functionals $H_j[u]$ whose Poisson brackets vanish. Their Hamiltonian vector fields then define commuting time evolutions. For example, a Poisson operator $J$ produces the field equation $u_t=J\delta H/\delta u$, where $\delta H/\delta u$ is the variational derivative.

A Lax equation packages many conservation laws into one operator identity. If $L_t=[A,L]$, cyclicity of an appropriate trace givesFor a formal pseudodifferential operator the relevant trace is the Adler trace; suitable fractional powers of a monic differential operator produce further conserved functionals. A Lenard-Magri recursion can prove that such functionals are Poisson-commuting functions, rather than merely conserved quantities. A Lax equation by itself does not establish all the independence and completeness properties needed for integrability.

The Korteweg-De Vries equation illustrates the mechanism. In one time normalization take $L={\partial }^2+u$ and $A={\partial }^3+\frac{3}{2}u\partial +\frac{3}{4}{u}_x$. Direct composition givesIts spectral problem, together with the inverse scattering transform for decaying data or suitable spectral coordinates for periodic data, turns nonlinear evolution into simple evolution of spectral data. Solitary waves arise from discrete spectral data, while continuous data describe dispersive radiation.

Thus infinite-dimensional Hamiltonian integrability is more than an infinite list of formulas: one wants sufficiently complete commuting invariants and a reconstruction mechanism that solves the evolution. Boundary conditions and the function space are essential. A formal hierarchy need not converge, and an infinite family of conserved quantities does not automatically give a compact infinite-dimensional torus or an unrestricted analogue of the Arnold-Liouville theorem.

First let $n\ge 0$. The ordinary Leibniz rule, applied to a test function $h$, givesEquivalently, the composition of differential operators isFor completeness, this formula follows by induction: applying $\partial$ differentiates either the coefficient or the remaining derivative of $h$, and the two contributions combine by $\left(\genfrac{}{}{0ex}{}{n}{r}\right)+\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{r}\right)$. The initial case is multiplication by $f$.

For all integers $n$, define $\left(\genfrac{}{}{0ex}{}{n}{0}\right)=1$ and $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=n(n-1)\cdots (n-r+1)/r!$. The formal pseudodifferential composition rule isFor $n=-s<0$, $\left(\genfrac{}{}{0ex}{}{-s}{r}\right)=(-1{)}^r(\genfrac{}{}{0ex}{}{s+r-1}{r})$, so this is usually an infinite series. In particularApplying $\partial$ on the left gives $m_f$: all later coefficient derivatives cancel in pairs. This uniquely determines the normal-ordered inverse expansion. Repeating the inverse construction, or using the same binomial recurrence downward in $n$, gives every negative power. More explicitly, applying $\partial$ to the proposed $n$-series yields coefficient $\left(\genfrac{}{}{0ex}{}{n}{r}\right)+\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{r}\right)$ at $f^{(r)}{\partial }^{n+1-r}$, so it is consistent with the already determined $(n+1)$-series. Negative powers here are formal; differentiation on all of $C^{\infty }(\mathbb{R})$ has a nontrivial kernel and has no genuine two-sided inverse there.

Use the canonical representatives of functionals of $u$: $P_f=f{\partial }^{-1}$, $P_g=g{\partial }^{-1}$. The formal pseudodifferential composition rule givesand the reversed product has $f,g$ exchanged. ThereforeMultiplying by $L={\partial }^2+u$ on the left, only the leading displayed commutator term can produce a residue. Differentiation of its coefficient produces powers at most ${\partial }^{-2}$, and multiplication by $u$ produces powers at most ${\partial }^{-3}$. Thusand, with the trace convention of part (i),There is no $u$ dependence in this bracket.

Moreover $\mathrm{res}(P_{f}L)=fu$, so $l_{P_f}=\int fudx$ and $\delta l_{P_f}/\delta u=f$. If $K(x,y)=\{u(x),u(y)\}$ is the distributional field bracket, it is characterized bySince $\int {\partial }_x\delta (x-y)g(y)dy=g^{\prime }(x)$, the answer isThe Dirac delta is periodic in the periodic setting, or the ordinary Dirac delta on the real line with admissible test functions. This is the first Hamiltonian structure of the KdV equation, with Poisson operator $-2{\partial }_x$. Its skewness follows by integration by parts; its coefficients are constant, so the functional Jacobi identity has no coefficient-variation terms.

A qualification is needed if the displayed trace formula is interpreted as a bracket on the reduced space for arbitrary operator representatives. Multiplication by a function $h$ gives $l_{m_h}({\partial }^2+u)=0$, yet direct expansion givesWith $h=g=\cos x$ on a period $[0,2\pi ]$, the latter is $\pi$. Hence the formula does not descend through every possible representative of a functional on this reduced affine space. The requested $f{\partial }^{-1},g{\partial }^{-1}$ representatives, used consistently for variational gradients, give the well-defined constant field bracket just computed; a general restriction from the ambient operator algebra needs an appropriate reduction.

Put every formal pseudodifferential operator in normal order, with all coefficient functions on the left. Its formal pseudodifferential residue is the coefficient of ${\partial }^{-1}$. The Adler trace isHere and below, use periodic coefficients and integrate over a period, or impose convergence and vanishing boundary terms on the real line. Equivalently, the algebraic integration functional must annihilate total derivatives. This convention is needed for the requested trace property; arbitrary elements of $C^{\infty }(\mathbb{R})$ do not supply it automatically.

It suffices to compute with monomials $A=a{\partial }^p$, $B=b{\partial }^q$. Write $r=p+q+1$. If $r<0$, no term in either product has power ${\partial }^{-1}$, so both residues are zero. If $r\ge 0$, the formal pseudodifferential composition rule givesSince $q=r-1-p$, reversal of the factors in the numerator provesAfter $r$ integrations by parts, $\int a{b}^{(r)}dx=(-1{)}^r\int b{a}^{(r)}dx$. Hence the two product traces agree. More explicitly, for $r\ge 1$ their residue difference is the total derivativeFor $r=0$ the residue difference is $ab-ba=0$. In products of operators bounded above in order, only finitely many monomial pairs can contribute to a residue: the required $r$ is nonnegative, so the two downward coefficient indices have a bounded sum. Summing the monomial result therefore proves

The boundary convention is substantive. Without it, take $A={\partial }^2$, $B=(\tanh x){\partial }^{-2}$. Both individual residues vanish, but $\mathrm{res}[A,B]=2(\tanh x{)}^{\prime }=2{\mathrm{sech}}^{2}x$ and its integral over $\mathbb{R}$ is $4$, not zero. The primitive has unequal limits at the two ends. Also a generic smooth residue need not have a convergent integral at all. Thus the proof supplies the intended cyclic trace under its necessary analytic or formal integration convention.

The intended construction is to solve the Lenard-Magri recursion one step at a time. With $H_0$ a Casimir function of the first Poisson structure, seek $H_n$ satisfyingHere $X_H^{(a)}f=\{f,H{\}}_a$, consistently with part 1(i). If every step can be solved by a globally defined smooth function, the formal series ${\sum }_{n\ge 0}{H}_n{\lambda }^n$ is a Casimir function for the Poisson pencil, and part (i) proves that all the $H_n$ are Poisson-commuting functions for every parameter value after complexification.

The recursive step is a concrete potential problem. Put $Y=-X_{H_{n-1}}^{(2)}$. It must first be tangent to the symplectic leaves of ${\Pi }_1$. On a regular leaf use canonical coordinates $(q_i,p_i)$. The required equations areThus the leaf one-formmust be exact. Where it is closed on a contractible coordinate domain, define $H_n$ by integrating ${\alpha }_Y$ from a base point; path independence follows from closedness. A function of the leaf labels can be added, corresponding to a Casimir function of the first Poisson structure. Global solvability also requires vanishing periods and smooth matching between leaves. This describes how to perform each solvable step without pretending that a degenerate Poisson structure can simply be inverted.

Compatibility alone does not ensure these solvability conditions. There is an immediate obstruction to a Lenard-Magri recursion even in two dimensions. On $M={\mathbb{R}}^2$ takeBoth are Poisson structures, and their linear combination is a scalar multiple of the canonical Poisson bracket, so they are compatible. Every function, including $H_0$, is a Casimir function for the first bracket. But the first recursion, evaluated at $f=y$, would requirea contradiction. An arbitrary initial Casimir need not extend even to $H_1$. Consequently the unconditional existence assertion requires an additional hypothesis: the indicated vector field must be Hamiltonian for ${\Pi }_1$ at every step. Under that hypothesis the recursive construction and involution proof above complete the intended result; without it the counterexample rules out the requested infinite extension.

The coefficients of the formal series must be smooth functions $H_n\in C^{\infty }(M)$, rather than points of $M$. Interpret complex parameter values in the complexified function algebra. The Casimir function condition for the Poisson pencil is a coefficientwise formal identity:Equality of the formal coefficients provesThese are the Lenard-Magri recursion relations, valid for every smooth function $f$.

For $i\ge 1$ and $j\ge 0$, use the recursion in the first argument, then antisymmetry and the recursion in the second:Each step lowers the first index while raising the second. Iterating $i$ times givesbecause $H_0$ is a Casimir function of the first Poisson structure. This includes $i=0$ directly. The recursion one step higher then givesThus the commuting coefficients of a Poisson pencil Casimir satisfyThey are therefore Poisson-commuting functions for every member of the Poisson pencil. The argument proves involution, not functional independence; an infinite sequence can contain repetitions or zero coefficients.

Choose $a_0\in A$ and put $R=\Vert a_0-y\Vert$. The intersection $K=A\cap \overline{B}(y,R)$ is nonempty, closed and bounded in ${\mathbb{R}}^n$, hence compact. The continuous distance function $a\mapsto \Vert a-y\Vert$ attains a minimum at some $x\in K$. For $a\in A\setminus K$ its value is strictly greater than $R$, while the value at $x$ is at most that at $a_0$, namely $R$. Thus this minimum on $K$ is a minimum on all of $A$:The set $A$ itself need not be bounded; only the compact portion needed to find the nearest point is bounded.

Close the real-line contour by a positively oriented upper semicircle of radius $R$, beyond every pole. The rational function is $O(R^{-2})$ uniformly on that arc, and $|e^{iz}|=e^{-\Im z}\le 1$. Its arc integral is therefore $O(R^{-1})$ and vanishes. The real integral is absolutely convergent, while no pole lies on it. The residue theorem givesOnly actual poles contribute; any removable zero of the denominator has zero residue. The same contour proof covers poles of any order.

For the particular denominator, the upper poles are $a=1+i$ and $b=-1+i$. They are simple, with residues $e^{ia}/(4a^3)$ and $e^{ib}/(4b^3)$. Sincetheir sum isMultiplication by $2\pi i$ gives the real answer. The sine part of the real-line integral is zero by oddness, so

Fixed endpoint variations yield the Euler–Lagrange equationWhen $F$ is independent of $z$, differentiation givesThis is the Beltrami identity. Denote the constant by $1/k$ when nonzero; for a general Lagrangian a zero constant is also possible.

The film area is $S=2\pi {\int }_{-H}^{H}r\sqrt{1+r^{\prime 2}}dz$. Suppressing its harmless positive factor $2\pi$, take $F=r\sqrt{1+r^{\prime 2}}$. The Euler–Lagrange equation reduces toThe first integral is $r/\sqrt{1+r^{\prime 2}}=1/k$ with $k>0$. Hence $1+r^{\prime 2}=k^2{r}^2$ and, using the differential equation without dividing by $r^{\prime }$, $r^{\prime \prime }=k^{2}r$. The positive solution obeying that first integral is $r=k^{-1}\cosh (k(z-z_0))$. Equal endpoint radii force $z_0=0$, givingThus every smooth positive stationary film in this axisymmetric class is a catenoid.

Put $u=kH>0$. The boundary equation is $R/H=f(u)=\cosh u/u$. The function tends to infinity at both endpoints of $(0,\infty )$, andIts minimum occurs when $u=\coth u$. The function $u-\coth u$ is strictly increasing, with derivative $1+{\mathrm{csch}}^{2}u>0$, and has limits $-\infty$ and $+\infty$. It therefore has exactly one positive zero $A$. At this point $1/A=\tanh A$, so $f(A)=\sinh A$. This proves the catenoid existence threshold for equal rings:At equality there is one stationary solution and above it two. The equation describes candidates for an area minimum, not a guarantee that both branches minimize. The thinner-neck branch is unstable to suitable axisymmetric variations; topology-changing collapse can also compete with connected films. In particular two separate planar disks have area $2\pi R^2$. The derived threshold concerns existence of the smooth annular stationary film; it does not exclude such disconnected competitors.