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The Arnold-Liouville theorem concerns a connected compact component $L$ of a regular common level of $n$ commuting independent smooth functions $F=(F_1,\dots ,F_n)$ on a $2n$-dimensional symplectic manifold. It states that $L$ is a Lagrangian torus, and that a neighbourhood of $L$ has action-angle variablesin which every $F_i$ depends only on $I$. If $H=F_1$ or, more generally, $H=h(F)$, its Hamiltonian flow satisfiesAngles have period $2\pi$ here. The theorem is local near this regular compact component; it does not assert globally defined action-angle variables across singular levels or over an entire base with monodromy.

First, $L$ has dimension of a manifold $n$ by the submersion theorem. The Hamiltonian vector fields $X_{F_i}$ are independent, tangent to $L$, and commute. They span $TL$. Moreoverso $L$ is a Lagrangian submanifold. Compactness makes these restricted vector fields complete. Their joint flow is an ${\mathbb{R}}^n$-action on $L$. Its orbits are open because the fields span $TL$, so connectedness gives one orbit. The stabilizer $\Lambda$ of a point is discrete by the inverse function theorem, andCompactness forces $\Lambda$ to be a full-rank Euclidean lattice: otherwise an unbounded linear coordinate transverse to its span would descend to the quotient. Thus $L$ is an $n$-torus.

Next choose a small ball of regular values near $F(L)$ and a neighbourhood $N$ of this component which is a product family of compact tori $L_c$. This local trivialization follows directly by choosing transverse vector fields $Z_j$ with $d{F}_i(Z_j)={\delta }_{ij}$ and lifting short radial paths in the base; compactness of $L$ gives a uniform neighbourhood where their flows exist. Hence $N$ deformation retracts onto $L$. The commuting joint flows on nearby fibres have smoothly varying full-rank Hamiltonian period lattices. A basis $t^{(j)}(c)\in {\mathbb{R}}^n$ of their periods can be chosen smoothly on this small ball: continue the return equations from a basis at the central fibre, using their nonsingular vertical flow derivatives and the implicit function theorem. No global choice of lattice basis is needed.

Because $\omega {|}_L=0$ and $N$ retracts onto $L$, its closed symplectic form is exact on $N$. Choose a one-form $\lambda$ with $d\lambda =-\omega$. Let ${\gamma }_j(c)$ be the smoothly continued cycles represented by the period basis and define the action integralsFor a transverse variation $v$ of the fibre, differentiating a cycle integral and using Cartan's magic formula eliminates the integral of the exact term. With the cycle parametrized by the joint-flow time $t^{(j)}$, this givesThusThe period matrix is nonsingular, so $I$ gives coordinates on the base. The corresponding Hamiltonian vector fields satisfyTheir time-$2\pi$ flows return along the basis cycles. They commute because each $I_j$ is a function of the commuting $F_i$. Consequently they give a free torus action on $N$.

Choose a section over the action-coordinate ball and use this torus action to define angles $\theta$. Since $X_{I_j}={\partial }_{{\theta }_j}$, the symplectic form takes the formThe closed base two-form $\beta$ is exact on the ball: write $\beta =d\alpha$, $\alpha ={\sum }_j{a}_j(I)d{I}_j$, by the Poincare lemma. Changing the angular origins by ${\theta }_j^{\prime }={\theta }_j+a_j(I)$ removes this term, since ${\sum }_{j}d{\theta }_j^{\prime }\wedge d{I}_j=\omega$. This completes the construction of action-angle variables. The functions $F_i$ are constant on the fibres, so are functions of $I$ alone, and Hamilton's equations give the claimed straight-line motion. This proves both the topological and symplectic parts of the Arnold-Liouville theorem.

For continuously differentiable coefficients on a simply connected domain, the exactness condition is $P_y=Q_x$. Locally the same condition suffices without a global topological assumption. An exact first-order ordinary differential equation has a potential $F$ with $F_x=P$ and $F_y=Q$, so $dF/dx=P+Q{y}^{\prime }=0$ along a solution. Thus its implicit solution is $F(x,y)=C$. One local construction, on a rectangle about $(x_0,y_0)$, isThe equality of cross derivatives verifies both required derivatives; on a simply connected domain one may use the corresponding path-independent integral of the exact differential form.

Here $P=4x+3y$, $Q=3x+3y^2$ and $P_y=Q_x=3$. Integrating $P$ with respect to $x$ gives $F=2x^2+3xy+K(y)$. Matching $F_y=Q$ gives $K^{\prime }=3y^2$, so take $K=y^3$. The initial value sets $C=F(1,2)=16$. ThereforeThis is the requested explicit relation between the coordinates. Since $F_y(1,2)=15\ne 0$, the implicit function theorem gives a unique local graph through the initial point, with slope $-10/15=-2/3$.

Along an isolated two-phase coexistence line, two distinct stable minima have equal free energy. If this line ends at an ordinary interior endpoint without encountering another phase, the distinction between the two phases must disappear: their minima and the intervening barrier merge. For separated nondegenerate minima, their stationary values are smooth functions of the controls. The difference $\Delta V=V(m_1)-V(m_2)$ has ${\partial }_h\Delta V=-m_1+m_2\ne 0$. The implicit function theorem therefore continues the equation $\Delta V=0$ as a local coexistence curve. It cannot end while those distinct stable minima persist.

At the merger the potential satisfiesThe third-derivative condition is essential: a stationary zero-curvature point with nonzero third derivative is an inflection, not a stable equilibrium. Shifting to $\psi =m-m_c$ gives an ordinary quartic Landau free energy with two independent control directions, temperature-like and field-like. The curvature and the order-parameter discontinuity vanish, the magnetic susceptibility diverges, and the correlation length diverges. The endpoint is therefore associated with a continuous transition. This is the critical endpoint of a quartic Landau free energy mechanism.

In the symmetric quartic example, the coexistence line is $h=0,r<0$. Its jump $2\sqrt{-r/u}$ vanishes at $(r,h)=(0,0)$, where $V=u{m}^4/4$. This supplies a concrete realization. The word “must” requires the stated ordinary-endpoint assumptions: a first-order line can instead meet a third phase, a multicritical point, a boundary of parameter space, or continue without a finite endpoint. A spinodal point alone is not an equilibrium critical endpoint.

The general-order Cauchy-Kovalevskaya theorem applies to analytic data on a non-characteristic hypersurface. In local coordinates $(t,z)$, a real analytic system of order $m$ in an unknown vector $U$ is in normal form whenwith $F$ real analytic near the initial jet. Analytic data ${\partial }_t^{j}U(0,z)=h_j(z)$ for $0\le j<m$ determine a unique real analytic solution near each initial point. More general systems, including different orders for the components, have the same conclusion after solving for the highest normal derivatives: the corresponding coefficient matrix, or highest-derivative Jacobian for a nonlinear system, must be invertible. This is the system's non-characteristic hypersurface condition. The uniqueness asserted by the Cauchy-Kovalevskaya theorem is initially uniqueness in the real analytic class.

A real analytic coordinate change flattens a real analytic initial hypersurface. Non-characteristicity permits solving for its highest normal derivatives by the real analytic implicit function theorem. Introduce all derivatives through order $m-1$ as additional unknowns. Their normal derivatives are either another jet variable, a tangential first derivative of a jet variable, or the right-hand side of the original equation. This gives a first-order real analytic system. The derivative-compatibility identities have zero initial data and are preserved by the system; equivalently, its real analytic coefficient recurrence reproduces the derivatives of $U$. This explains both reductions, rather than assuming arbitrary real analytic hypersurfaces are already flat.

For the first example, the principal symbol of the Laplace equation at the conormal $d{x}_1$ is $1$, so the plane is a non-characteristic hypersurface. The theorem therefore applies to real analytic $u{|}_{x_1=0}$ and ${\partial }_{1}u{|}_{x_1=0}$:This does not give Hadamard well-posedness in smooth or Sobolev space norms. For $n\ge 2$, the harmonic functionshave zero value data and normal derivative data $e^{-\sqrt{N}}\cos (N{x}_2)$ tending to zero with every fixed tangential derivative. At any fixed $x_1=\delta >0$, however, $u_N(\delta ,0)$ diverges. Thus the elliptic Cauchy problem for a partial differential equation is unstable despite real analytic solvability. The exceptional case $n=1$ is the elementary affine ordinary differential equation and has no such tangential high-frequency instability.

Set $F(z,w)=w^2-z^n+1$. Its complex gradient is $(-n{z}^{n-1},2w)$, never zero on $F=0$: simultaneous zero would give $z=w=0$, which is not on the curve. The complex implicit function theorem therefore gives a one-dimensional complex manifold, hence a Riemann surface. Where $w\ne 0$ use $z$ as coordinate, and where $z\ne 0$ use $w$; the transition maps and both projections are holomorphic.

For $\pi$, a ramification point of a holomorphic map has $w=0$ and $z=\zeta$ with ${\zeta }^n=1$. In the coordinate $w$, $z-\zeta =w^2/(n{\zeta }^{n-1})+O(w^4)$, giving ramification index $2$. Thus there are $n$ such points, and their branch points are the $n$th roots of unity.

For $\tau$, a ramification point of a holomorphic map has $z=0,w=\pm i$. In the coordinate $z$, $w-w_0=z^n/(2w_0)+O(z^{2n})$, giving ramification index $n$ at both points and branch points $\pm i$.

Write $n=2m$. At infinity set $u=1/z$, $v=w{u}^m$; then $v^2=1-u^n$. At $u=0$ there are two smooth points, $v=\pm 1$, and $u$ is a local coordinate. The extended $\pi$ has a simple pole at each, so neither is ramified. The degree-two Riemann-Hurwitz formula for the map to the Riemann sphere gives $2g-2=2(-2)+n$, henceThe list for $\tau$ above concerns the original affine Riemann surface; on the compactification its poles at the two added points have order $m$ and are additionally ramified if $m>1$.

Treat $n^2=N^2/{\Omega }^2$ as a signed small parameter, with $|n^2|\ll 1$, as printed in the original PDF. At $n^2=0$, the dispersion relation factors asThus the two epicyclic motion modes have $s_0=\pm i\Omega$, while the thermal energy mode of a shearing sheet hasIt decays on the thermal diffusion time ${\beta }^{-1}$. For completeness, its next correction is $s_E=-\beta +\beta N^2/({\beta }^2+{\Omega }^2)+O(\Omega n^4)$ at fixed $\beta /\Omega$.

For either epicyclic motion root, the implicit function theorem applied to the dispersion relation givesHenceSince $\beta >0$, the oscillations undergo overstability precisely whenThe convective overstability growth rate tends to zero both for very slow diffusion and for very fast diffusion. Differentiating $\beta /({\beta }^2+{\Omega }^2)$ shows that its maximum occurs at $\beta =\Omega$, corresponding to $k^2=\Omega /\xi$. ThereforeThis is the growth rate of the perturbation amplitude; a quadratic perturbation energy grows at twice that rate. The convective overstability arises when an adverse radial specific entropy gradient couples to epicyclic motion with a finite thermal conduction lag.

The formula $[X:Y:Z]\mapsto [X:Z]$ is initially undefined at the unique point $P_{\infty }=[0:1:0]$ of $X$ at infinity. Since $X$ is a smooth projective curve, the extension of a rational map from a smooth projective curve extends it uniquely to a morphismOn the affine chart it is the rational function $x$, and the equation $y^3=x^4+1$ shows that $\phi$ has degree three.

For each of the four roots $\alpha$ of $x^4+1$, let $P_{\alpha }=(\alpha ,0)$. Since the root is simple, $t=y$ is a uniformizer at $P_{\alpha }$, and the equation givesThus the ramification index of a holomorphic map at every $P_{\alpha }$ is three.

At $P_{\infty }$, use the chart $Y=1$ and put $u=X/Y$, $v=Z/Y$. The equation becomesThe implicit function theorem makes $u$ a uniformizer and gives $v=u^4$ times a local unit. Henceso $P_{\infty }$ also has ramification index three. These are all the ramification points of a holomorphic map because the resulting ramification degree is $5(3-1)=10$. Indeed, the Riemann-Hurwitz formula givesTherefore $g_X=3$, in agreement with the genus of a smooth plane curve, and the Euler characteristic of the underlying compact orientable surface is

The implicit function theorem states that if $F(x_0,y_0)=0$ and the partial derivative $D_{y}F(x_0,y_0)$ is an invertible linear map, then near $(x_0,y_0)$ the zero set of $F$ is uniquely the graph $y=g(x)$ of a continuously differentiable function.

If $f$ is a differentiable bijection with differentiable inverse $g$, the chain rule applied to $g\circ f$ and $f\circ g$ givesThus $D{f}_x$ is an isomorphism with inverse $D{g}_{f(x)}$.

If a continuously differentiable map $F:{\mathbb{R}}^n\to {\mathbb{R}}^n$ has invertible derivative everywhere, the inverse function theorem makes it a local diffeomorphism. In particular it is an open map, so its image is open. Its image need not be closed: $x\mapsto \arctan x$ has nonzero derivative everywhere and image $(-\pi /2,\pi /2)$.

For the given map of elementary symmetric polynomials,and direct evaluation of the determinant givesHence the critical set isIts complement is the Zariski-open set on which the three coordinates are pairwise distinct. Each point has one of the six possible strict coordinate orderings, and each ordering defines a nonempty convex open region. A continuous path cannot change an ordering without crossing $C$. Therefore ${\mathbb{R}}^3\setminus C$ has exactlyconnected components.

For $a\in U$, the function $f:U\to {\mathbb{R}}^n$ is differentiable at $a$ if there is a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$ such thatThe linear map is unique and is the Frechet derivative $D{f}_a=A$. The function is continuously differentiable at $a$ if it is differentiable on a neighbourhood of $a$ and the map $x\mapsto D{f}_x$, with values in the space of linear maps equipped with the operator norm, is continuous at $a$.

If $T:{\mathbb{R}}^m\to {\mathbb{R}}^n$ is linear, thenConsequently $D{T}_a=T$ at every $a$. The derivative is a constant function of $a$, hence is continuous, so every linear map is continuously differentiable everywhere.

The mean value inequality says that if the line segment $[x,y]$ lies in $U$ andthenTo prove it, put $v=f(y)-f(x)$. The claim is immediate if $v=0$. Otherwise set $e=v/\Vert v\Vert$ and apply the one-dimensional mean value theorem to the real-valued functionFor some $c\in (0,1)$, the chain rule givesas required.

Now suppose that $U$ is open and connected and that $D{f}_a=0$ for every $a\in U$. Every point has an open ball contained in $U$. The mean value inequality with $M=0$ shows that $f$ is constant on each such ball, so $f$ is locally constant. Fix $a_0\in U$. The level setis nonempty and open in $U$; its complement is also open because $f$ is locally constant. Since $U$ is connected, $E=U$. This proves the zero derivative on a connected open set result: $f$ is constant.

The inverse function theorem states that if $f:U\to {\mathbb{R}}^m$ is continuously differentiable, $a\in U$, and $D{f}_a$ is an invertible linear map, then there are open neighbourhoods $V$ of $a$ and $W$ of $f(a)$ such that $f{|}_V:V\to W$ is a bijection whose inverse is continuously differentiable.

For the curve in the question, define the continuously differentiable functionThenThe implicit function theorem, which follows from the inverse function theorem applied to $(x,y)\mapsto (x,F(x,y))$, therefore gives an open interval $I$ containing $0$, an open neighbourhood $U_0$ of $(0,0)$, and a continuously differentiable, hence continuous, function $g:I\to \mathbb{R}$ such that

Choose holomorphic coordinates centered at $p$. Locally, the blowup of a complex manifold at a point iswith $\sigma (z,[\ell ])=z$; away from $p$ this is an isomorphism, so it glues to $X\setminus \{p\}$. The exceptional divisor is $E={\sigma }^{-1}(p)\cong {\mathbb{CP}}^{n-1}$.

The proper transform is the closure of ${\sigma }^{-1}(Y\setminus \{p\})$. If $p\notin Y$, it is isomorphic to $Y$. If $p\in Y$, the holomorphic implicit function theorem supplies coordinates in which $Y=\{z_1=0\}$. In the blowup chart with $z_i=t$ and $z_j=t{u}_j$ for $j\ne i$, every chart with $i\ne 1$ describes the proper transform by $u_1=0$, while the chart $i=1$ does not meet it. These are smooth coordinate hypersurfaces, so $\tilde{Y}$ is smooth.

For a divisor $D$, the line bundle associated to a divisor $\mathcal{O}(D)$ consists locally of meromorphic functions $f$ such that $(f)+D\ge 0$. Pulling back a local defining function for $Y$ shows that its divisor iswhere $m$ is the order of vanishing at $p$ of a local defining function for $Y$. ThereforeBecause $Y$ is smooth, $m=1$ when $p\in Y$ and $m=0$ when $p\notin Y$.

Applying the definition with $D=E$ gives directlyThe map sends a section to its local meromorphic coefficient relative to the canonical meromorphic section of $\mathcal{O}(E)$; the divisor inequality is exactly the condition that these coefficients define a holomorphic section, and the inverse construction is local multiplication by that canonical section.

If $P({\lambda }_0,0)=0$ and $P_{\lambda }({\lambda }_0,0)\ne 0$, the implicit function theorem gives a local root branch. For analytic coefficients it has an ordinary power series in the perturbation parameter. Its corrections follow by substituting that series into the polynomial and comparing coefficients. A vanishing leading root may require its first nonzero correction to capture a slow exponential mode.