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With the convention that the residual term is a sum rather than an average, the penalised least squares criterion isMinimize over functions whose first derivative is absolutely continuous and whose second derivative is square-integrable, that is, the usual Sobolev space $H^2[x_1,x_n]$. The cubic smoothing spline ${\hat{g}}_{\lambda }$ is its minimizer, with natural boundary conditions and linear continuation outside the data range. It is a natural cubic spline with possible knots at the observed $x_i$, but usually does not interpolate the noisy $y_i$.

To see why the minimizer is a natural spline, for any $g$ take the natural cubic spline interpolant $s$ through its values $g(x_i)$ and write $h=g-s$. Then $h(x_i)=0$. Integrating $\int s^{\prime \prime }{h}^{\prime \prime }$ by parts on every knot interval twice gives zero: the terms involving $h$ vanish at knots, the terms involving $h^{\prime }$ cancel because $s^{\prime \prime }$ is continuous there, and the outer terms vanish by the natural boundary conditions. Hencewhile their residual sums of squares agree. Replacing $g$ by $s$ never increases the criterion. This proves the minimum roughness property of the natural cubic spline interpolant needed here.

For fitted values $u_i=s(x_i)$, write $\int (s^{\prime \prime }{)}^2=u^{T}Ku$, where $K$ is positive semidefinite and has the sampled affine functions as its two-dimensional nullspace. The finite-dimensional minimization becomes $\Vert y-u{\Vert }^2+\lambda u^{T}Ku$. Its strictly positive definite Hessian is $2(I+\lambda K)$, so it has the unique solutionThis is the roughness-matrix formula for a natural cubic smoothing spline. Larger $\lambda$ penalizes curvature more strongly. As $\lambda \downarrow 0$ the fit approaches interpolation; as $\lambda \to \infty$ it approaches the ordinary least-squares affine line.

A cubic spline belongs to $C^2[a,b]$ and is a polynomial of degree at most three on each interval separated by the spline knots. A natural cubic spline is additionally an affine function on the exterior intervals $[a,x_1]$ and $[x_n,b]$. On $[x_1,x_n]$ this is equivalent to the endpoint conditions $s^{\prime \prime }(x_1)=s^{\prime \prime }(x_n)=0$; these conditions remain part of the definition when an outer knot coincides with $a$ or $b$.

Let $s$ be the natural cubic spline interpolant of the prescribed values, and let $v$ be any other interpolating $C^2$ function. Set $u=v-s$, so $u(x_i)=0$. On the interval $[x_i,x_{i+1}]$, integration by parts twice givesbecause the fourth derivative of each cubic piece is zero. The second boundary term is zero because $u$ vanishes at the spline knots. After summing the first terms, the interior contributions cancel by continuity of $s^{\prime \prime }$ and $u^{\prime }$, and the two remaining terms vanish by the natural endpoint conditions. On the exterior intervals $s^{\prime \prime }=0$. Thus ${\int }_a^b{s}^{\prime \prime }{u}^{\prime \prime }=0$, and the second derivative roughness penalty satisfiesIf equality holds, the continuous nonnegative function $(u^{\prime \prime }{)}^2$ has zero integral, so $u^{\prime \prime }=0$ everywhere. Hence $u$ is an affine function, and its zeros at two distinct spline knots force $u=0$. This proves the minimum roughness property of the natural cubic spline interpolant and its equality case: the unique minimum-roughness interpolant is $s$.

For the fixed-design nonparametric regression fit, use the penalized least squares criterionLet $s_z$ be the natural cubic spline interpolant with value vector $z=(z_1,\dots ,z_n{)}^T$. Replacing $v$ by $s_z$ for $z_i=v(x_i)$ leaves the residual sum unchanged and reduces the second derivative roughness penalty, strictly unless $v=s_z$. It therefore remains to minimize the finite-dimensional quadratic formThe supplied spline roughness penalty matrix $K$ is positive semidefinite. For every nonzero vector $w$,Consequently $I+\lambda K$ is a positive-definite matrix, and $Q_{\lambda }$ is strictly convex. Its normal equations for linear least squares have the unique solutionThe preceding strict reduction also proves uniqueness over the whole class $C^2[a,b]$, not just over the natural cubic splines. This is the cubic smoothing spline. If the residual criterion is divided by $n$ instead, the corresponding formula is $(I+n\lambda K{)}^{-1}Y$, reflecting a change in the numerical convention for the smoothing parameter.

The independence and specified error mean and variance are not needed for existence or uniqueness of this minimizer. They do imply, with smoothing matrix $A_{\lambda }=(I+\lambda K{)}^{-1}$ and $g_X=(g(x_i){)}_i$, that $\mathbb{E}\hat{z}=A_{\lambda }{g}_X$ and $\mathrm{Cov}(\hat{z})={\sigma }^2{A}_{\lambda }^2$. The null space of $K$ consists of sampled affine functions: these have zero roughness, and conversely zero roughness forces their natural cubic spline interpolant to be an affine function. Thus these functions are reproduced without bias of an estimator, while other components are smoothed.

At a target $t$, local polynomial regression fits a polynomial of degree at most $p$ by weighted least squares:Define the response vector $Y=(Y_1,\dots ,Y_n{)}^T$, the design matrix $X$ by $X_{i,j+1}=(x_i-t{)}^j$ for $j=0,\dots ,p$, and the diagonal matrix $W$ by $W_{ii}=K((x_i-t)/h)$. The normal equations for linear least squares giveThis formula assumes the local polynomial Gram matrix $X^{T}WX$ is invertible. It is invertible whenever at least $p+1$ distinct sites have positive weights: a nonzero degree-$p$ polynomial cannot vanish at all of them. With fixed $p$, a nonzero continuous regression kernel, and $nh\to \infty$, this condition holds eventually at the target points considered here. Using weights $h^{-1}K((x_i-t)/h)$ instead gives exactly the same fit.

Use the usual unit-integral normalization ${\mu }_0(K)=1$. A regression kernel need not intrinsically have unit integral, but multiplying all weights by a positive constant leaves local polynomial regression unchanged. With unnormalized raw moments the interior coefficient below is ${\mu }_2(K)/{\mu }_0(K)$; the version written using ${\mu }_2(K)$ alone requires this normalization. A zero regression kernel cannot define an estimator.

Write the local moment sums asThe local constant estimator is ${\hat{m}}_h(t;0)={\sum }_{i}K((x_i-t)/h)Y_i/{\sum }_{i}K((x_i-t)/h)$. Since the errors have zero mean, its bias of an estimator is the corresponding weighted average of $m(x_i)-m(t)$. By the second-order Taylor expansion, uniformly for contributing sites with $|x_i-t|\le h$,where ${\eta }_n\to 0$. This uniform remainder follows from uniform continuity of $m^{\prime \prime }$ on $[0,1]$. Nonnegativity of the regression kernel therefore giveswhenever $A_0$ tends to a positive limit and $A_2=O(h^2)$.

For fixed interior $x$, the moment approximations give $A_0(x)=1+o(1)$, $A_1(x)=O(n^{-1})$, and $A_2(x)=h^2{\mu }_2(K)+o(h^2)$. The hypothesis $n{h}^2\to \infty$ makes $n^{-1}=o(h^2)$, so the odd-moment contribution is negligible at second order. ThusThe symmetry of the regression kernel is what removes the first-order interior bias of an estimator.

At $z_n=\alpha h$, the rescaled observation window is truncated to $[-\alpha ,1]$. Put ${\mu }_{r,\alpha }={\int }_{-\alpha }^1{u}^{r}K(u)du$ and ${\beta }_r={\mu }_{r,\alpha }/{\mu }_{0,\alpha }$. Symmetry, nonnegativity, and unit mass give ${\mu }_{0,\alpha }\ge 1/2$. The supplied moment approximations already imply the leading boundary bias of local constant regression:To retain the second-order term rigorously, we need slightly more precise control of the first moment. The rescaled grid $u_i=i/(nh)-\alpha$ has spacing $1/(nh)$. Each truncated function $u^{r}K(u)1_{[-\alpha ,1]}$ has bounded variation, because $K$ is continuously differentiable on its support. The Riemann sum error is bounded by this variation times the grid spacing: summing the oscillation bounds on the grid cells provesEndpoint cells, including the omitted site at zero, contribute the same error order. ConsequentlyBoth discretization errors are $o(h^2)$ under $n{h}^2\to \infty$. The preceding Taylor expansion therefore proves the second-order boundary expansion for local constant regression:Equivalently, expanding the derivatives about zero givesIn general the truncated first moment is nonzero, so the boundary bias of an estimator has order $h$, although it can be smaller when $m^{\prime }(0)=0$ or the omitted part of the regression kernel carries no mass.

For local linear regression, the polynomial reproduction property of local polynomial regression removes the constant and linear terms exactly. Let $\Delta =A_0{A}_2-A_1^2$. Its effective kernel weights areso ${\sum }_i{l}_i=1$, ${\sum }_i{l}_i(x_i-t)=0$, andAt either an interior or the specified boundary target, $\Delta$ is asymptotic to a positive constant times $h^2$. At the boundary that constant is ${\mu }_{0,\alpha }{\mu }_{2,\alpha }-{\mu }_{1,\alpha }^2>0$, the mass squared times the variance of the continuous normalized truncated regression kernel. Moreover ${\sum }_i|l_i|=O(1)$, as follows by bounding $|A_2-A_1(x_i-t)|=O(h^2)$ in the window. Thus the Taylor remainder still contributes $o(h^2)$, even though some effective kernel weights may be negative. In particular,At an interior point this coefficient reduces to ${\mu }_2(K)$ by symmetry and normalization. Hence local linear regression has bias $O(h^2)$ at both interior and boundary points; its linear reproduction removes the generic order-$h$ boundary term.

For an oriented Euclidean string worldsheet with metric $h_{ab}$, use the string nonlinear sigma model actionHere ${ϵ}^{ab}$ is an antisymmetric density with ${ϵ}^{12}=1$. The factor $i$ in the Euclidean Kalb–Ramond field term follows from continuing the real Lorentzian two-form coupling; reversing orientation or the convention for $B$ reverses its sign. The conventional positive Euclidean dilaton coupling is being used. With worldsheet boundaries its curvature coupling also needs ${\int }_{\partial \Sigma }\Phi kds/(2\pi )$, where $k$ is the boundary geodesic curvature, to have the proper topological normalization for constant $\Phi$.

All three terms are invariant under worldsheet diffeomorphisms. Target-coordinate changes act on $X,G,B,\Phi$ by their ordinary scalar and tensor transformation rules, making the target description covariant as well. For the two-form there is also the target gauge symmetry $B\mapsto B+d\Lambda$. Its change in the closed-worldsheet action is an integral of an exact two-form and hence zero. With boundaries it is instead $i{\int }_{\partial \Sigma }{X}^{\ast }\Lambda /(2\pi {\alpha }^{\prime })$; an endpoint gauge field with $\delta A=-\Lambda /(2\pi {\alpha }^{\prime })$ cancels it.

Under a local Weyl transformation $h_{ab}\mapsto e^{2\omega }{h}_{ab}$, the metric term is invariant because the two-dimensional factors $\sqrt{h}$ and $h^{ab}$ cancel. The $B$ term is independent of $h$ and is separately invariant. The curvature transforms asOn a closed surface the dilaton variation is consequentlyThus a general nonconstant dilaton term is not separately invariant under Weyl transformations. Its variation is the Weyl variation of a nonconstant dilaton coupling; it vanishes for constant $\Phi$ after the boundary completion where necessary.

For $\Phi ={\Phi }_0$, the Gauss-Bonnet theorem gives $S_{\Phi }={\Phi }_0\chi (\Sigma )$. In the Euclidean path integral the resulting weight isFor a connected orientable genus-$g$ surface with $b$ boundaries, $\chi =2-2g-b$, so the weight is $g_s^{2g+b-2}$. A closed-string handle therefore adds two powers of string coupling. External vertex normalizations add their own coupling factors; they are distinct from this dilaton Euler-characteristic weighting. The constant dilaton controls the string genus expansion, rather than adding a force to the classical embedding equation.

Quantization treats the target fields as position-dependent couplings of a two-dimensional interacting quantum field theory. Expanding $X$ around a background embedding gives fluctuation vertices involving target curvature and derivatives of $B$ and $\Phi$. Short-distance contractions require local counterterms, so the target fields become renormalized couplings. In particular a curvature vertex contracted once yields a Ricci tensor counterterm proportional to ${\alpha }^{\prime }{R}_{\mu \nu }$. Each additional sigma-model loop brings another power of ${\alpha }^{\prime }$ and more target derivatives. The expansion is useful when the target variation scale is large compared with $\sqrt{{\alpha }^{\prime }}$; it is not the separate expansion in powers of $g_s$.

After gauge fixing, a worldsheet Weyl anomaly would make amplitudes depend on the discarded local conformal factor. Quantum consistency therefore requires the trace anomaly to vanish, including the reparameterization worldsheet ghost field contribution. In a conventional renormalization scheme the resulting bosonic sigma-model Weyl anomaly coefficients start asThe barred quantities include the dilaton stress-energy tensor improvement; ordinary coupling sigma-model beta functions can differ by field redefinitions or target diffeomorphisms. Setting these coefficients to zero gives differential equations for the background fields order by order, including the metric gravitational equation and the antisymmetric-field equation. These conventions for the leading coefficients are described in the background-field lecture notes. Higher orders supply higher-curvature and higher-derivative corrections. This explains how spacetime dynamics follows from quantum gauge consistency of the worldsheet, not from asking only for a classical stationary worldsheet in an arbitrarily chosen background.

The apparently noninvariant dilaton term is essential outside the critical dimension because its stress-energy tensor improvement can cancel the quantum central charge mismatch. A flat constant-dilaton background has matter central charge $d$ and worldsheet ghost field central charge $-26$. If $d\ne 26$, changing the constant value of $\Phi$ only changes $g_s$ and cannot cure that mismatch.

A concrete solution is the linear dilaton conformal field theory with $G=\eta$, $B=0$, and $\Phi ={\Phi }_0+V_{\mu }{X}^{\mu }$. Its metric and antisymmetric-field anomaly coefficients vanish, while the scalar condition isOne can also see this cancellation directly from the improved holomorphic stress-energy tensor. With the positive Euclidean curvature convention above,The free-field contraction is $X^{\mu }(z)X^{\nu }(w)\sim -{\alpha }^{\prime }{\eta }^{\mu \nu }\log (z-w)/2$. Differentiating it twice at each point gives a fourth-order pole $3{\alpha }^{\prime }{V}^2/(z-w{)}^4$ from the improvement. Since the fourth-order stress-tensor pole is $c/2$, the matter central charge becomes $c=d+6{\alpha }^{\prime }{V}^2=26$. The worldsheet ghost field contribution then cancels it. This is why a classical Weyl variation need not mean inconsistency: it can cancel the quantum variation of the other terms. For $d<26$ the required gradient is spacelike; for $d>26$ it is timelike in the Lorentzian target theory. Such backgrounds are not flat constant-dilaton Minkowski vacua with full target Lorentz symmetry, nor does a central charge cancellation by itself prove stability or unitarity of every noncritical background. If the conformal factor is instead retained dynamically, its Liouville field theory supplies the corresponding noncritical description.

Use time dependence $e^{-i\omega t}$, positive wavenumber $k$ and aperture normal $\mathbf{n}={\mathbf{e}}_x$. Write $F(y^{\prime },z^{\prime })=E_{\mathrm{inc},y}(0,y^{\prime },z^{\prime })$ on the aperture and extend $F$ by zero outside it. DefineThe outgoing branch has $\kappa \ge 0$ for propagating components and $\mathrm{Im}\kappa >0$ for evanescent components.

Fix the Green function normalization by $G_0(\mathbf{r})=e^{ik|\mathbf{r}|}/(4\pi |\mathbf{r}|)$, so $(\Delta +k^2)G_0=-\delta$. Its transverse Fourier transform satisfies $({\partial }_x^2+{\kappa }^2){\hat{G}}_0=-\delta (x)$. An outgoing solution on both sides is proportional to $e^{i\kappa |x|}$, and its derivative jump fixes ${\hat{G}}_0=i{e}^{i\kappa |x|}/(2\kappa )$. Thus the Weyl plane-wave representation for observation at $x>0$ isSince $\mathbf{n}\times {\mathbf{E}}_{\mathrm{inc}}=F{\mathbf{e}}_z$, the given aperture integral becomes a superposition of curls of ${\mathbf{e}}_z{e}^{i\mathbf{q}\cdot \mathbf{r}}$. Each curl supplies $i\mathbf{q}\times {\mathbf{e}}_z=i(\nu ,-\kappa ,0)$. Therefore, with the unit prefactor in that integral,Every vector amplitude is transverse to $\mathbf{q}$, as Maxwell's equations require. The polarizer specifies the incident aperture field; diffraction can regenerate an $x$ component outside the aperture. In contrast, its $z$ component remains zero in this representation. Evanescent plane waves are needed for the complete near field, not just the propagating disk ${\nu }^2+{\eta }^2\le k^2$.

The factor $1/2$ follows explicitly from the standard outgoing Green function for the three-dimensional Helmholtz equation and the printed unit curl prefactor. If the intended kernel is the doubled half-space kernel $2G_0$, or the aperture formula carries the usual compensating factor $2$, delete that $1/2$. The latter normalization makes the tangential output trace equal to $F$. Keeping the convention explicit avoids silently changing the supplied representation; the angular dependence and transversality of an electromagnetic plane wave are the same in either normalization. Reversing the specified normal also reverses the overall sign.

Write the unit-amplitude incident plane wave as ${\psi }_i=e^{i{p}_{i}x-i{q}_{i}z}$, with $q_i=\sqrt{k^2-p_i^2}>0$. The flat Dirichlet reflected field is ${\psi }_s^{[0]}=-e^{i{p}_{i}x+i{q}_{i}z}$, so the flat total field vanishes at $z=0$ and has derivative ${\partial }_z{\psi }^{[0]}(x,0)=-2i{q}_i{e}^{i{p}_{i}x}$.

For a controlled height expansion take $f=\epsilon h$ with fixed regular bounded shape $h$. Taylor expansion at the moving boundary yieldsThus the first-order scattered trace is ${\psi }_s^{[1]}(x,0)=2i{q}_{i}f(x)e^{i{p}_{i}x}$, and its transform follows by the Translation property of the Fourier transform:The first formula is the roughness correction; the second includes the zeroth-order reflection. The small-height assumption controls the Taylor expansion for a fixed profile; arbitrarily fine surface scales can require additional derivative control, not merely a small height at fixed $k$.

If the mean roughness height is a constant $\bar{f}$, as for stationary roughness, then $⟨\hat{f}(\nu -p_i)⟩=2\pi \bar{f}\delta (\nu -p_i)$. Consequently the specular first-order mean reflection isIts tangential wavenumber is unchanged and its normal wavenumber is reversed relative to incidence, which is precisely specular reflection. For zero mean height the first-order mean correction is zero, but the full mean reflected field is not zero: it still contains the flat reflection.

Small height alone does not imply the statistical conclusion. For example, deterministic mean height $\bar{f}(x)=a\cos Kx$ with $|ka|\ll 1$ produces first-order mean sidebands at $p_i\pm K$ and is not purely specular. Constant mean is sufficient at first order; stationarity supplies it and, with suitable translation-invariant statistics, constrains higher-order coherent reflection as well.

For the cubic population iteration, put $g(u)=\lambda u(1-u^2)$. Its fixed points solveThus $u_{\ast }=0$ always exists. For $\lambda \ne 0$ there are alsowhenever the expression under the square root is positive, namely $\lambda <0$ or $\lambda >1$. At $\lambda =1$ these coincide with zero; at $\lambda =0$ only zero exists.

The criterion for fixed point stability for an iteration is $|g^{\prime }(u_{\ast })|<1$. To justify it, continuity of $g^{\prime }$ gives a small interval around the fixed point where $|g^{\prime }|\le q<1$. The mean value theorem then gives $|g(u)-u_{\ast }|\le q|u-u_{\ast }|$, so this interval is invariant and the error decreases geometrically. This proves local asymptotic stability rather than merely linear boundedness.

Here $g^{\prime }(u)=\lambda (1-3u^2)$. At zero its value is $\lambda$, so zero is locally asymptotically stable for $-1<\lambda <1$. At a nonzero fixed point it isThe strict inequality $|3-2\lambda |<1$ is equivalent to $1<\lambda <2$. Both nonzero real fixed points are therefore locally asymptotically stable on that interval. This proves the two requested existence ranges. The strict derivative criterion alone makes no conclusion at the endpoint multipliers of modulus one.

Measure temperature relative to ambient by ${\theta }_i=T_i-T_{\infty }$. Newton's law of cooling models ordinary cooling by ${\theta }_i^{\prime }=-a{\theta }_i$, since heat loss is proportional to the excess temperature. The additional negative forcing represents the cooling from the milk, with the temperature drop or rate normalized to one in the chosen units.

The Dirac delta function has unit integral and is supported at the instant of addition. It therefore imposes a sudden unit drop for cup 1. The difference $H(t-1)-H(t-2)$ of Heaviside step functions is one during the addition interval and zero outside it, giving a unit cooling rate for cup 2. The second temperature is continuous; only its derivative changes at the switching times. The value assigned to a Heaviside step function exactly at its jump does not change the integrated solution. This is the impulse versus finite-duration cooling comparison.

During the addition interval, subtracting the formulas in (ii) and (iii) cancels the common unforced temperature:This is zero precisely when $e^{-a(t-1)}=1/(1+a)$, soThe inequalities $0<\log (1+a)<a$ give $1<t^{\ast }<2$, so using the active-interval formula was justified. The difference has derivative $(1+a)e^{-a(t-1)}>0$, making this crossing unique within that interval. Immediately after time one it is minus one; after time two it is strictly positive by (iv). Hence there is no other equality at any finite time after time one. Both temperatures approach the same ambient limit, but that limiting equality is not an additional crossing.

For this quartic potential with fourfold symmetry, the first derivatives factor:The critical points are the origin, the four axis points $(\pm 1,0),(0,\pm 1)$, and four points with both coordinates nonzero. At the latter, $x^2+b{y}^2=y^2+b{x}^2=1$. Subtracting and using $b\ne 1$ gives $x^2=y^2=1/(1+b)$. Thus the complete list iswhere the two signs in the final group are chosen separately.

The Hessian isAt the origin it is $2I$, so there is a strict local minimum with value zero. At an axis point its eigenvalues are $-4$ and $2(1-b)$, giving a saddle point of a scalar function if $b<1$ and a strict local maximum if $b>1$. All four axis values are $1/2$.

At a diagonal point the Hessian has eigenvaluesHence these four points are strict local maxima for $b<1$ and saddle points of a scalar function for $b>1$. Their value is $1/(1+b)$. None of these critical points is degenerate when $b\ne 1$.

To describe the level curves precisely, use polar coordinates and writeFor a level $f=c$, the possible radii satisfyretaining only real, nonnegative values. This formula, together with reflection in each axis and interchange of the coordinates, determines the level curves. The radial maximum in a direction is $1/(2q)$, attained at $r^2=1/q$. It also proves that the local maxima identified above are global: the largest directional value is $1/(1+b)$ on the diagonals if $b<1$, and $1/2$ on the axes if $b>1$.

For negative $c$ only the plus sign gives a positive radius, producing one outer closed level curve. At $c=0$ there is the isolated origin and the outer curve $r^2=2/q$. For $0<c<c_s$, where $c_s$ is the saddle value, both radial branches exist in every direction, giving an inner loop around the origin and an outer loop. At $c=c_s$ they meet at the four saddle points of a scalar function. Above the saddle value, four separate loops surround the four maxima; they shrink to those points at the maximum value. There are no level curves above it.

For $0<b<1$, the saddle value is $c_s=1/2$ and the four high-level loops lie along the diagonals. For $b>1$, $c_s=1/(1+b)$ and the four high-level loops lie along the coordinate axes. The sketches show representative values in both regimes; the highlighted level curve is the saddle level.

Finally, at $b=1$ the expression depends only on radius:Consequently $\boxed{\max f=1/2\text{, attained exactly on }{x}^2+y^2=1.}$ The maximum set is the whole unit circle, not merely four isolated points.

Choose $0<R<R_1$, $0<S<S_1$ and $0<T<T_1$. On this smaller closed polydisc, the holomorphic function $A$ and its $w$-derivative are bounded; write $|A|\le M$ and $|{\partial }_{w}A|\le L$. Choose $0<r<R$ with $rM<S/2$ and $rL<1$, replacing a zero bound by any positive upper bound if necessary, and choose $0<\epsilon <T$.

Consider functions $u(z,\alpha )$ continuous on $\overline{D(0,r)}\times \overline{D(0,\epsilon )}$, jointly holomorphic in the interior, with $u(0,\alpha )=0$ and $\Vert u{\Vert }_{\infty }\le S/2$. This is a closed subset of a Banach space in the supremum norm. The integral operatoris well defined on the straight segment and is jointly holomorphic. Indeed, it equals $z{\int }_0^{1}A(tz,u(tz,\alpha ),\alpha )dt$, an integral of jointly holomorphic functions locally uniformly bounded in both variables. Convexity of the $w$-disc and the derivative bound giveThus the Banach fixed-point theorem gives a fixed point $f(z,\alpha )$. More explicitly, the successive integral iterates starting from zero converge uniformly, and locally uniform convergence of holomorphic functions makes their limit jointly holomorphic. Differentiating the integral equation gives the required complex ordinary differential equation and initial value. This proves local existence, uniqueness and joint holomorphic dependence on $(z,\alpha )$. Uniqueness among arbitrary local solutions follows first on a small disc where their values stay within $|w|<S$, by the same contraction mapping estimate, and then throughout their common connected domain by the identity theorem.

For holomorphic dependence of ordinary differential equations on parameters in the initial value, fix an admissible initial value $w_{\ast }$ and put $u=f-w_0$. The equation becomesIts right-hand side is jointly holomorphic in $(z,u,w_0)$ near $(0,0,w_{\ast })$. The preceding construction supplies a common disc and a jointly holomorphic solution for $w_0$ near $w_{\ast }$. Consequently $f=u+w_0$ is jointly holomorphic in $z$ and its initial value $w_0$.

A fixed singularity of a complex differential equation has a location determined by the coefficients, independently of the initial value. A movable singularity of a complex differential equation belongs to an individual continued solution and changes location when the initial value changes. For example, $w^{\prime }=w^2$ has the movable pole $z=c$ in $w=-1/(z-c)$. A fixed exceptional location need not be singular for every solution.

The precise first-order form of the Painlevé determinateness theorem needed here is as follows. Let $w^{\prime }=A(z,w)$ be rational in $w$, with coefficients meromorphic in $z$ on the chosen independent-variable surface. Algebraic coefficients can be made single valued on their own surface. Exclude the fixed coefficient singularities and exceptional fibres where a reduced numerator and denominator vanish together, working also in the dependent coordinate $v=1/w$. If a solution is continued along an arc $\gamma :[0,1)\to U$ with endpoint $z_{\ast }={\lim }_{t\to 1}\gamma (t)$ outside that exceptional set, then $w(\gamma (t))$ has a definite finite or infinite limit. Any singularity there is a pole or an algebraic branch point; there is no movable essential or logarithmic singularity. The standard rectifiable-arc formulation is included in this statement.

Here is a proof. First suppose a sequence along the arc has $(z_n,w_n)\to (z_{\ast },w_{\ast })$ at a point where the vector field is holomorphic. Choose a smaller closed bidisc about this point. Uniform bounds on the vector field and its dependent-variable derivative give, by the integral contraction mapping proof, one radius $r>0$ on which every solution starting sufficiently close to $(z_{\ast },w_{\ast })$ exists and remains bounded. For large $n$, the solution through $(z_n,w_n)$ is therefore defined on a fixed disc about $z_{\ast }$. The entire sufficiently late tail of $\gamma$ lies in that disc. Uniqueness identifies the original continued solution with this local solution throughout that tail, so the solution extends through $z_{\ast }$ and has a limit. The same argument applies to an ordinary point in the reciprocal coordinate $v=1/w$.

If the limit did not exist, consider the cluster set $K$ in the dependent-variable Riemann sphere:Each closed tail is compact and connected, and these sets are nested; hence $K$ is nonempty, compact and connected. The preceding paragraph excludes every ordinary point from $K$. At $z_{\ast }$ the rational vector field has only finitely many nonordinary dependent values, in both coordinate charts: an entire fibre of singular points was excluded among the fixed exceptions. Thus $K$ is a connected subset of a finite set, so it is a singleton. Compactness then forces convergence to its unique point, contradicting the assumed lack of a limit. This proves determinateness.

It remains to determine the local type rather than merely the limit. At a finite limiting value $w_{\ast }$ which is a pole of the vector field, write the reduced equation as $w^{\prime }=P/Q$. Since $z_{\ast }$ is not exceptional, $P(z_{\ast },w_{\ast })\ne 0$, and the inverse equationis holomorphic near $(z_{\ast },w_{\ast })$. The holomorphic dependence of ordinary differential equations on parameters makes its local solution curves jointly holomorphic in $w$ and their initial $z$ values. In particular, the value at the transverse line $w=w_{\ast }$ is a holomorphic first integral labelling these curves. The original graph has constant label; taking its endpoint limit shows that it lies on the inverse solution through $(z_{\ast },w_{\ast })$. That inverse solution cannot be the vertical curve $z\equiv z_{\ast }$: this would require $Q(z_{\ast },w)$ to vanish identically near $w_{\ast }$, an excluded fixed singular fibre. Consequently its power series has a first nonzero term,Here $k\ge 2$ at a vector-field pole. Factor the right side as $(w-w_{\ast }{)}^k$ times a nonvanishing holomorphic function and take its local $k$th root. This gives a new dependent coordinate with nonzero derivative, and the holomorphic inverse function theorem expresses $w$ as a convergent power series in $(z-z_{\ast }{)}^{1/k}$. Thus the singularity is an algebraic branch point. At an infinite limit, repeat the argument for $v=1/w$. An ordinary zero of $v$ gives a pole of $w$, while a finite ramification of $v$ gives an algebraic pole branch. This proves the asserted local types and the theorem.

For the given equation, the finite-coordinate rational field isThe whole fibre $z=-1$ is singular in the displayed coefficient, and at $(z,w)=(0,0)$ the reduced numerator and denominator vanish together. Hence the finite fixed exceptional set is $\{-1,0\}$. Away from these two locations, the only possible finite dependent limits at a singularity are $w_{\ast }=\sigma z_{\ast }$, with $\sigma =\pm 1$.

At such a point the inverse equation isIt is holomorphic when $z_{\ast }\ne 0,-1$ and $w_{\ast }=\sigma z_{\ast }$. Its inverse solution satisfies $z_w(w_{\ast })=0$, and differentiation givesThereforeandThese are genuine square-root algebraic branch points. The inverse solutions can be based at arbitrary $z_{\ast }\notin \{0,-1\}$ and then continued to nearby ordinary initial points. Their distinct branch locations require different initial data, so these branch points are movable. The signs specify the two inverse branches near each of the two values $w_{\ast }=\pm z_{\ast }$.