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The Painlevé property says that the general solution has no movable critical singularities: analytic continuation around a singularity whose location depends on the initial data must not create a different branch. A fixed singularity of a complex differential equation has its location prescribed by the coefficients; a movable singularity of a complex differential equation has a location which varies with the initial data. A commonly used stronger form requires movable singularities to be poles, so that the solution is locally a meromorphic function there. Movable branch points are excluded; a movable pole is permitted. We will prove the stronger form for this equation by finding its complete general solution family as rational functions.

Use the linearization of a Riccati equation. If $w^{\prime }=a{w}^2+bw+c$, put $w=-v^{\prime }/(av)$ wherever $a,v\ne 0$. Differentiating this expression and cancelling the quadratic terms yieldsConversely a nonzero solution $v$ of this linear ordinary differential equation gives a solution $w$, wherever the quotient is defined. Every local finite solution $w$ at an ordinary point arises this way: solve $v^{\prime }=-awv$ with a nonzero initial value and differentiate to get the second-order equation. Multiplying $v$ by a nonzero constant leaves $w$ unchanged, giving the one effective constant of integration expected for a first-order Riccati equation.

For the present coefficients, $a^{\prime }/a=1/z+1/(z+1)$, so the linear ordinary differential equation simplifies toTwo direct substitutions give solutions $v_1=z-1$ and $v_2=1/(z+1)$. Their Wronskian iswhich is nonzero at every ordinary point $z\notin \{0,-1\}$. They therefore form a fundamental pair there, and all local solutions have the formTaking its logarithmic derivative givesThus, for $C_1\ne 0$, writing $k=(C_2-C_1)/C_1$, the general family isThe remaining member, corresponding to $C_1=0$, isThe completeness of the linearizing pair proves that no other local solution is omitted. These are rational functions, single-valued meromorphic functions; hence their continuation has no branch points. Their only possible moving singularities are zeros of $z^2+k$. A root $z_0$ outside $\{0,-1\}$ is simple, and its residue isEquivalently, poles of a Riccati solution from zeros of its linearizing solution are simple at ordinary points, because a nontrivial solution of a linear ordinary differential equation cannot have both $v(z_0)=0$ and $v^{\prime }(z_0)=0$. The exceptional collisions $k=0$ or $k=-1$ occur at the fixed coefficient singularities, and still give rational functions. Consequently every movable singularity is a pole, establishing the Painlevé property in its stronger meromorphic sense.

For an invertible fundamental matrix, differentiating $Y_{\lambda }=AY$ in $z$ and $Y_z=BY$ in $\lambda$ gives $Y_{\lambda z}=(A_z+AB)Y$ and $Y_{z\lambda }=(B_{\lambda }+BA)Y$. Thus the necessary and sufficient compatibility identity isHere both derivatives must use the actual deformation variable $z$; the printed $t$ has not been defined. The isomonodromic deformation identity can be checked as a polynomial in $\lambda$, without assuming the conclusion.

First retain the coefficients actually printed in the PDF. Write $p=u^{\prime }$, $d=w+z/2$, and introduce the two residualsDirect matrix multiplication and differentiation giveFor example, the quadratic lower entry is the difference between $[D,B_0{]}_{21}=2w/u$ and the contribution $(A_1{)}_{21}=-zw/u$; it does not cancel. The constant lower entry follows by differentiating $2u^{\prime }w/u^2$ and collecting it as $2wH/u^2+2u^{\prime }K/u^2$. These entries also provide independent checks of the two most important cancellations.

Work on an open connected deformation domain where $u\ne 0$, as the matrices require. The quadratic entry forces $(2-z)w=0$ throughout the domain, hence $w\equiv 0$ by continuity, including a possible point $z=2$. All remaining entries then vanish exactly when $u^{\prime \prime }+(z/2)u=0$. Thus the literal answer isThis is a rescaled Airy function equation: with $\xi =-2^{-1/3}z$, the equation is $u_{\xi \xi }-\xi u=0$, so $u=c_1\mathrm{Ai}(\xi )+c_2\mathrm{Bi}(\xi )$ on a domain avoiding its zeros. In particular, choose the local solution with $u(1)=u^{\prime }(1)=1$ and $w=0$. It gives compatible matrices but violates the printed $w=u^{\prime }$. This proves that the requested assertion is false for the original coefficients, independently of the undefined $t/z$ in the printed differential equation.

For completeness, use the explicitly corrected lower coefficient $(A_1{)}_{21}=-2w/u$ from the preceding solution. The same multiplication gives the much simpler residualConsequently compatibility is equivalent to $K=H=0$. The first equation integrates without dividing by $w$:The constant of integration $C$ therefore satisfies $w=C{u}^2$, and the second equation becomesConversely these two equations give $K=H=0$, and substitution makes every residual entry vanish, proving sufficiency as well as necessity. This is the quadratic spectral Lax pair for zero-parameter Painlevé II. For $C\ne 0$, choose constants $b,a$ with $b^3=-2$ and $C{a}^2{b}^2=-2$, set $z=bx$, $u=aq(x)$, and obtain $q^{\prime \prime }=2q^3+xq$, the zero-parameter Painlevé II equation. For $C=0$ it reduces to the Airy function equation. Even with this correction, the conclusions are $w=C{u}^2$ and the coefficient $z/2$, rather than the incompatible expressions printed in the question.

Work locally in $z$ where $u\ne 0$, and take the matrix entries literally as they appear in the original PDF. At fixed $z$, abbreviateThen $A=D{\lambda }^2+\left(\begin{array}{cc}0& U\\ V& 0\end{array}\right)\lambda +\left(\begin{array}{cc}d& P\\ R& -d\end{array}\right)$. DefineThe literal coefficients giveIn general neither $h=z/2$ nor $\theta =0$ is valid for the printed system.

The normalized formal fundamental matrix iswith a chosen logarithm branch. In particular the diagonal normal form has ${\Lambda }_1=0$, ${\Lambda }_2=hD$, ${\Lambda }_3=\theta D$. Its first two prefactor coefficients areThese formulas include the diagonal terms required by the derivative of the prefactor. Setting every diagonal prefactor coefficient to zero while retaining only the displayed exponent would generally fail the differential equation.

Here is the coefficient calculation and the full recursion. Write $A_1=\left(\begin{array}{cc}0& U\\ V& 0\end{array}\right)$, $A_2=\left(\begin{array}{cc}d& P\\ R& -d\end{array}\right)$ and $G_0=I$, with negative-index $G_j=0$. Substitution into the gauge transformation of a linear differential system gives, for $m\ge 1$,At $m=1$, the commutator determines $(G_1{)}_{12}=-U/2$, $(G_1{)}_{21}=V/2$ and the absence of a linear diagonal term. The diagonal part at $m=2$ gives $h=d+UV/2$. The next diagonal part gives $\theta =(UR+VP)/2$. At $m=4$ diagonal solvability gives $-\rho =PR/2-UV(d+h)/4$; the next order fixes the two diagonal entries of $G_2$ above. The off-diagonal entries are fixed at each stage because $[D,X{]}_{12}=2X_{12}$ and $[D,X{]}_{21}=-2X_{21}$. Later diagonal solvability fixes the remaining normalization coefficients. Thus this is a complete formal solution, not a convergence assertion. It is the formal cubic-exponential two-by-two system expansion. The printed hint's second prefactor term must have denominator ${\lambda }^2$, and the remainder after the $1/\lambda$ term of the diagonal expansion must start at $1/{\lambda }^2$.

Now impose an isomonodromic deformation in the full irregular sense: its Stokes matrices and formal exponent are fixed in $z$, with the leading formal normalization fixed. Sectorial matrices satisfy $Y_{j+1}=Y_j{C}_j$ with $(C_j{)}_z=0$, so $B_j=(Y_j{)}_z{Y}_j^{-1}$ is the same on overlaps. Since the spectral system has no finite singularities, its fundamental matrix and inverse are entire in $\lambda$, making $B$ entire. Differentiating the formal expression, using ${\theta }_z=0$, givesBut $[G_1,D]=A_1$. The sectorial cover makes this a global growth bound, so Liouville's theorem applied to $B-h_z(\lambda D+A_1)$ gives the literal consequenceAn arbitrary parameter dependence of $u,w$ need not be isomonodromic; this analytic argument presupposes the stated constancy of generalized monodromy data.

The requested matrix would require $h_z=1/2$ from its diagonal and upper entry, and then its lower entry would require $(z-2)w=0$. On an open deformation domain this forces $w\equiv 0$. Thus that requested expression is not the general deformation matrix for the printed coefficients.

There is a consistent nearby correction: replace the lower entry of $A_1$ by $-2w/u$. Then $V=-2w/u$, so $h=z/2$ and $\theta =0$, and the same derivation proves exactlyFor this corrected system use the same prefactor formulas with $V=-2w/u$: $\rho =-w(w+z)/2-w(u^{\prime }/u{)}^2$ and both diagonal entries of $G_2$ are ${\rho }^2/2+w/4$. Compatibility determines the allowed $u,w$ in the next solution. This correction is stated explicitly rather than silently substituting it for the original PDF.

Use the local coordinate $t=1/\lambda$. The differential system becomesFor the nonzero quadratic leading term, the coefficient of the local linear ordinary differential equation system has a pole of order four, so infinity is an irregular singular point of Poincaré rank three in this gauge. The matrix-valued function $A(\lambda )$ itself has a pole of order two; the extra factor $d\lambda /dt=-t^{-2}$ is essential when classifying the equation. If $A_0$ vanishes the degree and these orders decrease. The specified distinct eigenvalues $1,-1$ give the nondegenerate unramified case.

The leading exponential function factors have actions $q_{\pm }=\pm {\lambda }^3/3$, so their difference is $2{\lambda }^3/3$. Equal exponential magnitudes occur onThese are commonly called Stokes rays in the sectorial ODE convention. With the action-phase convention used by the wiki's Stokes line article they are anti-Stokes lines; the same-phase rays instead haveGiving both defining equations removes the nomenclature ambiguity; both sets contain six directions.

The standard construction with distinct leading eigenvalues uses six sectorial fundamental solutions per turn, with basic angular spacing $\pi /3$. They are fundamental matrices, each containing two independent column solutions. On the logarithm cover choose six consecutive overlapping sectors of width $\pi /3+2\delta$, with $\delta >\epsilon /2$ small. Their union has angular opening $2\pi +2\delta$, covering the requested $2\pi +\epsilon$. One often also writes a seventh index to compare the closing sector with the first after a full turn; that is a closing copy, not a seventh independent sector in the six-sector construction. The number is a generic sectorial fundamental solution normalization count, not a universal minimum for exceptional systems with vanishing Stokes matrices: the exactly diagonal system already has one global elementary fundamental matrix.

There is a missing qualification in the requested product identity. Polynomial coefficients give an entire invertible fundamental matrix, hence trivial actual monodromy about a large circle: the circle contracts through ordinary points. But the formal matrix generally has the form $\hat{G}{\lambda }^{\Theta }{e}^Q$, and a positive turn changes its logarithm branch by the formal monodromy $M_f=e^{2\pi i\Theta }$. This factor cannot be discarded.

To specify the printed product order, let $S_j$ act on column vectors of solution coefficients across the $j$th sector boundary, so adjacent normalized matrices satisfy $Y_{j+1}=Y_j{S}_j^{-1}$. Six transitions give $Y_7=Y_1(S_6\cdots S_1{)}^{-1}$. The normalized formal branch after the turn is ${\hat{Y}}_1{M}_f$, so sectorial uniqueness gives $Y_7=Y_1{M}_f$. ConsequentlyUsing Stokes matrices for right changes of fundamental matrices instead reverses/inverts the corresponding convention; the formal monodromy factor remains necessary. The formal monodromy correction to an entire-system Stokes product shows that the printed identity holds when $M_f=I$, including the corrected Lax specialization discussed below.

It does not follow from polynomiality alone. For example takeThe coefficient matching in question 2 gives $\Theta =\frac{1}{2}\mathrm{diag}(1,-1)$, hence $M_f=-I$ and the ordered Stokes matrix product is $-I$, not $I$, despite having no finite singularity. This is an explicit counterexample to the unqualified final clause.