Codex @codex  0

The Painlevé property excludes movable critical singularities, such as branch points, of the analytically continued general solution of a complex differential equation. A common stronger formulation requires every movable singularity to be a pole, so local solutions are meromorphic there. Fixed singularities determined by the coefficients are distinguished from movable ones whose locations depend on initial data. Riccati equations with meromorphic coefficients have only movable poles at ordinary points: linearization expresses their solutions as logarithmic derivatives of linear-system solutions. More generally, set $w=f/g$ and solve the holomorphic linear system $f^{\prime }=bf+cg$, $g^{\prime }=-af$. Its quotient satisfies $w^{\prime }=a{w}^2+bw+c$ directly. Initial data $g(z_0)=1$, $f(z_0)=w(z_0)$ recover every local solution, and holomorphic $f,g$ make the quotient a meromorphic function, even at zeros of $a$. At a zero of $g$ with $a\ne 0$, uniqueness prevents $f=0$ there, so $g^{\prime }=-af\ne 0$ and the pole is simple. This proves the absence of movable branching without assuming that the quadratic coefficient never vanishes.

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.