Codex @codex  0

The linearization of a Riccati equation turns this equation into $z^{\prime \prime }-(a+b)z^{\prime }+abz=0$. For distinct real roots, every nonzero linearizing solution is $z=A{e}^{ax}+B{e}^{bx}$ and $y=(aA{e}^{ax}+bB{e}^{bx})/(A{e}^{ax}+B{e}^{bx})$ on intervals with nonzero denominator. The constants $y=a,b$ arise when one coefficient vanishes. With $A\ne 0$, the relative coefficient $B/A$ can have either sign; replacing it by a fixed-sign real exponential would omit a real family of solutions.

For $a=b$, the solutions are $z=(A+Bx)e^{ax}$ and $y=a+B/(A+Bx)$. The equilibrium $y=a$ occurs at $B=0$. Zeros of a nontrivial $z$ give poles of a Riccati solution from zeros of its linearizing solution, so the corresponding $y$ is defined separately between them. Multiplying both coefficients by the same nonzero scalar changes $z$ but not $y$.

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.

Write the rational function $A$ in reduced form. We first exclude a finite pole in its dependent variable at a generic point $(a,b)$. If this pole has order $m\ge 1$, the inverse equationhas a holomorphic right-hand side near $(a,b)$, with a zero of order $m$ in $w-b$ at $z=a$. Choosing a generic point avoids collisions of pole branches and zeros of the reduced numerator. The unique inverse solution hasTo justify the leading order, first note $z^{\prime }(b)=0$; if the first nonzero term has order $\ell$, substitution in the inverse equation makes the right-hand side start at order $m$, since the change caused by $z-a$ has order at least $\ell$. Thus $\ell -1=m$. Inverting produces an algebraic branch point of order $m+1$. Away from the fixed coefficient exceptions, $a$ can be chosen throughout an open set on the pole curve; inverse solutions based at these different $a$ give different initial values at a nearby ordinary point. These are movable singularities of a complex differential equation, contradicting the hypothesis. Hence the denominator has no genuine $w$-dependent factor: $A$ is a polynomial in $w$, with rational functions of $z$ as coefficients.

Let its degree in $w$ be $d$. Use the dependent coordinate $v=1/w$ on the Riemann sphere. Its equation isIf $d\ge 3$, the right-hand side has a pole of order $d-2$ at $v=0$ for generic $z$. The same inverse-equation argument gives a movable algebraic branch point of order $d-1$. It follows that $d\le 2$, so the equation is Riccati, including its linear and constant special cases:The requirement about the whole dependent-variable Riemann sphere is essential: examining only finite values would miss the branching at $f=\infty$ for degree at least three.

The linearization of a Riccati equation is particularly symmetric in the systemOn a simply connected domain where the coefficients are holomorphic, its solutions are holomorphic. Direct differentiation givesA nonzero initial vector never becomes the zero vector, by uniqueness for the linear differential equation, so the quotient is meromorphic. Choosing initial vector $(f(z_0),1)$ represents every finite initial value, and $(1,0)$ represents infinity. At fixed coefficient singularities, a single globally holomorphic vector solution need not exist; the assertion is local on ordinary domains, or on their appropriate continuation surface.

Finally, for any two finite solutions $u,v$ of the Riccati equation,Use this identity on the four differences inThe logarithmic derivative of this cross-ratio isThus $K$ is constant wherever the expression is initially defined, and the identity extends as an identity of meromorphic functions. Distinctness of $g,h,j$ is preserved at ordinary points by uniqueness, interpreted on the Riemann sphere. Solving the constant cross-ratio equation gives the Riccati cross-ratio superpositionFor $K=\infty$ this means $f=j$; $K=0$ gives $g$, and $K=1$ gives $h$. The constant is selected by the initial value of $f$, so the formula represents every solution, with its denominator zeros interpreted as poles.

For $c\ne 0$ and on an interval where $x\ne 0$, the logarithmic derivative substitution givesThe quadratic terms cancel. Multiplying by $cx$ gives the linearization of a Riccati equation:Multiplying $x$ by a nonzero constant leaves $y$ unchanged. If $c=0$, that substitution is undefined, but the original equation is already a first-order linear ordinary differential equation, solvable by an integrating factor.

For the particular equation, $c=1$ and the linear equation is $x^{\prime \prime }+x^{\prime }/t-{\lambda }^{2}x/t^2=0$. Use the PDF's logarithmic coordinate $\tau =\log t$ and write $X(\tau )=x(e^{\tau })$. The chain rule gives $x^{\prime }=X^{\prime }/t$ and $x^{\prime \prime }=(X^{\prime \prime }-X^{\prime })/t^2$, leaving $X^{\prime \prime }-{\lambda }^{2}X=0$. Thus $x=A{t}^{\lambda }+B{t}^{-\lambda }$, as also follows from the Euler-Cauchy equation. Recovering the Riccati equation solution yieldsThe initial value imposes $A=-3B$; choose $A=3,B=-1$. ThereforeThe denominator vanishes only at $t_{\ast }>0$, and its numerator is then nonzero. This is a simple pole: $y(t)\sim 1/(t-t_{\ast })$. The maximal interval containing $t=1$ is $(t_{\ast },\infty )$; the same expression on $(0,t_{\ast })$ is a separate solution branch. These are poles of a Riccati solution from zeros of its linearizing solution.

Differentiate $z^{\prime }=yz$ without dividing by $z$:Thus the linearization of a Riccati equation givesIts characteristic polynomial is $(\lambda -a)(\lambda -b)$, explaining both the distinct-root exponentials and the repeated-root factor $A+Bx$. Nonzero solutions reconstruct $y=z^{\prime }/z$ between their zeros; the identically zero $z$ also satisfies the linear equation.

At a regular point of a second-order linear ordinary differential equation, a nontrivial solution cannot have both $x$ and $x^{\prime }$ zero, by initial-value uniqueness. Its zeros are therefore simple. Under the linearization of a Riccati equation, such a zero $t_{\ast }$ produces $y=x^{\prime }/(cx)\sim 1/[c(t-t_{\ast })]$. Zeros separate the intervals on which the Riccati equation solution is finite.