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For the CIR model under its money-market risk-neutral measure, the Feynman-Kac formula gives $B^{\prime }=1-bB-{\sigma }^2{B}^2/2$ and $A^{\prime }/A=-aB$, with $B(0)=0$, $A(0)=1$. Solving this Riccati equation gives explicit exponential-affine zero-coupon bond prices. The positive root parameter is $\sqrt{b^2+2{\sigma }^2}$; the squared diffusion coefficient, not the volatility itself, enters this expression.

The conformal Friedmann equation is ${\mathcal{H}}^2=H_0^2[{\Omega }_m/a+1-{\Omega }_m]$. Differentiating and using $a^{\prime }=a\mathcal{H}$ eliminates the matter term and yields a first-order Riccati equation. The expanding Big-Bang branch for $0<{\Omega }_m<1$ has $\mathcal{H}=\alpha \coth (\alpha \tau /2)$.

For a perfect fluid in general relativity with constant equation-of-state parameter $w$, use conformal time $dt=ad\tau$ and the conformal Hubble parameter $\mathcal{H}=a^{\prime }/a=aH$. The Friedmann acceleration equation gives ${\mathcal{H}}^{\prime }=a\ddot{a}=-4\pi G{a}^2(1+3w)\rho /3$. The Friedmann equation eliminates $\rho$ and gives this Riccati equation. For positive energy density, $\alpha \ne 0$ and appropriately shifted conformal time, its usual branches are $\cot (\alpha \tau )$ for $k=1$, $1/(\alpha \tau )$ for $k=0$, and $\coth (\alpha \tau )$ for $k=-1$. Their scale factors are proportional to ${\sin }^{1/\alpha }(\alpha \tau )$, $(\alpha \tau {)}^{1/\alpha }$ and ${\sinh }^{1/\alpha }(\alpha \tau )$ respectively on intervals where the bases are positive. At $\alpha =0$ the conformal Hubble parameter is constant and the separate solution is $a=A{e}^{h\tau }$.

For $A_T=-\overline{A}B/2$ and $B_T=-A^2/4$, write $A=R{e}^{i\theta }$, $B=P{e}^{i\varphi }$ and $\delta =\varphi -2\theta$. The polar equations areWhere the polar phases are defined, direct differentiation gives the first integrals $R^2-2P^2$ and $R^{2}P\sin \delta$. In particular $R^2{\theta }_T$ and $P^2{\varphi }_T$ are constant. If $\delta =\pi$ and $R^2-2P^2=C>0$, then $P_T=P^2/2+C/4$, a Riccati equation whose positive solution develops a finite-time pole. This is finite-time blowup of the reduced amplitude system; it does not establish blowup of the full wave equation beyond the domain of the weakly nonlinear expansion.

For Ornstein-Uhlenbeck process volatility, matching powers of $\sigma$ reduces the transform PDE to a scalar Riccati equation for $R$, a linear equation for $Q$ and an integral for $P$. All coefficients start at zero. The representation exists up to the Riccati moment-explosion horizon; it need not remain finite on every horizon for arbitrary exponential powers.

A linear-quadratic optimal-control problem has linear state dynamics and running and terminal costs defined by quadratic forms. Its value function is quadratic in the state, the optimal feedback is linear, and its coefficient obeys a Riccati equation.

Writing $v=e^S$ in $v^{\prime \prime }=q(x)v$ produces the Riccati equationExpanding $S^{\prime }$ in inverse powers of $x$ determines the exponential phase and algebraic prefactor recursively.

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 given particular solution $y_p=x^2$ suggests the standard reduction of a Riccati equation. Substitution of $y=x^2+1/v$ givesOn either side of $x=0$, an integrating factor is $|x{|}^{5/2}$, so $v=(C-2\sqrt{|x|})/|x{|}^{5/2}$. Hencealong with the separately recovered particular solution $y=x^2$. The denominator must remain nonzero on the solution interval; $x=0$ is a singular point of the divided differential equation.

For the positive-$x$ branch, $y(1)=(C-1)/(C-2)$ and $y(4)=16(C-2)/(C-4)$. Nonsingular endpoint values exclude $C=2,4$, soThe particular solution $y=x^2$ gives the same value directly. These evaluations concern one solution branch, with a common integration constant, rather than independently chosen solutions at the two points.

Use the curvature convention $R(X,Y)={\nabla }_X{\nabla }_Y-{\nabla }_Y{\nabla }_X-{\nabla }_{[X,Y]}$ of the earlier curvature calculation. A geodesic variation of $\gamma :[a,b]\to M$ is a smooth map $F:(-\epsilon ,\epsilon )\times [a,b]\to M$ with $F(0,t)=\gamma (t)$ such that every $t\mapsto F(s,t)$ is an affinely parametrized geodesic. Its variation vector field $J={\partial }_{s}F{|}_{s=0}$ satisfies the Jacobi equationIndeed $D_{s}T=D_{t}J$ by vanishing torsion tensor; differentiating $D_{t}T=0$ in $s$ and commuting the covariant derivatives gives this equation. A Jacobi field is a field satisfying this equation. Conversely, any prescribed pair $J(a),D_{t}J(a)$ is obtained by smoothly varying the initial point and velocity of a geodesic; smooth dependence of the geodesic equation then constructs a geodesic variation on the compact parameter interval, whose variation vector field is the prescribed Jacobi field by uniqueness of this linear equation.

For continuous piecewise smooth vector fields along a curve $V,W$ along $\gamma$, the Riemannian index form is the symmetric bilinear formContinuity at the breakpoints is part of the admissibility condition; without it an arbitrary jump could destroy the claimed index inequality. The curvature symmetries make $V\mapsto R(V,T)T$ self-adjoint. For fields with zero endpoints, this form is the second variation of geodesic energy $E=\frac{1}{2}{\int }_a^b|\dot{\gamma }{|}^{2}dt$. More explicitly, twice differentiating energy, commuting $D_s,D_t$ and integrating the term $⟨D_t(D_s{\partial }_{s}F),T⟩$ by parts leaves the displayed integral; $D_{t}T=0$ and fixed endpoints remove the remaining terms.

The endpoints $p=\gamma (a)$ and $q=\gamma (b)$ are conjugate points along $\gamma$ when there is a nonzero Jacobi field $J$ with $J(a)=J(b)=0$. Equivalently, the Riemannian exponential map at the initial velocity for this segment has singular differential: variations of initial velocity give precisely the Jacobi fields with zero initial value. For a unit-speed geodesic starting at $p$, its cut time isIf finite and the geodesic is defined there, its endpoint $\gamma (c(v))$ is the Riemannian cut point along that geodesic. Before the cut time the segment minimizes, at that time it still minimizes by continuity, and every positive extension ceases to minimize. This is the Riemannian notion, independent of whether deleting the endpoint disconnects the space.

Let ${\mathcal{H}}_0$ denote the continuous piecewise smooth vector fields along a curve with zero endpoints. The precise characterization at conjugate endpoints isThus Jacobi fields form the radical of the fixed-endpoint index form. For a smooth field $J$, integration by parts givesA Jacobi field makes both terms vanish. Conversely, if $I(J,W)=0$ for all tests, first take tests supported inside each smooth subinterval. The fundamental lemma gives the Jacobi equation on each piece. Integrating by parts piecewise then leaves, at an interior breakpoint $t_i$, the term $⟨D_{t}J(t_i^{-})-D_{t}J(t_i^{+}),W(t_i)⟩$. The values of the tests there are arbitrary, so all derivative jumps vanish. The continuous field and its continuous first covariant derivative solve the smooth Jacobi equation globally, making $J$ smooth by uniqueness of its initial-value problem. In particular, conjugacy means that this radical is nonzero. Merely imposing $I(J,J)=0$ is insufficient when the Riemannian index form is indefinite; a zero value of an indefinite quadratic form need not mean membership in its radical.

Now suppose there are no conjugate points on $(0,1]$. In a parallel orthonormal frame along $\gamma$, write the Jacobi equation as $j^{\prime \prime }+K(t)j=0$, where the symmetric matrix $K$ represents $Z\mapsto R(Z,T)T$. LetEvery Jacobi field vanishing at zero has components $j(t)=A(t)c$. The nonconjugacy assumption means $A(t)$ is invertible for every $0<t\le 1$, and hence the unique Jacobi field with the requested endpoint values isPut $U=W-V$, so $U(0)=U(1)=0$. Since $V$ solves the Jacobi equation, integration by parts gives $I(V,U)=0$ and thereforeTo establish the needed strict positivity, use the Riccati factorization of the Riemannian index form. Differentiation givesand the initial data make this conserved quantity zero. Consequently $S=A^{\prime }{A}^{-1}$ is symmetric and obeys the matrix Riccati equation $S^{\prime }+S^2+K=0$. For the components $u$ of $U$,Near zero, the differential equation gives $A(t)=t{I}_n+O(t^3)$ and $S(t)=t^{-1}{I}_n+O(t)$. Since $U$ is piecewise smooth and $U(0)=0$, $u(t)=O(t)$, so $u^{T}Su\to 0$. At time one the same boundary term vanishes because $u(1)=0$. Integration, including cancellation at all interior breakpoints by continuity, therefore givesIf equality holds, $u^{\prime }=Su$ on each piece, so $(A^{-1}u{)}^{\prime }=0$ on $(0,1]$. Continuity across the breakpoints gives $u=Ac$ with one constant vector $c$; $u(1)=0$ and invertibility of $A(1)$ force $c=0$. HenceThe nonzero endpoint condition is compatible with this proof, but the minimizing property itself holds for arbitrary specified final value.

For the relation to the Riemannian cut point, let $a>0$ be the first conjugate time along a unit-speed geodesic, and let $J$ be a nonzero Jacobi field on $[0,a]$ vanishing at both endpoints. Necessarily $D_{t}J(a)\ne 0$, otherwise uniqueness of the Jacobi equation would give $J=0$. Fix any later time $b>a$ for which the geodesic exists. Extend $J$ by zero to a continuous piecewise smooth field $Z$ on $[0,b]$. Then $I(Z,Z)=0$, while for any endpoint-vanishing smooth field $P$,Choose $P(a)=-D_{t}J(a)$, using a smooth bump times parallel fields, so this equals $-|D_{t}J(a){|}^2<0$. For small $\epsilon >0$,If a smooth variation is desired, smooth the single corner of this field while preserving the endpoints. The field and its derivative converge in the integral defining the Riemannian index form, so strict negativity persists. The second variation of geodesic energy then gives a fixed-endpoint path of smaller energy. By the Cauchy-Schwarz inequality, $L^2\le 2bE$ for paths on $[0,b]$, with equality for the original unit-speed geodesic; the new path is therefore shorter. This proves loss of geodesic minimality beyond a conjugate point, and henceA missing conjugate time is interpreted as infinity. If a finite conjugate time exists and the geodesic continues past it, every later segment fails to minimize, so its first loss of minimality occurs no later than that conjugate time.

Finally, the cut-point dichotomy for complete Riemannian manifolds is the following characterization, stated without proof. On a complete connected Riemannian manifold, the cut time is the first time at which either the endpoint is conjugate to $p$ along the given geodesic, or a distinct minimizing geodesic from $p$ reaches the same endpoint with the same length. Thus a finite Riemannian cut point is either the first conjugate point or the meeting point of at least two distinct minimizing geodesics from the initial point; both alternatives may occur together. Completeness is the hypothesis guaranteeing that competing minimizing geodesics exist at a nonconjugate cut endpoint. It is not needed for the preceding local index-form argument on an existing segment.

Use natural units and conformal time $dt=ad\tau$. The conformal Hubble parameter is $\mathcal{H}=a^{\prime }/a=aH$. For the barotropic equation of state $P=(\gamma -1)\rho$, the Friedmann acceleration equation givesThe Friedmann equation gives $8\pi G{a}^2\rho /3={\mathcal{H}}^2+k$. Hence the constant-equation-of-state conformal Riccati equation isFor $\alpha \ne 0$, separation and integration of this Riccati equation yield, after choosing the origin of conformal time,The open positive-energy density branch uses the hyperbolic cotangent, rather than the cotangent. Integrating $a^{\prime }/a=\mathcal{H}$ giveswhere the positive constants are fixed by the energy density normalization. These formulas apply on branches where the bases are positive. For $\alpha >0$, take the Big Bang at $\tau =0$. The cosmic time is thenThe integrals also give the general answer after selecting a suitable reference time when a finite Big Bang endpoint does not exist. The exceptional coasting fluid $\gamma =2/3$ has $\alpha =0$, so $\mathcal{H}=h$ is constant: $a=A{e}^{h\tau }$ and $t=(A/h)e^{h\tau }+t_{\ast }$. If $h=0$, instead $a=A$ and $t=A\tau +t_{\ast }$. A flat fluid with $\gamma =0$ is the constant-density de Sitter spacetime case, for which the flat time integral is logarithmic.

During matter domination, $\gamma =1$ and $\alpha =1/2$. Absorbing numerical factors into separate positive constants gives the closed matter-dominated Friedmann solution, the flat Einstein-de Sitter universe and the open matter-dominated Friedmann solution:To compare the age of a dust universe without a cosmological constant, specify the normalization. At fixed present Hubble parameter $H_0$, with $x=a/a_0$ and present matter cosmological density parameter ${\Omega }_m>0$, the Friedmann equation givesThe integrand decreases strictly with ${\Omega }_m$ for $0<x<1$. Therefore age ordering of dust universes at fixed Hubble parameter gives, on the expanding branches,The final bound is approached by the empty Milne universe. This compares models with the same $H_0$; fixing instead the present scale factor and energy density gives a different comparison.

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.

Work under the money-market risk-neutral measure, with $r_0\ge 0$. The zero-coupon bond price is $P(0,T)=\mathbb{E}[\exp (-{\int }_0^T{r}_{s}ds)]$. For this CIR model, write $p(\tau ,r)=A(\tau )e^{-B(\tau )r}$. The Feynman-Kac formula givesMatching the constant and linear coefficients of $r$ yieldsFor CIR bond pricing, solving this Riccati equation and integrating the scalar equation for $A$, put $\gamma =\sqrt{b^2+2{\sigma }^2}$ and $D(\tau )=(\gamma +b)(e^{\gamma \tau }-1)+2\gamma$. ThenSubstitution verifies both equations and their initial conditions. The discounted candidate solves the martingale pricing equation; its boundedness on finite horizons, or Feynman-Kac formula, identifies it with the conditional expectation. The answer is $\boxed{P(0,T)=A(T)e^{-B(T)r_0}}$. More generally the same coefficients give $P(t,T)=A(T-t)e^{-B(T-t)r_t}$.

The CIR model mean reverts to $a/b$ and has volatility $\sigma \sqrt{r}$, so it remains nonnegative and its fluctuation size depends on the rate level. For positive initial rate, zero is inaccessible if $2a\ge {\sigma }^2$; below this threshold zero is reachable but the process remains nonnegative. Its stationary law is a gamma distribution, of shape $2a/{\sigma }^2$ and rate $2b/{\sigma }^2$. Explicit affine bond prices and nonnegative rates are useful features. Limitations include exclusion of negative rates, restricted volatility behavior and a single stochastic factor; the time-homogeneous parameter family cannot fit an arbitrary initial term structure exactly.

Set $\tau =T-t$, $\kappa =\lambda -\rho b\theta$ and $k=\theta (\theta -1)/2$. For the Ornstein-Uhlenbeck process volatility, the transformed partial differential equation isUse the Gaussian volatility exponential-quadratic transform ansatz $V=\exp (P+Q\sigma +R{\sigma }^2)$, where the coefficients depend on $\tau$. Its derivatives satisfyMatching the constant, linear and quadratic powers of $\sigma$ givesThe terminal condition requiresThese polynomial ordinary differential equations have a unique local solution by the Picard's theorem for ordinary differential equations. Substitution then proves the desired partial differential equation solution on every horizon for which the coefficient solution remains finite. The Riccati equation for $R$ is solved first; $Q$ subsequently solves a linear equation and $P$ is an integral of known coefficients.

Unrestricted global existence needs a qualification. Take $a=0$, $\lambda =b=1$, $\rho =0$ and $\theta =2$. Then $Q=0$ and the Riccati equation becomesThis solves the initial condition but explodes at $\tau =3\pi /4$. Therefore no finite real exponential-quadratic solution with the required terminal condition exists on an entire horizon $T\ge 3\pi /4$ for these allowed parameters. The correct general claim is local existence, or existence before the Riccati moment-explosion horizon.

A useful sufficient global condition is $0\le \theta \le 1$, so $k\le 0$. If $b\ne 0$, the lower equilibriumtraps the solution in $r_{-}\le R(\tau )\le 0$: the polynomial vector field points inward at the upper endpoint and vanishes at the lower endpoint. If $b=0$, the equation for $R$ is linear. In either case there is no finite-time explosion; $Q$ is then a linear equation with coefficients bounded on compact time intervals, and $P$ is finite on those intervals. This proves the intended ansatz globally under that sufficient parameter restriction, without asserting it for every real $\theta$.

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.

A homogeneous linear differential operator satisfies $L(e^{\epsilon }u)=e^{\epsilon }L(u)$, so the scaling $x\mapsto x$, $u\mapsto e^{\epsilon }u$ preserves every homogeneous linear equation. Its infinitesimal generator is $V=u{\partial }_u$.

On a patch with $u\ne 0$, take $s=\log |u|$, $t=x$. Then $V(s)=1$, $V(t)=0$. Set $w=ds/dt=u^{\prime }/u$. Since $u^{\prime \prime }/u=w^{\prime }+w^2$, division by $u$ reduces the second-order equation to the Riccati equationRecover a nonzero solution by $u=C\exp (\int wdt)$, allowing the sign in $C$. The zero solution is separate, and intervals containing zeros of other solutions require separate coordinate patches. The order reduction occurs because the transformed equation depends on $s^{\prime }$ but not on $s$ itself.

With the sign convention of the displayed operator, the Lax equation is $L_t=[L,A]$. Differentiating $L\phi =k^2\phi$ shows that ${\phi }_t+A\phi$ is another solution at the same spectral parameter. The normalization at $-\infty$ fixessince $A{e}^{-ikx}=4i{k}^3{e}^{-ikx}$ there. At $+\infty$, comparison of the two plane-wave coefficients givesThus the transmission denominator $a$ is time independent. This is the Time evolution of KdV scattering data for the stated normalization.

The logarithmic derivative $S={\phi }_x/\phi +ik$ obeys the Riccati equationIts formal large-$k$ expansion givesThere is an important qualification to the integral identity in the question. For real $k$ with nonzero reflection, $e^{ikx}\phi \sim a+b{e}^{2ikx}$ has no limit at $+\infty$, so the literal improper integral of the exact $S$ need not exist. Continue to $\mathrm{Im}k>0$, where the reflected exponential decays relative to $e^{-ikx}$, and take sufficiently large $k$ away from zeros and with a consistent logarithm. Then the normalization givesThe second equality is an asymptotic expansion, not an asserted convergent series. Time independence of $a$ fixes every coefficient, so each $\int S_{n}dx$ is a conserved quantity of the Korteweg-De Vries equation. The first two nontrivial examples are $-\int udx$ and $\int u^{2}dx$.

For the usual real-valued KdV potential and real formal $k$, write $S=X+iY$ with real coefficient functions. Its imaginary part of the Riccati equation is $Y^{\prime }-2kX+2XY=0$, henceIn the formal expansion, $X$ contains precisely the even-indexed $S_n$ and $Y$ the odd-indexed ones. Rapid decay of $u$ and its derivatives makes every coefficient of the logarithm vanish at both spatial ends. Therefore $\boxed{{\int }_{\mathbb{R}}{S}_{2j}dx=0(j\ge 1)}$. This is a coefficientwise formal identity; it does not claim that the exact real-axis reflected solution has a convergent integral of $X$. The analytically continued scattering identity and the formal expansion are the conventions needed to make the requested argument valid.

Put $q={\mu }^2-1/4$ and $S=\log y$. The differential equation becomes the Riccati equation $S^{\prime \prime }+(S^{\prime }{)}^2=1/4+q/x^2$. The inverse-power exponential hierarchy begins withso choose $S_0=\epsilon x/2$, $S_1=0$ with $\epsilon =\pm 1$; additive constants specify the overall normalization. At order $x^{-2}$, $2S_0^{\prime }{S}_2^{\prime }=q/x^2$, and at subsequent ordersThese equations are formally equivalent coefficient by coefficient; the order-zero transport term happens to be constant, so its chosen zero value is not interpreted as a literal denominator in a successive-ratio hypothesis. The ansatz $S_j=c_j{x}^{-(j-1)}$ is consistent for $j\ge 2$, withEmpty sums are zero. In particular $c_3=-q$, which agrees with direct expansion of the Riccati equation.

Exponentiating the inverse-power series gives $y_{\epsilon }\sim e^{\epsilon x/2}{\sum }_{j\ge 0}{A}_j^{\epsilon }{x}^{-j}$ with $A_0=1$. Substitute this amplitude series into $u^{\prime \prime }+\epsilon u^{\prime }=q{x}^{-2}u$ to obtainFormal matching alone does not prove existence of actual solutions with these asymptotics. Here actual solutions are obtained from the Modified Bessel differential equation and the large-argument asymptotic expansion of a modified Bessel function: substituting $y=\sqrt{x}w(x/2)$ yields solutions $\sqrt{\pi }\sqrt{x}{I}_{\mu }(x/2)$ and $\sqrt{x/\pi }{K}_{\mu }(x/2)$, using the Modified Bessel function of the first kind and Modified Bessel function of the second kind, whose positive-real-axis asymptotic expansions have precisely these normalized growing and decaying series. They are linearly independent. The series need not converge; it is an asymptotic expansion with a remainder after each fixed truncation. When $q=j(j+1)$ the amplitude recurrence terminates.

Completing the square in the Hamilton-Jacobi-Bellman equation shows that the minimizing feedback isThis is a linear-quadratic optimal control problem, so set $F(x,t)=a(t)x^2$. The equation and terminal condition becomeThis Riccati equation is separable and givesConsequently the general optimal feedback law isWhen $c_t=c$ is constant,Along the corresponding optimal trajectory, $x_s=x(c+h-s)/(c+h-t)$, so the open-loop control is constant: $u_s^{\ast }=-x/(c+h-t)$ for $s\in [t,h]$.