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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 $A_T=-\overline{A}B/2$ 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.

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 $s\in [t,h]$.