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Answer by Tom Copeland for Is there an underlying explanation for the magical powers of the Schwarzian derivative?

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The velocity for a solution $f(x-c\cdot t)$ of the one dimensional heat equation

$$\frac{\partial}{\partial t}f(x-ct)=-c\frac{\partial}{\partial x}f(x-ct)=\alpha\frac{\partial^{2}}{\partial x^{2}}f(x-ct)\;$$

is determined by $$c=\frac{-\alpha D_{x}^{2}\,f(x)}{D_{x\,}f(x)}=-\alpha D_{x}\ln[D_{x}f(x))]\,.$$

Similarly, the velocity for a soliton solution $u(x-ct)$of the KdV equation

$$u_{t}-u\cdot u_{x}+\frac{1}{12e_{2}}\,u_{xxx}=0\,$$

is determined by

$$c=\frac{D_{x}^{3}u(x)-6e_{2}D_{x}(u(x))^{2}}{12e_{2}D_{x}u(x)}=\frac{1}{6e_2}S_{x}\{h^{-1}(x)\}(g(x))^{2}=-\frac{1}{6e_2}S_{x}\{h(x)\}$$

$$=\frac{e_{2}}{3}\left(\frac{\omega_{2}-\omega_{1}}{2}\right)^{2},$$

where $\,u(x)=h'(x)$ and $\,g(x)=e_{2}(x-\omega_{1})(x-\omega_{2})=\frac{1}{d[h^{-1}(x)]/dx}=h^{'}(h^{-1}(x))$.

Note:

$$g(h(x)) = e_{2} \; (h(x)-\omega_{1})\;(h(x)-\omega_{2}) = h'(x)$$

is a Riccati equation, and the Schwarzian can be expressed as

$$S_{x}\{h(x)\}=D_{x}^{2}\ln[D_{x}h(x)]-\frac{1}{2}\left[D_x\ln[D_{x}h(x)]\right]^{2}=D_{x}^{2}\ln[u(x)]-\frac{1}{2}\left[D_x\ln(u(x))\right]^{2} \; .$$

(See Old and New on the Schwarzian Derivative by Osgood for an excellent survey on the derivative and Peter Michor's papers on the relationships among solutions of the KdV eqn., geodesics of the Virasoro-Bott group, the Schwarzian, and Moebius transformations--the homogeneous solns.)

(Explicit note on the Riccati equation added July 2, 2021,)


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