Conditional flow matching and transport paths — Examples

Apply [F1] with $d=1$ and $x_1=a$. It gives the law and density of $X_t$, the field $u_t(x\mid a)$, and the optimality of $T(z)=a+{\sigma }_{\min }z$ with cost $a^2+(1-{\sigma }_{\min }{)}^2$.

For fixed $z$, differentiation gives $\frac{d}{dt}({\sigma }_{t}z+ta)=({\sigma }_{\min }-1)z+a$. Substituting $x={\sigma }_{t}z+ta$ into the displayed field yields $u_t(x\mid a)=a+\frac{{\sigma }_{\min }-1}{{\sigma }_t}{\sigma }_{t}z=a+({\sigma }_{\min }-1)z,$ which is that particle derivative. Since ${\sigma }_t>0$, its dependence on $t$ is affine and the endpoints are $z$ and $a+{\sigma }_{\min }z$.

Apply [F1] to the two endpoints $x_1=a$ and $x_1=-a$. Their conditional densities are positive and continuous in $(t,x)$, as are their time derivatives, fluxes, and spatial flux derivatives; ${\sigma }_t>0$ throughout $[0,1]$. Hence for each compact $K\subset (0,1)\times \mathbb{R}$ the finitely many such functions are bounded on $K$ by [F4]. Taking a common bound as the constant function $G_K$ gives an element of $L^1(\pi )$ because $\pi$ has total mass one. Thus [F5] applies. The mixture density is normalized by [F3] and is strictly positive because it is a sum of two positive Gaussian densities.

Put $A_{+}(t,x):=\exp \text{\hspace{0.17em}⁣}(-\frac{(x-ta{)}^2}{2{\sigma }_t^2}),A_{-}(t,x):=\exp \text{\hspace{0.17em}⁣}(-\frac{(x+ta{)}^2}{2{\sigma }_t^2})\mathrm{.}$ The common Gaussian normalizing factor cancels in the endpoint-weighted mean, so [F2] gives $m_t(x):=\frac{\int x_1{\rho }_t(x\mid x_1)\pi (d{x}_1)}{p_t(x)}=a\frac{A_{+}(t,x)-A_{-}(t,x)}{A_{+}(t,x)+A_{-}(t,x)}\mathrm{.}$ Expanding the two squares gives $A_{+}\mathrm{/}{A}_{-}=\exp (2atx\mathrm{/}{\sigma }_t^2)$ by [F7] and [F9]. Multiplying numerator and denominator by $\exp (-atx\mathrm{/}{\sigma }_t^2)$ therefore yields $m_t(x)=a\tanh (atx\mathrm{/}{\sigma }_t^2)$.

For endpoint $x_1\in \{-a,a\}$, [F1] gives $u_t^c(x\mid x_1)=cx+(1-tc)x_1,c:=\frac{{\sigma }_{\min }-1}{{\sigma }_t}\mathrm{.}$ Taking the endpoint-weighted mean and using step 1.2 gives $u_t^m(x)=cx+(1-tc)m_t(x)$. Since ${\sigma }_t=1+t({\sigma }_{\min }-1)$, we have ${\sigma }_t-t({\sigma }_{\min }-1)=1$ and hence $1-tc=1\mathrm{/}{\sigma }_t$. Substitution gives the marginal velocity formula in the Example.

Fix $t\in (0,1)$ and put $\beta :=a\mathrm{/}{\sigma }_t>0$ and $k:=at\mathrm{/}{\sigma }_t^2>0$. The formula for $\tanh$, the addition formula [F7], and the reciprocal identity [F9] give $\tanh (-y)=-\tanh (y)$, so $u_t^m$ is odd. For $y>0$, [F7] gives $\exp (0)=\exp (0{)}^2$, and positivity [F9] therefore gives $\exp (0)=1$; strict increase [F8] now yields $r:=\exp (2y)>1$. Put $s:=(r-1)\mathrm{/}(r+1)\in (0,1)$. Rewriting $\tanh (y)$ and $\tanh (2y)$ using [F7, F9] gives $\tanh (y)=s,\tanh (2y)=\frac{r^2-1}{r^2+1}=\frac{2s}{1+s^2}<2s\mathrm{.}$ Thus for every $x>0$, step 2.1 gives $u_t^m(2x)-2u_t^m(x)=\beta (\tanh (2kx)-2\tanh (kx))<0.$ An odd affine function must be linear through the origin, and every such function has $u(2x)-2u(x)=0$. Hence $u_t^m$ is not affine and is nonlinear.