Existence, uniqueness, and stability of Wasserstein gradient flows — Examples

Every open ball centered at a point of $S^1$ intersects an arc of positive arc length, so [F1] gives $\mathrm{supp}\sigma =S^1$. Since every point of $S^1$ has norm $1<2$, the support lies in $B(0,2)$.

Let $\gamma :[0,1]\to {\mathbb{R}}^2$ be continuous with $\mathrm{\mid }\gamma (0)\mathrm{\mid }=1\mathrm{/}2$ and $\mathrm{\mid }\gamma (1)\mathrm{\mid }=2$. The scalar map $t\mapsto \mathrm{\mid }\gamma (t)\mathrm{\mid }$ is continuous because $\mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }-\mathrm{\mid }y\mathrm{\mid }\mathrm{\mid }\le \mathrm{\mid }x-y\mathrm{\mid }$. By [F2], it equals $1$ for some $t$, so $\gamma (t)\in S^1$ and the support separates the two spheres.

The support of ${\delta }_0$ is $\{0\}$ by [F1]. The radial segment $\gamma (t)=((1-t)\mathrm{/}2+2t,0)$ joins $(1\mathrm{/}2,0)$ to $(2,0)$ and has strictly positive norm throughout, so it avoids $\{0\}$. Thus this support does not separate the spheres. No conclusion about energy minimizers follows from either topological calculation.

For $a,b\in \mathbb{R}$ and $0\le t\le 1$, expansion gives $(1-t)a^2+t{b}^2-((1-t)a+tb{)}^2=t(1-t)(a-b{)}^2\ge 0$. Thus the real square obeys the convexity inequality, including $t=0,1$ and $a=b$, by core algebra alone.

For finite nonnegative measures ${\mu }_0,{\mu }_1$, write $a=\int \sin xd{\mu }_0$ and $b=\int \sin xd{\mu }_1$. Linearity of integration gives $F((1-t){\mu }_0+t{\mu }_1)=((1-t)a+tb{)}^2\le (1-t)a^2+t{b}^2=(1-t)F({\mu }_0)+tF({\mu }_1)$ by step 1.1. Thus $F$ is mixture-convex.

For ${\mu }_0={\delta }_0$ and ${\mu }_1={\delta }_{\pi }$, [F2] gives the geodesic ${\mu }_t={\delta }_{\pi t}$. Then $F({\mu }_0)=F({\mu }_1)=0$, whereas $F({\mu }_{1\mathrm{/}2})={\sin }^2(\pi \mathrm{/}2)=1$. The convexity inequality at $t=1\mathrm{/}2$ would require $1\le 0$, which is false. Hence mixture convexity does not imply displacement convexity.

The characteristic equation $\dot{x}=-x$ with $x(0)=x_0$ has the solution $x(t)=e^{-t}{x}_0$, as direct differentiation verifies. Pushing ${\mu }_0$ forward by this map gives a continuity-equation solution with velocity $v_t(x)=-x$. Its action on $[0,T]$ is finite because $\int \mathrm{\mid }x{\mathrm{\mid }}^{2}d{\mu }_t=e^{-2t}\int \mathrm{\mid }x{\mathrm{\mid }}^{2}d{\mu }_0$. For compactly supported initial laws it is the flow identified by the global uniqueness theorem. For arbitrary ${\mu }_0\in {\mathcal{P}}_2$, the same conclusion follows directly: every finite-action PDE solution with the prescribed smooth velocity $-x$ is represented by its unique characteristics via superposition, so it equals this pushforward, even when ${\mu }_0$ lies in no finite cutoff.

The metric $m^{-1}{\sum }_i{x}_i\cdot y_i$ gives the particle gradient of $m^{-1}{\sum }_i\mathrm{\mid }{x}_i{\mathrm{\mid }}^2\mathrm{/}2$ equal to $(x_1,\dots ,x_m)$, so the same scalar ODE gives $x_i(t)=e^{-t}{x}_i(0)$. Consequently ${\mu }_{m,t}=(e^{-t}\mathrm{id}{)}_{\mathrm{\#}}{\mu }_{m,0}$.

For $a>0$, pushing both marginals of any coupling forward by $x\mapsto ax$ multiplies its quadratic cost by $a^2$; applying the inverse scaling gives the reverse inequality for the infimum. Thus $W_2(a_{\mathrm{\#}}\alpha ,a_{\mathrm{\#}}\beta )=a{W}_2(\alpha ,\beta )$. Apply this with $a=e^{-t}$, $\alpha ={\mu }_{m,0}$, and $\beta ={\mu }_0$ to obtain the displayed identity. Since $e^{-t}\le 1$, its supremum for $0\le t\le T$ is at most the initial W2 error, proving compact-time uniform convergence.