so at time $t$ we interpret $\vec v^{(h, \varepsilon)} (t,y)$ as the Eulerian velocity vector field at $y$, while $V^{(h, \varepsilon)} (x,t)$ is the Lagrangian velocity with label $x$.
Hence we have the continuity equation

$${\frac{\partial \rho^{(h, \varepsilon)} (x,t)}{\partial t}}+\nabla\cdot \bigl(\vec v^{(h, \varepsilon )}(t,x)\rho^{(h, \varepsilon)} (x,t)\bigr)=0$$

in the weak sense, and we proceed to obtain a metric version. We have

$$\int_{{\mathbb{S}}^2}\Vert \vec v^{(h, \varepsilon )}(t,y)\Vert^2\rho^{(h, \varepsilon)} (y,t)m(dy)=\int_{{\mathbb{S}}^2}\Vert \vec v^{(h, \varepsilon )}(t,X^{(h, \varepsilon )}(x,t))\Vert^2\rho_0 (x)m(dx).$$

Next we compare $\rho_{jh}$ from the approximating path with the density $\rho^{(h, \varepsilon)} (x,t)$ that is  induced by from $\rho_0$ by $x\mapsto X^{(h, \varepsilon )}(t; x, v)$ where $v=\nabla q_0(x)$; we write $
X^{(h, \varepsilon )}(t; x)=X^{(h, \varepsilon )}(t; x, v)$ for $v=\nabla q_0(x)$. Suppose that $h=\tau /N$ for some integer $N$, and let  $t_0=0$ and $t_{j+1}=t_j+h$.
Also for all $\psi\in C({\mathbb{S}}^2; {\mathbb R})$ and $t_j=jh$, we take

\begin{align}\bigl\vert \int_{{\mathbb{S}}^2}&\psi (y)\rho^{(h, \varepsilon)} (y,t_j)m(dy)- \int_{{\mathbb{S}}^2}\psi (y)\rho_{jh} (y)m(dy)\Bigr\vert\nonumber\\
&=\Bigl\vert \int_{{\mathbb{S}}^2}\bigl( \psi (X^{(h, \varepsilon )}(x,t_j))-\psi (X_{jh}(x))\bigr)\rho_{0}(x)m(dx)\Bigr\vert\nonumber\\
&\leq \Vert\nabla\psi\Vert_{L^\infty} \int_{{\mathbb{S}}^2}\Vert X^{(h, \varepsilon )}(x,t_j)-X_{jh}(x)\Vert\rho_{0}(x)m(dx), \end{align}

$$W_1(\rho^{(h, \varepsilon)} (\cdot , t_j), \rho_{jh})\leq  \int_{{\mathbb{S}}^2}\Vert X^{(h, \varepsilon )}(x,t_j)-X_{jh}(x)\Vert\rho_{0}(x)m(dx).$$

The measures $\rho^{(h, \varepsilon )}(x,t_j)$ satisfy a weaker variant of (<ref>), so $t\mapsto \rho^{(h, \varepsilon )}(\cdot ,t)$ is $2$-absolutely continuous $[0, \tau ]\rightarrow {\mathcal W}^1( {\mathbb{S}}^2).$ From the integral equation (<ref>), we have

\begin{align}\sum_{j=1}^N {\frac{\Bigl\Vert \begin{bmatrix} X^{(h, \varepsilon )}(x, t_{j+1})-X^{(h, \varepsilon )}(x,t_j)\\ V^{(h, \varepsilon )}(x,t_{j+1})-V^{(h, \varepsilon )}(x,t_j)\end{bmatrix}\Bigr\Vert^2}{t_{j+1}-t_j}}&\leq 2\sum_{j=1}^N (t_{j+1}-t_j) \Bigl\Vert \begin{bmatrix} X^{(h, \varepsilon )}(x,t_j)\\ V^{(h, \varepsilon )}(x,t_j)\end{bmatrix}\Bigr\Vert^2\nonumber\\
&+2\int_0^\tau \Vert\nabla (\Theta_1\circ p^{(h, \varepsilon )}_u )(X^{(h, \varepsilon )}(x,u))\Vert^2 du\end{align}

so  integrating this against $\rho_0(x)m(dx)$, we obtain

\begin{align}\sum_{j=1}^N {\frac{W_1(\rho^{(h,\varepsilon )}(\cdot , t_{j+1}), \rho^{(h,\varepsilon )}(\cdot , t_{j}))^2}{t_{j+1}-t_j}}&\leq 2\sum_{j=1}^N (t_{j+1}-t_j) \int_{{\mathbb{S}}^2}\Bigl\Vert \begin{bmatrix} X^{(h, \varepsilon )}(x,t_j)\\ V^{(h, \varepsilon )}(x,t_j)\end{bmatrix}\Bigr\Vert^2\rho_0(x)m(dx)\nonumber\\
&+2\int_0^\tau\int_{{\mathbb{S}}^2}\Vert\nabla(\Theta_1\circ p^{(h, \varepsilon )}_u )(x)\Vert^2\rho^{(h, \varepsilon )}(x,u)m(dx)du\end{align}

The final step is to let $h,\varepsilon\rightarrow 0+$. The family of functions $[X^{(h, \varepsilon )}; V^{(h, \varepsilon )}]$ is uniformly equicontinuous, so by Arzela–Ascoli's theorem, there exists a sequence $h_\nu\rightarrow 0$ and $\varepsilon_\nu\rightarrow 0$ such that   $[X^{(h_\nu, \varepsilon_\nu )};V^{(h_\nu, \varepsilon_\nu )}]\rightarrow [X;V]$ uniformly on ${\mathbb {S}}^2\times [0, \tau ]$. Let $\rho (x,t)$ be the probability density that is induced from $\rho_0$ by $x\mapsto X(x,t)$.
We have

\begin{align} \Bigl\vert\int_{{\mathbb{S}}^2}&\psi (X(x,t))\rho_0(x)m(dx)-\int_{{\mathbb{S}}^2}\psi (X^{(h_\nu, \varepsilon_\nu )}(x,t)\rho_0(x)m(dx)\Bigr\vert\nonumber\\
&\leq \Vert\nabla\psi\Vert_{L^\infty} \int_{{\mathbb{S}}^2}\Vert X^{(h_\nu, \varepsilon_\nu)}(x,t)-X(x,t)\Vert\rho_{0}(x)m(dx), \end{align}

$$W_1(\rho (\cdot ,t),\rho^{(h_\nu, \varepsilon_\nu)}(\cdot , t))\leq \int_{{\mathbb{S}}^2}\Vert X^{(h_\nu, \varepsilon_\nu)}(x,t)-X(x,t)\Vert\rho_{0}(x)m(dx)$$

where the right-hand side goes to $0$ as $(h_\nu, \varepsilon_\nu)\rightarrow (0,0).$We can take the limit of (<ref>), and deduce that the weak continuity equation holds with velocity $\vec v(x,t)$ and density $\rho (x,t)$. We have

\begin{align}\int_0^\tau  \int_{{\mathbb{S}}^2}\psi (x,t) \rho^{(h_\nu ,\varepsilon_\nu )}(x,t)m(dx)dt&=\int_0^\tau  \int_{{\mathbb{S}}^2}\psi (X^{(h_\nu ,\varepsilon_\nu )}(x,t),t)\rho_0(x)m(dx)dt\nonumber\\
&\rightarrow \int_0^\tau  \int_{{\mathbb{S}}^2}\psi (x,t) \rho (x,t)m(dx)dt\end{align}

for all $\psi\in  C({\mathbb{S}}^2\times [0, \tau ];{\mathbb{R}})$, so $p(x,t)=\rho (x,t)$ almost everywhere, at least for the sequence $h_\nu\rightarrow 0$. Finally, one can easily check that $\delta\leq \rho (x,t)\leq 1/\delta$ for almost all $(x,t)\in {\mathbb{S}}^2\times [0, \tau ]$.

§.§ The acceleration equation

The function $T{\mathbb{S}}^2\rightarrow {\mathbb{R}}^4:$ $[X;V]\mapsto [V; -X-\nabla (\Theta_1\circ \rho_u)]$ on the right-hand side of (<ref>) has derivative

$$\begin{bmatrix} 0& -I-D^2(\Theta_1\circ \rho_u )\\ I&0\end{bmatrix},$$

where $D^2(\Theta_1\circ \rho )$ is a matrix of measures by (<ref>), and the diagonal terms are evidently zero, so the function is of bounded variation. In the following result, we hypothesize that the family of measures $D^2(\Theta_1\circ\rho^{(h, \varepsilon )}_t)$ is $L^2$. As in Ambrosio's stability Theorem 6.3 [1]  for Lagrangian flows, we impose additional assumptions on the vector fields in order to control the flow on the measures that the differential equation generates.

Suppose that there exist $\varepsilon_0,h_0>0$ such that the solution of (<ref>) gives a differentiabe function $\vec v^{(h, \varepsilon )}: {\mathbb{S}}\times [0, \tau ]\rightarrow {\mathbb R}^2$ such that

$$\int_0^\tau \int_{{\mathbb{S}}^2}\bigl\Vert D_y\bigl(\vec v^{(h,\varepsilon)}(y,t)\bigr)\bigr\Vert^2\rho^{(h, \varepsilon)}(y,t)m(dy)dt$$

are uniformly bounded for $h_0>h>0$ and $\varepsilon_0>\varepsilon>0$. (i) Then there exists a weak solution to the Euler equations.(ii) Suppose moreover that $\Vert D^2(\Theta_1\circ p_u^{(h,\varepsilon)})(x)\Vert\leq K$ for all $x\in {\mathbb{S}}^2$ and $h_0>h>0$ and $\varepsilon_0>\varepsilon>0$. Then as $(h_k, \varepsilon_k)\rightarrow (0,0)$, the approximate solution $p_t^{(h_k, \varepsilon_k)}$ converges to the weak solution $\rho (x,t)$ in ${\mathcal{W}}^2( {\mathbb{S}}^2).$
The acceleration equation leads to

\begin{align}\label{weakacceleration}\int_0^\tau\int_{{\mathbb{S}}^2}&\phi (X^{(h,\varepsilon)}(x,t),t)\cdot {\frac{\partial V^{(h,\varepsilon)}(x,t)}{\partial t}}\rho_0(x)m(dx)dt\nonumber\\
&= \int_0^\tau\int_{{\mathbb{S}}^2}\Bigl(-\phi (X^{(h,\varepsilon)}(x,t),t)\cdot
X^{(h,\varepsilon )}(x,t)\nonumber\\
&\quad -\phi (X^{(h,\varepsilon)}(x,t),t) \cdot\nabla (\Theta_1\circ p_t^{(
h, \varepsilon)})(X^{(h,\varepsilon)}(x,t)\Bigr)\rho_0(x)m(dx)dt;\end{align}

for all  $\phi\in  C^1({\mathbb{S}}^2\times [0, \tau ];{\mathbb{R}})$, which gives

\begin{align}\label{weakacceleration}\int_0^\tau\int_{{\mathbb{S}}^2}&\phi (y,t)\cdot {\frac{\partial \vec v^{(h,\varepsilon)}(y,t)}{\partial t}}\rho ^{(h, \varepsilon)}(y,t)m(dy)dt\nonumber\\
&= \int_0^\tau\int_{{\mathbb{S}}^2}\Bigl(-\phi (y,t)\cdot y-\phi (y,t) \cdot\nabla (\Theta_1\circ p_t^{(h, \varepsilon)})(y)\Bigr)\rho^{(h, \varepsilon)}(y,t)m(dy)dt;\end{align}

which is bounded above by

\begin{align} {}&\int_0^\tau\int_{{\mathbb{S}}^2}\Vert\phi (y,t)\Vert\rho ^{(h, \varepsilon)}(y,t)m(dy)dt+\Bigl( \int_0^\tau\int_{{\mathbb{S}}^2}\Vert\phi (y,t)\Vert^2\rho^{(h, \varepsilon)}(y,t)m(dy)dt\Bigr)^{1/2}\nonumber\\
&\times \Bigl( \int_0^\tau\int_{{\mathbb{S}}^2}\Vert\nabla (\Theta_1\circ p_t^{(h, \varepsilon)}) (y)\Vert^2\rho ^{(h,\varepsilon)}(y,t)m(dy)dt\Bigr)^{1/2}.\end{align}

Now the left-hand side of (<ref>) leads to

\begin{align}\label{split}\int_0^\tau \int_{{\mathbb{S}}^2}&\phi (y,t)\cdot {\frac{\partial \vec v^{(h,\varepsilon)}(y,t)}{\partial t}}\rho^{(h, \varepsilon)}(y,t)m(dy)dt\nonumber\\
&=-\int_0^\tau \int_{{\mathbb{S}}^2}
{\frac{\partial \phi (y,t)}{\partial t}}\cdot  \vec v^{(h,\varepsilon)}(y,t)\rho^{(h, \varepsilon)}(y,t)m(dy)dt\nonumber\\
&\quad-\int_0^\tau \int_{{\mathbb{S}}^2}\phi (y,t)\cdot  \vec v^{(h,\varepsilon)}(y,t){\frac{\partial\rho^{(h, \varepsilon)}(y,t)}{\partial t}}m(dy)dt,\end{align}

in which the first integral in (<ref>) is bounded in modulus by

\begin{align}\leq \Bigl(\int_0^\tau &\int_{{\mathbb{S}}^2}\Bigl\Vert {\frac{\partial \phi (y,t)}{\partial t}}\Bigr\Vert^2\rho^{(h, \varepsilon)}(y,t)m(dy)dt\Bigr)^{1/2}\Bigl(\int_0^\tau \int_{{\mathbb{S}}^2}\Vert \vec v^{(h,\varepsilon )}(y,t)\Vert^2\rho^{(h, \varepsilon)}(y,t)m(dy)dt\Bigr)^{1/2};\end{align}

while we use the continuity equation to replace the second integral in (<ref>) by

\begin{align}\int_0^\tau \int_{{\mathbb{S}}^2}&\phi (y,t)\cdot  \vec v^{(h,\varepsilon)}(y,t)\nabla \cdot\bigl( \vec v^{(h,\varepsilon)}(y,t)\rho^{(h, \varepsilon)}(y,t)\bigr)m(dy)dt\nonumber\\
&=-\int_0^\tau \int_{{\mathbb{S}}^2}\nabla\bigl(\phi (y,t)\cdot  \vec v^{(h,\varepsilon)}(y,t)\bigr)\cdot\bigl( \vec v^{(h,\varepsilon)}(y,t)\rho^{(h, \varepsilon)}(y,t)\bigr)m(dy)dt\nonumber\\
&\leq\Bigl( \int_0^\tau \int_{{\mathbb{S}}^2}\bigl\Vert\nabla\bigl(\phi (y,t)\cdot  \vec v^{(h,\varepsilon)}(y,t)\bigr)\bigr\Vert^2\rho^{(h, \varepsilon)}(y,t)m(dy)dt\Bigr)^{1/2}\nonumber\\
&\quad\times \Bigl( \int_0^\tau \int_{{\mathbb{S}}^2}\bigl\Vert \vec v^{(h,\varepsilon)}(y,t)\bigr\Vert^2\rho^{(h, \varepsilon)}(y,t)m(dy)dt\Bigr)^{1/2},\end{align}

where the last step is in accord with (<ref>). Hence both sides of (<ref>) define bounded linear functionals on $C^1({\mathbb{S}}^2\times [0, \tau ];{\mathbb{R}})$, and we obtain a weak solution in the limit as $(h_\nu , \varepsilon_\nu )\rightarrow (0,0)$ through some subsequence.(ii) We introduce

$$\chi (x,t)=\Bigl\Vert \begin{bmatrix} X^{(h_1, \varepsilon_1)}(x,v,t)-X^{(h_2, \varepsilon_2)}(x,v,t)\\  V^{(h_1, \varepsilon_1)}(x,v,t) -V^{(h_2, \varepsilon_2)}(x,v,t)\end{bmatrix}\Bigr\Vert$$

\begin{align}\label{eta}\eta (x,t)=\int_0^t \bigl\Vert \nabla (\Theta_1\circ p_u^{(h_1, \varepsilon_1 )})(X^{(h_1,\varepsilon_1)}(x,v,u)- \nabla (\Theta_1\circ p_u^{(h_2, \varepsilon_2 )})(X^{(h_1,\varepsilon_1)}(x,v,u)\bigr\Vert du,\end{align}

so that by (<ref>), we have

$$\chi (x,t)\leq \eta (x,t)+\int_0^t \bigl\Vert D^2(\Theta_1\circ p_u^{(h_1, \varepsilon_1)} )(y)\bigr\Vert_{L^\infty_y}\chi (x,u)du.$$

Then by Gronwall's inequality Theorem 12.3.3 of [25], we deduce

\begin{align}\label{Gron}\chi &(x,t)\leq \eta(x,t)\nonumber\\
&+\int_0^t  \bigl\Vert D^2(\Theta_1\circ p_s^{(h_1, \varepsilon_1)} )(y) \bigr\Vert_{L^\infty_y}\exp\Bigl( \int_s^t\bigl\Vert D^2(\Theta_1\circ p_u^{(h_1, \varepsilon_1)} )(y) \bigr\Vert_{L^\infty_y}du\Bigr) \eta (s,x)ds,\end{align}

\begin{align}\label{intchi}W_1\bigl(\rho^{(h_1, \varepsilon_1)}(\cdot , t), \rho^{(h_2, \varepsilon_2)}(\cdot , t)\bigr)\leq \int_{{\mathbb{S}}^2}\chi (x,t)\rho_0(x)m(dx)\end{align}

\begin{align}\label{inteta}&\int_{{\mathbb{S}}^2}\eta (x,t)\rho_0(x)m(dx)\nonumber\\
&=\int_0^t \int_{{\mathbb{S}}^2}\bigl\Vert \nabla (\Theta_1\circ p_u^{(h_1, \varepsilon_1 )})(y)- \nabla (\Theta_1\circ p_u^{(h_2, \varepsilon_2 )})(y)\bigr\Vert\rho^{(h_1, \varepsilon_1)}(y,u)m(dy)du.\end{align}

For $\varepsilon >0$, we deduce that $W_1((\rho^{(h_1, \varepsilon )}(\cdot , t), \rho^{(h_2, \varepsilon )}(\cdot , t))\rightarrow 0$ as $h_1,h_2\rightarrow 0$.We have $\nabla p_{t}^{(h, \varepsilon)}=
\nabla \psi_\varepsilon \ast p^{(h)}_{t}\rightarrow \nabla \rho_t$ in $L^2$ where $t=hj$ as $h, \varepsilon\rightarrow 0$, so $\nabla ( \Theta_1\circ p_{t}^{(h_1, \varepsilon_1)}(y)-  \Theta_1\circ p_{t}^{(h_2, \varepsilon_2)}(y))\rightarrow 0$ in $L^2$ as $(h_1, \varepsilon_1), (h_2, \varepsilon_2)\rightarrow (0,0)$. We deduce that (<ref>) converges to $0$, hence by (<ref>) and (<ref>), $\rho^{(h_1, \varepsilon_1)}(\cdot , t)$ converges to $\rho (\cdot ,t)$.

(i) In their solution of the compressible semigeostrophic equations in dual space, Cullen, Gilbert and Pelloni [18] (6.2) use a continuity equation in which the velocity vector field is divergence free, and they can therefore exploit directly the results of Ambrosio [1] on vector fields of bounded variation. In Propositions <ref> and <ref>, we have a vector field on $T{\mathbb{S}}^2$ which is given by velocity and acceleration on ${\mathbb{S}}$, and have the additional complication that the density appears in both differential equations.(ii) The estimates of Corollary <ref> seem too weak to force convergence of the $\rho_{hj}$ in $L^p$ spaces; see page 1067 of [24].
[1] L. Ambrosio, Transport equation and Cauchy problem for $BV$ vector fields, Invent. Math. 158 (2004), 227-260.[2] L. Ambrosio and W. Gangbo, Hamiltonian ODEs in the Wasserstein space of probability measures, Comm. Pure Appl. Math. 61 (2008), 18-53.

[4] L. Ambrosio, N. Gigli and G. Savare, Gradient flows: In metric spaces, and in the space of probability measures, Second Edition, (Springer, 2008).

[5] A. Bennett, Lagrangian Fluid  Dynamics, (Cambridge University Press, 2006).[6] G. Blower, The Gaussian isoperimetric inequality and concentration, Positivity 7 (2003), 203-224.[7] F. Bolley and I. Gentil, Phi-entropy inequalities for diffusion semigroups, J. Math. Pures Appl. (9) 93 (2010), 449-473.

[9] X. Cabre, Nondivergent elliptic equations on manifolds with nonnegative curvature, Comm. Pure Appl. Math. 50 (1997), 623-665.
[11] I. Chavel, Isoperimetric inequalities:  Differential geometric and analytic perspectives, (Cambridge University Press, 2001).[12] I. Chavel, Riemannian geometry: A Modern Introduction, (Cambridge University Press, 1993).[13] E. Chiodaroli, C. De Lewis, and O. Kremil, Global ill-posedness of the isentropic system of gas dynamics, Comm. Pure Appl. Math. 68 (2015), 1157-1190.

[15] D. Cordero-Erausquin, R.J. McCann and M. Schmückenschlager, Prekopa-Leindler inequalities on Riemannian manifolds, Jacobi fields, and optimal transport, Ann. Fac. Sci. Toulouse Math. (6) 15 (2006), 613-635.[16]M.J. P. Cullen, R.J. Douglas, I. Roulstone and M.J. Sewell, Generalized semi geostrophic theory on a sphere, J. Fluid Mechanics 531 (2005), 123-157.[17] M.J.P. Cullen and W. Gangbo, A variational approach to the $2$-dimensional semi geostrophic shallow water equations Arch. Ration. Mech. Anal. 156 (2001), 241-273.[18] M. Cullen, D.K. Gilbert and B. Pelloni, Solutions of the fully compressible semi-geostrophic system, Comm. Partial Differential Equations 39 (2014), 591-625.

[19] M.J.P. Cullen, T. Kuna, B. Pelloni, and M. Wilkinson, The stability principle and global weak solutions of the free surface semi-geostrophic equations in geostrophic coordinates, Proc. A. 475 (2019), no. 2229, 15pp.
[21] J.-D. Deuschel and D.W. Stroock, Hypercontractivity and spectral gap of symmetric diffusions with applications to the stochastic Ising models, J. Funct. Anal. 92 (1990), 30-48.[22] N. Dunford and J. Schwartz, Linear Operators: Part I General theory, (Wiley Interscience, 1957).
[24] W. Gangbo and M. Westdickenberg, Optimal transport for the system of isentropic Euler equations, Comm. Partial Differential Equations 34 (2009), 1041-1073.[25] E. Hille, Methods in classical and functional analysis, (Addison-Wesley, 1972).[26] R.A. Horn and C.A. Johnson, Matrix Analysis, (Cambridge University Press, 1985).

[28] M. Ledoux, On an integral criterion for hypercontractivity of diffusion semigroups and extremal functions, J. Funct. Anal. 105 (1992), 444-465.[29] M. Liero, A. Mielke and G. Savaré, Optimal entropy transport problems and a new Hellinger-Kantorovich distance between positive measures, Invent. Math. 211 (2018), 969-1117.[30] G. Loeper, On the regularity of solutions of optimal transportation problems, Acta Math. 202 (2009), 241-283.[31] G. Loeper, Regularity of optimal maps on the sphere: the quadratic cost and the reflector antenna. Arch. Ration. Mech. Anal. 199 (2011), 269-289.
[33] R.J. McCann, Polar factorization of maps on Riemannian manifolds, Geom. Funct. Anal. 11 (2001), 589-608.[34] C. Niculescu and L.-E. Persson, Convex functions and their applications, (Canadian Math. Soc., 2006).[35] E. Onofri, On the positivity of the effective action in a theory of random surfaces, Comm. Math. Phys. 86 (1982), 321-326.

[37] C. Villani, Topics in Optimal transportation, (American Mathematical Society, 2003).[38] C. Villani, Optimal Transport: Old and New, (Springer, 2009).[39] M. Westdickenberg and J. Wilkening, Variational particle schemes for the porous medium equation and the system of isentropic Euler equations, M2AN Math. Model. Numer. Anal. 44 (2010), 133-166.

Gordon Blower

Suppose that $(h_k)_{k=1}^\infty$ is a decreasing sequence such that $h_k\rightarrow 0$ as $k\rightarrow\infty$, and observe that $H_k(t)=\sum_{j=0}^{\lfloor \tau/h_k\rfloor} {\mathcal{H}}(\rho_{jh_k}, q_{jh_k}){\mathbb{I}}_{[jh_k, (j+1)h_k]}(t)$ gives a decreasing function of $t\in [0,\tau ]$. By Helly's selection theorem, there exists a subsequence, also denoted  $(h_k)_{k=1}^\infty$
and a decreasing function $H$, such that $H_k(t)\rightarrow H(t)$ as $k\rightarrow\infty$ for all $t\in [0,\tau ]$.

and the derivatives satisfy

$$ C_N:\qquad (1+1/N)r^2\Theta'(r)+r^3\Theta''(r)\geq 0\qquad (r>0),$$

or equivalently, $r\mapsto \Theta (r^{-N})$ is convex; or
The notation emphasizes the vector field since

\begin{align}\Bigl({\frac{d}{dt}}\Bigr)_{t=0}\int_M g(y)(T\circ \Phi_t)\sharp f(y)m(dy)&=\Bigl({\frac{d}{dt}}\Bigr)_{t=0}\int_M (g\circ T)(\Phi_t(y))f(y)m(dy)\nonumber\\
&= \int_M\langle \nabla (g\circ T)(y), V(y)\rangle f(y)m(dy)\end{align}

For $T\in C^1({\mathbb{R}}^n, {\mathbb{R}}^n)$, one can compute that

$$E[T]={\frac{1}{2}}\int_{{\mathbb{R}}^n}\Vert T(x)-x\Vert^2 f(x)dx$$

has first inner variation

$$\langle \delta_i E[T], V\rangle =  \int_{{\mathbb{R}}^n}\langle  T(x)-x, DT(x)V(x)\rangle f(x)dx.$$

For curved manifolds, it is much more difficult to compute the inner variation.

(vi) For $\Theta_0(X)=\Theta_1(\rho(X,t))$,

$$\Vert\nabla \Theta_1(\rho (x,t))\Vert^2={\frac{1}{\dot s^2}}\Bigl( {\frac{\partial \Theta_1(\rho(x,t))}{\partial t}}\Bigr)^2 +\kappa_g^2\dot s^4.$$
The intended application is to $\Theta_0 (X(x,t))=\rho (x,t)\Theta' (\rho (x,t))+\Theta (\rho(x,t))=\Theta_1(\rho (x,t))$, so in terms of the frame $\{ X,V, X\times V\}$, we have coordinates
$\nabla(\Theta_1\circ \rho)=\theta_VV+\theta_g(X\times V)$, which satisfy

$$\theta_g\dot s^2=-\nabla (\Theta_1\circ \rho )\cdot (X\times V)=V\cdot (X\times \nabla (\Theta_1\circ \rho ))=-\kappa_g\dot s^3,$$

$$\theta_VV\cdot V=V\cdot \nabla(\Theta_1\circ \rho )=-{\frac{\partial}{\partial t}}\Theta_1(\rho (x,t)),$$

by the continuity equation, so

$$\bigl\Vert \nabla (\Theta_1\circ \rho )\bigr\Vert^2=\theta_V^2\dot s^2+\theta_g^2\dot s^2={\frac{1}{\dot s^2}}\Bigl( {\frac{\partial\Theta_1(\rho )}{\partial t}}\Bigr)^2+\kappa_g^2\dot s^4.$$

An optimal transport map $x\mapsto X(x,t)$  admits of a polar factorization ${\mathbb{S}}^2\rightarrow [0, \pi]\times SO(3):$ $x\mapsto (r(x,t), \hat (X,t))$, where $X(x,t)=\hat X(x,t))x$ and $r(x,t)=d(X(x,t),x)$.
Given $x,y\in {\mathbb{S}}^2$ with $x\neq \pm y$, there exists $p\in {\mathbb{S}}^2$ such that $p\cdot x=p\cdot y=0$ and $v$ such that $y\in {\hbox{span}}\{ x,v\}$ and  $F=[x,v,p]\in SO(3)$ gives an orthonormal frame at $x$. Let $r =d(x,y)\in (0, \pi )$. Then $(\exp (tJ)F)_{t\in {\mathbb{R}}}$ is a coset such that $y=\exp (r J)x$ and $\exp (r J)F$ is an orthonormal frame at $y$. In particular, when $y=\exp_x\nabla \phi (x)$ arises from an optimal transport map, we can take $r =\Vert \nabla \phi (x)\Vert$ and $v=\nabla \phi (x)/r.$ Thus an optimal transport map $X(\cdot ,t): {\mathbb{S}}^2\rightarrow {\mathbb{S}}^2$ determines a map ${\mathbb{S}}^2\rightarrow [0, \pi]\times SO(3):$ $x\mapsto (r(x,t),\hat X(x,t))$, where  $r(x,t)=d(X(x,t),x)$ and $y=X(x,r)=\hat X(x,r)x$ with $\hat X(x,r)v=(d/dt)_{t=r} \exp_x(tv)$.

which by the continuty equation is

\begin{align}=-\int_0^\tau \int_{{\mathbb{S}}^2}{\frac{\partial \phi (y,t)}{\partial t}}\cdot
\vec v^{(h)}(y,t)\rho_t^{(h)}(y)m(dy)dt\nonumber\\
-\int_0^\tau \int_{{\mathbb{S}}^2}\vec v^{(h)}(y,t)\cdot\nabla\bigl( \phi (y,t)\cdot
\vec v^{(h)}(y,t)\bigr)\rho_t^{(h)}(y)m(dy)dt\end{align}

(iv) with  $\theta_1^2=\int_M \Theta_1(\rho_1(x))^2\rho_1(x)m(dx)$, the internal energy satisfies
\begin{align}{\mathcal U}(f)\geq{\mathcal U}(\rho_1)+{\frac{\lambda_1\Theta (1/m(M))^2}{4}}\Bigl( \int_M \Bigl\vert {\frac{\Theta_1^2(\rho_1(x))}{\theta_1^2}}-1\Bigr\vert \rho_1(x)m(dx)\Bigr)^{2}.\end{align}

(iv) Let $\theta_1^2=\int_M \Theta_1(\rho_1(x))^2\rho_1(x)m(dx)$ and $g(x)=\Theta_1(\rho_1(x))^2/\theta_1^2$, so that $g\rho_1$ is a probability density function; here $\Theta_1(r)\geq \Theta (r)$, so

$$\theta_1^2\geq \Bigl(\int_M \Theta (\rho_1 )\rho_1dm\Bigr)^{2}\geq \Theta (1/m(M))^2$$

by (5.25), since $m(M)\rho_1$ is a probability density function with respect to the probability measure $dm/m(M)$. Then by the logarithmic Sobolev inequality

\begin{align}\int_M \rho_1(x) \bigl\Vert  \nabla_M\Theta_1(\rho_1)\bigr\Vert^2 m(dx)&\geq \theta_1^2\int_M g(x)\log g(x) \, \rho_1(x)m(dx)\nonumber\\
&\geq {\frac{\lambda_1\theta_1^2}{4}}\Bigl( \int_M \vert g(x)-1\vert \rho_1(x)m(dx)\Bigr)^{2},\end{align}

where the final step follows from Csiszar's inequality.
This gives a quantitative measure of the $L^1$ distance between $g\rho_1$ and $\rho_1$. One can make the right-hand side of (6.14) equal to zero by choosing $\rho_1$ to be a multiple of an indicator function; whereas such a choice conflicts with Theorem 6.3(i).

We introduce the momentum $P=\rho u$ and consider the energy

$${\mathcal H}(P, \rho )={\frac{1}{2}}\int_M {\frac{\Vert P(x)\Vert^2}{\rho (x) }}m(dx)+\int_M \Theta (\rho (x) )\rho (x)m(dx).$$

Suppose that $\Theta (e^{r})$ is convex and increasing for $r\in {\mathbb{R}}$. Then ${\mathcal H}$ is a convex function of $(P, \rho )$ for the pointwise linear structure on $T_xM\times [0, \infty )$.
Proof.  Given $(P_0, \rho_0)$ and $(P_1, \rho_1)$ at $x\in M$, we introduce
$(P_\tau, \rho_\tau )=(1-\tau)(P_0, \rho_0)+\tau (P_1, \rho_1)$ for $\tau\in [0,1]$, and we find

$${\frac{d^2}{d\tau^2}}\Bigl({\frac{\Vert P_\tau\Vert^2}{2\rho_\tau}}+\Theta (\rho_\tau)\rho_\tau\Bigr)={\frac{\Vert P'_\tau \rho_\tau-P_\tau\rho'_\tau\Vert^2}{\rho_\tau^3}}+ \bigl(\rho_\tau \Theta''(\rho_\tau )+2\Theta'(\rho_\tau )\bigr) (\rho_\tau')^2\geq 0.$$

Let $\psi\in C^2( [0, \tau )\times {\mathbb{S}}^2;{\mathbb{R}})$ and observe that

\begin{align}-\int_{{\mathbb{S}}^2} &\psi (0,x)\rho_0(x)m(dx)\nonumber\\
&=\int_0^\tau\int_{{\mathbb{S}}^2}{\frac{\partial}{\partial t}}\bigl(\psi (t,X(t,x))\bigr)\rho_0(x)m(dx)dt\nonumber\\
&=\int_0^\tau\int_{{\mathbb{S}}^2}\Bigl({\frac{\partial}{\partial t}}\psi (t,X(t,x))+ \nabla \psi (t,X(t,x)) \cdot {\frac{\partial X(t,x)}{\partial t}}\Bigr)\rho_0(x)m(dx)dt\nonumber\\
\vskip.05in
\vfill
\eject
\end
\indent (iv) Suppose $\Theta (r)=r^{\gamma-1}/\gamma $ where $1<\gamma <3/2$. Then there exists $\kappa>0$ such that
\begin{align}\int_{{\mathbb{S}}^2} \rho(x)^{2\gamma -3}\Vert \nabla \rho (x)\Vert^2m(dx)\geq  {\frac{\kappa}{ (2\gamma -1)(\gamma -1)}}\Bigl(  \Bigl(\int_{{\mathbb{S}}^2} \rho(x)^\gamma m(dx)\Bigr)^{(2\gamma -1)/\gamma}-1\Bigr)\end{align}
\indent (iv) See [7], Theorem 2, and the discussion before the Corollary.

\par
\begin{prop}\label{Proposition1}Let $\gamma$ be any unit speed curve. Then $S$ satisfies the Serret-Frenet
differential equation $S'=\Omega S$, or equivalently
$F'=(U^{-1}\Omega U -U^{-1}U')F$, and the transportation cost for (\ref{defc}) satisfies
\begin{align}c_1((\gamma (0), \gamma'(0)), (\gamma (h), \gamma'(h))\leq {\frac{1}{2}}\int_0^h \Vert \Omega (s)\Vert_{{\mathcal L}^2}^2 ds\end{align}
\noindent where ${\mathcal L}^2$ denotes the Hilbert-Schmidt norm.\end{prop}
\vskip.05in
\noindent {\bf Proof.}  We have $S,F\in O(3)$, and $\Omega$ is a real skew symmetric matrix
$$\Omega =\begin{bmatrix}0&\kappa&0\cr -\kappa&0&\sigma\cr 0&-\sigma&0\end{bmatrix}$$
with Hilbert-Schmidt norm $\Vert \Omega\Vert_{{\mathcal L}^2}=\sqrt{2(\sigma^2+\kappa^2)},$ so $\Vert\gamma''\Vert^2\leq \Vert \Omega\Vert_{{\mathcal L}^2}^2/2$.
Observe that if $\gamma$ is a geodesic on $M$, then $F$ and $S$ both satisfy $F'=\Omega F$. \par
\vskip.05in
\noindent {\bf Example} Consider the case $\Theta (r)=r^2$ with solutions that are depend only upon the colatitude $\theta$, and are independent of the longitude $\phi$. We reduce to the Hamiltonian
$${\mathcal H}=\int_0^\pi \rho (\theta )\Bigl( {\frac {\partial q}{\partial\theta }}\Bigr)^2{\frac{d\theta}{2\pi}}+ \int_0^\pi \rho (\theta)^3 {\frac{d\theta}{6\pi}}$$
with canonical equations giving
$$\pi {\frac{\partial \rho}{\partial t}}= - {\frac{\partial (v\rho)}{\partial t}}$$
$$\pi {\frac{\partial v}{\partial t}}= - v{\frac{\partial v}{\partial t}}- \rho{\frac{\partial \rho}{\partial t}}.$$
Evidently, one can obtain simultaneous solutions with $v=\pm \rho $. We suppose that $\varphi : (0, \pi /2)\rightarrow [0, \infty )$ is a $C^2$ function such that
$\varphi (x)\rightarrow 0$ as $x\rightarrow 0+$ and $x\rightarrow \pi/2-$ and $\vert \varphi'(x)\vert \leq \kappa_1$; then let
$$\rho_0(\xi )=\varphi (\xi ){\mathbb{I}}_{[0, \pi /2]}(\xi )+ \varphi (\pi -\xi ){\mathbb{I}}_{[\pi /2,\pi]}(\xi ).$$
We choose $\tau>0$ such that $1-2\kappa_1\tau /\pi>0$, then for $0<t<\tau$ we observe that $X(\xi )=\xi +2\rho_0(\xi )t{\hbox{sgn}}(\pi /2-\pi )/\pi$ is a continuous and strictly increasing function $X : [0, \pi ]\rightarrow [0, \pi ]$ such that $X (0)=0$ and $X(\pi )=\pi$;  hence for all $\theta\in [0, \pi ]$ there exists a unique $\xi$ such that $\theta =X(\xi ).$ This shows that the characteristic lines $\theta=\xi +2\rho_0(\xi )t{\hbox{sgn}}(\pi /2-\pi )/\pi$  are straight lines, and they do not cross fror $t\in [0, \tau ]$, and $\rho$ is constant along characteristic lines. For larger $t>\tau$, the crossing characteristics produce conflicting values for $v=\pm\rho$.\par
\indent (iii) Suppose that $\Theta$ satisfies $C_\infty$ and $\Delta_2$; in particular, these hold for $\Theta (r)=r^{\gamma -1}$ with $\gamma >1$.Then there exist $K_1, K_2>0$ depending only upon $\Theta, n$ and $K$ such that
$${\mathcal U}(f_h)\leq e^{hK_1(K_2-1)/\log 2}{\mathcal U}(\rho_0).$$

\indent (iii) Now let $f_t$ be  a path of probability measures from $\rho_0$ to $f_h$ such that
$${\mathcal U}(f_t)=\int_M \Theta\Bigl( {\frac{\rho_0(x)}{\Delta (t,x)}}\Bigr)\rho_0(x)(m(dx);$$
$${\frac{d}{dt}}{\mathcal U}(f_t)=\int_M {\frac{\dot \Delta (t,x)}{\Delta (t,x)}} \Theta'\Bigl( {\frac{\rho_0(x)}{\Delta (t,x)}}\Bigr) {\frac{\rho_0(x)^2}{\Delta (t,x)}}m(dx),$$
\noindent where
$$ {\frac{\dot \Delta (t,x)}{\Delta (t,x)}}={\hbox{trace}}\bigl( (d/dt)A_x(t)\, A_x(t)^{-1}\bigr)={\hbox{trace}}\, U(x,t)$$
so $\vert \dot \Delta (t,x)/\Delta (t,x)\vert\leq K_1$ by (5.38), and by the $\Delta_2$ condition and convexity of $\Theta (e^{-r})$, we have
$$(K_2-1)\Theta (r)\geq \Theta(2r)-\Theta (r)\geq (\log 2) r\Theta'(r)$$
$${\mathcal U}(f_t)\leq {\mathcal U}(\rho_0)+{\frac{K_1(K_2-1)}{\log 2}}\int_0^t {\mathcal U}(f_\tau )d\tau$$
hence by Gronwall's inequality
$${\mathcal U}(f_h)\leq e^{K_1(K_2-1)h/\log 2}{\mathcal U}(\rho_0).$$
\indent In section 5 we have obtained information about $f_h$, which is an intermediate step in the algorithm essentially a predictor for $\rho_h$. In the following section 6, we derive corresponding results about $\rho_h$.\par

For any regular contour $\gamma$ bounding a region $B$ in $M$, there is a frame $F=\{ \gamma', N, N\times \gamma'\}$ where $\vec n =N\times\gamma'$ is the outward pointing normal to $B$ in $TM$. The divergence theorem gives
$$\int\!\!\!\int_{\partial B} \nabla_M\cdot v_M(x) m(dx)=\int_{\gamma} v_M(\gamma (s))\cdot \vec n(s)\, ds.$$
We can associate curves on $M$  with moving frames over ${\mathbb{S}}^2$, as follows. For any smooth surface $M$, there is the Gauss map $G: M\rightarrow {\mathbb{S}}^2$ $X\rightarrow N(X)$  so that any curve $\gamma$ on $M$  satisfies $G\circ \gamma (t)=N(\gamma (t))$ and $G$ has differential $dG: TM\rightarrow T{\mathbb{S}}^2$. This $dG$ is expressed in terms of the first and second fundamental forms of $M$. There is a natural map $TM\rightarrow T{\mathbb{S}}^2$ $(X,V)\mapsto (N(X), V)$ where $N\cdot V=0$, thus $(X,V)\mapsto [N;V; N\times V]$ gives an orthonormal frame for each unit tangent vector $V\in T_XM$. If $G$ is invertible, then we have an inverse $[N;V;N\times V]\mapsto (G^{-1}(N), V),$ so
$$\{ (X,V): X\in M, V\in {\mathbb{R}}^3;\Vert V\Vert =1; N(X)\cdot V=0\}\leftrightarrow \{ F=[N;V; N\times V]\in SO(3)\}.$$
This ODE is determined via the Gauss map $G:M\rightarrow {\mathbb{S}}^2:$ $X\mapsto N(X)$.\par

\indent The case of particular interest is when
$g(x,t)=-\nabla \gamma \rho (x,t)^{\gamma -1}$, and existence of the gradient is problematic. In Propositions 3.4 and 4.5, we give sufficient conditions for $\rho (x,t)$ to have a gradient such that $\int_M \rho (x,t)^{2\gamma -3} \Vert \nabla \rho (x,t)\Vert^2 m(dx)<\infty$. Uniqueness of solutions to (1.8)  is guaranteed by Lipschitz continuity of $x\mapsto g(x,t)$. Otherwise, there is a possible lack of uniqueness of solutions of (\ref{ODE}); the system
$${\frac{d}{dt}}\begin{bmatrix} X\cr V\end{bmatrix}=\begin{bmatrix} V\cr \beta (\gamma -1)X^{2\gamma -1}\end{bmatrix}$$
with $1<\gamma <2$ provides a cautionary example. \par
\indent (iii) Suppose that $\Theta (r)/r^\kappa$ is increasing for some $0<\kappa<1$ Then one can show that
$${\frac{\kappa}{1+\kappa}} \Theta_1(r)\leq \Theta_1(r)-\Theta (r)\leq \Theta_1(r),$$
so with $r=\chi (v)$, we have
$${\frac{\kappa}{1+\kappa}} v\chi (v)\leq (v-\Theta\circ \chi (v))\chi (v)\leq v\chi (v);$$
hence the Orlicz norm associated with $x\Theta (x)$ is equivalent to
\begin{align}\Vert u\Vert_{L_U}\sim \sup \Bigl\{ \int u(x)v(x)m(dx): \int v(x)\chi (v(x))m(dx)\leq 1\Bigr\}.\end{align}
\indent  (iv) {\sl Jacobi fields on the sphere} Let $\{ e_1, e_2, e_3\}$ be the usual orthogonal basis of ${\mathbb{R}}^3$ and
consider the tangent space to ${\mathbb{S}}^2$ at $e_3$.
\indent Now let $v\in T_{e_3}{\mathbb{S}}^2$  be a unit vector, so
$$\exp_{e_3}(tv)=\begin{bmatrix}\sin t\cos \psi\cr \sin t\sin\psi\cr\cos t\end{bmatrix}$$
and $e_3,\exp_{e_3}(tx), \exp_{e_3} (\xi +u\zeta )$ gives a spherical triangle, in which the distance $d=d(\exp_{e_3}(tx), \exp_{e_3} (\xi +u\zeta ))$
$$\cos d=\cos t\cos\theta+\sin t\sin\theta \cos (\phi-\psi )$$
by the cosine rule. Then
$${\frac{\partial^2\cos d}{\partial t^2}}=-\cos t\cos\theta -\sin t\sin\theta\cos (\phi-\psi )$$
\begin{align}{\frac{\partial^4\cos d}{\partial^2u \partial t^2}}&=\cos t\cos\theta \Bigl({\frac{\partial\theta}{\partial u}}\Bigr)^2+\cos t\sin\theta \Bigl({\frac{\partial^2\theta}{\partial^2u}}\Bigr)\nonumber\\
&\quad +\sin t\sin\theta\cos (\phi-\psi )\Bigl({\frac{\partial\theta}{\partial u}}\Bigr)^2-\sin t\cos\theta \cos (\phi -\psi )\Bigl({\frac{\partial^2\theta}{\partial u^2}}\Bigr)\nonumber\\
&\quad +2\sin t\cos\theta\sin (\phi-\psi )\Bigl({\frac{\partial\theta}{\partial u}}\Bigr)\Bigl({\frac{\partial\phi}{\partial u}}\Bigr)\nonumber\\
&\quad +\sin t\sin\theta\cos (\phi-\psi )\Bigl({\frac{\partial\phi}{\partial u}}\Bigr)^2+\sin t\sin\theta \sin (\phi-\psi )\Bigl({\frac{\partial^2\phi}{\partial u^2}}\Bigr),\end{align}
which at $t=0$ reduces to
$$\Bigl({\frac{\partial^4\cos d}{\partial^2u \partial t^2}}\Bigr)_{t=0}=\cos\theta \Bigl({\frac{\partial\theta}{\partial u}}\Bigr)^2+\sin\theta  \Bigl({\frac{\partial^2\theta}{\partial^2u}}\Bigr).$$
By differentiating the formula $\theta^2=\Vert \xi +u\zeta \Vert^2$ twice, we obtain
$$\Bigl({\frac{\partial^4\cos d}{\partial^2u \partial t^2}}\Bigr)_{t=0}=\cos\theta \Bigl\langle
{\frac{\xi +u\zeta }{\Vert \xi +u\zeta \Vert} }, \zeta \Bigr\rangle^2+{\frac{\sin\theta}{\theta}}\Bigl( \langle \zeta, \zeta \rangle -\Bigl\langle
{\frac{\xi +u\zeta }{\Vert \xi +u\zeta \Vert} }, \zeta \Bigr\rangle^2\Bigr),$$
\noindent which is nonnegative.\par
\noindent In the case of $M={\mathbb{S}}^2$, we compute the speed and acceleration of both these curves.\par
\indent (ii) In the case of $M={\mathbb{S}}^2$, we can convert \ref{ODE} into a differential equation for vectors in the tangent space. For $C^2$ curve $\xi :[0,1]\rightarrow T_y {\mathbb{S}}^2$, we introduce $X=\exp_y(\xi (t))$, and use the exponential map to change bases. Let $\lambda (t)(y\times \xi (t))=\xi '(t)-\xi (t)\cdot \xi '(t)\xi (t)/\Vert \xi (t)\Vert^2$ be the component of velocity in $T_y{\mathbb{S}}^2$ that is perpendicular to $\xi (t)$; then
$$\begin{bmatrix} X\cr V\cr X\times V\end{bmatrix}=\begin{bmatrix} \cos (\Vert \xi (t)\Vert) & {\frac{\sin (\Vert \xi (t)\Vert)}{\Vert \xi (t)\Vert}} &0\cr
{\frac{-\sin (\Vert \xi (t)\Vert )\xi (t)\cdot \xi '(t) }{\Vert \xi (t)\Vert}}&{\frac{\cos (\Vert \xi (t)\Vert )\xi (t)\cdot \xi '(t)}{\Vert \xi (t)\Vert^2}}&\lambda {\frac{\sin (\Vert \xi (t)\Vert)}{\Vert \xi (t)\Vert}}\cr
\lambda\sin^2 (\Vert \xi (t)\Vert) & -{\frac{\lambda \sin (\Vert \xi (t)\Vert) \cos (\Vert \xi (t)\Vert)}{\Vert \xi (t)\Vert}}& {\frac{\xi (t)\cdot \xi '(t)}{\Vert \xi (t)\Vert^2}}\end{bmatrix}\begin{bmatrix} y\cr \xi (t)\cr y\times \xi (t)\end{bmatrix}$$
so the squared speed on ${\mathbb{S}}^2$ is
$$\Bigl( {\frac{ds}{dt}}\Bigr)^2=\Bigl\Vert{\frac{dX}{dt}}\Bigr\Vert^2=\Bigl( {\frac{\xi (t)\cdot \xi '(t)}{\Vert \xi (t)\Vert }}\Bigr)^2+ \Bigl({\frac{\sin\Vert \xi (t)\Vert}{\Vert \xi (t)\Vert}}\Bigr)^2\Bigl\Vert \xi '(t)-{\frac{\xi (t)\cdot \xi '(t)}{\Vert \xi (t)\Vert^2}}\xi (t)\Bigr\Vert^2,$$
and  the determinant of this matrix is
$${\frac{(\xi (t)\cdot \xi'(t))^2}{\Vert \xi (t)\Vert^4}}+\lambda(t)^2 \Bigl({\frac{\sin(\Vert \xi (t)\Vert )}{\Vert \xi (t)\Vert}}\Bigr)^2={\frac{1}{\Vert \xi (t)\Vert^2}} \Bigl( {\frac{ds}{dt}}\Bigr)^2.$$
\noindent It follows that if $\Vert \xi (t)\Vert <\pi/2$, then the speed in the tangent space $ds_T/dt$ satisfies $ds_T/dt\geq ds/dt\geq (2/\pi )ds_T/dt$. Introducing $a(t)$ and $b(t)$ such that
\begin{align}{\frac{\xi ''(t)}{\Vert \xi (t)\Vert}}&-{\frac{2\xi (t)\cdot \xi '(t) \xi '(t)}{\Vert \xi (t)\Vert^3}}+{\frac{3(\xi (t)\cdot \xi '(t))^2 \xi (t)}{\Vert \xi (t)\Vert^5}}-{\frac{\xi '(t)\cdot \xi '(t) \xi (t)}{\Vert \xi (t)\Vert^3}}\nonumber\\
& = a(t)  {\frac{\xi '(t)\cdot \xi (t) \xi (t)}{\Vert \xi (t)\Vert^3}} +b(t)\Bigl( {\frac{\xi '(t)}{\Vert \$xi (t)\Vert}}-{\frac{\xi '(t)\cdot \xi (t) \xi (t)}{\Vert \xi (t)\Vert^3}}\Bigr),\end{align}
we obtain the acceleration in terms of the orthogonal basis
\begin{align}{} &{\frac{d^2}{dt^2}}\exp_y\xi\nonumber \\
&=\Bigl[ -\cos (\Vert \xi \Vert ){\frac{(\xi \cdot \xi ')^2}{\Vert \xi \Vert^2}}-\sin (\Vert \xi \Vert) \Bigl( {\frac{\xi '\cdot \xi '}{\Vert \xi \Vert}} +{\frac{\xi \cdot \xi ''}{\Vert \xi \Vert}} -{\frac{(\xi \cdot \xi ')^2}{\Vert \xi \Vert^3}}\Bigr)\Bigr] y\nonumber\\
&\quad + \Bigl[ \sin (\Vert \xi \Vert ) \Bigl( -{\frac{(\xi \cdot \xi ')^2}{\Vert \xi \Vert^3}} +a(t){\frac{\xi '\cdot \xi }{\Vert \xi\Vert^3}}\Bigr) +\cos (\Vert \xi\Vert) \Bigl( {\frac{\xi '\cdot \xi '}{\Vert \xi (\Vert^2}}+{\frac{\xi \cdot \xi ''}{\Vert \xi \Vert^2}}\Bigr)\Bigr] \xi \nonumber\\
& \quad + \Bigl[ 2\cos (\Vert \xi \Vert){\frac{\xi\cdot \xi '}{\Vert \xi \Vert^2}} +\sin (\Vert \xi \Vert ){\frac{b(t)}{\Vert \xi \Vert}}\Bigr] \Bigl( \xi '-{\frac{\xi \cdot \xi ' \xi }{\Vert \xi \Vert^2}}\Bigr).\end{align}
The following result  introduces a notion of local geodesic interpolation which may be compared with Definition 17.13 of [38].
We consider geodesic triangles within a cap $B_r(x_0)=\{ y\in {\mathbb{S}}^2: d(x_0,y)\leq r\}$. \par
\vskip.1in

\begin{prop}\label{proposition4}Suppose that the probability density functions $f, \rho_0, \rho_1$ are all supported on some cap $B_{\pi/8}(x_0)$ in ${\mathbb{S}}^2$.
Let $T_s:{\mathbb{S}}^2\rightarrow {\mathbb{S}}^2$ be a transport map inducing $\rho_s$ from $f$ for all $s\in [0,1]$ such that $T_0$ and $T_1$ are optimal transport maps, and $s\mapsto T_s(x)$ is a geodesic for all $x\in B_{\pi/8}(x_0)$. Then
\begin{align}(1-s)W_2^2(\rho_0,f)+sW_2^2(\rho_1,f)-W_2^2(\rho_s,f)\geq {\frac{\pi s(1-s)}{4}}W_2^2(\rho_0,\rho_1)\qquad (s\in [0,1]).\end{align}\end{prop}
\vskip.05in
\noindent {\bf Proof.}
Let $A,B,C$ be points on ${\mathbb{S}}^2$ and in the traditional notation let $A,B,C$ be the angles in the spherical triangle $\triangle ABC$. Then let $s\mapsto C(s)$ for $s\in [0,1]$ be the geodesic from $A$ to $C$ at constant speed $h=d(A,C)$, so $b(s)=d(A,C(s))=sh$, $a(s)=d(B,C(s))$ and $c=d(A,B)$ for $s\in [0,1]$. Then the cosine formula applied to $\triangle ABC(s)$ gives
$$\cos a(s)=\cos sh\cos c+\sin sh \sin c\cos A$$
hence the derivatives satisfy
$$-\dot a(s)\sin a(s)=(-\sin sh\cos c+\cos sh \sin c\cos A)h,$$
$$-\ddot a(s)\sin a(s)-\dot a(s)^2 \cos a(s)=-h\cos a(s);$$
hence $\dot a(s)^2\sin^2a(s)\leq h^2$ and the minimum value of $a(s)$ occurs, either when $\tan sh=\tan c\cos A$, or at the endpoints  $C(s)=A$ or $C(s)=C$. The diameter of $\triangle ABC$ is less than or equal to $\pi/4$, so $a(s)\leq a=d(C,B)\leq \pi/4$. Hence
\begin{align}{\frac{d^2}{ds^2}}\Bigl(d(C(s),B)^2/2\Bigr)&=\dot a(s)^2+\ddot a(s)a(s)\nonumber\\
&=a(s)\cot a(s)\, h^2+ \dot a(s)^2 \bigl(1-a(s)\cot a(s)\bigr)\geq \pi h^2/4,\end{align}
\noindent and since $a(s)\cot a(s)\geq \pi/4$ for all $a(s)\in (0, \pi/4 ].$
We can choose $A=T_0(x)$, $B=x$ and $C=T_1(x)$, where $h=d(T_0(x), T_1(x))$ depends upon $x$; indeed,
By the cosine rule for $T_0(x)=\exp_x\nabla \phi_0(x)$ and $T_1(x)=\exp_x\nabla\phi_1(x)$, we have
\begin{align}\cos \bigl( d(T_0(x), T_1(x))\bigr)&=\cos (\Vert\nabla\phi_0(x)\Vert )\cos (\Vert\nabla\phi_1(x)\Vert)\nonumber\\
&\quad  +{\frac{\sin (\Vert \nabla\phi_0(x)\Vert)}{ \Vert \nabla\phi_0(x)\Vert}}
{\frac{\sin (\Vert \nabla\phi_1(x)\Vert)}{\Vert \nabla\phi_1(x)\Vert}}\nabla\phi_0(x)\cdot \nabla\phi_1(x).\end{align}
Also $x, T_0(x), T_1(x)\in B_{\pi/8}(x_0)$.  Hence
\begin{align} {\frac{d^2}{ds^2}}{\frac{1}{2}}\int_{{\mathbb{S}}^2} \Bigl((1-s)d(T_0(x),x)^2&+sd(T_h(x),x)^2-d(T_s(x),x)^2\Bigr) f(x)m(dx)\nonumber\\
&\leq -{\frac{\pi}{4}}\int_{{\mathbb{S}}^2}d(T_0(x),T_1(x))^2f(x)m(dx).\end{align}
Since $T_0$ and $T_1$ are optimal, we have
\begin{align}(&1-s)W_2^2(\rho_0,f)+sW_2^2(\rho_1,f)-W_2^2(\rho_s,f)\nonumber\\
&\geq {\frac{1}{2}}\int_{{\mathbb{S}}^2} \Bigl((1-s)d(T_0(x),x)^2+sd(T_h(x),x)^2-d(T_s(x),x)^2\Bigr) f(x)m(dx)\nonumber\\
&\geq {\frac{\pi s(1-s)}{8}}\int_{{\mathbb{S}}^2} d^2(T_0(x),T_1(x))f(x)m(dx) \nonumber\\
&\geq {\frac{\pi}{4}}s(1-s)W_2^2(\rho_0, \rho_1).\end{align}

\vskip.05in
\noindent {\bf 3.2 Remark} I suspect that Proposition \ref{proposition4} holds for $f,\rho_0,\rho_1$ supported on a hemisphere.\par
From the continuity equation, we have a $2$-absolutely continuous path $[0, \tau ]\rightarrow {\mathcal W}^2({\mathbb S}^2):$ $t\mapsto \rho_t$ such that
$${\frac{W_2^2(\rho^{(h, \varepsilon )}_t, \rho^{(h, \varepsilon )}_s)}{t-s}}\leq \int_s^t \int_{{\mathbb{S}}^2} \Vert \vec v^{(h, \varepsilon )}(x,u)\Vert^2\rho^{(h, \varepsilon )}_u(x)m(dx)du;$$

We therefore consider continuous paths $t\mapsto \rho (\cdot , t)$ that are Lipschitz continuous $[0, \tau]\rightarrow {\mathcal W}$.

Using the Lipschitz condition, we deduce that

\begin{align}\Bigl\vert \int_0^\tau\int_{{\mathbb{S}}^2} \psi (t,x)\rho (t,x)m(dx)dt&-h\sum_j \int_{{\mathbb{S}}^2} \psi (t_j,x)\rho (t_j,x)m(dx)\Bigr\vert\nonumber\\
&\leq \tau h\Bigl\Vert{\frac{\partial \psi}{\partial t}}\Bigr\Vert_{L^\infty} +h\tau \Vert\nabla \psi\Vert_{L^\infty}{\hbox{Lip}}(t\mapsto \rho_t).\end{align}

\nonumber\\
&+\int_0^\tau \bigl\Vert \nabla \bigl(\phi (y,t)\cdot \vec v^{(h,\varepsilon)}(y,t)\bigr)\bigr\Vert_{L^\infty_y} \int_{{\mathbb{S}}^2} \Vert \vec v(x,t)\Vert^2\rho (x,t)\, m(dx) dt\end{align}

where we have used (<ref>). For $\varepsilon, h>0$, the solution determines a bounded linear functional on $H^1 ({\mathbb{S}}^2\times [0, \tau ]; {\mathbb{R}}^2)$; by hypothesis, the $\vec v^{(h, \varepsilon )}$ are uniformly Lipschitz, hence the norm of these functionals is bounded independent of $\varepsilon ,h>0$. Taking $h, \varepsilon\rightarrow 0$ through some subsequence, we obtain the existence of
a weak solution to the acceleration equation.
\begin{align}&\leq \tau\sum_{j=0}^{N}2h\int_{{\mathbb{S}}^2} \Vert\nabla q_{jh}\Vert^2\rho_{jh} m(dx)+ {\frac{\tau}{h}}\sum_{j=0}^{N}\bigl( {\mathcal{H}}(\rho_{jh}, q_{jh})-{\mathcal{H}}(\rho_{(j+1)h}, q_{(j+1)h})\bigr)\nonumber\\
&\leq 2\tau\int_0^\tau  \int_{{\mathbb{S}}^2} \Vert\nabla q_{t}(x)\Vert^2\rho_{t}(x) m(dx)dt+{\frac{\tau}{h}}\bigl({\mathcal H}(\rho_0, q_0)-{\mathcal H}(\rho_\tau, q_\tau )\bigr),\end{align}

For $\psi  \in C([0, \tau ]\times {\mathbb{S}}^2; {\mathbb R})$, we have

\begin{align}\psi& (X^{(h, \varepsilon )}(x,t_{j+1}),t_{j+1})-\psi (X^{(h, \varepsilon )}(x,t_{j}),t_{j})\nonumber\\
&=h{\frac{\partial\psi}{\partial t}}(X^{(h, \varepsilon )}(x,t_j))+(X(x,t_{j+1})-X^{(h, \varepsilon )}(x,t_j))\cdot \nabla\psi (X(x,t_j),t_j)+O(h^2)\end{align}

so integrating this against $\rho_0(x)m(dx)$, we obtain

\begin{align}\int_{{\mathbb{S}}^2}& \psi (x,t_{j+1})\rho(x, t_{j+1})m(dx)-\int_{{\mathbb{S}}^2} \psi (x,t_{j})\rho(x, t_{j})m(dx)\nonumber\\
&=h\int_{{\mathbb{S}}^2} {\frac{\psi (x,t_{j})}{\partial t}}\rho(x, t_{j})m(dx)\nonumber\\
&\quad +\int_{{\mathbb{S}}^2}(X^{(h, \varepsilon )}(x,t_{j+1})-X^{(h, \varepsilon )}(x,t_j))\cdot \nabla\psi (X^{(h, \varepsilon )}(x,t_j),t_j)\rho_0(x)m(dx)+O(h^2),\end{align}

so summing over $j$, we obtain

\begin{align}-\int_{{\mathbb{S}}^2}& \psi (x,0)\rho_0(x)m(dx)\nonumber\\&=h\sum_{j=0}^{\lfloor \tau/h\rfloor}\int_{{\mathbb{S}}^2} {\frac{\psi (x,t_{j})}{\partial t}}\rho(x, t_{j})m(dx)
\nonumber\\
&\quad +h\sum_{j=0}^{\lfloor \tau/h\rfloor}\int_{{\mathbb{S}}^2}{\frac{(X^{(h, \varepsilon )}(x,t_{j+1})-X^{(h, \varepsilon )}(x,t_j))}{h}}\cdot \nabla\psi (X^{(h, \varepsilon )}(x,t_j),t_j)\rho_0(x)m(dx)+O(h),\end{align}

By the hypotheses, both sides of  (<ref>) define uniformly bounded family of functionals on $C^1([0, \tau]\times {\mathbb{S}}^2; {\mathbb{R}}).$

$$\psi\mapsto h\sum_{j=0}^{\lfloor \tau/h\rfloor}\int_{{\mathbb{S}}^2} \psi (t_j,x)\rho (t_j,x)m(dx)\qquad (\psi  \in C([0, \tau ]\times {\mathbb{S}}^2; {\mathbb R}))$$

determines a bounded family of positive measures on $[0, \tau ]\times {\mathbb{S}}^2$, which has a weak${}*$ limit as $h\rightarrow 0+$ through some subsequence. We deduce that there
exists $\rho\in L^\infty([0, \tau ]\times {\mathbb{S}}^2; {\mathbb{R}})$ such that

$$ h\sum_{j=0}^{\lfloor \tau/h\rfloor}\int_{{\mathbb{S}}^2} \psi (t_j,x)\rho (t_j,x)m(dx)\rightarrow \int_0^\tau \int_{{\mathbb{S}}^2} \psi (t,x)\rho (x,t)m(dx)dt\qquad (\psi  \in C([0, \tau ]\times {\mathbb{S}}^2; {\mathbb R}))$$

as $h\rightarrow 0+$.where

$$\int_0^\tau \int_{{\mathbb{S}}^2}\phi (y,t)\cdot  \vec v^{(h,\varepsilon)}(y,t)\bigl(\nabla\cdot \vec v^{(h,\varepsilon)}(y,t)\bigr)\rho^{(h, \varepsilon)}(y,t)m(dy)dt$$

involves the bounded terms  $\vec v^{(h,\varepsilon)}$ and  $\rho^{(h, \varepsilon)}$, and the  measure $\nabla\cdot \vec v^{(h,\varepsilon)}$ from the bounded family  and

\begin{align}\Bigl\vert\int_0^\tau \int_{{\mathbb{S}}^2}&\phi (y,t)\cdot  \vec v^{(h,\varepsilon)}(y,t) \vec v^{(h,\varepsilon)}(y,t)\cdot \nabla \rho^{(h, \varepsilon)}(y,t)m(dy)dt\Bigr\vert\nonumber\\
&\leq \Vert\phi\Vert_{L^\infty}\Vert \vec v^{(h, \varepsilon)}\Vert^2_{L^\infty}\int_0^\tau \int_{{\mathbb{S}}^2} \Vert\nabla \rho^{(h, \varepsilon)} (x,t)\Vert m(dx)dt\end{align}

$${\mathcal U}(\psi_\varepsilon\ast \rho_1)+\int_M \bigl( \psi_\varepsilon\ast \rho_1\bigr) (x)\bigl\Vert  \nabla \bigl(\Theta_1(\psi_\varepsilon\ast \rho_1)(x)\bigr)\bigr\Vert^2 m(dx)\leq {\mathcal U}(f),$$

and by Fatou's Lemma, we can take $\varepsilon\rightarrow 0+$. If $f, \rho_1\in C^2({\mathbb{S}}^2)$, then $\phi\in C^3({\mathbb{S}}^2)$ and we can use the results of section 5.

(ii) We consider Proposition <ref>(iv), and make a discrete approximation to solutions of the ODE (<ref>) by making a predictor-corrector based on successive rotations. For the initial orthonormal frame $[X_0;V_0; X_0\times V_0]$ and time step $h>0$, we rotate about axis $X_0\times V_0$ by $\exp (h J)$ to give a new predictor frame $[Y; W;Y\times W] =\exp (h J)[X_0; V_0; X_0\times V_0]$; here $Y\times W=X_0\times V_0$.Recall that $-\nabla_{{\mathbb{S}}^2}\Theta_0(X)=\kappa_g X\times V$ in Proposition <ref>(iv). Choosing $\delta =h\Vert \nabla_{{\mathbb{S}}^2}\Theta_0(Y)\Vert$ and $S=-\nabla_{{\mathbb{S}}^2}\Theta_0(Y)/\Vert \nabla_{{\mathbb{S}}^2}\Theta_0(Y)\Vert$, so $S$ is a unit vector in ${\hbox{span}}\{ W, Y\times W\}$. Then we apply a rotation about the $Y$ axis through $\delta$ by letting $R[Y;W; Y\times W]=[W; Y\times W; Y]$ and applying $R^{-1}\exp (\delta J) R$ to give correction $W\mapsto (\cos \delta ) W+(\sin \delta ) Y\times W$, and finally we get our new frame

$$[X_h; V_h; X_h\times V_h]=R^{-1}\exp (\delta J)R\exp (hJ)[X_0;V_0; X_0\times V_0].$$

(iii) The acceleration cost function for ${\mathbb{R}}^d$ from [24] is

$$C_\tau\bigl((x, \xi ), (y, \eta )\bigr)=D_\tau\bigl((x+\tau\xi , (y,\eta )\bigr)$$

$$D_h\bigl( (x, \xi ), (y, \eta)\bigr) =3\Bigl\Vert {\frac{y-x}{\tau}}-{\frac{\eta-\xi}{2}}\Bigr\Vert^2+{\frac{\Vert \eta-\xi\Vert^2}{4}}\qquad (x, y,\xi,\eta \in {\mathbb{R}}^d) .$$

Here the geodesics are straight lines, and rectilinear motion at constant speed has zero acceleration, hence the journey from $(x,\xi )$ to $(x+\tau\xi ,\xi)$ incurs acceleration cost zero. This is quite different to the situation of motion on a sphere, where any movement between distinct points $x$ and $y$ involves a positive cost, by Proposition <ref>.
; so $\nabla\rho\in L^1({\mathbb{R}}^3)$ for $\rho\in C^1_0({\mathbb{R}}^3)$ ensures that $\rho\in L^{\gamma}({\mathbb{R}}^3).$